JAMB Past Questions and Answers - Page 1304

6,516.

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

6,517.

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.

6,519.

Determine the opening cash balance of the club?

A.

N3,070

B.

N2,220

C.

N5,710

D.

N4,320

Correct answer is B

No explanation has been provided for this answer.