1.Chemical Reactions and Equations

Class X
Chapter 1.
Chemical Reactions and Equations
Questions Answers (NCERT Textbook)

1.1 Chemical Equations

Q.No.1. Why should a magnesium ribbon be cleaned before burning in air ?
Answer:
Magnesium ribbon is very reactive metal so it easily reacts with air and make a layer of magnesium oxide. The layer of magnesium oxide is quite stable and prevents further reaction of magnesium with air or oxygen. Therefore, magnesium ribbon is cleaned by sand paper to remove the layer of magnesium oxide before burning in air. Magnesium ribbon burns with air and gives magnesium oxide.
The word equation for the reaction would be
Magnesium + Oxygen → Magnesium Oxide.
This word equation may be represented by following chemical equation.
2Mg + O₂ → 2MgO

Q.No.2. Write the balanced equation for the following chemical reactions.
(i) Hydrogen + Chlorine → Hydrogen chloride
Answer:
The above word equation may be represented by following chemical equation
H₂ + Cl₂ → 2HCl

(ii) Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride
Answer:
The above word equation may be represented by
3BaCl₂ + Al₂(SO₄)₃ → 3BaSO₄ + 2AlCl₃

(iii) Sodium + Water → Sodium hydroxide + Oxygen
Answer:
The above word equation may be represented by following chemical equation.
2Na + 2H₂O → 2NaOH + H₂

Q.No.3. Write a balanced chemical equation with symbols for the following reactions.
(i) Solution of barium chloride and sodium sulphate in water react to give insoluble barium sulphate and the solution of sodium chloride.
Answer:
The above equation may be represented by following chemical equation.
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

(ii) Sodium hydroxide solution (in water) reacts with hydrochloric acid solution (in water) to produce sodium chloride solution and water.
Answer:
The above equation may be represented by following chemical equation.
NaOH(aq) + HCl(aq) → NaCl(l) + H₂O(l)

3. Pair of Linear Equation in two variables

Class X
Exercise 3.3

Q.No. 1. Solve the following pair of linear equation by substitution method.
(i) 
x + y = 14
x - y = 4
(ii)
s - t = 3
s/3 + t/2 = 6
(iii)
3x - y = 3
9x - 3y = 9
(iv)
0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v)
√2x + √3y = 0
√3x - √8y = 0
(vi)
3x/2 + 5y/3 = -2
x/3 + y/2 = 13/6

Solution:

(i)
From equation no. (i)
x = 14 - y ------------- (iii)
Now, substitute the value of x in equation no. (ii)
14 - y - y = 4
or, 14 - 2y = 4
or, 2y = 14 - 4 = 10
∴ y  = 10/2 = 5
Now, putting the value of y in equation no. (iii)
 x = 14 - 5 = 9
Hence, the value of x = 5 and y = 9 Ans.

(ii)
From equation no. (i)
s = 3 + t ---------- (iii)
Now, from equation no. (ii)
(3 + t)/3 + t/2 = 6
or, 2(3 + t) + 3t = 6x6
or, 6 + 2t + 3t = 36
or, 6 + 5t = 36
or, 5t = 36 - 6 
or, t = 30/5 = 6
Now putting the value of t in equation no. (iii)
s = 3 + 6 = 9
Hence, the value of s = 9 ant t = 6 Ans.

(iii)
From equation no. (i)
x = (3 + y)/3 ---------------- (iii)
Now, substitute the value of x in equation no. (ii)
9(3 + y)/3 - 3y = 9
9 + 3y - 3y = 9
9 = 9
This is independent of y.
Hence, y has infinite many solution.
Therefore x has infinite many solution.

(iv)

From equation no. (i)
x = (1.3 - 0.3y)/0.2 ----------- (iii)
Now, substitute the value of x in equation no. (ii)
0.4(1.3 - 0.3y)/0.2 + 0.5y = 2.3
or, 2.6 - 0.6y + 0.5y = 2.3
or, 0.1y = 0.3
or, y = 3
Now, putting the value of y in equation no. (iii)
x = (1.3 - 0.9)/0.2
or, x = 2
Hence, x = 2 and y = 4.

(v)
From equation no. (i)
x = -√3y/√2 ------------ (iii)
Now substitute the value of x in equation no. (ii)
-√3x√3y/√2 - √8y = 0
or, 3y - 4y = 0
or, y = 0
Now, putting the value of y in equation no. (iii)
x = 0
Hence, x = 0 and y = 0. Ans.

(vi)
From equation no. (i)
x = (-12 - 10y)/9 -------------- (iii)
Now, substitute the value of x in equation no.(ii)
(-12 - 10y)/27 + y/3 = 13/6
or, -12 - 10y + 9y = 13x27/6
or, -12 - y = 39/2
y = (-24 - 39)/2
y = -63/2
Now, putting the value of y in equation no. (iii)
x = (-12 - 165)/9
x = - 177/9
x = -59/3
x = -59/3 and y = -63/2 Ans.

Exercise 3.4

Q.No.1. Solve the following pair of linear equations by the elimination methods and substitution method:

(i) x + y = 5 and 2x - 3y = 4
(ii) 3x + 4y = 10 and 2x - 3y = 4
(iii) 3x - 5y - 4 = 0 and 9x = 2y + 7
(iv) x/2 + 2y = -1 and x - y/3 = 3

(i) Solution:
Elimination Method
Multiply equation no. (i) by 3 and (ii) by 1
3x + 3y = 15 ------------ (iii)
2x - 3y = 4 --------------- (iv)
Now, add equation no. (iii) and (iv)
5x = 19
x = 19/5
Now, putting the value of x in equation no. (iii)
3x19/5 + 3y = 15
3y = 15 - 57/5
3y = (75 - 57)/5
y = 18/15 = 6/5
Required answer 
x = 19/5 and y = 6/5

(ii)
Solution:
Elimination Method
Multiply the equation no. (i) by 3 and (ii) by 4
9x + 12y = 30 ---------- (iii)
8x - 12y = 16 ----------- (iv)
Now, add the equations
17x = 46
x = 46/17
Now, putting the value of x in equation no.(ii)
92/17 - 3y = 4
3y = 92/17 - 4
3y = (92 - 68)/17
y = 24/51
Required answer 
x = 46/17 and y = 24/51 Ans.

(iii)
Solution:
Elimination Method
3x - 5y = 4 ------------ (i)
9x - 2y = 7 ----------- (ii)
Multiply equation no. (i) by 2 and (ii) by 5
6x - 10y = 8 ----------- (iii)
45x - 10y = 35 -------- (iv)
Now, subtract equation no.(iii) from (iv)
39x = 27
x = 27/39
x = 9/13
Now, substitute the value of x in equation no.(i)
27/13 - 5y = 4
5y = 27/13 - 4
5y = (27 - 52)/13
y = -25/65
y = -5/13
Required Answer
x = 9/13 and y = -5/13 Ans.

(iv)
Solution:
x/2 + 2y = -1 ------- (i)
x - y/3 = 3
Multiply equation no.(i) by 1/3 and (ii) by 2
x/6 + 2y/3 = -1/3 ---------- (iii)
2x - 2y/3 = 6 ---------- (iv)
Now, add equation no. (iii) and (iv)
x/6 + 2x = -1/3 + 6
13x/6 = 17/3
x = 34/13
Now, substitute the value of x in equation no.(ii)
34/13 - y/3 = 3
y/3 = 34/13 - 3
y/3 = (34 - 39)/13
y = -5/39
Required Answer
x = 34/13 and y = -5/39 Ans.
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