PHYSICAL SCIENCES
GRADE 12
PAPER 2 
NSC PAST PAPERS AND MEMOS
SEPTEMBER 2017

GUIDELINES FOR MARKING
This section provides guidelines for the way in which marks will be allocated. The broad  principles must be adhered to in the marking of Physical Sciences tests and examinations. 
1.1 MARK ALLOCATION
1.1.1 Definitions/Definisies: Two marks will be awarded for a correct  definition. No marks will be awarded for an incorrect or partially correct  definition. 
1.1.2 Calculations: 

• Marks will awarded for: correct formula, correct substitution,  correct answer with unit.
• No marks will be awarded if an incorrect or inappropriate formula  is used, even though there may be relevant symbols and  applicable substitutions. 

1.1.3 Explanations and interpretations:
Allocation of marks to questions requiring interpretation or explanation  e.g. AS 1.4, 2.2, 2.3, 3.1, 3.2 and 3.3, will differ and may include the  use of rubrics, checklists, memoranda, etc. In all such answers  emphasis must be placed on scientific concepts relating to the  question. 

1.2 FORMULAE AND SUBSTITUTIONS/FORMULES EN SUBSTITUSIE 
1.2.1 Mathematical manipulations and change of subjects of appropriate  formulae carry no marks, but if a candidate starts with the correct  formula and then changes the subject of the formula incorrectly, marks  will be awarded for the formula and the correct substitutions. The mark  for the incorrect numerical answer is forfeited. 
1.2.2 When an error is made during substitution into a correct formula, a  mark will be awarded for the correct formula and for the correct  substitutions, but no further marks will be given. 
1.2.3 Marks are only awarded for a formula if a calculation had been  attempted, i.e. substitutions have been made or a numerical answer  given. 
1.2.4 Marks can only be allocated for substitutions when values are  substituted into formulae and not when listed before a calculation  starts. 
1.2.5 All calculations, when not specified in the question, must be done to  two decimal places. 

1.3 UNITS
1.3.1 Candidates will only be penalised once for the repeated use of an  incorrect unit within a question or sub-question. 
1.3.2 Units are only required in the final answer to a calculation. 
1.3.3 Marks are only awarded for an answer, and not for a unit per se.  Candidates will therefore forfeit the mark allocated for the answer in  each of the following situations: 

• correct answer + wrong unit
• wrong answer + correct unit
• correct answer + no unit. 

1.3.4 SI units must be used except in certain cases, e.g. V.m^-1 instead of  N.C^-1, and cm.s^-1 or km.h^-1instead of m.s^-1 where the question warrants  this. (This instruction only applies to Paper 1). 

1.4 POSTIVE MARKING
Positive marking regarding calculations will be followed in the following cases:
1.4.1 Sub-question to sub-question: When a certain variable is calculated in  one sub-question (e.g. 3.1) and needs to be substituted in another (3.2  or 3.3), e.g. if the answer for 3.1 is incorrect and is substituted correctly  in 3.2 or 3.3, full marks are to be awarded for the subsequent sub questions. 
1.4.2 A multi-step question in a sub-question: If the candidate has to  calculate, for example, current in the first step and gets it wrong due to a  substitution error, the mark for the substitution and the final answer will  be forfeited. 
1.4.3 If a final answer to a calculation is correct, full marks will not automatically  be awarded. Markers will always ensure that the correct/ appropriate  formula is used and that workings, including substitutions, are correct. 
1.4.4 Questions where a series of calculations have to be made (e.g. a circuit  diagram question) do not necessarily always have to follow the same  order. FULL MARKS will be awarded provided it is a valid solution to the  problem. However, any calculation that will not bring the candidate closer  to the answer than the original data, will not count any marks. 
1.4.5 If one answer or calculation is required, but two given by the candidate,  only the first one will be marked, irrespective of which one is correct. If  two answers are required, only the first two will be marked, etc. 
1.4.6 Normally, if based on a conceptual mistake, an incorrect answer cannot  be correctly motivated. If the candidate is therefore required to motivate  in question 3.2 the answer given to question 3.1, and 3.1 is incorrect, no  marks can be awarded for question 3.2. However, if the answer for e.g.  3.1 is based on a calculation, the motivation for the incorrect answer for  3.2 could be considered. 
1.4.7 If instructions regarding method of answering are not followed, e.g. the  candidate does a calculation when the instruction was to solve by  construction and measurement, a candidate may forfeit all the marks  for the specific question. 
1.4.8 For an error of principle, no marks are awarded (Rule 1) e.g. If the  potential difference is 200 V and resistance is 25 Ω, calculate the current. 

1.5 GENERAL PRINCIPLES OF MARKING IN CHEMISTRY
The following are a number of guidelines that specifically apply to Paper 2. 
1.5.1 When a chemical FORMULA is asked, and the NAME is given as  answer, only one of the two marks will be awarded. The same rule  applies when the NAME is asked and the FORMULA is given. 
1.5.2 When redox half-reactions are to be written, the correct arrow should  be used. If the equation is the correct answer, the following marks will be given: 

H₂S → S + 2H⁺ + 2e^-(²/₂) 

• H₂S ⇋ S + 2H⁺ + 2e^-(¹/₂)
• H₂S ← S + 2H⁺ + 2e^-(⁰/₂)
• S + 2H⁺ + 2e^- ← H₂S (²/₂)
• S + 2H⁺ + 2e^- ⇋ H₂S (⁰/₂) 

1.5.3 When candidates are required to give an explanation involving the  relative strength of oxidising and reducing agents, the following is  unacceptable: 

• Stating the position of a substance on Table 4 only (e.g. Cu is above  Mg).
• Using relative reactivity only (e.g. Mg is more reactive than Cu).
• The correct answer would for instance be: Mg is a stronger reducing  agent than Cu, and therefore Mg will be able to reduce Cu²⁺ ions to Cu.  The answer can also be given in terms of the relative strength as  electron acceptors and donors. 

1.5.4 One mark will be forfeited when the charge of an ion is omitted per  equation. 
1.5.5 The error carrying principle does not apply to chemical equations or  half-reactions. For example, if a learner writes the wrong  oxidation/reduction half-reaction in the sub-question and carries the  answer to another sub-question (balancing of equations or calculations  of Eθcell) then the learner is not credited for this substitution. 
1.5.6 When a calculation of the cell potential of a galvanic cell is expected,  marks will only be awarded for the formula if one of the formulae  indicated on the data sheet (Table 2) is used. The use of any other  formula using abbreviations etc. will carry no marks. 
1.5.7 In the structural formula of an organic molecule all hydrogen atoms  must be shown. Marks will be deducted if hydrogen atoms are omitted. 
1.5.8 When a structural formula is asked, marks will be deducted if the  candidate writes the condensed formula. 
1.5.9 When an IUPAC name is asked, and the candidate omits the hyphen  (e.g. instead of 1-pentene the candidate writes 1 pentene), marks will  be forfeited. 

MEMORANDUM

QUESTION 1
1.1 B ✔✔ (2)
1.2 D ✔✔ (2)
1.3 A ✔✔ (2)
1.4 C ✔✔ (2)
1.5 C ✔✔ (2)
1.6 C ✔✔ (2)
1.7 A ✔✔ (2)
1.8 B ✔✔ (2)
1.9 C ✔✔ (2)
1.10 C ✔✔ (2) [20] 

QUESTION 2
2.1.1 B ✔ (1)
2.1.2 F ✔ (1)
2.2.1 pentan-2-one ✔✔ 
Accept  : 2-pentanone
IF/AS pentanone one mark (2)

2.2.2 3-bromo-3,4-dimethylhexane
Marking criteria:

• Correct stem i.e.hexane ✔
• First substituent, bromo, correctly identified ✔
• Second substituent, dimethyl, correctly identified ✔
• Subtract a mark for missing hyphens, commas ,incorrect numbering.  (3) 

2.3.1 Ester ✔ (1) 
2.3.2 
  (2) 
Marking criteria:

• Functional group correct. 1/2
• Whole structure correct. 2/2            

2.3.3
  (2) 
Marking criteria:

• Functional group correct. 1/2
• Whole structure correct. 2/2 

2.4.1 Organic molecules with the same molecular formula  but different structural  formulae. (2) 
2.4.2 Ethyl ✔ methanoate  (2)
2.5.1 Thermal  (1) 
2.5.2 C₄H₁₀ ✔ (1) [16] 

QUESTION 3
3.1 Secondary ✔ (1)
3.2
3.2.1 Substitution ✔ (1) 
3.2.2 Elimination ✔/dehydrohalogenation  (1) 
3.2.3 Addition ✔/hydration  (1) 
3.3 
(2) 

Marking criteria:

• Functional group correct 1/2
• Whole structure correct. 2/2 

3.4  (5) 
Accept H₂O  
3.5 Concentrated ✔  (1) [12] 

QUESTION  4 
4.1.1 The pressure exerted by a vapour at equilibrium with its liquid in a closed  system.✔✔ (2 OR0) (2) 
4.1.2 Molecular mass (size) ✔ (1) 
4.1.3 E has two sides for hydrogen bonding and F has one OR E forms dimers ✔ (1) 
4.2.1 From A to C/Van A na C 

• Chain length decreases/surface area decreases/More branches. ✔
• Strength of intermolecular forces/London forces/dispersion forces/induced dipole forces decreases. ✔
• Less energy needed to overcome/break intermolecular forces. ✔

OR

From C to A/Van C na A 

• Chain length increase/surface area increases/Less branches. ✔
• Strength of intermolecular forces /London forces/dispersion forces/induced dipole forces increases✔
• Less energy needed to overcome/break intermolecular forces increases. ✔ (3) 

4.2.2 Surface area ✔ /Position of side (methyl) chain  (1) 
4.2.3 n(O₂) = V/Vm 
 = 96/24 ✔
 = 4 mol 

n(C₆H₁₄) = 2/19 x 4 ✔
 = 0,42 mol 
Marking Criteria:  

• Divide by 22,4✔
• Use ratio ✔
• Multiply by 4163✔
• Final answer✔

Net energy released = 0,42 x - 4163  = -1 748 kJ/mol ✔
NOTE: Accept positive answer (4) [12] 

QUESTION  5 
5.1.1 (Use zinc) powder ✔ / Increase surface area (of zinc)  (1) 
5.1.2 Increase (temperature) ✔ / Heat  (1) 
5.2.1 Reaction is complete✔ / Reaction stops/Zinc is used up/ (1) 
5.2.2 t₁✔
 Gradient highest✔ / Steepest gradient  (2)
5.3.1 

Criteria for graph Mg

• Shape as shown with a steeper gradient below graph of Zn, starting from  the same point and ends at the same point.✔
• Graph becomes horizontal in less time ✔ (2) 

5.3.2 Cu²⁺ is a stronger oxidising agent than H⁺ ✔✔ OR H⁺is a weaker oxidising  agent than Cu²⁺  (2) 
5.4.1 Decrease in temperature ✔  (1) 
5.4.2

• Increase in temperature increases reaction rate. ✔
• Kinetic energy of particles increases as temperature increases. ✔
• More particles will have sufficient/enough (kinetic) energy/ Ek ≥ EA. ✔
• More effective collisions per unit time/second. ✔

OR

• Decrease in temperature decreases reaction rate ✔
• Kinetic energy of particles decreases as temperature decreases ✔
• Less particles will have sufficient/enough (kinetic) energy/Ek ≥ EA. ✔
• Less effective collisions per unit time/second. ✔ (4) [14] 

QUESTION 6 
6.1 A reaction is reversible when products can be converted back to reactants.✔✔. (2 or/of 0) (2) 
6.2.1 Decreases ✔  (1)
6.2.2 Increases ✔ (1)
6.3 10^-3 ✔ (1)
6.4 Marking Criteria:

• Equilibrium n(AX₂) = Equilibrium c(AX₂) x V ✔
• Change in n(AX₂) = equilibrium n(AX₂) – initial n(NH₃) ✔
• USE RATIO for change in n(AX₂) and change in n(X₂) change.✔ 
• n(equilibrium) = n(initial) – n (change) for n(X₂) ✔
• Divide equilibrium n(X₂) by V to calculate equilibrium c(X₂).✔
• Correct K_c expression (formulae in square brackets).✔
• Substitution of concentrations into K_c expression. ✔
• Final answer/Finale antwoord: 1,84 dm^-3 ✔

POSITIVE MARKING from QUESTION 6.3 
OPTION 1

• Ratio ✔

K_c = [AX₂]² / [X₂]³✔ 
10^-3 = (0,1)²/ (0,46 / V – 0,15)³ ✔
 V = 0,2 dm^-3✔

OPTION 2
 K_c = [AX₂]² / [X₂]³ ✔
 10^-3 = (0,1)²/ [X₂]³ 
 [X₂] = 2,15 mol.dm^-3 

2,15 V + 0,15 V = 0,46 ✔
 V = 0,2 dm³ ✔

• Ratio✔

OPTION 3
USING CONCENTRATIONS

•  K_c = [AX₂]² / [X₂]³ ✔
  10^-3 = (0,1)²/ [X₂]³ 
   [X₂] = 2,15 mol.dm^-3 

• Ratio✔

c = n/V ✔
2,3 = 0,46 /V ✔
V = 0,2 dm^-3 ✔ (8)

6.5 Decreases✔

• An increase in temperature causes a decrease in Kc. ✔
• When the temperature is increased, the reaction that will oppose this  increase in temperature, will be favoured. ✔
• Reverse reaction is favoured by a decrease in temperature. ✔

OR

• The forward reaction is exothermic. ✔
• An increase in temperature favours the endothermic reaction. ✔
• The reverse reaction is favoured. ✔ (4) [18] 

QUESTION 7
7.1.1 Hydrolysis ✔ (1) 
7.1.2 Weak ✔

• K_b value is < 1 ✔

OR

• K_b value is low (2) 

7.1.3 Acids ✔

• Both act as proton (H⁺)donors ✔ /Donate protons(H⁺)/Lose protons(H⁺). (2)

7.2.1 Burette ✔(1)
7.2.2 20 cm³ ✔ (1)
7.2.3 POSITIVE MARKING from QUESTION 7.2.2/ 

• n = Cv ✔
  n(NaOH) = 0,2 x 20/1000 ✔= 0,004 mol ✔ (4 x 10^-3 mol) (3) 

7.2.4 POSITIVE MARKING FROM QUESTION 7.2.3

• n(H₂) = ½ (0,004)✔ = 0,002 mol (2 x 10^-3 mol)
  c(H₂X)= n / V = 0,002/ (40/1000) ✔= 0,05 mol.dm^-3 
  pH = - log[H₃O⁺] ✔= - log(2 x 0,05) ✔= 1 ✔(5) 

Marking Criteria :

• Use of ratio NaOH : H₂X ✔
• Substitution into c = n/V to calculate n(H₂X).
• Use formule: pH = -log [H₃O⁺] ✔
• Substitution of 2 x c(H₂X) into [H₃O⁺]✔
• Final answer✔

7.2.5 Neutral ✔

• (Titration of) strong base with a strong acid. ✔  (2) [18] 

QUESTION 8
8.1 It is solution/liquid/dissolved substance  that conducts electricity through  the movement of ions. ✔. (2) 
8.2 Positive ✔

• It is the cathode. ✔

OR

• It is the electrode where reduction takes place.  (2) 

8.3.1 1 mol.dm^-3 ✔ (1) 
8.3.2        ✔                      ✔                  ✔
Ti (s)/ Ti³⁺(aq) // Ag⁺ (aq)/ Ag(s) Accept Ti/ Ti³⁺// Ag⁺/ Ag (3) 
8.4 E⁰_cell = E⁰_cathode  – E⁰_anode ✔
 2,43 = 0,80 – E0anode  

E⁰_anode = -1,63 V ✔

Notes: 

• Accept any other correct formula from the data sheet.
• Any other formula using unconventional abbreviations, e.g. E⁰_cell = E⁰_OA – E⁰_RA  3/4 (4) 

8.5 3 Mg + 2 Ti³⁺ ✔  ⇌ 3 Mg²⁺ + 2 Ti ✔  ✔ balancing 
Reactants ✔ Products ✔ Balancing ✔
Accept single arrow  (3) [17] 

QUESTION 9
9.1 Negative ✔
9.2 To improve (electrical) conductivity.  (1)
9.3 Decreases ✔

• Cu Cu²⁺+ 2e^- ✔✔
  Ignore phases (3) 

9.4 n(Cu) = ½ x 2✔                             Cu²⁺+ 2e^- → Cu 
 = 1 mol 
m(Cu) = nM = 1(63,5) ✔= 63,5 g 
% Purity = m(pure)/ m(impure) x 100 
m(impure) = 63,5 x 100 ✔
                          95,7 ✔
 = 66,35 g ✔ (5) [10] 

QUESTION 10 
10.1.1 Ammonia ✔(1)
10.1.2 Ostwald(process) ✔ (1)
10.1.3 NO ✔ (1)
10.1.4 HNO₃ ✔ and NO ✔ (2)
10.1.4 NH3(g) + HNO₃(aq) ✔ NH₄NO₃ ✔                  Bal. ✔ (3)
Notes:

• Reactants ✔ Products ✔ Balancing ✔
• Ignore  ⇌ and phases 
• Marking rule 6.3.10

10.2.1 The percentage of fertiliser in the bag ✔/ The percentage of the primary  nutrients (N:P:K) in the bag of fertilizers. (1) 
10.2.2 Dead zones ✔ (1) 
10.2.3 % N in (NH₄)₂SO₄ = 2(14) x 100 ✔ = 21,21% 
                                             132 
m(N) = 21,21 x 39,6 ✔ = 8,399 kg 
             100 
% N in bag = 7/15 x 36 ✔= 16,8% 
16,8 x m = 8,399 kg ✔
100  
m = 50 kg ✔ (5) [15] 

Marking criteria: 

• Calculate  % N in (NH₄)₂SO₄ ✔
• Calculate m(N) in (NH₄)₂SO₄ ✔
• % N in bag of fertiliser✔
• Calculate mass m ✔
• Final answer: 50 kg ✔

TOTAL 150