# Chemistry (2017) | Study Mode

Question 31
A

fused CaO

B

H2O

C

NaOH

D

Concentrated H2SO4

##### Explanation
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

Question 32
A
3-methybut-3-ene
B
2-methylbut-1-ene
C
2-ethylprop-1-ene
D
2-methylbut-2-ene
##### Explanation
CH2CH2 - CCH3CH2
Start the numbering from the terminal carbon.

Question 33
A
Fe3 + (H2O)6
B
FeO.H2O
C
Fe2O3.3H2O
D
Fe3O4.2H22O
##### Explanation
Iron [Fe] reacts with H2 in the presence of oxygen to form a rust.

4Fe + 3O2 ? 2 Fe2O2

Fe2O3 + H2O ? Fe2O3.H2O

Question 34
A
sp2 hybridized
B
sp3 hybridized
C
sp4 hybridized
D
sp hybridized
##### Explanation
Alkane's family are sp3 hybridized

Question 35
A

100c

B

273c

C

373c

D

0c

##### Explanation

Where -273c = 0K

i.e -273c + 273 = 0K

At zero kelvin, the volume of a gas becomes zero.

Question 36
A
20
B
32
C
14
D
12
##### Explanation
Molecular mass = vapour density X2

Mm of XO2 = x + 16(2) = x + 32

vapour density = 32

? x + 32 = 32 x 2

x + 32 = 64

x = 64 - 32

x = 32

? the relative molecular mass of X is 32
Relative molecular mass = vapour density x 2

Question 37
A
chlorine
B
sulphur (IV) oxide
C
carbon (IV) oxide
D
ammonia
##### Explanation
Upward delivery works well for hydrogen and ammonia, which are both less densed than air. Sometimes, they are collected over water.

Question 38
A
positive ion
B
neutral atom of a metal
C
neutral atom of a non-metal
D
negative ion
##### Explanation
protons = 9

neutrons = 10

electrons = 10

Electronic configuration = 2, 8

Ground state Electronic configuration=2, 7

It means that the atom has gained an electron thereby making it have a negative ion.

When an atom donates an electron, it becomes positively charged.

When an atom accepts an electron, it becomes negatively charged

Question 39
A

Reforming

B

Polymerization

C

Distillation

D

Cracking

##### Explanation

Ethene is produced from cracking which involves breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high temperatures and pressures without a catalyst.

C15H32 2C2H4 + C3H6 + C8H14
......heat...ethene........propene.........octane

Question 40
A
2.536
B
1.623
C
4.736
D
0.394
##### Explanation
For an ideal gas PV = nRT

Amount in moles = n

Volume v = 10.5dm3

Pressure P = 6atm

Temperature T = 30C + 273 = 303k

R, Gas constant = 0.082 atmdm3k-1 mol

Recall from ideal gas equation

pv = nRT

n = $$\frac{RV}{RT}$$

n = $$\frac{6 \times 105}{0.082 \times 303}$$

n= 2.536mol

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