JAMB Chemistry Past Questions & Answers - Page 20

96.

The pollutant usually presents in a city which generates its electricity from coal?

A.

fog

B.

carbon(ii)oxide

C.

smog

D.

sulphur(iv)oxide

Correct answer is D

In the majority of power plants, sulfur appears due to coal burning which is used to generate electricity.

97.

The reactions below represent neutralization reaction, in which of them is the value of ΔH highest?

A.

CH\(_3\)CH\(_2\)COOH + KOH → CH\(_3\)CH\(_2\)COOK + H\(_2\)O

B.

NH\(_4\)OH + HCL → NH\(_4\) + H\(_2\)O

C.

NaOH + HCL → NaCL + H\(_2\)O

D.

CH\(_3\)COOH + NaOH → CH\(_3\)COONa + H\(_2\)O

Correct answer is C

The enthalpy change of neutralisation involving weak acids are smaller in magnitude compared to that between strong acids and strong bases. This is because weak acids/bases are partially dissociated, which means not all hydrogen ions are free to react with hydroxide ions.

 For strong acids and bases, the enthalpy of neutralisation values are always closely similar with the values -57 and -58 KJ/mol. We know that strong acids completely ionize in water. In case of weak acids, only partial dissociation of its ions occurs.

98.

If the volume of a given mass of a gas at 0ºc is 29.5cm\(^3\). What will be the volume of the gas at 15ºc, given that the pressure remains constant. 

A.

31.6

B.

62.2

C.

32.7

D.

31.1

Correct answer is D

Charles' law is a special case of the ideal gas law in which the pressure of a gas is constant.

Charles's law defines the direct relationship between temperature and volume.

GIVEN DATA: V\(_1\) = 29.5cm\(^3\),  V\(_2\) = ? ,  T\(_1\) = 0ºc or 273k,  T\(_2\) = 15ºc or 288k

\(\frac{V_1}{T_1}\) = \(\frac{V_2}{T_2}\)

V\(_2\) = \(\frac{V_1 x T_2}{T_1}\) → \(\frac{29.5 x 288}{273}\)

V\(_2\) = 31.1cm\(^3\)

99.

In which of the following will hydrogen form ionic compound? 

A.

HCL

B.

NaH

C.

NH\(_3\)

D.

CH\(_4\)

Correct answer is B

NaH is a saline (salt-like) hydride, composed of Na+ and H− ions

100.

2H\(_2\) + O\(_2\) → 2H\(_2\)O

From the equation above, calculate the volume of unreacted oxygen gas if a mixture of 50cm\(^3\) of hydroden and 75cm\(^3\) of oxygen are involved

A.

85cm\(^3\)

B.

50cm\(^3\)

C.

125cm\(^3\)

D.

55cm\(^3\)

Correct answer is B

Since 2 mol H\(_2\) reacted with 1 mol of O\(_2\)

: 50cm\(^3\) of hydrogen reacted with 25cm\(^3\) of oxygen

Remainder oxygen = Supplied amount - Reacted amount → 75cm\(^3\) - 25cm\(^3\)

= 50cm\(^3\)