JAMB Chemistry Past Questions & Answers - Page 51

251.

A radioactive nucleus has a half-life of 20 years, starting with 100,000 particles, how many particles will be left exactly at the end of 40 years

A.

75,000 particles

B.

35,000 particles

C.

25,000 particles

D.

50,000 particles

Correct answer is C

Given:

t\(\frac{1}{2}\) = 20 years

After the first 20 years, half of the substance (\(\frac{1}{2} \times 100,000 = 50,000\)) will have decayed. Hence, we have 100,000 - 50,000 = 50,000 particles left.

After the second 20 years (being 40 years in all), half of the remaining substance (\(\frac{1}{2} \times 50,000 = 25,000\)) will have decayed. 

Remaining particles after 40 years = 50,000 - 25,000

= 25,000 particles.

252.

Na\(_2\)CO\(_3\) + 2HCl \(\to\) 2NaCl + H\(_2\)O + CO\(_2\)

The indicator most suitable for this reaction should have a pH equal to

A.

5

B.

7

C.

3

D.

9

Correct answer is C

Methyl orange is the best indicator for the reaction with range 3.1 - 4.4.

253.

The molecular shape and bond angle of water are respectively

A.

linear, 180°

B.

bent, 109.5°

C.

tetrahedral, 109.5°

D.

bent, 105°

Correct answer is D

The shape of water molecule = Bent/ V- shaped The bond angle of water = 104.5°/ 105°

254.

Which of the following properties increases from left to right along the period but decreases down the group in the Periodic Table?
i. Atomic Number   ii. Ionization energy  iii. Metallic character  iv. Electron affinity

A.

ii and iii only

B.

ii and iv only

C.

iii and iv only

D.

i and ii only

Correct answer is B

Ionization energy and electron affinity increase across a period, and decrease down a group.

255.

At 27°C, 58.5g of sodium chloride is present in 250cm\(^3\) of a solution. The solubility of sodium chloride at this temperature is?
(molar mass of sodium chloride = 111.0gmol\(^{-1}\))

A.

2.0 moldm\(^{-3}\)

B.

0.25 moldm\(^{-3}\)

C.

1.0 moldm\(^{-3}\)

D.

0.5 moldm\(^{-3}\)

Correct answer is A

Given the Mass of the salt = 58.5g

Volume = 250 cm\(^3\) = 0.25 dm\(^3\)

Mass concentration = \(\frac{Mass}{Volume}\)

= \(\frac{58.5}{0.25}\) = 234 gdm\(^{-3}\)

Solubility (in moldm\(^{-3}\) = \(\frac{234}{111} \)

= 2.11 moldm\(^{-3}\)

\(\approxeq\) 2.0 moldm\(^{-3}\)