Spotlight A+ Form 4 & 5 Chemistry KSSM

(b) Figure 8.3 shows the reactions carried out on compound K.
Heat
Compound K Black solid L + Gas M
+ Sulphuric acid, + Limewater
H SO
2 4
Blue solution Limewater
W formed turns cloudy

+ Ammonia, NH 3
solution in excess

©PAN ASIA PUBLICATIONS
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.
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 SPM MODEL PAPER
I



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

Excess zinc, Zn
II



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

Excess zinc, Zn
III




50 cm of 0.5 mol dm –3 70 ºC
3
sulphuric acid, H SO
2 4
Excess zinc, Zn

Table 9

535

(a) Referring to Experiments, I, II and III, state:
(i) The meaning of the rate of reaction. [1 mark]
(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 gas, H released in Experiment I. [3 marks]
2
[Molar gas volume at room conditions: 24 dm mol ]
–1
3
(d) Sketch the graph of the volume of hydrogen gas, H against time for Experiments I, II dan III on the
2
same axis. [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, AgNO
3
solution in a test tube.

Metal Q
powder Blue solution


Silver nitrate, AgNO
3 Shiny grey
solution (colourless)
solid
Figure 10.1
SPM MODEL PAPER ©PAN ASIA PUBLICATIONS
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
G
experiment to investigate a redox reaction.
Based on Figure 10.2, describe the oxidation

Carbon
and reduction that occurs. Your answer
must include the following:
FeSO solution
4
(i) The role of each reactant. [2 marks] Iron(II) sulphate, electrodes
Acidified potassium
(ii) The transfer of electron of each reactant. Dilute sulphuric acid, dichromate(VI),
K Cr O solution
[2 marks] H SO 2 2 7
(iii) The colour changes that can be 2 4
observed after 15 minutes. [2 marks] Figure 10.2
(c) Figure 10.3 shows two types of cells, P and Q.
V

A B C D
Magnesium, Copper, Copper, Cu
Mg Cu

Copper(II) sulphate,
CuSO solution
4
P Q
Figure 10.3
Compare between cells P and Q. Include in your answer the following:
(i) The change of energy. Given the following E value:
0
(ii) The flow of electron. Mg (aq) + 2e ⇌ Mg(s) E = –2.38 V
0
–
2+
(iii) The product formed at anode. 2H O(l) + 2e ⇌ H (g) + 2OH (aq) E = –0.83 V
0
–
–
2
2
(iv) Half equation for the discharge at cathode. 2H O(l) → O (g) + 4H (aq) + 4e – E = +1.23 V
+
0
2
2
(v) The colour change of the solution. Cu (aq) + 2e ⇌ Cu(s) E = +0.34 V
0
2+
–
[10 marks] 2SO (aq) → S O + 2e E = +2.01 V
0
2–
2–
–
4
8
2
536

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
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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
compound Properties

Miscible with water in all proportions. Burns with blue flame to form carbon
V
dioxide, CO and water, H O.
2
2
Soluble in water, H O. Reacts with calcium carbonate, CaCO to produce carbon
W 2 3
dioxide gas, CO .
2 SPM MODEL PAPER
Insoluble in water, H O. Decolourise the orange colour of acidified potassium
X 2
dichromate(VI), K Cr O solution.
8
2
2
Y Insoluble in water, H O. Sweet smell.
2
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 C C C C C C C H H C C C C C C C H

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 silver nitrate, AgNO solution
–3
3
L2 = 0.5 mol dm sodium chloride, NaCl solution
–3
L3 = 0.5 mol dm potassium chloride, KCl solution
–3
Carry out the experiment according to the following instructions:
1 Measure and pour 25 cm of L1 solution into a polystyrene cup.
3
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 cm of L2 solution into another polystyrene cup.
3
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 ©PAN ASIA PUBLICATIONS
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:
–3
–1
–1
[Specific heat capacity of solution: 4.2 J g °C ; density of solution: 1 g cm ]
(a) Silver nitrate, AgNO solution + sodium chloride, NaCl (L1 + L2)
3
(b) Silver nitrate, AgNO solution + potassium chloride, KCl (L1 + L3)
3
4 State the operational definition of the heat of precipitation of silver chloride, AgCl.
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 Pour the dilute sulfuric acid, H SO into the test tube and warm the mixture slowly.
2
4
3 Carry out the experiments in Table 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]
©PAN ASIA PUBLICATIONS
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 cm beaker.
3
2 Add distilled water until half full.
3 Stir until all D2 powder dissolves.
4 Pour 2 cm of the prepared solution into three test tubes, I, II and III.
3
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
drop by drop until in excess.

(b) Test tube II - Add ammonia solution drop by
drop until in excess. SPM MODEL PAPER
(c) Test tube III - Add 2 cm of sodium sulphate
3
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, AgNO + sodium chloride solution, NaCl)
3
Thermometer

40 40
30 30
3
25 cm of
20 solution L1 20
Solution L2
Polystyrene cup
White precipitate
(b) L1 + L3 (Silver nitrate solution, AgNO + potassium chloride solution, KCl)
3
Thermometer
40 40
30 30
3
25 cm of
20 solution L1 20
Solution L3
Polystyrene cup
White precipitate
539

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

No 'pop' sound
Lighted Lighted wooden
wooden splinter
splinter
Dilute sulphuric Dilute sulphuric
acid, H SO acid, H SO
2 4 2 4
D1 D1
Dilute sulphuric Dilute sulphuric
acid, H SO acid, H SO
2 4 2 4





Limewater Limewater
turns cloudy
D1 D1


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

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





White Soluble white
precipitate precipitate

Add a little ammonia, Add in excess ammonia,
Test tube 2
NH solution NH solution
3 3







White Insoluble white
precipitate precipitate
Add sodium sulphate,
Test tube 3
NaSO solution
4







White
precipitate

540

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




Form 4 (b) (i) Vinegar; salt; baking powder

(ii) Vinegar: preserves food
Salt: gives salty taste
Chapter 1 Introduction to Chemistry Baking powder: raises the dough
(c) Chemist; doctor
1.1 (d) Hydrogen peroxide waste with low concentration
6. B ©PAN ASIA PUBLICATIONS
can be poured directly into the sink. Concentrated
1. A field of science that studies the structures, properties, hydrogen peroxide wastes need to be diluted with
compositions and interactions between matter. water. Then, it is added with sodium sulphite for the
2. Herbicide; Hormone purpose of decomposition before being poured into the
3. With nanotechnology, sunscreens are no longer oily sink.
nowadays and are colourless when applied to the skin. 2. (a) (i) Presence of water and oxygen
4. (a) Cosmetic consultant (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)
Test tube Observation
1.2 A Iron nail does not rust.
1. A systematic scientific method used to solve science related B Iron nail rusts.
problems. C Iron nail does not rust.
2. (a) When the temperature increases, the mass of salt
dissolved also increases. (d) Oxygen and water must be present for the iron nail to
(b) Manipulated variable: Temperature rust.
Responding variable: Mass of salt dissolved
Section B
3. (a) The scientific method is a systematic approach to solve
1.3 problems in science. ANSWER FORM 4
1. • Do not eat, drink, chase or run in the laboratory. (b) Making an observation
• Do not pour the chemicals back to the reagent bottles.
2. • Keep flammable substances away from the heat source. Making an inference
• Do not point the mouth of the test tube at your face or
at other people.
3. (a) To carry out experiment that involves the release of Identifying the problem
toxic vapours, gases that can cause combustion or gases
with pungent smell. Making a hypothesis
(b) To remove dirt, oil, chemicals or microorganisms from
the hands. Identifying the variables
(c) To wash and clean the body if accident happens on
parts of the body. It is also used to put out fire at any Controlling the variables
part of the body if there is fire.
4. (a) Kept in paraffin oil to prevent reaction between this Planning an experiment
chemical with moisture, water and air.
(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 Writing a report
1. A 2. A 3. C 4. B 5. C
7. B 8. D 9. B 10. C Section C
11. B 12. C 13. A 14. C 15. D
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.
studies the structures, properties, compositions and • Try to stop the spill from spreading to other areas
interactions between matter. using sand to border it.
541

Chemistry Answer
Paper 2 Element Carbon, C Hydrogen, H Oxygen, O
Section A
1. (a) To remove the layer of the magnesium oxide on its Mass (g) 54.55 9.09 36.36
surface.
(b) To allow the oxygen to enter the crucible for the Number of moles 54.55 9.09 36.36
complete combustion of magnesium. of atoms 12 1 16
(c) (i) = 4.546 = 9.09 = 2.273
Element Magnesium, Mg Oxygen, O
22.30 – 20.50 23.50 – 22.30 4.546 9.09 2.273
Mass (g) Simplest ratio 2.273 2.273 2.273
=1.80 = 1.2
= 2 = 4 = 1
Number of
©PAN ASIA PUBLICATIONS
moles of 1.8 = 0.075 1.2 = 0.075 Thus, the empirical formula of compound Y is
atoms 24 16 C H O.
2
4
(ii) Assume that the molecular formula of Y
Simplest ratio 1 1 = (C H O) n
2
4
Thus the empirical formula of magnesium oxide is Given that the molecular mass of (C H O) = 88
2
n
4
MgO. n[2(12) + 4(1) + 16] = 88
(ii) 2Mg(s) + O (g) → 2MgO(s) 44n = 88
2
(d) (i) n = 88
44
Glass tube
Metal oxide W n = 2
Rubber Therefore, the molecular formula of compound Y
tubing
Ethanol, Glass tube = (C H O)
Air hole Spirit 2 4 2
C H OH = C H O
lamp 2 5 4 8 2
Glass tube
Hydrochloric Section B
Water acid 4. (a) (i) The molecular formula of a compound shows the
Zinc, Zn actual number of atoms of each element present in
a molecule of the compound.
granules (ii) The relative molecular mass of glucose
ANSWER FORM 4 2. (a) The empirical formula of a compound is the chemical (b) (i) Consider 100 g of compound Q. 2
Wooden block
= 6(12) + 12(1) + 6(16)



(ii) The hydrogen gas is allowed to flow through the
= 180
apparatus for 10 to 15 seconds to remove all the
A molecule of glucose consists of 6 carbon atoms,

air.
12 hydrogen atoms and 6 oxygen atoms.

The empirical formula of glucose is CH O.
formula that shows the simplest ratio of the number of
atoms of each element in the compound.
Element
hydrogen.
14.29%

85.71%
(ii) The heating, cooling and weighing steps are
(b) (i) Method I. Magnesium is more reactive than Percentage Carbon, C Hydrogen, H
repeated until a constant mass is obtained. Mass (g) 85.71 14.29
(c) (i) Mass of copper
= 61.71 – 55.31 = 6.4 g Number of 85 14.29
(ii) Number of moles of copper moles of atoms 12 1
= 6.4 = 0.1 mol = 7.1425 = 14.29
64
(iii) Mass of oxygen 7.1425 14.29
= 63.31 – 61.71 = 1.6 g Simplest ratio 7.1425 7.1425
(iv) Number of moles of oxygen = 1 = 2
= 1.6 = 0.1 mol Thus, the empirical formula Q is CH . 2
16
(v) 0.1 mol of copper atom combines with 0.1 mol (b) (ii) Assume that the empirical formula of Q = (CH )
2 n
of oxygen atom. Thus, 1 mol of copper atom = 42
combines with 1 mol of oxygen atom. Empirical Given that the relative molecular mass of
formula of copper oxide is CuO. (CH ) = 42
2 n
3. (a) The molecular formula is the chemical formula that n[12+ 2(1)] = 42
shows the actual number of atoms of each element 14n = 42
found in a molecule of a compound. n = 42
(b) C H 12 14
n = 3
6
(c) 6(12) + 12(1) = 84 Therefore, the molecular formula of compound Q

(d) CH 2
(e) (i) Consider 100 g of compound Y. = (CH )
2 3
= C H
3 6
546 547

Chemistry Answer
to achieve the stable octet electron arrangement, 2.8. A
Reinforcement & Assessment of Science Process Skill positive magnesium ion, Mg is formed. Mg → Mg
2+
2+
+ 2e . A chlorine atom has an electron arrangement of
–
Observation 2.8.7. Each of two chlorine atoms accepts one electron
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,
2+
–
–
–
–
acid, HNO 3 hydroxide, NaOH Cl is formed. Cl + e → Cl One Mg ion and two Cl
ions are attracted to each other by a strong electrostatic
solution
attraction force. An ionic compound magnesium
White powder dissolved chloride, MgCl is formed.
2
Oxide P White powder cannot (b)
(K1) dissolve. to form a colourless – 2+ –
solution.
Cl Mg Cl
White powder
Oxide Q White powder cannot
(K2) dissolved to form a dissolve.
colourless solution.
White powder White powder dissolved 5.3
Oxide R dissolved to form a to form a colourless
(K3) 1. (a) A single covalent bond
colourless solution. 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
solution/ nitric acid Formula: HCl
2. Amphoteric Base Acid (b) Four single covalent bonds

ANSWER FORM 4 3. Q, R, P ©PAN ASIA PUBLICATIONS
H
Oxide R
Oxide P
Oxide Q
Table 2
H
H
C
4. Aluminium oxide
Chapter 5 Chemical Bonds Formula: CH H 4
(c) Two double covalent bonds
5.1
1. Chemical bonds are forces that hold atoms together to make O C O
compounds or molecules through transfer of electrons or
sharing of electrons.
2. Neon atom has achieved the stable octet electron Formula: CO 2
arrangement. It does not need to gain, lose or share 2. (a) 2.5
electrons with other atoms. (b) Covalent compounds
3. No. The electron arrangement of magnesium atom is 2.8.2. (c) NH 3
It does not achieve the stable octet electron arrangement. (d)
4. The electron arrangement of sodium atom is 2.8.1. Sodium H
atom can donate the single valence electron to achieve a
stable electron arrangement, 2.8.
5. (a) Ionic bond H N H
(b) Covalent bond
(c) Covalent bond
(d) Ionic bond
(e) Covalent bond
(f) Ionic bond 5.4
1. A hydrogen bond is a type of attraction force between a
5.2 hydrogen atom which bonded to a strongly electronegative
atom such as fluorine, oxygen or nitrogen with another
1. (a) 2.8; Y 2– electronegative atom (fluorine, oxygen or nitrogen) in other
(b) 2, W molecule.
+
2. A positive ion; Q 3+ 2. Molecules of HCl are attracted by weak Van der Waals
3. (a) A magnesium atom has an electron arrangement of attraction forces only. Less heat energy is required to
2.8.2. A magnesium atom donates 2 valence electrons overcome the weak Van der Waals attraction forces. On
550 551

Answer Chemistry
(b) • Use small marble chips experiment II is higher. Frequency of effective collision
• Use higher concentration of hydrochloric between particles in experiment II is higher.
acid
(c) Graph of volume of carbon dioxide
gas against time
Paper 1
Volume of carbon dioxide (cm ) 3
1. D 2. C 3. D 4. C 5. A
6. B 7. A 8. C 9. A 10. A
40 11. D 12. B 13. D 14. A 15. D
Paper 2
30
Section A
©PAN ASIA PUBLICATIONS
1. (a)
Graph of volume of gas collected against time
20
Volume of gas (cm 3 )
40
10
Δy = 37 – 30
0 = 7
30 60 90 120 150 180 210
Time (s) 30 Δx = 114 – 54
= 60
(d) 48 cm . Some of the carbon dioxide gas dissolves in
3
water.
(e) Saturate the water with carbon dioxide gas before
collecting the gas in the burette. 20
Δy = 26.5 – 8
7.3 = 18.5
1. On a warm night, the surrounding temperature is higher.
The chemical reaction in the fireflies' body increases. Thus, 10
it flashes faster and more frequent. Δx = 51 – 9
2. Food kept in the kitchen cabinet is exposed to high = 42
room temperature. The bacterial action towards food is ANSWER FORM 4
faster. More toxic is released. Therefore, the decaying and Time (s)
decomposition of food is faster. 0 30 60 90 120 150 180 210 240
3. Powdered detergent has a larger total surface area that (b) The instantaneous rate of reaction at 30 seconds
act on the dirt. Hot water also provided a medium with a ∆y
higher temperature. The cleansing action become faster. = ∆x
–1
= 18.5 = 0.44 cm s
3
7.4 42
The instantaneous rate of reaction at 90 seconds
1. (a) Minimum energy that colliding particles of the = ∆y
reactants must possess to start a chemical reaction. ∆x
(b) Collision that causes chemical reaction. The particles = 7 = 0.12 cm s
3
–1
collide in the correct orientation and are able to achieve 60
the activation energy. (c) • The instantaneous rate of reaction at 30 seconds is
2. (a) The total energy content in the reactants is higher than higher than the instantaneous rate of reaction at 90
the total energy content in the products. seconds.
(b) & (c) • The amount of potassium chlorate(V) is greater at 30
Energy seconds.
(d) (i) Number of moles of oxygen gas
= 37
No catalyst 24 000
E = 0.0015 mol
a With catalyst
E '
Reactant a (ii) From the chemical equation, 2 mol of potassium
chlorate(V) produces 3 mol of oxygen
Product
x mol KClO → 0.0015 mol of oxygen
3
Reaction path x = 0.0015 × 2 3
3. (a) Use of catalyst = 0.001 mol
(b) 2NaOCl(aq) → 2NaCl(aq) + O (g) Mass of potassium chlorate(V)
2
(c) Time taken to collect 200 cm of gas in experiment = 0.001 × [39 + 35.5 + 3(16)]
3
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
556 557

Chemistry Answer
2. (a) There are some empty spaces between atoms in the orderly
arrangement of atoms. When a metal is knocked, atoms of
Set of experiment 1 2 3 4 5 the same size slide into new positions to fill the empty of
space. Therefore, metals are malleable.
2. (a) Iron
Concentration of atom
Na S O solution 0.1(50) 0.1(40) 0.1(30) 0.1(20) 0.1(10)
50
50
50
50
50
3
2 2
that reacts, M 2 Carbon
(mol dm ) = 0.1 = 0.08 = 0.06 = 0.04 = 0.02 Pure iron atom
–3
Steel
1 (s ) (b) The presence of carbon atoms which are of different
–1
Time 0.05 = 0.1 0.03 0.02 0.01 sizes with iron atoms disrupt the orderly arrangement

of iron atoms. Layers of iron atoms become more
Table 2 difficult to slide over one another when force is applied.
3. (a) Copper, aluminium, magnesium,
(b) Graph of a gainst concentration of thiosulphate manganese (any two answers).
1
—
Time
1 (b) Light, hard, resistant to corrosion.
–1
— (s )
Time 4. (a) Pewter
(b) Zinc, copper
0.05 (c) Manufacture of musical instruments; Manufacture of
electrical parts
0.04
8.2
1. Silicon and oxygen
0.03
2. (a) Silica
(b) Soda-lime glass
0.02 (c) Silica, sodium carbonate, calcium carbonate, aluminium
oxide, boron oxide 15 × 2500 g
ANSWER FORM 4 ©PAN ASIA PUBLICATIONS
(d) Lead cystal glass

0.01
3. Mass of boron oxide =
100
= 375 g


4. Lead crystal glass. It has high refraction index and is able to
0.10
0.08
0.06
0.02
0.04
refract the light from the bulb to the surroundings to make the
lobby look brighter and more beautiful.
Concentration (mol dm )
–3
(c) When the concentration of sodium thiosulphate,
1
solution increases, the value of Time also increases. 8.3
(d) Yellow precipitate is formed. 1. Extremely hard and strong; Brittle and easily break;
(e) The higher the concentration of sodium thiosulphate Chemically inert; Opaque
solution, the shorter the time for the ‘X’ mark to 2. Silicon, oxygen, metal
disappear from sight. 3. Oxide, carbide, nitride
(f) Add distilled water to sodium thiosulphate solution to 4. (a) Have semiconducting properties and can store charges.
change the concentration of sodium thiosulphate. (b) Withstand thermal shock and higher heat resistance.
Use the same size of conical flask, same concentration
and volume of dilute sulphuric acid.
8.4
Chapter 8 Manufactured Substances in Industry 1. Substance made from a combination of two or more
non-homogeneous materials, namely matrix substance and
8.1 strengthening substance.
2. Optical fibre, fibre glass.
1. Ductility of metals:
3. The concrete is strong but brittle with low tensile strength.
Therefore, steel should be added in concrete to strengthen the
Force Layers of concrete and produce a composite material which is strong,
atoms slide
Force over one 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
Malleability of metals: inside the condominium is neither too bright nor too hot.
When cloudy, the photochromic glass becomes transparent. The
condition inside the condominium is not dark and does not
Force The shape
of the metal need to switch on the light.
changes
560 561

Answer Chemistry
(iii) Dative bond. The nitrogen atom contributes a lone (b)
pair of electrons to share with the hydrogen ion, H . +
Acidified
8. (a) (i) Add water. potassium
(ii) Ethanoic acid is a weak acid that ionises partially in Reactant Iron(II) sulphate dichromate(VI)
water. Thus, the concentration of H in ethanoic acid solution
+
is low.
(iii) Number of moles of NaOH = 0.5 × 20 = 0.01 mol (i) Role Reducing agent Oxidising agent
1 000
Number of moles of H SO = 0.01 = 0.005 mol (ii) Transfer Donates electron. Accept/receive
2+
2 4 2 of electron Iron(II) ion, Fe electron.
2–
Molarity of H SO = 0.005 = 0.2 mol dm –3 donates electron Cr O ion
7
2
2 4 0.025 to produce accepts electrons
3+
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(b) (i) K: Copper(II) carbonate iron(III) ion, Fe . to produce
L: Copper(II) oxide chromate ion,
3+
M: Carbon dioxide Cr .
W: Copper(II) sulphate (iii) Colour Pale green to Orange to green
3
(ii) Add 2 cm of hydrochloric acid, HCl and follow by change brown/
2 cm of barium chloride, BaCl solution. A white yellowish-brown
3
precipitate is formed. (c)
Section B Matter Cell P Cell Q
9. (a) (i) Change in the volume of hydrogen gas per second
// The volume of hydrogen gas released per second. (i) The energy Chemical Electrical energy
(ii) Temperature; Concentration of hydrogen ion changes. energy to to chemical
(b) Zn + 2HCl → ZnCl + H 2 electrical energy energy
2
(c) Number of moles of HCl= 50 × 0.5 = 0.025 mol (ii) The electron Electrode A to Electrode C to
1 000 flow electrode B electrode
Number of moles H = 0.025 = 0.0125 mol through D through the
2 2 the external external
Volume of H = 0.0125 × 24 dm = 0.3 dm 3 circuit. circuit.
3
2
= 300 cm 3 (iii) The product Magnesium ion, Oxygen, O
3
(d) Volume of hydrogen gas (cm ) formed at Mg 2+ 2
anode
Experiment III ANSWER FORM 5
(iv) Half Cu (aq) + 2e – Cu (aq) + 2e –
2+
2+
Experiment II equation at → Cu(s) → Cu(s)
cathode
Experiment I (v) The colour Blue to The blue
change of the colourless solution remains
Time (s) solution unchanged.
(e) Experiments I and II:
The rate of reaction in Experiment II is higher. The Section C
temperature of HCL acid in Experiment II is higher. H + 11. (a) (i) C H
2n + 2
n
ions in Experiment II has more kinetic energy and move (ii) Prercentage of carbon in Q = 72 × 100%
+
faster. The frequency of collision between H ions and 86
zinc atoms in Experiment II is higher. The frequency of = 83.72%
effective collision in Experiment II is higher. Prercentage of carbon in R = 72 × 100%
(f) Experiments II and III: 84
= 85.71%
The rate of reaction in Experiment III is higher. The Prercentage of carbon by mass in R is higher.
concentration of H ions in Experiment III is higher (b) (i)
+
because H SO is a diprotic acid while HCl is a monoprotic
4
2
+
acid. The number of H ions per unit volume solution Compound Homologues series
in Experiment III is higher. The frequency of collision
between H ions and zinc atoms in Experiment III is V Alcohol
+
higher. The frequency of effective collision in Experiment W Carboxylic acid
III is higher.
10. (a) Metal Q: Copper X Alkene
Oxidation half-equation: Y Ester
Cu(s) → Cu (aq) + 2e –
2+
Reduction half-equation: (ii) Functional group V: Hydroxyl group
–
Ag (aq) + e → Ag(s) Functional group X: Carbon-carbon double bond
+
577