PS F4 Set 1

Set 1 Marking Scheme: The Structure of the Atom & Chemical Equation 2010
Chemistry Perfect Score Module Form 4 2010 Marking Scheme Set 1 1
THE STRUCTURE OF THE ATOMS
PAPER 2 : STRUCTURE
Question Explanation Marks
1(a)(i)
(ii)
(iii)
Diffusion
Molecules
- Made of tiny / discrete particles/ molecules
- Move faster / rapidly
- Between air / another particles
1
1
1+1+1
(b)(i)
(ii)
(iii)
83 o
C
Remain, because heat absorbed is used to overcome the forces between
particles.
Move faster
Particles gain kinetic energy.
1
1+1
1
1
2(a)(i)
(ii)
(iii)
White fume
Ammonium chloride
Diffusion
1
1
1
(b) Ammonia, because ammonia is lighter than hydrogen chloride. 1+1
(c) NH3 + HCl  NH4Cl 1
3(a)(i)
(ii)
Number of proton in the nucleus of an atom.
12
1
1
(b) X and Y. Because they have same proton number but different neutron or
nucleon number.
1+1
(c) 2.8.7 1
(d)(i)
(ii)
(iii)
7
Y-
1+1
1
1
(e) 35
X 1
4(a)(i)
(ii)
(iii)
(iv)
- Both axis are labeled with unit and consistent scales
- All points transfer correctly
- Size of graph at least ½ of the page of the graph paper
- Curve of the graph is correct and smooth
80o
C
Heat is absorbed is used to overcome the forces of attraction between particles.
3
1
2
1
Y
17

(b) To ensure uniform heating 1
(c) Solid X is highly flammable 1
5(a) Iodine and ethanol 2
(b) copper 1
(c) copper 1
(d) ion 1
PAPER : ESSAY
No. 6 Rubric Marks
(a)
Any 2 pairs
Particle Relative charge Relative mass
Proton +1 1
Neutron 0 (neutron) 1
Electron -1 1/1840
1+1
1+1
1+1
4
(b)(i) 1. Nucleus contains 7 protons and 7 neutrons.
2. Electrons moves around the nucleus.
3. Two shells filled with electrons.
4. 5 valance electrons / electrons arrangement 2.5
1
1
1
1
4
(ii)
Nucleon number
Proton number
1
1
2
(c)
Stage/Time State of matter Particle arrangement Change in
energy
to – t1 Solid Closely packed Kinetic energy
increases
t1 – t2 Solid –liquid Close together but not in
orderly arrangement
Higher kinetic
energy
t2 – t3 Liquid Further apart Highest kinetic
energy
1
1+1+1
3
1+1+1
3
1+1+1
3
10
Total 20
14
7
X

No. 7 Rubric Marks
7 (a) Melting point is the temperature at which a solid changes into a liquid at a
particular pressure.
Because the energy absorbed by the naphthalene is used to overcome the forces
of attraction between the molecules of naphthalene.
1
1
1
1 4
7(b) Before condensation
 The kinetic energy is high
 The particles are very far apart from each other
 The attraction forces between particles are very weak.
During condensation
 The kinetic energy decreases
 The particles begin to move closer toward one another
/ the distance between the particles decrease
 The attraction forces between particles become stronger.
After condensation
 The kinetic energy is low
 The particles are packed closely together NOT in an orderly manner
 The attraction forces between particles are strong.
1
1
1
1
1
1
1
1
1+1
1
10
7(c)(i) Proton number = 13
Nucleon number = 14 + 13 = 27
1
1 2
7(c)(ii)
Number of moles of Y =
27
8.10
= 0.4
Number of moles of Y2O3 = 0.4 ÷ 2 = 0.2
Relative formulas mass of Y2O3 = 2(27) + 3(16) = 102
Mass of Y2O3 = 0.2  102 = 20.4 g
1
1
1
1 4
Total 20
PAPER 3: STRUCTURE
8 Rubric Marks
(a) To study the rate of diffusion in the three states of matter. 3
(b) The rate of diffusion in matter decreases in the order of gas > liquid > solid 3
(c)(i)
(ii)
(iii)
Manipulated : medium of diffusion gas, liquid and solid
Responding : rate of diffusion
Fixed : temperature of each medium
3
(d) Solid , liquid, gas 3
(e) Solid – the particles are very closely pack
Liquid – the particles closely but there are more space between them
Gas – the particles far apart from each other
3
(f) Smell perfume / smell gas from leaked pipe or gas cylinder 3

CHEMICAL FORMULA AND EQUATION
1.
Ion Chloride Nitrate Hydroxide Sulphate Carbonate oxide
Sodium NaCl NaNO3 NaOH Na2SO4 Na2CO3 Na2O
Magnesium MgCl2 Mg(NO3)2 Ma(OH)2 MgSO4 MgCO3 MgO
Lead(II) PbCl2 Pb(NO3)2 Pb(OH)2 PbSO4 PbCO3 PbO
Copper (II) CuCl2 Cu(NO3)2 Cu(OH)2 CuSO4 CuCO3 CuO
Iron (II) FeCl2 Fe(NO3)2 Fe(OH)2 FeSO4 FeCO3 FeO
Iron (III) FeCl3 Fe(NO3)3 Fe(OH)3 Fe2 (SO4)3 Fe2( CO3)3 Fe2O3
Aluminium AlCl3 Al(NO3)3 Al(OH)3 Al2 (SO4)3 Al2( CO3)3 Al2O3
2. (a) CuCO3  CuO + CO2
(b) HNO3 + NaOH  NaNO3 + H2O
(c) 2HCl + Zn  ZnCl2 + H2
(d) Cu(NO3)2 + Mg  Mg(NO3)2 + Cu
(e) Cl2 + 2LiOH  LiCl + LiOCl + H2O
3. (a) 0.1 x 6.02 x 1023
= 6.02 x 1022
atoms
(b) 1.5 x 6.02 x 1023
= 9.05 x 1023
atoms
(c) 2.0 x 6.02 x 1023
= 1.2 x 1024
molecules
(d) 1.5 x 6.02 x 1023
= 9.05 x 1023
atoms
(e) 2.0 x 6.02 x 1023
= 1.2 x 1024
molecules
4. (a) 6.02 x 1023
÷ 6.02 x 1023
= 1.0 mol
(b) 1.8 x 1021
÷ 6.02 x 1023
= 30.0 mol
(c) 1.2 x 1023
÷ 6.02 x 1023
= 0.2 mol
(d) 2.4 x 1020
÷ 6.02 x 1023
= 4.0 x 10-4
mol
(e) 3.0 x 1023
÷ 6.02 x 1023
= 0.5 mol
5. (a) 1.5 x 71 = 106.5 g
(b) 2.5 x 32 = 80 g
(c) 2 x 98 = 196 g
(d) 0.5 x 17 = 8.5 g
(e) 2.5 x 267 = 667.5 g
(f) 0.5 x 188 = 94 g
6. (a) 0.5 mol x 22.4 dm3
mol-1
= 11.2 dm3
(b) 0.2 mo x 24 dm3
mol-1
= 4.8 dm3
(c) 1.5 mol x 24 dm3
mol-1
= 36 dm3
(d) 0.5 mol x 24 dm3
mol-1
= 12.0 dm3
(e) 2.5 mol x 22.4 dm3
mol-1
= 56 dm3
7. (a) 250 cm3
÷ 24000 cm3
mol-1
= 0.01 mol
(b) 500 cm3
÷ 22400 cm3
mol-1
= 0.02 mol
(c) 200 cm3
÷ 24000 cm3
mol-1
= 8.3 x 10-3
mol
(d) 750 cm3
÷ 24000 cm3
mol-1
= 0.03 mol
(e) 300 cm3
÷ 22400 cm3
mol-1
= 0.013 mol

8. (a) Mg + 2HCl  MgCl2 + H2
(b) (i) 1. Mol of Mg = 2.4 /24 = 0.1 mol
2. ration : 1 mol of Mg produced 1 mol MgCl2
0.1 mol Mg produced 0.1 mol MgCl2
3. mass of MgCl2 = 0.1 x 95 = 9.5 g
(ii) 1. Ratio : 1 mol of Mg produced 1 mol of H2
0.1 mol of Mg produced 0.1 mol of H2
2. volume of H2 = 0.1 mol x ÷ 24 dm3
mol-1
= 2.4 dm3
9. (a) C + 2PbO  CO2 + 2Pb
(b) 2 mol of PbO produced 2 mol of Pb
0.5 mol of PbO produced 0.5 mol of Pb
(c) 2 mol of PbO reacts with 1 mol of C
0.5 mol of PbO reacts with 0.25 mol of C
(d) (i) mol of PbO = 44.6 / 223 = 0.2 mol
(ii) mass of Pb produced = 0.2 x 207 = 41.4 g
10. (a) CuCl2 + Na2CO3  2NaCl + CuCO3
(b) (i) mol of CuCl2 = 0.5 x 50 / 1000 = 0.025 mol
(ii) 1 mol of CuCl2 produced 1 mol of CuCO3
0.025 mol of CuCl2 produced 0.025 mol of CuCO3
(iii) mass of salt = 0.025 x 124 = 3.1 g
11. (a) CaCO3 + 2HCl  CaCl2 + CO2 + H2O
(b) (i) mol CaCO3 = 5 /100 = 0.05 mol
(ii) 1 mol of CaCO3 produced 1 mol of CO2
0.05 mol of CaCO3 produced 0.05 mol of CO2
(iii) Volume of CO2 = 0.05 mol x 24 dm3
mol-1
= 1.2 dm3
12. (a) 2NaHCO3  Na2CO3 + CO2 + H2O
(b) (i) mol of NaHCO3 = 8.4 /84 = 0.1 mol
2 mol of NaHCO3 produced 1 mol of CO3
0.1 mol of NaHCO3 produced 0.05 mol of CO3
Volume of CO2 = 0.05 x 24 dm3
mol-1
= 1.2 dm3
(ii) Mass of Na2CO3 = 0.05 x 106 = 5.3 g
13. (i) Zn + 2HCl  ZnCl2 + H2
(ii) mol of HCl = 2 x 100 / 1000 = 0.2 mol
(iii) volume of H2 = 0.1 x 24 dm3
mol-1
= 2.4 dm3
14. (i) CaCO3 + 2HNO3  Ca(NO3)2 + CO2 + H2O
(ii) mol of CaCO3 = 5/100 = 0.05 mol
(iii) 1 mol of CaCO3 reacts with 2 mol of HNO3
0.05 mol of CaCO3 reacts with 0.1 mol of HNO3
(iv) 1 mol of CaCO3 produced 1 mol of CO2
0.05 mol of CaCO3 produced 0.05 mol of CO2
Volume of CO2 = 0.05 mol x 24 dm3
mol-1
= 1.2 dm3
15. (i) Na2CO3 + 2 HCl  2 NaCl + CO2 + H2O
(ii) mol of HCl = 1 x 100 /1000 = 0.1 mol
(iii) 2 mol of HCl reacts with 1 mol Na2CO3
0.1 mol of HCl reacts with 0.05 mol Na2CO3
mass of Na2CO3 = 0.05 x 106 = 5.3 g

16. (i) CuO + 2 HCl  CuCl2 + H2O
(ii) mol of CuO = 10 / 80 = 0.125 mol
1 mol CuO reacts with 2 mol HCl
0.125 mol CuO reacts with 0.25 mol HCl
Molarity of HCl = 0.25 x1000 / 100 = 2.5 mol dm-3
17. (a) mol of NaOH = 8/40 = 0.2 mol
Molarity of NaOH = 0.2 mol dm-3
(b) M1V1 = M2V2
0.2 x 50 = M2 x 150
M2 = 0.67 mol dm-3
18. (a) 2NaOH + H2SO4  Na2SO4 + 2H2O
(b) mol NaOH = 2 x 25 /1000 = 0.05 mol
(c) 2 mol of NaOH reacts with 1 mol of H2SO4
0.05 Mol NaOH racts with 0.025 mol of H2SO4
(d) molarity of H2SO4 = 0.025 x 1000 / 18.5 = 1.35 mol dm-3
19. (a) 2HCl + Na2CO3  2NaCl + CO2 + H2O
(b) mol of Na2CO3 = 1 x 25 /1000 = 0.025 mol
(c) (i) mol of HCl = 0.025 x 2 = 0.05 mol
(ii) volume of HCl = 0.05 x 1000 / 1.25 = 40 cm3
PAPER 2 :STRUCTURE
Question Explanation Mark 
1(a) Chemical formula that shows the simplest whole ratio of atom of each
elements in the compound.
(b)(i)
(ii)
(iii)
(iv)
Mass of Cu = 20.35 – 18.75 = 1.6 g
Mass of O = 20.75 – 20.35 = 0.4 g
Mol of Cu = 1.6 / 64 = 0.025 mol
Mol of O = 0.4 / 16 = 0.025 mol
Mol ratio : Cu : O = 0.025 : 0.025 = 1 : 1
Empirical formula CuO
CuO + H2  Cu + H2O
(c) 1.flow dry hydrogen
2.collect the gas
3. place lighted splinter at the mouth of the test tube.
2 (a) (i)
(ii)
46
189
1
1
(b) (i)
(ii)
2.408 × 1024
molecules
4.816 × 1024
atoms
1
1
(c) 0.0015 mol 1
(d) 434 g 1

(e)
276
28.8
= 0.03 mol Ag2CO3
2 mol of Ag2CO3 produce 2 mol of CO2
0.03 mol of Ag2CO3 produce 0.03 mol of CO2
0.03 × 24 dm3
= 0.72 dm3
1
1
1
Total 9
3(a) 1. green turns black
2. lime water turns cloudy
2
(b) CuCO3  CuO + CO2 1
(c) Mol of CuCO3 = 3.1 / 124 = 0.025 mol 1
(d) 1 mol of CuCO3 produced 1 mol of CO2
0.025 mol of CuCO3 produced 0.025 mol of CO2
Volume of CO2 = 0.025 mol x 22.4 dm3
mol-1
= 0.56 dm3
1
1
1
(e)(i)
(ii)
Black turns brown
CuO + H2  Cu + H2O
Mass of Cu = 0.025 x 64 = 1.6 g
1
1
1
10
ESSAY
4.
(a) (i) C H
mol 85.70 14.30
12 1
7.14 14.30
7.14 7.14
ratio 1 2
The empirical formula CH2
1
1
1………..3
(ii) ( CH2 )n = 56
[ 12 + 2(1) ]n = 56
56 // 4
14
The molecular formula C4H8
1
1
1………..3

(iii) Empirical formula Molecular formula
The formula shows that carbon
and hydrogen are present
The formula shows that carbon and
hydrogen are present
The formula shows that the
ratio of carbon to hydrogen is
1:2
The formula shows that one
molecule of X consists of 4 carbon
atoms and 8 hydrogen atoms
2
(b) 1 An empty crucible and its lid are weighed and the mass are recorded 1
2 Magnesium ribbon is cleaned with sandpaper, placed in crucible and
weighed again. The mass are recorded
1
3 The crucible and its contents are heated over a strong flame 1
4 The crucible lid opened once in a while during the experiment 1
5 When the magnesium does not burn anymore, the crucible and its contents
are cooled in room temperature,
1
6 and then weighed. The mass is recorded 1
7 The heating, cooling and weighing is repeated until the final mass becomes
constant
1
8 Result
Mass of crucible + lid = a g
Mass of crucible + lid + magnesium ribbon = b g
Mass of crucible + lid + magnesium oxide = c g
1
9 Mass of magnesium = (b – a) g
Mass of oxygen = (b – c) g
1
10 Mol of magnesium atom = b-a
24
Mol of oxygen atom = b-c
16
1
11 Simplest ratio mol of magnesium atom to mol of oxygen atom = x:y/ 1:1 1
12 Empirical formula : MgxOy / MgO 1
PAPER 3 : STRUCTURE

5(a)
Rubric Score
[Able to state three inferences according to the observations correctly]
Example:
Observations Inferences
i) white fumes is released. magnesium oxide is produced //
magnesium has been oxidised.
ii) Bright burning. magnesium is a reactive metal.
iii) The mass increases. magnesium has combined with oxygen.
3
5(b)
Rubric Score
[Able to record the data correctly]
Example:
Description Mass / g
The crucible and lid.
The crucible, lid and magnesium powder.
The crucible, lid and magnesium oxide.
123.36
128.16
131.36
3
5(c)
Rubric Score
[Able to calculate the mass of magnesium, mass of oxygen and show the steps to
determine the empirical formula of magnesium oxide accurately]
Example:
i) Mass of magnesium = 128.16 – 123.36 = 4.80 g
ii) Mass of oxygen = 131.36 – 128.16 = 3.2 g
iii)The empirical formula of magnesium oxide:
Element Mg O
No. of moles 4.8
24
= 0.2
3.2
16
= 0.2
Simplest ratio 1 1
Empirical formula : MgO
3
5(d)
Rubric Score
Cannot. Lead is less reactive metal 3

PS F4 Set 1

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