PHYSICAL SCIENCES PAPER 2
GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
MEMORANDUM
MAY/JUNE 2021

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

QUESTION 2
2.1

2.1.1 F (1)

2.1.2 B & F (1)

2.1.3 C (1)

2.2
2.2.1 Haloalkane / alkyl halide(1)

2.2.2  3,5-dibromooctane
Marking criteria:

• Octane
• Dibromo
• Substituents (dibromo) correctly numbered, hyphens, commas correctly used. (3)

2.3
2.3.1 Pentan-3-one 
OR
3-pentanone
Marking criteria

• Pentanone/pentanoon 
• Correct position of functional group (2)

2.3.2 3-methyl butan-2-one
OR
3-methyl butanone

OR
methyl butanone

OR
3-methyl- 2-butanone (2)

2.4
2.4.1 Hexyl methanoate  (2)

2.4.2 
(1)

2.5
2.5.1 Cracking/Elimination  (1)

2.5.2 C₇H₁₆(2)

2.5.3

Notes

• Functional group
• Whole structure correct (2)

[19]

QUESTION 3
3.1 Marking guidelines;
If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark
The pressure exerted by a vapour at equilibrium with its liquid in a closed system.(2)

3.2 Functional group/Type of intermolecular forces/Homologous series (1)

3.3 B (1)

3.4 Marking criteria

• State hydrogen bonding in A.
• State dipole-dipole forces in B
• Compare strengths of IMFs
• Compare energies required
• Compound A/butan-1-ol has hydrogen bonding (dipole-dipole and London forces) between molecules. 
• Compound B/butan-2-one has dipole-dipole forces (and London forces) between molecules. 
• Intermolecular forces in compound A/butan-1-ol are stronger than intermolecular forces in compound B/butan-2-one. 
• OR
  Intermolecular forces in compound B/butan-2-one are weaker than intermolecular forces in compound A/butan-1-ol. 
• More energy is needed to overcome/break intermolecular forces in compound A/butan-ol than in compound B/butan-2-one. (4)

3.5
3.5.1 Boiling point (of compound A/butan-1-ol)(1)

3.5.2 Gas (1)

3.5.3

Marking criteria:

• Curve C starts below curve A
• Curve C remains below curve A(2)

Accept

• If C is labelled as B
• If graph below graph A is unlabelled

Note:If both graphs unlabelled: 0 marks
[12]

QUESTION 4
4.1
4.1.1 Heat/sunlight/ultraviolet light/radiation/light(1)

4.1.2 HBr/hydrogen bromide(1)

4.1.3 Hydrolysis(1)

4.1.4 H₂O/water
Accept
hydrogen oxide

OR
NaOH/KOH/LiOH/sodium hydroxide/potassium hydroxide/lithium hydroxide (1)

4.1.5 2-bromo propane(2)

4.2 Marking criteria:
(Mark bullets independently)

• React chloroethane with (conc) NaOH or NaOH in ethanol. 
• Indicate heat/Δ (on the arrow) or as a reactant in the reaction of chloroethane. 
• Correct condensed formula for ethene as product.
• Product NaCℓ in the reaction of chloroethane. 
• Product H₂O in the reaction of chloroethane. 
• React ethene with H₂
• Indicate Pt on the arrow of / at the reaction of ethene with H₂ 
• Correct condensed formula of ethane as product. 

                     + NaOH (in ethanol)
CH₃CH₂Cℓ + (conc) NaOH → CH₂CH₂ + NaCℓ + H₂O
CH₂CH₂ + H₂ → CH₃CH₃ 
Note:
Any additional reactants or products: Deduct one mark per reaction (8)
[14]

QUESTION 5
5.1 NOTE
Give the mark for per unit time only if in context of reaction rate.
ANY ONE

• Change in concentration of products/reactants per (unit) time.
• 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.
• Rate of change in concentration/amount/number of moles/volume/mass. (2 or 0)(2)

5.2

• Time/tyd
• Volume of gas/CO₂/carbon dioxide (in gas syringe)

OR

• Time taken for Aℓ₂(CO₃)₃ to be used up.

Accept
Measure volume of gas/CO₂ at regular time intervals.(2)

5.3 Experiment II:

• More (HCℓ) particles per unit volume./More particles with correct orientation.
• More effective collisions per unit time./Higher frequency of effective collisions.
• Higher reaction rate.(3)

OR
Experiment I:

• Less (HCℓ) particles per unit volume.
• Less effective collisions per unit time./Lower frequency of effective collisions.
• Lower reaction rate.

5.4 OPTION 1
ave rate = - Δn 
                   Δt
4,4 x 10^-3 = -n_r - 0.016 
                       2,5 (0)
n[Aℓ₂(CO₃)₃] = 0,005 (mol)

Marking criteria:

• Substitute average rate and Δt
• Substitute Δn. 
• Final answer: 0,005 (mol) 

OPTION 2
ave rate = - Δn 
                   Δt
4,4 x 10^-3 = Δn 
                   2,5
n[Aℓ₂(CO₃)₃] = 0,016 – 0,011
= 0,005 mol

NOTE

• Accept negative answers when the negative sign in front of the formula is omitted
• Do not penalise if initial and final mole values or time values are swopped.

OPTION 3
With reference to CO₂
ave. rate =  Δn 
                   Δt
4,4 x 10^-3 = Δn 
                   2,5
Δn(CO₂) = 0,011 mol
n(CO₂) : n(Aℓ₂(CO₃)₃
    3       :       1
0,011 : 3,67 x 10^-3 mol 
n(Aℓ₂(CO₃)₃ left = 0,016 - 3,67 x 10^-3 = 1,23 x 10^-2 mol

OPTION 4
With reference to HCℓ
ave. rate = Δn 
                  Δt
4,4 x 10^-3 =  Δn 
                    2,5
Δn(HCℓ) = 0,011 mol
n[Aℓ2(CO₃)₃] = 0,011 = 0.0018 mol
                           6
n(Aℓ₂(CO₃)₃ left = 0,016 - 0,0055 = 0,0105 mol

OPTION 5
With reference to AℓCℓ₃ 
ave. rate= Δn 
                 Δt
4,4 x 10^-3 =  Δn 
                    2,5
Δn(HCℓ) = 0,011 mol
n[Aℓ₂(CO₃)₃] = 0,0055 mol
n[Aℓ₂(CO₃)₃] left = 0,016 – 0,0055 = 0,0105 mol (3)

5.5 Marking criteria
Use mol ratio: n(CO₂) : n(Aℓ₂(CO₃)₃) = 3 : 1
Substitute 24 000 cm³∙mol-¹/24 dm³∙mol^-1 in n = ^V/_VM or in ratio.

Final answer/Finale antwoord: 1 152 cm³ / 1,152 dm³

OPTION 1
n(CO₂) = 3n[Aℓ₂(CO₃)₃]
= 3(0,016) 
= 0,048 mol
n(CO₂) = V 
               V_M
0,048 =    V    
            24000
V(CO₂) = 1 152 cm³ (1,152 dm³) 

OPTION 2
n(CO₂) = 3n[Aℓ₂(CO₃)₃]
= 3(0,016) 
= 0,048 mol
1 mol ……………….24 000 cm³
0,048 mol ………….V
V(CO₂) =0,048 x 24000
                       1
= 1 152 cm³ (1,152 dm³) (3)
[13]

QUESTION 6
6.1 (The stage in a chemical reaction when the) rate of forward reaction equals the rate of reverse reaction.
OR
(The stage in a chemical reaction when the) concentrations of reactants and products remain constant. (2)

6.2
6.2.1 X 
ANY ONE

• The concentration of products increases (from 0 – 6 min.).
• The concentration of reactants decreases (from 0 – 6 min.).
• No products were present initially. 
• The curve begins at zero. (2)

6.2.2 Higher than (1)

6.3 CALCULATIONS USING NUMBER OF MOLES
Marking criteria

• Calculate mol HI: n(HI)ini. = 1(0,5).
• Use mol ratio: 2:1:1 / n(HI) = 2n(H₂) = 2n(I₂). 
• n(H2)equilibrium = n(H₂)formed
  n(I2)equilibrium= n(I2)formed
  Note: If Δn not shown award mark for equal nequilibrium
• n((HI)equilibrium = n(HI)initial - n(HI)change
• Divide n(HI)equil & n(H₂)equil & n(H₂)equil by 0,5 dm³
• Correct Kc expression (formulae in square brackets)
• Substitute 0,04 into Kc expression
• Substitute equilibrium concentrations in Kc expression.
• Final answer  0,07 mol
  Range: 0,07 – 0,072 mol

OPTION 1
n(HI) = 1(0,5) = 0,5 mol

Kc =[H₂][I₂ ]
          [HI]²
0,04 =(^x/_0.5)(^x/_0.5)
            (0,5 - 2x)²
                 0,5 
x = 0,071 mol

CALCULATIONS USING CONCENTRATION
Marking criteria:

• Use initial c(HI) = 1 mol∙dm^-3.
• Use mol ratio: 2 : 1: 1 / n(HI) = 2n(H₂) = 2n(I₂).
• c(H₂)equilibrium = c(H₂)formed
  c(I₂)equilibrium = c(I₂)formed
  Note: If Δc not shown award mark for equal cequilibrium
• c(HI)equilibrium = c(HI)initial - c(HI)change.
• Correct Kc expression (formulae in square brackets).
• Substitution of 0,04 into Kc expression. 
• Substitution of equilibrium concentrations into Kc expression. 
• Multiply concentration by 0,5 dm³.
• Final answer: 0,07 mol 
  Range: 0,07 to/tot 0,072 mol

OPTION 2

Kc =[H₂][I₂ ]
          [HI]²
0,04 =  (x)(x)  
          (1 - 2x)²
x = 0,143 mol∙dm^-3
n(I₂) = cV
= 0,143 x 0,5 
= 0,072 mol  (9)

6.4
6.4.1 Both forward and reverse(1)
6.4.2 Positive

• The forward reaction is favoured.
• An increase in temperature favours the endothermic reaction.
• The forward reaction is endothermic(4)

[19]

QUESTION 7
7.1 Standard solution(1)

7.2
7.2.1 Marking criteria

• Any one of the formulae c = ^m/_MV / n = ^m/_M/c = ⁿ/_V
• Substitution of 40 g∙mol-1 into correct formula.
• Substitution of 0,25 dm³ into correct formula.
• Final answer/Finale antwoord: 0,2 mol∙dm^-3

OPTION 1
c = ^m/_MV
=      2      
  40 x 0.25
= 0,20 mol∙dm^-3 

OPTION 2
n = ^m/_M
= ²/₄₀
= 0,05 mol
c = ⁿ/_V
= 0,05
   0,25
= 0,20 mol∙dm^-3 (4)

7.2.2 POSITIVE MARKING FROM 7.2.1.
OPTION 1
[H₃O+][OH-] = 1 x 10^-14
[H₃O+](0,2) = 1 x 10^-14
[H₃O+] = 5 x 10^-14 mol∙dm^-3
pH = -log[H₃O+] 
= -log(5 x 10^-14) 
= 13,30 

OPTION 2
pOH = -log[OH-] 
= -log(0,2) 
= 0,6989 (0,7)
pH + pOH = 14
pH = 14 – 0,6989 
= 13,30 (4)

7.3 POSITIVE MARKING FROM QUESTION 7.2.
Marking criteria:

• Substitution to calculate n(NaOH).
• Use mol ratio: n(HCℓ)excess: n(NaOH) = 1 : 1. 
• Substitute 100 g·mol-1 in n = ^m/_M
• Use mol ratio: n(HCℓ)reacted : n(CaCO₃) = 2 : 1.
• n(HCℓ)initial = n(HCℓ) excess + n(HCℓ) reacted 
• Substitute 0,05 dm3 to calculate either c(HCℓ)initial or c(HCℓ) reacted
• Final answer: 0,7 mol∙dm^-3 
  Range: 0,70 to 0,90 mol∙dm^-3

OPTION 1
n(NaOH)used = c_bV_b
= 0,2 x 0,025 
= 5 x 10^-3 mol

OPTION 2
n(NaOH)used =  25   x   2  
                          250      40
= 5 x 10^-3 mol
n(HCℓ)excess = n(NaOH) = 5 x 10^-3 mol
n(CaCO₃) = ^m/_M
= 1.5 
   100
= 0,015 mol (0,02 mol)
n(HCℓ)reacted= 2n(CaCO₃) = 0,03 mol (0,04 mol)
n(HCℓ)ini. = 5 x 10^-3 + 0,03
= 0,035 mol (0,045 mol)
c(HCℓ)ini = ⁿ/_V
= 0,035
    0,05
= 0,70 mol∙dm^-3 (0,90 mol∙dm^-3)

OPTION 3
caVa = na 
cbVb    nb
ca (0,025) =  1 
(0,2)(0,05)    1 
ca = c(HCℓ)excess
= 0,1 mol·dm^-3

OPTION 4
(NaOH)used = cbVb
= (0,2)(0,025)
= 0,005 mol
n(HCℓ)excess = n(NaOH)
= 0,005 mol
c(HCℓ)excess = 0,005 
                           0,05
= 0,1 mol·dm^-3

n(CaCO₃) = ^m/_M
= 1.5 
  100
= 0,015 mol
n(CaCO₃) : n(HCℓ) = 1 : 2
n(HCℓ)reacted = 2(0,015) 
= 0,03 mol
c(HCℓ)reacted = ⁿ/_V
= 0,03 
   0,05
= 0,6 mol·dm^-3
c(HCℓ)initial = c(HCℓ)reacted + c(HCℓ)excess
= 0,6 + 0,1 
= 0,7 mol·dm^-3 (8)
[17]

QUESTION 8
8.1
8.1.1 Gain of electrons. (2 or 0) (2)
8.1.2 2H2O(ℓ) + 2e─  H₂(g) + 2OH─(aq) 
Ignore phases.
Marking criteria:

• H₂(g) + 2OH─(aq) ← 2H₂O(ℓ) + 2e─ (²/₂)
  2H₂O(ℓ) + 2e─ ⇌ H₂(g) + 2OH─(aq) (½)
  H₂g) + 2OH─(aq) ⇌ 2H₂O(ℓ) + 2e─ (⁰/₂)
  2H₂O(ℓ) + 2e─ ← H₂(g) + 2OH─(aq) (⁰/₂)
• Ignore if charge omitted on electron.
• If charge (-) omitted on OH─
  Example: 2H₂O(ℓ) + 2e─ H₂(g) + 2OH(aq) Max.:½ (2)

8.1.3 2Na(s) + 2H₂O(ℓ) → H₂(g) + 2OH-(aq) + 2Na+(aq) Bal

OR
2Na(s) + 2H₂O(ℓ)  → H₂(g) + 2NaOH(aq) Bal 
Ignore phases/.
Marking criteria:

• Reactants   Products   Balancing 
• Ignore double arrows.
• Ignore phases
• Marking rule 6.3.10. (3)

8.1.4 Formation of hydroxide ions / OH- / sodium hydroxide/base/ alkaline/ pH > 7(1)

8.1.5 Cu is a weaker reducing agent than H₂ (and OH─) and H₂O will not be reduced (to H₂ and OH─).
OR
H₂ (and OH-) are stronger reducing agent than Cu and H₂O will not be reduced(to H2 and OH-). (3)

8.2
8.2.1 Phase separator/boundary/difference(1)

8.2.2 Chemical (energy) to electrical (energy)(1)

8.2.3 OPTION 1
Eθcell = Eθreduction - Eθoxidation
= 0,77 - (-0,13)
Eθcell = 0,90 V 

• Notes
  Accept any other correct formula from the data sheet
• Any other formula using unconventional abbreviations, e.g. Eθcell = EθOA - EθRA followed by correct substitutions:(4)

OPTION 2
Pb(s) → Pb²⁺(aq) + 2e- 0,13 (V) 
2Fe3+(aq) + 2e- → 2Fe²⁺(aq) 0,77 (V) 
Pb²⁺(aq) + 2Fe³⁺(aq) → Pb(s) + 2Fe²⁺(aq) 0,90 V 
[17]

QUESTION 9
9.1 Electrolytic (cell)
Cells have a battery/DC power source/ /Electrical energy is converted to chemical energy. (2)

9.2
9.2.1 2Cℓ─  Cℓ₂ + 2e─ 
Notes:
2Cℓ─ ⇌ Cℓ2 + 2e─ (½)   Cℓ₂ + 2e─ ← 2Cℓ─ (²/₂)
Cℓ₂ + 2e─ ⇌ 2Cℓ─ (⁰/₂ ) 2Cℓ─ ← Cℓ₂ + 2e─ (⁰/₂)

• Ignore if charge omitted on electron.
• If charge (-) omitted on Cℓ─

9.2.2 Aℓ³⁺ + 3e─ → Aℓ 
Notes/Aantekeninge
Aℓ3+ + 3e─ ⇌ Aℓ (½)   Aℓ ← Aℓ3+ + 3e─ (²/₂)
Aℓ ⇌ Aℓ3+ + 3e─ (⁰/₂)  Aℓ³⁺ + 3e ← Aℓ(⁰/₂)

• Ignore if charge omitted on electron.
• If charge (+) omitted on Aℓ3+ 
  Example: Aℓ3(aq) + 3e─ → Aℓ(s) Max. ½ (2)

9.2.3 Cu/copper(1)

9.3 ANY ONE

• The electrode/carbon/C reacts with oxygen. 
• C + O₂ → CO₂
• Oxidation takes place./Electrons are lost.
• Oxygen corrodes the carbon electrode. (1)

[8]

QUESTION 10
10.1
10.1.1 Sulphur dioxide/SO₂(1)

10.1.2 Sulphur trioxide/SO₃(1)

10.1.3 Vanadium pentoxide/V₂O₅/ Vanadium(V) oxide(1)

10.1.4 H₂SO4 + 2NH₃  → (NH4)2SO₄ bal
Marking guidelines:

• Reactants    Products    Balancing 
• Ignore → and phases 
• Marking rule 6.3.10 (3)

10.2
10.2.1 The ratio of nitrogen (N), phosphorous (P) and potassium (K) in a fertiliser./The ratio of the primary nutrients (1)

10.2.2 OPTION 1
Mass N in 4 kg NH₄NO₃ / Massa N in 4 kg NH₄NO₃
m(N) = ²⁸/₈₀ x 4 
= 1,4 kg
m(K) = 2m(N)
= 2,8 kg
m(P) = 3m(N)
= 4,2 kg
m(fertiliser) = 1,4 + 2,8 + 4,2
= 8,4 kg 

OPTION 2
Mass N in 4 kg NH₄NO₃ / Massa N in 4 kg NH₄NO₃
m(N) = ²⁸/₈₀ x 4 
= 1,4 kg
N : P : K
1 : 3 : 2
∴ m(fertiliser) = (6)(1,4)
= 8,4 kg 

OPTION 3
% N = (2)(14) x 100 = 35%
              80
Nitrogen in 4 kg = 35% of 4 = 1,4 kg 
N : P : K
1 : 3 : 2
1,4 : 4,2 : 2,8 
Total mass of fertiliser = 1,4 + 4,2 + 2,8
= 8,4 kg  (4)
[11]
TOTAL: 150