JAMB Chemistry Past Questions & Answers - Page 186

926.

Indicate which of the following statements is not true as we move from left to right along the periodic Table:

A.

atomic number of elements increases

B.

atomic mass of elements increases

C.

electropositive character of elements increases

D.

electronegative character of elements increases

E.

number of electrons in the outermost orbits of element increases

Correct answer is C

No explanation has been provided for this answer.

927.

When a piece of charcoal enclosed in a cylinder containing air is ignited

A.

the toatl volume of air is increased

B.

the relative amount of oxygen present is increased

C.

the relative amount of nitrogen present is decreased

D.

the relative amount of carbondioxide present is increased

E.

the ratio of oxygen to nitrogen in the system is increased

Correct answer is D

No explanation has been provided for this answer.

928.

The empirical formula of an oxide of nitrogen containing 30.4 per cent of nitrogen is (N=14.0,O=16.0)

A.

N2O2

B.

NO

C.

NO2

D.

N2O

E.

N2O3

Correct answer is C

No explanation has been provided for this answer.

929.

In the electrolysis of dilute sulphuric acid using platinum electrodes, the products obtained at the anode and cathode are:
Anode Cathode

A.

Sulphur hydrogen

B.

Hydrogen oxygen

C.

Oxygen Hydrogen

D.

Hydrogen sulphate ions

E.

Sulphur oxygen

Correct answer is C

Dissociation of sulfuric acid,

\(H_2SO_4→2H^+_{(aq)}+SO^{−2}_{4(aq)}\)

Dissociation of water,

\(H_2O→H^+_{(aq)} + OH^−_{(aq)}\)

At Anode : \(4OH^−→2H_2O_{(l)} + O_{2(g)} + 4e_{−}\)

At Cathode : \(2H^+_{(aq)}  + 2e^− → H_{2(g)}\)

930.

12 g of solid protassium chlorate (KCIO) were added to 40 g of water and heated to dissolve all the solid. As the solution cools, crystals of potassium chlorate at 66 oC. The solubility of potassium chlorate at 66 oC is therefore?

A.

66

B.

33

C.

10

D.

20

E.

30

Correct answer is E

solubility = \(\frac{solute x 100}{solvent}\)

The solubility of KClO3 at 66oC will be \(\frac{12}{40} \times \frac{100}{1}\)

= 30