FOCUS SPM Chemistry (2023)

Chemistry SPM Chapter 4 Polymer

SPM Practice 4



Objective Questions D H H I Natural rubber molecule
is made of many
Choose the correct answer. H C C H 2-methylbut-1,3-diene
CI CI units
4.1 Polymer II Natural rubber molecule
is not a hydrocarbon
3. Which of the following is a
1. Which polymer is made of natural polymer? III Natural rubber can be
monomer with empirical A Polyvinyl chloride strengthened through
formula CH 2 ? B Polyproprene vulcanisation process
A H CH IV Natural rubber has basic
3 C Polyisoprene property
C C D Polystyrene
A I and II
H H n 4. Which of the following B I and III
B H CH 3 H H monomers can undergo C III and IV
C C C C condensation polymerisation? D II and IV
I H CI H
H H n 7. Rubber hoses used in our
C C C C houses are made from
C
H H vulcanised rubber. What is
H the property of vulcanised
II H CH
C C 3 rubber that makes it suitable
C C to be used for making rubber
H H n
H COOCH 3 hoses?
D A Harder than unvulcanised
H III COOH rubber
C C H N C H B Stronger than
2
H CH 3 unvulcanised rubber
n IV O O C Resistant to oxidation
D Resistant to heat
CI C (CH ) 2 8 C CI
2. Figure 1 shows the structural 8. What is the advantage of the
formula of a polymer. A I and II vulcanised rubber produced
B I and III from vulcanisation process
H H H H H H
C II and IV using metal oxides or
C C C C C C D III and IV peroxides?
CI H CI H CI H 5. Thermoplastic polymers A More environmentally
Polyethene are polymers that can be friendly
Figure 1 B Biodegradable
remoulded and recycled C More durable
Which of the following is many times. Which of the D More elastic
the structural formula of its following is an example of a
monomer? thermoplastic? 9. W, X, Y and Z are stages in
Form
A CI H A Nylon latex coagulation.
C C B Bakelite W – Acid provides H ions
+
5
CI H C Polybutadiene
D Polyurethane X – Rubber particles collide with
B H CI each other
C C 4.2 Natural Rubber Y – Negatively-charged protein
CI H membrane are neutralised
6. Which of the following Z – Protein membrane breaks,
C H CI statements are true regarding rubber molecules are free
C C natural rubber? and entangled
H H




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Chemistry SPM Chapter 4 Polymer

Which of the following D Ammonia To add OH A I and II
–
arrangement of the stages is solution ions and B I and III
true? neutralise C II and IV
A W, Z, Y, X H ions from D III and IV
+
B W, Y, X, Z lactic acid
C Y, X, W, Z 4.3 Synthetic Rubber
D Y, W, Z, X 11. Figure 2 shows the structure
of two types of rubber, P and
10. A rubber tapper faces a Q. 12. Improper disposal of
problem of latex turning into synthetic rubber from vehicle
solid form when he collects C C C C C C tyres causes environmental
them on the second day S S S S pollution. Which of the
after tapping. So, he adds a C C C C following properties causes
substance into the latex to S S this problem?
solve the problem. S C C C S C C C A Non-biodegradable
Which of the following is the B Do not corrode
correct substance added and Rubber P C Resistant to oxidation
its explanation to solve the C C C C C C D Resistant to chemical
problem? C C C C solvents
Substance Explanation C C C C C C 13. Which of the following is not
added Rubber Q the method that can reduce
the disposal of rubber tyres?
A Water To dilute A Use discarded tyres to
latex Figure 2 generate electricity
Which of the following are the
B Sodium To preserve B Turn old tyres into stools
chloride latex and correct properties for rubber C Stick a new layer of
P and rubber Q?
solution maintain its rubber to old tyres
original state Rubber P Rubber Q D Burn discarded tyres
+
C Formic To add H I Less More elastic
acid ions and elastic
neutralise
the II Not sticky Become
negatively- when it is more sticky
charged heated when it is
heated
protein
membrane III Soft Hard
of rubber IV Not easy Easy to be
particles to be oxidised
oxidised



Subjective Questions

Section A Form
1. (a) Polystyrene is a compound commonly used to make food containers. The structural formula of styrene 5
is shown in Figure 1.1.


H
C C
H H
Figure 1.1

(i) State the source of raw material to make polystyrene. [1 mark]
(ii) State the type of polymerisation reaction used to produce polystyrene. [1 mark]



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Chemistry SPM Chapter 4 Polymer

(iii) Write a chemical equation for the polymerisation reaction in 1(a)(ii). [2 marks]
(iv) Describe a chemical test to confirm all the styrene molecules have been reacted to form
polystyrene. [2 marks]

(d) Nylon is a synthetic polymer used to make swimwear and fishing lines. Figure 1.2 shows the structural
formula of nylon.
H H H H H H H H O H H H H H H H H O
N C C C C C C N C C C C C C C C C C
H H H H H H H H H H H H H H n
Figure 1.2
(i) Draw the structural formulae of two monomers used to make nylon. [2 marks]
(ii) State the properties of nylon that make it suitable to be used for swimwear and fishing lines.
[2 marks]

Section B

2. (a) Mr Rosli is a chemistry laboratory assistant. He prepared latex in open beakers one day before the
experiment. He found out that the latex has turned into solid the next morning.
(i) Explain how the latex changed into a solid form. [4 marks]

(ii) Suggest a method to convert the solid rubber to latex again. [2 marks]
(b) Table 1.1 shows the results of latex coagulation.
Table 1.1
Procedure Initial time Latex coagulated time


Ethanoic acid is added into latex 8.00 a.m. 8.05 a.m.


Latex is left naturally in the laboratory 8.00 a.m. 1.30 p.m.

Explain why there is a difference in latex coagulation time. [6 marks]
(c) Table 1.2 shows the lengths of unvulcanised and vulcanised rubber strands in different stretching phases.
Table 1.2

Length of unvulcanised Length of vulcanised rubber
Stretching phases
rubber strand (mm) strand (mm)
Form
Before 100 100
5
During 120 110


After 105 100

Describe an experiment to compare the characteristic shown in Table 1.2 for both types of rubber.
[8 marks]





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Chemistry SPM Chapter 4 Polymer

Section C
3. (a) What is meant by polymer? [1 mark]
(i) Explain how the latex changed into a solid form. [4 marks]

(ii) Suggest a method to convert the solid rubber to latex again. [2 marks]
(b) Table 2 shows the three synthetic rubbers, X, Y and Z with their uses.
Table 2
Synthetic rubber Use


X To make engine hoses


Y To make medical implants


Z To make shoe soles

Based on Table 2, determine synthetic rubbers X, Y and Z.
Give one specific property of each synthetic rubber according to its use. [6 marks]

(c) Plastics are synthetic polymers which are very useful materials to improve our quality of life. However,
the widespread use of plastics results in large waste creation and eventually causes pollution problems.
Figure 2 shows a landfill full of plastic trash.
















Figure 2
(i) Name two examples of non-biodegradable synthetic polymer that can be recycled. State their
uses. [4 marks]

(ii) Improper disposal of synthetic polymer products can cause pollution. Describe three negative effects Form
of synthetic polymers on the environment. [3 marks]
(iii) Suggest three ways to overcome the environmental problems caused by synthetic polymer waste. 5
[3 marks]
(v) Although the use of synthetic polymers causes environmental problems, justify the use of synthetic
polymers. [3 marks]











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Chemistry SPM Answers
ANSWERS





mouth. A pipette should be used • Manipulated variable is
FORM 4 to suck up the solution/ liquid the factor that is purposely
chemical into the apparatus. changed in an experiment.
Chapter
1 Introduction to Chemistry Q3 He should remove his laboratory • Responding variable is the
coat immediately. If the acid wets his factor that changes with the
manipulated variable.
Checkpoint 1.1 clothes, he should rinse his whole • Fixed variable is the factor
body under the safety shower. At the
Q1 Chemistry is a branch of science same time, he should ask his friend that is kept constant in an
that studies the composition, to report the incident to the teacher. experiment.
structure, characteristics and (b) • List out all the materials and
interaction of matter. 1 apparatus needed to be
Q2 Iron: It is used to make nails, screws SPM Practice used.
and nuts; Chlorine: It is used to treat Objective Questions • Determine how to control
the manipulated and
tap water; Calcium carbonate: It is 1. D 2. B 3. C 4. B 5. A constant variables.
used as construction material/ to 6. D 7. C 8. C • Determine how to measure
manufacture cement. Subjective Questions the responding variable.
Q3 To carry out research and Section A • Determine how to collect,
development of food products/ To analyse and interpret data.
carry out analysis to check the 1. (a) (i) Long hair that is not tied up (c) (i) Manipulated variable –
quality of products/ To provide easily catches fire. Mass of salt added to pure
technical support to the marketing (ii) Closed-up shoes will be water.
section. able to protect the feet from Responding variable – Time
chemical spills or injury due taken to freeze the water.
Checkpoint 1.2
to glass apparatus falling (ii) The larger the mass of salt
Q1 Make observations, make an onto the feet or floor. added to pure water, the longer
inference, identify the problem, make (iii) If the excess solution is time it takes for the water to
a hypothesis, identify the variables, contaminated, the action freeze.
control the variables, plan and carry of pouring it back into (iii) Volume of pure water used.
out an experiment, collect data, its reagent bottle will Type of container used.
interpret data, make a conclusion, contaminate all the solution Type of freezer used.
write a report. in the bottle. (any two)
Q2 (a) The more water is added to the (b) I will report the incident to the (d) (i) The larger the mass of salt
acid, the higher the pH of the teacher immediately. Then, added to pure water, the
acid. using gloves, I will dispose higher is its density.
(b) Manipulated variable: Volume of of the broken test tube into (ii) • The manipulated variable
water added to the acid a specific container that is is the mass of salt added
Responding variable: pH value prepared and clean up the spill to pure water.
of acid as instructed by the teacher. • So, different mass of salt
Fixed variable: Type and volume (c) A fume chamber is a chamber is added to pure water, for
of acid, pH paper example 1 g, 2 g, 3 g, 4 g
where the air in it is always and 5 g.
Checkpoint 1.3 being sucked out. A fume • The responding variable is
chamber is used to carry
Q1 Goggles – to protect the eyes from the density of water.
chemicals out activities that involve • So, he can measure the
Safety shower – to remove volatile, inflammable or initial mass and final mass
chemicals that the body comes in toxic chemicals so that the of each mixture of water
contact by showering all parts of the vapour from the chemicals is and salt.
body disposed of immediately from The density of the mixture
Fire blanket – to put out the fire on a the fume chamber and does is calculated as follows:
victim’s body not contaminate the air in the Density =
laboratory.
Q2 A – Chemicals should not be sniffed Mass of mixture (g) 3
directly. Instead, the individual Section B Volume of mixture (cm )
should use his hands to waft the 2. (a) • Hypothesis is a statement • The fixed variables are
smell to his nose. that links the manipulated pure water and type of
variable to the responding salt used.
B – The solution/ liquid chemical variable. • So, pure water is used
should not be sucked using every time. Only one type



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Chemistry SPM Answers

of salt is used throughout Fixed variable: Presence of Checkpoint 2.2
the experiment, for water Q1 (a) Hydrogen atom has 1 proton
example sodium chloride. Procedure: with relative charge +1 and 1
Section C electron with relative charge –1.
Hence the net charge of H atom
3. (a) • We use various chemicals is (+1) + (–1) = 0
in many aspects of our Tap water (b) All the protons and neutrons
lives such as in foods, of an atom are found in the
agriculture, industries and Painted nucleus.
medicine. iron nail A proton or neutron is
• For example, we use 1) Two iron nails are cleaned approximately 2000 times
sodium chloride and with sand paper. heavier than an electron.
monosodium glutamate 2) One iron nail is painted with
(MSG) as flavourings in our a layer of paint and left to Q2 An atom is a discrete particle, very
small and indivisible.
dishes. dry.
• Sucrose or sugar is used Q3 (a) Electron
in various foods such 3) Both nails are placed (b) Thomson’s atom is a positively-
as desserts, cakes and in separate test tubes charged sphere embedded with
delicacies. containing tap water and are electrons.
• Acetic acid (ethanoic acid) left aside for three days. Q4 • Rutherford discovered the proton
is used as vinegar to pickle Observation: which is positively charged.
foods. The painted iron nail does not • He proposed the atomic model
• In agriculture, ammonia is rust while non-painted iron nail which consists of a positively-
used to make fertilisers for rusts. charged nucleus and negatively-
crops. Conclusion: charged electrons orbiting
• Slaked lime (calcium Painting prevents rusting of iron. around the nucleus.
hydroxide) is used to treat • He stated that the mass of an
acidic soil. Chapter atom is concentrated in the
• Hydrocarbons from 2 Matter and the Atomic Structure nucleus.
petroleum are separated
by fractional distillation and Checkpoint 2.1 Q5 • Bohr introduced the concept of
are important fuels. For shells.
example, petrol and diesel Q1 • The particles in a gas are very far • He stated that electrons orbit the
are fuels for vehicles. apart from each other and in random nucleus in fixed shells.
• We also use various motion. The particles can vibrate, Q6 (a) Neutron
petrochemical products rotate and move freely. They have (b) • Proton and neutron are
such as plastics, detergents, high energy content. found in the nucleus.
synthetic textiles and • When the gas is cooled, the particles • Neutrons contribute half the
synthetic rubber. of the gas lose energy and move overall mass of an atom.
• Metals such as copper, zinc, slower.
aluminium and iron are used • Eventually, the gas changes into Checkpoint 2.3
in various machines and a liquid. The particles of the liquid Q1 (a) Proton number = 9
important metal equipment are packed closely together but Nucleon number = 19
in industries. not in an orderly arrangement. The (b) F
19
• Various salts are used to particles can vibrate, rotate and 9
treat malnutrition diseases move throughout the liquid. They Q2 (a) 2.8.4
such as iron(II) sulphate is have low energy content. (b) (i) 14
used to treat lack of iron in Q2 (a) 0 °C (ii) silicon
anaemic patients. (b) This is because the heat energy (c) 29
• Barium sulphate salt is absorbed by the molecules of ice
used to check our digestive is used to overcome the forces Q3 (a) C
system using X-ray. Patients between molecules so that ice can (b) A and C
will have a drink of the salt turn into water. (c) B and D
before X-ray images are (c)
taken. Temperature (°C) (d) B; C; D and E
(any 10 examples of Checkpoint 2.4
other chemicals are also Q1 (a) Isotopes are atoms of the
accepted) 120 same element having the same
(b) Hypothesis: Painting prevents 100 number of protons but different
rusting of iron number of neutrons.
Manipulated variable: Presence (b) 82 electrons; 126 neutrons and
of a layer of paint on iron 15 S 82 protons
Responding variable: Rusting of Q
0
iron –3 P R
Time (min)


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Chemistry SPM Answers

Q2 RAM of Fe = [(5.85/100) × 54] + Temperature (°C)
[(91.75/100) × 56] + [(2.40/100) ×
57] = 55.91 T 1
Q3 Assume abundance of sulphur is T
S-32, x% and S-34, (100 – x)% 2
]
x
RAM of S = [( ) × 32 + T 3 Time (min)
100
100 – x × 34] = 32.1
100 Section B
32x + 3400 – 34x = 3210 2. (a) (i) Atoms with the same proton number but different nucleon numbers.
2x = 190 (ii) Chlorine
x = 95
∴ S-32, 95% dan S-34, 5% (iii) Isotope Chlorine-35 Chlorine-37
Q4 (a) Investigate uptake of
phosphorus in plants Proton number 17 17
(b) Detect and treat thyroid Nucleon number 35 37
(c) Measure the thickness of paper
and film Number of protons 17 17
(d) Detect leaks in underground Number of electrons 17 17
pipes
Number of neutrons 18 20
SPM Practice 2 Electron arrangement 2.8.7 2.8.7
Objective Questions Number of velance electrons 7 7
1. A 2. B 3. D 4. A 5. D
6. B 7. A 8. D 9. B 10. C
11. D 12. B 13. B 14. C 15. B
16. A 17. A 18. D 19. B 20. D (iv) Iodine-131: Treat thyroid diseases 3. Another gas jar filled with
21. A 22. B 23. D 24. C 25. C Carbon-14: Determine the age air is inverted on the jar
26. C of artifacts and fossils containing bromine vapour.
Subjective Questions (b) (i) Melting 4. The cover between the two
jars is removed.
Section A (ii) Kinetic energy of particles is
higher 5. The apparatus is set aside
1. (a) Sublimation (c) Chemical required: Liquid for a few minutes.
(b) T 2 °C bromine Observation:
(c) (i) Procedure: Red-brown vapour spreads
1. A few drops of liquid rapidly to both gas jars in a few
bromine are dropped into a minutes.
gas jar. Conclusion:
2. The jar is immediately Bromine gas is made up of tiny
closed and put aside for a discrete particles.
few minutes. Note: You can also choose
diffusion in liquid or solid.
Section C
(ii)
3. (a) Temperature T 1 T 2
Arrangement Particles are closely-packed Particles are closely-packed
of particles in an orderly manner. but not in an orderly manner.
Movement of Particles only vibrate and Particles can vibrate, rotate
particles rotate about fixed positions. and move from one point to
another point.

(d) The heat energy lost to the Attraction Strong forces of attraction Particles are held by weak
surrounding balances the heat force between between particles. forces of attraction.
energy given out when particles particles
of X form bonds with one
another to form a solid. Energy content Low energy content. Moderate energy content.
of particles
(e) (i) Particles move more slowly.
(ii) Particles lose its kinetic
energy.
(f)

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Chemistry SPM Answers

(b) (b) 4.515 × 10 atom (b) Carbon-12 can combine with
24
Thermometer (c) 60 dm 3 many elements/ Carbon-12
Q2 (a) 0.6 mol exists as solid at room
Test tube (b) 0.4 mol temperature and thus can
be handled easily during
Solid G Q3 (a) 160 g mol –1 investigations/ Carbon-12 is
Coconut oil (b) 80 g also used as standard in mass
Q4 4.8 g spectrometer / Carbon-12 is the
Q5 Z, X, Y most abundant carbon isotope,
thus the mass of 12 units
Procedure: Checkpoint 3.3 assigned to a carbon-12 atom is
1. A boiling tube is filled an accurate value. (any one)
with solid G to a depth of Q1 MgSO 4
3 cm and a thermometer Q2 (a) NO 2 (c) 48
is inserted into it. (b) N 2 O 4 ; Dinitrogen tetroxide (d) (i) Y + 2Br 2 → YBr 4
2. The boiling tube is Q3 (a) 12.8 g (ii) 1 mole of Y atoms reacts
clamped in a beaker half- (b) 1.5 mol with 2 moles of Br 2
filled with coconut oil molecules to produce 1
Q4 (a) K 2 SO 4
3. The level of solid G must (b) ZnCl 2 mole of YBr 4 units / 1 Y
be below the surface of (c) SnO atom reacts with 2 Br 2
the oil in the beaker. (d) (NH 4 ) 2 CO 3 molecules to produce 1
4. The coconut oil is heated (e) Mg(NO 3 ) 2 YBr 4 unit.
gently. (f) Na 3 PO 4 (iii) 0.15 mol of bromine gas
5. Solid G is stirred slowly Q5 (a) Aluminium hydroxide (iv)
with the thermometer. (b) Iron(II) sulphate
6. The temperature and (c) Ammonium chloride Element Y O
condition of solid G (d) Calcium nitrate Mass (g) 9.6 6.4
is recorded every half (e) Potassium carbonate
minute until solid G is (f) Zinc sulphide Number of 9.6 = 0.2 6.4 = 0.4
completely melted. moles of 48 16
• A graph of temperature against Q6 C, Mg(OH) 2 , FeBr 3 atoms
time is plotted for the heating Q7 Nitrogen monoxide; diboron trioxide; Simplest
of solid G. sulphur hexafluoride ratio of 1 2
Temperature (°C) Checkpoint 3.4 moles
Q1 (a) SO 3 (g) + H 2 O(l) → H 2 SO 4 (aq) Section B
(b) 2Mg(s) + CO 2 (g) → 2MgO(s) + 2. (a) (i) • RAM of P = 2 × RAM of Q
C(s) 28 = 2 × RAM of Q. So,
120 °C (c) N 2 (g) + 3H 2 (g) → 2NH 3 (g) 28
(d) 2AgNO 3 (aq) + Cu(s) → 2Ag(s) RAM of Q = 2 = 14
Time (min) + Cu(NO 3 ) 2 • 10 × RAM of P = 7 ×
Q2 (a) 2KClO 3 (s) → 2KCl(s) + 3O 2 (g) RAM of R
• Melting point of solid G is (b) 7.2 dm 3 10 × 28 = 7 × RAM of R.
120°C. So, RAM of
Q3 (a) 0.5 mol 10 × 28
(b) 1.204 × 10 atom R = = 40
23
Chapter 7
3 The Mole Concept, Chemical
Formula and Equation Q4 (a) Sodium hydroxide solution and • Ascending order of RAM
hydrogen gas = Q, P, R
Checkpoint 3.1 (b) 5 mol (ii) • Number of moles of
Q1 16 (c) 23 g atoms in the sample of
Q2 32 14 g of P
Q3 (a) 124 SPM Practice 3 = Mass of P
(b) 28 Objective Questions Molar mass of P
14 g
(c) 342 1. D 2. A 3. D 4. C 5. C = 28 g mol –1
(d) 122 6. A 7. B 8. C 9. B 10. D
11. D 12. C 13. B = 0.5 mol of atoms
Q4 (a) 62 • Number of moles of
(b) 189 Subjective Questions atoms in the sample 20 g
(c) 190 Section A of R
(d) 286 1. (a) Relative atomic mass of an Mass of R
Q5 24 element is the average mass = Molar mass of R
of an atom of the element
Checkpoint 3.2 1 = 20 g
24
Q1 (a) 1.505 × 10 molecules compared to 2 of the mass of 40 g mol –1
a carbon-12 atom. = 0.5 mol atom

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Chemistry SPM Answers

• As both samples have equal So, 13n = 26 Q2 (a) Helium. Helium has lowest
number of moles of atoms, 26 density hence, it can float in the
the samples have equal n = 13 = 2 air.
number of atoms. • Therefore, the molecular (b) Boiling point increases from
• Number of atoms in each formula of hydrocarbon W is helium to argon.
sample C 2 H 2 . Atomic size increases from
= number of moles × N A helium to argon. The van
= 0.5 × 6.02 × 10 atoms Chapter der Waals force of attraction
23
= 3.01 × 10 atoms 4 The Periodic Table of Elements between atoms become
23
(iii) – A crucible with its lid is stronger. More heat energy is
weighed and its mass is Checkpoint 4.1 needed to overcome the force of
recorded. Q1 (a) • Both of them arranged the attraction between the atoms.
– About 2 g of R powder elements in increasing order of
is placed in the crucible. relative atomic mass. Checkpoint 4.4
The crucible, its lid • Both of them grouped the Q1 (a) Group 1, all atom Li, Na dan K
and its content is elements with similar chemical has one valence electron.
weighed and its mass is properties into the same vertical (b) Li, Na, K
recorded. column named Group.
– The crucible is heated (b) • Left empty spaces in the (c) Atomic radius increases,
reactivity increases.
strongly without its lid. Periodic Table of Elements for
– When R starts to undiscovered elements Checkpoint 4.5
glow, the crucible is • He predicted the properties of the Q 1 (a) F: 2.7 Cl: 2.8.7
covered by its lid. The undiscovered elements based on
lid is removed a little at its position in the Periodic Table (b) Group 17
intervals. of Elements. (c) 7
– When the burning is • He mutually exchanged the (d) Yes. All atoms have the same
complete , the crucible position of nickel with cobalt number of valence electrons.
is opened and heated and iodine with tellurium so that
strongly. elements with similar chemical (e) Fluorine atom has smaller
– The crucible, its lid and properties were placed under the atomic size.
its content is weighed, same group. The outermost shell is closer
cooled and weighed. (c) He arranged the elements in to the nucleus. The nuclear
The mass reading is ascending order of their relative attraction on valence electrons
recorded. atomic mass instead of proton in fluorine atom is stronger.
– The heating, cooling number. Fluorine atom can attract one
and weighing are electon into its outermost shell
repeated a few times Checkpoint 4.2 easier.
until a constant mass is Q1 (a) (i) Nucleon number Q2 (a) Blue litmus paper turns red and
obtained. (ii) 2.8.1 then bleached.
(b) • Based on the given (iii) Group 1, Period 3 (b) The solution formed is acidic.
percentage composition, (b) Element X and Y. Atom X and Y (c) Cl 2 + H 2 O → HCl + HOCl
each 100 g of hydrocarbon have same number of valence (d) Blue litmus paper turns red and
W contains 92.3 g of carbon electrons. then bleached.
and 7.7 g of hydrogen.
(c) Element Y and Z. Atom Y and Checkpoint 4.6
Element C H Z have same number of shells Q1 (a) Period 2
filled with electrons.
Mass (g) 92.3 7.7 Q2 (a) Ascending order of proton (b) – Element X.
number – Atomic size of atom X is
Number 92.3 7.7 bigger than atom Y.
of moles 12 = 7.7 1 = 7.7 (b) (i) Alkali metal – The valence electron is
of atoms (ii) Halogen further from nucleus.
(iii) Transition elements – Force of attraction between
Simplest 1 1 (c) Period 2. Atom A, B and C have nucleus and valence
ratio of electrons is weaker.
moles two shells filled with electrons.
Checkpoint 4.3 – Atom X can donates
• So, the empirical formula of valence electron easier.
W is CH. Q1 (a) (i) X and Z Checkpoint 4.7
• Assume that the molecular (ii) Inert gas
formula of W is (CH) n (b) No, atom Z had achieved octet Q1 (a) Transition elements
• The molecular relative mass electron arrangement. Atom (b) – Form coloured ion or
of W Y does not donate, receive or compound
= n [12 + 1] share electron with atom Y. – Has more than one
= 13n oxidation number



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Chemistry SPM Answers

– Can act as catalyst Checkpoint 5.2 (b)
– Can form complex ions
Q1 (a) Lithium atom donates one
(c) Use as catalyst in Haber valence electron. Fluorine atom N
process. receive one electron into its
outermost shell. N
SPM Practice 4 (b) Cation Li , anion F – M N
+
(c) Li → Li + e ; F + e → F –
–
–
+
Objective Questions
1. C 2. D 3. B 4. D 5. A Q2 (a) Li 2 O N
6. D 7. A 8. D 9. C 10. C (b) LiCl
11. D 12. B 13. B (c) MgCl 2
(d) MgO
Subjective Questions (e) Al 2 O 3 (c)
Section A Q3 (a) Ionic bond N N
1. (a) X: chlorine, 2.8.7 (b)
Y : bromine, 2.8.8.7 + 2– +
(b) Group 17 because atom Y has M O M
7 valence electrons and Period Checkpoint 5.4
4 because it has 4 shells filled Q1 Hydrogen atom becomes partially
with electrons. (c) – Electron arrangement of positive charged because the
(c) (i) X is more reactive. atom M is 2.1. bonded pair electrons are pulled
(ii) Atomic size of atom X is – Atom M donates one closer to highly electronegative
smaller than atom Y. valence electron to achieve atom. The highly electronegative
The nuclear attraction on duplet electron arrangement atom in the molecule becomes
electrons is stronger. and forming cation M . + partially negative charged.
+
Atom X can receive one – M → M + e – Q2 Water is universal solvent because
electron in the outermost – Electron arrangement of water is a polar molecule which is
shell easier. atom O is 2.6 able to form hydrogen bonds. The
(iii) 2Fe + 3X 2 → 2FeX 3 – Oxygen atom receives 2 partial positive charged end of water
electrons from two M atoms molecule can form hydrogen bond
(d) Element Y is less reactive than to achieve stable octet
element X. with partial positive charge end from
electron arrangement and nearby ammonia, NH 3 , hydrogen
Section B forming anion O . chloride, HCl or ethanol, C 2 H 5 OH
2–
2. (a) (i) Group 1 because atom W – O + 2e → O 2– molecules.
–
has 1 valence electron, – Cation M and anion O Q3 Water is a bipolar molecule while
2–
+
Period 3 because W has 3 are attracted by strong carbon dioxide is a non-polar
shells filled with electron. electrostatic force and molecule. The hydrogen bonds
(ii) 2W + 2H 2 O → 2WOH + H 2 forming an ionic bond. between water molecules is stonger
(iii) Red litmus paper turns blue – An ionic compound, M 2 O than the Van der Waals attraction
(iv) Colourless solution / formed. between carbon dioxide molecules.
solution WOH is alkaline. Hence, more heat is needed to
Checkpoint
(b) (i) Y Q1 (a) XY 2 5.3 overcome the hydrogen bonds
(ii) 4Y + 3O 2 → 2Y 2 O 3 during boiling.
(c) W, X, Y, Z. (b) Y 2 Checkpoint 5.5
Proton number increases from Q2 (a) H O H
W, X, Y to Z. Q1 Dative bond is a type of covalent
The nuclear attraction on bond in which two electrons derive
electrons become stronger, (b) from one atom.
electrons are attracted closer H Cl Q2 Lone pair electrons
toward nucleus. Q3 (a)
(c) H H +
Chapter H C N
5 Chemical Bond H N + H + H N H
Q3 (a)
Checkpoint 5.1 H H
H
Q1 Inert gases have stable duplet or (b)
octet electron arrangement. Atoms of H H +
inert gases do not donate, receive or H M H
share electrons with other atom. H O + H + H O H
Q2 Ionic bond and covalent bond.
H




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Chemistry SPM Answers

+
Checkpoint 5.6 (b) (i) NH 3 + H 2 O → NH 4 + OH – Chapter
(ii) – Hydrogen atom shares 6 Acid, Base and Salt
Q1 (a) Metallic bond electrons with a more
(b) Metallic bonds are formed when electronegative nitrogen
metal cations are attracted by atom to form an Checkpoint 6.1
delocalised electrons in the sea ammonia molecule. Q1 Example of acid: hyrochloric acid, HCl
of electrons. – Electrons are not Hydrochloric acid, HCI ionises in
(c) Aluminium can conduct distributed evenly water to produce hydrogen ions, H .
+
electricity in solid state because because the shared Example of alkali: ammonia, NH 3
the delocalised electrons in the pair electrons are pulled Ammonia, NH 3 ionises in water to
sea of electrons can move freely closer to nitrogen atom. produce hydroxide ions, OH . –
and carry electrical charges. – Ammonia molecule Q2 H atoms in ammonia cannot ionise
becomes polar molecule to produce H ions.
+
SPM Practice 5 like water molecule. NH 3 molecule dissolves in water and
Objective Questions – Ammonia molecule form ionises to produce OH ion.
–
1. C 2. A 3. D 4. B 5. D hydrogen bond to water NH 3 + H 2 O → NH 4 + OH –
+
molecule.
6. C 7. C 8. D 9. A 10. B Dative bond Q3 Monoprotic acid: nitric acid and
11. A 12. C (iii) H + ethanoic acid
(iv)
Subjective Questions Diprotic acid: carbonic acid
Section A H N H Triprotic acid: phosphoric acid
1. (a) X H Q4 In water, ascorbic acid ionises to
produce H ion. H ion reacts with
+
+
(b) (i) YZ 2 reactive metal zinc to form salt and
(ii) Covalent – In ammonia molecule, hydrogen gas.
(iii) Molecule
NH 3 , nitrogen atom, N
(iv) has achieved an octet Hydrogen gas causes the
Z Y H electron arrangement and effervescence observed. In
hydrogen atoms, H have tetrachloromethane, ascorbic
acid does not ionise. Without
achieved duplet electron +
(c) (i) H ions, zinc does not react in
arrangements. tetrachloromethane.
+
– Hydrogen ion, H does
not have any electron in Q5 Moist red litmus paper contains
Z the shell. water. Ammonia ionises in the water
–
–
W W – The lone pair of electrons and produces OH ions. OH ion
exhibits alkaline properties and the
that are not involved
in covalent bond in red litmus paper turns blue.
ammonia molecule, NH 3
(ii) – B is a bipolar molecule Checkpoint 6.2
while A is non-polar are shared with hydrogen Q1 (a) pH value is used to measure
+
molecule. ion, H through the acidity and alkalinity of an
– Molecules B are attracted formation of dative bond. + aqueous solution.
by stronger hydrogen – Ion ammonium ion, NH 4 , +
bonds while molecules A nitrogen atom, N and (b) Definition: pH = –log [H ];
negative logarithm of the
are attracted by weaker hydrogen atoms, H have +
van der Waals attraction. achieved octet and duplet concentration of H ions in mol
–3
– More heat energy is electron arrangement dm .
needed to overcome the respectively. (c) At pH value = 7: +
Concentration of H ions is

hydrogen bonds between (c) – Valence electrons of copper –7 –3
molecules B during atoms are delocalised into 1 × 10 mol dm
boiling. sea of electrons. pOH value = 14 – 7 = 7
– Less heat is needed to – Electrons in the sea of Hence, water is neutral because
overcome the force of electron can move freely the concentration of H ions is
+
attraction between and carry electrical charges. the same as the concentration
molecule A during boiling. Hence, copper can conduct of OH ions.
–
electricity.
Section B – Metal cations are attracted Q2 (a) pH = –log [0.50] = 0.30
2. (a) – Substance Y has high melting by delocalised electrons (b) pH = –log [0.20] = 0.70
(c) pH = –log [0.04] = 1.40
point while substance X has strongly.
low melting point. – High heat energy is needed Q3 (a) pOH = –log [0.08] = 1.1;
– Substance X is a covalent to overcome the strong pH = 14 – 1.1 = 12.9
compound. metallic bonds during the (b) pOH = –log [0.001] = 3.0;
– Substance Y is an ionic melting. pH = 14 – 3.0 = 11
compound.
500 501
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Chemistry SPM Answers

(c) pOH = –log [0.05] = 1.30; (b) (i) Copper(II) chloride Mass of Na 2 CO 3
pH = 14 – 1.30 = 12.7 (ii) CuO + 2HCl → CuCl 2 + H 2 O = (number of moles) × (molar mass)
–1
Q4 P: 1.5 = –log [H ]; [H ] = 10 –1.5 = (0.625 mol) × (106 g mol )
+
+
= 0.0316 mol dm –3 Checkpoint 6.5 = 66.25 g
Q: 3.0 = –log [H ]; [H ] = 10 –3.0 Q1 (a) 3.2 = 12.8 g dm Q2 Number of moles of (COOH) 2 .2H 2 O
+
+
–3
= 0.001 mol dm 0.250 = (1.5) × (0.500) = 0.75 mol
–3
R: 4.7 = –log [H ]; [H ] = 10 –4.7 (b) 3.2 = Mass of (COOH) 2 .2H 2 O
+
+
= 2.0 × 10 mol dm (23 + 16 + 1) = (number of moles) × (molar mass)
–3
–5
S: 11.1 = –log [H ]; [H ] = 10 –11.1 0.250 = (0.75 mol) × (126 g mol )
+
+
–1
–3
= 7.9 × 10 mol dm 3.2 = 94.5 g
–12
T: 13.8 = –log [H ]; [H ] = 10 –13.8 40
+
+
–3
–3
= 1.6 × 10 mol dm 0.250 = 0.32 mol dm Q3 (2.0 mol dm ) × (V) = (0.50 mol
–3
–14
dm ) × (2.5 dm )
3
–3
Q2 (a) Al(NO 3 ) 3 → Al + 3NO 3 –
3+
Checkpoint 6.3 Molarity of Al 3+ V = (0.50 × 2.5)
–3
Q1 (a) HX ionises completely in water; = 1 × (0.30 mol dm ) 2.0
degree of ionisation is 100%. = 0.30 mol dm –3 = 0.625 dm 3
3
(b) HBr(aq) → H (aq) + Brˉ(aq) Molarity of NO 3 – –3 = 625 cm –3 3
+
Q2 (a) A weak acid is an acid that = 3 × (0.30 mol dm ) Q4 (1.8 mol dm ) × (888 cm ) 3
–3
–3
ionises partially in water. = 0.90 mol dm = (1.0 mol dm ) × (888 + V cm )
2+
(b) (i) HCOOH(aq) → HCOOˉ(aq) (b) MgCl 2 → Mg + 2Cl – 888 + V = 1.8 × 888
2+
+ H (ak) Molarity of Mg 1.0
+
–3
(ii) HCOOH; HCOOˉ; H = 1 × (0.90 mol dm ) V = 710.4 cm 3
+
–3

Q3 (a) 2K(s) + 2H 2 O(l) → 2KOH(aq) + = 0.90 mol dm –
H 2 (g) Molarity of NO 3 Checkpoint 6.7
–3
= 2 × (0.90 mol dm )
(b) KOH is a strong alkali because = 1.80 mol dm –3 Q1 (a) pH = 7
it ionises completely in water. (c) Na 2 SO 4 → 2Na + SO 4 (b) 5 cm 3
+
2–
KOH(aq) → K (aq) + OHˉ(aq) Molarity of Na + (c) (x mol dm ) × (5 cm ) = (0.20
+
–3
3
Q4 (a) A weak alkali is an alkali that = 2 × (1.80 mol dm ) mol dm ) × (25 cm )
3
–3
–3
ionises partially in water. = 3.60 mol dm –3 x = 1.0 mol dm –3
(b) Ammonium ion and hydroxide 2– Q2 Number of moles of HBr
ion. Molarity of SO 4 –3 0.240
= 1 × (1.80 mol dm )

(c) 0.5 × 200 = 1 = 1.80 mol dm –3 = 24 = 0.01 mol
100
Q3 (a) Number of moles of Pb(NO 3 ) 2 HBr + NH 3 → NH 4 Br
Checkpoint 6.4 = (0.50) × (0.500) 1 mol 1 mol
Q1 (a) 3H 2 SO 4 + 2Al(OH) 3 → Al 2 (SO 4 ) 3 = 0.25 mol Number of moles of NH 3 = 0.01 mol
+ 6H 2 O Mass of Pb(NO 3 ) 2 0.01 = (0.50 mol dm ) × (V dm ) 3
–3
3
(b) 2HNO 3 + PbO → Pb(NO 3 ) 2 + = (number of moles) × (molar V = 0.02 dm = 20 cm 3
H 2 O mass) Q3 Number of moles of K 2 CO 3 =
–1
(c) H 3 PO 4 + 3NH 3 → (NH4) 3 PO 4 = (0.25 mol) × (331 g mol ) 13.8 = 13.8 = 0.10 mol
(d) 2HCl + Zn → ZnCl 2 + H 2 O = 82.75 g 2(39) + 12 + 3(16) 138
(b) Number of moles of K 3 PO 4 2H 3 PO 4 + 3K 2 CO 3 → 2K 3 PO 4 + 3H 2 O
(e) 2CH 3 COOH + Na 2 CO 3 → = (1.0) × (1.5)
2CH 3 COONa + CO 2 + H 2 O + 3CO 2
= 1.5 mol Mole ratio: K 2 CO 3 :H 3 PO 4 = 3:2
(f) (NH 4 ) 2 SO 4 + 2NaOH → Na 2 SO 4 Number of moles of H 3 PO 4 required
+ 2NH 3 + 2H 2 O Mass of K 3 PO 4 2 2
= (number of moles) × (molar
(g) 2NaOH + MgCl 2 → Mg(OH) 2 + mass) = 0.10 × 3 = 30 mol
2NaCl = (1.5 mol) × (212 g mol ) Molarity of H 3 PO 4
–1
Q2 (a) (i) Hydrogen = 318 g = 49.0 = 49.0 = 0.5 mol dm –3
(ii) Zn + H 2 SO 4 → ZnSO 4 + H 2 (c) Number of moles of Na 2 S 2 O 3 3 + 31 + 64 98
(b) (i) Copper = (2.20) × (2.0) Volume of acid = 2/30 = 0.133 dm 3
(ii) Copper is an unreactive = 4.4 mol 0.5
3
metal and does not react Mass of Na 2 S 2 O 3 = 133 cm
with dilute acids. = (number of moles) × (molar Q4 % N in MAP, NH 4 H 2 PO 4 =
Q3 (a) (i) Alkali metal oxide: sodium mass) –1 1(14) × 100
oxide = (4.4 mol) × (158 g mol ) 14 + 4 + 2 + 31 +4(16)
(ii) Metal hydroxide: sodium = 695.2 g = 14 × 100
hydroxide Checkpoint 6.6 115
(iii) M 2 O + H 2 O → 2MOH = 12.2%
Or Q1 Number of moles of Na 2 CO 3
Na 2 O + H 2 O → 2NaOH = (2.5) × (0.250)
= 0.625 mol
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Chemistry SPM Answers

Checkpoint 6.8 • The salt solution should not Q4 (a) Silver nitrate solution is added to
be evaporated to dryness potassium chromate(VI) solution.
Q1 (a) Salt is an ionic compound because the salt might
produced when hydrogen ion in decompose by the heat. 2AgNO 3 (aq) + K 2 CrO 4 (aq) →
an acid is replaced by a metal Evaporate the salt solution Ag 2 CrO 4 (s) + 2KNO 3 (aq)
ion or ammonium ion. until it is saturated and The precipitate is removed by
(b) (i) CaCO 3 then allow it to cool for filtration, washed with distilled
(ii) Calcium ion and carbonate crystallisation to occur. water and dried between layers
ion (b) Equation for reaction: of filter papers.
(iii) Carbonic acid and calcium 2CH 3 COOH(aq) + PbO(s) → (b) Number of moles of Pb =
2+
hydroxide Pb(CH 3 COO) 2 (s) + H 2 O(l) MV
Q2 (a) Ammonium ethanoate Number of moles of PbO 1000 5
–3
(b) Potassium phosphate = mass of PbO (1.0 mol dm ) × 1000 dm 3
(c) Sodium bromide molar mass of PbO = 0.005 mol
4.46 g MV
Q3 (a) Food additives and food = (207 + 16) g mol –1 Number of moles = 1000
preservatives 4.46 -3 20 3
(b) Rising flour and antacid = 223 (0.5 mol dm ) × 1000 dm
Q4 The particles are closely packed = 0.020 mol = 0.010 mol
–
2+
and orderly. The strong forces of From equation: 1 mol PbO Simple mole ratio Pb : I
attraction between particles keep the produces 1 mol Pb(CH 3 COO) 2 → = 0.005:0.010 = 1:2
particles in position. 0.020 mol PbO produces 0.020 Ionic equation:
mol Pb(CH 3 COO) 2 Pb (aq) + 2I (aq) → PbI 2 (s)
2+
–
Checkpoint 6.9
Mass of Pb(CH 3 COO) 2 produced Checkpoint 6.10
Q1 (a) (i) Sodium nitrate; lead(II) = [number of moles
ethanoate; copper(II) Pb(CH 3 COO) 2 ] × [molar mass Q1 Silver nitrate; ammonium chloride;
chloride; ammonium Pb(CH 3 COO) 2 ] iron(II) sulfate; calcium carbonate
sulphate; zinc nitrate = 0.020 mol × [207 + 2 (2 × 12) Q2 (a) Ammonia, oxygen, hydrogen,
(ii) Iron(III) carbonate; silver + 3 × 1 + 2 × 16)] g mol carbon dioxide, sulphur dioxide,
–1
chloride = 0.020 × 325 = 6.5 g hydrogen chloride
(b) (i) Ammonia solution and Q3 (a) Sodium chloride (b) Nitrogen dioxide
sulphuric acid: (b) 2HCl(aq) + Na 2 CO 3 (aq) → (c) Chlorine; carbon dioxide;
2NH 3 (aq) + H 2 SO 4 (aq) → 2NaCl(aq) + H 2 O(l) + CO 2 (g) nitrogen dioxide; sulphur
(NH 4 ) 2 PO 4 (aq) (c) Volume of HCl used = 27.60 – dioxide; hydrogen chloride
(ii) Copper(II) carbonate and 1.10 = 26.50 cm (d) Ammonia
3
hydrochloric acid: Number of moles of HCl Q3 (a) Green solid turns black.
CuCO 3 (s) + 2HCl(aq) → = MV (b) Flow the gas through limewater.
CuCl 2 (aq) + H 2 O(l) + CO 2 (g) 1000 –3 26.5 3 Limewater turns cloudy (white
(iii) Iron(III) chloride and calcium (2.0 mol dm ) × 1000 dm precipitate is formed)
carbonate: = 0.053 mol The gas is carbon dioxide.
2FeCl 3 (aq) + 3K 2 CO 3 (aq) → From equation: 2 mol HCl reacts (c) CuCO 3 (s) → CuO(s) + CO 2 (g)
Fe 2 CO 3 (s) + 6KNO 3 (aq) with 1 mol Na 2 CO 3 → 0.053 mol
1
(iv) Zinc hydroxide and nitric HCl reacts with 0.053 × 2 = Q4 (a) 10NaCl + 8H 2 SO 4 + 2KMnO 4 →
5Na 2 SO 4 + K 2 SO 4 + 2MnSO 4 +
acid: 0.0265 mol Na 2 CO 3
Molarity of Na 2 CO 3 solution 8H 2 O + 5Cl 2
Zn(OH) 2 (aq) + 2HNO 3 (aq) number of moles (b) (i) Chlorine
→ Zn(NO 3 ) 2 (aq) + 2H 2 O(l) = volume (ii) Bring a piece of moist blue
Q2 (a) • The mixture of ethanoic 0.0265 litmus paper to the mouth of
acid and lead(II) oxide 25 the test tube.
must be stirred to speed = 1000 The blue litmus paper turns
up the reaction and ensure –3 red and then white.
complete reaction. = 1.06 mol dm (c) Number of moles of NaCl
• Adding acid to oxide will (d) From the equation: 2 mol HCl mass of NaCI
result in excess acid added. produce 1 mol NaCl = molar mass of NaCI
The excess acid cannot be Hence, 0.053 mol HCl produce 10 10
removed by filtration. The 0.053 mol NaCl = (23 + 35.5) = 58.5
salt becomes contaminated. Mass of NaCl produces
Hence, lead(II) oxide = (number of moles of NaCl) × = 0.1709 mol
powder must be added to (molar mass of NaCl) –1 10 mol NaCl produces 5 mol Cl 2
= 0.053 mol × (23 + 35.5) g mol
a fixed amount of acid until 0.1709 mol NaCl produces
the powder cannot dissolve = 0.053 × 58.5 = 3.10 g 0.1709 × 5 = 0.08545 mol Cl 2
anymore. 10



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Chemistry SPM Answers

Volume of Cl 2 liberated NaOH(aq) → Na (aq) + OH (aq) From the equation:
–
+
= (number of moles) × (molar A weak alkali such as ammonia 2 mol NaOH = 1 mol H 2 SO 4
volume) ionises partially in water to ∴ 0.0642 mol NaOH
–1
3
= 0.08545 mol × 24 dm mol produce a low concentration of 1
= 2.05 dm 3 hydroxide ions. = 0.0642 × mol H 2 SO 4
2
NH 3 (aq) + H 2 O(l) L Number of moles of H 2 SO 4
–
Checkpoint 6.11 NH 4 (aq) + OH (aq) MV
+
=
Q1 (a) (i) Magnesium oxide (b) (i) Sodium hydroxide solution 1000
(ii) Carbon dioxide reacts with lead(II) nitrate = 0.0321 mol
(iii) CaCO 3 solution to form a white 0.0321 = M × (0.025)
(iv) MgCO 3 (s) → MgO(s) + precipitate. The precipitate Concentration of sulphuric
CO 2 (g) is lead(II) hydroxide. acid
2+
–
Pb (aq) + 2OH (aq) → 0.0321
(b) (i) X = nitric acid; Pb(OH) 2 (s) = 0.025
Y = hydrochloric acid The white precipitate = 1.3 mol dm –3
(ii) Mg dissolves in excess sodium
2+
(iii) MgO(s) + 2H (aq) → hydroxide solution to form a Chapter
+
Mg (aq) + H 2 O(l) colourless solution. 7 Rate of Reaction
2+
(iv) Magnesium hydroxide (ii) Ammonia solution reacts
(v) Mg (aq) + 2OH (aq) → with lead(II) nitrate solution Checkpoint 7.1
2+
–
Mg(OH) 2 (s) to form a white precipitate. Q1 Fast: combustion of natural gas,
The precipitate is lead(II) neutralisation of acid in mouth with
SPM Practice 6 hydroxide. toothpaste.
2+
–
Objective Questions Pb (aq) + 2OH (aq) → Slow: browning of newspaper,
Pb(OH) 2 (s)
1. B 2. D 3. A 4. A 5. A The white precipitate does fermentation
6. D 7. B 8. B 9. B 10. D not dissolve in excess (any suitable answer)
11. B 12. C 13. D 14. A 15. B ammonia solution. Q2 (i)
16. D 17. D 18. C 19. A 20. B (c) (i) A standard solution is a Mass of CaCO 3 (g)
21. C 22. A solution which has a known
concentration.
Subjective Questions
Section A (ii) Calculate the mass of
NaOH required:
1. (a) X = calcium oxide; Number of moles of NaOH
Y = carbon dioxide; = 2.0 mol dm × 1.0 dm 3
–3
Z = limewater/ calcium hydroxide = 2.0 mol 0
solution Mass of NaOH Time (s)
(b) CaCO 3 (s) → CaO(s) + CO 2 (g) = 2.0 mol × (23 + 16 + 1) (ii)
–1
(c) Calcium oxide is a base. g mol Mass of CaSO 4 (g)
CaO dissolves in water and = 80 g
ionises to produce hydroxide Procedure:
ions. Hydroxide ion causes the • Weigh exactly 80 g of
solution to be alkaline. solid sodium hydroxide in
a weighing bottle.
(d) Carbon dioxide reacts with water • Transfer the solid into a
to form carbonic acid which 1.0 dm volumetric flask.
3
ionises to produce hydrogen • Add a little distilled water 0
ions. Hydrogen ion causes the to dissolve the solid. Time (s)
solution to be acidic. • Add distilled water to (iii)
(e) Dilute solid X with distilled make exactly 1 dm of Concentration of H 2 SO 4 (mol dm )
3
–3
water. Channel gas Y into the solution.
solution and a white precipitate • Close the flask and
(carbonate) is formed. shake to obtain a uniform
CaO(s) + H 2 O(l) → solution.
Ca(OH) 2 (aq) (iii) Number of moles of NaOH
Ca(OH) 2 (aq) + CO 2 (g) → MV
CaCO 3 (s) + H 2 O(l) = 1000
= 2.0 × 0.321
Section B 0 Time (s)
= 0.0642 mol
2. (a) A strong alkali such as sodium H 2 SO 4 (aq) + 2NaOH(aq) →
hydroxide ionises completely Na 2 SO 4 (aq) + 2H 2 O(l)
in water to produce a high
concentration of hydroxide ions.
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Chemistry SPM Answers

(iv) Q3 Porridge is a diluted rice. The (b)
Volume of CO 2 (cm ) number of rice per unit volume
3
in porridge is smaller. Amylase
enzyme take shorter time to digest. Oxygen
Q4 • Cut the chicken meats into smaller Sulphuric gas
acid
cubes and cooks it in a pressure
cooker. Calcium
• The temperature of the chicken carbonate Water
soup in pressure cooker is higher. chips
0 • Smaller cubes of chicken meat
Time (s) have larger total surface area.
Q3 More heat energy is absorbed in
shorter time. Sulphuric Calcium
Oxygen acid carbonate
chips
gas Checkpoint 7.4 100 g
Q1 Collision between the reactant (c) CaCO 3 + HCl → CaCl 2 + CO 2 +
particles in correct orientation, H 2 O
Hydrogen achieved the activation energy and (d) (i)
peroxide Water results products. Volume of carbon dioxide gas (cm )
3
Manganese(IV) oxide Q2 (a) Set I
13.6
Q4 (a) Overall rate of reaction = 60 2V
maximum volume of oxygen 3 –1 III II
= gas collected = 0.23 cm s
time taken of reaction to Set II V I
complete 32.6
46.50 =
= 60
180 = 0.54 cm s 0
3 –1
3
= 0.26 cm s –1 Volume of H 2 gas (cm ) Time (min)
3
(b) Average rate of reaction in the (ii) – Temperature of sulfuric
first minute 32.6 acid in set III is higher,
volume of oxygen gas thus particles have
= collected in first 60 s higher kinetic energy.
time taken 13.4 – Frequency of collisions
27.50 between CaCO 3 and H +
= 0 ion increases.
60 Time (s) – Frequency of effective
3
= 0.46 cm s –1 (c) Experiment II has higher rate of collisions increases.
(c) Average rate of reaction in the reaction. 2. (a) (i) Manganese(IV) oxide
first 3 minutes The size of magnesium powder (ii) 2H 2 O 2 → 2 H 2 O + O 2
volume of oxygen gas is smaller than magnesium (iii) Manganese(IV) oxide acts
= collected from 120 s to 180 s ribbon. as catalyst by lowering
time taken The total surface area of the the activation energy of
50.00 – 41.50 magnesium powder exposed to decomposition of hydrogen
=
60 hydrochloric acid is larger. peroxide. More hydrogen
3
= 0.14 cm s –1 Frequency of collision between peroxide molecules can
magnesium atom, Mg and achieve the activation
Checkpoint 7.2 hydrogen ion, H is higher. energy. The frequency of
+
Frequency of effective collision effective collision increases.
Q1 (a) Zn + 2HCl → ZnCl 2 + H 2
is higher. (b) Antacid B
(b) (i) Concentration of
hydrochloric acid Antacid B in powder form is
(ii) The rate of reaction of SPM Practice 7 smaller in size and it has larger
experiment II is higher than Objective Questions total surface area exposed
to acid. More H ions are
+
experiment I 1. A 2. A 3. D 4. D 5. B neutralises in short time.
6. D 7. C 8. B 9. B 10. C
Checkpoint 7.3 (c)
11. C 12. C 13. D
Q1 The temperature of hot frying oil is Fast reaction Slow reaction
higher than boiling water. Potato Subjective Questions
strips absorbed more heat in hot oil. Section A • Combustion of • Fermentation of
yeast
ethanol
Q2 Smaller sized charcoal has larger 1. (a) Volume of carbon dioxide gas • Displacement of • Electrolysis
total surface area exposed to evolved // mass of calcium metal from its salt
oxygen in air. Allowing smaller sized carbonate solution
charcoal to burn easily.
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Chemistry SPM Answers

(b) (i) Advanced ceramic Subjective Questions
Chapter Manufactured Substances in
8 (ii) • Lithium silicon oxide Section A
Industry
is used as heat shield 1. (a) Alloy is a mixture of two or
that can withstand high
Checkpoint 8.1 temperature and is used more elements where the main
Q1 (a) Alloy is a mixture of two or as heat insulator in rocket element is metal.
more elements where the main nozzle. (b) W: Carbon
element is a metal. • Alumina Al 2 O 3 and X: Pewter
(b) 90% Cu, 10% Sn silica SiO 2 is used Y: Aluminium
(c) in manufacture of Z: Stainless steel
microchips. (c) (i) To increase the strength and
(any suitable answers)
Zn resistance to corrosion.
(ii) Carbon atom has smaller
Checkpoint 8.4
atomic size. Carbon
Cu Q1 Composite material is a new atom disrupt the orderly
substance produced from combining
(d) The presence of smaller size arrangement of iron atoms.
zinc atoms disrupt the orderly two or more non-homogeneous The presence of carbon
arrangement of copper atoms in substances such as alloy, glass, atom prevents the iron
pure copper. The layer of metal polymer and ceramic. atoms from, sliding when a
atoms cannot slide easily if force Q2 force is applied.
is applied. Composite 2. (a) (i) Fibre glass
Q2 (a) Ductile and malleable materials Uses (ii) Glass fibre and plastic
(b) (i) Duralumin Fibre glass Make helmet (iii) Lighter and can be moulded
(ii) Lighter, stronger and Reinforced Building materials for into any shape
resistant to corrosion concrete tall building (iv) • Use of composite
(iii) Material used in aircraft material can improve
construction Optical fibre Cable for computer living standards because
networking composite material is
Checkpoint 8.2 superior to its original
Photochromic Optical lenses
Q1 (a) Transparent, hard but brittle, glass component.
chemically inert, electrical Superconduktor Components in • Composite material
insulator Magnetic Reson- produced through
(b) Silicon dioxide, SiO 2 once Imaging (MRI) chemical engineering can
reduce the exploitation
Q2 (a) Soda-lime glass. Soda- (any three answers) of natural resources and
lime glass has high thermal conserves a balanced
expansion coefficient and Q3 – Original component: silica glass ecosystem.
it cannot withstand sudden fibre and plastic polymer • Producing composite
temperature change. – Silica glass fibre breaks easily material which is more
but fibre glass is durable.
(b) Borosilicate glass. durable and long lasting
Borosilicate glass has zero – Plastic has low stretching can reduce the waste
thermal expansion coefficient strength but fibre glass has high generation by human.
thus it is resistant to thermal stretching strength. (b) (i) Advanced ceramic is
stress. produced by mixing additive
SPM Practice 8
Checkpoint 8.3 substances such as oxides,
Objective Questions carbides and nitrides into
Q1 (a) – Inert chemically to make ceramic.
dental enamel. 1. B 2. B 3. A 4. D 5. D (ii) Semiconductor, heat
– Hard and strong to make 6. D 7. A 8. D 9. A 10. C insulator which can
construction metals. 11. A 12. C 13. D 14. B 15. C withstand high heat.
– Heat insulator to make inner 16. C
wall of oven.
(any suitable answers)


Oxidising agent: CuO (g) Ba = +2
FORM 5 Reducing agent: H 2 (h) Fe = +3
2. (a) Pb = +2
Chapter 3. Reaction between hydrochloric acid
1 Redox Equilibrium (b) Cu = +1 and sodium hydroxide solution
(c) O = –1 HCl + NaOH → NaCl + H 2 O
Checkpoint 1.1 (d) V = +4
1. (a) CuO + H 2 → Cu + H 2 O (e) Cl = +1 +1 –1 +1–2 +1 +1 –1 +1 –2
(b) Oxidised substance: H 2 (f) H = –1 The oxidation number of element
before the reaction:
Reduced substance: CuO

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Answers Focus SPM Chemistry.indd 506 03/04/2023 2:18 PM

Chemistry SPM Answers

H = +1, Cl = –1, Na = +1, O = –2, Half equation for readuction: to +2. Copper undergoes
H = +1 Br 2 + 2e → 2Br – oxidation.
–
The oxidation number of element (c) Oxidising agent: Bromine – The oxidation number of
after the reaction: Reducing agent: Iron(II) sulphate silver decreases from +1
Na = +1, Cl = –1, H = +1, O = –2 solution // Solution X to 0. Silver ion undergoes
reduction.
The oxidation numbers of all 5. (a) Grey solid: Silver – Oxidation and reduction
elements remain unchanged. Cation in blue solution: occur simultaneously.
Copper(II) ion
4. (a) X: Iron(II) sulphate solution (or
2+
+
any soluble iron(II) salt solution) (b) Cu + 2Ag → Cu + 2Ag
(b) Half equation for oxidation: (c) – The oxidation number of
Fe → Fe + e – copper increases from 0
2+
3+
Checkpoint 1.2
1.
Half-cells Stronger oxidising agent Stronger reducing agent
(a) Cu (aq) + 2e L Cu(s) E = +0.34 V Cu 2+ Zn
o
–
2+
Zn (aq) + 2e L Zn(s) E = –0.76 V
–
2+
o
(b) Na (aq) + e L Na(s) E = –2.71 V
+
–
o
I 2 (s) + 2e L 2I (aq) E = +0.54 V I 2 Na
–
o
–
(c) Cl 2 (g) + 2e L 2Cl (aq) E = +1.36 V
–
–
o
Cl 2 Ca
Ca (aq) + 2e L Ca(s) E = –2.87 V
o
–
2+
2. (a) Ag, Pb, Mg (b) Fe(s) | Fe (aq) || Ag (aq) | Ag(s)
+
2+
(b) • Reaction occurs. (c) Voltage = (+0.80) – (–0.44)
o
• E value of magnesium, Mg is = +1.24 V
more negative than E value (d) The intensity of green colour increases. Iron atoms donate electrons to form
o
of silver, Ag. iron(II) ions. The concentration of iron(II) ions increases.
• Mg is a stronger reducing
agent compared to Ag. Checkpoint 1.4
• Mg atom is easier to donate 1. (a) P: E value
o
2+
electrons to form Mg ion.
• Mg undergoes oxidation. Q: Concentration of ion
+
• E value of silver ion, Ag is (b)
o
more positive than E value of
o
magnesium ion, Mg . Electrode P (Cathode) Q (Anode)
2+
• Ag ion is a strong oxidising Ions that are attracted Na , H + Cl , OH –
+
+
–
agent compared to Mg ion. Ions that are selectively Hydrogen ion Chloride ion
2+
• Ag ion is easier to receive discharged
+
electrons to form Ag atom.
+
3. (a) Reaction does not occur Reason Eo value of H ion is more Concentration of chloride
ions, Cl is higher than
positive than E value of
–
o
(b) Reaction occurs Na ion hydroxide ions, OH –
+
(c) Reaction does not occur Half equation 2H + 2e → H 2 2Cl → Cl 2 + 2e –
–
–
+
(d) Reaction occurs
2. (a) To make the iron spoon more (d) Silver nitrate solution
Checkpoint 1.3 attractive and resistant to 3. (a)
1. (a) Pb(s) | Pb (aq) || Ag (aq) | Ag(s) corrosion.
+
2+
(b) Al(s) | Al (aq) || Sn (aq) | Sn(s) (b) Connect the iron spoon to the A
2+
3+
2+
(c) Mg(s) | Mg (aq) || Cl 2 (g) | Cl – negative terminal of the batteries
(aq) | Pt(s) and connect the copper plate
2. (a) +3.18 V to the positive terminal of the Impure Pure
batteries. silver silver plate
(b) 0.63 V (c) 1. Rotate the iron spoon during plate Silver nitrate,
(c) +0.59 V the electroplating process to AgNO solution
3
3. (a) (i) Negative terminal: Iron get a uniform layer of plating.
Positive terminal: Silver 2. Use a low current during the (b) Anode: Impure silver plate
(ii) Half equation for oxidation: electroplating process to get becomes thinner
Fe → Fe + 2e – a strong layer of coating on Cathode: Pure silver plate
2+
Half equation for reduction: the iron spoon. becomes thicker
Ag + e → Ag
+
–
507
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Answers Focus SPM Chemistry.indd 507 03/04/2023 2:18 PM

Chemistry SPM Answers

(c) Half equation at anode: (ii) • Intensity of the blue colour
Ag → Ag + e – SPM Practice 1 decreases
+
Half equation at cathode: Objective Questions • Cu ion receives electrons to
2+
Ag + e → Ag form copper atom
–
+
1. A 2. B 3. B 4. C 5. D 2+
6. A 7. B 8. D 9. A 10. A • Concentration of Cu ion
Checkpoint 1.5 decreases
11. A 12. D 13. B 14. C 15. C
Q1 (a) Fe 2 O 3 16. A 17. C 18. C 19. C 20. D (c) • Metal X: Copper // Lead // Silver
(b) 2 Fe + 3(–2) = 0 21. D 22. C 23. B 24. C 25. D • Metal Y: Magnesium // Zinc //
2 Fe = +6 26. C 27. C 28. C 29. B 30. A Aluminium
Fe = +3 31. B 32. B 33. B 34. C 35. D • Procedure:
IUPAC nomenclature: Iron(III) oxide 36. A 1. Rub / Clean the iron nails,
(c) 2Fe 2 O 3 + 3C → 4Fe + 3CO 2 Subjective Questions X strip and Y strip with
sandpaper.
(d) Calcium carbonate decomposes to Section A 2. Coil the iron nails with metal
form calcium oxide. Calcium oxide 1. (a) X and metal Y respectively.
reacts with silicon(IV) dioxide in iron 3. Place the iron nails into two
ore to form slag or calcium silicate. different test tubes.
This can remove the impurities from A 4. Pour hot jelly solution
molten iron. Impure with potassium
(e) Aluminium is more reactive than Pure copper hexacyanoferrate(III) solution
carbon. Carbon cannot reduce copper plate and phenolphthalein into
aluminium oxide to aluminium. The plate 0.5 mol dm –3 each test tube.
reaction between aluminium oxide copper(II) 5. Leave the test tubes for
and carbon does not occur. sulphate three days and record the
(f) Tin(II) oxide solution observations.
(b) Mass of copper deposited (d) • No.
Checkpoint 1.6 = 0.05 × 64 • Electroplating of the iron ring
1. • Water and oxygen are present = 3.2 g does not occur.
during rusting of iron. Iron atom Remaining mass of impure • Silver is connected to the positive
donates electrons and oxidises to copper terminal of the battery / acts
Fe . = 5.0 – 3.2 as the anode while iron ring
2+
• Oxygen and water molecules = 1.8 g is connected to the negative
receive electrons and undergo (c) (i) A: Hydrochloric acid terminal of the battery / acts as
reduction to form OH ion. B: Sulphuric acid // Nitric the cathode.
–
• Fe ion reacts with OH ion to acid 3. (a) • Reactions which involve
–
2+
form iron(II) hydroxide, Fe(OH) 2 . (ii) P: 2Cl → Cl 2 + 2e – oxidation and reduction that
–
• Iron(II) hydroxide is oxidised to R: 4OH → O 2 + 2H 2 O + 4e – occur simultaneously // at the
–
hydrated iron(III) oxide. (d) • Replace 1.0 mol dm same time
–3
• The formation of brown solid can hydrochloric acid with 0.0001 • X: Iron(II) sulphate solution //
be prevented by applying spray mol dm hydrochloric acid / FeSO 4
–3
paint or grease. very dilute hydrochloric acid. • Y: Acidified potassium
2. Tin forms a protective oxide layer. • Hydroxide ion is discharged manganate(VII) solution //
KMnO 4 / H
+
The tin oxide layer protects the • because the E value of OH is • Oxidation: Fe → Fe + e –
o
–
3+
2+
steel from coming into contact with more negative / less positive • Reduction:
oxygen and water. Therefore, steel than the E value of Cl ion. MnO 4 + 8H + 5e → Mn +
–
o
–
–
2+
+
does not oxidise and rusting does 2. (a) • A chemical reaction where 4H 2 O
not occur. oxidation and reduction occur
3. Metal P is copper while metal Q is simultaneously / at the same (b) Number of moles of Al 2 O 3
magnesium. time. 2000 000
• Cl 2 + 2KBr → 2KCl + Br 2 = 102
Experiment I Experiment II (Any suitable pair of halogen = 19607.84 mol
• Iron rusts • Iron does not and halide solution) Mole ratio:
• Iron is more rust • Oxidation number of chlorine Al 2 O 3 : Al
electropositive • Iron is less decreases from 0 to –1.
than P electropositive Chlorine undergoes reduction. 2 mol : 4 mol
• Iron atom than Q • Oxidation number of iodine 19 607.84 mol : 39215.68 mol
donates • Oxygen and increases from –1 to 0. Mass of aluminium
electrons to water receive Potassium iodide undergoes = 39215.68 × 27
form Fe ion electrons to oxidation. = 1058823.36 g //
2+
–
• Half equation: form OH ion (b) (i) • Negative terminal: Magnesium = 1058.82 kg
Fe → Fe + 2e – • Half equation: • Positive terminal: Copper (c) • The grey solid is silver // Ag
2+
• Iron undergoes O 2 + 2H 2 O + • Mg(s) | Mg (aq) || Cu (aq) | • The cation present in the blue
2+
2+
–
–
oxidation 4e → 4OH Cu(s) solution is copper(II) ion / Cu 2+
• Metal Q • Voltage = +0.34 – (–2.38) • Ionic equation:
undergoes + 2+
oxidation = +2.72 V Cu + 2Ag → Cu + 2Ag
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Answers Focus SPM Chemistry.indd 508 03/04/2023 2:18 PM

Chemistry SPM Answers

• Copper / Cu is oxidised to form • When petroleum is heated, 14 unit (CH 2 )
copper(II) ion fractions with lowest boiling • Have similar chemical properties
• Copper / Cu atom loses electrons point will boil first to change • Physical properties differ gradually
• Silver ion / Ag is reduced to into vapour and then condenses with increasing number of carbon
+
silver back to liquid on cooling. atoms
• Silver ion / Ag+ gains electron • The boiling and condensation Q5 (a) Molecule A = Ethanol
(d) • Clean the rust with sandpaper processes are repeated as the • Alcohol has 1 O atom in the
• to remove the oxide layer. vapour rises up the fractionating functional group, –OH.
• Spray grease // spray paint column. • General formula of alcohol =
• to prevent the iron pipe in contact (b) (i) Petroleum gas C n H 2n + 1 OH
with oxygen and water. (ii) Fuel for cooking • n = 2;
Q6 (a) Cracking is a process to break up C 2 H 2(2) + 1 OH = C 2 H 5 OH
Chapter a large hydrocarbon molecule into
2 Carbon Compounds smaller hydrocarbon molecules. = C 2 H 6 O
• Carbon chain has no double
(b) C 8 H 18 : C 4 H 8 + C 4 H 10 bond; only carbon-carbon single
Checkpoint 2.1 bonds
Q1 (a) Carbon compounds are chemical Checkpoint 2.2 Molecule B = propanoic acid
compounds that contain carbon Q1 Homologous series is a series of • Carboxylic acid has 2 O atoms
atoms. compounds having the same functional in the functional group –COOH.
(b) Organic compounds are carbon group, similar chemical properties • General formula of carboxylic acid
compounds except oxides of and can be represented by a general = C n H 2n + 1 COOH
carbon, cyanides, carbides, formula. • n = 2;
carbonates and hydrogen Q2 (a) Propanoic acid C 2 H 2(2) + 1 COOH = C 2 H 5 COOH
carbonates. carbon dioxide is an Carboxyl group; –COOH
oxide of carbon. C 3 H 6 O 2 = C 2 H 6 O 2
(c) Limestone (calcium carbonate), (b) Ethanol Molecule C = Propyne (C 3 H 4 )
• Hydrocarbon; contains only C
CaCO 3 ; Carbon monoxide, CO; Hydroxyl group; –OH and H
potassium cyanide, KCN C 2 H 6 O
Q2 • Hydrocarbon compounds contain (c) Butane • n = 3:
carbon and hydrogen only. Methane Carbon-carbon single bond; C–C alkane: C n H 2n + 2 = C 3 H 2(3) + 2
is made up of carbon and hydrogen = C 3 H 8
only. C 4 H 10 alkene: C n H 2n = C 3 H 2(3) = C 3 H 6
• Non hydrocarbon compounds contain Q3 (a) Element: C H alkyne: C n H 2n – 2 = C 3 H 2(3) – 2
carbon, hydrogen and other elements Number of 92.31 7.69 = C 3 H 4
such as oxygen, nitrogen, halogens moles: 12 1 Molecule D = Propene (C 3 H 6 )
and others. Methanol is made up of Mole ratio: 1 : 1 • Hydrocarbon; contains only C
carbon, hydrogen and oxygen. Empirical formula: CH and H
Q3 • Saturated hydrocarbon molecules (b) Molecular formula: (CH) n • n = 3:
contain single covalent bonds (12 + 1)n = 26 alkane: C n H 2n + 2 = C 3 H 2(3) + 2
only. All covalent bonds in ethane 13n = 26
molecules are single bonds. n = 2 = C 3 H 8
• Unsaturated hydrocarbon molecules Molecular formula = C 2 H 2 alkene: C n H 2n = C 3 H 2(3) = C 3 H 6
contain at least one carbon-carbon (c) Structural formula: H!C#C!H alkyne: C n H 2n – 2 = C 3 H 2(3) – 2

double covalent bond; C"C. Ethene (d) Alkyne = C 3 H 4
molecule has one carbon-carbon Molecule E = Pentane (C 5 H 12 )
double bond. Q4 Characteristics of alcohol homologous • Hydrocarbon; contains only C
Q4 (a) Main sources: Petroleum, natural series: and H
gas and coal. • General formula: C n H 2n+1 OH • n = 5:
Alternative sources: Biogas, • Functional group: hydroxyl, –OH alkane: C n H 2n + 2 = C 5 H 2(5) + 2
bioalcohol (methanol and ethanol), • Molecular mass of successive = C 5 H 12
biodiesel members differ from each other by alkene: C n H 2n = C 5 H 2(5) = C 5 H 10
(b) • Fossil fuels are non-renewable – alkyne: C n H 2n – 2 = C 5 H 2(5) – 2
depleting resources; increasing = C 5 H 8
cost (b) H H
• Alternative sources obtain from H H O H
biomass – renewable; do not H C C O H
increase carbon content in H C C C O H H C C C H
atmosphere H H H H H
Q5 (a) • Petroleum is a mixture of
hydrocarbons with different A = Ethanol B = Ethanoic acid C = Propyne
boiling points. H H H H H H H H
• Different liquids boil at different
temperatures and vaporising at H C C C H H C C C C C H
different temperatures.
H H H H H H
D = Propene E = Pentane


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Chemistry SPM Answers

Q6 (a) • Water molecules form hydrogen Q2 (a) In an addition reaction, a bromine (b) (i) & (ii)
bonds between each other. molecule is added across the H H H
• Hydrocarbon molecules do not double bond in propene.
form hydrogen bonds with water (b) (i) 1,2-dibromopropene H H H H H C C C H
molecules. (ii) H H H
• Hence, hydrocarbon molecules H C C C C H H C H
H
H
are not miscible with water H C C C H
molecules. H H H H H
(b) Methanol and methanoic acid Br Br H Butane 2-methylpropane
forms hydrogen bonds with water (iii) C 3 H 6 + Br 2 → C 2 H 6 Br 2
molecules. (iv) Brown colour of bromine is (c) (i) 2C 4 H 10 + 13O 2 → 8CO 2 + 10H 2 O
H decolourised. (ii) Number of moles C 4 H 10
(c) In addition reactions, no atom is lost. = 2.9 g
[4(12) + 10(1)] g mol
H C O H Q3 (a) Liquid = 0.05 mol –1
H H O (b) • Effervescence Mole ratio :

• Potassium dissolves
H • Colourless solution is formed C 4 H 10 : CO 2
Hydrogen bonds (c) (i) • Ethanol burns when heated 2 : 8
O
between molecules in oxygen gas. 0.05 mol : Number of moles
C O H • Produces carbon dioxide and CO 2
water. Number of moles CO 2
8
H H O C 2 H 5 OH + 3O 2 → 2CO 2 + 3H 2 O = 2 × (0.05) mol
H (ii) • Ethanol decolourises the = 0.2 mol
(c) • Strength of van der Waals purple colour of acidified
attractive forces depend on potassium manganate(VII Volume of CO 2 3 –1
= 0.2 mol × 24 dm mol
molecule size. solution. = 4.8 dm 3
• Molecule size increases with • Produces ethanoic acid and
increasing number of carbon water Q2 (a) C 4 H 8
(b) Isomer I: but-1-ene
atoms. C 2 H 5 OH + 2[O] → CH 3 CO 2 H + H 2 O
• The larger the molecule, the (d) • Dehydration reaction Isomer II: 2-methylpropene
stronger the van der Waals • Alkene reacts with steam at (c) Both isomers:
forces of attraction. 300 °C and 60 atm in the • are gases at room temperature
• More energy is needed to presence of phosphoric acid and pressure
overcome these forces; boiling catalyst. • colourless
point increases. Q4 1 = Cu(CH 3 CH 2 COO) 2 ; 2 = H 2 O • do not conduct electricity
(d) • Hexane is a liquid insoluble in 3 = CH 3 CH 2 COOC 2 H 5 ; 4 = H 2 O • do not dissolve in water
water; immiscible with water. 5 = CH 3 CH 2 COOK; 6 = H 2 O • dissolve in organic solvents
• Hexane is less dense than water; 7 = Mg(CH 3 CH 2 COO) 2 ; 8 = H 2 ; 9 = H 2 O (d) • Isomer I and II have the same
its density is less than 1 g cm . functional group; a carbon-
–3
Q7 • A solution of ethanol in water does 10 = CO 2 ; 11 = Zn(CH 3 CH 2 COO) 2 carbon double bond, C"C.
not contain charged particles; i.e. Checkpoint 2.4 • Isomer I and II undergo
no free moving ions to conduct addition reactions; decolourise
electricity. Q1 (a) Isomers are molecules that have brown bromine water at room
• Ethanoic acid is an acid compound the same molecular formula but temperature.
that ionises in water. different structural formulae.
+
–
CHCOOH(aq) L CHCOO (aq) + H (aq)
• The presence of mobile ions makes C H + Br C H Br 2
8
4
4
2
8
the ethanoic acid solution to become H H H H H H H H
an electrical conductor.
H C C C C H + Br (aq) H C C C C H
Checkpoint 2.3 2
Q1 (a) In a substitution reaction, a H H Br Br H H
hydrogen atom from ethane
molecule is replaced by a bromine Isomer I
atom.
(b) (i) Bromoethane H H
(ii) H H H C H H C H
H H H
H C C Br
H C C C H + Br (aq) H C C C H
H H 2
(iii) C 2 H 6 + Br 2 → C 2 H 5 Br + HBr H H H Br Br H
(iv) Brown colour of bromine is
decolourised Isomer II
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Answers Focus SPM Chemistry.indd 510 03/04/2023 2:19 PM

Chemistry SPM Answers


SPM Practice 2 • % C in Q (C 3 H 6 )
36
Objective Questions 42 × 100 = 85.7%
1. A 2. C 3. B 4. D 5. B • Q has higher carbon content than P.
6. A 7. C 8. B 9. A 10. C (ii) C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
11. B 12. C 13. A 14. A 15. C (c)
16. A 17. D 18. D 19. B 20. B H H H
21. A 22. B
H C C C H
Subjective Questions
Section A OH H H
1. (a) (i) Hydrogenation Propan-1-ol
(ii) Hydrogen H H H
(b) (i) C 2 H 4 (g) + H 2 O(l) →
C 2 H 5 OH(l) H C C C H
(ii) Ethanol
(c) (i) Ethanoic acid H OH H
(ii) Add hot acidified potassium
manganate(VII) solution Propan-2-ol
(d) (i) H O H H Section C
& ' & &
C
H C O ! !C! H 1. (a) Number of H atom = 4: Number of H atom = 6: Number of H atom = 8:
! !C!
& & & (i) C 2 H 4 C 3 H 6 C 4 H 8
H H H (ii) H H H H H H H H H
(ii) Concentrated sulphuric acid
(e) C 2 H 5 OH(l) + 3O 2 (g) → H C C H H C C C H H C C C C H
2CO 2 (g) + 3H 2 O(l) H H H
2. (a) (i) C n H 2n+1 OH (ii) C 3 H 7 OH (iii) Ethene Propene But-1-ene
(iii) H H H (b) (i) Compound X:
& & &
H!C!C!C!OH H H H H H H H H H
& & &
H H H H C C H H C C C H H C C C C H
H H H
& & & H OH OH H H H OH H H
H!C!C!C!H Ethanol Propan-1-ol Butan-2-ol
& & &
H OH H Compound Y:
(b) (i) Liquid Y: Water H O H H O H H H O
(ii) The glass tube should be
inserted into the limewater. H C C OH H C C C OH H C C C C OH
(c) (i) Z = Ethanoic acid
(ii) H H H H H H
CH 3 COOH + CH 3 CH 2 CH 2 OH Ethanoic acid Propanoic acid Butanoic acid
→ CH 3 COOCH 2 CH 2 CH 3 + H 2 O
(iii) Ester is insoluble in water; (ii) Acidified potassium • Carefully heat the mixture for
two layers of liquids are manganate(VII) solution 5 minutes.
formed. or acidified potassium • Pour the contents in the boiling
Ester is less dense than dichromate(VI) solution; tube into a beaker of distilled
water, floats on water Heat under reflux. water.
surface. (c) Ester name: ethyl ethanoate Observations:
Chemical equation:
Section B CH 3 COOH + C 2 H 5 OH → • Oily layer floating on water.
• Fruity smell.
3. (a) (i) P = C 3 H 8 CH 3 COOC 2 H 5 + H 2 O
Q = C 3 H 6 Materials and apparatus: Chapter
Glacial ethanoic acid, absolute
R = C 3 H 6 O 2 ethanol, concentrated sulphuric 3 Thermochemistry
S = C 3 H 8 O acid, distilled water, boiling
(ii) P = Alkane tube, Bunsen burner, tongs and
Q = Alkene beaker. Checkpoint 3.1
R = Carboxylic acid Procedure: Q1 (a) Electrical energy is changed into
3
S = Alcohol • Add 2 cm of glacial ethanoic chemical energy.
3
(iii) R = –COOH ; S = –OH acid and 2 cm of absolute (b) (i) An endothermic reaction is a
chemical reaction that absorbs
ethanol in a boiling tube.
(b) (i) • Molecule Q • Carefully add 1 cm of heat from the surroundings.
3
• % C in P (C 3 H 8 ) concentrated sulphuric acid
36 into the boiling tube.
44 × 100 = 81.8%
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Answers Focus SPM Chemistry.indd 511 03/04/2023 2:19 PM

Chemistry SPM Answers

(ii) Electrical energy is required = 50.4 kJ mol –1 = Number of moles CH 3 COONa
to decompose sodium chloride ∴ ∆H = –50.4 kJ mol –1 = 1.0 × 100
into its elements, sodium and Q2 1. Calculate heat change, Q 1000
chlorine. Mass of water, m = 0.10 mol
2NaCl(l) → 2Na(l) + Cl 2 (g) = (volume of water) × (density of
Q2 (a) C 3 H 4 (g) + 4O 2 (g) → 3CO 2 (g) + water) (c) – Evaporate salt solution until
saturated.
3
2H 2 O(l) = 200 cm × 1 g cm –3
(b) = 200 g – Cool salt solution for
crystallisation to occur.
Energy Temperature change – Remove crystals formed by
= 61.5 – 31.5 filtration.
C H (g) + 4O (g) = 30.0 °C
3 4 2
Heat change, Q (d) 1. Dissolve sodium ethanoate
= mcθ crystals in water until saturated.
ΔH = –1466 kJ –1 –1
= 200 g × 4.2 J g °C × 30.0 °C 2. Pour saturated solution into a
= 25200 J plastic bag.
= 25.2 kJ 3. To activate the pack, add a grain
3CO (g) + 2H O(l) 2. Calculate number of moles of of sodium ethanoate crystal into
2
2
alcohol burnt. the plastic bag.
Number of moles of P
1
Checkpoint 3.2 = m Q3 (a) H 2 (g) + O 2 (g) → H 2 O(l)
Q1 (a) (i) Pb(NO 3 ) 2 (aq) + H 2 SO 4 (aq) M 2 ∆ H = –286 kJ mol –1
→ PbSO 4 (aq) + 2HNO 3 (aq) = 234.64 – 233.77
(ii) Number of moles PbSO 4 JMR (b) 2(1) = 2 g
precipitated = 0.90 g (c) Heat value of hydrogen
6.06 g JMR g mol –1
= 286 kJ
(207 + 32 + 64) g mol –1 = 0.90 mol = 2 g
= 0.02 mol JMR = 143 kJ g –1
Mole ratio: PbSO 4 : H 2 SO 4 3. Calculate heat of combustion of
1 : 1 alcohol (d) Heat change, Q
0.02 : 0.02 Heat of combustion, ∆H = heat absorbed by water
Number of moles H 2 SO 4 = Q = mcθ
–1
–1
required n heat change = 100 g × 4.2 J g °C × (50 °C
= 0.02 mol – 30 °C)
MV = number of moles of alcohol = 100 × 4.2 × 20
n = burnt
1000 25.2 kJ = 8400 J
M × 100 = = 8.4 kJ
0.02 = 0.90
1000 Combustion of 1 mole of H 2
∴ Molarity = 0.2 mol dm –3 = 28 M kJ mol –1 releases 286 kJ of heat.
Given, ∆H = –1286 kJ mol –1 Number of moles of H 2 that releases
(b) 1. Calculate heat change, Q
Mass of mixture solution, m ∴ –1286 = –28 M 8.4 kJ 8.4 kJ
= (volume of mixture solution) M = 46 = –1
× (density of water) 4. Determine molecular formula of 286 kJ mol
= (100 + 100) cm × 1 g cm –3 alcohol = 0.0294 mol
3
= 200 g General formula of alcohol Volume of gas H 2
Temperature change = C n H 2n + 1 OH = number of moles × molar volume
= 1.2 °C n(12) + 2n + 2(1) + 1(16) = 46 = 0.0294 mol × 24 dm mol –1
3
Heat change, Q 14n = 28 = 0.7056 dm 3
= mcθ ∴ n = 2 = 705.6 cm 3
= 200 g × 4.2 J g °C × 1.2 °C Formula of alcohol P = C 2 H 5 OH
–1
–1
= 1008 J SPM Practice 3
= 1.008 kJ Checkpoint 3.3
2. Calculate number of moles of Q1 (a) Anhydrous copper(II) sulphate Objective Questions
precipitate formed: (b) Exothermic reaction 1. A 2. A 3. C 4. B 5. D
Number of moles PbSO 4 H 2 O 6. D 7. C 8. B 9. D 10. B
2–
2+
precipitated = 0.02 mol (c) CuSO 4 (s) !: Cu (aq) + SO 4 (aq) 11. B 12. B 13. D 14. D 15. A
3. Calculate heat of precipitation: 16. D 17. C 18. B 19. C 20. C
Heat of precipitation, ∆H Q2 (a) Acid X: Ethanoic acid 21. C 22. B 23. D 24. B 25. B
Q Alkali Y: Sodium hydroxide 26. A
=
n (b) Chemical equation: Subjective Questions
heat change
= CH 3 COOH + NaOH → CH 3 COONa + H 2 O Section A
number of moles of alcohol Mole ratio:
burnt 1 1 1 1. (a) Fuel is a chemical substance
1.008 kJ that releases a lot of heat when
= Number of moles CH 3 CO 2 H
0.02 mol = Number of moles NaOH completely burnt in excess
oxygen.
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Chemistry SPM Answers

(b) C 3 H 8 (g) + 5O 2 (g) Bahagian B 2 × 50
→ 3CO 2 (g) + 4H 2 O(l) 3. (a) = 1000
(c) Number of moles C 3 H 8 burnt Heat of combustion = 0.10 mol
–1
= m (kJ mol ) Heat of neutralisation of acid
JMR P, ∆H
1.1 g
= 2800 5.04 kJ
3(12) + 8(1) g mol –1 = 0.10 mol
1.1 2600 –1
= = 50.4 kJ mol
44 2400
= 0.025 mol 2200 Set II:
Heat change, Q
Heat of combustion of propane, = mcθ
∆H 2000 = 100 g × 4.2 J g °C
–1
–1
heat change 1800 × (43.5 – 30 °C)
= number of moles burnt = 5670 J
55.5 kJ 1600 = 5.67 kJ
=
0.025 mol 1400 Number of moles of H 2 O
= 2220 kJ mol –1 formed
1200 MV
∴ ∆H = –2220 kJ mol –1 = 1000
1000
(d) (i) C 4 H 10 2 × 50
(ii) Carbon dioxide and water 800 = 1000
(iii) • 2880 kJ mol –1 = 0.10 mol
• Both isomers have the 600 1 2 3 4
same number of C and H Number of C atoms Heat of neutralisation of acid
atoms. per alcohol molecule Q, ∆H
• Total energy absorbed From graph, heat of combustion = 5.67 kJ
for breaking bonds is the of ethanol = 1370 kJ mol –1 0.10 mol
same. (b) (i) Heat change, Q = 56.7 kJ mol –1
• Total energy released is = mcθ (c) (ii) – Acid P and acid Q are
the same. = 200 g × 4.2 J g °C × θ acids of different strength.
–1
–1
= 840θ J – Acid P could be a weak
2. (a) Heat of displacement of copper = 0.84θ kJ acid and acid Q could be
is the heat change that occurs a strong acid.
when 1 mole of copper is Number of moles of – Heat of neutralisation of
displaced from its salt solution propan-1-ol, strong acid with strong
by a more reactive metal. CH 2 CH 2 CH 2 OH, burnt alkali is –57 kJ mol .
–1
(b) Plastic is a heat insulator that m – Heat of neutralisation of
can reduce the heat loss to the = JMR acid Q is closer to 57 kJ.
surroundings. 1.08 g
(c) Blue colour of copper(II) nitrate = (36 + 8 + 16) g mol –1 Section C
solution gradually fades and 1.08 4. (a) (i) Dissolving solid sodium
becomes colourless. = 60 hydroxide in water is highly
(d) Exothermic reaction = 0.018 mol exothermic. Ice absorbs the
(e) (i) Mg(p) + Cu (aq) heat liberated.
2+
→ Mg (aq) + Cu(s) Heat released from (ii) Heat released from the
2+
(ii) Heat change, Q combustion of 0.018 mole of solution of 0.50 mole of
propan-1-ol
= mcθ NaOH(s)
= 100 g × 4.2 J g °C = number of moles × heat of = (number of moles NaOH
–1
–1
× (44.2 – 32.2 °C) combustion –1 dissolved) × (45 kJ mol )
–1
= 100 × 4.2 × 12 = 0.018 mol × 2020 kJ mol = 0.50 mol × 45 kJ mol
–1
= 5040 J = 36.36 kJ = 22.5 kJ (absorbed by water)
= 5.04 kJ Hence, Heat absorbed by 800 cm
3
0.84θ kJ = 36.36 kJ
Number of moles of Cu water
∴ θ = = 43.3 °C
displaced (ii) C 4 H 9 OH(l) + 6O 2 (g) → = mcθ
–1
= MV = 800 g × 4.2 J g °C × θ °C
–1
1000 4CO 2 (g) + 5H 2 O(l) = 3360θ J
100 × 0.25 (c) (i) Set I: = 3.36θ kJ
= Heat change, Q
1000 3.36θ = 22.5
= 0.025 mol = mcθ –1 –1 22.5
= 100 g × 4.2 J g °C
Heat of displacement, ∆H × (42 – 30 °C) ∴ θ = 3.36 = 6.7 °C
5.04 = 5040 J
= (b) Reaction of Na 2 CO 3 with HCl:
0.025 = 5.04 kJ
= 201.6 kJ mol –1 Number of moles H 2 O – Exothermic reaction (∆H
(f) To ensure all copper is formed negative).
displaced from the copper(II) = MV – Temperature of surroundings
nitrate solution. 1000 increases.
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Chemistry SPM Answers

– Total energy absorbed for bond is extinguished, and
breaking in reactants is less the highest temperature Highest
than total energy released from achieved is recorded. temperature T 2 T 4
bond forming in the products. 10. Mass of lamp and its of water (°C)
– Total chemical energy of content is quickly weighed Initial mass
reactants is higher than total and recorded. of lamp (g) m 1 m 3
chemical energy of products. 11. Steps 1 to 10 are repeated Final mass
Reaction of NaHCO 3 with HCl: using diesel. of lamp (g) m 2 m 4
– Endothermic reaction (∆H Results:
positive). Fuel Petrol Diesel
– Temperature of surroundings
decreases. Initial
– Total energy absorbed for bond temperature T 1 T 3
breaking in reactants is greater of water (°C)
than total energy released from
bond forming in the products. Analysing data:
– Total chemical energy of 1. The change in temperature and mass of fuel burnt are calculated.
reactants is lower than total Fuel Petrol Diesel
chemical energy of products.
(c) Materials: Petrol, diesel and Temperature change, θ (°C) (T 2 – T 1 ) = θ p (T 4 – T 3 ) = θ d
distilled water Mass of fuel burnt (g) (m 2 – m 1 ) = m p (m 3 – m 4 ) = m d
Apparatus: Copper can, spirit
lamp, 100 cm 2. Fuel value of each fuel is calculated.
3
measuring cylinder, Petrol Diesel
electronic balance,
tripod stand, pipe 1. Calculate heat change, Q p 1. Calculate heat change, Q d
clay triangle, wind Mass of water, m p Mass of water, m d
shield and wooden = (volume of water) × (density = (volume of water) × (density
block. of water) of water)
3
Procedure: = 200 cm × 1 g cm –3 = 200 cm × 1 g cm –3
3
1. 200 cm of distilled water = 100 g = 100 g
3
is measured using a
measuring cylinder and Heat change, Q p Heat change, Q d
= mcθ
= mcθ
poured into a copper can. = 200 g × 4.2 J g °C × θ p °C = 200 g × 4.2 J g °C × θ d °C
–1
–1
–1
–1
2. Copper can is placed on a = 840θ p J = 840θ d J
tripod stand. = 0.840θ p kJ = 0.840θ d kJ
3. The initial temperature of
water is measured after a 2. Calculate number of moles of 2. Calculate number of moles of
few minutes. petrol burnt, n p diesel burnt, n d
4. A wind shield is set up as = mass of petrol burnt = mass of diesel burnt
shown in the diagram below. molar mass of petrol molar mass of diesel
m p g m d g
Thermometer = [(8)(12) + (18)(1)] g mol –1 = [(12)(12) + (23)(1)] g mol –1
Wind shield
m p
m d
Water Copper can = 114 mol = 167 mol
Tripod stand 3. Calculate heat of combustion, ∆H p 3. Calculate heat of combustion, ∆H d
Spirit = Q p = Q d
lamp Wooden n p n d
Petrol block = 0.840 θ p = 0.840 θ d
5. 50 cm of petrol is m p m d
3
measured using a 114 167
measuring cylinder and pour = 0.840 × 114 × θ p = 0.840 × 167 × θ d
into a spirit lamp. m p m d
6. Mass of spirit lamp and = 95.76 θ p kJ mol –1 = 140.28 θ d kJ mol –1
its content is weighed and m p m d
recorded.
7. The lamp is placed beneath Chapter (b) Chloroethene
the copper can and the wick 4 Polymer (c) H H
lighted.
8. Water in the copper can C C
is stirred continuously Checkpoint 4.1
throughout the experiment. 1. (a) Polymer is a long chain CI H
9. When the temperature molecule that is made from a (d) Addition polymerisation
of the water had raised combination of many repeating
about 35 C the flame basic units.
o
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Chemistry SPM Answers

(e) Similarity: Polymer A and its • Solubility of vulcanised rubber Thiokol rubber To make chemical
basic unit contains C, H and Cl in organic solvent is lower solvent and oil
Difference: Polymer A contains than unvulcanised rubber. storage tank
single covalent bonds between [Any two answers] coatings, petrol
carbon atoms, C–C whereas (c) Sulphur cross-links between pipes and sealants
its basic unit contains a double vulcanised rubber polymer
covalent bond between carbon chains enable the vulcanised Nitrile rubber To make gloves,
atoms, C=C. rubber to return to its original synthetic leather,
(f) Hydrogen chloride gas shape and size after it is gaskets, oil seals
and engine hoses
stretched.
2. (a) Boats // Decorative souvenirs (d) Alternative methods: [Any 2 types of synthetic rubbers]
(b) • Synthetic polymers are suitable • Using metal oxides
to be used • Using peroxides 4. (a) Styrene-butadiene rubber (SBR)
• They are lighter and more • Irradiation (e) Natural rubber shoe sole has
durable [Any two answers] higher tear resistance and low
• They can be innovated into Advantage: The product formed odour
various useful items is more environmentally friendly 3. • SBR tyres are suitable to be used.
• SBR tyres are more resistant
Checkpoint 4.2 Checkpoint 4.3 towards heat and abrasion.
1. (a) Structural formula: 3. (a) Synthetic rubber is synthetic Or
H CH H H elastomer polymer which is • SBR tyres are not suitable to be
3
elastic in nature. used.
H C C C C H (e) Elastic // Hard // Resistant to • SBR tyres are difficult to
IUPAC name: heat // Resistant to chemicals decompose naturally and lead to
2-methylbut-1,3-diene // Resistant to oxidation // pollution.
(b) Addition polymerisation Resistant to solvent // Good • SBR tyres needs to be disposed
electrical insulator systematically such as recycle and
H CH H H H CH H H [Any three answers] reuse to reduce pollution.
3
3
n C C C C C C C C (f) SPM Practice 4
Type of
H H H H n synthetic Use Objective Questions
(c) • Bacteria in the air produce rubber 1. A 2. C 3. C 4. D 5. A
lactic acid when they attack 6. B 7. C 8. A 9. B 10. D
the protein membrane of rubber Neoprene To make gloves 11. C 12. B 13. A 14. D
particles. (polychloroprene) / bearing pads /
• Hydrogen ions, H neutralise fire seals / firemen Subjective Questions
+
the negatively-charged protein suspenders / petrol Section A
membrane of rubber particles. rubber hoses / 1. (a) (i) Petroleum
• Neutral rubber particles collide gaskets and seals / (ii) Addition polymerisation
conveyor belts
with each other to break the
protein membranes. Styrene- Used for surfaces (iii)
• Rubber polymer chains or butadiene rubber corrosion protection
rubber molecules combine (SBR) // To make tyres H H
causing the formation of white or light automotive n C C
lumps and latex coagulates. parts / conveyor C C
belts and rubber
(d) Add ammonia solution, NH 3 into gaskets / shoe soles H H H H n
latex. Silicone rubber To make aircraft (iv) • 2 cm of polymerisation
3
2. (a) • Method: Vulcanisation components, sample is poured into a
• Dip the natural rubber strip automotive test tube. Bromine water
into disulphur dichloride components and is added.
solution, S 2 Cl 2 in a beaker for cooking utensils / • Bromine water remains
5 minutes. sealants / medical brown shows that all
• Dry the rubber strip in the air. implants / baby care styrene molecules have
(b) Similarity: Both are insoluble in products completely reacted.
water
Differences: H H H H H H H O H H H H H H H H O
• Vulcanised rubber is more H
elastic than unvulcanised N C C C C C C N C C C C C C C C C C
rubber.
• Vulcanised rubber is more H H H H H H H H CI H H H H H H H H CI
oxidation resistant than
unvulcanised rubber. (b) (i) Swimwear – Elastic // dry Section B
• Vulcanised rubber has quickly
higher heat resistance than Fishing lines – Strong // 2. (a) (i) • Bacteria in the air
unvulcanised rubber. damage resistant produced lactic acid.




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Chemistry SPM Answers

• Acid / H ions neutralise • with a low concentration of H (iii) The ways to overome the
+
+
the negatively-charged ion. environmental problems
protein membrane of rubber • The negatively charged caused by synthetic polymer
particles. protein membrane of rubber waste:
• Neutral rubber particles particles are neutralised • Recycle synthetic polymer
collide with one another slowly. waste
causing the breakage of the (c) • Vulcanised and unvulcanised • Reuse and reduce the use
protein membrane. rubber strands are cut into the of synthetic polymer
• The rubber molecules same size (100 mm 50 mm). • Use biodegradable
combine to form a solid. • The vulcanised and synthetic polymer
(ii) • Ammonia solution is added unvulcanised rubber strands • Replace synthetic polymer
into the solid rubber in a are hung onto two different with an environmental
beaker. retort stands. friendly product
• The mixture is stirred. • The initial lengths of both [Any three answers]
(b) • Ethanoic acid contains H rubber strands are measured. (iv) • Synthetic polymers are
+
ions. • 50 g weight is hung on each suitable to be used
• The negatively charged rubber strand and the length • Synthetic polymers are
protein membrane of is measured. cheaper
the rubber particles are • The weight is removed and • Synthetic polymers can be
neutralised quickly. the lengths of the rubber reused // innovated into
• There are bacteria in latex. strands are measured again. various useful items
• The action of bacteria on the • All the readings are recorded Or
protein membrane of rubber in a table. • Synthetic polymers are not
particles produces lactic acid suitable to be used
• Improper disposal of
Length of rubber Length of rubber synthetic polymers
Initial length strand when the strand after the • Can cause air / soil / water
Type of rubber pollution
(mm) weight is hung weight is removed
(mm) (mm)
Chapter
Unvulcanised rubber 5 Consumer and Industrial
Chemistry
Vulcanised rubber
• Conclusion: Vulcanised rubber is more elastic than unvulcanised rubber. Checkpoint 5.1
Section C 1. (a) Ester
3. (a) A polymer is a long chain molecule that is formed from a combination of many (b) • Saturated fats are the fats
repeating basic units known as monomers. that contain saturated fatty
(b) acids.
• Unsaturated fats are the fats
Synthetic rubber Use Name Specific property that contain unsaturated fatty
acids.
X To make engine Nitrile rubber Chemical solvent
hoses and oil resistance (c) Unsaturated fats can be
converted into saturated fats
Y To make medical Silicone rubber Inert chemically through hydrogenation reaction.
implants Hydrogen gas, H 2 is flowed
into hot unsaturated fat in the
Z To make shoe Styrene-butadiene Abrasion resistant
soles rubber (SBR) presence of nickel, Ni as a
catalyst at a temperature of
200 °C and pressure of 4 atm.
(c) (i) • Toxic chemicals from the 2. Similarities:
buried synthetic polymers • Oils and fats are made of glycerol
Non-biodegradable Use can leach into water and fatty acids
synthetic rubber catchment deep in the • Oils and fats are in homologous
Polyvinyl chloride To make water ground series of ester
(PVC) pipes • Many animals are killed • Oils and fats have same functional
because synthetic group of carboxylate
Polystyrene To make food polymers are mistaken for [Any two answers]
packaging food by them
materials Differences:
(ii) Negative effects of Aspect Oils Fats
synthetic polymers on the
environment: Source Plants Animals
• Open burning of synthetic Melting point Lower (< 20°C) Higher (> 20°C)
polymers releases
poisonous gases Physical state at room Liquid Solid
temperature



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Chemistry SPM Answers

3. (a)
Molecular size Smaller Bigger
Types
Saturation Unsaturated ester Saturated ester Food of food Function
additive
Content of fatty acids High percentage of High percentage of additive
unsaturated fatty acids saturated fatty acids
Monosodium Flavouring To bring
Chemical bond between At least one double Only single covalent bond glutamate out the
carbon atoms covalent bond between between carbon atoms, (MSG) flavour in
carbon atoms, C=C C–C the food
[Any two answers]
Sugar Flavouring To
sweeten
3. Uses of oils and fats: (c) Scum is insoluble salt formed the food
• As biofuel when soap anion reacts
• Source of energy and nutrition with calcium ion, Ca and Sunset Food dye To add
2+
• To manufacture soap and personal magnesium ion, Mg 2+ yellow yellow and
care products (d) Calcium ion, Ca and orange
2+
• To manufacture food products and magnesium ion, Mg 2+ colour to
animal feeds the food
[Any three answers] (e) Cleaning agent Y is
biodegradable and
4. Our body needs fats and oils: environmentally friendly Ascorbic Antioxidant To slow
• As a source of energy acid down the
• As solvents for vitamins A, D, E 3. (a) Detergent is sodium salt of oxidation
and K to help our body absorb sulphonic acid. of food
them (b) (i) Sulphonation and
• As a component of cell membrane (ii) Neutralisation prevent
• As heat insulator (c) Detergent anion reacts with rancidity
• To protect important organs hydrogen ion, H in acidic water (b) • Ingredient not suitable for a
+
to form soluble organic acid diabetic patient: Sugar
Checkpoint 5.2
1. (a) Soap is sodium or potassium (d) Type: Foam control agent • Replacement: Aspartame /
stevia / sorbitol
Example: Alkyl

salt of long chain fatty acid monoethanolamide, silicones
(b) Saponification Checkpoint 5.4
(c) To reduce solubility of soap in (e) • Lower the surface tension of
water water 1.
(d) Potassium hydroxide solution • To float the dirt removed Traditional Modern
(e) • Soap reduces the surface medicines medicines
tension of water and increases Checkpoint 5.3
the ability of water to wet the 1. Obtained from Produced by
surface of the cloth. herbal plants scientists in
• Hydrophobic part of soap Type of food Function or animals laboratories
anion dissolves in oil. additive without chemical come from the
Hydrophilic part dissolves in Preservatives To make the food last processing substances
water. longer found in plants,
• Water movement during microorganisms
scrubbing and agitation pulls Antioxidants To slow down the or synthetic
oily stain away from the oxidation of oils or fats chemicals
surface of the cloth. The oil in food
breaks into small droplets. Have not Have undergone
• The repulsion of negative Flavourings To replace the original undergone clinical clinical tests
charges of hydrophilic parts in flavour lost in food tests
the oil droplets surface makes processing Slower efficacy Faster efficacy
oil droplets do not reattach Stabilisers To give a uniform and
and clump together again onto smooth texture to food 2.
the cloth surface.
• The oil droplets are Emulsifiers To prevent water-oil Type of
suspended in water to form emulsion separation modern Function
emulsion. medicine
• Rinsing with water will remove Thickeners To thicken food Analgesics To relieve pain in
the oil droplets and clean the Dyes To restore food colour conscious without
surface of the cloth. causing numbness
2. (a) X: Detergent 2. (a) Preservative
Y: Soap (b) Common salt draws the water Antimicrobials To kill or retard
(b) X: ROSO 3 Na out of the microorganisms’ cells bacterial, fungus,
Y: RCOONa to delay their growth microorganisms or
pathogens growth




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Chemistry SPM Answers

nanotechnology has improved 5. Graphene has a high surface area
Psychotic To alter abnormal
drugs thinking, feelings or drug delivery to enable the direct and rapid electron movement. It
behaviour of psychotic treatment on diseased cells without is suitable to be used to make
patients to reduce destroying healthy somatic cells. bioelectronics such as sensors and
symptoms of mental In cosmetic fields, the usage of medicine delivery systems as well as
illnesses nanoparticles which can easily for tissue engineering.
penetrate the skin provides more
Anti allergies To relieve allergic satisfying outcomes. Checkpoint 5.6
symptoms such as
runny nose, itchiness, 3. (a) Production of electronics and 1. (a) Green Technology is a
sneezing, cough, semiconductors: technology and environmentally
watery eyes, rashes • To make smaller and friendly application of products
more efficient electronic or equipment and a system
Corticosteroids • To treat the infection components developed to minimise the
for arthritis, eczema, • Nanotechnology is used negative effects of harmful
psoriasis in high conductivity wiring human activities and to conserve
• To relieve the system the environment.
inflamed area (b) Medical:
• To reduce swelling • Nanotechnology is used (b) Energy supply // waste and
and joint or organ in highly sensitive testing wastewater management //
pain transportation // agriculture and
devices forestry // buildings // industry
3. (a) Student P: Antibiotics • To produce a more effective and manufacturing
Student Q: Psychotic drugs drug delivery system [Any three answers]
Student R: Analgesics (c) Energy and electricity: 2. (a) • Agree
(b) Finish the medicine • To manufacture • To prevent groundwater
(c) Can cause bleeding or ulcer in supercapacitors and provide sources from being polluted
the stomach efficient energy storage (b) Human: Brings negative effects
4. • To make smaller and more on human health and causes
efficient solar cell
diseases
Type of (d) Agriculture: Environment: Causes
cosmetics Function • Nanotechnology approaches eutrophication and pollution
are used to detect plant
diseases 3. • Use bacteria to decompose the
Makeup cosmetics To beautify the harmful substances in wastewater
users’ face • Nanotechnology is used for • Isolate solid waste from
highly efficient and thorough wastewater treatment through
fertilisation
Treatment To treat the body electrocoagulation
cosmetics (e) Textile: 3. The use of electrical energy in
• To make water, fire and dirt
resistance fabrics hybrid cars and the use of solar
Fragrances To provide • To manufacture anti-wrinkle energy as electrical energy in green
fragrant smells and ultraviolet ray protective buildings reduce the release of
fabrics carbon dioxide gas which contributes
5. • Agree (f) Food: to global warming.
• Commercial cosmetic products
contain harmful chemicals • To coat raw food packaging SPM Practice 5
• that lead to harmful side effects materials that enable
Or prolonged food shelf life Objective Questions
• Disagree • Nanoscale food additives 1. B 2. C 3. B 4. A 5. C
• High quality ingredients in are used for better nutrient 6. D 7. D 8. B 9. A 10. A
commercial cosmetic products delivery, improving flavours, 11. D 12. C 13. B 14. D 15. D
• can give the desired effects texture and colours 16. B 17. B 18. D 19. C 20. B
(Any three answers) 21. C 22. C 23. D 24. A 25. D
Checkpoint 5.5 4. Physical properties of graphene: 26. A
1. Nanoscience is a study about • Transparent Subjective Questions
the processing of substances at • Relatively very strong and hard Section A
nanoscales between 1 nanometre compared to diamonds and steel
to 100 nanometres while • Very good conductor of heat and 1. (a) (i) Oil Y
nanotechnology is the development electricity (ii) • Oil Y has a higher
of substances or gadgets using the • Very low electrical resistance percentage of saturated
properties of nanoparticles. • Impermeable fat
2. Nanoparticle size from 1 to 100 • More elastic than rubber • Less oxidation occurred
nm, extremely small size, enables • The thinnest substance with only (b) (i) Carboxylate group
various applications to be invented. one atom thick (ii) Saponification
For example, in the medical industry, • The lightest substance (iii) CH 3 (CH 2 ) 10 COONa; H 2 O
(Any three answers)



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Chemistry SPM Answers

(iv) • Total surface area of
Hydrophobic part nanogold is much bigger
than normal gold
• Atom on the material
CH CH CH CH CH CH
3 2 2 2 2 2 surface of nanogold is
more reactive than those
CH CH CH CH CH C O – in the centre, therefore
2 2 2 2 2
nanogold is more reactive
O than normal gold
(iii) • Silver ions can reduce the
ability of bacteria to take
Hydrophilic part up oxygen for respiration
• The larger the surface
(v) To float the dirt removed Function To prevent To area of nanosilver particles
(c) (i) Set I: Oil stain remains the growth of produce increases the anti-bacterial
Set II: Oil stain is removed microorganisms a smooth efficiency
(ii) Type of additive: Biological / uniform (b) (i) Atom R: Atom carbon
enzyme texture of Bond Q: Covalent bond
Example: Lipase / Amylase / food (ii) • Carbon atom has four
Protease / Cellulase valence electrons
(ii) Cancer / intestine ulcer / • Three valence electrons
stomach ulcer
Section B (iii) • Justification: Food are used to form covalent
bonds
2. (a) (i) additives can be used if • One valence electron is
they obey Food Act 1983 /
Food X Y Food Regulation 1985 free to move
additive • There are a lot of spaces
• Advantage: Food additives for electrons in the
Type Preservative Stabiliser can prolong the life span honeycomb structure to
of food / enhance the move rapidly.
Name Benzoic acid Lecithin / appearance of food / (c) (i) Green
/ sodium gelatine enhance the taste of food (ii) • The wastewater treatment
benzoate • Disadvantage: Food for the effluent of
additives can cause side chromium-electroplating
effect / cancer / allergy plants is needed
(b) • Chromium(III) ion is toxic
• Chromium(III) ion may
Student Name of medicine Type of medicine Correct usage cause respiratory cancer
Rahimah Barbiturate Psychotic drug / Taken based and kidney failure
Antidepressant on doctor’s (iii) • Treat wastewater via
prescription electrocoagulation
process.
Mary Penicillin / Antibiotics Complete full • Chromium(III) ions, Cr
3+
Streptomycin course as doctor’s react with hydroxide ions,
prescription OH in water to form solid
–
chromium(III) hydroxide,
flocs.
(c) (i) A: Treatment cosmetics Section C • Flocs are trapped in
B: Makeup cosmetics 3. (a) (i) A particle with the size hydrogen gas produced at
(ii) • Shift to homemade between 1 nanometre to the cathode and carried
cosmetics produced from 100 nanometres to the water surface to be
natural herbs (ii) • Normal gold is chemically removed.
• Reduce consumption of inert while nanogold is • Ionic equation:
cosmetic reactive Cr + 3OH → Cr(OH) 3
–
3+

















519





Answers Focus SPM Chemistry.indd 519 03/04/2023 2:19 PM

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Answers Focus SPM Chemistry.indd 520 03/04/2023 2:19 PM

Chemistry SPM Answers
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Answers Focus SPM Chemistry.indd 521 03/04/2023 2:19 PM

Chemistry SPM Answers

















































































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Answers Focus SPM Chemistry.indd 522 03/04/2023 2:19 PM

Format 190mm X 260mm Extent : 528pg (25.26mm) confirmed (All 4C/70gsm) Status CRC Date 3/4

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