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UTME Chemistry 1990 JAMB Past Questions and Answers, Jamb, Pre-degree, Nursing, Waec, Neco

UTME Chemistry 1990 JAMB Past Questions and Answers, Jamb, Pre-degree, Nursing, Waec, Neco

UTME Chemistry 1990 JAMB Past Questions and Answers

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1.  What volume of 11.0 M hydrochloric acid must be dilute to obtain 1 dm3 of 0.05 M acid?

Options

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A) 0.05 dm3

B) 0.10 dm3

C) 0.55 dm3

D) 11.0 dm3

The correct answer is A.

2. If 10.8 g of silver is deposited in a silver coulometer volume of oxygen liberated is?

Options

A) 0.56 dm3

B) 5.60 dm3

C) 11.20 dm3

D) 22.40 dm3

The correct answer is A.

Explanation:

Ag+ + e
H = 108 g Ag
= 0.11 10.8 Ag
4OH → O2 + 2H2O + 4e
4F ↔ 1mole O2
4F = 22.4 dm3 χ 0.1 = 0.56 dm3

3. 0.1 faraday of electricity deposited 2.95 g of nickel during electrolysis of an aqueous solution.

Calculate the number of moles of nickel that will be deposited by 0.4 faraday? (Ni = 58.7)

Options

A) 0.20

B) 0.30

C) 0.04

D) 5.87

The correct answer is A.

Explanation:

0.1F ↔ 2.95 g Ni = 0.4F ↔ 11.8 g Ni

No of moles = (11.8)/(58.2) = 0.20 moles

4. 10.0 dm3 of air containing H2S as an impurity was passed through a solution of Pb(NO3)2 until all the H2S had reacted. The precipitate of PbS was found to weigh 5.02 g. According to the equation: Pb(NO3)2 + H2S → PbS + 2HNO3 the percentage by volume of hydrogen sulphide in the air is?

(Pb = 207, S = 32, GMV at s.t.p = 22.4dm3)

Options

A) 50.2

B) 47.0

C) 4.70

D) 0.47

The correct answer is C.

Explanation:

Pb(NO3)2 + H2S → PbS + 2HNO3
34 g H2S → 239 g pbs
5.02g pbs → (34)/(237g) χ 5.02 g = 0.714 g
34g → 22.4 dm3
=0.714 → (22.4)/34g) χ 100 = 4.70

5. A blue solid, T, which weighed 5.0g was placed on a table. After 8 hours, the resulting pink solid was found to weigh 5.5g. It can be inferred that substance T?

Options

A) is deliquescent

B) is hydroscopic

C) has some molecules of water of crystallization

D) is efflorescent

The correct answer is A.

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