

@Hak cipta BPSBPSK/SBP/2013
1 Perfect Score & X A –Plus Module/mark scheme 2013
BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER
JAWAPAN MODUL PERFECT SCORE &
X A-PLUS 2013
CHEMISTRY
Set 1
Set 2 Set 3
Set 4 Set 5
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@Hak cipta BPSBPSK/SBP/2013
2 Perfect Score & X A –Plus Module/mark scheme 2013
MODULE PERFECT SCORE & X A-PLUS 2013
SET 1 :THE STRUCTURE OF ATOM, PERIODIC TABLE OF ELEMENTS AND CHEMICAL BONDS
Question No Mark schemes Mark
1 (a) (i) Melting 1
(ii) Molecule 1
(b) The heat energy absorbed by the particles is used to overcome the forces of attraction
between the naphthalene molecules / particles.
1
(c) The particles move faster 1
(d) (i) X : electron Y : nucleus 1
(ii) Electron 1
(e) (i) W and X 1
(ii) W and X atom have different number of neutrons but same number of protons
Atom// Element W and X has different nucleon number but same proton number
1+1
Σ 10
Question No Mark schemes Mark
2 (a) No of electrons = 18, No of neutrons = 22 1+1
(b) (i) The total number of protons and neutrons in the nucleus of an atom 1
(ii) 40
(c) (i) 2.1 1
(ii)
X
(d) (i) W and Y 1
(ii) Atom W and Y have the same number of valence electrons 1
(iii) To estimate the age of fossils /artefacts. 1
Σ 10
X
X
e
3p
4n X e
e
X
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@Hak cipta BPSBPSK/SBP/2013
3 Perfect Score & X A –Plus Module/mark scheme 2013
Question No. Mark Scheme Marks
3 (a) (i) Total number of protons and neutrons in the nucleus of an atom 1
(ii) 35 – 18 = 17 1
(iii) shows nucleus and three shells occupied with electron
Label 12 proton, 12 neutron
1 +1
(iv) Number of electrons = 2 1 ...5
(b) (i) Liquid 1
(ii)
Q R
1+1
...3
(c)
1st mark - Label X and Y axis with correct unit
2 nd mark - Correct shape of curve
1+1
10
Temperature/oC
Time/s
67
90

@Hak cipta BPSBPSK/SBP/2013
4 Perfect Score & X A –Plus Module/mark scheme 2013
4 a) (i) F 1
(ii) Atom F has achieve stable/octet electron arrangement // has 8 valence electron 1
b) (i) 2D + 2H2O 2DOH + H2
Correct reactant & correct product Balance equation
1
1
(ii) The nuclei attraction towards the valence electrons is weaker in atom G. More easier for atom G to lose / release an electron to form a positively charged ion.
1+1
c) (i) Covalent bond 1
(ii)
1 1
(iii) Cannot conduct electricity at any state/ low melting and boiling point/.... 1
(d) Show coloured ion//formed complex ion//has various oxidation number//act as
catalyst 1
11
5 (a) Increasing of proton number. 1
(b) (i) Na/sodium, Mg/magnesium .... 1
(ii) Atomic size decreases across the period // Period 3. 1
(iii) 1. Number of protons in atom increases when across the period.
2. Force of attraction between nucleus and electrons in the shell is stronger.
1+1
..4 (c) Chlorine more reactive than bromine
Size of chlorine atom is smaller than bromine atom
Chlorine atom is easier to receive one electron
1+1
(d) Al3+
1
(e) (i) Ionic compound 1
(ii)
1+1
11
Y Y x X x
x
x x x
E E Y

@Hak cipta BPSBPSK/SBP/2013
5 Perfect Score & X A –Plus Module/mark scheme 2013
6 (a)
P : liquid Q : solid R : gas 1 +1+1
(b) (i) 1. P can be change to Q through freezing process. 2. When the liquid cooled, the particles in liquid lose energy and move slower.
3. As temperature drops, the liquid particles attract tone another and change
into solid
1 1
1
(ii) 1. P can change to R through boiling. 2. When liquid is heated, the particles of the liquid gain kinetic energy and
move faster as the temperature increase
3. The particles have enough energy to overcome the forces between them and gas is formed
1 1
1
(iii) 1. R can be change to P through condensation process. 2. When the gas cooled, the particles in gas lose energy and move slower.
3. Particles attract one another and change into liquid
1 1
1
(c) (i) 1. Uniform scale for X-axis and Y-axis and labelled/size of graph plotted ¾ of
graph paper. 2. Tranfer of point
3. Smooth curve
1 1 1
(ii) 1. Dotted line on the graph from the horizontal line to Y-axis at 80oC.
2. Arrow mark freezing point at 80oC
1 1
(iii) 1. Heat released to sorrounding
2. Is balanced when particles comes together to form a solid
1 1
(iv) Supercooling 1
20

@Hak cipta BPSBPSK/SBP/2013
6 Perfect Score & X A –Plus Module/mark scheme 2013
Question No. Mark Scheme Mark 7 (a) (i) Atom R is located in Group 17, Period 3 1 + 1
(ii) Electron arrangement of atom R is 2.8.7.
Group 17 because it has seven valence electron. Period 3 because it has three shells filled with electron
1
1 1
(b) (i) Atoms P and R form covalent bond. To achieve the stable electron arrangement, atom P needs 4 electrons while atom R needs one electron. Thus, atom P shares 4 pairs of electrons with 4 atoms of R, forming a molecule with the formula PR4 // diagram
1 1 1
1
1
(ii) Atom Q and atom R form ionic bond. Electron arrangement for atom Q is 2.8.1 and electron arrangement for
atom R is 2.8.7// Atom Q has 1 valence electron while atow R has 7
valence electron To achieve a stable (octet ) electron arrangement, atom Q donates 1 electron to form a positive ion// equation Q Q
+ + e
Atom R receives an electron to form ion R
-//equation
and achieve a stable
octet electron arrangement. R + e R
-
Ion Q
+ and ion R
- are attracted together by the strong electrostatic forces
to form a compound with the formula QR// diagram
1
1
1
1 1
1
R
R
R
R
P
Q R
+ - -

@Hak cipta BPSBPSK/SBP/2013
7 Perfect Score & X A –Plus Module/mark scheme 2013
Question
No Mark scheme Mark
8 (a) 12 represents the nucleon number. 6 represents the proton number.
1 1
(b) Able to draw the structure of an atom elements X. The diagram should be able to show the following informations: 1. correct number and position of proton in the nucleus/ at the centre of the atom. 2. correct number and position of neutron in the nucleus/ at the centre of the atom. 3. correct number and position of electron circulating the nucleus 4. correct number of valence electrons Sample answer:
or
1
1
1 1
e-
e- e
-
e-
e-
e-
e- e
-
e-
e-
e-
11p
12n √1 √2
√3
√4
e-
e- e
-
e-
e-
e-
e- e
-
e-
e-
e-
11p + 12n

@Hak cipta BPSBPSK/SBP/2013
8 Perfect Score & X A –Plus Module/mark scheme 2013
(c) (i) Atoms W and Y form covalent bond. To achieve the stable electron arrangement, atom W contributes 4 electrons while atom Y contributes one electron for
sharing. Thus, atom W shares 4 pairs of electrons with 4 atoms of Y, forming a molecule with the formula WY4 // diagram
1 1 1 1 1
(ii) Atom X and atom Y form ionic bond. Electron arrangement for atom X is 2.8.1 and electron arrangement for atom Y is 2.8.7 To achieve a stable (octet )electron arrangement, atom X donates 1 electron to form a positive ion // equation X X
+ + e
Atom Y receives an electron to form ion Y-//equation
and achieve a stable octet
electron arrangement. Y + e Y
-
Ion X+ and ion Y
- are attracted together by the strong electrostatic forces to
form a compound with the formula XY// diagram
1
1
1
1 1
1
(d) The melting point of the ionic compound/ (b)(ii) is higher than that of the covalent
compound/ (b)(i) . This is because in ionic compounds oppositely ions are held by strong electrostatic forces. High energy is needed to overcome these forces. In covalent compounds, molecules are held by weak intermolecular forces. Only a little energy is required to overcome the attractive forces. OR The ionic compound/(b)(ii) conducts electricity in the molten or aqueous state whereas the covalent compound/(b)(i) does not conduct electricity. This is because in the molten or aqueous state, ionic compounds consist of freely
moving ions carry electrical charges. Covalent compounds are made up of molecules only
1
1
1 1 1
or 1 1 1 1 1
20
X Y
+ - -
Y
Y
Y
W Y

@Hak cipta BPSBPSK/SBP/2013
9 Perfect Score & X A –Plus Module/mark scheme 2013
9 (a) (i)
1. Correct number of shells and valence electrons
2. Black dot or label Q at the center of the atom
1 1
(ii) 1. Group 14
2. There are 4 valence electrons 3. Period 2
4. Atom consists of 2 shells occupied with electrons
1 1 1 1
(b) (i) 1. Floats and moves fast on the water
2. ‘Hiss’ sound occurs 3. Gas liberates / bubble
[any two]
1 1
(ii) 2Q + 2H2O 2QOH + H2 1. Correct reactant and product
2. Balanced equation
1 1
(c) (i) Compound X Sharing electron between atom B and A
1 1
(ii) Choose any one ionic compound and any one covalent compound.
Melting/boiling point
Ionic compound Covalent compound
1. 2.
3.
High force of attraction between oppositely charged ions are
strong. more heat energy needs to overcome the forces.
low force of attraction between molecules are weak. less heat energy needs to
overcome the forces.
1 1 1 1
Electrical conductivity
Ionic compound Covalent compound
4. 5.
Conduct in molten state
or aqueous solution. The free moving ions are
able to carry electrical
charges.
Not conduct electricity.
Neutral molecules are not
able to carry electrical
charges.
1 1 1
1
Solubility
Ionic compound Covalent compound
6
7
Soluble in water.
Water molecule is
polar solvent.
soluble in benzene/ toluene /
any organic solvents. The attraction forces between
molecules in solute and
solvent are the same.
20
Q

@Hak cipta BPSBPSK/SBP/2013
10 Perfect Score & X A –Plus Module/mark scheme 2013
10
(i)
Compound formed between X
and Y
Molecule formed between Z and
Y
Types of
chemical
bonds
Ionic bond is formed because X
atom donates electrons and Y
atom receives electrons to achieve stable octet electron
arrangement/involve transfer
electron
Covalent bond is formed because
Z and Y atoms share the
electrons to achieve stable electron arrangement //
Inovelve sharing of electron
Boiling
point and
melting point
High because a lot of heat
energy needed to overcome the
strong electrostatic forces between ions
Low because less heat energy is
needed to overcome the weak
forces of attraction between molecules
2
2
(b)
1.Correct electron arrangement of 2 ions 2.Correct charges and nuclei are shown
3. X atom with an electron arrangement of 2.8.2 donates 2 valence electrons to achieve the stable octet electron arrangement, 2.8. X
2+ ion is formed //
X X2+
+ 2e-
4. Y atom with an electron arrangement of 2.6 accept 2 electrons to achieve the
stable octet electron arrangement, 2.8. Y2-
ion is formed // Y + 2e
- Y
2-
5. The oppositely-charged ions, X2+
and Y2-
are attracted to each other by a strong electrostatic force.
6. An ionic compound XY is formed
1 1
1 1
1
1
1
X
X
X
X
X
X
X
X
X
X
X X
X
X
X
X
X
X
X
2+ 2-
X2+
Y2-
X
X
X
X
X
X
X
X
X
X
X X
X
X
X
X
X
X
X

@Hak cipta BPSBPSK/SBP/2013
11 Perfect Score & X A –Plus Module/mark scheme 2013
(c)
1. A crucible is filled with solid P until it is half full. 2. Two carbon electrodes are dipped in the solid P and connected to the batteries
using connecting wire.
3. Switch is turned on and observation is recorded. 4. The solid P is then heated until it melts completely.
5. The switch is turned on again and observation is recorded. 6. Steps 1 to 5 are repeated using solid Q to replace solid P.
7. Observations: P does not light up the bulb in both solid and molten states.
Q lights up the bulb in molten state only.
P: naphthalene // any suitable answer Q: lead(II) bromide // any suitable answer
1 1 1 1 1 1 1 1 1 1 1 1 1
20
11 (a) (i) Z : 2.8.7 X : 2.4
1 1 ..2
(ii)
Z atom has 7 valence electrons needs one electron X atom has 4 valence electrons ,hence it needs 4 more electron each atom achieves stable octet electron arrangement share electrons between them four Z atoms , each contributes 1 electron // [ diagram one X atom contributes 4 electrons //[diagram] - four single covalent bonds are formed - the molecular formula is XZ4
- diagram [ no. of electrons in all the occupied shells in the X and Z atoms - correct] [ sharing of 4 pairs of single covalent bonds between 1 X atom and 4 Z atoms ]
1 1 1 1 1
1 1 1 1
1
..10
(iii) Colourless liquid 1
b) [Procedures of the experiment] eg. 1. Add a quarter of spatula of YZ solid and add into a test tube. 2. Pour 2-5 cm
3 of distilled water into the test tube containing theYZ2
3. Stopper the test tube and shake well. 4. Repeat Steps 1 to 3 using [ named organic solvent eg ether ] 5. Observe the changes and record them in a table
1 1 1 1 1
. [Results] Eg Solvent Observation Distilled water Colourless solution obtained [named organic solvent] e.g ether
Solid crystals insoluble in
liquid
[Conclusion] eg ZY is insoluble in organic solvent/[named organic solvent] but soluble in water.
1
1
..7

@Hak cipta BPSBPSK/SBP/2013
12 Perfect Score & X A –Plus Module/mark scheme 2013
No Explanation Sub Total
12 (a)(i) Y more reactive 1 5 Atomic size of Y bigger than X // The number of shell occupied with
electron atom Y more than X. 1
The single valence electron becomes further away from the nucleus. 1 the valence electron becomes weakly pulled by the nucleus. 1 The valence electron can be released more easily. 1
(ii) Name : Sodium 4Na + O2 2Na2O Chemical formulae Balance equation
1 1 1
3
(b) Put group1 metal into bottle that contain paraffin oil Group 1 metal readily reacts with air/moisture in atmosphere/ water
1 1
2
(c) Name : Sodium/any group 1 element Material : group 1 elements, water, Apparatus : forceps , knife, filter paper, basin, litmus paper.
1
1
[procedure] 3. Pour some water into the basin 4. Group 1 metal is take out from paraffin oil using forceps 5. A small piece of group 1 element is cut using a small knife 6. Oil on group 1 element is dried using a filter paper 7. The group 1 element is placed in the basin contain water. 8. Dip a red litmus paper into water
1 1 1 1 1 1
Max 5 [observation] 9. Color of red litmus paper turn to blue
1
[chemical equation ] Sample answer 2 Na + 2 H2O 2NaOH + H2
Chemical formulae Balance equation
1 1
Total 20
No Explanation ` Total
13. (a) Glucose // naphthalene // any solid covalent compound covalent
Intermolecular forces are weak
Small amount of heat energy needed to overcomes the forces
1
1
1
1
4
(b) X = 2.1 X = 2.2
Y = 2.7 // Y = 2.6 //
1. Suitable electron aranggement
2. Ionic bond
3. to achieve octet electron arrangement
4. One atom of X donates 1 electron to form ion X+
5. One atom of Y receives an electron to form ion Y-
6. Ion X+ and ion Y
- are attracted together by the strong
electrostatic forces
1
1
1
1
1
1
1
7
(c) material and apparatus;
compound XY, Carbon electrode, cell, wire, crucible,
bulb/ammeter/galvanometer
1

@Hak cipta BPSBPSK/SBP/2013
13 Perfect Score & X A –Plus Module/mark scheme 2013
Procedure
A crucible is half fill with solid XY powder
Dipped two carbon electrode
Connect the electrodes with connecting wire to the battery and
bulb
Observed whether bulb glow
Heated the solid XY in the crucible
Observed whether bulb glow
Observation
Solid XY - bulb does not glow
Molten XY - bulb glow
Diagram
Functional diagram
Labeled
1
1
1
1
1
1
1
1
9
TOTAL 20
SET 1:CHEMICAL FORMULAE AND EQUATIONS
Question No Mark scheme Mark
1
(a)
Molar mass is the mass of a substance that contains one mole of the substance. Example : Molar mass of one mole of magnesium is 24gmol
-1 .
1
(b)
Substance Molar mass / gmol
-1
N2 14x2 = 28
CO2 12+2(16) = 44
H2S 2(1)+ 32 = 34
H2O 2(1)+16 = 18
4
(c)
Mole of water = 0.9/ 18 = 0.05
Number of molecules = 0.05 x 6.02 x 1023
= 0.3 x 1023
// 3 x 1022
Mole of carbon dioxide = 2.2 / 44 = 0.05
1
1
1 1

@Hak cipta BPSBPSK/SBP/2013
14 Perfect Score & X A –Plus Module/mark scheme 2013
Number of molecules = 0.05 x 6.02 x 1023
= 0.3 x 1023
// 3 x 1022
Number of molecule is simmilar
1
2 (a) (i) Volume CO2 = 0.1 mol x 24dm3mol
-1 = 2.4 dm
3 1
(ii) Mass of CO2 = 0.1 mol x 44 gmol-1
= 4.4 g 1
(iii) Number of molecules = 0.1 mol x 6.02 x 1023 1
(iv) Number of atoms = 6.02 x 1022
x 3 = 1.806 x 10
23 1+1
(b) (i)
Heating, cooling and weighing processes are repeated a few times until a
constant mass is obtained.
(ii)
Compound Anhydrous CoCl2 H2O
Mass/g (34.10-31.50)g = 2.60 g
(36.26-34.10)g = 2.16 g
Number of moles 2.60/130 = 0.02 2.16/18 = 0.12
Ratio of moles 0.02/0.02 = 1 0.12/0.02 = 6
Simplest ratio of moles 1 6
1 mole of CoCl2 combines with 6 moles of H2O Therefore, the molecular formula of hydrated cobalt(II) chloride crystal is CoCl2.6H2O. Hence, the value of x in CoCl2.xH2O is 6.
1
1
1
(iii) Percentage of water
= 6(18)2(35.5)59
)18(6
x 100%
= 108 x 100% = 45.4% 238
1
1
Total 10
3 (a) (i) concentrated sulphuric acid 1
(ii) zink and hydrochloric acid[ any suitable metal and acid ] 1
(iii) Zn + 2HCl ZnCl2 + H2
(b)
(i) Mole of oxygen = 46.35 - 45.15 16
= 1.2 = 0.075 16
1

@Hak cipta BPSBPSK/SBP/2013
15 Perfect Score & X A –Plus Module/mark scheme 2013
(ii) Mole of copper = 45.15 - 40.35 64 = 4.8 = 0.075 64
1
1
(iii) Empirical formula = CuO
1
(c) (i) Collect the hydrogen gas in a test tube Put a burning wooden splinter at the mouth of the test tube ‘No pop sound ‘ produced.
1 1 1
(ii) To avoid the hot copper react with oxygen/air 1
(iii) Repeat heating, cooling and weighing processes until a constant mass obtained. 1
Total 11
4 (a) (i) Pb(NO3)2 1
(ii) AgCl 1
(b) (i) Pb2+
+ 2 Cl- PbCl2
Correct formula for reactants and product Balance ionic equation
1+1
(ii) Qualitative aspect : Lead(II) nitrate and sodium chloride are the reactants and lead (II) chloride and sodium nitrate are the products // Lead(II) nitrate solution reacts with sodium chloride solution to form lead(II)
chloride precipitate and sodium nitrate solution.
Quantitative aspect : One mole of lead(II) nitrate reacts with 2 mole sodium chloride to produce 1 mole of lead(II) chloride and 2 mole of sodium nitrate.
1
1
(c) (i) 2 Pb(NO3)2 2 PbO + 4NO2 + O2 1
Compound Colour of the
residue when
hot
Colour of the
residue when
cold
PbO
Brown
Yellow
Gases
Colour of the gas released
NO2
Brown
O2
Colourless
1
1
1
Total 10

@Hak cipta BPSBPSK/SBP/2013
16 Perfect Score & X A –Plus Module/mark scheme 2013
No Explanation Mark
5 (a) (i) Al3+
, Pb4+ 1+ 1
(ii) Aluminium oxide Lead(IV) oxide
1 + 1
(b) (i) (CH2O)n = 60 12n + 2n + 16n = 60 n = 2 Molecular formula = C2H4O2//CH3COOH
1
1 1
(ii) CaCO3 + 2CH3COOH (CH3COO)2Ca + H2O + CO2
2
(c) (i) 1.Green solid turn Black 2. Lime water becomes cloudy
1 1
(ii) CuCO3 CuO + CO2
1 + 1
(iii) 1. 1 mol of copper(II) carbonate decomposed into 1 mol of copper(II) oxide and
1 mol of carbon dioxide 2. copper(II) carbonate is in solid state, copper(II) oxide is in solid state and
carbon dioxide is in gaseous state
1
1
(iv) 1. No. of mole for CuCO3 = 12.4 / 124 = 0.1 mol 2. 1 mol of CuCO3 produces 1 mol of CuO Therefor No. of mole for CuO = 0.1 mol 3. Mass of CuO = 0.1 mol X 80 g mol
-1 = 8 g
1 1
1
(v) Mass of oxygen is 0.8g Simplest mol ratio : Cu : O = 3.2/64 : 0.8/16 = 1 : 1
1 1
20
Mark
6
(a) (i) Empirical formula of a compound is a formula that shows the simplest whole number ratio of each atoms of each element in a compound.
1
(ii) (ii) Substance Empirical formula
C10H8 C5H4
H2SO4
H2SO4
1
1

@Hak cipta BPSBPSK/SBP/2013
17 Perfect Score & X A –Plus Module/mark scheme 2013
(b) Element Carbon Hydrogen Oxygen
Percentage (%) 62.07 10.34 27.59 Mass/ g 62.07 10.34 27.59 Mole 62.07/12
= 5.17 10.34/1 = 10.34
27.59/16 = 1.72
Simplest mole ratio
5.17/1.72 = 3
10.34/1.72 = 6
1.72/1.72 =1
Empirical formula = C3H6O
n [C3H6O ] = 116 [ 3(12) + 6(1) + 16 ] n = 116 58 n = 116 n = 2 Molecular formula = C6H12O2
1
1
1
1
1
(c) Procedure :
1. Clean magnesium ribbon with sand paper. 2.Weigh crucible and its lid. 3. Put magnesium ribbon into the crucible and weigh the crucible with its lid. 4. Heat strongly the crucible without its lid. 5. Cover the crucible when the magnesium starts to burn and lift/raise the lid a
little at intervals. 6. Remove the lid when the magnesium burnt completely. 7.Heat strongly the crucible for a few minutes. 8.Cool and weigh the crucible with its lid and the content. 9. Repeat the processes of heating, cooling and weighing until a constant mass is obtained. 10.Record all the mass.
Tabulation of result :
Description Mass/ g Crucible + lid a Crucible + lid + magnesium b Crucible + lid + magnesium oxide c
10
1
Element Magnesium Oxygen Mass / g b-a c-b Mole b-a/ 24 c-b / 16 Simplest ratio of
mole x y
Empirical formula = MgxOy
1 1
1
Max
11
Total 20

@Hak cipta BPSBPSK/SBP/2013
18 Perfect Score & X A –Plus Module/mark scheme 2013
No Sub T
7. (a)
1. Empirical formula is the chemical formula that shows the simplest ratio of
atoms of each element in the compound.
2. Molecular formula is the formula that shows the actual number of atoms of each element in the compound.
3. Example : empirical formula of ethene is CH2 and the molecular formula is
C2H4
1 1 1
3
(b)(i)
(ii)
Element Carbon Hydrogen Oxygen
Percentage 40.00 6.66 53.33
Number of moles
40
123.33
6.661
6.66
53.33
163.33
Ratio of moles 1 2 1
Empirical formula is CH2O
n(CH2O) = 180 12n + 2n + 16n = 180 30n = 180 n=6 molecular formula = C6H12O6
1
1
1
1 1
5
(c)(i)
Magnesium is more reactive than hydrogen//Position of magnesium is above
hydrogen in the reactivity series 1
(ii)
Lead(II) oxide / Stanum oxide / iron oxide / copper(II) oxide
(iii)
1. Clean [5 – 15] cm magnesium ribbon with sandpaper and coil it.
2. Weigh an empty crucible with its lid.
3. Place the magnesium in the crucible and weigh again. 4. Record the reading.
5. Heat the crucible very strongly. 6. Open and close the lid very quickly.
7. When burning is complete stop the heating 8. Let the crucible cool and then weigh it again
9. The heating, cooling and weighing process is repeated until a constant mass is
recorded. 10.
Description Mass(g)
Crucible + lid
Crucible + lid + Mg / Zn / Al
Crucible + lid + MgO / ZnO / Al2O3
10
Total 20

@Hak cipta BPSBPSK/SBP/2013
19 Perfect Score & X A –Plus Module/mark scheme 2013
SET 2 :ELECTROCHEMISTRY
Question
No Mark scheme Mark
1(a) Electrical to chemical energy / Tenaga elektrik kepada tenaga kimia 1 (b) Pure copper / Kuprum tulen 1 (c) Cu
2+ and H
+ 1
(d)(i) (ii)
Become thicker / brown solid formed Bertambah tebal / pepejal perang terbentuk Cu
2+ + 2e Cu
1 1
(e) Blue solution remain unchanged // the intensity of blue solution is the same. Larutan biru tidak berubah // keamatan warna biru larutan adalah sama.
(i) the concentration of Cu2+
ions remains the same. kepekatan ion kuprum(II) tidak berubah
(ii) the rate of ionized copper at the anode same as the rate of discharged copper(II)
ion at the cathode . kadar pengionan kuprum di anode sama dengan kadar ion kuprum(II) dinyahcaskan di katod
1
1
1
(f) Oxidation / pengoksidaan Copper atom released electron to form copper(II) ion. Atom kuprum menderMarkan / membebaskan elektron menghasilkan ion kuprum(II).
1 1
(g) Electroplating of metal // extraction of metal Penyaduran logam // pengekstrakan logam
1
Total 11
2(a)(i) Chloride ion / Cl-, hydroxide ion / OH
-, sodium ion / Na
+ and hydrogen ion / H
+
Ion klorida / Cl-, ion hidroksida /OH
-, ion natrium , Na
+ dan ion hidrogen / H
+ 1
(ii) Cl-. The concentration of chloride ion is higher than hydroxide ion.
Cl-. Kepekatan ion klorida lebih tinggi daripada ion hidroksida
1 + 1
(iii) 2Cl- Cl2 + 2e 1
(b)(i)
Functional – 1 Label - 1
1 1
(ii) - place lighted splinter at the mouth of the test tube containing hydrogen gas
- “pop” sound produced - Letakkan kayu uji menyala ke dalam tabung uji berisi gas hydrogen
- Bunyi “pop” terhasil
1
1
(iii) - Sodium ion and hydrogen ions move to the cathode, hydrogen ion is selectively
discharged
- hydrogen ion is lower than sodium ion in the Electrochemical Series. - Ion natrium dan ion hydrogen bergerak / tertarik ke katod, ion hidrogen terpilih
untuk nyahcas / discas
- Ion hidrogen terletak di bawah ion natrium dalam Siri Elektrokimia
1 1
Total 11
Carbon electrodes
Elektrod karbon
Sodium sulphate
solution
Larutan natrium
sulfat A
Oxygen gas
Gas oksigen
Hydrogen gas
Gas hidrogen
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20 Perfect Score & X A –Plus Module/mark scheme 2013
Question
No Mark scheme Mark
3(a) Cu2+
, H+
1 (b) Carbon electrode which connect to copper electrode in cell A.
Because oxidation takes place Elektrod karbon yang disambung kepada elektrod kuprum dalam sell A Kerana proses pengoksidaan berlaku
1 1
(c)(i) X – silver electrode / elektrod argentum Y – impure silver electrode / elektrod argentum tak tulen
1 1
(ii) Ag+
+ e Ag 1 (d)(i) - The electrode become thinner
- Silver atom ionized / silver atom oxidized to form silver ion - elektrod seMarkin nipis - atom argentum mengion / atom argentum dioksidakan membentuk argentum ion.
1 1
(ii) Y : Ag Ag+ + e
Z : Ag+
+ e Ag 1 1
(e) The waste chemicals emitted contain poisonous heavy metal ions and cyanide ions / alter
the pH of water. Bahan buangan kimia dibebaskan mengandungi logam berat yang beracun dan sianid / mengubah nilai pH air
1
11
Question
No Mark scheme Mark
4(a)(i) Lead(II) ion// Pb2+
, bromide ion// Br-
Ion plumbum(II)// Pb2+
, ion bromida// Br-
1
(ii) Sodium ion // Na+, hydrogen ion// H
+, sulphate ion// SO4
2-, hydroxide ion//OH
-
ion natrium // Na+, ion hidrogen// H
+, ion sulfat // SO4
2-, ion hidroksida //OH
- 1
(b)(i) Lead / Plumbum 1
(ii) Pb2+
+ 2e Pb 1
(iii) Brown gas / Gas berwarna perang 1
(c)(i) hydroxide ion / ion hidroksida 1
(ii) Anode : Oxygen gas anod : Gas oksigen Cathode : hydrogen gas Katod : gas hidrogen
1
1
(iii) Sodium nitrate solution // sulphuric acid Larutan natrium nitrat // asid sulfurik (Any suitable electrolyte)
1
9

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21 Perfect Score & X A –Plus Module/mark scheme 2013
Rubric Mark
5(a) (i) Q, R, S , Cu 1 …. 1
(ii) positive terminal : Cu Potential difference : 0.7 V S is higher than Cu in the Electrochemical Series
1 1 1 ..... 3
(b) (i) positive terminal : copper / Cu Negative terminal : Metal P
(ii) metal P : Zinc / Zn // Magnesium/Mg (any suitable metal) Solution Q : Zinc sulphate // magnesium sulphate (any suitable electrolyte)
1 1 1
1
..... 4
(c) (i) anode : greenish yellow gas cathode : colourless gas (bubbles)
1 1 ….. 2
(ii) gas X : hydrogen gas Y : chlorine
1 1 ….. 2
(iii)
Anode Cathode
Ions move to / ion
attracted to
Hydroxide ion/OH-
Chloride ion/Cl-
Hydrogen ion/H+ ,
Potassium ion/K+
Ions selectively
discharged Cl
- H
+
Reason
Concentration Cl- higher
than OH-
Position of hydrogen
ion/H+ is lower than
potassium ion/K+ in the
Electrochemical Series. Half equation
2Cl- Cl2 + 2e 2H
+ + 2e H2
1+1
1+1
1+1
1+1 …. 8
Total 20
Question
No Mark scheme Mark
6(a) (i) Substance R : Glucose / ethanol (any suitable covalent compound) Substance S : Sodium chloride solution ( any salt solution / acid / alkali)
1 1
….. 2
(ii) 1. S conducts electricity but R does not 2. S has free moving ions // ions free to move 3. R consists of molecules / no free moving ions
1 1 1
….. 3
(b) (i) negative terminal : zinc positive terminal : copper
1 1
….. 2
(ii) 1. zinc electrode become thinner 2. Zn Zn
2+ + 2e
1 1
….. 2
(iii) 1. the potential difference decreases 2. iron is lower than zinc in the Electrochemical Series // iron is less electropositive than zinc // distance between iron and
1 1

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22 Perfect Score & X A –Plus Module/mark scheme 2013
copper is shorter than distance between zinc and copper in the Electrochemical Series
….. 2
(c) (i) Sample answer Lead(II) bromide / lead(II) iodide /sodium chloride/sodium iodide (any suitable ionic compound) r : substance that decompose when heated. Example : lead(II) nitrate, lead(II) carbonate
1
(ii)
Diagram: Functional Label
Observation: Anode : brown gas Cathode: grey solid
Half equation: Anode : 2Br
- Br2 + 2e
Cathode : Pb2+
+ 2e Pb
Product: Anode : lead Cathode : bromine gas
1 1
1 1
1 1
1 1
….. 8
Total 20
Question
No Mark scheme Mark
7(a)
Sample answer Silver nitrate solution
Functional – 1 Label - 1
Anode : Ag Ag
+ + e
Cathode : Ag+
+ e Ag
1
1 1 1 1 ….. 5
(b) 1. metal X is more electropositive than copper // X is higher than copper in the
Electrochemical Series 1
Carbon electrodes
Elektrod karbon
PbI2 // PbBr2 //
NaCl
Heat
Panaskan
Note :
Observations and half-equations are
based on the substance suggested.
Silver nitrate solution Iron spoon
Silver

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2. atom X oxidises to X ion // atom X releases electron 3. copper(II) ion accepts electron to form copper 4. the concentration of copper(II) ion decreases 5. metal Y is less electropositive than copper // Y is lower than copper in the
Electrochemical Series
1 1 1 1 ….. 5
(c) Material 0.5 mol dm
-3 of P nitrate, Q nitrate, R nitrate, S nitrate solutions, metal P, Q, R and S
Apparatus Test tube, test tube rack, sand paper
Procedure
1. Clean the metal strips with sand paper
2. Pour 5 cm3 of P nitrate solution , R nitrate solution , S nitrate solution into different
test tubes.
3. Place a strip of metal P into each test tube 4. Record the observation after 5 minutes
5. Repeat steps 2 to 4 using strip of metal Q, R and S to replace metal P.
Observation
Metal Metal ion P Metal ion Q Metal ion R Metal ion S P X X X
Q / X X
R / / X
S / / /
1
1
1 1 1 1 1
1 1
Conclusion The electropositivity of metals increases in the order of P,Q,R,S
1 …..10
TOTAL 20
SET 2 :OXIDATION AND REDUCTION
Question
No Mark scheme Mark
1 (a) To allow the flow / movement / transfer of ions through it 1
(b) chemical energy to electrical energy 1
( c) mark at electrodes 1
Cell 1 Cell 2 Positive electrode
Negative electrode
Positive electrode
Negative electrode
Q P R S
(d)(i) magnesium more electropositive than copper // above copper in the Electrochemical Series
(ii) blue becomes paler / colourless 1
Concentration / number of Cu2+
ion decreases 1
(iii) Mg→ Mg2+
+ 2e 1
(iv) Oxidation 1
(e)(i) copper become thicker // brown solid deposited 1
(ii) zinc 1
(iii) zinc undergoes oxidation // zinc atom release electron to form zinc ion 1
11

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24 Perfect Score & X A –Plus Module/mark scheme 2013
Question
No Mark scheme Mark
2(a) A reaction which involves oxidation and reduction occur at the same time 1 (b) (i) green to yellow/brown 1
(ii) oxidation 1
(iii) Fe2+
→ Fe3+
+ e 1
(iv) 0 1 (c) (i) magnesium 1
(ii) Mg +Fe2+
→ Mg2+
+ Fe 1
(iii) +2 to 0 1
(d) 1. label for iron, water and oxygen
2. ionization of iron in the water droplet (at anode) 3. flow of electron in the iron to the edge of water droplet
Water droplet O2
Iron
1
1
1
11
3 (a) Reaction A : not a redox reaction Reaction B : a redox reaction
1 1
Reaction A: No change in oxidation number
Reaction B: Oxidation number of magnesium changes/increases from 0 to +2 // Oxidation number of zinc changes/decreases from +2 to 0
1
1.....4
(b) (i) Oxidation number of copper in compound P is + 2 Oxidation number of copper in compound Q is + 1
1 1.....2
(ii) Compound P : Copper(II) oxide Compound Q : Copper(I) oxide Oxidation number of copper in compound P is +2 Oxidation number of copper in compound P is +1
1 1 1 1.....4
(iii) Substance that is oxidised : H2
Substance that is reduced : CuO
Oxidizing agent : CuO
Reducing agent : H2
1 1 1 1.....4
(c) (i) X, Z, Y 1
Y : Copper Z : Lead X : Magnesium
1 1 1.....3
Fe Fe
2+ +2e
e e

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25 Perfect Score & X A –Plus Module/mark scheme 2013
2Mg + O2 → 2MgO // 2X + O2 → 2XO [Correct formulae of reactants and product] [Balanced equation]
1 1.....2
TOTAL 20
4 (a) (i) Iron(II) ion releases / loses one electron and is oxidised to iron(III) ion// Oxidation number of iron in iron(II) ion increases from +2 to +3. Iron(II) ion undergoes oxidation, Iron(II) ion acts as a reducing agent
1
1
(ii) Iron(II) ion receives/ gain one electron and is reduced to iron.// Oxidization number of iron in iron(II) iron decreases from +2 to 0. iron(II) ion undergoes reduction, Iron(II) ion acts as an oxidising agent
1
1
(b) eMgMg 22
Oxidation number of magnesium increases from 0 to +2 magnesium undergoes oxidation
CueCu 22 oxidation number of copper in copper(II) ion decreases from +2 to 0 copper(II) ion undergoes reduction
1 1 1
1 1 1
(c) At the negative terminal: Iron(II) ion release / lose one electron and is oxidised to iron(III) ion. Fe
2+ Fe
3+ + e
The green coloured solution of iron(II) sulphate turns brown. Fe
2+ act as a reducing agent.
At the positive terminal: Bromine molecules accepts electrons and is reduced to bromide ions, Br
- Br2 + 2e 2Br
- The brown colour of bromine water turns colourless. Bromine acts as an oxidising agent
1 1 1 1 1
1 1 1 1 1
20
Question No Mark scheme Mark
5 (a) 1. Mg/Al/Fe/Pb/Zn
2. Magnesium undergoes oxidation as oxidation number of magnesium
increases from 0 to +2 and 3. Copper (II) oxide undergoes reduction as oxidation number of copper in
copper(II) oxide decreases from +2 to 0 4. Oxidation and reduction occur at the same time.
1 1 1 1
(b) Experiment I
1. Fe
2+ ion present
2. Metal X lower than iron in the Electrochemical Series // Metal X is less electropositive than iron
3. Iron atoms releases electrons to form iron(II) ions
1 1 1

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26 Perfect Score & X A –Plus Module/mark scheme 2013
Experiment II 1. OH
- ion present
2. Metal Y higher than iron in the Electrochemical Series // Metal Y is more electropositive than iron
3. Atom Y releases electrons to form Yn+
ions 4. Water and oxygen gain electron to form OH
- ion //
2H2O + O2 + 4e → 4OH-
1 1 1 1
Max 3
(c) Procedure
1. One spatula of copper(II)oxide powder and one spatula of carbon powder is
placed into a crucible 2. The crucible and its content are heated strongly
3. The reaction and the changes that occur are observed
4. Steps 1 to 3 are repeated by replacing copper(II)oxide powder with zinc oxide powder and magnesium oxide powder.
Observation
Mixture Observation
Carbon and
copper(II)oxide The mixture burns brightly. The black powder turns brown
Carbon and zinc
oxide The mixture glows dimly. The white powder turns grey.
Carbon and
magnesium oxide No Changes
1
1
1
1
1+1
Explanation Carbon can react with copper(II)oxide and zinc oxide Carbon more reactive than copper and zinc / carbon is above copper and zinc in
the Reactivity Series Carbon cannot react with magnesium oxide Carbon less reactive than magnesium / carbon is below magnesium in the Reactivity Series
1
1
1
1
20
6 Sample answer
(a) Magnesium/Aluminium/zinc/iron/lead 1 Magnesium dissolve//The blue colour of copper(II)sulphate solution become paler // brown solid deposited 1 Mg→Mg
2+ + 2e 1
Cu2+
+ 2e→ Cu 1 Oxidising agent- Cu
2+ ion / copper(II) sulphate 1
Reducing agent- Mg 1..6 (b) sample answer
Pb(NO3)2 + 2KI Pbl2 + 2KNO3 1
Oxidation number: +2 +5 -2 +1 -1 +2 -1 +1 +5 -2 1

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27 Perfect Score & X A –Plus Module/mark scheme 2013
no changes of oxidation number of all elements in the compounds of reactants and products. 1
Neutralization 1...4 (c ) sample answer
[Material : Any suitable oxidizing agent (example : acidified potassium manganate(VII) solution, acidified potassium dichromate(VI) solution, chlorine water, bromine water), any suitable reducing agent (example : potassium iodide solution, iron(II) sulphate solution) and any suitable electrolyte] 1 [ Apparatus : U-tube , carbon electrodes , connecting wires and galvanometer] 1 Diagram Functional 1 Labelled 1
Procedure 1 Sulphuric acid is added into a U-tube until 1/3 full 1 2 Bromine water is added into one end of the U-tube while potassium iodide solution is added into the other end of the U-tube 1 3 carefully 1 4 Two carbon electrodes connected by connecting wires to a galvanometer are dipped into the two solution at the two ends of the U-tube. 1 Observation The colour of bromine water change from brown to colourless// The colour of potassium iodide solution change from colourless to yellow/brown// The needle of the galvanometer is deflected 1
Oxidation reaction : Br2 + 2e→ 2Br
- 1
Reduction reaction: 2I- → I2 + 2e 1
Max : 10 20

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28 Perfect Score & X A –Plus Module/mark scheme 2013
SET 3 :ACIDS, BASES AND SALTS
Question
No Mark scheme Mark
1 (a)(i) Propanone / Methylbenzene / [any suitable organic solvent] 1
(ii) Water 1
(b)(i) Molecule 1
(ii) Ion 1
(c) 1. Beaker A : No observable change Beaker B : Gas bubbles released 2. H
+ ion does not present in beaker A but H
+ ion present in beaker B //
Hydrogen chloride in beaker A does not show acidic properties but hydrogen chloride in beaker B shows acidic properties
1
1
(d)(i) 1. Correct formula of reactants and products 2. Balanced equation
Mg + 2HCl → MgCl2 + H2
1 1
(ii) 1. Mole of HCl 2. Mole ratio 3. Answer with correct unit
Mole HCl =
// 0.005
2 mol HCl reacts with 1 mol Mg 0.005 moles HCl reacts with 0.0025 moles Mg
Mass Mg = 0.0025 x 24 // 0.06 g
1 1 1
TOTAL 10
Question
No Mark scheme Mark
2 (a)(i) Substance that ionize / dissociate in water to produce H+ ion 1
(ii) 3 1
(iii) 1. Concentration of acid / H+ ion in Set II is lower than Set I
2. The lower the concentration of H+ ion the higher the pH value
1 1
(iv) 1. Ethanoic acid is weak acid while hydrochloric acid is strong acid
2. Ethanoic acid ionises partially in water to produce low concentration of H+ ion
while
3. hydrochloric acid ionises completely in water to produce high concentration of H+
ion
1 1
1
(b)(i) 1. The pH value of sodium hydroxide in volumetric flask B is lower than A 2. Concentration of sodium hydroxide / OH
- ion in volumetric flask B
is lower than A
1
1
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29 Perfect Score & X A –Plus Module/mark scheme 2013
(ii) 1. Mole of NaOH 2. Mass of NaOH with correct unit
Mole NaOH =
// 0.005
Mass NaOH = 0.005 x 40 g // 0.2 g
1 1
(iii) 0.01 x V = 0.002 x 100 // 20 cm3
TOTAL 10
Question
No Mark scheme Mark
3 (a) Pink to colourless
1
(b) Potassium nitrate
1
(c)(i) HNO3 + KOH → KNO3 + H2O
1
(ii) 1. Mole of HNO3 // Substitution 2. Mole ratio
3. Concentration of KOH with
Mole HNO3 =
// 0.01
0.01 mole HNO3 reacts with 0.01 mole KOH
Molarity KOH =
mol dm
-3 // 0.4 mol dm
-3
1 1 1
(d)(i) 10 cm3 1
(ii) 1. Sulphuric acid is diprotic acid but nitric acid is monoprotic acid // 1 mole of
sulphuric acid produce 2 moles of H+
ion but 1 mole of nitric acid produce 1 mole of H
+ ion
2. Concentration of H+ ion in sulphuric acid is double compare to nitric acid
3. Volume of sulphuric acid needed is half
1
1 1
TOTAL 10
Question
No Mark scheme Mark
4 (a) Ionic compound formed when H+
ion from an acid is replaced by a metal ion or
ammonium ion 1
(b) Pb(NO3)2 1
(c) To ensure all the nitric acid reacts completely 1
(d)(i) 1. Correct formula of reactants and products 2. Balanced equation 2H
+ + PbO → Pb
2+ + H2O
1 1

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30 Perfect Score & X A –Plus Module/mark scheme 2013
(ii) 1. Mole of acid 2. Mole ratio 3. Answer with correct unit
Mole HNO3 =
// 0.05
0.05 moles HNO3 produce 0.025 moles salt G Mass of salt G = 0.025 x 331 g // 8.275 g
1 1 1
(e) 1. Add 2 cm3 dilute sulphuric acid followed by 2 cm
3 of Iron(II) sulphate solution
Slowly add concentrated sulphuric acid by slanted the test tube. Then turn it upright. 2. Brown ring is formed.
1 1
TOTAL
Question
No Mark scheme Mark
5 (a)(i) Salt W : Copper(II) carbonate Solid X : Copper(II) oxide
1 1
(ii) 1. Flow gas into lime water 2. Lime water turns cloudy / chalky
3.
1 1
(iii) Neutralisation
(iv) 1. Correct formula of reactants and products 2. Balanced equation CuO + 2HCl → CuCl2 + H2O
1 1
(b) Cation : Cu2+
ion // copper(II) ion Anion : Cl
- ion // chloride ion
1 1
(c)(i) Ag+ + Cl
- → AgCl
1
(ii) Double decomposition reaction
1
TOTAL
Question
No Mark scheme Mark
6 (a)(i) Green 1
(ii) Double decomposition reaction 1
(b)(i) Carbon dioxide 1
(ii) CuCO3 → CuO + CO2 1
(iii) 1. Functional apparatus 2. Label
1 1
(c)(i) Sulphuric acid // H2SO4 1
Heat
Copper(II) carbonate
Lime
water

@Hak cipta BPSBPSK/SBP/2013
31 Perfect Score & X A –Plus Module/mark scheme 2013
(ii) 1. Mole of CuCO3 2. Mole ratio 3. Answer with correct unit
Mole CuCO3 =
// 0.1
0.1 moles CuCO3 produces 0.1 mole CuO Mass CuO = 0.1 x 80 g // 8 g
1 1 1
TOTAL
7 (a) 1. Vinegar 2. Wasp sting is alkali
3. Vinegar can neutralize wasp sting
1 1 1
(b) 1. Water is present in test tube X but in test tube Y there is no water. 2. Water helps ammonia to ionise // ammonia ionise in water
3. OH- ion present
4. OH- ion causes ammonia to show its alkaline properties
5. Without water ammonia exist as molecule // without water OH- ion does not
present
6. When OH- ion does not present, ammonia cannot show its alkaline properties
1 1 1 1 1 1
(c) 1. Sulphuric acid is a diprotic acid but nitric acid is a monoprotic acid
2. 1 mole of sulphuric acid ionize in water to produce two moles of H+ ion but 1 mole
of nitric acid ionize in water to produce one mole of H+ ion
3. The concentration of H+ ion in sulphuric acid is double / higher
4. The higher the concentration of H+ ion the lower the pH value
1 1
1 1
(d)(i) 1. Mole of KOH
2. Molarity of KOH and correct unit
Mole KOH =
// 0.25
Molarity =
mol dm
-3 // 1 mol dm
-3
1 1
(ii) 1. Correct formula of reactants 2. Correct formula of products
3. Mole of KOH // Substitution 4. Mole ratio
5. Answer with correct unit
HCl + KOH → KCl + H2O
Mole KOH =
// 0.025
0.025 mole KOH produce 0.025 mole KCl Mass KCl = 0.025 x 74.5 g // 1.86 g
1 1 1 1 1
TOTAL 20

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32 Perfect Score & X A –Plus Module/mark scheme 2013
Question
No Mark scheme Mark
8 (a)(i) 1. PbCl2
2. Double decomposition reaction
1 1
(ii) Copper (II) chloride : Copper(II) oxide / copper(II) carbonate , Hydrochloric acid Lead (II) chloride : Lead (II) nitrate solution , sodium chloride solution ( any solution that contains Cl
- ion)
1 + 1
1 + 1
(b)(i) 1. S = zinc nitrate 2. T = zinc oxide
3. U = nitrogen dioxide 4. W = oxygen
1 1 1 1
(ii) 2Zn(NO3)2 2ZnO + 4NO2 + O2 1+1
(c)(i) 1. Both axes are label and have correct unit
2. Scale and size of graph is more than half of graph paper 3. All points are transferred correctly
1 1 1
(ii)
1
(iii) Mole Ba2+
ion =
// 0.0025
Mole SO4 2-
ion =
// 0.0025
Ba
2+ ion : SO4
2- ion
0.0025 : 0.0025 // 1 : 1
1
1
1
(iv) Ba2+
+ SO42-
→ BaSO4 1
TOTAL 20
Question
No Mark scheme Mark
9 (a) 1. HCl // HNO3 2. 1 mole acid ionises in water to produce 1 mole of H
+ ion
3. H2SO4
4. 1 mole acid ionises in water to produce 2 moles of H+ ion
1 1 1 1
(b) 1. Sodium hydroxide is a strong alkali 2. Ammonia is a weak alkali 3. Sodium hydroxide ionises completely in water to produce high concentration of OH
-
ion 4. Ammonia ionises partially in water to produce low concentration of OH
- ion
5. Concentration of OH- ion in sodium hydroxide is higher than in ammonia
6. The higher the concentration of OH- ion the higher the pH value
1 1 1 1 1 1
5

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33 Perfect Score & X A –Plus Module/mark scheme 2013
(c) 1. Volumetric flask used is 250 cm3
2. Mass of potassium hydroxide needed = 0.25 X 56 = 14 g 3. Weigh 14 g of KOH in a beaker 4. Add water 5. Stir until all KOH dissolve 6. Pour the solution into volumetric flask 7. Rinse beaker, glass rod and filter funnel. 8. Add water 9. when near the graduation mark, add water drop by drop until meniscus reaches the
graduation mark 10. stopper the volumetric flask and shake the solution
1 1 1 1 1 1 1 1 1
1
TOTAL 20
Question
No Mark scheme Mark
10 (a)(i) Substance C : Glacial ethanoic acid Solvent D : Propanone [ or any organic solvent]
1 1
(ii) Solution E
1. Ethanoic acid ionises in water 2. Can conduct electricity because presence of freely moving ions
3. blue litmus paper turns to red because of H+ ions is present
Solution F
4. Ethanoic acid exist as molecules
5. Cannot conduct electricity because no freely moving ion 6. Cannot change the colour of blue litmus paper because no H
+ ion
1 1 1 1 1 1
(b) 1. Measure and pour [20-100 cm3] of [0.1-2.0 mol dm
-3]zinc nitrate solution into a
beaker
2. Add [20-100 cm3] of [0.1-2.0 mol dm
-3]sodium carbonate solution
3. Stir the mixture and filter 4. Rinse the residue with distilled water
5. Zn(NO3)2 + Na2CO3ZnCO3 + 2NaNO3
6. Measure and pour [20-100cm3]of [0.1-1.0mol dm
-3]sulphuric acid into a beaker
7. Add the residue/ zinc carbonate into the acid until in excess
8. Stir the mixture and filter 9. Heat the filtrate until saturated / 1/3 of original volume
10. Cool the solution and filter
11. Dry the crystal by pressing between two filter papers 12. ZnCO3 + H2SO4 ZnSO4 + H2O + CO2
1
1 1 1 1 1 1 1 1 1 1 1
TOTAL 20

@Hak cipta BPSBPSK/SBP/2013
34 Perfect Score & X A –Plus Module/mark scheme 2013
SET 3 :RATE OF REACTION
Question
No Mark scheme Mark
1(a)(i) Set II 1
(ii) Able to draw the graph with these criterion:
1 Labelled axis with correct unit 2. Uniform scale for X and Y axis & size of the graph is at least half of the graph
paper 3. All points are transferred correctly 4. Curve is smooth.
1 1
1 1
(b)(i) Set I : 1.Tangen shown in graph correctly 2.Rate of reaction = 0.19 cm
3s
-1 ( +- 0.05)
Set II : 1.Tangen shown in graph correctly 2.Rate of reaction = 0.23 cm
3s
-1 (+- 0.05)
1 1
1 1
(ii) Add catalyst Increase the temperature Use smaller size/ metal powder Increases the concentration of acid// Double the concentration of acid but half volume
[Any two]
1 1
Question
No Mark scheme Mark
2 (a) 1. Correct formulae of reactants and product 2. Balanced equation CaCO3+ 2HNO3 → Ca(NO3) 2+ CO2 + H2O
1 1
(b)
Functional diagram
Label
1 1
(c) 1. Mole of nitric acid 2. Mole ratio 3. Answer with correct unit Number of moles of HNO3 = 0.2 X 50 = 0.01 mol // 1000 2 mol of HNO3 produce 1 mol of CO2 0.01 mol of HNO3 produce 0.005 mol of CO2
1 1 1
Nitric acid
Calcium carbonate
Water
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@Hak cipta BPSBPSK/SBP/2013
35 Perfect Score & X A –Plus Module/mark scheme 2013
Maximum volume of CO2 = 0.005 x 24 = 0.12 dm
3 // 120 cm
3
(d) Experiment I = 0.12 X 1000 // 0.2 cm3 s
-1 //
10 X 60 //0.12 //0.012 dm
3 min
-1 10 Experiment II = 0.12 X 1000 // 0.4 cm
3 s
-1 //
5 X 60 // 0.12 // 0.024 dm
3 min
-1 5
1
1
(e)(i) Rate of reaction in Experiment II is higher than I 1
(ii) - The size of calcium carbonate in Experiment II is smaller than Experiment I // calcium carbonate powder in Experiment II has a larger total surface area exposed to
collision than Experiment I. - The frequency of collision between between calcium carbonate and hydrogen ion in Experiment II is higher than Experiment I. - The frequency of effective collision s in Experiment II is higher than Experiment I
1
1
1
Question
No Mark scheme Mark
3 (a)
-Total surface area of smaller pieces wood is larger/bigger/ greater than the bigger
pieces of wood - More surface area exposed to air for burning
1 1
(b)(i)
1. Experiment II 2. Present of catalyst /manganase(IV) oxide in Experiment I
1 1
(ii) 1.Correct formulae of reactants and product 2.Balanced equation
2H2O2 → 2H2O + O2
1 1
(iii)
1. Arrow upward with energy label ,two levels and position of reactant and products are correct
2. Curve of Experiment I and experiment II are correct and label
3. Activation energy of experiment I and experiment II are shown and labelled
1 1 1
(c)(i) 1.Correct formulae of reactants and product 2.Balanced equation Zn + 2HCl ZnCl2 + H2
1 1
(ii) No. of mol HCl = 50 X 0.5 // 0.025 1000
1
Energy
2H2O2
Ea
2 H2O +
O2
Ea’

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2 mol HCl : 1 mol H2 0.025 mol HCl : 0.0125 mol H2 Volume of H2 = 0.0125 x 24 // 0.3dm
3 // 300 cm
3
1
1
(iii) 1. Add excess zinc powder with 12.5 cm3 of 1 mol dm
-3hydrochloric acid .
2. At the same temperature
OR
1. Add excess zinc powder with 25 cm3 of 0.5 mol dm
-3hydrochloric acid
2. At the higher temperature //present of catalyst
1
1
1
1
(iv) 1. Rate of reaction using sulphuric acid is higher 2. The concentration of H
+ ion in sulphuric acid is higher
3. Maximum volume of gas collected is double
4. The number of mole of H+ ion in sulphuric acid is double
1 1 1 1
20
Question
No Mark scheme Mark
4 (a) 1. Temperature in refrigerator is lower than in cabinet
2. The activity of microorganisme (bacteria) in refrigerator is lower than in
refrigerator 3. The amount of toxin produced in the refrigerator is less then in the kitchen
cabinet.
1 1
1
(b)(i) 1. Correct formula of reactants and products 2. Mol of sulphuric acid
3. Mole ratio 4. Volume and ratio
Zn + H2SO4 ------- ZnSO4 + H2 No. Of mol H2SO4 = 1 X 50/1000 // 0.05 1 mol of H2SO4 : 1 mol of H2 0.05 mol of H2SO4 : 0.05 mol of H2
Volume of H2 = 0.05 x 24 dm
3 //1.2 dm
3 //0.05 x 24000//1200 cm
3
1
1
1
1
(ii) Experiment I = 1200 // 15 cm3 s
-1 80 Experiment II = 1200 // 7.5 cm
3 s
-1 160 Experiment III = 600 // 2.5 cm
3 s
-1 240
1
1
1
(iii) Exp I and II 1.Rate of reaction of Expt I is higher 2.The size of zinc in Expt I is smaller 3.Total surface area of zinc in Expt I is bigger/larger 4.The frequency of collision between zinc atom and hydrogen ion/H
+ in Expt I is
higher 5. The frequency of effective collision in Exp I is higher
1
1 1 1 1

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Exp II and III 1. Rate of reaction in Expt II is higher 2.The concentration of sulphuric acid/ H
+ ion in Exp II is higher
3. The no. of H+ per unit volume in Expt II is higher/greater in Expt II//
4. The frequency of collision between zinc atom and H+ in Expt II is higher
5. The frequency of effective collision in Expt II is higher
1 1 1 1 1
20
Question
No Mark scheme Mark
5.(a) (i)
N2 + 3H2 ------- 2NH3 1 + 1
(ii) Temperature : 450 – 550 ˚ C Pressure : 200 – 300 atm Catalyst : Powdered iron// Iron filling
[ Any two]
1
1
(b)(i) Example of acid Sample answer : Hydrochloric acid / HCl// Sulphuric acid // Nitric acid Correct formula of reactant and product Balance Sample answer 2HCl + Mg → MgCl2 + H2
1
1
1
(ii) 1. Experiment I : 20 cm3 / 60 s // 0.33 cm
3s
-1 2. Experiment II : 20 cm
3 / 50 s // 0.4 cm
3s
-1 1
1
(iii) (Catalyst) Experiment 1:
1.Pour /measure (50-100) cm3 of (0.1-2 mol dm
-3 ) hydrochloric acid .
2.Add excess zinc powder/granules 3.Add a (2-5 cm
3 ) of copper(II) sulphate solution
4.At the same temperature Experiment II :
1. Pour /measure (50-100) cm3 of (0.1-2 mol dm
-3 ) hydrochloric acid .
2. Add excess zinc powder/granule 3. At the same temperature
OR (Temperature) Experiment 1:
1. Pour /measure (50-100) cm3 of (0.1-2 mol dm
-3 ) hydrochloric acid
2. Heat acid to (30-80OC)
3. Add excess zinc powder/granule
Experiment II :
1. Pour /measure (50-100) cm3 of (0.1-2 mol dm
-3 ) hydrochloric acid .
2. Without heating
3. Add excess zinc powder/granules
OR (Concentration) Experiment 1:
1.Pour /measure (50-100) cm3 of (0.2-2 mol dm
-3 ) hydrochloric acid .
2. Add excess zinc powder/granules 3.At the same temperature
1
1
1
1
1
1
1
1
1
1
1
1
1

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Experiment II :
1. Pour /measure (50-100) cm3 of (0.1-1 mol dm
-3 ) hydrochloric acid .
2. Add excess zinc powder/granules
3. At the same temperature
OR (Size) Experiment 1:
1.Pour /measure (50-100) cm3 of (0.1-2 mol dm
-3 ) hydrochloric acid .
2. Add excess zinc powder 3.At the same temperature
Experiment II :
1. Pour /measure (50-100) cm3 of (0.1-2 mol dm
-3 ) hydrochloric acid .
2. Add excess zinc granule 3. At the same temperature
1
1
1
1
1
1
1
1
(iv) (Catalyst) 1.Catalyst/copper(II) sulphate is used in Experiment I 2. Catalyst/(copper(II) sulphate) lower activation energy (and provide an alternative
path) 3. More colliding particles / ions are able to achieve that lower activation energy. 4.The frequency of effective collision between magnesium atoms and hydrogen ion increases. 5. The rate of reaction of Experiment I is higher.
(Any 4) (Temperature)
1. Rate of reaction in Experiment I is higher. 2. The temperature of reaction in Experiment I is higher
3. The kinetic energy of particles increases in Experiment I // The particles move faster
4. Frequency of collision between magnesium atom and H+ ion in Experiment I is
higher 5. Frequency of effective collision in Experiment I is higher
(Any 4)
(Concentration)
1. Rate of reaction in Experiment II is higher 2. The concentration of acid in Experiment I is higher
3. The number of hydrogen ion per unit volume in Experiment II is higher
4. Frequency of collision between magnesium atom and H+ ion in Experiment I is higher
5. Frequency of effective collision in Experiment II is higher (Any 4) (Size) 1.Rate of reaction in Experiment I is higher 2.The size of magnesium in Experiment I is smaller 3.Total surface area of magnesium in Experiment I is bigger/larger 4.The frequency of collision between magnesium atoms and hydrogen ions in Experiment I higher 5.The frequency of effective collision between in Experiment I is higher (Any 4)
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
(v) The number of mol are same // The concentration and volume of acid are same 1

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Question
No Mark scheme Mark
6.(a) (i)
1. First minute = 24/60 =0.4 cm3 s
-1 // 24 cm
3 min
-1
2. 2 nd
minute = 34-24/60 =0.167 cm3 s
-1 // 10 cm
3 min
-1
1
1
(ii) 3. rate in 1 st minute higher than 2
nd minute (vice versa)
4. concentration of sulphuric acid / mass of zinc decreases 1
1
(iii) All hydrogen ion from acid was completely reacts 1
(iv) A catalyst lower activation energy provide an alternative path More colliding particles /zinc atoms and hydrogen ions are able to overcome the lower
activation energy. The frequency of effective collisions between zinc atom and hydrogen ion in is higher.
(any 2 )
1
1
(b) - hydrogen and oxygen molecules collide - with correct orientation -total energy of particles higher or equal to activation /minimum energy
1 1 1
(Temperature)
Materials: 0.2 mol dm
-3 sodium thiosulphate, 1.0 mol dm
-3 sulphuric acid, a piece of white paper
marked ‘X’ at the centre. Apparatus: 150 cm
3 conical flask, stopwatch, 50 cm
3 measuring cylinder, 10 cm
3 measuring
cylinder, thermometer, Bunsen burner, wire gauze.
Procedure:
1.Using a measuring cylinder, 50 cm
3 of 0.2 mol dm
-3 sodium thiosulphate solution is
measured and poured into a conical flask.
2.The conical flask is placed on top of a piece of white paper marked ‘X’ at the centre. 3.5 cm
3 of 1.0 mol dm
-3 sulphuric acid is measured using another measuring cylinder.
4.The sulphuric acid is poured immediately and carefully into the conical flask. At the
same time, the stop watch is started 5.The mixture in a conical flask is swirled.
6.The ‘X’ mark is observed vertically from the top of the conical flask through the
solution. 7.The stopwatch is stopped once the ‘X’ mark disappears from view.
8.Step 1 – 7 are repeated using 50 cm
3 of 0.2 mol dm
-3 sodium thiosulphate solution at
40oC, 50
oC, 60
oC by heating the solution before 5 cm
3 of sulphuric acid is added in.
(Max 7) Conclusion When the temperature of sodium thiosulphate solution is higher , the rate of reaction is
higher
1
1
1
1
1
1
1
1
1
1
1

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(Temperature)
Materials: 0.2 mol dm
-3 sodium thiosulphate, 1.0 mol dm
-3 sulphuric acid, water, a piece of white
paper marked ‘X’ at the centre.
Apparatus: 150 cm
3 conical flask, stopwatch, 50 cm
3 measuring cylinder, 10 cm
3 measuring
cylinder, wire gauze.
Procedure:
1.Using a measuring cylinder, 50 cm
3 of 0.2 mol dm
-3 sodium thiosulphate solution is
measured and poured into a conical flask.
2.The conical flask is placed on top of a piece of white paper marked ‘X’ at the centre.
3.5 cm
3 of 1.0 mol dm
-3 sulphuric acid is measured using another measuring cylinder.
4.The sulphuric acid is poured immediately and carefully into the conical flask. At the same time, the stop watch is atarted
5.The mixture in a conical flask is swirled. 6.The ‘X’ mark is observed vertically from the top of the conical flask through the
solution. 7.The stopwatch is stopped once the ‘X’ mark disappears from view. 8.Step 1 – 7 are repeated by adding 5 cm
3, 10 cm
3, 15 cm
3, 20 cm
3 and 40 cm
3 of
distilled water .(at the same time) maintaining the total volume of solution at 50 cm3
after dilution//table of dilution (Max 7)
Conclusion When the temperature of sodium thiosulphate solution is higher , the rate of reaction is
higher
1
1
1
1
1
1
1
1
1
1
1
SET 3 :THERMOCHEMISTRY
Question No
Mark scheme Mark
1 (a) Heat change /released when 1 mol copper is displaced from copper (II)
sulphate solution by zinc 1
(b) Blue to colourless 1
(c) (i) 50 X 4.2 X 6 J // 1260 J 1
(ii) (1.0 )(50) 1000
1
(iii) 1260 0.05
1
// 0.05
J // 25200 J mol-1

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= - 25.2 kJ mol-1 1
(d) 1. Correct reactant and product 2. Correct two energy level for exothermic reaction 3. Correct value heat of displacement and unit
Sample answer Energy
1 1 1
(e) (i) 3°C 1
(ii)
Number of mole copper displaced is half Heat released is half / 1260 2
1 1
TOTAL 12
Question No Mark scheme Mark
2 (a)
Heat of precipitation is the heat change when one mole of a precipitate is
formed from its solution. 1
(b)
To reduce heat loss to the surrounding. Reject : prevent
1
(c) Ag+
+ Cl- → AgCl 1
(d) (i) The heat released =(50 + 50) x 4.2 x 3.5 =1470 J
1
(ii) Number of moles of Ag+
= (50 x 0.5) = 0.025 mol 1000 Number of moles of Cl
- = (50 x 0.5) = 0.025 mol 1000
1
1
(iii)
0.025 mole of Ag+ reacts with 0.025 mole of Cl
- to form 0.025 mole of
AgCl Number of moles of AgCl = 0.025 mol
1
(iv) =
x 1470 J
=58 800 J Heat of precipitation of AgCl = -58.8 kJ mol
-1
1
1
(e)
(i)
Ag+ + Cl
-→AgCl ∆H = -58.8kJmol
-1 // AgNO3 + NaCl →AgCl + NaNO3 ∆H = -58.8kJmol
-1 1
J // 630 J
Zn + CuSO4 //Zn + Cu2+
∆H = - 25.2 kJmol-1
ZnSO4 + Cu //Zn2+
+ Cu

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(ii)
1. Label axes 2. Energy levels of reactants and products correct with formula of
reactants and products 3. Heat of precipitation written
1 1
1
Total
Question No Mark scheme Mark
3. (a) (i) Ethanol
1
(ii) 1260 kJ of heat energy is released when one mole of ethanol is burnt
completely in excess oxygen
1
(b)
(i) No of moles of alcohol = 0.23 / 46 = 0.005 mol
1 mol of alcohol burnt released 1260 kJ
Thus, 0.005 mol of alcohol burnt released 6.3 kJ
1
1
(ii) mc = 6.3 kJ
mc = 6.3 x 1000
= 6300/ 200 x 4.2
= 7.5 0 C
1
1
( c) Heat is lost to the surrounding // Heat is absorbed by the apparatus or
containers // Incomplete combustion of alcohol
1
(d)
(i)
1. Label axes 2. Energy levels of reactants and products correct with formula of reactants
and products
3. Heat of combustion written
1
1
1
C2 H5 O H + 3 O2
2 CO2 + H2 O
∆ H = - 1260 kJmol-1
Ag+ + Cl
-
∆H = -58.8kJmol-1
AgCl
Energy
Energy

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(ii)
1. Label
2. Functional
1
1
(e) (i) - 2656 kJmol
-1 // 2500-2700 kJmol
-1
1
(ii) 1. The molecular size/number of carbon atom per molecule propanol is
bigger/higher methanol
2. Combustion of propanol produce more carbon dioxide and water molecules
3. More heat is released during formation of carbon dioxide and water
molecules
1
1
1
Total marks
Question No Mark scheme Mark
4 (a) (i) Characteristic Diagram 4.1 Diagram 4.2 Change in
temperature Increase Decrease
Type of chemical
reaction
Exothermic reaction
Endothermic reaction
Energy content of reactants
and products
The total energy content of the reactants more than
the energy content of the
products
The total energy content of the reactants less than the
energy content of the
products
Amount of
heat absorbed
/realeased during
breaking of
bonds
Amount of heat absorbed
for the breaking of bond in
the reactant is less than heat released during
formation of bond in the
products
Amount of heat absorbed for
the breaking of bond in the
reactant is more than heat released during formation of
bond in the products
1 1
1+1
1+1
(ii) Number of moles of FeSO4 = MV 1000 = (0.2)(50) = 0.01 mol 1000 Heat change = 0.01 x 200 kJ = 2 kJ // 2000 J Heat change = mcθ θ = 2000 (50)(4.2) θ = 9.5
oC
1
1
1

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(b) 1. Number of mole of Ag+ ion in both experiment
= 25 x 0.5 // 0.0125 mol
1000
2. Number of mole of Cl- ion in both experiment
= 25 x 0.5 // 0.0125 mol
1000
3. Number of mole of silver chloride formed is the same
4. Na+
ion and K
+ ion not involved in the reaction // Ag
+ ion
and Cl
- involved in the
reaction
1
1
1 1
(c) (i) Heat change = mcθ = (100)(4.2)(42.2 – 30.2) = 5040 J / 5.04 kJ
Number of moles of HCl / H
+ ion = (50)(2 = 0.1 mol
1000 Number of moles of NaOH / OH
- ion = (50)(2) = 0.1 mol
1000 The heat of neutralization = 5.04 0.1 ΔH = - 50.4 kJ mol
-1
1
1
1 1
(ii) Temperature change is 12.0 oC // same
Number of moles of sodium hydroxide reacted when hydrochloric acid or
sulphuric acid is used is the same // 0.01 mol Number of mole of water formed when hydrochloric acid or sulphuric acid used
is the same // 0.01 mol H
+ ion in excess when sulphuric acid is used
1
1
1 1
Total marks 20
Question No Mark scheme Mark
5 (a) (i) Neutralisation//Exothermic reaction 1
(ii) Total energy content of reactant is higher than total energy content in
product 1
(iii) 1. The heat of neutralization of Experiment 1 is higher than Experiment 2
2. HCl is strong acid while ethanoic acid is weak acid 3. HCl ionises completely in water to produce high concentration of H
+ ion
4. CH3COOH ionizes partially in water to produce low concentration of H+
ion and most of ethanoic acid exist as molecules
5. In Expt 2,Some of heat given out during neutralization reaction is used to
dissociate the ethanoic acid molecules completely in water//part of heat that is released is used to break the bonds in the molecules of ethanoic
acid that has not been ionised
1 1 1
1
1
(b)
(i) No of mol acid/alkali= 50 X 1 /1000= 0.05 Q = ∆ H X no of mol = 57.3 X 0.05 = 2.865 kJ // 2865 J
1
1 1
(ii) 2865 = 100 X 4.2 X 0
θ = 2865 ÷ 420 1 1

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= 6.8 oC ( correct unit) 1
(iii) 1. Some of heat is lost to the sorrounding
2. Heat is absorbed by polystyrene cup
1 1
(c )
A B The reaction is exothermic// Heat is released to the surrounding during the
reaction
The reaction is endothermic// Heat is absorbed from the surrounding
during the reaction
Heat released is x kJ when 1 mol
product is formed Heat absorbed is y kJ when 1 mol
product is formed. The total energy content in reactant is
higher than total energy content in
product
The total energy content in
reactant is lower than total energy
content in product The temperature increases during the reaction
The temperature decreases during the reaaction
Heat released during the formation of bond in product is higher than heat
absorbed during the breaking of bond
in reactant
Heat absorbed during the breaking of bond in reactant is higher than
heat released during the formation
of bond in product
1
1
1
1
1
TOTAL 20
6
(a) (i)
1. Y-axes : energy
2. Two different level of energy
1 1
(ii) 1. reactants have more energy // products have less energy 2.energy is released during the experiment // this is exothermic reaction
1 1
(b) No. of mol of H+ ion/OH
- = 1x50/1000// 0.05
Heat change = 100x 4.2 x7//2940 Joule//2.94 kJ Heat of neutralization= -2940/0.05 = -58800 J mol
-1//-58.8 kJ mol
-1
1 1 1 1
(c) 1. Heat of combustion of propane is higher 2. The molecular size/number of carbon atom per molecule propane is
bigger/higher 3. Produce more carbon dioxide and water molecules//released more heat energy
1 1 1
1. Methanol/ethanol/ propanol, Diagram: 2. -labelled diagram 3. -arrangement of apparatus is functional
1
1
1..3
energy
Zn + CuSO4
ZnSO4 + Cu
∆H = -152 kJmol-1

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1. (100-250 cm
3 )of water is measured and poured into a copper can and the
copper can is placed on a tripod stand 2. the initial temperature of the water is measured and recorded 3. a spirit lamp with ethanol is weighed and its mass is recorded 4. the lamp is then placed under the copper can and the wick of the lamp is lighted
up immediately 5. the water in the can is stirred continuously until the temperature of the water
increases by about 30oC.
6. the flame is put off and the highest temperature reached by the water is
recorded 7. The lamp and its content is weighed and the mass is recorded
…. 8 max 4 Data
The highest temperature of water = t2 The initial temperature of water = t1 Increase in temperature, = t2 - t1 =
Mass of lamp after burning = m2
Mass of lamp before burning = m1
Mass of lamp ethanol burnt, m = m1 - m2 = m …..1
Calculation : Number of mole of ethanol, C2H5OH, n = m
46 ……1 The heat energy given out during combustion by ethanol = the heat energy absorbed
by water= 100x x c x J Heat of combustion of ethanol = m c KJ mol
-1
n
= -p kJ/mol …1
..4
..3
Total marks 20

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Question No Mark scheme Mark 7 (a) (i) Heat change = mc = (25+25)(4.2)(33-29) = 445 J
Heat of precipitation of AgCl = - 445 / 0.0125 = -35600 J mol
-1 // 35.6 kJ mol
-1
1. The position and name /formulae of reactants and products are correct. 2. Label for the energy axis and arrow for two levels are shown.
1
1
1 1
(b) (i)
(ii)
1. HCl is a strong acid // CH3COOH is a weak acid. 2. HCl ionised completely in water to produce higher concentration of H
+
ion. // 3. CH3COOH ionised partially in water to produce lower concentration of
H+ ion.
4. during neutralisation reaction, some of the heat released are absorbed by
CH3COOH molecules to dissociate further in the molecules.
1. H2SO4 is a diprotic acid// HCl is a monoprotic acid. 2. H2SO4 produced two moles of hydrogen ion/H
+ when one mole of the acid
ionised in water // 3. HCl produced one mole of hydrogen ion/ H
+ when one mole of the acid
ionised in water. 4. When one mole of OH
- reacts with two moles of H
+ will produce one
mole of water, the heat of neutralisation is still the same as Experiment I because the definition of heat of neutralisation is based on the formation
of one mole of water.
4Max
3
4Max
3
(c) - apparatus and material : 2 marks - procedures : 5 marks - Table : 1 mark - Calculation : 2 marks
Sample answer: Apparatus : Polystyrene cup, thermometer, measuring cylinder. Materials : Copper (II) sulphate, CuSO4 solution, zinc powder.
Procedures :
1. Measure 25 cm3 of 0.2 mol dm
-3 copper (II) sulphate, CuSO4 solution and pour it
into a polystyrene cup.
2. Put the thermometer in the polystyrene cup and record the initial temperature of
the solution. 3. Add half a spatula of zinc powder quickly and carefully into the polystyrene cup.
4. Stir the reaction mixture with the thermometer to mix the reactants.
5. Record the highest temperature reached.
1 1
1 1 1 1 1
Energy
AgNO3 + NaCl
AgCl + NaNO3*
H = -35.6 kJ mol-1
* Accept ionic equation

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SET 4 :CARBON COMPOUNDS
Question No Mark scheme Mark 1 (a)
Or
1
(b) C3H7OH + 9/2O2 3CO2 + 4H2O 1
(c) (i) Sweet/ pleasant smell /// fruity smell 1
(ii) Methanoic acid 1
(iii)
1+1
(d) (i) Oxidation 1
(ii) Orange colour of acidified potassium dichromate (VI) solution turns green 1
(iii) C3H7OH + 2[O] C2H5COOH + H2O 1
(e) C3H7OH C3H6 + H2O
(ii)
1+1
Tabulation of data: Initial temperature of CuSO4 solution (
oC) 1
Highest temperature of the reaction mixture (oC) 2
Temperature change (oC) 2 - 1
....1 Calculation : Number of mole of CuSO 4 = MV/1000 = (0.2)(25)/1000 = 0.005 mol ……1
Heat change = mc(2 - 1) = x J Heat of displacement = x / 0.005 kJ mol
-1 = y kJ mol
-1 …….1
TOTAL 20
propene
propanol
O H H H
H C O C C C H
H H H
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Question No Mark scheme Mark 2 (a) (i) Fermentation 1
(ii) Ethanol 1
(iii)
1
(b) C2H5OH + 3O2 → 2CO2 + 3H2O 1+1
(c) (i) Ethene 1
(ii)
1
(d) Purple to colourless 1
(e) (i) Ethyl ethanoate 1
(ii) CH3COOH + C2H5OH CH3COOC2H5 + H2O 1+1
Question
No Mark scheme Mark
3 (a) Characteristics Explanation
Same general formula CnH2n + 1OH
successive member is different from each other by – CH2
Relative atomic mass is different by 14
Gradual change in physical
properties // Melting / boiling point increase
Number of carbon atom per
molecules increase // size of molecule increase
Similar chemical properties // oxidation produce carboxylic acid
Have same chemical/similar functional group
Can be prepared by similar method // can be prepared by hydration of
alkene
Have same chemical properties // have same functional group
1+1
1+1
1+1
1+1
1+1
(b) (i) (CH2O)n = 60 (12 + 2 + 16)n = 60 n = 2
1
C2H4O2 1
(ii) Carboxylic acid 1
React with carbonate to produce carbon dioxide 1
OH
C
H
H C H
H
H
H H
৷ ৷
C - C
৷ ৷
H H n

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(iii) 2 CH3COOH + CaCO3 → (CH3COO)2Ca + H2O + CO2
Correct formula of reactants and products Balanced equation
1 1
(c) Compound P Q The number of carbon atom 2 2
The number of hydrogen atom 4 6 number of hydrogen atom Q is higher
Type of covalent bond
between // carbon/ Type of
hydrocarbon
Double bond / /
Unsaturated Single bond/ /
Saturated
Type of homologous series // // Name of compound
Alkene// Ethene //
Alkane // Ethane
General formula// Molecular formula of the
compound
CnH2n // C2H4
CnH2n+2 // C2H6
1 1
1
1
1
Max 4
20
Question No Mark scheme Mark 4 (a) (i) 14.3 % 1
(ii)
Element C H
Mass/ % 85.7 14.3
No. of moles
12
7.85 = 7.14
1
3.14 = 14.3
Ratio of moles/ Simplest ratio 14.7
14.7= 1
14.7
3.14= 2
Empirical formula = CH2
RMM of (CH2)n = 56 .............1
[(12 + 1(2)]n = 56
14n = 56
n = 14
56
= 4 ………..1 Molecular formula : C4H8 ………………..1
1
1
1
6 max
5
(iii)
[any 2]
1+1
1+1
Max 4
But-1-ene
But-2-ene
2-methylpropene

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(iv) Compound M (Butene, C4H8) has a higher percentage of carbon atom in their
molecule than butane, C4H10 …………….1
% of C in C4H8 = 8)12(4
)12(4
x 100%
= 56
48 x 100%
= 85.7% …………1
% of C in C4H10 = 10)12(4
)12(4
x 100%
= 58
48 x 100%
= 82.7% ………..1
.....3
(b) (i) Starch Protein / natural silk
1 1
(ii) H H CH3 H I I I I C = C – C = C I I H H
2-methylbut-1,3-diene or isoprene
1 1..2
(c) (i) Rubber that has been treated with sulphur 1
(ii)
In vulcanised rubber sulphur atoms form cross-links between the rubber molecules These prevent rubber molecules from sliding too much when stretched
1
1
TOTAL 20
Question No Mark scheme Mark
5 (a) (i)
Hydrocarbon Type of
bond Homologous
series General
formula
A covalent alkane CnH2n+2
B covalent alkene CnH2n
3
3
(ii) Carbon dioxide 2C4H10 + 13O2 → 8CO2 + 10H2O [Chemical formulae of reactants and products] [Balanced]
1
1 1
(iii) Hydrocarbon B. Hydrocarbon B is an unsaturated hydrocarbon which react with bromine. Hydrocarbon A is a saturated hydrocarbon which do not react with bromine.
1
1
1

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(iv) Hydrocarbon B more sootiness. B has higher percentage of carbon by mass. % of carbon by mass ; Hydrocarbon A : 4(12) × 100 // 82.76 % 4(12) + 10(1)
Hydrocarbon B : 4(12) × 100 // 85.71 % 4(12) + 8(1)
1 1
1
1
(b) Carboxylic acid X :
Propanoic acid Alcohol Y:
Ethanol
1
1
1
1
TOTAL 20
Question No Mark scheme Mark 6 (a) (i) X - any acid – methanoic acid
Y - any alkali – ammonia aqueous solution
1 1
(ii) 1. Methanoic acid contains hydrogen ions 2. Hydrogen ions neutralise the negative charges of protein membrane 3. Rubber particles collide, 4. Protein membrane breaks 5. Rubber polymers combine together
1 1 1 1 1
5 max 4
(iii) Ammonia aqueous solution contains hydroxide ions Hydroxide ions neutralise hydrogen ions (acid) produced by activities of bacteria
1
1
(b) (i) Alcohol 1
(ii) Burns in oxygen to form carbon dioxide and water Oxidised by oxidising agent (acidified potassium dichromate (VI) solution) to
form carboxylic acid
1 1
(iii) Procedure: 1. Place glass wool in a boiling tube
2. Soak the glass wool with 2 cm3 of ethanol
3. Place pieces of porous pot chips in the boiling tube
4. Heat the porous pot chips strongly
5. Heat glass wool gently

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6. Using test tube collect the gas given off
Diagram:
[Functional diagram] [Labeled – porcelain chips, water, named alcohol, heat] Test: Add a few drops of bromine water Brown colour of bromine water decolourised
6 max
5
1 1
1 1
Total 20
Question
No Mark scheme Mark
7
(a) Carbon dioxide/ CO2 and water/ H2O Any one correct chemical equation Example 2C4H10 + 13O2 → 8CO2 + 10H2O Chemical formula of reactants Balanced
1
1 1
(b) Compound B & Compound D Same molecular formula / C4H8 Different structural formula
1 1 1
(c) Pour compound A/B into a test tube Add bromine water to the test tube and shake Test tube contain compound A unchanged Test tube contain compound B brown colour turn colourless
or Pour compound A/B into a test tube Add acidified Potassium manganate(VII) solution to the test tube and shake Test tube contain compound A unchanged Test tube contain compound B purple colour turn colourless
1 1 1 1
(d)
(i) Any members of carboxylic acid and correct ester Example [Methanoic acid] [Propylmethanoate]
1 1
1
1
Heat Heat
Glass wool
soaked with ethanol
Porcelain chips
Water

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(d)
(ii) Pour 2 cm
3 of [methanoic acid] into a boiling tube
Add 2 cm3 of propanol/compound E into the boiling tube
Slowly/carefully/drop 1 cm3 of concentrated sulphuric acid
Heat the mixture gently Pour the mixture in a beaker that contain water Observation : Colorless liquid with fruity smell is formed / Colorless liquid float on
water surface
1 1 1 1 1 1
TOTAL 20
Question
No Mark scheme Mark
8(a)
But-2-ene
2-methylpropene
1+1
1+1
(b)
(i)
(ii)
Propanoic acid Ethanol Chemical properties for propanoic acid:
1. React with reactive metal to produce salt and hydrogen gas 2. React with bases/alkali to produce salt and water
3. React with carbonates metal to produce salt, carbon dioxide gas and water 4. React with alcohol to produce ester
[any three]
Chemical properties for ethanol: 1. Undergo combustion to produce carbon dioxide and water
2. Burnt in excess oxygen to produce CO2 and H2O 3. Undergo oxidation to produce carboxylic acid / ethanoic acid
4. React with acidified K2Cr2O7 /KMnO4 to produce carboxylic acid / ethanoic acid
5. Undergo dehydration to produce alkene / ethene. [Any three answers]
1 1
1 1 1 1 1
1 1 1 1
1
(c) (i) P : Hexane Q : Hexene // Hex-1-ene
(ii) Reaction with bromine // acidified potassium manganate(VII) solution
Procedure:
1 1
1
1
C C C C H
H H H H
H
H H
C
C C C
H
H
H
H
H
H
H
H

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1. Pour about [2 -5 cm3] of P into a test tube.
2. Add 4-5 drops of bromine water / acidified potassium manganate(VII) solution
and shake.
3. Observe and record any changes. 4. Repeat steps 1 to 3 by replacing P with Q
Observation: P : Brown/ Purple colour remains unchanged. Q : Brown/ Purple colours decolourise / turn colourless.
1
1 1
1 1
Max 6
20
SET 4 :MANUFACTURED SUBSTANCE IN INDUSTRY
Question No Mark scheme Mark 1
(a) (i)
Contact process
1
(ii)
Ammonia 1
(iii) Vanadium(V) oxide, 450 oC - 500
oC 1
(iv)
Ammonium sulphate 1
(v) 2NH3 + H2SO4 (NH4)2SO4 1+1
(b) (i) Composite material 1
(ii) Correct arrangement Correct label
1 1
(iii)
nC2H3Cl --( C2H3Cl )n 1
(iv)
It has low thermal expansion coefficient // resistant to thermal shock 1
TOTAL 11
Question No Mark scheme Mark
2 (a) (i)
(ii)
SO2 + H2O H2SO3
Corrodes buildings
Corrodes metal structures
pH of the soil decreases
Lakes and rivers become acidic [Able to state any three items correctly]
1
3 4
Tin atom
Copper atom
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56 Perfect Score & X A –Plus Module/mark scheme 2013
(b) (i)
(ii) (iii)
Oleum
2SO2 + O2 2SO3
Moles of sulphur = 48 / 32 =1.5
Moles of SO2 = moles of sulphur
= 1.5
Volume of SO2 = 1.5 24 dm3
= 36 dm3
1 1 1
1 1 1 6
(c) (i) Pure metal are made up of same type of atoms and are of the same size.
The atoms are arranged in an orderly manner.
The layer of atoms can slide over each other.
Thus, pure copper are ductile.
There are empty spaces in between the atoms.
When a pure copper is knocked, atoms slide.
Thus, pure copper are malleable.
1
1 1
1 1 1 1 Max:5
(ii) Zinc.
Zinc atoms are of different size,
The presence of zinc atoms distrupt the orderly arrangement of copper atoms.
This reduce the layer of atoms from sliding.
Arrangement of atoms – 1; Label - 1
1 1 1 1
1
1 Max: 5
Total 20
Question No Mark scheme Mark 3 (a) Haber process
Iron
N2 + 3H2 2NH3
1 1
1+1
(b) Pure copper Bronze
Bronze is harder than pure copper
Tin atoms are of different size
The presence of tin atoms distrupt the orderly arrangement of copper
1
1+1
1 1
Zinc atom
Copper atom
Tin atom
Copper atom

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57 Perfect Score & X A –Plus Module/mark scheme 2013
atoms.
This reduce the layer of atoms from sliding.
1
1
MAX 6
Procedure: 1. Iron nail and steel nail are cleaned using sandpaper.
2. Iron nail is placed into test tube A and steel nail is placed into test tube
B. 3. Pour the agar-agar solution mixed with potassium
hexacyanoferrate(III) solution into test tubes A and B until it covers
the nails. 4. Leave for 1 day.
5. Both test tubes are observed to determine whether there is any blue spots formed or if there are any changes on the nails.
6. The observations are recorded
Results:
Test tube The intensity of blue spots A High B Low
Conclusion: Iron rust faster than steel.
1 1+ 1
1
1 1
1 1
1
TOTAL 20
SET 4 :CHEMICALS FOR CONSUMERS
Question No Mark scheme Mark 1 (a) (i) To improve the colour of food
1
(ii) Absorbs water /inhibits the growth of microorganisms 1
(iii)
1. Preservative
2. Flavouring 1 1
(b) (i) Analgesic 1
(ii) To relieve pain 1
(c) (i) Saponification // alkaline hydrolysis 1
(ii)
Hydrophobic hydrophilic
1+1
(iii) Soap form scum/insoluble salts in hard water. 1
TOTAL 10

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58 Perfect Score & X A –Plus Module/mark scheme 2013
Question No Mark scheme Mark
2 (a) Examples of food preservatives and their functions:
Sodium nitrite – slow down the growth of microorganisms in meat
Vinegar – provide an acidic condition that inhibits the growth of microorganisms in pickled foods
1+1
1+1
(b) (i) No // cannot Because aspirin can cause brain and liver damage if given to children with
flu or chicken pox. // It causes internal bleeding and ulceration
1 1
(ii) Paracetamol
Codeine 1 1
(iii) 1. If the child is given a overdose of codeine, it may lead to addition. 2. If the child is given paracetamol on a regular basis for a long time, it
may cause skin rashes/ blood disorders /acute inflammation of the
pancreas.
1
1
(c)
Type of food
additives Examples Function
Preservatives Sugar, salt To slow down the growth
of microorganisms Flavourings Monosodium
glutamate, spice,
garlic
To improve and enhance the taste of food
Antioxidants Ascorbic acid To prevent oxidation of
food Dyes/ Colourings Tartrazine
Turmeric To add or restore the
colour in food Disadvantages of any two food additives: Sugar – eating too much can cause obesity, tooth decay and diabetes Salt – may cause high blood pressure, heart attack and stroke. Tartrazine – can worsen the condition of asthma patients
- May cause children to be hyperactive MSG – can cause difficult in breathing, headaches and vomiting.
2
2
2
2
1 1
TOTAL 20
Question No Mark scheme Mark 3 (a) (i) Traditional medicines are derived from plants or animals.
Modern medicines are made by scientists in laboratory and based on
substances found in nature.
1 1
(ii) Type Modern medicine
Analgesics Aspirin Paracetamol Codein
Antibiotics Penicillin Psychotherapeutic Chloropromazin
Caffeina
1 1 1 1
1 1
MAX
5
(iii) Penicillin Cause allergic reaction, diarrhoea, difficulty breathing and easily bruising
1

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Codeine Cause addiction, drowsiness, trouble sleeping, irregular heartbeat and
hallucinations. Aspirin Cause brain and liver damage if given to children with flu or chicken pox. Cause internal bleeding and ulceration
1
1
(b) Hard water contains calcium ions and magnesium ions. Example : sea water
Procedure 1. 20cm
3 of hard water (magnesium sulphate solution) is poured into two
separate beakers X and Y.
2. 50 cm3 of soap and detergent solutions are added separately in beaker X
and beaker Y.
3. A small piece of cloth with oily stains is dipped into each beaker. 4. Each cloth is washed.
5. The cleansing action of the soap and detergent is observed.
Results
Beaker Observation X The cloth is still dirty.
Y The cloth becomes clean.
Conclusion The cleansing action of detergent is more effective than soap in hard water
1 1
1
1 1 1 1
1 1
1

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60 Perfect Score & X A –Plus Module/mark scheme 2013
SET 5 :PAPER 3 SET 1
Rubric Score
1(a)(i) Able to give correct observation Sample answer: Colourless solution formed//Aluminium oxide powder dissolved in nitric acid/sodium hydroxide solution.
3
Rubric Score
1(a)(ii) Able to give the correct inference.
Sample answer Aluminium oxide is soluble in nitric acid/sodium hydroxide solution//Aluminium oxide shows basic/acidic properties
3
Rubric Score
1(a) (iii) Able to give the correct property of aluminium oxide.
Answer: amphoteric
3
Rubric Score
1(b) Able to state the hypothesis correctly. Sample answer: When aluminium oxide dissolves in nitric acid, it shows basic properties,
when aluminium oxide dissolves in sodium hydroxide solution, shows
acidic properties.
3
Rubric Score
1(c) Able to state all the variables correctly.
Answer: Manipulated variable: type of solutions // nitric acid and sodium hydroxide solution Responding variable: solubility of aluminium oxide in acid and alkali//property of aluminium oxide Fixed variable: aluminium oxide
3
Rubric Score
1(d) Able to state the operational definition correctly.
Sample answer. When aluminium oxide solid is added into sodium hydroxide solution, the
solid dissolved.
3
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61 Perfect Score & X A –Plus Module/mark scheme 2013
Rubric Score
1(e)(i) Able to give the correct observations for both experiments. Red litmus paper turns blue Blue litmus paper turns red
3
Rubric Score
1(e)(ii) Able to classify all the oxides correctly. Acidic oxide Basic axide Carbon dioxide Phosphorous pentoxide
Magnesium oxide Calcium oxide
3
Rubric Score
2(a) Able to state the observation Sample Answer: 1. Iron glowed brightly 2. Iron ignited rapidly with bright flame. 3. Iron glowed dimly
3
Rubric Score
2(b) Able to state the observation and the way on how to control variable
Sample Answer : 1. change bromine with chlorine and iodine 2. Ignition or glowing of halogen 3. Use the same quantity of iron wool in each experiment.
3
Rubric Score
2(c) Able to state the correct hypothesis by relating the manipulated variable
and responding variable Sample Answer : 1. The higher the position of halogen in group 17 the higher the reactivity towards iron. 2. The higher the position of halogen in group 17 the greater the ignition
or glowing reaction with iron.
Rubric Score
2(d) Able to state the inference correctly. Sample answer: The solid of Iron(lll) bromide formed//Bromine combined with iron //Iron is oxidized by bromine//Bromine is reduced by iron
3
Rubric Score
2(e) Able to arrange the three position of halogen based on the reactivity
toward iron in ascending order Answer : Iodine. Bromine, Chlorine,
3
Rubric Score
3(a) Able to give the correct arrangement of the metals Answer: Magnesium, Y, copper
3

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62 Perfect Score & X A –Plus Module/mark scheme 2013
Rubric Score
3(b) Able to give the name of metal Y correctly.
Answer: Zinc//Iron//Lead
3
Rubric Score
3 (c) Able to give the three observations correctly. Answer:
1. Brown solid deposited
2. Blue solution turns light blue
3. Zinc strip becomes pale blue.
3
Rubric Score
4(a)
Able to give the problem statement correctly.
Sample answer: How is the effect of other metals on the rusting of iron when the metals
are in contact with iron.
3
Rubric Score
4(b) Able to state the three variables correctly.
Answer: Manipulated variable: Type of metals//Zinc and copper Responding variable: Rusting of iron Fixed variable: iron nail
3
Rubric Score
4(c) Able to state the hypothesis correctly.
Sample answer: When iron is in contact with a more electropositive metal/zinc, rusting
will not occur, when iron is in contact with less electropositive
metal/copper, rusting will occur.
3
Rubric Score
4(d) Able to list the apparatus and materials needed for the experiment. Apparatus: two test tubes, test-tube rack, Materials: hot agar-agar solution added with phenolphthalein and
potassium hexacyanoferrate(III) solution, iron nails, zinc strip, copper strip, sand paper.
3
Rubric Score
4(e) Able to give the procedures correctly
Sample answer:
1. Clean 2 pieces of iron nails, zinc strip and copper strip with sand
paper. 2. Coil the iron nails with zinc strip and copper strip each.
3. Put the iron nails into two different test tubes
4. Pour hot agar into each test tube until the iron nail is immersed. 5. Leave the apparatus for about 1 day and record the observations.
3

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63 Perfect Score & X A –Plus Module/mark scheme 2013
Rubric Score
4(f) Able to tabulate the data correctly
Answer:
Experiment Observation Iron nail coiled with zinc Iron nail coiled with copper
2
PAPER 3 SET 2
Rubric Score
1(a) Able to construct the table correctly with the following aspects:
Experiment Ammeter reading/A I 0.0 II 0.5 III 0.0
3
Rubric Score
1(b) Able to state the inference correctly.
Sample answer: Lead(II) bromide can conduct electricity in molten state//Naphthalene/Glucose
cannot conduct electricity in molten state
3
Rubric Score
1(c) Able to state the type of compound correctly
Answer: ionic compound
3
Rubric Score
1(d) Able to state all the three variables correctly:
Answer: Manipulated variable: type of compound Responding variable: ammeter reading//conductivity of electricity Fixed variable: state of compound//ammeter
3
Rubric Score
1(e) Able to state the hypothesis correctly.
Sample answer: Molten ionic compound can conduct electricity but molten covalent compound
cannot conduct electricity.
3
Rubric Score
1(f) Able to state the operational definition correctly. Sample answer: When carbon electrodes are dipped into molten lead(II) bromide, ammeter
shows a reading/ammeter needle deflects
3
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64 Perfect Score & X A –Plus Module/mark scheme 2013
Rubric Score
1(g) Able to explain the difference in conductivity of electricity in Experiment I and
II. Sample answer: In Experiment II, molten lead(II) bromide consists of free moving ions that
carry the electrical current, In Experiment I molten naphthalene consists of neutral molecules.
3
Rubric Score
1(h) Able to classify the substances correctly. Answer:
Substance can conduct electricity Substance cannot conduct electricity Carbon rod Copper(II) sulphate solution
Glacial ethanoic acid Molten polyvinyl chloride
3
Rubric Score
2(a) Able to give the correct value of the reading.
Answer: Final burette reading = 40.20 cm
3 Initial burette reading = 47.20 cm
3 X = 5.0 cm
3
3
Rubric Score
2(b) Able to draw the correct graph with the following aspects.
1. X –axis and y-axis with label and unit 2. Correct scale 3. Correct shape of graph
3
Rubric Score
2(c) Able to determine the correct mole ratio.
Answer: Ag
+ : Cl
- 1.0 x 5 : 1.0 x 5 1000 1000 0.005 : 0.005 1 : 1
3
Rubric Score
2(d) Able to write the ionic equation correctly. Answer: Ag
+ + Cl
- → AgCl
3
Rubric Score
2(e) Able to sketch the correct curve: Graph constant at V = 10 cm
3 3
Rubric Score

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65 Perfect Score & X A –Plus Module/mark scheme 2013
2(f) Able to classify the salts correctly.
Soluble salt Insoluble salt Potassium chloride Nickel nitrate Ammonium carbonate
Barium sulphate
3
Rubric Score
3. (a) Able to state the problem statement correctly.
Sample answer: What is the effect of size of zinc on the rate of reaction with sulphuric acid?
3
Rubric Score
3(b) Able to state the hypothesis correctly
Sample answer: When size of zinc is smaller, the rate of reaction is higher.
3
Rubric Score
3(c) Able to state the all the variables correctly
Answer: Manipulated variable: big sized granulated zinc and small sized granulated zinc Responding variable: rate of reaction Fixed variable: volume and concentration of sulphuric acid
3
Rubric Score
3(d) Able to list the necessary materials and apparatus needed.
Sample answer: Materials: big sized granulated zinc, small sized granulated zinc, 0.1 mol dm
-3
sulphuric acid, water. Apparatus: burette, conical flask, delivery tube with stopper, basin, retort, basin, weighing balance, stop watch, measuring cylinder.
3
Rubric Score
3(e) Able to list procedures for the experiment Sample answer.
1. [5-10] g of big sized granulated zinc is weighed and put into the conical flask.
2. Half filled a basin with water.
3. Fill burette with water and invert into the basin and record the initial reading.
4. Measure 50 cm3 of sulphuric acid and pour into the conical flask.
5. Stopper the conical flask and immediately start the stop watch.
6. Record the burette reading every 30 s intervals for 5 minutes.
7. Repeat the experiment by replacing the big sized granulated zinc with small sized granulated zinc.
3

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66 Perfect Score & X A –Plus Module/mark scheme 2013
Rubric Score
3(f) Able to tabulate the data with the following aspects:
Time/s 0 30 60 90 120 150 180 210 Burette
reading/cm3
Volume of
gas/cm3
2
PAPER 3 SET 3
RUBRIC SCORE
1(a) Able to record all the temperature accurately Sample answer :
Experiment 1
Initial temperature = 28.0 Highest temperature = 40.0 Change of temperature = 12.0 Experiment II
Initial temperature = 28.0 Highest temperature = 38.0 Change of temperature = 10.0
3
RUBRIC SCORE
1(b) Able to construct table accurately with correct title and unit Sample answer :
Temperature Experiment I Experiment II Initial temperature of mixture,
oC 28.0 28.0
Highest temperature of mixture, oC 40.0 38.0
Change of temperature, oC 12.0 10.0
3
RUBRIC SCORE
1(c) Able to state the relationship between manipulated variable and responding variable
with direction correctly
Sample answer :
Manipulated variable : type of acid Responding variable : heat of neutralisation Direction : ?
The reaction between a strong acid and strong alkali produce a greater heat of
3

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67 Perfect Score & X A –Plus Module/mark scheme 2013
neutralization than the reaction between a weak acid and strong alkali.// The reaction between hydrochloric acid and sodium hydroxide produce a greater heat of neutralization than the reaction between ethanoic acid and sodium hydroxide// The heat of neutralization between a strong acid and a strong alkali is greater than the
heat of neutralization between a weak acid and a strong alkali
RUBRIC SCORE
1(d) Able to explain with two correct reasons
Sample answer :
This is to enable the change in temperature to be measured.
The change of temperature is needed to calculate the heat of neutralization
3
RUBRIC SCORE
1(e) Able to state the formula accurately
Sample answer :
Change in temperature = Highest temperature of mixture - initial temperature of
mixture
3
RUBRIC SCORE
1(f) Able to state three observation correctly
Sample answer :
1. A colourless mixture of solution is obtained 2. The vinegar smell of ethanoic acid disappears
3. The polystyrene cup becomes warmer
3
RUBRIC SCORE
1(g) Able to state three constant variables correctly
Sample answer :
1. The volumes and concentration of the acid and the alkali 2. The type of cup used in the experiment
3. The type of alkali
3
RUBRIC SCORE
1(h) Able to calculate the heat of neutralisation for experiment I and II correctly
Sample answer :
Experiment I
Heat released = mcƟ = 50 x 4.2 x 12 = 2520 J
3

@Hak cipta BPSBPSK/SBP/2013
68 Perfect Score & X A –Plus Module/mark scheme 2013
Number of mole of sodium hydroxide = MV = 2.0 x 25/1000 = 0.05 mol 0.05 mole of sodium hydroxide releases 2520 J heat energy
1.0 mole of sodium hydroxide releases = heat released / number of mole = 2520 / 0.05 = 50400 J Heat of neutralisation = - 50.40 kJ/mol Experiment II
Heat released = mcƟ = 50 x 4.2 x 10 = 2100 J
Number of mole of sodium hydroxide = MV = 2.0 x 25/1000 = 0.05 mol 0.05 mole of sodium hydroxide releases 2100 J heat energy
1.0 mole of sodium hydroxide releases = heat released / number of mole = 2100 / 0.05 = 42000 J
Heat of neutralisation = - 42.0 kJ/mol
RUBRIC SCORE
1(i) Able to write the operational definition for the heat of neutralisation correctly. Able to
describe the following criteria
(i) What should be done (ii) What should be observed
Sample answer :
The heat of neutralization is defined as the temperature rises when one mole of water is
produced from reaction between acid and alkali
3
RUBRIC SCORE
1(j) Able to state the relationship between type of acid and value of heat of neutralization and
explain the difference correctly.
Sample answer :
1. The heat of neutralization of a weak acid by a strong alkali is less than the heat of neutralization of a strong acid by a strong alkali.
Explanation :
3

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69 Perfect Score & X A –Plus Module/mark scheme 2013
2. Experiment I uses a strong acid whereas Experiment II uses a weak acid.
3. During neutralization of a weak acid such as ethanoic acid, small portion of the heat released in experiment II is absorbed to help the dissociation of the ethanoic acid
molecules
RUBRIC SCORE
1(k) Able to predict the temperature change accurately
Sample answer :
Lower than 10
oC
3
RUBRIC SCORE
1(l) Able to classify the acids as strong acid or weak acid.
Sample answer :
Name of acid Heat of neutralization /kJmol
-1 Type of acid
Ethanoic acid - 50.3 Weak acid
Hydrochloric acid - 57.2 Strong acid
Methanoic acid - 50.5 Weak acid
3
RUBRIC SCORE
2(a) Able to record all the temperature accurately one decimal places.
Time 55.0 s at 30
oC
Time 48.0 s at 35oC
Time 42.0 s at 40oC
Time 37.0 s at 45oC
Time 33.0 s at 50oC
3
RUBRIC SCORE
2(b) Able to construct table accurately with correct title and unit
Sample answer :
Temperature/
oC 30 35 40 45 50
Time/s 55.0 48.0 42.0 37.0 33.0 1/time / s
-1 0.018 0.021 0.024 0.027 0.030
3
RUBRIC SCORE
2(c)(i) Able to draw the graph of temperature against 1/time correctly
i) Axis x : temperature / 0C and axis y : 1/time /1/s
ii) Consistent scale and the graph half of graph paper
iii) All the points are transferred correctly iv) Correct curve
3

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70 Perfect Score & X A –Plus Module/mark scheme 2013
RUBRIC SCORE
2(c)(ii) state the relationship between the rate of reaction and temperature correctly
The rate of reaction increases with the increase in temperature
3
RUBRIC SCORE
2(d
) Able to predict the time taken
From the graph, when temperature = 55
oC,
1/time = 0.033 s-1
Time = 1/0.033 = 30.3 s
3
RUBRIC SCORE
2(e)(i) Able to state all variables correctly Manipulated variable : Temperature of sodium thiosulphate solution Responding variable : Rate of reaction between sodium thiosulphate and hydrochloric
acid//time taken for the sign X disappear Constant variable : Concentration and volume of sodium thiosulphate solution and
hydrochloric acid
3
RUBRIC SCORE
2(e)(ii) Able to state how to manipulate one variable while keeping the other variables
constant.
Temperature is the manipulated variable. Heating sodium thiosulphate with several different temperatures by remaining the
3

@Hak cipta BPSBPSK/SBP/2013
71 Perfect Score & X A –Plus Module/mark scheme 2013
concentration and volume of sodium thiosulphate and hydrochloric acid constant helps
maintain the responding variable.
RUBRIC SCORE
2(f) Able to give the hypothesis accurately
Manipulated variable : Temperature of sodium thiosulphate solution Responding variable : Rate of reaction between sodium thiosulphate and hydrochloric
acid//time taken for the sign X disappear The higher the temperature, the higher the rate of reaction is
3
RUBRIC SCORE
2(g) Able to state the relationship between temperature and the rate reaction in our daily lives correctly
The lower the temperature, the lower the rate of food turns bad
3
RUBRIC SCORE
3(a) Able to Marke a statement of the problem accurately and must be in question form
Does concentration of ions affect the product of electrolysis process at the anode?
3
RUBRIC SCORE
3(b) Able to state the relationship between manipulated variable and responding variable
correctly
The higher the concentration of ions at the anode, the higher its tendency to be
discharge.
3
RUBRIC SCORE
3(c) Able to state all the three variables correctly
Manipulated variables : concentration of sodium chloride solution Responding variables : product formed at anode Controlled variables : quantity of current, carbon electrodes
3
RUBRIC SCORE
3(d) Able to state the list of substances and apparatus correctly and completely Materials : 0.0001 mol dm
-3 sodium chloride solution, 2.0 mol dm-3 sodium chloride
solution. Apparatus : carbon electrode, electrolytic cell, test tubes, dry cell, blue litmus paper,
wooden splinter, Bunsen burner.
3

@Hak cipta BPSBPSK/SBP/2013
72 Perfect Score & X A –Plus Module/mark scheme 2013
RUBRIC SCORE
3(e) Able to state a complete experimental procedure
1. Fill electrolytic cell with 0.0001 mol dm-3
sodium chloride solution. 2. Connect carbon electrodes to the power supply and ammeter. 3. Switch on the circuit for half hour. 4. Collect the gas at the anode and test with a glowing wooden splinter and a damp
blue litmus paper.
5. Repeat the step 1 to 4 by replacing 0.0001 mol dm-3 sodium chloride solution with
2.0 mol dm-3 sodium chloride solution.
3
RUBRIC SCORE
3(f) Able to draw a suitable table with title correctly
Solution Observation Product formed at
anode 0.0001 mol dm-3sodium chloride solution
2.0 mol dm-3sodium chloride solution
3
RUBRIC SCORE
4 (a)
Able to give the statement of problem correctly.
Sample answer: Does the type of electrode/anode affect the choice of ions to be discharged?
3
RUBRIC SCORE
4 (b) Able to state all variables correctly.
Sample answer:
Manipulated variable : Type of electrode/ anode Responding variable : Product formed at anode Controlled variable : Electrolyte
3
RUBRIC SCORE
4(c) Able to give the hypothesis accurately Sample answer:
Type of electrode/anode will influence the choice of ion to be discharged// type of
electrode/anode will produce different product.
3
RUBRIC SCORE
4(d) Able to list completely the materials and apparatus.
Sample answer: Materials:
3

@Hak cipta BPSBPSK/SBP/2013
73 Perfect Score & X A –Plus Module/mark scheme 2013
1. copper(II) sulphate solution, (0.5 – 2.0) mol dm-3
//any suitable solution that match with metal plate used.
2. carbon rod
3. copper plate// any metal plate that match with a solution used. 4. wooden splinter// any suitable material used for testing a gas or any product at anode.
Apparatus: 1. electrolytic cell
2. battery 3. connecting wire
4. test tube
RUBRIC SCORE
4(e) Able to state all procedures completely and correctly.
Sample answer:
1. Fill the electrolytic cell (beaker) with half full of copper (II) sulpahate solution
(any suitable electrolyte that match with metal plate used).
2. A test tube filled with copper (II) solution is inverted on the anode carbon electrode.
3. Complete the circuit. 4. Electricity is flowed.
5. Record observation at anode..
6. Step 1-5 is repeated using copper plate
3
RUBRIC SCORE
4(f) Able to exhibit the tabulation of data correctly.
Sample answer: Type of electrode Observation Carbon Copper/any metal
2
PAPER 3 SET 4
Rubric Score
1(a) Able to state all the observations and inferences correctly
Sample answers:
Observations Inferences
1. Zinc electrode become thinner Zinc atom ionised to zinc ions//zinc atom
ionises 2. Brown deposite is formed at copper electrode//thicker
Copper atom is formed
3. Blue solution turn to colourless/ become paler // The intensity of blue solution decrease
Copper(II) ions is discharged to copper atom//concentration of copper(II) ion
decreases
6
Rubric Score

@Hak cipta BPSBPSK/SBP/2013
74 Perfect Score & X A –Plus Module/mark scheme 2013
1(b) Able to state all the voltmeter readings accurately with unit
Sample answer:
Zinc and copper : 1.4 V P and copper : 0.8 V Q and copper : 2.8 V R and copper : 0.4 V
3
Rubric Score
1(c) Able to construct a table to record the voltmeter reading for each pair of metals
accurately Sample answer:
Pairs of metals Voltage / V
Zinc and copper 1.4 P and copper 0.8 Q and copper 2.8 R and copper 0.4
3
Rubric Score
1(d) Able to arrange all the metals in ascending order in electrochemical series
Sample answer: Copper, R, P, Zinc, Q
3
Rubric Score
1(e) Able to state the relationship between the manipulated variable and the responding variable with direction.
Sample answer: The further the distance between two/pair of metals in the electrochemical series the
higher/larger/bigger the voltage value.
3
Rubric Score
1(f) Able to state all the three variables correctly Sample answer: Manipulated variable : Pairs of metals Responding variable :Voltmeterreading/voltage/potential difference Constant variable : copper electrode, copper(II) sulphate solution
3
Rubric Score
1(g) Able to state the operational definition for the potential difference accurately Sample answer: The potential difference is the voltmeter reading when two different metals are dipped
in an electrolyte.
3

@Hak cipta BPSBPSK/SBP/2013
75 Perfect Score & X A –Plus Module/mark scheme 2013
Rubric Score
1(h) Able to classify the cations and anions in copper(II)sulphate solution correctly
Sample answer:
Cations Cu
2+, H
+ anions SO4
2-, OH
-
3
Rubric Score
1(i) Able to predict the positive terminal and the voltage value correctly Sample answer:
Positive terminal Voltage /V
P 2.0
3
Rubric Score
1(i) Able to explain the relationship between the time for negative terminal to corrode and
the position in electrochemical series accurately
Sample answer: The distance between magnesium and copper in electrochemical series further//the
distance between zinc and copper in electrochemical series is closer
3
Rubric Score
2 (a) Able to state the inference correctly.
Sample answer: The reactivity (of alkali metals with oxygen) increase from lithium to potassium. //
Lithium, sodium and potassium / alkali metals show similar chemical in their reactions with oxygen.
3
Rubric Score
2 (b) Able to state the three variables correctly:
1. Method to manipulate variable. 2. The responding variable.
3. The controlled variable. Sample answer: (i) Use different types of (alkali) / (group 1) metals (ii) Reactivity of metals with oxygen // Vigorousness of the reaction between
metals and oxygen. (iii) Oxygen gas // size / mass of metal
3
Rubric Score
1 (c) Able to state the relationship correctly between the manipulated variable and the
responding variable. Sample answer: (The lower/higher the position of metal in)/(Going down/up) Group 1, the more/less
3

@Hak cipta BPSBPSK/SBP/2013
76 Perfect Score & X A –Plus Module/mark scheme 2013
reactive is the metal in reaction with oxygen. // The lower/higher the metal in Group 1 the more/less reactive the reaction with oxygen.
Rubric Score
2 (d) Able to give the operational definition accurately by stating the following three
information.
- alkali metals - vigorously / more vigorous / reactive with oxygen
- more / highly reactive
Sample answer: An alkali metal that reacts more vigorously with oxygen is a more reactive metal.
3
Rubric Score
2 (e)(i) Able to state the position of metal X in Group 1 accurately.
Sample answer: Period 5/6/7
3
Rubric Score
2 (e)(ii) Able to arrange the metals in ascending order based on their reactivity.
Sample answer: Lithium, Sodium, Potassium, X // Li, Na, K, X
3
Rubric Score
2 (f) Able to state the relationship between the mass of sodium and the time taken for the metal to burn completely in oxygen gas.
- the higher the mass / the bigger the size - the longer the time taken
- burn completely Sample answer: The higher the mass of metals, the longer the time taken to burn completely. // The bigger the size of metals, the longer the time taken to burn completely.
3
Rubric Score
2 (g) Able to record all the readings with one decimal place accurately.
Sample answer:
10.1 , 10.6, 10.9
3

@Hak cipta BPSBPSK/SBP/2013
77 Perfect Score & X A –Plus Module/mark scheme 2013
Rubric Score
2 (h) Able to state observations for blue and red litmus paper correctly. Sample answer: Solutions Red litmus paper Blue litmus paper Gas Jar I Turns blue No change Gas Jar II Turns blue No change Gas Jar III Turns blue No change
3
Rubric Score
2 (i) Able to write the two balanced chemical equations for the reaction accurately.
Sample answer : i. 4Na + O2 2Na2O and ii. Na2O + H2O 2NaOH Notes: Sodium can be replaced with any alkali metals from Table 1.
3
Rubric Score
2 (j) Able to classify all alkaline solutions into strong alkali and one weak alkali
correctly. Sample answer: Strong alkali : Sodium hydroxide / NaOH, Potassium hydroxide / KOH Calcium hydroxide / Ca(OH)2 Weak alkali : Ammonia solution/ NH3
3
Rubric Score
3(a) Able to give the statement of the problem accurately. Response is in question form.
Sample answer: Does the temperature of sodium thiosulphate solution affect the rate of reaction
between sodium thiosulphate solution and sulphuric acid? // How does the temperature of sodium thiosulphate solution affect the rate of reaction? between sodium thiosulphate solution and sulphuric acid?
3
Rubric Score
3 (b) Able to state the three variables correctly Sample answer: Manipulated variable: Temperature of sodium thiosulphate solution Responding variable : Rate of reaction // Time taken for mark ‘X’ to become invisible /disappear Constant variable: Volume and concentration of sodium thiosulphate/ sulphuric acid
/ size of conical flask
3
Rubric Score
3 (c) Able to state the relationship correctly between the manipulated variable and the
responding variable with direction. Sample answer:
3

@Hak cipta BPSBPSK/SBP/2013
78 Perfect Score & X A –Plus Module/mark scheme 2013
The higher/lower the temperature of sodium thiosulphate solution, the higher/lower
the rate of reaction. // The higher/lower the temperature of sodium thiosulphate solution, the
shorter/longer the time taken for mark ‘X’ to disappear from sight/view //
Rubric Score
3(d) Able to give complete list of materials and apparatus Sample answer: Materials : Sodium thiosulphate solution, sulphuric acid. Apparatus : Conical flask, ,bunsen burner, measuring cylinder, stop-watch,
filter paper.
3
Rubric Score
3(e) Able to list all the steps correctly Sample Answer:
1. ‘X ‘mark is drawn on a piece of white/filter/ cardboard paper.
2. 50 cm3 of sodium thiosuphate solution [(0.01-1.0) mol dm
-3] is
measured with a measuring cylinder and is poured into a conical flask.
3. The solution is slowly heated until 30 oC.
4. 5 cm3 of hydrochloric acid [(0.1- 2.0) mol dm
-3] is measured with a
measuring cylinder and is added to the conical flask. A stop-watch is
started immediately.
5. The conical flask is swirled and is placed on a filter paper with a mark ‘X’.
6. The ‘X’ mark is observed vertically from the top through the solution.
7. The stop-watch is stopped immediately when the ‘X’ mark cannot be
seen. Time is recorded. 8. The experiment is repeated by using the sodium thiosuphate solution at
40 oC, 50
oC, 60
oC and 70
oC respectively.
3

@Hak cipta BPSBPSK/SBP/2013
79 Perfect Score & X A –Plus Module/mark scheme 2013
Rubric Score
3 (f) Able to tabulate the data with following aspects
1. Correct titles with units 2. Complete list of temperatures
Sample answer:
Temperature (oC) Time (s)
30 40 50 60 70
2
PAPER 3 SET 5
Rubric Score
1(a) Able to state four observations correctly Sample answers:
Observations at anode Blue litmus paper : turn red then bleached / decolourise Glowing splinter : no change Blue litmus paper : no change Glowing splinter : is rekindled / relighted
3
Rubric Score
1(b) Able to state the colour change in the copper (II) chloride solution correctly Sample answer:
The intensity of the blue solution decreases / reduced // Blue colour of solution fades gradually // Blue solution becomes light blue
3
Rubric Score
1(c) Able to state all the variable and the action to be taken correctly
Sample answer:
Name of variables Action to be taken Concentration of copper (II) chloride
solution Change the concentration from 1.0 mol
dm-3
to 0.001 mol dm-3
Gas collected at anode
The change of damp blue litmus paper
and glowing splinter
Type of solution
Use the same copper (II) chloride
solution
6
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@Hak cipta BPSBPSK/SBP/2013
80 Perfect Score & X A –Plus Module/mark scheme 2013
Rubric Score
1(d) Able to state the relationship between the manipulated variable and the responding
variable with direction. Sample answer: The higher the concentration of ion in the solution in the electrolyte, the higher the chance the ion discharged at anode
3
Rubric Score
1(e) Able to classify the ions correctly -write the name or symbols of the ions. Sample answer:
Cations Anions Copper (II) ions, Cu
2+ Hydrogen ions, H
+ Hydroxide ions, OH
- Chloride ions, Cl
-
3
Rubric Score
2(a)
Able to state all the observation and inference correctly.
Sample answer:
Observation inference White fume is released White solid is formed The mass of crucible and its content
increases
Magnesium oxide is formed Magnesium reacts wth oxygen
6
Rubric Score
2 (b) Able to state all the masses accurately
Sample answer: The crucible and lid = 25.35 g The crucible, lid and magnesium ribbon = 27.75 g The crucible, lid and magnesium oxide when cooled = 29.35 g
3
Rubric Score
2 (c) (i) The mass of Mg = 27.75 – 25.35 = 2.4 g
(ii) The mass of Oxygen = 29.35 – 27.75 = 1.6 g
(iii) The number of mole of Mg = 2.4/24 = 0.1 mole The number of mole of O = 1.6/16 = 0.1 mole The ratio of Mg : O = 1 : 1 The empirical formula is MgO
3

@Hak cipta BPSBPSK/SBP/2013
81 Perfect Score & X A –Plus Module/mark scheme 2013
Rubric Score
2 (d) 0.1 mole of Mg reacts with 0.1 mole of oxygen// 1 mole of Mg reacts with 1 mole of oxygen
3
Rubric Score
2 (e) Able to predict and give a reason for the prediction Sample answer:
Cannot because copper is a less electropositive metal. Copper cannot reacts with
oxygen gas to produce copper (II) oxide.
3
Rubric Score
2 (f) Able to classify the oxides into two groups, those which are basic oxides and those
which are acidic oxides correctly
Sample answer:
Basic oxides Acidic oxides
Magnesium oxide
Copper (II) oxide
Sulphur oxide
Carbon dioxide
3
Rubric Score
3(a)
Able to give the statement of the problem accurately. Response is in question form.
Sample answer How does ethanoic acid and ammonia solution affects the coagulation of latex?
3
Rubric Score
3(b) Able to state the three variables correctly Sample answer: Manipulated : ethanoic acid and ammonia solution Responding : coagulate / coagulation of latex Fixed : latex
3
Rubric Score
3(c) Able to state the relationship correctly
Ethanoic acid coagulates the latex while ammonia solution does not coagulate the
latex.
3
Rubric Score
3(d) Able to state the complete list of apparatus and material as follows.
Materials: ethanoic acid 0.5 mol dm
-3 and ammonia solution
Apparatus: Beaker, measuring cylinder, glass rod, dropper
3

@Hak cipta BPSBPSK/SBP/2013
82 Perfect Score & X A –Plus Module/mark scheme 2013
Rubric Score
3(e) Able to list all the steps correctly
1. 10 cm3 of latex is poured into a beaker.
2. Ethanoic acid is added into the beaker using a dropper. 3. The mixture is stirred using glass rod.
4. The beaker is left aside.
5. The observation is recorded 6. Experiment is repeated using ammonia solution to replace ethanoic
acid.
3
Rubric Score
3(f) Able to tabulate the data correctly
Mixture Observation Latex + ethanoic acid Latex + ammonia solution
2
PAPER 3 SET 6
RUBRIC SCORE
1(a)(i) Able to record all reading accurately with units Sample answer :
Experiment Copper Bronze I 1.35 cm 1.20 cm II 1.60 cm 1.00 cm III 1.50 cm 1.20 cm
3
RUBRIC SCORE
1(a)(ii) Able to construct the table with correct label and unit
Sample answer :
Type of
blocks Diameter of dents (cm) Average diameter
of dents (cm) I II III Copper 1.35 1.60 1.50 1.48 Bronze 1.20 1.00 1.20 1.13
3
RUBRIC SCORE
1(b) Able to state the observation correctly and accurately
Sample answer : The average diameter of dents on bronze block is 1.13 cm and the average
diameter of dents on copper block is 1.48 cm// The size / diameter of dents on bronze block is smaller than size / diameter of
dents on copper block//
3
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@Hak cipta BPSBPSK/SBP/2013
83 Perfect Score & X A –Plus Module/mark scheme 2013
1(c) RUBRIC SCORE
Able to state the inference correctly and accurately Sample answer : Bronze is harder than copper // Copper is less harder than bronze
3
1(d) RUBRIC SCORE
Able to state operational definition correctly Sample answer : The smaller dent produced when 1 kg weight is dropped on the block.
3
1(e) RUBRIC SCORE
Able to explain the arrangement of particles in the materials correctly
Sample answer :
1. The atomic size of tin is bigger than copper // the atomic size of tin and copper are different.
2. The presence of tin atoms in bronze disrupts the orderly arrangement
of copper atoms. 3. Reduces / prevent the layers of atoms from sliding over each other
easily
3
1 (f) RUBRIC SCORE
Able to state the hypothesis correctly
Sample answer : Bronze is harder than copper // Copper is less harder than bronze
3
1 (g) RUBRIC SCORE
Able to state all three variables and all three action correctly
Sample answer :
Name of variables Action to be taken
(i) Manipulated variable:
Type of materials //
copper and bronze
(i) The way to manipulate variable:
Replace copper with bronze
(ii) Responding variable:
Diameter of dent
(ii) What to observe in the responding
variable:
The diameter of the dent formed on copper block and bronze block
(iii) Controlled variable:
Mass of weight // height
of the weight // size of steel ball bearing
(iii) The way to maintain the control variable:
Uses same mass of weight // same height of the weight // same size of ball
bearing
3

@Hak cipta BPSBPSK/SBP/2013
84 Perfect Score & X A –Plus Module/mark scheme 2013
RUBRIC SCORE
2(a) Able to state 5 correct observations.
Sample answer
Test tube Observation 1 blue colour /solutions 2 High intensity of pink colour/ solutions 3 High intensity of blue colour /solutions 4 Low intensity of pink colour/ solutions 5 Low intensity of blue colour /solutions
3
RUBRIC SCORE
2(a) Able to state 5 correct inferences.
Sample answer
Test tube Inference 1
Iron(II) / Fe2+
ions formed / produced in the solutions // Iron / Fe rusted/corroded/oxidised
2 Iron(II) / Fe
2+ ions are not formed /produced in the solutions //
Iron / Fe does not rust/ corrode/oxidised Magnesium/Mg rusted/corroded /oxidised
3 Iron(II) / Fe
2+ ions formed / produced in the solutions //
Iron / Fe rusted/ corroded/ oxidised
4 Iron(II) / Fe
2+ ions are not formed /produced in the solutions //
Iron / Fe does not rust/ corrode/oxidised // Zinc/Zn rusted/ corroded / oxidised
5 Iron(II) / Fe
2+ ions formed / produced in the solutions //
Iron / Fe is rusted / corroded/ oxidised
3
RUBRIC SCORE
2(b) Able to explain a difference in observation correctly between test tube 2 and 3 Sample answer Iron/Ferum/Fe in test tube 2 does not rust/ corrode/ oxidised because ferum is
in contact with a more electropositive metal, but iron/Ferum/Fe in test tube 3 rusts/ corrodes/ is oxidised because ferum is in contact with a less
electropositive metal. // In test tube 2, magnesium is more electropositive than ferum/iron and and in test tube 3, copper is less electropositive the ferum/iron.
3
RUBRIC SCORE
2(c) Able to state the hypothesis correctly.
Sample answer When a more/less electropositive metal is in contact with iron/ferum/Fe, the
metal inhibits/(speeds up) rusting/corrosion of iron // If the metal in contact with iron is higher/lower than iron/ferum/Fe in
electrochemical series, the rusting/corrosion of iron is slower/faster //
3

@Hak cipta BPSBPSK/SBP/2013
85 Perfect Score & X A –Plus Module/mark scheme 2013
RUBRIC SCORE
2(d) Able to state all the variables in this experiment correctly.
Sample answer (i) Manipulated variables : Type/different metal // position of metal in
electrochemical series (ii) Responding variable : Rusting / corrosion // presence of blue/pink colour (iii) Constant variable : Size/mass of iron nail // type of nail // clean iron nails //
temperature // medium in which the iron nail are kept
3
RUBRIC SCORE
2(e) Able to state the operational definition for the rusting of iron nail correctly. Sample answer Rusting occurs when iron nail is in contact with copper/tin /less electropositive metal and form blue colouration in potassium hexacyanoferrate(III) solution
3
RUBRIC SCORE
2(f) Able to classify all the metals correctly.
Sample answer
Metals that inhibit rusting Metals that speed up rusting Magnesium/Mg
Zinc/Zn Tin/Sn
Copper/Cu
3
RUBRIC SCORE
2(g)(i) Able to state the relationship between the time taken and the amount of rust
formed correctly.
Sample answer The longer the time taken, the greater/bigger/larger the rust formed // The
longer the time taken, more rust is formed // The rust formed is greater/bigger/larger, when the time taken is longer.
3
RUBRIC SCORE
2(g)(ii) Able to predict the time taken for the iron nail to completely rust correctly.
Answer Less than 5 days
3
RUBRIC SCORE
2(h)(i) Able to record the voltmeter readings correctly in one decimal place.
Answer
Pairs of metal
Positive terminal
Voltmeter reading
(V) Magnesium and iron Iron 2.0 Iron and copper Copper 0.8 Iron and zinc Iron 0.4 Iron and tin Tin 0.2
3

@Hak cipta BPSBPSK/SBP/2013
86 Perfect Score & X A –Plus Module/mark scheme 2013
RUBRIC SCORE
2(h)(ii) Able to draw a labelled diagram accurately.
Sample Answer
3
RUBRIC SCORE
3 (a) Able to Marke a statement of the problem accurately and must be in
question form
Sample Answer Able to How do the heat of neutralisation for reactions between acids and
alkalis of different strengths differ?
3
RUBRIC SCORE
3 (b) Able to state all the three variables correctly
Manipulated Variable : different strength of acid // hydrochloric acid and
ethanoic acid Responding variable : the value of heat of neutralisation Fixed variable : volume and concentration of acid // volume and
concentration of alkali // polystyrene cup
3
RUBRIC SCORE
3 (c) Able to state the relationship between manipulated variable and responding
variable correctly
The value of heat of neutralisation for reaction between strong acid and
strong alkali is higher than of reaction between weak acid and strong alkali// The value of heat of neutralisation for reaction between hydrochloric acid
and sodium hydroxide is higher than of reaction between ethanoic acid and
sodium hydroxide
3
RUBRIC SCORE
3 (d) Able to state the list of substances and apparatus correctly and completely
Apparatus : Measuring cylinders, polystyrene cup with covers, thermometer Material : 2.0 mol dm
3 sodium hydroxide, 2.0 mol dm
3 ethanoic acid, 2.0
mol dm3 hydrochloric acid
3
V
Magnesium/Mg Iron/Fe
Dilute sulphuric
acid /H2SO4
Voltmeter

@Hak cipta BPSBPSK/SBP/2013
87 Perfect Score & X A –Plus Module/mark scheme 2013
RUBRIC SCORE
3 (e) Able to state a complete experimental procedure
1. Measure 50 cm3 of 2.0 mol dm
-3 sodium hydroxide solution, NaOH
solution using a measuring cylinder. Pour it into a polystyrene cup
with a cover.
2. Measure 50 cm3 of 2.0 mol dm
-3 hydrochloric solution, HCl solution
using another measuring cylinder. Pour it into another polystyrene
cup with a cover.
3. Leave both the polystyrene cups on the table for 5 minutes. After 5 minutes, measure and record the initial temperatures of both the
solution.
4. Pour the hydrochloric acid, HCl quickly and carefully into the polystyrene cup containing sodium hydroxide solution.
5. Stir the mixture using the thermometer. 6. Record the highest temperature of the reaction mixture.
7. Repeat steps 1 to 5 using ethanoic acid to replace the hydrochloric
acid.
3
RUBRIC SCORE
3 (f) Able to tabulate the data correctly
Hydrochloric acid Ethanoic acid
Initial temperature of alkali, oC
Initial temperature of acid, oC
Highest temperature of the
reaction mixture, oC
3
PAPER 3 SET 7
RUBRIC SCORE
1 (a) Able to record all the readings accurately to two decimal points with units.
Sample answer: Activity I : 26.05 cm
3, 26.90 cm
3, 30.05 cm
3 Activity II : 13.30 cm
3, 25.85 cm
3, 38.45 cm
3
3
RUBRIC SCORE
1(b)
Able to construct a table containing the following information:
1. Headings in the table 2. Transfer all data from 1(a) correctly
3. With units Sample answer:
Titration
number Initial burette reading /
cm3
Final burette reading /
cm3
Volume of acid /
cm3
3

@Hak cipta BPSBPSK/SBP/2013
88 Perfect Score & X A –Plus Module/mark scheme 2013
RUBRIC SCORE
1 0.80 13.30 12.50 2 13.40 25.85 12.45 3 25.90 38.45 12.55
RUBRIC SCORE
1(c) Able to show all the steps to calculate the concentration of sulphuric acid correctly. Sample answer: Step 1: Write the chemical equation: 2NaOH + H2SO4 Na2SO4 + 2H2O Step 2: Calculating the number of moles of sodium hydroxide Number of mol of NaOH : 0.1 x 25 // 0.0025 1000 Step 3: Calculating the concentration of sulphuric acid Concentration of H2SO4 : ( 0.0025 x 1000 ) // 0.1 mol/dm
3 12.50 x 2
3
RUBRIC SCORE
1(d) Able to state the colour change correctly Sample answer: Activity I : Pink change to colourless Activity II : Yellow change to orange
3
1(e) Able to state the correct type of acid in activity I and II and give the correct reason.
Sample answer: Type of acid : Activity I use monoprotic acid and Activity II use diprotic acid. Reason : The volume of acid used in activity I is twice with the volume of acid used in activity II.
3
RUBRIC SCORE
1(f) Able to state the colour change correctly Sample answer: Yellow change to orange and finally change to red
3
RUBRIC SCORE
1(g)
Able to predict the volume with the unit Sample answer: More than 25.00 cm
3 // 25.05 – 50.00 cm
3
3
1(h)
Able to state all the variable correctly Manipulated Variable : Type of acid uses // type of indicator Responding Vvariable : Volume of acid to neutralize 25.0 cm
3 of mol dm
-3 sodium
hydroxide solution // Change in the colour of the indicator. Fixed Variable : Concentration and volume of sodium hydroxide solution.
3
RUBRIC SCORE
1(i) Able to state the hypothesis (relate the manipulated variable with the responding
variable) correctly. Sample answer: If use different type of acid to neutralize 25.0 cm
3 of 1.0 mol dm
-3 sodium hydroxide
solution, the volume of acid use also different// Different indicator used in the titration create different colour.
3

@Hak cipta BPSBPSK/SBP/2013
89 Perfect Score & X A –Plus Module/mark scheme 2013
RUBRIC SCORE
RUBRIC SCORE
1(j) Able to give the operational definition for the end-point of titration in activity I correctly. Able to describe the following criteria
(i) What should be done
(ii) What should be observed Sample answer: When hydrochloric acid is added to sodium hydroxide solution with phenolphthalein,
pink turns to colourless.
3
RUBRIC SCORE
1(k)
Able to classify all the acids into strong acid and weak acid correctly.
Sample answer:
Strong acid Weak acid Nitric acid Ethanoic acid
Phosphoric acid Ascorbic acid
3
RUBRIC SCORE
2(a) Able to state the inference accurately Sample answer When alcohol react with carboxylic acid, ester is formed//Esters have sweet pleasant smell property
3
RUBRIC SCORE
2(b) Able to construct a table correctly with the following information:
1. Columns with titles for alcohol, carboxylic acid, Ester
2. Name of all alcohols, carboxylic acid and ester
Alcohol Carboxylic acid Ester Methanol Ethanoic acid Methyl ethanoate Ethanol Propanoic acid Ethyl propanoate Propanol Methanoic acid Propyl methanoate
3
RUBRIC SCORE
2(c) Able to name the alcohol and carboxylic acid correctly. Alcohol: Propanol Carboxylic acid: Butanoic acid
3
RUBRIC SCORE
2(d)(i) Able to state the three variables correctly.
Sample answer Manipulated variable : Hexane and hexene Responding variable : Colour change of bromine water // colour change of
potassium manganate (VII) solution Fixed variable : Bromine water//acidified potassium manganate (VII) solution
3

@Hak cipta BPSBPSK/SBP/2013
90 Perfect Score & X A –Plus Module/mark scheme 2013
RUBRIC SCORE
2(d) (ii)
Able to state the hypothesis accurately
Sample answer: Hexene declourised the brown colour of bromine water, hexane does not// Hexene
declourised the purple colour of acidified potassium manganate(VII) solution, hexane does not
3
RUBRIC SCORE
2(d)(iii) Able to predict and Marke explanations accurately
Answer
1. Hexene 2. Percentage of carbon atoms per molecule hexene is higher than hexane
3. Percentage of carbon in hexane = 72 x 100 84 = 85.71 %
4. Percentage of carbon in hexane = 72 x 100 86 = 83.72 %
3
RUBRIC SCORE
3(a) Able to state the problem statement accurately
Sample answer Are the effectiveness of the cleansing action of soap and detergent in hard water
different?
3
RUBRIC SCORE
3(b) Able to state the three variables accurately.
Answer Manipulated variable: Soap and detergent Responding variable: Effectiveness of cleansing action // the ability to remove
the oily stains on cloth Fixed variable: cloth with oily stains, hard water
3
RUBRIC SCORE
3(c) Able to state the hypothesis accurately with direction
Sample answer The cleansing action of a detergent is more effective in hard water than a soap
3
RUBRIC SCORE
3(d) Able to state the complete list of apparatus and material as follows
List of apparatus : 2 beakers, , glass rod List of material : Hard water, soft water, soap and detergent solution, 2 pieces of
cloths stained with oil
3
RUBRIC SCORE
3(e) Able to state procedures correctly as follows
1. [50 - 200] cm3 of hard water is poured into a beaker
2. Soap is added into the beakers
3

@Hak cipta BPSBPSK/SBP/2013
91 Perfect Score & X A –Plus Module/mark scheme 2013
3. A piece of cloth stained with oil is immersed in the solution 4. The cloth is shaken/rubbed/stirred
5. Observation is recorded
6. Repeat steps 1 – 6 by using detergent .
RUBRIC SCORE
3(f) Able to tabulate the data correctly Sample answer
Type of cleaning agent Observation
Soap Detergent
3
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