14 Spotlight A+1 Form 4.5 Chemistry

(iii) Sodium chloride, NaCl cannot conduct electricity in solid state but it can conduct electricity in
aqueous state. Explain why. [2 marks]
(b) Figure 7.2 shows the symbol of two elements, N and H. Element N reacts with element H to form a
compound.

N H14 1

71
Figure 7.2
(i) State the type of particle in the compound formed. [1 mark]
(ii) Draw a diagram to show the electron arrangement in the compound formed between atoms N
and H. [2 marks]
(iii) The molecule formed in (b)(ii) can combine with hydrogen ion, H+ to form ammonium ion,
NH4+. S tate the type of chemical b ond and exp lain briefly how the chemical bond is fo[2rmmeadr.ks]

8. (a) Figure 8.1 shows atchirdeeantdesHt YtubaceisdcaornetastirnondgiluatceidHs.2X acid, HY acid and dry glacial ethanoic acid,
CH3COOH. H2X

PQR

0.1 mol dm–3 0.1 mol dm–3 Anhydrous
dilute H2X acid dilute HY acid ethanoic acid,
CH3COOH

SPM MODEL PAPER Calcium carbonate, Calcium carbonate, Calcium carbonate,
CaCO3 CaCO3 CaCO3

Figure 8.1

(i) There are no gas bubbles produced in test tube R. Suggest a method that can make calcium
carbonate, CaCO3 reacts with ethanoic acid, CH3COOH to produce gas bubbles. [1 mark]
(ii) Table 8 shows the pH value of aqueous solution of ethanoic acid, CH3COOH and dilute HY acid
of the same molarity.

Type of acid Ethanoic acid Dilute HY acid

Molarity 0.1 mol dm–3 0.1 mol dm–3

pH value 4.0 1.0

Table 8 higher than the
Explain why the pH value of aqueous solution of ethanoic acid, CH3COOH is [2 marks]
pH value of dilute HY acid.
(iii) Figure 8.2 shows the apparatus set up for the titration of 20 cm3 of 0.5 mol dm–3 sodium
hydroxide, NaOH solution with H2X acid of an unknown concentration. Phenolphthalein is
used as the indicator.

H2X acid

Sodium hydroxide, NaOH
solution + phenolphthalein

Figure 8.2
The chemical equation for the reaction is as the following.

H2X(aq) + 2NaOH(aq) → Na2X (aq) + 2H2O(l)

If 25 cm3 of Hca2lXcualactied is required to neutralise the sodium hydroxide, NaOH solution in the
conical flask, the concentration of the H2X acid solution in mol dm–3. [3 marks]

534

(b) Figure 8.3 shows the reactions carried out on compound K.

Compound K Heat Black solid L + Gas M

+ Sulphuric acid, + Limewater
H2SO4

Blue solution Limewater
W formed turns cloudy

+ Ammonia, eNxHce3ss
solution in

Dark blue
solution formed

Figure 8.3

(i) Based on Figure 8.3, identify K, L, M and W. [4 marks]
(ii) Describe a chemical test to identify the anion present in solution W. [2 marks]

Section B

[20 marks]

Answer any one question. SPM MODEL PAPER

9. Three sets of experiments, I, II and III are carried out to investigate the factors affecting the rate of reaction.
Table 9 shows the reactants and the conditions of the reaction involved.

Set of experiment Reactants Condition

I

50 cm3 of 0.5 mol dm–3 Room temperature
hydrochloric acid, HC

Excess zinc, Zn

II

50 cm3 of 0.5 mol dm–3 70 ºC
hydrochloric acid, HCl

Excess zinc, Zn

III

50 cm3 of 0.5 mol dm–3 70 ºC
sulphuric acid, H2SO4 535

Excess zinc, Zn

Table 9

(a) Referring to Experiments, I, II and III, state: [1 mark]
(i) The meaning of the rate of reaction.
(ii) Two factors that affect the rate of reaction. [2 marks]
(b) Write a balanced chemical equation for the reaction in Experiment I. [2 marks]
(c) Calculate the total volume of hydrogen g2a4sd, Hm23 released in Experiment I. [3 marks]
[Molar gas volume at room conditions: mol–1]
(d) sSakmetechaxthise. graph o f the volum e of hydrogen gas, H2 agains t time for Experiments I, II dan III on the
[2 marks]
(e) Compare the rate of reaction between Experiment I and Experiment II. Explain your answer using
the collision theory. [5 marks]
(f) Compare the rate of reaction between Experiment II and Experiment III. Explain your answer using
the collision theory. [5 marks]

10. (a) Figure 10.1 shows the displacement reaction. Metal Q powder is added to silver nitrate, AgNO3
solution in a test tube.

Metal Q Blue solution
powder

Silver nitrate, AgNO3 Shiny grey
solution (colourless) solid

SPM MODEL PAPER Figure 10.1
Based on Figure 10.1, state the identity of metal Q. Write the half equations to represent the oxidation
and reduction. State the change in the oxidation number of silver, Ag. [4 marks]
(b) Figure 10.2 shows an apparatus set up for an
experiment to investigate a redox reaction. G

Based on Figure 10.2, describe the oxidation Carbon
and reduction that occurs. Your answer electrodes
must include the following: Iron(II) sulphate, Acidified potassium
(i) The role of each reactant. [2 marks] FeSO4 solution dichromate(VI),
(ii) The transfer of electron of each reactant. Dilute sulphuric acid, K2Cr2O7 solution
[2 marks] H2SO4
(iii) The colour changes that can be
observed after 15 minutes. [2 marks] Figure 10.2

(c) Figure 10.3 shows two types of cells, P and Q.

V

AB CD

Magnesium, Copper, Copper, Cu
Mg Cu

Copper(II) sulphate,
P CuSO4 solution
Q

Figure 10.3

Compare between cells P and Q. Include in your answer the following:
536 (i) The change of energy.
(ii) The flow of electron. Given the following E0 value: E0 = –2.38 V
(iii) The product formed at anode. Mg2+(aq) + 2e– ⇌ Mg(s)
(iv) Half equation for the discharge at cathode. 2C2HHu222+OO(a((qll)))+→+22Oee–2–(⇌g⇌) +H42(Hg)+(+aq2)O+H4–e(–a q) EE00 = –0.83 V
(v) The colour change of the solution. Cu(s) E0 = +1.23 V
= +0.34 V
[10 marks] 2SO42–(aq) → S2O82– + 2e– E0 = +2.01 V

Section C

[20 marks]

Answer the question.
11. (a) Liquid of compounds Q and R are hydrocarbons. An experiment is carried out to investigate the

quantity of soot produced when compounds Q and R are burnt in oxygen. The results are shown in
Table 11.1

Compound Q Compound R

Burn in the air producing yellow sooty flame. Burn in the air producing yellow and more sooty flame.

Table 11.1
(i) State the general formula of compound Q. [1 mark]
(ii) Explain the differences in the observations of the combustion of compounds Q and R by
comparing the percentage of carbon by mass. [3 marks]
[Relative molecular mass: Q = 86; R = 84; Relative atomic mass C = 12]
(b) Table 11.2 shows the properties of four organic compounds. Each compound has three carbon atoms
per molecule.

Organic Properties
compound

V Miscible with water in all proportions. Burns with blue flame to form carbon SPM MODEL PAPER
dioxide, CO2 and water, H2O.

W Soluble ignasw, aCtOer,2.H2O. Reacts with calcium carbonate, CaCO3 to produce carbon
dioxide

X Idnicsholruobmleatien(VwIa)t,eKr, 2HC2rO2O. D8 seocloultoiounri.se the orange colour of acidified potassium

Y Insoluble in water, H2O. Sweet smell.

Table 11.2
Reffering to Table 11.2, state each of the following:
(i) Name of the homologues series for compounds V, W, X and Y. [4 marks]
(ii) Functional group of compound V and X. [2 marks]
(c) Figure 11.1 shows the structural formulae of hydrocarbon A and hydrocarbon B. Both hydrocarbons
have different chemical properties.

H H H H H H H H H H H H H H H H

HCCCCCCC HCCCCCCC

H H H H H H H H H H H H

Hydrocarbon A Hydrocarbon B

Figure 11.1

(i) Draw structural formula for two isomers of hydrocarbon A. [2 marks]
(ii) Describe a laboratory experiment to differentiate both hydrocarbons. In your description,
include the following:
• Procedure
• Observations
• Chemical equation involved [8 marks]

537

Paper 3

1. In this experiment you are required to investigate the effect of different solutions of chloride salts on the heat
of precipitation of silver chloride, AgCl.
You are provided with the following materials:
L1 = 0.5 mol dm–3 ssiolvdeiurmnitcrhaltoer, iAdge,NNOa3Cslosloultuiotinon
L2 = 0.5 mol dm–3
L3 = 0.5 mol dm–3 potassium chloride, KCl solution

Carry out the experiment according to the following instructions:
1 Measure and pour 25 cm3 of L1 solution into a polystyrene cup.
2 Put the thermometer into the polystyrene cup containing L1 solution and leave it for 1 minute. Record

the initial temperature of L1 solution in Table 1.
3 Measure and pour 25 cm3 of L2 solution into another polystyrene cup.
4 Put the thermometer into the polystyrene cup containing L2 solution and leave it for 1 minute. Record

the initial temperature of L2 solution in Table 1.
5 Pour the L1 solution carefully and quickly into the polystyrene cup containing L2 solution.
6 Stir the mixture using a thermometer and record the highest or lowest temperature.
7 Repeat steps 1 to step 6 by replacing L2 solution with L3 solution.

Chloride salt solution L2 L3

Initial temperature of L1 solution (°C)

SPM MODEL PAPER Initial temperature of chloride salt solution (°C)

Highest/Lowest temperature of the mixture (°C)

Table 1
[**Complete Table 1 with reference to the Simulation experiments and sample results given]

Based on the experiment carried out:
1 State an inference based on the change in temperature in this experiment.
2 State the variable involved in this experiment:
(a) Manipulated variable
(b) Responding variable
(c) Fixed variable
3 Calculate the heat of precipitation of argentum chloride in both reactions:
[Specific heat capacity of solution: 4.2 J g–1 °C–1; density of solution: 1 g cm–3]
S((bata)) t eSStiihllvveeeorrpnneiirttarraatittoeen,, AAalggdNNeOOfin33 isstooiolluunttiiooofnnth++epshooedtaaitusosmifupcmrhelccohirplioditreai,tdiNoe,naKCoClf(lsLi(l1Lv1e+r+Lch2L)l3o)ride, AgCl.

4
5 Based on the results of the experiment, state the effect of different chloride salt solution on the
precipitation of argentum chloride, AgCl. Explain your answer.

2. In this experiment, you are provided with substances D1 and D2.

Part A Carry out the experiment according to the following instructions:

1 Put the appropriate quantity of D1 into a test tube.
2 PCoaurrrythoeutdtihlueteexspuelfruimricenatcsidin, HT2aSbOle4 into the test tube and warm the mixture slowly.
3 2 on the released gas.
4 Write the observations and inferences in Table 2.

538

Experiment Observation Inference
(a) Place the lighted wooden

splinter near the mouth of
the test tube.
(b) Bubble the gas released
into the limewater.

Table 2
[**Complete Table 2 with reference to the Simulation experiments and sample results given]
5 Based on the experiments conducted, name the anions found in D1.

Part B Carry out the experiment according to the following instructions:
1 Put a spatula of D2 powder into a 50 cm3 beaker.
2 Add distilled water until half full.
3 Stir until all D2 powder dissolves.
4 Pour 2 cm3 of the prepared solution into three test tubes, I, II and III.
5 Carry out the experiments in Table 3 on the solution in the test tube.
6 Write the observations and inferences in Table 3.

Experiment Observation Inference

(a) Test tube I - Add sodium hydroxide solution SPM MODEL PAPER
drop by drop until in excess.

(b) Test tube II - Add ammonia solution drop by
drop until in excess.

(c) Test tube III - Add 2 cm3 of sodium sulphate
solution.

Table 3
[**Complete Table 3 with reference to the Simulation experiments and sample result given]
7 Based on the experiments conducted, name the cations found in D2.

Note: Simulation experiments and sample results of Question 1

(a) L1 + L2 (Silver nitrate solution, AgNO3 + sodium chloride solution, NaCl)

Thermometer

40 40

30 25 cm3 of 30
20 solution L1 20
Polystyrene cup
Solution L2 White precipitate

(b) L1 + L3 (Silver nitrate solution, AgNO3 + potassium chloride solution, KCl)

Thermometer

40 40

30 30
20
25 cm3 of
20 solution L1

Solution L3 Polystyrene cup White precipitate

539

Note: Simulation experiments and sample results of Question 2 (Part A)

Lighted No 'pop' sound Lighted wooden
wooden splinter
splinter Dilute sulphuric
acid, H2SO4
Dilute sulphuric
acid, H2SO4 D1
Dilute sulphuric
D1 acid, H2SO4

Dilute sulphuric
acid, H2SO4

Limewater Limewater
D1 turns cloudy
D1

Note: Simulation experiments and sample results of Question 2 (Part B)

SPM MODEL PAPER Test tube 1 Add a little sodium hydroxide, Add in excess sodium hydroxide,
NaOH solution NaOH solution

White Soluble white
precipitate precipitate

Test tube 2 Add a little ammonia, Add in excess ammonia,
NH3 solution NH3 solution

White Insoluble white
precipitate precipitate

Test tube 3 Add sodium sulphate,
NaSO4 solution

White
precipitate

540

ANSWERS Complete Answers
https://bit.ly/3sn6L0o

Form 4 (b) (i) Vinegar; salt; baking powder
(ii) Vinegar: preserves food
Chapter 1  Introduction to Chemistry Salt: gives salty taste
Baking powder: raises the dough
1.1 (c) Chemist; doctor
1. A field of science that studies the structures, properties, (d) Hydrogen peroxide waste with low concentration

compositions and interactions between matter. can be poured directly into the sink. Concentrated
2. Herbicide; Hormone hydrogen peroxide wastes need to be diluted with
3. With nanotechnology, sunscreens are no longer oily water. Then, it is added with sodium sulphite for the
purpose of decomposition before being poured into the
nowadays and are colourless when applied to the skin. sink.
4. (a) Cosmetic consultant 2. (a) (i) Presence of water and oxygen
(ii) Rusting of iron nail
(b) Dietitian (iii) Type of nail
(c) Pathologist (b) Water and oxygen are needed for iron rusting.
(d) Veterinarian (c)

1.2 Test tube Observation
1. A systematic scientific method used to solve science related
A Iron nail does not rust.
problems.
2. (a) When the temperature increases, the mass of salt B Iron nail rusts.

dissolved also increases. C Iron nail does not rust.
(b) Manipulated variable: Temperature
Responding variable: Mass of salt dissolved (d) Oxygen and water must be present for the iron nail to ANSWER FORM 4
rust.
1.3
1. • Do not eat, drink, chase or run in the laboratory. Section B
• Do not pour the chemicals back to the reagent bottles. 3. (a) The scientific method is a systematic approach to solve
2. • Keep flammable substances away from the heat source. problems in science.
• Do not point the mouth of the test tube at your face or (b)
Making an observation
at other people.
3. (a) To carry out experiment that involves the release of Making an inference

toxic vapours, gases that can cause combustion or gases Identifying the problem
with pungent smell.
(b) To remove dirt, oil, chemicals or microorganisms from Making a hypothesis
the hands.
(c) To wash and clean the body if accident happens on Identifying the variables
parts of the body. It is also used to put out fire at any
part of the body if there is fire. Controlling the variables
4. (a) Kept in paraffin oil to prevent reaction between this
chemical with moisture, water and air. Planning an experiment
(b) Kept in dark bottles to avoid the exposure of sunlight.
5. Mercury poisoning is a phenomenon when a person is Collecting data
exposed to mercury in a certain amount. Two symptoms of
mercury poisoning are vomiting and difficult in breathing. Interpreting data

Making a conclusion

Paper 1 2. A 3. C 4. B 5. C Writing a report
1. A 7. B 8. D 9. B 10. C
6. B 12. C 13. A 14. C 15. D Section C
11. B 4. (a) (i) Ammonium nitrate / Urea
(ii) Occupation A: Pharmacist
Paper 2 Occupation B: Pathologist
Section A (b) • Inform the accident to the teacher immediately.
1. (a) Chemistry is defined as a field of science that • Make the spill area as a restricted area for students.
• Try to stop the spill from spreading to other areas
studies the structures, properties, compositions and
interactions between matter. using sand to border it.

541

Chemistry   Answer

Paper 2 Element Carbon, C Hydrogen, H Oxygen, O
9.09 36.36
Section A
1. (a) To remove the layer of the magnesium oxide on its 54.55 9.09 36.36
surface. Mass (g) 1 16
= 9.09 = 2.273
(b) To allow the oxygen to enter the crucible for the Number of moles 54.55
complete combustion of magnesium. of atoms 12
(c) (i) Element = 4.546
Magnesium, Mg Oxygen, O
4.546 9.09 2.273
Mass (g) 22.30 – 20.50 23.50 – 22.30 Simplest ratio 2.273 2.273 2.273
=1.80 = 1.2 =2 =4 =1

Number of 1.8 1.2 Thus, the empirical formula of compound Y is
moles of 24 16
atoms = 0.075 = 0.075 (ii) CA2sHsu4mO.e that the molecular formula of Y

Simplest ratio 1 1 G= i(vCe2nHt4hOa)tnthe
n[2(12) + 4(1)
Thus the empirical formula of magnesium oxide is molecular mass of (C2H4O)n = 88
+ 16] = 88
MgO. 44n = 8888
(ii) 2Mg(s) + O2(g) → 2MgO(s) n = 44
(d) (i)

Glass tube Metal oxide W n = 2
Ethanol,
Rubber C2H5OH Therefore, the molecular formula of compound Y
tubing
Glass tube = (CC4H2H8O4O2 )2
Air hole Spirit =
Hydrochloric
Glass tube lamp acid
Zinc, Zn
granules Section B
4. (a) (i) The molecular formula of a compound shows the
Water actual number of atoms of each element present in

ANSWER FORM 4 Wooden block a molecule of the compound.
(ii) The relative molecular mass of glucose
(ii) The hydrogen gas is allowed to flow through the = 6(12) + 12(1) + 6(16)
apparatus for 10 to 15 seconds to remove all the = 180
air. A molecule of glucose consists of 6 carbon atoms,
2. (a) The empirical formula of a compound is the chemical 12 hydrogen atoms and 6 oxygen atoms.
formula that shows the simplest ratio of the number of The empirical formula of glucose is CH2O.
atoms of each element in the compound. (b) (i) Consider 100 g of compound Q.
(b) (i) Method I. Magnesium is more reactive than
hydrogen. Element Carbon, C Hydrogen, H

(ii) The heating, cooling and weighing steps are Percentage 85.71% 14.29%
repeated until a constant mass is obtained.
(c) (i) Mass of copper Mass (g) 85.71 14.29

= 61.71 – 55.31 = 6.4 g Number of 85 14.29
(ii) =Nu66m.44be=r of moles of copper moles of atoms 12 1
0.1 mol = 7.1425 = 14.29

(iii) Mass of oxygen 7.1425 14.29
= 63.31 – 61.71 = 1.6 g
(iv) =Nu11m.66be=r of moles of oxygen Simplest ratio 7.1425 7.1425
0.1 mol =1 =2

AThssuusm, the etheamt pthireiceaml fpoirrmicaullafoQrmisuClaHo2f. Q =
(v) 0.1 mol of copper atom combines with 0.1 mol (b) (ii) = (4C2H2)n
of
of oxygen atom. Thus, 1 mol of copper atom Given that the relative molecular mass
combines with 1 mol of oxygen atom. Empirical
formula of copper oxide is CuO. n(C[1H22+)n2 =(14)2] = 42
3. (a) The molecular formula is the chemical formula that
shows the actual number of atoms of each element 14n = 4422
found in a molecule of a compound. n = 14
(b) 6C(61H21)2
(c) + 12(1) = 84 n=3
Therefore, the molecular formula of compound Q
(d) C(i)HC2 onsider = C(C3HH62)3
(e) 100 g of compound Y. =

546

Chemistry   Answer

Reinforcement & Assessment of Science Process Skill to achieve the stable octet electron arrangement, 2.8. A
positive magnesium ion, Mg2+ is formed. Mg → Mg2+
+ 2e–. A chlorine atom has an electron arrangement of
2.8.7. Each of two chlorine atoms accepts one electron
Observation from a magnesium atom to achieve the stable octet

Oxide Reaction with nitric Reaction with sodium electron arrangement, 2.8.8. Two negative chloride ions,
acid, HNO3 hydroxide, NaOH Cl– is formed. Cl + e– → Cl– One Mg2+ ion and two Cl–
solution ions are attracted to each other by a strong electrostatic
attraction force. An ionic compound magnesium
chloride, MgCl2 is formed.
Oxide P White powder cannot White powder dissolved (b) – 2+
(K1) dissolve. to form a colourless –
solution.
Oxide Q White powder White powder cannot Cl Mg Cl
(K2) dissolved to form a dissolve.
colourless solution. White powder dissolved 5.3
Oxide R White powder to form a colourless 1. (a) A single covalent bond
(K3) dissolved to form a solution.
colourless solution.

Table 1

1. Variables: H Cl
(a) Manipulated: Types of oxide of elements in Period 3
(b) Responding: Solubility in acid and alkali.
(c) Fixed: Volume and concentration of sodium hydroxide Formula: HCl
(b) Four single covalent bonds
solution/ nitric acid
2.

ANSWER FORM 4 Amphoteric Base Acid H

Oxide R Oxide Q Oxide P

Table 2 HCH

3. Q, R, P H
4. Aluminium oxide (c) TFowromduolau:bCleHc4ovalent bonds

Chapter 5  Chemical Bonds

5.1 OC O
1. Chemical bonds are forces that hold atoms together to make

compounds or molecules through transfer of electrons or Formula: CO2
sharing of electrons. 2. (a) 2.5
2. Neon atom has achieved the stable octet electron (b) Covalent compounds
arrangement. It does not need to gain, lose or share ((cd)) NH3
electrons with other atoms.
3. No. The electron arrangement of magnesium atom is 2.8.2. H
It does not achieve the stable octet electron arrangement.
4. The electron arrangement of sodium atom is 2.8.1. Sodium HNH
atom can donate the single valence electron to achieve a
stable electron arrangement, 2.8. 5.4
5. (a) Ionic bond 1. A hydrogen bond is a type of attraction force between a
(b) Covalent bond
(c) Covalent bond hydrogen atom which bonded to a strongly electronegative
(d) Ionic bond atom such as fluorine, oxygen or nitrogen with another
(e) Covalent bond electronegative atom (fluorine, oxygen or nitrogen) in other
(f) Ionic bond molecule.
2. Molecules of HCl are attracted by weak Van der Waals
5.2 attraction forces only. Less heat energy is required to
1. (a) 2.8; Y2– overcome the weak Van der Waals attraction forces. On
(b) 2, W+
2. A positive ion; Q3+
3. (a) A magnesium atom has an electron arrangement of

2.8.2. A magnesium atom donates 2 valence electrons

550

(b) • Use small marble chips Answer Chemistry
• Use higher concentration of hydrochloric experiment II is higher. Frequency of effective collision
between particles in experiment II is higher.
acid
(c) Graph of volume of carbon dioxide Paper 1 2. C 3. D 4. C 5. A
1. D 7. A 8. C 9. A 10. A
gas against time 6. B 12. B 13. D 14. A 15. D
Volume of carbon dioxide (cm3) 11. D

40

30 Paper 2

20 Section A
1. (a)
Graph of volume of gas collected against time

Volume of gas (cm3)

10 40

0 30 60 90 120 150 180 210 Δy = 37 – 30
Time (s) =7

(d) 48 cm3. Some of the carbon dioxide gas dissolves in 30 Δx = 114 – 54
water. = 60

(e) Saturate the water with carbon dioxide gas before 20
collecting the gas in the burette. Δy = 26.5 – 8
= 18.5
7.3
1. On a warm night, the surrounding temperature is higher. 10 ANSWER FORM 4
Δx = 51 – 9
The chemical reaction in the fireflies' body increases. Thus, = 42
it flashes faster and more frequent.
2. Food kept in the kitchen cabinet is exposed to high 0 Time (s)
room temperature. The bacterial action towards food is 30 60 90 120 150 180 210 240
faster. More toxic is released. Therefore, the decaying and
decomposition of food is faster. (b) The instantaneous rate of reaction at 30 seconds
3. Powdered detergent has a larger total surface area that ∆y
act on the dirt. Hot water also provided a medium with a = ∆x
higher temperature. The cleansing action become faster.

= 18.5 = 0.44 cm3 s–1
42
7.4 The instantaneous rate of reaction at 90 seconds
∆y
1. (a) Minimum energy that colliding particles of the = ∆x
reactants must possess to start a chemical reaction.
= 7 = 0.12 cm3 s–1
(b) Collision that causes chemical reaction. The particles 60
collide in the correct orientation and are able to achieve (c) • The instantaneous rate of reaction at 30 seconds is
the activation energy. higher than the instantaneous rate of reaction at 90
seconds.
2. (a) The total energy content in the reactants is higher than • The amount of potassium chlorate(V) is greater at 30
the total energy content in the products. seconds.
(d) (i) Number of moles of oxygen gas
(b) & (c)
Energy

No catalyst = 37
Ea' With catalyst 24 000
Ea = 0.0015 mol
Reactant Product
(ii) From the chemical equation, 2 mol of potassium
chlorate(V) produces 3 mol of oxygen
mol KClO3 0.0015 mol of oxygen
x = 0.0015 × →
Reaction path x 2
3
3. (a) Use of catalyst = 0.001 mol
((cb)) 2TNimaOe tCalk(eanq)to→co2lNleacCt l2(0a0q)cm+ 3Oo2f(gg)as Mass of potassium chlorate(V)
in experiment = 0.001 × [39 + 35.5 + 3(16)]
II is shorter. Hence, experiment II has a higher rate = 0.1225 g
of reaction. Manganese(IV) oxide acts as a catalyst 2. (a) Total surface area / Size of marble
that provide an alternative path with lower activation (b) Carbon dioxide gas
energy. Frequency of collision between particles in

557

Chemistry   Answer There are some empty spaces between atoms in the orderly
2. (a) arrangement of atoms. When a metal is knocked, atoms of
the same size slide into new positions to fill the empty of
Set of experiment 1 2 3 4 5 space. Therefore, metals are malleable.

Concentration of 2. (a) Iron
Na2S2O3 solution
that reacts, M2 0.1(50) 0.1(40) 0.1(30) 0.1(20) 0.1(10) atom
(mol dm–3) 50 50 50 50 50
Pure iron Carbon
= 0.1 = 0.08 = 0.06 = 0.04 = 0.02 atom

Ti1me (s–1) 0.05 = 0.1 0.03 0.02 0.01 Steel
(b) The presence of carbon atoms which are of different
Table 2
sizes with iron atoms disrupt the orderly arrangement
(b) Graph of—Tim1 e a gainst concentration of thiosulphate of iron atoms. Layers of iron atoms become more
difficult to slide over one another when force is applied.
—Tim1 e (s–1) 3. (a) Copper, aluminium, magnesium,
manganese (any two answers).
0.05 (b) Light, hard, resistant to corrosion.
4. (a) Pewter
(b) Zinc, copper
(c) Manufacture of musical instruments; Manufacture of
electrical parts

0.04

8.2

0.03 1. Silicon and oxygen
2. (a) Silica
(b) Soda-lime glass
(c) Silica, sodium carbonate, calcium carbonate, aluminium
0.02 oxide, boron oxide

ANSWER FORM 4 0.01 (d) Lead cystal glass 15
100
3. Mass of boron oxide = × 2500 g

= 375 g
4. Lead crystal glass. It has high refraction index and is able to
0 0.02 0.04 0.06 0.08 0.10 refract the light from the bulb to the surroundings to make the

Concentration (mol dm–3) lobby look brighter and more beautiful.

(c) sWYoellhuloetnwiontphrieenccciorpeniatcasetenest,irstahfteoiorvnmaloeudfe.soofdiTuim1methailossouilnpchraetaes,es. 8.3
(d) 1. Extremely hard and strong; Brittle and easily break;
(e) The higher the concentration of sodium thiosulphate
solution, the shorter the time for the ‘X’ mark to Chemically inert; Opaque
disappear from sight. 2. Silicon, oxygen, metal
(f) Add distilled water to sodium thiosulphate solution to 3. Oxide, carbide, nitride
change the concentration of sodium thiosulphate. 4. (a) Have semiconducting properties and can store charges.
Use the same size of conical flask, same concentration (b) Withstand thermal shock and higher heat resistance.

and volume of dilute sulphuric acid.

Chapter 8  Manufactured Substances in Industry 8.4
1. Substance made from a combination of two or more
8.1
1. Ductility of metals: non-homogeneous materials, namely matrix substance and
strengthening substance.
Force Layers of 2. Optical fibre, fibre glass.
atoms slide 3. The concrete is strong but brittle with low tensile strength.
Force Therefore, steel should be added in concrete to strengthen the
over one concrete and produce a composite material which is strong,
high tensile strength and high compression strength.
another 4. Silica glass fibre is a strengthening substance, glass or plastic
coating are matrix substances.
Equal size of atoms are orderly arranged. The layers of atoms 5. The condominium is prone to intense sunlight. Photochromic
can slide over one another when a force is applied. Therefore, glass can turn dark when exposed to light. As the sun becomes
metals are ductile. brighter, the photochromic glass becomes darker. So, the space
inside the condominium is neither too bright nor too hot.
Malleability of metals: When cloudy, the photochromic glass becomes transparent. The
condition inside the condominium is not dark and does not
Force The shape need to switch on the light.
of the metal
changes

560