JAMB Chemistry Past Questions & Answers - Page 80

396.

The reaction of halogens with alkanes in the presence of sunlight is an example of

A.

oxidation reaction

B.

addition reaction

C.

hydrogenation reaction

D.

substitution reaction

Correct answer is D

Alkanes undergoes substitution reaction and it is an example of halogenation substitution reaction.

Reaction of halogen with alkane in the presence of sunlight (ultraviolet) is termed halogenation. Halogenation is an example of substitution reaction. In addition, alkanes only undergo substitution reaction but not addition reaction.

397.

Calculate the amount in moles of silver deposited when 9650C of electricty is passed through a solution of silver salt [= 96500 Cmol-1]

A.

0.05

B.

10.80

C.

10.00

D.

0.10

Correct answer is D

m = \(\frac{Mm \times Q}{96500n
}\)

where Q = IT

M = Mm × \(\frac{Q}{96500n}\)

where m = mass

Mm = Molar mass

Q = Quantity of electricity

n = number of change= +1

\(\frac{M}{Mm}\) = mole = \(\frac{mass}{Molarmass}\)

\(\frac{M}{Mm}\) = \(\frac{Q}{96500n}\)

= \(\frac{9650}{96500n}\) × 1

= \(\frac{1}{10}\) = 0.1mol

398.

Tin is unaffected by air at ordinary temperature due to its?

A.

Low melting point

B.

Weak electropositive character

C.

High boiling point

D.

White lustrous appearance

Correct answer is A

Tin has a melting point of 232° which enables it to be unaffected by air at ordinary temperature coupled with the fact that it also helps in making a good metal for alloying.

Tin is relatively unaffected by both water and oxygen at room temperature due to its low melting point. It does not rust, corrode, or react in any other way. This explains one of its major uses: as a coating to protect other metals.

399.

The mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s is – (Ag = 108 F = 96500(mol-1)

A.

54.0g

B.

27.0g

C.

13.5g

D.

108.0gQ

Correct answer is A

Recall that

mass deposited = \(\frac{MmIt}{96500n}\)

Mm =108, t = 4830s

I = 10A, n = 1

m = 108 × 10 × (\(\frac{4830}{96500}\)) × 1

m = 54.0g

400.

The shape of ammonia molecules is

A.

trigonal planar

B.

octahedral

C.

square planar

D.

tetrahedral

Correct answer is A

Ammonia = NH3

It has three bonds of hydrogen to the Nitrogen.