JAMB Chemistry Past Questions & Answers - Page 185

921.

An anhydride is?

A.

a compound wich has no water of crystallization

B.

an oxide whose solution in water has a PH greater than 7

C.

an oxide whose solution in water has a PH less than 7

D.

an oxide that has hydrogen atoms

E.

an amphoteric oxide

Correct answer is A

No explanation has been provided for this answer.

922.

In the redox reaction\( 2Fe^{2+} + CI_2 → 2Fe^{3+} + 2Cl^-\)

A.

Cl2 is reduced because it has lost electrons

B.

Cl2 is reduced because its oxidation number has decreased

C.

Cl2 is reduced because its molecule is change to two ions

D.

Fe2+ is reduced because it has lost electrons

E.

Fe2+ is reduced because it has gained electrons

Correct answer is B

decrease in oxidation number indicates reduction while an increase in oxidation number indicates oxidation hence chloride gas is reduced because its oxidation number changed from o to -1 

923.

In the extraction of aluminum from purified bauxite by electrolysis, cryolite is used because

A.

it makes bauxite a better conductor of electricity

B.

it makes bauxite melt at a lower temperature

C.

it makes aluminium purer

D.

it prevents aluminium from getting oxidized

E.

it protects the carbon electrodes used in the process

Correct answer is B

The bauxite is purified to yield a white powder, aluminium oxide, from which aluminium can be extracted. The extraction is done by electrolysis. ... Instead, it is dissolved in molten cryolite, an aluminium compound with a lower melting point than aluminium oxide.

924.

Which one of these compounds will NOT give an oxygen gas on heating?

A.

Manganese dioxide

B.

Hydrogen peroxide

C.

Zinc nitrate

D.

Sodium nitrate

E.

Ammonium nitrate

Correct answer is E

No explanation has been provided for this answer.

925.

0.1 Faraday of electricity was passed through a solution of copper (ll) sulphate. The maximum weight of copper deposited on the cathode would be [Cu = 64]

A.

64.0g

B.

32.0g

C.

16.0g

D.

6.4g

E.

3.2g

Correct answer is E

CuSO4 \(\to\) Cu2+ + SO42+

Cu2+ + 2e \(\to\) Cu

1 Faraday will deposit \(\frac{64}{2}\)gms of copper

0.1 faraday will deposit \(\frac{64}{2} \times \frac{1}{10} = \frac{64}{20}\)

= \(\frac{32}{10}\) = 3.2gms