Srinivas

Well-designed AP Board Solutions Class 7 Maths Chapter 5 Lines and Angles InText Questions offers step-by-step explanations to help students understand problem-solving strategies.

AP 7th Class Maths 5th Chapter Data Handling InText Questions

Try These (Page No. 150)

Question 1.
List ten figures around you and identify the acute, obtuse and right angles found in them.
Solution:
Class room activity.

Think discuss and write (Page No: 152)

Question 1.
Can two acute angles be complement to each other?
Solution:
Yes, two acute angles can be complementary. Acute angles are angles less than 90°.
Ex: 30° and 60° angles. Their sum is 90°, so, they complement each other.

Question 2.
Can two obtuse angles be complement to each other?
Solution:
Angles measuring than 90° are called obtuse angles. Ex : ∠100° and ∠120° are obtuse angle and their sum is 220°. Thus, they are not complement to each other.

Question 3.
Can two right angles be complement to each other?
Solution:
Two right angles cannot be complement to each other.

Try These (Page No. 152)

Question 1.
Which pairs of following angles are complementary? (Fig.)
i)
Solution:
70° + 20° = 90°
∴ The angles are complementary.

ii)
Solution:
75° + 25° = 100° > 90°
∴ The angles are not complementary.

iii)
Solution:
48° + 52° = 100° > 90°
∴ The angles are not complementary.

iv)
Solution:
35° + 55° = 90°
∴ The angles are complementary.

Question 2.
What is the measure of the complement of each of the following angles?
i) 45°
Solution:
Complement of 45° = 90° – 45° = 45°

ii) 65°
Solution:
Complement of 65° = 90° – 65° = 25°

iii) 41°
Solution:
Complement of 41° = 90° – 41° = 49°

iv) 54°
Solution:
Complement of 54° = 90° – 54° = 36°

Question 3.
The difference in the measures of two complementary angles is 12°. Find the measures of the angles.
Solution:
Method 1 :
x + y = 90°
\(\frac{x-y=12^{\circ}}{2 x=102}\)
x = \(\frac{102^{\circ}}{2}\) = 51°
y = 90° – 51° = 39°
∴ 51° and 39° on the two complementary angles.
(OR)
Method 2:
Let the first angle = x
Second angle = 90 – x
According to the problem
x – (90 – x) = 12
x – 90 + x = 12
2x = 12 + 90
2x = 102
x = \(\frac { 102 }{ 2 }\) ⇒ x = 51
90 – x = 90 – 51° = 39°
∴ The angle are 51° and 39°

Think discuss and write (Page No. 154)

Question 1.
Can two obtuse angles be supplementary?
Solution:
No, sum will exceed 180°

Question 2.
Can two acute angles be supplementary?
Solution:
No, sum will be less than 180°

Question 3.
Can two right angles be supplementary?
Solution:
Yes, sum will be 180°

Try These (Page No. 154)

Question 1.
Find the pairs of supplementary angles in Fig.
i)
Solution:
110° + 50° = 160°
Not Supplementary angles

ii)
Solution:
105° + 65° = 170°
Not Supplementary angles

iii)
Solution:
50° + 130° = 180°
Supplementary angles

iv)
Solution:
45° + 45° = 90°
Not Supplementary angles

Question 2.
What will be the measure of the supplement of each one of the following angles ?
i) 100°
Solution:
The measure of supplementary angle of 100° = 180° – 100° = 80°

ii) 90°
Solution:
The measure of supplementary angle of 90° = 180° – 90° = 90°

iii) 55°
Solution:
The measure of supplementary angle of 55° = 180° – 55° = 125°

iv) 125°
Solution:
The measure of supplementary angle of 125° = 180° – 125° = 55°

Question 3.
Among two supplementary angles the measure of the larger angle is 44° more than the measure of the smaller. Find their measures.
Solution:
Let the smaller angle = x°
Larger angle = x° + 44°
According to the problem
The angles are supplementary
∴ x° + x° + 44° = 180°
2x° + 180° – 44°
2x° = 136°
x° = \(\frac{136^{\circ}}{2}\)
x° = 68°
x° + 44° = 68° + 44° = 112°
∴ The angles are 68° and 112°

Think, discuss and write (Page No. 158)

Question 1.
In Fig. AC and BE intersect at P.

AC and BC intersect at C, AC and EC intersect at C.
Try to find another ten pairs of intersecting line segments.
Should any two lines or line segments necessarily intersect ? Can you find two pairs of non-intersecting line segments in the figure?
Can two lines intersect in more than one point? Think about it.
Solution:
Pairs of intersecting line segments.
BC and BE – B
BC and AC – C
AC and EC – C
AC and BE – P
BC and CE – C
BE and CE – E
Total 6 pairs are intersecting.
Yes, any two lines are intersecting in this figure.
Any two lines are necessarily intersecting only when they are not parallel.
No, two lines cannot intersect at two or more points.
This is possible only when both lines coincide. These lines intersect at infinite number of points

Try These (Page No. 160)

Question 1.
Find examples from your surroundings where lines intersect at right angles.
Solution:
Class room activity.
Ex : Edge of paper sheet.

Question 2.
Find the measures of the angles made by the intersecting lines at the vertices of an equilateral triangle.
Solution:

Each angle will be equal to other
∴ Measure of each angle = \(\frac{180^{\circ}}{3}\) = 60°

Question 3.
Draw any rectangle and find the measures of angles at the four vertices made by the intersecting lines.
Solution:

Question 4.
If two lines intersect, do they always intersect at right angles ?
Solution:
No, it need not be that they intersect at right angles.

Try These (Page No. 162)

Question 1.
Suppose two lines are given. How many transversals can you draw for these lines?
Solution:
Infinite.

Question 2.
If a line is a transversal to three lines, how many points of intersections are there?
Solution:

Question 3.
Try to identify a few transversals in your surroundings.
Solution:
Class room activity.
Ex: Mesh in Tennis bat.

Try These (Page No. 164)

Question 1.
Name the pairs of angles in each figure:

Solution:

• ∠1 and ∠2 are corresponding angles.
• ∠3 and ∠4 are alternate interior angles.
• ∠5 and∠6 are interior angles on the same side of transversal.
• ∠7 and∠8 are corresponding angles.
• ∠9 and ∠10 are alternate interior angles.
• ∠11 and ∠12 are linear pair of angles.

Do This (Page No. 164)

Question 1.
Take a ruled sheet of paper. Draw (in thick colour) two parallel lines I and m. Draw a transversal l to the lines l and m . Label ∠1 and ∠2 as shown fig. (i). Place a tracing paper over the figure drawn. Trace the lines l, m and t. Slide the tracing paper along t, until l coincides with m .
You find that ∠1 on the traced figure coincides with ∠2 of the original figure.
In fact, you can see all the following results by similar tracing and sliding activity.
i) ∠1 = ∠2
ii) ∠3 = ∠4
iii) ∠5 = ∠6
iv) ∠7 = ∠8

Solution:
Student Activity.

Do This (Page No. 168)

Question 1.
Draw a pair of parallel lines and a transversal. Verify the above three statements by actually measuring the angles.
Solution:
Students self activity.

Try these (Page No. 170)

Question 1.
i)
Lines l // m; t is a transversal ∠x = ?
Solution:
l // m
t is the tranversal
∠x = 60°
If a transversal intersects two parallel lines, then alternate interior angles are equal.

ii)
Lines a //b; c is a transversal ∠y =?
Solution:
a//b
c is a transversal
If a transversal intersects two parallel lines, then alternate interior angles are equal.

iii)
l₁, l₂ be two lines t is a transversal. Is ∠1 = ∠2?
Solution:
l₁, l₂ be two lines
t is a transversal
l₁ & l₂ are not parallel lines
∴ ∠1 ≠ ∠2

iv)
Lines l //m; t is a transversal ∠z= ?
Solution:
l//m
t is a transversal
If a transversal intersects two parallel lines, then sum of interior angles in the same side of the transversal is supplementary.

v)
Lines l//m; t is a transversal ∠x =?
Solution:
l//m

t is a transversal
If a transversal intersects two parallel lines, then corresponding angles are equal
x° = 120°

vi)
Lines l//m, p//q, Find a, b, c, d
Solution:
l // m
p || q
l //m, then p & q are transversals
p // m and l is the transversal
a + 60° = 180° (Sum of interior angles on the same side of transversal)
a = 180° – 60°
a = 120°
l // m, and q is the transversal.
∠a = ∠c = 120° (Alternate interior angles)
∴ ∠c = 120°

∠b = ∠c = 120°(Vertically opposite angles)
∴ ∠b = 120°
∠c + ∠d = 180° (Linear pair)
120 + ∠d = 180°
∠d = 180° – 120°
∠d = 60°

Try these (Page No. 172)

Question 1.
i)
Is l || m? Why?
Solution:
50° = 50°
l and m are two lines
t is the transversal
Then alternate interior angles are equal
∴ l || m

ii)
Is l || m? Why?
Solution:

Let us take the figure is the following way l and m are two lines
t is the transversal let x = 50° (Alternate angles)
x° + 130° = 50° + 130° = 180°
∴ l || m

iii)
Is l || m, what is ∠x?
Solution:
l || m
t is a transversal
70° + x° = 180°
(Interior angles on the same side of the transversal are supplementary)
x° = 180° – 70°
x° = 110°

Well-designed AP Board Solutions Class 7 Maths Chapter 5 Lines and Angles Exercise 5.2 offers step-by-step explanations to help students understand problem-solving strategies.

Lines and Angles Class 7 Exercise 5.2 Solutions – 7th Class Maths 5.2 Exercise Solutions

Question 1.
State the property that is used in each of the following statements?

i) If a || b, then ∠1 = ∠5
Solution:
a|| b, then ∠1 = ∠5
corresponding angles property.

ii) If ∠4 = ∠6, then a || b.
Solution:
∠4 = ∠6, then a || b
Alternate interior angle property.

iii) If ∠4 + ∠5 = 180°, then a || b.
Solution:
∠4 + ∠5 = 180°, then a || b
Interior angles on the same side of the transversal are supplementary.

Question 2.
In the below figure, identify

i) the pairs of corresponding angles.
Solution:
Pairs of corresponding angles
∠1, ∠5; ∠2, ∠6; ∠3, ∠7; ∠4, ∠8.

ii) the pairs of alternate interior angles.
Solution:
Pairs of Alternate interior angles
∠2, ∠8; ∠3, ∠5

iii) the pairs of interior angles on the same side of the transversal.
Solution:
Pairs of interior angles on the same side of the transversal.
∠2, ∠5; ∠3,∠8

iv) the vertically opposite angles.
Solution:
Vertically opposite angles
∠1, ∠3; ∠2, ∠4; ∠5, ∠7; ∠6,∠8.

Question 3.
In the below figure, p || q. Find the unknown angles.

Solution:
In the figure, given p // q
125° + ∠e° = 180° (Linear pair)
∠e = 180° – 125°
∠e = 55°
∠e = ∠f = 55° (Vertically opposite angles)
∠f = 55°
125° = ∠d (Corresponding angles)
∴ ∠d = 125°
∠d = ∠b = 125° (Vertically opposite angles)
∠b = 125°
∠e = ∠a = 55° (Corresponding angles)
∴ ∠a = 55°
∠c = ∠a = 55° (Vertically opposite angles)
∴ ∠c = 55°

Question 4.
Find the value of x in each of the following figures if l || m.
i)
Solution:
Let us draw the figure in the following was

y = 110° (Vertitically opposite angles)
given l || m
∴ x + y = 180° (Interior angles on the same side of the transversal are supple-mentary)
x° + 110° = 180°
x° = 180° – 110°
x° = 70°

ii)
Solution:
l || m
a is the transversal.
∴ x = 100° (Corresponding angles)

Question 5.
In the given figure, the arms of two angles are parallel.

If ∠ABC = 70°, then find
i) ∠DGC
ii) ∠DEF
Solution:
Given in the figure, the arm of two angles are parallel
given ∠ABC = 70°
i) ∠DGC = 70° (Corresponding angles)
(AB || ED, BC is the transversal)
∴ ∠DGC = 70°

Question 6.
In the given figures below, decide whether l is parallel to m.
i)
Solution:
126° + 44° = 170° ≠ 180°

ii)
Solution:
Let us take the figure in the following way.

Let x =75° (Vertically opposite angles)
x + 75° = 75° + 75° = 150° ≠ 180°

iii)
Solution:
Let us take the figure in the following way.

x = 57° (Vertically opposite angles)
x° + 123° = 57° + 123° = 180°
∴ l || m

iv)
Solution:
Let us take the figure diagram.

x = 72° (Vertically opposite angles)
x° + 98° = 72° + 98° = 170° ≠ 180°
∴ l || m

Well-designed AP Board Solutions Class 7 Maths Chapter 5 Lines and Angles Exercise 5.1 offers step-by-step explanations to help students understand problem-solving strategies.

Lines and Angles Class 7 Exercise 5.1 Solutions – 7th Class Maths 5.1 Exercise Solutions

Question 1.
Find the complement of each of the following angles :
i)
Solution:
Complement of 20° = 90° – 20° = 70°

ii)
Solution:
Complement of 63° = 90° – 63° = 27°

iii)
Solution:
Complement of 57° = 90° – 57° = 33°

Question 2.
Find the supplement of each of the following angles:
i)
Solution:
Supplement of 105° = 180° – 105° = 75°

ii)
Solution:
Supplement of 87° = 180° – 87° = 93°

iii)
Solution:
Supplement of 154° = 180° – 154° = 26°

Question 3.
Identify which of the following pairs of angles are complementary and which are supplementary.
i) 65°, 115°
Solution:
65°, 115°
65° + 115° = 180°
65° and 115° are supplementary angles.

ii) 63°, 27°
Solution:
63°, 27°
63° + 27° = 90°
63° and 27° are complementary angles.

iii) 112°, 68°
Solution:
112°, 68°
112° and 68° are supplementary angles.

iv) 130°, 50°
Solution:
130°, 50°
130° + 50° = 180°
130°, 50° are supplementary angles.

v) 45°, 45°
Solution:
45°, 45°
45° + 45° = 90°
45°, 45° are complementary angles.

vi) 80°, 10°
Solution:
80°, 10°
80° + 10° = 90°
80° and 10° are complementary angles.

Question 4.
Find the angle which is equal to its complement.
Solution:
Let the angle be x°
According to the problem
x° + x° = 90°
2x° = 90°
x° = \(\frac{90^{\circ}}{2}\)
x° = 45°
∴ 45° is the angle which is equal to its complement.

Question 5.
Find the angle which is equal to its supplement.
Solution:
Let the angle be x°
According to the problem
x° + x° = 180°
2x° = 180°
x° = \(\frac{180^{\circ}}{2}\)
x° = 90°
∴ 90° is the angle which is equal to its supplement.

Question 6.
In the given figure, ∠1 and ∠2 are supplementary angles.

If ∠1 is decreased, what changes should take place in ∠2 so that both the angles still remain supplementary.
Solution:
Given ∠1 and ∠2 are supplementary
∴ ∠1 + ∠2 = 180°
Let ∠1 = 120°
then, 120° + ∠2 = 180°
∠2 = 180° – 120°
∠2 = 60°
If ∠1 is decreased by 10°, then
1 = 120° – 10° = 110°
then, ∠2 = 180° – 110° = 70°
Finally if ∠1 is decreased then ∠2 will be increased and both will again form supplementary angles.

Question 7.
Can two angles be supplementary if both of them are :
i) acute?
ii) obtuse?
iii) right?
Solution:
i) Acute :
Let 60° and 70° are acute angles
60° + 70° = 130° ≠ 180°
∴ If both angles are acute, they cannot form supplementary angles.

ii) Obtuse :
Let 100° and 110° are two obtuse angles
100° + 110° = 210° ≠ 180°
∴ If both angles are obtuse, the sum of the two angles will be more than 180°.
∴ They can’t form supplementary angles.

iii) Right :
90° + 90° = 180°
Yes, if both angles are 90° (i.e., right angle),
They form supplementary angles.

Question 8.
An angle is greater than 45°. Is its complementary angle greater than 45° or equal to 45° or less than 45°?
Solution:
Let the angle greater than 45° is 46°
Complementary angle of 46° = 90° – 46° = 44°
∴ An angle is greater than 45°, then its complementary angle is less than 45°.

Question 9.
Fill in the blanks:
i) If two angles are complementary, then the sum of their measures is _____
ii) If two angles are supplementary, then the sum of their measures is: _____
iii) If two adjacent angles are supplementary, they form a _____
Solution:
i) 90°
ii) 180°
iii) a linear pair.

Question 10.
In the below figure, name the following pairs of angles.

i) Obtuse vertically opposite angles
ii) Adjacent complementary angles
iii) Equal supplementary angles
iv) Unequal supplementary angles
v) Adjacent angles that do not form a linear pair
Solution:
i) ∠BOC and ∠AOD obtuse vertically opp osite angles.
ii) ∠AOB and ∠AOE one adjacent complementary angles.
iii) ∠EOB and ∠EOD are equal supplementary angles.
iv) ∠EOA and ∠EOC are unequa! supplementary angles.
v) ∠AOB and ∠AOE, ∠AOE, and ∠EOD, ∠EOD and ∠COD, are adjacent angles but not form a linear pair.

Well-designed AP Board Solutions Class 7 Maths Chapter 4 Simple Equations InText Questions offers step-by-step explanations to help students understand problem-solving strategies.

AP 7th Class Maths 4th Chapter Simple Equations InText Questions

Try These (Page No. 120)

Question 1.
The value of the expression (10y – 20) depends on the value of y. Verify this by giving five different values to y and finding for each y the value of ( 10 y 20). From the different values of (10y -20) you obtain, do you see a solution to 10y – 20 = 50? If there is no solution, try giving more values to y and find whether the condition 10y – 20 = 50 is met.
Solution:
Let us find the value of the expression (10y – 20) for different values of y.
Let y = 2.
We have : 10y – 20 = 10(2) – 20 = 0
Let y = 3.
We have : 10y – 20 = 10(3) – 20 = 10
Let y = 4.
We have : 10y – 20 = 10(4) – 20 = 20
Let y = 5.
We have: 10y – 20 = 10(5) – 20 = 30
Let y = 6.
We have : 10y – 20 = 10(6) – 20 = 40
Hence 10y – 20 depends upon y
Now, consider 10y – 20 = 50
Transpose – 20 to the RHS :
10y = 50 + 20 = 70
or y =7

Try These (Page No. 124)

Question 1.
Write atleast one other form for each equation (ii), (iii) and (iv).
Solution:
Other forms are as unde r:
i) p multiplied by 5 gives 20
ii) 7 added to thrice of n gives 1
iii) one-fifth of m less 2 gives 6 .

Try These (Page No: 140)

Question 1.
i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is ?
Solution:
Let the number required be x
x is multiplied by 6 , the product = 6x.
5 is subtracted from 6x, then difference = 6x – 5
Result = 7
6x – 5 = 7 ⇒ 6x = 7 + 5
6x = 12 ∴ x = \(\frac{12}{6}\) = 2
∴ Required number = 2

ii) What is that number one third of which added to 5 gives 8 ?
Solution:
Let the number required be x
One third of number = \(\frac{x}{3}\)
According to the problem
\(\frac{x}{3}\) + 5 = 8 ⇒ \(\frac{x}{3}\) = 8 – 5
\(\frac{x}{3}\) = 3 ⇒ x= 3 × 3 ⇒ x = 9
∴ The required number = 9

Try These (Page No : 142)

Question 1.
There are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of the smaller type. Each larger box contains 100 mangoes. Find the number of mangoes contained in the smaller box?
Solution:
Let the number of mangoes in the smaller box be x
According to the question
8x + 4 = 100 ⇒ 8x = 100 – 4
8x = 96 ⇒ x = \(\frac{96}{8}\) ∴ x = 12
∴ The number of mangoes in the smaller box = 12

Well-designed AP Board Solutions Class 7 Maths Chapter 4 Simple Equations Exercise 4.3 offers step-by-step explanations to help students understand problem-solving strategies.

Simple Equations Class 7 Exercise 4.3 Solutions – 7th Class Maths 4.3 Exercise Solutions

Question 1.
Set up equations and solve them to find the unknown numbers in the following cases :
a) Add 4 to eight times a number; you get 60.
Solution:
Let the number required be x
Eight times of the number = 8x
When 4 is added to 8x, the sum = 8x + 4
According to the problem
8x + 4 = 60 ⇒ 8x = 60 – 4
8x = 56 ⇒ x = \(\frac { 56 }{ 8 }\) ⇒ x = 7
∴ The number required = 7

b) One-fifth of a number minus 4 gives 3.
Solution:
Let the number required be x
One-fifth of the number = \(\frac { x }{ 5 }\)
According to the problem
\(\frac { x }{ 5 }\) – 4 3 ⇒ \(\frac { x }{ 5 }\) = 3 + 4
\(\frac { x }{ 5 }\) = 7 ⇒ x = 7 × 5 ⇒ x = 35
∴ The number required = 35

c) If I take three-fourths of a number and add 3 to it, I get 21.
Solution:
Let the number required be x
Three-fourths of the number = \(\frac { 3x }{ 4 }\)
According to the problem
\(\frac { 3x }{ 4 }\) + 3 = 21 ⇒ \(\frac { 3x }{ 4 }\) = 21 – 3
\(\frac { 3x }{ 4 }\) = 18 ⇒ 3x = 18 × 4
x = \(\frac{18 \times 4}{3}\) ⇒ x = 6 × 4 ⇒ x = 24
∴ The required number = 24

d) When I subtracted 11 from twice a number, the result was 15.
Solution:
Let the number required be x
Twice of the number = 2x
According to the problem
2x – 11 = 15 ⇒ 2x = 15 + 11
⇒ 2x = 26 ⇒ x = \(\frac { 26 }{ 2 }\) ⇒ x = 13
∴ The required number = 13

e) Munna subtracts thrice the number of notebooks he has from 50 , he finds the result to be 8.
Solution:
Let the number required be x
Thrice of the number = 3x
According to the problem
50 – 3x = 8 ⇒ 50 – 8 = 3x
⇒ 3x = 42 ⇒ x = \(\frac { 42 }{ 3 }\) ⇒ x = 14
∴ The required number = 14

f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5 , she will get 8 .
Solution:
Let the number required be x
When 19 is added to x then the sum = x + 19
According to the problem
\(\frac { x+19 }{ 5 }\) = 8 ⇒ x + 19 = 40
x = 40 – 19 ⇒ x = 21
∴ The required number = 21.

g) Anwar thinks of a number. If he takes away 7 from \(\frac { 5 }{ 2 }\) of the number, the result is 23 .
Solution:
Let the number required be x\(\frac { 5 }{ 2 }\) of a number = \(\frac { 5x }{ 2 }\)
According to the problem
\(\frac { 5x }{ 2 }\) – 7 = 23 ⇒ \(\frac { 5x }{ 2 }\) = 23 + 7
\(\frac { 5x }{ 2 }\) = 30 ⇒ 5x = 30 × 2
5x = 60 ⇒ x = \(\frac { 60 }{ 5 }\) ⇒ x = 12
∴ The required number = 12

Question 2.
Solve the following :
a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87 What is the lowest score?
Solution:
Let the lowest score be x
Twice the lowest score = 2x
Highest marks = 87
According to the problem
2x + 7 = 87
2x = 87 – 7 (Transposing 7 to RHS)
2x = 80
x = \(\frac { 80 }{ 2 }\) (divide both sides by 2)
x = 40
∴ Lowest score = 40

b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle ? (Remember, the sum of three angles of a triangle is 180°).
Solution:
Let the equal angle is Isosceles triangle = a
Vertex angle = 40°

We know that the sum of the angles of a triangle = 180°
40° + a° + a° = 180°
2a = 180° – 40°
(Transposing 40° to RHS)
2a = 140°
a = \(\frac{140^{\circ}}{2}\) (divide both sides by 2 )
a = 70°
∴ Base angles of the triangle = 70°

c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Solution:
Let the score of Rahul = x
Score of Sachin = 2x
Two short of double century = 200 – 2 = 198
According to the problem
2x + x = 198
3x = 198
x = \(\frac{198}{3}\) (divide by 3 on both sides)
x = 66
∴ Rahul’s score = 66
Sachin’s score = 2 × 66 = 132

Question 3.
Solve the following :
i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
Solution:
Let the number of marbles of Parmit has = x
5 times of marbles of Parmit = 5x
According to the problem
5x + 7 = 37
5x = 37 – 7 (Transposing 7 to RHS)
5x = 30
x = \(\frac{30}{5}\) (Divide both sides by 5)
x = 6
∴ There are 6 marbles with Parmit.

ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age ?
Solution:
Let the age of Laxmi be x years
3 times the age of Laxmi = 3x years
According to the problem
3x + 4 = 49
3x = 49 – 4 (Transposing 4 to RHS)
3x = 45
x = \(\frac{45}{3}\) (Divide both sides by 3 )
x = 15
∴ Age of Laxmi = 15.

iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77 ?
Solution:
Let the number of fruit trees be x
3 times the number = 3x
According to the problem
3x + 2 = 77
3x = 77 – 2 (Transposing 2 to RHS)
3x = 75 (Divide both sides by 3)
x = \(\frac{75}{3}\)
x = 25
∴ The number of fruit trees = 25.

Question 4.
Solve the following riddle:
I am a number,
Tell my identity !
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty !
Solution:
Let the number = x
7 times the number = 7x
7 x is added to 50 , then sum = 7x + 50.
According to the problem
7x + 50 + 40 = 300
7x + 90 = 300
7x = 300 – 90 (Transposing 90 to RHS)
7x = 210
x = \(\frac { 210 }{ 7 }\) (Divide both sides by 7 )
x = 30
∴ The number required = 30

Well-designed AP Board Solutions Class 7 Maths Chapter 4 Simple Equations Exercise 4.2 offers step-by-step explanations to help students understand problem-solving strategies.

Simple Equations Class 7 Exercise 4.2 Solutions – 7th Class Maths 4.2 Exercise Solutions

Question 1.
Give first the step you will use to separate the variable and then solve the equation :
a) x – 1 = 0
Solution:
x – 1 = 0
Adding ‘ 1 ‘ on both sides
x – 1 + 1 = 0 + 1 ⇒ x = 1

b) x + 1 = 0
Solution:
x + 1 = 0
Adding ‘ -1 ‘ on both sides
x + 1 – 1 = 0 – 1 ⇒ x = – 1

c) x – 1 = 5
Solution:
x – 1 = 5
Adding ‘ 1 ‘ on both sides
x – 1 + 1 = 5 + 1 ⇒ x = 6

d) x + 6 = 2
Solution:
x + 6 = 2
Adding ‘ -6 ‘ on both sides
x + 6 – 6 = 2 – 6 ⇒ x = – 4

e) y – 4 = – 7
Solution:
y – 4 = – 7
Adding 4 on both sides
y – 4 + 4 = – 7 + 4 ⇒ y = – 3

f) y – 4 = 4
Solution:
y – 4 = 4
Adding 4 on both sides
y – 4 + 4 = 4 + 4 ⇒ y = 8

g) y + 4 = 4
Solution:
y + 4 = 4
Adding -4 on both șides
y + 4 – 4 = 4 – 4 ⇒ y = 0

h) y + 4 = – 4
Solution:
y + 4 = – 4
Adding -4 on both sides
y + 4 – 4 = – 4 – 4 ⇒ y = – 8.

Question 2.
Give first the step you will use to separate the variable and then solve the equation :
a) 3l = 42
Solution:
3l = 42
Divide by 3 on both sides
\(\frac { 3l }{ 3 }\) = \(\frac { 42 }{ 3 }\) ⇒ l = 14

b) \(\frac { b }{ 2 }\) = 6
Solution:
\(\frac { b }{ 2 }\) = 6
Multiplying by 2 on both sides
\(\frac { b }{ 2 }\) × 2 = 6 × 2 ⇒ b = 12

c) \(\frac { p }{ 7 }\) = 4
Solution:
\(\frac { p }{ 7 }\) = 4
Multiply by 7 on both sides
\(\frac { p }{ 7 }\) × 7 = 4 × 7 ⇒ p = 28

d) 4x = 25
Solution:
4x = 25
Divide by 4 on both sides
\(\frac { 4x }{ 4 }\) = \(\frac { 25 }{ 4 }\) ⇒ x = \(\frac { 25 }{ 4 }\)

e) 8y = 36
Solution:
8y = 36
Divide by 8 on both sides
\(\frac { 8y }{ 8 }\) = \(\frac { 36 }{ 8 }\) ⇒ y = \(\frac { 36 }{ 8 }\) = \(\frac { 18 }{ 4 }\) = \(\frac { 9 }{ 2 }\)

f) \(\frac { z }{ 3 }\) = \(\frac { 5 }{ 4 }\)
Solution:
\(\frac { z }{ 3 }\) = \(\frac { 5 }{ 4 }\)
Multiplying both sides by 3
\(\frac { z }{ 3 }\) × 3 = \(\frac { 5 }{ 4 }\) × 3 ⇒ z = \(\frac { 15 }{ 4 }\)

g) \(\frac { a }{ 5 }\) = \(\frac { 7 }{ 15 }\)
Solution:
\(\frac { a }{ 5 }\) = \(\frac { 7 }{ 15 }\)
Multiplying both sides by 5
\(\frac { a }{ 5 }\) × 5 = \(\frac { 7 }{ 15 }\) × 5 ⇒ a = \(\frac { 7 }{ 3 }\)

h) 20t = – 10
Solution:
20t = – 10
Divide both sides by 20
\(\frac { 20t }{ 20 }\) = \(\frac { -10 }{ 20 }\) ⇒ t = \(\frac { -1 }{ 2 }\)

Question 3.
Give the steps you will use to separate the variable and then solve the equation :
a) 3n – 2 = 46
Solution:
3n – 2 = 46
Adding 2 on both sides
3n – 2 + 2 = 46 + 2 ⇒ 3n = 48
Divide both sides by 3
\(\frac { 3n }{ 3 }\) = \(\frac { 48 }{ 3}\) ⇒ n = 16

b) 5m + 7 = 17
Solution:
5m + 7 = 17
Adding -7 on both sides
5 m + 7 – 7 = 17 – 7 ⇒ 5m = 10
Divide both sides by 5
\(\frac { 5m }{ 5 }\) = \(\frac { 10 }{ 5 }\) ⇒ m = 2

c) \(\frac { 20p }{ 3 }\) = 40
Solution:
\(\frac { 20p }{ 3 }\) = 40

Multiply both sides by 3
\(\frac { 20p }{ 3 }\) × 3 = 40 × 3 ⇒ 20p = 120
Divide both sides by 20
\(\frac { 20p }{ 20 }\) = \(\frac { 120 }{ 20 }\) ⇒ p = 6

d) \(\frac { 3p }{ 10 }\) = 6
Solution:
\(\frac { 3p }{ 10 }\) = 6
Multiply both sides by 10
\(\frac { 3p }{ 10 }\) × 10 = 6 × 10 ⇒ 3p = 60
Divide both sides by 3
\(\frac { 3p }{ 3 }\) = \(\frac { 60 }{ 3 }\) ⇒ p = 20.

Question 4.
Solve the following equations :
a) 10p = 100
Solution:
10p = 100
Divide both sides by 10
\(\frac { 10p }{ 10 }\) = \(\frac { 100 }{ 10 }\) ⇒ p = 10

b) 10p + 10 = 100
Solution:
10p + 10 = 100
Subtract 10 on both sides
10p + 10 – 10 = 100 – 10 ⇒ 10p = 90
Divide both sides by 10
\(\frac { 10p }{ 10 }\) = \(\frac { 90 }{ 10 }\) ⇒ p = 9

c) \(\frac { p }{ 4 }\) = 5
Solution:
\(\frac { p }{ 4 }\) = 5
Multiply both sides by 4
\(\frac { p }{ 4 }\) × 4 = 5 × 4 ⇒ p = 20

d) \(\frac { -p }{ 3 }\) = 5
Solution:
\(\frac { -p }{ 3 }\) = 5
Multiply both sides by 3
\(\frac { -p }{ 3 }\) × 3 = 5 × 3 ⇒ – p = 15
Multiply both sides by -1
-1 × – p = 15 × – 1 ⇒ p = – 15

e) \(\frac { 3p }{ 4 }\) = 6
Solution:
\(\frac { 3p }{ 4 }\) = 6
Multiply both sides by 4
\(\frac { 3p }{ 4 }\) × 4 = 6 × 4 ⇒ 3p = 24
Divide both sides by 3
\(\frac { 3p }{ 3 }\) = \(\frac { 24 }{ 3 }\) ⇒ p = 8.

f) 3s = – 9
Solution:
3s = – 9
Divide both sides by 3
\(\frac { 3s }{ 3 }\) = \(\frac { -9 }{ 3 }\) ⇒ s = -3

g) 3s + 12 = 0
Solution:
3s + 12 = 0
Adding – 12 on both sides
3s + 12 – 12 = 0 – 12
Divide both sides by 3
\(\frac { 3s }{ 3 }\) = \(\frac { -12 }{ 3 }\) ⇒ s = – 4

h) 3s = 0
Solution:
3s = 0
Divide both sides by 3
\(\frac { 3s }{ 3 }\) = \(\frac { 0 }{ 3 }\) ⇒ s = 0

i) 2q = 6
Solution:
2q = 6
Divide both sides by 2
\(\frac { 2q }{ 2 }\) = \(\frac { 6 }{ 2 }\) ⇒ q = 3

j) 2q – 6 = 0
Solution:
2q – 6 = 0
Adding 6 on both sides
2q – 6 + 6 = 0 + 6 ⇒ 2q = 6
Divide both sides by 2
\(\frac { 2q }{ 2 }\) = \(\frac { 6 }{ 2 }\) ⇒ q = 3

k) 2q + 6 = 0
Solution:
2q + 6 = 0
Subtract 6 from both sides
2q + 6 – 6 = 0 – 6 ⇒ 2q = – 6
Divide both sides by 2
\(\frac { 2q }{ 2 }\) = \(\frac { -6 }{ 2 }\) ⇒ q = -3

l) 2q + 6 = 12
Solution:
2q + 6 = 12
Subtract 6 from both sides.
2q + 6 – 6 = 12 – 6 ⇒ 2q = 6
Divide both sides by 2
\(\frac { 2q }{ 2 }\) = \(\frac { 6 }{ 2 }\) ⇒ q = 3

These AP 10th Class Physics Important Questions 3rd Lesson Metals and Non-metals will help students prepare well for the exams.

AP 10th Class Physical Science 3rd Lesson Metals and Non-metals Important Questions

Class 10 Physical Science Chapter 3 Important Questions – 1 Mark

Question 1.
What are metals ?
Answer:
Those elements which can lose electrons easily and from positive ions. They are mostly solids, posses high density.

Question 2.
Write the name of an allotrope of carbon.
Answer:
Diamond, Graphite and Fullerences are the allotropes of carbon.

Question 3.
Name the metal which reacts with a very dilute HNO₃ to evolve hydrogen gas.
Answer:
Magnesium

Question 4.
What do you mean by thermal conductivity ?
Answer:
It is the property due to which metals can conduct heat. e.g. copper, silver, aluminium, gold and iron are good conductors of heat.

Question 5.
What is metalic lustre ?
Answer:
Most of metals have shiny surface i.e. they show metalic lustre. Eg. Au, Ag, Pt are lustrous.

Question 6.
Name any one metal which reacts neither with cold water nor with hot water, but reacts with heated stem to produce hydrogen gas.
Answer:
Iron, 3 Fe_(s) + 4H₂O_(g) → Fe₃O_{4 (s)} + 4H_2(g)

Question 7.
Arrange the following metals in the decreasing order of reactivity.
Na, K, Cu, Ag
Answer:
K > Na > Cu > Ag

Question 8.
Metals generally occur in solid state. Name and write symbol of a metal that exists in liquid state at room temperature.
Answer:
Mercury (Hg) exists in liquid state at room temperature.

Question 9.
Name two metals that are soft and can be cut with knife.
Answer:
Na (Sodium) and K (Potassium)

Question 10.
Which of the following two metals will melt at body temperature (37° C) ?
Gallium, Magnesium, Cesium, Aluminium.
Answer:
Gallium and Cesium.

Question 11.
How many valence electrons do metals have ?
Answer:
They have 1 to 3 valence electrons.

Question 12.
Why are metals lustrous ?
Answer:
They absorb light and radiate a part of it.

Question 13.
Name the property due to which metal produces sound ?
Answer:
Sonorous

Question 14.
Reaction of metal with dilute acid is exothermic or endothermic process ?
Answer:
Exothermic process.

Question 15.
Which metal does not react with water at all ?
Answer:
Copper because it is less reactive than hydrogen.

Question 16.
Why do we use copper and aluminium wire for transmission of electric current ?
Answer:
It is because copper and Al are good conductor of electricity and ductile, i.e. drawn into wires easily.

Question 17.
Name a non-metal which is lustrous and a metal which is non-lustrous.
Answer:
Iodine is a non-metal which is lustrous, lead is a non-lustrous metal.

Question 18.
Name one metal and one non-metal in liquid state at room temperature.
Answer:
Mercury is a liquid metal and bromine is liquid non-metal.

Question 19.
Name the ion made by non-metalls : cations / anion.
Answer:
Anions (negative ions) because non-metals can gain electrons easily.

Question 20.
An element forms an oxide, A₂O₃ which is acidic in nature. Identify A as a metal or non-metal.
Answer:
‘A’ is non-metal as non-metallic oxides are acidic in nature.

Question 21.
What happens Zn reacts with NaOH ?
Answer:
H₂ gas is liberated and sodium zincate is formed.

Question 22.
Write the chemical equation for the reaction taking place when steam is passed over hot aluminium.
Answer:
2Al_(s) + 3H₂O_(g) → Al₂O_3(s) + 3H_2(g)

Question 23.
What is name given to electrons lost or gained by an element ?
Answer:
Electrovalency

Question 24.
What happens when carbon dioxide is compressed in water at higher pressure ?
Answer:
Carbonic acid is formed CO₂ + H₂O → H₂CO₃

Question 25.
Choose the amphoteric oxide amongst the following : Na₂O, ZnO, Al₂O₃, CO₂, H₂O.
Answer:
Al₂O₃ and ZnO are amphoteric oxides, Na₂O is basic, H₂O is neutral, CO₂ is acidic oxide.

Question 26.
What is metallurgy ?
Answer:
All the process involved in the extraction of metals from their ores and refining them for use, is called metallurgy.

Question 27.
How are sulphide ores concentrated ?
Answer:
By froth floatation process.

Question 28.
How is hematite ore concentrated ?
Answer:
By hydraulic washing.

Question 29.
Why sulphide and carbonate ores are converted into oxide ?
Answer:
It is because it is easier to reduce oxides as compound to sulphide and carbonate ores.

Question 30.
Why oxide of highly reactive metals can not be reduced by carbon ?
Answer:
It is because highly reactive metals themselves are stronger reducing agents than carbon.

Question 31.
What is formula of rust ?
Answer:
Fe₂O₃. XH₂O is formula of rust.

Question 32.
A green layer is gradually formed on a copper plate left exposed to air for a week in a bathroom. What could this green substance be ?
Answer:
It is due to the formation of basic copper carbonate (CuCO₃, Cu(OH)₂)

Question 33.
Define amalgam.
Answer:
When a metal is alloyed with mercury, it is called amalgam.

Question 34.
Do we use 24 carat gold for ornaments ?
Answer:
No, it is very soft and changes its shape.

Question 35.
Which metal is present both in solder and bronze ?
Answer:
Tin.

Question 36.
Why is solder used for welding electrical wires ?
Answer:
It has low melting point.

Question 37.
WTiat is the process of depositing zinc on iron called ?
Answer:
It is called galvanisation.

Question 38.
What is galvanisation ?
Answer:
Galvanisation is a method of protecting steel and iron from rusting by coating them with a thin layer of zinc.

Question 39.
Name two metals which react with dil. HNO₃ to evolve H₂ gas.
Answer:
Magnesium (Mg) and manganese (Mn).

Question 40.
Why oxides of high reactive metals cannot be reduced by carbon ?
Answer:
This is because these metals have more affinity for oxygen than carbon.

Question 41.
What is formed when sodium absorbs moisture from air ? Give equation also.
Answer:
It forms sodium hydroxide
2Na + 2H₂O → 2 NaOH + H₂

Question 42.
Why the sulphide and carbonate ores are converted into oxides ?
Answer:
It is because it is easier to obtain a metal from its oxide, as compared to its sulphide and carbonates.

Question 43.
From amongst the metals sodium, calcium, aluminium, copper and magnesium, name the metal
i) which reacts with water only on boiling and
ii) another which does not react even with steam ?
Answer:
i) Magnesium
ii) Copper

Question 44.
Which metals are mixed with iron to get stainless steel ?
Answer:
Nickel and chromium.

Question 45.
Why is stainless steel preferred for making household utensils ?
Answer:
Stainless steel is preferred as it is non-reactive and so the milk or food is not spoiled in it.

Question 46.
Write balanced chemical equation for the reaction.
Aluminium when heated in air. Write the name of the product.
Answer:

Question 47.
Why do ionic compounds not conduct electricity to the solid state ?
Answer:
Movement of ions in the solid state is not possible due to their rigid structure.

Question 48.
Name the most ductile metal.
Answer:
Gold.

Question 49.
Name any two metals that have low densities.
Answer:
Lithium, sodium.

Question 50.
How can the pure iron which is very soft be made useful ?
Answer:
Iron is mixed with about 0.05% carbon to make in hard and strong. Thus, it can be made useful.

Question 51.
What happens when calcium is treated with water ?
Answer:
Calcium reacts less violently with water and bubbles of hydrogen gas stick to its surface.

Question 52.
What is the nature of non-metal oxide ?
Answer:
Non metal oxides are acidic or neutral in nature.

Question 53.
What is the nature of metal oxide ?
Answer:
Metal oxides are basic in nature.

Question 54.
What is flux ?
Answer:
Flux is a substance which is added to the ore, during reduction, for removing the non- fusible impurities.

Question 55.
Why cannot aluminium be obtained by reduction of its oxide with carbon ?
Answer:
Aluminium has more affinity for oxygen than carbon.

Question 56.
Metals are refined by using different methods. Name two metals refined by electrolytic refining.
Answer:
Copper and gold.

Question 57.
What is rust ?
Answer:
The coating of brown, flaky substance on the surface of iron when it is kept exposed in moist air is called rust.

Question 58.
What is corrosion ?
Answer:
When the surface of a metal is attracted by air, water and some other substances, it is said to be corroded. This phenomenon is known as corrosion.

Question 59.
Observe the experimental setup carefully and give correct order of reactivity of these metals with dil. HCl:

Answer:
C > B > A > D.

Question 60.
An element ‘M’ has 50% of the electrons filed in the 3^rd shell as in the 2^nd shell. WThat is the atomic number of ‘M’ ?
Answer:
14

Question 61.
A student while burning a magnesium ribbon in air, collected the products in a wet watch glass. What was the new product obtained ?
Answer:
Magnesium hydroxide

Question 62.
Study the diagram given below and identify the gas formed in the reaction.

Answer:
Hydrogen which while burning produces a popping sound.

Question 63.
The diagram shows the reaction between metal and dil. acid.

What is the reason for different behaviour of Mg in test tube B?
Answer:
Mg reacts with dil. HCl to produce H₂ gas which helps in floating.

Question 64.
Two statements are given below as Assertion (A) and Reason (R).
Read the statements and choose the correct option :
Assertion (A) : Alloys are commonly used in electrical heating devices like electric iron and heater.
Reason (R) : Resistivity of an alloy is generally higher than that of its constituent . metals but the alloys have low meting points than their constituent metals.
a) Both (A) and (R) are true and (R) is the correct explanation of (A).
b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
c) (A) is true but (R) is false.
d) (A) is false but (R) is true.
Answer:
c) (A) is true but (R) is false.

Question 65.
Two statements are given below as Assertion (A) and Reason (R).
Read the statements and choose the correct option :
Assertion (A) : Rusting of Iron is endothermic in nature.
Reason (R) : As the reaction is slow, the release of heat is barely evident.
a) Both (A) and (R) are true and (R) is the correct explanation of (A).
b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
c) (A) is true but (R) is false.
d) (A) is false but (R) is true.
Answer:
d) (A) is false but (R) is true.

Question 66.
Assertion (A) : The solution of ionic compounds are good conductors of electricity.
Reason (R): Movement of atoms of elements take place in solution.
a) Both (A) and (R) are true and (R) is the correct explanation of (A).
b) Both (A) and (R) are true but (R) is not the correct explanation of (A).
c) (A) is true, but (R) is false.
d) (A) is false, but (R) is true.
Answer:
c) (A) is true, but (R) is false.

Question 67.
Metals like calcium and magnesium are never found in their free state in nature. Explain.
Answer:
Metals like calcium and magnesium are highly reactive metals therefore they never found in free state in nature. They always found in combined state.

Question 68.
Carbonate of metal ‘X’ is abundant in earth crust and its hydroxide is used in ‘white washing’. Identify metal ‘X’.
Answer:
Calcium carbonate is abundant in earth’s crust and calcium hydroxide is used in white washing. Hence, metal X is calcium (Ca).

Question 69.
Write a chemical equation to illustrate the use of aluminium for joining cracked railway lines.
Answer:

Metals and Non-metals Class 10 Important Questions – 2 Marks

Question 1.
Why are ionic compounds usually hard ?
Answer:
In all ionic compounds, their positive and negative ions are attracted to each other by a strong ionic bond. So, they are rigid and hard solids.

Question 2.
Why does aluminium not react with water under ordinary conditions ?
Answer:
Aluminium does not react with water under ordinary conditions because of the presence of a thin layer of aluminium oxide on its surface.

Question 3.
In nature, metal A is found in a free state while metal B is found in the form of its compounds which of these two will nearer to the top of the activity series of metals?
Answer:
Metal B will be nearer to the top of the activity series of metals as it is so reactive that it is formed in combined state.

Question 4.
‘X’ is amphoteric oxide. It is reduced to metal by using suitable reducing agents such as carbon. Identify ‘X’. Write its reaction with acid as well as base.
Answer:
‘X’ is zinc oxide (ZnO)

Question 5.
A metal ‘X’ when added to a solution containing ZnSO₄ shows no change in the colour of the solution. The metal ‘X’ is also used to join railway tracks.
i) Identify the metal ‘X’.
ii) What is the order reactant used in the reaction with ‘X’ to join railway tracks ?
Answer:
i) X is aluminium
ii) FeO₃

Question 6.
The following reaction takes place when aluminium powder is heated with MnO₂
3MnO_2(s) + 4Al_(s) → 3Mn_(l) + 2Al₂O_3(l) + Heat
i) Is aluminium getting reduced ?
ii) Is MnO₂ getting oxidised ?
Answer:
i) No, because oxygen is added to aluminium therefore, it is getting oxidised,
ii) No, since manganese has lost oxygen therefore, it is getting reduced.

Question 7.
Generally, when metals are treated with mineral acids, hydrogen gas is liberated but when metals (except Mn and Mg) are treated with HNO₃, hydrogen is not liberated, why ?
Answer:
It is because HNO₃ is a strong oxidising agent. It oxidises the H₂ produced to H₂O and itself get reduced to any of the oxides of the nitrogen, like NO₂, NO, etc.

Question 8.
When calcium metal is added to water the gas evolved does not catch fire but the same gas evolved on adding sodium metal to water catches fire. Why is it so ?
Answer:
In both cases, the gas evolved is H₂, when calcium reacts with water the heat evolved is not sufficient for hydrogen to catch fire. On the other hand, sodium metals reacts with water violently and in this case a lot of heat is evolved which is sufficient for hydrogen to catch.

Question 9.
Which of the following reactions will not accur ? Give reason.
i) MgSO_{4 (aq)} + Fe_(s) → FeSO_{4 (aq)} + Mg_(s)
ii) MgSO_{4 (aq)} + Cu_(s) → CuSO_{4 (aq)} + Mg_(s)
iii) CuSO_{4 (aq)} + Fe_(s) → FeSO_{4 (aq)} + Cu_(s)
Answer:
Reaction (i) will not occur because Fe is less reactive than Mg. Reaction (ii) will not occur because Cu is less reactive than Mg.

Question 10.
What happens when
i) ZnCO₃ is heated in the absence of oxygen ?
ii) a mixture of Cu₂O and Cu₂S is heated ?
Answer:
i) It undergoes calcination. The chemical reaction can be given as

ii) It undergoes and reduction forming copper and sulphur dioxide

Question 11.
Give the reaction involved during extraction of zinc from its ore by.
i) roasting of zinc ore
ii) calcination of zinc ore
Answer:

Question 12.
Why are food cans tin – plated instead of zinc plated through zinc is cheaper than tin ?
Answer:
Tin is less reactive than zinc. It is less likely to dissolve in the liquid stored in the food cans. Tin reacts only with powerful acids whereas zinc can easily react even with tomatoes, so it is not safe to store food in zinc plated cans.

Question 13.
When a metal X is treated with cold water, it gives a basic salt Y with molecular formula XOH (Molecular mass = 40) and liberates a gas Z which easily catches fire? Identify X, Y and Z and also write the reaction involved.
Answer:
Sodium (Na) and Potassium (K) react with cold water to form basic salt NaOH and KOH respectively. The molecular mass of NaOH is 40. So, X is Na and Y is,NaOH. The gas liberated during the reaction is hydrogen (H₂. So Z is H₂.
2Na + 2H₂O → 2NaOH + H₂ + Heat energy

Question 14.
The three metals X, Y and Z, X reacts with cold water, Y with hot water and Z with steam only. Identify X, Y and Z and also arrange them in order of increasing reactivity.
Answer:
X is an alkali metal, Na or K
Y is an alkali earth metal, Mg or Ca
Z is Fe, Al or Zn
Therefore increasing order of reactivity Fe < Mg < Na.

Question 15.
An element A burns with golden flame in air. It reacts with another element B, atomic number 17 to give a product C. An aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen. Identify A, B, C and D. Also write down the equation for the reactions involved.
Answer:

Question 16.
Compound X and aluminium are used to join railway tracks,
i) Identify the compound X
ii) Name the reaction
iii) Write down its reaction.
Answer:
i) The compound, X is Fe₂O₃
ii) Thermite reaction
iii) Fe₂O_3(s) + 2Al_(s) → 2Fe_(l) + Al₂O_3(s) + Heat

Question 17.
During extraction of metals, electrolytic refining is used to obtain pure metals.
i) Which material will be used as anode and cathode for refining of silver metal by this process ?
ii) Suggest a suitable electrolyte also.
iii) In this electrolytic cell, where do we get pure silver after passing electric current?
Answer:
i) Anode : Impure silver
Cathode: Pure silver
ii) Electrolyte : Silver salt, such as AgNO₃, AgCl, etc.
iii) We get pure silver at cathode.

Question 18.
Iqbal treated a lustrous, divalent element M with sodium hydroxide. He observed the formation of bubbles in reaction mixture. He made the same observations when this element was treated with hydrochloric acid. Suggest how can he identify the produced gas. Write chemical equation for both reactions.
Answer:
The element is a metal
M + 2NaOH → Na₂MO₂ + H₂↑
M + 2 HCl → MCl₂ + H₂↑
The produced gas can be identified by bringing a burning match-stick near the reaction vessel the gas burns with a pop sound. Thus confirms that the gas is H₂.

Question 19.
Write the names and symbols of the constituents present in the following alloys.
a) Brass
b) Bronze
c) Solder
Answer:
a) Brass -Copper (Cu), Zinc (Zn)
b) Bronze – Copper (Cu), Tin (Sn)
c) Solder – Lead (Pb), Tin (Sn)

Question 20.
Electrical wires are coated with plastic. Why ?
Answer:
Electrical wires are made up of copper. Copper reacts with moist carbon dioxide in the air and slowly loses its shiny brown surface and gains a green coat. So, these are coated with plastic.

Question 21.
Why is potassium kept immersed in kerosene ?
Answer:
Potassium is a very reactive metal and combines explosively with air at room temperature., Hence, it catches fire if kept in open. Therefore, to prevent accidental fires and accidents, potassium is kept immersed in kerosene.

Question 22.
What is electrical conductance ?
Answer:
It is the property due to which electric current can pass through the metal. It is due to presence of free electrons or mobile electrons. Eg. Copper, silver, gold, aluminium are good conductors of electricity.

Question 23.
A non-metal ‘X’ exists in two different forms Y and Z, Y is the hardest natural substance, whereas Z is a good conductor of electricity. Identify X, Y and Z.
Answer:
‘X’ is carbon, ‘Y’ is diamond as it is the hardest natural substance and ‘Z’ is graphite as it is good conductor of electricity.

Question 24.
When metal reacts with nitric acid, H₂ is not evolved why ?
Answer:
HNO₃ is good oxidising agent, therefore, it oxides H₂ to H₂O and itself gets reduced to N₂O, NO and NO₂ depending upon concentration of acids.

Question 25.
From amongst the metals sodium, calcium, aluminium, copper and magnesium, name the metal.
a) which reacts with water only on boiling, and
b) another which does not react even with steam ?
Answer:
a) Magnesium reacts with water only on boiling.
b) Copper does not react even with steam.

Question 26.
Why are electrical wires covered with PVC ?
Answer:
PVC acts as insulator, therefore, electric wires are coated with plastic so that we donot get electric shock on touching wires.

Question 27.
Which one of the methods gi ven in column I are used for extraction of each of the metals given in column II.

Column – I	Column – II
i) Electrolytic reduction	Al, Zn
ii) Reduction with carbon	Na, Fe
iii) Reduction with Al	Mn, Sn

Answer:
i) Electrolytic reduction is used in case of Al, Na.
ii) Reduction with carbon is done in case of Zn, Fe, Sn.
iii) Reduction with Al is carried out in case of Mn.

Question 28.
Give reason for the following
a) School bells are made up of metals
b) Electric wire are made up of copper.
Answer:
a) It is because metals are sonorous, i.e., produce sound when struck with a hard substance.
b) It is because copper is good conductor of substance.

Question 29.
Give two examples each of the metals that are good conductors and poor conductors of heat respectively.
Answer:
Good conductors of heat are copper and silver.
Poor conductors of heat are lead and mercury.

Question 30.
X + YSO₄ → XSO₄ + Y
Y + XSO₄ → No reaction
Out of the two elements, ‘X’ and ‘Y’, which is more reactive and why ?
Answer:
‘X’ is more reactive than ‘Y’ because it displaces ‘Y’ from its salt solution.

Question 31.
Why does galvanised iron remain protected against rusting even if the zinc coating is broken ?
Answer:

1. Galvanised iron remains protected against rusting because of a phenomenon called cathodic protection.
2. When the zinc coating is broken, exposing the iron beneath, the zinc acts as a sacrificial anode.
3. The zinc corrodes instead of the iron, preventing rust from forming on the iron surface.

Question 32.
A teacher asks her students to identify a metal, M. She gives them the following clues to help them.
P) Its oxide reacts with both HCl and NaOH
Q) It does not react with hot or cold water but reacts with steam.
R) It can be extracted by electrolysis of its ore.
a) Identify the metal.
b) Write the chemical equations for the reaction of the metal with HCl and NaOH respectively.
c) What would happen if the metal is reacted with iron oxide ?
Answer:
a) Aluminium
b) Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O
Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
c) It would displace iron to form aluminium oxide.

Question 33.
a) Name a non-metallic element which conducts electricity.
b) Give reasons for the following :
i) Metals conduct electricity.
ii) Reaction of nitric acid with metals generally does not evolve hydrogen gas.
Answer:
a) Carbon in the form of graphite conducts electricity, used as electrodes in electrolytic cells and dry cells.

b) i) Metals conduct electricity due to presence of free electrons which can easily move.
ii) It is because nitric acid good oxidising agent and it oxidises H₂ to H₂O.

Question 34.
Write combination reactions that occur when the metal barium reacts with the fol¬lowing non-metals.
a) Hydrogen
b) sulfur
c) nitrogen
Answer:
a) Ba (s) + H₂ (g) → BaH₂ (s)
b) 8 Ba (s) + S (g) → 8 BaS (s)
c) 3 Ba (s) + N₂ (g) → Ba₃N₂ (s)

Question 35.
a) Name the chief ore of iron. Write its formula.
b) How is an iron ore concentrated ? Describe it briefly.
Answer:
a) Hematite is chief ore of iron. Its formula is Fe₂O₃. xH₂O
b) It is concentrated by hydraulic washing. The ore is washed with stream of water under high pressure. Lighter impurities are washed away whereas heavier minerals are left behind.

Question 36.
Name the anode, the cathode and the electrolyte used in the electrolytic refining of impure copper.
Answer:
Anode – Impure copper
Cathode – Strip of pure copper
Electrolyte – Acidified copper sulphate solution

Question 37.
Complete and balance the following chemical equations :
i) Al₂O₃ + HCl →
ii) K₂O + H₂O →
iii) Fe + H₂O →
Answer:
i) Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
ii) K₂O + H₂O → 2KOH
iii) Fe + 2H₂O → Fe(OH)₂ + H₂

Question 38.
How are alloys better than metals ? Give composition of solder and amalgam.
Answer:

1. Alloys are stronger than the metals from which they are made, more resistant to corrosion, have lower melting point, have lower electrical conductivity.
2. Solder is an alloy of lead and tin. An amalgam is cm alloy of mercury with another metal.

Important Questions on Metals and Non-metals Class 10 – 4 Marks

Question 1.
Name a metal / non-metal.
i) Which makes iron hard and strong ?
ii) Which is alloyed with any other metal is make an amalgam ?
iii) Which is used to galvanise iron articles ?
iv) Whose articles when exposed to air form a black coating ?
Answer:
i) Carbon makes iron hard and strong. Tungsten can also make iron hard strong.
ii) Mercury
iii) Zinc
iv) Silver.

Question 2.
State reasons for the following facts :
a) Sulphur is a non-metal
b) Magnesium is a metal.
One of the reason must be supported with a chemical equation.
Answer:

a) Sulphur is a non-metal	b) Magnesium is a metal
1) Poor conductor of heat and electricity	1) Good conductor of heat and electricity.
2) Neither malleable nor ductive	2) Malleable and ductile
3) S + O2 → SO2 Sulphur dioxide is oxide acidic oxide.	3) 2Mg + O2 → 2MgO Magnesium oxide is basic in nature
4) SO2 + H2O → H2SO3 (Sulphurous acid)	4) MgO + Hv2O → Mg(OH)2 (Magnesium hydroxide)

Question 3.
a) What are amphoteric oxides ? Choose the amphoteric oxides from amount the following oxides
Na₂O, ZnO, Al₂O₃, CO₂, H₂O
b) Why is that non-metals do not displace hydrogen from dilute acids ?
Answer:
a) Those oxides which are both acidic as well as basic are called amphoteric oxides, e.g. Al₂O₃, ZnO.
b) It is because non-metals do not supply electrons to change H⁺ ions into hydrogen gas.

Question 4.
An ore on heating in air produces sulphur dioxide. Which process whould you suggest for its concentration ? Describe briefly any two steps involved in the convertion of this concentrated ore into related metal.
Answer:
It is concentrated by froth – floatation process.
i) Roasting : The concentrated sulphide ore is heated strongly in the presence of oxygen to convert it into its oxide.
2ZnS_(s) + 3O_2(g) → 2ZnO_(s) + 2SO_2(g)

ii) Reduction : This oxide of metal is reduced with suitable reducing agent to get free metal.

Question 5.
Suggest a method of reduction for the following metals during their metallurgical processes
a) Metal ‘A’ which is one of the last second or third position in the reactivity.
b) Metal ‘B’ which gives vigorous reaction even with water and air.
c) Metal ‘C’ which is kept in the middle of activity series.
Answer:
a) ‘A’ can be obtained by chemical reduction by using carbon or carbon monoxide as reducing agent.
b) ‘B’ can be obtained by electrolytic reduction
c) ‘C’ can be reduced by reducing agent like ‘Al’.

Question 6.
a) Differentiate between roasting and calcination. Explain the two with the help of suitable chemical equations. How is zinc extracted from its ore ?
b) Name two metals that can be used to reduce metal oxides to metals.
Answer:

Calcination	Roasting
1) It is process in which carbonate ore is heated in absence of air to form oxide.	1) It is process in which sulphide ore is heated in presence of oxygen to convert into oxide.
2)	2) 2ZnS + 3O2 → 2ZnO + 2SO2

By reduction process. Zn can be extracted from its ore.
Reduction : ZnO + C → Zn + CO₂
b) Aluminium, Magnesium.

Question 7.
Two ores ‘A’ and ‘B’ were taken as heating ore A gives CO₂ where as, ore B gives SO₂, What steps will you take to convert them into metals ?
Answer:
Since ore A gives CO₂ and ore B gives SO₂, therefore, ores are MCO₃ and MS
From A

Question 8.
Explain the following:
a) Reactivity of Al decreases if it is dipped in HNO₃
b) Carbon cannot reduce the oxides of Na or Mg
c) Iron articles are galvanised
d) Metals like Na, K, Ca and Mg are never found in their free state in nature.
Answer:
a) Due to the formation of a layer of oxide i.e., Al₂O₃
b) Na or Mg are more reactive metals as compared to carbon
c) To protect from corrosion
d) They are highly reactive.

Question 9.
Here are some general statements. Give one exception for each statement.
i) Metals are lustrous
ii) Metals are solids at room temperature
iii) Metals have high melting points
iv) Non-metals are non-lustrous.
Answer:
i) Sodium
ii) Mercury
iii) Gallium
iv) Iodine

Question 10.
Write chemical equation when zinc granules react with :
i) Sulphuric acid
ii) Hydrochloric acid
iii) Aluminium chloride
iv) Sodium hydroxide
Answer:
i) Zn + H₂SO₄ → ZnSO₄ + H₂
ii) Zn + 2HCl → ZnCl₂ + H₂
iii) Zn + AlCl₃ → No reaction
iv) Zn + 2NaOH → Na₂ZnO₂ + H₂

Question 11.
Name the metal which
a) becomes blakish after certain time
b) develops green coating on exposure to air
c) does not corrode at all
d) is used in making steel
Answer:
a) Silver
b) Copper
c) Platinum
d) Nickel, Chromium

Question 12.
A metal ‘E’ is stored under kerosene oil. When a small piece of it is left open in the air, it catches fire. When the product formed is dissolved in water it turns red litmus blue.
i) Name the metal E
ii) Write the chemical equation for the reaction when it is exposed to air and when the product is dissolved in water
iii) Explain the process by which the metals is obtained from its molten chloride.
Answer:
i) Sodium
ii) 4Na + O₂ → 2Na₂O
Na₂O + H₂O → 2NaOH
iii) Sodium is obtained from sodium chloride by the process of electrolytic reduction Reactions :
At cathode : Na⁺ + e^– → Na
At anode : 2Cl^– → Cl₂ + 2e^–

Question 13.
Give a reason to explain
i) Why do aluminium sheets not corrode easily ?
ii) Why is copper vessel covered with a green coating in rainy season ?
Answer:
i) Aluminium does not corrode easily because on exposure to air, its surface is coated with a thin layer of aluminium oxide which is protective and non-reactive.

ii) It is because copper reacts with moist carbon dioxide in the air and slowly loses its shiny brown surface and gains a green coat.

Question 14.
A compound ‘Z’ is formed by the transfer of electrons from a metal ‘X’ to a nonmetal ‘Y’. Identify the type of bond formed in the compound. List three general properties of the compounds formed by such type of bonds.
Answer:
Ionic compounds are formed in these compounds. The general properties of ionic compounds are:
i) Solid
ii) High melting and boiling point
iii) Soluble in water
iv) Good conductor of electricity

Question 15.
Give reasons for the following
i) Metals conduct electricity
ii) Non-metals in general do not displace hydrogen from dilute acids
iii) Aluminium easily combines with oxygen but still it can be used for making kitchen utensils.
Answer:
i) Metals have electrons which are free to move
ii) A non-metal is an electron acceptor. It cannot supply electrons to H⁺ ions. Hydrogen can only be displaced from dilute acids if electrons are supplied to the H⁺ ions of the acid.
iii) The layer of aluminium oxide formed prevents the metal from further oxidation.

Question 16.
How metals can be differentiated from non-metals on the basis of any three chemical properties ?
Answer:

Metals	Non Metals
1) Burn in air to form metal oxides which are basic in nature.	1) Burn in air to form non-metal oxides which are acidic in nature.
2) React with acids to form salt and hydrogen.	2) They are electron acceptors so cannot produce hydrogen from acids.
3) Metals form compounds with non-metals which are ionic.	3) Non-metals form compounds with non-metals which are covalent.
4) Example : NaCl	4) Example : CCl4

Question 17.
What is thermit reaction ? How is it used to join the railway tracks or cracked machine parts ?
Answer:
The reaction which is highly exothermic and in which metal is produced in molten state is know as thermit reaction. Consider the following displacement reaction.

In this reaction, when iron (II) oxide (Fe₂O₃) is heated with aluminium powder (Al), the amount of heat evolved is so large that the metal iron (Fe) is produced in molten state. Hence, this reaction is used to join railway tracks or craked machine parts.

Question 18.
a) Show the formation of MgO by the transfer of electrons.
b) Name the cation and anion.
c) Write three properties of ionic compounds.
Answer:

b) i) Cation is positively charged (Mg²⁺)
ii) Anion is negatively charged (O^2-)

c) A solution of ionic compounds contain ions since electricity conduction requires movement of charged particles they conduct electricity.
i) Soluble in water
ii) Good conductor of electricity in molten state and aqueous solution
iii) High melting and boiling point.

Question 19.
Mention the metals which are used for the following purpose
i) Domestic utensils and factory equipments.
ii) Jewelleries
iii) Packaging food materials
iv) Transport of electricity
Answer:
i) Iron, aluminium and copper
ii) Gold and silver
iii) Thin foils of aluminium
iv) Copper and Aluminium

Question 20.
Define the term ‘alloy’. State any two advantages of making alloys.
Answer:
An alloy is a uniform mixture of one metal with one, or more metals, or non-metals.
Advantages of making alloys :
i) An alloy increases resistance to corrosion than the combining substances.
ii) An alloy has better tensile strength than the combining substances.

Question 21.
Describe the properties of ionic compounds.
Answer:

1. Physical state : Ionic compounds are solid at room temperature.
2. Hardness : They are somewhat hard and brittle due to strong ion-ion attraction.
3. Melting and boiling points : They have high melting and boiling points.
4. Solubility : Ionic compounds are generally soluble in water but insoluble in kerosene or petrol.
5. Conductivity: They conduct electricity when dissolved in water or in molten form.

Question 22.
Observe the following figure and answer the following questions.

i) What does Figure illustrate ?
Answer:
Figure illustrates a setup for an activity to test the conductivity of heat in metals.

ii) What can be concluded from Figure ?
Answer:
Based on Figure, it can be concluded that metals are good conductors of heat.

iii) What is the significance of Figure ?
Answer:
Figure highlights the ability of metals to conduct heat effectively, which is-important , in various applications where heat transfer is required, such as in electrical wires and cooking utensils.

Question 23.
How does Figure demonstrate that metals are good conductors of electricity ?

Answer:

1. Figure shows an electric circuit set up with a bulb, terminals A and B, and a metal to be tested.
2. When the metal is placed between terminals A and B in the circuit, the bulb glows. This indicates that the electric current is able to flow through the metal.
3. Since the bulb requires electricity to light up, the glowing bulb confirms that the metal is conducting electricity.
   Therefore, Figure demonstrates that metals are good conductors of electricity.

Question 24.
What does Figure illustrate about the reaction of metals with steam ?

Answer:
Figure shows the action of steam on a metal.

1. It demonstrates the reaction of certain metals with steam.
2. It helps to observe and understand the reactivity of metals with steam.
3. By arranging the apparatus as shown in the figure, the reaction of metals with steam can be observed.
4. It allows us to determine which metals react with steam and which do not.

Question 25.

i) What is the purpose of heating a salt sample on a spatula in Figure ?
Answer:
The purpose of heating a salt sample on a spatula in Figure is to observe whether the sample imparts any colour to the flame.

ii) What can be inferred from the observation of heating a salt sample on a spatula in Figure ?
Answer:
From the observation of heating a salt sample on a spatula in Figure, it can be inferred whether the sample is capable of melting or not.

iii) How does conducting the heating of a salt sample on a spatula in Figure help in understanding the nature of the compound?
Answer:
Conducting the heating of a salt sample on a spatula in Figure helps in understanding the nature of the compound by observing its behaviour when subjected to heat. This observation provides information about the properties such as solubility and conductivity of the compound, which can assist in determining its overall nature.

Question 26.

What are the observations and conclusions from the investigation on the conditions under which iron rusts in tubes A, B, and C ?
Answer:
Tube A contains both air and water.
Tube B contains water, but no air dissolved in it.
Tube C contains dry air.
Observation : The iron nails in tube A rusted, while the nails in tube B and C did not rust.
Conclusion : The presence of both air and water is necessary for iron to rust. If either of them is absent, the rusting process is prevented.

Question 27.
i) Which metal is the most reactive according to the activity series table ?

Answer:
Potassium is the most reactive metal according to the activity series table.

ii) Which metal is the least reactive according to the activity series table ?
Answer:
Gold is the least reactive metal according to the activity series table.

iii) What is the order of reactivity of zinc, iron, and lead according to the activity series table ?
Answer:
According to the activity series table, the order of reactivity from most to least reac-tive is : Zinc, Iron, and Lead.

Question 28.

Ionic compound	Melting point (K)	Boiling point (K)
NaCl	1074	1686
LiCl	887	1600
CaCl2	1045	1900
CaO	2850	3120
MgCl2	981	1685

i) What is the melting and boiling point of NaCl (sodium chloride) ?
Answer:
The melting point of NaCl is 1074 K (degree Kelvin) and the boiling point is 1686 K.

ii) What are the melting and boiling points of LiCl (lithium chloride) ?
Answer:
The melting point of LiCl is 887 K and the boiling point is 1600 K.

iii) What are the melting and boiling points of CaCl₂ (calcium chloride) ?
Answer:
The melting point of CaCl₂ is 1045 K and the boiling point is 1900 K.

Question 29.
The physical and chemical properties of metals and non-metals are explored in this document. The classification of elements as metals or nonmetals is discussed, along with their respective properties. The physical properties of metals, such as metallic luster, malleability, ductility, hardness, and heat and electricity con- ductivity, are explained through various activities. The chemical properties of metals, including their reactions with oxygen, water, and acids, are also examined. Nonmetals are discussed in terms of their production of acidic oxides when dissolved in water. Overall, this document provides a comprehensive overview of the classification and properties of metals and nonmetals.

1. What are the physical properties of metals ?
Answer:
The physical properties of metals include metallic luster, malleability, ductility, hardness, and conductivity of heat and electricity.

2. How can metals be categorized ?
Answer:
Elements can be classified as metals or non-metals based on their properties.

3. What are the chemical properties of metals ?
Answer:
The chemical properties of metals include their reaction with oxygen, water, and acids. Metals react with oxygen to form metal oxides, react with water to produce metal oxides and hydrogen gas, and react with acids to form salts and hydrogen gas.

4. What are the properties of non-metals ?
Answer:
Non-metals produce acidic oxides when dissolved in water, in contrast to metals that produce basic oxides.

Question 30.
The easiest way to start grouping substances is by comparing their physical properties. Let us study this with the help of the following activities. For performing Activities 3.1 to 3.6, collect the samples of following metals – iron, copper, aluminium, magnesium, sodium, lead, zinc and any other metal that is easily available.Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. Clean the surface of each sample by rubbing them with sandpaper and note their appearance again. Metals, in their pure state, have a shining surface. This property is called metallic luster.

i) Which property of metals is described as having a shining surface ?
Answer:
Metallic luster

ii) What are the metals used in Activity to show the luster property?
Answer:
Iron, copper, aluminium, magnesium

iii) What is the term for the process of rubbing the metal samples with sandpaper to clean their surface?
Answer:
Polishing

iv) Which of the following metals has the most metallic luster ?
Answer:
Copper Iron, Copper Aluminium, Magnesium.

Question 31.
A pale green solution of ferrous sulphate was taken in four separate test tubes marked I, II, III and IV. Pieces of Cu, Zn and Al were dropped in test tubes II, III and IV respectively. In which case(s)
a) the colour of ferrous sulphate solution will match with the colour in test tube (I)? Give reason.
b) the colour of ferrous sulphate solution will fade and black mass will be deposited on the surface of the metal ?
Answer:
a) The colour of the ferrous sulphate solution in test tube (1) will match the colour of the solution in test tube II when a piece of copper is dropped in it. This is because copper is less reactive than iron and will not displace iron from ferrous sulphate solution, resulting in no change in the colour of the solution.

b) The colour of the ferrous sulphate solution will fade and a black mass will be deposited on the surface of the metal when a piece of aluminium or zinc is dropped in test tube III or IV. This is because aluminium and zinc are more reactive than iron and will displace iron from the ferrous sulphate solution, forming iron oxide (black mass) on their surfaces.

Question 32.
Reverse of the following chemical reaction is not possible :
Zn_(s) + CuSO_4(aq) → ZnSO_4(aq) + Cu_(s)
Justify this statement with reason.
Answer:

1. The reverse of the given chemical reaction is not possible because it violates the principle of conservation of mass.
2. In the forward reaction, zinc (Zn) displaces copper (Cu) from copper sulphate (CuSO₄) to form zinc sulphate (ZnSO₄) and copper.
3. However, in the reverse reaction, copper cannot displace zinc from zinc sulphate, as copper is less reactive than zinc.
4. Hence, the reverse of this reaction is not feasible.

Question 33.
Name a metal which :
a) is the best conductor of heat.
b) has a very low melting point.
c) does not react with oxygen even at high temperature.
d) is most ductile.
Answer:
a) Metal which is the best conductor of heat is silver.
b) Gallium has a very low melting point.
c) Silver and gold do not react with oxygen even at high temperature.
d) Gold is the most ductile metal.

Question 34.
A clear solution of slaked lime is made by dissolving Ca(OH)₂ in an excess of water. This solution is left exposed to air. The solution slowly goes milky as faint faint white precipitate forms. Explain why a white precipitate forms, support your response with the help of a chemical equation. (OR)
Keerti added dilute Hydrochloric acid to four metals and recorded her observations as shown in the table given below :

Metal	Gas Evolved
Copper	Yes
Iron	Yes
Magnesium	No
Zinc	Yes

Select the correct observation (s) and give chemical equation(s) of the reaction involved.
Answer:
Calcium hydroxide reacts with Carbon dioxide present in the atmosphere to form Calcium carbonate which results in milkiness/white ppt / Formation of Calcium carbonate Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
(OR)
Fe + HCl → FeCl₂/ FeCl₃ + H₂
Zn + HCl → ZnCl₂ + H₂

Question 35.
A student added few pieces of aluminium metal to two test tubes A and B containing aqueous solutions of iron sulphate and copper sulphate. In the second part of her experiment, she added iron metal to another test tubes C and D containing aqueous solution of aluminium sulphate and copper sulphate.
In which test tube Or test tubes will she observe colour change ? On the basis of this experiment, state which one is the most reactive metal and why ?
Answer:

Colour change occurs in test tube A, B and D, no colour change will be observed in test tube (c) because there is no reaction.
Al is most reactive because it displaces copper, zinc and iron from their sulphate salts.

Question 36.
Give reasons for the following:
a) Metals are regarded as electro positive elements.
b) When a piece of Copper metal is added to a solution of zinc sulphate, no change takes place, but the blue colour of copper sulphate fades away when a piece of – zinc is placed in its solution.
Answer:
a) It is because metals can lose electrons easily to form positive ions, therefore, regarded as electropositive elements.

b) It is because copper is less reactive than Zn, therefore, it cannot displace Zn from ZnSO₄ solution.
Cu + ZnSO₄ → No reaction
But when Zn is dipped in copper sulphate solution, it displaces copper from CuSO₄ to form ZnSO₄ which is colourless. Reddish brown copper metal gets precipitated because Zn is more reactive than Cu.

Question 37.
By the transfer of electrons, illustrate the formation of bond in magnesium chlo¬ride and identify the ions present in this compound.
Answer:
Bond of Magnesium chloride –

In MgCl₂, Cation is Magnesium (Mg²⁺)
Anion is Chloride (Cl^–)

Question 38.
An element ‘M’ with electronic configuration 2, 8, 3 combines separately with Cl^–, SO₄^-2 anions. Write the chemical formulae of the compounds formed. Predict with the suitable reason the nature of the bond formed by element ‘M’ in general. How will the electrical conductivity of the compounds formed vary with respect to ‘M’ ?
Answer:
MCl₃; M₂(SO₄)₃
M in general forms Ionjc bond. It can acquire a stable electronic configuration of neon (2, 8) by losing its three valence electrons to form M³⁺ cation.
Compounds formed will conduct electricity in liquid/molten state but in solid state in contrast to ‘M’.

Question 39.
Three pieces of a rust free iron rod are completely coated with following:
i) plastic
ii) oil paint
iii) zinc
An identical scratch is made on each piece, thus exposing the iron. The pieces of. iron are kept exposed to moist air for 10 days and then checked for rust formation.
a) State if rusting will be observed at the point of the scratch on the three iron pieces.
b) Give reasons for your answer in each case.
c) Name the process of applying a protective zinc coating to steel or iron.
Answer:
a) i) Rust will be seen on the plastic coated iron piece.
ii) Rust will be seen on the painted iron piece.
iii) No rust will be seen on the zinc coated iron piece.

b) i) The iron rod is in contact with air and moisture.
ii) The iron rod is in contact with air and moisture.
iii) Zinc is more reactive than iron and gets oxidised in preference to the iron object.

c) Galvanisation.

Question 40.
A piece of iron rusts when it comes in contact with air and moisture. Prakash had two identical shiny iron places P and Q. To prevent the pieces from rusting, he coated piece P with oil paint and he galvanized piece Q with a coat of zinc metal, He noticed that the coatings were not complete and that a small part of the iron was exposed in both the pieces.

What is Prakash likely to observer about the exposed parts of the two iron pieces after some days ? Explain why.
Answer:
a) The exposed part of piece P is rusted.
b) The exposed part of piece Q not rusted.
c) Oil painting prevents rusting only by preventing contact of iron with moist air.
d) Galvanising also protects by zinc getting oxidised in preference to iron as it is more reactive than iron.

Question 41.
Compare in tabular form the reactivities of the following metals with cold and hot water :
a) Sodium
b) Calcium
c) Magnesium
Answer:

Metal	Reaction with Water
a) Sodium	Reacts violently with cold water 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
b) Calcium	Reacts less violently with cold water Ca(s) + 2H2O(l) → CaOH2(aq) + H2(g)
c) Magnesium	Reacts less violently with cold water Mg(s) + 2H2O(l) → MgOH2(aq) + H2(g)

Question 42.
State what would happen if:
i) some zinc pieces are placed in blue copper sulphate solution.
ii) some copper pieces are placed in green ferrous sulphate solution.
in) an iron nail is dipped in a solution of copper sulphate for some time.
Answer:
i) When zinc pieces are put into blue colour copper sulphate solution, Zinc displaces copper from copper sulphate solution. In this process, blue copper sulphate solution turns in to colourless.
Zn(s) + CuSO₄(aq) → Cu(s) + ZnSO₄(aq).

ii) Cu + FeSO₄ → No Reaction

iii) When an iron nail is dipped in copper sulphate solution, then the blue colour of copper sulphate fades gradually and a reddish brown coating is formed on the iron nail.

As iron is more reactive than copper, it displaces copper from copper sulphate solution. Cu is less reactive than Fe, thus, it cannot displace Fe from FeSO₄ solution.

Question 43.
A metal ‘X’ combines with a non-metal ‘Y’ by the transfer of electrons to form a compound Z.
i) State the type of bond in compound Z.
ii) What can you say about the melting point and boiling point of compound Z?
iii) Will this compound dissolve in kerosene or petrol?
iv) Will this compound be a good conductor of electricity?
Answer:
i) The type of bond in compound Z is an ionic bond.

ii) Ionic compounds generally have high melting and boiling points due to the strong electrostatic forces of attraction between the positive and negative ions.

iii) Ionic compounds like compound Z do not dissolve in non-polar solvents like kerosene or petrol. They are only soluble in polar solvents.

iv) No, compound Z will not be a good conductor of electricity in solid state because the ions are held in a fixed position and cannot move. However, it may conduct electricity when dissolved in water or in molten state as the ions become free to move and carry electric charge.

Question 44.
Explain the following :
a) Reactivity of Al decreases if it is dipped in HNO₃
b) Carbon cannot reduce the oxides of Na or Mg
c) NaCl is not a conductor of electricity in solid state whereas it does conduct electricity in aqueous solution as well as in molten state.
d) Iron articles are galvanised.
e) Metals like Na, K, Ca and Mg are never found in their free state in nature.
Answer:
a) Due to the formation of a layer of oxide ie., Al₂O₃
b) Na or Mg are more reactive metals as compared to carbon.
c) In solid NaCl, the movement of ions is not possible due to its rigid structure but in aqueous solution or molten state, the ions can move freely.
d) To protect from corrosion.
e) They are highly reactive.

Question 45.
Distinguish between ‘roasting’ and ‘calcination’. Which of these two is used for sulphide ores and why ?
Answer:
Roasting : It is the process in which sulphide ores of the metals are converted into oxides by heating them in the presence of excess air.
For example : zinc sulphide is converted into zinc oxide by roasting.

Calcination : It is the process in which carbonate ores of the metals are decomposed into oxides by heating them in the absence or limited air.
For example : Zinc carbonate is decomposed into zinc oxide and carbon dioxide by calcination.

Out of roasting and calcination, only roasting is used for sulphide ores. This is because it is easier to obtain metal from its oxide as compared to its sulphide.

Question 46.
On the basis of reactivity metals are grouped into three categories.
i) Metals of low reactivity,
ii) Metals of medium reactivity
iii) Metals of high reactivity
Therefore metals extracted in pure form from their ores on the basis of their chemical properties.
Metals of high reactivity are extracted from their ores by electrolysis of the molten ore.
Metals of low reactivity are extracted from their sulphide ores, which are converted into their oxides. The oxides of these metals are reduced to metals by simple heating.
a) Name the process of reduction used for a metal that gives vigorous reaction with air and water both.
b) Carbon cannot be used as a reducing agent to obtain aluminium from its oxide? Why?
c) Describe briefly the method to obtain mercury from cinnabar. Write the chemical equation for the reactions involved in the process.
Answer:
a) Highly reactive metals like Na and obtained by ‘Electrolytic reduction’.
The metal is deposited at cathode of the electrolytic cell.

b) Carbon cannot be used as a reducing agent to obtain aluminium from its oxide because Al lies up in the reactivity series of metals and for such metals, electrolytic reduction in a suitable method.

c) Cinnabar (HgS) is a sulphide ore of Hg.
Hg lies low in the reactivity series. First, HgS is heated to give HgO which upon further heating, gives Hg.

Question 47.
Draw a labelled diagram of electrolytic refining of copper.
Answer:

Question 48.
Name the following :
i) Metal that can be cut by knife
ii) Lustrous non-metal
iii) Metal that exists in liquid state at room temperature
iv) Most malleable and ductile metal
v) Metal that is best conductor of electricity
vi) Non-metal that can exist in different forms
Answer:
i) Sodium
ii) Iodine
iii) Mercury
iv) Gold
v) Silver
vi) Carbon

Question 49.
A student, took four metals P, Q, R and S and carried out different experiments to study the properties of metals. Some of the observations were.
• All metals could not be cut with knife except metal R.
• Metal P combined with oxygen to form an oxide M₂O₃ which reacted with both acids and bases.
• Reaction with water.
P – Did not react either with cold or hot water but reacted with steam.
Q – Reacted with hot water and the metal started floating.
R – Reacted violently with cold water
S – Did not react with water at all

Based on the above observations answer the following :
i) Out of the given metals, the one which needs to be stored used Kerosene is
a) P
b) R
c) S
d) Q
Answer:
b) The metal which reacted violently with cold water needs to be stored in kerosene.

ii) Out of the given metals, the metal Q is
a) iron
b) zinc
c) potassium
d) magnesium
Answer:
d) Magnesium reacts with hot water and the metal started floating.
Mg + 2H₂O → Mg(OH)₂ + H₂↑
Mg does not react with cold water.

iii) Metal which forms amphoteric oxides is
a) P
b) Q
c) R
d) S
Answer:
a) Metal P combines with oxygen to form an oxide M₂O₃ which reacted with both acids and bases.
Hence, acts as an amphoteric oxide.

iv) The increasing order of the reactivity of the four metals is :
a) P < Q < R < S
b) S < R < Q < P
c) S < P < Q < R
d) P < R < Q < S
Answer:
c) The increasing order of reactivity of metal is S < P < Q < R.

Question 50.
Read the case given below carefully and answer the questions that follow :
Krunal connected a copper plate and an iron plate to the positive and negative terminals of a battery respectively along with a switch. He immersed the plates into beaker containing acidified copper sulphate solution.

Questions:
i) After a few minutes, even before he turned the switch on, he noticed that copper was deposited on the iron plate. This could have been due to
a) electrolysis
b) electroplating
c) a combination reaction
d) a displacement reaction
Answer:
d) a displacement reaction

ii) Which of the following is likely to happen when the current is started ?
a) Iron will be deposited on the copper plate
b) Copper will continue to be deposited on the iron plate.
c) No reaction will occur at the iron plate or at the copper plate.
d) The copper already deposited on the iron plate will go back into the solution,
Answer:
b) Copper will continue to be deposited on the iron plate.

iii) Krunal now replaces the iron plate with a silver plate. He sees that there is no deposition of copper on the silver plate before starting the current.
Which of the following could be the reason ?
a) Silver is more reactive than iron
b) Silver is less reactive than copper
c) Silver is a poorer conductor of electricity than iron
d) Silver is a better conductor of electricity than copper
Answer:
b) Silver is less reactive than copper

iv) What is likely to happen to the concentration of copper sulphate in the solution on passing electric current through the solution in the set up with silver plate ?
a) it will increase
b) it will decrease
c) it will remain the same
d) cannot say without knowing the amount of current passed.
Answer:
c) it will remain the same

v) Which of the following will happen to the weights of the silver and copper plates
after passing the current for same time ?
a) The weight of the silver plate will increase and that of the copper of the copper plate will decrease
b) The weight of the copper plate will increase and that of the silver plate will decrease.
c) Both the plates will decrease in weight.
d) Both the plates will increase in weight.
Answer:
a) The weight of the silver plate will increase and that of the copper of the copper plate will decrease

Question 51.
Two students decided to’investigate the effect of water and air on iron object under identical experimental conditions. They measured the mass of each object before placing it partially immersed in 10 mL of water. After a few days, the objects were removed, dried and their masses were measured. The table shows their results.

Student	Object	Mass of Object before Rusting in g	Mass of the coated object in g
A	Nail	3.0	3.15
B	Thin plate	6.0	6.33

a) What might be the reason for the varied observations of the two students ?

b) In another set up the students coated iron nails with zinc metal and noted that, iron nails coated with zinc prevents rusting. They also observed that zinc initially acts as a physical barrier, but an extra advantage of using zinc is that it continues to prevent rusting even if the layer of Zinc is damaged. Name this process of rust prevention and give any two other methods to prevent rusting.

c) In which of the following applications of Iron, rusting will occur most? Support your answer with valid reason.

A) Iron Bucket electroplated with Zinc
B) Electricity cables having iron wires covered with aluminium
C) Iron hinges on a gate
D) Painted iron fence
Answer:
a) Rusting occurs in both A and B so there is an increase in mass. As the surface area of B is more, extent of rusting is more.
b) Galvanisation Oiling/ greasing/ painting/ alloying/ chromium plating or any other
c) C – Iron hinges on a gate : Iron is in contact with both atmospheric oxygen and moisture/water vapour.

Extra Questions on Metals and Non-metals Class 10 – 8 Marks

Question 1.
Give reasons why :
i) Ionic compounds have high melting and boiling points
ii) Ionic compounds are soluble in water
iii) Ionic compounds are solid and some what hard
iv) The metals have tendency to lose electrons
Answer:
i) Ionic compounds have high melting and boiling points because a considerable amount of energy is required to break the strong inter-ionic attraction.

ii) Ionic compounds are soluble in water because their oppositely charged ion break by the opposite pole of water molecules.

iii) They are solids and are some what hard because of the strong force of attraction between the positive and negative ions.

iv) Metals lose electrons as a tendency to attain a completely filled valence shell.

Question 2.
What happens when
i) A piece of calcium is dropped in water
ii) Sodium is kept exposed in air
iii) Iron nail is dipped in copper sulphate solution
iv) Potassium is dropped in cold water.
Answer:
i) Calcium starts floating because the bubbles of hydrogen gas formed stick to its surface
Ca_(s) + 2H₂O_(l) → Ca(OH)_2(aq) + H_2(g)

ii) In dry air, sodium forms sodium peroxide

If the air is moist and contains a lot of water, sodium reacts violently with the water content of air and forms sodium hydroxide and hydrogen gas, producing a lot of heat
2Na_(s) + 2H₂O_(cl) → 2NaOH_(s) + H_2(g) + Heat

iii) Iron displaces copper from copper sulphate solution
Fe + CuSO₄ → FeSO₄ + Cu

iv) Potassium reacts violently with cold water
2K_(s) + 2H₂O_(l) → 2KOH_(aq) + H_2(g) + Heat energy

Question 3.
What is the purpose of making alloys ?
Answer:
Purpose of making Alloy are as follows:

1. Alloys do not get corroded or corroded to very less extent.
2. They are harder and stronger than pure metal, e.g. gold mixed with copper is harder than pure gold.
3. They have less conductance than pure metals, e.g. copper is good conductor of heat and electricity whereas brass and bronze are not good conductors.
4. Some alloys have lower melting point than pure metals, e.g. solder is an alloy of lead.

Question 4.
a) In the formation of compound between two atoms A and B, A loses two electrons and B gains one electron.
i) What is the nature of bond between A and B ?
ii) Suggest the formula of the compound formed between A and B.
b) Explain the formation of MgCl₂ molecule.
c) Common salt conducts electricity only in the molten state. Why?
d) Why is melting point at NaCl high?
Answer:
a) i) Ionic bond
ii)

b) Mg → Mg²⁺ + 2e^–
2Cl + 2e^– → 2Cl^–

c) Na⁺ and Cl^– are free to move in molten state but not in solid state.
d) It is due to strong force of attraction between Na⁺ and Cl^–

Question 5.
a) Write the steps involved in the extraction of pure metals in the middle of the activity series from their carbonate ores.
b) How is copper extracted from its sulphide ore ?
Explain the various steps supported by chemical equations. Draw labelled diagram for the electrolytic refining of copper.
Answer:
a) i) Concentration of ore
ii) Calcination
iii) Reduction
iv) Purification

b) i) Ore of copper is concentrated by forth floatation process
ii) Roasting
2CuFeS₂ + O₂ → Cu₂S + 2FeS + SO₂
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂
2FeS + 3O₂ → 2FeO + 2SO₂
FeO + SiO₂ → FeSiO₃

iii) Smelting 2Cu₂O + Cu₂S → 6Cu + SO₂

iv) Electrolytic refining : Impure copper is taken as anode, pure copper is taken as cathode. Acidified CuSO₂ is taken as electrolyte. Impure copper changes into Cu²⁺ which gain electron at cathode forming pure Cu. Impurities are left behind as anode mud.
At anode Cu (impure) → Cu²⁺ + 2e^–
At cathode Cu²⁺ + 2e^– → Cu (pure)

Question 6.
Distinguish between metals and non-metals on the basis of (i) two physical and (ii) three chemical properties.
Answer:

Metals	Non-metals
Physical 1) Metals are malleable and ductile.	1) Non-metals are brittle.
2) Metals are good conductors of heat and electricity.	2) Non-metals do not conduct heat and electricity.
Chemical 1) Reactive metals displace hydrogen from dilute acids.	1) Non-metals do not displace H, from dilute acids.
2) Metallic oxides are basic in nature.	2) Non-metallic oxides are acidic in nature.
3) Metal can lose electron to form positive ions.	3) Non-metals can gain electrons to form negative ions.

Question 7.
Describe the reaction of metals with dilute acids.
Answer:

1. When metals react with dilute acids, they produce a salt and hydrogen gas.
2. Some metals react more vigorously than others.
3. Magnesium is highly reactive with dilute hydrochloric acid, producing lots of bubbles and heat.
4. The reactivity decreases in the order of magnesium > aluminium> zinc > iron.
5. Copper does not react with dilute hydrochloric acid, and no bubbles or temperature change is observed.
6. In the case of nitric acid, hydrogen gas is not evolved because HNO₃ is a strong oxidizing agent.
7. Magnesium and manganese react with very dilute nitric acid to evolve hydrogen gas.

Question 8.
Explain the reactivity series of metals. Provide a clear answer in 8 to 10 simple bullet points.
Answer:

1. The reactivity series is a list of metals arranged in order of decreasing reactivity.
2. At the top of the series are the most reactive metals, while the least reactive ones are at the bottom.
3. Potassium and sodium are among the most reactive metals and can even catch fire when exposed to air or water.
4. Metals like calcium and magnesium are also highly reactive.
5. Aluminium, zinc, and iron follow, showing moderate reactivity.
6. Metals like copper, mercury, silver, arid gold are at the bottom of the series and are the least reactive.
7. The reactivity series helps determine which metal can displace another from its solution in a displacement reaction.
8. It provides valuable information about the relative reactivities of metals.

Question 9.
Discuss the reactions of metals with oxygen.
Answer:

1. Most metals combine with oxygen to form metal oxides when heated.
2. The reaction between a metal and oxygen is a combustion reaction.
3. Magnesium bums in air with a dazzling white flame.
4. Some metals, like potassium and sodium, are So reactive that they catch fire when exposed to air.
5. Metals like aluminium, zinc, and lead develop a thin layer of oxide on their surface, preventing further oxidation.
6. Iron does riot burn on heating, but iron filings can burn vigorously.
7. Copper does not burn but forms a black layer of copper(II) oxide when hot.
8. Silver and gold do not react with oxygen even at high temperatures.

Question 10.
Explain the properties of amphoteric oxides with examples.
Answer:

1. Amphoteric oxides are metal oxides that exhibit both acidic and basic properties.
2. They can react with both acids and bases to form salts and water.
3. Aluminium oxide (Al₂O₃) is an example of an amphoteric oxide.
4. When reacted with hydrochloric acid, it forms aluminium chloride (AlCl₃ and water.
5. When reacted with sodium hydroxide, it forms sodium aluminate (NaAlO₂ and water.
6. Zinc oxide (ZnO) is another example of an amphoteric oxide.
7. These oxides are versatile and can participate in various chemical reactions.
8. They show characteristics of both acids and bases, depending on the reaction’ conditions.

Question 11.
Explain the methods used for preventing the corrosion of iron.
Answer:

1. Painting : Applying a protective layer of paint on iron surfaces prevents direct contact with air and moisture, thus inhibiting rust formation.
2. Oiling and greasing : Coating iron with oil or grease creates a barrier that prevents air and moisture from coming into contact with the metal.
3. Galvanising : Coating iron or steel with a thin layer of zinc (galvanisation) provides a sacrificial layer that corrodes instead of the underlying metal. Even if the zinc coating is damaged, it continues to protect the iron or steel.
4. Chrome plating : Electroplating iron or steel with chromium creates a corrosion- resistant surface.
5. Anodising : This process forms an oxide layer on the surface of aluminium, protecting it from corrosion.
6. Making alloys : Mixing iron with other elements, such as carbon, nickel, and chromium, can alter its properties. For example, stainless steel, formed by mixing iron, nickel, and chromium, is both strong and rust-resistant.

Question 12.
Explain why alloys like brass and bronze have different electrical conductivities compared to their constituent metals.
Answer:

1. Alloys like brass (copper and zinc) and bronze (copper and tin) have different electrical conductivities compared to their constituent metals due to changes in their atomic structures.
2. The introduction of additional elements disrupts the regular arrangement of atoms present in pure metals.
3. This disruption hinders the flow of electrons through the alloy, reducing its electrical conductivity.
4. In the case of brass and bronze, the added zinc and tin atoms create irregularities in the copper lattice, making them less effective conductors of electricity.
5. Copper, in its pure state, has excellent electrical conductivity, which is why it is used for making electrical circuits.
6. In contrast, alloys like brass and bronze are chosen for their combination of specific properties, such as strength and corrosion resistance, rather than their electrical conductivity.

Question 13.
Metals and non-metals have different physical and chemical properties. Metals are characterized by their metallic luster, malleability, ductility, hardness, and conductivity of heat and electricity. Non-metals, on the other hand, have proper¬ties such as dullness, brittleness, low conductivity, and low melting points. These properties play a significant role in the uses and applications of metals and non-metals in our daily lives.

i) What are some physical properties of metals ?
Answer:
Metallic luster, malleability, ductility, hardness, and conductivity of heat and electricity.

ii) What are some physical properties of non-metals ?
Answer:
Dullness, brittleness, low conductivity, and low melting points.

iii) How do the physical properties of metals and non-metals differ ?
Answer:
Metals have metallic luster, are malleable and ductile, hard, and have high conductivity, while non-metals are dull, brittle, and have low conductivity and low melting points.

iv) What are the main differences in the physical and chemical properties of metals and non-metals?
Answer:
Metals have properties such as metallic lustre, malleability, ductility, hardness, and conductivity of heat and electricity. Non-metals, on the other hand, have properties such as dullness, brittleness, low conductivity, and low melting points. These differences in properties contribute to the various uses and applications of metals and non-metals in our daily lives.

Question 14.
a) Write electron dot structures of Ca (At. No. 20) and O (At. No. 8)
b) Show the formation of calcium oxide by transfer of electrons.
c) Name the ions present in these compounds.
d) List four important characteristics of these compound.
Answer:

c) The ions present are:
Ca²⁺ – Calcium ion which is a cation.
O^2- – Oxide ion which is an anion.

d) CaO is an ionic compound.
It properties:

1. High melting and boiling points
2. Imparts specific colours to the flame
3. Insoluble in organic C solvents like petrol, kerosen
4. Conducts electricity in molten and aqueous state.

Question 15.
Name the following :
a) A metal that catches fire in open air and gives off white fumes.
b) A metal that forms two types of oxides and rusts in moisture : Write their formulae also
c) A metal used in stainless steel
d) A metal used in long distance cables wires.
e) A metal added to gold to harden it.
Answer:
a) The metal that catches fire in open air and gives off white fumes is sodium.

b) The metal that forms two types of oxides and rusts in moisture is iron. The formulas of its oxides are FeO; Fe₃O₃

c) The metal used in stainless steel is chromium besides iron and carbon as it increases corrosion resistance character of steel.

d) The metal used in long distance cables wires is aluminium, because it,is a light metal and a very good conductor of electricity.

e) The metal added to gold to harden it is copper.

Question 16.
Write balanced chemical equations to explain what happens, when
i) Mercuric oxide is heated.
ii) Mixture of cuprous oxide and cuprous sulphide is heated.
iii) Aluminium is reacted with manganese dioxide.
iv) Ferric oxide is reduced with aluminium.
v) Zinc carbonate undergoes calcination.
Answer:
i) Mercuric oxide decomposes to mercury and oxygen gas on heating.
2 HgO → 2 Hg + O₂

ii) When cuprous oxide is heated with cuprous sulphide, copper metal is obtained.
2Cu₂O + Cu₂S → 6Cu + SO₂

iii) When manganese dioxide reacts with aluminium then it results into the formation of aluminium trioxide and manganese.
4Al + 3MnO₂ → 2Al₂O₃ + 3Mn

iv) When Ferric oxide is heated with aluminium, iron metal is obtained.
Fe₂O₃ + Al → 2Fe + Al₂O₃

v) When zinc carbonate is heated in calcination process it converted to zinc oxide then it can be easily converted to metal by heating with e.
ZnCO₃ → ZnO + CO₂

Question 17.
Draw a schematic diagram of the various steps involved in the extraction of metals from ores for metals of medium reactivity and for metals of low reactivity.
Answer:
Various steps involved in the extraction of a metal from its ore followed by refining of the metal is called ‘metallurgy’. The steps involved are summarised as follows :

Question 18.
How is copper obtained from its ore (Cu₂S) ? Write only the chemical equations. How is copper thus obtained refined? Name and explain the process along with a labelled diagram.
(OR)
Copper produced by heating the ore in air is not very pure. Describe the method used for refining impure copper. Draw labelled diagram of the process.
Answer:
Copper glance (Cu₂S) when heated in air gets partially oxidised to copper oxide which further reacts with the remaining copper glance to give copper metal.

(OR)
Copper obtained is refined by electrolytic refining.
Electrolytic refining of crude copper : Thick block of impure metal acts as anode and a thin strip of pure copper metal acts as cathode. The electrolyse used is aqueous solution of copper sulphate containing a small amount of sulphuric acid. On passing electric current through the electrolyte, the metal from the anode dissolves into the electrolyte. An equivalent amount of copper metal from copper sulphate solution gets deposited on cathode.

Question 19.
The melting points and boiling points of some ionic compounds are given below:

Ionic compound	Melting point (K)	Boiling point (K)
NaCl	1074	1686
LiCl	887	1600
CaCl2	1045	1900
CaO	2850	3120
MgCl2	981	1685

These compounds are termed ionic because they are formed by the transfer of electrons from a metal to a non-metal. The electron transfer in such compounds is controlled by the electric configuration of the elements involved. Every element tends to attain a completely filled valence shell of its nearest noble gas or a stable octet.
i) Show the election transfer in the formation of magnesium chloride.
ii) List two properties of ionic compounds other than their high melting and boiling points.
iii) a) While forming an ionic compound say sodium chloride how does sodium atom
attain its stable configuration ?
b) Give reasons:
i) Why do ionic compounds in the solid state not conduct electricity ?
ii) What happens at the cathode when electricity is passed through an aque¬ous solution of sodium chloride ?
Answer:

ii) Properties of ionic compounds :
a) Ionic compounds are generally soluble in water.
b) Their aqueous solutions conduct electricity.

iii) a) Sodium atom has one electron in its valence shell. Na loses that one valence electron to form the Na” cation and attains a stable configuration of 2,8

b) i) Ionic compounds do not conduct electricity in the solid state because the move¬ment of ions is not possible due to the rigidity of the structures of the solid.
ii) Sodium Chloride (NaCl) is electrolysed in its aqueous solution.
At cathode, reduction of H+ and Na* ions takes place.

Hydrogen gas and sodium metal are obtained at cathode.

Question 20.
Aluminium is a reactive metal but is still used for packing food articles. Give reason.
Answer:

1. Aluminium is a strong and cheap metal.
2. It is also a good conductor of heat.
3. But it is highly reactive.
4. When it is pxposed to moist air, its surface is covered with a thin impervious layer of aluminium oxide (Al₂O₃).
5. This layer does not allow moist air to come in contact with the fresh metal and hence, protects the metal underneath from further damage or corrosion.
6. Thus, after the formation of this protective layer of Al₂O₃, aluminium becomes re-sistant to corrosion. .
7. It is because of this reason that although aluminium is a highly reactive metal, it is still used in food packaging.

Class 10 Science Chapter 13 MCQ Our Environment

MCQ on Our Environment Class 10

Question 1.
Among the following pairs, which one is a mis-matched pair ?
A) Plant – Producer.
B) Rabbit – Primary consumer.
C) Lion – Carnivore.
D) Giraffe – Secondary consumer.
Answer:
D) Giraffe – Secondary consumer.

Question 2.
The correct sequence of a food chain
A) Plants → Grasshopper → Frog → Snake → Hawk.
B) Plant → Frog → Grasshopper → Hawk → Snake.
C) Plant → Hawk → Snake → Frog → Grasshopper.
D) Plant → Snake → Frog → Hawk → Grasshopper.
Answer:
A) Plants → Grasshopper → Frog → Snake → Hawk.

Question 3.
Identify the correct one from the following food chains.
A) Eagle → Rat → Snake → Grains.
B) Grains → Rat → Snake → Eagle.
C) Grains → Eagie → Rat → Snake.
D) Rat → Snake → Grains → Eagle.
Answer:
B) Grains → Rat → Snake → Eagle.

Question 4.
The region of the world of living things is termed as
A) Biosphere
B) Ecosystem
C) Food chain
D) Food web
Answer:
A) Biosphere

Question 5.
The arrows between each Item in a food chain always point from
A) Feeder to food
B) Food to feeder
C) Animals to animals
D) Carnivores to herbivores
Answer:
B) Food to feeder

Question 6.
First tropic level of a ecological pyramid includes
A) Producers
B) Primary consumers
C) Secondary consumers
D) Tertiary consumers
Answer:
A) Producers

Question 7.
This pyramid shows the population of organisms at each tropic level in a food chain
A) Pyramid of biomass
B) Pyramid of number
C) Pyramid of Egypt
D) Pyramid of energy
Answer:
B) Pyramid of number

Question 8.
In a food chain, the third trophic level is always occupied by
A) Producers
B) Primary consumers
C) Carnivores
D) Decomposers
Answer:
C) Carnivores

Question 9.
These organisms helps in recycling of materials in an ecosystem
A) Producers
B) Saprophytes
C) Omnivores
D) Herbivores
Answer:
B) Saprophytes

Question 10.
In the food chain comprising of a snake, grass, insect and frog, the secondary consumer is
A) Insect
B) Frog
C) Snake
D) Grass
Answer:
B) Frog

Question 11.
Arrange the following in sequential order.
1) Rat
2) Snake
3) Grain
4) Eagle
A) 3,2,1,4
B) 3,1,2,4
C) 4,1,2,3
D) 1,3,2,4
Answer:
B) 3,1,2,4

Question 12.
Complete the flow chart.

A) Rat
B) Wolf
C) Mosquito
D) Fly
Answer:
B) Wolf

Question 13.
Proper order of food chain is ……,…….. ( )
A) Eagle → rat → snake → grain
B) Grain → rat → snake → eagle
C) Snake → eagle → rat → grain
D) Rat → snake → grain → eagle
Answer:
B) Grain → rat → snake → eagle

Question 14.
In this food chain, the organism which belongs to the second trophic level is

A) Snake
B) Plant
C) Grasshopper
D) Frog
Answer:
C) Grasshopper

Question 15.
Identify the mismatched pair.
A) Producers i Plants
B) Herbivores – Cow, Rabbit
C) Predators – Lion, Tiger
D) Ecological pyramids – A.G. Tansley
Answer:
D) Ecological pyramids – A.G. Tansley

Question 16.
The primary consumer in the food chain comprising of a snake, insect, frog and grass is
A) Grass
B) Frog
C) Insect
D) Snake
Answer:
C) Insect

Question 17.
Arrange the following in the sequential order with reference to biomass.
1) Small fish
2) Zooplanktons
3) Phytoplanktons
4) Big fish
5) Man
A) 1,3,2,5,4
B) 3,2,1,4,5
C) 1,2,4,3,5
D) 2,1,3,5,4
Answer:
B) 3,2,1,4,5

Question 18.
Complete the flow chart.

A) Tertiary consumers
B) Primary producers
C) Planktons
D) Decomposers
Answer:
A) Tertiary consumers

Question 19.
Plants : Producers :: Frog : …..
A) Primary Consumer
B) Secondary Consumer
C) Tertiary Consumer
D) Top Carnivore
Answer:
B) Secondary Consumer

Question 20.
Arrange the food chain in the correct order.

A) 2,4,5,3,1
B) 1,4,5,2,3
C) 2,5,4,3,1
D) 2,4,5,1,3
Answer:
D) 2,4,5,1,3

Question 21.
Which one of the following acts as primary as well as secondary consumer in the forest ecosystem ?
A) Rabbit
B) Deer
C) Tiger
D) Man
Answer:
D) Man

Question 22.

This is an example for pyramid of biomass of ……….
A) Forest ecosystem
B) Desert ecosystem
C) Grassland ecosystem
D) Aquatic ecosystem
Answer:
D) Aquatic ecosystem

Question 23.
One of the following is not a consumer
A) Deer
B) Kangaroo
C) Dolphin
D) Diatoms
Answer:
D) Diatoms

Question 24.

In the above food chain, fish occupies
A) 1st trophic level
B) 2nd trophic level
C) 3rd trophic level
D) 4th trophic level
Answer:
C) 3rd trophic level

Question 25.

Which of the following has much accumulation of pollutants in this pyramid?( )
A) Diatoms
B) Zooplanktons
C) Fish
D) Man
Answer:
D) Man

Question 26.
Observe the food chain.

A) 15 kJ
B) 20 kJ
C) 10,000 kJ
D) 0.1 kJ
Answer:
D) 0.1 kJ

Question 27.

In the above food chain, this represents the second trophic level
A) Tiger
B) Tree
C) Fish
D) Deer
Answer:
D) Deer

Question 28.
In the above food relationship, which one represents the secondary consumer ?( )

A) Grass
B) Snake
C) Rabbit
D) Eagle
Answer:
B) Snake

Question 29.

This figure represents the following.
A) Pyramid of energy
B) Food web
C) Pyramid of numbers
D) The pyramid of biomass
Answer:
B) Food web

Question 30.
What percent of energy loss can be observed at each trophic level ?
A) 90 %
B) 70%
C) 60%
D) 10%
Answer:
D) 10%

Question 31.
The flow of energy in a food chain is minimum at
A) Herbivores
B) Primary consumers
C) Producers
D) Top carnivores
Answer:
D) Top carnivores

Question 32.
If the energy available at the producer level in a food chain is 200J, how much energy will be available at tertiary consumer level? ( 10% Law)
A) 20 J
B) 2 J
C) 0.2 J
D) 200 J
Answer:
C) 0.2 J

Question 33.
If the energy available at teritiary consumer level is 20 J in a food chain, how much energy will be available at the primary consumer level ? ( 10% Law)
A) 2000 J
B) 1000 J
C) 200 J
D) 20 J
Answer:
A) 2000 J

Question 34.
Arrange the following in the correct order with reference to biomass.
1) Tiger -0.1 KJ
2) Rabbit – 100 KJ
3) Grass – 1000 KJ
4) Fox – 10 KJ
A) 3,2,4,1
B) 1,3,2,4
C) 1,2,4,3
D) 4,1,2,3
Answer:
A) 3,2,4,1

Question 35.
Complete the flow chart.

A) Producers
B) Predators
C) Parasites
D) Trophic level
Answer:
A) Producers

Question 36.
Write the proper order of food chain.
1) Grass
2) Goat
3) Lion
A) 3,2,1
B) 1,3,2
C) 1,2,3
D) 3,1,2
Answer:
C) 1,2,3

10th Class Biology Bits

Well-designed AP Board Solutions Class 7 Maths Chapter 4 Simple Equations Exercise 4.1 offers step-by-step explanations to help students understand problem-solving strategies.

Simple Equations Class 7 Exercise 4.1 Solutions – 7th Class Maths 4.1 Exercise Solutions

Question 1.
Complete the last column of the table.

Solution:
i) x + 3 = 0
∵ x = 3 ⇒ 3 + 3 = 6 ≠ 0 No

ii) x + 3 = 0
∵ x = 0 ⇒ 0 + 3 = 3 ≠ 0 No

iii) x + 3 = 0
∵ x = – 3 ⇒ – 3 + 3 = 0 Yes

iv) x – 7 = 1
∵ x = 7 ⇒ 7 – 7 = 0 ≠ 1 No

v) x – 7 = 1
∵ x = 8 ⇒ 8 – 7 = 1 Yes

vi) 5x = 25
∵ x = 0 ⇒ 5 × 0 = 5 No

vii) 5x = 25
∵ x = 5 ⇒ 5(5) = 25 Yes

viii) 5x = 25
∵ x = – 5 ⇒ 5(-5) = – 25 ≠ 25 No

ix) \(\frac { m }{ 3 }\) = 2
∵ m = – 6 ⇒ \(\frac { -6 }{ 3 }\) = – 2 ≠ 2 No

x) \(\frac { m }{ 3 }\) = 2
∵ m = 0 ⇒ \(\frac { 0 }{ 3 }\) = 0 ≠ 2 No

xi) \(\frac { m }{ 3 }\) = 2
∵ m = 6 ⇒ \(\frac { 6 }{ 3 }\) = 2 Yes

Question 2.
Check whether the value given in the brackets is a solution to the given equation or not :
a) n + 5 = 19(n = 1)
Solution:
n + 5 = 19
∵ n = 1 ⇒ 1 + 5 = 6 ≠ 19
n = 1 is not the solution.

b) 7n + 5=19(n = – 2)
Solution:
7n + 5 = 19 ∵ n = – 2
7n + 5 = 7(- 2) + 5
= – 14 + 5 = – 9 ≠ 19
∴ n = – 2 is not the solution.

c) 7n + 5 = 19 (n = 2)
Solution:
7n + 5 = 19
∵ n = 2
7(2) + 5 = 14 + 5 = 19
∴ n = 2 is the solution.

d) 4p – 3 = 13 (p = 1)
Solution:
4p – 3 = 13 ∵ p = 1
4p – 3 = 4(1) – 3 = 4 – 3 = 1 ≠ 13
∴ p = 1 is not the solution.

e) 4p – 3 = 13(p = – 4)
Solution:
4p – 3 = 13
4p – 3 = 4(-4) – 3 = – 16 – 3 = – 19 ≠ 13
∴ p = – 4 is not the solution.

f) 4p – 3 = 13(p = 0)
Solution:
4p – 3 = 13 ∵ p = 0
4p – 3 = 4(0) – 3 = – 3 ≠ 13
∴ p = 0 is not the solution.

Question 3.
Solve the following equations by trial and error method:
i) 5p + 2 = 17
Solution:
Trial and error method
Let p = 1
LHS 5p + 2 = 5(1) + 2 = 5 + 2 = 7
RHS = 17
∴ LHS ≠ RHS
p = 1 is not the solution

Let p = 2
LHS 5p + 2 = 5(2) + 2 = 10 + 2 = 12
RHS = 17
LHS ≠ RHS
p = 2 is not the solution

Let p = 3
LHS 5p + 2 = 5(3) + 2 = 15 + 2 = 17
RHS = 17
∴ LHS = RHS
∴ p = 3 is the solution.

ii) 3m – 14 = 4
Solution:
Let m = 1
LHS = 3m – 14 = 3(1) – 14 = 3 – 14 = – 11
RHS = 4
LHS ≠ RHS
∴ m = 1 is not the solution

Let m = 3
LHS = 3m – 14 = 3(3) – 14 = 9 – 14 = – 5
RHS = 4
∴ LHS ≠ RHS
∴ m = 3 is not the solution

Let m = 6
LHS 3m – 14 = 3(6) – 14 = 18 – 14 = 4
RHS = 4
∴ LHS = RHS
∴ m = 6 is the solution

Question 4.
Write equations for the following statements :
i) The sum of numbers x and 4 is 9.
Solution:
x + 4 = 9

ii) 2 subtracted from y is 8.
Solution:
y – 2 = 8

iii) Ten times a is 70.
Solution:
10 a = 70

iv) The number b divided by 5 gives 6 .
Solution:
\(\frac { b }{ 5 }\) = 6

v) Three-fourth of t is 15.
Solution:
\(\frac { 3 }{ 4 }\)(t) = 15

vi) Seven times m plus 7 gets you 77 .
Solution:
7m + 7 = 77

vii) One-fourth of a number x minus 4 gives 4.
Solution:
\(\frac { 1 }{ 4 }\)x – 4 = 4

viii) If you take away 6 from 6 times y, you get 60.
Solution:
6y – 6 = 60

ix) If you add 3 to one-third of z, you get 30 .
Solution:
\(\frac { 1 }{ 3 }\)z + 3 = 30

Question 5.
Write the following equations in statement forms:
i) p + 4 = 15
Solution:
4 is added to p gives 15

ii) m – 7 = 3
Solution:
7 is subtracted from m, the difference is 3 .

iii) 2m = 7
Solution:
Two times of m is 7

iv) \(\frac { m }{ 5 }\) = 3
Solution:
m is divided by 5 gives quotient 3

v) \(\frac { 3m }{ 5 }\) = 6
Solution:
3 times of m divided by 5 gives 6.

vi) 3p + 4 = 25
Solution:
3 times of p is added to 4 to give result 25

vii) 4p – 2 = 18
Solution:
2 is subtracted from 4 times of p gives 18

viii) \(\frac { p }{ 2 }\) + 2 = 8
Solution:
p is divided by 2 and 2 is added the result is 8.

Question 6.
Set up an equation in the following cases :
i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)
Solution:
Let m be the number of Parmit marbles
5 times of m = 5m
7 more than 5m = 5m + 7
According to the problem
5m + 7 = 37

ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)
Solution:
Let Laxmi’s age to be ‘y’ years 3 times Laxmi’s age = 3y years
4 year older than 3y years = 3y + 4
According to the problem 3y + 4 = 49

iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87 . (Take the lowest score to be l)
Solution:
Let the lowest score = l
Twice of l = 2l
Twice of l plus 7 = 2l + 7
According to the problem 2l + 7 = 87

iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be ‘b’ in degrees. Remember that the sum of angles of a triangle is 180 degrees).
Solution:
Let the base angle = b°
Twice the base angle = 2b°

We know that the sum of the angles in a triangle = 180°
2b + b + b = 180°
4b = 180°
b = \(\frac{180}{4}\)
b = 45°