2017 - JAMB Chemistry Past Questions and Answers - page 4
31
A
fused CaO
B
H2O
C
NaOH
D
Concentrated H2SO4
correct option: d
In the preparation of sulphur dioxide by the action of dilute acids on sulphates and bisulphites. conc H2SO4 helps to release SO2 from the mixture.
The setup represents the production of sulphur dioxide
Users' Answers & CommentsThe setup represents the production of sulphur dioxide
32
Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol-1]
A
0.300g
B
0.250g
C
0.2242g
D
0.448g
correct option: d
M = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)
= \(\frac{MmIT}{96500n}\)
Copper II Chloride = CuCl2
CuCl2 → Cu2+ + 2Cl2
Mass of compound deposited = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)
Q = IT
I = 0.5A
T = 45 × 60
T = 2700s
Q = 0.5 × 2700
= 1350c
Molarmass = 64gmol-1
no of charge = + 2
Mass = \(\frac{64 \times 1350}{96500 \times 2}\)
Mass = 0.448g
Users' Answers & Comments= \(\frac{MmIT}{96500n}\)
Copper II Chloride = CuCl2
CuCl2 → Cu2+ + 2Cl2
Mass of compound deposited = \(\frac{\text{Molar mass × Quantity of Electricity}}{\text{96500 × no of charge}}\)
Q = IT
I = 0.5A
T = 45 × 60
T = 2700s
Q = 0.5 × 2700
= 1350c
Molarmass = 64gmol-1
no of charge = + 2
Mass = \(\frac{64 \times 1350}{96500 \times 2}\)
Mass = 0.448g
33
C3H5 - COH3 =CH2
The IUPAC nomenclature of the structure above is
The IUPAC nomenclature of the structure above is
A
3-methybut-3-ene
B
2-methylbut-1-ene
C
2-ethylprop-1-ene
D
2-methylbut-2-ene
correct option: b
CH2CH2 - CCH3CH2
Start the numbering from the terminal carbon.
Users' Answers & CommentsStart the numbering from the terminal carbon.
34
The reddish–brown rust on ion roofing sheets consists of
A
Fe3 + (H2O)6
B
FeO.H2O
C
Fe2O3.3H2O
D
Fe3O4.2H22O
correct option: c
Iron [Fe] reacts with H2 in the presence of oxygen to form a rust.
4Fe + 3O2 → 2 Fe2O2
Fe2O3 + H2O → Fe2O3.H2O
Users' Answers & Comments4Fe + 3O2 → 2 Fe2O2
Fe2O3 + H2O → Fe2O3.H2O
35
The densities of two gases, X and Y are 0.5gdm-3 and 2.0gdm-3 respectively. What is the rate of diffusion of X relative to Y?
A
0.1
B
0.5
C
2.0
D
4.0
correct option: c
The rate of dimension of a gas inversely proportional to the square root of its molecular mass or its density, which is Graham's Law of diffusion of gas.
R ∝ \(\frac{1}{\sqrt{Mm}}\) or R ∝ \(\frac{1}{\sqrt{D}}\)
Dx = 0.5gdm-3, Dy = 2gdm-3
R= \(\frac{K}{\sqrt{D}}\)
R\(\sqrt{D}\) = k
R1\(\sqrt{D_1}\) = R1\(\sqrt{D_2}\)
Rx\(\sqrt{D_x}\) = Ry\(\sqrt{D_y}\)
\(\frac{R_x}{R_y}\) = \(\frac{\sqrt{D_y}}{\sqrt{D_x}}\)
= \(\frac{\sqrt{2}}{\sqrt{0.5}}\)
= 2.0
Users' Answers & CommentsR ∝ \(\frac{1}{\sqrt{Mm}}\) or R ∝ \(\frac{1}{\sqrt{D}}\)
Dx = 0.5gdm-3, Dy = 2gdm-3
R= \(\frac{K}{\sqrt{D}}\)
R\(\sqrt{D}\) = k
R1\(\sqrt{D_1}\) = R1\(\sqrt{D_2}\)
Rx\(\sqrt{D_x}\) = Ry\(\sqrt{D_y}\)
\(\frac{R_x}{R_y}\) = \(\frac{\sqrt{D_y}}{\sqrt{D_x}}\)
= \(\frac{\sqrt{2}}{\sqrt{0.5}}\)
= 2.0
36
The carbon atoms on ethane are
A
sp2 hybridized
B
sp3 hybridized
C
sp4 hybridized
D
sp hybridized
37
According to Charle's law, the volume of a gas becomes zero at
A
−100°c
B
−273°c
C
−373°c
D
0°c
correct option: b
Where −273°c = OK
i.e −273°c + 273 = OK
At zero kelvin, the volume of a gas becomes zero.
Users' Answers & Commentsi.e −273°c + 273 = OK
At zero kelvin, the volume of a gas becomes zero.
38
An oxide XO2 has a vapour density of 32. What is the atomic mass of X?
A
20
B
32
C
14
D
12
correct option: b
Molecular mass = vapour density X2
Mm of XO2 = x + 16(2) = x + 32
vapour density = 32
∴ x + 32 = 32 × 2
x + 32 = 64
x = 64 - 32
x = 32
∴ the relative molecular mass of X is 32
Relative molecular mass = vapour density × 2
Users' Answers & CommentsMm of XO2 = x + 16(2) = x + 32
vapour density = 32
∴ x + 32 = 32 × 2
x + 32 = 64
x = 64 - 32
x = 32
∴ the relative molecular mass of X is 32
Relative molecular mass = vapour density × 2
39
The gas that can be collected by downward displacement of air is
A
chlorine
B
sulphur (IV) oxide
C
carbon (IV) oxide
D
ammonia
correct option: d
Upward delivery works well for hydrogen and ammonia, which are both less densed than air. Sometimes, they are collected over water.
Users' Answers & Comments40
In the laboratory preparation of trioxonitrate (V) acid the nitrogen(iv) oxide formed as a by-product is removed by
A
further heating
B
adding concentrated H2SO4
C
cooling the acid solution with cold water
D
bubbling air through the acid solution
correct option: d
Bubbling of air through the acid solution removes deposited oxides of nitrogen.
Nitric acid is prepared in the laboratory by heating a nitrate salt with the concentrated acid.
NaNO3 + H2SO4 → NaHSO4 + HNO3
Vapours of nitric acid are condensed to a brown liquid in a receiver cooled under cold water. "Dissolved oxides of nitrogen" e.g NO2 are removed by redistillation or blowing a current of carbondioxide or dry air through the warm acid
Users' Answers & CommentsNitric acid is prepared in the laboratory by heating a nitrate salt with the concentrated acid.
NaNO3 + H2SO4 → NaHSO4 + HNO3
Vapours of nitric acid are condensed to a brown liquid in a receiver cooled under cold water. "Dissolved oxides of nitrogen" e.g NO2 are removed by redistillation or blowing a current of carbondioxide or dry air through the warm acid