# Physical Sciences Grade 12 Paper 2 Limpopo Mock 2025 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the Limpopo Mock Exam 2025, 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physics-grade-12-nsc-p2-memo-september-2025-limpopo/46f1e839-8ea6-431d-897b-3ce2ee9b8779
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Limpopo Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 10

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## About this document

Physical Sciences · Grade 12 · Limpopo Mock Exam · 2025. Memorandum, 10 pages. Read online or download the PDF.

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## Document text

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### Page 1

REPUBLIC OF SOUTH AFRICA

DEPARTMENT OF EDUCATION

NATIONAL
SENIOR CERTIFICATE

GRADE 12

PHYSICAL SCIENCES PAPER 2

MEMO

PREPARATORY 2025

ee ee ee ee

MARKS: 150
TIME: 3 Hours

This Marking Guidelines consist of 10 pages including the cover page

### Page 2

QUESTION 1
14 cvy
1.2 AVY
1.3 BYv
1.4 Avy
1.5 Byv
1.6 avy
1.7 Avy
1.8 Byy
1.9 Byy
1.10 Dyv
QUESTION 2

24

2.3

2.4
2.5

2.2.1
2.2.2

2.2.3

2.2.4

DoE/Preparatory 2025

Chain isomers are compounds with the same molecular formula but

different types of chains. vv
Cv

4 — bromo - 2,2 — dimethyipentanev vv,

Notes:

1. Bromo, dimethyl identifiedy
2. Pentanev
3. Whole structure correctY

Pent — 1 —- enevv Notes:

1. Correct functional groupY
2. Whole structure correctY

CnHonY

HH

\ | id

H-—-C—C H

Pld
Tay
H

. Correct functional groupy
2. Whole structure correctv

CsH12 + 8O2v — 5CO2 + 6H20v (“ balance)

Aldehydes

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(2)
(2)
(2)
(2)
(2)
(2)
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(2)
(2)
[20]

(2)
(1)

(3)

(2)
(1)

(2)
(3)
(1)

### Page 3

DoE/Preparatory 2025

2.6 20% C — atoms in comp.C = od x Sv
=v

Compound E has 6 C — atoms
4
“xX = 6¥ and y = 14v OR (x = 6 and y= 14vv) (4)

[19]

QUESTION 3

3.1 Melting point is the temperature at which the solid and the liquid phases
of a substance are at equilibrium. vv (2)

3.2 As the chain length/molar mass/number of carbon atoms increases, the
boiling/melting point increases. v ¥ (2)

3.3 Stronger than. ” (1)

3.4 Fair. Y The compounds belong to the same homologous/intermolecular
force/functional group/there is only one independent variable. v (2)

3.5 Gas. v (1)

3.6 e Propane has a longer chain length/larger/ surface area/ bigger
molecular size/more carbon atoms than ethane.v
e Both have London forces/dispersion/dipole-dipole which are stronger
between propane molecules than in ethane./the intermolecular
forces of propane are stronger
e More energy is needed to overcome the intermolecular force in
propane than in ethane.v

OR

e Ethane has shorter chain length/smaller surface area/smaller
molecular size/less carbon atoms than propane.“

e Both have London forces/dispersion/dipole-dipole which are weaker
between ethane molecules than in propane.”

e Less energy in needed to overcome the intermolecular forces in
ethane than in propane.v (3)

[11]

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### Page 4

QUESTION 4
4.44
4.1.2

4.21

4.2.2

4.3.1

4.3.2

4.4.1

4.4.2

4.5.1

4.5.2

DoE/Preparatory 2025

Substitution/hydrolysis~

Butan — 2 - ol’ “ Accept 2-butanol (1 mark for butanol, 1 mark for
position)

Ai H
\\ i
v Ct+-CTH Notes:
YW) \ Vv
ce ri 1. Correct functional groupY
We 2. Whole structure correctv
H
(Strong) Heatv’

H
v| |
rH

H clH H v

Lf tm “ iH Vv
H—C—C—C—C—H org

ro] tod eee H~ \ ws v

H HH H conc. KOH in ethanol H + 4H H + K—Cl

Positive marking form 4.3.1
But-2-ene.v v fAccept 2-Butene (1 mark for butene, 1 mark for position)

Addition/hydrogenation. ~
Ni/Pt/Pd/Nickel/Platinum/Palladium.”

Cracking is the chemical process in which longer chain hydrocarbon
molecules are broken down to shorter more useful molecules. v ¥

Catalyst (Pt, Pd and Ni).”

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(2)

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(5)

(2)
(1)
(1)

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[18]

### Page 5

DoE/Preparatory 2025

QUESTION 5
5.4 Rate of reaction is the change in concentration of reactants or products
per unit time.v Vv (2)
5.2 Neutralisation reaction.v /Acid-base reaction (1)
5.3 State of division/surface area.” (1)
5.4 Rate of reaction.” (1)
5.5 n(COz) = wy
=e tt
22400
= 0,002366 mol
n(HCl) _ 2(0,002366) /
“22.400
= 0,004732 mol
Rate = “v
At
= 0,004732 /,
60 . - : (5)
= 0,0000789 mol.s™!/(7,89 x 1075mol. s~1)v
[10]

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### Page 6

QUESTION 6

6.1.1

6.1.2

6.2.1

6.2.2

6.2.3

DoE/Preparatory 2025

Chemical equilibrium is a dynamic equilibrium when the rate of the

forward reaction equals to the rate of the reverse reaction.v Vv

| SO2 O22 | SO |

Ratio 2 1 2
Initial amount(mol) x 0.5 0
Change -0,75 -0,375Y 0,75
Equilibrium x-—0,75 0,125” 0,75
Concentration =—= 0,0625 | 0,375

S03]
K.= [ |

[SO2]?[02]

10V = (0,375) v

(£278)" 0625 fi

x= 1,70 molv

Positive marking from 6.1.2

m
=—Vv
n M

m
= —VvY
1,70 r3

m= 108, 8gv
Decreases.v

Stays the same.v

(2)

¥ Ratio

(8)

(3)
(1)
(1)

Decreases.” A decrease in pressure favours the reaction that produces

more moles of gas/more volume.v The reverse reaction is favoured.”

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(3)
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### Page 7

DoE/Preparatory 2025

QUESTION 7
7.4 A weak base is a base that ionises/dissociates incompletely in water to

form a low concentration of hydroxide ions.v¥ (2)
7.2 Option 1:

[ H30*] = 2 x 0,02= 0,04 mol.dm?
Kw = [ Hs0*][OH]¥
1x 10-4 = 0,04 [OH ]v [OH] = 2,5 x 10°73 mol.dm3v
Option 2:
pH = -log[Hs0*]v
= -log{0,04) v
= 1,398
pH + pOH = 14
1,398 + pOH = 14v
pOH = 12,602
pOH = -log[OH]
12,602 = -logjOH]
[OH7] = 2,50 x 10°3mol.dm3v

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### Page 8

7.3.1

7.3.2

7.4.1
7.4.2

7.4.3

DoE/Preparatory 2025

n(H2SO4) = cvv

= (0,02)(0,025)v
= 06,0005 mal
n(VH3) = 2(0,0005)”
= 0,001 mol
[NH] in diluted solution = ;
_ 0,001
0,0312v

= m0,032051 mol
n(NHs) in 250 cm$ = cv
= 0,032051 x 0,25v
= 0,0080128 mol
n(NHs) in 250 cm? = 7(NH3) in 20 cm3 of undiluted solution

NH;] ==
[NH] = 5
= 0,080128 /,

0,02

= 0,4 mol.dm=3v

Below 7v

NHa4* + HoOvw ———-» NHs3 + Hs0*tvv

An Ampholyte is a substance that can act as either an acid or a basevv
HC204-v

H2C204v and H3O*v

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(4)

(1)

(2)
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### Page 9

DoE/Preparatory 2025

QUESTION 8
8.1 A galvanic cell is a cell in which chemical energy is converted to electrical
energy.vv (2)
8.2 Concentration of 1mol.dm3v
Temperature of 25°C/298Kv (2)
8.3.1 Pty (1)
8.3.2 Crv (1)
8.4 2 mol.dm3v (2)
8.5 3Sn4* + 2Crv ——> 3Sn2 + 2Cr3+v (vbalancing) (3)
8.6 E® cet = E® cathode - E®anode”
= +0,15v- (-0,74)v
= +0,89Vv (4)
8.7 Decreases.v (1)
[16]

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### Page 10

DoE/Preparatory 2025

QUESTION 9

9.4 Electrolysis is the chemical process in which electrical energy is
converted to chemical energy.v v (2)

9.2 Cu———* Cut + 2e vv

Notes/Notas:
Ignore phases/ignoreer fases

Cu?* + 2e° — Cu (2/2) Cu** + 2e° @ Cu (0/2)
Cu © Cu?* + 2e (1/2) Cu?* + 2e° > Cu (0/2) (2)
9.3 Impure copper.v (1)
9.4.1 Zinc/Nickel/Zn/Ni.v (1)
9.4.2  Zinc/Nickel is a stronger reducing agentv than copper.” (2)
9.4.3 They will fall to the bottom.” (1)
9.5 mass of copper + impurutues = 600 — 390V
= 210g
mass of pure copper = 320 — 150V
= 170g
% purity = = x 100V
= 80,95%Y (4)
9.6 It remains constant.v The rate of oxidation of copper is equal to the rate
of reduction of copper ions.v v (3)
[16]

TOTAL: 150

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