JAMB Chemistry Past Questions & Answers - Page 58

286.

3H\(_{2(g)}\)+ N\(_{2(g)}\)⇔ 2NH\(_{3(g)}\) ; H= -ve
In the reaction above, lowering of temperature will

A.

have no effect on the equilibrium position

B.

increase the rate of both the forward and reverse reactions equally

C.

favour the forward reaction

D.

favour the reverse reaction

Correct answer is C

N\(_{2(g)}\) + 3H\(_{2(g)}\) ⇔ 2NH\(_{3(g)}\) In the Haber process, the forward reaction is exothermic and the backward reaction is endothermic. If the temperature is decreased, the yield from the exothermic direction is increased i.e. the forward reaction is increased.

287.

Sieving is a technique used to separate mixtures containing solid particles of

A.

small sizes

B.

large sizes

C.

different sizes

D.

the same size

Correct answer is C

Sieving is a simple and convenient technique of separating particles of different sizes.

288.

If the molecular mass of tetraoxosulphate (VI) acid is 98, calculate its vapour density.

A.

196

B.

49

C.

106

D.

82

Correct answer is B

Molecular mass = Vapour Density x 2

  Vapour Density = \(\frac{\text{Molecular Mass}}{2}\)

  Where Molecular mass = 98g/mol

  Vapour Density = \(\frac{98}{2}\)

  Vapour Density = 49g/mol

289.

Which of these alloys contains copper?

A.

solder

B.

steel

C.

permallory

D.

bronze

Correct answer is D

Bronze is an alloy consisting primarily of copper, commonly with about 12% tin and often with the addition of other metals (such as aluminium, manganese, nickel or zinc) and sometimes non-metals or metalloids such as arsenic, phosphorus or silicon.

290.

Calculate the percentage composition of oxygen in calcium trioxocarbonate(IV) [Ca=40, C=12, O=16]

A.

16

B.

48

C.

40

D.

12

Correct answer is B

Calcium trioxocarbonate (IV) = CaCO\(_3\)

  Percentage of Oxygen = \(\frac{\text{Molar mass of 3O}}{\text{Molar mass of} CaCO_3}\) × 100%

  Percentage of Oxygen = \(\frac{(3 \times 16)}{(40 + 12 + 48)}\)  x 100%

  Percentage of Oxygen = \(\frac{48}{100}\) x 100%

  Percentage of Oxygen = 48%