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
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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.
Physical Sciences · Grade 12 · Limpopo Mock Exam · 2025. Memorandum, 10 pages. Read online or download the PDF.
- Subject
- Physical Sciences
- Grade
- Grade 12
- Document type
- Memorandum
- Year
- 2025
- Exam period
- Limpopo Mock Exam
- Paper
- 2
- Curriculum
- CAPS
- Pages
- 10
- File size
- 1.4 MB
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This exam
Limpopo Mock Exam 2025
- Paper 1Question paperMemo
- Paper 2Question paperMemo
Questions in this paper
- Question 1Organic Compounds and Macromolecules20 marks
- Question 2Organic Compounds and Macromolecules19 marks
- Question 3Organic Compounds and Macromolecules11 marks
- Question 4Organic Compounds and Macromolecules18 marks
- Question 5Rate and Extent of Reactions10 marks
- Question 6Chemical Equilibrium18 marks
Document textSearch extracted text and jump to a page.
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)
(2)
(2)
(2)
[20]
(2)
(1)
(3)
(2)
(1)
(2)
(3)
(1)
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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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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(1)
(2)
(2)
(1)
(5)
(2)
(1)
(1)
(2)
(1)
[18]
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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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)
[18]
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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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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)
[22]
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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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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