PHYSICAL SCIENCES: CHEMISTRY
PAPER 2
GRADE 12 
NSC PAST PAPERS AND MEMOS
FEBRUARY/MARCH 2017

MEMORANDUM 

QUESTION 1
1.1 B ✔✔ (2)
1.2 B ✔✔ (2)
1.3 A ✔✔ (2)
1.4 A ✔✔ (2)
1.5 C ✔✔ (2)
1.6 D ✔✔ (2)
1.7 C ✔✔ (2)
1.8 B ✔✔ (2)
1.9 C ✔✔ (2) 
1.10 A ✔✔ (2)

[20] 

QUESTION 2
2.1 

2.1.1 B ✔ (1)
2.1.2 D OR/OF E ✔ (1)
2.1.3 F ✔ (1) 

2.2 

2.2.1 Butanal ✔ (1) 
2.2.2 2,3,3-trimethyl✔but-1-ene ✔ 
Accept
2,3,3- trimethyl ✔-1- butene

(3) 

(2) 
2.4 

2.4.1 Esterification / Condensation ✔   (1) 
2.4.2 Propan-1-ol ✔ ✔ 
If propanol (1 mark)  (2) 
2.4.3  

(2) 

2.4.4 Propyl ✔ butanoate ✔(2)

[16] 

QUESTION 3
3.1

• The temperature at which the vapour pressure equals atmospheric (external)  pressure. ✔✔ (2 or 0)  (2) 

3.2

• Flammable / Catch fire easily. / Volatile ✔ (1) 

3.3 
3.3.1

• Use straight chain✔ primary alcohols ✔ (2) 

3.3.2. OPTION 1

• Structure
  Chain length / more C atoms in chain / molecular size / molecular mass / surface area increases from top to bottom / butan-1-ol to hexan-1-ol.✔
• Intermolecular forces
  Intermolecular forces / Van der Waals forces / London forces / dispersion  forces increases from top to bottom / butan-1-ol to hexan-1-ol. ✔
• Energy
  Energy needed to overcome / break intermolecular forces increases from  top to bottom / butan-1-ol to hexan-1-ol. ✔

OPTION 2

• Structure
  Chain length / number of C atoms in the chain / molecular size / molecular  mass/surface area decreases from bottom to top / hexan-1-ol to  butan-1-ol. ✔
• Intermolecular forces
  Intermolecular forces / Van der Waals forces/London forces / dispersion  forces decreases from bottom to top/hexan-1-ol to butan-1-ol. ✔
• Energy
  Energy needed to overcome / break intermolecular forces decreases from  bottom to top / hexan-1-ol to butan-1-ol. ✔  (3) 

3.4 Remains the same / Bly dieselfde ✔ (1) 
3.5 

3.5.1 Functional group / Type of homologous series ✔ (1) 
3.5.2

• Type of intermolecular forces
  Between molecules of aldehyde / hexanal are dipole-dipole forces. ✔
• Between molecules of alcohols / hexan-1ol are (in addition to dipole-dipole  forces and London forces) hydrogen bonds. ✔
• Strength of intermolecular forces
  Dipole-dipole forces are weaker than hydrogen bonds. ✔
  OR
  Hydrogen bonds are stronger than dipole-dipole forces.
• Energy
  More energy needed to overcome / break intermolecular forces in  hexan-1-ol. ✔
  OR
  Less energy needed to overcome / break intermolecular forces in  hexanal.✔  (4)

[14]

QUESTION 4
4.1 

4.1.1 Substitution / hydrolysis ✔ (1)
4.1.2 H₂O/water ✔
OR
Dilute sodium hydroxide /NaOH(aq) 
OR
Dilute potassium hydroxide/KOH(aq)(1)
4.1.3 Tertiary ✔ (1) 
4.1.4 Elimination / dehydrohalogenation / dehydrobromination ✔ (1) ✔✔
4.1.5 2-methylprop-1-ene / methylpropene / 2-methylpropene  (2) 
4.1.6 Halogenation / bromination ✔ (1) 
4.1.7   (4) 

4.2 

4.2.1 Monomers  ✔ (1)
4.2.2 Alkenes ✔ (1) 
4.2.3 Addition (polymerisation) ✔ (1)

[14] 

QUESTION 5/VRAAG 5 
5.1 ANY TWO:

• Increase temperature of HCℓ. ✔
• Add a catalyst ✔
• Increase the concentration of HCℓ. ✔
• Increase the state of division of CuCO₃ ✔
• Agitation / Stirring  ✔ (2)

5.2 Accepted range : 42 s to 50 s ✔ (1) 
5.3 
5.3.

• average = - Δ m
                      Δ t
  = - (169,76 - 170,000)✔
               (20 - 0) ✔
  = 0,012(g.s^-1)✔

If answer is negative (minus 1 mark)  (3) 
5.3.2 Pure sample:
m(CO₂)_formed = 170,00 – 169,73 ✔
 = 0,27 g  
Impure sample:
m(CO₂)_formed = 170,00 - 169,78 ✔
 = 0,22 g  
%Purity = 0,22 × 100  ✔
                0,27 
= 81,48% ✔ (4) 

5.3.3 POSITIVE MARKING FROM QUESTION 5.3.2. 

• n(CO₂) formed  = m 
                              M
  = 0,27
      44
  = 6,13   ×  10^-3 mol
  n(CO₂) formed  = V 
                              V_M
  6,13 × 10^-3 =      V  
                          22,4
  V = 0,137 dm³    (3) 

5.4 POSITIVE MARKING FROM QUESTION 5.2. 

(2) 

[15] 

QUESTION 6
6.1 Amount / number of moles / volume of (gas) reactants equals amount/number  of moles/volume of (gas) products. ✔
OR
A change in pressure will change the concentration of the reactants and  products equally. (1) 
6.2 

OPTION 1
K_C =    [HI]²   
         [H₂][I₂]
∴55,3 =         [H_I]²               
             (0,014)(0,0085)
∴[HI] = 0,08112 mol.dm^-3

OR
K_C =    [HI]²   
         [H₂][I₂]
∴55,3 =         X²               
             (0,014)(0,0085)
∴X = 0,08112 mol.dm^-3

Initial quantity I₂(mol) (mol) = 0,08112 + 0,017 
= 0,09812 mol 
m(I₂) = nM 
 = (0,09812)(254) ✔
= 24,92 g ✔

OPTION 2

n(HI at equilibrium) = (0,08112)(2) = 0,1622 mol 
n(HI formed) = n(HI at equilibrium) = 0,1622 mol
n(I₂ reacted) = ½n(HI formed) = 0,08112 mol 
n(I₂ initial) = n(I₂ reacted) + n(I2 equilibrium)  
= 0,08112 + 0,017 
 = 0,09812 mol 
m(I₂ initial) = nM 
 = (0,09812)(254) 
 = 24,92 (g)✔

OPTION 3

6.3 (Chemical/dynamic) equilibrium 
OR
The rate of the forward reaction equals the rate of the reverse reaction. ✔  (1) 
6.4

• Addition of a catalyst.
• Increase in pressure. (2) 

6.5.1 Endothermic 

• The rate of the forward reaction decreases more. / The rate of the reverse  reaction decreases less.
• A decrease in temperature favours the exothermic reaction. (3)

6.5.2 Decreases  (1) 
6.6 Reactants / H2 / I2 removed

[18] 

QUESTION 7
7.1 A substance that ionises incompletely/to a small extent.(2) 
7.2

• Oxalic acid
• Higher K_a value
  OR
• Carbonic acid has a lower Ka value .(2) 

7.3 

• H₂O
• (COO)2⁻₂ (2) 

7.4 (4) 
7.5 
7.5.1 (5)

7.5.2

• C / phenolphthalein
• Titration of weak acid and strong base.
  OR
• The endpoint will be at pH > 7 which is in the range of the indicator.  (2)

[17] 

QUESTION 8
8.1 

8.1.1 Salt bridge  (1) 
8.1.2 Voltaic / Galvanic cell (1) 

8.2 

8.2.1 Decreases(1)
8.2.2 Increases (1) 

8.3 

8.3.1 Y(s) → Y²⁺(aq) + 2e^- Ignore phases
OR
Mg(s) → Mg²⁺(aq) + 2e ^-

(2) 

8.3.2 Y(s) |Y²⁺(aq) ||Aℓ³⁺(aq) | Aℓ(s) OR/OF Mg(s) |Mg²⁺(aq) ||Aℓ³⁺(aq) | Aℓ(s 
OR
Y(s) | Y²⁺ (1 mol∙dm^-3) || Aℓ³⁺(1 mol∙dm^-3) | Aℓ(s) 
Accept
Y | Y²⁺ || Aℓ³⁺ | Aℓ (3) 
8.4 (5) 

[14] 

QUESTION 9
9.1 Bauxite (1)
9.2 Oxidation  (1)
9.3 Reduce melting point .
OR
To lower the temperature / energy needed to melt the Aℓ₂O₃. (1) 

9.4 Aℓ³⁺(aq) + 3e^- → Aℓ(s)
Ignore phases

(2) 
9.5

• C + O₂  → CO₂  Bal
  OR
• 2Aℓ₂O₃ + 3C  → 4Aℓ + 3CO₂  Bal  

(3) 

[8]

QUESTION 10/VRAAG 10 
10.1 
10.1.1 Ostwald (process) (1) 
10.1.2 Catalyst/Speeds up the rate of the reaction (1)
10.1.3 Nitrogen dioxide  (1)
10.1.4 3NO₂ + H₂O ⇌ 2HNO₃(aq) + NO Bal.  (2) 

10.1.5

• Decrease pressure / Increase volume
• Decrease temperature \ (2) 

10.2 
10.2.1 (Ratio of the) nitrogen, phosphorous and potassium in the fertiliser.  (1)
10.2.2 (6) 

Marking criteria: Use ratio 3/8 x 50 kg x 25 / 25 % Divide previous answer by  39 Multiply by 74,5  Final answer 8,94 kg

OPTION 4 %K =3/8 x 25 = 9,38%  m(K) = 9,38  x 50 = 4,69 kg              100  %K in KCℓ =   39   x 100 = 52,35%                      74,5  52,35% KCℓ: 4,69 kg  m(100% KCℓ) = 4,69  x 100                           52,35   = 8,96 kg

[14] 
TOTAL: 150