PHYSICAL SCIENCES: CHEMISTRY
PAPER 2
GRADE 12 
AMENDED SENIOR CERTIFICATE EXAMS
PAST PAPERS AND MEMOS
MAY/JUNE 2018

MEMORANDUM 

QUESTION 1
1.1 D ✓✓ (2)
1.2 A ✓✓ (2)
1.3 B ✓✓ (2)
1.4 B ✓✓ (2)
1.5 D ✓✓ (2)
1.6 C ✓✓ (2)
1.7 B ✓✓ (2)
1.8 D ✓✓ (2)
1.9 D ✓✓ (2) 
1.10 C ✓✓ (2) [20] 
QUESTION 2
2.1 
2.1.1 A ✓ (1)
2.1.2 D ✓ (1)
2.1.3 B ✓ (1)
2.1.4 E ✓ (1)
2.1.5 B ✓ (1) 
2.2 
2.2.1 

Accept

• Any correct arrangement of correct number of  atoms (2) 

2.2.2 ANY ONE: 

• Methyl ✓ethanoate ✓
  OR/OF
• Ethyl ✓methanoate ✓ (2) 

2.3 
2.3.1 A large molecule ✓composed of smaller monomer units covalently bonded to  each other in a repeating pattern. ✓  (2) 
2.3.2 Polyethene ✓ 
Accept:

• Polyethylene/polythene  (1) 

2.3.3 
Accept as reactant:

Accept as product:
 (3) 
2.4 Hydrolysis/Substitution ✓  (1) 

2.5

• Use concentrated strong base/NaOH/KOH/LiOH OR ethanolic/alcoholic  strong base/NaOH/KOH/LiOH. ✓/Use ethanol instead of water./No water. 
• Heat strongly ✓ 

Accept: Increase temperature (2) [18] 

QUESTION 3 
3.1

• Structure:
  The chain length/molecular size /molecular structure/molecular mass/  surface area increases. ✓ 
• Intermolecular forces:
  Increase in strength of intermolecular forces/induced dipole /London/  dispersion /Van der Waals forces/momentary dipoles. ✓ 
• Energy:
  More energy needed to overcome/break intermolecular forces. ✓
  OR
  Structure:
  From 4 C atoms to 1 C atom/bottom to top the chain length/molecular  size/molecular structure/molecular mass/surface area decreases. ✓ 
• Intermolecular forces:
  Decrease in strength of intermolecular forces/ induced dipole forces/  London forces/dispersion forces. ✓ 
  Energy:
  Less energy needed to overcome/break intermolecular forces. ✓  (3) 

3.2

• Alkanes have London/dispersion/induced dipole forces. ✓ 
• Alcohols have hydrogen bonding (in addition to London/dispersion/ induced dipole forces and dipole dipole forces). ✓ 
• Hydrogen bonding are stronger intermolecular forces than London/ dispersion/ induced dipole forces. ✓
  OR/OF
  More energy needed to overcome/break intermolecular forces in alcohols 
• Alcohols have higher boiling points than alkanes. ✓  (4)

3.3 Decrease ✓ (1)
3.4 Lower than✓ -  2-methylpropane/It is more branched/has a smaller surface area/has a  shorter chain length (than butane/chain isomer) ✓ 
OR
Butane/chain isomer is less branched /has larger surface area/longer chain  length (than 2-methylpropane). (2) [10] 

QUESTION 4
4.1 
4.1.1 Substitution/halogenation/bromonation✓  (1) 
4.1.2 Elimination/dehydration ✓  (1) 
4.1.3 Esterification/condensation ✓ (1) 
4.1.4 Addition/hydrohalogenation/hydrobromonation ✓ (1) 
4.2 
4.2.1 Catalyst/dehydrating agent/speeds up reaction ✓  (1)
4.2.2 Propyl ✓ ethanoate ✓/Propieletanoaat (2) 
4.2.3 (2) 

4.3  (5) 

[14] 

QUESTION 5
5.1 ONLY ANY ONE OF:

• Change in concentration of products/reactants ✓ per (unit) time. ✓ 
• Rate of change in concentration. ✓✓ 
• Change in amount/number of moles/volume/mass ✓ of products or  reactants per (unit) time. ✓ 
• Amount/number of moles/volume/mass (of products) formed/(reactants) used✓ per (unit) time.✓  (2)

5.2 
5.2.1 Surface area/State of division ✓  (1) 
5.2.2 ANY ONE:

• Amount/mass of magnesium ✓ 
• Concentration of HCℓ/acid
• (Initial) temperature (1)

5.3 
5.3.1  (5)

Marking criteria: Calculate change in m(Mg) or n(Mg) ✓ Substitute/Vervang 24 g∙mol-1 in n =  m  ✓                                                             M Use mol ratio: n(Mg) = n(H2) = 1:1 ✓ Substitute 25 dm3 in n =   V  ✓                                          Vm Final answer: 2,5 dm3 ✓

5.3.2 

ave rate / tempo = ∆n  ✓ 
                              ∆t
(5) 
5.4

• Larger surface area/state of division. ✓ 
• More particles (per volume) with correct orientation ✓
  OR
• More contact points./Meer kontakpunte. 
• More effective collisions per (unit) time./Frequency of effective collisions  increases./More particles collide with sufficient kinetic energy & correct  orientation per (unit) time.✓✓ (3) [17] 

QUESTION 6
6.1 The stage in a chemical reaction when the rate of forward reaction equals the  rate of reverse reaction./Both forward and reverse reactions take place at  same rate. ✓✓ 
OR
The stage in a chemical reaction when the concentrations of reactants and  products remain constant. ✓✓  (2) 
6.2 
6.2.1 2 ✓ (1)
6.2.2 1 ✓ (1)
6.2.3 3 ✓ (1) 
6.3 POSITIVE MARKING FROM QUESTION 6.2. 

6.4 Endothermic ✔ 

• (An increase in temperature) favours the reverse reaction. ✔ 
• An increase in temperature favours an endothermic reaction. ✔ (3) [15] 

QUESTION 7
7.1 Titration/Volumetric analysis ✓  (1) 
7.2 To measure the (exact) volume of acid needed to reach endpoint/to neutralise  the base. ✓ (1) 
7.3 Acids produce hydrogen ions (H⁺)/hydronium ions (H₃O⁺) in solution/when  dissolved in water. ✓✓ 
IF:

• Acids produce hydrogen ions (H⁺)/hydronium ions (H₃O⁺). ✓ (2)

7.4 H₂SO₄ ionises completely.✓ (1)
7.5 Blue to yellow✓ (1) 
7.6 (4)

7.7 POSITIVE MARKING FROM QUESTION 7.6. 

Marking guidelines: Formula: c =  n  ✓                                     V Substitution of : (0,1)(0,005)/0,0175 in n = cV ✓ Substitute V = 0,0425 dm3 ✓ Use [H3O+] : [H2SO4] = 2 : 1 ✓ Formula: pH = -log[H3O+] ✓ Substitute [H+] ✓ Final answer: 1,63 ✓

(7)

[17]

QUESTION 8
8.1 
8.1.1 Galvanic (cell)/Voltaic (cell) ✓  (1) 
8.1.2 Indicates phase boundary./Interphase /phase separator✓ (1)
8.1.3 Fe²⁺ → Fe³⁺ + e^- ✓✓  (2) 

8.1.4  (5) 

8.2
8.2.1 Pt ✓ (1) 
8.2.2 Iron(III) (ions)Ferric ions✓  (1)
8.2.3 2Fe³⁺ + Cu ✓→ 2Fe²⁺ + Cu²⁺ ✓ Bal. ✓ (3) 

[14]

QUESTION 9
9.1 
9.1.1 Electrolyte ✓ (1) 
9.1.2 Conduct electricity/Carry charges ✓  (1)
9.2 Cu(NO₃)₂ ✓ (1) 
9.3 Iron rod✓ - Reduction takes place. ✓ (2)
9.4 Cu → Cu²⁺ + 2e^-✓✓ (2) 

9.5 
9.5.1 Copper(II) (ions)/Cu²⁺ ✓and silver (ions)/Ag⁺ ✓ 
Accept

• Cu (ions) and Ag (ions) (Ions are stated in the question.)  (2) 

9.5.2 Ag⁺/silver(I) ions is a stronger oxidising agent ✓ than Cu²⁺/Copper(II) ions and will be reduced (more readily) ✓ to form silver/Ag on the iron rod.  (2 ) [11] 

QUESTION 10
10.1 
10.1.1 (Catalytic) oxidation (of ammonia)✓ (1) 
10.1.2 Neutralisation/acid-base reaction ✓  (1) 
10.2 
10.2.1 Nitrogen/N₂✓ (1)
10.2.2 NO₂/nitrogen dioxide✓ (1)
10.2.3 Nitric acid/HNO₃✓ (1)  
10.3 
10.3.1 2NH₃ + H₂SO₄ ✓ → (NH₄)₂SO₄ ✓ Bal. ✓  (3) 

10.3.2 4NH₃ + 5O₂ ✓ → 4NO + 6H₂O ✓ Bal. ✓  (3) 

10.4 % N = 28 × 100 ✓
                   80 ✓ 
= 35% ✓ (3) [14] 

TOTAL: 150