PHYSICAL SCIENCES PAPER 2
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
SEPTEMBER 2018

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: 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 be 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
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 has 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-1 instead of m.s-1 where the question warrants this. (This instruction only applies to Paper 1).
1.4 POSITIVE MARKING
Positive marking regarding calculations will be followed in the following cases:
1.4.1 Subquestion to subquestion: When a certain variable is calculated in one subquestion (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 subquestion: 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.
1.5.1When 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
H₂S → S + 2H+ + 2e- (2/2)
is the correct answer, the following marks will be given:
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 subquestion and carries the answer to another subquestion (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.

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

QUESTION 2
2.1.1 Alkanes ✓(1)
2.1.2 

(3)

2.1.3 CO₂ ✓ and H₂O ✓ (2)
2.2.1 Compounds with the same molecular formula, ✓ but different positions of the functional group. ✓(2)
2.2.2

3-methyl✓pentan-2-one ✓/ 3-methyl-2- pentanone (4)
2.3.1 2-chloro-3,4-dimethylhexane 

(3)
2.3.2 SECONDARY ✓ (1)
[16]

QUESTION 3
3.1 They are organic compounds that contain hydrogens and carbons only.✓✓. (2 or 0) (2)
3.2 53,3 ✓ (kPa)
From A to 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
Chain length increases/surface area increases/less branches. ✓
Strength of intermolecular forces/London forces/dispersion forces/induced-dipole forces increases.✓
More energy needed to overcome/break intermolecular forces. ✓ (4)
3.3 E ✓ (1)
3.3.1 Compound with lowest vapour pressure ✓ (1)
3.3.2

• Compound D(ethanol) has one site for hydrogen bonding, whereas compound E(methanoic acid) has two sites for hydrogen bonding. ✓
• Compound E has stronger hydrogen bonds than compound D. ✓(2)

[10]

QUESTION 4
4.1 Substitution ✓(1)
4.2.1 Waft your hand across the beaker/test tube toward your nose and sniff/smell (cautiously).✓
OR
Pour the ester into a bowl of water and this will help you to identify the smell of the ester.(1)
4.2.2 Esterification ✓(1)
4.2.3 Ethanoic acid ✓✓(2)
4.2.4

(2)
4.3

Accept: H₂O (4)
4.4.1

(2)
4.4.2 Concentrated H₂SO₄/Sulphuric acid ✓(1)
4.4.3 (Excess) water/H₂O ✓/(1)
[15]

QUESTION 5
5.1 ONLY ANY ONE OF:

• Change in concentration ✓ of reactants/products per (unit) time. ✓
• Rate of change in concentration. ✓✓
• Change in the 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 EQUAL TO ✓ (1)
5.3 Ave rate/Gem. tempo = ΔV = 28–0 ✓ = 1,87 (cm³.s^-1) ✓
                                            Δt     15–0
(2)
5.4

Examples:

• What is the effect of a catalyst on the rate of the reaction?
• What is the relationship between a catalyst and the rate of reaction? (2)

5.5 EXOTHERMIC ✓
(Net) Energy is absorbed./More energy is released than absorbed. ✓/ Energy of reactants > Energy of products.(2)
5.6 Experiment II ✓ (1)
5.7

• A catalyst provides an alternative pathway of lower activation energy. ✓
• More particles will have sufficient/enough (kinetic) energy / Ek ≥ EA. ✓
• More effective collisions per unit time /second. /Frequency of effective collisions increases.(3)

[13]

QUESTION 6
6.1 The stage in a chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. ✓✓ (2 marks or no marks)
OR
The state where the concentrations/quantities of reactants and products remain constant. ✓✓ (2 marks or no marks)(2)
6.2.1 DECREASES✓(1)
6.2.2 NO EFFECT✓
Kc will only change with a change in temperature.✓/Kc is temperature dependent(2)
6.3.1

Related Items

OPTION 1 
Kc = [NO₂]²✓
         [N₂O₄]
= (0,0125)² ✓
0,0336
= 0,00465

Kc = [NO₂]²
        [N₂O₄]
0,00465✓ = (0,0156)²✓
                     [N₂O₄]
[N₂O₄]eq = 0,052 mol.dm^-3

OR

OPTION 2
Kc = [NO₂]²
        [N₂O₄]
= (0,0107)²
0,0246
= 0,00465

OPTION 3 
K_c = [NO₂]²✓
        [N₂O₄]
= (0,0125)² ✓
0,0336
= 0,00465
K_c = [NO₂]²
         [N₂O₄]
0,00465✓= (0,0156)²✓
[N₂O₄]
[N₂O₄]eq = 0,052 mol.dm^-3

OR

OPTION 4
K_c = [NO₂]²
        [N₂O₄]
= (0,0107)²
0,0246
= 0,00465

n(N₂O₄)_eq = cV = (0,052)(2)✓= 0,1046 mol
n(NO₂)_eq = cV = (0,0156)(2)✓= 0,0312 mol
Δn(NO₂) = n(NO₂)eq - n(NO₂)i= 0,0312 – 0 = 0,0312 mol✓
Δn(N₂O₄) = ½(0,0312)✓= 0,0156 mol Ratio = 1:2
n(N₂O₄)i = n(N₂O₄)_eq + Δn(N₂O₄)
= 0,1046 + 0,0156
= 0,12 mol✓ (9)
6.3.2 EQUAL TO ✓ (1)
6.3.3 Decrease in concentration of N₂O₄. ✓/ Decrease in [ N₂O₄ ] (1)
[16]

QUESTION 7
7.1.1 WEAK ACID✓
It has a low Ka value ✓/ The ionisation is NOT complete/incomplete.(2)
7.1.2 C₂O4^2-✓ (1)
7.2.1 Hydrolysis is the reaction (of a salt) with water✓✓(2)
7.2.2 CH₃COOH ✓and OH-✓ (2)
7.3.1 pH = - log[H₃O⁺]✓
1 = - log[H₃O⁺]
[H₃O+] = 0,1 mol.dm^-3✓ (3)
7.3.2 POSTIVE MARKING FROM QUESTION 7.3.1
OPTION 1
n(NaOH) = cV ✓
= (0,025)(0,0322) ✓
=8,05 x 10^-4 mol
n(HCℓ)_excess = n(NaOH)
= 8,05 x 10^-4mol✓

OPTION 2
na = c_aV_a ✓
nb    c_bV_b
1  =   (0,1)Va    
1   (0,025)(32,2) ✓
V(HCℓ)_excess = 8,05 cm³
n(HCℓ)_excess = cV
=0,1(0,00805)
= 8,05 x 10^-4 mol✓ (3)
7.3.3 POSTIVE MARKING FROM QUESTION 7.3.2
OPTION 1
V(HCℓ)_excess = n/c
= 0,000805 / 0,1
= 0,00805 dm³
V(HCℓ)_reacted
= 0,050 – 0,00805✓
= 0,04195 dm³
n(HCℓ) _reacted = cV
= (0,1)(0,04195) ✓
= 0,004195 mol
n(CaCO₃) = ½ n(HCℓ) _reacted
= 0,0020975 mol✓
m(CaCO₃) = nM
= (0,0020975 )(100) ✓
= 0,20975 g
% purity/suiwerheid = 0,20975 x100✓
                                        0,3
= 69,92%✓

Marking guidelines

• Subtract 0,00805 from 0,005✓
• Substitution into n(HCℓ) reacted =cV ✓
• Use ratio 1 : 2✓
• Substitution into m(CaCO₃)= nM✓
• Calculate/Bereken: % purity✓
• Final answer: 69,92%✓

RANGE (69,92 TO 70%)

OPTION 2
[HCl] = [H₃O⁺] = 0,1 mol.dm^-3
nHCl initial = cV = 0,1 x 50/1000 ✓
= 5 x 10^-3 mol
nHCl excess = 8,05 x 10 -4 mol (From 7.3.2)
nHCl reacted = 5 x 10^-3 - 8,05 x 10^-4✓
= 4,195 mol
nCaCO₃ = ½ x 4,195✓
= 2,097 x 10^-3 mol
m = nM= 2,097 x 10^-3 x 100✓
= 0,20975 g
%Purity = m_pure x 100
= 0,20975/0,3 x100 ✓
= 69,92% ✓ (6)
[19]

QUESTION 8
8.1 A solution / liquid / dissolved substance that conducts electricity through the movement of ions ✓✓ (2)
8.2.1 Pressure: 1 atmosphere (atm) ✓/101,3 kPa / 1,013 x 105 Pa (1)
8.2.2 Platinum is a conductor of electricity ✓ (1)
8.3 ANODE ✓ (1)
8.4 

(5)
8.5 BECOMES ZERO✓
The electrical circuit is incomplete✓(2)
8.6 3Mg(s) + 2Aℓ +3(aq) ✓ → 3Mg+2(aq) + 2Aℓ (s) ✓ Bal ✓

(3)
[15]

QUESTION 9
9.1.1 Battery ✓/Energy source/Power source (1)
9.1.2 Chlorine ✓/Cℓ₂ (1)
9.1.3 2H₂O(ℓ) + 2e- → H₂(g) + 2OH-(aq) ✓✓
Ignore phases
Notes:
H₂(g) + 2OH-(aq)← 2H₂O(ℓ) + 2e- (2/2)          2H2O(ℓ) + 2e- ⇌ H₂(g) + 2OH-(aq) (1/2)
H₂(g) + 2OH-(aq) ⇌2H₂O(ℓ) + 2e- (0/2)           2H₂O(ℓ) + 2e- ← H₂(g) + 2OH-(aq) (1/2) (2)
9.1.4 Cℓ ✓ OR NaCℓ (1)
9.2 [H₃O+] > 1 x 10^-7✓
Solution is alkaline✓/NaOH is formed (2)
9.3 Cu⁺² is a stronger oxidising agent ✓(than Na⁺) and will be reduced ✓ to Cu.✓(3)
[10]

QUESTION 10
10.1.1 Vanadium pentoxide V₂O₅ ✓ (1)
10.1.2 H₂SO₄ ✓ (1)
10.1.3 Water ✓ (1)
10.1.4 Contact (process) ✓ (1)
10.1.5 NH₃(g) + H₂SO₄(aq) ✓ → ( NH₄)₂SO₄ ✓ Bal. ✓ (3)
Notes:

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

10.1.6 HIGHER THAN 450ºC ✓
Mark independently:

• For all pressure values ✓
• Yield is lower for graph Q ✓
• Reverse reaction which is endothermic will be favoured ✓ (4)

10.2.1 Ensures quality seeds, fruit, vegetables and flowers. ✓
OR
Help plants fight frost and resist diseases. ✓ (1)
10.2.2 OPTION 1
% fertiliser = 100 – 76 = 24%✓
% K = 24 – (4+8) ✓= 12
m(K) = 12
x m(bag) = 6 kg ✓
100
m(bag) = 50 kg ✓

OPTION 2
% fertiliser = 100 – 76 = 24%✓
% K = 24 – (4+8) ✓= 12
m(K) = 12 x m(fertiliser) = 6 kg ✓
            24
m(fertiliser) = 12 kg ✓
m(filler) = 12 x 76 = 38 kg (24 : 76)
                24
m(bag) = 12 + 38 = 50 kg

(4)
[16]

TOTAL: 150