PHYSICAL SCIENCES PAPER 2 GRADE 12 MEMORANDUM - NSC PAST PAPERS AND MEMOS NOVEMBER 2017

More in this category: « GRADE 12 MATHEMATICAL LITERACY PAPER 1 ADDENDUM - NSC PAST PAPERS AND MEMOS NOVEMBER 2017 PHYSICAL SCIENCES PAPER 2 GRADE 12 QUESTIONS - NSC PAST PAPERS AND MEMOS NOVEMBER 2017 »

Share via Whatsapp Join our WhatsApp Group Join our Telegram Group

PHYSICAL SCIENCES: CHEMISTRY (PAPER 2)
GRADE 12
NOVEMBER 2017
MEMORANDUM
NATIONAL SENIOR CERTIFICATE

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

QUESTION 2
2.1
2.1.1 Esters ✓ (1)
2.1.2 Ethyl ✓ butanoate ✓ (2)
2.1.3 Butanoic acid ✓ (1)
2.1.4 (2)
2.2 (3)
2.3
2.3.1 CnH2n-2 ✓ (1)
2.3.2 5-ethyl-2,6-dimethylhept-3-yne/5-ethyl-2,6-dimethyl-3-heptyne (3)
[13]

QUESTION 3
3.1 ANY ONE:

They have ONLY single bonds. ✓
They have single bonds between C atoms.
They have no double OR triple bonds OR multiple bonds.
They contain the maximum number of H atoms bonded to C atoms.
Each C atom is bonded to four other atoms. (1)

3.2 The pressure exerted by a vapour in equilibrium with its liquid ✓ in a closed system. ✓ (2)
3.3
3.3.1 Increases ✓ (1)
3.3.2 Q ✓
It is the temperature where the graph intercepts the dotted line. ✓
OR
It is the temperature where the vapour pressure of compound Q equals atmospheric pressure/is equal to 760 mmHg. (2)
3.3.3 S ✓

At a given temperature, S has the lowest vapour pressure/highest boiling point. ✓ B
Strongest intermolecular forces/London forces. ✓
Highest energy needed to overcome/break the intermolecular forces. ✓ (4)

3.4
3.4.1 OR(3)
3.4.2 Higher than ✓ (1)
[14]

QUESTION 4
4.1 Secondary ✓✓
The C atom bonded to the –OH group is bonded to TWO other C atoms. ✓(2)
4.2
4.2.1 Dehydration ✓ (1)
4.2.2 Hydration ✓ (1)
4.2.3 Dehydrohalogenation/dehydrobromination ✓ (1)
4.3
4.3.1 Substitution ✓ (1)
4.3.2

Dilute base/sodium hydroxide/NaOH ✓
Moderate temperature/(mild) heat ✓ (2)

4.3.3 2-✓bromobutane ✓(2)
4.4 NaOH/KOH ✓ (1)
4.5 (3)
4.6 Butane(1)
[15]

QUESTION 5
5.1 ANY ONE:

Change in concentration of products/reactants per (unit) time. ✓✓
Rate of change in concentration.
Change in amount/number of moles/volume/mass of products or reactants per (unit) time.
Amount/number of moles/volume/mass of products formed or reactants used per (unit) time. (2)

5.2

Marking criteria:
Dependent and independent variables correctly identified.	✓
Ask a question about the relationship between the independent and dependent variables.	✓

Examples:

What is the relationship between concentration and reaction rate?
How does the reaction rate change when the concentration changes/increases/decreases? (2)

5.3 Q 

Smaller gradient./Less steep. ✓
Reaction I has the lowest HCℓ concentration and will take longer to reach completion/for the maximum volume of gas to be formed. ✓ (3)

5.4

OPTION 1 Ave rate = ΔV Δt 15 = ΔV 30-0 ✓ V(H₂)produced = 450 cm³ n(H₂)produced = V V_m = 450 24000 ✓ =0,0188 mol n(Zn) = n(H₂) = 0,0188 mol ✓ n(Zn)_used = m M ∴ 0,0188 = m 65 ✓ ∴ m(Zn) = 1,22 g ✓	OPTION 2 Ave rate = 15 24000 ✓ = 6,25 × 10^-4 mol·s^-1 V(H₂)_produced= 6,25 × 30 ✓ = 0,0188 mol ✓ n(Zn) = n(H2) = 0,0188 mol ✓ n(Zn)_used = m M 0,0188 = m 65 ✓ ∴ m(Zn) = 1,22 g 
	OPTION 3 Ave rate = ΔV Δt 15 = ΔV ✓ 30-0 V(H₂)produced/berei = 450 cm³ 65 g ✓ Zn ……… 24 000 cm³ ✓ x g Zn ……………450 cm³ ✓ x = 1,22 g ✓

(5)
5.5
5.5.1 Equal to ✓ (1)
5.5.2 Equal to ✓ (1)
5.6

At higher temperature the average kinetic energy of particles is higher. ✓
More molecules gain sufficient/enough kinetic energy OR more molecules have kinetic energy equal to or greater than the activation energy. ✓
More effective collisions per unit time./Frequency of effective collisions increases. ✓ (3)

[17]

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

OPTION 1
n = m
M
=1,12
28 ✓
= 0,04 mol

	COBr₂	CO	Br₂
Initial quantity (mol)		0	0
Change (mol)	0,04	0,04	0,04 ✓
Quantity at equilibrium (mol)		0,04	0,04
Equilibrium concentration		0,02	0,02

Divide by 2 ✓
K_c = [CO][Br₂] ✓
[COBr₂]
0,19 ✓ = (0,02) ² ✓
[COBr₂]

OPTION 2
n = m
M
=1,2
28 ✓
= 0,04 mol
n(CO)_formed = n(Br₂)_formed ✓
= 0,04 mol
c(CO)eq = c(Br₂)eq
= n
V
= 0,04
2 ✓
= 0,02 mol·dm^-3
K_c = [CO][Br₂]✓
[COBr₂]
0,19 = (0,02) ²
[COBr₂] ✓
[COBr₂] = 2,11 x 10^-3 mol∙dm^-3 ✓

(7)
6.2.2

OPTION 1
n(COBr₂)_eq = cV
= 2,11 ×10^-3 × 2 ✓
= 4,22 × 10^-3 mol
n(COBr₂)_initial
= 0,04 + 4,22 × 10^-3 ✓
= 0,044 mol
% decomposed = 0,04 ×100 ✓
0,044
= 90,46% ✓
Range: 90,46 – 90,9%

OPTION 2
n(COBr₂)_eq = cV
= 2,11 ×10^-3 × 2 ✓
= 4,22 × 10^-3 mol
n(COBr₂)_initial
= 0,04 + 4,22 × 10^-3 ✓
= 0,044 mol
m(COBr₂)_initial = nM
= 0,044 × 188
= 8,27 g
m(COBr2)_reacted = 0,04 × 188
= 7,52 g
% decomposed = 7,52 × 100 ✓
8,27
= 90,9% ✓

(4)
6.3 Kc < 0,19 (1)
6.4 Decreases ✓
A decreases in pressure favours the reaction that produces the larger number of moles of gas.
The forward reaction will be favoured. ✓ (3)
[17]

QUESTION 7
7.1
7.1.1 Weak ✓
Dissociates/Ionises incompletely (in water) ✓ (2)
7.1.2 NH⁺₄ ✓ (1)
7.1.3 H₂O/water OR NH₃ ✓ (1)
7.2
7.2.1 Acidic ✓
pH < 7 ✓ (2)
7.2.2

OPTION 1
pH = -log[H₃O⁺] ✓
6 ✓ = -log[H₃O⁺]
[H₃O⁺] = 1 × 10^-6 mol∙dm^-3
[H₃O⁺][OH-] = 10^-14 ✓
[OH-] = 1 × 10-8 mol∙dm^-3 ✓

OPTION 2
pH + pOH = 14 ✓
6 + pOH = 14
pOH = -log[OH^-] ✓
8 = -log[OH^-]
[OH-] = 1 × 10^-8 mol∙dm^-3 ✓

(4)
7.3

OPTION 1
n(Na₂CO₃) = m
M ✓
= 0,29
106 ✓
= 2,74 × 10^-3 mol
n(HCℓ) = 2n(Na₂CO₃) ✓
= 5,47 × 10^-3 mol
c(HCℓ)dilute = n
V
= 5,74 x 10^-3
0,05 ✓
= 0,1094 mol∙dm^-3
cV(HCℓ)_dilute = cV(HCℓ)_conc
0,1094 × 500  = (HCℓ)_conc × 5 ✓
∴ c(HCℓ)_conc = 10,94 mol∙dm^-3 ✓

OPTION 2
n(Na₂CO₃) = m
M ✓
= 0,29
106 ✓
= 2,74 × 10^-3 mol
n(HCℓ) = 2n(Na₂CO₃) ✓
= 5,47 × 10^-3 mol
In 500 cm³:
n(HCℓ) = 5,47 × 10^-3 mol✓
In 500cm³
n(HCℓ) = 500 (5,47 × 10^-3)✓
50
= 0,547 mol
c(HCℓ)_conc = 0,547 × 1000
5 ✓
= 10,94 mol∙dm^-3 ✓

(7)
[17]

QUESTION 8
8.
8.1.1 Voltmeter ✓ (1)
8.1.2 Anode ✓ (1)
8.1.3 3Ag+(aq) + Aℓ(s) ✓ → 3Ag(s) + Aℓ 3+(aq) ✓ Bal. ✓ (3)
8.1.4

OPTION1
E^θ_cell = E^θ_reduction - E^θ_oxidation✓
= +0,80 ✓- (- 1,66) ✓
= 2,46 V ✓

OPTION 2
Ag+(aq) + e- → Ag(s) 0,80 V ✓
Aℓ(s) → Aℓ ³⁺(aq) + 3e- 1,66 V ✓
3Ag⁺(aq) + Aℓ(s) → 3Ag(s) + Aℓ ³⁺(aq) 2,46 V ✓ (4)

8.2
8.2.1 Platinum/Pt/Carbon/C ✓ (1)
8.2.2 ANY TWO:
Concentration: 1 mol∙dm^-3 ✓
Temperature: 25°C/298 K ✓
Pressure: 101,3 kPa/1,01 x 10⁵ Pa/1 atm (2)
8.2.3 Zinc/Zn ✓ (1)
8.2.4 PQ ✓ (1)
[14]

QUESTION 9
9.1 DC ✓ (1)
9.2 Cathode ✓
Cu²⁺(aq) + 2e^- → Cu ✓✓
9.3 Cu2+ is a stronger oxidising agent ✓ than Zn2+ ions ✓ and therefore Zn2+ ions will not be reduced (to Zn). ✓ (3)
9.4
9.4.1 (Chlorine) gas/bubbles is/are formed. ✓(1)
9.4.2 Decreases ✓ (1)
[9]

QUESTION 10
10.1
10.1.1 Ammonia ✓ (1)
10.1.2 NO₂ ✓ (1)
10.1.3 Catalytic oxidation of ammonia ✓ (1)
10.1.4 Platinum/Pt ✓ (1)
10.1.5 Ostwald (process)✓ (1)
10.1.6 Haber (process)✓ (1)
10.1.7 NH₃ + HNO₃ ✓ → NH₄NO₃✓ Bal. ✓ (3)
10.2
10.2.1

OPTION 1
N : P : K
10 : 5 : 15
m(fertiliser) = 30 x 15
100
= 4,5 kg
m(P) = 5 x 4,5 ✓
30
= 0,75 kg ✓

OPTION 2
m(fertiliser) = 5 x 15 ✓
100
= 0,75 kg ✓

(2)
10.2.2 %fertiliser = 10 + 5 + 15 = 30%
%filler = 100 – 30 = 70%
m_(filler) = 70 ✓ x 15 ✓
100
= 10,5 kg ✓ (3)
[14]

TOTAL: 150

Last modified on Monday, 16 August 2021 13:15

Published in Grade 12 NSC Exam Past Papers and Memos November 2017

PHYSICAL SCIENCES PAPER 2 GRADE 12 MEMORANDUM - NSC PAST PAPERS AND MEMOS NOVEMBER 2017

Related items