Cambridge IGCSE™ Chemistry
Cambridge IGCSETM Chemistry
Workbook answers
Cambridge Assessment International Education bears no responsibility for the example answers to
questions taken from its past question papers which are contained in this publication.
1 States of matter
Core
1
In ice, I (the water molecule) am bonded to other water molecules and vibrating in a fixed
position. As the temperature rises, there is enough energy for some of the bonds between us to
weaken and begin to break due to the increase in vibrations. This allows me to move further
from my neighbouring molecules and I am able to move past them. As the temperature
continues to rise, more of the bonds between us begin to break and I move away from my
neighbouring molecules and move much more quickly. At 100 °C I have so little attraction to
my neighbouring molecules that I am able to break away and move a long way away from
them.
2
a
i
When a substance changes from a solid to a liquid
ii
When substances are changed into different substances by chemical reaction
iii When a gas changes into a liquid
iv When a liquid changes into a gas
v
When a liquid changes into a solid
vi When heating of a liquid causes it to vaporize
b
3
i
Condensation
ii
Melting
a
Diffusion cannot occur in solids because the particles are bonded to one another more
strongly than those in liquids and gases, which do diffuse. This prevents them from being
able to move relative to one another.
b
Gases can be compressed more than liquids because the particles in the gas are much more
spread out than those in a liquid. This means that the gas particles can be squeezed together
more than those in a liquid, which are much closer to one another.
c
Liquids can flow because their particles have much weaker bonds between them than those
in a solid. This means that the particles in a liquid can move over one another.
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Supplement
4
5
a
[Correct scale, axis labels and units, plotting of points, and line]
b
5 °C
c
82 °C
a
Ammonia
b
Point B. Both HCl and NH3 particles move randomly along the tube, but the ammonia
particles are lighter and so will move faster and further along the tube than the HCl
particles.
c
The particles of hydrogen chloride and ammonia move along the tube by randomly
colliding with other particles and the sides of the tube; the particles of ammonia move
quicker as they are lighter.
Exam-style questions
Core
1
a
i
Solid – any two of:
cannot be compressed, has a fixed shape, expands on heating [2]
Liquid – any two of:
will flow and take up the shape of the container, can only be compressed a small
amount, its volume increases when heated [2]
Gas – any two of:
is easily compressed, flows easily, will fill the whole volume of its container [2]
ii
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[1 for each]
b
i
oxygen [1]
ii
phosphorus and iron [1]
iii bromine [1]
2
a
Gas particles from the coffee are moving randomly from the coffee shop, colliding with
other particles in the air, until they reach you. [1] They are diffusing. [1]
b
When the temperature rises, the steel tracks will expand. [1] The gaps allow the tracks to
expand without buckling the railway lines. [1]
c
The particles of tea move randomly and collide with water molecules in the water. [1]
They diffuse from the tea bag until eventually all of the water in the cup changes colour as
the tea dissolves. [1]
d
Water vapour inside the house, caused by central heating, hot water and people breathing
[1], condenses onto the glass of the windows, which is colder. [1]
e
At the bottom of the ocean the bubble has a large amount of pressure all around it due to
the depth of water, and has an equal interior pressure caused by methane molecules. [1] As
the bubble rises, the external pressure decreases and the relative greater interior pressure of
the gas in the bubble causes the volume of the methane bubble to increase. [1]
Supplement
3
a
i
The external pressure decreases. [1]
ii
The volume of the balloon will increase. [1]
iii As the external pressure decreases the particles of helium gas on the inside are still
colliding with the inside surface [1] of the balloon, causing the balloon to increase in
volume. [1].
b
i
The external temperature decreases. [1]
ii
As the external temperature decreases this will have an effect of cooling down the gas
in the balloon and the particles will lose kinetic energy [1] and will cause them to
collide with the inside surface of the balloon less violently [1] and this will lead to the
balloon’s decreasing in volume. [1]
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2 Atoms, elements and compounds
Core
1
a
The elements in a compound, e.g. iron sulfide, cannot be separated by physical means,
whilst a mixture of iron filings and sulfur powder can be separated using a magnet.
b
i
carbon monoxide, sulfuric acid, methane, sodium hydroxide, limestone
ii
stainless steel, lemonade, cement, beer, brass
c
Accept any sensible answer with a reasonable explanation, for example:
i
water, water is a compound
ii
chromium, chromium is an element
iii F2, F2 is an element
2
d
A group of atoms chemically joined/bonded together.
a
i
Use eye protection e.g. safety spectacles to protect eyes.
ii
A new substance is being produced.
iii zinc sulfide
iv zinc + sulfur → zinc sulfide
Zn(s) + S(s) → ZnS(s) [Correct reactants, correct products]
b
v
Zinc and sulfur can be separated by physical means (e.g. using an organic solvent to
dissolve the sulfur) whereas the compound zinc sulfide needs chemical means to
separate the elements.
i
copper, oxygen, carbon
ii
copper sulfide, copper oxide, sulfur dioxide, carbon dioxide
iii copper sulfide + oxygen → copper oxide + sulfur dioxide
2CuS(s) + 3O2(g) → 2CuO(s) + 2SO2(g) [Correct reactants, correct products, correct
balancing]
copper oxide + carbon → copper + carbon dioxide
2CuO(s) + C(s) → 2Cu(s) + CO2(g) [Correct reactants, correct products, correct
balancing]
3
a
Formula of substance
LiNO3
Elements present
Total number
of atoms
Symbol
Name
Number of atoms
Li
Lithium
1
N
Nitrogen
1
O
Oxygen
3
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CaCO3
Mg3N2
Ag2CrO4
b
Ca
Calcium
1
C
Carbon
1
O
Oxygen
3
Mg
Magnesium
3
N
Nitrogen
2
Ag
Silver
2
Cr
Chromium
1
O
Oxygen
4
i
2Pb(s) + O2(g) → 2PbO(s) [Correct products, correct reactants]
ii
2H2(g) + O2(g) → 2H2O(l) [Correct products, correct reactants]
5
5
7
iii C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) [Correct products, correct reactants]
4
a
Element
Proton number
Lithium
3
Boron
Aluminium
a
6
5
20
i
23
11Na: 2, 8, 1
ii
1632 32
16S: 2, 8, 6
iii
39
19K: 2, 8, 8, 1
Nucleon number
27
7
Calcium
5
Number of
electrons
3
13
Nitrogen
b
Number of
neutrons
7
20
12 × mass of 1 atom of carbon-12 Ar = average mass of the isotopes of an element
compared to 1/12th of the mass of an atom of 12C.
b
Average mass =
Ar =
c
(1.10 × 13) + (98.90 × 12)
= 12.01
100
12.01
= 12.01
1
These are atoms of the same element, C. They have different nucleon numbers. C-12 has 6
neutrons and C-14 has 8 neutrons. Both atoms have 6 protons, hence they are isotopes of
carbon.
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d
These are atoms of the same element with the same chemical properties because they have
the same number of electrons and therefore the same electronic configuration.
Supplement
6
a
An oxidising agent is a substance that oxidises another substance and is itself reduced. A
reducing agent is a substance that reduces another substance and is itself oxidised.
b
i
ZnO or zinc oxide
ii
C or carbon
Exam-style questions
Core
1
a
Element
Metal or nonmetal?
W
Shiny?
Conductor of
electricity?
Melting point
Yes
X
Y
Metal
Z
No
b
high density
c
i
iron or oxygen
ii
iron oxide
iii 4Fe(s) + 3O2(g) → 2Fe2O3 (s) [Correct reactants, correct products, correct balancing]
Supplement
2
a
i
C(s) + O2(g) → CO2(g) [Correct reactants, correct products]
ii
CO2(g) + C(s) → 2CO(g) [Correct reactants, correct products, correct balancing]
b
carbon
c
carbon dioxide [1]
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d
Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g) [Correctly balanced reactants, correctly balanced
products]
e
reducing agent CO, oxidising agent Fe2O3
3 Bonding and structure
Core
1
2
a
false
b
true
c
false
d
true
e
true
f
true
a
A diamond, B graphite
i
covalent
b
ii
giant covalent
c
i
4
ii
3
d
graphite – lubricant or electrode, diamond – cutting tools
e
Substance
Electrical conductivity Melting point
Hardness
A
Poor
High
High
B
Good
High
Low
3
a
Ionic bonds are usually found in compounds that contain metals combined with nonmetals. When this type of bond is formed, electrons are transferred from the metal atoms
to the non-metal atoms during the chemical reaction. In doing this positive metal ions and
negative non-metal ions are formed. The positive ions are known as cations. The negative
ions are known as anions. When oppositely charged ions form a crystal structure they
have a strong attractive force between the them called an electrostatic force of attraction.
This is known as the ionic bond.
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b
[Each metal and non-metal ion shown correctly, including charge]
i
ii
4
LiCl
CaS
[In each, correct numbers of electrons, and correct sharing in overlap areas]
a
HF
b
NCl3
Supplement
5
a
i
CuF2
ii
Na2CO3
iii Ag3PO4
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iv (NH4)2SO4
v
Mg3(PO4)2
vi Al(OH)3
vii FeBr3
6
Metals are good conductors of electricity and heat, because the free electrons from the outer
energy levels of metal atoms carry a negative charge or heat energy through the metal. The
free electrons are often described as delocalised. The free electrons allow metal ions to slide
over each other, so metals are malleable and ductile. They have high melting and boiling
points due to the strong attractive forces within the structure of the metal.
Exam-style questions
Core
1
a
covalent (single bonds) [1]
b
electrons [1]
c
In the molecule, the carbon atom has eight electrons in its outer energy level. [1] It
achieves this by sharing electrons with four hydrogen atoms. [1]
d
i
helium [1]
ii
neon [1]
iii The only shell in the hydrogen atom is the first, which is full when it contains only two
electrons. [1]
Supplement
2
a
Na: 2,8,1 [1], Na+: 2,8 [1]
b
Cl: 2,8,7 [1], Cl–: 2,8,8 [1]
c
ionic [1]
d
i
Each Na+ ion is surrounded by six Cl– ions [1] and vice versa. [1]
ii
There are strong electrostatic forces between the oppositely charged ions. [1] A lot of
energy is therefore needed to separate the ions and melt the substance. [1]
iii In the solid state the ions are not free to move [1] but in the molten state the ions are
free to move to the oppositely charged electrodes. [1]
4 Stoichiometry – chemical calculations
Core
1
a
The molecular formula of a compound is the number and type of different atoms in one
molecule.
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b
A CH4;
B C2H6O;
2
g/dm3
mol/dm3
3
a
i
46
ii
88
C C2H4O2
iii 74
iv 42
b
i
106
ii
74
iii 132
Supplement
4
a
i
Fe(s) + 2HCl(aq) → FeCl2(aq) + H2(g) [Correct reactants, correct products, correct
balancing]
ii
moles of Fe used = 5.6/56 = 0.1 mole
iii moles of FeCl2 = 0.1 mole; mass of FeCl2 = 0.1 × (56 + 35.5 + 35.5) = 12.7 g
iv percentage yield = (9.17/12.7) × 100 = 72.2%
b
1 mole of Fe2O3 = (56 × 2) + (16 × 3) = 160 g
number of moles of Fe2O3 = 100 × 106/160 = 625 000 moles
moles of Fe formed, if 100% yield = 2 × 625 000 = 1 250 000 moles
mass of Fe formed if 100% yield = 1 250 000 × 56 = 70 000 000 g (70 tonnes)
percentage yield = (7/70) × 100 = 10%
5
6
7
8
a
i
2 moles
ii
0.01 mole
b
0.5 mole
c
3.01 × 1023 sodium ions and 3.01 × 1023 chloride ions = 6.02 × 1023 ions
a
i
200g
ii
4.5g
b
100
c
56 g/mol
a
The empirical formula of a compound is the simplest whole number ratio of the different
elements in a compound.
b
2.16/12 = 0.18 mole carbon, 0.36/1 = 0.36 mole hydrogen, 1.44/16 = 0.09 mole
oxygen, so two C and four H for every one O, C2H4O
c
Every 100 g has 27.4/23 = 1.19 moles sodium, 1.2/1 = 1.2 moles hydrogen, 14.3/12 =
1.19 moles carbon, 57.1/16 = 3.56 moles oxygen; approximately a ratio of 1 : 1 : 1 : 3,
so NaHCO3
a
i
0.125 mole
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b
9
ii
0.05 mole
i
1 mol/dm3
ii
1 mol/dm3
a
2HCl + Na2CO3 → 2NaCl + H2O + CO2 [Correct reactants, correct products, correct
balancing]
b
[five (or more) from the following]
c
•
25 cm3 of the sodium carbonate solution is placed in a conical flask using a pipette and
safety filler.
•
3 or 4 drops of thymolphthalein (or methyl orange) indicator are added to the sodium
carbonate solution.
•
A burette is filled with the hydrochloric acid solution, ensuring that some runs through
the valve, and the initial burette reading is taken.
•
The acid is added from the burette into the flask, with swirling, until the colour of the
indicator just changes.
•
The final burette reading is taken and the volume of acid needed to neutralise the
sodium carbonate is found.
•
The process is repeated to obtain three concordant results (within 0.10 cm3 of one
another).
moles of Na2CO3 used = 0.1 × 25/1000 = 2.5 × 10–3 mole
moles of HCl = 2 × 2.5 × 10–3 = 5 × 10–3 mole
concentration of HCl = 5 × 10–3 × 1000/18.95 = 0.26 mol/dm3
Exam-style questions
Supplement
1
a
i
moles of CuO used = 8/(64 + 16) = 0.1 mole [1]
moles of H2 gas needed = 0.1 mole [1]
volume of H2 gas = 0.1 × 24 = 2.4 dm3 (2400 cm3) [1]
ii
moles of Cu obtained = 0.1 mole [1]
mass of Cu = 0.1 × 64 = 6.4 g [1]
iii percentage yield = 5.8 × 100/6.4 [1] = 90.6% [1]
b
i
moles of propane = 10/24 = 0.417 mole [1]
moles of O2 needed = 5 × 0.417 = 2.085 mole [1]
volume of O2 needed = 2.085 × 24 = 50 dm3 [1]
ii
moles of O2 = 10/24 = 0.417
volume of CO2 = 0.417/5 × 3 × 24 = 6 dm3 [1]
volume of H2O = 0.417/5 × 4 × 24 = 8 dm3 [1]
total volume = 6 + 8 = 14 dm3 [1]
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2
a
Rough
1
2
3
21.75
22.25
22.35
22.30
Final burette reading/cm3
Initial burette reading/cm3
Volume of sulfuric acid used/cm3
[1]
b
average volume = (22.25 + 22.35 + 22.30)/3 [1] = 22.30 cm3 [1]
c
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) [1 for reactants, 1 for products, 1 for
correct balancing]
d
moles of NaOH = 0.25 × 25/1000 [1] = 6.25 × 10–3 mole [1]
e
moles of H2SO4 neutralised = 0.5 × 6.25 × 10–3 = 3.125 × 10–3 mole [1]
f
concentration of H2SO4 = 3.125 × 10–3 × 1000/22.30 [1] = 0.14 mol/dm3 [1]
5 Electrochemistry
Core
1
a
substances T, U, W and Z
b
substances X and Y
c
substance U
d
substance X
e
substance V
f
any dilute acid or dilute alkali
g
copper(II) oxide
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2
Material
electrodes
Substance
of Substance formed at Substance formed at
the cathode
the anode
Lead
Chlorine
Calcium
Bromine
Dilute sulfuric acid
Carbon
Molten sodium chloride
Carbon
3
Electroplating is the process that uses electrolysis to plate (or coat) one metal with another.
Often the purpose of electroplating is to give a protective coating to the surface beneath. For
example, bath taps are chromium plated to prevent corrosion, and at the same time are given a
shiny, more attractive finish. The electroplating process is carried out in a cell. This is often
known as the ‘plating bath’ and it contains a suitable electrolyte, usually a solution of a metal
salt. For silver plating, the electrolyte is a solution of a silver salt. The article to be plated is
made the cathode in the cell so that the ions move to it when the current is switched on. [8]
4
a
This increases the conductivity of the water.
b
Chlorine and sodium hydroxide will react together and so the products would not be as
wanted.
c
The ions will be mobile in the liquid state.
Supplement
5
6
[Correct ionic charge, correct molecules, correct numbers of electrons]
a
Na+ + e– → Na
b
2Br– → Br2 + 2e–
c
Ca2+ + 2e– → Ca
d
Cu2+ + 2e– → Cu
e
2I– → I2 + 2e–
f
4OH– → 2H2O + O2 + 4e–
a
[Correct balancing of ions and electrons]
b
i
2H+(aq) + 2e– → H2(g)
ii
2Cl–(aq) → Cl2 + 2e–
i
1 mole of NaCl is 23 + 35.5 = 58.5 g; so moles of NaCl electrolysed
= 234/58.5 = 4 moles
From the equation, 2 moles NaCl produces 2 moles NaOH, so 4 moles NaCl produces
4 moles NaOH, or 4 × (23 + 16 + 1) g, i.e. 160 g of NaOH
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ii
2 moles of NaCl are needed to produce 1 mole Cl2, so 4 moles NaCl produce 2 moles
Cl2 , i.e. 2 × (35.5 + 35.5) g = 142 g of Cl2
iii 2 moles of NaCl are needed for 1 mole H2, so 4 moles NaCl produce 2 moles H2, i.e.,
2 × (1+1) g H2 = 4 g of H2
Exam-style questions
Supplement
1
a
The temperatures required for chemical reduction are too high [1] and therefore energy
costs are too high. [1]
b
The mixture of cryolite and aluminium oxide [1] has a much lower melting point, of about
1000°C. [1]
c
i
cathode [1]
ii
Al3+ + 3e– → Al [1]
i
carbon [1]
ii
2O2– → O2 + 2e–
i
The carbon anodes react with oxygen at high temperatures [1], producing carbon
dioxide. [1]
ii
C(s) + O2(g) → CO2(g) [1 for reactants, 1 for products]
d
e
2
f
It is cheaper to recycle the aluminium than to get it from the ore. [1] It is environmentally
more sensible – less mining is required and there is less waste. [1]
a
i
increases [1]
ii
Metallic tin is deposited at the cathode. [1]
i
+2 [1]
ii
The electrolyte is tin(II) sulfate. [1]
i
No [1]
ii
It stays the same because the tin taken out of the solution [1] is replaced by tin from
the anode. [1]
i
Sn2+(aq) + 2e– → Sn(s) [1 for reactants, 1 for products, 1 for correct balancing]
ii
Sn(s) → Sn2+(aq) + 2e– [1 for reactants, 1 for products, 1 for correct balancing]
b
c
d
e
Mild steel is mainly iron, which would corrode when in contact with the food and any
solutions in the can. [1]
6 Chemical energetics
Core
1
a
Ease of importing oil by sea
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b
i
The crude oil is heated to about 400°C. This vaporises most of the different substances
in the crude oil mixture.
ii
naphtha
iii surfacing roads
iv diesel oil
c
2
i
bitumen
ii
refinery gas
d
gasoline or petrol
e
fractionally distilled
a
A substance that releases energy when it is combusted.
b
i
one from: gasoline/petrol, diesel or other liquid fuel from the fractional distillation of
crude oil
ii
e.g. coal
iii e.g. natural gas
c
e.g. methane + oxygen → carbon dioxide + water
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) [Correct reactants, correct products, correct
balancing]
Supplement
3
a
i
2 × –728 = –1456 kJ
ii
0.25 × –728 = –182 kJ
iii (8/16) × –728 = –364 kJ
iv (64/16) × –728 = –2912 kJ
4
b
[Correctly shown axes, correct formulae and balancing, reactants shown higher than
products]
a
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Breaking: 2 C—C, + 8 C—H, + 5 O═O: (2 × 347) + (8 × 413) + (5 × 498) = 6488 kJ/mol
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Forming: 6 C═O, + 8 O—H: (6 × 805) + (8 × 464) = 8542 kJ/mol
∆H = 6488 – 8542 = –2054 kJ/mol
b
c
i
[Correctly shown axes, correct formulae and balancing, reactants shown higher than
products]
ii
The activation energy, Ea, is the minimum energy that colliding particles must have in
order to react.
i
0.5 × –2054 = –1027 kJ
ii
5 × –2054 = –10 270 kJ
iii (11/44) × –2054 = –513.5 kJ
5
a
2H2(g) + O2(g) → 2H2O(l) (or H2(g) + ½O2(g) → H2O(l) ) [Correct reactants, correct
products, correct balancing, correct state symbols]
b
endothermic, energy has to be supplied to break chemical bonds
c
If the first equation is given in part a:
Breaking: 2 H—H + 1 O═O, (2 × 436) + (1 × 498) = 1370 kJ
Forming: 4 O—H, 4 × 464 = 1856 kJ
∆H = 1370 – 1856 = –486 kJ (for 2 moles H2O)
Or, if the second equation is given in part a:
Breaking: 2 H—H + 1 O═O, (1 × 436) + (½ × 498) = 685 kJ
Forming: 2 O—H, 2 × 464 = 928 kJ
∆H = 685 – 928 = –243 kJ (for 1 mole H2O)
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6
a
[Correct scale, axis labels, plotting of points]
b i
ii
c
–3300 kJ/mol (±150 kJ/mol) [Correct value, correct ‘–' sign]
Extrapolated the graph to a relative molecular mass of 88, assuming that the change in
enthalpy of the next alcohol which has one extra –CH2 unit would have the same effect
on the trend shown.
The amount of energy released per mole gets progressively larger.
Exam-style questions
Core
1
a
i
An exothermic reaction is one which transfers thermal energy to the surroundings [1]
leading to an increase in the temperature of the surroundings. [1]
ii
A, B, D [1]
iii B [1]
iv It has the biggest energy gap between the reactants and products. [1]
b
i
An endothermic reaction is one which takes in thermal energy from the surroundings
[1] leading to a decrease in the temperature of the surroundings. [1]
ii
E [1]
iii It has the biggest energy gap between the reactants and products. [1]
Supplement
2
a
1.91 g [1]
b
20.8 °C [1]
c
energy transferred = 500 [1] × 4.2 × 20.8 [1] = 43 680 J [1]
d
relative molecular mass of butan-1-ol = 74 [1]
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moles of butan-1-ol burned = 1.91/74 [1] = 0.026 mole [1]
e
enthalpy of combustion of butan-1-ol = 43 680/0.026 [1] = 1 680 000 J/mol (1680 kJ/mol)
[1 for value, 1 for units]
7 Chemical reactions
Core
1
2
a
chemical change
b
physical change
c
physical change
d
chemical change
a
The volume of hydrogen gas collected in a burette/measuring cylinder/gas syringe could be
recorded at set time intervals.
b
i
A
ii
Steepest line at the beginning of the curves, or flattens soonest.
iii Yes because all the reactions stop, shown by the flattened lines, to produce the same
final volume of hydrogen gas.
3
4
a
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g) [Correct reactants, correct
products, correct balancing]
b
To prevent the loss of acid spray from the flask, which would have led to increased values
for the loss in mass of the flask and its contents.
c
Carbon dioxide gas was being lost from the flask.
d
[Correct scale, correct axis labels, key, points plotted correctly (each graph), each correct
best-fit line]
e
The reaction with the smaller chips. The line is steeper at the beginning.
a
It indicates the reaction is reversible.
b
Water could be added.
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c
Hydrated copper(II) sulfate contains water of crystallisation within its structure, anhydrous
copper(II) sulfate has no water in its structure.
Supplement
5
a
In all experiments, moles of Mg used = 2/24 = 0.083 mole.
Maximum number of moles of H2SO4 used = 0.1 × 40/1000 = 0.004 mole.
There is a 1 : 1 mole relationship in the balanced equation, so magnesium is in excess.
b
6
Line
A
B
C
D
E
Experiment
II
III
I
V
IV
c
The reaction shown by line B uses 2 g of powdered magnesium, which has a larger surface
area than the magnesium ribbon used in the reaction shown by line C. Hence more
collisions occur in the same time, so successful collisions occur more frequently and this
gives a faster rate of reaction.
d
The reaction shown by line D is carried out at a higher temperature. So the particles have
more energy and move faster, causing successful collisions to occur more frequently and
hence giving rise to a faster rate of reaction.
a
2CO(g) + 2NO(g) → 2CO2(g) + N2(g) [correct reactants, correct products, correct
balancing]
b
i
carbon monoxide
ii
nitrogen monoxide
i
platinum (alloyed with small amounts of palladium and rhodium)
ii
A catalyst increases the rate of reaction by providing an alternative reaction path with a
lower activation energy, so more of the collisions are successful, so increasing the rate
of the reaction.
i
C8H18(l) + 12½O2(g) → 8CO2(g) + 9H2O(g) (or 2C8H18 + 25O2 → 16CO2 + 18H2O)
[Correct reactants, correct products, correct balancing]
ii
Mass of octane burned = 5000 × 0.70 = 3500 g
c
d
Moles of octane burned = 3500/114 = 30.7 moles
Mass of carbon dioxide produced = 30.7 × 44 × 8 = 10 806 g (10.806 kg)
iii Volume of carbon dioxide produced = 30.7 × 8 × 24 = 5894.4 dm3
iv 100 g of CO is 100/28 = 3.57 moles; produces 3.57 moles of CO2, this is
3.57 × 44 = 157 g
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Exam-style questions
Core
1
a
b
i
Yes. [1]
ii
More oxygen gas is produced, more quickly when more black powder is used. [1] The
mass of black powder remains unchanged. [1]
i
At the beginning [1], the line is steepest here. [1]
ii
Maximum concentration of reactants. [1]
iii The reactants are being used up therefore less oxygen gas is produced in a given time.
[1]
iv 43 cm3 [1]
v
48 seconds [1]
Supplement
2
a
To ensure a fair test [1]; different volumes would lead to different concentrations of
hydrochloric acid. [1]
b
Solid yellow sulfur is formed in the reaction [1] so it becomes impossible to see the cross.
[1]
c
Experiment
Rate of reaction / s
1
2
2.3 × 10–2
3
1.5 × 10–2
4
1.0 × 10–2
5
0.6 × 10–2
[1 for each]
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d
[Correct scale, correct axis labels, points plotted correctly, correct best-fit line]
e
i
0.15 mol/dm [1]
ii
45 cm3 [1]
iii 83 s [1]
3
a
Reaction A [1], steepest line [1]
b
i
Reaction A [1]
ii
Reaction C [1]
c
By using powdered zinc or more concentrated acid. [1]
d
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) [1]
Moles of H2 produced = 25/24 000 = 1.04 × 10–3 mole [1]
Volume of H2SO4 = 1.04 × 10–3 × 1000/0.05 = 20.8 cm3 [1]
e
Moles of H2 produced = 50/24 000 = 2.08 × 10–3 mole [1]
Mass of H2 produced = 2.08 × 10–3 × 2 = 4.16 × 10–3 g [1]
4
a
A reversible reaction, in a closed system reaches an equilibrium when the rate of the
forward reaction is equal to the rate of the reverse reaction [1]. The concentrations of the
reactants and products no longer change. [1]
b
i
Increasing the pressure increases the yield of ammonia [1] at any temperature. [1]
ii
Decreasing the temperature increases the yield of ammonia. [1]
c
As the pressure is increased, the position of equilibrium will move to the right [1] as this
produces fewer moles of gas [1], lowering the pressure of the system. [1]
d
The production of ammonia is an exothermic process (negative sign). [1] A decrease in
temperature would favour the exothermic reaction. [1] As the yield of ammonia increases
with decreasing temperature, the forward reaction must be exothermic. [1]
e
25%
f
At lower temperatures, although more ammonia would be produced [1], the rate would be
too slow for the process to be economic. [1]
g
It would be expensive [1] and dangerous. [1]
h
If more nitrogen gas was introduced into the reaction vessel its concentration would be
increased [1] and so the equilibrium would move to the right, producing more ammonia.
[1]
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5
i
The iron catalyst would not affect the position of the equilibrium, [1] but it would increase
the rate of both the forward and reverse reactions equally. [1]
a
i
2SO2(g) + O2(g) ⇌ 2SO3(g) [1 for correctly balanced equation, 1 for reversible
arrows]
ii
If the temperature is increased, the position of equilibrium will move to the left. [1]
This is because the back reaction is endothermic and this is favoured by higher
temperatures. [1]
If the pressure is increased, the position of equilibrium will move to the right. [1]
This is because there are 3 moles of gas on the left and 2 moles of gas on the right, so
moving to the right will form fewer moles of gas, lowering the pressure. [1]
b
By burning sulfur in air, OR from the roasting of sulfide ores. [1]
c
Pressure 200 kPa; [1] Temperature 450oC; [1] Catalyst vanadium(V) oxide. [1]
8 Acids, bases and salts
Core
1
Indicator
Red litmus
Colour with hydrochloric acid
Red
Colour with sodium hydroxide
Blue
Blue litmus
Red
Blue
Thymolphthalein
Colourless
Blue
Methyl orange
Red
Yellow
2
ammonium chloride + calcium hydroxide → calcium chloride + ammonia + water [Correct
reactants, correct products]
3
a
sodium carbonate + nitric acid → sodium nitrate + water + carbon dioxide
Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2 [Correct reactants, correct products, correct
balancing]
b
magnesium + hydrochloric acid → magnesium chloride + hydrogen
Mg + 2HCl → MgCl2 + H2 [Correct reactants, correct products, correct balancing]
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4
Substances used to make the salt
Salt prepared
Other products
Calcium chloride
Water
Sodium hydroxide/oxide
Nitric acid
Zinc
Lead chloride
Barium sulfate
5
a
Potassium chloride
i
Add a small amount of dilute nitric acid followed by a few drops of silver nitrate
solution. A white precipitate will be formed.
ii
Add a small amount of dilute nitric acid followed by a few drops of silver nitrate
solution. A cream (off-white) precipitate will be formed.
iii Add a small amount of dilute nitric acid followed by a few drops of silver nitrate
solution. A yellow precipitate will be formed.
6
b
Add some dilute hydrochloric acid. Fizzing (effervescence) will be observed, caused by the
production of carbon dioxide (which would turn limewater cloudy).
c
Add a small amount of dilute hydrochloric acid followed by some barium chloride. A white
precipitate will be formed.
d
i
Add some dilute sodium hydroxide solution. A green precipitate will be formed.
ii
Add some dilute sodium hydroxide solution. An orange-brown precipitate will be
formed.
a
ammonium
b
soluble, silver
c
soluble
d
sulfates lead
e
insoluble
Supplement
7
a
P is iron(III) chloride, FeCl3
b
Q is iron(III) hydroxide, Fe(OH)3
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8
c
R is iron(II) chloride, FeCl2
d
S is hydrogen, H2
e
T is silver chloride, AgCl
a
Substance
0.1 mol/dm3 HCl
pH
1.0
0.1 mol/dm3 NaOH
13.0
0.1 mol/dm3 CH3COOH
2.9
Pure H2O
7.0
0.1 mol/dm3 NH3 solution
11.0
b
Although they have the same concentration, HCl has a lower pH because it is a strong acid
and completely dissociates to produce more H+ ions than CH3COOH, which is a weak acid
and only partially dissociates to give fewer H+ ions. The higher the concentration of H+
ions in the solution the lower the pH.
c
i
A proton is a hydrogen ion, H+.
ii
An acid is a proton donor.
iii A base is a proton acceptor.
9
a
i
silver nitrate, sodium chloride (or other soluble chloride)
ii
barium chloride or nitrate, sodium sulfate (or other soluble sulfate)
iii calcium nitrate or chloride, sodium or potassium carbonate
b
Mix together solutions of lead(II) nitrate and sodium or potassium iodide to give a yellow
precipitate of lead(II) iodide. Filter the mixture obtained. Wash with distilled water. Dry
the yellow solid.
Pb2+(aq) + 2I–(aq) → PbI2(s)
10 a
b
c
d
It produces hydrogen ions (H+(aq)) when added to water.
i
an acid that only partially ionises (dissociates) when added to water
ii
an acid that completely ionises (dissociates) when added to water
i
HCl(g) → H+(aq) + Cl–(aq)
ii
CH3COOH(l) ⇌ CH3COO–(aq) + H+(aq)
A strong acid is one that completely ionises (dissociates) when added to water, whereas a
weak acid ionises (dissociates) only partially. A concentrated acid contains a high
concentration of acid particles, whereas a dilute acid contains a much lower concentration
of acid particles and lots of water. [1]
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Exam-style questions
Core
1
a
Acids dissolve in water to produce hydrogen ions, which can be written as H+(aq). Bases
are oxides or hydroxides of metals. Alkalis are soluble bases. They dissolve in water to
produce hydroxide ions, which can be written as OH–(aq). Acids and alkalis react together
to produce solutions with a pH of 7; these are called neutralisation reactions. [1 for each]
b
H+(aq) + OH–(aq) → H2O(l) [1 for reactants, 1 for products, 1 for correct balancing]
c
i
2HCl(aq) + K2CO3(s) → 2KCl(aq) + H2O(l) + CO2(g) [1 for reactants, 1 for products,
1 for correct balancing]
ii
To ensure that all the hydrochloric acid had been neutralised. [1]
iii potassium chloride solution [1]
iv a saturated solution [1]
Supplement
2
a
Seven (or more) from:
•
25 cm3 of the dilute sodium hydroxide solution is placed in a conical flask using a
pipette and safety filler. [1]
•
Three or four drops of thymolphthalein indicator are added to the sodium carbonate
solution. [1]
•
A burette is filled with the dilute sulfuric acid solution, ensuring that some runs
through the valve, and the initial burette reading is taken. [1]
•
The acid is added from the burette into the flask, with swirling, until the colour of the
indicator just changes from blue to colourless. [1]
•
The final burette reading is taken and the volume of acid needed to neutralise the
sodium hydroxide is found. [1]
•
The process is repeated, this time without an indicator but using exactly the same
volumes of dilute sodium hydroxide and dilute sulfuric acid as found in the first
experiment. [1]
•
The neutralised solution from the repeat experiment is poured into an evaporating
basin and the solution is heated until half of the volume has evaporated. [1]
•
b
The solution is then left to cool and the crystals will form slowly. [1]
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) [1 for reactants, 1 for products, 1 for
correct balancing]
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9 The Periodic Table
Core
1
2
a
chemical change
b
60–100°C 350 – 400 K
c
The melting points increase with increasing atomic number.
d
F and Cl
a
aluminium + chlorine → aluminium chloride
2Al(s) + 3Cl2(g) → 2AlCl3(s)
b
Chlorine is a toxic gas.
c
fluorine
d
any suitable metal, e.g. sodium, potassium, magnesium or gallium
3
The modern Periodic Table has been credited to the work of the Russian chemist Dmitri
Mendeleev. After many years of chemists across the world trying to classify the elements in a
useful way, he came up with the table that we have been using for nearly 150 years. He
arranged the elements in order of increasing atomic weight. Occasionally he had to swap
elements around so that they were in the same group as other elements with similar properties,
for example swapping tellurium (Te) and iodine (I). The major change that he introduced to
his classification was that he left spaces/gaps for elements that had not been discovered at the
time. Today, the elements in the modern Periodic Table are arranged in order of increasing
atomic number. Going across a period, from left to right, in the modern Periodic Table
elements change from being metallic to non-metallic.
4
a
i
19 electrons, 19 protons, 20 neutrons
ii
2,8,8,1
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iii Group I, it has one electron in the outer energy level of the atom
iv Y+
b
7
c
i
potassium bromide
ii
potassium + bromine → potassium bromide
2K + Br2 → 2KBr
iii Both chlorine and bromine gain an electron, from the potassium, when they react.
Chlorine is more reactive than bromine as it is a smaller atom, which attracts the
incoming electron more strongly as its nucleus is closer to its outer energy level.
5
6
a
Mn
b
C
c
Br
d
Na
e
Ne or Kr
f
P
g
Na
h
F
a
Group I metals
Transition metals
Density
Low
High
Melting points
Low
High
Colour of solid compounds
White
Coloured
Good catalysts?
No
Yes
b
Higher density than Group I metals, harder/stronger than Group I metals.
c
i
metallic bonding
ii
d
i
It speeds up the chemical reaction and allows lower temperatures/less energy to be
used.
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ii
Haber process – iron catalyst, or
Contact process – vanadium(V) oxide catalyst
Supplement
7
a
b
i
element C: 2,8,3; element D: 2,6
ii
17
i
element A
ii
element B
iii element E
iv element C
c
i
element C
ii
element D
iii element A
Exam-style questions
Core
1
a
potassium bromide + chlorine → potassium chloride + bromine [1]
2KBr(aq) + Cl2(aq) → 2KCl(aq) + Br2(aq) [1 for reactants, 1 for products, 1 for correct
balancing]
2
3
b
Chlorine is a smaller atom than bromine. [1] Therefore it attracts the incoming electron
from the potassium more strongly as its nucleus is closer to the outer energy level than
bromine’s. [1]
c
fluorine [1]
d
No [1], bromine is less reactive than fluorine [1]
a
Na 2,8,1 [1]
b
Potassium. [1] Both potassium and sodium have one electron in their outer energy level,
which is lost when they react with the water. [1] In potassium (the bigger atom), the
electron in the outer energy level is further from the nucleus and is less tightly held in the
atom and so is lost more easily, making it more reactive. [1]
c
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) [1 for reactants, 1 for products, 1 for correct
balancing]
d
Francium. [1] It is at the bottom of the group, is the largest atom and loses its outer
electron most easily. [1]
a
chlorine: gas, pale yellow-green [1]
bromine: liquid, red-brown [1]
iodine: solid, green-black [1]
b
7 [1]
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c
gain one electron [1]
d
–1 [1]
e
Fluorine. [1] It is the smallest of all the halogen atoms [1] and attracts the incoming
electron most strongly, as its nucleus is closest to the outer energy level into which the
electron is coming. [1]
f
i
black [1] solid [1]
ii
Anywhere between –67 and –150 oC [1]
Supplement
4
Formula of compound Oxidation state Name of compound
FeCl2
+2 [1]
iron(II) chloride [1]
Fe2O3
+3 [1]
iron(III) oxide [1]
CuO
+2 [1]
copper(II) oxide [1]
CoCl2
+2 [1]
cobalt(II) chloride [1]
10 Metals
Core
1
2
Physical property
Metals
Non-metals
Thermal conductivity
High
Low
Electrical conductivity
High
Low
Malleability and ductility
Very
Brittle
Melting and boiling points
High
Low
a
Iron: displacement using carbon. Aluminium: electrolysis.
Iron is a moderately reactive metal. Aluminium is a more reactive metal and carbon would
not be able to displace aluminium from its ore so a more vigorous method is needed.
b
To produce heat required by reacting with oxygen; to produce carbon monoxide.
c
After decomposing, to form a slag with the sandy impurities from the hematite.
d
[Correct reactants, correct products, for each part]
i
calcium carbonate → calcium oxide + carbon dioxide
ii
carbon (coke) + oxygen → carbon dioxide
iii carbon (coke) + carbon dioxide → carbon monoxide
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iv iron(III) oxide + carbon monoxide → iron + carbon dioxide
calcium oxide + sand (silicon dioxide) → slag (calcium silicate)
v
e
[Correct reactants, correct products, correct balancing, for each part]
i
CaCO3(s) → CaO(s) + CO2(g)
ii
C(s) + O2(g) → CO2(g)
iii C(s) + CO2(g) → 2CO(g)
iv Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
CaO(s) + SiO2(s) → CaSiO3(s)
v
3
f
reducing agent
a
i
carbon
ii
It is reacted with oxygen gas, through a water-cooled lance, to produce oxides that are
lost as gases.
i
alloy
ii
They can be harder and stronger than the pure metals and so are more useful.
b
iii hard, resists corrosion
c
Stainless steel is too expensive and too dense.
d
4
Object
Properties
Steel
Car body
Easily shaped, not brittle
Mild steel
Axe
Tough
Hard steel
Surgical
knife
Tough, sharp-edged, non-reactive
Stainless steel
e
A
a
i
It is a drying agent and removes the water vapour from the air.
ii
Boiling the water removes any oxygen gas from the water because gases are less
soluble in hot water.
iii Oil is less dense than water and floats on its surface. It prevents oxygen gas from the
air above re-dissolving into the water.
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b
Tube
Water
Oxygen
A
✓
✓
✓
B
C
✓
D
✓
c
d
e
f
✓
i
tubes B and C
ii
In tube B there is no water and in tube C there is no oxygen. Both water and oxygen
are needed for iron to rust.
i
tube D
ii
There is a higher concentration of oxygen gas in tube D than in tube A.
i
A barrier method by painting/greasing/coating with plastic/(galvanising).
ii
Barrier methods prevent rusting by excluding oxygen or water.
i
Zinc is a more reactive metal than iron and it would react rather than the iron.
ii
Galvanising. It acts as a barrier method in that whilst the thin layer is intact it prevents
air and water getting to the metal. Because it is a more reactive metal than iron, and
loses electrons more easily, the zinc, rather than the iron, reacts with air and water,
sacrificially protecting the iron.
Supplement
5
6
a
MgCO3(s) + H2SO4(aq) → MgSO4(aq) + H2O(l) + CO2(g)
b
2Ca(s) + O2(g) → 2CaO(s)
c
Mg(s) + ZnSO4(aq) → MgSO4(aq) + Zn(s)
d
2Mg(s) + O2(g) → 2MgO(s)
e
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
a
2K(s) + 2H2O(l) → 2KOH(aq) + H2(g) [Correct reactants, correct products, correct
balancing]
b
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) [Correct reactants, correct products, correct
balancing]
c
Mg(s) + CuO(s) → MgO(s) + Cu(s) [Correct reactants, correct products, correct
balancing]
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7
8
d
Mg(s) + H2O(g) → MgO(s) + H2(g) [Correct reactants, correct products, correct
balancing]
e
Mg(s) + 2H2O(l) → Mg(OH)2(aq) + H2(g) [Correct reactants, correct products, correct
balancing]
a
a change in colour of the solution and/or the solid
b
displacement
c
E, F, B, A, D, C
a
copper(II) oxide + zinc → copper + zinc oxide
CuO(s) + Zn(s) → Cu(s) + ZnO(s)
b
copper
c
zinc oxide
d
i
zinc
ii
Zinc is gaining oxygen, or the zinc is losing electrons.
iii The copper is being reduced.
Exam-style questions
Core
1
a
A, zinc oxide, ZnO; B, magnesium oxide, MgO; C, zinc, Zn; D, zinc chloride, ZnCl2; E,
hydrogen gas, H2; F, magnesium, Mg; G, oxygen gas, O2; H, copper, Cu; I, magnesium
sulfate, MgSO4 [1 for each]
b
i
zinc oxide + magnesium → magnesium oxide + zinc [1 for reactants, 1 for products]
ii
zinc + hydrochloric acid → zinc chloride + hydrogen [1 for reactants, 1 for products, 1
for correct balancing]
c
Anode: 2O2– → O2(g) + 4e– [1 for reactants, 1 for products, 1 for correct balancing]
Cathode: Mg2+ + 2e– → Mg(s) [1 for reactants, 1 for products, 1 for correct balancing]
Supplement
2
a
CuO + Zn → ZnO + Cu [1 for reactants, 1 for products]
b
Cu: +2, 0 [1]
Zn: 0, +2 [1]
c
0 [1]
d
Cu2+ + 2e- → Cu [1]
e
reduction [1] it decreases [1]
f
Zn → Zn2+ + 2e- [1]
g
oxidation [1] it increases [1]
h
redox reaction [1]
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3
a
[1 for each]
b
In the pure metal all the atoms are of the same size, so when a force is applied, the layers of
atoms can easily move over one another. [1] In an alloy the atoms are not all of the same
size [1] so when a force is applied the layers of atoms can no longer slide over one another,
making the alloy harder and stronger than the pure metal. [1]
11 Chemistry of the environment
Core
1
2
3
a
false
b
true
c
false
d
true
e
false
f
false
g
false
h
true
a
Excess fertiliser dissolves and runs off fields into streams and rivers during wet weather.
b
ammonium, NH4+, nitrate, NO3–
c
ammonium nitrate, NH4NO3
d
i
algae and aquatic plants
ii
As algae and aquatic plants die and decay, oxygen is removed from the water. This
leaves insufficient oxygen for fish and other organisms to survive.
i
These are gases produced by industries or vehicles and they would not naturally be
found in the air.
ii
Any two of these gases:
a
Nitrogen monoxide from the reaction of nitrogen and oxygen, from the air, in the car
engine.
Nitrogen dioxide from the oxidation of nitrogen monoxide by oxygen from the air.
Carbon monoxide from in incomplete combustion of the fuel.
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b
i
Fossil fuels, e.g. oil, coal and natural gas
ii
64 kg
iii sulfurous acid, H2SO3, sulfuric acid, H2SO4
iv Extensive damage to forests, increased corrosion of exposed metals; and damage to
buildings and statues made from limestone and marble
4
a
v
flue gas desulfurisation units or FGD units
i
The water is passed through screens.
ii
To filter out floating debris.
b
To kill any remaining bacteria in the water.
c
hydrochloric acid, HCl
d
To neutralise the acidic solution.
e
Removes colour and smells.
f
fluoride, F–
Supplement
5
a
i
100 × 14/(23 + 14 + 48) = 16.47%
ii
100 × (14 × 3)/[3(14 + 4) + (31+64)] = 28.19%
iii 100 × (14 × 2)/[2(14 + 4) + (32+64)] = 21.21%
6
b
potassium (K) and phosphorus (P)
c
Ca3(PO4)2
a
i
1 mole CH4 is 12 + (4 × 1) = 16 g, and produces 1 mole CO2, 12 + (2 × 16) = 44 g
32 g of CH4 = 2 moles; produces 2 moles, i.e., 88 g of CO2
ii
88 g of CO2 = 2 moles of CO2; it occupies 2 × 24 dm3 = 48 dm3
b
C8H18 contains 8 times as much carbon as is found in methane.
Exam-style questions
Core
1
a
i
a gas that contributes to the greenhouse effect [1] by absorbing infrared radiation [1],
leading to atmospheric warming
ii
a general warming [1] across the surface of the Earth caused by the greenhouse effect
[1]
b
climate changes [1], melting of the ice caps [1], flooding caused by rise in sea levels [1]
c
Methane [1], CH4 [1]
d
Plant more trees [1], halt deforestation [1], use less fossil fuel for the production of
electricity [1], use fewer/no petrol or diesel vehicles by using electric ones [1], increase the
use of alternative energy sources [1], build houses that are more energy-efficient, [1] or
other suitable ways.
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Supplement
2
a
i
96 dm3 [1]
ii
2NO(g) + O2(g) → 2NO2(g)
iii catalytic converter [1]
b
i
nitric acid [1], HNO3 [1]
ii
nitrogen(IV) oxide + water → nitric acid + nitrogen monoxide [1]
3NO2(g) + H2O(g) → 2HNO3(aq) + NO(g) [1 for reactants, 1 for products, 1 for
correct balancing]
12 Organic Chemistry 1
Core
1
2
a
The process of breaking large molecules into smaller, more useful molecules.
b
A substance that alters the rate of a chemical reaction without itself being chemically
changed at the end of the reaction.
c
The chemical breakdown of a substance under the influence of heat.
d
A family of saturated hydrocarbons with the general formula CnH2n+2.
e
Molecules that possess only single covalent bonds.
f
Molecules that contain one or more double covalent bonds.
g
An atom or group of atoms that determine the chemical properties of a homologous series.
h
Compounds that contain carbon and hydrogen only.
a
When small molecules such as ethene join together to form long chains of atoms, called
polymers, the process is called polymerisation. These small molecules that can join
together in this way are called monomers. The polymer formed with ethene is an addition
polymer.
b
Single C—C bonds and shows at least 3 molecules.
3
c
The colour changes from orange to colourless.
d
Some plastics chemically break down during the recycling process. Hence they cannot be
recycled.
a
i
addition polymerisation
ii
poly(ethene)
iii CnH2n
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b
i
The plastic does not react with substances in the environment and is not decomposed
by bacteria, so it will not rot away. The bags will therefore stay unchanged over many,
many years. They create a polluted environment and are a danger to animals.
ii
In recycling, new bags are made from the existing plastic, saving the costs of energy
and raw materials.
iii Not all plastics just melt, some break down when heated.
iv Landfill sites are getting full.
Plastics are accumulating in the oceans.
Plastics form toxic gases when burned.
4
a
i
Homologous series: a homologous series is a family of similar compounds, with
similar chemical properties, due to the presence of the same functional group, with the
same general formula, differing by CH2.
Hydrocarbons: molecules that contain hydrogen and carbon atoms only.
ii
b
Combustion and substitution with chlorine.
i
Alkane
Formula
Methane
CH4
Structure
Ethane
Propane
C3H8
Butane
C4H10
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ii
covalent bonding
iii
iv Decane is likely to be a liquid.
v
CnH2n+2
Supplement
5
a
complete combustion
b
i
It is an exothermic reaction.
ii
1 mole
iii moles of methane = 64/16 = 4 moles
mass of carbon dioxide = 44 × 4 = 176 g
iv volume of CH4 = volume of CO2. Therefore 100 dm3 of methane would give 100 dm3
of CO2.
6
a
a carbon–carbon double bond
b
7
c
one
a
chloromethane
b
ethanol
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c
ethane
d
1,2 dibromoethane
e
ethene
Exam-style questions
Core
1
a
A, paraffin-soaked mineral wool; B, hard glass boiling tube; C, water; D, gaseous alkene [1
for each]
b
as a hot surface [1] for the alkane molecules to break up over [1]
c
The substance collected over water is an unsaturated hydrocarbon [1] and reacts with the
bromine dissolved in the organic solvent. [1] The substance produced, 1,2-dibromoethane,
is colourless. [1]
d
A star should be drawn near the delivery tube on the diagram, between the bung and the
crystallising dish. [1]
e
hydrogen, H2
f
i
C12H26 → C4H8 + C8H18 [1]
ii
butene, octane [2]
g
Larger less useful molecules [1] may be broken down into smaller more useful molecules
[1] such as octane and ethene. [1]
Supplement
2
a
Isomers are compounds that have the same molecular formula [1] but different structural
formulae [1], e.g. butane [1] and 2-methylpropane. [1]
[1 for each correct diagram]
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b
i
The isomer with the side chain will have lower melting and boiling points. [1]
ii
The isomer with the side chain has less surface area, so the forces of attraction are less
strong. [1] Hence the heat energy required to separate the molecules (to melt or boil
the substance) is not so great. [1]
13 Organic chemistry 2
Core
1
2
a
false
b
true
c
true
d
false
e
true
CH3COOH – Ethanoic acid
C2H5OH – Has an Mr of 46
CO2 – Produced when ethanol burns completely in air
(CH3COOH)2Mg – A salt of ethanoic acid
Phosphoric acid – Catalyst used in the manufacture of ethanol
CnH2n+1OH – General formula for alcohols
3
a
CaCO3 + 2HCOOH → Ca(HCOO)2 + CO2 + H2O
b
2HCOOH + Mg → (HCOO)2Mg + H2
Supplement
4
a
b
butanol + oxygen → carbon dioxide + water
CH3CH2CH2CH2OH(l) + 6O2(g) → 4CO2(g) + 5H2O(g)
5
a
The ethanol in the alcoholic drink is oxidised to ethanoic acid, which gives the sour taste.
ethanol + oxygen (from air) → ethanoic acid + water
CH3CH2OH(l) + O2(g) → CH3COOH(aq) + H2O(l)
6
b
esters
a
covalent
b
i
the part containing the carbon–carbon double bond
ii
the part containing the –OH group
c
the part containing the –OH group
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Cambridge IGCSE Chemistry Workbook answers
7
a
condensation polymerisation
b
i
1,6-diaminohexane, hexanedioic acid
ii
ethane-1,2-diol, benzene-1,4-dicarboxylic acid
c
water, H2O
d
i
amide
ii
ester
i
making ropes, woven into fabric
ii
woven into fabric, plastic bottles
e
f
8
In condensation polymerisation a small molecule (water) is produced during the process,
whereas addition polymerisation is the addition of monomers without producing another
product.
a
b
i
The isomer with the side chain –OH will have a lower boiling point.
ii
The isomer with the side chain –OH has less surface area, so the forces of attraction are
less strong. Hence the heat energy required to separate the molecules (to melt or boil
the substance) is not so great.
Exam-style questions
Core
1
a
Ethanol
[1]
b
i
C6H12O6(aq) → 2C2H5OH(l) + 2CO2(g) [2]
ii
glucose [1]
iii 180 [1]
c
i
carboxylic acids [1]
ii
CnH2n+1COOH [1]
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Cambridge IGCSE Chemistry Workbook answers
Supplement
2
c
a
amine [1], carboxylic acid [1]
b
[6 for correct single bonds (1 each); 3 for non-bonding pairs, on O of OH, on CO and on N
(1 each), 1 for CO double bond]
i
condensation polymerisation [1]
ii
nylon [1]
iii
[1]
14 Experimental techniques and chemical analysis
Core
1
a
How close each measurement is to the true value. It depends on the quality of the
measuring apparatus (e.g. the thermometer or electronic balance) and on the skill of the
scientists taking the measurement.
b
i
stopwatch
ii
thermometer
iii balance
iv burette, pipette, measuring cylinder
c
v
gas syringe
i
one hundredth of a second
ii
one tenth of a degree
iii one hundredth of a gram
iv one tenth of a cm3
2
a
1, thermometer; 2, cold water out; 3, cold water in; 4, fractionating column; 5, heat; 6,
distillate; 7, mixture of liquids; 8, Liebig condenser
b
petroleum, the fractions within the mixture have different boiling points; liquid air, the
liquid gases in the mixture that is liquid air all have different boiling points
c
1, chromatography paper; 2, beaker; 3, pencil line; 4, solvent; 5, samples
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3
4
5
6
a
electrolysis
b
carbon or platinum
c
chlorine, Cl2
d
Ignite the gas with a lit splint – gives a ‘pop’ if it is hydrogen.
a
top left – liquid paraffin, right – aluminium oxide or broken pot
b
one arrow pointing at mineral wool, one arrow pointing at aluminium oxide or broken pot
c
to stop sucking back of water
d
Shake with orange bromine water – turns colourless if there is unsaturation
a
oxygen, O2
b
hydrogen, H2
c
sulfur dioxide, SO2
d
ammonia, NH3
e
chlorine, Cl2
a
i
calcium
ii
Chloride ions are present.
iii Ag+(aq) + Cl–(aq) → AgCl(s)
b
i
potassium
ii
sulfate
iii Ba2+(aq) + SO42–(aq) → BaSO4(s)
c
d
i
copper
ii
carbonate
i
CaCl2
ii
K2SO4
iii CuCO3
7
8
a
G = zinc carbonate, ZnCO3; H = carbon dioxide, CO2
b
I = zinc oxide, ZnO; J = zinc chloride, ZnCl2
c
K = zinc hydroxide, Zn(OH)2
d
L = silver chloride, AgCl
a
Compare with genuine banknote
Dissolve ink from supposed forgery and genuine banknote
Apply spots of inks to paper
Set up apparatus shown in the diagram
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Cambridge IGCSE Chemistry Workbook answers
Place in solvent or organic solvent or water
The spots rise up paper
Compare spot heights of samples on chromatogram
b
by use of a locating agent
c
Rf =
distance travelled by substance
distance travelled by solvent
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