HKDSE CHEMISTRY – A Modern View
(Second Edition)
(Reprinted with minor amendments 2019)
(Chemistry and Combined Science)
Coursebook 1
Suggested answers
Chapter 1
Fundamentals of Chemistry
Page
Number

Class practice
1

Think about
2

Chapter exercise
2
Chapter 2
The atmosphere

Class practice
4

Think about
5

Chapter exercise
5
Chapter 3
The ocean

Class practice
9

Think about
10

Chapter exercise
10
Chapter 4
Rocks and minerals

Class practice
14

Think about
14

Chapter exercise
15
© Aristo Educational Press Ltd. 2019

Part exercise
Chapter 5
18
Atomic Structure

Class practice
22

Self-test
24

Think about
24

Try it now
25

Chapter exercise
25
Chapter 6
The Periodic Table

Class practice
28

Think about
29

Chapter exercise
29
Chapter 7
Chemical bonding: ionic bonding and metallic bonding

Class practice
32

Think about
33

Try it now
34

Chapter exercise
34
Chapter 8
Chemical bonding: covalent bonding

Class practice
39

Think about
40

Try it now
40

Chapter exercise
41
Chapter 9
Structures and properties of substances

Class practice
45

Think about
46
© Aristo Educational Press Ltd. 2019

Chapter exercise
47

Part exercise
51
Chapter 10 Occurrence and extraction of metals

Class practice
56

Think about
57

Chapter exercise
57
Chapter 11 Reactivity of metals

Class practice
60

Think about
61

Try it now
62

Chapter exercise
62
Chapter 12 Reacting masses

Class practice
65

Self-test
69

Think about
72

Try it now
72

Chapter exercise
72
Chapter 13 Corrosion of metals and their protection

Class practice
80

Think about
80

Chapter exercise
81

Part exercise
84
© Aristo Educational Press Ltd. 2019
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 1
Coursebook 1
Fundamentals of Chemistry
Class Practice
A1.2 (p.1-8)

Student A should not run in the laboratory as she may knock on the other
students who are doing experiments and cause accidents. Besides, she should
wear safety spectacles.

Student B should not smell the gas directly as the gas may be poisonous.

Student C should wear safety spectacles when doing experiment as the chemical



may spill into her eyes. In addition, the flammable chemical should not be placed
near the flame as it catches fire easily.
Student E should not throw rubbish into the sink as the solid waste may block the
drain of the sink. In addition, he should wear safety spectacles.
Student F should not touch the hot beaker with bare hands as she would get heat
burns.
Student G should not eat in the laboratory as the food and drinks may be
contaminated with chemicals. Besides, he should wear safety spectacles.
A1.3 (p.1-12)
Name and vertical section diagram
Name and vertical section diagram
(a)
Test tube
(b)
Conical flask
(c)
Glass rod
(d)
Dropper
(e)
Tripod
(f)
Wire gauze
© Aristo Educational Press Ltd. 2019
1
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
(g)
Evaporating dish
(h)
Coursebook 1
Watch glass
Think about
Think about (p.1-2)
1. Chemistry is the study of substances, including their compositions, structures,
properties and the changes among them.
2.
3.
Many new substances can be made by applying the knowledge of Chemistry.
Hence, Chemistry helps provide us with a good standard of living and is
important for our survival.
We have to observe carefully and fully when doing experiment, report
experimental results clearly and accurately, analyse the results, try to interpret
them and finally draw conclusions for the experiment.
Chapter exercise (p.1-17)
1. matter; space; mass
2.
3.
4.
substances; compositions, changes
feeling, smelling; hearing
laboratory safety rules
5.
D
Refer to p.3 to 5 of chapter 1 for details.
6.
C
7.
C
Broken glass pieces should be cleaned up with a broom and the glass pieces
should be disposed of in a waste bin which is used for collecting broken glass.
8.
C
Hydrogen is flammable.
9.
D
Concentrated hydrochloric acid is corrosive.
10. D
© Aristo Educational Press Ltd. 2019
2
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
11. Any FIVE:
She did not stopper the reagent bottles after taking the chemicals out from them.
She did not tie up her long hair.
She did not wear safety spectacles (and laboratory coat).
She should NOT have heated the ethanol directly as ethanol is flammable.
She should not have pointed the test tube to herself.
She should not have placed the bottle of ethanol near the Bunsen flame as
ethanol is flammable.
12. (a)
(b) Wear safety spectacles.
Avoid contact with the eyes and skin. When accidentally spilt into the eyes
or onto the skin, wash the affected area with a lot of water.
(c) Colourless gas bubbles evolve from the egg shell pieces.
The egg shell pieces dissolve in the acid.
13. (a) A.
B.
C.
D.
E.
F.
G.
H.
(b) (i)
(ii)
(iii)
(iv)
Beaker
Test tube/boiling tube
Bunsen burner
Wire gauze
Tripod
Heat-resistant mat
Test tube holder
Spatula
H
B
B, C, F, G
A, C, D, E, F
© Aristo Educational Press Ltd. 2019
3
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 2
Coursebook 1
The atmosphere
Class practice
A2.1 (p.2-8)
(a) Element
(b) Compound. Hydrogen sulphide is made up of hydrogen and sulphur.
(c) Element
(d) Compound. Nitrogen dioxide is made up of nitrogen and oxygen.
A2.2 (p.2-10)
(a) Common salt is a compound because its appearance is different from that of their
constituent elements. Common salt is a white solid while sodium is a shiny
silvery solid and chlorine is a greenish yellow gas. Besides, sodium chloride has
no reaction with water while sodium reacts vigorously with water. Sodium
chloride is non-toxic while chlorine is toxic.
(b) Distillation
A2.3 (p.2-12)
1. (a) Chemical property
(b) Physical property
2.
State of
Temperature
nitrogen
argon
oxygen
(C)
(m.p. = 210C;
b.p.= 196C)
(m.p. = 189C;
b.p.= 186C)
(m.p. = 219C;
b.p.= 183C)
–205
liquid
solid
liquid
–188
gas
liquid
liquid
–185
gas
gas
liquid
–182
gas
gas
gas
A2.4 (p.2-13)
Elements: nitrogen, oxygen, argon, helium, neon, krypton and xenon
Compounds: water vapour and carbon dioxide
A2.5 (p.2-15)
(a) Fractional distillation of liquid air
(b) Nitrogen, argon, oxygen, krypton, xenon
© Aristo Educational Press Ltd. 2019
4
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
A2.6 (p.2-19)
1. No. This is because oxygen is colourless and many gases are also colourless.
2. (a) Test the three gases with a glowing splint. The gas which can relight the
glowing splint is oxygen.
(b) The gas which burns with a ‘pop’ sound is hydrogen.
The gas which does not show any observable change is oxygen.
The gas which extinguishes the burning splint is nitrogen.
Think about
Think about (p.2-2)
1. Nitrogen can be obtained from the atmosphere by fractional distillation of liquid
air.
2. Apart from nitrogen, oxygen, water vapour, carbon dioxide and noble gases such
as helium, neon and argon are present in the atmosphere.
Think about (p.2-4)
Pure substances that can be found in sea water include water, sodium chloride,
magnesium chloride, oxygen, carbon dioxide, etc.
Rocks have varied chemical compositions. But calcium carbonate is a pure substance
that is commonly found in rocks.
Chapter exercise (p.2-24)
1. Earth’s crust; ocean
2. mixture
3. element; compound; mixture
4. chemically; hydrogen; oxygen
5.
6.
7.
8.
9.
10.
chemical
sulphur; compound
retains; different
atmosphere
nitrogen; oxygen
(a) filtered; Carbon dioxide; water vapour
(b) compressed, liquid
(c) Fractional distillation; different
11. burning; glowing
12. D
(1): milk tea is a mixture of water, milk, tea and sugar.
(2): petroleum is a mixture of hydrocarbons.
© Aristo Educational Press Ltd. 2019
5
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(3): stainless steel is a mixture of iron, chromium, nickel and carbon.
13. A
(1): calcium oxide is a compound of calcium and oxygen.
(2): ethanol is a compound of carbon, hydrogen and oxygen.
14. C
Compound is a pure substance. Air is a mixture of compounds and elements.
Elements cannot be broken down into anything simpler by chemical methods.
15. D
Carbon dioxide is a compound of carbon and oxygen.
16. B
When a compound forms, heat is usually released or absorbed.
17. C
The reaction of nitrogen with oxygen to give nitrogen dioxide at high
temperature is a chemical property of nitrogen.
18. C
There is only one type of particles which are composed of particles of one
element in Diagram C.
19. B
There is only one type of particles which are composed of particles of two
different elements in Diagram B.
20. D
There are two types of particles in Diagram D. Each of these particles is
composed of particles of one element.
21. A
There are two types of particles in Diagram A. Each of these particles is
composed of particles of two different elements.
22. B
Refer to p.13 of chapter 2 for details.
© Aristo Educational Press Ltd. 2019
6
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
23. B
The respective boiling points of nitrogen and oxygen are 196C and 183C.
Nitrogen has a lower boiling point so it boils off before oxygen.
24. B
25. C
Nitrogen can be used as a refrigerant. Carbon dioxide can be used to make fire
extinguishers. Oxygen can be used in oxy-acetylene torch to cut and weld metals.
26. C
Oxygen is not flammable. It only supports burning.
27. (a) Oxygen, chlorine, sodium, hydrogen, iron, sulphur and mercury
(b) An element is a pure substance that cannot be broken down into anything
simpler by chemical methods.
(c) Ammonia, sodium chloride and water
(d) A compound is a pure substance made up of two or more elements
chemically combined together.
(e) A mixture consists of two or more pure substances which have not
chemically combined together.
(f) Sodium chloride solution is a mixture (because a solution is a mixture of
solute(s) and solvent).
28. (a) No. At room temperature and pressure, both hydrogen and oxygen are gases
while glucose is a solid. Carbon is black in colour while glucose is white.
(b) Glucose solution is a mixture. This is because there is no chemical reaction
taking place between glucose and water.
(c) New substances (carbon dioxide and water) form when glucose burns.
Hence, burning of glucose is a chemical property of glucose.
29. (a) The volume of argon, nitrogen and oxygen obtained are 940 litres, 78 000
litres and 21 000 litres respectively.
(b) Fractional distillation of liquid air
(c) No. Oxygen is the most reactive gas in air whereas nitrogen is unreactive. If
there were more oxygen in air, metals would be oxidized and corroded
faster. Things would also burn easier so there would be a higher risk of fire.
© Aristo Educational Press Ltd. 2019
7
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
30. (a) (i) Fractional distillation of liquid air
(ii) These gases have different boiling points. The gas with a lower boiling
point will boil off first.
(b) To remove the dust particles.
(c) By repeated cooling and compression of the purified air in the liquefaction
unit.
(d) Fractionating column
(e) (i) A, B, C
(ii) A: nitrogen; B: argon; C: oxygen
(f)
31. –
A: used to fill food packets/as a refrigerant/making ammonia (Any ONE)
B: to fill light bulbs
C: to support breathing for divers, fire-fighters, etc./to help patients with
breathing difficulties/to support burning of fuels (Any ONE)
The air is first purified to remove dust particles. The filtered air is then
cooled to –80C. At this temperature, carbon dioxide and water vapour
become solid and are removed from the filtered air.
–
–
–
The air is then liquefied by repeated cooling and compression.
The liquid air is warmed up bit by bit very slowly.
Different gases in air boil at different temperatures, so they can be collected
one by one.
–
The one boiling off first is nitrogen (boiling point: –196C). The second one
to be collected is argon (boiling point: –186C). Finally, oxygen (boiling
point: –183C) is collected.
32. (a) Fractional distillation of liquid air
(b) Oxidizing
(c) Physical property of oxygen:
Oxygen is colourless/odourless/has a boiling point of –183C. (Any ONE.
Accept other correct physical properties of oxygen.)
Chemical property of oxygen:
Oxygen reacts with hydrogen to give water/reacts with carbon to give
carbon dioxide. (Any ONE. Accept other correct chemical properties of
oxygen.)
(d) We can test oxygen using a glowing splint. Oxygen relights the glowing
splint.
© Aristo Educational Press Ltd. 2019
8
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 3
Coursebook 1
The ocean
Class practice
A3.1 (p.3-4)
1. (a) Sugar is the solute and water is the solvent.
(b) Magnesium chloride is the solute and water is the solvent.
(c) Iodine is the solute and alcohol is the solvent.
2. Solution A is more concentrated than solution B.
This is because solution A and solution B have the same volume but solution A
contains a larger amount of solute than solution B.
A3.2 (p.3-8)
Set-up for performing filtration:
filter paper
sand
filter funnel
salt and water
Set-up for performing evaporation:
evaporating dish
wire gauze
salt and water
heat
tripod
A3.3 (p.3-15)
(a) (i) Moisten a clean platinum wire with concentrated hydrochloric acid. Then,
dip the wire into a crushed sample (or solution) of the unknown salt. After
that, heat the end of the wire strongly in a non-luminous flame.
(ii) The unknown salt contains potassium.
(b) (i) Dissolve the unknown salt sample in deionized water. Then, add excess
dilute nitric acid to the salt solution. After that, add a few drops of silver
nitrate solution to the salt solution.
(ii) The unknown salt contains chloride.
© Aristo Educational Press Ltd. 2019
9
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
Think about
Think about (p.3-2)
1. Sea water contains water and about 3.5% by mass of dissolved substances. Most
of the dissolved substances are salts, e.g. sodium chloride, magnesium chloride,
sodium sulphate, etc.
2. Common salt can be obtained from sea water by filtration, followed by
evaporation or crystallization.
3. Uses of hydrogen: to make margarine, as rocket fuel, to make ammonia, etc.
Uses of chlorine: to sterilize swimming pool water, to make polyvinyl chloride
(PVC) and solvents such as the thinner used in the correction fluids, etc.
Uses of sodium hydroxide: to make soaps and drain cleaner, to neutralize acidic
effluents from factories, etc.
Think about (p.3-7)
Dip the glass rod to the hot solution and then take it out. If the immersed end becomes
cloudy within a few seconds, the solution is saturated enough to form crystals.
Think about (p.3-11)
Not all water vapour can be condensed into pure water.
Chapter exercise (p.3-22)
1. solute(s); solvent
2. saturated solution
3. filtration; crystallization
4. filtrate; residue
5. boiling; condensation
6.
7.
8.
9.
distillate; residue
lilac; golden yellow; brick-red; bluish green
white precipitate
water; white; blue; blue; pink
10. B
Refer to p.3 of chapter 3 for details.
11. C
Refer to p.3 of chapter 3 for details.
© Aristo Educational Press Ltd. 2019
10
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
12. B
Refer to p.5 of chapter 3 for details.
13. A
Sand can be removed from the mixture by filtration as sand is insoluble in water
and sugar is soluble in water. Then, water can be removed from the sugar
solution (filtrate) by evaporation. Sugar will be left behind as residue.
14. D
Since X contains sodium, it would give a golden yellow flame in the flame test.
Since X does not contain chloride, it would not show any observable change in
the silver nitrate test.
15. C
Refer to p.15 of chapter 3 for details.
16. A
Electrolysis of sea water produces hydrogen, chlorine and sodium hydroxide.
17. C
Refer to p.17 of chapter 3 for details.
18. Dissolve the mixture in distilled water. Filter the mixture and sand can be
separated out from the mixture as residue. After that, evaporate the filtrate
(sodium chloride solution) to dryness. The solid obtained is sodium chloride.
19. (a) Pure water can be obtained from sucrose solution by distillation. During
distillation, sucrose solution is heated to boil and the water changes into
steam. Then the steam is cooled and condenses into water, which is
collected as distillate.
(b) The direction of water flow in the condenser is wrong. The water should
enter the condenser from the lower opening.
The thermometer should not be dipped into the sea water. It should be
placed near the side-arm of the still head.
There are no anti-bumping granules in the round-bottomed flask. The
student should add a few anti-bumping granules to the sea water to prevent
bumping due to overheating.
© Aristo Educational Press Ltd. 2019
11
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
20. (a) Filtration
(b)
filter paper
sand
filter funnel
sodium chloride solution
(c) Distillation
(d)
clamp
thermometer
r
delivery tube
boiling tube
heat
anti-bumping
granule
sodium
chloride
solution
test tube
(receiver)
water
pure water
(e) Test for the presence of sodium: flame test
The sample gives a golden yellow flame if sodium is present.
Test for the presence of chloride: silver nitrate test
The appearance of a white precipitate indicates the presence of chloride.
21. (a) This is because some metals and metal compounds can give a characteristic
coloured flame when they are heated strongly.
(b) Moisten a clean platinum wire with concentrated hydrochloric acid. Then,
dip the wire into the sample to be tested. After that, heat the end of the wire
strongly in a non-luminous flame.
(c) Potassium: lilac; calcium: brick-red; copper: bluish green
22. (a) No. This is because the liquid may be unclean, harmful or even poisonous.
(b) Flame test
(c) Add excess dilute nitric acid to a sample of the liquid. Then, add silver
nitrate solution to the sample. If chloride is present, a white precipitate
forms.
© Aristo Educational Press Ltd. 2019
12
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(d) Add a few drops of the liquid to anhydrous copper(II) sulphate powder. If
water is present, the colour of the powder changes from white to blue. OR
Add a few drops of the liquid to a piece of dry cobalt(II) chloride paper. If
water is present, the colour of the paper changes from blue to pink.
(e) No. Even if the tests showed that sodium, chloride and water were present,
the liquid might not necessarily be sea water. For example, it might be just a
sodium chloride solution, without any other salts naturally present in sea
water.
23. (a)
(b)
(c)
(d)
Electrolysis means decomposition by electricity.
Gas X is hydrogen and gas Y is chlorine.
Sodium hydroxide solution
Uses of gas X: to make margarine/as a rocket fuel (Accept other correct
uses)
Use of gas Y: to sterilize swimming pool water/to make polyvinyl chloride
(PVC) (Accept other correct uses)
© Aristo Educational Press Ltd. 2019
13
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 4
Coursebook 1
Rocks and minerals
Class practice
A4.1 (p.4-8)
1. (a) Physical change. This is because no new substances are produced during the
process.
(b) Chemical change. This is because new substances, such as carbon, carbon
dioxide, etc. are produced when paper burns.
(c) Physical change. This is because no new substances are produced during the
process.
(d) Chemical change. This is because a new substance, copper, is produced
during the process.
heat
2.
(a) silver oxide  silver + oxygen
heat
(b) lead(II) oxide + carbon  lead + carbon dioxide
electrolysis
(c) aluminium oxide  aluminium + oxygen
A4.2 (p.4-14)
heat
(a) (i) calcium carbonate  calcium oxide + carbon dioxide
(ii) calcium oxide + water  calcium hydroxide
(iii) carbon dioxide + calcium hydroxide  calcium carbonate + water
(b) The rock dissolves and colourless gas bubbles evolve. (Magnesium carbonate
can also react with dilute hydrochloric acid. Carbon dioxide is produced in the
reaction.)
A4.3 (p.4-16)
(a) The white solid sample contains calcium and carbonate.
(b) Pass the gas into a test tube of limewater. If the gas is carbon dioxide, the
limewater turns milky.
(c) No. The results of the tests only show that the sample contains calcium and
carbonate.
Think about
Think about (p.4-2)
1. A rock is a solid mass of a mineral or a mixture of minerals.
2. Rocks have varied chemical compositions.
© Aristo Educational Press Ltd. 2019
14
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
3.
Coursebook 1
Common methods used to extract metals from ores are: mechanical separation,
heating the metal ore alone, heating the metal ore with carbon and electrolysis of
the molten ore.
Chapter exercise (p.4-22)
1. mineral; mixture
2. elements
3. physical
4. new substances
5.
6.
7.
8.
9.
heating
galena; carbon
bauxite; electrolysis
chalk; marble
Weathering
10. quicklime; slaked lime
11. milky
12. brick-red
13.
14.
15.
16.
17.
18.
19.
silver; oxygen
lead; carbon dioxide
calcium oxide; carbon dioxide
calcium hydroxide
calcium carbonate; water
calcium hydrogencarbonate
calcium chloride; carbon dioxide; water
20. D
A physical change is a change in which no new substances are produced.
21. D
Metal
Usual extraction method
Copper
Heating the copper ore with carbon
Iron
Heating the iron ore with carbon
Magnesium
Electrolysis of its molten ore
22. D
23. B
(2): Powdered limestone can be used to neutralize acidic soil.
© Aristo Educational Press Ltd. 2019
15
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
24. D
Refer to p.10 and 11 of chapter 4 for details.
25. D
Refer to p.12 of chapter 4 for details.
26. C
Refer to Figure 4.20 on p.14 of chapter 4 for details.
27. A
Limestone is insoluble in water and has no reaction with water.
28. D
When calcium carbonate is heated strongly, it decomposes to calcium oxide and
carbon dioxide. When silver oxide is heated strongly, it decomposes to silver and
oxygen.
29. A
The sample may contain calcium as it gives a brick-red flame in the flame test.
30. (a) Rainwater is slightly acidic because carbon dioxide in air dissolves slightly
in it, forming carbonic acid.
carbon dioxide + water  carbonic acid
The carbonic acid formed reacts with calcium carbonate in the rock,
forming calcium hydrogencarbonate.
calcium carbonate + carbonic acid  calcium hydrogencarbonate
Calcium hydrogencarbonate is soluble in water and thus the rock is slowly
worn away.
(b) To make cement/to make glass/to build statue/as the construction material
for some footpaths/ to neutralize acidic soil (Any THREE)
31. (a) Calcium carbonate
heat
(b) calcium carbonate  calcium oxide + carbon dioxide
© Aristo Educational Press Ltd. 2019
16
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(c)
limestone
limewater
heat
(d) When the gas is passed into the limewater, the limewater turns milky.
32. (a) Limestone/marble/chalk (Any ONE)
(b) A is calcium oxide and B is carbon dioxide.
(c) (i) Quicklime
(ii) Limewater
(d) The calcium hydroxide solution turns milky.
calcium hydroxide + carbon dioxide  calcium carbonate + water
(e) The calcium hydroxide solution turns milky and then becomes colourless
again. This is because the insoluble calcium carbonate reacts with excess
carbon dioxide and water to form soluble calcium hydrogencarbonate.
33. Moisten a clean platinum wire with concentrated hydrochloric acid. Then, dip the
wire into the sample of crushed chalk. After that, heat the end of the wire
strongly in a non-luminous flame. If calcium is present in the sample, a brick-red
flame is seen.
© Aristo Educational Press Ltd. 2019
17
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
Part I Planet Earth
Part exercise (p.4-26)
1. A
(3): Some compounds form naturally on the Earth.
2.
A
(2): The composition by mass of a mixture is variable. The substances in the
mixture can be mixed together in any proportions.
(3): It is not easy to separate the components of some mixtures. For example, it is
not easy to separate iron and carbon in steel.
3.
B
Sublimation of dry ice is a physical change.
4.
C
(1): Nitrogen is an element, which is a pure substance.
5.
D
Refer to p.14 of chapter 2 for details.
6.
A
Hydrogen, but not oxygen, burns with a ‘pop’ sound when a burning splint is
brought near to it.
7.
A
If 100 cm3 of water can dissolve a maximum of 30 g of salt Y, in other words, 50
cm3 of water can dissolve a maximum of 15 g of salt Y and 200 cm3 of water can
dissolve a maximum of 60 g of salt Y.
8.
A
Sea water is first filtered to remove any insoluble substances. Common salt can
then be separated from the filtered sea water by evaporation.
9.
B
Refer to Figure 3.11 on p.10 of chapter 3 for details.
© Aristo Educational Press Ltd. 2019
18
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
10. B
Potassium-containing compounds give a lilac flame and copper(II)-containing
compounds give a bluish-green flame in the flame test.
11. C
The presence of chloride in a sample can be tested by using acidified silver
nitrate solution. If chloride is present in the sample, a white precipitate forms.
12. D
Haematite is the main ore of iron. It is mostly iron(III) oxide.
13. D
A chemical change is a change in which one or more new substances are
produced.
14. B
Copper is extracted by heating its ore with carbon.
15. C
Refer to Figure 4.20 on p.14 of chapter 4 for details.
16. B
Refer to p.12 of chapter 4 for details.
17. C
The carbonate in a sample reacts with dilute hydrochloric acid to give carbon
dioxide.
18. B
Solid P gives a golden yellow flame in the flame test. This shows that P contains
sodium.
Solid P has no reaction with dilute hydrochloric acid. This shows that P does not
contain carbonate.
The positive result of silver nitrate test shows that P contains chloride.
19. A
(1): Carbon dioxide reacts with calcium hydroxide solution to give a white
precipitate of calcium carbonate.
© Aristo Educational Press Ltd. 2019
19
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(2): Silver nitrate solution reacts with the chloride in sea water to give a white
precipitate of silver chloride.
(3): Water only turns white anhydrous copper(II) sulphate to blue. No white
precipitate would form.
20. (a) Dead Sea water has a higher concentration of dissolved substances.
(b) Sodium chloride/magnesium chloride/calcium chloride, etc. (Any ONE)
(c) No. Dead Sea water contains other salts apart from sodium chloride. These
salts will crystallize out together with common salt from Dead Sea water.
21. (a) X: hydrogen; Y: chlorine; Z: sodium hydroxide solution
(b) The process is a chemical change because new substances (hydrogen,
chlorine and sodium hydroxide solution) are produced during the process.
electrolysis
(c) sea water  hydrogen + chlorine + sodium hydroxide solution
(d) X: flammable; Y: toxic; Z: corrosive
22. (a) Nitrogen: used in food packaging/as a refrigerant/making ammonia (Any
ONE)
Oxygen: to support breathing for divers, fire-fighters, etc./to help patients
with breathing difficulties/to support burning of fuels (Any ONE)
(b) (i) No. Oxygen and nitrogen inside the container mix to form a gaseous
mixture. A mixture is an impure substance.
(ii) It is a physical change as no new substances are produced during the
process.
(c) (i) The product is water. It is a compound.
(ii) It is a chemical change as a new substance, water, forms during the
process.
(d) (i) Fractional distillation of liquid air
(ii) This property is boiling point, which is a physical property of
substances.
23. (a) (i) Calcium carbonate
(ii) Calcium oxide
heat
(b) Calcium carbonate  calcium oxide + carbon dioxide
(c) carbon dioxide
(d) It can be used to neutralize acidic soil.
© Aristo Educational Press Ltd. 2019
20
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
24. (a) Since the sample gave a brick-red flame in the flame test, the sample
contains calcium.
Since the sample gave colourless gas bubbles when reacted with dilute
hydrochloric acid, the sample contains carbonate.
Since the sample gave a positive result in the silver nitrate test, the sample
contains chloride.
Thus, the sample should be a mixture of calcium carbonate and calcium
chloride.
(b) Add distilled water to the mixture to dissolve the calcium chloride.
Filter the mixture. The residue left on the filter paper is calcium carbonate.
filter paper
calcium carbonate
filter funnel
calcium chloride solution
Evaporate the filtrate to dryness. The residue collected is calcium chloride.
evaporating dish
wire gauze
calcium chloride
solution
heat
tripod
(c) (i) Pass the gas into limewater.
(ii) If the gas is carbon dioxide, the limewater turns milky.
carbon dioxide + calcium hydroxide  calcium carbonate + water
25. Perform flame test on the two samples separately. The one which gives a lilac
flame is potassium chloride. The one which gives a brick-red flame is calcium
carbonate.
Add the two samples into water separately. The one which dissolves in water is
potassium chloride. The one which does not dissolve in water is calcium
carbonate.
Add dilute hydrochloric acid to the two samples separately. The one which gives
colourless gas bubbles is calcium carbonate. The one which has no observable
change is potassium chloride.
© Aristo Educational Press Ltd. 2019
21
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 5
Coursebook 1
Atomic Structure
Class practice
A5.1 (p.5-7)
1. (a) P is a non-metal because it is brittle and does not conduct electricity.
Although Q conducts electricity, it is brittle. Hence, Q is a non-metal. R is a
semi-metal because it conducts electricity only when heated or slightly
impure. S is a metal because it is hard and strong. Moreover, it conducts
electricity.
2.
(b) Q could be graphite.
(a) Mercury. All are metals. Mercury is a liquid, while others are solids at room
conditions.
(b) Sulphur. Sulphur is a non-metal, while others are metals.
(c) Iodine. All are non-metals. Iodine is a solid, while others are gases at room
conditions.
(d) Graphite. All are non-metals. Graphite conducts electricity, while others are
non-conductors of electricity.
(Accept other reasonable answers.)
A5.2 (p.5-10)
1. A, B, D
2. (a) D
(b) A
3. Diameter of an atom = 2 × 10–10 × 2 × 1000 mm = 4 × 10–7 mm
Number of atoms =
A5.3 (p.5-12)
(a) (i) Mg
(b) (i) N
(c) (i) fluorine
1 mm
= 2.5 × 106
4  10 7 mm
(ii) O
(ii) Na
(ii) chlorine
(iii) He
(iii) Br
(iii) mercury
A5.4 (p.5-15)
(a) Hydrogen atom
(b) 91 electrons. The number of neutrons cannot be predicted from the given data.
(c) It is not an atom. The numbers of protons and electrons are not equal.
© Aristo Educational Press Ltd. 2019
22
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
A5.5 (p.5-19)
1. (a) Aluminium
(b)
27
13Al
(c) (i) 13
(ii) 13
(iii) 27 – 13 = 14
2.
3.
4.
Number of
Atomic
Mass
number
number
protons
neutrons
electrons
Neon
10
20
10
10
10
Potassium
19
39
19
20
19
Gold
79
197
79
118
79
Iron
26
56
26
30
26
Element
B
D
A5.6 (p.5-22)
1. (a) 3
(b)
2.
16
8O
A and D are isotopes because they have the same number of protons but different
numbers of neutrons.
A5.7 (p.5-26)
1. (a) 1
(b) 2
(c) 4
(d) 235
(e) We cannot tell from the given data because the mass number is not given.
2. Relative atomic mass of magnesium
= 24 × 78.6% + 25 × 10.1% + 26 × 11.3% = 24.3
3. Let the relative abundance of 85Q and 87Q be y% and (100 – y)% respectively.
85.5 = 85 × y% + 87 × (100 – y)%
8550 = 85y + 8700 – 87y
y = 75
Thus, the relative abundance of 85Q is 75% and that of 87Q is 25%.
© Aristo Educational Press Ltd. 2019
23
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
A5.8 (p.5-29)
1. (a) 2
(b) 2,5
(c) 2,8,7
(d) 2,8,8,1
2.
(a)
(b)
(c)
(d)
Self-test
Self-test 5.2 (p.5-25)
Relative atomic mass of boron = 10 × 19.7% + 11 × 80.3% = 10.8
Self-test 5.3 (p.5-25)
Let the relative abundance of 63X be y% and that of 65X be (100 – y)%.
63.5 = 63 × y% + 65 × (100 – y)%
6350 = 63y + 6500 – 65y
∴ y = 75
(100 – y) = 100 – 75 = 25
Thus, the relative abundance of 63X is 75% and that of 65X is 25%.
Think about
Think about (p.5-2)
1. Yes. The smaller particles present in an atom are protons, neutrons and electrons.
2. Refer to Section 5.2.
3. Refer to Section 5.3.
Think about (p.5-4)
The brown solid is copper. It is a metal.
Try it now (p.5-19)
1
○
Atomic number of
40
18Ar
= 18; mass number of
© Aristo Educational Press Ltd. 2019
40
18Ar
= 40
24
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
2
○
Number of protons in
40
18Ar
Coursebook 1
= 18
3
○
Number of electrons in
40
18Ar
= 18
4
○
Number of neutrons in
40
18Ar
= 40 – 18 = 22
Chapter exercise (p.5-35)
1. metals; semi-metals
2.
3.
4.
5.
6.
bromine; mercury
metals; graphite
atom
element
atoms
7.
8.
9.
symbol
nucleus; neutrons; electrons
proton; electron; neutron
10.
11.
12.
13.
14.
15.
16.
atomic
mass
Isotopes; protons
carbon-12
relative isotopic masses
electron shells
electronic arrangement
17. (a)
Atom
7
24
Number of
Atomic
Mass
number number protons neutrons electrons
Electronic
arrangement
Li
3
7
3
4
3
2,1
Mg
12
24
12
12
12
2,8,2
18
40
18
22
18
2,8,8
19
39
19
20
19
2,8,8,1
40
Ar
39
K
(b)
© Aristo Educational Press Ltd. 2019
25
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
18. B
19. A
20. D
21. A
The atom of 11H has no neutron.
22. C
An atom of the
13
6C
isotope contains 6 electrons.
23. C
24. B
In the nucleus of an oxygen atom, there are 8 protons and 8 neutrons.
In the nucleus of an atom of 11H, there is 1 proton but no neutron.
25. C
Let the relative abundance of 69
31X be y% and that of
69.7 = 69 × y% + 71 × (100 – y)%
6970 = 69y + 7100 – 71y
y = 65
Thus, the relative abundance of
69
31X
71
31X
is 65% and that of
be (100 – y)%.
71
31X
is 35%.
26. B
The innermost shell holds two electrons and the outermost shell holds five
electrons. The atomic number of Q can be 2 + 5 = 7.
27. (a) True. This is because there is no gaseous metal or semi-metal under room
conditions.
(b) False. This is because mercury is a liquid metal under room conditions.
(c) False. This is because carbon (graphite) is a non-metal which can conduct
electricity. /This is because pure semi-metals cannot conduct electricity
under room conditions.
(d) False. This is because some non-metals (e.g. diamond and graphite) have
high melting points and boiling points.
(e) True. This is because non-metals are not malleable and not ductile.
© Aristo Educational Press Ltd. 2019
26
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
28. (a) Q and R
(b) Carbon
(c) Carbon-13 and carbon-14
(d)
13
13
14
14
5 P, 6 Q, 6 R, 7 S
29. (a) 235
(b) Neutron
(c) It has 56 protons, 56 electrons and 88 neutrons.
(d) Relative atomic mass of uranium
= 234 × 0.0055% + 235 × 0.72% + 238 × 99.27% = 238
30. (a) Making transistors/computer chips.
(b) The electrical conductivity of silicon can be increased by heating it.
(c) (i)
Isotopes are different atoms of the same element, with the same
number of protons but different numbers of neutrons.
(ii) 28
(iii) Relative atomic mass of silicon
= 28 × 92.23% + 29 × 4.68% + 30 × 3.09% = 28.1
(d) (i) Atomic number of an atom is the number of protons in the atom.
Mass number of an atom is the sum of the numbers of protons and
neutrons in the atom.
(ii)
Atom
Number of
protons
Number of
neutrons
Electronic
arrangement
28
14
Si
14
14
2,8,4
29
14
Si
14
15
2,8,4
30
14
Si
14
16
2,8,4
© Aristo Educational Press Ltd. 2019
27
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 6
Coursebook 1
The Periodic Table
Class practice
A6.1 (p.6-5)
(a) Period 4, Group VII
(b) Halogens
(c) Bromine
A6.2 (p.6-10)
1. (a) 2,8,8,2
(b) Yes, because it is a metal.
(c) 12R
2. (a)
(b) 3X has two occupied electron shells and
11Y
has three occupied electron
shells.
(c) Yes, this is because they have the same number of outermost shell electrons.
A6.3 (p.6-19)
(a) (i) They both have one (the same number of) outermost shell electron.
(ii) Lithium atom has two occupied electron shells while sodium atom has three
/ they have different numbers of occupied electron shells.
(b) Both of them are soft metals. They can be cut with a knife.
Both of them have low densities.
(c) The reactivity of Group II elements increases down the group.
(d) Potassium. Group I elements are generally more reactive than Group II elements.
Moreover, the reactivity of Group I elements increases down the group.
(e) Both of them react readily with dilute hydrochloric acid to form hydrogen.
(f) Fluorine. The reactivity of Group VII elements decreases down the group.
(g) This is because Group 0 elements have a stable electronic structure/a duplet or an
octet of electrons.
A6.4 (p.6-20)
(a) 2
(b) They have the same number of outermost shell electrons in their atoms.
(c) The reactivity of Group II elements increases down the group.
© Aristo Educational Press Ltd. 2019
28
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(d) (i) Beryllium has no reaction with cold water.
(ii) Barium reacts vigorously with cold water.
(e) Hydrogen
(f) Barium is more reactive than strontium. It should be stored under paraffin oil.
Think about
Think about (p.6-2)
1. They are classified in categories according to their uses.
2. It would be very difficult for customers to find what they want in a short period
3.
of time.
Yes. Elements with similar chemical properties are arranged in the same group in
the Periodic Table.
Chapter exercise (p.6-25)
1. atomic number
2. period; group
3. metals; semi-metals; non-metals
4.
5.
6.
7.
8.
9.
electrons; outermost
alkali; one; increases
alkaline earth; two; increases
seven; halogens; decreases
eight; noble gases
octet rule
10. (a)
Electronic
Period
arrangement
number
5
2,3
2
III
Q
8
2,6
2
VI
R
10
2,8
2
0
S
11
2,8,1
3
I
T
14
2,8,4
3
IV
U
20
2,8,8,2
4
II
Element
Atomic Number
P
Group number
(b) Metals: S, U
Non-metals: Q, R
Semi-metals: P,T
(c) R
© Aristo Educational Press Ltd. 2019
29
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
11. B
Refer to p.3 of chapter 6 for details.
12. B
The electronic arrangement of W is 2,5 and that of Y is 2,8,5.
13. C
The electronic arrangement of R is 2,8,1 and that of S is 2,8,8. They have the
same number of occupied electron shells in their atoms.
14. A
Elements of the same period have the same number of occupied electron shells in
their atoms.
15. C
The electronic arrangement of calcium is 2,8,8,2. Calcium has the same number
of outermost shell electrons as element X.
16. C
Helium atom has two electrons in the outermost shell.
17. C
Helium, instead of argon, is used to fill balloons and airships.
18. B
(1): Cl2(g), Br2(l) and I2(s) belong to the same group, but they have different
physical states.
(3): The reactivity of Group VII elements decreases down the group.
19. C
Potassium is more reactive than lithium. Fluorine is more reactive than bromine.
20. (a) Non-metal
(b) 2,8,5
(c) 5
(d) Group V, Period 3
21. (a) r = 18; s = 4
© Aristo Educational Press Ltd. 2019
30
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(b) 50
(c) Period 5
(d) Any TWO:
‒
It is a shiny solid at room temperature and pressure.
‒
It is hard and strong.
‒
It is malleable and ductile.
‒
It is a conductor of heat and electricity.
(Accept other reasonable answers.)
22. (a)
(b)
(c)
(d)
(e)
Magnesium, silicon and chlorine. They are in Period 3.
Chlorine, bromine and iodine. They are in Group VII.
Magnesium and helium
Oxygen, chlorine and helium
Rubidium
(f) Iron and copper
(g) Helium
(h) Bromine and iodine
23. (a) Group 0
(b) Noble gases
(c) (i) 2
(ii) 8
(d) Xenon is a gas under room conditions.
(e) Xenon is unreactive. This is because it has a stable electronic structure/has
an octet of electrons/has 8 electrons in the outermost shell.
(f) The balloon falls to the ground because xenon is denser than air.
24. (a) 7
(b) Halogens
(c) The colours of elements become darkened down the group. They change
from greenish yellow gas (chlorine) to dark red liquid (bromine) and then
black solid (iodine).
(d) (i) Black solid
(ii) There is no reaction between iodine and hydrogen. The reactivity of
Group VII elements decreases down the group. Hence astatine should
have no reaction with hydrogen.
© Aristo Educational Press Ltd. 2019
31
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 7
Coursebook 1
Chemical bonding: ionic bonding and metallic
bonding
Class practice
A7.1 (p.7-3)
Conductors: calcium, lithium
Electrolytes: calcium chloride, sodium bromide
Non-conductors: nitrogen, glucose
A7.2 (p.7-8)
(a) Z
(b) W and X
(c) (i) X and Y
(ii) X:
Y:
A7.3 (p.7-12)
1
(a) Simple ions: H+, Mn2+, O2, Cu2+
Polyatomic ions: NH4+, OH, MnO4
(b) H+: hydrogen ion, NH4+: ammonium ion, OH: hydroxide ion,
Mn2+: manganese(II) ion, O2: oxide ion, Cu2+: copper(II) ion,
MnO4: permanganate ion
2.
Group
Charge of ions
3.
I
II
III
V
VI
VII
+1
+2
+3
3
2
1
(a) strontium ion: +2; astatide ion: 1
(b) Sr2+, At
A7.4 (p.7-15)
(a)
(b)
(c)
(d)
© Aristo Educational Press Ltd. 2019
32
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
A7.5 (p.7-18)
1
Cation
(a)
2.
NH4
+
2+
Anion
Cl
–
–
Name of the compound
Formula of the compound
Ammonium chloride
NH4Cl
(b)
Mg
OH
Magnesium hydroxide
Mg(OH)2
(c)
Li+
H–
Lithium hydride
LiH
Zinc sulphate
ZnSO4
Ammonium nitrate
NH4NO3
2+
(d)
Zn
(e)
NH4+
(a)
(b)
(c)
(d)
SO4
2–
NO3–
Calcium nitrate
Iron(III) oxide
Aluminium hydroxide
Magnesium sulphide
A7.6 (p.7-21)
(a) Colourless
(b) Purple
(c) Orange
(d) Pale green
A7.7 (p.7-24)
(a) Potassium ions migrate towards the negative electrode. Potassium ions are
positively charged and therefore they are attracted towards the negative
electrode.
(b) No. Potassium ions are colourless.
(c) A green spot would move slowly towards the negative electrode. Chromium(III)
ions are green in colour and positively charged. They are attracted towards the
negative electrode.
A7.8 (p.7-25)
(a) 3
(b) 3
(c) Metallic bonding
Think about
Think about (p.7-2)
1. An ion is an atom or a group of atoms having an overall electric charge.
2. Unlike ion, an atom is overall electrically neutral.
3. No. Ions can be produced by either losing or gaining electron(s).
© Aristo Educational Press Ltd. 2019
33
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
Think about (p.7-24)
Ionic bonds are the attractions between oppositely charged ions. Metallic bonds are
the attractions between delocalized electrons and positively charged metal ions.
Try it now (p.7-18)
1
○
Al3+ O2 –
2
○
Al3+ O2 –
3
○
Al3+ O2 –
= Al2
O3
4
○
Al2O3
Chapter exercise (p.7-31)
1. conductors; electrolytes
2. ionic; covalent; metallic
3. electrons; noble gas
4. cations; anions
5. simple; polyatomic
6. group
7. 8
8.
9.
transferred; Calcium (Ca2+); oxide (O2); ionic bonds
formula
10. colourless
11. migration
12. delocalized; ions
13. (a)
(b)
(c)
(d)
© Aristo Educational Press Ltd. 2019
34
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(e)
14. (a)
(b)
(c)
(d)
(e)
15.
Name of compound
Formula of compound
(a) Lithium chloride
LiCl
(b) Copper(II) oxide
CuO
(c) Mercury(I) nitrate
HgNO3
(d) Potassium dichromate
K2Cr2O7
(e) Iron(III) hydroxide
Fe(OH)3
(f)
Silver oxide
Ag2O
(g) Barium sulphate
BaSO4
(h) Aluminium hydride
AlH3
(i)
Zinc nitride
Zn3N2
(j)
Copper(I) bromide
CuBr
16.
Formula of compound
Name of compound
Colour of solution
potassium permanganate
purple
(b) FeCl2
iron(II) chloride
pale green
(c) NiSO4
nickel(II) sulphate
green
iron(III) nitrate
yellow or brown
cobalt(II) chloride
pink
ammonium chloride
colourless
(g) KNO2
potassium nitrite
colourless
(h) Na2CrO4
sodium chromate
yellow
(a) KMnO4
(d) Fe(NO3)3
(e) CoCl2
(f)
NH4Cl
© Aristo Educational Press Ltd. 2019
35
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
(i)
CuSO3
copper(II) sulphite
Coursebook 1
blue or green
17. A
Mg2+ ion has 12 protons and 10 electrons.
18. B
(1): Electronic arrangement of K+: 2,8,8
(2): Electronic arrangement of Al: 2,8,3
(3): Electronic arrangement of Cl–: 2,8,8
19. B
In particle X, the number of protons is larger than that of electrons.
20. A
21. A
Fe2O3 is the formula of iron(III) oxide.
The formula of potassium permanganate is KMnO4.
CuOH is the formula of copper(I) hydroxide.
22. C
The electronic arrangements of X and Y are 2,8,8,1 and 2,6 respectively.
23. D
K+(aq) is colourless.
24. C
25. B
26. C
27. (a) copper, gold, lithium, mercury
(b) iodine, water, oil, sugar
(c) sodium chloride, lead(II) bromide, copper(II) nitrate, calcium fluoride
28. (a) Charge on the ion of X = +2
(b) Charge on the ion of Y = 3
© Aristo Educational Press Ltd. 2019
36
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(c) X atom has 2 more electrons than X2+ ion, hence its electronic arrangement
is 2,8,2. Y atom has 3 less electrons than Y3 ion, hence its electronic
arrangement is 2,5.
(d) X is magnesium. Y is nitrogen.
29. (a) Calcium atom:
Chlorine atom:
(b) Metallic bonding
(c) Calcium can conduct electricity/is a shiny silvery solid under room
conditions. (Accept other appropriate answers.)
(d)
(e) Ionic bonding
(f) Calcium chloride solution is colourless. Calcium chloride solid is white in
colour.
30. (a) To increase the electrical conductivity of the filter paper.
(b) Permanganate ion
(c) Electrode X is the positive electrode because negatively charged
permanganate ions are attracted towards the positive electrode.
(d) The purple spot would move towards electrode Y because negatively
charged permanganate ions are attracted towards electrode Y, which is now
the positive electrode.
31. (a)
(b)
(c)
(d)
Calcium sulphate
CaSO4
Cation: calcium ion; Anion: sulphate ion
Ionic bonding
(e) The coagulant is white in colour.
(f) Polyatomic ion. This is because it is derived from a group of atoms.
32. (a) Ionic bonding
© Aristo Educational Press Ltd. 2019
37
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(b) A strontium atom has 2 outermost shell electrons while a chlorine atom has
7 outermost shell electrons. To get the electronic arrangement of the nearest
noble gas, a strontium atom loses 2 electrons and each of the two chlorine
atoms gains 1 electron. By transfer of electrons, strontium chloride is
produced.
(c)
(d) SrCl2
(e) (i) Colourless
(ii) Colourless
© Aristo Educational Press Ltd. 2019
38
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 8
Coursebook 1
Chemical bonding: covalent bonding
Class practice
A8.1 (p.8-5)
1. CH3OH, I2, HCl
2. (a) 1
(c) 1
(e) 5
(b) 2
(d) 2
A8.2 (p.8-11)
1.
(a)
(b) Covalent bonding. It forms when each atom of X contributes two outermost
shell electrons for sharing.
(c) There are two bond pairs and two lone pairs on each atom of X.
2.
(a)
(b) There are three bond pairs and one lone pair on the nitrogen atom.
(c) Molecular formula: NCl3
Structural formula:
A8.3 (p.8-13)
1.
A8.4 (p.8-14)
(a) CF4
2.
(b) H2S
A8.5 (p.8-15)
(a) Hydrogen chloride
© Aristo Educational Press Ltd. 2019
(c) PH3
(d) SiCl4
(b) Carbon monoxide
39
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
(c) Carbon dioxide
(e) Sulphur trioxide
Coursebook 1
(d) Sulphur dioxide
A8.6 (p.8-18)
1. (a) 12.0 + 1.0 × 4 = 16.0
(b) 12.0 × 2 + 1.0 × 6 = 30.0
(c) 12.0 × 12 + 1.0 × 22 + 16.0 × 11 = 342.0
2. (a) 23.0 + 35.5 = 58.5
(b) 12.0 × 2 + 1.0 × 6 = 30.0 (same as relative molecular mass)
(c) 12.0 + 16.0 × 3 = 60.0
(d) 63.5 + (14.0 + 16.0 × 3) × 2 + 3 × (1.0 × 2 + 16.0) = 241.5
A8.7 (p.8-19)
(a) Molecule
(b) Molecule
(c) Ion
(e) Molecule
(g) Molecule
(d) Atom
(f) Ion
(h) Atom
Think about
Think about (p.8-2)
1. A covalent compound is a compound in which the atoms are held together by
covalent bonds.
2. An ionic compound is usually produced by combining a metal with a non-metal.
It consists of positive ions and negative ions held together by ionic bonds. In a
covalent compound, non-metal atoms are held together by covalent bonds.
3. Refer to Table 8.2 on p.9 of this chapter for more examples.
Try it now (p.8-14)
1
○
C
Cl
2
○
C
Cl
2,4
3
○
C
4
4
○
C
2,8,7
Cl
1
Cl
© Aristo Educational Press Ltd. 2019
40
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
4
= C1
Coursebook 1
1
Cl4
5
○
CCl4
Chapter exercise (p.8-25)
1. molecule
2. atomicity
3. share; covalent
4.
5.
6.
7.
8.
two; double
three; triple
molecular
structural
dative
9. Relative molecular mass
10. Formula mass
11. electrostatic; non-directional; directional
12. molecules; ions
13. (a)
(b)
(c)
(d)
(e)
(f)
14. Electron diagram of NH4Cl:
© Aristo Educational Press Ltd. 2019
Electron diagram of CHCl3:
41
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
15.
Molecular formula
Name of the
of the compound
compound
(a) Carbon and oxygen
CO2
carbon dioxide
(b) Nitrogen and hydrogen
NH3
ammonia
(c) Carbon and fluorine
CF4
carbon tetrafluoride/
tetrachloromethane
(d) Nitrogen and chlorine
NCl3
nitrogen trichloride
(e) Carbon and hydrogen
CH4
methane
(f)
PCl3
phosphorus trichloride
Constituent elements
Phosphorus and chlorine
16. C
X is a metal. It reacts with W and Y to form ionic compounds instead of covalent
compounds.
Z is a noble gas. It does not form compounds with other elements.
17. D
(A): An ionic compound, NaF, forms when sodium reacts with fluorine.
(B): A covalent compound, CS2, forms when carbon reacts with sulphur.
(C): A covalent compound, NO2, forms when nitrogen reacts with oxygen.
(D): A covalent compound, Cl2O, forms when oxygen reacts with chlorine.
© Aristo Educational Press Ltd. 2019
42
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
18. B
No. of bond pair of electrons
N2
3
H2
1
O2
2
NH3
3
19. A
No. of lone pair of electrons
HCl
3
NH3
1
N2
2
H2O
2
20. C
(A): It is the chemical symbol of hydrogen.
(B): It is the structural formula of hydrogen.
(D): It is the electron diagram of hydrogen.
21. C
22. A
An atom of element X needs three electrons to attain the stable electronic
arrangement while an atom of Y needs one.
23. B
Both relative molecular mass and formula mass carry no units.
24. (a) A: 2,1; B: 2,4; C: 2,6; D: 2,8; E: 2,8,2; F: 2,8,7
(b) Element D
(c) (i)
(ii)
(d) BC2; C=B=C
(e) A diatomic molecule refers to a molecule of an element or a compound
which consists of 2 atoms (i.e. its atomicity is 2).
© Aristo Educational Press Ltd. 2019
43
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
(f)
25. (a)
(b)
(c)
(d)
(e)
(i)
Coursebook 1
(ii)
Chlorine, carbon monoxide, nitrogen, water, oxygen, helium
Carbon monoxide, water
Ammonium chloride, chlorine, carbon monoxide, nitrogen, water, oxygen
Ammonium chloride
Ammonium chloride, sodium chloride
26. (a)
(b) Covalent bonding is present in an ammonia molecule.
(c) There are three bond pairs and one lone pair of electrons in an ammonia
molecule.
(d) (i) Molecular formula: NH3; structural formula:
(ii) Molecular formula: HCl; structural formula: HCl
(e) Ammonium chloride
(f)
In ammonium chloride, there are four NH covalent bonds, in which three
are normal covalent bonds and one is dative covalent bond.
Ionic bond is present between NH4+ and Cl ions.
Electron diagram of ammonium chloride:
© Aristo Educational Press Ltd. 2019
44
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 9
Coursebook 1
Structures and properties of substances
Class practice
A9.1 (p.9-8)
1. Iodine, bromine, chlorine and fluorine have simple molecular structures and their
molecules are held together by weak van der Waals’ forces. The larger the
molecular size, the stronger are the van der Waals’ forces. The molecular sizes of
the substances are: I2 > Br2 > Cl2 > F2. Therefore, the strength of the van der
Waals’ forces are: I2 > Br2 > Cl2 > F2.
2.
(a) Covalent bonding
(b) Intermolecular forces/van der Waals’ forces
(c) Sulphur is a low-melting solid. This is because it has a simple molecular
structure. The sulphur molecules are held together by weak van der Waals’
forces. Only a small amount of heat energy is needed to separate the
molecules during melting.
(d) No. This is because there are no mobile ions or delocalized electrons in the
structure.
(e) No
A9.2 (p.9-12)
(a) Giant covalent structure
(b) Covalent bonding
(c) Quartz is a high-melting solid. This is because a lot of heat energy is needed to
break a large number of strong covalent bonds in the structure.
(d) Quartz is neither soluble in water nor heptane.
(e) No. This is because there are no mobile ions or delocalized electrons in the
structure.
A9.3 (p.9-15)
(a) Giant ionic structure
(b) Ionic bonding
(c) Caesium chloride is a high-melting solid. This is because a lot of energy is
needed to break the strong ionic bonds between the ions during melting.
(d) Caesium chloride is soluble in water. This is because when dissolved in water,
strong attraction exists between ions in caesium chloride and water molecules.
However, there is no such attraction between ions in caesium chloride and
heptane molecules. Hence, caesium chloride is insoluble in heptane.
(e) No
© Aristo Educational Press Ltd. 2019
45
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
A9.4 (p.9-17)
(a) Tungsten has a very high melting point.
(b) This is because there are delocalized electrons in the structure.
(c) When a force is applied to a piece of tungsten metal, the layers of tungsten ions
can slide over one another without breaking the metallic bonds.
A9.5 (p.9-20)
(a) B. This is because it conducts electricity in the solid state.
(b) D. This is because it does not conduct electricity in the solid state but conducts
electricity when molten.
(c) A. This is because it has a low melting point and a low boiling point. Besides, it
does not conduct electricity no matter it is in the solid state or when molten.
(d) C. This is because it has a high melting point and a high boiling point. Besides, it
does not conduct electricity no matter it is in the solid state or when molten.
(e) A. This is because substances with simple molecular structure are usually soluble
in non-aqueous solvent.
A9.6 (p.9-21)
(a) (i) MgBr2
(ii) It has a giant ionic structure.
(iii) Its physical properties are: (1) High melting point and boiling point (2)
Solid at room temperature and pressure (3) Soluble in water but insoluble in
non-aqueous solvents (4) Non-conductor of electricity in the solid state;
conductor when molten or in aqueous solution
(b) (i) PCl3
(ii) It has a simple molecular structure.
(iii) Its physical properties are: (1) Low melting point and boiling point (2) Gas
at room temperature and pressure (3) Insoluble in water but soluble in
non-aqueous solvents (4): Non-conductor of electricity no matter it is in the
solid or liquid state.
Think about
Think about (p.9-2)
1. Graphite has lubricating property while diamond does not.
2.
3.
Refer to Section 9.3 on p.8 to 10 of this chapter for details.
The physical properties of a substance are closely related to its structure.
Substances composed of different elements may show similar physical properties
if their structures are similar.
© Aristo Educational Press Ltd. 2019
46
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
Think about (p.9-10)
Diamond is commonly used for making jewellery, diamond cutter, diamond-tipped
drill, etc. Graphite is commonly used for making ‘wet’ lubricants, electrodes, pencil
lead, etc.
Chapter exercise (p.9-29)
1. giant
2. molecules; covalent; intermolecular
3. giant lattice (or network)
4.
5.
6.
7.
8.
low; non-conductor; water; soluble
solids; high; graphite; non-conductor; insoluble
high; molten; aqueous solution; soluble; non-aqueous
giant metallic
delocalized; malleable; high
9. structure
10. A
(2): Iodine has a low boiling point because iodine molecules are held together by
weak van der Waals’ forces.
(3): I2 has a simple molecular structure.
11. A
Substances with a giant covalent structure (except graphite) cannot conduct
electricity because there are no mobile ions or delocalized electrons in their
structures. Besides, they are neither soluble in water nor non-aqueous solvents.
12. C
Substance
Structure
SiO2
Giant covalent structure
NO
Simple molecular structure
SO2
Simple molecular structure
13. B
In this structure, each ion of X is surrounded by 8 ions of M and vice versa.
14. C
X shows the properties of ionic compounds.
15. B
© Aristo Educational Press Ltd. 2019
47
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
P and S show the properties of ionic compounds or substances with giant
covalent structures.
Q shows the properties of substances with simple molecular structures.
R shows the properties of metals.
16. (a) I2
(b)
iodine molecule
(c) It has a simple molecular structure.
(d) Iodine atoms are linked together by covalent bond within each iodine
molecule. Iodine molecules in the lattice are held together by weak van der
Waals’ forces.
(e) The attractive forces between water molecules are quite strong. The weak
attractive forces between iodine and water molecules are not strong enough
to overcome the attractive forces between water molecules. On the other
hand, the attractive forces between ethanol molecules are similar to that
between iodine molecules in strength. Thus, iodine is more soluble in
ethanol than in water.
17. (a) Fractional distillation of liquid air
(b) Covalent bonding
(c) A nitrogen atom has five electrons in its outermost shell. To attain an
electronic arrangement of the nearest noble gas, each nitrogen atom shares
three of its outermost shell electrons with another nitrogen atom. As a result,
a triple covalent bond forms.
(d)
(e) There are three bond pairs and two lone pairs of electrons in a nitrogen
(f)
molecule.
This is because to react with other substances, the strong triple covalent
bond between the nitrogen atoms has to be broken first, which requires a lot
of energy.
© Aristo Educational Press Ltd. 2019
48
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(g) Intermolecular forces/van der Waals’ forces
18. (a) B
(b) (i) E. This is because it conducts electricity in the solid state.
(ii) A. This is because it does not conduct electricity in the solid state but
conducts electricity when molten.
(iii) B and C. This is because both of them do not conduct electricity no
matter in the solid state or when molten. Besides, they have low
melting points.
(iv) D. This is because it does not conduct electricity no matter in the solid
state or when molten. However, it has a very high melting point.
(c) B and C
(d) C. Wax does not conduct electricity no matter it is in the solid state or when
molten. Besides, it is a low-melting solid.
19. (a) W: giant covalent structure; X: giant ionic structure; Y: giant covalent
structure; Z: simple molecular structure
(b) W: graphite; X: sodium chloride; Y: diamond; Z: iodine
(c) X: ionic bonding; Y: covalent bonding
(d) In solid W, the atoms within each layer are linked by strong covalent bonds.
The layers are held together by weak intermolecular forces. In solid Z, the
atoms within each molecule are linked by strong covalent bonds. The
molecules are held together by weak intermolecular forces.
(e) X has a giant ionic structure while Z has a simple molecular structure. A
large amount of energy is needed to break the strong ionic bonds between
the ions during the melting of X. On the other hand, to melt Z, only a small
(f)
amount of energy is needed to separate the molecules. Hence, X has a higher
melting point than Z.
In Y, each carbon atom is covalently bonded with four other carbon atoms.
All electrons are localized. In W, each carbon atom is covalently bonded
with only three other carbon atoms in its layer. Each carbon atom has one
delocalized electron. These delocalized electrons can move in the direction
of electric field.
20. (a) Covalent bonding
(b) Giant covalent structure
© Aristo Educational Press Ltd. 2019
49
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(c) Structure of diamond:
carbon atom
Structure of quartz:
silicon atom
oxygen atom
(d) Diamond. This is because it has a shiny beautiful appearance.
(e) Diamond. This is because it is very hard.
© Aristo Educational Press Ltd. 2019
50
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Part II
Coursebook 1
Microscopic World I
Part exercise (p.9-33)
1. B
The mass number of Y is 18.
2.
B
Relative atomic mass of copper = 63 × 69.09% + 65 × 30.91% = 63.62
3.
C
The most reactive metals are found in the bottom left-hand corner of the table.
4.
D
Metals do not decompose when conducting electricity.
5.
D
Copper(II) sulphate conducts electricity when molten or in aqueous solution and
is decomposed at the same time.
6.
A
(3): Electrons are responsible for the electrical conduction of metals.
7.
A
The formula of this compound can be worked out as follows:
X 2+
Y 3
= X3
Y2
8.
A
9.
C
The correct electron diagram of a NO2 molecule is:
10. D
Ionic bond exists between NH4+ and Cl ions. Covalent bond and dative covalent
bond exist in NH4+ ion.
© Aristo Educational Press Ltd. 2019
51
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
11. A
Refer to p.5 of chapter 9 for details.
12. A
Metallic bonds exist between mercury(II) ions and the ‘sea’ of delocalized
electrons in mercury. Covalent bonds exist between silicon and oxygen atoms in
quartz.
13. (a) (i)
Isotopes are different atoms of the same elements, with the same
number of protons but different numbers of neutrons.
(ii)
Number of
Isotope
proton(s)
neutron(s)
electron(s)
Protium
1
0
1
Deuterium
1
1
1
Tritium
1
2
1
(iii) 2
(b) (i)
20
21
22
10 Ne, 10 Ne, 10 Ne
(ii) Relative atomic mass of neon
= 20 × 90.48% + 21 × 0.27% + 22 × 9.25% = 20.2
14. (a) (1)
(2)
(3)
(4)
B
K
C
E
(b) This is because it has a stable electronic arrangement/an octet of electrons.
(c) This is because A has delocalized electrons in its structure while G does not
have any delocalized electrons or mobile ions for conducting electricity.
(d) K < L< M
15. (a) A purple colour gradually appears in the solution around the positive
electrode because negatively charged permanganate ions migrate towards
the positive electrode.
(b) There is no observable change at the negative electrode. Though the
positively charged ammonium ions migrate towards the negative electrode,
they cannot be seen as they are colourless.
(c) The gel slows down the mixing of dilute sulphuric acid and ammonium
© Aristo Educational Press Ltd. 2019
52
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
permanganate.
16. (a) When a force is applied to a piece of strontium, the layers of strontium ions
can slide over one another without breaking the metallic bonds.
(b) Yes. This is because strontium has delocalized electrons in its structure.
(c) Giant metallic structure
(d) Alkaline earth metals
(e) Strontium reacts more vigorously with water than calcium does to give
hydrogen and strontium hydroxide.
17. (a) The electronic arrangements of C and D are 2,6 and 2,8,7 respectively.
Their atoms can attain electronic arrangements of the nearest noble gases by
sharing electrons with another atom of their own.
Electron diagram of C2:
Electron diagram of D2:
(b) A. The atom of A can attain the electronic arrangement of a helium atom by
losing one outermost shell electron.
(c) (i) A2C; ionic bonding
(ii) AD; ionic bonding
(iii) BC2; covalent bonding
(d) A2C and AD have giant ionic structures while BC2 has a simple molecular
structure. When melting or boiling A2C or AD, a large amount of heat
energy is needed to break the strong ionic bonds between the ions in them.
On the other hand, only a small amount of heat energy is needed to
overcome the weak van der Waals’ forces between the molecules in BC2
during melting or boiling. Hence, BC2 has the lowest melting point and
boiling point.
(e) No. This is because the atom of E has a stable electronic arrangement.
18. (a) (NH4)2CO3
(b) Both of them are colourless.
(c) Ionic bonding
(d) (i)
© Aristo Educational Press Ltd. 2019
53
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(ii) Covalent compound
(iii) The molecular size of ammonia is small. The intermolecular forces
between the molecules are weak.
19. (a) KCl
(b) CCl4
(c) Electron diagram of potassium chloride:
Electron diagram of carbon tetrachloride:
(d) Potassium chloride has a giant ionic structure while carbon tetrachloride has
a simple molecular structure. A large amount of heat energy is needed to
break the strong ionic bonds between the ions in potassium chloride during
melting. On the other hand, only a small amount of heat energy is needed to
overcome the weak van der Waals’ forces between the molecules in carbon
tetrachloride during melting.
(e) The ions in potassium chloride become mobile when potassium chloride is
molten or in aqueous solution.
(f) The attractive forces between water molecules are quite strong. The weak
attractive forces between carbon tetrachloride and water molecules are not
strong enough to overcome the attractive forces between water molecules.
Thus, carbon tetrachloride is insoluble in water.
20. (a) A: giant ionic structure; B: giant covalent structure; C: simple molecular
structure
(b) A: caesium chloride; B: silicon dioxide; C: carbon dioxide
(c) (i) A
(ii) C
(iii) B
(iv) A
© Aristo Educational Press Ltd. 2019
54
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
21. (a) Carbon
(b) Structure of graphite:
carbon atom
covalent bond
van der Waals’ forces
Structure of diamond:
carbon atom
(c) Giant covalent structure
(d) In diamond, each carbon atom is covalently bonded with four other carbon
atoms. All electrons are localized. In graphite, each carbon atom is
covalently bonded with only three other carbon atoms in its layer. Each
carbon atom has one delocalized electron. These delocalized electrons can
move in the direction of electric field.
(e) This is because a lot of energy is needed to break the large number of strong
covalent bonds between the carbon atoms in graphite and diamond during
melting.
22. (a) SiO2
(b) Giant covalent structure
(c)
silicon atom
oxygen atom
(d) It is transparent. It has a very high melting point (At high temperatures,
silicon dioxide can be drawn into fibres.). When light travels from air to
silicon dioxide, total internal reflection occurs.
© Aristo Educational Press Ltd. 2019
55
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 10
Coursebook 1
Occurrence and extraction of metals
Class practice
A10.1 (p.10-7)
(a) Copper. This is because it is non-poisonous, strong, malleable, ductile and
corrosion resistant.
(b) Aluminium. This is because it is strong and corrosion resistant.
(c) Mercury. This is because it is a liquid under room conditions. It expands on
heating.
A10.2 (p.10-10)
heat
Step 1: zinc sulphide + oxygen  zinc oxide + sulphur dioxide
heat
Step 2: zinc oxide + carbon  zinc + carbon dioxide
A10.3 (p.10-12)
(a) Heating the metal ore alone
heat
silver sulphide + oxygen  silver + sulphur dioxide
(b) Electrolysis of the molten ore
electricity
sodium chloride  sodium + chlorine
(c) Heating the metal ore with carbon/carbon reduction
heat
iron(III) oxide + carbon monoxide  iron + carbon dioxide
A10.4 (p.10-17)
1. (a) Q, R, P
(b) Q, R, P. The more easily a metal can be extracted, the earlier it was
discovered.
2. (a) Gold is less reactive than iron. It can be easily obtained by physical
methods.
(b) This is because gold is very rare.
A10.5 (p.10-20)
(a) This is because aluminium has low density, non-toxic, very malleable and
corrosion resistant.
(b) Recycling metals means melting down used metals and using them again.
© Aristo Educational Press Ltd. 2019
56
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(c) Any TWO of the following:

It saves metal resources.

It saves energy and other resources such as electricity, water and fuels.

It reduces metal waste and land used for waste disposal.

It reduces pollution arising from the mining and extraction of metals.
Think about
Think about (p.10-2)
1.
2.
3.
Silver comes from its ore  argentite. Gold exists in nature in free state.
Copper, tin and zinc can be obtained by heating their ores with carbon.
Refer to Section 10.1 on p.4 to 6 of this chapter for details.
Think about (p.10-7)
Gold and platinum are unreactive. They do not combine with other elements to form
compounds easily.
Think about (p.10-17)
The demand of a metal will also affect the price of that metal.
Think about (p.10-19)
Steel cans can be separated from aluminium cans by using magnets.
Chapter exercise (p.10-26)
1. compounds
2. (a) Heating
(b) carbon
3.
4.
5.
6.
(c) Electrolysis
extraction
(a) Abundance
(b) mining
non-renewable; conserving
(a) Reusing
(b) Reducing
(c) Recycling
7.
D
8.
C
© Aristo Educational Press Ltd. 2019
57
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
Mercury is used to make thermometers. Aluminium is used to make overhead
power cables and aircraft bodies.
9.
A
The main ore of mercury is cinnabar, which mainly consists of mercury(II)
sulphide. Copper pyrite is the main ore of copper. It mainly consists of copper
iron sulphide.
10. C
Refer to p.10 of chapter 10 for details.
11. D
Refer to p.8 to 9 of chapter 10 for details.
12. D
Reactive metals such as aluminium, calcium, sodium, etc. can be extracted from
the molten ores by electrolysis.
13. D
The extraction of aluminium by electrolysis is an expensive method. It is because
a large amount of electricity is used.
14. D
Refer to p.20 of chapter 10 for details.
15. (a) This is because aluminium has a low density/is a good conductor of
electricity/ductile. (Any TWO)
(b) This is because aluminium has delocalized electrons in the structure to
conduct electricity.
(c) This is because copper has a higher density than aluminium. The overhead
power cables made of copper may be quite heavy.
(d) This is because silver is much more expensive than copper.
16. (a) (i)
Aluminium. This is because aluminium is corrosion resistant, strong
and has a low density (hence convenient to carry).
(ii) Iron. This is because iron is strong, cheap, malleable and ductile.
(iii) Gold. This is because gold is extremely corrosion resistant, malleable
and ductile.
© Aristo Educational Press Ltd. 2019
58
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(b) This is because gold exists as free element in nature and can be extracted
easily by physical method.
(c) This is because iron is much more abundant than copper in the Earth’s crust.
17. (a) Lead metal
(b) Carbon dioxide
heat
(c) lead(II) oxide + carbon  lead + carbon dioxide
(d) (i) Yes
heat
copper(II) oxide + carbon  copper + carbon dioxide
(ii) No
This is because magnesium is a reactive metal.
18. (a) (i)
Electrolysis of molten ore
electricity
(ii) aluminium oxide  aluminium + oxygen
(iii) Potassium/sodium/calcium/magnesium (Any ONE)
(b) (i) Word equation for the extraction of mercury from cinnabar:
heat
mercury(II) sulphide + oxygen  mercury + sulphur dioxide
Word equation for the extraction of zinc from zinc blende:
heat
Step 1: zinc sulphide + oxygen  zinc oxide + sulphur dioxide
heat
Step 2: zinc oxide + carbon  zinc + carbon dioxide
(ii) Aluminium, zinc, mercury
(iii) Mercury, zinc, aluminium
19. (a) Tin is malleable/corrosion resistant/non-toxic. (Any ONE)
(b) They will be disposed of in the landfill site.
(c) Economic importance of recycling metals:
It saves energy and other resources such as electricity, water and fuels.
Environmental importance of recycling metals (Any ONE):

It saves metal resources.

It reduces metal waste and land used for waste disposal.

It reduces pollution arising from the mining and extraction of metals.
(d) Reuse these mooncake containers to store things.
Recycle these mooncake containers by putting them into metal recycling
bins.
© Aristo Educational Press Ltd. 2019
59
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 11
Coursebook 1
Reactivity of metals
Class practice
A11.1 (p.11-6)
(a) Sodium burns vigorously with a golden yellow flame to produce a white powder.
sodium + oxygen  sodium oxide
(b) Zinc burns to give out some heat; a powder (yellow when hot, white when cold)
forms.
zinc + oxygen  zinc oxide
(c) The surface of copper turns black.
copper + oxygen  copper(II) oxide
A11.2 (p.11-10)
(a) Yes. This is because the reactivity of Group I elements increases down the group.
(b) Rubidium floats on the water surface during the reaction.
(c) rubidium + water  rubidium hydroxide + hydrogen
(d) The resultant solution is alkaline because the rubidium hydroxide formed is
alkaline.
A11.3 (p.11-11)
(a) Yes
calcium + dilute hydrochloric acid  calcium chloride + hydrogen
(b) Yes
zinc + dilute sulphuric acid  zinc sulphate + hydrogen
(c) No
A11.4 (p.11-13)
(a) A, C, B
B is the most reactive because it reacts explosively with dilute hydrochloric acid.
C is more reactive than A because C reacts more rapidly with dilute hydrochloric
acid than A.
(b) A: iron; B: sodium; C: zinc
A11.5 (p.11-15)
1.
2.
(a) 2Ca(s) + O2(g)  2CaO(s)
(b) Fe(s) + 2HCl(aq)  FeCl2(aq) + H2(g)
1 molecule of SO2 reacts with 2 formula units of NaOH to produce 1 formula
unit of Na2SO3 and 1 molecule of H2O. In addition, SO2 is a gas. NaOH and
Na2SO3 are aqueous solutions. H2O is a liquid.
© Aristo Educational Press Ltd. 2019
60
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
A11.6 (p.11-18)
1.
2.
3.
(a)
(b)
(c)
(d)
(a)
(b)
(a)
(b)
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
PbO(s) + CO(g)  Pb(s) + CO2(g)
4K(s) + O2(g)  2K2O(s)
3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g)
2Ag2O(s)  4Ag(s) + O2(g)
2Fe2O3(s) + 3C(s)  4Fe(s) + 3CO2(g)
2Zn(s) + O2(g)  2ZnO(s)
Ca(s) + 2H2O(l)  Ca(OH)2(s) + H2(g)
(c) 2Al(s) + 3H2O(g)  Al2O3(s) + 3H2(g)
(d) Mg(s) + H2SO4(aq)  MgSO4(aq) + H2(g)
A11.7 (p.11-24)
1.
2.
(a) 3Mg(s) + 2Al3+(aq)  3Mg2+(aq) + 2Al(s)
(b) Ca(s) + 2H+(aq)  Ca2+(aq) + H2(g)
(c) Pb2+(aq) + SO42(aq)  PbSO4(s)
(d) Br2(aq) + 2OH(aq)  Br(aq) + OBr(aq) + H2O(l)
x = 1; y = 3; z = 4
A11.8 (p.11-26)
(a) Hydrogen
(b) Displacement reaction
(c) Copper
(d) 2ZO(s) + C(s)  2Z(s) + CO2(g) OR
2CuO(s) + C(s)  2Cu(s) + CO2(g)
(e) Z, X, Y. Y is the most reactive among the three metals because only it can react
with cold water. X is more reactive than Z because it can displace copper metal
from copper(II) sulphate solution.
Think about
Think about (p.11-2)
1. Reactivity of a metal refers to the readiness of it to react with other substances.
2. Water, acids, aqueous solution of another metal ion (another metal which is less
reactive)
Think about (p.11-14)
A word equation cannot tell the physical states of the substances involved in the
reaction.
© Aristo Educational Press Ltd. 2019
61
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
Try it now (p.11-16)
1 sodium + oxygen  sodium oxide
○
2 Na + O2  Na2O
○
3 4Na + O2  2Na2O
○
4 4Na(s) + O2(g)  2Na2O(s)
○
Try it now (p.11-23)
1
○
Fe(s) + CuSO4(aq)  FeSO4(aq) + Cu(s)
2
○
Fe(s) + Cu2+(aq) + SO42(aq)  Fe2+(aq) + SO42(aq) + Cu(s)
3
○
Fe(s) + Cu2+(aq) + SO42(aq)  Fe2+(aq) + SO42(aq) + Cu(s)
4
○
Fe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)
Chapter exercise (p.11-32)
1. (a) hydroxide; hydrogen
(b) oxide; hydrogen
2. metal; reactivity series
3. electrons; positive/metal; electrons
4. displace
5.
6.
ionic; equation; spectator
lower
7.
(a)
(b)
(c)
(d)
(e)
(f)
8.
(a) Magnesium dissolves and colourless gas bubbles evolve.
4Al(s) + 3O2(g)  2Al2O3(s)
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(l)
2Cu(NO3)2(s)  2CuO(s) + 4NO2(g) + O2(g)
3CuO(s) + 2NH3(g)  3Cu(s) + N2(g) + 3H2O(l)
4FeS2(s) + 11O2(g)  2Fe2O3(s) + 8SO2(g)
2Al(s) + 3Fe2+(aq)  2Al3+(aq) + 3Fe(s)
Mg(s) + H2SO4(aq)  MgSO4(aq) + H2(g)
(b) Calcium burns quite vigorously with a brick-red flame to produce a white
© Aristo Educational Press Ltd. 2019
62
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
powder.
2Ca(s) + O2(g)  2CaO(s)
(c) Potassium melts to form a silvery ball. The ball moves about very quickly
on the water surface with a hissing sound. It burns with a lilac flame.
2K(s) + 2H2O(l)  2KOH(aq) + H2(g)
(d) Magnesium gives an intense white light. A white solid is produced.
Mg(s) + H2O(g)  MgO(s) + H2(g)
(e) Zinc dissolves and colourless gas bubbles evolve.
(f)
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
Copper slowly dissolves and some shiny silvery deposits form on the
copper surface. The solution gradually turns blue.
Cu(s) + 2AgNO3(aq)  Cu(NO3)2(aq) + 2Ag(s)
9.
(a)
(b)
(c)
(d)
H+(aq) + OH(aq)  H2O(l)
Cl(aq) + Ag+(aq)  AgCl(s)
Mg(OH)2(s) + 2H+(aq)  Mg2+(aq) + 2H2O(l)
Pb2+(aq) + 2I(aq)  PbI2(s)
(e) 2H+(aq) + CaCO3(s)  Ca2+(aq) + CO2(g) + H2O(l)
10. B
Refer to p.4 of chapter 11 for details.
11. A
(B): Copper has no reaction with hydrochloric acid.
(C): The insoluble lead(II) sulphate formed prevents lead from further reaction
with the acid.
(D): Magnesium has no reaction with cold water.
12. A
For a balanced equation, the number of each type of atoms is the same on both
reactant and product sides.
13. D
14. D
As zinc is more reactive than copper, zinc can displace copper from copper(II)
sulphate solution.
© Aristo Educational Press Ltd. 2019
63
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
15. C
From the observations of the addition of nickel to solutions of compounds X, Y
and Z, the order of metal reactivity is: X, Y > Ni > Z.
From the observations of the addition of iron to solutions of compounds X, Y and
Z, the order of metal reactivity is: X > Fe > Y, Z.
16. B
17. C
Ionic equations include only those ions that are produced or changed during the
reaction.
18. D
The ores of Al, Ca and Mg are stable. It is difficult to extract the metals from
them.
19. (a) Magnesium oxide
(b) Magnesium burns with a very bright white light.
(c) 2Mg(s) + O2(g)  2MgO(s)
(d) When calcium burns with oxygen, a brick-red flame instead of a very bright
white light is produced. Hence, calcium is not suitable to replace
magnesium in the flashbulb.
20. (a) Silver
(b) Copper(II) ion
(c) Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s)
(d) Copper is less reactive than zinc as it cannot displace zinc from aqueous
zinc nitrate solution. On the other hand, copper is more reactive than silver
as it can displace silver from aqueous silver nitrate solution. Hence, the
ascending order of reactivity is: silver, copper, zinc.
21. (a) Oxygen
(b) Silver
(c) Hydrogen
(d) Potassium or sodium
(e) (i) Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
(ii) Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
(f) A, C, B
© Aristo Educational Press Ltd. 2019
64
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
22. (a) Lead
(b) 2Al(s) + 3PbO(s)  Al2O3(s) + 3Pb(s)
(c) (i) Yes.
2Al(s) + Fe2O3(s)  Al2O3(s) + 2Fe(s)
(ii) No. As aluminium is less reactive than magnesium, it cannot displace
magnesium from magnesium oxide.
© Aristo Educational Press Ltd. 2019
65
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 12
Coursebook 1
Reacting masses
Class practice
A12.1 (p.12-4)
1. (a) Number of oxygen molecules = 0.5 × 6.02 × 1023 = 3.01 × 1023
(b) As there are two oxygen atoms in each oxygen molecule, number of oxygen
atoms = 3.01 × 1023 × 2 = 6.02 × 1023
2.
Number of moles of sodium atoms =
1.204  10 24
mol = 2 mol
6.02  10 23
A12.2 (p.12-5)
1.
2.
(a)
(b)
(c)
(d)
(e)
Molar mass of Ag = 107.9 g mol1
Molar mass of F2 = (19.0 × 2) g mol1= 38.0 g mol1
Molar mass of NH3 = (14.0 + 1.0 × 3) g mol1 = 17.0 g mol1
Molar mass of C2H5OH = (12.0 × 2 + 1.0 × 6 + 16.0) g mol1 = 46.0 g mol1
Molar mass of Fe2(SO4)3
= 55.8 × 2 + 3 × (32.1 + 16.0 × 4) g mol1 = 399.9 g mol1
(a) Mass of 1 mole of Na2SO4 = (23.0 × 2 + 32.1 + 16.0 × 4) g = 142.1 g
(b) Mass of 0.5 mole of CCl4 = 0.5 × (12.0 + 35.5 × 4) g = 77.0 g
A12.3 (p.12-7)
1. (a) Mass of 0.200 mole of Cl atoms = 0.200 × 35.5 g = 7.1 g
(b) Mass of 0.200 mole of Cl2 molecules = 0.200 × (35.5 × 2) g = 14.2 g
(c) Mass of Cl2 = 1.20 × (35.5 × 2) g = 85.2 g
2.
Substance
No. of
No. of
(g)
moles
(mol)
molecules/
formula units
Molar mass
Mass
(g mol1)
(a)
Sodium hydroxide
40.0
10.0
0.250
1.51 × 1023
(b)
Helium
4.0
0.20
0.05
3.01 × 1022
(c)
Sulphur dioxide
64.1
320.5
5
3.01 × 1024
(d)
Compound X
46.0
23.0
0.5
3.01 × 1023
A12.4 (p.12-10)
1. Let the relative atomic mass of M be a.
79.87% =
79.87 =
a
× 100%
a  16.0
100 a
a  16.0
© Aristo Educational Press Ltd. 2019
66
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
a = 63.5
∴the relative atomic mass of M is 63.5.
2.
Let the relative atomic mass of M be a.
5.68
35.5
=
26.88 a  35.5
a = 132.5
∴the relative atomic mass of M is 132.5.
3.
Number of moles of Na =
100
mol = 4.35 mol
23.0
Since 1 formula unit of NaNO3 contains 1 Na, number of moles of NaNO3
= 4.35 mol
Mass of NaNO3 = 4.35 × (23.0 + 14.0 + 16.0 × 3) g = 369.75 g
Percentage by mass of N in NaNO3 =
14.0
× 100% = 16.5%
23.0  14.0  16.0  3
Mass of N in the NaNO3 sample = 369.75 g × 16.5% = 61.0 g
4.
Number of moles of Na =
4 .6
mol = 0.2 mol
23.0
Since 1 formula unit of Na2CO3 • 10H2O contains 2 Na, number of moles of
Na2CO3 • 10H2O =
0 .2
mol = 0.1 mol
2
Molar mass of Na2CO3 • 10H2O
= [23.0 × 2 + 12.0 + 16.0 × 3 + 10 × (1.0 × 2 + 16.0)] g mol1 = 286.0 g mol1
Mass of Na2CO3 • 10H2O = 0.1 × 286.0 g = 28.6 g
Percentage by mass of H2O in Na2CO3 • 10H2O
=
10  (1.0  2  16.0)
×100% = 62.9%
286.0
Mass of H2O in the Na2CO3 • 10H2O sample = 28.6 g × 62.9% = 18.0 g
A12.5 (p.12-11)
Substance
Empirical formula
Molecular formula
Structural formula
Oxygen
/
O2
O=O
Water
H2O
H2O
HOH
Ethane
CH3
C2H6
© Aristo Educational Press Ltd. 2019
67
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
But-1-ene
CH2
Coursebook 1
C4H8
A12.6 (p.12-15)
Extension 1.
Mass (g)
Mg
O
28.698  28.092 = 0.606
29.103  28.698 = 0.405
0.606
= 0.0249
24.3
0.405
= 0.0253
16.0
0.0249
=1
0.0249
0.0253
= 1.02 ≈ 1
0.0249
Number of moles of
atoms (mol)
Simplest whole number
mole ratio of atoms
∴the empirical formula of the oxide of magnesium is MgO.
2.
Mass of C in the compound = 1.173 ×
12.0
g = 0.320 g
12.0  16.0  2
Mass of H in the compound = 0.240 ×
1 .0  2
g = 0.0267 g
1.0  2  16.0
Mass of O in the compound = (1.200  0.320  0.0267) g = 0.853 g
C
H
O
0.320
0.0267
0.853
0.320
= 0.0267
12.0
0.0267
= 0.0267
1 .0
0.853
= 0.0533
16.0
0.0267
=1
0.0267
0.0267
=1
0.0267
0.0533
=2
0.0267
Mass (g)
Number of moles of
atoms (mol)
Simplest whole number
mole ratio of atoms
∴the empirical formula of the compound is CHO2.
3.
C
H
0.857
1.000  0.857 = 0.143
0.857
= 0.0714
12.0
0.143
= 0.143
1.0
0.0714
=1
0.0714
0.143
=2
0.0714
Mass (g)
Number of moles of
atoms (mol)
Simplest whole number
mole ratio of atoms
∴the empirical formula of the compound is CH2.
4.
Assume that there are 100 g of X. Then, there are 26.95 g of sulphur, 13.44 g of
oxygen and 59.61 g of chlorine.
S
© Aristo Educational Press Ltd. 2019
O
Cl
68
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Mass (g)
Number of moles of
atoms (mol)
Coursebook 1
26.95
13.44
59.61
26.95
= 0.840
32.1
13.44
= 0.84
16.0
59.61
= 1.68
35.5
0.840
=1
0.840
0.84
=1
0.840
1.68
=2
0.840
Simplest whole number
mole ratio of atoms
∴the empirical formula of the compound is SOCl2.
A12.7 (p.12-19)
1.
Mass of C in the compound = 1.32 ×
12.0
g = 0.36 g
12.0  16.0  2
Mass of H in the compound = 0.45 ×
1 .0  2
g = 0.05 g
1.0  2  16.0
Mass of O in the compound = (0.81  0.36  0.05) g = 0.40 g
C
H
O
0.36
0.05
0.40
Number of moles of
atoms (mol)
0.36
= 0.03
12.0
0.05
= 0.05
1 .0
0.40
= 0.025
16.0
Simplest whole number
mole ratio of atoms
0.03
= 1.2
0.025
0.05
=2
0.025
0.025
=1
0.025
1.2 × 5 = 6
2 × 5 = 10
1×5=5
Mass (g)
∴the empirical formula of the compound is C6H10O5.
Let the molecular formula of the compound be (C6H10O5)n.
320.0 = n × (12.0 × 6 + 1.0 × 10 + 16.0 × 5)
n = 1.98 ≈ 2
∴the molecular formula of the compound is C12H20O10.
2.
Assume that there are 100 g of the compound. Then, there are 40.00 g of carbon,
6.67 g of hydrogen and 53.33 g of oxygen.
C
H
O
40.00
6.67
53.33
40.00
= 3.33
12.0
6.67
= 6.67
1.0
53.33
= 3.33
16.0
3.33
=1
3.33
6.67
=2
3.33
3.33
=1
3.33
Mass (g)
Number of moles of
atoms (mol)
Simplest whole
number mole ratio of
atoms
∴the empirical formula of the compound is CH2O.
© Aristo Educational Press Ltd. 2019
69
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
Let the molecular formula of the compound be (CH2O)n.
60.0 = n × (12.0 + 1.0 × 2 + 16.0)
n=2
∴the molecular formula of the compound is C2H4O2.
3.
Assume that there are 100 g of Epsom salt. Then, there are 51.22 g of water of
crystallization and (100  51.22) g = 48.78 g of MgSO4.
MgSO4
H2O
48.78
51.22
24.3 + 32.1 + 16.0 × 4
= 120.4
1.0 × 2 + 16.0
= 18.0
Number of moles of
formula units (mol)
48.78
= 0.4051
120.4
51.22
= 2.85
18.0
Simplest whole
number mole ratio of
formula units
0.4051
=1
0.4051
2.85
= 7.04 ≈ 7
0.4051
Mass (g)
Formula mass
∴the value of n is 7.
A12.8 (p.12-26)
1.
2Ag2O(s)  4Ag(s) + O2(g)
Number of moles of Ag2O used =
6.96
mol = 0.0300 mol
107.9  2  16.0
From the equation, mole ratio of Ag2O to Ag is 1 : 2.
∴number of moles of Ag produced = 0.0300 × 2 mol = 0.0600 mol
Mass of Ag produced = 0.0600 × 107.9 g = 6.47 g
2.
Number of moles of Mg used =
5.42  10 6
mol = 223 045 mol
24.3
1.77  10 7
mol = 93 207 mol
47.9  35.5  4
From the equation, mole ratio of TiCl4 to Mg is 1 : 2.
Number of moles of TiCl4 used =
∴TiCl4 is the limiting reactant.
From the equation, mole ratio of TiCl4 to Ti is 1 : 1.
∴number of moles of Ti formed = 93 207 mol
Mass of Ti formed = 93 207 × 47.9 g = 4 464 615 g
© Aristo Educational Press Ltd. 2019
70
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
3.
Coursebook 1
(a) Ca(s) + 2H2O(l)  Ca(OH)2(s) + H2(g)
(b) Number of moles of Ca used =
1.50
mol = 0.0374 mol
40.1
From the equation, mole ratio of Ca to Ca(OH)2 = 1 : 1.
∴number of moles of Ca(OH)2 formed = 0.0374 mol
Theoretical mass of Ca(OH)2 formed
= 0.0374 × [40.1 + (16.0 + 1.0) × 2] g = 2.77 g
(c) Possible reasons:
The calcium used was impure.
Some calcium hydroxide was lost during filtration.
Self-test
Self-test 12.1 (p.12-6)
(a) Molar mass of magnesium hydroxide (Mg(OH)2)
= (24.3 + 16.0 × 2 + 1.0 × 2) g mol1 = 58.3 g mol1
10.21 g
Number of moles of Mg(OH)2 =
= 0.175 mol
58.3 g mol1
(b) Since 1 formula unit of Mg(OH)2 contains 2 OH ions,
number of moles of OH ions = 0.175 × 2 mol = 0.350 mol
Number of OH ions = 0.350 mol × 6.02 × 1023 mol1 = 2.11 × 1023
Self-test 12.2 (p.12-7)
24.3 g mol1
(a) Mass of 1 Mg atom =
= 4.04 × 10–23 g
6.02  10 23 mol1
126.9  2 g mol1
(b) Mass of 1 I2 molecule =
= 4.22 × 10–22 g
6.02  10 23 mol1
(40.1  12.0  16.0  3) g mol1
(c) Mass of 1 formula unit of CaCO3 =
6.02  10 23 mol1
=1.66 × 10–22 g
Self-test 12.3 (p.12-8)
Formula mass of NaOH = (23.0 + 16.0 + 1.0) g mol–1 = 40.0 g mol–1
Percentage by mass of Na in NaOH =
23.0
× 100% = 57.5%
40.0
Self-test 12.4 (p.12-9)
Formula mass of K2Cr2O7 = (39.1 × 2 + 52.0 × 2 + 16.0 × 7) g mol–1 = 294.2 g mol–1
© Aristo Educational Press Ltd. 2019
71
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Percentage by mass of K in K2Cr2O7 =
Coursebook 1
39.1  2
× 100% = 26.6%
294 .2
Mass of K in 7.91 g of K2Cr2O7 = 7.91 g × 26.6% = 2.10 g
Self-test 12.5 (p.12-9)
Let the relative atomic mass of X be a.
25.6% =
25.6 =
a
× 100%
a  79.9  2
100 a
a  159.8
a = 55.0
∴the relative atomic mass of X is 55.0.
Self-test 12.11 (p.12-22)
Molar mass of PbO = (207.2 + 16.0) g mol–1 = 223.2 g mol–1
Number of moles of PbO =
10.55
mol = 0.0473 mol
223.2
From the equation, mole ratio of Mg to PbO is 1 : 1.
∴number of moles of Mg required = 0.0473 mol
Mass of Mg required = 0.0473 × 24.3 g = 1.15 g
Self-test 12.12 (p.12-22)
Number of moles of Na reacted =
8.51
mol = 0.37 mol
23.0
From the equation, mole ratio of Na to H2 is 2 : 1.
∴number of moles of H2 formed =
0.37
mol = 0.185 mol
2
Mass of H2 produced = 0.185 × 1.0 × 2 g = 0.37 g
Self-test 12.13 (p.12-24)
Molar mass of NO = (14.0 + 16.0) g mol–1 = 30.0 g mol–1
Number of moles of NO =
26.58
mol = 0.886 mol
30.0
Molar mass of O2 = 16.0 × 2 g mol–1 = 32.0 g mol–1
Number of moles of O2 =
8.06
mol = 0.252 mol
32.0
© Aristo Educational Press Ltd. 2019
72
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
From the equation, mole ratio of NO to O2 = 2 : 1.
∴O2 is the limiting reactant.
Molar mass of NO2 = (14.0 + 16.0 × 2) g mol–1 = 46.0 g mol–1
From the equation, mole ratio of O2 to NO2 = 1 : 2.
∴number of moles of NO2 formed = 0.252 × 2 mol = 0.504 mol
Mass of NO2 formed = 0.504 × 46.0 g = 23.2 g
Self-test 12.14 (p.12-25)
(a) Number of moles of H2 =
430
mol = 215 mol
1.0  2
Molar mass of CH3OH = (12.0 + 1.0 × 4 + 16.0) g mol–1 = 32.0 g mol–1
From the equation, mole ratio of H2 to CH3OH = 2 : 1.
∴number of moles of CH3OH produced =
215
mol = 107.5 mol
2
Theoretical yield of CH3OH =107.5 × 32.0 g = 3440 g
(b) Actual yield of CH3OH = 3440 g × 45% = 1548 g
Think about
Think about (p.12-2)
1. We can count the number of nitroglycerin molecules by weighing.
2. The reactant and any of the products in the above reaction are related by a ratio,
as given by the stoichiometric coefficients in the chemical equation.
Think about (p.12-4)
Relative molecular mass of chlorine = 35.5 × 2 = 71.0
Hence, the mass of one mole of chlorine is 71.0 g.
Try it now (p.12-20)
1
○
CuO(s) + H2(g)  Cu(s) + H2O(l)
2
○
Molar mass of CuO = (63.5 + 16.0) g mol–1 = 79.5 g mol–1
Number of moles of CuO =
15.9
mol = 0.2 mol
79.5
3
○
From the equation, mole ratio of CuO to Cu is 1 : 1.
© Aristo Educational Press Ltd. 2019
73
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
∴number of moles of Cu = 0.2 mol
4
○
Mass of Cu produced = 0.2 × 63.5 g = 12.7 g
Chapter exercise (p.12-32)
1. relative atomic masses
2. 6.02 × 1023; Avogadro constant
3. molar mass
4.
5.
6.
7.
8.
Relative atomic mass; number
empirical; composition
relative molecular mass
Limiting reactant
actual yield; theoretical yield
9.
(a) Number of sodium atoms = 2 × 6.02 × 1023 = 1.204 × 1024
(b) Number of moles of oxygen molecules =
2
mol = 0.0625 mol
16.0  2
Number of oxygen atoms = 0.0625 × 2 × 6.02 × 1023 = 7.525 × 1022
(c) Number of atoms in 1.5 moles of nitrogen dioxide gas
= 1.5 × 3 × 6.02 × 1023 = 2.709 × 1024
(d) Number of atoms in 0.5 mole of sodium carbonate-10-water
= 0.5 × 36 × 6.02 × 1023 = 1.084 × 1025
(e) Number of moles of aluminium sulphate
22
=
mol = 0.0643 mol
27.0  2  (32.1  16.0  4)  3
Number of atoms in 0.0643 mol of aluminium sulphate
= 0.0643 × 17 × 6.02 × 1023 = 6.58 × 1023
10. (a) Formula mass of CH4 = 12.0 + 1.0 × 4 = 16.0
Percentage by mass of C in CH4 =
12.0
×100% = 75%
16.0
(b) Formula mass of anhydrous Na2SO4 = 23.0 × 2 + 32.1 + 16.0 × 4 = 142.1
Percentage by mass of S in anhydrous Na2SO4 =
32.1
×100% = 22.6%
142.1
(c) Formula mass of Na2CO3 • 10H2O
= 23.0 × 2 + 12.0 + 16.0 × 3 + 10 × (1.0 × 2 + 16.0) = 286.0
© Aristo Educational Press Ltd. 2019
74
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
Percentage by mass of H2O in Na2CO3 • 10H2O =
10  18.0
×100% = 62.9%
286.0
(d) Formula mass of FeSO4 • 7H2O
= 55.8 + 32.1 + 16.0 × 4 + 7 × (1.0 × 2 + 16.0) = 277.9
Percentage by mass of O in FeSO4 • 7H2O =
16.0  11
×100% = 63.3%
277.9
11. (a) Formula mass of CH4 = 12.0 + 1.0 × 4 = 16.0
Percentage by mass of H in CH4 =
1.0  4
×100% = 25%
16.0
Mass of H in 10 g of CH4 = 10 g × 25% = 2.5 g
(b) Formula mass of anhydrous Na2SO4 = 23.0 × 2 + 32.1 + 16.0 × 4 = 142.1
Percentage by mass of Na in anhydrous Na2SO4 =
23.0  2
×100% = 32.4%
142 .1
Mass of Na in 50 g of anhydrous Na2SO4 = 50 g × 32.4% = 16.2 g
(c) Formula mass of FeCl3 • 6H2O
= 55.8 + 35.5 × 3 + 6 × (1.0 × 2 + 16.0) = 270.3
Percentage by mass of Cl in FeCl3 • 6H2O =
35.5  3
×100% = 39.4%
270 .3
Mass of 2 moles of FeCl3 • 6H2O = 2 × 270.3 g = 540.6 g
Mass of Cl in 2 moles of FeCl3 • 6H2O = 540.6 × 39.4% = 213 g
(d) Formula mass of CaCl2 • 6H2O
= 40.1 + 35.5 × 2 + 6 × (1.0 × 2 + 16.0) = 219.1
Percentage by mass of H2O in CaCl2 • 6H2O =
6  18.0
×100% = 49.3%
219 .1
Mass of 1.25 moles of CaCl2 • 6H2O = 1.25 × 219.1 g = 273.9 g
Mass of H2O in 1.25 moles of CaCl2 • 6H2O = 273.9 × 49.3% = 135.0 g
12. (a) Assume that there are 100 g of the compound.
Mass (g)
Number of moles of atoms
(mol)
Simplest whole number
mole ratio of atoms
C
H
75
25
75
= 6.25
12.0
25
= 25
1 .0
6.25
=1
6.25
25
=4
6.25
∴the empirical formula of the compound is CH4.
(b) Assume that there are 100 g of the compound.
© Aristo Educational Press Ltd. 2019
75
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Pb
O
86.6
13.4
86.6
= 0.418
207.2
13.4
= 0.838
16.0
0.418
=1
0.418
0.838
=2
0.418
Mass (g)
Number of moles of atoms
(mol)
Coursebook 1
Simplest whole number
mole ratio of atoms
∴the empirical formula of the compound is PbO2.
(c) Assume that there are 100 g of the compound.
Na
S
O
36.5
25.4
38.1
36.5
= 1.59
23.0
25.4
= 0.79
32.1
38.1
= 2.38
16.0
1.59
= 2.01 ≈ 2
0.79
0.79
=1
0.79
2.38
= 3.01 ≈ 3
0.79
Mass (g)
Number of moles of atoms
(mol)
Simplest whole number
mole ratio of atoms
∴the empirical formula of the compound is Na2SO3.
(d) Assume that there are 100 g of the compound.
Mass (g)
Number of
moles of
atoms (mol)
Simplest
whole number
mole ratio of
atoms
C
N
O
H
40.67
23.73
27.13
8.47
40.67
= 3.39
12.0
23.73
= 1.70
14.0
27.13
= 1.70
16.0
8.47
= 8.47
1 .0
1.70
=1
1.70
1.70
=1
1.70
3.39
1.70
= 1.99 ≈ 2
8.47
1.70
= 4.98 ≈ 5
∴the empirical formula of the compound is C2NOH5.
(e) Assume that there are 100 g of the compound.
Mass (g)
Number of moles
of formula units
(mol)
Simplest whole
number mole ratio
Cu
Cl
H2O
37.11
41.68
(100  37.11  41.68)
= 21.21
37.11
= 0.584
63.5
41.68
= 1.174
35.5
21.21
=1.178
18.0
0.584
=1
0.584
1.174
= 2.01 ≈ 2
0.584
1.178
= 2.02 ≈ 2
0.584
© Aristo Educational Press Ltd. 2019
76
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
of formula units
∴the empirical formula of the compound is CuCl2•2H2O.
13. B
(1): The molar mass of a substance is the mass in gram of one mole of it.
(2): Both molecules of oxygen and nitrogen are diatomic.
(3): Molar mass of oxygen is (16.0 × 2) g mol–1 = 32.0 g mol–1; molar mass of
nitrogen is (14.0 × 2) g mol–1 = 28.0 g mol–1
14. A
(A): Number of moles of H2 molecules =
2.0
mol = 1 mol
1.0  2
Number of moles of H atoms = 1 × 2 mol = 2 mol
(B): Number of moles of Mg atoms =
24.3
mol = 1 mol
24.3
(C): Number of moles of Ne atoms =
30.3
mol = 1.5 mol
20.2
(D): Number of moles of Cu atoms =
31.8
mol = 0.501 mol
63.5
15. C
Number of helium molecules present in 2 g of helium gas = y =
2
× Avogadro
4 .0
constant
∴Avogadro constant = 2y
Number of moles of fluorine molecules present in 38 g of fluorine gas
=
38
38
× Avogadro constant =
× 2y = 2y
19.0  2
19.0  2
16. D
Percentage by mass of water of crystallization in Na2CO3 • H2O
=
1.0  2  16.0
×100% = 14.5%
23.0  2  12.0  16.0  3  1.0  2  16.0
17. C
Mass of Fe3O4 in 100 g of iron ore = 100 g × 70% = 70 g
© Aristo Educational Press Ltd. 2019
77
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Mass of Fe in 70 g of Fe3O4 = 70 ×
Coursebook 1
55.8  3
g = 50.6 g
55.8  3  16.0  4
18. A
Mass of metal X in the oxide = (8.42  2.40) g = 6.02 g
Mass of O in the oxide = 2.40 g
Number of moles of X : number of moles of O
=
6.02 2.40
:
= 0.150 : 0.150 = 1 : 1
40.1 16.0
19. D
2H2O(l)  2H2(g) + O2(g)
Atoms cannot be created or destroyed in a reaction. Hence, the mass of products
(H2(g) and O2(g)) is the same as that of the reactant, i.e. 20.0 g.
20. A
R(s) + Cl2(g)  RCl2(s)
Number of moles of Cl2 used =
5.33
mol = 0.0751 mol
35.5  2
From the equation, mole ratio of R to Cl2 is 1 : 1.
∴number of moles of R used = 0.0751 mol
Let the molar mass of R be y g mol1.
4.76
0.0751 =
y
y = 63.4
21. B
CaCO3(s)  CaO(s) + CO2(g)
Number of moles of CaCO3 used =
10.01
mol = 0.1 mol
40.1  12.0  16.0  3
From the equation, mole ratio of CaCO3 to CaO = 1 : 1.
∴mass of CaO produced = 0.1 × (40.1 + 16.0) g = 5.61 g
22. C
Percentage yield of the reaction =
© Aristo Educational Press Ltd. 2019
4.31 g
× 100% = 76.8%
5.61 g
78
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
23. (a) SnF2
(b) Formula mass of SnF2 = (118.7 + 19.0 × 2) = 156.7
(c) Percentage by mass of F in SnF2 =
19.0  2
×100% = 24.3%
156 .7
(d) Mass of F in 1.50 g of SnF2 = 1.50 g × 24.3% = 0.365 g
24. Assume that the mass of the sodium to be 1 g.
Oxide of sodium formed in air:
Mass (g)
Number of moles of
atoms (mol)
Simplest whole number
mole ratio of atoms
Na
O
1
0.35
1
= 0.0435
23.0
0.35
= 0.0219
16.0
0.0435
= 1.99 ≈ 2
0.0219
0.0219
=1
0.0219
∴the empirical formula of the oxide of sodium formed in air is Na2O.
Oxide of sodium formed in pure oxygen:
Mass (g)
Number of moles of
atoms (mol)
Simplest whole number
mole ratio of atoms
Na
O
1
0.70
1
= 0.0435
23.0
0.70
= 0.0438
16.0
0.0435
=1
0.0435
0.0438
=1
0.0435
∴the empirical formula of the oxide of sodium formed in pure oxygen is NaO.
25. Assume that there are 100 g of paracetamol.
Mass (g)
Number of moles
of atoms (mol)
Simplest whole
number mole
ratio of atoms
C
H
N
O
63.58
5.96
9.27
21.19
63.58
= 5.30
12.0
5.96
= 5.96
1 .0
9.27
= 0.662
14.0
21.19
= 1.32
16.0
5.30
=8
0.662
5.96
=9
0.662
0.662
=1
0.662
1.32
=2
0.662
∴the empirical formula of paracetamol is C8H9NO2.
© Aristo Educational Press Ltd. 2019
79
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
Let the molecular formula of paracetamol be (C8H9NO2)n.
n × (12.0 × 8 + 1.0 × 9 + 14.0 + 16.0 × 2) = 151.0
n=1
∴the molecular formula of paracetamol is C8H9NO2.
26. (a) Mass of anhydrous Na2CO3 = (13.07  8.23) g = 4.84 g
(b) Formula mass of anhydrous Na2CO3 = (23.0 × 2 + 12.0 + 16.0 × 3) = 106.0
(c) Number of moles of anhydrous Na2CO3 in the sample
=
4.84
mol = 0.0457 mol
106.0
(d) Number of moles of H2O in the sample =
8.23
mol = 0.457 mol
18.0
(e)
Na2CO3
H2O
Number of moles of formula
units (mol)
0.0457
0.457
Simplest whole number mole
ratio of formula units
0.0457
=1
0.0457
0.457
= 10
0.0457
∴the value of n is 10.
27. (a) Number of moles of C2H2 =
Number of moles of Br2 =
2.00
mol = 0.0769 mol
(12.0  2  1.0  2)
5.20
mol = 0.0325 mol
79.9  2
From the equation, mole ratio of C2H2 to Br2 is 1 : 2.
∴Br2 is the limiting reactant.
Number of moles of C2H2Br4 formed =
0.0325
mol = 0.0163 mol
2
Theoretical yield of C2H2Br4 = 0.0163 × (12.0 × 2 + 1.0 × 2 + 79.9 × 4) g
= 5.63 g
5.02 g
(b) Percentage yield of the reaction =
×100% = 89.2%
5.63 g
28. (a) The whole process may be represented by a sequence of steps:
© Aristo Educational Press Ltd. 2019
80
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
Fe  Fe2+  Fe3+ Fe(OH)3  Fe2O3
And the whole process can be represented by the overall equation:
2Fe  Fe2O3 (the ‘2’ is added to balance the number of Fe atoms)
Thus, mole ratio of Fe to Fe2O3 = 2 : 1.
Number of moles of Fe =
5.91
mol = 0.106 mol
55.8
Number of moles of Fe2O3 formed =
0.106
mol = 0.053 mol
2
Molar mass of Fe2O3 = (55.8 × 2 + 16.0 × 3) g mol1 = 159.6 g mol1
Theoretical yield of Fe2O3 = 0.053 × 159.6 g = 8.46 g
7.95 g
(b) (i) Percentage yield of iron(III) oxide =
× 100% = 94.0%
8.46 g
(ii) The actual yield is smaller than the theoretical yield. The possible
reasons for the difference (Any TWO):



The reaction was incomplete.
The iron used might be impure.
There was a loss of materials during various experimental
processes, e.g. filtration.
© Aristo Educational Press Ltd. 2019
81
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Chapter 13
Coursebook 1
Corrosion of metals and their protection
Class practice
A13.1 (p.13-8)
(a) (i) The positive ion: iron(II) ion; the negative ion: the hydroxide ion
(ii) 4Fe(s) + 3O2(g) + 2nH2O(l)  2Fe2O3 • nH2O(s)
(b) In tube 3, dissolved air in distilled water had been driven off by boiling. Besides,
the oil layer on top prevented air from dissolving in water again. Because there
was no air (oxygen), the iron nail in tube 3 did not rust. In tube 4, iron nail was
immersed in oil. Air does not dissolve in oil. Because there were no air and water,
the iron nail in tube 4 did not rust.
(c) Copper is less reactive than iron. If the iron nail were wrapped with a copper
wire, iron would lose electrons more readily. As a result, rusting would occur
faster.
A13.2 (p.13-16)
(a) Magnesium, aluminium and zinc
(b) Galvanizing/sacrificial protection
(c) The tin layer protects iron from rusting by preventing it from contacting air and
water.
(d) This is because tin ions are not poisonous while zinc ions are poisonous.
A13.3 (p.13-21)
(a) The rust formed is just loosely attached to the surface of the iron-made object. It
falls from the iron surface easily. When the fresh iron surface is exposed to the
environment, it reacts with air and water. Thus, rusting continues until the iron
piece corrodes completely.
(b) A layer of protective oxide layer forms on the aluminium surface. This oxide
layer is tough and is impermeable to air and water. It can protect aluminium
underneath from further corrosion.
(c) Anodization
Think about
Think about (p.13-2)
1. Corrosion of metals results in changes in properties and subsequent malfunction
2.
of the metal objects.
Common protection methods include applying a protective layer, cathodic
protection, sacrificial protection or using alloys of iron. (Accept other reasonable
answers.)
© Aristo Educational Press Ltd. 2019
82
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
Think about (p.13-4)
No. This is because there is no air on the Moon.
(In March 1998, NASA revealed that there was strong evidence for the existence of a
large quantity of ice at the poles of the Moon. Thus, the old idea that there was no
water on the Moon might have to be changed.)
Think about (p.13-6)
Rusting should be faster in Hong Kong, where humidity is high. The Sahara Desert is
very dry with almost no rainfall all year round. Rusting does not occur where there is
no water. OR
Hong Kong has serious air pollution. The acid rain would speed up rusting.
Chapter exercise (p.13-25)
1. Corrosion; air (oxygen); water
2.
3.
4.
5.
6.
Rusting
iron(III) oxide
(a) acidic pollutants
(b) soluble; ionic
(c) High
(d) less reactive
(e) Scratched; bent; sharp
potassium hexacyanoferrate(III); phenolphthalein
(a) coating
(b) Cathodic
(c) Sacrificial; galvanizing
(d) alloys
7.
aluminium oxide/oxide; anodization
8.
D
Refer to p.5 of chapter 13 for details.
9.
C
When an iron nail is in contact with a less reactive metal, it would rust faster.
10. C
The iron nail in (C) is protected from rusting by sacrificial protection. Zinc
corrodes instead of iron.
© Aristo Educational Press Ltd. 2019
83
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
11. A
Tin is lower than iron in the metal reactivity series.
12. B
13. D
Zinc offers sacrificial protection even when the zinc coating is damaged. Thus,
galvanized iron can be used for making objects that are often scratched or
knocked about during use.
14. B
Metal object
Usual corrosion prevention method
(A)
Head of electric drill
Oiling/greasing
(C)
Paper clip
Coating with plastic/tin-plating
(D)
Ship body
Sacrificial protection
15. C
Refer to p.20 of chapter 13 for details.
16. D
Distilled water contains dissolved oxygen. Hence, an iron nail rusts in a test tube
of distilled water.
17. (a) It can absorb water (moisture) from the air.
(b) It can prevent air from dissolving in water.
(c) (i) Tubes 2, 3, 4 and 6. This is because both air (oxygen) and water are
present in these test tubes.
(ii) Tube 6. This is because the presence of soluble ionic compounds in sea
water can speed up the rusting of iron.
(d) Tubes 1 and 5. This is because tube 1 does not contain water, and the boiled
distilled water in tube 5 does not contain air.
(e) 4Fe(s) + 3O2(g) + 2nH2O(l)  2Fe2O3 • nH2O(s)
18. (a) (i)
To detect the presence of iron(II) ions.
(ii) To detect the presence of hydroxide ions.
(b) The appearance of blue colour around the heads and tips of iron nails
indicates the presence of iron(II) ions. Rusting occurs faster in these regions
as these regions are sharp.
© Aristo Educational Press Ltd. 2019
84
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(c) Dish 1: The iron nail rusts in the presence of air and water. Iron loses
electrons to form iron(II) ions which turn potassium hexacyanoferrate(III)
blue. The hydroxide ions formed turn phenolphthalein pink.
Dish 2: Copper is less reactive than iron. The iron nail wrapped with copper
wire loses electrons to form iron(II) ions more readily. The iron(II) ions
formed turn potassium hexacyanoferrate(III) blue. The hydroxide ions
formed turn phenolphthalein pink.
Dish 3: Magnesium is more reactive than iron. It loses electrons more
readily than iron, preventing iron from losing electrons. Thus, the iron nail
does not rust and no iron(II) ions form. Hydroxide ions form when
magnesium corrodes. Hence, a pink colour appears around the head and tip
of the iron nail.
(d) Magnesium, iron, copper
19. (a) The presence of soluble ionic compounds in sea water speeds up the rusting
of iron pylons.
(b) The iron pylons can be protected from rusting by attaching some zinc
blocks to the surface. Zinc is more reactive than iron. It will corrode instead
of iron.
(c) (i) Stainless steel is an iron alloy produced by mixing carbon and other
metals such as chromium, nickel and manganese with iron.
(ii) The cost of using stainless steel pylons is high.
20. (a) Aluminium is very malleable/non-poisonous/corrosion resistant. (Any
TWO)
(b) Anodization
(c) During anodization, a layer of aluminium oxide forms on the surface of the
milk bottle cap. As a result, the oxide layer is thickened.
2Al(s) + 3H2O(l)  Al2O3(s) + 3H2(g)
(d) The layer of aluminium oxide is tough and impermeable to air and water. It
can prevent the aluminium underneath from further corrosion.
© Aristo Educational Press Ltd. 2019
85
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Part III
Coursebook 1
Metals
Part exercise (p.13-29)
1.
D
In general, the more easily the metal can be extracted, the earlier it was
discovered.
2.
D
Magnesium can be extracted by electrolysis of its molten ore.
3.
B
Calcium hydroxide, instead of calcium oxide, forms when calcium reacts with
cold water. Hydrogen, instead of water, forms when calcium reacts with dilute
hydrochloric acid. Calcium is not a transition metal.
4.
B
A more reactive metal can displace a less reactive metal from its compound. The
ease of extracting a metal from its metal ore is related to the stability of the metal
ore. The more stable the metal ore, the more difficult the metal is extracted.
5.
B
A more reactive metal can displace a less reactive metal from its compound.
6.
A
Number of moles of atoms in 28.0 g of N2 =
Number of moles of atoms in 2.0 g of H2 =
28.0
× 2 mol = 2 mol
14.0  2
2.0
× 2 mol = 2 mol
1.0  2
Number of moles of atoms in 18.0 g of O2 =
18.0
× 2 mol = 1.13 mol
16.0  2
Number of moles of atoms in 34.0 g of Cl2 =
Number of moles of atoms in 40.0 g of Ar =
7.
34.0
× 2 mol = 0.958 mol
35.5  2
40.0
mol = 1 mol
40.0
C
Assume that there are 100 g of compound X. Then, there are 72 g of carbon, 12 g
of hydrogen and 16 g of oxygen.
© Aristo Educational Press Ltd. 2019
86
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Mass (g)
Number of moles of
atoms (mol)
Simplest whole number
mole ratio of atoms
Coursebook 1
C
H
O
72
12
16
72
=6
12.0
12
= 12
1 .0
16
=1
16.0
6
12
1
∴the empirical formula of compound X is C6H12O.
Let the molecular formula of compound X be (C6H12O)n.
200.0 = n × (12.0 × 6 + 1.0 × 12 + 16.0)
n=2
∴the molecular formula of compound X is C12H24O 2.
8.
C
Mass of water of crystallization in 5.0 g of CuSO4•5H2O
5  (1.0  2  16.0)
= 5.0 ×
g = 1.8 g
63.5  32.1  16.0  4  5  (1.0  2  16.0)
9.
A
Let the relative atomic mass of metal M be a.
For oxide X,
Mole ratio of M to O =
3.76 ( 4.38  3.76)
:
=1:1
a
16.0
a = 97.0
For oxide Y,
Mole ratio of M to O =
3.76 ( 4.07  3.76)
:
=2:1
97.0
16.0
∴the formula of Y is M2O.
10. A
(2): Once the tin coating is partly damaged, the iron object will corrode faster
than expected.
(3): Tin-plated iron is not a kind of iron alloy.
11. A
© Aristo Educational Press Ltd. 2019
87
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
12. (a) Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
(b) (i) It can remove the impurities in the haematite.
(ii) CaCO3(s)  CaO(s) + CO2(g)
(c) Economic importance of recycling of iron:
It saves energy and other resources such as electricity, water and fuels.
Environmental importance of recycling of iron (Any ONE):
It saves metal resources. OR
It reduces metal waste and land used for waste disposal. OR
It reduces pollution arising from the mining and extraction of metals.
13. (a) Copper can be extracted from chalcocite by heating the ore with carbon,
which is a cheap material. On the other hand, the extraction of aluminium
from bauxite involves the use of a large amount of electricity. The cost of
using electricity is very high.
(b) Electrolysis of the molten ore
electricity
(c) 2Al2O3(l)  4Al(l) + 3O2(g)
3
(d) Cu2S(s) + O2(g)  Cu2O(s) + SO2(g)
2
2Cu2O(s) + C(s)  4Cu(s) + CO2(g)
(e) (i) (Any ONE):
The cost of this metal extraction method is lower. OR
The burning of plants gives out energy which can be used for heating.
(ii) The growth of plants is very slow.
14. (a) (i)
Moisten a clean platinum wire with concentrated hydrochloric acid.
Then, dip the wire into a crushed sample (or the solution) of the
substance to be tested. After that, heat the end of the wire strongly in a
non-luminous flame.
(ii) P is sodium and Q is calcium.
(b) This is because P reacts with air very readily.
(c) (i) P melts to form a silvery ball. The ball moves about very quickly on
the water surface with a hissing sound. It burns with a golden yellow
flame.
(ii) The resultant solution is alkaline. The red litmus paper would turn
blue.
© Aristo Educational Press Ltd. 2019
88
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
(d) (i)
hydrogen
dilute hydrochloric acid
inverted filter funnel
metal Q
(ii) Test the gas with a burning splint. If the gas burns with a ‘pop’ sound,
it should be hydrogen.
(e) (i) Mercury
(ii) It can be used to make thermometers.
(f) P, Q, R
(g)
Metal
Metal extraction method
P
Electrolysis of the molten metal oxide
Q
Electrolysis of the molten metal oxide
R
Heating the metal oxide in air
15. (a) To burn away excess town gas as town gas is flammable.
(b) The total mass of the combustion tube and the copper formed would be
larger than 22.16 g.
(c) Mass of oxide of copper used = (22.68  20.10) g = 2.58 g
(d) Mass of copper formed = (22.16  20.10) g = 2.06 g
(e) Mass of oxygen in the oxide of copper = (2.58  2.06) g = 0.52 g
(f)
Number of moles of copper formed =
2.06
mol = 0.0324 mol
63.5
(g) Number of moles of oxygen in the oxide of copper =
0.52
mol = 0.0325 mol
16.0
(h)
Cu
O
0.0324
0.0325
0.0324
=1
0.0324
0.0325
=1
0.0324
Number of moles of atoms (mol)
Simplest whole number mole
ratio of atoms
∴the empirical formula of the oxide of copper is CuO.
(i)
CuO(s) + H2(g)  Cu(s) + H2O(l) OR
© Aristo Educational Press Ltd. 2019
89
HKDSE CHEMISTRY  A Modern View (Second Edition)
(Reprinted with minor amendments 2019)
Coursebook 1
CuO(s) + CO(g)  Cu(s) + CO2(g)
16. (a) 2NaN3(s)  2Na(s) + 3N2(g)
(b) Molar mass of NaN3 = (23.0 + 14.0 × 3) g mol1 = 65.0 g mol1
Number of moles of NaN3 decomposed =
130
mol = 2 mol
65.0
From the equation, mole ratio of NaN3 to N2 is 2 : 3.
∴number of moles of N2 formed = 2 mol ×
3
= 3 mol
2
Mass of N2 formed = 3 × 14.0 × 2 g = 84.0 g
(c) (i)
Sodium is very reactive and it reacts with air vigorously. OR
It is a flammable metal.
(ii) 2KNO3(s) + 10Na(s)  5Na2O(s) + K2O(s) + N2(g)
(iii) From the equation in (a), mole ratio of NaN3 to Na = 1 : 1.
∴number of moles of Na produced = 2 mol
From the equation in (c)(ii), mole ratio of KNO3 to Na = 1 : 5.
∴number of moles of KNO3 required =
2
mol
5
2
Mass of KNO3 required = × (39.1 + 14.0 + 16.0 ×3) g = 40.44 g
5
17. (a) The essential conditions for the rusting of iron are water and air (oxygen).
(b) Painting provides a protective layer which prevents iron from contacting air
and water.
(c) (i) (Any ONE):
It lasts longer than some other rust prevention methods such as
painting or oiling/greasing. OR
The coating of plastic can be used for decoration purpose.
(ii) It is more expensive than using painting.
(d) By oiling/greasing. This is because the oil/grease would not be scratched off
easily like paint or plastic. Besides, it can serve as a lubricant for the chain.
(e) This is because the water film on the surface of bicycle usually contains
dissolved soluble ionic compounds which would speed up the rusting of
iron.
© Aristo Educational Press Ltd. 2019
90