PHYSICAL SCIENCES PAPER 2
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
JUNE COMMON PAPER 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.
1.1MARK 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 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 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-1 instead 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.
CORRECT      ANSWER (1)       POSSIBLE       ANSWER (2)    POSSIBLE
I =  ✓                  R =    ✓             R = ✓               R =   ✓            I = V✓
     R                               I                             I                            I                        R
= 200   ✓                = 200  ✓                = 200  ✓          I = R ✓                       = 8A✓
    25                         25                           25                      V  
   =8A  ✓                   =8A                     =8A                   = 25
                                                                                        200
= 0.125A
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
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 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.

QUESTION 1
1.1 C ✓✓
1.2 A ✓✓
1.3 C ✓✓
1.4 B ✓✓
1.5 A ✓✓
1.6 D ✓✓
1.7 A ✓✓
1.8 C ✓✓
1.9 A ✓✓
1.10 B ✓✓ (10 x 2)
[20]

QUESTION 2
2.1
2.1.1 G ✓ (1)
2.1.2 F ✓ (1)
2.2
2.2.1 CnH2n+2 ✓ (1)
2.2.2 Carboxylic group ✓(1)
2.2.3 CH2 ✓ (1)
2.3
2.3.1 4-bromo-3-methylheptane

Marking criteria:

  • Stem i.e. heptan. ✓
  • Bromo and methyl ✓
  • Substituents correctly numbered, hyphens, commas and sequence correctly used. ✓(3)

2.3.2 pentan-3-one ✓✓
Accept:3-pentanone

IF:
pentanone (1/2) 

(2)
2.3.3 methyl ✓ methanoate ✓ (2)
2.4

2.4
 Marking criteria:
Whole structure correct. (2/2)
TWO methyl groups (1/2)

(2)
2.5
2.5.1 Sulphuric acid/H2SO4 ✓(1)
2.5.2

2.52
Marking criteria:
Whole structure correct/ (2/2)
Functional group correct (1/2)
Notes:
Condensed structural formula or semi structural formule (1/2) 

(2)
2.5.3 Flammable/Volatile ✓(1)

2.6
2.6

Marking criteria:
Whole structure correct.
Functional group correct (2/2)
Notes:
Condensed structural formula or semi structural formule(1/2)

2-methyl ✓propan-2-ol✓
Accept: 2-methyl-2-propanol

Marking criteria
2-methyl/2-metiel
Propan-2-ol 

(2)
2.7 CO2, ✓ H2O ✓ (2)
[24]

QUESTION 3
3.1 The temperature at which the vapour pressure equals the atmospheric (external) pressure. ✓✓(2)
3.2.1 Chain length/surface area/molecular size(mass)/number of C atoms. ✓(1)
3.2.2 Thermometer ✓(1)
3.3 C/ CH3CH2CH2OH ✓ (1)
3.4 From A to C

  • Chain length/surface area/molecular size/molecular mass/ number of C’s in the chain increases. ✓
  • Strength of intermolecular forces/London forces/induced dipole/dispersion forces increases. ✓
  • More energy needed to overcome/break intermolecular forces. ✓

From C to A

  • The chain length/surface area/molecular size/molecular mass/number of C’s in the chain decreases. ✓
  • Strength of intermolecular forces/London forces/induced dipole/dispersion forces decreases. ✓
  • Less energy needed to overcome/break intermolecular forces. ✓ (3)

3.5
3.5.1 A series of organic compounds with the same general formula. ✓✓ OR in which one member differs from the next with a CH2 group.(2 or 0)(2)
3.5.2 Between molecules of D there are hydrogen bonds. ✓
Between molecules of E there are dipole-dipole forces. ✓
Hydrogen bonds are stronger than dipole-dipole forces. ✓
More energy needed to overcome/break intermolecular forces.in D. ✓
OR
Between molecules of D there are hydrogen bonds. ✓
Between molecules of E there are dipole-dipole forces. ✓
Dipole-dipole forces are weaker than hydrogen bonds. ✓
Less energy needed to overcome/break intermolecular forces in E. ✓ (4)
3.5.3 E ✓
Higher boiling point ✓(2)
[16]

Related Items

QUESTION 4
4.1
4.1.1

  1. Substitution/halogenation/bromination. ✓(1)

4.1.2 Elimination/dehydrohalogenation ✓(1)
4.2
4.2.1 Bromine/Br2 ✓(1)
4.2.2 Heat/Ultraviolet ✓(1)
4.2.3 Butane ✓✓ (2)
4.3 
4.3

Marking criteria:
Whole structure (2/2)
Only functional group correct. (1/2) 

(2)
4.4
4.4

Marking criteria:
Whole structure(2/2)
Only functional group correct (1/2) 

(2)
4.5 Dilute KOH ✓ and mild heat ✓(2)
[12]

QUESTION 5
5.1 The change in concentration/mass/number (amount) of moles/volume ✓ of reactants or products per unit time. ✓(2)
5.2 Experiment 1 ✓

  • Highest temperature ✓
  • Largest surface area ✓ (3)

5.3
5.3.1 Temperature✓/Surface area/State of division.(1)
5.3.2 Equal to ✓
Number/amount of mol/mass of Mg used is the same. ✓(2)
5.4 More than ONE independent variable. ✓(1)
5.5

Marking criteria:

  • Formula : n = cV ✓
  • Substitution into: n = cV ✓
  • Divide by 24 ✓
  • Use of ratio n(HCℓ) = 2n(Mg) ✓
  • n(HCℓ) remaining = n(HCℓ)initial - n(HCℓ)reacting n(HCℓ) ✓
  • Substitution of 36,5 into n = m/M ✓
  • Final answer. ✓

n( HCℓ) initial/aanvanklik = c.V ✓
= 2 x 80/1 000 ✓ = 0,16 mol
n( Mg )reacting = m/M
= 0.24/24 ✓
= 0.01 mol
n(HCℓ)reacting = 2 x 0.01✓ = 0.02 mol
n (HCℓ)remaining = 0.16 – 0.02 ✓= 0.14 mol
m( HCℓ)remaining/= nM = 0.14 x 36,5 ✓= 5,11 g ✓ (7)
5.6
5.6.1 Number/Fraction of molecules/particles ✓(1)
5.6.2 P ✓ (1)
5.6.3 Increase in temperature

  • Increases (average) kinetic energy of particles. ✓
  • More particles have enough/sufficient (kinetic) energy to react. EK ≥ EA ✓
  • More effective collisions per unit time/second. ✓

OR
Rate/Frequency of effective collisions increases. ✓

  • Increases reaction rate ✓(4)

[22]

QUESTION 6
6.1 The stage in a chemical reaction when the rate of the forward reaction is equal to the rate of the reverse reaction. ✓✓(2)
6.2
6.2.1 Increases ✓(1)
6.2.2 Decreases ✓(1)
6.3 Change is opposed/Increase in pressure is opposed/The reaction that will oppose this increase in pressure/decrease the pressure will be favoured✓
Increase in pressure favours the reaction which produces less gas moles/volume. ✓
The reverse reaction is favoured. ✓ (3)
6.4
6.4

Notes:

  • Curves starts at same concentration and intercept at the same final concentration. ✓
  • Broken line on graph is below solid line and levels off faster than the solid line. ✓

(2)
6.5 
6.5.1 T = 23 °C ✓
Largest KC value. ✓ (2)
6.5.2

Marking criteria:

  • Correct KC expression (formulae in square brackets). ✓
  • Substitution of 0,32 for KC value. ✓
  • Substitute 0,5 as [N2O4] ✓
  • [NO2]equilibrium✓
  • Change in concentration of NO2 ✓
  • Using ratio: N2O4: NO2 = 1:2 ✓
  • Initial[N2O4] = Equilibrium[N2O4] + Change in [N2O4] ✓

KC = [NO2]2/[N2O4] ✓
0,32 ✓ = [NO2]2/0,5 ✓
[NO2] = 0,4 mol.dm-3

  N2O4  NO2 
Initial concentration (mol.dm-3)   ✓ 0,7  0
Change in concentration (mol.dm-3)  0,2  0,4 ✓
Equilibrium concentration (mol.dm-3)   0,5  0,4 ✓

Ratio ✓ (7)
6.5.3 Exothermic. ✓
When temperature increases, KC decreases ✓ therefore [products] decreases.
Reverse reaction is favoured. ✓
Increase in temperature favours the endothermic reaction (reverse reaction). ✓(4)
[22]

QUESTION 7
7.1
7.1.1 An acid that donates one proton (H+) ✓ per molecule.(2)
7.1.2 CH3COO ✓ (1)
7.1.3 Weak acid ✓
Low Ka value ✓ (2)
7.2
7.2.1 OPTION 1
KW = [OH-][H3O+]
10-14 = 1 x 10-5 [H3O+] ✓
[H3O+] = 10-9 mol.dm-3
pH = -log [H3O+] ✓
= -log 10-9 ✓
= 9 ✓

OPTION 2
pOH = -log [OH-] ✓
= -log(1 x 10-5) ✓
= 5✓
14 = pH +pOH
14 = pH + 5 ✓
pH = 9 ✓ (4)
7.2.2 c1V1 = c2V2
(1 x 10-5)(10) ✓ = (1 x 10-6)V2 ✓
V2 = 100 cm-3 ✓ (3)
7.3
7.3.1 n = cV ✓
= (0,8(100/1 000) ✓
= 0,08 ✓ (mol) (3)
7.3.2

Marking criteria:

  • n (KOH) = cV ✓
  • Substitution of c and V to find n(KOH) ✓
  • Use MOL RATIO: n(H2SO4) : n(KOH) = 1 : 2 ✓
  • n(H2SO4)reacted = n(H2SO4) initial – n(H2SO4)excess ✓
  • Use MOL RATIO: n(H2SO4) : n(Na2CO3) = 1 : 1 ✓
  • Substitution into m = nM to calculate m(Na2CO3) ✓
  • Final answer: 7.92 g/ 7.95 g ✓

POSITIVE MARKING from QUESTION 7.3.1
n(KOH) = cV ✓ = (0,3)(0,035) ✓ = 0,01 mol
n(H2SO4) = 1/2n(KOH) = 0,5(0,01) ✓ = 0,005 mol
n(H2SO4)reacted = 0,08 – 0,005✓= 0,075 mol
n(Na2CO3) = n(H2SO4)reacted = 0,075 mol ✓
m(Na2CO3) = nM = (0,075)(106) ✓ = 7,95 g ✓ (8)
7.4 CO32- + H2O ✓✓ HCO3 + OH- ✓✓ bal
OR CO32- + H2O ⇌ H2CO3 + OH-
OR Na2CO3 + H2O ⇌ NaHCO3 + OH-
Accept: Single arrow

Marking guidelines:

  • Reactants ✓ Products ✓ Balancing ✓
  • Marking rule 6.3.10 6.3.10 

(3)
[26]

QUESTION 8
8.1
8.1.1 Percentage yield decreases as the temperature increases. ✓
OR
Percentage yield increases as the temperature decreases. ✓(2)
8.1.2 50 ✓ (%) (1)
8.1.3 200 ✓ (atmosphere/atmosfeer) (1)
8.2 n(N2) = m/M = 112/28 ✓= 4 mol
n(NH3) = 2n(N2) = 2(4) ✓ = 8 mol
% Yield = Actual yield/Theoretical yield x 100
60 = Actual yield/Werklike opbrengs x 100/8 ✓ (4)
[8]

TOTAL: 150

Last modified on Monday, 12 July 2021 09:27