Chemistry questions and answers

Chemistry Questions and Answers

Learn more about the properties, composition, and structure of substances (elements and compounds) with these Chemistry questions and answers. This Test can be used by students preparing for Chemistry in JAMB, WAEC, NECO or Post UTME.

1,301.

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.

1,302.

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.

1,303.

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.

1,304.

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.

1,305.

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\(_2\)