WAEC Past Questions and Answers - Page 1422

7,106.

A metal M forms two types of chlorine, MCl\(_2\) and MCl\(_3\) which of the following laws best explains the relationship between the chlorides? Law of

A.

conservation of mass

B.

definite proportion

C.

multiple proportion

D.

reciprocal proportion

Correct answer is C

Law of multiple proportions, statement that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.

7,107.

A sample of a gas may be identified as chlorine if it turns

A.

damp blue litmus paper red

B.

lime water miky

C.

lead ethanoate paper black

D.

starch iodide paper blue-black

Correct answer is D

A test for chlorine is to place damp starch-iodide paper into a test tube of chlorine. A positive result for chlorine is indicated by the paper turning from white to blue-black.

7,108.

The formula of mercury (l) dioxonitrate (III) is

A.

HgNO3

B.

Hg2NO2

C.

Hg2(NO2)2

D.

Hg(NO3)2

Correct answer is C

No explanation has been provided for this answer.

7,109.

Consider the reaction represented by the following equation: xCH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + zH\(_2\)O. The values of x,y and z respectively, are

A.

2,3 and 4

B.

2, 4 and 3

C.

1,2 and 3

D.

1, 3 and 5

Correct answer is A

xCH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + zH\(_2\)O

Balancing the equation, we have

C = x (on the left) and 2 (on the right)

⇒ x = 2

2CH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + zH\(_2\)O

On the left, H = 8 and on the right = 2z

⇒ 2z = 8 ; z = 4

2CH\(_3\)OH + yO\(_2\) → 2CO\(_2\) + 4H\(_2\)O

On the right, O = 4 + 4 = 8.

On the left, O = 2 + 2y

⇒ 2 + 2y = 8

2y = 6 ; y = 3

x, y, z = 2, 3, 4.

7,110.

The number of sulphur atoms in 3.20g of SO\(_{2(g)}\) is [O = 16.0;S = 32.0; Avogadro constant = 6.02 x 10\(^{23}\)]

A.

3.01 x 1022

B.

6.02 x 1022

C.

6.02 x1023

D.

1.20 x 1024

Correct answer is A

32g of sulphur = 64g / 6.02 x 10\(^{23}\) atoms of SO\(_2\)

xg of sulphur = 3.2g of SO\(_2\) 

: 1.6g of sulphur is present in 3.2g / 3.01 x 10\(^{23}\) atoms of SO\(_2\)