The document discusses a laboratory experiment to determine the empirical formula of metal oxide X. The procedure involves weighing a crucible before and after heating a sample of metal X inside it to form an oxide. The mass changes are used to calculate the empirical formula as XpOq by determining the ratio of metal X to oxygen in the oxide.

1. PANITIA KIMIA
SEKOLAH MENENGAH SAINS TAPAH
JALAN PAHANG, 35000 TAPAH,
PERAK DARUL RIDZUAN
PECUTAN AKHIR SPM 2015

2. RATE OF REACTION
• A group of students carried out three experiments to investigate the factors affecting the rate of reaction between
hydrochloric acid and zinc. Table shows the results of the experiments.
i) Calculate the average rate of reaction for Experiment I and Experiment II. [2 marks]
ii) On the same axis , sketch the graph for the three sets of experiments for the liberation of
40 cm³ of hydrogen gas [3 marks]
iii) Write the ionic equation for the reaction between zinc and hydrochloric acid. [2 marks]
iv) Based on Table above, compare the rate of reaction between  Experiment I and Experiment II
 Experiment II and Experiment III
Explain the difference in the rate of reaction based on the Collision Theory. [10 marks]
Experiment Reactants
Time taken to collect 40 cm³
of hydrogen gas (s)
I
50 cm³ of 1.0 mol dm¯³ hydrochloric acid + zinc granule + a few
drops of copper(II) sulphate solution
90
II 50 cm³ of 1.0 mol dm¯³ hydrochloric acid + zinc granule 150
III 50 cm³ of 0.5 mol dm¯³ hydrochloric acid + zinc granule 270

3. ANSWER
i) Average rate of reaction experiment I = 40/90 = 0.44 cm³ s¯¹
Average rate of reaction experiment II = 40/150 = 0.27 cm³ s¯¹
ii)
iii) Zn + 2H⁺  Zn²⁺ + H₂

4. iv) Experiment I and II
1. Rate of reaction of Experiment I is higher than Experiment II
2. Copper(II) sulphate solution acts as catalyst
3. Which provide an alternative path with lower activation energy
4. More colliding particles able to achieve the activation energy
5. The frequency of effective collision between zinc atom and hydrogen ions is higher in Experiment I than in
Experiment II
Experiment II and III
1. Rate of reaction in experiment II is higher than Experiment III
2. The concentration of hydrochloric acid in experiment II is higher than in Experiment III
3. The number of hydrogen ions per unit volume is higher in Experiment II
4. The frequency of collision between zinc atom and hydrogen ions is higher in Experiment II than Experiment III
5. The frequency of effective collision between zinc atom and hydrogen ions is higher in Experiment II than in
Experiment III

5. MANUFACTURED OF SUBSTANCES IN INDUSTRY
Table shows three substance, example and their components respectively
i) Name substance P and suggest how the strength of concrete can be increased to be used as pillars of
building. [2 marks]
ii) Polymer R is formed through polymerization process.
Write the chemical equation to produce the polymer R and give a name for the polymer. [3 marks]
iii) Q is one type of alloy. Compare and explain the hardness between alloy Q and its pure metal. [5 marks]
Substance Example Component
P Reinforced concrete cement, sand, small pebbles and steel
Alloy Q copper and zinc
Polymer R ethene

6. ANSWER
i) Composite material - Adding steel rod into the concrete
i) nC2H4  [C2H4 ]
Polyethene
iii) - Alloy Q/brass is harder than its pure metal/copper
- the presence of zinc atom in alloy Q disrupts the orderly arrangement of copper atom
- These make the atomic layers of atoms harder to slide over on another
- in pure metal/copper the atoms are arranged packed closely and in orderly manner.
- this allow the layers of atoms are easily to slide one another
n

7. CHEMICAL FOR CONSUMERS
Diagram shows the apparatus used to investigate the cleaning action of cleaning agent X and Y to remove oily
stain from the cloth.
• Based on Diagram compare and explain the effectiveness of cleaning action between experiment I and II.
Identify the cleaning agent X and Y.

8. ANSWER
- Cleaning agent Y in experiment II is more effective than cleaning agent X
- Cleaning agent Y do not form scum in hard water therefore it can remove oily stain from
the cloth
- Cleaning agent X in experiment I is not effective in hard water because hard water
contain high calcium ion and magnesium ion
- These ions will react with cleaning agent X to formed an insoluble precipitate/scum
- The formation of scum will reduces the number of cleaning agent A
• Cleaning agent X is soap
• Cleaning agent Y is detergent

9. CHEMICAL BOND
By using the element in Diagram above, explain the formation of compound between
element R and Q , also formation of compound between element L and T. The two
compounds should have different bond types.
[10 marks]

10. ANSWER
 R and Q
1. Electron arrangement of atom R is 2.8.2
2. Electron arrangement of atom Q is 2.6.
3. Atom R donates 2 electrons to achieve stable octet electron arrangement to form R²⁺ ion
4. Atom Q receive 2 electrons to achieve stable octet electron arrangement to form Q²¯ ion
5. R²⁺ ion and Q²¯ ion are attracted by a strong electrostatic force to form ionic bond
 L and T
1. Electron arrangement of atom L is 2.4
2. Electron arrangement of atom T is 2.8.7
3. Atom L contribute 4 electron and atom T contribute 1 electron for sharing
4. To achieve stable octet electron arrangement
5. 1 atom L share 4 pairs of electron with 4 atom T to form covalent bond

11. ELECTROCHEMISTRYDiagram shows two type of cells.
Compare and contrast cell X and cell Y in terms of:
• Type of cell
• The energy change
• The terminals of the cells
• Ions presence in the electrolyte
• Observation
• Half equation for both electrodes
• Name of the processes occurred at the positive terminal of each cell [ 10 marks]
Cell X Cell Y

12. ANSWER
Cell X Cell Y
Type of cell Electrolytic cell Voltaic cell
The energy change Electrical energy to chemical energy Chemical energy to electrical energy
The terminal of the cell
Positive terminal / anode: Copper
Negative terminal / cathode: copper
Positive terminal / cathode: copper
Negative terminal / anode: aluminium
Ions present in the electrolyte
Cu 2+, H+
SO4
2- , OH-
Observation
Anode: Electrode become thinner
Cathode: Brown solid is deposited//thicker
Negative terminal/Aluminium plate:
Electrode become thinner
Positive terminal/Copper plate:
Brown solid is deposited//copper plate
become thicker
Half equation for both
electrodes
Anode:
Cu  Cu 2+ + 2e
Cathode:
Cu 2+ +2e  Cu
Al plate/- terminal:
Al  Al 3+ + 3e
Cu plate//+ terminal:
Cu 2+ +2e  Cu
Name of the process occurred
at both electrodes/terminal
Anode/Al plate: Oxidation
Cathode/Copper plate//negative terminal: Reduction

13. EMPIRICAL FORMULA
Metal X is more reactive than hydrogen.
[Relative atomic mass: O = 16 ; X = 24 , ionic formula : X²⁺ ]
Describe a laboratory experiment to determine the empirical formula of oxide X. Your answer should
consist of the following:
 Procedure of the experiment
 Calculation involved
[ 10 marks ]

14. ANSWER
• Procedure:
1. A crucible and its lid are weighed
2. 10 cm of cleaned X ribbon is coiled loosely and placed in the crucible.
3. The crucible with its lid and content are weighed again.
4. The crucible is heated strongly without its lid.
5. Using a pair of tongs, the lid is lifted at intervals.
6. When the burning is completed, the lid is removed and the crucible is heated strongly
for 2 minutes.
7. The crucible is allowed to cool to room temperature.
8. The crucible and its lid and content are weighed again

15. ANSWER
• Calculation:
Mass of crucible + lid = a g
Mass of crucible + lid + magnesium = b g
Mass of crucible + lid + X oxide = c g
Empirical formula is XpOq

16. REDOX
Diagram 10.2 shows the reactions involving Fe2+ ion and Fe3+ ion
i) State the suitable examples for reagent A and reagent B.
ii) Explain the oxidation and reduction in terms of electron transfer in Reaction I.
iii) Write the half equation involved in Reaction II and state the observation occurred. [7 marks]

17. ANSWER
i) A : Bromine/chlorine water // acidified KMnO4 solution
B : Zinc
ii) Fe2+ undergoes oxidation reaction because Fe2+ releases electron.
Bromine water undergoes reduction because Bromine water gains electron
iii) Fe3+ + e → Fe2+
Zn → Zn2+ + 2e
Observation : Brown solution turns to green