Perfect Score & X A-Plus 2013 Chemistry Module Solutions

@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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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

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/o
C
Time/s
67
90

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 Yx X x
x
x
x x
E EY

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 80o
C.
2. Arrow mark freezing point at 80o
C
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

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
+
- -

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

(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
WY

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
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
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

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. X2+
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-
Y2-
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
XX
X
X
X
X
X
X
X
2+ 2-
X2+
Y2-
X
X
X
X
X
X
X
X
X
X
XX
X
X
X
X
X
X
X

(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 cm3
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

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

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

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 24dm3
mol-1
= 2.4 dm3
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 1023
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

(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

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

(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

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
12
3.33
6.66
1 6.66 53.33
16
3.33
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

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) Cu2+
and H+
1
(d)(i)
(ii)
Become thicker / brown solid formed
Bertambah tebal / pepejal perang terbentuk
Cu2+
+ 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

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

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  Zn2+
+ 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

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 solutionIron spoon
Silver

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

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  Fe2+
+2e
ee

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.
Fe2+
 Fe3+
+ e
The green coloured solution of iron(II) sulphate turns brown.
Fe2+
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
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. Fe2+
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

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→Mg2+
+ 2e 1
Cu2+
+ 2e→ Cu 1
Oxidising agent- Cu2+
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

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

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
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

(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
0.01mole 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
2H+
+ PbO → Pb2+
+ H2O
1
1

(ii) 1. Mole of acid
2. Mole ratio
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 cm3
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
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

(ii) 1. Mole of CuCO3
2. Mole ratio
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
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

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
Ba2+
ion : SO4
2-
ion
0.0025 : 0.0025 //
1 : 1
1
1
1
(iv) Ba2+
+ SO4
2-
→ 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

(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 + Na2CO3ZnCO3 + 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

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 cm3
s-1
( +- 0.05)
Set II :
1.Tangen shown in graph correctly
2.Rate of reaction = 0.23 cm3
s-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
CaCO3+ 2HNO3 → Ca(NO3) 2+ CO2 + H2O
1
1
(b)
Functional diagram
Label
1
1
(c) 1. Mole of nitric acid
2. Mole ratio
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

Maximum volume of CO2 = 0.005 x 24 = 0.12 dm3
// 120 cm3
(d) Experiment I = 0.12 X 1000 // 0.2 cm3
s-1
//
10 X 60
//0.12 //0.012 dm3
min-1
10
Experiment II = 0.12 X 1000 // 0.4 cm3
s-1
//
5 X 60
// 0.12 // 0.024 dm3
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
’

2 mol HCl : 1 mol H2
0.025 mol HCl : 0.0125 mol H2
Volume of H2 = 0.0125 x 24 // 0.3dm3
// 300 cm3
1
1
(iii) 1. Add excess zinc powder with 12.5 cm3
of 1 mol dm-3
hydrochloric acid .
2. At the same temperature
OR
1. Add excess zinc powder with 25 cm3
of 0.5 mol dm-3
hydrochloric 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 dm3
//1.2 dm3
//0.05 x 24000//1200 cm3
1
1
1
1
(ii) Experiment I = 1200 // 15 cm3
s-1
80
Experiment II = 1200 // 7.5 cm3
s-1
160
Experiment III = 600 // 2.5 cm3
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

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 cm3
s-1
2. Experiment II : 20 cm3
/ 50 s // 0.4 cm3
s-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 cm3
) 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
2. Add excess zinc powder/granule
OR
(Temperature)
Experiment 1:
of (0.1-2 mol dm-3
) hydrochloric acid
2. Heat acid to (30-80O
C)
3. Add excess zinc powder/granule
Experiment II :
of (0.1-2 mol dm-3
2. Without heating
3. Add excess zinc powder/granules
OR
(Concentration)
Experiment 1:
of (0.2-2 mol dm-3
1
1
1
1
1
1
1
1
1
1
1
1
1

Experiment II :
of (0.1-1 mol dm-3
OR
(Size)
Experiment 1:
of (0.1-2 mol dm-3
2. Add excess zinc powder
Experiment II :
of (0.1-2 mol dm-3
2. Add excess zinc granule
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

Question
No
Mark scheme Mark
6.(a)
(i)
1. First minute = 24/60 =0.4 cm3
s-1
// 24 cm3
min-1
2. 2 nd
minute = 34-24/60 =0.167 cm3
s-1
// 10 cm3
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 cm3
conical flask, stopwatch, 50 cm3
measuring cylinder, 10 cm3
measuring
cylinder, thermometer, Bunsen burner, wire gauze.
Procedure:
1.Using a measuring cylinder, 50 cm3
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 cm3
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 cm3
of 0.2 mol dm-3
sodium thiosulphate solution at
40o
C, 50o
C, 60 o
C by heating the solution before 5 cm3
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

(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 cm3
conical flask, stopwatch, 50 cm3
measuring cylinder, 10 cm3
measuring
cylinder, wire gauze.
Procedure:
1.Using a measuring cylinder, 50 cm3
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 cm3
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 cm3
, 10 cm3
, 15 cm3
, 20 cm3
and 40 cm3
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

= - 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
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

(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
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

(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
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 o
C
1
1
1

(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+
ionand K+
ion not involved in the reaction // Ag+
ionand 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 o
C // 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
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

= 6.8 o
C ( 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

1. (100-250 cm3
)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 30o
C.
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

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

SET 4 :CARBON COMPOUNDS
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 (o
C) 1
Highest temperature of the reaction mixture (o
C) 2
Temperature change (o
C) 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

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
HCH
H
H
H H
৷ ৷
C - C
৷ ৷
H H n

(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
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-eneBut-2-ene 2-methylpropene

(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
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

(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
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

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

(d)
(ii)
Pour 2 cm3
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 CH
H H H H
H
H H
C
C C C
H
H
H
H
H
H
H
H

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
1 (a) (i) Contact process 1
(ii) Ammonia 1
(iii) Vanadium(V) oxide, 450 o
C - 500o
C 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
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

(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
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

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
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

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
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

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. 20cm3
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

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

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

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

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

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 cm3
Initial burette reading = 47.20 cm3
X = 5.0 cm3
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 cm3
3
Rubric Score

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

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, o
C 28.0 28.0
Highest temperature of mixture, o
C 40.0 38.0
Change of temperature, o
C 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

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

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

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 10o
C
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 30o
C
Time 48.0 s at 35o
C
Time 42.0 s at 40o
C
Time 37.0 s at 45o
C
Time 33.0 s at 50o
C
3
RUBRIC SCORE
2(b) Able to construct table accurately with correct title and unit
Sample answer :
Temperature/o
C 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 / 0
C 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

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 = 55o
C,
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

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

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

Perfect Score & X A-Plus 2013 Chemistry Module Solutions

Perfect Score & X A-Plus 2013 Chemistry Module Solutions

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Perfect Score & X A-Plus 2013 Chemistry Module Solutions