PHYSICAL SCIENCES: CHEMISTRY (PAPER 2)
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
NOVEMBER 2020

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

QUESTION 2
2.1.1 Ketones✓(1)

2.1.2 Pentanal✓✓
ACCEPT
2,2-dimethylpropanal
2-methylbutanal
3-methylbutanal

Marking criteria

• Correct functional group,i.e. –al 
• Whole name correct (2)

2.2.1 5 – bromo-2,3 – dimethylhexane
Marking criteria:

• Correct stem i.e. hexane
• All substituents (bromo and dimethyl) correctly identified.
• IUPAC name completely correct including numbering, sequence, hyphens and commas. (3)

2.2.2

Marking criteria

• Whole structure correct ²/₂
• Only functional group correct: ½

IF
More than one functional group ⁰/₂ (2)
IF
Molecular formula ⁰/₂
Condensed structural formula ½

2.3.1 The C atom bonded to the hydroxyl group is bonded to only one other C-atom. ✓✓ (2 or 0)
OR
The hydroxyl group/-OH/ is bonded to a C atom which is bonded to two hydrogens atoms. (2 or 0)
OR
The hydroxyl group/functional group/-OH is bonded to:
a primary C atom / the first C atom (2 or 0)
OR
(2)
2.3.2 Esterification/condensation ✓(1)

2.3.3 Butanoic acid ✓(1)
[12]

QUESTION 3
3.1 Marking criteria
If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark.
The temperature at which the vapour pressure equals atmospheric (external) pressure. ✓✓ (2)
3.2
(1)
3.3

• Increase in the number of C-atoms increases molecular mass/size/chain length/surface area. ✓
• Strength of the intermolecular forces increases/More sites for London forces. ✓
• More energy is needed to overcome/break intermolecular forces. ✓ (3)

3.4.1 C ✓ (1)

3.4.2 B ✓
Marking criteria

• Compare strength of intermolecular forces of A, B and C. ✓✓
• Compare boiling points/energy required to overcome intermolecular forces of alcohols/A and aldehydes/B. ✓
  OR
  Alcohols have the highest boiling point.
• Compare boiling points/ energy required to overcome intermolecular force of aldehydes/B and alkanes/C .✓
  OR
  Alkanes have the lowest boiling point.

Aldehydes/B have (in addition to London forces) dipole-dipole forces which are stronger than London forces, but weaker than hydrogen bonds. ✓
Therefore aldehydes/B have lower boiling points/require less energy to overcome intermolecular forces than alcohols/A,✓but higher boiling points / require more energy to overcome intermolecular forces than alkanes/C. ✓
OR
Aldehydes/B have stronger intermolecular forces than alkanes, but weaker intermolecular forces than alcohols/A. ✓
Therefore aldehydes/B have higher boiling points/ more energy required to overcome intermolecular forces than alkanes/C, ✓ but lower boiling points/ less energy to overcome intermolecular forces than alcohols/A.✓ (4)

3.5 Butanal ✓✓
Marking criteria

• Correct stem, i.e. but ✓
• Whole name correct(2)

3.6 Pentan-1-ol ✓✓
OR
1-pentanol ✓✓(2)
[15]

QUESTION 4
4.1 Marking criteria

• Addition reaction / reaction of alkene / reaction of C – C double bond /reaction of unsaturated hydrocarbon✓
• (Addition of) hydrogen halide/HX/ hydrogen and halide. ✓
  The addition of a hydrogen halide/HX✓to an alkene. (2)

4.2 

Marking criteria

• Whole structure correct ²/₂
• Only functional group correct: ½

(2)
4.3.1 Cracking ✓ (1)
4.3.2 C₈H₁₈ ✓ (1)
4.4 1,2–dibromo ✓ propane ✓
1,2-dibromopropaan (2)
4.5.1

Marking criteria for the alcohol

• Whole structure correct ²/₂
• Only functional group correct: ½

Notes:

• If 1-chloropropane used as reactant, 2 marks for the primary alcohol.
• Condensed or semi-structural formula: Max. ⁴/₅
• Molecular formula:²/₅
• Any additional reactants or products: Max.⁴/₅
• If arrow in completely correct equation omitted: Max.⁴/₅
• The product NaCℓ/KCℓ/HCℓ must be marked in conjunction with reactant NaOH/KOH/H₂O. (5)

4.5.2

• (Mild) heat
• Dilute strong base/NaOH/LiOH/KOH OR water/H₂O ✓ (2)

[15]

QUESTION 5
5.1.1 (Reaction) rate(1)

5.1.2 Surface area/state of division /particle size (1)

5.2.1 (Decreasing gradient indicates) rate of reaction is decreasing. ✓(1)

5.2.2 (Gradient is zero, indicates) reaction rate is zero ✓(1)

5.3
ave rate =^ΔV/_Δt
= 500 (-0) = 8,33 (cm³∙s-1)
    60 (-0)
(3)

5.4 Equal to (1)

5.5 Greater than
Experiment C:

• Surface area of CaCO₃ powder is greater than that of CaCO₃ granules./
  More particles are exposed /More particles with correct orientation ✓
• More effective collisions per unit time/Higher frequency of effective collisions. ✓
• Increase in reaction rate.✓

OR
Experiment A:

• Surface area of CaCO₃ granules is smaller/Fewer particles are exposed
  (than that of powdered CaCO₃). Less particles with correct orientation ✓
• Less effective collisions per unit time./Lower frequency of effective collisions. ✓
• Decrease in reaction rate.✓ (4)

5.6 Marking criteria:

• Divide volume by 25,7 in n  = ^V/_VM
  If no substitution step shown, award mark for answer: 0,0195 mol
• Ratio: n(CO₂) = n(CaCO₃).
• Substitute 100 in n = ⁿ/_M or in ratio
• Final answer: 1,95 g to 2 g. 

OPTION 1
n(CO₂) = ^V/_Vm = ^0,5/_25,7
= 0,0195 mol
n(CaCO₃) = n(CO₂) = 0,0195 mol ✓
m(CaCO₃) = nM
= 0,0195(100)
= 1,95 g ✓

OPTION 2
25,7 dm³ .........1 mol
0,5 dm³ ..........0,0195 mol ✓
100 g ✓...........1 mol
x ....................0,0195 mol ✓
x = m(CaCO₃) = 1,95 g ✓

OPTION 3
n(CO₂) = ^V/_Vm = ^0,5/_25,7
= 0,0195 mol
0,0195 mol CO₂ ≡ 0,856 g CO₂ ✓
m(CO₂) produced : m(CaCO₃)
44 g : 100 g 
0,856 : x
x = 1,95 g CaCO₃ (4)
[16]

QUESTION 6
6.1 Products can be converted back to reactants. ✓
OR
Both forward and reverse reactions can take place.
OR
A reaction which can take place in both directions. (1)

6.2.1 Remains the same(1)

6.2.2 Increases (1)

6.3

• (When pressure is increased) the reaction that leads to the smaller amount of gas / side with less molecules/number of moles is favoured.✓
• The reverse reaction is favoured. (2)

6.4 Endothermic

• Kc decreases with decrease in temperature. ✓
• Reverse reaction is favoured. / Concentration of reactants increases. / Concentration of products decreases./Yield decreases✓
• Decrease in temperature favours an exothermic reaction. ✓
  OR
• Kc increases with increase in temperature. ✓
• Forward reaction is favoured. / Concentration of reactants decreases. / Concentration of products increases./Yield increases ✓
• Increase in temperature favours an endothermic reaction.✓ (4)

6.5 CALCULATIONS USING NUMBER OF MOLES
Mark allocation

• Correct Kc expression (formulae in square brackets).✓
• Substitution of equilibrium concentrations into Kc expression. ✓
• Substitution of Kc value. ✓
• Multiply equilibrium concentrations of I₂ and I by 12,3 dm³. ✓ (OPTION 1)
• Multiply equilibrium concentrations of I by 12,3 dm³ and divide equilibrium mol of I₂ by 12,3 dm³. ✓(OPTION 2)
• Change in n(I) = n(I at equilibrium). ✓
• USING ratio: I₂ : I = 1 : 2 ✓
• Initial n(I2) = equilibrium n(I₂) + change in n(I₂).✓
• Substitute 254 g·mol-1 as molar mass for I₂.✓
• Final answer: (26 g - 27,94 g). ✓

OPTION 1
K_c =  [I] ²
         [I₂]
3,76x10^-3 =(4,79 x 10^-3)²
                           [I₂]
[I₂] = 6,102 x 10^-3 mol∙dm^-3

OPTION 2

K_c =  [I] ²
         [I₂]
3,76x10^-3 =(4,79 x 10^-3)²
                    x - 0.0295
                        12.3
x = 0,1045 mol
m = nM
= (0,1045)(254)
= 26,543 g ✓
Wrong Kc expression ⁶/₉
No Kc expression, correct substitution ⁸/₉

CALCULATIONS USING CONCENTRATION
Mark allocation

• Correct Kc expression (formulae in square brackets). ✓
• Substitution of equilibrium concentrations into Kc expression. ✓
• Substitution of Kc value ✓
• Change in n(I) = n(I at equilibrium). ✓
• USING ratio: I2 : I = 1 : 2 ✓
• Initial [I₂] = equilibrium [I₂] + change in [I₂]. ✓
• Divide by 12,3 dm3. ✓
• Substitute 254 g·mol-1 as molar mass for I₂.✓
• Final answer 26,543 g. ✓

OPTION 3
K_c =  [I] ²
         [I₂]
3,76x10^-3 =(4,79 x 10^-3)²
                           [I₂]
[I₂] = 6,102 x 10^-3 mol∙dm^-3

c =  m  
      MV
8.497 x 10^-3 =       m        
                      (254)(12.3)
m = 26.546g
(9)
[18]

QUESTION 7
7.1.1 Weak
Ionises/Dissociates incompletely/partially (in water) (2)
7.1.2 OPTION 1
pH = -log[H₃O+] 
3,85  = -log[H₃O⁺]
[H₃O+] = 1,41 x 10^-4 mol∙dm^-3 

OPTION 2
[H₃O+] = 10-pH 
= 10-3,85 
= 1,41 x 10-4 mol∙dm^-3 (3)

7.1.3 Greater than (1)

7.1.4 CH₃COO-(aq) + H₂O(ℓ) ⇌ CH₃COOH(aq) + OH-(aq) 
OR
CH₃COONa(aq) + H₂O(ℓ) ⇌ CH₃COOH(aq) + NaOH(aq) 
Due to formation of hydroxide/OH- / the solution is basic/alkaline /pH > 7. (3)

7.2.1 Marking criteria
Substitute/vervang: 1 x 0,0145 OR 1 x 14,5 in c = n  / ca x Va  = na  
                                                                               V    cb x Vb     nb

• Use: n(CH₃COOH) : n(NaOH) = 1:1 
• Final answer/Finale antwoord: 0,0145 mol 

OPTION 1
n(NaOH)_reacted = cV
= 1(0,0145) 
= 0,0145 mol
n(CH₃COOH)_diluted = n(NaOH) 
= 0,0145mol (3)

7.2.2 POSITIVE MARKING FROM 7.2.1.
Marking criteria

• Calculate mass CH₃COOH in 25 cm³ (1,13 g). ✓
• Formula: n = ^m/_M
• Substitute: M = 60 g∙mol^-1. 
• n(CH₃COOH)reacted = ninitial- nunreacted 
• USE mol ratio: n(CaCO₃) : n(CH₃COOH) = 1 : 2.✓
• Substitution of 100 g∙mol-1 in m = nM. ✓
• Calculate percentage: 0,217 x 100
                                        1,2
• Final answer: 18,08% (17,92 – 22,92)

m(CH₃COOH) = 4,52 x 25 = 1,13 g
                           100
n(CH₃COOH)ini. = ^m/_M
=1,13= 0,01883 mol
   60
n(CH₃COOH)rea = 0,01883 – 0,0145 = 0,0043 mol
n(CaCO₃) = ½n(CH₃COOH)
= 0,5(0,0043) 
= 0,00217 mol
m(CaCO₃) = nM
= 0,00217(100) = 0,217 g
% CaCO₃= 0,217 x 100
                    1,2
= 18,08 %  (8)
[20]

QUESTION 8
8.1 Provides path for movement of ions./Ensures(electrical)neutrality in the cell. (1)

8.2 (The electrode) where oxidation takes place/electrons are lost.(2)

8.3 Mg/Magnesium (1)

8.4.1 2H⁺+ 2e^- → H₂ 
Marking criteria
H₂ ← 2H+ + 2e- (²/₂)   2H+ + 2e- ⇌ H₂ (½)
H₂ ⇌ 2H+ + 2e- (⁰/₂)    2H+ + 2e ← H₂ (⁰/₂)

• Ignore if charge omitted on electron.
• If charge (+) omitted on H+:
  Example: 2H+ 2e- → H₂ Max.:½ (2)

8.4.2 Magnesium/Mg ✓ (1)

8.5 OPTION 1
E^θ_cell = E^θ_reduction - E^θ_oxidation
= 0 - (- 2,36)
E^θ_cell  = 2,36 V

Notes

• Accept any other correct formula from the data sheet.
• Any other formula using unconventional abbreviations, e.g.E^θ_cell= E°_OA - E°_RA followed by correct substitutions:

OPTION 2
2H⁺+ 2e^- → H₂                                          E^θ = 0 V 
Mg(s) → Mg²⁺(aq) + 2e-                           E^θ = +2,36 V 
Mg(s) + 2H⁺(aq)→ Mg2+(aq) + H2(g)       E^θ = +2,36 V  (4)

8.6 H₂ is a stronger reducing agent than Cu and therefore Cu2+/Cu ions are reduced/H₂ is oxidised Electrons flow from H₂ to Cu. (3)
[14 ]

QUESTION 9
9.1 ANY ONE:

• The chemical process in which electrical energy is converted to chemical energy. ✓✓ (2 or 0)
• The use of electrical energy to produce a chemical change. (2 or 0)
• Decomposition of an ionic compound by means of electrical energy. (2 or 0)
• The process during which and electric current passes through a solution/ionic liquid/molten ionic compound. (2 or 0) (2)

9.2 Battery/cell/ power source ✓(1)

9.3 Silver nitrate/AgNO₃/ Silver ethanoate/CH₃COOAg / Silver fluoride /AgF/ Silver perchlorate AgCℓO₄. (1)

9.4 Remains the same
Rate of oxidation is equal to the rate of reduction. (2)

9.5 Ag → Ag⁺ + e^- 
Notes
Ag+ + e- ← Ag (²/₂)   Ag ⇌ Ag+ + e- (½)
Ag ← Ag+ + e- (⁰/₂)   Ag+ + e- ⇌ Ag (⁰/₂)

• Ignore if charge omitted on electron.
• If charge (+) omitted on Ag+:
  Example: Ag → Ag + e- ✓Max./Maks: (2)

[8]

QUESTION 10
10.1.1 (Liquid) Air (1)

10.1.2 Natural gas/methane/oil/coal/coke✓(1)

10.1.3 Iron/iron oxide/Fe/FeO (1)

10.1.4 NH₃/Ammonia/Ammoniak ✓ (1)

10.1.5 Ostwald (process)/Ostwald(proses) ✓(1)

10.1.6 NH₃ + HNO₃ → NH₄NO₃ Bal 
Marking criteria

• Reactants  Products  Balancing 
• Ignore double arrows
• Marking rule 6.3.10.

10.2.1 NPK ratio/Ratio of primary nutrients ✓ (1)
10.2.2 OPTION 1
⁴/₉  x  ^X/₁₀₀  x 20 = 2,315 kg
X = 26 (26,04)

OPTION 2
m(P) = 2,315 kg
Mass of 1 part P =2,315= 0,57575
                                4
Mass of N = (0,57575)(2) = 1,1575 kg
Mass of K = (0,57575)(3) = 1,73625 kg
Total mass of fertiliser:
1,1575 + 2,315 + 1,73625 = 5,20875 kg ✓
X =5,20875 x 100= 26,04 (3)
           20
[12]
TOTAL: 150