Class 10

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables

Short/Long Answer Type Questions

Question 1.
Solve the following pair of linear equation by graphical method : 4x – 5y = 20, and 3x + 5y = 15, with the help of this find the value of m while 4x + 3y = m.
Solution :
The given equations are:
4x – 5y = 20 …………(1)
And 3x + 5y = 15 …………..(2)
For representation of these equations graphically, we draw the graphs of these equations as follows:
4x – 5y = 20
y = \(\frac{20-4 x}{-5}\)
y = \(\frac{4 x-20}{5}\)
We put the different values of x in this equation, then we get different values of y and we prepare the table of x, y for the equaiton 4x – 5y = 20.

And 3x + 5y = 15
⇒ y = \(\frac{15-3 x}{5}\)
We put the different values of x in this equation, then we get different values of y and we prepare the table of x, y for the equation 3x + 5y = 15.

Now, we plot the values of x hind y from tables 1 and 2 on the graph paper and we draw the graphs of the equations 1 and 2, those passes through these values.
Observe that we get two straight lines which intersect each other at point (5, 0). Hence, x = 5, y = 0 is the required solution put these values of x and y in the equation 4x + 3y = m, we get
4 × 5 + 3 × 0 = m
⇒ 20 + 0 = m
⇒ m = 20

Question 2.
Solve the following pair of linear equations by graphical method :
3x – 5y = – 1 and 2x – y = – 3. Thus find the value of A in the relation (x + y)² = A
Solution :
The given equation are :
3x – 5y = – 1
And 2x – y = – 3 …………(2)
For representation of these equations graphically, we draw the graphs of these equations are follows:
3x – 5y = – 1
⇒ y = \(\frac{-1-3 x}{-5}\)
⇒ y = \(\frac{3 x+1}{5}\)
We put the different values of x in this equation, then we get different values of y and we prepare the table of x and y for the equation 3x – 5y = – 1.

And 2x – y = – 3
y = \(\frac{-3-2 x}{-1}\)
= 3 + 2x
We put the different values of x in this equation, then we get different values of y and we prepare the table of x and y for the equation
2x – y = – 3

Now, we plot the values of x and y from tables 1 and 2 on the graph paper and we draws the graphs of equations 1 and 2 those passes through these values.

Observe that we get two straight lines which intersects each other at point (-2, -1) Hence, x = -2, y = -1 is the required solution put the values of x and y in the equation (x + y)² = A, we get
[- 2 + (-1)]² = A
(-3) = A
A = 9

Question 4.
Solve the following pair of linear equations by graphical method 2x + y = 6 and 2x – y = 2. Thus find the value of P in the relation 6x + 7y = P.
Solution :
The given equations are :
2x + y = 6 ………..(1)
And 2x – y = 2 …………(2)
For representation of these equations graphically, we draw the graphs of these equations as follows
2x + y = 6
⇒ y = 6 – 2x
We put the different values of x in this equation, then we get different galues of y and we prepare the table of x, y for the equation 2x + y = 6.

And 2x – y = 2
⇒ y = \(\frac{2-2 x}{-1}\)
⇒ y = 2x – 2
We put the different values of x in this equation, then we get different values of y and we prepare the table of x and y for the equation
2x – y = 2

Now, we plot the values of x and y from tables 1 and 2 on the graph paper and we draw the graphs of the equations 1 and 2, those passes through these values.
Observe that we get two different straight lines which intersect each other at point (2, 2).
Hence, x = 2, y = 2 is the required solution. Put the values of x and y in the relation
6x + 7y = P we get
6 × 2 + 7 × 2 = P
⇒ 12 + 14 = P
⇒ P = 26

Question 5.
For ultarakhand Flood victims two reactions A and B of class 10 contributed ₹ 1500. If the contribution of class 10, A was ₹ 100 less than that of class 10 B, find the graphically the amount contributed by both reactions.
Solution :
Let contribution of A be ₹ x and contribution of B be ₹ y.
According to question
x + y = 1500 ………….. (1)
Contribution of A is ₹ 100 less than that of B
∴ x + 100 = y ………… (2)
For representation of these equations graphically, we draw the graphs of these equations as follows:
x + y = 1500
⇒ y = 1500 – x
We put the different values of x in this equation, then we get different values of y and we prepare the table of x, y for the equation
x + y = 1500

And x + 100 = y
⇒ y = x + 100
We put the different values of x in this equation, then we get different value of y and we prepare the table of x, y for the equation
x + 100 = y

Now, we plot the values of x and y from tables 1 and 2 on the graph paper and we draw the graphs of the equations 1 and 2 those passes through these values.
Observe that we get two different straight lines which intersect each other at point (700, 800)
Hence,
x = ₹ 700,
y = ₹ 800
Therefore, contribution of A is ₹ 700 and contribution of B is ₹ 800.

Question 6.
If 2x + y = 23 and 4x – y = 19, find the value of (5y – 2x) and (\(\frac {y}{x}\)– 2)
Solution :
The given equations are :
2x + y = 23 ………(1)
4x – y = 19 ……..(2)
Adding equation (1) and (2) we get
6x = 42
⇒ x = \(\frac {42}{6}\) = 7
Substituting the value of x in equation (1), we get
2 × 7 + y = 23
⇒ y = 23 – 14 = 9
So, x = 7, y = 9
Putting the values of x and y in 5y – 2x, we get 5 × 9 – 2 × 7 = 45 – 14 = 31.
And putting the values of x and y in the \(\frac {y}{x}\) – 2, we get
\(\frac {9}{7}\) – 2 = \(\frac{9-14}{7}=\frac{-5}{7}\)
Hence, 5y – 2x = 31 and \(\frac {y}{x}\) – 2 = – \(\frac {5}{7}\)

Question 7.
A father’s age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father.
Solution :
Let the age of father be x years and the sum of ages of his children be y years.
According to questions,
x = 3y …….. (1)
After 5 years
Age of father = (x + 5) years
And sum of ages of his children = (y + 10) years
According to the question,
x + 5 = 2(y + 10)
⇒ x + 5 = 2y + 20
⇒ x – 2y = 20 – 5
⇒ x – 2y = 15 ……..(2)
From equation (1), substituting the value of x in the equation, we get
3y – 2y – 15
⇒ y = 15 years
Putting the value of y in the equation we get
x = 3 × 15 = 45 years
Hence, present age of father = 45 years.

Question 8.
Raghav scored 70 marks in a test, getting 4 marks for each right answer and losing 1 mark for each wrong answer. Had 5 marks been awarded for each correct answer and 2 marks been deducted for each wrong answer, then Raghav would have scored so marks. How many questions were there in the test.
Solution :
Let the number of right answers be x and number of wrong answers be y.
According to question,
4x – y = 70 ………… (1)
And 5x – 2y = 80 ………..(2)
Multiplying equation in (1) by 2 and substracting equ. (2) from equ. (1), we get

⇒ x = \(\frac {60}{3}\) = 20
Substituting the value of x in the equ. (1) we
4 × 20 – y = 70
⇒ – y = 70 – 80 = – 10
⇒ y = 10.
Hence, total number of questions = x + y = 20 + 10 = 30.

Question 9.
In a painting competition of a school a child made Indian national flag whose perimeter was 50 cm. Ita area will be decreased by 6 square em, if length is decreased by 3 cm and breadth is increased by 2 cm then find the dimension of flag.
Solution :
Let the length of the flag be x cm and its breadth be y cm,
so, its area = xy cm² ………..(1)
According to condition (i)
2(x + y) = 50
⇒ x + y = 25 …….(2)
If length is decreased by 3 cm. then new length is (x – 3) cm. And breadth is increased by 2 cm then new breadth is (y + 2) cm.
New area = (x – 3) (y + 2) cm²
According to condition (ii)
(x – 3) (y + 2) = xy – 6 [Using equ. (1)]
⇒ xy + 2x – 3y – 6 = xy – 6
⇒ 2x – 3y = 0 ……(3)
So, the equation of pair is
x + y – 25 = 0 …….(2)
2x – 3y + 0 = 0 …….(3)
By cross multiplication method, we get

⇒ x = 15 cm and y = 10 cm.
Hence, length of flag is 15 cm and its breadth is 10 cm.

Question 10.
Sumit is 3 times as old as his son. Five years later, he shall be two and half times as old as his son. How old is sumit at present age ?
Solution :
Let present age of sumit be x years and present age of his son be y years.
According to question
x = 3y
x – 3y = 0 …….. (1)
After 5 years later
Age of sumit = (x + 5) years
And age of his son = (y + 5) years
According to question,
x + 5 = 2\(\frac {1}{2}\)(y + 5)
x + 5 = \(\frac {5}{2}\)(y + 5)
2x + 10 – 5y + 25
2x – 5y = 25 – 10
2x – 5y = 15
Multiplying equation (1) by 2 and substracting eqn. (2) from equ. (1), we get

⇒ y = 15
y = 15 Substituting the value of y in the equ. (1) we
x – 3 × 15 = 0
⇒ x = 45 years
Hence, present age of sumit = 45 years.

Question 11.
For what value of k, will the following pair of equations have infinitely many solutions :
2x + 3y = 7 and (k + 2) x – 3 (1 – k)y = 5k + 1.
Solution :
The given equations are :
2x + 3y – 7 = 0
And (k + 2)x – 3 (1 – k)y – (5k + 1) = 0
The condition for infinitely many solutions is :
⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
⇒ \(\frac{2}{k+2}=\frac{3}{-3(1-k)}=\frac{-7}{-(5 k+1)}\)
⇒ – 6 (1 – k) = 3k + 6
and – 10k – 2 = – 7k – 14
⇒ – 6 + 6k = 3k + 6
and – 10k + 7k = – 14 + 2
⇒ 3k = 12 and – 3k = – 12
⇒ k = 4 and k = \(\frac {-12}{-3}\) = 4
Hence k = 4

Question 12.
For what values of m and n the following system of linear equations has infinitely many solutions:
3x + 4y = 12 and (m + n)x + 2 (m -n)y = 5m – 1.
Solution :
The given equations are :
3x + 4y – 12 = 0 ………(1)
(m + n)x + 2(m – n)y – (5m – 1) = 0
The condition for infinitely many solution is :

⇒ 6m – 6n = 4m + 4n
and 15m – 3 = 12m + 12n
⇒ 6m – 4m = 6n + 4n
and 15m – 12m – 12n = 3
2m = 10n and 3m – 12n = 3
⇒ m = 5n and m – 4n = 1
Putting the value of m in (m – 4n = 1) we get
5n – 4n = 1
n = 1
Putting the value of n is m = 5n, we get
m = 5 × 1 = 5
Hence, m = 5, n = 1.

Fill in the Blanks

Question 1.
If \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) the pair of equations a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 is …………
Solution :
Inconsistent

Question 2.
The graph of a pair of linear equations in two variables is represented by two ……………. lines.
Solution :
Straight

Question 3.
Every solution of a linear equation in two variables is a ………. on the line representing it.
Solution :
Point

Question 4.
A pair of linear equation in two variables can be represented and solved by the algebraic and ………….. methods.
Solution :
Graphical

Question 5.
If the lines are ……….. then the pair of linear equations in two variables has no solution.
Solution :
Parallel

Question 6.
The general form of a pair of ……………… equations in two variables x and y is a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 where \(a_1^2+b_1^2\) ≠ 0, \(a_2^2+b_2^2\) ≠ 0.
Solution :
Linear.

Multiple Choice Questions

Question 1.
The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is :
(a) – \(\frac {14}{3}\)
(b) \(\frac {2}{5}\)
(c) 5
(d) 10
Solution :
(d) 10

Given equation are :
x + 2y – 3 = 0 And
5x + ky + 7 = 0
The condition for inconsistant is:
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Taking first two order, we get
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}\)
⇒ \(\frac{1}{5}=\frac{2}{k}\)
⇒ k = 10

Question 2.
The value of k for which the system of equations.
2x + 3y = 5
4x + ky = 10 has infinite number of solutions, is :
(a) 3
(b) 6
(c) 0
(d) 1
Solution :
(b) 6

For inifinite number of solutions
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
\(\frac{2}{4}=\frac{3}{k}=\frac{5}{10}\)
⇒ k = \(\frac {12}{2}\) = 6
So correct option (b).

Question 3.
The value of k for which the given system of equations has a unique solution is :
kx + 2y – 5 = 0
x + 3y – 2 = 0
(a) ≠ \(\frac {3}{2}\)
(b) ≠ \(\frac {2}{3}\)
(c) ≠ \(\frac {2}{3}\)
(d) ≠ \(\frac {3}{2}\)
Solution :
(b) ≠ \(\frac {2}{3}\)

kx + 2y – 5 = 0
x + 3y – 2 = 0
For unique solution
\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
⇒ \(\frac{k}{1} \neq \frac{2}{3}\)
⇒ k ≠ \(\frac {2}{3}\)
So correct option (b).

Question 4.
The value of k for which the system of equations
2x + 4y = 5
6x + ky = 9 has no solution, is :
(a) 10
(b) 9
(c) 12
(d) 11.
Solution :
(c) 12

2x + 4y = 5
6x + ky = 9
For no solution
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}\)
⇒ \(\frac{2}{6}=\frac{4}{k}\)
⇒ k = \(\frac{6 \times 4}{2}\) = 12
So correct option (c).

Question 5.
The values of a and b for which the following system of equations
2x – 3y = 7
(a + b)x – (a + b – 3)y = 4a + b has infinitely many solutions, are :
(a) a = -5, b = 2
(b) a = -2, b = -5
(c) a = -5, b = -1
(d) a = -1, b = -5.
Solution :
(c) a = -5, b = -1

2x – 3y = 7
(a + b)x – (a + b – 3)y
= 4a + b
For inifinitily many solutions 2 -3
\(\frac{2}{a+b}=\frac{-3}{-(a+b-3)}=\frac{7}{4 a+b}\)
By solving a = -5, b = – 1 so correct choice is (c).

Question 6.
For which values of P, will the lines represented by the following pair of linear equations be parallel 3x – y – 5 = 0, 6x – 2y – P = 0.
(a) all real values expect 10
(b) 10
(c) \(\frac {5}{2}\)
(d) \(\frac {1}{2}\)
Solution :
(a) all real values expect 10

The given equation are :
3x – y – 5 = 0 …………….(1)
6x – 2y – P = 0 …………….(2)
The condition for parallel lines is

So, P can have any real values other than 10.

Haryana State Board HBSE 10th Class Science Solutions Chapter 15 Our Environment Textbook Exercise Questions and Answers.

Haryana Board 10th Class Science Solutions Chapter 15 Our Environment

HBSE 10th Class Science Our Environment Textbook Questions and Answers

Question 1.
Which of the following groups contain only biodegradable items?
(a) Grass, flowers and leather
(b) Grass, wood and plastic
(c) Fruit-peels, cake and lime-juice
(d) Cake, wood and grass
Answer:
(a) (c) and (d) are all biodegradable items.

Question 2.
Which of the following constitute a food-chain?
(a) Grass, wheat and mango
(b) Grass, goat and human
(c) Goat, cow and elephant
(d) Grass, fish and goat
Answer:
(b) Grass, goat and human

Question 3.
Which of the following are environment-friendly practices?
(a) Carrying cloth-bags to put purchases in while shopping
(b) Switching-off unnecessary lights and fans
(c) Walking to school Instead of getting your mother to drop you on her scooter
(d) All of the above
Answer:
(d) All of the above

Question 4.
What will happen if we kill all the organisms in one trophic level?
Answer:
Nature has created a finely balanced ecosystem with right proportion of individuals at each trophic level.
1. If all the organisms of one trophic level are killed the organisms in the next level would also die out of starving. This will in turn affect other levels too.
2. The number of individuals in the lower trophic level will rise drastically because there would be no one to feed on them.
3. Thus, by killing all the organisms in any one trophic level would imbalance the entire ecosystem.

Question 5.
Will the Impact of removing all the organisms in a trophic level be different for different trophic levels?
Can the organisms of any trophic level be removed without causing any damage to the ecosystem?
Answer:
1. Removing all the organisms of a trophic level will have the same impact as removing any other organism completely from any other trophic level.
Example: Grass → Grasshopper → Frog → Snake

2. In the above example, whether you remove grasshopper or remove frog completely, both will have same impact on the balance of ecosystem because the next level individuals will starve to death.

Question 6.
What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?
Answer:
1. Pesticides and other chemicals are sprinkled on crops to protect them from diseases or pests.
2. There is no control over the amount of pesticides used. Hence, gradually, with the passage of time, the amount of these chemicals increase in the soil and water bodies.
3. The pesticides then enter the body of plants through soil or water.
4. Later, when herbivores consume these plants, pesticides enter their bodies too; and when carnivores eat affected herbivores, the pesticides enter the bodies of carnivores too.
5. These pesticides and other harmful chemicals are not degradable. So they get accumulated at each trophic level of a food chain.
6. This phenomenon of progressive accumulation of pesticides at each level in a food chain is called biological magnification.
7. Humans occupy the top level of any food chain. Hence, the maximum concentration of these chemicals get accumulated in human bodies.

Question 7.
What are the problems caused by the non-biodegradable wastes that we generate?
Answer:

• The non-biodegradable waste is inert and so it does not break down into simpler forms. Hence, such waste keeps on accumulating in the nature.
• Accumulation of such wastes pollutes soil, air and water.
• It also leads to increased level of bio-magnification.

Question 8.
If all the waste we generate is biodegradable, will this have no impact on the environment?
Answer:
If all the generated waste is biodegradable and if the authorities properly manage for its disposal or recycle, it will not cause any harmful impact on the environment.

Question 9.
Why is damage to the ozone layer a cause for concern? What steps are being taken to limit this damage?
Answer:
1. Ozone layer protects us from harmful UV rays.
2. If the ozone layer gets damaged, the UV radiations will fall directly on the earth and cause several diseases.
3. Diseases such as skin cancer, cataract, etc. would be rampant.
4. To reduce the damage to the ozone layer, use of CFC has been minimized.
5. In this regard, In 1987, the United Nations Environment Programme (UNEP) successfully got an agreement done to freeze CFC production at the base year level of 1986.

HBSE 10th Class Science Our Environment InText Activity Questions and Answers

Textbook Page no – 257

Question 1.
Why are some substances biodegradable and some non-biodegradable?
Answer:
Many substances are such that they get broken down into simpler substances by decomposers like bacteria and fungi present in the environment. Hence they get classified as biodegradable substances. However, several substances are such which the decomposers cannot convert into simpler substances. As a result, such substances remain in the environment and get termed as non-biodegradable substances.

Question 2.
Give any two ways in which biodegradable substances would affect the environment.
Answer:
(i) Presence of biodegradable substances would pollute the environment and also cause foul smell.
(ii) The location where biodegradable substances are present becomes breeding centre for insects, flies, mosquitoes, etc.

Question 3.
Give any two rays in which non-biodegradable substances would affect the environment.
Answer:
(i) The non-biodegradable substances remain present in the environment and cause pollution.
(ii) These substances remain in the environment and so enters into the ecosystem leading to death and diseases in plants, animals and humans.

Textbook Page no – 261

Question 1.
What are trophic levels? Give an example of a food chain and state the different trophic levels in it.
Answer:
1. No living organism is independent.
2. Each organism has to depend on other for its food requirement. Thus, organisms form a chain in which they depend on other for their food. This is termed as food chain.
3. Each step or level of a food chain forms a trophic leveL
4. The producers form the first trophic level, the herbivores form the second and the carnivores form the third level.

An example of food chain has been given below:
Grass →  Insects → Grasshopper → Rats → Snakes → Hawks
5. A food chain generally starts with primary producers and ends with carnivores.
6. The autotrophs or the producers are at the first trophic level. They produce energy for themselves.
7. The autotrophs are consumed by herbivores i.e. the primary consumers. The carnivores (the secondary consumers) consume herbivores.
8. This way, the producers are at the first trophic level, the primary consumers (herbivores) at second, small carnivores i.e. secondary consumers at the third level and large carnivores or tertiary consumers at the fourth trophic level.

Question 2.
What Is the role of decomposers In the ecosystem?
Answer:
Decomposers and transformers:

• Decomposers are the last type of consumers.
• The decomposers obtain their food by decomposing dead bodies of plants and animals.
• The decomposers convert complex organic matter into simple organic constituents and then transform these constituents into inorganic ones and consume them as food.

Textbook Page no – 264

Question 1.
What is ozone and how does it affect any ecosystem?
Answer:
1. Ozone (O₃) is a type of gas formed by three oxygen atoms.
2. Ozone is a deadly poisonous gas, however it helps us too.
3. Ozone gas forms a layer of protective shield. This layer prevents the damage to living organisms from the harmful ultraviolet radiations.

Question 2.
How can you help in reducing the problem of waste disposal? Give any two methods.
(1) Minimizing the use of such substances which are not biodegradable.
(2) Avoiding use of disposable items such as disposable crock&y, disposable pen, razors, etc. and produce such products which can be re-used.

Activities

Activity 1.

1. Collect waste material from your homes. This could include all the waste generated during the day, like kitchen waste (spoilt food, vegetables peels, used tea leaves, milk packets and empty cartoons), waste paper, empty medicine bottles/strips/bubble packs, old and torn clothes and broken footwear.

2. Bury this material in a pit in the school garden or if there is no space available, you can collect the material in an old bucket/flower pot and cover with at least 15 cm of soil,

3. Keep this material moist and observe at 15 days intervals.

Question 1.
What are the materials that remain unchanged over long periods of time?
Answer:
Materials such as plastic bags, broken melamine crockery, empty cartons, etc. are all non-biodegradable and hence remain unchanged for very long period.

Question 2.
What are the materials which change their form and structure over time?
Answer:
Biodegradable material such as left over food, fruit and vegetable peel, used tea leaves, newspaper etc. change their form and structure over time.

Question 3.
Of the materials that are changed, which one changes the tastest?
Answer:
Materials such as fruit and vegetable peel, tea-leaves and rotten food change their form quite fast.

Activity 2.

Use the library or internet to find out more about biodegradable and non-biodegradable substances.

Question 1.
How long are various non-biodegradable substances expected to last in our environment?
Answer:
Non-biodegradable substances remain present en the environment ranging from few to several decades as well as centuries.

Question 2.
These days, new types of plastics which are said to be biodegradable are available. Find out more about such materials and whether they do or do not harm the environment.
Answer:
Students need to find out answer for this question on their own.

Activity 3.

You might have seen an aquarium. Let us try to design one. An aquarium is an example of a human made eco-system.

Question 1.
What are the things that we need to keep in mind when we create an aquarium?
Answer:
An aquarium is a small and artificial ecosystem. So, it does not have sufficient amount of decomposers to clean the waste and dirt created. Hence, we need to clean the aquarium after few days.

Question 2.
If we add a few aquatic plants and animals it can become a self-sustaining system. Can you think how this happens?
Answer:
Aquatic plants are producers. They will make their own food through the available light and grow.
This food will be then available to herbivore fishes. This is how the food chain will be formed.

Question 3.
Can we leave the aquarium as such after we set it up? Why does It have to be cleaned once in a while? Do we have to clean ponds or lakes in the same manner? Why or why not?
Answer:
Ponds and lakes are natural entities. They get sufficient sunlight and atmospheric gases. Also, there are several decomposers and transformers in very large number in waste generated in the water bodies.

Activity 4.

Question 1.
While creating an aquarium did you take care not to put an aquatic animal which would eat others? What would have happened otherwise?
Answer:
If we put a carnivore fish, it would eat other herbivore fish of the aquarium. Since the aquarium is not a natural habitat, new fishes would not generate on their own. This would kill the aquarium ecosystem soon.

Question 2.
Make groups and discuss how each of the above groups of organisms is dependent on each other.
Answer:
In the aquarium, plants produce food which will be eaten by herbivores. The carnivores will eat herbivores. When any of these entities dies or releases waste, the decomposers will convert all such material into simpler substances. This simple substance will again be used by the plants to grow and the cycle will continue.

Question 3.
Write the aquatic organisms in order of who eats whom and form a chain of at least three steps.
Answer:

Question 4.
Would you consider any one group of organisms to be of primary importance? Why or why not?
Answer:
All groups of organisms are almost equally important because each group plays its unique role in maintaining the ecosystem.

Activity 5 to Activity 10.

To be done by students on their own.

Haryana State Board HBSE 10th Class Science Solutions Chapter 14 Sources of Energy Textbook Exercise Questions and Answers.

Haryana Board 10th Class Science Solutions Chapter 14 Sources of Energy

HBSE 10th Class Science Sources of Energy Textbook Questions and Answers

Question 1.
A solar water heater cannot be used to get hot water on ……………..
(a) a sunny day
(b) a cloudy day
(c) a hot day
(d) a windy day
Answer:
(b) a cloudy day

Question 2.
Which of the following is not an example of a bio.mass energy source?
(a) wood
(b) gobar-gas
(c) nuclear energy
(d) coal
Answer:
(c) nuclear energy

Question 3.
Most of the sources of energy we use represent stored solar energy. Which of the following is not ultimately derived from the Sun’s energy?
(a) geothermal energy
(b) wind energy
(c) nuclear energy
(d) bio-mass
Answer:
(c) nuclear energy

Question 4.
Compare and contrast fossil fuels and the Sun as direct sources of energy.
Answer:

Question 5.
Compare and contrast bio-mass and hydro electricity as sources of energy.
Answer:

Question 6.
What are the limitations of extracting energy from:
(a) the wind? (b) waves? (c) tides?
(a) Limitations of wind energy:
Answer:
(a) Wind: Moving air is called wind.

(b) Wave energy:
The biggest limitation of this energy ¡s that it can be only obtained where waves are very strong.

(a) Tidal energy:

• Tidal energy can be obtained only in coastal areas.
• The electricity generated is in lesser quantity. So, its commercial use is not possible.

Question 7.
On what basis would you classify energy sources as:
(a) renewable and non-renewable?
(b) exhaustible and inexhaustible?
Are the options given in (a) and (b) the same?
Answer:
We can classify an energy resource as renewable or non-renewable on the basis of following difference.

The options given in (a) and (b) are same Le. renewable can be called inexhaustible and non renewable, exhaustible.

Question 8.
What are the qualities of an ideal source of energy?
Answer:
Qualities of an ideal source of energy:

• The source of energy should be available In good quantity.
• It should have good output i.e. it should be able to do a large amount of work per unit volume or mass.
• It should be easily accessible.
• It should be easy to store and transport.
• It should be economical.

Question 9.
What are the advantages and disadvantages of using a solar cooker? Are there places where solar cookers would have limited utility?
Answer:
Advantages of solar cooker:

• No fuel is required for combustion.
• Maintenance is negligible.
• It is pollution free.
• The solar cooker conserves all the nutrients and vitamins and hence the natural taste of food is maintained.
• No personal attention is needed while preparing food and so time is saved.

Disadvantages :

• Food cannot be prepared on a cloudy as well as a rainy day.
• Cooking in solar cooker consumes more time.

Question 10.
What are the environmental consequences of the Increasing demand for energy? What steps would you suggest to reduce energy consumption?
Answer:
Impact on environment due to Increasing demand of energy:

• More the use of energy, more the pollution
• Natural resources will deplete at a faster rate
• Rise in global warming
• Increase in acid rain which then damages crops, monuments, iron bridges and structures, etc.
• Rise of sea-level

Steps to reduce energy consumption:

• Use public transport as much as possible
• Use bicycle or go walking to nearby places
• Energy efficient lightening system, gadgets, buildings, etc. should be developed
• Emphasis should be given on developing renewable energy sources

HBSE 10th Class Science Sources of Energy InText Activity Questions and Answers

Textbook Page no – 243

Question 1.
What is a good source of energy?
Answer:
Source of energy:

• A source from which useful energy can be extracted either directly or indirectly by means of a conversion or transformation is known as the source of energy.
• For example, sources of energy that provide us heat for cooking are LPG, kerosene, sunlight, etc.

Factors to consider while selecting an energy source:

• The source of energy should be available in good quantity.
• It should have good output i.e. it should be able to do a large amount of work per unit volume or mass.
• It should be easily accessible.
• It should be easy to store and transport.
• It should be economical.

Question 2.
What is a good fuel?
Answer:
Characteristics of a good fuel:

• The fuel should be cheap.
• It should be easily available and in good quantity.
• It should neither produce a lot of smoke nor leave a lot of residue.
• It should have good thermal capacity.

Question 3.
If you could use any source of energy for heating your food, which one would you use and why?
Answer:
LPG and PNG are the best fuels when it comes to cooking.
Reason:

• Both LPG and PNG are easily available and economical.
• They have good calorific value.
• They can be easily transported.
• They do not cause pollution and also do not leave any residue after burning.
•  In villages, gobar gas is considered as the best source of energy for cooking food. Gobar gas helps in getting rid of the waste generated, is a clean fuel and the residue works as an excellent manure.

Textbook Page no – 248

Question 1.
What are the disadvantages of fossil fuels?
Answer:
Disadvantages of fossil fuels:

• Burning fossil fuels create smoke. This causes respiratory problems.
• Fossil fuels such as coal and petroleum cause severe air pollution.
• Acidic oxides of carbon, oxygen and sulphur are released on burning fossil fuels. These substances cause acid rain which affects our water and soil resources.
• Release of carbon dioxide leads to the greenhouse effect which causes global warming.

Question 2.
Why are we looking at alternate sources of energy?
Answer:
1. Since several years the most convenient sources of energy for man are coal and petroleum. All these fuels are fossil fuels.
2. It takes millions of years for the formation of fossil fuels. But, man has almost emptied these fuels from the earth within few centuries.
3. Now, there is no possibility that man will be able to regenerate them. However, the demand is ever increasing due to rise in technology, population, inventions, etc. Hence, man is left with no option but to look for alternate sources of energy.

Question 3.
How has the traditional use of wind and water energy been modified for our convenience?
Answer:
1. Traditionally, wind and water energies were used to carry out basic tasks such as sailing boats, separating husk from wheat, drawing water from the earth using horses bullocks, watering farms using water wheels, etc.
2. In today’s time with advent of technology we use these sources of energy in several modern ways.
3. We have constructed dams to convert the energy of flowing water into electricity we have also set-up wind farms to produce electricity on large-scale commercial basis.

Textbook Page no – 253

Question 1.
What kind of mirror – concave, convex or plain – would be best suited for use in a solar cooker? Why?
Answer:
A plane mirror would be best suited in solar cooker. The reason for this is that plane mirror is a converging mirror. It reflects all the light falling on it and so more heat can be trapped in the cooker.

Question 2.
What are the limitations of the energy that can be obtained from the oceans?
Answer:
Energy from ocean is available in three forms. The limitations of each form are listed below.
(a) Tidal energy:

• Tidal energy can be obtained only in coastal areas.
• The electricity generated is in lesser quantity. So, its commercial use is not possible.

(b) Wave energy:

• The biggest limitation of this energy ¡s that it can be only obtained where waves are very strong.

(c) Ocean thermal energy:

• To obtain this energy it is necessary to have a temperature difference of 20° C or more between the surface water and water upto depth of 2 km. This increases the cost of production.

Question 3.
What is geothermal energy?
Answer:
Geothermal energy:
The deep interior region of the earth where magma is found is very hot. The energy utilized from this heat is called geothermal energy.

Question 4.
What are the advantages of nuclear energy?
Answer:
1. When the nucleus of a heavy atom (such as uranium, plutonium or thorium) is bombarded with low-energy neutrons, it gets split into lighter nuclei. This process is called nuclear fission.
2. During the splitting of nucleus, tremendous amount of energy is released. This energy is called nuclear energy.
3. The mass of the original nucleus which is bombarded is slightly more than the sum of the masses of individual nucleus formed.
4. The released energy can be used to produce steam and hence generate electricity.

Advantages:

• The atomic fission of uranium produces 10 million times the energy produced by the combustion of an atom of carbon from coal.
• The nuclear fuel can itself go on chain reaction and release energy at a controlled rate.

Textbook Page no – 253

Question 1.
Can any source of energy be pollution-free? Why or why not?
Answer:
1. Non-renewable sources cause a direct impact on environment that too at quite a fast pace.
2. In contrast to this we prefer alternative sources such as solar, CNG, hydro energy, etc and call them clean fuels.
3. Although these fuels are clean but in fact they are not fully clean. They are just cleaner than the renewable sources.
4. Generating solar power needs solar cells for which we extract silicon from earth. Similarly, we need materials like plastic, glass, etc. to make solar devices. Manufacturing or extracting these things causes damage to the environment.
5. Similarly, for constructing dams thousands of trees are cut and ecosystems are destroyed. Moreover, cement, iron and steel and several such materials are needed to build up hydropower plant. All these things cause environmental pollution or degradation.

Question 2.
Hydrogen has been used as a rocket fuel. Would you consider It a cleaner fuel than CNG? Why or why not?
Answer:
1. CNG releases carbon dioxide on burning but hydrogen does not. Moreover, hydrogen combusts completely. In these two aspects, hydrogen is a cleaner fuel than CNG.
2. On the flip side, we do not have proper technology to make hydrogen a domestic fuel like CNG. Hence, hydrogen cannot replace CNG in current scenario.

Textbook Page no – 254

Question 1.
Name two energy sources that you would consider to be renewable. Give reasons for your choices.
Answer:
(i) Energy obtained from biomass can be considered as a renewable source of energy because a lot of farm waste, plant waste and animal waste is produced continuously on a daily basis. All these wastes can be used to obtain biomass energy.

(ii) Energy derived from sun, flowing water, wind and ocean are also considered renewable because energy from these sources can be utilized as long as the solar system exists i.e. almost perpetual.

Question 2.
Give the names of two energy sources that you would consider to be exhaustible. Give reasons for your choices.
Answer:
1. Coal and petroleum are two energy resources that can be considered exhaustible.
2. These sources take millions of years for their formation and are likely to exhaust within next 200 years.

Activities

Activity 1.

Question 1.
List four forms of energy that you use from morning, when you wake up, till you reach the school.
Answer:
Although we see several forms of energies before we even reach our school, four of them are:
(i) Light energy (sunlight),
(ii) Heat energy (cooking),
(iii) Muscular energy (getting ready, bathing, etc.) and
(iv) Mechanical and electrical energy (Moving fan, mixer grinder, vehicle, etc.)

Question 2.
From where do we get these different forms of energy?
Answer:
Light energy from sun as well as electricity, heat energy from LPG/PNG, muscular energy from body, mechanical and electrical energy from various fossil fuels, etc.

Question 3.
Can we call these ‘sources’ of energy? Why or why not?
Answer:
Only fuel and sunlight can be called a source of energy. Rest all other are energies derived from some source.

Activity 2.

Question 1.
Consider the various options we have when we choose a fuel for cooking our food.
Answer:
We can either use heat energy of LPG/PNG to cook food or heat energy generated by electrical appliance such as induction.

Question 2.
What are the criteria you would consider when trying to categorize something as a good fuel?
Answer:
A good fuel should –

• Have good calorific value
• Be economical
• Cause no or least pollution
• Have easy accessibility and continuous supply
• Be easy to handle and transport

Question 3.
Would your choice be different if you lived
(a) In a forest?
(b) In a remote mountain village or small island?
(C) In New Delhi?
(d) Five centuries ago?
Answer:
(a) In case of forests, our source of energy would be wood and leaves.
(b) In rural regions it would be cattle dung, wood from trees and twigs.
(c) Since New Delhi is a city, the source of fuel would be LPG or PNG.
(d) Before five centuries, the only fuel source known to man was wood.

Question 4.
How are the factors different in each case?
Answer:
Which fuel should be selected is dependent upon the availability of the fuel and its several characteristics such as calorific value, feasibility, etc.

Activity 3.

• Take a table-tennis ball and make three slits into it.
• Put semicircular fins cut out of a metal sheet into these slits.
• Pivot the tennis ball on an axle through its centre with a straight metal wire fixed to a rigid support. Ensure that the tennis ball rotates freely about the axle.
• Now connect a cycle dynamo to this.
• Connect a bulb in series.

Question 1.
Direct a Jet of water or steam produced in a pressure cooker at the fins as shown in the figure. What do you observe?
Answer:
Observation:
We can see that the bulb starts glowing. The bulb keeps on glowing as long as the source of fuel i.e. steam runs the turbine and dynamo.

Activity 4.

Question 1.
Find out from your grand-parents or other elders —
(a) How did they go to school?
(b) How did they get water for their daily needs when they were young?
(c) What means of entertainment did they use?
Answer:
(a) They used to walk, cycle or use bullock/horse cart for going to school.
(b) They used to draw water from the well, rivers and ponds.
(c) The means of entertainment they had were fairs, circus, stage drama, dance programmes, etc.

Question 2.
Compare the above answers with how you do these tasks now.
Answer:
(a) Today over and above bicycles, we also make use of fuel and even electric powered vehicles for going to school.
(b) We get water right in our bathrooms and kitchens through a property developed plumbing network.
(c) For entertainment we today have TV, mobile phones, theaters and malls, etc.

Question 3.
Is there a difference? If yes, ¡n which case more energy from external sources is consumed?
Answer:
Yes, there is a major difference. The energy consumed today is much higher than it was in the older times.

Activity 5.

• Take two conical flasks and paint one white and the other black colour. Fill both with water.
• Place the conical flask in direct sunlight for half an hour to one hour.

Question 1.
Touch the conical flasks. Which one is hotter? You should also measure the temperature of the water in the two conical flasks with a thermometer.
Answer:
The black flask is hotter than the white.

Question 2.
Can you think of ways In which this finding could be used In your daily life?
Answer:
From this activity we conclude that black ( and dark) coloured objects absorb more heat as compared to white or light colored objects. Because of these reasons we wear light coloured clothes in summer to reflect heat. Same principle is applied in the home interior. We use light colours on the walls so that more light is reflected and house looks spacious.

Activity 6.

Study the structure and working of a solar cooker and/or a solar water-heater, particularly with regard to how it is insulated and maximum heat absorption is ensured. Design and build a solar cooker or water-heater using low cost material available and check what temperatures are achieved in your system.

Construction :

• The body of the solar. cooker is made up of wood, bad conductor material such as plastic or fibre.
• The external surface of the solar cooker is coated with an insulated material to prevent loss of heat energy. The internal surface of the cooker and the containers used in it are of black colour to absorb maximum heat.
• An adjustable plane mirror is fixed on the top of the box and it is adjusted in such a way that it can reflect maximum sunlight into the box. Moreover, the box is covered with another glass to retain the heat that goes into the container.
• By exposing solar cooker under Sunlight for 2-3 hours continuously one can obtain a temperature of 100 °C to 140°  inside the cooker.

Uses:
Solar cookers are used to prepare food items such as rice, dal, pulses and vegetables.

Question 1.
Discuss what would be the advantages and limitations of using the solar cooker or water- heater.
Answer:
Advantages of solar cooker:

• No fuel is required for combustion.
• Maintenance is negligible.

It is pollution free.

• The solar cooker conserves all the nutrients and vitamins and hence the natural taste of food is maintained.
• No personal attention is needed while preparing food and so time is saved.

Disadvantages :

• Food cannot be prepared on a cloudy as well as a rainy day.
• Cooking in solar cooker consumes more time.

Activity 7.

Question 1.
Discuss in class the questions of what is the ultimate source of energy for bio-mass, wind and ocean thermal energy.
Answer:
Sun is the ultimate source of energy for biomass, wind and ocean thermal energy because –
(i) Plants i.e. the producers of ecosystem produce food through sunlight. Animals and humans get their food as fuel from plants and other animals.
(ii) Wind blows due to sun.
(iii) Ocean thermal energy can be harnessed because of temperature difference at the sea surface and deep water. The water on the surface is hot whereas deep water is quite cold.

Question 2.
Is geothermal energy and nuclear energy different in this respect? Why?
Answer:
1. Geothermal energy is developed due to presence and heat inside the earth. In this regard, geothermal energy is not directly dependent on solar energy.
2. Nuclear energy is also not dependent on sun because this energy is obtained due to nuclear reaction of radioactive substances.

Question 3.
Where would you place hydroelectricity and wave energy?
Answer:
Hydroelectricity and wave energy under renewable energy sources.

Activity 8 and 9.

Students should do these activities on their own.

Haryana State Board HBSE 10th Class Science Solutions Chapter 12 Electricity Textbook Exercise Questions and Answers.

Haryana Board 10th Class Science Solutions Chapter 12 Electricity

HBSE 10th Class Science Electricity Textbook Questions and Answers

Question 1.
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R / R is ………
(a) 1/25
(b) 1/5
(c) 5
(d) 25
Answer:
(d) Resistance of each part = \(\frac{R}{5}\)
When the five parts are connected in parallel, the equivalent resistance R’ is given by
\(\frac{1}{R^{\prime}}=\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}+\frac{5}{R}=\frac{25}{R}\)
∴ \(\frac{\mathrm{R}}{\mathrm{R}^{\prime}}=25\)

Question 2.
Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V²/R
Answer:
(b) IR²

Question 3.
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be ………………
(a) 100W
(b) 75W
(c) 50W
(d) 25W
Answer:
Resistance \(R=\frac{V^2}{P}=\frac{(220)^2}{100}=484 \Omega\)
When operated on 110 V, the power consumed will be \(P^{\prime}=\frac{V^{\prime 2}}{R}=\frac{(110)^2}{484}=25 \mathrm{~W}\)

Question 4.
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be …………….
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
OR
Two Identical wires are first connected in series and then In parallel to a source of supply. Find the ratio of the heat produced in two cases.
Answer:
(c) 1:4
Let R be the resistance of each wire. In senes combination, the total resistance will be 2R. Heat produced,

In parallel combination, the total resistance is R/2. Heat produced.

Question 5.
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
A voltmeter is always connected in parallel connection with the circuit.

Question 6.
A copper wire has a diameter 0.5 mm and a resistivity of 1.6 x 10^-8 Ωm. What will be the length of this wire to make its resistance 10Ω? How much does the resistance change if the diameter is doubled?
Answer:
Given the diameter of the wire,
d = 0.5mm = 0.5 x 10m=5 x 10m
Resistivity of copper p = 1.6 x 10 Qm
Required resistance R = 10
Length I = ?

Conclusion: When diameter d is doubled, then resistance R becomes one-fourth of its original value.

Question 7.
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistors are given below.

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:
1. The graph between V and I plotted by using the given data is shown here.
2. Since the graph is a straight line, the slope of the graph will be reciprocal of resistance. \(\left(\text { i.e., } \frac{1}{R}\right)\), because the slope of the graph \(=\frac{\Delta \mathrm{I}}{\Delta \mathrm{V}}=\frac{1}{\mathrm{R}}\)

Question 8.
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer:
Here,v=12V, l=2.5 mA = 2.5 x 10^-3A; R=?
The resistance of the resistor R = \(\mathrm{R}=\frac{\mathrm{V}}{\mathrm{l}}=\frac{12}{2.5 \times 10^{-3}} \)
= 4800 Ω = 4.8 kΩ

Question 9.
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Answer:
Since all the given resistors are connected in series, their equivalent resistance
Rs = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4Ω
The current through the circuit,
\(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{S}}}=\frac{9}{13.4}=0.67 \mathrm{~A}\)
In a series combination, the equal amount of current I flows through all the resistors, so the current flowing through 12 Ω resistor = 0.67 A

Question 10.
How many 176 Ω resistors (In parallel) are required to carry 5 A on a 220 V line?
Answer:
Here, I = 5A; V = 220 V
∴ The total resistance of the given circuit is Rtotal = \(\frac{\mathrm{V}}{\mathrm{l}}=\frac{220}{5}=44 \Omega\)
So, when 44 Ω resistance is connected with 220 v line, 5A current would flow through the given circuit. Now, suppose ‘n’ resistors, each of resistance R, are required to be connected in parallel, so that the total resistance Rtotal becomes 44 Ω.

Thus, when we connect 4 resistors each of 176 in parallel, it will give the total resistance of 44. This will allow 5A current to flow when connected to 220 V line.

Question 11.
Show how you would connect three resistors, each of resistance 6Ω, so that the combination has a resistance of
(i) 9Ω, (ii) 4Ω.
Answer:
(i) For getting total resistance of 9Ω from three resistors each of 6Ω, we
(a) First connect two 6Ω resistors in parallel.
(b) Then, we connect this parallel combination in series with the third 6Ω resistors as shown in
For resistors connected in parallel,
\(\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\)
∴ Rp = 3Ω

Now, total resistance HaryanaBoardSolutio
= Resistance in parallel + 6 Ω of resistance connected in series.
= Rp+6Ω
= 3Ω + 6Ω = 9Ω

(iii) For getting total resistance of 4Ω from three resistors each of 6Ω, we
(a) First connect two 6Ω resistors in series
(b) Then, we connect this senes combination in parallel to the third 6Ω resistor as shown in the diagram.
For resistors connected in series,
For resistors connected in series,
Rs = 6 + 6 = 12 Ω
Now, for total resistance

Question 12.
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected In parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer:
Voltage rating of each bulb = 220 V
Power rating of each bulb = 10 W
∴ The resistance of each bulb
\(R=\frac{V^2}{P}=\frac{220^2}{10}=\frac{220 \times 220}{10}=4840 \Omega\)
Now, V = 220 v and I = 5A
∴  The total resistance of the given circuit is
\(\text { Rtotal }=\frac{V}{l}=\frac{220}{5}=44 \Omega\)
Thus, when 44Ω resistance is connected with 220 V line, 5A current flows through the given circuit.
So, when ‘n’ bulbs each having resistance R, are connected in parallel, their equivalent resistance

Thus, 110 bulbs each having resistance 4840Ω connected in parallel will give total resistance of 44 Q causing a current of 5 A to flow when we connect them to 220 V line.

Question 13.
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24Ω resistance, which may be used separately, In series, or in parallel. What are the currents In the three cases?
Answer:
Potential difference V = 220 V
Resistance of each coil R_A = R_B 24 Ω

Case-1: When coil A or B is connected separately, the current through each coil is
\(I=\frac{V}{R_A} \text { or } \frac{V}{R_B}=\frac{220}{24}=9.166 \mathrm{~A}\)

Case-2: When the coils A and B are connected in series, the equivalent resistance of the circuit
R_S = R_A + R_B = 24 + 24 = 48Ω
Now, the current through the series combination
\(\text { Is }=\frac{V}{R_s}=\frac{220}{48}=4.58 A \approx 4.6 \mathrm{~A}\)

Case-3: When the coils A and B are connected in parallel, the equivalent resistance R_P of the circuit is given by

Question 14.
Compare the power used In the 2Ω resistor in each of the following circuits:
(i) a 6 V battery in series with 1Ω and 2Ω resistors, and
(ii) a 4 V battery in parallel wIth 12Ω and 2Ω resistors.
Answer:
(i) 1Ω resistor and 2Ω resistor are connected in series.
∴ The equivalent resistance Rs = 1 + 2 = 3 Ω
Now, the voltage of the battery V = 6V
So, the current flowing through the circuit,
\(I s=\frac{V}{R_s}=\frac{6}{3}=2 A\)
In a series combination equal current 2 A flows through each resistor.
Hence, the current flowing through 2Ω resistor is also 2A.
∴ Power used in 2Ω resistor
P1 =I²sR
= (2)² x 2 = 8W

(ii) 12Ω resistor and 2Ω resistor are connected in parallel. A 4 V battery is connected in parallel with this parallel combination of resistors. The potential difference across 2Ω resistor will also be 4V.
∴ Power used in 2Ω resistor –
\(P 2=\frac{V^2}{R}=\frac{4^2}{2}=\frac{16}{2}=8 \mathrm{~W}\)
For comparing the power used in 2Ω resistor in two different circuits, we find the ratio of P1 and P2.
∴ \(\frac{P_1}{P_2}=\frac{8}{8}=1\)
∴ P1 = P2
Hence, we conclude that 2Ω resistor uses equal power i.e. 8W in both the circuits.

Question 15.
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line If the supply voltage is 220 V?
Answer:
Resistance of 100 w lamp,

When, these lamps are connected in parallel, their equivalent resistance Rp would be

Question 16.
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer:
(a) For TV set:
P=250W = 25\(\frac{\mathrm{J}}{\mathrm{s}}\)
Time t = 1h = 3600s
Electric energy (used by TV set)
= P x t
= 250\(\frac{\mathrm{J}}{\mathrm{s}}\) x 3600s
= 900000 = 900kJ

(b) For toaster:
Power P = 1200 W = 1200\(\frac{\mathrm{J}}{\mathrm{s}}\)
Time t = 10 minute = 10 x 60 = 600 s
Electric energy (used by toaster) = P x t
= 1200\(\frac{\mathrm{J}}{\mathrm{s}}\) x 600 s = 720000 720 kJ
Thus, we can say a 250 W TV set in 1 h consumes more electric energy than a 1200W toaster in 10 minute.

Question 17.
An electric heater of resistance 8 Q draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed In the heater.
Answer:
Here, I = 15A; R = 8Ω; t = 2h
The rate at which heat is developed in the heater means its electric power,
P =I²R=(15)² x 8 = 225 x 8=1800W=1800\(\frac{\mathrm{J}}{\mathrm{s}}\)

Question 18.
Explain the following.
(a) Why is tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why copper and aluminium wires are most widely employed for electricity transmission?
Answer:
(a) Tungsten metal has a very high melting point of 3380°C and so it does not melt even at very high temperature.

• Tungsten has the ability to retain much of the heat generated. So, when it becomes very hot it emits light without getting melted.
• Tungsten is very flexible and can be easily molded into filament.
• Owing to all these reasons, tungsten is the only metal used in filament of electric lamps.

(b) Compared to metals, alloys have very high resistance. This helps in better heat generation. Hence,

(c) In a series connection, the voltage gets divided. Due to this, the appliances give less output.
If fault occurs in one appliance or at one part of the circuit, the current stops flowing in the entire circuit and other appliances also stop working.

• All the electrical appliances connected in a series are operated by one single switch. So, we cannot control individual appliance.
• Owing to all these reasons, we cannot use series connection for domestic use.

(d) The resistance R of a wire is inversely proportional to its cross-section area A i.e. R α \(\frac{1}{A}\)

• So, thicker the wire (i.e. larger the cross-section area) lesser the resistance and vice-versa.

(e) Copper and aluminium have low electric resistivity. So, they can conduct electric current without much heat loss.

• Compared to other metals copper and aluminium are cheaper. Moreover, wires can be easily drawn from these metals.

HBSE 10th Class Science Electricity InText Activity Questions and Answers

Textbook Page no – 200

Question 1.
What does an electric circuit mean?
Answer:
A continuous and closed path along which electric current flows is called an electric circuit.

Question 2.
Define the unit of current.
Answer:
Electric current:
1. The rate of flow of electric charge is known as electric current. In other words, the net quantity of electric charge that flows through any cross-section of a conductor is known as the electric current.

2. Thus, electric current \(=\frac{\text { Quantity of electric charge }}{\text { Time }}\)

3. Thus, if Q is the amount of electric charge passing through any cross-section of a conductor in time t then, electric current (I) =\(\frac{Q}{t}\)

4. If a quantity of one Coulomb (1Q) electric charge passes through the conductor in 1 second, then we can say that an electric current of one ampere (1 A) is flowing through the conductor.

5. SI unit of electric current is Coulomb/second (C/s).

6. Electric current is also measured in Ampere (A), after the French scientist Andre-Marie Ampere.

7. Milliampere (mA) and microampere (μA) are smaller units to measure electric current.

8. 1 mA = 10^-3 A and 1 μA = 10^-6A.

9.  An instrument called Ammeter ¡s used to measure the electric current.

Question 3.
Calculate the number of electrons constituting one coulomb of charge.
Answer:
Charge on 1 electron = 1.6 x 10^-19C
Also, total charge Q = 1C
But, Q = ne
∴ Number of electrons

Thus, 625 x 10¹⁸ electrons will constitute 1 coulomb of charge.

Textbook Page no – 202

Question 1.
Name a device that helps to maintain a potential difference across a conductor.
Answer:
An electric cell or battery is a device that helps to maintain potential difference across a conductor.

Question 2.
What is meant by saying that the potential difference between two points is 1 V?
Answer:
If the work done to bring 1 coulomb electric charge from one point to another is 1 joule, then the potential difference between these two points is called 1 volt.
Thus, 1 volt = \(\frac{1 \text { joule }}{1 \text { Coulomb }}\)

Question 3.
How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer:
Each coulomb means every 1 coulomb i.e.
Q = 1 coulomb
Potential difference V = 6 volt
Energy = work done i.e. W =?
Work done W = V.Q.
= 6 V.1C
= 6J
Thus, 6 J energy is given to each coulomb of charge passing through a 6v battery.

Textbook Page no – 209

Question 1.
On what factors does the resistance of a conductor depend?
Answer:
The resistance of a conductor depends

• on its length,
• on its area of cross-section,
• its temperature and
• on the nature of its material

Question 2.
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer:
The current will flow more easily through a thick wire as compared to a thin wire.

Reason:

• Resistance \(\mathrm{R} \propto \frac{1}{\text { Cross – section A }}\) This means larger the value of A, lesser will be the value of resistance R.
• Thick wire will have more cross-section area compared to the thin-wire and so its resistance will be lower.

Question 3.
Let the resistance of an electrical component remain constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer:
According to Ohm’s law, I = \(\frac{\mathrm{V}}{\mathrm{R}}\). It is given that R = constant. ∴  I α V
Since I is directly proportional to potential difference V, if the value of V is halved, the value of current I will also become half.

Question 4.
Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer:
The coils of electrical heating devices such as toaster and electric iron are made of alloys such as nichrome and not of pure metals due to following reasons:
(i) Alloys have higher resistivity as compared to pure metals. This helps in easily controlling the heat
to be produced.
(ii) Alloys do not oxidize i.e. they burn readily at high temperature.
(iii) Alloys have high melting point compared to several metals.

Question 5.
Use the data of Table 12.2 to answer the following.
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor?
Answer:
(a) The electrical resistivity of iron is 10.0 x 10^-8 Ω m whereas that of mercury is 94 x 10^-8 Ωm.
Lesser the resistivity, more will be the conduction of electricity. Hence, iron (Fe) is a better conductor than mercury (Hg).

(b) In the entire table, silver has the lowest electrical resistivity of just 1.60 x 10^-8 Ωm. Hence, silver metal is the best conductor of electricity.

Textbook Page no – 213.

Question 1.
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5Ω resistor, an 8Ω resistor, and a 12Ω resistor, and a plug key, all connected in series.
Answer:
The required circuit diagram is as shown in the adjoining figure.

Question 2.
Redraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12Ω resistor. What would be the readings in the ammeter and the voltmeter?
Answer:
The required circuit diagram is as shown here:

Total voltage i.e. Potential difference V = 3 x 2 = 6V
Total resistance, Rs = 5 + 8 + 12 = 25Ω
Reading of ammeter, I =\(\frac{6}{25}\)= 0.24 A.
Reading of voltmeter,
V = l x R = 0.24 x 12 = 2.88V.

Textbook Page no – 216.

Question 1.
Judge the equivalent resistance when the following are connected in parallel –
(a) 1Ω and 106Ω,
(b) 1Ω and 10³ Ω,and 10⁶ Ω.
Answer:
(a) In both cases equivalent resistance will be less than 1Ω but approximately 1Ω.

Reason:

• As per the characteristic of parallel connection, when resistors are connected in parallel, the equivalent resistance is less than the least resistance.
• Here, in case (a) as well as (b), the lowest resistance value is of 1Ω. So, naturally, the value of equivalent resistance can be judged as lesser than 1Ω.

Question 2.
An electric lamp of 100Ω, a toaster of resistance 50Ω, and a water filter of resistance 500Ω are connected In parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Answer:
Resistance of electric lamp, R₁ = 100Ω
Resistance of toaster, R₂ = 50Ω
Resistance of water filler, R₃ = 500Ω
Equivalent resistance Rp of the three appliances connected in parallel is given by

\(=\frac{500}{16}=\frac{125}{4}\)
= 31.25Ω
Resistance of electric iron = Equivalent resistance of the three appliances connected in parallel
= 31.25Ω
Applied voltage, V = 220 V
Current, \(\mathrm{I}=\frac{\mathrm{V}}{\mathrm{R}}=\frac{220 \mathrm{~V}}{31.25 \Omega}=7.04 \mathrm{~A}\)

Question 3.
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer:

Question 4.
How can three resistors of resistances 2Ω, 3Ω, and 6Ω be connected to give a total resistance of  (a) 4Ω (b)1Ω?
Answer:
(a) First connect resistors of 3Ω and 6Ω in parallel connection.
Then connect 2Ω resistor in series connection with the above-mentioned two resistors.
The connection is shown in the diagram below:

Resistance in parallel connection, Rp
\(=\frac{1}{R_1}+\frac{1}{R_2}=\frac{R_1 R_2}{R_1+R_2}=\frac{3 \times 6}{3+6}=2\)
Resistance between parallel connection R_p and third resistor R₃ will be
R_s= R_p+R₃  = 2 + 2 = 4Ω
Thus, RXY gives us a total resistance of 4Ω using resistors of 2Ω 3Ω and 6Ω.

(b) On connecting all the three resistors, the diagram we get is as follows:

Total resistance in this parallel connection \(\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)

Question 5.
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4Ω, 2Ω, 8Ω, 12Ω, 24Ω?
Answer:
(a) The highest resistance can be obtained by connecting all the four coils in series. In this case,
Rs =R1+R2+R3+R4 = 4 + 8 + 12 + 24 = 48
Thus, the highest resistance which can be secured is 48Ω.

(b) The lowest resistance can be secured by connecting all the four coils in parallel. In this case,

Thus, the lowest resistance which can be secured is 22.

Textbook Page no – 218.

Question 1.
Why does the cord of an electric heater not glow while the heating element does?
Answer:
1. The heating element of the electric heater is made up of an alloy such as nichrome. It has high resistance. So, it becomes red hot on heating and starts glowing,
2. The cord is made of pure metal like copper which has very low resistance. So when electricity is passed through it, it does not become red hot and so it does not glow.

Question 2.
Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50V.
Answer:
Here, charge, Q = 96000 C,
Time t =1 h=60 x 60 = 3600s.
Potential difference V = 50 V
Now, Heat generated H = Vlt

= 50 x 96000 = 4800000 J = 4.8 X 10⁶ J
Or
=4.8 x 10³ x 10³J=4.8 x 10³ kJ = 4800 kJ

Question 3.
An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Answer:
Here, current I = 5A
Resistance R = 20 , time t = 30 s
Now,
Heat produced H = I²Rt
= (5)² x 20 x 30 = 25 x 20 x 30
=15000J = 15kJ

Textbook Page no – 220.

Question 1.
What determines the rate at which energy is delivered by a current?
Answer:
The electric current remains the same. So, it is the type of resistance placed in the circuit that determines the rate at which energy is delivered by the current.

Question 2.
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
Here, I = 5A,V = 220 V
T= 2h = (2 x 60 x 60) s= 7200 s
Power P = VI = 220 x 5 = 1100W = 1100J/s
Now,
Energy consumed W = Pt = 1100 J/s x 7200 s
= 7920000 J = 7.92 x 10⁶ j

Activities

Activity 1.

Discuss the factors on which strength of electric current in a given conductor depends.
OR
How can you say that the current flowing in a circuit is strong or weak? Discuss.
Answer:
Ohm s law says, current (I) = \((\mathrm{I})=\frac{\mathrm{V} \text { (Potential difference) }}{\mathrm{R} \text { (Resistance) }}\)

From this relation we draw two conclusions. They are –
(i) I α V (It R = constant) (ii) I \(\frac{1}{R}\)(if V = constant)

From 1st:

• Since, current (I) is directly proportional to potential difference V, it value of V is doubled, the current (I) flowing through it will also get doubled.
• It V is halved, the current will also get halved.

From 2nd:

• Current (I) is inversely proportional to resistance (R). So, if the value of resistance is doubled, the current (I) flowing will get halved.
• If resistance is halved, the current (I) will get doubled.

Conclusion:
1. The strength of electric current in a given conductor depends on  –
(i) Potential difference (V) across the ends of the conductor (ii) Resistance (R) of the conductor

Activity 2.

Aim: To show that the strength of an electric current in a circuit depends on the resistance used in the circuit.

Answer:
1. Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0 – 5A range), a plug key and some connecting wires.
2. Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter leaving a gap
XV in the circuit, as shown in Fig.
3. Complete the circuit by connecting the nichrome wire in the gap XV. Plug the key. Note down the ammeter reading. Take out the key from the plug. [Note: Always take out the key from the plug after measuring the current through the circuit.
4. Replace the nichrome wire with the torch bulb in the circuit and find the current through It by measuring the reading of the ammeter.
5. Now repeat the above step with the lo W bulb in the gap XY.
6. Are the ammeter readings different for different components connected in the gap XV? What do the above observations indicate?
7. You may repeat this activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyze the observations.

Observation:

• Yes, the ammeter readings are different for different components (Nichrome wire, a torch bulb and a low bulb) when connected in the gap XY.
• This happened because certain components offer an easy path for the flow of electric current while the others resist the flow. This means that the current through an electric component depends on its resistance.

Activity 3.

Procedure: To study the factors on which resistance of conducting wire depends.
Procedure:
Complete an electric circuit consisting of a cell, an ammeter, a nichrome wire of length I [say, marked (1)] and a plug key, as shown in figure.

1. Now, plug the key. Note the current in the ammeter.

2. Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is 21 [marked (2) in the Fig.

3. Note the ammeter reading.

4. Now replace the wire by a thicker nichron wire, of the same length I [marked (3)]. A thicker wire has larger cross-sectional area. Again note down the current through the circuit.

5. Instead of taking a nichrome wire, connect copper wire (marked (4) in Figure] in the circuit.

6. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire [marked (1)1. Note the value of the current.

7. Notice the difference in the current in all cases.

8. Does the current depend on the length of the conductor?

9. Does the current depend on the area of cross section of the wire used?

Observation and Conclusion:
From the above activity, we can note the following observations:

• When the length of the wire is doubled, the ammeter reading decreases to one-half of its initial value.
• When we use a thicker wire of the same material and of the same length, the current in the circuit increases.
• When we use copper wire of similar dimensions in place of nichrome wire, the current is the circuit increases.

Thus we conclude that the resistance of a conductor depends

• on its length,
• on its area of cross-section and
• on the nature of its material.

Activity 4.

Aim: To study the series combination of resistors.

1. Join three resistors of different values in series. Connect them with a battery, an ammeter and a plug key, as shown in figure. You may use the resistors of values like 1Ω, 2Ω, 3Ω etc., and a battery of 6 V for performing this activity.
2. Plug the key. Note the ammeter reading.
3. Change the position of the ammeter to anywhere in between the resistors. Note the ammeter reading each time.
4. Do you find any change in the value of current through the ammeter?

Observation and conclusion:

• The ammeter reading is nearly lA. This value remains same irrespective of the position of the ammeter.
  This means the current passing through each and every part of circuit is same.
• We conclude that in a series connection the current flowing is same throughout the circuit and through each resistor.

Activity 5.

Aim: To show that in a series combination of resistors, the total potential difference across the combination divides itself across the individual resistors.
Answer:
1. In Activity 124, insert a voltmeter across the ends X and Y of the series combination of three resistors, as shown in Fig. 12.16 of text-book.
2. Plug the key in the circuit and note the voltmeter reading. It gives the potential difference across the series combination of resistors. Let it be V. Now measure the potential difference across the two terminals of the battery. Compare the two values.
3. Take out the plug key and disconnect the voltmeter. Now insert the voltmeter across the ends X and P of the first resistor, as shown in Fig. 12.16 of text-book.
4. Plug the key and measure the potential difference across the first resistor. Let it be V₁.
5. Similarly, measure the potential difference across the other two resistors, separately. Let these values be V₂ and V₃, respectively.
6. Deduce a relationship between V, V₁, V₂ and V₃

Observation and Conclusion:
1. The potential difference across the series combination is equal to the potential difference across the two terminals of the battery.

2. Potential difference across the first resistor R₁ is found out to be V₁.

3. Similarly, potential difference across resistors R₂: and R₃ separately are found out to be V₂ and V₃ respectively. Moreover, V = V₁ + V₂ + V₃.

4. In a series combination of different resistors, the potential difference across different resistors is different,
i.e. If R₁ ≠ R₂ ≠ R₃, V₁ ≠ V₂ ≠ V₃.

5. The total potential difference across the combination Is equal to the sum of potential difference across each of the resistors, i.e., total potential difference across the combination divides itself across the individual resistors,
i.e. V is equal to V₁ + V₂ + V₃.

Activity 6.

Aim: To study the parallel combination of resistors.

Procedure:
1. Make a parallel combination, XV, of three re&stors having resistances R₁, R₂, and R₃, respectively. Connect it with a battery, a plug key and an ammeter, as shown in Fig. 12.10. Also connect a voltmeter in parallel with the combination of resistors.

2. Plug the key and note the ammeter reading. Let the current be I. Also, take the voltmeter reading. It gives the potential difference V, across the combination. The potential difference across each resistor is also V. This can be checked by connecting the voltmeter across each individual resistor (see Fig. 12.11).

3. Take out the plug from the key. Remove the ammeter and voltmeter from the circuit. Insert the ammeter in series with the resistor R₁, as shown in Fig. 12.11. Note the ammeter reading, l.

4. Similarly, measure the currents through R₂ and R₃. Let these be I₂ and I₃, respectively. What is the relationship between
I, I₁ , I₂ , and I₃

Observation and conclusion:

1. Ammeter showed the reading I when connected. The voltmeter showed reading V when connected.

2. Potential difference across each resistor is same (i.e. V), as across the combination of resistors, i.e., V₁ = V₂ = V₃ = V.

3. Ammeter showed reading li, when connected with the resistor R1.

4. Similarly ammeter showed reading I₂ and I₃ measured separately through R₂ and R₃ respectively.

5. The relationship between I, I₁, I₂ and I₃ is I = I₁ + I₂ + I₃
Where, I = Total current through the combination of resistors in parallel I₁, I₂, I₃ = Currents through R₁, R₂ and R₃ respectively.

6. In parallel combination of resistors, the potential difference across different resistances will be same i.e. V = V₁ = V₂ = V₃

7. Total current I in the given parallel circuit is divided amongst different resistors i.e. I = I₁ + I₂ + I₃

Haryana State Board HBSE 10th Class Science Solutions Chapter 7 Control and Coordination Textbook Exercise Questions and Answers.

Haryana Board 10th Class Science Solutions Chapter 7 Control and Coordination

HBSE 10th Class Science Control and Coordination Textbook Questions and Answers

Question 1.
Which of the following is a plant hormone?
(a) Insulin
(b) Thyroxin
(c) Estrogen
(d) Cytokinin
Answer:
(d) Cytokinin

Question 2.
The gap between two neurons is called a ……………
(a) dendrite
(b) synapse
(c) axon
(d) impulse
Answer:
(b) Synapse

Question 3.
The brain Is responsible for ………
(a) thinking
(b) regulating the heart beat
(c) balancing the body
(d) all of the above
Answer:
(d) All of the above

Question 4.
What is the function of receptors in our body? Think of situations where receptors do not work properly. What problems are likely to arise?
Answer:
1. Receptors are the special type of nerve endings which can detect the specific information from the environment.

2. Generally, they are in the sensory organs such as ears, nose, tongue, eye, etc. If receptors do not work properly, the environmental stimuli will not cause nerve impulse and hence body will not be able to respond.

Question 5.
Draw the structure of a neuron and explain its function.
Answer:
Nerve cell:
Nerve cells or neurons are the building as well as functional units of the nervous system. They carry information from one part of the body to another. Nerve cell has three components:

• Cell body,
• Dendrites and
• Axon

Cell body: The cell body of a neuron is like a typical animal cell which contains cytoplasm and a nucleus. A number of short or long fibers stretch out from the cell body. These fibers are called nerve fibers.

Dendrites: The short fibers on the cell body are known as dendrites.

Axon: The longest fiber on the cell body is called axon.

Movement of Impulse (working of nerve cell):
1. The end of the dendritic tip receives message from receptors. It then initiates a neuro-chemical reaction due to which an electric impulse known as nerve impulse is produced.

2. The dendrites pass the impulse to the cell body and then to axon upto the end of axon i.e. axon terminal bud.

3. Synapse — The gap between nerve ending of an axon of one neuron and dendrite of the next neuron is called synapse.

4. At the end of the axon terminal bud, the electrical impulse releases certain chemicals.

5. These chemicals cross the gap or synapse and start similar electrical impulse in the dendrite of the next neuron. This is how the impulse or message travels from one neuron to another in the body.

6. Finally with the help of similar synapse, the impulses are delivered from neurons to other cells such as muscle cells or glands.

Question 6.
How does phototropism occur in plants?
Answer:
1. Movement of plant in response to light is called photoperiodism.

2. The shoot apex releases auxin. a growth-promoting plant hormone. When the light comes from the side of the plant, auxin diffuses towards the shady side of the shoot and helps to grow. So, the plants appear to bend towards light.

Question 7.
Which signals will get disrupted in case of a spinal cord injury?
Answer:

• In case of injury of spinal cord, the reflex action will be disturbed.
• Due to this, quick or emergency response are affected which may harm the body.

Question 8.
How does chemical co-ordination occur in plants?
Answer:
1. Various plant hormones are important to co-ordinate plant growth and development. In addition to this, the plant hormones also play an important role to response to the environmental stimuli.
2. The plant hormones diffuse towards the location of action, from where they are secreted.
3. For example, auxin is secreted from the shoot tip. Then, it diffuses to other cells and induces growth.

Question 9.
What is the need for a system of control and coordination in an organism?
Answer:
1. Multicellular organisms are made-up of various organs and organ system.
2. It is a basic need of the body to control and co-ordinate these organs and organ systems, so that they can function properly and accomplish the voluntary as well as involuntary functions.

Question 10.
How are involuntary actions and reflex actions different from each other?
Answer:

Question 11.
Compare and contrast nervous and hormonal mechanism for control and co-ordination in animals.
Answer:

Question 12.
What is the difference between the manner in which movement takes place in a sensitive plant and the movement in our legs?
Answer:
Movement in the sensitive plants:
1. When we touch a mimosa plant (a touch-sensitive plant) with our fingers, the leaves of plant fold-up.
2. In this plant, the ‘touch’ of our fingers is the stimulus where as the ‘folding-up’ of the leaves is the nastic movement.
3. When the leaves of sensitive plants mimosa are touched with a finger, an electric impulse is generated which travels through the cells.
4. The electrical impulses act on a plant hormone.
5. Due to the impulse, the plant hormone makes the water migrate from the cells of one-half of a pulvinus to the intercellular spaces in the other half of the pulvinus.
6. This loss of water from one-half of pulvinus causes the pulvinus to lose its firmness which in turn makes the leaf fold.

Movement in our legs:

• Leg movement may be voluntary or even due to a reflex action.
• Voluntary movement of leg: Under the wish of the organism, the brain orders the leg muscles (voluntary muscles) to act. So under the control of our nervous system, the leg muscles contract and relax.
• For example, when we walk, run, play garba, dance, etc. the movement of leg muscles is under our wish.

Reflex action of leg:

1. When we encounter a stimulus that can cause harm to us, we respond through reflex action. For example, if our feet by chance touches floor which is quite hot due to exposure of the sun, we quickly lift it (i.e. move). Such movements of legs are involuntary and are done under the order from spinal cord.

2. This stimulus reaches to the spinal cord and there is an order for the immediate movement. Here, the movement is under the control of the spinal cord.

HBSE 10th Class Science Control and Coordination InText Activity Questions and Answers

Textbook Page no – 119.

Question 1.
What is the difference between a reflex action and walking?
Answer:

Question 2.
What happens at the synapse between two neurons?
Answer:
Nerve cell:
Nerve cells or neurons are the building as well as functional units of the nervous system. They carry information from one part of the body to another. Nerve cell has three components:

• Cell body,
• Dendrites and
• Axon

CEO body: The cell body of a neuron is like a typical animal cell which contains cytoplasm and a nucleus. A number of short or long fibers stretch out from the cell body. These fibers are called nerve fibers.

Dendrites: The short fibers on the cell body are known as dendrites.

Axon: The longest fiber on the cell body is called axon.

Question 3.
Which part of the brain maintains posture and equilibrium of the body?
Answer:
Hind brain is responsible for maintaining posture and equilibrium of the body.

Question 4.
How do we detect the smell of an agarbatti (incense stick)?
Answer:
1. On lightning an agarbatti, the molecules responsible for fragrance of agarbatti (incense stick) sproad in the air.

2. On Inhaling this air, the fragrance comes in contact of the factory receptor cells present in the nose.

3. From these cells, this stimuli or information reaches i.e. is transmitted to the olfactory lobe in the fore-brain.

4. The brain interprets the information and so we are able to smell the agarbatti and also identify the type of fragrance such as daisy, rose or sandalwood.

5. This entire physiological process occurs within a fraction of a second.

Question 5.
What is the role of the brain in reflex action?
Answer:
1. The task done under reflex action is done through reflex arc in co-ordination with the spinal cord.

2. In reflex action, the signal is not transmitted to the brain, it is sent only to the spinal cord and the immediate response is made.

3. After the action is completed, the brain is informed or say reported about the action. The brain then keeps that action in its memory for analyzing and responding to similar stimuli in the future.

Textbook Page no – 122

Question 1.
What are plant hormones?
Answer:
1. Plant hormones are special chemical compounds which are synthesized at one place/organ of the plants and migrate to the target organ for acting.

2. They play an important role in control and co-ordination activities of plants, as well as in growth, development and response to the environment and reproduction.

Question 2.
How is the movement of leaves of the sensitive plant different from the movement of a shoot towards light?
Answer:
1. The movement of leaves of the sensitive plants is in response to the touch stimulus. This movement is based on the difference of water molecules in that specific area. Hence, it is not a directional movement.

2. The movement of a shoot towards light due to growth promotes hormone auxin. It is against gravity.

Question 3.
Give an example of a plant hormone that promotes growth.
Answer:
Auxin, gibberellins and cytokinins are the growth promoter hormones. They promote cell division and cell growth. As a result, growth of plant organs occur.

Question 4.
How do auxin promote the growth of a tendril around a support?
Answer:
1. Auxins are growth promoting plant hormones.

2. Tendril is a thread like structure of the plant, which is highly sensitive to touch.

3. When the tendril comes in contact with any object/support, particularly at the apical parts of the plants, the growth hormone auxin diffuses. This promotes the growth of the tendril which then circles around or wraps to the support and grows further on the support.

Question 5.
Design an experiment to demonstrate hydrotropism.
Answer:
The movement of plant in response to water is known as hydrotropism.

• Take a rectangular transparent plastic container and fill it with dry soil.
• Plant a small plant on one side of the container.
• Take water in a clay glass and place it in the soil on the other end of the container as shown in the figure.
• Water will slowly move into the soil below the clay vessel and make it moist.

Observation:

We can observe that the root of the plant does not grow deep towards the soil but rather grows in the direction of moist soil. This means the root tend to grow in the direction where water is available. This property of growth is called hydrotropism.

Conclusion:

We can conclude from this activity that root grows and elongates in the direction of source of water. This means the root shows positive hydrotropism.

Textbook Page no – 125

Question 1.
How does chemical co-ordination takes place in animals?
Answer:
1. In animals, the chemical co-ordination is done by various hormones secreted by endocrine glands.
2. Endocrine glands are ductless glands.
3. The animal hormones are directly poured into blood. These hormone molecules reach the target organs and perform the task of control and co-ordination for various body functions.

Question 2.
Why is the use of iodized salt advisable?
Answer:
Adrenal gland:

• The adrenal gland releases a hormone called adrenaline. Adrenaline hormone is also called fight or flight (run-away) hormone.
• This hormone is secreted directly into the blood and carried to different parts of the body.

Effects of adrenaline:

• When adrenaline is released the heart beats become fast and so, more oxygen gets supplied to our heart.
• The muscles around the small arteries of heart and other organs contract due to which the digestive system and skin receive less blood. This diverts the blood to our skeletal muscles.
• Moreover, the breathing rate increases due to contractions of the diaphragm and the rib muscles.
• All these responses together enable the animal and human body to be ready for dealing with the situation of fight or flight i.e. fight or run-away immediately from the situation.

Question 3.
How does our body respond when adrenaline is secreted Into the blood?
Answer:
When adrenaline hormone is released into the body, the body shows the following responses:
(1) Increase in heart-rate
(2) Increase in breathing-rate
(3) The muscles of the ribs and diaphragm contract
(4) Eyes get opened widely

Question 4.
Why are some patients of diabetes treated by giving Insulin?
Answer:
1. In human bodies, pancreas release a hormone called insulin which regulates the level of sugar in the blood.

2. When insulin is not secreted in proper amounts, the sugar level in the blood rises and the person may suffer from a disease called diabetes.

3. Diabetes can be harmful for heart, brain, eyes, kidneys, etc.

4. Hence, in order to save the diabetic person from such harmful effects, he is given the injections of insulin which regulates his sugar level in the blood.

Activities

Activity – 1

Aim: To detect the taste bud (gustatory receptor) and its functions

Question 1.
Put some sugar in your mouth. How does it taste?
Answer:
Sugar is sweet in taste.

Question 2.
Block your nose by pressing it between your thumb and index finger. Now eat sugar again. Is there any difference in its taste?
Answer:
When the nose is blocked, even though the sugar taste ¡s sweet, it is somewhat different than the previous sweetness.

Question 3.
While eating lunch, block your nose in the same way and notice if you can fully appreciate the taste of the food you are eating.
Answer:
Generally, the person cannot appreciate the taste of food fully when the nose is closed/blocked.

Activity 2.

Perform an activity to study the response of plants towards light (Photoperiodism).
Answer:
Procedure:
1. Fill a conical flask with water.
2. Cover the neck of the flask with a wire mesh.
3. Keep two or three freshly germinated bean seeds on the wire mesh.
4. Take a cardboard box which is open from one side.
5. Keep the flask in the box in such a manner that the open side of the box faces light coming from a window as shown in the figure.
6. After two or three days, you will notice that the shoots bend towards light and roots away from light.
7. Now turn the flask so that the shoots are away from light and the roots towards light. Leave it undisturbed in this condition for a few days.

Observation:
The shoot apex releases auxin, a growth promoting plant hormone. When the light comes from the side of the plant, auxin diffuses towards the shady side of the shoot and helps to grow. So, the plants appear to bend towards light.

Question 1.
Have the old parts of the shoot and root changed direction?
Answer:
The older part of the shoot shows a slight movement towards light. Moreover, here the root grows slightly away from the light.

Question 2.
Are there differences in the direction of the new growth?
Answer:
The upcoming or young parts of the shoot bend more towards the light and the root apex moves away from the light.

Question 3.
What can we conclude from this activity?
Answer:
We can conclude that shoot always shows positive phototropism where as root shows negative phototropism.

Activity 3.

(i) Look at Figure 7 of text-book — Endocrine glands in human beings.
(ii) Identify the endocrine glands mentioned in the figure.
(iii) Some of these glands .have been listed in Table and discussed in the text. Consult books in the library and discuss with your teachers to find out about other glands.

Name of glands shown in the figure:

• Hypothalamus,
• Pineal gland,
• Pituitary gland,
• Thyroid gland,
• Parathyroid gland,
• Thymus gland,
• Pancreas gland,
• Adrenal gland,
• Testis (in male),
• Ovaries (in female)

Activity 4.

Hormones are secreted by endocrine glands and have specific functions. Complete Table 1 based on the hormone, the endocrine gland or the functions provided. Table 1: Some important hormones and their functions

Haryana State Board HBSE 10th Class Maths Notes Chapter 2 Polynomials Notes.

Haryana Board 10th Class Maths Notes Chapter 2 Polynomials

Introduction
We have studied about polynomials in one variable and their degrees, factors, multiples etc., in previous classes. Recall that p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial. For example, 4y² – 3y + 8 is a polynomial in y of degree 2: In this chapter we shall extend our knowledge about zeroes of a polynomial, relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials and quadratic polynomials.

Polynomial
1. Polynomial: An algebraic expression of the form a_nxⁿ + a_n-1x^n-1 + a_n-2x^n-2 + ……. + a₁x + a₀ is called a polynomial in x of degree x. Where a₀, a₁……, a_n are real numbers and coefficients of each term of polynomial, n is a non-negative integer and a_n ≠ 0.
For example :
(i) 4x + 9 is a polynomial in x of degree 1.
(ii) 7y² + 5y + 9 is polynomial in y of degree 2.
(iii) 2p³ + 1/2P² + 3P + 8 is a polynomial in P of degree 3.
The expressions like \(\frac{1}{x+2}\), \(\sqrt{x}\) + 1, \(\frac{1}{x^2+2 x+1}\) etc. are not polynomials.
2. Degree of a polynomial: The highest power of the variable in a polynomial is called the degree of the polynomial.
3. Monomials, binomials, trinomials etc. are the classifications of polynomials on the bases of number of terms.
4. Monomials, binomials and trinomials have 1, 2 and 3 terms respectively.
Degrees of linear, quadratic, cubic, biquadratic and constant polynomials are 1, 2, 3, 4 and 0 respectively.
5. Zero of a polynomial: A real number is said to be a zero of the polynomial p(x) if p(k) = 0.
6. The zeroes of polynomial p(x) are the x-coordinates of the points, where the graph of y = p(x) intersects the x-axis.
7. A linear polynomial has one and only one zero, a quadratac polynomial can have at most 2 zeroes, a cubic polynomial can have at most 3 zerors and a biquadratic polynomial can have at most 4 zeroes.
8. 0 (zero) is a constant polynomial and is also called zero polynomial. The degree of the zero polynomial is not defined.
9. A non-zero constant polynomial has no zero and every real number is a zero of the zero polynomial.
10. Division algorithm for polynomials: If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then p(x) = g(x) × q(x) + r(x)
where r(x) = 0 or degree of r(x) < degree of g(x).

Types of Polynomial
(a) Constant polynomial: A polynomial of degree zero is called a constant polynomial and it is of the form p(x) = k.
For example:
f(x) = 7, g(x) = 13, p(x) = –\(\frac{7}{3}\)
(b) Linear Polynomial: A polynomial of degree 1, is called linear polynomial.
It is of the form p(x) = ax + b, where a ≠ 0 and a and b are real numbers.
For example:
(i) p(x) = 3x + 9
(ii) f(x) = \(\frac{4}{5}\)x – 5
(c) Quadratic Polynomial: A polynomial of degree 2 is called quadratic polynomial. It is of the form p(x) = ax² + bx + c, where a and a,b,c are real numbers.
For example:
(i) p(x) = 2x² + 4x – 7
(ii) f(x) = x² – 3x + \(\sqrt{5}\)
(iii) g(y) = \(\frac{y}{3}\) – 4y² + 5
(d) Cubic Polynomial: A polynomial of degree 3 is called a cubic polynomial.
It is of the form p(x) = ax³ + bx² + cx + d.
where a ≠ 0. and a, b, c, d are real numbers.
For example:
(i) p(z) = 2z³ + z² – 4 + 9
(ii) f(x) = 4x³ + \(\sqrt{3}\)x² – 8x + 4

(e) Biquadratic polynomial: A polynomial of degree 4 is called a biquadratic polynomial. It is of the form p(x) = ax⁴ + bx³ + cx² + dx + e, where a ≠ 0 and a, b, c, d, e are real number.
For example:
(i) p(y)= y⁴ – 5y³ + 2y² + 9y + 5
(ii) f(x) = 3r⁴ – 5x² + 2.

Value of a Polynomial
In a polynomial p(x), the real number obtained by replacing x by a in p(x), is called the value of p(x) at x = a and it is denoted by p(a).
For example: Find the value of polynomial
p(x) = 4x² – 5x + 6 at
x = -1
Solution:
p(x) = 4x² – 5x + 6
putting x = -1
p(-1) = 4 (-1)² – 5 × (-1) + 6
4 + 5 + 6 = 15.

Graph of the Polynomial
(a) Graph of a linear polynomial is a straight line.
(b) For a quadratic polynomial ax² + bx + c, where a ≠ 0, the graph of the corresponding equation y = zx² + bx + c is a parabola. It has one of the two shapes either opens upward like ∪ (where a > 0) or opens downward linen ∩ (where a < 0).

Number of zeroes of a polynomial in a graph
In a polynomial p(x) of degree, the graph of y = p(x) intersects the x-axis at most points. Therefore, a polynomial p(x) of degree n has at most zeroes.

Relationship between zeroes and coefficients of a polynomial
(A) Relationship between reroes and coefficients of a quadratic polynomial:
Let α and β are zeroes of quadratic polynominl P(x) = ax² + bx + c, where a ≠ 0, then (x – α) and (x – β) are factors of P(x).
Therefore, ax² + bx + c = k(x – α)(x – β), where k is constant
= k[x² – (α + β)x + α × β]
= kx² – k(α + β)x + kαβ
On comparing the coefficients of x², x and constant terms on both sides, we get,
a = k, b = -k(α + β) and c = kαβ.
b = -k(α + β)
b = -a(α + β) [Put k = a]
(α + β) = \(-\frac{b}{a}\)
and c = kαβ
c = aαβ
αβ = \(\frac{c}{a}\)
∴ Sum of zeroes = \(-\frac{b}{a}=-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^2}\)
Product of zeroes = \(\frac{c}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}\)

(B) Relationship between zeroes and coefficients of cubic polynomial:
If, α, β and γ are zeroes of the cubic polynomial p(x) = ax³ + bx² + cx + d, where a ≠ 0.
Then by factor theorem, x – α, x – β and x – γ are the factors of p(x).
p(x) = k(x – α) (x – β) (x – γ)
ax³ + bx² + cx + d = k(x – α) (x – β) (x – γ)
= k[(x² – xβ – xα + xβ) (x – γ)]
= k[x³ – x²γ – x²β + xβγ – x²α + xαγ + xαβ – αβγ]
= k[x³ – (α + β + γ)x² + (αβ + βγ + αγ)x – αβγ]
= kx³ – k(α + β + γ)x² + k(αβ + βγ + αγ)x – kαβγ
Equating the coefficient of x³, x², x and the constant terms on the both sides, we get
a = k, …..(i)
b = -k(α + β + γ) …….(ii)
c = k(αβ + βγ + αγ) …….(iii)
d = -kαβγ ……..(iv)
From (ii),
– k(α + β + γ) = b
α + β + γ = \(-\frac{b}{k}\)
α + β + γ = \(-\frac{b}{a}\) [a = k from (i)]
from (iii)
k(αβ + βγ + αγ) = c
αβ + βγ + αγ = \(\frac{c}{k}=\frac{c}{a}\) [k = a from (i)]
From (iv),
-kαβγ = d
αβγ = \(\frac{d}{-k}=-\frac{d}{a}\) [k = a from (i)]
Hence,
(1) Sum of zeroes = \(-\frac{b}{a}=-\frac{\text { Coefficient of } x^2}{\text { Coefficient of } x^3}\)
(2) Sum of the products of the zeroes taken two at a time = \(\frac{c}{a}=\frac{\text { Coefficient of } x}{\text { Coefficient of } x^3}\)
(3) Product of zeroes = \(-\frac{d}{a}\)
= \(-\frac{\text { Constant term }}{\text { Coefficient of } x^3}\)

(C) Relationship between Zeroes and Coefficients of Biquadratic Polynomial :
If α, β, γ and δ are zeroes of Biquadratic polynomial p(x) = ax⁴ + bx³ + cx² + dx + e, where a ≠ 0.
(1) α + β + γ + δ = \(-\frac{b}{a}=-\frac{\text { Coefficient of } x^3}{\text { Coefficient of } x^4}\)
(2) αβ + βγ + γδ + δα + δβ + αγ = \(\frac{c}{a}\)
\(=\frac{\text { Coefficient of } x^2}{\text { Coefficient of } x^4}\)
(3) αγδ + αβδ + αβγ + βγδ = \(-\frac{d}{a}\) = \(-\frac{\text { Coefficient of } x}{\text { Coefficient of } x^4}\)
(4) αβγδ = \(\frac{e}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^4}\)

(D) Find a quadratic polynomial when its zeroes a and are given:
p(x) = [x² – (α + β)x + α × β]
or p(x) = [x² – (sum of zeroes)x + product of zeroes]

(E) Find a cubic polynomial when its zeroes, α, β and γ are given:
p(x) = [x³ (α + β + γ)x² + (αβ + βγ + γα)x – αβγ]
or p(x) = [x³ – (sum of zeroes)x² + (sum of products of zeroes taken two at a time) x – product of zeroes]

Division Algorithm for Polynomials
In chapter 1, we have studied Euclid’s division algorithm. Recall when a positive integer is divided by another positive integer we obtain quotient and remainder. It can be expressed in the form
a = bq + r Where 0 ≤ r < b.
Where, a is dividend, is quotient, b is divisor and r is remainder.

Here, we study the division of polynomials where a polynomial f(x) is divided by an another polynomial g(x), we get quotient q(x) and remainder r(x). We can express it in the form of long division as

By Euclid division algorithm it follows that f(x) = g(x) × q(x) + r(x)
Where r(x) = 0 or degree of r(x) < degree of g(x).
This rule is known as the division algorithm for polynomials.
Remarks: (i) If r(x) = 0, then polynomial g(x) is a factor of f(x).
(ii) If α is a zero of the polynomial f(x), then (x – α) is a factor of f(x).

For example : Divide polynomial f(x) = 4x³ – 3x² + 2x – 4 by other polynomial g(x) = -1. Find the quotient and remainder.

Here, quotient q(x) = 4x² + x + 3
and remainder r(x) = -1
∵ f(x) = g(x) × q(x) + r(x)
∴ 4x³ – 3x² + 2x – 4 = (x – 1)(4x² + x + 3) + (-1)

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

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

HBSE 10th Class Science Carbon and Its Compounds Textbook Questions and Answers

Question 1.
Ethane, with the molecular formula C₂H₆ has ……………..
(a) 6 covalent bonds
(b) 7 covalent bonds
(c) 8 covalent bonds
(d) 9 covalent bonds.
Answer:
(b) 7 covalent bonds

Question 2.
Butanone is a tour-carbon compound with the functional group
(a) carboxylic acid
(b) aldehyde
(c) ketone
(d) alcohol
Answer:
(c) ketone

Question 3.
While cooking, if the bottom of the vessel is getting blackened on the outside, it means that
(a) the food is not cooked completely.
(b) the fuel is not burning completely.
(c) the fuel is wet.
(d) the fuel is burning completely.
Answer:
(b) the fuel is not burning completely.

Question 4.
Explain the nature of the covalent bond using the bond formation in CH₃Cl.
Answer:
The atomic numbers of carbon, hydrogen and chlorine are 6, 1 and 17 respectively. Their electronic configurations are: Carbon C (2, 4), Hydrogen H (1), Chlorine Cl (2, 8, 7)
1. As can be seen, carbon requires 4 electrons to complete its octet, hydrogen requires 1 electron to complete its duplet and chlorine requires 1 electron to complete its octet.

2. Therefore, carbon shares its four electrons, one electron each with three hydrogen atoms and one electron with a chlorine atom to form four covalent bonds and complete its octet.

3. Thus, carbon attains the stable electronic configuration of nearest noble gas neon, hydrogen attains the electronic configuration of noble gas helium, while chlorine attains the electronic configuration of noble gas argon.

4. Chloromethane forms three C – H covalent bonds and one C – Cl covalent bond.

Question 5.
Draw the electron dot structures for
(a) ethanoic acid
(b) H₂S
(c) propanone
(d) F₂
Answer:

Question 6.
What is a homologous series? Explain with an example.
Answer:
Homologous series:

• 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.
• Each compound of the homologous series differs from its previous or later compound by (CH₂).
• For example, the alkanes namely methane (CH₄), ethane (C₂H₆), propane (C₃H₈) and so on form a homologous series with a definite difference of CH₂.
• 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 7.
How can ethanol and ethanoic acid be differentiated on the basis of their physical and chemical properties? Difference between ethanol and ethanoic acid:

Question 8.
Why does micelle formation take place when soap ¡s added to water? Will a micelle be formed in other solvents such as ethanol also?
(a) Micelle formation:
(b) Solubility in ethanol:
Answer:
1. Micelle formation does not take place in all types of solvents. Micelles are formed only in those solvents in which soap is soluble.
2. Since soap is not soluble in ethanol, micelles will not be formed.

Question 9.
Why are carbon and its compounds used as fuels for most applications?
Answer:
Carbon and its compounds have high thermal capacity. On burning them, a large amount of heat is used which can be put to several use, Hence, carbon and its compounds are used as fuels for most applications.

Question 10.
Explain the formation of scum when hard water is treated with soap.
Answer:
1. Hard water contains salts of calcium and magnesium.
2. When soap molecules come in contact with these salts, they form a curdy white precipitate, insoluble in water. This is called scum.

Question 11.
What change will you observe you test soap with litmus paper (red and blue)?
Answer:
Soap is alkaline (basic) in nature. Hence, it does not show any effect on the blue litmus paper. However, soap will turn red litmus paper blue.

Question 12.
What is hydrogenation? What is its Industrial application?
Answer:
1. A reaction in which adding one molecule to an organic compound gives a new but single organic compound is called an addition reaction.

2. For example, on adding hydrogen to an unsaturated (alkene or alkyne) hydrocarbon in the presence of catalysts such as palladium or nickel gives a single but saturated (alkane) product. This reaction is called an addition reaction.

Question 13.
Which of the following hydrocarbons undergo addition reactions?
C₂H₆, C₃H₈, C₃H₆, C₂H₂ and CH₄.
Answer:
1. Addition reaction occurs in unsaturated hydrocarbons.
2. Out of the given compounds only C₃H₆ and C₂H₂ are unsaturated hydrocarbons. Hence, only these compounds Will undergo additional reaction.

Question 14.
Give a test that can be used to differentiate between saturated and unsaturated hydrocarbons.
Answer:
1. Alkaline potassium permanganate (KMnO₄) can be used to test if the compound is saturated or unsaturated.
2. Put some butter in one test tube and cooking oil in another. Add KMnO₄ to both.
3. The pink colour of KMnO₄ will disappear in oil but not in butter. Hence, we can conclude that cooking oil is unsaturated whereas butter is not.

Question 15.
Explain the mechanism of the cleaning action of soaps.
Answer:
Detergents:

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

HBSE 10th Class Science Carbon and Its Compounds InText Activity Questions and Answers

Textbook Page no – 61

Question 1.
What would be the electron dot structure of carbon dioxide which has the formula CO₂?
Answer:
The electronic configuration of carbon is (2, 4) and that of oxygen is to 6. Hence, the electron dot structure of carbon dioxide will be as shown below.

Question 2.
What would be the electron dot structure of a molecule of sulphur which is made up of eight atoms of sulphur?
(Hint – The eight atoms of sulphur are joined together in the form of a ring.)
Answer:
The electronic configuration of sulphur is (2, 8, 6). Hence, the electron dot structure of sulphur will be as shown below, It is a wavy structure.

Textbook Page no – 68-69

Question 1.
How many structural isomers can you draw for pentane?
Answer:
We can draw three structural isomers of pentane (C₅H₁₂). They are:

Question 2.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer:
(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 3.
What will be the formula and electron dot Structure of cyclopentane?
Answer:

Question 4.
Draw the structures for the following compounds.
(i) Ethanoic acid
(ii) Bromopentane
(iii) Butanone
(iv) Hexanal
Are structural isomers possible for bromopentane?

Yes, structural isomers for bromopentane are possible.

Isomers of bromopentane (C₅H₁₁Br):

Question 5.
How would you name the following compounds?
Answer:

The name of the compounds are (i) Bromoethane, (ii) Methanal and (iii) 1, Hexyne

Textbook Page no – 71.

Question 1.
Why is the conversion of ethanol to ethanoic acid an oxidation reaction?
Answer:
For conversion of ethanol to ethanoic acid, oxygen is added to ethanol by oxidizing agents like alkaline potassium permanganate or acidified potassium dichromate. Hence, it is called an oxidation reaction.

Question 2.
A mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?
Answer:
Over and above oxygen, air also contains other gaseous contents. So, sufficient oxygen will not be available for wielding and it will result in incomplete combustion with a sooty flame. Using pure oxygen will give a complete combustion with blue flame. Hence, oxygen and not air s used for welding.

Textbook Page no –  74

Question 1.
How would you distinguish experimentally between an alcohol and a carboxyllc acid?
To experimentally distinguish alcohol and a carboxylic acid we can conduct the following tests:
Answer:
(a) Acid test:

• Take few drops of alcohol (let us say ethanol) in one test tube and few drops of carboxylic acid in another test tube.
• Add sodium hydrogen carbonate to each test tube and observe.
• The test-tube which produces brisk effervescence (= quick bubbling or fizzing in solution) of CO₂ gas will be carboxylic acid.
• You will see that ethanoic acid produces brisk effervescence of CO₂  gas.

(b) Alcohol test:

• Take small amount of ethanol and ethanoic acid in test tube A and B respectively.
• Add 5% solution of alkaline potassium permanganate drop by drop to this solution and warm the test tubes.
• The colour of potassium permanganate will disappear in test tube containing alcohol and not in test-tube containing acid.

Question 2.
What are oxidizing agents?
Answer:
Oxidizing agents are substances which add oxygen to other substances.
Example:
Alkaline potassium permanganate or acidified potassium dichromate can give oxygen to alcohols to convert them into carboxylic acids.

Textbook Page no – 76

Question 1.
Would you be able to check if water is hard by using a detergent?
Answer:
Detergent is equally effective in both hard and soft water. Hence, we cannot check if the water is hard or soft using detergent.

Question 2.
People use a variety of methods to wash clothes. Usually after adding the soap, they ‘beat’ the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture Is agitated in a washing machine. Why is agitation necessary to get clean clothes?
Answer:
1. The dirt that sticks to our clothes is oily in nature. We know that oil is not soluble water. So, if we simply rinse the dirty cloth in water, the cloth will not become clean because the oily dirt will not dissolve in water and remain stuck to the cloth. So, to remove the dirt one has to make some action that can help to break the dirt or make it thin.

2. Agitation i.e. stirring the clothes is also one such act that helps in making the dirt layer thin. Later, the action of soap easily cleans the clothes.

Activities

Activity 1.

Students should do this activity on their own.

Activity 2.

Aim: To Study homologous series

Question 1.
Calculate the difference in the formula and molecular masses for
(a) CH₃OH and C₂H₅OH,
(b) C₂H₅OH and C₃H₇OH, and
(c) C₃H₇OH and C₄H₉OH.
Answer:

Question 2.
Is there any similarity in these three?
Answer:
All the compounds belong to the same functional group of alcohol ( – OH). Hence, all the compounds have similar chemical properties.

Question 3.
Arrange these alcohols In the order of increasing carbon atoms to get a family. Can we call this family a homologous series?
Answer:
CH₃OH, C₂H₅OH, C₃H₇OH, C₄H₉OH. Yes this is a homologous series of alcohols.

Question 4.
Generate the homologous serles for compounds containing up to four carbons for the other functional groups given in table 4.3 of textbook.
Answer:

Activity 3.

Aim: To study the flame obtained by burning carbon compounds.
Caution: This activity needs the teacher’s assistance.

• Take some carbon compounds (naphthalene, camphor, alcohol) one by one on a spatula and burn them.
• Observe the nature of the flame and note whether smoke is produced.
• Place a metal plate above the flame.
• Is there a deposition on the plate in case of any of the compounds?

Answer:
The nature of flames obtained by burning various carbon compounds is given in the table below:

Observation and conclusion:
1. Camphor and alcohol burns with a blue non-sooty flame without any residue. This means complete combustion takes place. As a result, alcohol is a saturated compound.

2. Napthalene burns with a yellow sooty flame leaving carbon deposits. This means napthalene undergoes incomplete combustion and is an unsaturated hydrocarbon.

Activity 4.

Aim: To study the type of flame formed by complete combustion and incomplete combustion.
Procedure:
Light a Bunsen burner.
(i) When do you get a yellow, sooty flame
(ii) When do you get a blue flame?
Answer:
(i) If some of the holes of the burner are blocked, the burner receives insufficient oxygen and hence we get a yellow sooty flame.
(ii) If all the holes of the burner are clean and open, the burner gets sufficient oxygen and the flame is blue.

Activity 5.

Aim:
1. To study the oxidation of alcohols with alkaline KMnO₄.
2. Take about 3 mL of ethanol in a test tube. Warm it gently in a water bath. Add to it 5% alkaline potassium permanganate solution slowly drop by drop.
3. Does the colour of potassium permanganate persist when it is added initially?
4. Why does the colour of potassium permanganate not disappear when excess is added?
Answer:
1. No, initially the colour of potassium permanganate disappears. This happens because potassium permanganate (KMnO₄) oxidizes ethanol to ethanoic acid and itself gets reduced to MnO₂.

2. When excess potassium permanganate is added, the colour does not disappear because due to excess KMnO₄ there is no alcohol left in the test-tube and hence no reaction takes place.

Activity 6.

Aim: To study the reaction of sodium metal with ethanol.

Answer:
Teacher’s demonstration:
Drop a small piece of sodium, about the size of a couple of grains of rice, into ethanol (absolute alcohol). What do you observe? How will you test the gas evolved?

Observation:
(i) Sodium reacts vigorously with ethanol and hydrogen gas is released.
(ii) On taking a lighted match stick near the product, the gas burns with a pop sound. This confirms that hydrogen gas is present as one of the product.

Activity 7.

Aim: To study the strength of dilute acetic acid and dilute hydrochloric acid.

1. Compare the pH of dilute acetic acid and dilute hydrochloric acid using both litmus paper and universal indicator.
2. Are both acids indicated by the litmus test?
3. Does the universal indicator show them as equally strong acids?
Answer:
1. Acetic acid as well as dilute hydrochloric acid turns litmus paper red. This confirms their acidity by the litmus test.
2. Universal indicator shows different colours for both these tests. The colour clearly tells that hydrochloric acid is stronger than acetic acid.

Activity 8.

Aim: To study the preparation of ester and its properties.

(i) Take 1 mL ethanol (absolute alcohol) and 1 mL glacial acetic acid along with a few drops of concentrated sulphuric acid in a test tube.
(ii) Warm in a water bath for at least five minutes as shown in figure.
(iii) Pour into a beaker containing 20-50 mL of water and smell the resulting mixture.
Answer:

When the mixture of absolute alcohol and glacial acetic acid is heated in the presence of concentrated sulphuric acid, it produces a sweet smelling ester called ethyl acetate.

Activity 9.

Aim: To show evolution of carbon dioxide gas when carboxylic acid reacts with sodium carbonate or sodium hydrogen carbonate.

• Take a spatula full of sodium carbonate Rubb in a test tube and add 2 mL of dilute ethanoic acid.
• What do you observe?
• Pass the gas produced through freshly el prepared lime-water. What do you observe?
• Can the gas produced by the reaction between ethanoic acid and sodium carbonate be identified by this test?
• Repeat this activity with sodium hydrogen carbonate instead of sodium carbonate.

Answer:
1. A brisk effervescence appears due to the evolution of CO₂ gas.
2. The gas turns lime water milky due to formation of insoluble calcium hydroxide.
3. Yes, the CO₂ gas produced is identified in the test.
4. Similar observations as above will be seen when activity is done with sodium hydrogen carbonate.

Activity 10.

Aim: To study the solubility of oil in soap solution.

1. Take about lo mL of water each in two test tubes.
2. Add a drop of oil (cooking oil) to both the test tubes and label them as A and B.
3. To test tube B, add a few drops of soap solution.
4. Now shake both the test tubes vigorously for the same period of time.
5. Can you see the oil and water layers separately in both the test tubes immediately after you stop shaking them?
6. Leave the test tubes undisturbed for some time and observe. Does the oil layer separate out? In which test tube does this happen first?
Answer:
(i) No, oil and water do not separate immediately in any of the test-tubes.
(ii) Two separate layers of oil and water are formed in test-tube A while in test-tube B separate layers are not formed.

Activity 11.

Aim: To study how soap reacts with hard and soft water.

1. Take about lo mL of distilled water (or rain water) and 10 mL of hard water (from a tube well or hand-pump) in separate test tubes.
2. Add a couple of drops of soap solution to the test tubes.
3. Shake the test tubes vigorously for an equal period of time and observe the amount of foam formed.

• In which test tube do you get more foam?
• In which test tube do you observe a white curdy precipitate?

Answer:
1. The test-tube containing distilled water produces more foam.
2. The test-tube which contains hard water shows curd-like white precipitate.

Activity 12.

Aim: To study the solubility of soap and detergent in hard water.

1. Take two test tubes with about 10 mL of hard water in each.
2. Add five drops of soap solution to one and five drops of detergent solution to the other.
3. Shake both test tubes for the same period.

• Do both test tubes have the same amount of foam?
• In which test tube is a curdy solid formed?

Answer:
1. The test-tube which contains detergent solution produces more foam.
2. Curdy solid is formed in test-tube which contain soap solution.

Haryana State Board HBSE 10th Class Science Solutions Chapter 1 Chemical Reactions and Equations Textbook Exercise Questions and Answers.

Haryana Board 10th Class Science Solutions Chapter 1 Chemical Reactions and Equations

HBSE 10th Class Science Chemical Reactions and Equations Textbook Questions and Answers

Question 1.
Which of the statements about the reaction below are incorrect?
2 PbO(s) +C(s) →2Pb (s)+CO₂(g)
(a) Lead is getting reduced.
(b) Carbon dioxide is getting oxidized.
(c) Carbon is getting oxidized.
(d) Lead oxide is getting reduced.
(i) (a) and (b)
(ii) (a) and (c)
(iii) (a), (b) and (c)
(iv) all
Answer:
(i) (a) and (b)

Question 2.
Fe₂O₃+2Al → Al₂O₃, +2Fe
The above reaction is an example of a ………..
(a) combination reaction.
(b) double displacement reaction.
(c) decomposition reaction.
(d) displacement reaction.
Answer:
(d) displacement reaction.
Reason: Aluminium (Al) displaces iron (Fe).

Question 3.
What happens when dilute hydrochloric acid is added to iron fillings? Tick the correct answer.
(a) Hydrogen gas and iron chloride are produced
(b) Chlorine gas and iron hydroxide are produced
(c) No reaction takes place
(d) Iron salt and water are produced
Answer:
(a) Hydrogen gas and iron chloride are produced.

Question 4.
What is a balanced chemical equation? Why should chemical equations be balanced?
Answer:
The process of adding atoms to the elements on one or both sides i.e. reactant and product side so that the numbers of atoms of elements on each side becomes equal is known as balancing a chemical equation. Such an equation is called a balanced equation.

Need of balancing:
As per the universal law of conservation of mass, mass can neither be created nor destroyed in a chemical reaction. In other words, the total mass of the elements present in the products of chemical reaction should be equal to the total mass of the elements present in the reactants.

Now atoms have mass. Hence, in order to fulfill the above condition we need to make sure that the number of atoms in each element remains same on both the sides.
(a) Mg + O₂ → MgO Unbalanced. Hence, wrong chemical equation
(b) 2Mg + O₂ → 2MgO Balanced. Hence, correct chemical equation

Question 5.
Translate the following statements into chemical equations and then balance them.
(a) Hydrogen gas combines with nitrogen to form ammonia.
(b) Hydrogen sulphide gas burns in air to give water and sulphur dioxide.
(c) Barium chloride reacts with aluminium sulphate to give aluminium chloride and a precipitate of barium sulphate.
(d) Potassium metal reacts with water to give potassium hydroxide and hydrogen gas.
Answer:
(a) 3H₂(g) + N₂(g) → 2NH₃(g)
(b) 2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(l)
(c) 3BaCl₂(aq) + Al₂(SO₄)₃(aq) → 2AlCl₃(aq) + 3BaSO₄(s)
(d) 2K(s) + 2H₂O(l) → 2KOH(aq) + H₂(g)

Question 6.
Balance the following chemical equations.
(a) HNO₃ + Ca(OH)₂ → Ca(NO₃)₂ + H₂O
(b) NaOH + H₂SO₄ → Na₂SO₄ + H₂O
(c) NaCl + AgNO₃ → AgCl + NaNO₃
(d) BaCl₂ + H₂SO₄ → BaSO₄ + HCl
Answer:
(a) 2HNO₃ + Ca(OH)₂→ Ca(NO₃)₂ + 2H₂O
(b) 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
(c) NaCl + AgNO₃ → AgCl + NaNO₃
(d) BaCl₂ + H₂SO₄ → BaSCU + 2HCl

Question 7.
Write the balanced chemical equations for the following reactions.
(a) Calcium hydroxide + Carbon dioxide → Calcium carbonate + Water
(b) Zinc + Silver nitrate → Zinc nitrate + Silver
(c) Aluminium + Copper chloride → Aluminium chloride + Copper
(d) Barium chloride + Potassium sulphate → Barium sulphate + Potassium chloride
Answer:
(a) Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
(b) Zn + 2AgNO₃ → Zn(NO₃)₂ + 2Ag
(c) 2Al + 3CuCl₃ → 2AlCl₃ + 3Cu
(d) BaCl₂ + K₂SO₄ → BaSO₄ + 2KCl

Question 8.
Write the balanced chemical equation for the following and identify the type of reaction In each case.
(a) Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(s)
(b) Zinc carbonate(s) → Zinc oxide(s) + Carbon dioxide(g)
(c) Hydrogen(g) + Chlorine(g) → Hydrogen chloride(g)
(d) Magnesium(s) + Hydrochloric acid(aq) → Magnesium chlorlde(aq) + Hydrogen(g)
Answer:
(a) 2KBr(aq) + Bal(aq) → 2Kl(aq) + BaBr₂(s) — Type: Double displacement reaction
(b) ZnCO₃(s) → ZnO(s) + CO₂(g) — Type: Decomposition reaction
(c) H(g) + Cl(g) → 2HCl(g) — Type: Combination reaction
(d) Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g) — Type: Displacement reaction

Question 9.
What does one mean by exothermic and endothermic reactions? Give examples.
Answer:
A chemical reaction in which heat is released (evolved) along with the formation of products is called an exothermic reaction. (Note: Exothermic is not a major type of reaction unlike combination reaction, displacement reaction, etc.)

Example:
(i) Burning of natural gas –

(ii) Respiration
(iii) Decomposition of vegetable matter into compost.

Endothermic reaction:
A reaction in which heat is absorbed or say required is called endothermic reaction.
Example:

When calcium carbonate is supplied heat, calcium oxide and carbon dioxide are formed.

Question 10.
Why is respiration considered an exothermic reaction? Explain.
Answer:

• Food gives us energy which helps us to survive.
• When we eat food, our body starts the digestion process. During digestion, the food gets broken into simple substances. For example, carbohydrate present in rice, potato, etc. breaks down to form glucose.
•  The glucose then combines with oxygen present in the body cells and provides energy. This reaction is known as respiration.

• Since heat energy is released during respiration, it is known as exothermic reaction.

Question 11.
Why are decomposition reactions called the opposite of combination reactions? Write equations for these reactions.
Answer:
In a decomposition reaction, a single compound breaks down to produce two or more simpler substances.
For example, mercuric oxide on heating decomposes to mercury and oxygen.
2HgO → 2Hg + O₂

In a combination reaction, two or more substances simply combine to form a new substance. For example, magnesium combines with oxygen to give magnesium oxide. This process is opposite of decomposition reaction.
2Mg + O₂ → 2MgO

Hence, decomposition reactions are called opposite of combination reaction.

Question 12.
Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.
Answer:

Question 13.
What is the difference between displacement and double displacement reactions? Write equations for these reactions.
Answer:
In displacement reactions, a more reactive metal displaces a less reactive metal from its solution
while in double displacement reactions, two reactants in solution exchange their ions. For example,
1. Fe(s) + CuSO₄(aq) → Cu(s) + FeSO₄(aq)
This is a displacement reaction because iron displaces copper from its solution.

2. AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃
This is a double displacement reaction. In this reaction silver nitrate and sodium chloride exchange CF and NO ions.

Question 14.
In the refining of silver, the recovery of silver from silver nitrate solution involved displacement by copper metal. Write down the reaction involved.
Answer:

Question 15.
What do you mean by a precipitation reaction? Explain by giving examples.
Answer:
When two reactants in a solution react and one or more of the products formed is insoluble or forms a precipitate, the reaction is called a precipitation reaction.

When the aqueous solution of sodium sulphate reacts with aqueous solution of barium chloride a white precipitate of BaSO₄ is formed.

Question 16.
Explain the following in terms of gain or loss of oxygen with two examples each.
(a) Oxidation, (b) Reduction
Answer:
(a) Oxidation: A chemical reaction in which a substance gains oxygen (or loses hydrogen) is called oxidation.

(b) Reduction: A chemical reaction in which a substance loses oxygen (or gains hydrogen) is called reduction.
Example:
ZnO + O₂ → Zn + CO (Here ZnO is reduced to Zn)
CuO + H₂ → Cu+ H₂O (Here CuO is reduced to Cu)

Question 17.
A shiny brown coloured element ‘X’ on heating in air becomes black in colour. Name the element ‘X’ and the black coloured compound formed.
Answer:

Question 18.
When iron objects come in contact of moist air they get corroded.
Answer:
By applying paint the iron surface gets a protective layer. This layer of paint works as a barrier and so the surface does not comes in direct contact of moist air. This prevents the corrosion of iron.

Question 19.
Oil and fat containing food items are flushed with nitrogen. Why?
Answer:
1. Food items such as snacks like pun, chakri, etc. containing oil react with oxygen and become rancid. They then develop bad taste and smell.

2. To prevent them from getting spoiled they are flushed with unreactive gases such as nitrogen.

Question 20.
Explain the following terms with one example each (a) Corrosion, (b) Rancidity.
Answer:
(a) Corrosion

• When a metal comes in contact with humid air, moisture or a chemical such as acid, the surface of metal starts getting eaten up. This is called corrosion.
• Corrosion is mainly caused by the oxidation of metals in the humid air. Rusting of iron is the most common form of corrosion.
• Due to corrosion, silver ornaments become black and a green coating gets deposited on copper vessels.
• Corrosion damages vehicles, bridges, iron rails, ships and other metal structures.
• Corrosion of iron is a serious problem. It causes huge sum to be spent on maintaining iron structures and replacing corroded parts.

(b) Rancidity.

• Oxidation affects food that contains fats and oils.
• When food items (such as snacks like pun, chakni, chavana, etc) prepared using fat and oils are kept for longer period, they develop an unpleasant smell and taste. We then say that the food item has become rancid.
• One can reduce the rate of rancidity by keeping the food items in air-tight containers. This, slows down oxidation of food.
• Packets of chips are flushed with nitrogen to prevent chips from becoming rancid.

HBSE 10th Class Science Chemical Reactions and Equations InText Activity Questions and Answers

Textbook Page no – 6

Question 1.
Why should a magnesium ribbon be cleaned before burning in air?
Answer:
Magnesium ribbon reacts with oxygen present in air to form a protective and inert layer of magnesium oxide on its surface. This layer is unreactive and prevents the ribbon from burning. Hence, it needs to be cleaned with sandpaper before burning in air.

Question 2.
Write the balanced equation for the following chemical reactions:
(i) Hydrogen + Chlorine → Hydrogen chloride
(ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride
(iii) Sodium + Water → Sodium hydroxide + Hydrogen
Answer:
(i) H₂ + Cl₂ → HCl
H₂ + Cl₂ → 2HCl

(ii) BaCl₂ + Al₂(SO₄)₃ → BaSO₄ + AlCl₃
∴ 3BaCl₂ + Al₂(SO₄)₃ -4 3BaSO₄ + 2AlCl₃

(iii) Na + H₂O → NaOH + H₂
2Na + 2H₂O → 2NaOH + H₂

Question 3.
Write a balanced chemical equation along with symbols for the following reactions:
1. Solutions of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.
2. Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.
Answer:

• BaCl₂(aq) + Na₂SO₄(aq) → BaSO(s) + 2NaCl(aq)
• NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

Textbook Page no – 10

Question 1.
A solution of a substance ‘X’ Is used for white washing.
1. Name the substance ‘X’ and write its formula.
2. Write the reaction of the substance ‘X’ named in (i) above with water.
Answer:
1. Substance ‘X’ is quick lime. Its formula is CaO.

2. Write the reaction of the substance ‘X’ named in (i) above with water.

Question 2.
Why is the amount of gas collected in one of the test tubes in Activity 1.7 double of the amount collected in the other? Name this gas.
Answer:
In activity 7. water is electrolysed to give H₂ gas at one electrode and O₂ gas at the other electrode. The chemical reaction for this is:
2H₂O(l) → 4 2H₂(g) + O₂(g)
Thus, two molecules of water on electrolysis give 2 molecules of hydrogen gas and 1 molecule of oxygen gas. Hence, the amount of hydrogen gas collected will be double than that of oxygen gas.

Textbook Page no – 13

Question 1.
Why does the colour of copper sulphate solution change when an Iron nail is dipped In It?
Answer:

Iron is more reactive than copper. Hence, it displaces copper from the solution of copper sulphate. As a result, the blue colour of copper sulphate fades away when it becomes green coloured solution of ferrous sulphate.

Question 2.
Give an example of a double displacement reaction other than the one given in Activity 1.10.
Answer:
When hydrochloric acid is added to lead nitrate, a white precipitate of lead chloride is formed.

PB⁺⁺ ions from lead nitrate combine with 2Cl^– ions from hydrochloric acid and a precipitate of lead chloride is formed. Thus, there is an exchange of ions between the two reactants i.e., lead nitrate and hydrochloric acid, and hence this reaction is a double displacement reaction.

Question 3.
Identify the substances that are oxidized and the substances that are reduced in the following reactions.
Answer:
(i) 4Na(s) + O₂(9) → 2Na₂O(s)
(i) 4Na(s) + O₂(g) → 2Na₂O(s)
Na has gained oxygen and forms Na₂O.
So, Na is oxidized and O₂ is reduced.

(ii) CuO(s) + H₂(g) → Cu(s) + H₂O(I)
(ii) CuO(s) + H₂(g) → Cu(s) + H₂O(I)
CuO has lost oxygen and forms Cu.
So, Cu is reduced while H₂ has gained oxygen, hence, it is oxidized.

Activities

Activity 1.
Perform an activity of burning magnesium ribbon.
Answer:
1. Clean a magnesium ribbon about 3-4 cm long by rubbing it with sandpaper.
2. Hold it with a pair of tongs. Burn it using a spirit lamp or burner and collect the ash formed in a watch-glass as shown in figure. Burn the magnesium ribbon keeping it away as far as possible from your eyes.

Observation:
The magnesium (Mg) ribbon starts burning with a dazzling white flame and changes into white powder. The powder formed is called magnesium oxide (MgO).

Conclusion:
On burning magnesium (Mg) in the presence of oxygen (O₂) produces magnesium oxide (MgO) through a chemical reaction between magnesium and oxygen.

Activity 2.
Take lead nitrate solution in a test tube.
Answer:
Add potassium iodide solution to this. What do you observe?

Observation:
A yellow precipitate of lead iodide is formed.

Conclusion:
Adding potassium iodide to solution of lead nitrate produces yellow precipitate of lead iodide

Activity 3.
Perform an activity to produce hydrogen gas from zinc.
Answer:
1. Take a few zinc granules in a conical flask or a test tube.
2. Add dilute hydrochloric acid or sulphuric acid to this.

• Do you observe anything happening around the zinc granules?
• Touch the conical flask or test tube. Is there any change in its temperature?

Observation:
Bubbles of hydrogen gas get formed around zinc granules.
The conical flask becomes hot.

Conclusion:
On adding dilute hydrochloric or sulphuric acid to zinc granules, hydrogen gas is produced. It also increases the temperature of the solution. This shows two characteristics,
(1) Evolution of gas and
(2) Change in temperature.

Activity 4.
Perform an activity with calcium oxide to demonstrate endothermic reaction.
Answer:
Take a small amount of calcium oxide (quick lime) in a beaker.
Now slowly add water in the beaker.
Touch the beaker.

Observation:
You will observe that the beaker has become hot. This means that the reaction has released heat and hence it is exothermic reaction.

Reaction:
When calcium oxide and water are mixed, calcium oxide reacts vigorously with water to produce slaked lime (calcium hydroxide) and release a large amount of heat.

Activity 5.
Explain thermal decomposition of ferrous sulphate with suitable diagram and chemical equation.
Answer:
1. Take about 2g ferrous sulphate crystals in a dry boiling tube.
2. Note the colour of the ferrous sulphate crystals.
3. Heat the boiling tube over the flame of a burner or spirit lamp as shown in figure.
4. Observe the colour of the crystals after heating.

Observation:
When ferrous sulphate is heated strongly, it looses water and decomposes to form ferric oxide, sulphur dioxide gas and sulphur trioxide gas.

Reaction:

The green coloured ferrous sulphate changes to brown coloured ferrous oxide.
The odour of burning sulphur can be smelled in the test tube.

Activity 6.
Perform an activity to show thermal decomposition reaction of lead nitrate.
Answer:

Take about 2g lead nitrate powder in a boiling tube.
Hold the boiling tube with a test-tube holder and heat it over a flame, as shown in figure.

Observation:
On heating lead nitrate, gets decomposed to produce lead oxide nitrogen dioxide and oxygen.
The reaction emits brown fumes of nitrogen dioxide.

Reaction:

Electrical decomposition/Electrolysis of water

Activity 7.

Explain electrolysis of water through electrical Decomposition
Answer:
Take a plastic mug. Drill two holes at its base and fit rubber stoppers in these holes. Insert carbon electrodes in these rubber stoppers as shown in figure.

1. Connect these electrodes to a 6-volt battery.
2. Fill the mug with water such that the electrodes get immersed. Add a few drops of dilute sulphuric acid to the water.
3. Take two test tubes filled with water and carefully invert them over the two carbon electrodes.
4. Switch on the current and leave the apparatus undisturbed for some timer
5. You will observe the formation of bubbles at both the electrodes. These bubbles are formed by decomposition of water on passing electricity.

Observation:
1. On passing electricity, the water decomposes into hydrogen gas and oxygen gas. The formation of bubbles at both the electrodes is due to these gases.

2. As can be seen, more gas is produced on the test-tube on the right hand side. This is hydrogen gas. The hydrogen gas is produced in double volume as compared to oxygen produced in the left test tube.

Testing the presence of gases:

•  Take out the test-tube on the left side i.e. the one which contains less gas and bring it near a burning candle. You will see that the gas burns brightly. This means the gas is oxygen.
• Similarly, the gas in the right test-tube burns with a popping sound which means it is hydrogen gas.

Activity 8.
Perform an activity to demonstrate decomposition through light. OR Perform and activity to show the decomposition of
silver chloride
Answer:
1. Take about 2g silver chloride (AgCI) in a china dish. Observe its colour.
2. Place the china dish under sunlight for some time. Again observe its colour.

Observation:
Initially the colour of silver chloride was white. After keeping it under sunlight for some time its Fig colour changed to grey.

Reaction:
When white coloured silver chloride is kept under sunlight it gets decomposed and produces white coloured silver metal and chlorine.

Activity 9.
Perform an activity to show the displacement of copper.
Answer:
1. Take three iron nails and clean them by rubbing with sand paper.

2. Take two test tubes marked as (A) and (B). In each test tube, take about 10 mL copper sulphate solution.

3. Tie two iron nails with a thread and immerse 04: them carefully In the copper sulphate solution in test tube B for about 20 minutes. Keep one iron nail aside for comparison.

4. After 20 minutes, take out the iron nails from the copper sulphate solution.

Question 1.
Compare the intensity of the blue colour of copper sulphate solutions in test tubes (A) and (B).
Answer:
Observation:
The colour of copper sulphate in test-tube A is Figur blue. The blue colour of copper sulphate fades in test-tube B and the solution becomes light blue.

Question 2.
Also, compare the colour of the iron nails dipped in the copper sulphate solution with the one kept aside [Fig. 1.8 (b).
Answer:
Observation:
iron nail dipped in copper-sulphate solution becomes shinning brown as compared to an ordinary iron nail.

Reaction:

Conclusion:
Iron is a more reactive element. So, it displaces copper from copper sulphate solution.

(b) Iron nails and copper sulphate solutions compared before and after the experiment.

Activity 10.
Perform an activity to show double displacement reaction.
Answer:
1. Take about 3 mL of sodium sulphate solution in a test tube.
2. In another test tube, take about 3 mL of barium chloride solution.
3. Mix the two solutions.

Observation:
1. When aqueous solution of sodium sulphate reacts with an aqueous solution of barium chloride, a white precipitate of BaSO₄ is formed.
2. The precipitate is insoluble whereas the other product i.e. sodium chloride remains in the solution.
Reaction:

Activity 11.
Perform an activity to study oxidation reaction.
Answer:
1. Take about 1 gm of copper powder in a china dish. Copper is reddish-brown coloured.
2. Strongly heat the china dish over a burner.

Observation:
The surface of copper gets coated with a black powder. This black substance ¡s copper oxide (CuO).

Oxidation reaction:
The reaction of formation of copper oxide by the addition of oxygen is an oxidation reaction

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

Haryana Board 10th Class Science Solutions Chapter 6 Life Processes

HBSE 10th Class Science Life Processes Textbook Questions and Answers

Question 1.
The kidneys in human beings are a part of the system for ……….
(a) nutrition
(b) respiration
(c) excretion
(d) transportation
Answer:
(c) excretion

Question 2.
The xylem in plants is responsible for ……….
(a) transport of water
(b) transport of food
(c) transport of amino acids
(d) transport of oxygen
Answer:
(a) transport of water

Question 3.
The autotrophic mode of nutrition requires ……….
(a) carbon dioxide and water
(b) chlorophyll
(c) sunlight
(d) all of the above
Answer:
(d) all of the above.

Question 4.
The breakdown of pyruvate to give carbon dioxide, water and energy takes place in ……….
(a) cytoplasm
(b) mitochondria
(c) chloroplast
(d) nucleus
Answer:
(b) mitochondria

Question 5.
How are fats digested in our bodies? Where does this process take place?
Answer:
1. Fat digestion occurs in small intestine.

2. First of all the large tat molecules gets converted into small molecules. This helps the fat-digesting enzymes to act on the fat molecules efficiently.

3. The bile juice which comes from the gall bladder (produced by liver) contains bile salts. These bile salts emulsify the fat molecules. Here, the surface tension of fat molecules decrease and they become smaller

4. As the surface area of these smaller molecule is more, it helps in effective action of enzymes.

5. The enzyme lipase which does this task comes from pancreas as well as intestinal glands. This lipase converts the fats into fatty acids and glycerol and this is how fats get digested.

Question 6.
What is the role of saliva in the digestion of food?
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.
The human digestive system consists of the alimentary canal and its associated glands.

2. 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.

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 7.
What are the necessary conditions for autotrophic nutrition and what are Its by-products?
Answer:
(a) Necessary conditions:

• Presence of chlorophyll in the plants
• Absorption of light energy
• Splitting of water molecules
• Reduction of carbon dioxide to carbohydrates

(b) By-products:
Oxygen is the by-product and carbohydrate (glucose) is the main product of the autotrophic nutrition.

Question 8.
What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration.
Answer:
1. Yeast, fungi. endoparasites (tape-worm, ascans), some bacteria are examples of some organisms who show anaerobic mode of respiration.

2.

Question 9.
How are the alveoli designed to maximize the exchange of gases?
Answer:
1. Alveoli are balloon-like structures. They can be also compared with a bunch of grapes. Their spacious structure facilitates the gaseous exchange with increased surface area.

2. Additionally, they are rich in blood supply and are thin-walled which helps in maximizing the exchange of gases.

Question 10.
What would be the consequences of deficiency of hemoglobin In our bodies?
Answer :
The desirable value of hemoglobin in males is 13-16 gibo mL blood and in females 12-15 gibo mL blood.
1. It hemoglobin is less, the oxygen-carrying capacity of blood decreases. It leads to lack of supply of oxygen to the body organs.

2. So, the heart has to pump the blood rapidly. As a result, a person gets tired soon and breathes faster than normal.

Question 11.
Describe double circulation of blood in human beings. Why Is it necessary?
Answer:
Double circulation:
In higher animals, the blood goes twice in the heart during each cardiac cycle. Hence, it is known as double circulation. The two circulations are called:

• Systemic circulation and
• Pulmonary circulation.

The systemic circulation occurs between the body cells (except lungs) and the heart while in pulmonary circulation, the circulation occurs only between the heart and the lungs.

Need of double circulation:
The need of oxygen is more in higher animals. In double circulation, the oxygenated and deoxygenated bloods do not mix. So, the body organs can get pure blood. It is helpful to maintain their metabolic needs.

Question 12.
What are the differences between the transport of materials In xylem and phloem?
Answer:

Question 13.
Compare the functioning of alveoli In the lungs and nephrons In the kidneys with respect to their structure and functioning.
Answer:

HBSE 10th Class Science Life Processes InText Activity Questions and Answers

Textbook Page no – 95

Question 1.
Why is diffusion insufficient to meet the oxygen requirements of multi-cellular organisms like humans?
Answer:
1. In multicellular organisms such as humans, all the cells of, body are not in direct contact with surrounding environment.

2. Moreover, the size of human body is quite large as well as complex. So, each and every cell cannot perform gaseous exchange with environment by diffusion.

Question  2.
What criteria do we use to decide whether something is alive?
Answer:
1. Movement is one of the basic criterions of life to decide if something is alive.

2. Animals show visible movement called locomotion, while plants show invisible movement of biomolecules with themselves.

Question 3.
What are outside raw materials used for by an organism?
Answer:
Chief raw materials used by an organism from outside are:
(a) Food (Carbon based food source):
Constituents: Carbohydrates. proteins, fats etc.
Chief function: Providing energy, photosynthesis in plants
(b) Oxygen: For respiration
(c) Water:
Constituents: H₂O and many water soluble minerals and salts.
Chief function: Universal medium for various metabolic processes in the body.

Question 4.
What processes would you consider essential for maintaining life?
Answer:
Processes essential for maintaining life include —

• Nutrition,
• Respiration,
• Transportation,
• Excretion,
• Control and Coordination,
• Movement,
• Reproduction.

Textbook Page no – 101.

Question 1.
What are the differences between autotrophic nutrition and heterotrophic nutrition?
Answer:

Question 2.
Where do plants get each of the raw materials required for photosynthesis?
Answer:
Plants need water. CO₂, chlorophyll and sunlight for photosynthesis. They get these raw material from the following sources:

• Water: Plants (Terrestrial plants) absorb water from the soil through roots.
• CO₂ : Plants get CO₂ from the environment through the stomata.
• Chlorophyll: Chlorophyll is a pigment found in the organelle called chloroplast of the cells of green parts of the plants.
• Sunlight: The sun gives sunlight.

Question  3.
What is the role of the acid In our stomach?
Answer:
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:

• Dilute hydrochloric acid,
• An enzyme called pepsin and
• 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 hydrochloric add may cause to the inner lining of the stomach.

Question  4.
What is the function of digestive enzymes?
Answer:
1. The digestive enzymes play an important role in digestion.
2. The food we eat contains complex macromolecules.
3. The digestive enzymes break down these complex molecules into smaller and simpler molecules.
4. These molecules are then easy to absorb by intestine.

Question 5.
How is the small intestine designed to absorb digested food?
Answer:
1. The process of digestion of food is almost completed in small intestine.
2. The small intestine is designed to provide maximum surface area for absorption of digested food.
3. To increase the surface area, the inner lining of the small intestine has numerous finger-like projections called ýilli.
4. These villi have a rich blood supply which helps to absorb food effectively.

Textbook Page no – 105

Question 1.
What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration?
Answer:
In aquatic environment. i.e. in water, the dissolved O₂ content is quite low compared to the oxygen-rich atmosphere of the terrestrial animal. So. the aquatic organisms have to breath at a much faster rate compared to the terrestrial organisms.

Question 2.
What are the different ways in which glucose Is oxidized to provide energy in various organisms?
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 3.
How is oxygen and carbondioxide transported in human beings?
Answer:
(a) Transport of oxygen:

• In human body, hemoglobin has a very high affinity for oxygen.
• Hemoglobin is present in the red blood cells. It carries oxygen from the respiratory surface of the lungs to the body cells.

(b) Transport of CO₂:
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 4.
How are the lungs designed in human beings to maximize the area for exchange of gases?
Answer:
1. Lungs have balloon-like structure.
2. The windpipe gets divided into left and right bronchi.
3. The bronchi enters into the lungs and further divides into narrower tubes called bronchioles.
4. The bronchioles terminate into balloon-like structure known as alveoli.
5. These alveoli provide a large surface area for maximum exchange of respiratory gases.

Textbook Page no – 110

Question 1.
What are the components of the transport system In human beings? What are the functions of these components?
Answer:
The transport system of human body contains three main components. They are –
(i) Heart,
(ii) Blood and
(iii) Blood vessels and
(iv) Lymph

(i) Functions of heart:

• The human heart is a muscular pump which keeps blood flowing in the body till the person ¡s alive.
• The heart receives purified (oxygenated) blood from the lungs and pumps it towards the body cells.
• The heart receives deoxygenated blood from the body parts and sends it towards the lungs for purification.

(ii) Function of various blood cells:

• Red blood cells (RBCs): They contain hemoglobin. Hence, they transport O₂ from lungs to the body cells.
• White blood cells (WBCs): They play an important role in providing immunity to the body. They protect the body from micro-organisms.
• Blood platelets: When any blood vessel gets cut, the platelets clot the blood and prevent bleeding.
• Blood plasma: Blood plasma transport nutrients, excretory substances, CO₂, hormones, enzymes, etc.

(iii) Functions of blood vessels:

• Arteries carry blood from heart to the different organs of the body.
• Exchange of materials between blood and its surrounding takes place through these capillaries.
• Veins collect blood from different parts of the body and bring it back to the heart.

(iv) Functions of lymph: Lymph carries digested and absorbed fat from intestine and drain excess fluid from intercellular spaces back into blood.

Question 2.
Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds?
Answer:
1. Human heart (and also birds) require high amount of energy. They 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 te heart. As a result, the oxygenated blood and deoxygenated blood remain in separate chambers and do not mix.

3. Such a separation enables a very efficient supply of oxygen to the body cells.

Question 3.
What are the components of the transport system in highly organized plants?
Answer:
Xylem tissue and phloem tissue are the components of the transport system in highly organized plants.

Question 4.
How are water and minerals transported In plants?
Answer:
Transport of water in higher plants:

• Higher plants possess xylem for transporting water to all its parts.
• Xylem contains tracheids and vessels.
• 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 concentrations 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.

Transpiration:

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

Textbook Page no – 112.

Question 1.
Describe the structure and functioning of nephrons.
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 walled 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:

1. The lower end of the Bowman’s capsule is called tubule.
2. The part of tubule which is near the Bowman’s capsule is quite small and is called the neck.
3. After the neck, the tubule becomes very narrow and coiled.
4. This region of tubule consists of

• A proximal convoluted tubule,
• A Henle’s loop and
• A distal convoluted tubule.

5. The posterior end of nephron is known as collecting tube.
6. Collecting tubule opens into renal pelvis.
7. The renal pelvis opens into ureter.

Question 2.
What are the methods used by plants to get rid of excretory products?
Answer:
Excretion in Plants:

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

Question 3.
How is the amount of urine produced In the body, regulated?
Answer:
The amount of urine formed in the body depends on —

• Amount of excess water present in the body
• Amount of nitrogenous wastes produced in the body
• Amount of re-absorption in kidneys
• Certain hormones like ADH (Anti Diuretic Hormone)
• Surrounding temperature

Activities

Activity 1.

Aim: To demonstrate that chlorophyll is essential for photosynthesis.

Procedure:

• Take a potted plant with variegated leaves — for example, money-plant or croton.
• Keep the plant in a dark room for three days so that all the starch gets used up.
• Now keep the plant in sunlight for about six hours.
• Pluck a leaf from the plant. Mark the green areas in it and trace them on a sheet of paper.
• Dip the leaf in boiling water for a few minutes.
• After this, immerse it in a beaker containing alcohol.
• Carefully place the above beaker in a water-bath and heat till the alcohol begins to boil.
• Now, dip the leaf in a dilute solution of iodine for a few minutes, Fig
• Take out the leaf and rinse off the iodine solution.

Question 1.
What happens to the colour of the leaf? What is the colour of the solution?
Answer:
The leaf becomes colourless because on boiling the leaf, it loses the pigmentation. The water becomes green in colour due to presence of chlorophyll that was present in the leaf.

Question 2.
Observe the colour of the leaf and compare this with the tracing of the leaf done In the beginning.
Answer:
The leaf shows blue-black and colourless patches.

Question 3.
What can you conclude about the presence of starch in various areas of the leaf?
Answer:
The green patched areas become blue-black when they are tested with iodine. While the chlorophyll-

Activity 2.

Aim: To demonstrate that carbon dioxide is necessary for photosynthesis.

1. Take two healthy potted plants which are nearly the same size.
2. Keep them in a dark room for three days.
3. Now place each plant on separate glass plates. Place a watch-glass containing potassium hydroxide by the side of one
of the plants. The potassium hydroxide is used to absorb carbon dioxide.
4. Cover both plants with separate bell-jars as shown in figure 2.
5. Use Vaseline to seal the bottom of the jars to the glass plates so that the set-up is air-tight.
6. Keep the plants in sunlight for about two hours.
7. Pluck a leaf from each plant and check for the presence of starch as in the above activity.

Question 1.
Do both the leaves show the presence of the same amount of starch?
Answer:
The leaves of plant ‘a’ show absence of starch, while the leaves of plant ‘b’ show presence of starch.

Question 2.
What can you conclude from this activity?
Answer:
1. Starch is not formed in the leaf of the plant kept under the bell-jar along with potassium hydroxide.
This is so because potassium hydroxide absorbs carbon dioxide from air. Hence, the plant does not get carbon dioxide and it can not undergo photosynthesis.
2. From the above activity, we can conclude that a good amount of CO₂ is necessary to carry out photosynthesis at proper rate.

Activity 3.

Aim: To check the effect of saliva on starch.
1. Take 1 mL starch solution (1%) in two test tubes (A and B).
2. Add 1 mL saliva to test tube A and leave both test tubes undisturbed for 20-30 minutes.
3. Now, add a few drops of dilute iodine solution to the test tubes.

Question 1.
In which test tube do you observe a colour change?
Answer:
The solution of test tube B will turn bluish black.

Question 2.
What does this indicate about the presence or absence of starch in the two test tubes?
Answer:
It shows that starch is present in test tube ‘B’ while test tube ‘A’ does not have starch.

Question 3.
What does this tell us about the action of saliva on starch?
Answer:
The result shows that the saliva of test tube ‘A’ acts upon starch. So, starch gets converted into sugar and the solution does not show reaction to iodine test.

Activity 7.

Aim: To find out the haemoglobin content in human beings and in animals such as buffalo and cow.

Question 1.
Visit a health centre in your locality and find out what is the normal range of haemoglobin content in human beings.
Answer:
It is 13 — 18 g/dL (Note: dL = decilitre which means 1/10th of 1 litre or say loo mL)

Question 2.
Is It the same for children and adults?
Answer:
No, it is 12.5 g/dL for children

Question 3.
Is there any difference in the haemoglobin levels for men and women?
Answer:
In males, it is 13 — 18 g/dL and in females, it is 12 -10 g/dL.

Question 4.
Visit a veterinary clinic In your locality. Find out what is the normal range of hemoglobin content in an animal like buffalo or cow.
Answer:
It Is 10.4 — 16.4 g/dL in such animals.

Question 5.
Is this content different in calves, male and female animals?
Answer:
Yes, the haemoglobin level of calves is more than the adult male and female animals.

Question 6.
How would the difference, if any, be explained?
Answer:
Hemoglobin content depends on the personal health, genetic map and environment.

Activity 8.

Aim : To demonstrate physiological process of transpiration in plants.

1. Take two small pots of approximately the same size and having the same amount of soil. One should have a plant in it. Place a stick of the same height as the plant in the other pot.
2. Cover the soil in both pots with a plastic sheet so that moisture cannot escape by evaporation.
3. Cover both sets, one with the plant and the other with the stick, with plastic sheets and place in bright sunlight for half an hour.

Question 1.
Do you observe any difference in the two cases?
Answer:
The plant which is covered with a plastic sheet, shows some water droplets inside the plastic sheet. This demonstrates that there is water loss from the aerial parts of a plant. This process is called transpiration.

Haryana State Board HBSE 10th Class Science Solutions Chapter 2 Acids, Bases and Salts Textbook Exercise Questions and Answers.

Haryana Board 10th Class Science Solutions Chapter 2 Acids, Bases and Salts

HBSE 10th Class Science Acids, Bases and Salts Textbook Questions and Answers

Question 1.
A solution turns red litmus blue, its pH is likely to be ……………
(a) 1
(b) 4
(c) 5
(d) 10
Answer:
(d) 10

Question 2.
A solution reacts with crushed egg-shells to give a gas that turns lime-water milky. The solution contains ………….
(a) NaCI
(b) HCI
(c) LiCI
(d) KCI
Answer:
(b) HCI

Question 3.
10 mL of a solution of NaOH is found to be completely neutralized by 8 mL of a given solution of HCI. If we take 20 mL of the same solution of NaOH, the amount HCI solution (the same solution as before) required to neutralize it will be ………………..
(a) 4 mL
(b) 8 mL
(c) 12 mL
(d) 16 mL
Answer:
(d) 16 mL
Here, 10 mL NaOH neutralizes 8 mL HCI
∴ 20 mL NaOH is neutralized by = \(\frac{20 \mathrm{~mL} \times 8 \mathrm{~mL}}{10 \mathrm{~mL}}=16 \mathrm{~mL}\)

Question 4.
Which one of the following types of medicines is used for treating indigestion?
(a) Antibiotic
(b) Analgesic
(c) Antacid
(d) Antiseptic
Answer:
(c) Antacid

Question 5.
Write word equations and then balanced equations for the reaction taking place when –
(a) dilute sulphuric acid reacts with zinc granules.
(b) dilute hydrochloric acid reacts with magnesium ribbon.
(c) dilute sulphuric acid reacts with aluminium powder.
(d) dilute hydrochloric acid reacts with iron filings.
Answer:

Question 6.
Compounds such as alcohols and glucose also contain hydrogen but are not categorized as acids. Describe an activity to prove it.
Answers:
1. Take a cork and fix two nails on it.
2. Keep the cork inside a beaker.
3. Connect the nails to the battery, bulb and key.
4. Take some solution of alcohol say ethanol and solution of glucose.
5. First put ethanol in the beaker and see f the bulb glows.
6. Remove ethanol and put glucose in the beaker and observe.

Observation:
You will observe that the bulb does not glow with any of Ethar these solutions. This means that these solutions do not release H⁺ ions. H⁺ ions conduct electricity and are produced by ions. Hence, none of these solutions is acidic.

Question 7.
Why does distilled water not conduct electricity, whereas rain water does?
Answer:
1. Distilled water is pure and so it does not form ions. Hence, it does not conduct electricity.
2. Rain water contains impurities such as acidic components. So, rain water release ions and hence conducts electricity.

Question 8.
Why do acids not show acidic behaviour In the absence of water?
Answer:
1. Acids cannot release H⁺ ions in absence of water.
2. Only when acids are dissolved in water, they release H ions which then enable conduction of electricity.

Question 9.
Five solutions A,B,C,D and E when tested with universal Indicator showed pH as 4,1,11,7 and 9, respectively.
(i) Which solution is
(a) neutral?
(b) strongly alkaline?
(c) strongly acidic?
(d) weakly acidic?
(e) weakly alkaline?
(ii) Arrange the pH in increasing order of hydrogen-ion concentration.
Answer:
(i)

• Solution ‘D’ with pH = 7 is neutral
• Solution ‘C’ with pH = 11 is strongly alkaline
• Solution ‘B’ with pH = 1 is strongly acidic
• Solution ‘A’ with pH = 4 is weakly acidic
• Solution ‘E’ with pH = 9 is weakly alkaline

(ii) Arrangement of these solutions in increasing order of hydrogen ion concentration:
C = 11 < E = 9< D = 7< A = 4< B = 1

Question 10.
Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCI) is added to test tube A, while acetic acid (CH₃COOH) is added to test-tube B. Amount and concentration taken for both the acids are same. In which test tube will the fizzing occur more vigorously and why?
Answer:
Hydrochloric acid (HCI) is a strong acid as compared to acetic acid (CH₃COOH). Hence, the fizzing will occur more vigorously in test-tube A which contains HCI.

Question 11.
Fresh milk has a pH of 6. How do you think the pH will change as it turns into curd? Explain your answer.
Answer:
Formation of curd from milk means milk becoming acidic. Hence, pH of milk will decrease when it turns to curd due to the formation of lactic acid. Formation of acid will make curd sour.

Question 12.
A milkman adds a very small amount of baking soda to fresh milk.
(a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline?
(b) Why does this milk take a long time to set as curd?
Answer:
(a) By shifting the pH of milk a bit towards alkaline the milkman is preventing the milk from becoming sour.

(b) Baking soda was added to the milk. Baking soda is a base substance and so it neutralizes the formation of lactic acid. Hence, the milk takes more time to set as curd because it will now take more time to become acidic.

Question 13.
Plaster of Paris should be stored in a moisture-proof container. Explain why?
Answer:
If plaster of Paris is kept in a moist environment it will absorb moisture and get converted to hard gypsum.

Question 14.
What is a neutralization reaction? Give two examples.
Answer:
The reaction between an acid and a base to give salt and water is called neutralization reaction.
Example:

Question 15.
Give two Important uses of washing soda and baking soda.
Answer:
Uses of baking soda:

• In a bakery for making bread, cakes, etc.
• In fire extinguisher

Uses of washing soda:

• For manufacturing soap and detergent
• For removing the permanent hardness of water.

HBSE 10th Class Science Acids, Bases and Salts  InText Activity Questions and Answers

Text Book Page no – 18.

Question 1.
You have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution, respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?
Answer:
1. Cut the litmus paper into three parts. Dip each part in a test tube and observe the change in colour.
2. The test-tube in which red litmus paper turns blue contains a basic solution. Now dip this blue litmus paper in remaining two test-tubes.
3. The solution that turns the blue litmus paper again red is an acidic solution.
4. The tube in which there occurs no change in the colour of blue litmus contains water.

Text Book Page no – 22.

Question 1.
Why should curd and sour substances not be kept in brass anal copper vessels?
Answer:
1. Both curd and sour substances such as lemon juice are acids whereas brass and copper are highly reactive metals.
2. When acids come in contact with metals they react and form salts and release hydrogen gas. Consuming such substances causes harm to the body.
3. Hence, food items acidic in nature should not be kept in copper and brass vessels.

Question 2.
Which gas is usually liberated when an acid reacts with a metal? Illustrate with an example. How will you test for the presence of this gas?
Answer:
Hydrogen (H₂) gas is liberated when an acid reacts with a metal.
Example: Take some zinc granules in a test-tube. Add 5 mL dilute hydrochloric (HCI) acid in it. Following reaction will occur.
Zn + 2HCI → ZnCI₂ + H₂

Testing evolution of gas:
Hydrogen (H₂) gas is insoluble in water. If you pass the gas in soap solution, it will form bubbles. The bubbles will burn with a popping sound when match stick is brought near the solution. This proves presence of gas.

Question 3.
Metal compound A reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.
Answer:
1. The gas extinguishes the burning candle, so it must be carbon dioxide gas. Moreover, since the end product is calcium chloride, the metal compound A must be calcium carbonate.
2. The reaction between calcium carbonate and hydrochloric acid which produces calcium chloride and carbon dioxide gas is:

Text Book Page no – 25.

Question 1.
Why do HCI, HNO₃, etc., show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?
Answer:
1. If on adding a substance to an aqueous solution, H⁺ ions get/eleased it means the substance added is acid.
2. On adding HCI and HNO₃, H⁺ ions are released in the aqueous solution and so they show acidic characteristic whereas C₂H₅OH and glucose do not release H⁺ ions. Hence they do not show acidic character.

Question 2.
Why does an aqueous solution of an acid conduct electricity?
Answer:
When an acid is added to an aqueous solution, it releases H⁺ ions. H⁺ ions are responsible for conducting electricity. Hence, aqueous solution of an acid conducts electricity.

Question 3.
Why does dry HCI gas not change the colour of the dry litmus paper?
Answer:
Dry hydrochloric (HCI) gas does not release H⁺ ions. So it does not possess acidic character Hence, it does not change the colour of dry litmus paper.

Question 4.
While diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?
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 5.
How is the concentration of hydronium ions (H₃O⁺) affected when a solution of an acid is diluted?
Answer:
1. There exists a fix number of hydronium ions per unit volume of the acidic solution.
2. On dilution, the number of hydronium ions decreases and the solution becomes less acidic.

Question 6.
How is the concentration of hydroxide ions (OH^–) affected when excess base is dissolved in a solution of sodium hydroxide?
Answer:
When the excess base is dissolved in the solution of sodium hydroxide, the amount of hydroxide (H^–) ions per unit volume increases. Hence, the concentration of hydroxide ions increases. This happens only when base added dissolves in water.

Text Book Page no – 28.

Question 1.
You have two solutions, A and B. The pH of solution A is 6 and pH of solution B is 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic?
Answer:
The pH of solution A is 6 which means it is an acidic solution and has higher concentration of hydrogen ions. On the other hand, pH of solution B is 8 which means it is basic and so has lower concentration of hydrogen ions.

Question 2.
What effect does the concentration of H⁺(aq) ions have on the nature of the solution?
Answer:
As the concentration of H⁺(aq) ions increase, the solution goes on becoming more and more acidic.

Question 3.
Do basic solutions also have H⁺(aq) ions? If yes, then why are these basic?
Answer:
Yes, basic solutions have H⁺(aq) ions, but their number is quite less as compared to OH^– ions. This make a solution basic. Hence, solutions that have much higher number of OH^– compared to H⁺ are basic.

Question 4.
Under what soil condition do you think a farmer would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)?
Answer:
1. Good soil for farming should have pH range of 6 to 8.
2. When the soil becomes acidic i.e. when the pH of soil becomes lesser than 7, the farmers would add quick lime (calcium oxide) or slaked lime (calcium hydroxide) or such base substances to maintain the pH level.

Text Book Page no – 33.

Question 1.
What is the common name of the compound CaOCl₂?
Answer:
The common name of CaOCl₂ is bleaching powder.

Question 2.
Name the substance which on treatment with chlorine yields bleaching powder.
Answer:
Calcium hydroxide (Dry slaked lime)

Question 3.
Name the sodium compound which is used for softening hard water.
Answer:
Sodium carbonate (Na₂CO₃)

Question 4.
What will happen if a solution of sodium hydrocarbonate is heated? Give the equation of the reaction involved.
Answer:
When sodium hydrocarbonate is heated, sodium carbonate, water and carbon dioxide gas are obtained.

Question 5.
Write an equation to show the reaction between plaster of Paris and water.
Answer:

Activities

Activity 1.

Aim:
To test certain acids and bases in the laboratory.
Answer:
1. Collect the following solutions from the science laboratory — hydrochloric acid (HCI), sulphuric acid (H₂SO₄), nitric acid (HNO₃). acetic acid (CH₃COOH), sodium hydroxide (NaOH). calcium hydroxide [Ca(OH)₂], potassium hydroxide (KOH), magnesium hydroxide [Mg(OH)₂] and ammonium hydroxide (NH₄OH).

2. Put a drop of each of the above solutions on a watch-glass one by one and test with a drop of the indicators shown in table.

3. What change in colour did you observe with red litmus, blue litmus, phenolphthalein and methyl orange solutions for each of the solutions taken?

4. Tabulate your observations in Table

The observations noted during various tests are tabulated in the table below:

Activity 2.

Aim:
To test acids and bases using olfactory indicators.

Question 1.
Take some finely chopped onions in a plastic bag along with some strips of clean cloth. Tie up the bag tightly and leave overnight in the fridge. The cloth strips can now be used to test for acids and bases. Take two of these cloth strips and check their odour.
Answer:
When cloth strips are kept in a bag with finely chopped onions for overnight the strips emit smell of onion.

Question 2.
Keep them on a clean surface and put a few drops of dilute HCI solution on one strip and a few drops of dilute NaOH solution on the other. Rinse both cloth strips with water and again check their odour. Note your observations.
Answer:
1. Cloth strip + dilute HCI solution = Smell of onion ÷ colour of strip changes to pale red.
2. Cloth strip + dilute NaOH solution = Red coloured cloth strip changes to green.

Question 3.
Now take some dilute vanilla essence and clove oil and check their odour. Take some dilute HCI solution in one test tube and dilute NaOH solution in another. Add a few drops of dilute vanilla essence to both test tubes and shake well. Check the odour once again and record changes in odour, if any.
Answer:
On adding vanilla essence —
1. In dilute NaOH: No smell,
2. In dilute HCI: Smell of vanilla is still present

Question 4.
Similarly, test the change in the odour of clove oil with dilute HCI and dilute NaOH solutions and record your observations.
Answer:
On adding clove oil:
1. In dilute HCI: Smell of clove is present.
2. In dilute NaOH: Smell of clove is absent

Activity 3.

Perform an activity to study how metal (zinc) reacts with acid (sulphuric acid) and products formed.
Answer:
Set the apparatus as shown in
1. Set the apparatus as shown in figure.
2. Take about 5 mL of dilute sulphuric acid in a test tube and add a few pieces of zinc granules to it.

Observation:
1. We can observe that bubbles form on the surface of zinc granules. The bubbles indicate formation of gas.

2. When we pass this gas through the soap solution, the gas gets trapped in soap molecules and bubbles are formed.

3. When a burning candle is brought near the gas tilled bubbles, the gas bubbles break up with a pop
sound. This indicates that the gas formed is hydrogen (H₂) gas.

Reaction:

We can record similar observations even when we repeat this activity with some more acids like HCI, HNO₃ and CH₂COOH.

Conclusion:
When a metal reacts with acid, it gives salt and hydrogen gas.

Activity 4.

Perform an activity to study how metal (zinc) reacts with base (sodium hydroxide) and products formed.
Answer:
1. Place a few pieces of granulated zinc metal in a test tube.
2. Add 2 mL of sodium hydroxide (NaOH) solution and warm the contents of the test tube.

Observation:

• We can observe that bubbles form on the surface of zinc granules. The bubbles indicate formation of gas.
• When we pass this gas through the soap solution, the gas gets trapped In soap molecules and bubbles are formed.
• When a burning candle is brought near a gas filled bubble, the gas bubbles break up with a pop sound. This indicates that the gas formed is hydrogen (H₂) gas.

Reaction:

Conclusion: When a metal reacts with a base, it gives salt and hydrogen gas.

Activity 5.

Perform an activity to study how acid reacts with metal carbonates and metal hydrogen
Answer:
1. Take two test tubes and label them as A and B.
2. Take about 0.5 g of sodium carbonate (Na₂CO₃) in test tube A and about 05 g of sodium hydrogencarbonate (NaHCO₃) in test tube B.
3. Add about 2 mL of dilute HCI in both the test tubes.

Observation:
When acids react with metal carbonates and metal hydrogencarbonates, they produce salt, water and carbon dioxide gas. This happens in both the test tubes.

Reaction

4. On passing carbon dioxide produced through lime water (calcium hydroxide solution), a milky precipitate of calcium carbonate is formed.

5. On passing excess carbon dioxide through above solution, precipitate of calcium carbonate dissolves due to formation of soluble calcium bicarbonate.

Activity 6.

Perform an activity to study how acids and bases react with each other?
Answer:
1. Take about 2 mL of dilute NaOH solution in a test tube and add two drops of phenolphthalein solution.
2. The solution will become pink.
3. Add dilute HCI solution to the above solution drop by drop and record your observation.

Observation:

• On adding dilute HCI to the solution of phenolphthalein and sodium hydroxide, the pink colour of the solution slowly disappears. This suggests that NaOH (base) gets neutralized by acid.
• If we add a few drops of NaOH to the above mixture we will see that the solution again becomes pink. This suggests that the solution becomes alkaline again due to excess of NaOH over HCI. Hence, the pink colour reappears.

Conclusion:
When acid and base react with each other, the solution becomes neutral. The reaction that takes place is called neutralization reaction.

Activity 7.

Perform an activity to study reaction of metallic oxides with acids.
Answer:
1. Take a small amount of copper oxide in a beaker. it is black in colour.
2. Add dilute hydrochloric acid slowly while stirring.

Observation:
Copper oxide reacts with dilute hydrochloric acid. This gives a blue coloured solution of copper chloride salt.

Conclusion:
When acid reacts with metallic oxides, salt and water are produced.

Activity 8.

Perform an activity to study conduction of electricity by acids and bases.
Answer:
1. Take solutions of glucose, alcohol, hydrochloric acid, sulphuric acid, sodium hydroxide, etc.
2. Fix two nails on a cork, and place the cork in a 100 mL beaker.
3. Connect the nails to the two terminals of a 6 volt battery through a bulb and a switch, as shown in fig.

4. Now pour some dilute HCI in the beaker and switch on the current. Note down what happens.
5. Repeat the steps with dilute sulphuric acid (H₂SO₄), glucose and alcohol.

Observation:

• We can observe that the bulb glows when it is passed through acids and bases. This happens because the ions present in the acid and base enable flow of electric current.
• Glucose and alcohol solutions do not have ions. So, electricity is not conducted and the bulb does not glow.

Conclusion:
Acids have H⁺ ions and bases have OH^– ions. As a result. they are able to conduct electricity.

Activity 9.

Perform an activity to show that HCI solution is acidic but its dry form is not.
Answer:
1. Take about 1 g solid NaCI in a clean and dry test tube and set up the apparatus as shown in figure.
2. Add some concentrated sulphuric acid to the test tube.

Observation:

• On adding concentratec sulphuric acid reaction takes place and hydrochloric acid gas is produced.
• NaCI(s) + H₂SO₄ → Na₂SO₄(aq) + 2HCI(g)

The gas produced has no effect on dry litmus paper. However, the gas changes the colour of wet litmus paper into red.

Conclusion:
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 HCI solution is acidic whereas dry HCI is not.

Note for teachers: If the climate is humid, you will have to pass the gas produced through a guard tube (drying tube) containing calcium chloride to dry the gas.

Activity 10.

Perform an activity to demonstrate that reaction between acid (or base) and water is an exothermic reaction.
Answer:
1. Take 10 mL water in a beaker.
2. Add a few drops of concentrated H₂SO₄ (or sodium hydroxide) to it and swirl the beaker slowly.
3. Touch the base of the beaker.

Observation:
On touching the beaker it can be felt that the temperature of the solution has increased. This means the reaction is exothermic.

Dissolution of acid (or base) in water releases heat and so this is an exothermic process.

Activity 11.

1. Test the pH values of solutions given in table 2.
2. Record your observations.

Q. What is the nature of each substance on the basis of your observations?
Answer:
Observations and answers are recorded in the table below.

Activity 12.

Aim:
To find out pH of the soil.
Answer:
1. Put about 2 gm soil in a test tube and add 5 mL water to it.
2. Shake the contents of the test tube.
3. Filter the contents and collect the filtrate in a test tube.
4. Check the pH of this filtrate with the help of universal indicator paper.

Q. What can you conclude about the ideal soil pH for the growth of plants in your region?
Answer: If the pH of soil is near 7, plants will grow well. If the pH <6.5, the soil will be acidic and hinder plant growth.

Activity 13.

Question 1.
Write the formula of the salts given below:
Potassium sulphate, sodium sulphate, calcium sulphate, magnesium sulphate, copper sulphate sodium chloride, sodium nitrate, sodium carbonate and ammonium chloride.
Identify the acids and bases from which the above salts may be obtained.
Answer:

Question 2.
Salts having the same positive or negative radicals are said to belong to a family. For example, NaCI and Na₂SO₄ belong to the family of sodium salts. Similarly, NaCI and KCI belong to the family of chloride salts. How many families can you identify among the salts given in this activity?
Answer:
The following families can be identified:
Family of sulphate salts: K₂SO₄, Na₂SO₄, Ca₂SO₄, MgSO₄, CuSO₄
Family of chloride salts: NaCI NH₄CI
Family of sodium salts: NaCI, Na₂SO₄, NaNO₃, Na₂CO₃

Activity 14.

Aim: To find the pH and test the solubility of soil in water.
Answer:
1. Collect the following salt samples – sodium chloride, potassium nitrate, aluminium chloride, zinc sulphate, copper sulphate, sodium acetate, sodium carbonate and sodium hydrogen carbonate (some other salts available can also be taken).

2. Check their solubility in water (use distilled water only).

3. Check the action of these solutions on litmus and find the pH using a pH paper.

• Which of the salts are acidic, basic or neutral?
• Identify the acid or base used to form the salt.
• Report your observations in table below.

Give an idea about the pH of the salt formed with acid and base pH ot salt:

Activity 15.

Perform an activity to demonstrate that crystals of salt are not dry.
Answer:
1. Heat a few crystals of copper sulphate in a dry boiling tube. The colour of copper sulphate will be blue.
2. Heat the tube and observe the colour. Then add 2-3 drops of water and again check the colour.

Observation:

• On heating the tube you will observe that the colour of the copper sulphate changes to white.
• Water droplets can be seen in the boiling tube due to condensation of water of crystallization of copper sulphate.
• On adding 2-3 drops of water on the sample of copper sulphate obtained after heating the blue colour of copper sulphate is restored.

Conclusion:
This experiment suggests that there are a fixed number of water of crystallization in the molecule of copper sulphate salt. When we heat this salt, the water gets removed and so it becomes colourless.

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

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

HBSE 10th Class Science Metals and Non-metals Textbook Questions and Answers

Question 1.
Which of the following pairs will give displacement reactions?
(a) NaCI solution and copper metal
(b) MgCI₂ solution and aluminium metal
(C) FeSO₄ solution and silver metal
(d) AgNO₃ solution and copper metal
Answer:
(d) AgNO₃ solution and copper metal

Question 2.
Which of the following methods is suitable for preventing an iron frying pan from rusting?
(a) Applying grease
(b) Applying paint
(c) Applying a coating of zinc
(d) All of these
Answer:
(c) Applying a coating of zinc
Note: Although all options can be used to prevent rusting of an ordinary iron. But, the frying pan is used for cooking so the only best option is to coat it with zinc.

Question 3.
An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble In water. The element is likely to be ……
(a) Calcium
(b) Carbon
(c) Silicon
(d) Iron
Answer:
(a) Calcium

Question  4.
Food cans are coated with tin and not with zinc because
(a) Zinc is costlier than tin.
(b) Zinc has a higher melting point than tin.
(c) Zinc Is more reactive than tin.
(d) Zinc is less reactive than tin.
Answer:
(c) Zinc Is more reactive than tin.

Question  5.
You are given a hammer, a battery, a bulb, wires and a switch.
(a) How could you use them to distinguish between samples of metals and non-metals?
(b) Assess the usefulness of these tests in distinguishing between metals and non-metals.
Answer:
1. (a) Take the given samples of metals and non-metals and strike them with the hammer. If a sample converts into a sheet, it is a metal and if not, it is non-metal. Metals are malleable, non-metals are not.

(b) If a sample produces sound when struck with the hammer, it is a metal. Metals are sonorous. But if it does not produce a round, it is non-metal.

(c) Now arranging the given objects to form an electric circuit. Insert any one sample between clips A and B, lithe bulb glows, it is a metal (good conductor of electricity). If the bulb does not glows, it is a non-metal.

2. From the above tests, it is clear that metals are generally malleable, sonorous and good conductors of electricity, while non-metals are generally non-malleable/brittle, non-sonorous and poor conductors of electricity.

Question 6.
What are amphoteric oxides? Give two examples of amphoteric oxides.
Answer:
Some metal oxides react with both acids and bases to produce salt and water. Such metal oxides are called amphoteric oxides.
Example:

Question 7.
Name two metals which will displace hydrogen from dilute acids, and two metals which will not.
Answer:
1. Sodium and magnesium will displace hydrogen from dilute acids because they lie above hydrogen in the activity series.

2. Metals such as copper and silver lie below hydrogen in the activity series and hence will not displace hydrogen from dilute acids.

Question 8.
In the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte?
Answer:
(a) Cathode (negatively charged) — Pure metal M
(b) Anode (positively charged) — Impure metal M
(c) Electroyte — Aqueous solution of a salt of metal M

Question 9.
Pratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it, as shown in figure below.
(a) What will be the action of gas on  :
(i) Dry litmus paper?
(ii) Moist litmus paper?
(b) Write a balanced chemical equation for the reaction taking place.
Answer:
(a) (i) There will be no effect of gas on dry litmus paper. Hence, no change in colour will take place in case of dry litmus paper.
(ii) Moist blue litmus paper will turn red because sulphur is non-metal and non-metal oxides are acidic in nature.

(b) (i) S + O₂ → SO₂
(ii) SO₂ + H₂O → H₂SO₃

Question 10.
State two ways to prevent the rusting of iron.
Answer:
Two ways to prevent rusting of iron:
(1) Paint the iron articles.
(2) Galvanize iron article with the protective layer of zinc metal.

Question 11.
What type of oxides are formed when non-metals combine with oxygen?
Answer:
When non-metals combine with oxygen they form acidic oxides. SO₂, CO₂, etc. are some examples.

Question 12.
Give reasons
(a) Platinum, gold and silver are used to make jewellery.
(b) Sodium, potassium and lithium are stored under oil.
(c) Aluminium is a highly reactive metal, yet it Is used to make utensils for cooking.
(d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction.
Answer:
(a) Metals such as gold, platinum and silver are rare as well as non-reactive metals.

These metals have very attractive luster and so catches the attention of people. Moreover, these metals are malleable and ductile and molded into desired shape and intricated into exquisite jewellery. Another major advantage is that these metals do not corrode when exposed to air. Hence, they can be kept and worn for years without fearing the loss of luster.

(b) 1. Sodium (and potassium) is a highly reactive metal. If kept open it reacts with oxygen at room temperature. The reaction is extremely exothermic i.e. a lot of heat is produced. This even causes burning and dangerous accidents.
2. Sodium does not react with kerosene and hence it is preserved in kerosene.

(c) Using metal vessels that can react with food can lead to severe harm to the body.

Although aluminium is quite reactive, if forms a protective layer of aluminium oxide on its surface. This protects it from getting corroded and hence reacting with food. Hence, even though aluminum is very reactive metal, it is used to make utensils for cooking.

(d) It is easier and a cheaper way to reduce metal oxide to metal. Hence, carbonate and sulphide ores are first converted into their oxides and then reduced into metals.

Question 13.
You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.
Answer:
1. On using copper metal, it gets corroded to form copper carbonate and get tarnished i.e. copper gets decoloured due to oxidation.

2. Copper carbonate dissolves in mild acids. Lemon or tamarind juice are mild acids which when rub on copper vessels dissolve copper carbonate and make copper vessels shiny again.

Question 14.
Differentiate between metal and non-metal on the basis of their chemical properties.
Answer:

Question 15.
A man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can
you play the detective to find out the nature of the solution he had used?
Answer:
1. The crooked goldsmith fooled her by using a specific solution. The name of this solution is aqua regia.
2. This solution contains 1 part of concentrated nitric acid (HNO₃) and 3 parts of hydrochloric acid (HCl). On putting gold Ornaments in this solution, the solution dissolves gold.

Question 16.
Give reasons why copper is used to make hot water tanks and not steel (an alloy of Iron).
Answer:
1. The advantages of using copper over steel for making hot water tanks are many.
2. Copper does not react with any form of metal, not even steam.
3. Copper is cheap.
4. Copper is easily available and is also a good conductor of heat which enables heating water.

HBSE 10th Class Science Metals and Non-metals  InText Activity Questions and Answers

Textbook Page no – 40

Question 1.
Give an example of a metal which
(i) is a liquid at room temperature
(ii) can be easily cut with a knife
(iii) is the best conductor of heat
(iv) is a poor conductor of heat
Answer:
(i) Mercury
(ii) Sodium
(iii) Silver
(iv) Lead

Question 2.
Explain the meanings of malleable and ductile.
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.

Textbook Page no – 46

Question 1.
Why Is sodium kept Immersed in kerosene oil?
Answer:
1. Sodium (and potassium) is a highly reactive metal. If kept open it reacts with oxygen at room temperature, The reaction is extremely exothermic i.e. a lot of heat is produced. This even causes burning and dangerous accidents.
2. Sodium does not react with kerosene and hence it is preserved in kerosene.

Question 2.
Write equations for the reactions of ……….
(i) Iron with steam,
(ii) Calcium and potassium with water
Answer:

Question 3.
Samples of four metals A, B, C and D were taken and added to the following solution one by one. The results obtained have been tabulated as follows.

Use the table above to answer the following questions about metals A, B, C and D.
(i) Which Is the most reactive metal?
(ii) What would you observe if B is added to a solution of copper (Il) sulphate?
(iii) Arrange the metals A, B, C and D in the order of decreasing reactivity.
Answer:
(i) B is the most reactive metal because it displaces iron from its salt solution.
(ii) On adding ‘B’ to copper sulphate solution, blue colour of copper sulphate solution disappears and reddish brown coloured copper metal gets deposited on metal B. This happens because B will displace Cu from CuSO₄.
(iii) The order of reactivity is B > A > C > D.

Question 4.
Which gas is produced when dilute hydrochloric acid is added to a reactive metal? Write the chemical reaction when iron reacts with dilute H₂SO₄ :
Answer:
Hydrogen gas is produced when dilute hydrochloride acid is added to a reactive metal.

Question 5.
What would you observe when zinc Is added to a solution of iron(lI) sulphate? Write the chemical reaction that takes place.
Answer:
Zinc is more reactive than iron. Hence, when zinc is added to iron (II) sulphate, zinc displaces iron metal and the colour of solution fades from green to colourless.

Textbook Page no – 49

(i) Write the electron-dot structures for sodium, oxygen and magnesium.
(ii) Show the formation of Na₂O and MgO by the transfer of electrons.
(iii) What are the ions present in these compounds?
Answer:

(iii) In Na₂O ions present are Na⁺ and O^2- — In MgO, ions present are Mg₂ and

Question 2.
Why do ionic compounds have high melting points?
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.

Textbook Page no – 53

Question 1.
Define the following terms.
(i) Mineral (ii) Ore (iii) 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 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 2.
Name two metals which are found in nature in the free state.
Answer:
Gold and silver.

Question 3.
What chemical process is used for obtaining a metal from 4ts oxide?
Answer:
Metal oxides can be converted to metal through the process of reduction using carbon (coke).
Example: Zinc oxide is reduced to obtain zinc metal by heating it with carbon.
ZnO_(s) + C_{(s) →} Zn_(s) + CO_(s)

Over and above carbon, highly reactive metals such as sodium, calcium, aluminium are also used as reducing agents.
Example: Fe₂O_3(s) + 2Al(s) → 2Fe₍₁₎ + Al₂O₃(s) + Heat

Textbook Page no – 55

Question 1.
Metallic oxides of zinc, magnesium and copper were heated with the following metals.

In which cases will you find displacement reactions taking place?
Answer:

Answer:
As per the reactivity series, magnesium is most reactive, then comes zinc and finally copper i.e. Mg > Zn > Cu. So, in the first column, zinc cannot displace itself. It also cannot displace magnesium because magnesium is more reactive than zinc. However, since zinc is more reactive than copper, it can displace copper. Similar events happen in the next two columns.

Question 2.
Which metals do not corrode easily?
Answer:
Gold and platinum

Question 3.
What are alloys?
Answer:
An alloy is a homogeneous mixture of two or more metals or metal and non-metal.

Activities

Activity 1.

Perform an activity to study that metals are lustrous.
Answer:
Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample.
1. Clean the surface of each sample by rubbing them with sand paper. Note their appearance again.
2. Initially the surface of these metals look dull. On rubbing with sand paper, the surface starts shinning This happened because pure metals have shiny surface.

Activity 2.

Perform an activity to study that metals are hard.
Answer.
1. Take small pieces of iron, copper, aluminium and magnesium. Try to cut these metals with a sharp knife. What do you observe?
2. Hold a piece of sodium metal with a pair of tongs. (Caution: Always handle sodium metal with care.) Dry it by pressing between the folds of a filter paper. Now, put it on a watch-glass and try to cut it with a knife. What do you observe?

Observation:
1. Pieces of iron, copper and aluminium are very hard and hence it is difficult to cut them with a knife. However, magnesium turned out to be soft and so could be cut
2. We can see that sodium metal gets cut easily with the knife.

Activity 3.

Perform an activity to study that metals are malleable.
1. Take pieces of iron, zinc, lead and copper.
2. Place any one metal on a block of iron and strike it four or five times with a hammer. What do you observe?
3. Repeat with other metals and record the change in the shape of these metals.
Answer:
1. On striking hammer, the metal becomes flat. This means metal is malleable on hammering.
2. Pieces of iron, zinc, lead and copper also become flat i.e. get converted into sheets on hammering.
3. This property of metals under which they can be hammered and made flat is called the property of malleability.

Activity 4.

Perform an activity to study that metals are ductile. Consider some metals such as iron, copper, aluminium, lead, etc. Which of these metals are also available in the form of wires?
Answer:
All the four metals mentioned here are available in the form of wire also. This property of metals where in they can be drawn into wires is called the property of ductility.

Activity 5.

Perform an activity to study that metals are good conductors of heat.
Answer:

1. Take an aluminium or copper wire. Clamp this wire on a stand, as shown in figure 1.
2. Fix a pin to the free end of the wire using wax.
3. Heat the wire with a spirit lamp, candle or a burner near the place where it is, clamped.

Activity (i).
What do you observe after sometime? Does the metal melt?
Answer:
1. On heating the wire on one end, soon the entire wire gets hot. The heat melts the wax and so the pin attached to it falls down. This means the heat given at one end of the metal got transferred to the entire metal wire. So, we conclude that aluminium and copper are good conductors of heat.
2. The heat melted the wax but not the metal wire because the temperature of the heat is not high enough.

Activity 6.

Perform an activity to study that metals are good conductors of electricity.
Set up an electric circuit as shown in figure.
Place the metal to be tested in the circuit between terminals A and B as shown. Does the bulb glow? What does this indicate?

Answer:
On joining the metal such as iron, aluminium or copper with two terminals namely A and B, the bulb starts glowing. This means that electricity flows through the metals. Hence, metals are good conductor of electricity.

Activity 7.

Aim: To study the properties of non-metals
Collect samples of carbon (coal or graphite), sulphur and iodine.

Question 1.
Carry out activities 1 to 6 with these non-metals and record your observations.
Answer:
Samples of carbon, sulphur and iodine do not show the properties of metals. However, carbon in the form of graphite is a good conductor of electricity and iodine has a shining surface.

Activity 8.

Perform an activity to study if the oxides of magnesium and sulphur are basic or acidic.
Answer:
Take a magnesium ribbon and some sulphur powder.
1. Burn the magnesium ribbon. Collect the ashes formed and dissolve them in water.
2. Test the resultant solution with both red and blue litmus paper. Is the product formed on burning magnesium acidic or basic?
3. Now burn sulphur powder. Place a test tube over the burning sulphur to collect the fumes produced. Add some water to the above test tube and shake. Test this solution with blue and red litmus paper.
Is the product formed on burning sulphur acidic or basic?

Q. Can you write equations for these reactions?
Answer:
On burning magnesium, magnesium oxide (MgO) is formed. The solution of magnesium oxide turns red litmus paper into blue. This means that magnesium oxide is basic.

1. When sulphur powder is burnt it produces sulphur dioxide (SO₂). Solution of SO₂ turns blue litmus into red. Hence, SO₂ is acidic.
The reactions of both these processes are as follows:
(i) 2Mg_(s)+ O_2(g) → 2MgO₍₅₎
(ii) S_(s) + O_2(g) →  SO_2(g)

Conclusion:
Oxide of magnesium is basic whereas oxide of sulphur is acidic.

Activity 9.

Aim: To study how metals burn in air
Caution: The following activity needs the teacher’s assistance. It would be better if students wear eye protection.
1. Collect samples of the following metals — Aluminium, copper, iron, lead magnesium, zinc and sodium.
2. Hold any of these samples with a pair of tongs and try burning over a flame. Repeat with the other metal samples.
3. Collect the product, if formed.
4. Let the products and the metal surface cool down.

Question 1.
Which metals burn easily?
Answer:
Sodium, potassium and magnesium burn easily.

Question 2.
What flame colour did you observe when the metal burnt?
Answer:
Refer table given below.

Question 3.
How does the metal surface appear after burning?
Answer:
Refer table given below.

Question 4.
Arrange the metals In the decreasing order of their reactivity towards oxygen.
Answer:
The reactivity of metals with oxygen decreasing order is –
Na > Ca > Mg > Al> Zn > Fe > Cu

Question 5.
Are the products soluble in water?
Answer:
Refer table given below.

Activity 10.

Aim: To study how metals react with water.
Answer:
Caution: This activity needs the teacher’s assistance.
1. Collect samples of the following metals — Aluminium, copper, iron, lead magnesium, zinc, calcium, sodium
and potassium.
2. Put small pieces of the samples separately in beakers half-filled with cold water.

Question 1.
Which metals reacted with cold water? Arrange them in the increasing order of their reactivity with cold water.
Answer:
Na, K and Ca reacted with cold water. The reactivity of metals with cold water in increasing order is: Mg < Ca < Na < K

Question 2.
Did any metal produce fire on water?
Answer:
Sodium (Na) and potassium (K) catch fire on water.

Question 3.
Does any metal start floating after some time?
Answer:
Calcium and magnesium start floating after some time.
1. Put the metals that did not react with cold water in beakers half-filled with hot water.
2. For the metals that did not react with hot water, arrange the apparatus as shown in the figure and observe their reaction with steam.

Question 4.
Which metals did not react even with steam?
Answer:
Metals namely copper, silver and gold did not react even with steam.

Question 5.
Arrange the metals, In the decreasing order of reactivity with water.
Answer:
Metals such as lead (Pb), copper (Cu), silver (Ag) and gold (Au) did not react with water. The reactivity of remaining metals with water in decreasing order is: K> Na > Ca > Mg > Al> Zn > Fe

Activity 11

Aim: To study how metals react with acids.
Answer:
Collect the sample of metal pieces such as magnesium, alumimnium, zinc, iron and copper. If the samples are tarnished, rub them clean with sand paper.

Caution: Do not take sodium and potassium as they react vigorously even with cold water.
1. Put the samples separately in test tubes containing dilute hydrochloric acid.
2. Suspend thermometers in the test tubes, so that their bulbs are dipped in the acid.

Question 1.
Observe the rate of formation of bubbles carefully.
Answer:
The rate of formation of bubbles is fastest in magnesium.

Question 2.
With which metal did you record the highest temperature?
Answer:
With magnesium (Mg).

Question 3.
Which metals reacted vigorously with dilute hydrochloric acid?
Answer:
Metals like Mg, Al, Zn and Fe react vigorously with dilute hydrochloric acid.

Question 4.
Arrange the metals in the decreasing order of reactivity with dilute acids.
Answer:
Mg > AI > Zn> Fe

Activity 12.

Aim: To study the reaction of metals with solution of other metals.
Answer:
1. Take a clean piece of copper wire of copper and an iron nail.
2. Put the copper wire in a solution of iron sulphate and the iron nail in a solution of copper sulphate taken in test tubes
3. Record your observations after 20 minutes.

Question 1.
In which test tube did you find that a reaction has occurred?
Answer:
The reaction took place in the test tube in which the iron nail is kept in copper sulphate (CuSO₄) solution.

Question 2.
On what basis can you say that a reaction has actually taken place?
Answer:
The blue colour of the copper sulphate solution starts fading. This assures occurrence of some chemical reaction.

Question 3.
Can you correlate your observations for activities 3.9, 3.10 and 3.11?
Answer:
This activity can be correlated with activities 3.9, 3.10 and 3.11 in the sense that all these activities affirm that iron (Fe) is more reactive than copper (Cu).

Question 4.
Write a balanced chemical equation for the reaction that has taken place.
Answer:
Chemical reaction:
Fe(s) + CUSO_4(aq) → CU(g) + FeSO_4(aq)

Question 5.
Name the type of reaction.
Answer:
The reaction is displacement reaction.

Activity 13.

Aim: To study the properties of ionic compounds.
Take samples of sodium chloride, potassium iodide, barium chloride or any other salt from the science laboratory.

1. What is the physical state of these salts?
Take a small amount of a sample on a metal spatula and heat directly on the flame. Repeat with other samples.

2. What did you observe? Did the samples impart any colour to the flame? Do these compounds melt?

3. Try to dissolve the samples in water, petrol and kerosene. Are they soluble?

4. Make a circuit as shown in figure and insert the electrodes into a solution of one salt. What did you observe? Test the other salt samples too in this manner.

5. What is your inference about the nature of these compounds?
Answer:
The answers of all these questions are tablulated below:

Activity 14.

Perform an activity to demonstrate that both air and water are needed for rusting of Iron.
Answer:
1. Take three test tubes and place clean iron nails in each of them.
2. Label these test tubes A, B and C.
3. Pour some water in test tube A and cork it. The nail should not completely submerge.
4. Pour boiled distilled water in test tube B, add about 1 mL of oil and cork it. The oil will float on water and prevent the air from dissolving in the water.
5. Put some anhydrous calcium chloride in test tube C and cork it. Anhydrous calcium chloride will absorb the moisture, if any, from the air. Leave these test tubes for a few days and then observe.

Observation:

• Nail in test tube A is in contact with water as well as air and so it will get rusted.
• Nail in test tube B is in contact with water but not air and so it will not rust.
• Nail in test tube C is in contact with air but not water and so it will not rust.

Conclusion :

• The nail in test-tube A got rusted due to oxidation reaction.
• The activity proves that rusting of iron can take place only if iron is in contact with (1) water as well as (2) air.