Answers to exam-style questions
Answers to exam-style questions
3
Chapter 1
1
Solid [1] because both the melting point and boiling
point are above 50°C. [1]
2
a
Solids have particles that are very close
together, are ordered and that vibrate about a
fixed position. Answer: C
b
Liquids have particles that are fairly close
together, are irregularly arranged and that
move slowly. Answer: B
c
Gases have particles that are very far apart,
are arranged randomly and that move at high
speeds. Answer: D
d
3
[1]
[1]
[1]
Particles that are very far apart must be in a
gas. Gases have particles arranged randomly
and moving at high speeds, which means they
cannot be ordered or vibrating about a fixed
position. Answer: A
[1]
a
Condensation
[1]
b
Diffusion
[1]
c
Freezing
[1]
d
Evaporation
[1]
4
4
a
C
[1]
b
E
[1]
c
Separation: the particles get closer.
[1]
Arrangement: particles are arranged in
fixed positions.
[1]
[1]
[1]
c
TiCl4(l) + 4Na(s) → Ti(s) + 4NaCl(s)
[1]
d
4KO2(s) + 2CO2(g) → 2K 2CO3(g) + 3O2(g)
[1]
e
2Al(s) + 6HCl(aq) → 2AlCl3 + 3H2(g)
[1]
a
Atomic number = number of protons = 20
[1]
b
Nucleon number = number of protons +
number of neutrons = 20 + 21 = 41
[1]
c
29
[1]
d
Number of neutrons = nucleon number –
atomic number = 63 − 29 = 34
[1]
e
From the Periodic Table, element with atomic
number 29 = copper
[1]
f
29
[1]
g
29 + 36 = 65
[1]
h
From the Periodic Table, zinc has an atomic
number of 30.
[1]
i
30
[1]
j
30 + 35 = 65
[1]
3p
4n
lithium
Nucleus containing 3 protons [1] and 4 neutrons [1]
Three electrons outside the nucleus arranged 2,1 [1]
6
a
Isotopes are atoms of the same element with
the same number of protons but different
numbers of neutrons.
b
[1]
Consider 100 boron atoms:
a
Iron(iii) oxide has a formula: compound.
[1]
b
Hematite is an ore: mixture.
[1]
c
Iron appears in the Periodic Table: element.
[1]
d
Stainless steel is an alloy: mixture.
[1]
80 boron atoms have mass of 11 relative mass
units each
[1]
e
Air is a mixture of gases.
[1]
Total mass = (20 × 10) + (80 × 11) = 1080
f
Oxygen appears in the Periodic Table: element. [1]
g
Natural gas is mainly methane but also contains
other gases: mixture.
[1]
Methane has a formula: compound.
[1]
Consider 100 copper atoms:
So (100 − x) = number of copper atoms with mass
number 65
[1]
H = 2, S = 1, O = 4
[1]
C = 2, O = 1, H = 6 (5 from H5 and 1 from OH)
[1]
c
5H2O = 10 × H and 5 × O
Total mass of 100 copper atoms = (63x) + 65(100 − x)
[1]
Multiplying the brackets gives MgN2O6.
Mg = 1, N = 2, O = 6
Relative atomic mass (Ar) of boron = 10.80
7
x = number of copper atoms with mass number 63
a
Cu = 1, S = 1, H = 10, O = 5 + 4 = 9
[1]
Average mass = 1080 ÷ 100 = 10.80
[1]
b
d
20 boron atoms have mass of 10 relative mass
units each
It is illegal to photocopy this page
h
2
[1]
Li
Chapter 2
1
CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g)
2KOH(aq) + H2 SO 4(aq) → K 2 SO 4(aq) + 2H2O(l)
5
Motion: particles vibrate but do not move from
place to place.
[1]
Forces of attraction: particles are more
strongly attracted.
a
b
= 63x + 6500 – 65x
= 6500 − 2x
[1]
[1]
© David Besser 2022
1
Answers to exam-style questions
Ar of copper = 63.5
Therefore, total mass of 100 copper atoms
= 100 × 63.5 = 6350
So 6500 − 2x = 6350
2x = 150
x = 150 ÷ 2 = 75% with mass number 63
[1]
(100 − x) = 25% with mass number 65
[1]
Chapter 3
1
Number
of
Particle protons
Number
Charge
of
Electronic
on
electrons configuration particle
A
20
18
2,8,8 [1]
2+ [1]
B
9
10
2,8
1− [1]
C
10 [1]
10
2,8
0
D
8
10
2,8
2− [1]
2
Giant ionic structures
Giant covalent structures
Simple molecules
Example
Any ionic substance, e.g.
sodium chloride [1]
Any giant covalent
substance, e.g. diamond [1]
Any simple molecular
substance, e.g. iodine [1]
Type of particle present
Ions [1]
Atoms [1]
Molecules [1]
Type of bonding between
particles
Ionic [1]
Covalent [1]
Intermolecular [1]
Melting point and boiling
point
High [1]
High [1]
Low [1]
Electrical conductivity
when solid
Poor [1]
Poor (except graphite) [1]
Poor [1]
Electrical conductivity
when aqueous
Good [1]
Insoluble in water
Poor [1]
Malleability and ductility
Not malleable or ductile [1]
Not malleable or ductile
Not malleable or ductile
F
It is illegal to photocopy this page
3
Cl
Si
[1]
−2x = 6350 − 6500
4
B
[1]
b
E
[1]
c
D
[1]
d
A
[1]
e
D
[1]
f
B
[1]
a
[1]
H
F
F
F
b
[1]
Cl
F
Cl
2
Cl
a
F
Si
Cl
H
S
H
Cambridge IGCSE
Chemistry Study and Revision Guide Third Edition
Cl
H
S
H
H
F
F
Answers to exam-style questions
F
c
[1]
Cl
Cl F
F
Si
F
Cl
H
S
H
Cl
7
Cl
Si
d
e
Pb(NO3)2
[1]
f
CaCO3
[1]
g
Al(NO3)3
[1]
h
K 2 SO3
[1]
i
ZnSO 4
[1]
j
(NH4)2 SO 4
[1]
a
Oxidation: Mg → Mg2+ + 2e −
2+
[1]
Cl
H
S
b
H
Cl
c
5
[1]
a
[1]
i
0
[1]
ii
+2
[1]
iii 0
[1]
iv +2
[1]
i
Mg
[1]
H
H
C
O
i
ii
H
H
[1]
Reduction: Cu + 2e → Cu
ii
d
−
Mg loses two electrons.
[1]
Mg increases in oxidation number
from 0 to +2.
[1]
Cu2+
[1]
2+
Cu gains two electrons.
[1]
Cu2+ decreases in oxidation number
from +2 to 0.
[1]
Chapter 4
1
[1]
b
a
(6 × 12) + (1 × 12) + (16 × 6) = 180
b
Na2 SO 4 = (23 × 2) + 32 + (16 × 4) = 142
[1]
10H2O = 10 × [(2 × 1) + 16] = 180
H
C
N
2
H
[1]
(39 × 2) + (52 × 2) + (16 × 7) = 294
[1]
a
More than enough oxygen than is required
to react with all the magnesium.
[1]
b
[1]
c
142 + 180 = 322
c
1.0 g of magnesium forms 10.0 ÷ 6.0 g of
magnesium oxide.
H
P
6.0 g of magnesium forms 10.0 g of magnesium
oxide.
(There is no need to evaluate 10.0 ÷ 6.0.)
18.0 g of magnesium forms 18.0 times as much
magnesium oxide as 1.0 g.
H
Therefore, mass of magnesium oxide formed
= (10.0 ÷ 6) × 18.0 = 30.0 g.
c
[1]
d
Cl
It is illegal to photocopy this page
1.0 g of magnesium oxide is formed from
6.0 ÷ 10.0 g of magnesium.
C
(There is no need to evaluate 6.0 ÷ 10.0.)
O
0.24 g of magnesium oxide are formed from
0.24 times as much magnesium as 1.0 g.
Cl
6
[1]
10.0 g of magnesium oxide are produced from
6.0 g of magnesium.
Therefore, mass of magnesium
= (6.0 ÷ 10.0) × 0.24 = 0.144 g.
a
Ca(OH)2
[1]
b
MgCl2
[1]
c
(NH4)3PO 4
[1]
d
Li2 S
[1]
3
[1]
Mr of H2 = (1 × 2) = 2
Moles of Al = 8.1 ÷ 27 = 0.30
[1]
Mole ratio from the equation:
2 moles Al : 3 moles H2
0.30 moles Al : 0.30 × 3/2 = 0.45 moles H2
[1]
© David Besser 2022
3
Answers to exam-style questions
Mass of H2 = number of moles × molar mass
= 0.45 × 2 = 0.90 g
4
Mass of Ti = 0.002 × 48 = 0.096 g
[1]
0.096 g is the theoretical yield but the actual yield
is only 0.024 g
Mr of CaC2 = 40 + (12 × 2) = 64
Percentage yield = (actual yield ÷ theoretical yield)
× 100
Moles of C2H2(g) = 120 ÷ 24 000
= 0.005 moles C2H2(g)
[1]
= (0.024 ÷ 0.096) × 100 = 25.0% [1]
(The volumes must both be in the same units so,
as the volume in the question is given in cm3, the
volume of one mole of a gas must be converted
from 24 dm3 to 24 000 cm3. Alternatively, 120 cm3
could be converted to 0.120 dm3.)
8
a
The Cl2 is in excess and Na is the limiting
reagent.
Mr of Cl2 = 35.5 × 2 = 71
Moles of Cl2 = 7.1 ÷ 71 = 0.1
0.005 moles C2H2 : 0.005 moles CaC2
[1]
The equation states that 2 moles of Na react
with 1 mole of Cl2.
[1]
Therefore, 0.1 moles of Na react with
0.1 ÷ 2 = 0.05 moles of Cl2.
[1]
Since there are 0.1 moles of Cl2, there is more
than enough Cl2 to react with all the Na.
[1]
Mass of CaC2 = number of moles × molar mass
= 0.005 × 64 = 0.32 g
35.0 × 0.20
= 0.007
1000
Mole ratio from equation:
1 mole H2 SO 4 : 2 moles KOH
Moles of H2 SO 4 =
b
As all the Na reacts, we use the moles of Na
to calculate the moles of NaCl followed by its
mass.
0.007 moles H2 SO 4 : 2 × 0.007 = 0.014 moles KOH[1]
a
Concentration of KOH =
moles ×1000
( )
3
volume cm
=
According to the equation, 2 moles of Na
produce 2 moles of NaCl.
0.014 ×1000
20.0
0.014 ×1000
=
3
20.0
volume cm = 0.70 mol/dm3
( )
Mr: NaCl = 23 + 35.5 = 58.5
Therefore, the mass of 0.1 moles of NaCl
= 0.1 × 58.5 = 5.85 g
[1]
Mr of KOH = 39 + 16 + 1 = 56
Mass = moles × molar mass
Chapter 5
= 0.70 × 56
6
[1]
Calculate the number of moles of atoms of each
element:
Hydrogen, H: 9.1 ÷ 1 = 9.1
a
[1]
Divide all of the above by the smallest:
C: 4.54 ÷ 2.275 = 2
H: 9.1 ÷ 2.275 = 4
O: 2.275 ÷ 2.275 = 1
Empirical formula = C2H4 O
It is illegal to photocopy this page
b
[1]
Mr of the compound = 44
Mr of C2H4 O = (12 × 2) + (1 × 4) + 16 = 44
1
Electrolyte
Carbon, C: 54.5 ÷ 12 = 4.54
Oxygen, O: 36.4 ÷ 16 = 2.275
[1]
n = Mr of the compound ÷ Mr of empirical
formula
7
Bromine [1]
Potassium [1]
Molten sodium
chloride
Chlorine [1]
Sodium [1]
Concentrated
aqueous sodium
chloride
Chlorine [1]
Hydrogen [1]
Iodine
Lead
Molten lead
iodide [1]
2
[1]
3
Mr of TiCl4 = 48 + (35.5 × 4) = 190
Moles of TiCl4 = 0.38 ÷ 190 = 0.002
[1]
Mole ratio from equation:
1 mole of TiCl4 : 1 mole of Ti
Therefore, 0.002 moles of TiCl4 : 0.002 moles of Ti [1]
4
Name of product Name of product
at anode (+)
at cathode (−)
Molten
potassium
bromide
n = 44 ÷ 44 = 1
Therefore, molecular formula = C2H4 O × 1
= C2H4 O
[2]
(5.9 g would be an acceptable answer.)
To convert concentration in mol/dm3 to
concentration in g/dm3, use:
= 39.2 g/dm3
[1]
Therefore, 0.1 moles of Na produce 0.1 moles
of NaCl.
moles ×1000
b
[1]
Moles of Na = 2.3 ÷ 23 = 0.1
Mole ratio from the equation:
1 mole C2H2 : 1 mole CaC2
5
[1]
4
a
i
The anode = positive electrode
ii
The cathode = negative electrode
[1]
[1]
iii The electrolyte = the liquid in the beaker
[1]
b
Dissolved in water
[1]
c
Ions [1] that are moving carry the charge [1].
a
Bauxite
[1]
b
Aluminium is above carbon in the reactivity
series.
[1]
Electrolysis is the process in which an electrolyte
is decomposed. The products of decomposition are
Cambridge IGCSE Chemistry Study and Revision Guide Third Edition
Answers to exam-style questions
formed at the electrodes. The positive electrode is
called the anode and the negative electrode is called
the cathode.
2
a
[3]
Electroplating means covering a metal object with
a thin layer of another metal. One of the reasons for
electroplating is to improve appearance. Another
reason is to resist corrosion.
[7]
a
Nickel
[1]
b
Aqueous nickel nitrate (accept an aqueous
solution of any soluble nickel salt)
[1]
The knife
[1]
c
6
7
a
Iodine b
+
2H + 2e → H2
[1]
c
Oxidation
[1]
energy/kJ
5
reactants
products
[1]
−
d
Potassium hydroxide
[1]
e
Electrons
[1]
f
Ions
[1]
progress of reaction
3
Name of
product at
anode (+)
Name of
product at
cathode (−)
Aqueous copper(ii)
sulfate
Oxygen [1]
Copper [1]
Concentrated aqueous
lithium bromide
Bromine [1]
Hydrogen [1]
Dilute aqueous sodium
chloride
Oxygen [1]
Hydrogen [1]
b
Exothermic – the reactants have more energy
than the products.
[1]
a
i
Volume of water
[1]
ii
So the temperature is the same throughout
the water.
[1]
b
a
Electrolyte
4
Petroleum spirit
[1]
Biggest temperature rise
[1]
a
[2]
H
H
H
C
C
H
H
H
C
H
O
O
H + O
O
O
O
O
O
O
O
O
C
O H
O
H
H
O
H
O
C
O
H
O
H
O
C
O H
O
H
It is a common error to include three C–C
bonds because the formula is C 3H8.
b
Bonds are: 2 C–C, 8 C–H, 5 O=O
[1]
c
Energy = (2 × 347) + (8 × 435) + (5 × 497)
= 6659 kJ
[1]
b
From blue to colourless
c
i
The copper anode goes into solution as
Cu2+ ions.
[1]
d
Bonds are: 6 C=O, 8 O–H
[1]
ii
There is no change in colour/stays blue. [1]
e
Energy = (6 × 803) + (8 × 464) = 8530 kJ
[1]
f
6659 − 8530= –1871 kJ/mol
[2]
The amount of energy given out when new
bonds form in the products (8530 kJ) is
bigger than the amount of energy taken
in to break bonds in the reac tants (6659 kJ),
so overall energy is given out and the
reaction is exothermic.
[3]
Chapter 6
a
Boiling point
[1]
b
i
Viscosity decreases.
[1]
ii
Volatility increases.
[1]
iii Chain length decreases.
c
[1]
g
ΔH = −1871 kJ/mol
[8]
Use
Lubricating oil [1]
Lubricants, waxes or
polishes
Refinery gas
Heating and cooking [1]
Bitumen [1]
Making roads
Naphtha
Chemical feedstock [1]
Fuel oil [1]
Fuel in ships or home
heating systems
Gasoline or petrol
Fuel in cars [1]
Diesel oil or gas oil [1]
Fuel for diesel engines
Kerosene or paraffin
Jet fuel [1]
5
a
[4]
It is illegal to photocopy this page
Fraction
[2]
(The negative sign shows that the reaction is
exothermic.)
products
Ea
energy/kJ
1
energy change
ΔH
reactants
progress of reaction
b
See diagram
[2]
© David Besser 2022
5
Answers to exam-style questions
Chapter 7
1
2
6
a
This is making a mixture that can be easily
separated later: physical change.
[1]
b
New substances are made so it is a chemical
change.
[1]
c
New substances are made so it is a chemical
change.
[1]
d
This is separating a mixture – a physical change.
[1]
e
This is separating a mixture – a physical change.
[1]
If the temperature of an equilibrium system is
decreased, the equilibrium shifts in the exothermic
direction. (The enthalpy change is the only factor
that needs to be considered. The actual equations
do not matter.)
7
In Experiment 2, the concentration of the acid
increases so the rate increases and the graph is
steeper than in Experiment 1. Twice as many moles of
acid are present, therefore the volume of hydrogen is
doubled. Answer: D
[1]
a
Exothermic direction is to the right.
[1]
b
Exothermic direction is to the left.
[1]
a
The rate of the forward reaction and the
rate of the reverse reaction are equal [1] and
concentrations of the reactants and
products are no longer changing (they
become constant).
[1]
b
i
In Experiments 3, 4 and 5, the number of moles of the
acid are the same as in Experiment 1. Therefore, the
volume of hydrogen is the same.
c
In Experiment 5, using a higher temperature means
that the rate increases and the graph is steeper.
Answer: B
[1]
3
a
Time, stopwatch
[2]
b
Carbon dioxide escapes from the apparatus. [1]
(‘Carbon dioxide is formed/given off’ is not
sufficient for the mark.)
c
Temperature, concentration of the acid
d
i
The steepest graph/graph that levels off first
shows the fastest reaction.
[1]
ii
3 top box, 2 middle box, 1 bottom box
Pushing in the plunger increases the
pressure, so equilibrium moves in the
direction of fewer gas molecules. There
are fewer gas molecules on the left-hand
side of the equilibrium. The gas on this
side is colourless, so the colour of the
mixture becomes paler.
[1]
i
Equilibrium shifts to the right.
[1]
ii
Forward reaction rate increases
and reverse reaction rate increases.
[1]
[1]
iii Graphs level off
4
Chapter 8
1
b
[1]
[1]
l the gas molecules move faster OR the gas
[1]
It is illegal to photocopy this page
b
c
6
Fewer gas molecules on the left so shift is
to the left.
Hydrochloric acid produces chlorides so
use (dilute) hydrochloric acid.
[1]
iii CoCO3 + 2HCl → CoCl2 + CO2 + H2O
[2]
i
Method 2
[1]
ii
Nitric acid produces nitrates so use (dilute)
nitric acid.
[1]
[1]
l Add magnesium carbonate.
[1]
l Stir or warm.
[1]
l Filter off excess magnesium carbonate.
[1]
l Heat the filtrate until crystals form on a glass
rod placed in the solution and withdrawn.
[1]
l Leave the hot saturated solution to cool slowly. [1]
Crystals should then form.
If pressure is increased, the equilibrium shifts
in the direction of fewer gas molecules.
Number of gas molecules on both sides is
the same so no change.
ii
solid remains undissolved OR no more bubbles
of gas are evolved.
[1]
energy greater than or equal to the activation
energy, therefore, a greater percentage of
collisions are successful collisions.
[1]
a
[1]
l Stop adding magnesium carbonate when some
[1]
l a greater percentage of molecules have
5
Method 1
l Pour dilute sulfuric acid [1] into a beaker.
l the frequency of collisions between
molecules increases
i
iii KOH + HNO3 → KNO3 + H2O
2
As the temperature increases:
molecules gain kinetic energy
a
[2]
iv Calcium carbonate is in excess, so the reaction
stops when all the hydrochloric acid is used.[1]
[1]
ii
In Experiment 3, using a powder means that the rate
increases and the graph is steeper. Answer: B
[1]
In Experiment 4, using a lower temperature means
that the rate decreases and the graph is less steep.
Answer: E
[1]
The NO2 molecules are pushed closer
together OR the concentration of NO2
increases. As NO2 is the coloured gas,
the colour gets darker.
l Remove crystals (by filtration if there is any
liquid left).
l Dry the crystals in a low oven or on a warm
[1]
[1]
Fewer gas molecules on the right so shift is
to the right.
[1]
[1]
windowsill.
3
[1]
l MgCO 3 + H2 SO 4 → MgSO 4 + CO2 + H2O
[1]
a
A: burette
[1]
B: conical flask
[1]
i
Indicator
[1]
ii
Methyl orange OR thymolphthalein
[1]
b
Cambridge IGCSE Chemistry Study and Revision Guide Third Edition
Answers to exam-style questions
c
d
pH starts above 7 OR 11 > value > 7
[1]
pH decreases to 7 (or below 7)
[1]
Ammonia + hydrochloric acid →
ammonium chloride
4
a
b
[1]
l Stir/warm to dissolve scandium oxide.
[1]
l Filter off copper(ii ) oxide.
[1]
l Wash with distilled water.
[1]
in a low oven.
l Add water to both salts.
5
[1]
l Filter the lead(ii ) iodide.
[1]
l Wash the residue of lead(ii ) iodide with distilled
2+
6
[1]
[1]
–
solid floats
–
solid disappears
–
bubbles
[2]
2Na + 2H2O → 2NaOH + H2
a
Copper(i) refers to Cu+ so the formula is Cu2O [1]
b
Copper(ii) refers to Cu2+ so the formula is
Cu(NO3)2
[1]
c
Iron(ii) refers to Fe2+ so the formula is FeCl2 [1]
ii
Proton donor
[1]
d
Iron(iii) refers to Fe3+ so the formula is
Fe2(SO 4)3
i
Oxidising agent
Oxygen
iii C2H5OH + O2 → CH3COOH + H2O
7
a
[1]
Cl2(g) + 2KI(aq) → 2KCl(aq) + I2(aq)
Correct formulae [1], balancing [1], state
symbols [1]
[1]
[1]
Cl2(g) + 2I– (aq) → 2Cl– (aq) + I2(aq)
b
Group VII elements become more reactive up
the group. Answer: D
[1]
c
Transition elements appear in the elongated
section between Groups II and III.
Answer: F OR G
[1]
d
Periods are the horizontal rows. Answer: A
e
Correct formulae [1], balancing [1],
state symbols [1]
Br2(l) + 2I– (aq) → 2Br– (aq) + I2(aq)
[1]
Chapter 10
a
(Dilute) nitric acid + magnesium →
magnesium nitrate + hydrogen
[1]
[1]
b
Atoms of Group IV elements have four
electrons in their outer shells. Answer: A
Chlorine + (aqueous) potassium bromide →
bromine + (aqueous) potassium chloride
[1]
[1]
c
Reaction occurs:
[1]
Iron(iii) oxide + carbon monoxide →
(molten) iron + carbon dioxide
[1]
F2 + 2KCl →2KF + Cl2
[1]
b
No reaction
[1]
Silicon dioxide + calcium oxide →
calcium silicate
[1]
c
Reaction occurs:
[1]
i
Hematite
[1]
Br2 + 2KAt → 2KBr + At2
[1]
ii
d
No reaction
[1]
Air (not oxygen; using oxygen instead of air
would be expensive and is unnecessary)
[1]
a
Any two from:
a
–
coloured compounds
–
high density
–
(compound acts as a) catalyst
1
[2]
d
2
a
It is illegal to photocopy this page
Group I elements become more reactive down
the group. Answer: B
[1]
[1]
Br2(l) + 2KI(aq) → 2KBr(aq) + I2(aq)
Chapter 9
a
[1]
I2(s) is acceptable in both equations because some
of it may form as a precipitate.
b
3
–
[1]
ii
2
solid moves
Partially dissociated
All formulae correct [1],  (indicates
acid is weak) [1], state symbols [1]
1
–
i
iii CH3COOH(aq)  CH3COO – (aq) + H+(aq)
b
Any two from:
The roman numerals used to represent oxidation
states are the same as the number of positive
charges on the cations.
Correct formulae [1], balancing [1], state
symbols [1]
a
[1]
Atoms of the noble gases have a full outer shell
of electrons. [1] This does not require sharing of
electrons (to form covalent bonds) or loss or gain of
electrons (to form ionic bonds). [2]
l Pb (aq) + 2I (aq) → PbI2(s)
6
[2]
Formulae [1], equation completely correct [1]
[1]
l Mix the two aqueous solutions.
l Dry (e.g. with filter paper).
(forms a) basic oxide
ii
[1]
or deionised water.
(forms) soluble salts
–
i
[1]
l Stir to dissolve both salts to make aqueous
solutions.
good conductor of electricity
–
Reactivity increases down the group
l Dry the copper(ii ) oxide on a warm windowsill/
5
–
(‘Reactivity decreases’ = 0 marks)
l Add aqueous sodium hydroxide or potassium
hydroxide.
Any two from:
[1]
NH3 [1] + HCl → NH4 Cl
4
b
iii Coke burns exothermically to provide the
high temperature required to carry out
endothermic reactions. [1] Coke forms
carbon monoxide, which reduces the iron(iii)
oxide to iron.
[1]
© David Besser 2022
7
Answers to exam-style questions
b
3
i
Calcium oxide [1], carbon dioxide [1]
ii
Slag or calcium silicate
3
[1]
a
Iron
[1]
b
Oxygen and water
[1]
c
Any two from:
d
4
5
–
paint
–
oil
–
plastic
[3]
adding carbon
undissolved particles sink
to bottom
chlorination
removes floating debris
filtration
removes unwanted
tastes
sedimentation
kills microbes
[2]
They prevent oxygen and water from coming into
contact with the steel.
[1]
a
B, C, A, D
[1]
b
B + A(NO3)2 → A + B(NO3)2
[1]
c
C + D2+ → D + C2+
[1]
a
Any three from:
b
–
solid moves around
–
solid floats
–
bubbles
–
solid disappears
[3]
4
2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g)
a
Correct formulae [1], balancing [1], state
symbols [1]
6
Yellow
a
Magnesium acts as a barrier that prevents
oxygen and water [1] from reaching the steel.[1]
i
ii
[1]
c
Galvanising
[1]
d
Copper is less reactive than iron.
[1]
5
a
It is illegal to photocopy this page
b
[2]
[2]
b
i
6CO2 + 6H2O → C6H12O6 + 6O2
[2]
ii
CO2 + C → 2CO
[2]
c
The greenhouse gases carbon dioxide and
methane cause global warming by the
absorption of thermal energy, reducing
the loss of thermal energy to space.
A substance that improves plant growth.
b
[4]
[1]
(NH4)3PO 4
Octane + oxygen → carbon dioxide + water [1]
ii
Calcium carbonate →
calcium oxide + carbon dioxide
[1]
iii Iron(iii) oxide + carbon monoxide →
iron + carbon dioxide
[1]
iv Calcium carbonate + dilute hydrochloric
acid →
calcium chloride + carbon dioxide + water
The percentage of nitrogen is
42
× 100 = 28.2%
149
[1]
An answer of 28% is acceptable.
i
Water
[1]
ii
Glucose [1], oxygen [1]
Multiplying out the brackets gives N3H12PO 4.
Mr = (14 × 3) + (1 × 12) + 31 + (16 × 4) = 149
[1]
Of which N = (14 × 3) = 42
[1]
Chapter 12
1
a
Combustion of fossil fuels containing sulfur
compounds as impurities
[1]
b
Acid rain
[1]
c
Using low-sulfur fuels
[1]
Using catalytic converters
[1]
Flue-gas desulfurisation by calcium oxide
[1]
2
8
[1]
[2]
2NO + 2CO → N2 + CO2
i
iii Chlorophyll [1], sunlight [1]
2
CaCO3 → CaO + CO2
v
a
Chapter 11
1
ii
[2]
iv Na2CO3 + 2HCl → 2NaCl + CO2 + H2O
[1]
Magnesium loses electrons instead of iron
[1] and forms positive magnesium ions. [1]
Sacrificial protection
2C10H22 + 31O2 → 20CO2 + 22H2O
iii C6H12O6 → 2C2H5OH + 2CO2
c
b
i
a
Ethene
b
Ethane [1], methane [1]
[1]
c
Poly(ethene)
[1]
d
Ethene
[1]
e
Ethene
[1]
f
Ethane [1], methane [1]
g
Ethane
a
Ethene [1], methane [1]
b
Methane
Cambridge IGCSE Chemistry Study and Revision Guide Third Edition
[1]
[1]
Answers to exam-style questions
c
H
H
C
3
H
a
Addition
[1]
b
The bromine changes colour from yellow–
brown [1] to colourless.
[1]
Ethene changes the colour of aqueous bromine
from orange to colourless. [1] Ethane gives no
colour change/solution stays yellow–brown. [1]
e
Carbon dioxide [1], methane [1]
f
i
Polymerisation
[1]
ii
Monomers
[1]
i
A substance which contains two or more
elements chemically combined in fixed
proportions by mass.
[1]
Not reactive
[1]
a
ii
iii A substance which increases the rate of
a chemical reaction [1] and is chemically
unchanged at the end of the reaction.
b
It is essential to give the initial and final colours
for both marks. Note that if excess propene
is not used, there may be some bromine left
and so the colour of the bromine may still be
visible.
c
8
CS2 + 3Cl2 → CCl4 + S2Cl2
a
Addition polymer
H CH3CH2
H
C
C
C
C
C
C
H
H
H
H
H
H
[1]
ii
Two hydrogen atoms are added to the
propene, giving C 3H8.
[1]
[1]
Nickel
a
U
sing n = 9 in the general formula for alkanes
Cn H2n+2 gives C9H20.
[1]
b
Answer may be structural formulae or
displayed formulae of but-1-ene or but-2-ene
from Table 12.6 (page 126) because ‘showing
all the atoms and bonds’ is not requested.
The C=C double bond must be shown.
[2]
c
i
[1]
H CH3CH2
Two bromine atoms are added to the
propene, giving C 3H6Br2.
d
b
CH3CH2
i
iii A water molecule (i.e. two hydrogen atoms
and one oxygen atom) is added to the
propene, giving C 3H8O.
[1]
[1]
Formulae correct [1], equation fully correct
4
7
C
H
d
[1]
ii
i
H
ii
5
H
H
H
H
H
C
C
C
C
H
H
H
H
6
a
H
H
H
C
C
C
C
H
H
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
H
H
C
C
C
C
C
H
H
C 8H18 → C5H10 + C 3H6 + H2
OR C 8H18 → C2H4 + 2C 3H6 + H2
[2]
There are other acceptable answers
with hydrogen, an alkyne (Cn H2n–2) and an
alkane as the products.
Chapter 13
1
H
H
H
[2]
[1]
But-1-ene
H
C 8Hl8 → C 4H10 + 2C2H4
OR C 8H18 → C2H4 + C6H12 + H2
The circle should be drawn around two
consecutive carbon atoms in the main chain
and all the atoms and groups of atoms
joined to them. One example is given.
[1]
c
[1]
[1]
H
H
H
H
H
C
C
C
C
C
H
H
a
H
H
H
2
Ultraviolet light
Chloroethane
Some or all the hydrogen atoms in ethane
can be substituted by chlorine atoms. The
total number of atoms bonded to the two
carbon atoms must be six in each case, so
the possible answers are:
C2H4 Cl2 / C2H3Cl3 / C2H2Cl4 / C2HCl5 / C2Cl6
(only one is required)
[1]
b
D
[1]
c
C
[1]
d
B
[1]
e
D
[1]
f
B
[1]
g
A
[1]
a
i
Zinc ethanoate
[1]
ii
[1]
Zinc + ethanoic acid →
zinc ethanoate + hydrogen
[1]
[1]
iii Solid dissolves [1], bubbles [1]
[2]
b
i
Magnesium ethanoate
[1]
ii
Magnesium carbonate + ethanoic acid →
magnesium ethanoate + carbon dioxide +
water
[1]
It is illegal to photocopy this page
b
c
A
iii Solid dissolves [1], bubbles [1]
[1]
3
a
The formula that shows the number of atoms
of each element in one molecule of an
element or compound.
[1]
b
Structural isomers (must include the word
‘structural’)
[1]
© David Besser 2022
9
Answers to exam-style questions
c
d
e
f
This formula cannot be simplified any further.
Therefore, the molecular formula and the
empirical formula are the same: C5H10O2
[1]
Esterification (condensation is also
acceptable)
Chapter 14
1
a
Dissolving (sugar in water) [1], filtration (to
remove sand) [1], crystallisation [1]
b
(Simple) distillation (not fractional distillation,
as this mixture is a solid and a liquid)
[1]
c
This is a mixture of two liquids, so use
fractional distillation.
[1]
The precipitate is an insoluble/undissolved
solid so start with filtering. [1] The solid will
be contaminated with a small amount of the
solution it was separated from, so wash
with distilled water [1] and dry on a warm
windowsill or in a low oven.
[1]
[1]
Heat [1] and a catalyst of concentrated
sulfuric acid
[1]
d
2C 3H8O + 9O2 → 6CO2 + 8H2O
Correct formulae [1], balancing [1]
Fractions or multiples are accepted in
equations.
g
A
H
H
H
B
H
C
C
C
H
O
H
H
H
H
H
H
H
C
C
C
H
H
H
2
O
H
H
H
C
C
H
H
(must state slowly). [1] Crystals should then
form.
H
O
C
l Remove crystals (by filtration if there is any
H
liquid left).
H
Condensation polymerisation is the formation
of a long-chain molecule (the polymer) from
small molecules (monomers). [1] A simple
molecule, such as water, is eliminated as the
monomers join together. [1]
b
A polyamide
[1]
[1]
l Dry the crystals in a low oven or on a warm
[3]
a
windowsill.
[1]
If a heat source is mentioned in the final step, it must
be stated that it is set on low, as high temperatures
will cause the crystals to lose water of crystallisation,
leaving anhydrous powder.
3
c,d [1 each]
linkage
[1]
l Leave the hot saturated solution to cool slowly
4
in the solution and withdrawn (must state
when to stop heating).
C
H
Steps:
l Heat until crystals form on a glass rod placed
repeat unit
l Add the mixture to the acid and stir OR shake
OR heat/warm/boil.
[1]
l Until the bubbling stops.
[1]
l Filter off the carbon.
[1]
l Wash (the residue/carbon) with distilled water. [1]
O
C
e
C6H4
O
H
C
N
C6H4
H
O
N
C
C6H4
O
H
C
N
A protein
[1]
f
[2]
H
N
H
It is illegal to photocopy this page
5
a
b
10
l Dry in a low oven or between filter papers.
4
H
C6H4
N
O
H
O
C
O
C6H4
C
O
H
H
CH3COOCH3 methyl ethanoate
[2]
Ethyl groups can be written as C2H5 - OR
CH3CH2- so HCOOCH2CH3 OR HCOOC2H5
are acceptable for ethyl methanoate.
[2]
Propyl groups can only be written as
CH3CH2CH2-. This is because C 3H7- can
be a straight chain or a branched chain.
HCOOCH2CH2CH3 propyl methanoate
[2]
CH3COOCH2CH3 OR CH3COOC2H5
ethyl ethanoate
[2]
CH3CH2COOCH3 OR C2H5COOCH3
methyl propanoate
[2]
Test
Observation
Conclusion
Aqueous ammonia
is added
Green
precipitate [1]
R contains
Cr3+ or Fe2+
An excess of
aqueous ammonia
is added
Green precipitate
remains [1]
R contains
Cr3+ or Fe2+
Aqueous sodium
hydroxide is added
Green
precipitate [1]
R contains
Cr3+ or Fe2+
Excess aqueous
sodium hydroxide
is added
Green precipitate
dissolves [1]
R contains
Cr3+
The mixture from
row above is
warmed and the
gas given off is
tested with damp
red litmus paper
Litmus turns
blue [1],
ammonia gas [1]
Contains
ammonium
ion [1]
Dilute nitric acid
[1], aqueous
barium nitrate [1]
White
precipitate [1]
R contains
sulfate ion
Cambridge IGCSE Chemistry Study and Revision Guide Third Edition
[1]
Answers to exam-style questions
5
a
b
A: fractionating column
[1]
B: Liebig condenser [1]
Water in and water out are the wrong way round. [1]
There should not be a bung in the conical flask. [1]
c
Fractional distillation
[1]
d
Pentane and heptane are both flammable.
[1]
e
Pentane has a lower boiling point.
[1]
6
Steps:
l Carry out paper chromatography using a
suitable solvent and allow solvent to reach the
top of the chromatography paper.
[1]
Use a simple diagram to describe how the
apparatus is set up even if the question does
not ask for one. Take care not to draw the
solvent level above the starting line on the
chromatography paper.
[1]
l Remove the chromatography paper and
allow to dry.
[1]
l Spray with locating agent.
[1]
l Determine Rf values using:
Rf =
distance travelled by component
distance travelled by solvent
[1]
l Compare Rf values with data book values to
identify amino acids.
[1]
It is illegal to photocopy this page
© David Besser 2022
11