2019 - JAMB Chemistry Past Questions and Answers - page 7

61

The oxidation state(s) of nitrogen in ammonium nitrite is/are

A

-3, +3

B

+5

C

-3, +5

D

-3

correct option: a

Ammonium nitrite = NH\(_4\)NO\(_2\)

NH\(_4 ^+\): Let the oxidation number of Nitrogen = x

x + 4 = 1 \(\implies\) x = 1 - 4

x = -3

NO\(_2 ^-\): x - 4 = -1

x = -1 + 4 \(\implies\) x = +3.

hence, 

the oxidation numbers for Nitrogen in Ammonium Nitrite are -3, +3.

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62

Hydrogen bond is a sort of

A
van der waals force
B
dative bond
C
ionic bond
D
a covalent bond
correct option: d

Hydrogen bond is a covalent intermolecular bond that exists between hydrogen and highly electronegative elements like nitrogen, oxygen and fluorine.

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63

Methane is prepared in the laboratory by heating a mixture of sodium ethanoate with soda lime. The chemical constituent(s) of soda lime is/are

A

Ca(OH)\(_2\) and NaOH

B

Na\(_2\)CO\(_3\)

C

NaCl

D

NaOH

correct option: a

Soda-lime is a mixture of caustic soda (NaOH) and lime water (Ca(OH)\(_2\)).

Soda-lime, solid is generally a white to greyish-white coloured solid. It is the mixture of calcium hydroxide and sodium or potassium hydroxide, both corrosive materials. It is noncombustible and soluble in water with release of heat. It is corrosive to metals and tissue.

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64

Which of the following sets of operation will completely separate a mixture of sodium chloride, sand and iodine?

A

addition of water, filtration, evaporation to dryness and sublimation

B

filtration, evaporation to dryness, addition of water and sublimation

C

sublimation, filtration, evaporation and addition of water

D

sublimation, addition of water, filtration and evaporation to dryness

correct option: a

The Process involves:

  • Add water to the mixture and filter it. This will result in a salt solution as the filtrate; a mixture of iodine and sand as the residue.
  • Evaporate the filtrate to dryness to get sodium chloride (salt).
  • Heat the residue to sublime the iodine from the sand.

That is,

Addition of water \(\to\) Filtration \(\to\) Evaporation \(\to\) Sublimation.

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65

200cm\(^3\) of 0.50mol/dm\(^3\) solution of calcium hydrogen trioxocarbonate (IV) is heated. The maximum weight of solid precipitated is

A

10g

B

25g

C

20g

D

15g

correct option: a

Equation: Ca(HCO\(_3\))\(_2\) \(\to\) CaCO\(_3\) (precipitate) + H\(_2\)O + CO\(_2\) (in the presence of heat)

V = 200 cm\(^3\) = 0.2 dm\(^3\)

C = 0.5 mol/dm\(^3\) = 0.5M

N = CV \(\implies\) N = 0.5 \(\times\) 0.2

= 0.1 mole

From the above equation,

1 mole of Ca(HCO\(_3\))\(_2\) gives 100g of CaCO\(_3\)

(1 mole CaCO\(_3\) = 40 + 12 + (3 x 16) = 100g\)

0.1 mole of Ca(HCO\(_3\))\(_2\) gives x g of CaCO\(_3\).

\(\frac{1}{0.10} = \frac{100}{x}\)

\(x = 100 \times 0.10 = 10 g\)

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66

Which of the following conditions will most enhance the spontaneity of a reaction?

A
\(\Delta\)H is negative and greater than T\(\Delta\)S
B
\(\Delta\)H is negative and \(\Delta\)S = 0
C
\(\Delta\)H is positive and less than T\(\Delta\)S
D
\(\Delta\)H is positive and equal to T\(\Delta\)S
correct option: a
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67

X is a substance which liberates CO\(_2\) on treatment with concentrated H\(_2\)SO\(_4\). A warm solution of X can decolorize acidified KMnO\(_4\). X is

A

HCOOH

B

NaHCO\(_3\)

C

FeCO\(_3\)

D

H\(_2\)C\(_2\)O\(_4\)

correct option: d

For X to liberate CO\(_2\), X must be carbonate or an oxalate. Since X decolorizes KMnO\(_4\), X must be an oxalate. 

Hence,

X is H\(_2\)C\(_2\)O\(_4\)

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68

The velocity, V of a gas is related to its mass, M by (k = proportionality constant)

A
V = \(\frac{k}{M}\)
B
V = \(\frac{k}{M^{\frac{1}{2}}}\)
C
V = \(kM^2\)
D
V = \((\frac{k}{M})^{\frac{1}{2}}\)
correct option: b

Recall:

V = \(\sqrt{\frac{3RT}{M}}\)

\(\therefore V \propto \frac{1}{\sqrt{M}}\)

\(V = \frac{k}{\sqrt{M}}\)

V = \(\frac{k}{M^{\frac{1}{2}}}\)

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69

The emission of two successive beta particles from the nucleus \(^{32} _{15} P\) will produce

A
\(^{30} _{14}\)Si
B
\(^{32} _{13}\)Al
C
\(^{32} _{17}\)Cl
D
\(^{32} _{16}\)S
correct option: c

\(^{32} _{15}\)P \(\to\) 2\(^{0} _{-1}\)e + \(^{a} _{b}\)X

32 = 2(0) + a

a = 32

15 = 2(-1) + b

b = 15 + 2

b = 17

\(\therefore\) \(^{a} _{b}\)X = \(^{32} _{17}\)Cl

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70

Hydrogen diffused through a porous plug

A

thrice as fast as oxygen

B

at the same speed as oxygen

C

four times as fast as oxygen

D

twice as fast as oxygen

correct option: c

By applying Graham's Law of Diffusion,

\(\frac{R_1}{R_2} = \sqrt{\frac{m_2}{m_1}}\)

\(\frac{R_H}{R_O} = \sqrt{\frac{m_O}{m_H}}\)

\(\frac{R_H}{R_O} = \sqrt{\frac{32}{2}}\)

\(\frac{R_H}{R_O} = \sqrt{16} = 4\)

\(\therefore R_H = 4 \times R_O\)

Therefore, hydrogen diffuses 4 times faster than oxygen.

Graham's law of effusion was formulated by Scottish physical chemist Thomas Graham in 1848. Graham found experimentally that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles. Wikipedia

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