Author name: Prasanna

Haryana State Board HBSE 10th Class Science Important Questions Chapter 6 Life Processes Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 6 Life Processes

Question 1.
Only what is visible is alive does not hold true for a life. Explain.
Answer:
1. When we see an animal moving we call it alive. Similarly when we see a green plant we call it alive.
2. Actions that we can see such as running, breathing, laughing, changing colours of leaves, etc. also make us believe that the organism is alive.
3. So, we believe some sort of movement, either growth-related or not as an evidence of being alive.
4. But, a plant which we cannot see growing and animals that breathe without showing any movement are also alive. Hence, it would be wrong to say that what is visible is only alive.

Question 2.
How one can differentiate living organisms and non-living things? What are the exceptions here?
Answer:
1. Movement is one of the basic criterions to identify life. All living organisms show movement without any external help.
2. Animals show clear visible movements.
3. Plants show invisible movements inside their body (i.e., movements of various elements and compounds). External movement of plants is quite minute and very slow and so it is difficult to notice.
4. Viruses are an exception here.

Question 3.
What do you mean by molecular movement? Explain.
Answer:
1. The movements of molecules within living organisms which we cannot see through our naked eyes but are critical for carrying life processes such as photosynthesis, respiration, digestion, excretion, etc. are called molecular movements.
2. Molecular movement includes movement of molecules such as oxygen for respiration, movement of compounds such as enzymes, nutrients, hormones, etc.

Need of molecular movement:

• Living organisms are highly complex, but very well organized structures. Going down they are made of organs, tissues, cells and cell organelles.
• Therefore it is necessary that molecules move and reach up to cell level so that functions such as growth, maintenance and survival are maintained.
• For example, oxygen inhaled move throughout the body via, the process of breathing.

Question 4.
What is the connecting link between living organisms and non-living things? Why?
Answer:
Viruses are the connecting link between living organisms and non-living things.

Reason:
1. It is said that molecular movement is necessary in living organisms. Non-living things do not show any molecular movement.
2. In a free environment, viruses do not show any molecular movement. But, when they come in the contact of the host cells they start acting as living organisms in the form of obligate parasite.
3. Thus, viruses on one side do not show molecular movement i.e. behave as non-living things, but under favourable environment they start showing molecular movement and behave like living organisms. Hence, viruses are called a connecting link between living organisms and non-living things.

Question 5.
What are life processes? State the important life processes that perform the job of maintaining these processes.
Answer:
Life processes:
1. All living organisms perform certain important functions to maintain their survival. These main functions are called life processes.
2. In other words, the processes which together perform the job of maintenance of the body are known as life processes.
3. Maintenance process requires energy. The energy is used for—

• Growth and maintenance of organisms and
• Preventing break-down.

The energy needed to the organism comes from external sources such as food, sun, atmosphere, etc.

The main life processes are:
(1) Nutrition,
(2) Respiration,
(3) Transportation,
(4) Excretion,
(5) Control and co-ordination,
(6) Movement and
(7) Reproduction.

Question 6.
Define and explain each life process very briefly.
Answer:
1. Nutrition:

• The process of transferring a source of energy which we call food, from outside the body of the organism to the inside is called nutrition.
• The food consumed is then converted into smaller and smaller units so that it can be absorbed by the body.
• Most of the food sources are carbon-based. Depending upon the complexity of these carbon sources, different organisms can then use different kind of nutritional processes.

2. Respiration:

• The process of acquiring oxygen from outside the body, and it in the process of break-down of food-sources for cellular needs, is known as respiration.
• Through respiration, organisms break down food and release energy.

3. Transportation:

• The process through which absorbed substances are transported to various parts of the body is called transportation.
• Transportation system is important to carry food and oxygen from one place to another.
• This system is also important for carrying waste products from body cells to excretory organs.

4. Excretion:
Waste materials produced in various cells of the body are removed by the process of excretion.

5. Control and co-ordination (Chapter 7):
Control and co-ordination make the living organisms adapt to the changing environment and survive there in.

6. Movement (Chapter 7):
The process of movement makes the living organism move from one place to another.

7. Reproduction (Chapter 8):
Reproduction involves multiplication of existing organisms. This enables them to maintain the existence of their species on the earth.)

Question 7.
What are nutrients? Give examples.
Answer:
1. The substances obtained by an organism from the surroundings and used them as a source of energy are called nutrients.
2. For example, carbohydrates, lipids, proteins, mineral, salts, etc.

Question 8.
Why a unicellular organism does not need specific organs for nutrition, respiration and transportation?
Answer:
In unicellular organisms, entire body surface of the organism is in the direct contact with the environment and transfer can take place through entire body. So, a unicellular organism does not need specific organs for nutrition, respiration and transportation.

Question 9.
What is nutrition? Classify the types of nutrition.
Answer:
1. Nutrition:

• The process of transferring a source of energy which we call food, from outside the body of the organism to the inside is called nutrition.
• The food consumed is then converted into smaller and smaller units so that it can be absorbed by the body.
• Most of the food sources are carbon-based. Depending upon the complexity of these carbon sources, different organisms can then use different kind of nutritional processes.

2. Respiration:

• The process of acquiring oxygen from outside the body, and it in the process of break-down of food-sources for cellular needs, is known as respiration.
• Through respiration, organisms break down food and release energy.

3. Transportation:

• The process through which absorbed substances are transported to various parts of the body is called transportation.
• Transportation system is important to carry food and oxygen from one place to another.
• This system is also important for carrying waste products from body cells to excretory organs.

4. Excretion:
Waste materials produced in various cells of the body are removed by the process of excretion.

5. Control and co-ordination (Chapter 7):
Control and co-ordination make the living organisms adapt to the changing environment and survive there in.

6. Movement (Chapter 7):
The process of movement makes the living organism move from one place to another.

7. Reproduction (Chapter 8):
Reproduction involves multiplication of existing organisms. This enables them to maintain the existence of their species on the earth.)

Types of nutrition:
(1) Autotrophic nutrition,
(2) Heterotrophic nutrition:

• Saprophytic nutrition
• Parasitic nutrition
• Holozoic nutrition

Question 10.
Define autotrophic nutrition and give examples of autotrophes.
Answer:
Autotrophic nutrition:
1. ‘Auto’ means ‘Self’ and ‘Trophe’ means ‘Nutrition’. So, the mode of ‘nutrition’ in which the organism ‘itself’ synthesizes its own food from the simple inorganic materials like carbon dioxide and water with the help of sunÍight, is called autotrophic nutrition.
2. All green plants and some bacteria show autotrophic mode of nutrition.

Question 11.
How do autotrophs fulfill their need of carbon and energy?
Answer:
1. Carbon dioxide (CO₂), (2) Water (H₂O) and (3) Nutrients are the main food materials that the autotrophic organisms need for their growth and maintenance. These materials fulfill the carbon and energy requirement of the autotrophs.
2. Autotrophs obtain these materials through the process called photosynthesis.
3. During photosynthesis plants take up carbon dioxide from the atmosphere, water from the ground and in the presence of sunlight convert these materials into carbohydrates. The carbohydrate then provides energy to the plants.
4. The excess carbohydrate remains stored as starch within the plants.
5. Autotrophs fulfill their needs of minerals such as nitrogen, phosphorus, iron and magnesium from the soil.

Question 12.
Write a note on photosynthesis. OR Define photosynthesis and enlist Its main events.
Answer:
Photosynthesis:

• The process by which the green plants make their own food by converting carbon dioxide and water into carbohydrates in the presence of sunlight and chlorophyll is called photosynthesis.
• Thus, the basic materials used in photosynthesis are carbon dioxide and water.

Chemical equation of photosynthesis:

Following events take place during photosynthesis:
(1) Absorption of light energy by chlorophyll.
(2) Conversion of light energy to chemical energy and splitting of water molecule into hydrogen and oxygen.
(3) Reduction of carbon dioxide to carbohydrate.

Question 13.
Is it necessary that all events of photosynthesis take place one after the other Immediately?
Answer:
Give an example to explain.
No, it is not necessary that all the events of photosynthesis occur one after the other.

Example:
1. The desert plants absorb CO₂ at night from the surrounding and repare the intermediate compound before the sun rises.
2. This intermediate compound is then acted upon after the chlorophyll absorbs solar radiation in the morning.

Question 14.
How do plants obtain carbon dioxide (CO₂) from the atmosphere?
Answer:
A large number of pores are present on the leaves of plants. These pores are known as stomata. These stomata are responsible for gaseous exchange.
1. During the gaseous exchange, a large amount of water is lost. To control the water loss, the stomata is surrounded by guard cells.
2. The guard cells control the opening and closing of the stomatal pore according to the need of CO₂ in the plant.
3. As the need of CO₂ arises, the guard cells absorbs water and so due to turgidity (swelling up) the pores get opened.
4. When the need gets satisfied, the guard cells lose the water. So, they tend to shrink. As a result, the pores are closed.

Question 15.
Draw a labeled diagram of cross-section of a leaf.
Answer:

Question 16.
How do plants obtain water (H₂O)?
Answer:
1. Roots of plants have specialized system to absorb underground or land water.
2. The water which is absorbed by roots reaches to the organs of the plants through xylem tissue.
3. In addition to this, the entire surface of plants is also able to absorb water Generally, plants absorb water through a process called osmosis.

Question 17.
State important functions performed by roots of terrestrial plants.
Answer:
1. In terrestnal plants, the root plays a crucial role of absorbing water and salts dissolved in the land water.
2. The root system also absorbs salts of nitrogen, phosphorus iron, magnesium, etc. from the soil.

Question 18.
What is the importance of nitrogen in plants? In which form do plants absorb nitrogen?
Answer
1. In plants, nitrogen is important for protein synthesis as well as for the production of other important substances.
2. Generally, plants absorb nitrogen in form of nitrates [N0O₃] or nitrite [NO₂] compounds.
3. Sometimes, the plants also take up atmospheric nitrogen prepared by symbiotic bacteria, in the form of organic compounds.

Question 19.
How is chlorophyll important for photosynthesis?
Answer:
1. Photosynthesis occurs in the chloroplast of a plant cell.
2. The chloroplast contains a pigment called chlorophyll to trap light energy.
3. The light energy of the sun is absorbed by the chlorophyll in the form of photons.

Question 20.
Define heterotrophic nutrition, and explain it.
Answer:
1. The term, ‘Hetero’ means ‘others’ and ‘Trophe’ means ‘nutrition’.
2. The mode of nutrition in which the living organisms cannot synthesis their own food material from the simple inorganic material they have and hence have to depend on other organisms for food is known as heterotrophic nutrition.

Question 21.
Describe the types of heterotrophic organisms on the basis of the type of food they consume.

Answer:
(a) Herbivores: These organisms feed on plants.
– Generally, the food source for organisms is stationary. For example, grass and plants. Herbivores such as cow and goat have to go to these sources to obtain nutrition.

(b) Carnivores: These organisms feed on other animals.

• The food source is mobile. For example, a lion has to hunt deer which is a moving source of nutrition.

Example: Frog, tiger, lizard, etc.

(c) Omnivores: These organisms feed on plants as well as animals.
Example: human beings, dog, ant, etc.

(d) Scavengers: These organisms feed on dead materials.
Example: Vulture

Question 22.
Describe the type of heterotrophic organisms on the basis of the mode of feeding. OR Discuss the strategies used to take up the nutrition.

(a) Saprophytic nutrition:

• Some organisms break down the complex food material outside their bodies and then absorb it. Such a mode of nutrition is called saprophytic nutrition.
• Saprophytes absorb soluble organic nutrients from the dead parts of the animals, plants, dung, etc.
  Example: Bacteria, fungi, yeast, mushrooms.

(b) Holozoic nutrition:
Some organisms undertake either a part or whole of animals or plants and then break down such sources inside their bodies. Such nutrition is known as holozoic nutrition. Example: Human beings, dog.

(c) Parasitic nutrition:
The organisms that live inside or outside of other organisms and obtain nutrition from them are known as parasites. The mode of their nutrition is called parasitic nutrition.

Example: Several bacteria, lice, tape warm, ascaris, cuscuta (plant).
Parasitic organisms derive nutrition from plants or animals without killing them.

Question 23.
Describe the process of nutrition in amoeba with the help of a figure.
Answer:
Amoeba is a unicellular organism. Its mode of nutrition is holozoic.
The various processes involved in nutrition are as follows :

(a) Ingestion: When amoeba comes in contact with food particle, it ingests the food by forming temporary finger-like extensions of the cell surface called pseudopodia.

• The food is then encaptured into a bag called food vacuole.

(b) Digestion: The complex food particles are broken down into simpler forms and digested in food vacuole with the help of digestive enzymes.
(c) Absorption: The digested food is absorbed directly from food vacuole into cytoplasm by the process of diffusion.
(d) Egestion: The undigested food is moved to the surface of the cell and thrown out.

Question 24.
Differentiate between autotrophic and heterotrophic nutrition.
Answer:

Question 25.
Describe the process of nutrition In paramoecium.
Answer:
1. Paramoecium is a unicellular organism whose cell has a definite shape. There is a specific spot in its cell. The entire body is covered with cilia which helps in bring the food to this specific spot.
2. The ingested (= entered) food material is then digested by paramoecium.

Question 26.
Differentiate between saprophytic and parasitic nutrition.
Answer:

Question 27.
Define digestion. What are the main functions of the human digestive system?
Answer:
In animals, the process of breaking-down the large and complex food material into small and simple absorbable molecules is known as digestion. The main functions of human digestive system are:

• Ingestion: To take in food
• Digestion: Converting food into small, simple and absorbable molecules
• Absorption: To absorb the digested food
• Egestion: To remove the undigested food from the body

Question 28.
Draw a labeled diagram of human digestive system and list out the main parts associated with the process of digestion.
Answer:
Human digestive system:
1. The function of digestive system is to break down larger molecules of food into smaller forms so that they can be absorbed easily into the body.
2. The human digestive system consists of alimentary canal and its associated glands.

The main organs of the human digestive system are:
Mouth, oesophagus, stomach, small intestine and large intestine.

The main glands associated with digestion are:
Salivary gland, liver and pancreas.

Question 29.
Explain the process of digestion in humans.
Answer:
Process of digestion In humans:
(1) Buccal cavity

(a) Mouth:

• Food is ingested through mouth.
• Humans put food in the mouth through their hands and the digestion starts simultaneously.
• The mouth (buccal) cavity contains teeth, salivary glands and tongue.

(b) Teeth:

• Teeth cut the food into small pieces and chew and grind it.

(c) Salivary gland:

• The salivary gland secretes saliva ¡n our mouth. It is a watery liquid and so it wets (lubricates) the food in mouth. It is easy to swallow wet food.
• The salivary gland also secretes an enzyme called amylase. Amylase breaks the complex molecule called starch present in the food into sugar.

(d) Tongue:

• The tongue does the job of properly mixing the food with saliva.
• The food remains for a very short time in the mouth and hence only a part of food gets digested here.

(2) Oesophagus:

• The partly digested food comes down to oesophagus or say food-pipe.
• The main function of the oesophagus is to push the food down to stomach.

(3) Stomach:

• Stomach is a large organ which expands when the food enters it. Moreover, the stomach releases certain juices that help in digestion.
• The muscular walls of the stomach churn the food thoroughly with these digestive juices.
• During this process, the food is broken down into smaller pieces and is converted into a semi-solid paste.
• The wall of stomach contains three tubular glands which secrete gastric juice.

The gastric juice contains:
(A) Dilute hydrochloric acid, (B) An enzyme called pepsin and (C) Mucus

(a) Dilute hydrochloric acid:

• Since the stomach releases dilute hydrochloric acid, the digestive juices are acidic in nature.
• The presence of acid enables the enzyme pepsin to digest protein present in the food.
• Therefore, the function of hydrochloric acid is to create an acidic medium in the stomach.
• It also kills the bacteria that enter the stomach through food.

(b) Pepsin:
Pepsin is an enzyme that helps in digesting protein and converting food into smaller molecules.

(c) Mucus:
The mucus prevents the damage that the hydrocholic acki may cause to the inner lining of the stomach.

(4) Small Intestine:

• The partly digested food then moves from stomach to small intestine with the help of sphincter muscles. These muscles release the food in the small intestine in small parts.
• The small intestine is the longest part of the alimentary canal and hence an extensive coiled structure. Owing to this structure it can fit in a very compact space.
• The small intestine is the main site for complete digestion of carbohydrates, proteins and fats. It receives secretions from liver and pancreas for this purpose (whose functions are discussed below).

(a) Function of bile juice:

• Liver secretes bile juice.
• Bile is alkaline in nature and so it converts the acidic food coming from stomach into alkaline form so that enzymes of pancreas can act on it.
• In addition, the bile salts break the fats present in the food into small globules (droplets), which makes it easy for the enzymes to act and digest them.

(b) Function of pancreatic juice:
The pancreas secrete pancreatic juice, which contains enzymes such as amylase for digesting carbohydrates, trypsin for digesting protein and lipase for digesting fats.

(c) Function of intestinal juice:

• The glands of the wall of the small intestine secrete intestinal juice.
• The intestinal juice contains various enzymes, which complete the digestion of carbohydrates into glucose, proteins into amino acids and fats into fatty acids and glycerol.

Question 30.
Explain the process of absorption of food in human body.
Answer:
1. Absorption of food takes place in the small intestine after the food has been digested completely.
2. The inner wall of small intestine contains millions of small, finger-like projections called villi.
3. The presence of villi gives the inner walls of small intestine a very large surface area. This helps in the rapid absorption of digested food.
4. The villi possess numerous blood-vessels which take the absorbed food to each and every cell of the body.
5. The food absorbed in the cells is then used for obtaining energy, building new tissues and repairing old tissues.

Question 31.
Explain the process of assimilation of food and its egestion from the human body
Answer:
Assimilation :

• Blood receives food which is absorbed from the walls of small Intestine.
• Blood then carnes this food to all the parts of the body where it is assimilated by the cells.
• This assimilated food is used by all the cells for obtaining energy, growth arid repair of the body tissues.

Egestion:

• The unabsorbed food is sent to the large intestine.
• In the large intestine, most of the water of the undigested food is absorbed again by villi. This makes the undigested food almost solid.
• This solid form called faeces or stool is removed from the body through the anus.
• Sphincter muscles present in the anus regulate the removal of this waste material from the body.

Question 32.
What is the function of saliva?
Answer:
Salivary gland:
1. The salivary gland secretes saliva in our mouth. It is a watery liquid and so it wets (lubricates) the food in mouth. It is easy to swallow wet food.
2. The salivary gland also secretes an enzyme called amylase. Amylase breaks the complex molecule called starch present in the food into sugar.

Question 33.
Which gastric juices are released in our stomach? State their roles. OR Why does the stomach release hydrochloric acid?
Answer:
Stomach:
1. Stomach is a large organ which expands when the food enters it. Moreover, the stomach releases certain juices that help in digestion.
2. The muscular walls of the stomach churn the food thoroughly with these digestive juices.
3. During this process, the food is broken down into smaller pieces and is converted into a semi-solid paste.
4. The wall of stomach contains three tubular glands which secrete gastric juice.

Question 34.
Discuss the functions of Intestinal and pancreatic juices.
Answer:
Small intestine:
1. The partly digested food then moves from stomach to small intestine with the help of sphincter muscles. These muscles release the food in the small intestine in small parts.
2. The small intestine is the longest part of the alimentary canal and hence an extensive coiled structure. Owing to this structure it can fit in a very compact space.
3. The small intestine is the main site for complete digestion of carbohydrates, proteins and fats. It receives secretions from liver and pancreas for this purpose (whose functions are discussed below).

(a) Function of bile juice:

• Liver secretes bile juice.
• Bile is alkaline in nature and so it converts the acidic food coming from stomach into alkaline form so that enzymes of pancreas can act on it.
• In addition, the bile salts break the fats present in the food into small globules (droplets), which makes it easy for the enzymes to act and digest them.

(b) Function of pancreatic juice:
The pancreas secrete pancreatic juice, which contains enzymes such as amylase for digesting carbohydrates, trypsin for digesting protein and lipase for digesting fats.

(c) Function of intestinal juice:

• The glands of the wall of the small intestine secrete intestinal juice.
• The intestinal juice contains various enzymes, which complete the digestion of carbohydrates into glucose, proteins into amino acids and fats into fatty acids and glycerol.

Question 35.
What are villi (Singular — vlllus, Plural — vllli)?
Answer:
The Finger like projections or structures present on the inner surface of the small intestine are known as villi.
Function: Villi increases the surface area of intestine for better absorption of digested food.

Question 36.
How does food move In the alimentary canal?
Answer:
1. The food moves in the alimentary canal by a process called ‘peristalsis’.
2. The circular and longitudinal muscles of the alimentary canal undergo involuntary contraction and relaxation. This results in the wavy movement which is known as ‘peristalsis’.
3. Due to peristalsis, the food is pushed ahead and the passage behind the food gradually closes. This also prevents the backward flow of food.

Question 37.
Differentiate between gastric juice and bile.
Answer:

Question 38.
Explain how different organisms adopt different ways to release energy obtained from food material. OR Explain break-down of glucose by various pathways.
Answer:
1. The food material taken in during the process of nutrition is used in cells to provide energy for various life processes. The process of converting food into energy varies among organisms.

2. In any case, the first step is to break-down glucose which is a 6-carbon molecule, into two 3-carbon  molecule called pyruvate. This process takes place in the cytoplasm.

Further conversion of pyruvate:
(1) In absence of oxygen (anaerobic respiration):

• Conversion of pyruvate in absence of oxygen is called anaerobic respiration.
• In the absence of oxygen, the conversion of pyruvate may take place in two ways. They are —

(a) The pyruvate may be converted into ethanol and carbon dioxide.
This process takes place in yeast during fermentation.

(b) When our muscles lack oxygen, the pyruvate gets converted into lactic acid which is also a 3-carbon molecule.
When we over-stress our muscles they build up lactic acid which then results in cramps.

(2) In presence of oxygen (aerobic respiration):

• The conversion of pyruvate in the presence of oxygen is called aerobic respiration.
• In aerobic respiration, pyruvate breaks down in the mitochondria. This process breaks up the three-carbon pyruvate molecule to give three molecules of carbon dioxide. Water is also formed in this process.
• Aerobic process releases much greater energy as compared to aerobic process.
• The energy released during cellular respiration is immediately used to synthesize a molecule called ATP which is used to fuel all other activities in the cell.

Question 39.
What is respiration? What are its types?
Answer:
Respiration:
The process of releasing energy from food substances such as glucose, amino acids and fats under the control of enzymes so as to carry Out various life processes by an organism is known as respiration. There are two types of respiration:

• Aerobic respiration
• Anaerobic respiration

Question 40.
Explain anaerobic respiration.
Answer:
Anaerobic respiration:

• The respiration which takes place in the absence of oxygen is known as anaerobic respiration.
• It is seen in microorganisms like bacteria, yeast, fungi, endoparasites and muscle cells.

(a) In yeast (plant):

During anaerobic respiration in yeast (plants), carbon dioxide and ethanol are formed as end products.

(b) In animal medium (In muscle cells):

Question 41.
Explain aerobic respiration.
Answer:
Aerobic respiration :

• The respiration which takes place in the presence of oxygen is known as aerobic respiration.
• Most of the organisms show aerobic respiration.
•  It takes place in the cells and so it ¡s also called cellular or internal respiration.
• During aerobic respiration, digested food (glucose), in the presence of oxygen, is completely broken down into carbon dioxide and water.
• The energy released during this process is stored in the form of ATP.

The overall equations is as under:

Question 43.
Explain: Respiration In plants.
Answer:
Respiration in plants:
1.  All plants need energy for their physiological processes. Like animals, plants also show gaseous exchange (or say respiration).
2. In plants, gaseous exchange takes place through stomata and the large inter-cellular spaces.
3. The gaseous exchange takes place through the process called diffusion. Diffusion mainly occurs through the surface of the plant leaves which is in direct contact with atmosphere.
4. The gases that are exchanged by diffusion are

• oxygen and
• carbon dioxide.

5. These gases may enter into plant cells or may move out of the plant cells into atmosphere. The direction of diffusion (i.e. movement of gases) depends upon

• environmental conditions and
• requirement of the plant.

6. In general, at night, when there is no photosynthesis, the plants eliminate maximum carbon dioxide (CO₂) whereas during daytime plants release oxygen (O₂).

Question 44.
Why do plant release oxygen during day time and carbon dioxide at night?
Answer:
Exchange of gases or say diffusion of gases is the process of respiration in the plants.

• Plants perform photosynthesis during day time. So, whatever carbon dioxide (CO₂) is generate due to respiration gets used in the photosynthesis. Hence, there is no carbon dioxide remaining to release. As a result, plants only release oxygen.
• Photosynthesis does not occur at night and hence plants release carbon dioxide.

Question 45.
How do aquatic animals respire?
Answer:
1. For respiration, aquatic animals use oxygen dissolved in water.
2. The amount of oxygen dissolved in water is quite low as compared to amount of oxygen present in air. As a result, the aquatic animals have to put more effort to obtain oxygen required for respiration. So, the rate of breathing in aquatic organisms is quite fast as compared to terrestrial organisms.
3. Fishes take in water through their mouths and force it past the gills where the dissolved oxygen is taken up by the blood.

Question 46.
What are the basic characteristics of respiratory surface in terrestrial animals?
Answer:
Important characteristics of the respiratory surface of terrestrial animals:

• Terrestrial animals have large surface area of their bodies which remains in contact with oxygen rich atmosphere.
• This surface is very thin, fine and delicate. The thin walls facilitates diffusion of respiratory gases O₂ and CO₂ and also H₂O in gaseous form.
• Usually, this surface lies within the body so that it can be properly protected. As a result there are passages that carry the oxygen to this surface from outside.
• There is also a mechanism for moving the air in and out of this area where oxygen is absorbed.

Question 47.
List out few organisms and their respiratory organs.
Answer:
Few organisms and their respiratory organs:

Question 48.
Explain the process of respiration in human beings along with the functions of the respiratory organs and a labelled diagram.
Answer:
Human respiratory system consists of nostril, nasal cavity, pharynx, Iaryngopharynx, trachea, bronchi, lungs, and diaphragm.
The process of respiration and functions of various respiratory organs of human beings are as follows:

1. Nose and nostrils:

• Human respiratory system starts with nose.
• Air is taken into the body from nostrils.

2. Nasal cavity:

• Nostril opens in the nasal cavity which is lined by fine nair and mucus.
• When air passes through the nasal cavity, the cavity traps the dust particles and microbes of the air and hence prevent them from entering the nose. Thus, filtered air moves further.

3. Throat:

• The air then passes through the throat and reaches the lungs.
• Cartilaginous rings are present in the throat. These rings ensure that the air passage remains open.

4. Lungs:

• Lungs are the chief respiratory organs.
•  In the lungs, air passage divides into smaller and smaller tubes which finally end in balloon like structures called alveoli (or alveolar sacs).
• The walls of alveoli are thin and covered by blood capillaries. This provides a large surface for exchange of gases.

Question 49.
Explain the mechanism of breathing.
Answer:
1. The function of human respiratory system is to breathe in O₂ and to breathe out CO₂.
2. The alternate process of inspiration (inhalation) and expiration (exhalation) is known as breathing.

(A) Inspiration:

• During this process, we lift our ribs and flatten our diaphragm. As a result, the chest cavity expands.
• Due to this, the oxygen rich air from the atmosphere rushes into the lungs and fills the expanded alveoli sacs.
• The blood then takes this oxygen in the alveolar blood vessels and transports it to all the cells of the body.

(B) Expiration:

• The blood brings carbon dioxide from all the body cells and release into alveoli. The diaphragm relaxes i.e. moves up and the air containing carbon dioxide is pushed out of the lungs into the atmosphere through nostrils.
• This process is known as exhalation or expiration.

Question 50.
What is a respiratory pigment? Explain human respiratory pigment. Also state its importance.
Answer:
1. When the body surface of an animal is large, only diffusion pressure is not sufficient to transport oxygen to all the parts of the body. So, a specialized pigment called respiratory pigment does this work.
2. In human body, hemoglobin (Hb) which has a very high affinity for oxygen works as the respiratory pigment.
3. Hemoglobin is present in the red blood cells. It carries oxygen from the respiratory surface of the lungs to the body cells.

Question 51.
How is carbon dioxide transported towards lungs in human beings?
Answer:
Carbon dioxide is a respiratory gas to be exhaled out of the body. CO₂ is more soluble in water than oxygen. So, majority of CO₂ is transported in the dissolved form in blood plasma.

Question 52.
How are the lungs designed to increase the surface area for exchange of gases?
Answer:
Lungs:

• Lungs are the chief respiratory organs.
•  In the lungs, air passage divides into smaller and smaller tubes which finally end in balloon like structures called alveoli (or alveolar sacs).
• The walls of alveoli are thin and covered by blood capillaries. This provides a large surface for exchange of gases.

Question 53.
Differentiate between breathing and respiration.
Answer:

Question 54.
State the role of blood in transportation. What are the basic requirements to run this transport system?
Answer:
Role of blood in transportation:

• Blood is a red coloured connective tissue in liquid form.
• Blood consists of fluid medium called plasma in which the cells remain in suspended form.
• Plasma does the work of transporting food, carbon dioxide, salts and nitrogenous wastes in dissolved form.
• The red blood cells transport oxygen.

Basic requirements to run this system:

• A pumping organ — To push blood around the body
• A network of tubes — To reach all the tissues
• A proper system —To ensure that this network can run properly and be repaired if damaged

Question 55.
Explain the structure of human heart along with a figure.
Answer:
1. Human heart is a muscular pumping organ to push blood around the body.
2. Its size is just as much as our fist.

Structure of human heart:

3. A human heart contains four chambers.
4. Both oxygen and carbon dioxide are transported by blood. So, in order to prevent the mixing of oxygen rich blood with blood containing carbon dioxide, there are four chambers in the heart.

The upper two chambers:

• These chambers are known as atria (singular : atrium).
• The left chamber is known as the left atrium while the right chamber is known as right atrium.
• The walls of atria are thin.

The lower two chambers:

• The lower chambers are known as ventricles.
• The left chamber is known as the left ventricle while the right chamber is known as the right ventricle.
• The walls of the ventricles are thick.
• All the four chambers are separated from each other by partitions called septa.

Valves:

• There is a bicuspid valve that allows the oxygenated blood to flow from left atrium to the left ventricle.
• Also, there is a tricuspid valve that allows deoxygenated blood to flow from right atrium to right ventricle.
• These valves prevent backward flow of blood from ventricles to the atria i.e. they are one-way valves.

Question 56.
Explain the flow of blood in human heart.
(Note : Although all the points of this answer are to be written as printed, but to understand the flow properly, read all the points marked as a’ together and then points marked as ‘b’ together.]
Answer:

Initially, all the tour chambers of the heart are in relaxed state. This state of the heart is known as the diastolic stage.

• A-1. During this stage, the deoxygenated blood from various organs (except lungs) passes through superior and inferior venacava and reaches the right atrium.
• B-1. At the same time, the oxygenated blood from the lungs passes through pulmonary veins and reaches left atrium.

Now, both the atria contract and the following process takes place:
A-2. The tricuspid valve opens and deoxygenated blood from the right atrium is poured Into the right ventricle.
B-2. The bicuspid valve opens and oxygenated blood form the left atrium is poured into the left ventricle.

Now both the ventricles contract and the following process takes place:

A-3. Deoxygenated blood from the right ventricle passes from pulmonary valve and pulmonary arteries and reaches lungs.

• In lungs, CO₂ is removed from blood and O₂ from air is added into it.

B-3. Oxygenated blood from the left ventricle passes through aortic valve, aorta and pulmonary arteries and gets distributed to all the parts of the body (except lungs).

• Thus, the blood flows from right ventricle towards lungs and that from left ventricle towards all the parts of the body, expect lungs.
• Since the blood circulates twice through the heart, it is called double circulation.
• The separation of both types of blood in the heart allows a highly efficient oxygen supply system to the body.

Question 57.
With the help of a schematic diagram, show the transport and exchange of oxygen and carbon dioxide.
Answer:

Schematic representation of transport and exchange of oxygen and carbon dioxide

Question 58.
Explain the evolution of heart starting from fishes to mammals. OR Explain the structure of hearts of lower animals (Note: For this question, write point (a) and (b) only).
Answer:
(a) 2-Chambered heart (Fishes):

• The heart of fishes have only two chambers.
• The blood is pumped to the grills. It is oxygenated there itself and passed directly to the rest of the body.
• Thus in fishes, the blood passes only once through the heart during one cycle of passage through the body.

(b) 3-Chambered heart (Amphibians and reptiles):

• In amphibians and many reptiles, there are three chambers in the heart. The oxygenated and deoxygenated blood gets mixed in the heart of such animals.
• Such organisms do not need high amount of energy like humans and so they can tolerate some Schematic representation of transport and exchange of oxygen and carbon dioxide

(c) 4-Chambered heart (Birds and mammals including humans):

• Birds and mammals have a four chambered heart.
• The separation of both types of blood i.e. oxygenated and deoxygenated in the heart allows a highly efficient oxygen apply system to the body.
• Such a system of separation is extremely useful for birds and mammals because they require very high energy for maintaining their body temperature.

Question 59.
Write a short note on blood vessels.
Answer:
Blood vessels:
Closed hollow tubes which transport blood from the heart to different organs and from different organs to the heart are called blood vessels. There are three types of blood vessels. They are:
(1) Arteries,
(2) Veins and
(3) Blood capillaries

(1) Arteries:

• Arteries carry blood from heart to all the organs of the body.
• For this, heart pumps blood into the arteries under high pressure.
• To sustain this pressure, walls of arteries are thick.
• After reaching the organs and tissues, arteries divide into many smaller vessels known as arterioles and finally into capillaries in order to make the blood reach to all the cells.

(2) Veins:

• Veins collect blood from different parts of the body and bring it back to the heart.
• Since blood is not under pressure while getting collected from the organs, the walls of veins are thin.
• Veins have valves to prevent backward flow of blood to ensure that the blood flows only in one direction.

(3) Capillaries:

• Capillaries are blood vessels having just a single-layered thick cell wall.
• Exchange of materials between blood and its surrounding talks place through these capillaries.
• These capillaries join to form venules and then veins.

Question 60.
What role do platelets play when we are Injured and bleeding? OR How does blood clot durIng an injury?
Answer:
1. If we get injured and our blood vessels rupture, the blood starts leaking form these vessels. This is called bleeding.
2. Due to leakage, the pressure of the blood will decrease. This will in turn decrease the efficiency of the pumping system i.e. the heart.
3. To avoid such a situation, blood contains platelet cells. The platelet cells keep on circulating around the body and plug the leaks in the vessels, if any. This clots the blood.

Question 61.
What is lymph? Write a short note on lymphatic system.
Answer:
1. Lymph is a colourless fluid which is also involved in transportation. It is also called tissue fluid. It consists of plasma and proteins.
2. Lymph does not contain red blood cells and so it is pale and colourless. Also, it contains less protein compared to blood.

Lymphatic system:

• The lymphatic system is a part of the immune system as well as circulatory system.
• Lymphatic system consists of lymph, lymph vessels, lymphatic capillaries and lymphatic nodes.

Flow of lymphatic system:

• Lymphatic flow begins in the areas around blood capillaries. In these areas, small amount of tissue fluid drains into lymphatic capillaries through its pores.
• The lymphatic capillaries then drain into larger lymph vessels that look like thin, transparent veins. Finally these veins open in large veins.

Functions of the lymphatic system:

• Lymph vessels carry digested and absorbed fats from intestine.
• It collects intercellular fluid through the medium of lymph vessels and returns ¡t to blood circulation.
• The system also protects against diseases.

Question 62.
Differentiate: Blood and lymph
Answer:

Question 63.
Differentiate: Atria and ventricles
Answer:

Question 64.
Differentiate: Arteries and veins
Answer:

Question 65.
State the need of transportation system in plants.
Answer:
Need of transportation in plants:
1. Different types of substances that are absorbed or synthesized in one part of the body are transported to the various parts of the body. This is known as transportation.
2. Under photosynthesis, plants convert solar energy into chemical energy by utilizing carbondioxide and water.
3. Plants also need other substances which they take up separately by the means of roots. Roots absorb these substances from soil.
4. If the distance between roots and leaves is small, the raw material and energy that roots and leaves possess can easily reach all the parts of the body through simple process called diffusion.
5. But If the distance is long, process of diffusion becomes inefficient in transportation. In such a situation, a proper transport system is required. However, since plants do not move from one place to another, their energy requirement is lesser compared to animals.

Plants have two transport systems. They are:
(A) Xylem — For transporting water and minerals
(B) Phloem — For transporting food material produced by the plants

Question 66.
Explain transport of water In higher plants.
Answer:
Transport of water In higher plants:
1. Higher plants possess xylem for transporting water to all its parts.
2. Xylem contains tracheids and vessels.
3. Xylem tissues of all the organs of a plant are connected end to end with each other to form a network of conducting tubes.

Process of transport:
(a) Through diffusion:

• The cells of the roots are in direct contact with the soil and they take up ions from the soil.
• Due to this, a difference is created between the concentration of ions in the roots and that of ions in the soil.
• The ions present in the soil water are at higher concentration and so the water moves up from soil to the roots through osmosis.
• This water movement creates a water column under which water is steadily pushed upwards.

(b) Through conduction (transpiration):

• In higher plants transporting water till the highest point of the plant with this system is inefficient.
  As a result, higher plants take help of a process called transpiration.
• The transpiration or evaporation of water from the leaves create a suction which pulls up water from the xylem vessels.
• Thus the process of transpiration helps in the upward movement of water from roots to the leaves through the stem.
• Since the stomata are open during the daytime, the transpiration pull becomes a major driving force in the movement of water in the xylem.

Question 67.
Explain transpiration in brief.
Answer:
Transpiration:
1. The loss of water in the form of water vapour from the aerial parts of the plant is known as transpiration.
2. Plants absorb the water through the roots and transport it to all its parts.
3. This absorption is in fact driven by transpiration that occurs through the tiny pores on the leaves called stomata.
4. Transpiration creates a suction which pulls the water from the xylem cells of roots.
5. Thus, we can say transpiration plays an important role in making it possible for all the parts of the plant to receive water.
6. Transpiration also helps in regulating temperature.

Question 68.
What is translocation? Explain translocation of substances through the phloem tissue. OR How are food and other substances are transported In plants?
Answer:
Transportation of food and other substances:
1. The transportation of food and other substances i.e. the soluble products of photosynthesis is known as translocation.
2. Phloem is the chief tissue for conducting transportation.
3. Phloem transport the products of photosynthesis, some amino acids and other substances.
These products are transported from leaves to the other plant organs, while amino acids and the other products are transported towards the storage organs of roots, fruits, seeds and also to the growing parts of the plant.
4. Sieve tubes and companion cells are important components of the phloem. They can transport the food material in both directions i.e. upward and downward.
5. Sometimes translocation needs energy in the form of ATR in general, molecules like sucrose are transferred into phloem with the energy obtained from ATP. Here, the osmotic pressure in the cell increases, which pushes the food to the adjacent cell.
6. Interestingly in the spring season, the stored food in root or stem is transported to the buds to help them grow.

Question 69.
Differentiate: Xylem and phloem
Answer:

Question 70.
Differentiate: Transportation and translocation
Answer:

Question 71.
What is excretion? Discuss its need.
Answer:
Excretion:

• The harmful substances produced during several biochemical reactions are known as excretory substances.
• The biological process of removing these harmful nitrogen-based metabolic wastes in liquid form is known as excretion.
• Most excreted substances leave the body in the form of urine. Some of them are also lost in sweat and some in air that we breathe out.

Need for excretion:

• Body cells perform biochemical processes for sustaining life.
• During these processes, both useful as well as harmful toxic substances are produced.
• Accumulation of harmful substances in the body can harm the body and so they need to be removed from time to time.
• The process of removing excretory substances is simple in unicellular organisms.
• Unicellular organisms remove the excretory substances by simple diffusion from the body surface in the surrounding water.
• However, excretion is a complex process in multi-cellular organisms and so they possess special organs and even organ-system to perform excretion.

Question 72.
Explain excretory system of humans with a labeled diagram. OR Discuss the removal of urine from the body.
Answer:
Human excretory system is made up of the following organs:
(a) A pair of kidneys,
(b) A pair of ureter,
(c) A urinary bladder and
(d) A urethra

• Kidneys are located in abdomen on either side of abdomen. Kidney is bean shaped.
• Urine produced in the kidneys passes through the ureters into urinary bladder.
• The urinary bladder opens into urethra. Finally, urethra removes the urine outside the body.

Question 73.
Describe the structure of nephron, with labeled diagram.
Answer:
Nephron:

• Nephron is the main functional unit of kidney.
• Each human kidney possesses about 10 lakh nephrons.

Structure of nephron:
Bowman’s capsule:

• Each nephron has a double wailed cup-shaped bag at its upper end which is called Bowman’s capsule.
• The Bowman’s capsule contains a mass (bundle) of blood capillaries which is called glomerulus.

Tubule:

• The lower end of the Bowman’s capsule is called tubule.
• The part of tubule which is near the Bowman’s capsule is quite small and is called the neck.
• After the neck, the tubule becomes very narrow and coiled.
• This region of tubule consists of
  (A) A proximal convoluted tubule,
  (B) A Henle’s loop and
  (C) A distal convoluted tubule.
• The posterior end of nephron is known as collecting tube.
• Collecting tubule opens into renal pelvis.
• The renal pelvis opens into ureter.

Question 74.
Explain urine formation in human kidneys.
Answer:
1. The purpose of making urine is to filter out waste products from the blood.
2. Urine contains nitrogenous wastes such as urea or uric acid which are removed from blood in the kidneys.
3. The renal arteries bring the waste material and the blood from the body to the kidney.
4. This blood is filtered out from the blood capillaries into the owman’s capsule.
5. This filtration takes place under heavy pressure and is known as ultrafiltration.
6. Each kidney has large numbers of filtration units called nephrons packed close together. The filtrate passes through the tubular parts of nephron.
7. During this, useful substances such as glucose, amino acid, salts and a major amount of water are reabsorbed by blood capillaries that surround the nephron.
8. The amount of water re-absorbed depends on how much excess water does the body contains and how much of dissolved waste is yet to be excreted.
9. The remaining fluid contains excretory substances called urine.
10. The urine passes from ureter to urinary bladder and gets stored there.
11. When the bladder is filled with urine it expands and creates a pressure which then creates an urge to pass the urine out of the body via, urethra.

Question 75.
How is urine produced?
Answer:
1. The purpose of making urine is to filter out waste products from the blood.
2. Urine contains nitrogenous waste such as urea or uric acid which are removed from blood in the kidneys.
3. It is then no surprise that the basic filtration unit in the kidneys, like in the lungs, is a cluster of very thin-walled blood capillaries.
4. Each capillary cluster in the kidney is associated with the cup-shaped end of a coiled tube called Bowman’s capsule that collects the filtrate
5. Each kidney has large numbers of these filtration units called nephrons packed close together.
6. Some substances in the initial filtrate, such as glucose, amino acids, salts and a major amount of water, are selectively re-absorbed as the urine flows along the tube.
7. The amount of water re-absorbed depends on how much excess water there is in the body and on how much of dissolved waste there is to be excreted.
8. The urine forming in each kidney eventually enters a long tube, the ureter, which connects the kidneys with the urinary bladder.
9. Urine is stored in the urinary bladder until the pressure of the expanded bladder leads to the urge to pass it out through the urethra.
10. The bladder is muscular, so it is under nervous control, as we have discussed elsewhere.
11. As a result, we can usually control the urge to urinate.

Question 76.
Explain excretion in plants.
Answer:
Excretion in plants:
1. For plants, oxygen is one of the end products of photosynthesis and can also be considered as a waste product.
2. Plants emit this oxygen into atmosphere through diffusion process.
3. Plants get rid of excess water through transpiration.
4. For other waste products, many of the plant tissues contain dead cells in themselves.
5. Some of the material is also lost by falling leaves.
6. In the plant cell, vacuoles are the excretory organelle.
7.  Around the root system in the soil, plant excretes some of the waste products.
8. Peeling of the bark is also the example of plant excretion.
9. Resins and gums are excretions of plants.

Question 77.
Life on the earth depends on the sun. Give reason.
Answer:
1. All living beings need energy for their survival. They constantly need energy.
2. Food is the source of energy. This food directly or indirectly comes through the green plants.
3. All green plants trap solar energy or light energy and convert it in the form of food through the process of photosynthesis. Thus, the sun is the base of life on the earth.

Question 78.
Two green plants are kept separately in oxygen free containers, one In the dark and the other in continuous light. Which one will live longer? Why?
Answer:
1. The plant which is kept in dark will not be able to conduct photosynthesis. So, the container will soon be filled with the carbon dioxide released by the plant. Hence, the plant will die.
2. On the other hand, the plant kept in light would be able to carry out photosynthesis and thus convert the produced carbon dioxide into oxygen. Hence, this plant would live longer.

Question 79.
Leaves of a healthy potted plant were coated with Vaseline. Will this plant remain healthy for long? Give reasons for your answer.
Answer:
The plant covered with Vaseline will not remain healthy for long because Vaseline would make an impervious layer on the leaves. This will cause the following effects:
(a) Plant will not get oxygen for respiration.
(b) It will not get carbon dioxide for photosynthesis.
(c) Upward movement of water and minerals will be affected due to lack of transpiration.

Question 80.
Why mucus is secreted along with HCl in the stomach?
Answer:
Mucus makes a protective layer on the innermost layer of the stomach and protects it from the effect of HCl.

Question 81.
Why is small intestine in herbivores longer than in carnivores?
Answer:
1. Herbivores mainly eat green plants. So, cellulose found in green plants forms the largest part of an herbivore’s food.
2. Digesting cellulose takes quite a long time. Hence, herbivores have longer small intestine in which the food stays for a longer duration in order to get digested properly.
3. Carnivores do not eat plants and so their diet does not contain cellulose. So, the small intestine is small in them.

Question 82.
Why does absorption of digested food occur mainly in the small intestine?
Answer:
Maximum absorption of digested food occurs In small Intestine due to following reasons:
(a) Digestion is completed in small intestine.
(b) Inner lining of small intestine is provided with villi which increase the surface area to ensure better absorption.
(c) Wall of intestine is richly supplied with blood vessels which take the absorbed food to each and every cell of the body.

Question 83.
Hydrochloric acid Is an important constituent of gastric juice. Give reason.
Answer:
1. The stomach secretes gastric juices from its gastric glands for the chemical digestion of food.HCl is one of the constituents of gastric juice.
2.  Hydrochloric acid destroys the bacteria and other microorganisms that enter along with the food and thus prevents the decay of food in stomach.
3. HCl creates acidic medium in the stomach so that the gastric enzyme can act properly.

Question 84.
Name the following —
(a) The process in plants that links light energy with chemical energy
(b) Organisms that can prepare their own food
(c) The cell organelle where photosynthesis occurs
(d) Cells that surround a stomatal pore
(e) Organisms that cannot prepare their own food
(f) An enzyme secreted from gastric glands in stomach that acts on proteins.
Answer:
(a) Photosynthesis
(b) Autotrophs
(c) Chloroplasts
(d) Guard cells
(e) Heterotrophs
(f) Pepsin

Question 85.
What is a residual volume of air in lung? What is its importance
Answer:
1. During the breathe-in cycle, when air is taken in and let out, some amount of air remains in the lungs. This is known as a residual volume of air in lungs.
2. The residual volume of air in lungs is important to provide sufficient time for oxygen to be absorbed and for carbon dioxide to be released. This way the exchange of respiratory gases can be carried out continuously.

Question 86.
Name the energy currency ¡n the living organisms. When and where is It produced?
Answer:
1. ATP (Adenosine Triphosphate) is the energy currency in the living organisms. It Is produced at the end of respiration in mitochondria.
2. The energy released during respiration is used to make an ATP molecule from ADP and inorganic phosphate.

Question 87.
The rate of breathing in aquatic organisms much faster than in terrestrial organisms. Give reason.
Answer:
1. Aquatic organisms like fishes obtain oxygen from water present in dissolved state through their gills.
2. The amount of dissolved oxygen is quite low in water as compared to the amount of oxygen in the air. So, the breathing-rate of aquatic organisms is much higher.
3. Terrestrial organisms use the oxygen present in the atmosphere for respiration. Oxygen is available in plenty in the atmosphere and so terrestrial organisms do not need to breathe very fast.

Question 88.
Give the main points of difference between respiration in plants and respiration in animals.
Answer:

Question 89.
Differentiate: Inhalation and Exhalation
Answer:

Question 90.
The walls of ventricles are thicker than atria. Give reason.
Answer:
1. The human heart is a constantly pumping organ.
2. The blood flows in very high force in ventricles as compared to atria. Hence, the walls of ventricles are thicker than atria.

Question 91.
What is the need of transportation in human beings? Why circulatory system is a need of human being?
Answer:
1. The circulatory system is very important to transport various substances required for life processes.
2. Circulatory system is important for transporting respiratory gases between lungs and body cells.
3. It is important to transport nutrients (digested) from alimentary canal to body cells.
4. It is important to transport excretory substances towards kidneys.
5. It is also important to transport some of the enzymes and various hormones.

Question 92.
Describe the functions of various blood cells.
Answer:
Functions of various blood cells:
(1) Red blood cells: They contain hemoglobin. Hence, they transport O₂ from lungs to the body cells.
(2) White blood cells: They play an important role to provide immunity to the body. They protect the body from micro-organisms.
(3) Blood platelets: When any blood vessel gets cut, the platelets clot the blood and prevent bleeding.

Question 93.
What is the importance of the structure of septum in human heart?
Answer:
1. Human heart (and also birds) require high amount of energy. They also need to maintain their body temperature constantly. So, they need constant energy supply and hence constant supply of oxygen.
2. The septum separates the right side and the left side of the heart. As a result, the oxygenated blood and deoxygenated blood remain in separate chambers and do not mix.
3. Such a separation made by septum enables a very efficient supply of oxygen to the body cells.

Very Short Answer Type Question:

Question 1.
What is the Importance of life processes?
Answer:
To maintain the body functions of the living organism and to reproduce.

Question 2.
Name important life processes to maintain life.
Answer:
(i) Nutrition
(ii) Respiration
(iii) Transportation
(iv) Excretion
(v) Control and Co-ordination
(vi) Movement and
(vii) Reproduction.

Question 3.
What happens when the guard cells lose water?
Answer:
They shrink and cause the pore to close

Question 4.
In which from is carbohydrate stored?
Answer:
In plants, the carbohydrates are stored in starch form, while in animals, the carbohydrates are stored in glycogen form.

Question 5.
Which is the most common source to provide energy to the living organisms?
Answer:
Carbohydrates are the most common source for providing energy to the living organisms.

Question 6.
Give the summary of photosynthesis in terms of a reaction.
Answer:
6CO₂ + 12H₂O + Chlorophyll + Sunlight → C₆H₁₂O₆ + 6O₂+ 6H₂O

Question 7.
Name the raw materials of photosynthesis.
Answer:
CO₂. Water, Sunlight and Chlorophyll (a plant pigment)

Question 8.
Give any two examples of plant parasites.
Answer:
Cascuta and loxanthus

Question 9.
Give any two examples of animal parasites.
Answer:
Liverfiuke and plasmodium

Question 10.
What is the source of oxygen liberated during pbtosynthesis?
Answer:
Water

Question 11.
Give examples of plant with variegated leaves.
Answer:
The examples of plant with variegated leaves are money plant and crotons.

Question 12.
What is the site of photosynthesis?
Answer:
Chloroplast found in plant leaf

Question 13.
Define Translocation.
Answer:
Transfer of products of photosynthesis and some other materials through phloem is called as translocation.

Question 14.
A few drops of Iodine solution were added to rice water. The solution turned blue-black in colour. What does this indicate about rice water?
Answer:
The blue-black colour of rice water confirms the presence of starch

Question 15.
Which is the first enzyme to mix with food in the digestive tract?
Answer:
Amylase

Question 16.
What Is the importance of assimilated food in human body?
Answer:
(i) As a fuel to get energy,
(ii) As a material for growth and repair of the body.

Question 17.
Define Ingestion.
Answer:
The process by which the organisms take in the food is called ingestion.

Question 18.
Define Egestion.
Answer:
The process of removal of undigested food from the body is called egestion.

Question 19.
Define Peristalsis.
Answer:
The muscles of alimentary canal contract in a designed rhythm to push the food forward, This is called peristalsis.

Question 20.
In nasal cavity, which substance traps the dust particles and microbes present in the air?
Answer:
Mucus

Question 21.
What is the reason for muscular cramps after hard work?
Answer:
During hard work (or exercise) muscles undergo partial breakdown of glucose and forms lactic acid. This lactic acid is accumulated in the muscles and causes muscular cramps.

Question 22.
Give two examples of animals who breathe through their cell membranes?
Answer:
Amoeba. planarian and paramecium.

Question 23.
What is the average breathing rate in an adult man?
Answer:
The average breathing rate In an adult man is 15 to 18 breathe per minute.

Question 24.
What is the name of the area through which the exchange of respiratory gases occurs in a woody stem?
Answer:
Woody stems of plants/trees have lenticels for exchanging respiratory gases.

Question 25.
Which valve regulates the flow of blood from left atrium to left ventricle?
Answer:
Bicuspid value

Question 26.
Define Transpiration.
Answer:
The loss of water in the form of vapour through the aerial parts of the plants is called transpiration.

Question 27.
Give the location and function of a companion cell.
Answer:
The companion cells are located just next to sieve tube cells n a phloem tissue. They have nucleus and many other important organelles. They control the function of the sieve tube cell.

Question 28.
Define Pulmonary cIrculation.
Answer:
In human circulatory system, the blood circulatory system which transports blood between the lungs and the heart Is called as pulmonary circulation.

Question 29.
Define Systemic circulation.
Answer:
In human circulatory system, the transportation pathway of blood between various organs (except lungs) and heart is known as systemic circulation.

Question 30.
What is the pulse rate of a healthy adult person?
Answer:
The pulse rate of a healthy adult person is 72 per minute in resting position.

Question 31.
What are the mediums of circulation In human body?
Answer:
Blood and lymph are the mediums of circulation in human body.

Question 32.
Arrange these body parts starting from the lower side of the body.
I. Urter, II. Urethra, Ill. Kidney, IV. Renal vein
Answer:
III, I. Il, IV

Question 33.
Define Osmoregulatlon,
Answer:
The process of regulating waste contents and ion concentration in the body is called osmooegulation.

Question 34.
Name two chemical waste products of the human body.
Answer:
The two chemical waste products of the human body are urea and uric acid.

Question 35.
What is Urethra?
Answer:
Urethra is a tube through which urine is passed out from the body.

Question 36.
Define Dialysis.
Answer:
The artificial procedure used for cleaning the blood of a person by separating the waste substance (urea) from it is called dialysis.

Fill in the Blanks

1. Plants store food in the form of ……………..
Answer:
Starch

2. Each stoma consists of minute pore surrounded by a pair of …………..
Answer:
Guard cells

3. Guard cells are present in ……………. of leaves.
Answer:
Stomata

4. In amoeba, the process of obtaining food is called ………….
Answer:
Phagocytosts

5. The digested food found in food vacuoles of amoeba is absorbed directly into cytoplasm through ………….
Answer:
Diffusion

6. ………….. is the last step in the respiration process of human beings.
Answer:
Release of carbon

7. The process of food ingestion in amoeba is called …………
Answer:
Phagocytosis

8. ……………. is the largest gland of the body.
Answer:
Liver

9. Pepsin digests………….
Answer:
Proteins

10. In human beings ……….. is a respiratory pigment.
Answer
Hemoglobin

11. When air is blown from mouth into a test-tube containing lime water, the lime water turns milky due to the presence of …………
Answer:
Carbon dioxide

12. During respiration exchange of gases take place in ……………..
Answer:
Alveoli of lungs

13. Lack of oxygen in muscles often leads to cramps among cricketers. This results due to…………..
Answer:
Non-conversIon of glucose to pyruvate

14. The scientific name of voice box of humans is……………..
Answer:
Larynx

15. Blood is a …… liquid tissue.
Answer:
Connective

16. The ………. blood from various organs of the body is received by the right atrium.
Answer:
De-oxygenated

17. Capillaries join together to form ………….
Answer:
Veins

18. The normal rate of heart beat is ………………..
Answer:
72/mm

19. The normal blood pressure is …………..
Answer:
120/80 mmHg

20. The instrument to measure blood pressure is called ……………..
Answer:
Sphygmomanometer

21. The extracellular fluid which always flows front body tissues to the heart is called ………………..
Answer:
Lymph

22. At night ……………. is important for water transportation in plants.
Answer:
Root pressure

23. In nephrons, the structure of tubule after the neck is ………………..
Answer:
narrow and coiled

24. The chief nitrogenous waste products in the human beings are and ………………..
Answer:
Urea; uric acid

25. Oxygen liberated during photosynthesis comes from………………..
Answer:
water

26. Life on the earth depends on based molecules.
Answer:
Carbon

27. Plants excretory product(s) other than O₂ and CO₂ is/are …………………
Answer:
Resins and Gums

True Or False

1. A plant which is not growing visibly is dead. — False
2. Most of the food sources on earth are oxygen based. — False
3. Bacteria can produce their own food. — True
4. The opening and closing of the stomatal pore depends upon oxygen in guard cells. — False
5. The green dots on plant leaves are chloroplasts. — True
6. In amoeba, the undigested food moves to the surface of the cell. — True
7. Lipase is present in pancreatic juice. — True
8. During respiration in humans glucose is oxidized. — True
9. Pepsin is a protein digesting enzyme — True
10. Exit of food from stomach into small intestine as well as from anus to outside the body is regulated by sphincter muscles. — True
11. The internal (cellular) energy reserve in autotrophs is glycogen. — False
12. During deficiency of oxygen in tissues of human beings, pyruvic acid is converted into lactic acid in the chloroplast. — False
13. The direction of diffusion is not fixed. It depends upon the environmental conditions and the requirements of plants. — True

14. Among plants, humans and fishes, plants use oxygen at the fastest rate whereas fishes at the lowest rate. — False
15. Plants have specific respiratory pigments to in take oxygen from the atmosphere. — False
16. In pisces, the heart does not pump oxygenated blood to different parts of the body. — True
17. The blood leaving the tissues becomes richer in oxygen. — False
18. The left atrium receives oxygenated blood from the lungs through pulmonary vein. — True
19. There is independent pathway for transportation in plants. — False
20. Movement of material in xylem occurs through physical fores whereas movement of material in phloem occurs by utilizing energy. — True
21. Complex multi-cellular organisms remove the excretory substances by diffusion. — False
22. Bowman’s capsule brings the waste material along with the blood to the kidneys. — False

Match the Following :

Question 1.

Answer: (A – R), (B – P), (C – Q)

Question 2.

Answer: (A – Q), (B – R), (C – P)

Question 3.

Answer: (A – R), (B – Q), (C – P)

Question 4.

Answer: (A – R), (B – Q), (C – S), (D – P)

Haryana State Board HBSE 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 5 Periodic Classification of Elements

Question 1.
How many elements are known to man? Why did he classify them?
Answer:
1. Today, man has discovered 118 elements.
2. All the elements have different properties. These elements are either used individually or in combination with other elements to form an endless variety of substances.
3. Classifying the elements help us to understand their properties and produce various products.

Question 2.
Explain Dobereiner’s classification of elements or Dobereiner’s triads.
Answer
1. In 1817, German chemist Johann Wolfgang Dobereiner started classifying elements on the basis of their chemical properties.
2. He discovered that there exists several ‘triads’ or ‘groups of three elements each’ that appeared to have similar chemical properties.
3. Law of Triads: If three elements are arranged in the increasing order of their atomic masses, the atomic mass of the intermediate (i.e. second) element would be almost equal to the average of atomic masses of first and third elements. This is known as Law of Triads. It was given by Dobereiner.

Based on this principle he identified the following three triads:

• Lithium, Sodium and Potassium,
• Chlorine, Bromine and Iodine,
• Calcium, Strontium and Barium.

Example:

• Atomic mass of Lithium (Li) is 6.9 u, Sodium (Na) is 23.0 u and that of potassium (K) is 39.0 u.
• Here, the atomic mass of sodium is the average of lithium and potassium i.e. = ((6.9 + 39)/2) = 23. Hence, these three elements form a triad.

Question 3.
The elements in the Dobereiner’s table were arranged in such a manner that the atomic mass of the intermediate (i.e. second) element would be almost equal to the average of atomic masses of first and third elements. Demonstrate this with the help of Dobereiner’s triads.
Answer:
1. Table given below shows Dobereiner’s triacis.
2. We can see that in all these three triads, the atomic mass of the second element is equal or nearly equal to the average of atomic mass of first and third element.

Question 4.
State the limitations of Doberelner’s classification of elements.
Answer:
Limitation of Dobereiner’s classification :

• Under Dobereiner’s classification, overall only a limited number of elements could be classified into triads.
• After arranging the elements in triads, it was found that there were certain other elements which could not be classified by Dobereiner’s method.

Question 5.
Explain Newlands’ method of classifying elements (or Newlands’ Law of Octaves)
Answer:
Newlands’ Law of Octaves :
1. When elements are arranged in the increasing order of their atomic masses, properties of every 8th element are found to be similar to the properties of the first element.
2. In 1866, a scientist named John Newlands arranged the elements in the increasing order of their atomic masses.
3. Newlands was the first person to arrange elements in their increasing order of atomic masses.
4. During this arrangement, he found properties of every 8th element to be similar i.e. property of 1st and then 8th element would be similar. Similarly property of 2nd and 9th element would be similar and so on.
5. He called this periodicity pattern as the Law of Octaves (octaves = eight). The law can be compared with the octaves or say 8 notes found in music where in the 8th note is similar to the 1st note.

Question 6.
Discuss limitations of Newlands’ Law of Octaves. OR How can you say that Newlands’ table worked well with lighter elements only?
Answer:
Limitation of Newlands’ Law of Octaves:
(1) The law of octaves was applicable only upto calcium. After calcium, every 8th element did not possess properties similar to that of 1st
(2) Newlands thought that there were only 56 elements in nature. He also thought that no more elements would be discovered in the future. However, later, several new elements were discovered that could not be arranged in the table as per Newlands’ law.
(3) In order to fit elements any how into his table, Newlands adjusted two elements in the slot even if the properties of elements did not match with other elements.

Question 7.
How did Mendeleev drill to arrange the elements into his Periodic Table? OR How did Mendeleev construct his Periodic Table?
Answer:
When in 1869, Russian chemist Dmitri Ivanovich Mendeleev started his work of classifying elements, 63 elements were known to man.
Mendeleev started by examining the relationship of
(a) ‘Atomic mass of an element with
(b) ‘Physical property of the element’ and
(c) Chemical property of the element.

(i) Classification on the basis of chemical properties:

• For classifying the elements on the basis of chemical properties, Mendeleev started studying the compounds that a particular element formed with oxygen as well as hydrogen.
•  A compound formed when an element combines with oxygen is called oxide whereas with hydrogen is called hydride. (For example, CaO, NaH, K₂H, etc.) The formulae of such oxides and hydrides were taken as ‘one of the basis of classification’.
• Mendeleev selected hydrogen and oxygen because they are very reactive and form compounds with most elements.

(ii) Classification on the basis of atomic mass:

• Mendeleev took 63 cards (1 card per element) and on each card he wrote the ‘atomic mass’ as well as ‘properties’ of a particular element.
• Mendeleev arranged the cards in the increasing order of atomic masses of elements starting from atomic number 1 to 63. Then he put the cards that showed similar properties into single group. This way he arranged elements into 8 groups. Each of these 8 groups were further divided in two sub-blocks, A and B.
• Hence, while arranging the elements in the increasing order of their atomic masses, he also arranged them in groups on the basis of their chemical properties.
• Through this classification, he concluded that “The properties of elements are periodic function (i.e. periodic in nature) of their atomic mass.” This law came to be known as Mendeleev’s Periodic Law.

Mendeleev’s observation:

1. Mendeleev observed that most of the elements got a place in his periodic table.
2. The elements got arranged in the order of increasing atomic masses.
3. Elements having similar physical and chemical properties occur periodically.

Question 8.
What observations did Mendeieev make while classifying the elements into his Periodic Table?
Answer:
Mendeleev’s observation:

1. Mendeleev observed that most of the elements got a place in his periodic table.
2. The elements got arranged in the order of increasing atomic masses.
3. Elements having similar physical and chemical properties occur periodically.

Question 9.
Which criteria did Mendeleev use for developing his periodic table?
Answer:
Criteria used by Mendeieev for developing periodic table:

• The properties of elements are the periodic function of their atomic masses. Hence, arranging elements in the increasing order of their atomic masses.
• Elements with similar properties are arranged in the same group.
• The formula of oxides and hydrides formed by an element.

Question 10.
Discuss the anomalies (irregularities) of Mendeleev’s Periodic Table.
Answer:
Anomalies (irregularities) of Mendeleev’s Periodic Table:
(1) Sequence of few elements was inverted:

•  In Mendeleev’s table, the elements were arranged in increasing order of their atomic masses. However, at few instances Mendeleev did not follow this arrangement.
• At few places he placed elements with slightly higher atomic mass before elements with lesser atomic mass. This means he inverted his sequence. He did this so that elements with similar properties could be grouped together.

(2) Gaps were kept at few places:

• Mendeleev had left some blank spaces in his Periodic Table. He was very sure that there exists some elements that would fit these spaces.
• Mendeleev named those blank places with a Sanskrit numeral prefix called ‘Eka

Example:
‘Eka’ means one place below something. So, Eka-aluminium means one space (left blank) below aluminium. In this manner scandium (Sc), gallium (Ga) and germanium (Ge) discovered later had properties similar to Eka-boron, Eka-aluminium and Eka-silicon respectively.

Question 11.
Discuss the achievements of Mendeleev’s Periodic Table.
Answer:
Achievements of Mendeleev’s Periodic Table:
(1) The gaps that Mendeleev had left in his table got properly filled when new elements were discovered.
(2) Chemists not only accepted Mendeleev’s table but also conferred him as the ‘originator of the concept’ on which Modern Periodic Table is based.
(3) Noble gas elements such as helium (He), neon (Ne), etc. were not known during Mendeleev’s time. So, those elements were not present in his table. But, Mendeleev’s table was so precisely designed that when these gases were discovered they were easily placed in a new separate group without disturbing the existing order.

Question 12.
Elements in Mendeleev’s Periodic Table were arranged in increasing order of their atomic masses.
Then why did Mendeleev place cobalt Co (atomic mass 58.93) before element nickel Ni (atomic mass 58.71) when In fact atomic mass of nickel was lesser than cobalt?
Answer:
1. Although mass of cobalt (58.93) was slightly more than Nickel (58.71), even then cobalt was placed before Nickel. Mendeleev did this so that he could maintain the properties of element placed in one group.
2. The properties of Co, Rh and Ir are similar and that of Ni, Pd and Pt are similar. Hence, Mendeleev placed Co before Ni so that elements Co, Rh and Ir having similar properties could be arranged in the same group.

Question 13.
Compare the properties of Eka-aluminium and gallium to sho the brilliant prediction Mendeleev’s table had.
Answer:
1. During Mendeleev’s time, gallium was not discovered. But, Mendeleev proposed properties of elements that could be filled in blank spaces that he left in his table. One such element was eka-aluminium.
2. The properties given by Mendeleev for Eka-aluminium and the actual element gallium that later replaced Eka-aluminium were quite similar. This can be seen in the table below.

Question 14.
Why did Mendeleev leave gaps in the Periodic Table?
Answer:
Gaps were kept at few places:

• Mendeleev had left some blank spaces in his Periodic Table. He was very sure that there exists some elements that would fit these spaces.
• Mendeleev named those blank places with a Sanskrit numeral prefix called ‘Eka

Example:
‘Eka’ means one place below something. So, Eka-aluminium means one space (left blank) below aluminium. In this manner scandium (Sc), gallium (Ga) and germanium (Ge) discovered later had properties similar to Eka-boron, Eka-aluminium and Eka-silicon respectively.

Question 15.
Discuss limitations of Mendeleev’s classification.
Answer:
Limitation of Mendeleev’s classification:
(1) Position of hydrogen:
Mendeleev had placed hydrogen (H) element in group 1 i.e. group of alkalis. He did so because like alkali metals, hydrogen combines with halogens and oxygen and sulphur to form compounds having similar formulae.

Contrary to this hydrogen also showed properties similar to elements of group 17 i.e. group of halogens. Thus, hydrogen could be placed in group 17 as well. Due to this confusion hydrogen could not be assigned a fixed place in Mendeleev’s Periodic Table.

(2) Position of isotopes:

• Isotopes are atoms of the same element having similar chemical properties but different atomic masses.
• Several elements have isotopes. For example, hydrogen has three isotopes and all have separate masses. Even then all these three were placed in one single position as hydrogen element H.

(3) Wrong order of some elements:

Mendeleev had arranged the elements on the basis of increasing order of atomic masses however, he broke this rule for certain elements. For example. he placed cobalt (Co) having larger mass before nickel (Ni) having lesser mass. This made it difficult to predict how many elements can be discovered between such two elements.

Question 16.
State the three limitations of Mendeleev’s Periodic Table.
Answer:
(1) The position of hydrogen could not be correctly assigned.
(2) The table could not assign proper position for the isotopes of various elements.
(3) Some elements were not arranged on the basis of their increasing atomic mass. This posed a question as to how many elements could still take the place between two elements.

Question 17.
What is a periodic table?
Answer:
Periodic table:

• The Periodic Table is a chart in which all the elements known to us are arranged in a systematic manner.
• A Periodic Table is divided into rows (periods) and columns (groups).
• Elements having similar properties are placed in the same group (i.e. vertical column).

(For Information only: Why is Periodic Table called so?

• During classification of elements it was found that the elements show similar physical and chemical properties after a fix interval or say ‘period’ i.e. the elements are periodic in their nature.
• The table was then prepared based on the periodic nature or periodicity of elements and so the table is called Periodic Table.)

Question 18.
How was Periodic Law of Mendeleev Improved Into Modern Periodic Law?
Answer:
1. Mendeleev had arranged the elements in his table on the basis of increasing atomic masses. In 1913, Henry Moseley showed that, rather than ‘atomic mass’, ‘atomic number’ is a ‘better fundamental property’. Henry said, atomic number and not atomic mass determine the chemical properties of elements.
2. Looking to the facts, the Periodic Law given by Mendeleev was modified into Modern Periodic Law which states “Properties of elements are a periodic function of their atomic number”.

Question 19.
State Mendeleev’s Periodic Law and Modern Periodic Law.
Answer:
(a) Mendeleev’s Periodic Law:
Properties of elements are the periodic function of their atomic masses.

(b) Modern Periodic Law:
Properties of elements are a periodic function of their atomic numbers.

Question 20.
On what basis were the elements arranged in the Modern Periodic Table? Why?
Answer:
1. The elements in the Modern Periodic Table were arranged on the basis of Modern Periodic Law which states that elements are a periodic function of their atomic numbers.
2. Atomic number (Z) tells us the number of protons in the nucleus of the atom of an element.
3. This number i.e. the atomic number goes on increasing as one moves from one element to the next.
4. Thus, in Modern Periodic Table the elements were arranged in the increasing order of their atomic number Le. Z.

Question 21.
Discuss the arrangement of elements In groups and periods in Modern Periodic Table.
Answer:
The Modern Periodic Table has 18 vertical columns known as ‘groups’ and 7 horizontal rows known as ‘periods
(a) Placement of an element in a group:
(i) Elements are placed in a particular group on the basis of the ‘valence electrons’ i.e. elements having same number of valence electrons will be placed same group.

Example:

• The electronic configuration of fluorine (f) is 2, 7 and that of chlorine (Cl) is 2, 8, 7. This means both these elements have valency = 1. Hence, both these elements are placed in same group i.e. group no. 17.
• Note that as we go down in a group, the number of shells increase. Fluorine has only 2 shells whereas chlorine has 3.

(b) Placement of an element In a period:
(i) Within a horizontal period, as one moves from left to right, the ‘elements have same number of shells’ but, ‘different valence electrons
(ii) Moreover, on moving left to right, the number of electrons of valence shell increase by 1 unit because the elements are arranged in the increasing order of their atomic number by 1 unit.
(iii) Thus, elements having same number of shells are placed in the same period. For example, Na, Mg, Al, Si, etc. have 3 shells and hence are placed In the 3rd period.

Question 22.
How are electrons arranged in various shells? How do we determine the number of elements in each period on the basis of number of elements in a shell?
Answer:
Arrangement of electrons in a shell:
1. The maximum number of electrons that can be accommodated in a shell depends on the formula 2n² (where n = the number of given shells from the nucieusý’
2. Letter ‘K’ denotes the first shell, ‘L’ second, M’ third and so on. Thus for shell ‘K, n = 1, for ‘L, n = 2 and for ‘M’, n = 3.
3. By knowing the values of ‘n’ for K, L and M we can derive the number of electrons that each of these shells can hold. It is:

• K shell = 2(n)² = 2(1)² = 2 electrons
• L shell = 2(n)² = 2(2)² 2(4) = 8 electrons
• M shell = 2(n)² = 2(3)² = 2(9) = 18 electrons

4. Based on this calculation we say that the 1st period has 2 elements and period has 8 elements.
5. Although M shell can hold 18 electrons but third period can hold only 8 electrons. Therefore, 3 period can hold only 8 elements.

Question 23.
What do you mean by periodic properties?
Answer:
1. The properties which are determined by the electronic configuration of elements or which depend on the electronic configuration of elements are known as periodic properties.
2. Also, the properties which show a recurring gradation within the same group or along a period are known as periodic properties.
Example: Valency, atomic radii (atomic size), metallic property, electronegativity, etc. exhibit periodic properties.

Question 24.
Bring out the trends of change In valency, atomic size and metallic and non-metallic properties in groups and periods.
Answer:

Question 25.
What is valency? How is it calculated?
Answer:
1. Relative ability of an element to combine with other element is known as a valency. OR Valency is the combining capacity of an atom of an element to acquire noble gas configuration.
2. Valency depends on the number of valence electrons that an atom of an element has.
3. The valency of an element is

• Either equal to the number of electrons in the valence shell OR
• Equal to eight minus the number of electrons in the valence shell.

Question 26.
Explain the trend of valency within a period and a group. OR How does the valency vary in a period on going from left to right and while going down In a group?
Answer:
1. As you move in the period from left to right, the valency first increases from 1 to 4. Then it decreases from 4 to 0.
2. All the elements in a given group possess equal number of electrons. Hence, valency does not change within a group.

Question 27.
Explain atomic size (atomic radius).
Answer:
Atomic size:

• The radius of the atom is called the atomic size of the atom.
• Atomic size (or radius) can be visualized as the distance between the centre of the nucleus and the outermost shell of an isolated atom.
• Atomic radius is expressed in angstrom (A°), or centimeter (cm) or picometer (pm). 1 pm = 10 12m.
• For example, the atomic radius of hydrogen atom is 37 pm.

Question 28.
Why does atomic radii increase as we go down in a group, while decrease as we move from left to right in a period? OR Explain the trends in atomic size of elements In a period and a group.
Answer:
Situation in a group :
1. In a group as we move from top to down, new orbits get added in the elements.
2. For example, in the first group, Lithium (Li) has 2 orbits, Sodium (Na) has 3 orbits, Potassium (K) has 4 orbits and so on.
3. Since the orbits increase as we move down, naturally their radii also increase.

Situation in a period :

• In a period, as we move form left to right, no new orbits are added unlike in group.
• Secondly, in the period, the positive electric charge of nucleus attracts more electrons and so the radii of atoms decrease.
• Thus, as we move from left to right in a period, the atomic radii decreases.

Question 29.
Explain the trends of the metallic character in a period and a group.
Answer:
Trend of the metallic character in a group:
On moving down in a group, the effective nuclear charge experienced by valence electrons decreases because the outermost electrons are far away from the nucleus. Therefore, tendency of the element to lose electrons increases and electrons can be lost easily. Hence, the metallic character increases on moving down in a group.

Trend of the metallic character In a period:

• On moving from left to right in a period the effective nuclear charge experienced by valence electrons increases.
• Therefore, the tendency to lose electrons will decrease.
• Thus metallic character decreases in a period on moving from left to right.

Question 30.
Explain the trends of the non-metallic character in a period and a group.
Answer:
Trend of the non-metallic character in a group:
On moving down within a group, the non-metallic character decreases.

Trend of the non-metallic character in a period:

• On moving from left to right in a period, the nuclear charge experienced by valence electrons increases. So, the tendency to attract electrons increase.
• As a result, the non-metallic character increases in a period on moving from left to right.

Question 31.
What are metalloids or semi-metallic elements? Give example.
Answer:
1. Elements which possess properties of both metals and non-metals are known as metalloids or semi-metallic elements.
2. In the Modern Periodic Table, a zig-zag line separates metals and non-metals. The border line elements such as boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te) and polonium (Po) on this zig-zag line are known as metalloids or semi-metals.

Question 32.
Give difference between ‘Elements of a group’ and ‘Elements of a period’
Answer:

Question 33.
Possibility of groups instead of triads arose. Explain.
Answer:
1. The method of classifying elements on the basis of triads was given by Dobereiner.
2. Under this method, Dobereiner stated that if three elements having similar chemical properties are arranged in their increasing order of atomic masses, the atomic mass of the intermediate (second) element would be similar to the average of atomic masses of first and third element.
3. However, under this classification, overall only a limited number of elements could be classified and so this method failed later on.
4. It was also discovered that more elements could be added to these triads.
5. For example, fluorine could be added to the triad of chlorine, bromine and iodine. Similarly magnesium could be added to the triad of calcium, strontium and barium.
6. Hence, scientists thought that rather than classifying elements only in triads, perhaps they could be arranged in larger groups.

Question 34.
Newlands’ law of arranging elements came to be known as ‘Law of Octaves’. Give reason.
Answer:
Newlands’ Law of Octaves :
1. When elements are arranged in the increasing order of their atomic masses, properties of every 8th element are found to be similar to the properties of the first element.
2. In 1866, a scientist named John Newlands arranged the elements in the increasing order of their atomic masses.
3. Newlands was the first person to arrange elements in their increasing order of atomic masses.
4. During this arrangement, he found properties of every 8th element to be similar i.e. property of 1st and then 8th element would be similar. Similarly property of 2nd and 9th element would be similar and so on.
5. He called this periodicity pattern as the Law of Octaves (octaves = eight). The law can be compared with the octaves or say 8 notes found in music where in the 8th note is similar to the l note.

Question 35.
Differentiate between groups and periods of Modern Periodic Table.
Answer:

Question 36.
Two elements X and Y belong to groups 1 and 3 respectively In the same period. Compare these elements with respect to their —
Answer:
(a) Metallic character (b) Size of atoms (c) Formulae of their oxides and chlorides
(a) As one moves in a period from left to right, the metallic character decreases. Hence, the metallic character decreases while moving from X to Y.
(b) As one moves in a period from left to right, the atomic radius decreases. Hence, atoms of element Y are smaller than that of element X.
(c) Oxides of element X and Y: X₂O, X₂O₃, Chlorides of element X and Y: XCl, YCl₃

Question 37.
Atomic number of magnesium is 12. What information can you get from this? (Hint: Keep In mind the Modern Periodic Table.)
Answer:
1. The atomic number of magnesium is 12. So, its electronic configuration is (2, 8, 2).
2. From this, we can say that it is positioned in the third period arid second group in the Periodic Table.
3. There are 3 orbits (2, 8, 2) and there are 2 electrons in its outermost orbit.
4. It has 2 electrons in the last orbit and so it will lose these 2 electrons and hence become mg²⁺ ion. Thus, its vaiency is +2.

Question 38.
Elements of which groups of the Periodic Table can easily lose electrons and elements of which groups of the Periodic Table can easily gain electrons?
Answer:
1. Elements of group IA, IIA, IIIA and that of group IB. IIB and IIIB will easily lose electrons.
2. Elements of group IVA, IVA, VIIIA and VIIIB will neither lose nor gain electrons.
3. Elements of group VA, VIA, VIIA and that of group VB, VIB and VIIB will easily gain electrons.

Question 39.
The position of three elements A, B and C in the Modern Periodic Table is shown below:

Giving reason for the following statements in one or two sentences.
(a) Element A is a metal.
(b) Atom of element C has a larger size than atom of element B.
(c) Element B has a valency of 3.
Answer:
(a) Element A has 1 valence electron. So it can lose this electron and become electropositive. Hence, we can consider element A as metal.
(b) Element B belongs to 2nd period and so has 2 shells whereas element C belongs to 3rd period and so it has 3 shells. As a result, atoms of element C are bigger than atoms of element B.
(c) Element B belongs to group 13 and so its valency is 3.

Question 40.
The atomic number of elements A,B, C, D and E are given below:

From the above table, answer the following questions:
(a) Which two elements are chemically similar? (b) Which is an inert gas? (c) Which elements belong to 31 d period of Periodic Table? (d) Which element among these is a non-metal?
Answer:
(a) Element C and D have similar electronic configuration, i.e. 2, 8, 2 and 2, 2. Hence, these two elements are similar.
(b) Electronic configuration of element B is 2, 8. Hence, it has 8 electrons in its outermost shell. So, element B is an inert gas.
(c) Atomic number of C is 12 and so its electronic configuration is (2, 8, 2) i.e. it has 3 shells. Hence, element C belongs to 3rd period of the Modern Periodic Table.
(d) Element A (2, 5) is non-metal.

Question 41.
What is meant by periodicity? Why are the properties of elements in the same group similar?
Answer:
1. The properties of the elements n the Modern Periodic Table depend on regular changes that are seen in the electronic configuration of elements arranged in groups and periods. This law is known as periodicity.
2. Valency decides the properties that an element possesses.
3. The electronic configuration of the valence orbit of all the elements in a group is same.
4. Thus, in the same group, properties of an element remains same.

Question 42.
The electronic configuration of an element ‘X’ is 2, 8, 8, 2. To which (a) period and (b) group of the Modern Periodic Table does ‘X’ belong? State its valency. Justify your answer in each case.
Answer:
(a) The electronic configuration of element X states that the element has four shells namely K, L, M and N. Hence, element X belongs to 4th period.
2. There are 2 electrons in the outermost i.e. N shell and so element X belongs to group 2.

Question 43.
Atomic number is considered to be a more appropriate parameter than atomic mass for classification of elements in a periodic table. Why?
Answer:
1. Atomic mass does not properly tell us the properties of elements. Moreover, the elements cannot be arranged in the increasing order of atomic mass because of certain anomalies.
2. The properties of elements depend upon the number of electrons present in the valence shell. The number of electrons can be known by its atomic number. In this regard, an element can be classified in a better way through atomic number.
3. Hence, atomic number is considered to be a more appropriate parameter than atomic mass for classification of elements in the Periodic Table.

Question 44.
From the part of a Periodic Table, answer the following questions:

(a) Atomic number of oxygen is 8. What would be the atomic number of Fluorine?
(b) Out of ‘X’ and ‘Q’ which element has larger atomic size. Give reason for your answer.
(c) Out of ‘Y’ and ‘Z’ which element has smaller atomic size. Give reason for your answer.
Answer:
(a) In the Modern Periodic Table, elements are arranged on the basis of increasing order of their atomic numbers. Fluorine comes after oxygen and so atomic number of fluorine is 9.
(b) As one moves from, left to right in a period, the atomic size of the element goes on decreasing. Hence, atomic size of element Q is smaller than that of element X.
(c) As one moves down from one period to another, the numbers of shells increase. Element Y lies above element Z. So, element Y will have lesser shells than Z and hence Y’s atomic size will be smaller.

Question 45.
The atomic number of an element ‘X’ is 20. Write —
(a) Its valency
(b) Whether it Is a metal or non-metal
(c) The formula & compound formed when the element ‘W reacts with an element ‘Y’ of atomic number 8.
Answer:
(a) Since the atomic number of X is 20, its electronic configuration will be 2, 8, 8, 2. Hence, its valency is 2.
(b) Its valency is 2, so it belongs to group 2 i.e. group of metals. Moreover, since its valency is 2. it will lose 2 electrons which means it shows the property of metal.
(c) Atomic number of Y is 8. So its electronic configuration is (2, 6). In other words, valency of Y is 2. Moreover valency of X is also 2. So, the compound formed by sharing 2 electrons will be compound ‘XV’.

Question 46.
An element X belongs to 13th group of periodic table. State its valency. What will be the formula of its sulphate?
Answer:
Element X belonging to group 13th has 3 valence electrons and so its valency is 3. Sulphate of X = X₂(SO₄)₃.

Question 47.
The elements Li, Na and K, each having one valence electron, are in period 2,3 and 4 respectively of Modern Periodic Table.
(a) In which group of the periodic table should they be?
(b) Which one of them is least reactive?
(c) Which one of them has the largest atomic radius? Give reason to justify your answer in each case.
Answer:
(a) All these elements have 1 valence electron and so they belong to Group 1.
(b) Li is the least reactive. Since Li is 2nd period it has only 2 shells. So, its outermost orbit is very near the nucleus. As a result, is difficult to remove the electron. This makes it least reactive of the three elements.
(c) Element K lies in 4th perioded has 4 shells. Hence, it has largest atomic radius of the three.

Question 48.
Sudha madam asked students to place three elements namely lithium, sodium and potassium in one group or say triad. Now, answer the following questions.
(a) How could she place all the three elements in one group?
(b) What is the name of this group? State the properties of the elements of this group.
Answer:
(a) Sudha madam placed the three elements in the same group because all three elements have similar properties.
(b) The name of this group is ‘alkali metal group’. Properties of the elements of this group are-

• All these elements are metals.
• Valency of each element is 1.
• These metals readily react with water and form alkalis and hydrogen gas.

Question 49.
In the Science class this morning, Prakash sir taught students how Newlands classified the elements known to him. He then gave the students an assignment to role-play Newland. Prakash sir gave them some elements along with their atomic weights which are as follows.

The students were to answer the following questions. Study them and draft your answer.
(a) Which two elements will have similar properties on the basis of Newland’s law of octaves? Justify the reason for your selection.
Answer:
Newlands had arranged the elements in the increasing order of their atomic weights. So, first we arrange the elements.

As per Newland’s law of Octaves, if elements are arranged in the increasing order of the atomic weights then every 8th element will have properties similar to the just one. Hence, element ‘a’ having atomic mass 2 will have properties similar to element ‘d’ having atomic mass 23. Similarly, element ‘b’ and ‘g’ will can be grouped together.

Question 50.
Introduction to the groups of elements of the Modern Periodic Table

Introduction:

118 elements are known to us. Majority of these elements are metals. These metals are not very reactive. They are malleable and ductile and good conductors of heat and electricity. The remaining elements are non-metals and semi-metals (metalloids).

Brief explanation of each group of the Modern Periodic Table is as follows:

(1) Group 1 elements (Alkali metals): Elements of this group are known as ‘Alkali metals’. There are in total 6 alkali metals from Lithium to Francium. Alkali metals are very soft and highly reactive metals. They are so reactive that they have to be stored in substances like oil because if kept open they will start reacting with atmospheric gases.

If they are not found freely then how do we obtain them? Well, chemists extract them from their compounds. Can explode when exposed to air The metals are malleable, ductile and good conductors. They have only 1 valence electron. They are always eager to lose this 1 electron and join with other electron to form a positive ion or say cation.

(2) Group 2 elements (Alkaline earth metals): Elements of this group are known as ‘alkaline earth metals’. There are in total 6 elements from Beryllium to Radium. Alkaline metals are also reactive metals but not as reactive as alkali metals. They have 2 electrons in their outer shell. So, they tend to lose 2 electrons to form positive ions.

(3) Group 3 to Group 12 (Transition metals (elements)): Elements of this group are known as ‘Transition metals’. They are moderately reactive. They are malleable and ductile. They have high melting and boiling points

(4) Group 13 to 16: Group 13 to 16 contains a mix of metals, metalloids and non-metals.

Metalloids: Metalloids are elements that have few properties of metals and few of non-metals.
Metalloids are the smallest class of elements (the other two classes of elements are metals and nonmetals). There are just seven rnetalloids namely boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb). tellurium (Te) and polonium (PO).

(5) Group 17 (Halogens): Halogens are non-metals. Examples of halogens are Fluorine (F), chlorine (Cl), Bromine (Br), Iodine (I), etc. The term “halogen” means salt-former” and compounds containing halogens are called “salts”.All halogens have 7 electrons in their last shell. So, they gain 1 electron They are highly reactive with alkali metals (group 1) arid alkaline metals (group 2).

(6) Group 18 (Noble gases): The 18th group (extreme right group) consists of non-metal elements known as noble gases or inert gases. The valence (outermost) shells of all noble gases are completely filled with electrons (2 for Helium,8 for all others). Hence, the noble gases are very stable and extremely less reactive.There are 6 noble gases that occur naturally. They are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and the radioactive radon (Rn).

(7) Inner-transItion metals: Inner-transition metals are the elements listed below the Modern Periodic Table. These are 30 rare earth metals.

Why are they placed below the table and not In the table?
If you look at the Modern Periodic Table, you will notice that after the element Lanthanum (atomic number 57), the next element that appears is Hafnium (atomic number 72). This means that the elements with atomic number 58 to 71 are missing. Similarly, after actinium (atomic number 89) the next element that occurs is Rutherfordium (atomic number 104).

The missing elements having atomic number 58 to 71 are placed below the table and are called lanthanides because they exhibit properties similar to element Lanthanum (atomic number 57). Similarly, the missing elements having atomic number 90 to 103 are placed below the table and are called actinides because they exhibit properties similar to element actinium (atomic number 89)

The inner-transition metals were discovered quite recently. Scientists and chemists are yet to determine their exact characteristics and properties. Hence, it was decided to keep these elements separate below the Modern Periodic Table.

(8) The case of hydrogen — an important point: Hydrogen is placed in group 1. However, its position is a matter of discussion. The reason for it follows.
Hydrogen shows similarity to group 1 elements i.e. alkalis as well as to group 17 elements i.e. halogens.

(a) Hydrogen’s similarity with alkalis: Hydrogen has only 1 electron in its valence shell. All the electrons that have 1 electron in their outer shells are placed in group 1 For example, lithium (2, 1), sodium (2, 8, 1) and so on. This means, hydrogen’s electronic configuration resembles that of alkali metals and hence it is placed in group 1. Hence, both alkalis and hydrogen have the tendency to lose 1 electron and become positive ion. The way alkali metals react with halogens (i.e. elements of group 17), oxygen and sulphur, hydrogen also reacts in a similar way. Hence, hydrogen is placed in the grotip of alkalies.

(b) Hydrogen’s simIlarity with halogens: The hydrogen also shows properties to halogens i.e. elements of group 17. All the elements of halogen group are 1 electron short of attaining their octet configuration. For example, Fluorine (2, 7),chlorine (2, 8, 7), etc. Hydrogen with electronic configuration of (1) is also 1 short of attaining octet configuration. So, the way the halogens require 1 electron to complete their configuration and become stable, hydrogen also require 1 electron to complete its configuration.

Hence, both halogens and hydrogen have the tendency to accept 1 electron and become negative ion. Halogens exist as diatornic molecules i.e. the molecule of halogens is made up of 2 atoms. For example, fluorine exists as, F₂, chlorine as C₁₂ and so on. Similarly, hydrogen also exists as a diatomic molecule i.e. as H₂. Just like halogens, hydrogen also combines with metals and non-metals to form covalent compounds. Considering the above points we conclude that hydrogen also resembles halogens.

(c) Conclusion: Since hydrogen shows properties similar to alkanes (group 1 elements) and halogens (group 2 elements). its position in the periodic table is till date contradictory and hence not fixed. However, we place hydrogen in group I simply for the purpose of convenience (technically, it can be placed in halogen group as well.

Very Short Answer Type Question :

Question 1.
What is the need to classify elements?
Answer:
We can learn a lot about the properties and characteristics of elements when they are systematically classified. Hence, there arose a need to classify them.

Question 2.
What were elements first classified into?
Answer:
When elements were classified for the first time they were classified as metals and non-metals.

Question 3.
Who was Dobereiner?
Answer:
John Wolfgang Dobereiner was a German chemist who in 1817 tried to arrange the elements with similar properties into groups.

Question 4.
What are Dobereiners Triads?
Answer:
Dobereiner identified some elements that could be grouped into groups of three on the basis of the atomic mass of elements. Such groups were called Dobereiner’s Triads.

Question 5.
State the principle of Dobereiner’s Triads.
Answer:
When three elements are arranged in the order of increasing atomic masses, the atomic mass of middle element is roughly the average of the atomic masses of other two elements.

Question 6.
The three elements A, B and C with similar properties have atomic masses X, Y and Z respectively. The mass of Y is approximately equal to the average mass of X and Z. What is such an arrangement of elements called as? Give one example of such a set of elements.
Answer:
Such an arrangement of elements is called Dobereiner’s triads.

Example:
Lithium (ll): atomic mass = 6.9, Sodium (Na): atomic mass = 23, Potassium (K): atomic mass = 39
Atomic mass of Na =\(\frac{39+6.9}{2}\)= 23

Question 7.
Can Na, Si, Cl form Dobereiner’s triad?
Answer:
Although the atomic mass of Si Is approximately the average of atomic masses of Na and Cl. still these three elements cannot form Dobereiner’s triads because the properties of these elements are different.

Question 8.
What can be considered as Newland’s biggest discovery?
Answer:
Newland’s thought that on arranging elements on the basis of increasing atomic masses, every 8th element has properties similar to the 1st can be considered his biggest discovery.

Question 9.
Fill the table given below with first seven elements as arranged by Newland’s.

Answer:

Question 10.
State Newland’s Law of Octaves.
Answer:
When elements are arranged in the order of increasing atomic masses, the properties of the eight elements (starting from a given element) are repetitions of the properties of the first element.

Question 11.
As per Patel Sir, Newland Law of Octaves was applicable only to tighter elements. Did he pass a correct statement? Why?
Answer:
Yes, Reason: Newland’s Law of Octaves is applicable only to lighter elements having mass upto element calcium. After calcium, the 1st and 8th element do not have similar properties.

Question 12.
Define Mendeleev’s Periodic Law.
Answer:
The properties of elements are a periodic function of their atomic masses,

Question 13.
What was the base of classification adopted by Mendeleev?
Answer:
(a) Arranging elements on the basis of increasing atomic masses.
(b) Grouping elements that have similar properties.

Question 14.
Why did Mendeleev leave some gaps in his periodic table?
Answer:
Mendeleev predicted that the gaps will be occupied by elements that would be discovered in some time future.

Question 15.
Magnesium and calcium are metals. Is this enough to classity both these elements into same group? Why?
Answer:
No, Reason: Elements should be put into same group on the basis of some fundamental properties that are common to such elements.

Question 16.
Mendeleev had knowledge about how many elements?
Answer:
63

Question 17.
What did Mendeleev focus on while considering chemical properties as one of the bases of classification?
Answer:
While considering chemical properties as one of the bases of classification, Mendeleev concentrated on compounds that the elements formed by combining with oxygen and hydrogen.

Question 18.
Why did Mendeleev consider compounds that elements formed with hydrogen and oxygen as a base of classification?
Answer:
Hydrogen and oxygen are very reactive and they form compounds with most elements. This provides a great help in classifying the elements. Hence,……….

Question 19.
Why were noble gases discovered very late?
Answer:
Elements are discovered on the basis of their reactivity with oxygen. Noble gases are inert i.e, not reactive. Hence, it was quite difficult to imagine existence of such elements.

Question 20.
Why hydrogen dId not find a suitable position in Mendeleev’s Periodic Table?
Answer:
Hydrogen is an element which shows properties similar to the elements of group 1 as well as to the elements of group 17 ie. halogens. Hence, till date where should hydrogen be placed is not correctly determined even in the Modern Periodic Table.

Question 21.
What are isotopes?
Answer:
Atoms of an element having same atomic number but different atomic masses are called isotopes.

Question 22.
Define valency.
Answer:
The combining capacity of an atom of an element to acquire noble gas configuration is called valency.

Question 23.
State Modern Periodic Law.
Answer:
Properties of elements are a periodic function of their atomic number.

Question 24.
What are groups in a Modern Periodic Table?
Answer:
The 18 vertical columns of the Modern Periodic Table are called groups.

Question 25.
What are periods?
Answer:
The 7 horizontal rows of the Modern Periodic Table are called periods.

Question 26.
Look at the table given below.

To which group do these elements belong? Wily?
Answer:
All these elements belong to Group 1 because every element has one valence electron in its outermost shell.

Question 27.
Loot at the table below.
Answer:

State the electronic configuration of each element. To which period will these elements belong? Why?
Answer:
1. Mg – 2, 8, 2, Si – 2, 8, 4, S – 2, 8, 6 Ar – 2, 8, 8
2. Electrons of all these elements are arranged in 3 shells and so all lie in 3rd period.

Question 28.
An element Q lies in group 2 whereas element R lies in group 15 of the periodic table. Now fill the table.

Answer:

Question 29.
Why all the elements in spite of belonging to one same period have different properties?
Answer:
Within a period the number of valence electrons is different for each element. So, the tendency to lose or gain electrons for each element varies. Hence …………

Question 30.
Define atomic size (radius).
Answer:
The distance between the centre of the nucleus and the outermost shell of an isolated atom is called atomic radius or atomic size.

Question 31.
What are metalloids?
Answer:
Elements that have some properties of metals and some of non-metals are known as metal bids.

Question 32.
State the number of valence electrons and valency of most of the elements of group 18.
Answer:
Generally, the elements of group 18 have 8 valence electrons and their valency is O (zero).

Question 33.
An element Q lies in group 2 of the periodic table. State the formula of its chloride and oxide.
Answer:
Chloride — XCl₂; Oxide XC.

Question 34.
State the formula of the compound formed when an element X of group 14 combines with element Y of group 16.
Answer:
Electronic configuration of X : 2, 8, 4
Electronic configuration of Y : 2, 8, 6
∴ X will combine with 2 atoms of Y.
∴ The formula of compound formed will be XY₂.

Question 35.
Write the positions of metallic, non-metallic and metalloid elements in the Modern Periodic
Table.
Answer:
In the Modern Periodic Table, metallic elements are on the left side while non-metallic elements are on the right side. Metalloids or semi-elements are in the middle of the Periodic Table.

Question 36.
The construction of Modern Periodic Table owes to research of which scientist?
Answer:
(A) Sir Ramsay
(B) Moseley
(C) Lord Rayleigh
(D) Mendeleev

True or False

1. ………… elements are known to us and out of them are available naturally.
Answer:
118; 98

2. Elements with similar properties were arranged in groups for the first time in the year …………
Answer:
1817

3. It was ………… who found that properties of elements are a periodic function of their atomic masses.
Answer:
Newlands

4. As per the ‘Law of Octaves’, lithium shows properties similar to …………
Answer:
Sodium

5. Newland’s Periodic Table ended at element number named ………… named …………
Answer:
56; Thorium

6. The main credit for classifying elements goes to …………  (write full name).
Answer:
Dmitri Ivanovich Mendeleev

7. While considering chemical properties as one of the parameter for classifying the elements, Mendeleev focused on compounds that the elements formed with ……………
Answer:
Hydrogen and oxygen

8. Element X has atomic number 20. It should find its place in ………… period.
Answer:
Fourth

9. Element that took place of Eka-silicon was …………
Answer:
Germanium

10. Atomic mass of Eka-aluminium: 68; Atomic mass of Eka-gallium……………
Answer:
69.7

11. Hydrogen’s properties resemble ………… to ……….. and (Name of groups).
Answer:
Alkalis; halogens

12. In 1913 …………….. showed that the atomic number of an element is a more fundamental property than its atomic mass.
Answer:
Henry Moseley

13. Atomic number is denoted as ……………….
Answer:
‘Z’

14. The Modern Periodic Table takes care of limitations of Mendeleev’s Periodic Table. (state the number)
Answer:
Three

15. With respect to group number, hydrogen can be placed in
Answer:
Group number 1 and also 7.

16. The number of shells that can be accommodated in a shell can be calculated using the formula ………..
Answer:
2n²

14. Atomic radius is measured in ………….
Answer:
Pico meter (pm)

18. 1 pm = ………………………
Answer:
10^-12m

19. Noble gases are also known as ……………..
Answer:
Inert gases

20. Metallic elements have …………….. electrons n their outermost orbit.
Answer:
1 to 3

21. If you move from right to left in a period, the atomic radius will ……………..
Answer:
Increase

22. Boron, silicon and germanium are examples of
Answer:
Metalloids (or semi-metals)

23. Oxides of elements tend to be basic.
Answer:
Metallic

24. Non-metals are electro-
Answer:
Negative

True or False

1. Triads are written in the increasing order of their atomic number. — False
2. Look at the table below.

Do these elements satisfy even to be a triad? (Yes = True, No = False) — True (Yes)
3. Dobereiner could identify only three triads and so his system of classification was not useful. — True
4. Newlands’ law failed after calcium — True
5. Newlands thought that after 63rd element no more elements will be discovered — False
6. Newlands put nickel and carbon in same slot — False
7. Newlands’ law worked well for heavier elements — False
8. Mendeleev knew about 7 more elements as compared to Newlands. — True
9. Nickel (atomic mass 58.93) appeared before cobalt (atomic mass 58.71) in Mendeleev’s periodic table. So as to group them together on the basis of similar properties. — False
10. Mendeleev’s Table did not consist of noble gases. — True
11. Halogens have valency of 7. — True
12. The position of hydrogen in the Modern Periodic Table is till date a matter of argument. — True
13. An element having atomic number 25.5 can be easily placed between Mn and Fe. — False
14. Each period signifies a new electronic shell getting filled. — True
15. As we go down in a group, the distance between the outermost electron and the nucleus decreases. — False
16. A zigzag line separates metals from non-metals within the Modern Periodic Table. — True
17. As the effective nucleus charge acting on the valence shell electrons increases across a period, the tendency to lose electrons decrease. — True
18. Metals are electro-positive. — True

Match the Following:

Question 1.

Answer:
(1-b), (2-c)

Question 2.

Answer:
(1-c), (2-a), (3-b), (4-d)

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 10 Circles Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 10 Circles

Short / Long Answer Type Questions

Question 1.
Two concetric circles of raddi a and b (a > b) are given. Find the length of chord of the larger circle which touches the smaller circle.
Solution :
Let two concetric circles of centre O with raddi a and b (a > b)
AO = OB = a and OM = b
OM ⊥ AB (By theorem 10.1)

In right triangle AMO, we have
AO² = AM² + OM²
(By Pythagoras theorem)
⇒ a² = AM² + b²
⇒ AM² = a² – b²
⇒ AM = \(\sqrt{a^2-b^2}\) units
we know that perpendicular drawn from centre to chord bisect the chord
∴ AM = MB
AB = 2 × AM
= 2 × \(\sqrt{a^2-b^2}\) units
Hence, length of chord of larger circle
= 2\(\sqrt{a^2-b^2}\) units

Question 2.
How many tangents can be drawn on the circle of radius 5 cm from a point lying outside the circle at distance 9 cm from the centre.
Solution :
Draw a circle of radius 5 cm with centre O. Let P be the point outside 9 cm from the centre O.

From external point P, we can draw two tangents PA and PB only.

Question 3.
In the given figure, two circles touch each other at the point C. Prove that common tangent to the circles at C, bisects the common tangents at P and Q.

Solution :
PT and CT are tangents at circle with centre A from external point T
∴ PT = CT …… (1) [by theorem 10.2]
Similarly
QT = CT ……… (2)
From equ. (1) and (2)
PT = QT
Now, PQ = PT + QT
⇒ PQ = 2 PT
⇒ PT = QT = \(\frac {1}{2}\)PQ
Hence, common tangent CT bisects the common tangents at P and Q. Hence proved

Question 4.
In the given figure ΔABC is circumscribing a circle, the length of BC is ….. cm.

Solution :
BQ = BP = 3 cm
[by theorem 10.2]
AR = AP = 4 cm
[by theorem 10.2]
Now,
CR = AC – AR
= 11 – 4 = 7
CR = QC
⇒ QC = 7 cm
BC = BQ + QC
= 3 + 7 = 10 cm

Question 5.
In the given figure, PQ and PR are tangents to the circle with centre O such that ∠QPR = 50°, then find ∠OQR.

Solution :
Join QR
PQ = PR by theorem 10.2]
⇒ ∠PQR = ∠PRQ [Angles Opp. to equal sides are equal] … (1)
In ΔPQR, we have
∠PQR + ∠PRO + ∠QPR = 180°
⇒ ∠PQR + ∠PRO + 50° = 180° [using equ. (1)]
⇒ 2∠PQR = 180° – 50°
⇒ ∠PQR = \(\frac {130°}{2}\)
⇒ ∠PQR = 65°
Now, OQ ⊥ PQ
[By theorem 10.1]
⇒ ∠PRO = 90°
⇒ ∠OQR = ∠PQO – ∠PQR
= 90° – 65° = 25°
Hence, ∠OQR = 25°

Question 6.
In the given figure PQ is a chord of length 6 cm of the circle of radius 6 cm. TP and TQ are tangents to the circle at points P and Q respectively. Find the ∠PTQ.

Solution :
Here PQ = 6 cm
PO = OQ = 6 cm
[equal raddi]
∴ PO = OQ = PQ
⇒ POQ is a equilateral triangle
∠POQ = 60°
Now, OP ⊥ PT and OQ ⊥ TQ
[By theorem 10.1]
In quadrilateral POQT, we have
∠POQ + ∠OPT + ∠PTQ + ∠OQT = 360°
⇒ 60° + 90° + ∠PTQ + 90° = 360°
⇒ 240° + ∠PTQ = 360°
⇒ ∠PTQ = 360° – 240°
⇒ ∠PTQ = 120°

Question 7.
In the figure, AB and CD are common tangents to two circles of unequal raddi. Prove that AB = CD.

Solution :
Produced AB and CD to meet at P
Now, PA = PC
(By theorem 10.2) …. (1)

PB = PD
(By theorem 10.2) … (2)
Subtracting equ. (2) from equ. (1), we get
PA – PB = PC – PD
⇒ AB = CD Hence proved

Question 8.
In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q.

If PA = 12 cm, QC = DQ = 3 cm, then find PC + PD.
Solution :
AC = CQ
(By theorem 10.2) ……. (1)
BD = DQ
(By theorem 10.2) …….(2)
But CQ = DQ …….(3)
By equ. (1). (2) and (3), we get
AC = BD ……….(4)
Now,
AP = PB
(by theorem 10.2)
⇒ CP + AC = PD + BD
⇒ CP + AC = PD + AC
[using equ. (4)]
⇒ CP = PD
CP = AP – AC = 12 – CQ [∴ CQ = AC]
⇒ CP = 12 – 3 = 9 cm
[CQ = DQ = 3 cm]
∴ DP = CP = 9 cm
PC + PD = 9 + 9 = 18 cm

Question 9.
In the given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the area of ΔABC is 54 cm², then find the lengths of sides AB and AC.

Solution :
Join OA, OB and OC
Draw OE ⊥ AC and OF ⊥ AB
Area of ΔABC = 54 cm² (given)
⇒ ar (ΔBOC) + ar (ΔAOC) + ar (ΔAOB) = 54
⇒ \(\frac {1}{2}\)BC × OD + \(\frac {1}{2}\)AC × OE + \(\frac {1}{2}\)AB × OF = 54
⇒ \(\frac {1}{2}\)[BC × 3 + AC × 3 + AB × 3] = 54
⇒ \(\frac {1}{2}\)[BC + AC + AB] = 54
⇒ AB + BC + AC = \(\frac{54 \times 2}{3}\)
⇒ AB + BC + AC = 36 cm ……….(1)
AF = AE [by theorem 10.2]
Let AF = AE = x cm
BD = BF and CD = CE [by theorem 10.2]
⇒ BF = 6 cm and CE = 9 cm

Now AB + BC + AC = 36
⇒ AF + BF + BD + CD + CE + AE = 36
⇒ x + 6 + 6 + 6 + 9 + 9 + 1 = 36
⇒ 2x + 30 = 36
⇒ 2x = 6
⇒ x = 3
∴ AB = 6 + 3 = 9 cm, AC = 9 + 3 = 12 cm.

Question 10.
Prove that the parallelogram circular scribing a circle is rhombus.
Solution :
Let ABCD be a parallelogram
∴ AB = CD and AD = BC
we know that length of tangents drawn to a circle from an exterior point are equal in length.
AP = AS …(1)
PB = BQ …(2)
CR = CQ …(3)
DR = DS ……(4)

Adding equations (1), (2), (3) and (4), we get
AP + PB + CR + DR = AS + BQ + CQ + DS
⇒ AB + CD = AD + BC
⇒ AB+ AB = BC + BC
⇒ 2 AB = 2 BC
⇒ AB = BC
since, in a parallelogram ABCD adjacent sides AB and BC are equal.
So, ABCD is a rhombus.

Fill in the Blanks

Question 1.
Tangent is perpendicular to the …….. through the point of contact.
Solution :
radius

Question 2.
Only …….. tangents can be drawn to a circle from an external point.
Solution :
two

Question 3.
Lengths of tangents from an external point to a circle are ……..
Solution :
equal

Question 4.
The line containing the radius through the point of contact is also sometimes called the …….. to the circle at the point.
Solution :
normal

Question 5.
Circles having the same …….. are called concentric circles.
Solution :
centre

Question 6.
The word ……. to circle has been deribed from the latin word “……..”
Solution :
tangent

Question 7.
The point at which the tangent line meets the circles is called the ……..
Solution :
point of contact

Multiple Choice Questions

Choose the correct answer each of the following :

Question 1.
In the adjoining figure, if PA and PB are tangents to the circle with centre such that ∠APB = 50°, then ∠OAB is equal to :

(a) 25°
(b) 30°
(c) 40°
(d) 50°
Solution :
From figure
∠1 + ∠P = 180°

[Opposite angles of cyclic ∠1]
⇒ ∠1 + 50° = 180°
⇒ ∠1 = 180° – 50° = 130°
In ΔOAB,
OA = OB (radius)
⇒ ∠2 = ∠3 (Isosceles Δprop)
∠2 + ∠3 + ∠AOB = 180°
(Angle sum property of Δ)
⇒ ∠2 + ∠2 + 130° = 180°
⇒ 2∠2 = 180° – 130° = 50°
⇒ ∠2 = ∠OAB = \(\frac {50°}{2}\) = 25°
Hence correct choice is (a).

Question 2.
In the adjoining figure, if AP = 4 cm, CR = 5 cm and BQ = 6 cm the perimeter of ΔABC is:

(a) 15 cm
(b) 30 cm
(c) 25 cm
(d) 20 cm
Solution :
From Fig. BQ = BP = 6 cm
CQ = CR = 5 cm
PA = AR = 4 cm
The perimeter of ΔABC = BQ + CQ + CR + AR + AP + BP
= 60 + 5 + 5 + 4 + 4 + 6
= 30 cm
Hence correct choice is (b).

Question 3.
In the adjoining figure, if ∠AOB = 125°, then ∠COD is equal to :
(a) 62.5°
(b) 45°
(c) 35°
(d) 55°

Solution :
From fig. ∠AOB + ∠COD = 180°
125° + ∠COD = 180°
⇒ ∠COD = 180° – 125°
= 55°
Hence correct choice is (d).

Question 4.
In the adjoining figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to :
(a) 4 cm
(b) 2 cm
(c) 2\(\sqrt{3}\) cm
(d) 4\(\sqrt{3}\) cm

Solution :
∠OAT = 90° (Angle between radius and tangent)
Cos 30° = \(\frac {AT}{OT}\)

\(\frac{\sqrt{3}}{2}\) = \(\frac {AT}{OT}\)
⇒ AT = 2\(\sqrt{3}\)cm
Hence correct choice is (c).

Question 5.
In the given figure QR is a common tangent to the given circle, which touch externally at P. If QP = 3.8 cm, the length of QR is :
(a) 38 cm
(b) 7.6 cm
(c) 5 cm
(d) 1.9 cm

Solution :
From Fig QP = PT = PR = 3.8 cm
So, QR = QP + PR
= 3.8 + 3.8
= 7.6 cm
Hence correct choice is (b).

Question 6.
In the given figure, if ∠APO = 40°. Then ∠AOB is :
(a) 100°
(b) 80°
(c) 50°
(d) 40°

Solution :
Given ∠APO = 40°
and ∠OAP = 90°
(radius is ⊥ to tangent)
∴ In ΔPAQ
∠AOP = 180 – (40 + 90)
= 50°
∴ ∠AOB = 2 × ∠AOP
= 2 × 50°
= 100°
So correct choice is (a)

Question 7.
In the adjoining figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR = 70°, then ∠AQB is equal to
(a) 20°
(b) 40°
(c) 35°
(d) 45°

Solution :
∠DQR = 90°
(Angle between radius and tangent)
∠BQD = ∠DQR – ∠BQR
= 90° – 70°
= 20°
Similarly ∠AQD = 20
∠AQB = 20° + 20°
So correct choice is (b).

Question 8.
If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm then length of each tangent is equal to :
(a) \(\frac {3}{2}\)\(\sqrt{3}\) cm
(b) 6 cm
(c) 3 cm
(d) 3\(\sqrt{3}\) cm
Solution :
∠OBA = 90°
(Angle between radius and tangent)
In right ΔOBA
Tan 30° = \(\frac {OB}{AB}\)
\(\frac{1}{\sqrt{3}}\) = \(\frac {3}{AB}\)

AB = 3\(\sqrt{3}\)
∴ AC = AB = 3\(\sqrt{3}\) (tangents from same external point)
So correct choice is (d).

Haryana State Board HBSE 10th Class Science Important Questions Chapter 4 Carbon and Its Compounds Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 4 Carbon and Its Compounds

Question 1.
Give a brief introduction of carbon.
Answer:
1. Carbon is a non-metallic element. Its symbol is C.

2. The amount of carbon in the atmosphere and earth’s crust is very less. Earth’s crust contain 0.02% carbon in the form of minerals such as carbonates, hydrogen carbonates, coal and petroleum. The atmosphere contains only 0.03% of carbon in the form of carbon dioxide.

3. In spite of being present in such a small quantity, carbon is an extremely important element. All the long things, plants and animals are made up of carbon based compounds known as organic compounds.

Atomic number and electron sharing:

1. The atomic number of carbon is 6. So, it has 2 electrons in Its first shell and 4 in the second or say outermost shell.

2. This makes the valency of carbon as 4 (i.e. 4 electrons in the outermost shell). This makes carbon quite a unique element. To complete the octet configuration, carbon either needs to lose 4 electrons or gain 4 electrons, both of which are not possible. Hence, carbon joins with other elements by sharing electrons and forming covalent bonds.

Question 2.
Explain electronic configuration, valency and bonding of carbon with other elements. OR ‘Carbon has a unique way of bonding’. Explain.
Answer:
1. The atomic number of carbon is 6. Hence, there are 2 electrons in its first (K) shell and 4 in second (L) i.e. outermost shell. Thus, carbon has 6 protons and 6 electrons.

2. The reactivity of an element is explained by its tendency to attain a completely filled outer shell to attain noble gas configuration.

3. Elements forming ionic compounds achieve noble gas configuration by either losing or gaining electrons from the outermost shell. The case of carbon is diUerent since it has 4 electrons in its outermost shell.

4. Carbon has to either gain or lose 4 electrons to attain noble gas configuration. The problem in doing this is discussed below:

• Carbon can gain 4 electrons to form C anion. If carbon does this, it will be difficult for the nucleus with 6 protons to hold 10 electrons (6 existing + 4 borrowed) Le. 4 extra electrons.
• Carbon can lose 4 electrons to form C4-’ cation. This would require a large amount of energy to remove 4 electrons leaving behind a carbon cation with 6 protons in its nucleus holding on to just 2 electrons.

Solution:
1. To overcome these problems, carbon neither accepts, nor gains but shares its valence electrons with other atoms of carbon or with atoms of other elements. The shared electrons belong to the outer shells of both the atoms. This way both the atoms attain noble gas configuration.

2. The bond formed by sharing of electrons in this manner is known as covalent bond.

3. Not only carbon but many other elements form molecules by sharing electrons and forming covalent bonds.

Question 3.
What do you mean by covalent bond and covalent compounds? Explain briefly with one example.
Answer:
Covalent bond and compounds :
1. Covalent compounds consist of molecules which are groups of atoms in which one or few pairs of atoms share the electrons by bonding.

2. In ionic bond, elements develop bonds by gaining or losing electrons but, in covalent bonds. elements develop bonds by sharing electrons.

3. For example, two atoms of hydrogen, each of them having one electron, share their electrons of the outermost orbit by forming a covalent bond and attain dual closed shell configuration of their nearby inert element, helium.

4. Here, both the hydrogen atoms jointly share the electrons for becoming inert and stable.

5. Only those electrons that are present in the outermost orbit of the atoms take part in bond formation.

6. The electron pair that takes part in sharing is known as bonding electron pair or bond electron pair.

Question 4.
What is an ionic bond and a covalent bond? Explain giving differences. OR State the key differences between an IonIc bond and a covalent bond.
Answer:

Question 5.
Explain covalent bonding In hydrogen molecule. OR Explain covalent single bond with the help of an example.
Answer:
1. The atomic number of hydrogen is 1 and so hydrogen atom possesses one electron.

2. Hydrogen atom requires one more electron to achieve the closed shell configuration of near by inert element, helium.

3. Hence, two atoms of hydrogen, each having one electren, will share their one electron by forming a covalent bond, thus giving rise to hydrogen (H₂) molecule.

4. Both these atoms will attain dual closed shell configuration like that of helium.

5. The electron pair that takes part in sharing is called bonding electron pair or bond electron pair.

6. The single line between two hydrogen atoms represent single covalent bond.

Question 6.
Explain covalent bonding in chlorine molecule.
Answer:
1. The atomic number of chlorine is 17 and so its electronic configuration is (2, 8, 7).
2. The chlorine atom will share the electron of its outermost orbit to attain the closed octet configuration of nearby inert gas.
3. Each chlorine atom requires one electron to attain octet configuration.
4. Hence, two chlorine atoms will share one-one electron with each other, form a single covalent bond, will attain octet configuration and form chlorine molecule.

Question 7.
Explain covalent bonding in oxygen. OR Explain covalent double bond with the help of an example.
Answer:
1. The atomic number of oxygen is 8 and so its electronic configuration is (2, 6).
2. Since the valency of oxygen is 2, each oxygen atom will share 2 electrons from its outermost orbit to complete the octet configuration.
3. Hence, two oxygen atoms will share two-two electrons with each other, form double covalent bond, will attain octet configuration and will form one oxygen (O₂) molecule.
4. Because of the double covalent bond, this compound is called divalent compound.

The double line between two oxygen atoms represent double covalent bond.

Question 8.
Explain covalent bonding in nitrogen. OR Explain covalent triple bond with the help of an example.
Answer:
1. The atomic number of nitrogen is 7 and so its electronic configuration is (2, 5).
2. Since the valency of nitrogen is 3, each nitrogen atom will share 3 electrons from its outermost orbit to complete octet configuration.
3. Hence, two nitrogen atoms will share three-three electrons with each other, form triple covalent bond, attain octet configuration and form nitrogen molecule.
4. Because of the triple covalent bond, this compound is called trivalent compound.

Question 9.
Explain the formation of covalent bonds in water molecule (H₂O).
Answer:
1. Oxygen is a central atom in the molecule of water. The atomic number of oxygen is 8,
so its electronic configuration is (2, 6)
2. Thus, oxygen has 6 electrons in its L shell and it needs two more electrons to fill the L shell.
3. The atomic number of hydrogen is 1 and so hydrogen atom possesses 1 electron in its K shell.
4. Oxygen shares two of its valence electrons with one electron each of K shell of two hydrogen atoms to form a molecule of water (H₂O).
5. This way oxygen atom of water attains the electronic configuration of its nearest noble gas neon (Ne), while hydrogen atom attains the electronic configuration of its nearest noble gas helium (He) which has two electrons in its K shell.

Question 10.
Explain covalent bonding in ammonia (NH₃) molecule.
Answer:
1. Nitrogen is a central atom in the molecule of ammonia. The atomic number of nitrogen is 7, so its electronic configuration is (2, 5).
2. Nitrogen has five electrons in its L shell and so it needs three more electrons to fill the L shell.
3. Nitrogen shares three of its valence electrons with one electron each of K shell of three hydrogen atoms to form a molecule of ammonia (NH₃).
4. Thus, nitrogen atom of ammonia attains the electronic configuration of its nearest noble gas neon (Ne), which has eight electrons in its L shell, while hydrogen atom attains the electronic configuration of its nearest noble gas helium (He), which has two electrons in its K shell.

Question 11.
Explain covalent bonding in methane (CH₄) molecule.
Answer:
1. Carbon is a central atom in the molecule of methane. The atomic number of carbon is 6, so its electronic configuration is (2, 5).
2. Thus, carbon has four electrons in its L shell and it needs four more electrons to fill the L shell.
3. Carbon shares its four valence electrons with one electron each of K shell of four hydrogen atoms to form a molecule of methane (CH₄).
4. This way carbon atom of methane attains the electronic configuration of its nearest noble gas neon (Ne), which has eight electrons in its L shell, while hydrogen atom attains the electronic configuration of its nearest noble gas helium (He), which has two electrons in its K shell.

Question 12.
State the properties of covalent compounds (or carbon compounds).
Answer:
Properties of covalent compounds (or carbon compounds):

• Covalent compounds exists in all the three forms i.e. solid, liquid and gas.
• They have weak force of attraction between the molecules.
• They have lower melting and boiling points.
• They are non-conductors of electricity.
• Generally they are insoluble in water but soluble in organic solvents.

Question 13.
How does carbon bond with other atoms of carbon? OR Explain catenation.
Answer:
(i) Catenation:

• Carbon has a unique ability to bond with other atoms of carbon and form long chain. This unique
  property of carbon is called catenation.
• Catenation results in formation of large molecules. Moreover, the ability of carbon to bond with several elements results in formation of a large number of carbon based compounds.

(ii) Bonding by carbon:

• Carbon atom bonds with the help of three types of covalent bonds namely, single bond, double bond and triple bond.

(a) Single bond compound (Saturated compound):

•  If carbon atom joins with another carbon atom with the help of only single bond then the compound formed is called a saturated compound.

(b) Double or triple bond (Un-saturated compound):

• If carbon atom joins with other carbon atoms via, double or triple bond, then the compounds formed are called unsaturated compounds.
• The carbon-carbon bond is very strong and hence stable. This gives rise to a large number of compounds with several carbon atoms linked with each other.

(iii) The structures formed by the three types of covalent bonds of carbon can be of the following types:

• Normal chain,
• Branched (Iso) chain or
• Cyclic chain

Question 14.
Why is carbon called tetravalent? How does It help carbon to bond?
Answer:
The valency of carbon is 4, ie. carbon has 4 electrons in its outermost shell. Hence, carbon is called tetravalent.
1. Since the valency of carbon is 4, it is capable of bonding with four other atoms of carbon or some mono-valent atoms i.e. atoms having one valency.
2. This way, carbon forms compounds with oxygen, hydrogen, nitrogen, sulphur, chlorine and many other elements and gives rise to several compounds.

Question 15.
State the two important properties of carbon that help it to form a large number of compounds.
Answer:
The two important properties of carbon that help it to form a large number of compounds are:
(1) Catenation:

• Carbon has a unique ability to bond with other atoms of carbon and form long chain. This unique property of carbon is called catenation.
• Catenation results in formation of large molecules. Moreover, the ability of carbon to bond with several elements results in formation of a large number of carbon based compounds.
• Carbon atom bonds with the help of three types of covalent bonds namely, single bond, double bond and triple bond.

(2) Tetravalency:

• The valency of carbon is 4, i.e. carbon has 4 electrons in its outermost shell. Hence, carbon is called tetravalent.
• Since the valency of carbon is 4, it is capable of bonding with four other atoms of carbon or some mono-valent atoms i.e. atoms having one valency.
• This way, carbon forms compounds with oxygen, hydrogen, nitrogen, sulphur, chlorine and many other elements and gives rise to several compounds.

(3) Other reasons:

• Carbon forms very strong bonds with other elements and so the compounds formed are extremely stable.
• No other element shows the property of catenation to the extent of carbon.

Question 16.
What are hydrocarbons? How are they classified?
Answer:
Hydrocarbons:

• Compounds containing hydrogen and carbon are called hydrocarbons.
• In organic chemistry, hydrocarbons are considered to be the simplest organic compounds.

Classification of hydrocarbons:

• Hydrocarbons are classified on the basis of the number of covalent bonds between carbon-carbon atoms.
• Thus, on the basis of covalent bonds, hydrocarbons can be classified as—

(I) Saturated hydrocarbons:

• Hydrocarbons having single covalent bonds between their carbon atoms are called saturated hydrocarbons.
• Alkanes are the main class of saturated hydrocarbons.

Question 17.
Give a classification of hydrocarbons. OR Differentiate between saturated and unsaturated hydrocarbons.
Answer:
Hydrocarbons can be classified into —
(i) Saturated hydrocarbons and
(ii) Unsaturated hydrocarbons

Question 18.
Give the molecular, electronic and structural formula of methane
Answer:
1. Methane is the first member of the alkane series hydrocarbon.
2. Molecular formula, electronic formula and structural formula of methane are as under.

Question 19.
With the help of the example of ethane, show how to draw the structure of a hydrocarbon.
Answer:
1. Ethane is a hydrocarbon formed by joining carbon and hydrogen.
2. The molecular formula of ethane is C₂H₆.

Steps to draw the structure:

Step-1:
1. Since ethane is a saturated hydrocarbon, it will have a single covalent bond. First we link two carbon atoms by single bond.
2. Here, one valency of carbon is used. Now, three valencies are to be filled.

C – C

Step-2:
1. We now attach hydrogen atoms with carbon to satisfy remaining three valencies of carbon. This gives us the following structure.
2. Each carbon atom is bonded by a single covalent bond with carbon as well as hydrogen. So, all the valencies are filled and the structure is completed.

Question 20.
Draw the electron dot structure of ethane.
Answer:

Question 21.
Give the molecular, electronic and structural formula of ethene.
Answer:
Molecular, electronic and structural formula of ethene are as under —

Question 22.
Give the molecular formula of ethyne and draw its electronic structure and bond structure.
Answer:
Ethyne (Acetylene):
1. Ethyne is the first member of the alkyne series.
2. Its molecular, electronic and structural formula are as follows —

Question 23.
Give formula and structure of first six alkanes
Answer:
Formula and structure of first six alkanes

Question 24.
Explain Isomers and Isomerism.
Answer:
Isomers and Isomerism:
1. Property of catenation possessed by carbon gives rise to a large number of compounds with different structural formula and different physical properties.

2. Organic compounds that have same molecular formula but different structural formula are called isomers and the phenomenon is called isomerism.

3. ‘Iso’ indicates a branched chain whereas normal (-n) Indicates a straight chain structure.

4. As the number of carbon atoms increase in a chain, the number of isomers also increase.

5. For example, butane (C₄H₁₀) has only two isomers whereas hexane (C₆-H₁₄) has five isomers.

Example of ‘n’ i.e. normal and ‘iso’ i.e. branched structured hydrocarbons:

Question 25.
State the characteristics of isomers.
Answer:
Characteristics of isomers:

• Isomers have same molecular formula but different structural formula.
• Isomers have different melting and boiling points.
• They have different chemical properties.

Question 26.
Explain the classification of hydrocarbon compounds based on their different structures.
Answer:
On the basis of their structures (arrangement of carbon atoms), hydrocarbon compounds can be classified into following three categories:
(a) Straight chain compounds,
(b) Branched chain compounds
(c) Ring structures OR Cyclic compounds

(a) Straight chain compounds:
Hydrocarbon compounds in which all carbon atoms are arranged linearly i.e. in a straight chain are called straight chain compounds.

For example, propane (C₃H₈):

Branched chain compounds:

A hydrocarbon compound in which carbon atoms are arranged in straight chain, as well as possess one or more branches is called a branched chain compound.
For example, Carbon-skeleton of four carbon atoms of butane can be arranged in two different possible ways as follows:

Now, when we satisfy the remaining valencies of carbon with hydrogen we get the following structures:

(c) Ring structure OR Cyclic compounds:
Hydrocarbon compounds in which the first carbon atom is directly linked with the end (last) carbon atom, forming ring or cyclic structures are called ring structures or cyclic compounds. For example, cyclohexane (C₆H₁₂).

Question 27.
Is cyclohexane an isomer of hexane? Explain with reason.
Answer:
1. No. The molecular formula of hexane is C₆H₁₄ whereas that of cyclohexane is C₁₆H₁₂.
2. Since the molecular formula of cyclohexane is different than hexane it, is not the isomer of hexane.

Question 28.
Draw the structure of benzene.
Answer:
The molecular formula of benzene is C₆H₆.

Question 29.
What are cyclic hydrocarbons? Draw structure of two cyclic hydrocarbons.
Answer:
Ring structure OR Cyclic compounds:
Hydrocarbon compounds in which the first carbon atom is directly linked with the end (last) carbon atom, forming ring or cyclic structures are called ring structures or cyclic compounds. For example, cyclohexane (C₆H₁₂).

The molecular formula of benzene is C₆H₆.

Question 30.
What are heteroatoms? Explain.
Answer:
Heteroatom:
1. Apart from hydrogen, carbon forms bonds with other elements as well.

2. The atom or group of atoms of an element that replaces hydrogen in a hydrocarbon is called an heteroatom.

3. Oxygen (O), nitrogen (N), sulphur (S), halogens such as fluorine (F), chlorine (Cl), bromine (Br) and iodine (I)) are typical heteroatoms that replace hydrogen.

Question 31.
What are functional groups? Explain.
Answer:
Functional groups:
1. A heteroatom (i.e. an atom or a group of atoms) which imparts specific properties to the organic compound they are attached to is called a functional group. (Note: When an heteroatom attaches to an organic compound, the physical and chemical properties of that compound changes. The heteroatom which Is responsible for changes in these properties Is called functional group.)

2. Thus, the functional group decides the physical and chemical properties of the carbon compound, irrespective of the length of the carbon chain.

3. There are several functional groups. Some of them are listed in the table below.

Question 32.
Explain how does the name of an organic compound changes when a functional got is attached to it
Answer:

Question 33.
“The number of carbon atoms in ethane, ethanol and ethanoic acid are same, yet all the three compounds show different physical and chemical properties”. Explain.
Answer:
1. Ethane is an alkane and its formula is C₂H₆.

2. When functional group hydroxyl – OH attach to ethane, the compound formed is ethanol or ethyl alcohol and its formula is C₂H₅OH. Similarly, when functional group carboxylic acid – COOH attach to ethane, the compound formed is ethanoic acid which is an acid and its formula is CH₃COOH.

3. In a hydrocarbon, the functional group changes the physical and chemical properties of the compound.

4. Now, although the number of carbon atoms in all the three compounds i.e. ethane, ethanol and ethanoic acid are same, yet all the three compounds show different physical and chemical properties because ethane has no functional group, ethanol contains functional group alcohol whereas ethanoic acid consists of another functional group called carboxylic acid.

Question 34.
What do you mean by homologous series? Give its characteristics.
Answer:
Homologous series:
1. The series of organic compounds in which a particular functional group attaches to the carbon chain in place of hydrogen atom is called a homologous series.
2. Each compound of the homologous series differs from its previous or later compound by (CH₂).
3. For example, the alkanes namely methane (CH₄), ethane (C₂H₆), propane (C₃H₈) and so on form a homologous series with a definite difference of CH₂.
4. Similarly, CH₃OH, C₂H₅OH, C₃H₇OH is the homologous series of alcohols with functional group – OH and with a definite difference of CH₂ between each compound.

Question 35.
State the characteristics of homologous series.
Answer:
Characteristics of homologous series:
1. Each member of the homologous series contains same elements as well as same functional group.

2. Each member of the homologous series can be expressed by a general formula. For example, general formula of alkane series of hydrocarbc,s is CnH_2n+2.

3. The difference in molecular formula between two successive members of the series of homologous compounds is CH₂.

4. Same prefix or suffix is applied to the nomenclature of each member of the series. For e.g., suffix sane’ is put to each member of the alkane series.

5. Difference between molecular masses of any two successive members of the series is 14u.

6. As the number of carbon and hydrogen increases in the series, the molecular mass of the compound also increases.

• As the molecular mass of the compound increases, the physical properties such as boiling point, melting point, solubility, etc. also change gradually.

7. The chemical properties of each compound of a homologous series remain same.

Question 36.
What is IUPAC name?
Answer:
IUPAC names:
1. Organic compounds or hydrocarbons have two names, (a) Common name and (b) IUPAC name.

2. IUPAC names are names given by IUPAC (Note: IUPAC = International Union of Pure and Applied Chemistry. These names are also called JUPAC nomenclature.)
Example: Methyl alcohol is an alcohol formed from methane. Methyl alcohol is a general (common) name where as methanol is its IUPAC name.

Question 37.
State the method to name the carbon compounds (hydrocarbons). OR Explain nomenclature of organic compounds.
Answer:
Nomenclature of organic compounds:
(Note: The number of carbon atoms in a hydrocarbon (or any other organic compound) is indicated by using following steps. The table is only for understanding purpose. Actual answer starts after the table.)

Steps for naming:
The names of compounds in homologous series are based on the name of the basic carbon chain which are modified by either a ‘prefix’ or a ‘suffix’ of the functional group.

(i) Identify the ‘number of carbon atoms’ in the compound. (Note: Refer the note above. Based on it, lets say a carbon compound has 3 carbon atoms, then its prefix would be ‘prop’.)

• If the compound is a saturated hydrocarbon with 3 carbon atoms and without any functional group, it will belong to alkane series and its name would be prop + ane = propane.

(ii) If there is a functional group present, it will be indicated with either a prefix or a suffix. For example, C₃H₇OH has 3 carbon atoms and has a functional group ‘-OH’ i.e. alcohol. So name of the compound is propanol.

(iii) As discussed in point (ii) if the name of the functional group is to be given as suffix, the name of the carbon chain is modified by replacing the last letter ‘e’ with proper suffix. For example, a 3-carbon chain with a ketone functional group would be named in the following manner:
Propane = Propan + ‘one’ = Propanone.

(iv) If the carbon chain is unsaturated (i.e. having double or triple covalent bond), the final ‘ane’ in the name of carbon chain is replaced by ‘ene’ or ‘yne’.
For example, a 3-carbon chain with a propene bond would be called propene and if it has a triple bond, it will be called propyne.

Nomenclature of Organic Compounds based on Functional Groups:

Question 38.
Discuss chemical properties of carbon compounds briefly.
(Note: Each property can be asked as a separate question.)
Answer:
Chemical properties of carbon compounds:
(i) Combustion:
Carbon present in all its allotropes burn in sufficient amount of oxygen. On burning, it produces carbondioxide and water and liberate heat and light. These reaction are oxidation reactions.

(ii) Oxidation:

• Oxidation is the reaction in which carbon compounds take up oxygen in the presence of oxidizing agents to give another compound.
• There are certain substances which are capable of adding oxygen to others i.e. adding oxygen to reactants. Such materials are called oxidizing agents.
• 

Example:
Ethanol (ethyl alcohol) gets oxidized into ethanoic acid (i.e. a carboxylic acid) in the presence of oxidizing agent alkaline potassium permanganate or acidified potassium dichromate.

(iii) Addition (Hydrogenation) reaction:

• A reaction in which adding one molecule to an organic compound gives a new but single organic compound is called addition reaction.
• For example, on adding hydrogen to an unsaturated (alkene or alkyne) hydrocarbon in the presence of catalyst such as palladium or nickel gives a single but saturated (alkane) product. This reaction is called addition reaction.

(1) Addition reaction In alkenes:

(2) Addition reaction in alkynes:

(iv) Substitution reaction:

• A reaction in which one or more hydrogen atoms of a hydrocarbon are substituted (replaced) by some other atom(s) (such as chlorine) is called a substitution reaction.
• Actually saturated hydrocarbons are quite unreactive and remain inert in the presence of most reagents. However, in the presence of sunlight, these compounds ie. alkanes undergo substitution reaction.

Example:
Methane (an alkane) reacts with chlorine in the presence of sunlight to form chloromethane and hydrochloric acid. In this reaction, one hydrogen (H) atom of methane gets substituted by a chlorine (Cl) atom. This converts CH₄ into CH₃Cl.

The way chlorine replaced one hydrogen atom from methane, it rapidly replaces each hydrogen atom with chlorine atom one by one.

Question 39
Give an example of addition reaction along with necessary chemical reaction.
Answer:
When hydrogen is added to ethene (an unsaturated hydrocarbon) and heated in the presence of nickel catalyst it gives a single saturated product ethane.

Question 40.
State the non difference between combustion and oxidation. OR ‘All combustion reactions are also oxidation reactions but, the reverse is not true.
Answer:
1. During combustion, a carbon compound is burnt in the air to give out carbon dioxide and water and liberate heat and light.
2. During oxidation the carbon compounds take up oxygen in the presence of oxidizing agents to give another carbon compound. Hence, all combustion reactions are also oxidation reactions but, the reverse is not true.

Question 41.
What is hydrogenation of oils? OR How ¡s vegetable ghee prepared from vegetable oil? OR Explain addition reaction giving example of vegetable oil.
(Note: Hydrogenetaion = Addition of hydrogen)
Answer:
1. Vegetable oils have long unsaturated fats having double bonds between sorne of their carbon atoms.
2. When a vegetable oil (like groundnut oil) is heated with hydrogen gas in the presence of nickel catalyst, the oil turns into a saturated fat called vegetable ghee or say naspati ghee. This reaction is known as hydrogenation of oils.

Vegetable oil is in liquid form which after hydrogenation turns into semi-solid ghee.

Question 42.
What is ethanol? State its properties.
Answer:
Ethanol: Ethanol is the second member of the homologous series of alcohol. Its formula is C₂H₂OH.
Ethanol or ethyl alcohol is the most common and most widely used alcohol and hence is also simply called alcohol.

Properties:

• At room temperature it exists in the liquid form.
• Ethanol is a very good solvent. It is soluble in water in any proportion.

Question 43.
State the uses and abuses of ethanol.
Answer:
Uses of ethanol:

• Ethanol is the active ingredient of all alcoholic drinks.
• Ethanol is a good solvent and so it is also used to make medicines such as tincture iodine, cough syrups and several other tonics.

Abuses:

• Consuming pure (undiluted) ethanol even in a small quantity can prove lethal.
• Consuming diluted ethanol that too in small quantity causes drunkenness.
• People fall prey and addicted to alcoholic dnnks. This ruins individual health, family and society at large.

Question 44.
State reactions (chemical properties) of ethanol.
Answer:
Reactions of ethanol:
(i) Reactions with sodium:
Ethanol reacts with sodium and produces sodium ethoxide along with evolution of hydrogen gas.

(ii) Dehydration:

• When ethanol is heated with excess concentrated sulphuric acid at 443 K, it gets dehydrated to form ethene.
• The concentrated sulphuric agent works as a dehydrating agent which removes water from ethanol.

Question 45.
What is ethanoic acid? State its physical properties.
Answer:
Ethanoic acid:

• Ethanoic acid is the second member of the homologous series of carboxylic acids.
• Its formula is CH₃COOH. The common name of ethanoic acid is acetic acid.

Properties of ethanoic acid:

• Ethanoic acid is a weak acid.
• It is sour in taste. 5-8% solution of ethanoic (acetic) acid in water is called vinegar. Vinegar is widely used as preservative in pickles and in preparation of certain food items.
• The melting point of pure ethanoic acid is 290 K i.e. just nearly 17° Celsius. As a result, it often freezes in winter.
• The frozen form of ethanoic acid looks like solid ice or say ‘glacier’. Hence, it is also called ‘glacial acetic (or ethanoic) acid’

Question 46.
Discuss the chemical properties (reactions) of ethanoic acid.
Answer:
Chemical properties (reactions) of ethanoic acid:
(i) Esterification reaction:

• When acid reacts with alcohols in the presence of little amount of concentrated sulphuric acid, the reaction produces esters. This reaction is called esterification reaction.
• Ester is a sweet smelling substance.
• When ethanoic acid reacts with ethanol in the presence of concentrated sulphuric acid, ester is produced.

(ii) Reaction with base Le. alkali (Saponification reaction):
When the ester formed in above reaction is heated with sodium hydroxide (a base) solution then the ester breaks down to give back original alcohol Le. ethanol and sodium salt of the carboxylic acid.

This reaction is called saponification because ills used in making soap.

(iii) Reaction with carbonates and hydrogencarbonates:
Ethanoic acid reacts with carbonates and hydrogen carbonates to produce salt, carbon dioxide and water. The salt produced is commonly called sodium acetate.

Question 47.
What is esterification?
Answer:
Esterification reaction:

• When acid reacts with alcohols in the presence of little amount of concentrated sulphuric acid, the reaction produces esters. This reaction is called esterification reaction.
• Ester is a sweet smelling substance.
• When ethanoic acid reacts with ethanol in the presence of concentrated sulphuric acid, ester is produced.

Question 48.
What Is an ester? State Its properties.
Answer:
1. Ester is a sweet smelling substance.
2. It is used in making perfumes and as a flavouring agent.
3. On treating ester with sodium hydroxide (a base alkali), the ester converts back to alcohol and sodium salt of carboxylic acid. This reaction is called the saponification process.

Question 49.
How does ethanoic acid react with alkali? OR What is saponification? State its reaction.
Answer:
Reaction with base Le. alkali (Saponification reaction):
When the ester formed in above reaction is heated with sodium hydroxide (a base) solution then the ester breaks down to give back original alcohol Le. ethanol and sodium salt of the carboxylic acid.

This reaction is called saponification because ills used in making soap.

Question 50.
State and explain chemical reaction of ethanoic acid with carbonates and hydrogencarbonates.
Answer:
Reaction with carbonates and hydrogencarbonates:
Ethanoic acid reacts with carbonates and hydrogen carbonates to produce salt, carbon dioxide and water. The salt produced is commonly called sodium acetate.

Question 51.
What is soap? Draw and explain the structure of its molecule.
Answer:
Soap:
1. A molecule of soap is a sodium or potassium salt of long chain carboxylic acid.
2. Each long chain soap molecule is made up of two parts. They are —

• a polar head and
• a polar tail.

The polar head (Hydrophilic end): It is made up of functional group sodium carboxylate ( – COONa). It is ionic ¡n nature. Moreover, it is hydrophilic which means it attracts water (or say dissolves in water).

The non-polar tail (Hydrophobic end): It is a long hydrocarbon chain. It is not ionic. It is hydrophobic which means it repels water (but dissolves in oil) or say dirt.

Question 52.
What are detergents? Explain their molecular structure.
Answer:
Detergents:

• Detergent is a chemical substance used for cleaning purposes.
• A molecule of detergent is ammonium or sulphonate salt of long chain carboxylic acid.
• In detergent, the functional group sodium sulphonate (-SO₃Na) is attached to the long chain of hydrocarbon.

Question 53.
How does a soap cleans the dirt from clothes? Explain.
Answer:
Cleansing action of soap:
1. When soap is added in water that has dirty clothes —

• The polar head (hydrophilic or ionic end) of the soap molecule dissolves in water whereas the
• The non-polar tail (hydrophobic end) dissolves in oily dirt.

2. The soap molecule dissolves and get arranged in a systematic orientation where in the head portion faces towards the water and the tail faces the dirt. Due to this orientation, the oily dirt will get trapped at the centre of the micelle. Such patterns having ring like structures are called micelles and the process is known as micelle formation.

3. The micelles capture the dirt particles within their rings and suspend them in water. Finally, when we brush and rub the clothes, the dirt gets removed from the clothes.

Question 54.
How are micelles formed?
Answer:
1. When soap is added in water that has dirty clothes —

• The polar head (hydrophilic or ionic end) of the soap molecule dissolves in water whereas the
• The non-polar tail (hydrophobic end) dissolves in oily dirt.

2. The soap molecule dissolves and get arranged in a systematic orientation where in the head portion faces towards the water and the tail faces the dirt. Due to this orientation, the oily dirt will get trapped at the centre of the micelle. Such patterns having ring like structures are called micelles and the process is known as micelle formation.

Question 55.
What is the advantage of detergent over soap? OR Use of detergent has increased compared to washing soap. Give reason. OR The limitations of soap ¡s overcome by a detergent. Explain.
Answer:
1. Hard water contains calcium (Ca) and magnesium (Mg) salts.
2. When hard water is used during washing, the salts of Ca and Mg react with soap to form insoluble salts or say precipitates of Ca and Mg. Also called scum. The scum does not dissolve in water. Hence, more soap is used for cleaning.
3. On the other hand, detergent reacts with Ca and Mg salts of hard water and forms soluble salts of Ca and Mg. These salts remain in water and so very less amount of detergent is needed for washing.
4. Hence, the use of detergent has increased compared to washing soap.

Question 56.
Differentiate between soap and detergent.
Answer:

Question 57.
Functional groups play a key role in organic compounds. Give reason.
Answer:
1. An atom or a group of atoms responsible for chemical behaviour of the parent molecule is called a functional group.
2. Different molecules that contain same kind of functional group or groups undergo similar reactions.
3. By learning the properties of these functional groups, we can study and understand properties of many organic compounds.
4. As a result, functional groups play a key role in organic compounds.

Question 58.
In a homologous series, as you progress in the series, the physical properties change but chemical properties do not. Give reason.
Answer:
1. In a homologous series, as the series progresses, the number of carbon and hydrogen atoms increase. This means the molecular mass of every next compound in the series is more than the previous compound.

2. Physical properties such as boiling point, melting point, density, solubility, etc. are dependent on the molecular mass. Since, the molecular mass gradually increase in the series, the compounds show change in physical properties.

3. The chemical properties of any homologous series are determined by the functional group of the series. Since, all the compounds of a given series have the same functional group, the chemical properties of all the compounds remain same.

Question 59.
Write the names of the following compounds.

Answer:
(a) This compound contains five carbon C atoms and  – COOH functional group. Hence, its name is Pentanoic acid.
Derivation of name: Remove ‘e’ from Pentane and add ‘oic acid’ for functional group – COOH = Pentanoic acid.

(b) This corn pound contains five carbon C atoms along with one triple bond. Hence, its name is Pentyne.
Derivation of name: Remove ‘e’ from Pentane and add ‘yne’ = Pentyne.

(c) This compound contains seven carbon C atoms and – CHO functional group. Hence, its name is Heptanal.
Derivation of name: Remove ‘e’ from Heptane and add ‘al’ for functional group – CHO = Heptanal.

(d) This compound contains five carbon C atoms and  – OH functional group. Hence, its name is Pentanol.
Derivation of name: Remove ‘e’ from Pentane and add ‘ol’ for functional group – OH = Pentanol.

Question 60.
Identify and name the functional groups present in the following compounds.

Answer:

Question 61.
How can we differentiate between saturated and unsaturated hydrocarbons on the basis of combustion?
Answer:
1. Saturated hydrocarbons burn with a blue flame i.e. a clean flame which neither emits smoke nor leaves sooty deposit.
2. In unsaturated hydrocarbons, the percentage of carbon is higher. Hence, they burn in air producing a yellow sooty flame.
3. Thus, by studying the type of flame and the residue we can differentiate between saturated and unsaturated hydrocarbons.

Question 62.
Why kerosene stove burns with a blue flame but a kerosene lantern with a yellow?
Answer:
1. The construction of stove is such that it allows oxygen-rich air to enter into the stove in sufficient quantity. So, the stove burns with a blue flame.
2. The lantern is covered with glass to prevent the flame from getting extinguished. Hence, there is limited supply of oxygen available to lantern and so it burns with a yellow flame.

Question 63.
A compound X ¡s formed by the reaction of a carboxylic acid C₂H₄O₂ and an alcohol in presence of a few drops of H₂SO₄. The alcohol on oxidation with alkaline KMnO₄ followed by acidification gives the same carboxylic acid as used in this reaction. Give the names and structures of (a) carboxylic acid, (b) alcohol and (c) the compound X. Also write the reaction.
Answer:
(a) Carboxyllc acid having molecular formula C₂H₄O₂ is acetic acid (or ethanoic acid) Its structure is

(b) Since, an alcohol which on oxidation with alkaline KMnO₄ followed by acidification gives ethanoic acid. it must be ethanol. Its structure is CH₃CH₂ – OH.

Question 64.
Which oil and ghee are good for health and which should be avoided?
Answer:
1. Vegetable oils containing unsaturated fat are good for health and hence should be used for cooking. Sun-flower oil, groundnut oil, etc. are some of them.
2. Vegetable ghee contains saturated fat and hence is not good for health.
3. Animal fats found in butter and desi ghee also contain saturated fats. But, they can be eaten in moderate quantity to maintain good health.

Question 65.
What happens when hydrogen atoms of methane are one by one replaced with chlorine atoms in the presence of sunlight?
Answer:
1. Formula of methane is CH₄.
2. When one hydrogen is replaced by chlorine, we get chioromethane (CH₃Cl), when two hydrogen are replaced by two chlorine atoms we get dichioromethane (CH₂Cl₂), when three hydrogen atoms are replaced by three chlorine atoms we get chloroform or carbon trichloride (C₃Cl₄). Finally, when four hydrogen atoms are replaced by four chlorine atoms, we get carbon tetrachioride (CCl₄).

Question 66.
Consumption of alcohol should be avoided. Give reason.
Answer:
1. Consuming pure (undiluted) ethanol e. alcohol even in a small quantity can prove lethal.
2. Consuming diluted ethanol that too in small quantity causes drunkenness.
3. People fall prey and addicted to alcoholic drinks, This wins individual health, family and society at large.
4. Due to all these harmful effects, it is said that consumption of alcohol should be avoided.

Question 67.
Ethene is formed when ethanol at 443 K is heated with excess of concentrated sulphuric acid. What is the role of sulphuric acid in this reaction? Write the balanced chemical equation of this reaction.
Answer:
When ethanol is heated with excess of concentrated sulphuric acid at 443 K, it gets dehydrated to form ethene.

Here, concentrated sulphuric acid acts as a dehydrating agent which removes water molecule from the ethanol molecule.

Question 68.
What is the role of metal or reagents written on arrows in the given chemical reactions?

Answer:
(a) Nickel (Ni) acts as the catalyst during the reaction.
(b) Conc. H₂SO₄ increases the rate of the forward reaction. In other words, conc. H₂SO₄ acts as a catalyst which works as a dehydrating agent.
(c) Alkaline KMnO₄ acts as an oxidizing agent and oxidizes ethanol to ethanoic acid.

Question 69.
Explain the given reactions with the examples.
(a) Hydrogenation reaction
(b) Oxidation reaction
(c) Substitution reaction
(d) Saponification reaction
(e) Combustion reaction
Answer:
(a) Hydrogen reaction:
Addition of hydrogen to the unsaturated molecule for making it saturated is known as hydrogenation.
Example:

(b) Oxidation reaction:
The reactions in which an oxidizing agent supplies nascent oxygen for oxidation are called oxidation reactions.

(c) Substitution reaction:
When one atom or a group of atoms replaces or substitutes another atom or a group of atoms from the molecule, It is known as substitution reaction.

(d) Saponification reaction:
When esters are hydrolyzed in the presence of a base (NaOH) then the reaction is called saponification reaction.
Example: CH₃COOCH₃ + NaOH → CH₃COONa + CH₃OH

(e) Combustion reaction:
Organic compounds burn readily in air to form CO₂ and water vapour along with lot of heat. Such a reaction is known as combustion reaction. Example: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O + Energy

Very Short Answer Type Question :

Question 1.
State the atomic number of carbon and its electronic configuration.
Answer:
Atomic number Z = 6. Electronic configuration: K – 2, L – 4.

Question 2.
What defines the reactivity of an element?
Answer:
The tendency to lose or gain the electrons so as to complete the outer shell to attain noble gas configuration decides how reactive an element would be.

Question 3.
How does carbon bonds?
Answer:
Carbon shares its valence electrons with other carbon atoms or with atoms of other elements for joining purpose and attaining noble gas configuration. The bond formed while sharing is covalent bond.

Question 4.
What is a covalent bond?
Answer:
A chemical bond formed between two or more atoms by mutual sharing of valence electrons is known as a covalent bond.

Question 5.
Why is carbon called tetravalent?
Answer:
Since the valency of carbon is 4, it is called tetravalent (tetra = 4, valent = valency)

Question 6.
Which two important characteristics of carbon are responsible for formation of a very large number of carbon compounds?
Answer:
Catenation and tetravalency

Question 7.
In which forms is carbon available?
Answer:
In free state carbon occurs as diamond and graphite. In combined state, carbon occurs in the form of compounds such as CO₂ in the air, carbonates, fossil fuels, etc.

Question 8.
Why covalent compounds have low melting and boiling point?
Answer:
Covalent bonds of compounds are weaker as compared to ionic bonds. Hence, in general, covalent compounds have lower melting and boiling points.

Question 9.
Why covalent compounds are bad conductors of electricity?
Answer:
Compounds having covalent bonds do not have ions or free electrons that conduct electricity. Hence, covalent compounds do not form strong electrolytes and so they are not very good conductors of electricity.

Question 10.
What is the problem with silicon exhibiting catenation property?
Answer:
Silicon can show catenation property by forming compounds with hydrogen which have chains upto seven or eight atoms, but the problem is that these compounds are very reactive.

Question 11.
Draw the electron bond between two oxygen atoms.
Answer:

Question 12.
In electron dot structure, the valence shell electrons are represented by crosses or dots.
(a) The atomic number of chlorine is 17. Write its electronic configuration
(b) Draw the electron dot structure of chlorine molecule.
Answer:
(a) Electronic configuration of Cl (17):

(b) Electron dot structure of chlorine molecule:

Question 13.
Define hydrocarbon. Give two examples.
Answer:
A compound macle up of only hydrogen and carbon is called hydrocarbon. Methane (CR₄), ethane (C₂H₄), etc.

Question 14.
Define saturated hydrocarbons.
Answer:
Hydrocarbons having single covalent bonds between the carbon atoms are called saturated hydrocarbons. For example, methane, ethane, etc.

Question 15.
Define unsaturated hydrocarbons.
Answer:
Hydrocarbons having double or triple covalent bonds between the carbon atoms are called unsaturated hydrocarbons. For example, ethene and ethyne.

Question 16.
What are alkanes? Give one example.
Answer:
Saturated hydrocarbons have single covalent bonds arid are called alkanes. For example, methane, ethane, etc.

Question 17.
Draw electron dot structure of ethane.
Answer:
Ethane (C₂H₆):

Question 18.
Draw electron dot structure if ethene.
Answer:
Ethene (C₂H₄):

Question 19.
What are alkenes and alkynes? Give one example.
Answer:
Unsaturated hydrocarbons have double or triple covalent bonds and are called alkenes and alkynes respectively. Ethane is alkene and ethyne is alkyne.

Question 20.
Which compound is shown here?
Answer:
Benzene

Question 21.
Name the following structure.
Answer:

Question 22.
What is heteroatom?
Answer:
In an organic compound, any atom other than carbon or hydrogen is known as heteroatom.

Question 23.
What Is a functional group?
Answer:
An atom or a group of atoms (i.e. heteroatoms) responsible for chemical behavior of the parent molecule is called a functional group.

Question 24.
Mention the functional group containing oxygen.
Answer:

Question 25.
What is homologous series?
Answer:
The series of organic compounds in which each compound differs from its previous or later compound by (CH₂) is called homologous series.

Question 26.
What is the difference between molecular masses of any two successive members of the homologous series? Why?
Answer:
Reason:The difference between two successive members of any homologous series is CH₂. The atomic mass of carbon is 12 u and that of hydrogen is 1 u. So, the molecular mass of CH₂ = 14 (12 x 1+1 x 2).

Question 27.
Why the physical properties of a compounds change as one moves in a homologous series?
Answer:
Physical properties are dependent on the molecular mass. Since, the molecular mass gradually increase in the series, the compounds show change in physical properties.

Question 28.
What do you mean by IUPAC names?
Answer:
Organic compounds or hydrocarbons have two names, (a) Common name and (b) IUPAC name. IUPAC names are names given by IUPAC i.e. International Union of Pure and Applied Chemistry. These names are also called IUPAC nomenclature.

Question 29.
Give common name and IUPAC name of any two compounds.
Answer:

Question 30.
How is the alcohol group represented?
Answer:
R-OH

Question 31.
What are aldehydes and ketones?
Answer:
Aldehydes and ketones are organic compounds  having ( – CHO) and ( – C = O) respectively as their functional groups.

Question 32.
What are carboxylic acids?
Answer:
Organic compounds containing carboxyl (- COOH) groups as their functional group are called carboxylic acids.

Question 33.
Which compound is this:

Answer:
Propanone

Question 34.
List four important properties of carboncorn pounds.
Answer:
(1) Combustion
(2) Oxidation
(3) Addition reaction arid
(4) Substitution reaction

Question 35.
Define complete combustion as a property of carbon compounds.
Answer:
Carbon and all its allotropes burn completely insufficient amount of oxygen. This is known as complete combustion.

Question 36.
Give the reaction when carbon burns insufficient supply of oxygen.
Answer:

Question 37.
Define oxidation of carbon compounds.
Answer:
Oxidation is the reaction in which carbon compounds take up oxygen in the presence of oxidizing agents to give another compound.

Question 38.
Define addition reaction.
Answer:
The reaction in which an unsaturated (alkene or alkyne) hydrocarbon combines with another substance to give a single but saturated (alkane) product is called addition reaction. Undergoing addition reaction is one of the properties of carbon compounds.

Question 39.
What is substitution reaction?
Answer:
The reaction in which one or more hydrogen atoms of a hydrocarbon are replaced by some other atoms (like chlorine) is called a substation reaction.

Question 40.
What is hydrogenation or hydrogenation of oils?
Answer:
When a vegetable oil (like groundnut oil) is heated with hydrogen in the presence of nickel catalyst, the oil turns into a saturated fat called vegetable ghee or say vanaspati ghee. This reaction is known as hydrogenation of oils.

Question 41.
State two uses of ethanol (alcohol).
Answer:
1. Ethanol is the active ingredient of all alcoholic drinks.
2. Ethanol is a good solvent and so it is also used to make medicines such as tincture iodine, cough syrups and several other tonics.

Question 42.
State the reaction of ethanol with sodium.
OR
A gas is evolved when ethanol reacts with sodium. Name the gas evolved and also write the balanced chemical equation of the reaction involved.
Answer:
Ethanol reacts with sodium and produces sodium ethoxide along with evolution of hydrogen gas.

Question 43.
What is ethanoic acid?
Answer:
1. Ethanoic acid is the second member of the homologous series of carboxylic acids.
2. Its formula is CH₃COOH. The common name of ethanoic acid is acetic acid.

Question 44.
Why in few countries a mixture of water and alcohol is used instead of simply water in car radiators?
Answer:
Water freezes at O °C. So, in countries where temperature falls below O°C mixing alcohol with water brings down the freezing point and car radiator can function property.

Question 45.
What is esterification?
Answer:
When ethanoic acid reacts with alcohols in the presence of little amount of concentrated sulphuric acid, the reaction produces esters. This reaction is called esterification reaction.

Question 46.
What is saponification?
Answer:
When the ester (CH₃COOC₂H₅) formed in above reaction is heated with sodium hydroxide (a base) solution then the ester breaks down to give back original alcohol Le. ethanol and sodium salt of the carboxylic acid. This reaction is called saponification because it is used in making soap.

Question 47.
What is a soap molecule formed of?
Answer:
A molecule of soap is a sodium or potassium salt of long chain carboxiylic acid.

Question 48.
Give an idea of the hydrophlllc end of the soap.
Answer:
Hydrophilic end i.e. the polar head of a soap molecule is made up of functional group sodium carboxylate (-COONa). It is ionic in nature. Since it is hydrophilic it attracts water (or say dissolves in water).

Question 49.
Give a brief description of the non-polar tall of the soap.
Answer:
The non-polar tail is a long hydrocarbon chain. It is not Ionic. It is hydrophobic which means it repels water (i.e. it dissolves in oil). The hydrophobic carbon chain dissolves In oil or say dirt.

Question 50.
What are micelles?
Answer:
The soap molecule dissolves and get arranged in a systematic orientation where in the head portion faces towards the water and the tail faces the dirt. Due to this orientation, the oily dirt will get trapped at the centre of the micelle. Such patterns having ring like structures are called micelles. water brings down the freezing point and car radiator can function property.

Fill in the Blanks:

1. Acetic acid melting point : 290 K; Boiling point …………..
Answer:
391 K

2. The correct electron dot structure of a water molecule is …………..
Answer:

3. ………….. the chief constituent of natural gas.
Answer:
Methane

4. Carbon forms four covalent bonds by sharing its four valence electrons with four univalent atoms, e.g. hydrogen. After the formation of four bonds. carbon attains the electronic configuration of ……………
Answer:
Neon

5. Covalently bonded molecules have ………….. intermolecular force.
Answer:
Weak

6. Covalent compounds have ……………. melting and …………… boiling points.
Answer:
Low; Low.

7. One of the allotrope of carbon is diamond. The other two are ………… and …………..
Answer:
Graphite; Buckminster fullerene.

8. As per an estimate there are about carbon compounds.
Answer:
3 million

9. Unsaturated compounds have bonds.
Answer:
Double or triple.

10. The first and simplest member of saturated hydrocarbon group is ……………
Answer:
Methane

11. Isomers are organic compounds having same formula but different formula.
Answer:
Molecular, structural

12. Almost all types of organic compounds have series.
Answer:
Homologous

13. Which out of alkyne, alkane and alkene is unsaturated?
Answer:
Alkene and alkyne

14. Chlorine, bromine and iodine belong to the functíonal group
Answer:
Halogen

15. The name of the compound CH₃ — CH₂ — CHO is …………….
Answer:
Propanal

16. Atomic mass of carbon is ……………….
Answer:
12u

17. The general formula of alkene is ……………….
Answer:
CnH2n

18. Kerosene will bum with ………………. type of flame.
Answer:
Yellow and sooty

19. We get a clean blue flame in our home stoves because of ……………….
Answer:
Burning saturated hydrocarbons In the presence of sufficient oxygen.

20. Substances capable of adding oxygen to others are called …………….
Answer:
Oxidizing agents

21. ….. and …… work as oxidizing alcohols to acids.
Answer:
Alkaline potassium permanganate; Acidified potassium dichromate.

22. Palladium and nickel work as ……………… in addition reaction.
Answer:
Catalysts

23. ………….. are substances that cause a reaction to occur or proceed at a different rate without affecting the reaction.
Answer:
Catalysts

24. Ideally, oils containing should be used for cooking.
Answer:
Unsaturated fatty acids.

25. ……………. an important compound is formed when sodium reacts with ethanol.
Answer:
Sodium ethoxide

26. Heating ethanol in at 443 K with excess concentrated sulphuric acid results in to give ethane.
Answer:
Dehydration of ethanol.

27. Because ethanoic acid freezes in winter it is also called …………..
Answer:
Glacial acid

28 hydrocarbons have a sweet fruity smell.
Answer:
Ester

29. Ester can be converted back into alcohol by treating with ……………
Answer:
Sodium hydroxide

30. Saponification reaction is used to prepare …………..
Answer:
Soap

31. Mast dirt of clothes is in nature.
Answer:
Oily

True Or False

1. Ethanol boils at 156 K. — False
2. Methanol boil at 111 K. — True
3. Nitrogen bonds with nitrogen through triple bond. — True
4. Carbon compounds are found in three shapes namely long chain, branched chain and elliptical rings.– False
5. The compounds of a homologous series show similar physical properties but different chemical properties. — False
6. For IUPAC nomenclature of Ketone compounds, the last alphabet ‘e’ is removed from the name of the hydrocarbon and the suffix ‘one’ is added. — True
7. The hydrocarbons in which any two nearby carbon atoms are combined by a double bond unsaturated hydrocarbons are called alkenes. — True
8. A molecule of detergent is ammonium or sulphonate salt of long chain carboxylic acki. — True
9. Soaps react with calcium and magnesium of hard water and then clean the clothes. — False

Match the Following

Question 1.

Answer:
(a-iv) (b-ii) (c-i) (d-iii)

Question 2.

Answer:
(a-iv) (b-i) (c-ii) (d-iii)

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 9 Some Applications of Trigonometry Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 9 Some Applications of Trigonometry

Short/Long Answer Type Questions

Question 1.
From a point on a bridge across a river the angles depression of the banks on opposite sides of the river are 30° and 45° respectively. If the bridge is at a height of 4m from the banks. Find the width of river.
Solution :
Let A and B represent points on the bank on opposite sides of the river so that AB is the width of river. Pis a on the bridg at a height of 4 m is DP = 4m. We are required to determine the width of the river, which is the length of side AB of ΔAPB
∴ AB = AD + BD

Angles of depression of the banks on opposite sides of the river are
⇒ ∠APR = 30°
⇒ ∠PAD = 30°
[alternate interior angles]
and ∠BPQ = 45°
∠PBD = 45°
In right triangle APD, we have
tan 30° = \(\frac {PD}{AD}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac {4}{AD}\)
⇒ AD = 4\(\sqrt{3}\) …………..(1)
In right triangle BPD, we have
tan 45° = \(\frac {PD}{BD}\)
⇒ 1 = \(\frac {4}{BD}\)
⇒ BD = 4 ………….(2)
Adding equ. (1) and (2), we get
AD + BD = 4\(\sqrt{3}\) + 4
= 4 × 1.732 + 4
⇒ AB = 6.928 + 4 = 10.928 m
Hence, width of river = 10.928 m.

Question 2.
As observed from the top of a 100 m high lighthouse from the sea level, the angles of depression of two ships are 30 and 45°. If one ship is exactly behind the other on same side of lighthouse, find the distance between two ships.
Solution :
Let the AB be the lighthouse of height of 100 m. Let D and C be position of two ships. The angle of depression of two ships are 30° and 45°

⇒ ∠DAP = 30°
⇒ ∠ADB = 30°
[Alternate interior angles]
and ∠CAP = 45°
⇒ ∠ACB = 45°
In right triangle ADB, we have
tan 30° = \(\frac {AB}{BD}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac {100}{BD}\)
⇒ BD = 100\(\sqrt{3}\)
⇒ 10 × 1.732 = 173.2 m
In right triangle ACB, we have
tan 45° = \(\frac {AB}{BC}\)
⇒ 1 = \(\frac {100}{BC}\)
⇒ BC = 100 m
⇒ DC = BD – BC
173.2 – 100 = 73.2 m
Hence distance between two ships
= DC = 73.2 m

Question 3.
The angle of elevation of the top of a hill from the foot of a tower is 60° and the angle of depression from the top of the tower of the foot of the hill is 30°. If tower is 50 m high, find the height of hill.
Solution :
Let AB be the hill of height hm and CD be the tower of height 50 m.

Angle of elevation of the top of a hill from the foot of a tower is 60° i.e. ∠ACB = 60° and angle of depression from top of the tower of the foot of the hill is 30 i.e. ∠BDE = 30°.
⇒ ∠DBC = 30°
[Alternate interior angles]
In right triangle BCD, we have
tan 30° = \(\frac {CD}{BC}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac {50}{BC}\)
⇒ BC = 50\(\sqrt{3}\) ……..(1)
In right triangle ABC, we have
tan 60° = \(\frac {AB}{BC}\)
⇒ \(\sqrt{3}\) = \(\frac {h}{BC}\)
⇒ h = \(\sqrt{3}\)BC ………..(2)
substituting, the value of BC in equ. (2), we get
h = \(\sqrt{3}\) × 50\(\sqrt{3}\)
= 50 × 3 = 150 m
Hence height of hill = 150 m

Question 4.
A vertical tower stands on a horizontal plane and is sur mounted by a vertical flag staff of height 6 m. At a point on the plane, the angle of elevation of the bottom and top of the flag staff are 30° and 45° respectvely. Find the height of the tower. Take \(\sqrt{3}\) = 1.73]
Solution :
Let AB be tower of height h m and BC be vertical flag staff of height 6m. At point D on the plane angle of elevation of the bottom and top of the flag staff are 30° and 45° respectively i.e. ∠ADB = 30° and ∠ADC = 45°
In right triangle ABD, we have

⇒ h = \(\frac{6 \times(1.73+1)}{(\sqrt{3})^2-(1)^2}\)
⇒ h = \(\frac{6 \times 2.73}{3-1}\)
⇒ h = \(\frac{6 \times 2.73}{2}\)
⇒ h = 3 × 2.73
⇒ h = 8.19 m
Hence, height of the tower h = 8.19 m

Question 5.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 48 meters high. Find the height of the building.
Solution :
Let AB be tower of the height 48 m and CD be building of height h m. Angle of elevation of the top of a building from the foot of the tower is 30° Le. ∠CBD = 30° and angle of elevation of top of the tower from the foot of the building is 60° L.E. ∠ACB = 60°,

Question 6.
An aeroplane when flying at a height of 4000 m. from the ground passes vertically above another aeroplane at an instant when the angles of the elevation of two planes from the same point on the ground are 60° and 45° repectively. Find the vertical distance between the aeroplanes at that instant.
Solution :
Let A and B be position of two aeroplanes. When A be the vertical above B and AC = 4000 m.
Let the angles of elevation of two nlanes from the point A be 60 and 45° respectively i.e. ∠ADC = 60° and ∠BDC = 45°

Hence, vertical distance between two aeroplanes = 1690.53 m.

Question 7.
Two men on either side of the cliff 80 m high observes the angles of the elevation of the top of the cliff to be 30° and 60° respectively. Find the distance between the two men.
Solution :
Let the CD be cliff of height 80 m. Let A abd B be position of eyes of two men. The angles of elevation of eyes of two men from top of the cliff be 30° and 60° respectively i.e. ∠CAD = 30° and ∠CBD = 60°.

Hence, distance between two men = 184.80 m

Question 8.
At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30°. The angle of depression of the reflection of the cloud in lake, at A is 60°. Find the distance of cloud from A.
Solution :
Let BR be surface of lake and A be point of observation P be position of cloud and C be the position of reflection of cloud in the lake. Draw AQ ⊥ PR. Angle of elevation of cloud from point A is 30° i.e. ∠PAQ = 30° and angle of depression of reflection of cloud in the lake is 60° i.e. ∠CAQ = 60°
Let PQ be h m, then PR = (h + 20) m and
RC = PR = (h + 20) m

substituting the value of h in equ. (1), we get
AQ = \(\sqrt{3}\) × 20 = 20\(\sqrt{3}\)m
In right ΔAPQ, we have
⇒ AP² = AQ² + PQ²
⇒ AP² = (20\(\sqrt{3}\))² + (20)²
⇒ AP² = 1200 + 400
⇒ AP² = 1600
⇒ AP = \(\sqrt{1600}\)
⇒ AP = 40
Hence, distance of cloud from point A = 40 m.

Question 9.
Amit standing on a horizontal plane, find a bird flying at a distance of 200m. From him at an elevation of 30°. Deepak standing on the roof of a 50m hight building, find the angle of elevation of the same bird to be 45°. Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak.
Solution :
Let A and B be two positions of Amit and Deepak. Let P be position of bird. Bird flying at a distance from A 200 m i.e. AP = 200 m. Angle of elevation of P from Amit is 30° i.e. ∠PAR = 30° and Angle of elevation of P from Deepak is 45° i.e.
∠PBQ = 45°

In right ΔAPR, we have
sin 30° = \(\frac {PR}{AP}\)
⇒ \(\frac {1}{2}\) = \(\frac{P Q+Q R}{200}\)
⇒ PQ + QR = \(\frac {200}{2}\)
⇒ PQ + 20 = 100 [∴ QR = BS]
⇒ PQ = 100 – 50 = 50 m
In right ΔPBQ, we have
sin 45° = \(\frac {PQ}{PB}\)
⇒ \(\frac{1}{\sqrt{2}}\) = \(\frac {50}{PB}\)
⇒ PB = 50\(\sqrt{2}\) m
Hence, distance of the bird from Deepak is 50\(\sqrt{2}\) m.

Question 10.
A vertical tower stands on a horizontal plane and is sur mounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and top of the flag staff are λ and β respectively. Prove that the height of the tower is \(\frac {h tan α}{tan β – tan α}\)
Solution :
Let AB be the vertical tower of height H and AD be flag staff of height h. The angle of elevation of bottom and top of flag staff are λ and β respectively i.e.
⇒ ∠ACB = α and
∠DCB = β

Hence proved.

Fill in the Blanks

Question 1.
……… is the height to which something is raised above a point of reference.
Solution :
Elevation

Question 2.
………. is the depth to which something is lowered below point of reference.
Solution :
depression

Question 3.
Numerically the angle of elevation is… to the angle of depression.
Solution :
equal

Question 4.
The angle of elevation and angle of depression both are measured with the ………. .
Solution :
horizontal

Question 5.
If two lines are ………, then the alternate interior angles formed are equal.
Solution :
parallel

Question 6.
Two angles having a sum of ………… are called complementary angles.
Solution :
90°

Question 7.
As the ……….. of an object increases, the angles of elevation of its top with the horizontal level also increases.
Solution :
height

Multiple Choice Questions

Choose the correct answer for each of the following:

Question 1.
A 20 m long ladder rests against a wall. If the feet of the ladder is 10 m away from the wall, then angle of the elevation is:
(a) 30°
(b) 60°
(c) 90°
(d) 45°
Solution :

Question 2.
When the sun is 30° above the horizontal, the length of the shadow casts by a 50 m high building is :
(a) 50\(\sqrt{3}\) m
(b) 25 m
(c) 25\(\sqrt{3}\) m
(d) \(\frac{50}{\sqrt{3}}\)m

Solution :
∵ Tan 30° = \(\frac {BC}{AB}\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac {50}{AB}\)
⇒ Length of shadow AB = 50\(\sqrt{3}\) m
Hence correct choice is (a)

Question 3.
The angle of elevation of a tower from a distance 100 m from its foot is 30°. Height of tower is :
(a) 50\(\sqrt{3}\)
(b) 100\(\sqrt{3}\)
(c) \(\frac{100}{\sqrt{3}}\)
(d) \(\frac{50}{\sqrt{3}}\)
Solution :
In ΔABC

Hence, height of tower = \(\frac{100}{\sqrt{3}}\)m
Hence correct choice is (c).

Question 4.
A kite is flying at a height of 30 m from the ground. The length of string from the kite to the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is :
(a) 45°
(b) 30°
(c) 60°
(d) 90°
Solution :
In ΔABC

sin θ = sin 30°
⇒ θ = 30°
Hence angle of elevation 30°
So correct choice is (b).

Question 5.
If the elevation of the Sun changed from 30° to 60°, then difference between the length of shadows of a pole of height 15 m is :
(a) 7.5 m
(b) 15 m
(c) 103 m
(d) 573 m
Solution :
In ΔBCD

⇒ x + 5\(\sqrt{3}\) = 15\(\sqrt{3}\)
⇒ x = 15\(\sqrt{3}\) – 5\(\sqrt{3}\)
⇒ x = 10\(\sqrt{3}\) m
Hence required difference x = 10\(\sqrt{3}\) m
So correct choice is (c)

Question 6.
From a window 15 above the ground, the angle of elevation of the top of a tower is α and angle of depression of the foot of the tower is β. Then height of the tower is :
(tan α = \(\frac {5}{3}\), tan β = \(\frac {2}{3}\))
(a) 50 m
(b) 52.5 m
(c) 55 m
(d) 53.5 m
Solution :
Tan α = \(\frac {5}{3}\), Tan β = \(\frac {2}{3}\)
In ΔАВС Tan β = \(\frac{2}{3}=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{15}{\mathrm{BC}}\)

= 37.5 m
∴ Height of tower = CD + DE
= 15 + 37.5 m
= 52.5 m
So correct choice is (b).

Question 7.
A, B, C are three collinear points on the ground such that B lies between A and Cand AB = 10 m. If the angles of elevation of the top of a vertical tower at C are respectively 30° and 60° as seen from A and B, then the height of tower is :
(a) 5\(\sqrt{3}\) m
(b) 5 m
(c) \(\frac{10}{\sqrt{3}}\)
(d) \(\frac{20}{\sqrt{3}}\)
Solution :
Let CD = x, and BC = y
In ΔBCD,

So correct choice is (a).

Question 8.
A person of height 2 m wants to get a fruit which is on a pole of height \(\frac {10}{3}\)m. If he stands at a distance of \(\frac {4}{3}\)m from the foot of the pole, then the angle at which he should throw the stone, so that it hits the fruits, is :
(a) 15°
(b) 30°
(c) 45°
(d) 60°
Solution :
From the figure

= Tan 45°
∴ θ = 45°
So correct choice is (c)

Haryana State Board HBSE 10th Class Science Important Questions Chapter 3 Metals and Non-metals Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 3 Metals and Non-metals

Question 1.
What are metals?
Answer:
1. Those elements which lose electrons to become positive ions are called metals. For example, magnesium (Mg) is a metal since it loses 2 electrons to become positive ion Mg²⁺.
2. Metals conduct heat and electricity.
3. They are hard, shiny, heavy and sonorous (i.e. make sound on banging).
4. Iron, copper, gold, aluminium, zinc, lead, etc. are all metals.

Question 2.
Discuss the physical properties of metals.
Answer:
Physical properties of metals :
1. Luster: Metals in their pure state have a shining surface. This property is called metallic luster. For e.g., gold, silver, etc.

2. Hardness: Most of the metals are hard. The hardness of various metals are different.
→ Interestingly metals like sodium and potassium are relatively soft and can be easily cut with a knife.

3. Malleability: Some metals can be hammered and turned into thin sheets. This property of metals is known as malleability. For e.g., gold, silver and aluminium.

4. Ductility: The ability of metals to be drawn into thin wires is called ductility. Metals are generally ductile.

5. Conductivity of heat: Metals are good conductors of heat.

6. Melting and boiling points: Metals have high melting and boiling points.

7. Electrical conductivity: Metals are good conductors of electricity. The power of conduction of heat and electricity is very high for copper, silver and gold metals.

8. Sonorous: Metals produce ringing sound on striking which means they are sonorous.

9. Alloys: Alloys can be prepared by adding one metal to the other. Brass, gold ornaments, stainless steel, etc. are the examples of alloys.

Question 3.
What is malleable and ductile? Explain.
Answer:
1. Some metals can be hammered and turned into thin sheets. This property of the metals is known as malleability.
2. This property is specially found in metals like gold, silver and aluminium. Hence, thin strips can be prepared from gold and silver and very thin paper-like foil can be prepared from aluminium.
3. Some metals are ductile i.e. we can draw thin wires from them. For e.g. gold and silver.
4. For example, about 2 kilometer long wire can be drawn from one gram gold. Wires can also be prepared from metals like copper and aluminium, by drawing.

Question 4.
Several metals conduct electricity even then mainly aluminium and copper are used for making electric wires. Why so?
Answer:
1. Almost all metals conduct electricity, but all metals are not available easily.
2. Metals like gold are very costly and metal like iron, rusts.
3. Aluminium and copper are two such metals which are easily available, cheap, and also do not rust. Hence, they are widely used in making electric wires.

Question 5.
What are non-metals?
Answer:
1. Elements that form negative ions by gaining electrons are called non-metals. For example, oxygen forms oxide ion O^-2 by gaining electron and hence oxygen is a non-metal.
2. Carbon, sulphur, hydrogen are other examples of non-metals.

Question 6.
State physical properties of non-metals.
Answer:
Physical properties of non-metals:

• Non-metals are neither ductile nor malleable.
• Solid non-metals except iodine, are lusterless.
• Non-metals except diamond, are soft.
• Non-metals are bad conductors of heat and electricity. However, graphite is a good conductor of electricity.

Question 7.
“We cannot classify elements only on the basis of their physical properties.” Explain. OR State exceptions of metallic and non-metallic elements.
Answer:
(A) Exceptions found in metals:

• Although metals are solid but mercury exists as liquid at room temperature.
• Metals have high melting points but metals namely gallium and caesium have very low melting points.
• Alkali metals (lithium, sodium and potassium) are so soft that they can be cut with knife. Also, they have low densities and melting point.

(B) Exceptions found in non-metals:

• Non-metals exists as solids or gases at room temperature but bromine exists as a liquid.
• Non-metals do not possess lustre but iodine does.
• Non-metals have low melting point, but diamond, an allotrope of carbon has very high melting and boiling point.
• Generally, non-metals do not conduct electricity, but graphite, an allotrope of carbon is the only non-metal which conducts electricity.
• Owing to such exceptions we cannot classify metals or non-metals only on the basis of physical properties. Hence, they can be classified more clearly on the basis of their chemical properties.

Question 8.
What is allotrope? Give an example.
Answer:
Some elements possess the property of existing in two or more different forms in the same physical state. Such property is called allotropy. The different forms of the elements are called allotropes.

Example:

• Carbon, a non-metal exists in different forms and each of its form is called an allotrope.
• Diamond and graphite are two well-known allotropes of carbon.

Question 9.
Differentiate between metals and non-metals on the basis of their physical properties.
Answer:

For information only:

Before studying the chemical properties of metals, let us take a basic idea of the reactivity series. The chemical properties of metals are determined by the way they react with oxygen, water, acid, etc. All metals do not react with a given element or a compound in the same manner. Some metals react vigorously, some react less vigorously and some do not react at all.

The question then is how do we determine if a metal will react with an element/compound? Or if it will react then in which manner? For this having a basic idea about the reactivity series of metals will make understanding the properties of metals quite easy.

Activity (reactivity) series :

The arrangement of metals in decreasing order of their reactivities is called the activity or reactivity series of metals. The series is given in the adjacent column. Metals above hydrogen are more reactive whereas those below hydrogen are less reactive.

Potassium (K) and sodium (Na) are most reactive metals. Silver (Ag) and gold (Au) are least reactive metals. As we move down from K to Au, the reactivity decreases. This fact will play a very important role in determining the type of reaction that will take place between the metal and the element/compound.

Now let us understand the chemical properties of metals. Note that the points discussed in each property have been arranged in the decreasing order of reactivity i.e. from most reactive metal to the least reactive metal.)

Question 10.
How does metal reaòt with oxygen of air?
Answer:
Reaction of metals with oxygen (O₂):

• Metals can easily give electrons to oxygen.
• Hence, metals combine with oxygen to form metal oxides.

Metal + Oxygen gas → Metal oxides

(1) Potassium (K) and sodium (Na) are so reactive that even if they are simply kept open in the room they react with oxygen (O₂) of air. On reacting they start burning very vigorously and catch fire. Hence, they are stored immersed in kerosene.

(2) Metals such as magnesium, aluminium, zinc, etc. are covered with a thin layer of oxide. Hence, they do not react directly with air. So, we need to provide heat to for making them react with oxygen.

(3) Iron on heating in air reacts with the oxygen and forms oxides. Iron does not bum in air the way potassium and sodium does. However, iron fillings do burn vigorously.

(4) When copper s heated in air it combines with oxygen to form black coloured copper (II) oxide.

(5) Silver and gold are least reactive. They do not react with oxygen under any case.

Question 11.
What are amphoterlc oxides? State examples
Answer:
Amphoteric oxides:
1. Some metal oxides react with both acids and bases to produce salt and water. Such metal oxides are called amphoteric oxides.
Example:
(i) Reaction with acid:

(ii) Reaction with base:

Question 12.
How do metal oxides react with water?
Answer:
Reaction of metal oxides with water:
Most metal oxides are insoluble in water. However, some dissolve in water and form alkalis.
Example: Sodium oxide and potassium oxide dissolve in water to produce alkalis.
(i) Na₂O_(s) + H₂O_(l) → 2NaOH_(aq)
(ii) K₂O_(s) + H₂O_(l) → 2KOH_(aq)

Question 13.
Discuss how metal reacts with water along with necessary chemical reactions.
Answer:
Reaction of metal with water (H₂O):
1. Metals on reaction with water form metal oxides and produce hydrogen gas. Metal oxides that are soluble in water dissolve in it to further form metal metal hydroxide.
Metal + Water → Metal oxide + Hydrogen + Heat gas
Metal oxide + Water → Metal hydrxide (on dissolving)

Question 14.
State and explain how metals react with dilute acid.
Answer:
Reaction of metals with dilute acid:

• All metals do not react with dilute acids.
• Those metals which react with dilute acids produce metal salt and hydrogen gas.

Hydrogen gas is liberated if the metal reacts with dilute hydrochloric acid (HCl), dilute sulphuric acid (H₂SO₄), etc. but, not with dilute nitric acid (HNO₃). However, magnesium and manganese metals can react with very dilute nitric acid and produce hydrogen gas.

Question 15.
Why hydrogen gas is liberated when metals react with dilute hydrochloric acid and dilute sulphuric acid but not with nitric acid?
Answer:
1. When metals react with dilute nitric acid (HNO₃), they do not produce hydrogen gas.
2. The reason behind this is that nitric acid is a strong oxidizing agent and so it oxidizes H₂ produced during the reaction to H₂O and its get reduced to any of the nitrogen oxides such as N₂O, NO, NO₂.

Question 16.
How will be the rate of liberation of hydrogen gas among various metals when they react with dilute acids?
Answer:
1. Metals above hydrogen in the reactivity series liberate hydrogen gas when they react with dilute mineral acid like hydrochloric acid, sulphuric acid. etc. (But not nitric acid).
2. Higher a metal in the reactivity series, the fastest will it liberate the hydrogen gas.
Example:
1. In the reactivity series, the elements magnesium (Mg), aluminium (Al), zinc (Zn), and iron (Fe) are placed above hydrogen in the sequence. Hence, the rate of liberation of hydrogen gas in the descending order will be Mg > Al > Zn > Fe.

Question 17.
Write equations for the reactions of magnesium, aluminium, zinc and iron with dilute hydrochloric acid.
Answer:
(i) Mg_(s) + 2HCl_(aq) → MgCl_2(aq) + H_2(g)
(ii) 2Al_(s) + 6HCl_(aq) → 2AlCl₃(_(aq) + 3H_2(g)
(iii) Zn_(s)+ 2HCl_(aq) →  ZnCl_2(aq) + H_2(g)
(iv) Fe_(s) + 2HCl_(aq) → FeCl_2(aq) + H_2(g)

Question 18.
State and explain how metals react with solutions of other metal salts.
Answer:
Reaction of metals with solutions of other metal salts:
1. When metals react with solutions of salts of other metals, the more reactive metals displace less reactive metals from their salt solutions. This can be stated as follows:

Example: When a strip of iron metal is put into solution of copper sulphate CuSO₄, the more reactive Fe metal displaces less reactive copper Cu metal.

Question 19.
State important chemical properties of metals without writing chemical equations.
Answer:
Chemical properties of metals:
1. Reaction of metals with oxygen (O₂):

• Metals can easily give electrons to oxygen.
• Hence, metals combine with oxygen to form metal oxides.
  Metal + Oxygen gas → Metal oxides

2. Reaction of metal with water (H₂O):

• Metals on reaction with water form metal oxides and produce hydrogen gas. Metal oxides that are soluble in water dissolve in it to further form metal metal hydroxide.

Metal + Water → Metal oxide + Hydrogen + Heat gas
Metal oxide + Water → Metal hydroxide (on dissolving)

3. Reaction of metals with dilute acid:

• All metals do not react with dilute acids.
• Those metals which react with dilute acids produce metal salt and hydrogen gas.

4. Reaction of metals with solutions of other metal salts:
When metals react with solutions of salts of other metals, the more reactive metals displace less reactive metals from their salt solutions. This can be stated as follows:

Question 20.
What is displacement reaction? Out of copper and iron, which is more reactive? Why iron is placed above copper in the reactivity series?
Answer:
1. The reaction in which a more reactive element displaces (removes) less reactive element from its compound is called a displacement reaction.
2. Iron (Fe) is more reactive than copper (Cu) which means iron will react easily compared to copper. Hence, iron is place above copper in the reactivity series.

Question 21.
Write a detailed note on activity (reactivity) series of metals.
Answer:
Activity (reactivity) series:
The arrangement of metals in decreasing order of their reactivities is called the activity or reactivity series of metals.
Explanation:

• All metals do not react at same rate.
• Some metals are more reactive while others are less reactive.
• Those metals which lose electrons easily and form ions are called more reactive metals. For example, calcium (Ca). On the other hand, less reactive metals do not lose electrons easily. For example, gold (Au).

Question 22.
What Information can one gain from reactivity series of metals?
Answer:
One can gain the following Information from reactivity series of metals:
(1) A more reactive metal will displace a less reactive metal (i.e. a metal below it in the reactivity series) from its solution.
(2) Metals present at the top of the activity series are less electropositive and do not occur freely In nature. On the other hand, metals at the bottom of the series are more electropositive and generally occur freely in nature.

Question 23.
Give two differences between chemical properties of metals and non-metals.
Answer:

Question 24.
What is chemical bond? Give Its types along with one example of each.
Answer:
Chemical bond: The phenomenon through which atoms of a molecule of a compound attract each other and combine is known as a chemical bond.
There are two type of chemical bonds. They are :

(i) Ionic bond : The bond formed between a metal and a. non-metal is called an ionic bond. For example, sodium (Na) metal and chlorine (Cl) non-netal bond with each other through ionic bond and form NaCl (common salt).

(ii) Covalent bond : The bond formed between two non-metals is called a covalent bond. For example, two hydrogen (H) atoms join with each other through a covalent bond and form a hydrogen molecule (H₂).

Question 25.
Explain how Na and Cl bond to form NaCl (Sodium chloride). OR Give an example to demonstrate how metals and non-metals react.
Answer:
1. Formation of NaCl:
Sodium metal (Na) and non-metal chlorine (Cl) combine with each other through ionic bond.

2. As can be seen in the table above, there is 1 electron in the outermost shell (orbit) of sodium (Na) metal.
3. The sodium metal will tend to lose 1 electron from its M shell. On doing so, ‘L’ will become the outermost shell with a stable octet.
4. The nucleus of this atom will still have 11 protons but the number of electrons after giving 1 electron will become 10. Thus, there is a net positive charge with sodium Na⁺.
5. Chlorine has 7 electrons ¡n its outermost. i.e. ‘M’ shell. It requires only 1 electron to complete its octet and become stable.

Reaction between sodium (Na) and chlorine (Cl):

1. If sodium and chlorine are reacted, sodium will lose its one electron which chlorine will take-up.
2. On gaining the electron, chlorine atom will get a negative charge, because its nucleus has 17  protons and 18 electrons. This will give us chloride anion Cl^–.

Both, Na and Cl perform a give and take relation as follows:

3. Sodium and chloride ions are oppositely charged and hence they attract each other and hold each other by strong electrostatic force of attraction.
4. Sodium chloride does not exist as an atom but as oppositely charged ions.

Question 26.
Explain the formation of ionic bond In magnesium chloride.
Answer:
1. Since magnesium’s atomic number is 12, its electronic configuration is (2, 8, 2).

2. Magnesium has 2 valence electrons in it and so it will tend to lose them, achieve octet configuration of nearby inert gas and form magnesium ion (Mg²⁺)

3. On the other hand, two chlorine (2,8,1) atoms will accept each of these electrons liberated by magnesium and form two Cl ions. Thus, chlorine will attain an octet configuration of (2, 8, 8).

4. Thus, the two electrons lost by magnesium atom are gained by two chlorine atoms (each one gets one electron) and forms magnesium chloride.

Question 27.
Give the properties of ionic compounds.
Answer:
Properties of Ionic compounds :
(1) Physical nature :

• Ionic compounds are obtained in solid form.
• The ionic compounds are hard due to strong force of attraction between positive and negative ions. Also, they are brittle and break into pieces on applying pressure.

(2) Melting point and boiling point :
In ionic compounds, positive and negative ions are joined strongly due to inter-ionic attraction. As a result, more energy is required to break such compounds and hence their melting and boiling points are high.

(3) Solubility :
Ionic substances (electrovalent compounds) are soluble in water but insoluble in organic solvents such as kerosene, petrol, etc.

(4) Conduction of electricity:
An ionic compound cannot conduct electricity in solid form. However, an ionic compound can conduct electricity when

• It is dissolved in water or
• When is in molten state.

5. The necessary condition for conducting electricity through a solution is that the charged particles should move in the solution.

Question 28.
Explain the conduction of electricity in ionic compounds.
Answer:
An ionic compound cannot conduct electricity in solid form. However, an ionic compound can conduct electricity when

• It is dissolved in water or
• When it is in molten state.

The necessary condition for conducting electricity through a solution is that the charged particles should move in the solution.

(a) in solids:
The ions cannot move and hence the electricity is not conducted. For example, in solid NaCl (common salt).

(b) ionic compound dissolved in water:
When ionic compounds dissolve in water, the solution so formed contain free-floating ions. When electricity is passed in this solution, the free-floating ions move towards the opposite electrodes. This confirms that ionic compound dissolved in water conducts electricity. For example, NaCl dissolved in water.

(c) Molten ionic compound:
To melt an ionic compound we need to supply heat. The heat overcomes the electrostatic force of attraction between the oppositely charged ions. As a result, the ions start moving freely and conduct electricity. For example, on heating NaCl (common salt). its molten form will conduct electricity.

Question 29.
Ionic compounds have high melting and boiling points. Give reason.
Answer:
Melting point and boiling point :
In ionic compounds, positive and negative ions are joined strongly due to inter-ionic attraction. As a result, more energy is required to break such compounds and hence their melting and boiling points are high.

Question 30.
Solid ionic compounds are bad conductor of electricity, but in aqueous solution they conduct electricity. Give reason.
Answer:
Molten ionic compound:
To melt an ionic compound we need to supply heat. The heat overcomes the electrostatic force of attraction between the oppositely charged ions. As a result, the ions start moving freely and conduct electricity. For example, on heating NaCl (common salt). its molten form will conduct electricity.

Question 31.
What is mineral ore and gangue?
Answer:
(i) Mineral:

• The elements or compounds that occur naturally in the earth’s crust are called minerals.
• Sea-water also contains salts of metals such as sodium chloride, magnesium chloride, etc.

(ii) Ores:
Those minerals from which the metals can be extracted, conveniently and profitably are called ores. (Note: Although all minerals contain metals, they are not termed as ‘ores’. The reason behind this is that it is not always practically possible to extract metals from these minerals. Hence, only those minerals in which metals is present in large quantity and from which metals can be extracted conveniently and reasonably are called ores.)

(iii) Gangue:
Impurities such as sand, mud, etc. present in the ore are called gangue.

Question 32.
All ores are minerals but all minerals are not ores. Give reason.
Answer:
(i) Mineral:

• The elements or compounds that occur naturally in the earth’s crust are called minerals.
• Sea-water also contains salts of metals such as sodium chloride, magnesium chloride, etc.

(ii) Ores:
Those minerals from which the metals can be extracted, conveniently and profitably are called ores. (Note: Although all minerals contain metals, they are not termed as ‘ores’. The reason behind this is that it is not always practically possible to extract metals from these minerals. Hence, only those minerals in which metals is present in large quantity and from which metals can be extracted conveniently and reasonably are called ores.)

Question 33.
How do metals occur On earth?
Answer:
(1) in free state (Least reactive metals):

• Only those metals which are very less reactive are found in the free state (or say native form) in nature. For example, silver (Ag), gold (Au) and platinum (Pt).
• Copper and silver are found in the form of oxides and sulphides also.

(2) in the form of compounds (Highly reactive metals):
Highly reactive metals such as potassium (K), sodium, Magnesium (Mg), Aluminium (Al), etc. are so reactive that they are never found in their free form in the nature. They are only found in the form of their compounds.

(3) In the form of oxides/sulphates/carbonates (Moderately reactive metals):

Moderately reactive metals such as zinc (Zn), Iron (Fe), Lead (Pbs, eth that lie at the ‘middle of the reactivity series are found in the earth in the form of oxides, sulphides and carbonates.
(Note: Textbook says that highly reactive minerals are found in the form of oxides, suiphides and carbonates. It is a mistake. This can be easily verified by studying section 3.4.4 Extracting Metals in the Middle of the Activity series of the textbook.)

In general —

• All the metals placed above copper (Cu) are found in nature only in the form of their compounds.
• Ores of many metals are found in the form of their oxides ie. the metal ore is oxide ore. This is because oxygen is a very reactive element and also it is found in abundance in the earth.
• Since, the reactivity of metals is different, the method to extract them from nature is also different.
• Extracting pure metals from their ores require several steps.

Question 34.
Explain In brief how are metals having different reactivities extracted from nature?
Answer:
(1) Least reactive metals:
Metals such as mercury (Hg) and copper (Cu) that lie in the lower section of the activity series are very unreactive.

They are extracted as follows:
(a) Since these metals are very less reactive they get extracted from their oxides simply by heating.
(b) Then, the metals are refined using electrolytic refining.

(2) Moderately reactive metals:

• Metals such as iron (Fe), zinc (Zn), lead (Pb), etc. lie at the middle of reactivity series and are moderately reactive.
• In nature, these metals occur in the form of suiphides and carbonates.

They are obtained in pure form as follows:
(a) The sulphide or carbonate ores are first heated (through roasting/calcination) to convert them into oxides.
(b) The metal oxides so formed are then reduced to their corresponding metals using reducing agents such as carbon (C).
(c) Finally, the metals are refined using electrolytic refining.

(3) Highly reactive metals:
Metals such as potassium (K), sodium (Na), calcium (Ca), magnesium (Mg), aluminium (Al), etc. are highly reactive and are placed at the top of the reactivity series.

They are obtained in pure form as follows:
(a) Since these metals are highly reactive, they are extracted using the method of electrolytic reduction or electrolysis. For example, sodium, magnesium and calcium are obtained by the electrolysis of their molten chlorides.

(b) Then, the metals are refined using electrolytic refining.
(Note: All metals extracted have one last common step which is ‘electrolytic refining’. This step removes all the impurities and finally gives us metal in its purest form. This method is discussed in text-book in section 3.4.6 Refining of Metals.)

Question 35.
List out steps involved in extracting metal from its ore.
Answer:
The chart below shows steps for extracting metals from their ores.

Question 36.
How are metals low in activity series extracted from their ores? OR How are less reactive metals extracted from their ores?
Answer:
Metals that lie in the lower section of the activity series are very unreactive. As a result, they easily get extracted from their oxides simply by heating.
Example:
(i) Mercury sulphide (HgS) also known as cinnabar is an ore of mercury.
On heating it, it first gets converted into mercuric oxide (HgO). On heating (HgO), we can obtain mercury (Hg) in pure form.

Question 37.
How copper is extracted from Its ore?
Answer:
1. Copper is a very less reactive metal. Hence, it is obtained from its ore simply by heating.
2. The chart below shows steps for extracting metals from their ores.

Question 38.
How are metals lying in the middle of activity series extracted?
Answer:
1. Metals such as iron (Fe), zinc (Zn), lead (Pb), etc. lie at the middle of reactivity series and are moderately reactive.
2. In nature, these metals occur in the form of sulphide and carbonates.
3. It is easier to obtain metals from their oxides rather than sulphides and carbonates. Hence, the moderately reactive metals are first converted into oxides. This is done as follows:

(a) Roasting:
1. This process is used for suiphide ore. Roasting is the process in which the sulphide ore is heated strongly in the presence of excess air to convert it into metal oxide.
Example:

(b) Calcination:
1. This process is used for carbonate ore.
2. Calcination is the process in which the carbonate ore is heated strongly in the limited air to convert it into metal oxide.

3. The metal oxides obtained by calcination or roasting are then reduced (converted) into free metal by using reducing agents such as carbon, aluminium, sodium or calcium. The reducing agent to be used again depends on the chemical reactivity of the metal to be extracted.

(i) Reduction of metal oxide using carbon:
The oxides of comparatively less reactive metals like zinc, iron, lead, etc. are reduced using carbon as the reducing agent.

(ii) Reduction using aluminium: If the metal oxide is of more reactive metal, then aluminium can be used as a reducing agent to extract metal from its metal oxide. The reason for using aluminium is that aluminium is a more reactive metal and hence it can displace a comparatively less reactive metal from its metal oxide to give free metal.

(iii) Manganese and iron (Fe) metal is extracted from its oxide using aluminium.

(iv) Note that this reaction is a displacement reaction as well as reduction and oxidation reaction.

(v) Such displacement reactions are highly exothermic.

Question 39.
What s thermite reaction? Give one example.
Answer:
Thermite reaction:
1. The process or reaction in which a metal is reduced from its oxide using aluminium powder as a reducing agent is called thermite reaction (or thermite process).

2. Thermite reactions are highly exothermic and produce so much heat that the metals are obtained in molten state.
Example:
Iron oxide is reacted with aluminium to obtain iron metal in molten state. The molten iron is then used to weld the railway tracks and machine parts.

Question 40.
What is roasting?
Answer:
Roasting:
1. This process is used for sulphide ore. Roasting is the process in which the sulphide ore is heated strongly in the presence of excess air to convert it into metal oxide.
Example:

Question 41.
What is calcination?
Answer:
Calcination:
1. This process is used for carbonate ore.
2. Calcination is the process in which the carbonate ore is heated strongly in the limited air to convert it into metal oxide.

3. The metal oxides obtained by calcination or roasting are then reduced (converted) into free metal by using reducing agents such as carbon, aluminium, sodium or calcium. The reducing agent to be used again depends on the chemical reactivity of the metal to be extracted.

Question 42.
Explain method to obtain metal using carbon.
Answer:

Reduction of metal oxide using carbon:
The oxides of comparatively less reactive metals like zinc, iron, lead, etc. are reduced using carbon as the reducing agent.

Question 43.
Explain method to obtain metal using aluminium.
Answer:
Reduction using aluminium: If the metal oxide is of more reactive metal, then aluminium can be used as a reducing agent to extract metal from its metal oxide. The reason for using aluminium is that aluminium is a more reactive metal and hence it can displace a comparatively less reactive metal from its metal oxide to give free metal.

Question 44.
How are metals present in the top of activity series extracted?
Answer:
1. Metals such as potassium (K), sodium (Na), calcium (Ca), magnesium (Mg), aluminium (Al), etc. are highly reactive and are placed at the top of the reactivity series.

2. The oxides of these metals have more affinity i.e. attraction towards oxygen rather than carbon.

3. Hence, a method called ‘electrolytic reduction or electrolysis’ is used to extract highly reactive metals from their molten chlorides or oxides.

4. Sodium, magnesium and calcium are obtained by the electrolysis of their molten chlorides. Aluminium is obtained by the electrolytic reduction of aluminium oxide.

5. The metal obtained will be at cathode (negative electrode) because metals are positively charged and get attracted to negatively charged electrode (the cathode).

Following reactions take place during electrolysis of sodium chloride to obtain sodium metal:

Question 45.
How is sodium metal extracted? OR State the reactions taking place during extraction of sodium.
Answer:
1. Sodium is a very reactive metal. Hence, it is extracted with the help of ‘electrolytic reduction (electrolysis)’ method.

2. Metals such as potassium (K), sodium (Na), calcium (Ca), magnesium (Mg), aluminium (Al), etc. are highly reactive and are placed at the top of the reactivity series.

3. The oxides of these metals have more affinity i.e. attraction towards oxygen rather than carbon.

4. Hence, a method called ‘electrolytic reduction or electrolysis is used to extract highly reactive metals from their molten chlorides or oxides.

5. Sodium, magnesium and calcium are obtained by the electrolysis of their molten chlorides. Aluminium is obtained by the electrolytic reduction of aluminium oxide.

6. The metal obtained will be at cathode (negative electrode) because metals are positively charged and get attracted to negatively charged electrode (the cathode).

Question 46.
Explain the procedure to perform electrolysis (electrolytic refining) of metal.
Answer:
Electrolytic refining:
Refining metals using the method of electrolysis is called electrolytic refining.

Electrolysis:
1. A thick rod of impure metal is taken as anode. It is connected to the +ve terminal of the battery.
2. A thin rod of pure metal is taken as cathode. It is connected to the  – ve terminal of the battery.
3. Aqueous acidified solution of salt of metal to be refined (i.e. impure metal) is taken as an electrolyte and filled in the electrolytic vessel.

Procedure:

• After setting up the apparatus as shown in the figure, electricity is supplied.
• On passing the electric current through electrolyte, pure metal from anode dissolves in the solution i.e. the electrolyte. An equivalent amount of pure metal from the solution will get deposited at cathode.

The impurities that move out of the anode rod get divided Into following two parts :
(A) Soluble impurities go into the aqueous solution and
(B) Insoluble impurities get collected below the anode. It is then called anode mud or anodic mud.

Application :
To refine metals such as copper, zinc, gold and silver.

Question 47.
Explain the electrolysis of copper. OR explain procedure to obtain pure copper from its ore using electrolytic reduction.
Answer:
Electrolysis of Copper

• A thick rod of copper (Impure metal) to be purified is taken as anode and it is connected with the positive terminal of the battery.
• AcomparatlvelythlnrodofpurecOpperlstaken as cathode and it is connected with the negative
  terminal of the battery
• Aqueous solution of copper sulphate (salt of copper) is taken as electrolyte. Also, a dilute sulphuric acid Is added to It
• The copper sulphate solution then becomes acidified copper sulphate solution.

Procedure:

• After setting up the apparatus as shown in the figure, electricity is supplied.
• The positively charged copper ions i.e. Cu⁺² will get attracted towards negatively charged electrodes i.e. cathode.
• This way pure metal cathode will become thick.
• The impure copper goes into the solution in the form of ions.

The Impurities that move out of anode of copper gets divided Into following two parts:
(A) Soluble impurities go into the aqueous solution
(B) Insoluble impurities get collected below the anode. It is then called anode mud or anodic mud.
The copper deposited at cathode is almost hundred percent pure.

Question 48.
What is corrosion ? State Its advantages and disadvantages.
Answer:
Corrosion:

• The erosion reaction that takes place between a metal and water or moisture when they come in contact with each other called metallic erosion or corrosion.
• We can see the effect of air and water i.e. corrosion on surface of many metals.

Examples:
1.  If objects made out of iron remain exposed to moist air for a longer period, their surface becomes brown and flaky. This is nothing but corrosion.
2. Silver articles or ornaments become black after sometime due to exposure to air.
3. A green coloured layer of copper carbonate forms on the surface of copper objects when they remain exposed in air for a long time.
4. Inert metal such as gold, never get corroded.

Disadvantages of corrosion:

• Every year, corrosion causes loss of crores of rupees in the world.
• Corrosion also reduces life of objects.

Question 49.
Enlist ways to prevent corrosion of iron. Explain the methods of galvanizing and making alloys for preventing corrosion.
Answer:
Corrosion (rusting) of iron can be prevented by painting, oiling, greasing, galvanizing, anodizing or making alloys.
(a) Galvanizing:

• Rusting of iron can be prevented by applying a coat of very fine layer of zinc metal on it.
• The process of applying zinc is called galvanizing and the iron on which it is applied is then called galvanized iron.
•  For example, iron sheets used in the roofs of house are galvanized iron sheets.

(b) Making alloys:

• Another effective way to prevent corrosion is to change the properties of metals and non—metals. This can be done by mixing different metals and non-metals.
• For example, stainless steel is an alloy which is formed by adding chromium and nickel to iron.
  This alloy does not get affected by air, water or alkali and it does not even get corroded.
• Hence, utensils used in kitchen, instruments used in surgery, big vessels used in industries, etc. are prepared from stainless steel.

Question 50.
What is an alloy? State its preparation and some examples.
Answer:
Alloys:

• An alloy is a homogeneous mixture of two or more metals or metal and non—metal.
• There are several advantages of using alloys over metals.

Preparation:

• For making alloy, first of all the chief metal e. the metal whose alloy is to be made is melted.
• Then the substance which is to be mixed is added in definite proportion and the mixture is melted again.
• Then this molten mixture is allowed to cool to obtain alloy.

Examples:

• On adding a very small amount of carbon (about 0.05%) to iron, it becomes quite hard and strong.
• Stainless steel is obtained on adding nickel and chromium to iron. This steel is strong and does not corrode.
• The alloy prepared by adding zinc metal to copper is known as brass.
• Amalgam is an alloy made by mixing mercury with any metal.

Question 51.
State advantages of alloys with the help of examples.
Answer:
Advantages of alloys:
1. On adding a very small amount of carbon (about 0.05%) to iron, it becomes quite hard and strong.
2. Stainless steel is obtained on adding nickel and chromium to iron. This steel is strong and does not corrode.
3. The alloy prepared by adding zinc metal to copper is known as brass.
4. Brass is used in making cooking vessels, parts of machinery, musical instruments, etc.
5. An alloy which contains mercury is called amalgam.
6. The electrical conductivity of an alloy is less than the pure metal. For example, impure copper has a lesser conductivity compared to pure copper.
7. Melting point of an alloy is less than those of component elements. For example, melting point of the alloy prepared from lead and tin is less and so it is used in soldering electric wires

Question 52.
State important differences between diamond and graphite.
Answer:

Question 53.
A non-metal X exists In two different forms Y and Z. Y Is the hardest natural substance, whereas Z is a good conductor of electricity. Identify X, Y and Z.
Answer:
1. The non-metal X is carbon (C). Carbon exists in two different forms called the allotropes of carbon. These allotropes are diamond and graphite.
2. The hardest natural substance is diamond. Hence, Y is diamond. Graphite is good conductor of electricity. Hence, Z is graphite.

Question 54.
Individually, hydrogen gas lacks the capability of combustion. Give reason.
Answer:
1. If you put an ignited match stick into glass jar, it extinguishes. But, if you allow oxygen to mix with hydrogen gas and then try to ignite t, it will ignite with a blue flame.
2. This suggests that hydrogen becomes combustible in presence of oxygen where as individually it is non-combustible.

Question 55.
A metal A, which is used in thermite process, when heated with oxygen gives an oxide B, which is amphoteric in nature. Identify A and B. Write down the reactions of oxide B with HCl and NaOH.
Answer:
1. Metal A is aluminium (Al) which is used in thermite process.
2. Aluminium reacts with oxygen to form aluminium oxide Al₂O₃ i.e. B which is amphoteric in nature.
4Al + 3O₂ → 2Al₂O₃(s)

Reactions of oxide B i.e. Al₂O₃ with HCl and NaOH:

Question 56.
lqbal treated a lustrous, divalent element M with sodium hydroxide. He observed the formation of bubbles in reaction mixture. He made the same observations when this element was treated with hydrochloric acid. Suggest how he can identify the produced gas. Write chemical equations for both the reactions.
Answer:
The divalent element M that lqbal used is zinc Zn.
(a) Reaction of zinc with sodium hydroxide:

The gas produced gas can be identified by bringing splinter near the gas. The gas will start burning with a pop sound. This suggests that it is hydrogen gas.

Question 57.
A solution of CuSO₄ was kept in an iron pot. After few days the iron pot was found to have a number of holes in it. Explain the reason in terms of reactivity. Write the equation of the reaction involved.
Answer:
The solution of CuSO₄ means solution of copper sulphate. Iron is a more reactive metal than copper. Hence, it displaces copper from copper sulphate solution.
Reaction:

Since, iron takes part in this reaction, it produces holes at places where iron metal has reacted to form iron (II) sulphate.

Question 58.
The principle ‘Less active metals can be displaced from the solution of its salt by the more active metal’ is used ¡n developing the activity series of metals. Explain.
Answer:
1. For preparing the activity series of metals, the metals need to be arranged in decreasing order of their reactivities.

2. For this, it is necessary to determine which metal is more reactive or less reactive than others.

3. Certain reactions are carried out to determine the reactivity level.

4. The reactions can be done either with oxygen, water or even acid. But, all metals cannot undergo reactions using these methods.

5. To overcome the above problem, displacement reaction is carried out on the metals and then their reactivities are
compared.

6. Displacement reaction works on the principle ‘Less active metals can be displaced from the solution of its salt by more active metal’. As a result, this principle is used to determine the activity series.

Question 59.
‘Hydrogen is not a metal even then it is placed in the reactivitý series of metals’. Why?
Answer:
1. The reactivity of metals depends on the ease with which they can lose electrons.

2. Hydrogen is not a metal but still it can lose one electron quite easily compared to many other metals and form positive ions, Hence, hydrogen is placed in the reactivity series of metals.

Question 60.
A non-metal A is an important constituent of our food and forms two oxides B and C. Oxide B is toxic whereas C causes global warming.
(a) Identify A, B and C (b) To which group of Periodic Table does A belong?
Answer:
(a) Non-metal A is carbon. Its two oxides are carbon monoxide (CO) and carbon dioxide. Oxide B is carbon monoxide (CO) as it is toxic while C is carbon dioxide (CO₂), because it is responsible for global warming.

(b) Atomic weight of non-metal A is 6. So, its electronic configuration is 2, 4. It is present in 14th group (10 + valence electrons), i.e. group IV of the Periodic Table.

Question 61.
‘Bonding alters the properties of elements.’ Explain this statement by giving examples.
Answer:
1. Sodium metal is highly reactive. It easily reacts with oxygen (O₂) of the air and forms sodium oxide (Na₂O). Na₂O further reacts with water and gives sodium hydroxide (NaOH).

2. Sodium metal is highly active because its configuration is (2, 8, 1) and so it has a tendency to lose one electron from its valence orbit and thus form sodium ion (Na⁺).

3. Na⁺ attracts chloride ion (Cl^–) which has a negative charge. Thus, Na⁺ and CF^– combine and form NaCl.

4. Even though individually Na⁺ and Cl^– are reactive and non-edible, when they combine by bonding, they lose their original individual properties and form NaCl i.e. common salt which can be consumed without any harm.

5. Thus, bonding alters the properties of elements.

Question 62.
Why should the metal suiphides and carbonates be converted to metal oxides in the process of extraction of metal from them?
Answer:
1. In metallurgy, it is easier to obtain metals from their oxides (by reduction) rather than from carbonate or suiphide ores.
2. As a result, before reduction the ore must be converted into metal oxide and then be reduced.

Question 63.
A metal that exists as a liquid at room temperature is obtained by heating its sulphide in the presence of air. Identify the metal and its ore and give the reaction involved.
Answer:
1. The only metal that exists in liquid state is mercury (Hg). Mercury is obtained from the sulphide ore called cinnabar ore (HgS).

2. Mercury sulphide (HgS) also known as cinnabar is an ore of mercury. On heating it, it first gets converted into mercuric oxide (HgO). On heating (HgO), we can obtain mercury (Hg) in pure form.

Question 64.
Compound X and aluminium are used to join railway tracks. (a) Identify the compound X, (b) Name the reaction and (c) Write down its reaction.
Answer:
(a) Compound X is iron (lll) oxide, Fe₂O₃.
(b) The reaction is known as a thermite reaction.
(c) The reaction that takes place is as follows

Question 65.
Switching over to alloys is an effective way to prevent corrosion. Give reason.
Answer:
1. Metals like iron, copper, etc. start getting corroded when they come in contact with air and water.
2. We can improve the anticorrosion properties of the metals by mixing them with certain other metals or non-metals.
3. For example, stainless steel which is an alloy made up of iron, chromium and nickel does not get affected by air, water or alkali and so it does not corrode.
4. Similarly, brass alloy which is made up of copper and zinc prevents corrosion in copper.
5. Thus, switching over to alloys is an effective way of preventing corrosion.

Question 66.
Explain the following
(a) Reactivity of Al decreases if it is dipped in HNO₃
(b) Carbon cannot reduce the oxides of Na or Mg
(c) NaCl is not a conductor of electricity in solid state whereas it does conduct electricity in aqueous solution as well as In molten state
Answer:
(a) Nitric acid (HNO₃) is an oxidizing agent. When aluminium is dipped in HNO₃, an oxide layer of aluminium is formed on the surface of the metal. This prevents it from further reaction. As a result, reactivity of Al decreases.

(b) Na, Mg. etc., metals are quite reactive and are present towards the top portion of the reactivity series. As a result, these metals have more affinity with oxygen than carbon. Therefore, their oxides are stable. To reduce them with carbon, very high temperature is required and at that temperature they will form their corresponding carbides. Hence, their oxides cannot be reduced by carbon.

(c) One of the conditions for conduction of electricity is movement of ions. Ions of solid or dry NaCl cannot move to carry the charge. However, they are free in molten state and in aqueous solution to carry the charge and hence, NaCl conducts electricity in molten state.

Very Short Answer Type Questions

Question 1.
What are metals?
Answer:
The elements which lose electrons to become positively charged Ions are called metals.

Question 2.
State two physical properties of metals.
Answer:
Metals conduct heat and electricity. They are hard, shiny, heavy and sonorous (i.e. make sound on banging).

Question 3.
Define: Thermal conductivity
Answer:
The ability of a material to conduct heat from one end to another is called thermal conductivity.

Question 4.
What is sonority?
Answer:
The characteristic, of metal to produce a deep resonant sound on striking is called the property of sonority.

Question 5.
What makes aluminium and copper the ideal metals for making wires?
Answer:
Both aluminium and copper are easily available, cheap, conduct electricity and do not rust. Hence

Question 6.
Name two metals that do not react with oxygen.
Answer:
Gold and silver

Question 7.
What are non-metals?
Answer:
Elements that form negative ions by gaining electrons are called non-metals.

Question 8.
State two physical properties of non-metals.
Answer:
Non-metals are dull aid are poor conductor of heat and electricity.

Question 9.
How can you say that magnesium oxide is basic?
Answer:
The solution of MgO turns red litmus paper blue. This suggests that

Question 10.
What is an allotrope?
Answer:
Some elements possess the property of existing in two or more different forms in the same physical state. Such a property is called allotropy. The different forms of the elements are called allotropes.

Question 11.
A non-metal exists in two different forms Y and Z. Y Is the hardest natural substance, whereas Z is a good conductor of electricity Identify Y and Z.
Answer:
Y is diamond whereas Z is graphite.

Question 12.
Name the metal that has such a less melting point that It can melt with the heat of your palm.
Answer:
Gallium

Question 13.
Which metal oxides dissolve In water to form alkali? Write their balanced chemical equations.
Answer:
Oxides of reactive metals such as sodium and potassium dissolve in water to form alkali.
Reactions:
Na₂O_(s) + H₂O_(l)→ 2NaOH_(aq)
K₂O_(s) + H₂O_(l) → 2KOH_(aq)

Question 14.
Give chemical reaction when an iron strip is put in the solution of copper sulphate.
Answer:
Fe + CuSO₄ → FeSO + Cu

Question 15.
What happens when metals react with acids?
Answer:
When metals react with acids they give salt and produce hydrogen gas.

Question 16.
Which metal out of calcium, aluminium, iron and zinc will perform the fastest reaction with acid to liberate hydrogen gas? Arrange the metals in the descending order with respect to the rate of liberation of hydrogen gas.
Answer:
1. Out of the given elements, calcium will liberate hydrogen gas at the fastest rate followed by aluminium, zinc and iron.
2. Arrangement of metals: Ca > Al > Zn > Fe.

Question 17.
Why copper does not react with dilute hydrochloric acid (HCl)?
Answer:
Since copper is not an active metal like Na, K.Zn. etc. it does not react with dilute hydrochloric acid (HCl).

Question 18.
Name two metals which react with dilute HNO₃ to evolve hydrogen gas.
Answer:
Manganese (Mn) and magnesium (Mg)

Question 19.
What is the reactivity (activity) series of metals?
Answer:
The arrangement of metals in decreasing order of their reactivities is called the reactivity series of metals

Question 20.
How can you call a metal as more reactive metal?
Answer:
If a metal loses electrons rapidly it forms a positive ion and reacts quickly with other substance such metal is then termed as a more reactive metal.

Question 21.
The following reaction takes place when aluminium powder is heated with MnO₂
3MnO₂(s) + 4Al(s) → 3Mn(l) + 2Al₂O₃(l) + Heat
(a) Is aluminium getting reduced?
(b) Is MnO₂ getting oxidized?
Answer:
(a) No, aluminium is not getting reduced. it is getting oxidized because there is addition of oxygen.
(b) No, MNO₂ is getting reduced because there is removal of oxygen.

Question 22.
What is an ion?
Answer:
An ions an electrically charged atom (or group of atoms)

Question 23.
How is an ion formed?
Answer:
When an atom gains or loses electrons, it forms an ion.

Question 24.
What is cation? OR How is positive ion formed?
Answer:
If an element has 1,2 or 3 electrons in the outermost shell of its atoms, then it loses these electrons and forms a positively charged ion or cation.

Question 25.
What is anion? OR How is negative Ion formed?
Answer:
If an element has 5.6 or 7 electrons In the outermost shell of its atoms, then it gains these electrons and forms a negatively charged ion or anion.

Question 26.
Define: ionic bond.
Answer:
The chemical bond formed by the transfer of electrons from one atom to another is known as an ionic bond.

Question 27.
Show how electrons transfer in the formation of MgCl₂ from Its elements.
Answer:
The transfer of electron in the formation of MgCl₂ from its elements is shown below:

Question 28.
How do metals and non-metals combine?
Answer:
Metals and non-metals combine through the transfer of electrons from metals to non-metals to form ionic bonds.

Question 29.
What is gangue?
Answer:
Impurities such as soil, sand, etc. present in the ore is called gangue.

Question 30.
What is concentration of ore?
Answer:
The removal of impurities from the ore is called concentration of ore?

Question 31.
What is mineral?
Answer:
The natural materials in which the metals or their compounds are found in earth are called minerals.

Question 32.
What is ore?
Answer:
Those minerals from which the metals can be extracted conveniently and profitably are called ores.

Question 33.
What is roasting?
Answer:
Roasting is the process in which sulphide ore is heated strongly in the presence of air to convert it into metal oxide.

Question 34.
What is calcination?
Answer:
Calcination is the process in which carbonate ore is heated strongly in the absence of air to convert it into metal oxide.

Question 35.
What is rust?
Answer:
The rust-brown flaky layer that develops on iron when iron is exposed to moist air is called rust.

Question 36.
What is corrosion?
Answer:
The erosion reaction that takes place between a metal and water or moisture when they come in contact is called metallic corrosion.

Question 37.
Can rusting of Iron nail occur In distilled water?
Answer:
Distilled water does not contain dissolved oxygen. Hence, rusting will not occur in it.

Question 38.
Which two things are necessary for corrosion?
Answer:
Presence of air and water

Question 39.
What will you do to prevent corrosion on iron sheets used in the roof of the factories?
Answer:
I would either use galvanized sheets or paint the sheets.

Question 40.
What are the constituents of solder alloy? Which property of solder makes It suitable for welding electrical wires?
Answer:
Solder is an alloy made from 50% lead and 50% tin. Solder has a low melting point and so it is used for soldering i.e. welding electrical wires.

Fill in the Blanks:

1. …………………..is a liquid non-metal.
Answer:
Bromine

2. …………………….. and ………………… are two aliotropes of carbon.
Answer:
Diamond and graphite

3. ……………….. metals (type) are so soft that they can be cut with knife.
Answer:
Alkali

4. Most metals produce ………………… when dissolved in water.
Answer:
Basic oxides

5. When sodium dissolves in water it produces (name of compound).
Answer:
Sodium hydroxide

6. On heating copper, it gets coated with ……………
Answer:
Black coloured layer of copper (II) oxide

7. When calcium reacts with water it produces hydrogen but the hydrogen does not catch tire because ………………..
Answer:
The heat evolved is not sufficient enough to catch fire

8. Although when metal reacts with acids hydrogen gas is evolved, it is not the case when metal reacts with ……………. acid.
Answer:
Nitric

9. On putting pieces of copper in dilute HCl, neither bubbles are produced nor temperature changes. This means that copper …………………………..
Answer:
Does not react with dilute HCl

10. Ionic compounds are also called ……………….
Answer:
Electrovalent compounds

11. The most reactive metals are placed at the ………………….. in the reactivity series.
Answer:
Top

12. In a reactivity series, metals that lie above ………………. are more reactive than the metals that lie below it.
Answer:
H

13. Ores of many metals are found in the form of oxides because ………………
Answer:
Oxygen is very reactive metal and is found in abundance in the earth.

14. Highly reactive metals are mainly extracted with ………………… process.
Answer:
Electrolysis

15. A carbonate ore will undergo while a sulphide ore ………………. before oxidizing.
Answer:
Calcination; Roasting

16. Mercury ore is also called ………………
Answer:
Cinnabar

17. Metal carbonates and metal sulphide must be first reduced to their corresponding oxides because ………
Answer:
It is easier to obtain a metal from Its oxides.

18. Is used as an electrolyte in electrolysis of copper……………………
Answer:
Copper sulphate

19. The unit of for measuring the purity of gold is ………………
Answer:
Carat

20. What is common among magnesium, zinc and iron?
Answer:
They react with vapour to produce hydrogen gas

21. The most common reducing agent for reducing metal oxides is …………………………
Answer:
Carbon

22. Highly reactive metals such as sodium, calcium and aluminium are used as reducing agents because………………………
Answer:
They can displace metals of lower reactivity from their compounds

23. …………….. and ……………….. are reacted to join the railway tracks.
Answer:
Iron (III) oxide; aluminium

24. Carbon cannot reduce very reactive metals such as sodium, magnesium and aluminium because these metals ……….
Answer:
Have more affinity for oxygen rather than carbon

25. In electrolytic refining, the insoluble impurities ……………..
Answer:
Settle at the bottom of anode

26. The erosion reaction of a metal with water or moisture is called …………….
Answer:
Erosion

27. …………………… percentage of carbon added to iron makes it very strong and hard.
Answer:
0.05

28. Steel is formed by mixing ………………. to iron.
Answer:
Nickel and chromium

29. In an alloy, It one of the metals is mercury, the alloy will be known as ………………..
Answer:
Amalgam

True Or False

1. Although iodine is a non-metal, it is lustrous. — True
2. Metalloids are called so because they exhibit the properties of metals as well as non-metals. — True
3. Iron does not burn on heating but iron fillings burn vigorously when sprinkled in the flame. — True
4. Magnesium reacts vigorously with cold water. — False
5. Metals like aluminium, iron and zinc neither react with cold water, nor with hot nor with steam. — False
6. Ionic compounds have low melting and boiling points. — False
7. Ionic compounds have the ability to conduct electricity in their solid forms. — False
8. Electrovalent compounds are soluble in organic solvents such as petrol and kerosene. — False
9. In electronic refining, pure metal is taken as anode. — False

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 8 Introduction to Trigonometry Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 8 Introduction to Trigonometry

Short/Long Answer Type Questions

Question 1.
If tan²A = cot (A – 24°). Find A.
Solution :
We have,
tan 2A = coat (A – 24°)
tan 2A = tan (90° – (A – 24))
[∴ tan (90 – θ) = cotθ]
⇒ 2A = 90° – (A – 24°)
⇒ 2A = 90° – A + 24°
⇒ 2A + A = 114°
⇒ 3A = 114°
⇒ A = \(\frac {114}{3}\) = 38°

Question 2.
If cos³A = sin (A – 34°), where A is an acute angle, find the value of A.
Solution :
We have,
cos 3A = sin (A – 34°)
⇒ sin (90° – 3A) = sin (A – 34°)
[∴ sin (90 – θ) = cos θ]
⇒ 90° – 3A = A – 34°
⇒ 90° + 34° = A + 3A
⇒ 124° = 4A
⇒ A = \(\frac {124°}{4°}\)
⇒ A = 31° Hence proved

Question 3.
The value of (tan 1° tan 2° …. tan 89°) is equal to ?
Solution :
We have, tan 1° tan 2° tan 3° ….. tan 43° tan 44° tan 45° tan 46° tan 47° ….. tan 87° tan 88° tan 89° = tan 1° tan 2° tan 3° ….. tan 43° tan 44°
1. tan (90° – 44°) tan (90° – 43°) ….. tan (43° – 3°). tan (90° – 2°). tan (90° – 1°).
= tan 1° tan 2° tan 3° ….. tan 43° tan 44° : 1. cot 44° cot 43°….. cot 3° cot 2° cot 1
= tan 1° × \(\frac {1}{tan 1°}\) . tan 2° × \(\frac {1}{tan 2°}\) . tan 3° × \(\frac {1}{tan 3°}\) ……… tan 43° × \(\frac {1}{tan 43°}\) . tan 44° × \(\frac {1}{tan 44°}\) . 1
= 1 × 1 × 1 × 1 × 1 × 1 = 1

Question 4.
The value of (sin²θ + \(\frac{1}{1+\tan ^2 \theta}\)) = _____
Solution :
We have,
sin²θ + \(\frac{1}{1+\tan ^2 \theta}\) = sin²θ + \(\frac{1}{\sec ^2 \theta}\)
[∴ 1 + tan²θ = sec²θ)
= sin²θ + cos²θ
= 1 Hence proved

Question 5.
(1 + tan²θ) (1 – sinθ) (1 + sinθ)
Solution :
(1 + tan²θ) (1 – sinθ) (1 + sinθ)
= sec²θ (1 – sin²θ)
= sec²θ × cos²θ
= \(\frac{1}{\cos ^2 \theta}\) × cos²θ
= 1 Hence proved

Question 6.
Prove that : \(\frac{\tan A-\sin A}{\tan A+\tan A}=\frac{\sec A-1}{\sec A+1}\)
Solution :
We have,

Question 7.
Prove the following identify, where the angles involved are acute angles for which the expression is defined.
\(\frac{1+\cot ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}\) = (\(\frac{1-\cot A}{1-\tan A}\))²
Solution :
We have,

Question 8.
Prove that

Solution :

Question 9.
Prove that :

Solution :
We have

Question 10.
Prove that:

Solution :
We have,

= R.H.S
L.H.S. = R.H.S. Hence proved

Question 11.
Prove that:

Solution :
We have,

Question 12.
If sin θ + cos θ = \(\sqrt{3}\), then prove that tan θ + cot θ = 1
Solution :
We have,
sin θ + cos θ = \(\sqrt{3}\)
⇒ (sin θ + cos θ)² = \(\sqrt{3}\)²
⇒ sin²θ + cos²θ + 2sin θ cos θ = 3
⇒ 1 + 2 sinθ cosθ = 3
⇒ sinθ cosθ = \(\frac {2}{2}\) = 1 ………..(1)
Now L.H.S. = tanθ + cotθ
= \(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\)
= \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}\)
= \(\frac {1}{1}\)[using equ. (1)]
= 1 = R.H.S.
L.H.S. = R.H.S. Hence proved

Question 13.
If sin A + sinA² = 1, then find the value of the expression cos²A + cost⁴A.
Solution :
We have, sin A + sin²A = 1
⇒ sin A = 1 – sin²A
⇒ sin A = cos²A
squaring both sides, we get
sin²A = cos⁴A
⇒ 1 – cos²A = cos⁴A
⇒ 1 = cos²A + cos⁴A
⇒ cos²A + cos⁴A = 1.
Hence proved

Fill in the Blanks

Question 1.
____ is the study of relationship between the sides and angles of a triangle.
Solution :
Trigonometry

Question 2.
An equation involving trignometric ratios of an angle is called a trignometeric.
Solution :
identity

Question 3.
The word is derived from the Greek words tri (means three), gon (means sides), metron (means measure).
Solution :
Trigonometry

Question 4.
In a right triangle, aide opposite to right angle is known as _______ .
Solution :
hypotenuse

Question 5.
cot θ is the abbrevation used for ________ of angle θ.
Solution :
cotangent

Question 6.
The ratio of the sides of a triangle with respect to its acute angle are called trigonometric ____ of the angle.
Solution :
ratios

Question 7.
If ΔABC is right, right angled at C, then value of cos (A + B) is ______.
Solution :
0 (zero).

Multiple Choice Questions

Question 1.
If cos A = \(\frac {12}{13}\), then cot A is :
(a) \(\frac {12}{5}\)
(b) \(\frac {13}{12}\)
(c) \(\frac {13}{5}\)
(d) \(\frac {5}{12}\)
Solution :
(a) \(\frac {12}{5}\)

Let us draw the triangle ABC in which ∠B = 90°
Then cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{12}{13}\)
Let AB = 12k and AC = 13k
where k is a positive integer
By using Pythagoars Theorem, we have
AC² = AB² + BC²
(13k)² = (12k)² + BC²
BC² = (13k)² – (12k)²
BC² = 169k² – 144k²
BC² = 25k²
\(\sqrt{\mathrm{BC}^2}\) = 5k
cot A = \(\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{12 k}{5 k}=\frac{12}{5}\)

Question 2.
The rod AC of TV disc antena is fixed at right angles to wall AB and a rod CD is supporting the disc as shown in given figure. If AC = 1.5 cm long and CD = 3m, then sec θ + cosec θ is :
(a) \(\frac {41}{12}\)
(b) \(\frac {41}{13}\)
(c) \(\frac {26}{12}\)
(d) \(\frac {15}{41}\)
Solution :
(b) \(\frac {41}{13}\)

The rod AC of TV disc antena is fixed at right angles to wall AB and rod CD is supporting the disc as shown. In right triangle ACD, we have
CD² = AD² + AC²
[By Phythogoras Theorem]
⇒ 3² = AD² + 1.5²
⇒ 3² – (1.5)² = AD²
⇒ AD² = 9 – 2.25
⇒ AD² = 6.75
⇒ AD = \(\sqrt{6.75}\)
⇒ AD = 2.6 om (approx)
⇒ sec θ + cosec θ = \(\frac{C D}{A D}+\frac{C D}{A C}\)

Question 3.
If 4 tan θ = 3, then [latex]\frac{4 \sin \theta-\cos \theta+1}{4 \sin \theta+\cos \theta+1}[/latex] is :
(a) \(\frac {13}{5}\)
(b) \(\frac {5}{21}\)
(c) \(\frac {13}{21}\)
(d) \(\frac {12}{13}\)
Solution :
(c) \(\frac {13}{21}\)

We have, 4 tan θ = 3
⇒ tan θ = \(\frac{3}{4}=\frac{\mathrm{BC}}{\mathrm{AB}}\)
consider a triangle ABC in which ∠B = 90°
Let BC be 3k and AB be 4k wherek is positive integer.
In right triangle ABC, we have
AC² = AB² + BC²
= (4k)² + (BC)²
= 16k² + 9k²
= 25k²
AC = \(\sqrt{25 k^2}\)
= 5k
sin θ = \(\frac{3 k}{5 k}=\frac{3}{5}\)
and cos θ = \(\frac{4 k}{5 k}=\frac{4}{5}\)

Question 4.
If sin (A + 2B) = \(\frac{\sqrt{3}}{2}\) and cos (A + 4B) = 0, A > B and A + 4B ≤ 90°, then A and B is :
(a) 30°, 45°
(b) 15°, 45°
(c) 60°, 15°
(d) 30°, 45°
Solution :
(d) 30°, 45°

We have,
sin (A + 2B) = \(\frac{\sqrt{3}}{2}\)
⇒ sin (A + 2B) = sin 60°
⇒ A + 2B = 60° ………(1)
And cos (A + 4B) = 0
⇒ Cos (A + 4B) = cos 90°
⇒ A + 4B = 90° ……(2)
Subtracting equation (2), from (1), we get

⇒ B = \(\frac{\sqrt{-30}}{-2}\) = 15°
Substituting the value of B in equ. (1), we get
A + 2 × 15 = 60°
⇒ A + 30° = 60°
⇒ A = 60 – 30 = 30°
Hence, A = 30° and B = 15°

Question 5.
The value of sin 23° cos 67° + cos 23° sin 67° is:
Solution :
We have, sin 23° cos 67° + cos 23° sin 67°
= sin 23° cos (90° – 23°) + cos23° sin (90° – 23°)
= sin 23° sin 23° + cos 23° cos 23°
= sin² 23° + cos²23°
= 1 [sin²θ + cos²θ = 1]

Question 6.
The value of sin 32° cos 58° + cos 32° sin 58° is:
(a) 0
(b) 1
(c) 2
(d) 3
Solution :
(b) 1

We have sin 32° cos 58° + cos 32° sin 58°
= sin 32° cos (90° – 32°) + cos 32° sin (98° – 32°)
= sin 32° sin 32° + cos 32° cos 32°
= sin²32° + cos² 32°
= 1 [∴ sin²θ + cos²θ = 1]

Question 7.
\(\frac{\cos 80^{\circ}}{\sin 10^{\circ}}\) + cos 59° cosec 31° = :
(a) 0
(b) 1
(c) 2
(d) 3
Solution :
(c) 2

We have, \(\frac{\cos 80^{\circ}}{\sin 10^{\circ}}\) + cos 59° cosec 31°
= \(\frac{\cos \left(90^{\circ}-10^{\circ}\right)}{\sin 10^{\circ}}\) + cos 59° cosec (90° – 5°)
= \(\frac{\sin 10^{\circ}}{\sin 10^{\circ}}\) + cos 59° sec 59°
= \(\frac{\sin 10^{\circ}}{\sin 10^{\circ}}+\frac{\cos 59^{\circ}}{\cos 59^{\circ}}\)
= 1 + 1 = 2

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 7 Coordinate Geometry Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 7 Coordinate Geometry

Short/Long Answer Type Questions

Question 1.
The distance between the point A(5, -3) and B(13, m) is 10 units. Calculate the value of m.
Solution :
Here, x₁ = 5, y₁ = – 3, x₂ = 13, y₂ = m
Distance (AB) = 10
⇒ \(\sqrt{(13-5)^2+(m+3)^2}\) = 10
⇒ \(\sqrt{8^2+m^2+9+6 m}\) = 10
⇒ \(\sqrt{64+m^2+9+6 m}\) = 10
⇒ \(\sqrt{73+m^2+6 m}\) = 10
⇒ [Squaring both sides]
⇒ m² + 6m + 73 – 100 = 0
⇒ m² + 6m – 27 = 0
⇒ m² + (9m – 3m) – 27 = 0
⇒ m² + 9m – 3m – 27 = 0
⇒ m(m + 9) – 3(m + 9) = 0
⇒ (m + 9) (m – 3) = 0
⇒ m + 9 = 0 or m – 3 = 0
⇒ m = – 9 or m = 3

Question 2.
If the distances of P(x, y) from A(5, 1) and B(-1, 5) are equal, then prove that 3x
Solution :
Since, of from A(5, 1) and B(-1, 5) are equal

So, PA = PB
⇒ PA² = PB²
⇒ (5 – x)² + (1 – y)² = (-1 – x)² + (5 – y)²
⇒ 25 + x² – 10x + 1 + y² – 2y = 1 + x² + 2x + 25 + y² – 10y
⇒ 26 – 10x – 2y = 26 + 2x – 10y
⇒ – 10x – 2x = – 10y + 2y
⇒ – 12x = – 8y
⇒ 3x = 2y Hence proved.

Question 3.
It k(5, 4) is the mid point of line segment PQ and coordinates of Q are (2, 3) then find the coordinates of point P.
Solution :
Let coordinate of point P be (x, y)

Since k(5, 4) is mid point of P(x, y) and Q(2, 3)
∴ 5 = \(\frac{x+2}{2}\) and 4 = \(\frac{y+3}{2}\)
⇒ 10 = x + 2 and 8 = y + 3
⇒ x = 10 – 2 and y = 8 – 3
⇒ x = 8 and y = 5
∴ coordinates of point P are (8, 5)

Question 4.
If the midpoint of two points A(2, 5) and B(-5, y) is (-\(\frac {7}{2}\), 3). then find the distance between points A and B.
Solution :
Let point C(-\(\frac {7}{2}\), 3) is the midpoint of line AB.

∴ 3 = \(\frac{5+y}{2}\)
⇒ 6 = 5 + y
⇒ y = 6 – 5 = 1
Coordinates of point is (-5, 1)
Length of AB = \(\sqrt{(-5+2)^2+(1-5)^2}\)
= \(\sqrt{(-3)^2+(-4)^2}\)
= \(\sqrt{9+16}\) = \(\sqrt{25}\) = 5
Hence, length of AB = 5 units.

Question 5.
In a parallelogram ABCD A(3, 1), B(5, 1), C(a, b) and D(4, 3) are the vertices. Find vertex C(a, b).
Solution :
We know that diagonals of a parallelogram bisects each other so, O is the mid point of AC as well as that of BD

∴ Coordinates of mid point of AC = coordinates of mid point of BD
⇒ (\(\left(\frac{3+a}{2}, \frac{1+b}{2}\right)\)) = (\(\left(\frac{5+4}{2}, \frac{1+3}{2}\right)\))
⇒ (\(\left(\frac{3+a}{2}, \frac{1+b}{2}\right)\)) = (\(\frac {9}{2}\), \(\frac {4}{2}\))
⇒ \(\frac{3+a}{2}\) = \(\frac {9}{2}\) and \(\frac{1+b}{2}\) = \(\frac {4}{2}\)
⇒ 3 + a = 9 and l + b = 4
⇒ a = 9 – 3 and b = 4 – 1
⇒ a = 6 and b = 3
Hence, coordinates of vertex C is (6, 3)

Question 6.
If the coordinates of two adjacent vertices of a parallelogram are (3, 2), (1, 0) and diagonals bisect each other at (2, – 5), find the coordinates of the other two vertices.
Solution :
Diagonals of a parallelogram bisect each other.
Let coordinates of vertices C be (x₁, y₁) and d be (x₂, y₂) O is the mid point of AC as well as that of BD.

∴ coordinates of mid point of AC = (2, – 5)
⇒ (\(\frac{3+x_1}{2}, \frac{2+y_1}{2}\)) = (2, – 5)
⇒ \(\frac{3+x_1}{2}\) = 2, \(\frac{2+y_1}{2}\) = – 5
⇒ 3 + x₁ = 4, 2 + y₁ = – 10
⇒ x₁ = 1, y₁ = – 12
And coordinates of mid point of BD = (2, – 5)
⇒ (\(\frac{1+x_2}{2}, \frac{0+y_2}{2}\)) = (2, – 5)
⇒ \(\frac{1+x_2}{2}\) = 2, \(\frac{0+y_2}{2}\) = – 5
⇒ 1 + x₂ = 4, y₂ = – 10
⇒ x₂ = 3, y₂ = – 10
Coordinates of vertice Care (1, -12) and vertice D are (3, – 10).

Question 7.
If the midpoint of the line segment joining the points A(3, 4) and B(k, 6) is P(x, y) and x + y – 10 = 0. Find the value of k.
Solution :
Since, mid point’s coordinates are C(x, y)
∴ x = \(\frac{3+k}{2}\) and y = \(\frac{4+6}{2}\) = 5
Since (x, y) lies the equation x + y – 10, so

substituting the value, A(3, 4) of x, y in this equ., we get
\(\frac{3+k}{2}\) + 5 – 10 = 0
⇒ \(\frac{3+k}{2}\) – 5=0
⇒ \(\frac{3+k-10}{2}\) = 0
⇒ k – 7 = 0
⇒ k = 7
Hence, value of k is 7.

Question 8.
The coordinate of the vertices of ΔABC are A (7, 2), B (9, 10) and C(1, 4). If E and F are the mid points of AB and AC respectvely, prove that EF = \(\frac {1}{2}\)BC
Solution :
Since, E and f are the mid points of AB and AC respectvely coordinates of points are
\(\frac{9+7}{2}, \frac{10+2}{2}\) = (8, 6)
Coordinates of points Fare
\(\frac{1+7}{2}, \frac{4+2}{2}\) = (4, 3)

Question 9.
If the point C(-1, 2) divides in ternally the time segments joining A(2, 5) and B(x, y) in the ratio 3 : 4, find the coordinates of B.
Solution :
Hence, x₁ = 2, y₁ = 5, x₂ = x, y₂ = y, x = – 1, y = 2, m₁ = 3, m₂ = – 4

⇒ 3x + 8 = – 7 and 3y + 20 = 14
⇒ 3x = – 7 – 8 and 3y = 14 – 20
⇒ 3x = -15 and 3y = -6
⇒ x = –\(\frac {15}{3}\) and y = – \(\frac {6}{3}\)
⇒ x = – 5 and y = – 2
Hence, coordinate of point Bare (-5, -2)

Question 10.
In what ratio does the point P (\(\frac {24}{11}\), y) divide the line segment joining the points P(2, -2) and Q(3, 7) ? Also find the value y.
Solution :
Let the required ratio be k : 1
Here, x₁ = 2, y = – 2, x₂ = 3, y₂ = 7, x = \(\frac {24}{11}\), y = y
m₁ = k, m₂ = 1
By section formula, we have

Question 11.
Find the value of P, if the points A(2, 3) B(4, P) and C(6, – 3) are collinear.
Solution :
The given points are collinear So, the area of ΔABC will be zero.
Here, x₁ = 2, y₁ = 3, x₂ = 4, y₂ = P, x₃ = 6, y₃ = – 3
Area of ΔABC = 0
⇒ \(\frac {1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ \(\frac {1}{2}\)[2 (P + 3) + 4 (-3 – 3) + 6 (3 – P)]= 0
⇒ \(\frac {1}{2}\)[2P + 6 – 24 + 18 – 6P] = 0
⇒ \(\frac {1}{2}\)[-4P + 0] = 0
⇒ – 4P = 0
⇒ P = 0

Question 12.
Find the area of triangle ABC with A(1, – 4) and the mid points of sides through A being (2, – 1) and (0, – 1).
Solution :
Let the coordinates of vertices B be (x₂, y₂) and C be (x₃, y₃)

Let E (2, – 1) and f(0, – 1) are mid points of AB and AC respectively,
∴ 2 = \(\frac{1+x_2}{2}\) and – 1 = \(\frac{-4+y_2}{2}\)
⇒ 4 = 1 + x₂ and – 2 = – 4 + y₂
⇒ x₂ = 3 and y₂ = 2
∴ (x₂, y₂) = (3, 2)
Again 0 = \(\frac{1+x_3}{2}\) and – 1 = \(\frac{-4+y_2}{2}\)
⇒ 0 = 1 + x₃ and – 2 = – 4 + y₃
⇒ x₃ = – 1 and y₃ = 2
∴ (x₃, y₃) = (-1, 2)
Here, x₁ = 1, y₁ = – 4, x₂ = 3, y₂ = 2, x₃ = -1, y₃ = 2
Area of ABC
= \(\frac {1}{2}\)[x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)]
= \(\frac {1}{2}\)[1(2 – 2) + 3 (2 + 4) + (-1)(-4 – 2)]
= \(\frac {1}{2}\)[1 × 0 + 3 × 6 + (-1) × (- 6)]
= \(\frac {1}{2}\)[0 + 18 + 6]
= \(\frac {1}{2}\) × 24 = 12
Hence, area of ABC = 12sq. units.

Question 13.
Prove that the points (2, -2), (2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the area of this trian.
Solution :
Let the points A (2, -2), B (-2, 1) and C (5, 2) are the vertices of right angled triangle.
AB² = (-2 – 2)² + (1 + 2)²
= 16 + 9 = 25
BC² = (5 + 2)² + (2 – 1)² = 49 + 1 = 50
And AC² = (5 – 2)² + (2 + 2)²
= 9 + 16 = 25
AB² + AC² = 25 + 25 = 50 = BC²

Since, BC² = AB² + AC², So by converse of phythagores theorem ∠A = 90° there fore point A (2, -2), B (-2, 1) and C (5, 2) are the vertices of a right angled triangle
Here, x₁ = 2, y₁ = -2, x₂ = -2, y₂ = – 1, x₃ = 5, y₃ = 2
Area of ΔABC
= \(\frac {1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\)[2(1 – 2) + (-2)(2 + 2) + 5(-2 – 1)]
= \(\frac {1}{2}\)[2 × (-1) + (-2) × 4 + 5 × (-3)]
= \(\frac {1}{2}\)[- 2 – 8 – 15]
= \(\frac {1}{2}\)[-25]
= \(\frac {-25}{2}\)
Hence, area of Δ = \(\frac {25}{2}\)sq. units.

Question 14.
Find the area of that triangle whose vertices are (-3, -2), (5, -2) and (5, 4). Also prove that it is a right-angled triangle
Solution :
Let the points of vertices be A(-3, – 2), B (5, -2) and C (5, 4)
Here, x₁ = – 3, y₁ = – 2, x₂ = 5, y₂ = – 2, x₃ = 5 and y₃ = 4
Area of D ABC
= \(\frac {1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\)[(-3)(-2 – 4) + 5 (4 + 2) + 5(-2 + 2)]
= \(\frac {1}{2}\)[(-3) × (-6) + 5 × 6 + 5 × 0)]
= \(\frac {1}{2}\)[18 + 30 + 0]
= \(\frac {1}{2}\) × 48 = 24 sq. units
AB² = (5 + 3)² + (-2 + 2)² = 8² + 0² = 64
BC² = (5 – 5)² + (4 + 2)² = 0² + 6² = 36
AC² = (5 + 3)² + (4 +2)² = 8² + 6² = 64 + 36 = 100
AB² + BC² = 64 + 36 = 100 = AC²

By converse of phythagoras theorem ∠B = 90° Hence, ΔABC is a right angled triangle.

Question 15.
In a ΔABC, A is (1, – 4). If E (0, -1) and Δ(2,-1) are the mid points of AB and AC. Calculate the area of ΔABC.
Solution :
Let the coordinates of vertices B is (x₂, y₂) and C is (x₃, y₃) since, E is the mid point of AB

∴ 0 = \(\frac{1+x_2}{2}\) and – 1 = \(\frac{-4+y_2}{2}\)
⇒ 0 = \(\frac{1+x_2}{2}\) and – 2 = – 4 + y₂
⇒ x₂ = -1 and y₂ = – 2 + 4 = 2
(x₂, y₂) = (-1, 2)
Since, D is the midpoint of AC
∴ 2 = \(\frac{1+x_3}{2}\) and – 1 = \(\frac{-4+y_2}{2}\)
⇒ 0 = 1 + x₂ and – 2 = – 4 + y₂
⇒ x₂ = -1 and y₂ = – 2 + 4 = 2
∴ (x₃, y₃) = (-1, 2)
Since, D is the midpoint of AC.
∴ 2 = \(\frac{1+x_3}{2}\) and – 1 = \(\frac{-4+y_3}{2}\)
4 = 1 + x₃ and -2 = – 4 + y₃
x₃ = 3 and y₃ = 4 – 2 = 2
∴ (x₃, y₃) = (3, 2)
Here, x₁ = 1, y₁ = – 4, x₂ = – 1, y₂ = 2, x₃ = 3, y₃ = 2
Area of ΔABC
= \(\frac {1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\)[1(2 – 2) + (-1) (2 + 4) + 3(-4 – 2)]
= \(\frac {1}{2}\)[1 × 0 + (-1) × 6 + 3 × (-6)]
= \(\frac {1}{2}\)[0 – 6 – 18]
= \(\frac {1}{2}\) × (-24) = – 12
Hence, area of ΔABC = 12 sq. units.

Question 16.
Find the area of the triangle formed by joining the mid points of the sides of a triangle, whose coordinates of vertices are (0, -1), (2, 1) and (0, 3).
Solution :
Let vertices of triangle be A (0, -1), B (2, 1) and (0, 3) in which of sides AB, BC and AC respectvely.

∴ Coordinates of point
(D) = \(\frac{1+0}{2}, \frac{1-1}{2}\) = (1, 0)
Coordinates of point
E = \(\frac{2+0}{2}, \frac{1+3}{2}\) = (1, 2)
And Coordinates of point
F = \(\frac{0+0}{2}, \frac{3-1}{2}\) = (0, 1)
Here, x₁ = – 1, y₁ = 0, x₂ = 1, y₂ = 2, x₃ = 0, y₃ = 1
Area of ΔDEF
= \(\frac {1}{2}\)[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]
= \(\frac {1}{2}\)[1 (2 – 1) + 1(1 – 0) + 0 (0 – 2)]
= \(\frac {1}{2}\)[1 × 1 + 1 × 1 + 0]
= \(\frac {1}{2}\) × (1 + 1)
= \(\frac {1}{2}\) × 2 = 1 sq. unit.
Hence, area of ΔDEF = 1sq. unit.

Fill in the Blanks

Question 1.
The point of intersection of two axes is called ……….
Solution :
origin

Question 2.
The distance of a point from the y axis is known its ………….
Solution :
abscissa

Question 3.
The ……… of a line segment divides the line segment in the ratio 1 : 1
Solution :
midpoint

Question 4.
If the area of a triangle is ……… sequare unit, then its vertices will be collinear.
Solution :
zero

Question 5.
The distance of a point from the x axis is called its …….
Solution :
ordinate

Question 6.
The distance of a point P(x, y) from the ………. is \(\sqrt{x^2+y^2}\)
Solution :
origin.

Multiple Choice Questions

Question 1.
The coordinates of the point which is reflection of point (-3, 5) in x axis are :
(a) (3, 5)
(b) (3, -5)
(c) (-3, -5)
(d) (-3, 5)

Solution :
(d) (-3, 5)

In graph paper we observe reflection of (-3, 5) in axis is (-3, -5).

Question 2.
The point P on x axis equidistant from the A(-1, 0) and B(5, 0) is:
(a) (2, 0)
(b) (0, 2)
(c) (3, 0)
(d) (2, 2)
Solution :
(a) (2, 0)

Since point P on x axis. So the coordinates of P(x, 0)

∵ Point P is equidistant from point A and point B.
∴ PA = PB
⇒ PA² = PB²
⇒ (-1 – x)² + (0 – 0)² = (5 – x)² + (0 – 0)²
⇒ 1 + x² + 2x + 0 = 25 + x² – 10x + 0
⇒ x² + 2x – x² + 10x = 25 – 1
⇒ 12x = 24
⇒ x = \(\frac {24}{12}\) = 2
Coordinates of point P(2, 0).

Question 3.
The distance between the points (acosθ + b sinθ, 0) and (0, asinθ – bcosθ), is :
(a) a² + b²
(b) a² – b²
(c) \(\sqrt{a^2+b^2}\)
(d) \(\sqrt{a^2-b^2}\)
Solution :
(c) \(\sqrt{a^2+b^2}\)

Let the given points A (a cosθ + b sinθ, 0) and (o, a sinθ – b cosθ)
The distance AB

Question 4.
If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3 : 1, then the value of y is:
(a) 4
(b) 3
(c) 2
(d) 1
Solution :
(d) 1

The point P(6, 2) divides the line segment joining A(6, 2) and B(4, y) in the ratio 3 : 1

Here x₁ = 6, y₁ = 5, x₂ = 4, y₂ = y, m₁ = 3, m₂ = 1, x = 6, y = 2
By section formula

Question 5.
If the point P(k, 0) divides the line segment joining the points A(2, – 2) and B(-7, 4) in the ratio 1 : 2 then the value of k is :
(a) 1
(b) 2
(c) – 2
(d) – 1
Solution :
(d) – 1

Here, x₁ = 2, y₁ = – 2, x₂ = – 7, y₂ = 4, x = k, y = 0, m₁ = 1, m₂ = 2.
By section formula,

Question 6.
The value of P, for which the points A(3, 1), B(5, P) and C (7, – 5) are collinear is :
(a) – 2
(b) 2
(c) – 1
(d) 1.
Solution :
(a) – 2

the given points are collinear. So, area of ΔABC is zero.
Here, x₁ = 3, y₁ = – 1, x₂ = 5, y₂ = P, x₃ = 7, y₃ = – 5
Area of ΔABC = 0
⇒ \(\frac {1}{2}\)[x₁ (y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)] = 0
⇒ \(\frac {1}{2}\)[3(P + 5) + 5(-5 – 1) + 7(1 – P)] = 0
⇒ \(\frac {1}{2}\)[3P + 15 + 5(-6) + 7 – 7P] = 0
⇒ \(\frac {1}{2}\)[3P + 15 – 30 + 7 – 7P] = 0
⇒ \(\frac {1}{2}\)[-4P – 8] = 0
⇒ -4P = 8
⇒ P = \(\frac {8}{- 4}\)
⇒ P = – 2

Haryana State Board HBSE 10th Class Science Important Questions Chapter 2 Acids, Bases and Salts Important Questions and Answers.

Haryana Board 10th Class Science Important Questions Chapter 2 Acids, Bases and Salts

Question 1.
Chemically, how can the elements found in nature be divided?
Answer:
1. In present times, 114 elements are known to us.
2. These elements combine in several ways and give rise to a very large number of compounds.
3. On the basis of their chemical properties aU the compounds can be added to three groups. They are:

• Acids,
• Bases and
• Salts.

4. Thus, all the compounds of this world belong to one or the other group.
5. In this sense, a compound may be acidic, basic or a salt.

Question 2.
What is an acid?
Answer:
Acids: An acid is a compound having hydrogen which when dissolved in water releases ie. dissociates hydrogen ions (H⁺) (to be specific (H₃O⁺ ions).
Example:

Hydrochloric acid (HCl), sulphuric acid (H₂SO₄,) nitric acid (HNO₃), etc. are all examples of acids.
(Note: Water H₂O also has hydrogen in it but when you add water to water, it does not release hydrogen ions (H⁺). Hence, water is not acid. Similarly, NaOH has hydrogen but it does not release hydrogen ions (H⁺) when dissolved in water. NaOH is a base. So, only those substance which on dissolving to water release hydrogen ions (H⁺) are called acids.)

Testing of acid:
If you put acid on blue litmus paper, the blue litmus paper will turn red. This means the substance put is acid.

Question 3.
What is a base?
(a) Bases:
1. A base is a metal hydroxide substance which when dissolved in water release hydroxide (OH) ions. (Second definition: A metal hydroxide or a metal oxide substance when dissolved in acid produces salt and water and hence is called a base.)

2. Sodium hydroxide (NaOH), calcium hydroxide (Ca(OH)₂, potassium hydroxide (KOH), Calcium oxide (CaO), etc. are bases

Testing of base:
If you put base on red litmus paper, the red litmus paper will turn blue. This means the substance put is base.

Question 4.
What is a salt?
Answer:
A salt is an ionic compound which is formed from the neutralization reaction of an acid and a base. Thus, we get salt when we react an acid with a base.
Note:
(1) When an acid and a base are mixed in right proportion, both acid and base lose their properties. In other words, they neutralize each other. Such a reaction is called neutralization reaction.
(2) When salt is dissolved in water it gets ionized into anions and cations)

Question 5.
State two ways to test if the substance is acid.
Whether a substance is an acid or not can be tested with any of the given methods:
Answer:
1. Litmus paper: If you place a drop of a given substance on a moist litmus paper and it turns red, then the substance is acid.
2. Test with metals:  If you put metal in an aqueous substance and the reaction releases hydrogen gas then it means that the aqueous substance is acid.

Question 6.
What are olfactory indicators?
Answer:
1. Olfactory means ‘relating to the sense of smell’. Those substances whose smell changes in acidic or basic solutions are called olfactory indicators.
2. When an acid or a base is added to an olfactory substance, the characteristic smell of that substance cannot be detected.

Question 7.
How does onion and vanilla essence help in finding if a substance is acid or a base?
Answer:
1. If you add onion juice (or vanilla essence) to a base, the onion juice (or vanilla) will lose its smell. However, it will not lose the smell if added to acid.
2. Thus change in smell will help us to find out if the substance added to onion (or vanilla) is acidic or basic.

Question 8.
State three properties of acids and bases.
Answer:
Properties of acid:

• Acids are sour in taste
• They turn blue colour litmus paper into red
• Acids react with base and form salt and water

Properties of bases:

• Bases are bitter in taste
• They change red litmus paper to blue
• Bases react with acids to form salt and water

Question 9.
What happens when an acid reacts with metal? Give one example.
Answer:
When acid reacts with metal, metallic salt of that metal and hydrogen gas are produced.
Acid + Metal →  Salt of Metal + Hydrogen gas
Example:
1. When zinc metal is added to sulphuric acid the reaction gives out zinc sulphate which is a salt and hydrogen gas.

Question 10.
What happens when a base reacts with a metal? Give one example.
Answer:
When a strong base reacts with certain metals, it produces salt and hydrogen gas.
Base + Metal → Salt + Hydrogen gas
Example:
1. When sodium hydroxide (NaOH) reacts with certain metals like zinc Zn, salt and hydrogen gas are produced.

Question 11.
When zinc metal is treated with dilute HCl or dilute H₂SO₄, hydrogen gas is evolved, but with dilute HNO₃ hydrogen gas is not evolved. Explain.
Answer:
Zinc metals reacts with dilute HCl and dilute H₂SO₄ evolving hydrogen gas. Since Zn metal is more reactive than H₂ gas, Zn can displace H₂ gas from dilute HCl and dilute H₂SO₄ solutions.
Reaction:

Hydrogen gas is not evolved by the reaction of Zn metal with dilute HNO₃, because HNO₃ is a strong oxidizing agent and H₂ gas evolved during reaction is oxidized to H₂O. Therefore, H₂ gas is not obtained during the reaction of Zn with HNO₃.

Question 12.
What happens when an acid reacts with metal carbonate or metal hydrogen carbonate? Give one example.
Answer:
Reaction of acid with metal carbonate or metal hydrogen carbonate:
When acids react with metal carbonate or metal hydrogen carbonate, most acids produce salt, water and carbon dioxide gas.

Question 13.
What happens when carbon dioxide in less proportion and excess proportion is passed through the solution of lime water? OR State the reactions that take place when carbon dioxide is passed through lime water (calcium hydroxide solution).
Answer:
On passing carbon dioxide in less proportion through lime water (calcium hydroxide solution), the solution turns milky because a white milky precipitate of calcium carbonate is formed.

On passing excess carbon dioxide through lime water, precipitate of calcium carbonate dissolves due to formation of soluble calcium bicarbonate.

Question 14.
What happens when an acid and a base react? Give one example.
Answer:
Reaction of acid with base:

• When acid reacts with base, salt and water are produced.
• Since base neutralizes the effect of acid, this reaction is called neutralization reaction.
  Acid → Base → Salt + Water

Example:
When sodium hydroxide, a base reacts with hydrochloric acid, it produces salt and water.

Question 15.
What is formed when copper oxide reacts with dilute hydrochloric acid? State the change in colour that you will observe along with the balanced chemical reaction. OR What happens when a metal oxide reacts with acid? Give one example.
Answer:
Reaction of acid with metal oxide:
1. When acid reacts with metal oxide, salt and water are produced.
Acid + Metal oxide → Salt + Water

2. When hydrochloric acid reacts with copper oxide, a salt of copper (Il) chloride is formed.
Example:

The colour of the solution is bluish-green due to the formation of copper chloride.

Question 16.
What happens when a non-metallic oxide reacts with a base?
Answer:
When a non-metallic oxide reacts with base, the reaction gives out salt and water.
Example:

Question 17.
How are hydrogen ions produced when an acid is dissolved in water? Explain with necessary chemical equations. Also mention method of writing the hydrogen ions so formed.
Answer:
1. When an acid (or an acidic substance) dissolves in water, it produces hydrogen (H⁺) ions.
2. This happens because all acids contain positively charged hydrogen (H⁺) ions.
Example:

Thus, hydrogen ion must always be written as H⁺(aq) or hydronium ion (H₃O⁺).

Question 18.
Write a brief note on base and alkali.
Answer:
Base:

• Chemical or substances which are hydroxides of metal and have a bitter taste are called bases.
• All metal oxides and metal hydroxide are bases.
• Fo example, sodium hydroxide (NaOH), calcium oxide (CaO), calcium hydroxide Ca(OH)₂, etc. are bases.
• Washing soda (Na₂CO₃ 10H₃O), baking soda(NaHCO₃), etc. are also bases or basic substances.

Alkali:

• Those bases which can dissolve in water are called alkalis.
• Sodium hydroxide (NaOH). potassium hydroxide (KOH), etc. are alkalis or say water soluble bases.
• When a base (or a basic substance) is dissolved in water, it always produces hydroxide (OH^–) ions.
• Thus base is a substance which dissolves in water to produce hydroxide (OH^–) ions in solution.

Example:

As shown in the above reaction, when sodium hydroxide which is a base, is dissolved in water it produces hydroxide (OH) ions over and above sodium (Na⁺) ions.

Here, when potassium hydroxide is dissolved in water, it gives hydroxide (OH) ions and potassium ions.

Question 19.
State the ions responsible for acidic and basic behaviour. Explain the acidic and basic behaviour by reaction with water.
Answer:
H⁺ or H₃O⁺ ions in aqueous solution are responsible for acidic character and OH ions in an aqueous solution are responsible for basic character.

H⁺(aq) ions are formed in aqueous solution of HCl. So it can be said that HCl is an acid.

OH^– ions are formed in an aqueous solution of NaOH. So NaOH is a base.

Question 20.
One should dissolve acid in water cautiously. Give reason. OR During dilution, acid must be added to water, not vice-versa.
Answer:
1. The process of dissolving an acid or a base is highly exothermic.
2. If one adds acid to water suddenly and in large amount or if one adds water to acid, the reaction will emit a lot of heat. This can even break the glass container.
3. The hot contents may come out of the container with an explosion and burn our eyes and body.
4. Hence, while making dilute solution of acid, acid must only be added to water that too slowly and by stirring continuously.

Question 21.
What Is dilution? How do you prepare a dilute acid?
Answer:
1. The process of mixing an acid or base with water decreases the concentration of ions (H₃O/OH) per unit volume. The acid/base so formed is called dilute acid/base.
2. For making dilute acid one should slowly add concentrated acid in small amounts to water and stir continuously.

Question 22.
Write a short note on strong and weak acids.
Answer:
Strong acids:
1. An acid which gets completely ionized completely in water or say which completely dissociates in water and produce a large amount of hydrogen [H⁺] ions (or say hydronium [H₃O⁺] ions) is called a strong acid.
2. Depending upon their structures, different acids produce different number of H⁺ ions. For example, 1 mole of hydrochloric acid will produce more hydrogen [H⁺] ions as compared to 1 mole acetic acid.

Weak acids:
1. An acid which does not ionize completely (i.e. does not dissociate completely in water) and thus produce a small amount of hydrogen [H⁺] ions (or say H₃O⁺ ions) is called a weak acid.
2. For example, when acids such as acetic acid, lactic acid, citric acid, tartaric acid, etc. are dissolved in water, they do not completely ionize and so are called weak acids.

Question 23.
Write a short note on strong and weak bases.
Answer:
Strong base:

• A base, which completely ionizes in water and thus produces a large amount of hydroxide (OH) ions, is called a strong base or a strong alkali.
• For example, when sodium hydroxide (NaOH) and potassium hydroxide (KQH) are dissolved in water, they completely get ionized and so are called strong bases.

Weak base:

• A base, which does not ionize completely in water and thus produces a small amount of hydroxide (OH) ions, is called a weak base or a weak alkali.
• For example, ammonium hydroxide (NH₄OH) and calcium hydroxide Ca(OH)₂ when dissolved in water do not completely ionize in water and so are considered weak bases.

Question 24.
Explain the methods of measuring the strength of an acid or a base. Strength (or weakness) of an acid or a base can be measured through following methods:
Answer:
(a) Through universal indicator :
1. Universal indicator is a mixture of many different indicators or say dyes which when added to a solution, changes the colour of the solution and thus indicate its pH value.

(b) Through pH scale:
1. To measure the acidic/basic nature of an aqueous solution, a scale called pH scale has been developed.
2. The pH scale measures concentration of hydrogen [H⁺] ions in the solution.
3. In German language, P of pH stands for ‘potenz’ i.e. power.
4. The scale points range from O to 14. 0 means very acidic and 14 means very alkaline. Scale point means neutral solution.
5. A lower pH value means there is higher concentration of H⁺ ions and hence the solution is acidic. So, as one moves towards O from 7, the solution goes on becoming more and more acidic.
6. Similarly, a higher pH value means there is lower concentration of H⁺ (and higher concentration of OH^–) ions and hence the solution is basic. So, as one moves from 7 to 14, the basicity of the solution increases.

Question 25.
Write a note on pH scale.
Answer:
Through pH scale:
1. To measure the acidic/basic nature of an aqueous solution, a scale called pH scale has been developed.
2. The pH scale measures concentration of hydrogen [H⁺] ions in the solution.
3. In German language, P of pH stands for ‘potenz’ i.e. power.
4. The scale points range from O to 14. 0 means very acidic and 14 means very alkaline. Scale point means neutral solution.
5. A lower pH value means there is higher concentration of H⁺ ions and hence the solution is acidic. So, as one moves towards O from 7, the solution goes on becoming more and more acidic.
6. Similarly, a higher pH value means there is lower concentration of H⁺ (and higher concentration of OH^–) ions and hence the solution is basic. So, as one moves from 7 to 14, the basicity of the solution increases.

Question 26.
How do you measure the strength of an acid or a base?
Answer:
1. The strength of an acid is determined by the number of H ions it produces, where as strength of a base is determined by the number of OH ions the base produces.

2. If we take 1 molar concentration (1 mole or 1 m acid dissolved in 1 litre of solution) of hydrochloric acid and acetic acid, then the acid which produces more number of H⁺ ions will be considered the stronger among the two. In this case, hydrochloric is a strong acid whereas acetic acid is weak.

3. Using the same method one can find the strength or weakness of a base. However, in base, the OH^– ions are released and hence they are counted (instead of H⁺).

Question 27.
Discuss the importance of pH In daily life.
Answer:
Importance of pH In everyday life:
(1) Importance of pH in existence of living beings:

• The physiological reactions occurring in our body takes place between a narrow range of 7.0 to 7.8 pH. If due to any reason the pH range gets disturbed, we may face several problems in the body.
• Even other living beings cannot tolerate more changes in their pH level.
• For example, when acid rain (i.e. rain having pH level of 5.6) falls into water reservoirs like rivers, ponds, etc., it decreases pH level of these water bodies and makes them highly acidic.
• Under such circumstances, existences of aquatic organisms like fish, microorganisms and vegetation is risked.

(2) Importance of pH in soil:

• Shrubs grow well if the pH level of soil is maintained between 6.5 to 7.3.
• The soil having pH <6.5 is called acidic soil, and that having pH > 7.3 is called alkaline.

(3) Importance of pH indigestion of food:

• Stomach plays an important role in digesting food.
• When we eat food, stomach secretes hydrochloric acid, The pH of this acid ¡s between 1 and 3.
• At such a low value of pH, an enzyme called pepsin becomes active which helps in digesting food.

(4) Importance of pH in stopping tooth decay:

• The outer layer of the teeth is made up of calcium phosphate. This layer does not dissolve in water but gets decayed when pH of inner side of mouth becomes less than 5.5. This causes tooth decay.
• When we eat food, bacteria decomposes the food particles that remain in the mouth and produce acid. This decreases the pH in the mouth.

(5) Self-defence by animals and plats through chemical warfare:

• When a honey bee bites a human being, pain, irritation and swelling is felt at the site of bite.
• This occurs because the honey bee releases acidic poison into human body during the bite.
• To neutralize the effect of this acid to some extent, substances like baking soda are applied around the bite. Stinging hair of nestle leaves injects methanoic acid and causes burning pain.

Question 28.
Indigestion causes pain and irritation. Suggest how to cure this. OR How does baking soda help in relieving stomach pain and Irritation?
Answer:
1. During indigestion, the stomach produces excessive acid. This causes pain and irritation.
2. To neutralize the effect of the acid, bases must be used. Such bases are called antacids i.e. anti-acids.
3. Magnesium hydroxide (Milk of magnesia) is one such antacid. Baking soda is also a mild base which helps in neutralizing excess acid. (Note: Eno available in medical store is also one type of antacid.)

Question 29.
When tooth pH goes below 5.5 it leads to decay. How would you prevent it?
Answer:
1. When pH in the mouth goes below 5.5, the bacteria of the mouth start producing acids and decaying the teeth.
2. The best way to prevent this is to clean the mouth properly after eating.
3. Toothpastes are basic in nature. Hence, cleaning the teeth with tooth paste neutralizes excess acid and prevents tooth decay.

Question 30.
What is acid rain? How does it affect aquatic life?
Answer:
1. If the pH of rain water becomes less than 5.6, then such rain is called acid rain.
2. When such acid water flows in water bodies such as rivers, ponds and lakes, it increases the acidity of these water sources.
3. The living organisms including humans have quite a narrow pH band of 7.0 to 7.8. Hence, even slightest change in water source makes survival difficult.

Question 31.
Why distilled (pure) water is used as a solvent in laboratories ? How does distilled water self – ionise?
Answer:
1. Distilled (pure) water is neutral. So in laboratories a solution is made using distilled water in order to get a correct pH value.
2. When an acid or a base is added to distilled water, the solution produces hydronium (H₃O⁺) and hydroxide (OH^–) ions respectively through self ionization of water. The self-ionization reaction is as under:

Question 32.
The aqueous solution of the salt produced by neutralization of weak acid and strong base possesses basic nature, while aqueous solution of salt produced by neutralization reaction of weak base and strong acid possess acidic nature – Explain.
Answer:
1. When the pH level of an aqueous solution increases more than 7, the solution goes on becoming basic.
2. In a neutralization reaction, when a weak acid and a strong base react, the salts obtained in the reaction hydrolyse to produce hydroxide [OH] ions.
3. As the level of [OH] ions increases in the solution, the solution becomes basic in nature.
4. In case, when in neutralization reaction, reaction takes place between a weak base and a strong acid, the salt obtained in the reaction hydrolyses in water to produce hydronium [H₃O⁺] ions.
5. The increased level of [H₃O⁺] ions in the aqueous solution makes the solution acidic.
6. Thus, the aqueous solution of the salt produced by neutralization reaction of weak acid and strong base possesses basic nature, while aqueous solution of salt produced by neutralization reaction of weak base and strong acid possess acidic nature.

Question 33.
What is neutralization reaction?
Answer:
Neutralization reaction:
1. When an acid and a base are mixed in right proportion, both acid and base lose their properties. Such a reaction is called neutralization reaction.

2. Thus, the reaction of acid and base is called a neutralization reaction. Moreover, reaction of acid and base gives us ‘salt’ and ‘water’.

Neutralization reaction can be generalized as follows:

Where, H refers to hydrogen ion
OH refers to hydroxide ion
MX refers to Salt
HOH refers to water (H₂O)

Question 34.
What s a salt?
Answer:
1. A salt is an ionic compound which is formed by the neutralization reaction of an acid and a base. Thus, we get salt when we react an acid with a base.
2. When we dissolve a salt in water it will get ionized and release cation (i.e. the positive +ve ion) and anion (i.e. the negative -ve ion).

3. ‘Salt’ is a general name and it does not refer only to NaCl i.e. sodium chloride. There exists a huge number of salts other than NaCl.
4. NaCl i.e. common salt so formed can be further used to produce several other products.
5. Sodium hydroxide (NaOH), baking soda (NaHCO₃), washing soda (Na₂CO₃ 10H₂O), bleaching powder (CaOCl₂) are all examples of salts.

Question 35.
How does a salt gets its name?
Answer:
The general form of naming a salt is ‘cation anion’. This means the name of cation (i.e. the positive +ve ion) of the salt will be put first and then name of its anion (i.e. the negative -ve ion) will be put second.
Example:
(a) NaCl: In NaCl salt, sodium Na⁺ has positive ions i.e. cations and chlorine CF has negative ions i.e. anions. Hence, we call NaCl salt as sodium chloride.

(b) K₂SO₄: Applying the same rule, the name of this salt is potassium sulphate.

Question 36.
What is a family of salts?
Answer:
1. There are several types of salts. The properties of all the salts are not completely different. In other words, properties of several salts are similar.
2. In general, salts having same type of cations (+ve ions) or anions (-ve) belong to the same family.
Example:
(a) Family of sodium salts (salts having Na ion): Na₂SO₄, NaCl, NaNO₃, Na₂CO₃
(b) Family of chloride salts (salts having C ion): NaCl NH₄Cl
(c) Family of sulphate salts (salts having SO₄ ion): K₂SO₄, Na₂SO₄, Ca₂SO₄, MgSO₄, CuSO₄

Preparation of Important Salts
List of important salts that we will study in this section.

Question 37.
What is brine? State its two important uses.
Answer:
1. The concentrated solution of sodium chloride i.e. concentrated solution of NaCl + H₂O is called brine.
2. Brine is used for preparing many compounds, however two products are prepared directly using brine.
They are:

• Caustic soda (sodium hydroxide) and
• Baking soda (sodium hydrogen carbonate).

Question 38.
Write a note on causltc soda (sodium hydroxide). OR Write a note on chlor-alkali process. Chemical name of caustic soda: Sodium hydroxide
Answer:
Chemical formula: NaOH
Preparation:
1. When electricity is passed through brine, it gets decomposed and produces three products. They are:

• Sodium hydroxide,
• Chlorine gas and
• Hydrogen gas.

Reaction:

2. The chlorine gas is produced at anode (+ve electrode) and hydrogen at cathode. Sodium hydroxide is formed near the cathode.

3. The process of preparing NaOH is also called chlor-alkali process because the process gives out sodium hydroxide which is an alkali and chlorine.

Question 39.
State uses of the products formed in the chior-alkali process.
Answer:
Uses of products formed in chlor-aikali process:

• NaOH: Making soap and detergent, paper, artificial fibres, de-greasing metals, etc.
• Chlorine gas: To disinfect water, added in swimming pools, making PVC, CFCs and pesticides
• Hydrogen gas: As a fuel, for making fertilizers and in making margarine

Question 40.
Write a note on preparation of baking soda.
Answer:
Chemical name of baking soda: Sodium hydrogen carbonate
Chemical formula: NaHCO₃

Preparation:
When brine reacts with ammonia in the presence of carbon dioxide gas, it produces sodium hydrogen carbonate and ammonium chloride.

Baking soda is a mild, non-corrosive basic salt.

Question 41.
Give an idea and preparation about soda ash. OR What happens when you heat baking soda? OR State and explain the reaction of baking soda when it Is heated during cooking. OR State the preparation of sodium carbonate.
Answer:
Chemical name of soda ash: Sodium carbonate
Chemical formula: Na₂CO₃
Preparation:
When baking soda gets heated during cooking, it produces sodium carbonate (soda ash) along with carbon dioxide and water.

Question 42.
What happens when baking soda is heated or mixed with water? When baking soda is heated or mixed with water than following action takes place:
Answer:

Carbon dioxide produced during this reaction causes bread or cake to rise making them soft and spongy.

Question 43.
Mention uses of baking soda:
Answer:
1. Baking socia is a soda (or a salt) commonly used in kitchen for making cakes, crispy pakoras, khaman, etc.
2. It is used for making baking powder commonly used in kitchen.
3. It is also used in making antacids. Antacids cure acidity of stomach.
4. To make soda-acid fire extinguishers
5. To make several industrial products

Question 44.
How is washing soda prepared?
Answer:
Chemical name of washing soda: Sodium carbonate decahydrate
Chemical formula: Na₂CO₃ 10H₂O
Preparation:
When sodium carbonate (i.e. soda ash) is dissolved in water, and recrystallized it gives washing soda crystals containing 10 molecules of water of crystallization.

Washing soda is a basic salt.

Question 45.
What does the number 10 represents In the formula Na₂CO₃ 10H₂O i.e. the formula of washing soda?
Answer:
The number 10 in the formula Na₂CO₃ 10H₂O tells us that there are lo water molecules in washing soda. Hence, it is also called decahydrate.

Question 46.
State the uses of washing soda.
Answer:
Uses of washing soda:

• It is used in making glass and paper.
• It is used for making sodium based compounds such as borax.
• It is used as a cleaning agent such as washing powder and soap.
• For removing permanent hardness of water.

Question 47.
Write a note on preparation of bleaching powder and its uses.
Answer:
Chemical name of bleaching powder: Calcium oxy-chloride
Chemical formula: CaOCl₂

Preparation:
1. It is prepared by the action of chlorine on slaked lime (Calcium hydroxide).
2. On passing chlorine gas over dry slaked lime (CaCOH₂), bleaching powder is obtained.

Uses of bleaching powder:

• For bleaching cotton and linen clothes in textile industry, for bleaching wood pulp in paper-making factories and bleaching washed clothes in laundries.
• As an oxidizing agent in many chemical industries.
• For disinfecting drinking water to make it free of germs.

Question 48.
State the uses of bleaching powder.
Answer:
Uses of bleaching powder:

• For bleaching cotton and linen clothes in textile industry, for bleaching wood pulp in paper-making factories and bleaching washed clothes in laundries.
• As an oxidizing agent in many chemical industries.
• For disinfecting drinking water to make it free of germs.

Question 49.
What is plaster of Paris? State its chemical name and preparation.
Answer:
Chemical name of plaster of Paris (POP): Calcium sulphate hemihydrate (hemi = half or say 1/2)
Chemical formula: CaSO₄ \(\frac{1}{2}\)H₂O (OR 2CaSO₄ . H₂O)

Preparation:
CaSO 2H₂O refers to a compound called gypsum. Its chemical name is calcium sulphate dehydrate (de =2).

• On heating gypsum at 373 K, it loses water molecules and forms calcium sulphate hemihydrate (Plaster of Paris) and \(1 \frac{1}{2}\) molecule of water.
• In CaSO₄ . \(\frac{1}{2}\)H₂O, the half molecule of water is attached as water of crystallization.
• Plaster of Paris is a white powder. On adding water, it again turns to gypsum having hard solid mass.

Question 50.
What do doctors use to join the fractured bones?
Answer:
1. Doctors use white powder of plaster of Paris to join the fractured bones.
2. On adding water to this powder, it forms a paste which can be easily applied on the fractured area. When the paste becomes dry it becomes solid hard substance called gypsum.

Question 51.
What is water of crystallization? Give examples.
Answer:
(a) The fix number of water molecules present in one formula unit of salt is called water of crystallization.
Examples:
(1) Calcium suphate hemihydrate (CaSO₄. H₂O) contains 1/2 i.e. half molecule of water of crystallization.
(2) Sodium carbonate decahydrate (Na₂CO₃ . 10H₂O) contains 10 molecules of water of crystallization.

Question 52.
List out two differences between acids and base on the basis of their chemical properties.
Answer:

Question 53.
State two difference between organic acid and mineral acid.
Answer:

Question 54.
What will be the action of the following substances on litmus paper? Dry HCl gas, Moistened NH₃ gas, Lemon juice, Carbonated soft drink, Curd, Soap solution.
Answer:
1. Dry HCl gas: No change on litmus paper
2. Moistened NH₃ gas: Red litmus will turn blue.
3. Lemon juice: Blue litmus will turn red.
4. Carbonated soft drinks: They contain carbonic acid. So blue litmus will turn red.
5. Curd: It contains lactic acid. So, blue litmus will turn red.
6. Soap solution: It is basic. So, red litmus will turn blue.

Question 55.
Name the acid present in ant sting and give its chemical formula. Also give the common method to get relief from the discomfort caused by the ant sting.
Answer:
When ant stings, it releases formic acid (or methanoic acid) in our body. Its chemical formula is HCOOH. Mix some baking soda with water and apply it on the sting. Baking soda is basic and so it will neutralize the acidic effect and provide relief.

Question 56.
What happens when nitric acid is added to egg shell?
Answer:
Egg shells contain calcium carbonate (CaCO₃). When nitric acid (HNO₃) is added to it, it results in brisk effervescence due to the formation of CO₂ gas. The reaction is as follows.

Question 57.
A student prepared solutions of (i) an acid and (ii) a base in two separate beakers. She forgot to label the solutions and litmus paper is not available In the laboratory. Since both the solutions are colourless, how will she distinguish between the two?
Answer:
Since the student does not have litmus, she can use any other indicator like methyl orange, phenolphthalein, etc. She can also use a natural indicator such as turmeric.

Question 58.
When zinc metal is treated with a dilute solution of a strong acid, a gas s evolved, which is utilized in the hydrogenation of oil. Name the gas evolved. Write the chemical equation of the reaction involved and also write a test to detect the gas formed.
Answer:
When zinc reacts with dilute solution of strong acid (like hydrochloric acid HCl), it forms salt and hydrogen gas is evolved which is used in hydrogenation of oil.

To test the presence of H₂ gas take a burning candle near the mouth of the test tube. The gas burns with a pop sound indicating presence of hydrogen gas.

Question 59.
How would you distinguish between baking powder and washing soda by heating?
Answer:
On heating baking soda (NaHCO₃), carbon dioxide (CO₂) gas is produced. It turns lime water milky.

If you heat washing soda (Na₂CO₃ . 10H₂O), water of crystallization is given out and the sail becomes anhydrous. The presence of water of crystallization given as product can be tested b treating it with anhydrous CuSO₄. It is white coloured which turns blue when it comes in contact ol anhydrous CuSO₄.

Question 60.
Salt A commonly used in bakery products on heating gets converted Into another salt B Which itself Is used for removal of hardness of water and a gas C is evolved. The gas C when passed through lime water, turns It milky. Identify A, B and C.
Answer:
Salt A: It is used in bakery and also gives Na₂CO₃ . Hence it is sodium bicarbonate NaHCO₃.
Salt B: It is sodium carbonate Na₂CO₃ because it is used for removal of hardness of water.
Gas C: It turns lime water milky and hence gas C is carbon dioxide CO₂.

Question 61.
A compound X of sodium is commonly used In kitchen for making crispy pakoras. It is also used for curing acidity in the stomach. Identify X. What is its chemical formula? State the reaction which takes place when it is heated during cooking.
Answer:
Compound X: Sodium bicarbonate or sodium hydrogen carbonate or baking soda
Formula: NaHCO₃
Chemical reaction:

Question 62.
For making cake, baking powder is taken. If at home your mother uses baking soda instead of baking powder in cake,
(a) How will it affect the taste of the cake and why?
(b) How can baking soda be converted into baking powder?
(c) What is the role of tartaric acid added to baking soda?
Answer:
(a) Baking powder contains tartaric acid. This acid reacts with sodium carbonate (Na₂CO₃) produced during decomposition of NaHCO₃ and neutralizes it.
If only sodium hydrogen carbonate (baking soda) is used in making cake, then sodium carbonate formed from it by the action of heat (during baking) will give a bitter taste to cake.
(b) By adding tartaric acid to baking soda we can convert baking soda into baking powder.
(c) Tartaric acid neutralizes the sodium carbonate formed during decomposition of NaHCO₃. Hence, it makes the cake tasty and prevents it from becoming bitter in taste.

Question 63.
“The aqueous solution of the salt produced by neutralization of weak acid and strong base possesses basic nature, while aqueous solution of salt produced by neutralization of weak base and strong acid possesses acidic nature.” Explain.
Answer:
1.

2. When we react carbonic acid which is a weak acid with a strong base of sodium hydroxide what we get is sodium carbonate.
3. The aqueous solution of sodium carbonate contains higher concentration of OH ions. Hence, the solution possesses basic nature.

4. When we react hydrochloric acid which is a strong acid with a weak base, we get ammonium chloride.
5. This aqueous solution contains higher concentration of H⁺(aq) ions compared to the concentration of OH⁺ (aq) ions. Therefore solution possesses acidic nature.

Question 64.
Identify compound X on the basis of the reactions given below. Also, write the name and chemical formulae of A, B and C.
Answer:

Question 65.
While eating food, you spill some curry on your white shirt. You immediately scrub It with soap.
What happens to its yellow colour on scrubbing with soap? Why? What happens to this stain. when the shirt is washed with plenty of water?
Answer:
When we scrub the shirt with soap, its colour changes from yellow to reddish brown. This happens because soap is basic in nature and the colour of turmeric changes from yellow to reddish brown in basic medium. If we wash the shirt with plenty of water, the stain will again turn yellow.

Very Short Answer Type Questions

Question 1.
How can we broadly classify all the compounds known to us?
Answer:
All the compounds that exist can be classified as

• Acids,
• Bases or
• Salts.

Question 2.
What are acids? Name two acids.
Answer:
Substances which when dissolved in water release hydrogen ions (H⁺) are called acids. Hydrochloric acid and sulphuric acid are two common acids.

Question 3.
State two properties of acids.
Answer:
Acids are sour in taste, Acids conduct electricity when dissolved in water, Acids react with metals to form salt and hydrogen gas.

Question 4.
What are bases?
Answer:
Bases are hydroxides of metals which give hydroxide (OH^–) ions when mixed with water.

Question 5.
State two properties of bases.
Answer:
Bases have a bitter taste. They change red litmus paper to blue.

Question 6.
What is salt?
Answer:
A salt is an ionic compound which is formed from the neutralization reaction of an acid and a base. Thus, we get salt when we react an acid with a base.

Question 7.
Read the paragraph and point out the erroneous statement.
Answer:
Acids are sour and they release H⁺ ions. Bases are bitter and they released H₃O⁺ ions. Also, bases turn litmus paper blue. Litmus is an artificial indicator to test presence of acids.

• Bases are sour
• Litmus is natural
• Bases release H⁺ ions Statement (i) and (iii) are erroneous.

Question 8.
Name two natural and synthetic indicators.
Answer:
1. Natural: Litmus and turmeric,
2. Synthetic: Methyl orange and phenolphthalein.

Question 9.
A knife, which is used to cut a fruit, was immediately dipped into water containing drops of blue litmus solution. If the colour of the solution changes to red, what inference can be drawn about the fruit and why?
Answer:
Since the colour of blue litmus turned red, the fruit is acidic.

Question 10.
What is an olfactory indicator?
Answer:
A substance whose smell changes in acidic or basic solution is called an olfactory indicator.

Question 11.
Name two olfactory indicators.
Answer:
Onion and vanilla extract.

Question 12.
What is a concentrated acid?
Answer:
An acid that contains minimum amount of water is called a concentrated acid.

Question 13.
What is a dilute acid?
Answer:
An acid obtained by mixing concentrated acid with large amount of water is called a dilute acid.

Question 14.
Name a substance that does not contain hydroxide ion yet it acts as a base. Also give its formula.
Answer:
Ammonia (NH₃)

Question 15.
You are given gold, silver and platinum. Which of these will not react with acid easily? Why?
Answer:
None. Because all are noble metals.

Question 16.
What will happen if you add zinc granules to dilute sulphuric acid?
Answer:
Bubbles containing H₂ gas will be formed on zinc granules.

Question 17.
Rasika took some dilute sulphuric acid in a test tube and added a few pieces of zinc granules to it. Her friend Shyam was supposed to write the observation in the journal but he was absent that day. As a science student write the observation that must have taken place in this experiment.
Answer:
We can observe that bubbles form on the surface of zinc granules indicating formation of gas. When we pass this gas through the soap solution, the gas gets trapped in soap molecules and bubbles are formed. The gas is hydrogen and it burns with a pop sound when a burning candle is brought near it.

Question 18.
What will be produced when you pass carbon dioxide gas through lime water?
Answer:
A milky precipitate of calcium carbonate (CaCO₃).

Question 19.
Take a small amount of copper oxide in a beaker and add dilute hydrochloric acid in it. State the chemical reaction.
Answer:
CuO + 2HCl CuCl₂ + H₂O

Question 20.
You might have seen lemon or tamarind juice being used to clean tarnished surface of copper vessels. Explain why these sour substances are effective in cleaning the vessels?
Answer:
Copper is metal whereas lemon juice and tamarind are acidic. When acid reacts with metal oxides, salt and water is formed. This makes the metal vessels shiny again

Question 21.
Why lemonade should not be prepared in copper vessel?
Answer:
Lemon is highly acidic and reacts vigorously with copper metal and causes copper poisoning. Hence………..

Question 22.
Take about 0.5 g of sodium carbonate (Na₂CO₃) in test tube A and about 0.5 g of sodium hydrogencarbonate (NaHCO₃) in test tube B. Add about 2 mL of dilute HCl in both the test tubes. State the observation of this experiment.
Answer:
When acids react with metal carbonates and metal hydrogen carbonates, they produce salt, water and carbon dioxide gas. This happens in both the test-tubes.

Question 23.
Look at the reaction given below and mention what will happen if you pass excess carbon dioxide from it. Also state the reaction.

Answer:
On passing excess carbon dioxide through the given solution, precipitate of calcium carbonate dissolves due to formation of soluble calcium ‘ bicarbonate.

Question 24.
What is neutralization reaction?
Answer:
When acid and base mix in right proportions, both of them lose their properties and produce salt and water. Such a reaction is called neutralization : reaction.

Question 25.
What will be the colour of solution if you add two drops of phenolphthalein solution in dilute NaOH solution?
Answer:
The solution will become pink.

Question 26.
Why does the colour of phenolphthalein
Answer:
NaOH gets neutralized by acid. Hence, the colour of phenolphthalein changes.

Question 27.
Give the neutralization reaction of potassium hydroxide with sulphuric acid.
Answer:

Question 28.
Write one word for the following:
(a) Water soluble base
(b) A substance which dissociates on dissolving in water to produce hydroxide ions.
Answer:
(a) Alkali
(b) Base

Question 29.
Write one word for the following:
(a) A substance which dissociates on dissolving in water to produce hydrogen ions.
(b) A reaction between an acid and a base to form salt and water.
Answer:
(a) Acid,
(b) neutralization

Question 30.
Give reason in one sentence: ‘Solution of sulphuric acid conducts electricity whereas alcohol does not’.
Answer:
Solution of sulphuric acid contains charged ions H⁺ and SO₄^-2 which helps in conducting electricity whereas alcohol does not. Hence,………..

Question 31.
Although compounds like alcohol and glucose contain hydrogen, they are not acids. Why?
Answer:
Although alcohol and glucose contain hydrogen, they do not give hydrogen ions in water and hence are not categorized as acids.

Question 32.
What is happening in this reaction?
Answer:
Here, when potassium hydroxide, a base is subjected to water, it generates hydroxide (OH^–) ions. Which out of the solutions of glucose,

Question 33.
Which out of the solutions of glucose, alcohol, hydrochloric acid, sulphuric acid and sodium hydroxide, will not conduct electricity?
Answer:
Glucose and alcohol will not conduct electricity because they will not release ions.

Question 34.
What will happen it you add water to strong acid for producing a dilute acid?
Answer:
A lot of heat will be produced. This may splash out the acid and burn our bodies. Even the glass container may break.

Question 35.
What is dilution?
Answer:
Mixing acid or base with water results in decrease in the concentration of H₃O⁺ or OH ions per unit volume. Such a process ¡s called dilution and the acid/base is called diluted.

Question 36.
An experiment was done in which about 1g solid NaCl was taken in a clean and dry test tube. Then some concentrated sulphuric acid was added to the test tube. The reaction produced hydrochloric acid. On the basis of the above activity, what do you infer about the acidic character of : (i) Dry HCl gas, (ii) HCl solution?
Answer:
The experiment suggests that hydrochloric acid produces hydrogen ions in the presence of water. But, dry hydrochloric acid does not release hydrogen ions. Thus, only HCl solution is acidic whereas dry HCl is not.

Question 37.
What is a universal Indicator?
Answer:
An indicator which can pass through a series of colour changes over a wide range of H₃O⁺ ion concentration is called universal indicator. It is a mixture of several indicators.

Question 38.
What is pH scale?
Answer:
A scale that measures the concentration of hydrogen ion in a solution is called a pH scale.

Question 39.
Arrange the following in an increasing order of their pH values: NaOH solution blood, lemon juice.
Answer:
NaOH < Blood < Lemon juice

Question 40.
Why 1M HCl solution will have a higher concentration of H ions compared to 1M CH₃COOH solution?
Answer:
1. HCl i.e. hydrochloric acid is a strong acid. Hence, HCl molecules dissociate completely into H⁺ ions and Cl^– ions and produce more H⁺ ions.
2. CH₃COOH i.e. acetic acid is a weak acid and so it does not dissociate completely. Hence, it produces less H⁺ ions. As a result

Question 41.
A student added a few drops of liquid P into distilled water. He observed that the pH of the water decreased. Can you guess the pH of the liquid P?
Answer:
We know that pH of distilled water is 7. On adding liquid P, the pH of the water decreased which means that P could be any acid such as HCl or H₃SO₄, etc.

Question 42.
Which type of substances are taken for getting relief from acidity?
Answer:
Basic substances or say antacids

Question 43.
Separate the following acids into strong acids and weak acids. Hydrochloric acid, citric acid, acetic acid, nitric acid, formic acid, sulphuric acid.
Answer:
Strong acid: Hydrochloric acid, nitric acid and sulphuric acid:
Weak acid: Citric acid acetic acid and formic acid

Question 44.
What do you mean by family of salts?
Answer:
1. There are several types of salts. The properties of all the salts are not completely different, In other words, properties of several salts are similar.
2. In general, salts having same type of cations (+ve ions) or anions (-ve) belong to the same family.

Question 45.
Which salt do you obtain by the reaction of hydrochloric acid and sodium hydroxide?
Answer:
Sodium chloride (NaCl)