# Physical Sciences Grade 12 P2 NSC November 2026 Memo (Eng)

> Grade 12 Physical Sciences Paper 2 memorandum from the NSC November Exam 2026 (English), 15 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-physical-sciences-2026-nsc-november-exam-unofficial-memorandum-paper-2/134837e3-5c8a-4d52-b478-70cc67085e8a
- Updated: 2026-11-02
- Subject: Physical Sciences
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 15

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## About this document

Physical Sciences · Grade 12 · NSC November Exam · 2026. Unofficial memorandum, Paper 2, 15 pages. Unofficial marking guidelines with full worked answers by Hlayiso for the Grade 12 Physical Sciences Paper 2 NSC November 2026 exam, plus how our prediction compared. Not the official DBE memorandum. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

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### Page 1

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Unofficial memo
hlayiso mock exams
Practice Series:
Grade 12 NSC Preparation
WWW.HLAYISO.COM
NATIONAL
SENIOR CERTIFICATE
GRADE 12
PHYSICAL SCIENCES: CHEMISTRY (P2)
NOVEMBER 2026 — UNOFFICIAL MARKING GUIDELINES
MARKS: 150
These marking guidelines consist of 15 pages.

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Physical Sciences/P2
2
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
IMPORTANT: These are unofficial marking guidelines prepared by Hlayiso for revision. They
are not the DBE’s official memorandum; the official marking guidelines may differ in mark
allocation and accepted alternatives.
GENERAL MARKING GUIDELINES
• Calculations: a tick (✓) shows one mark. Marks are given for the formula, the substitution, the
use of mole ratios and the final answer WITH the unit. A formula on its own (no substitution) earns
no mark. Accept answers rounded to a minimum of TWO decimal places; ranges are given where
rounding affects the answer.
• Units: penalise a missing or wrong unit in a final answer once per question only. Accept mol.dm–3
or mol/dm3 for mol·dm–3.
• Positive marking: an incorrect answer carried into a later calculation is marked positively.
Negative marking: if the choice in a “choose and explain” item is wrong, no marks are given for
the explanation.
• Definitions: deduct 1 mark for each underlined key word/phrase that is omitted or used in the
wrong context. “(2 or 0)” means no part marks.
• Equations (rule 6.3.10): marks for reactants, products and balancing. Reactants followed by an
arrow, or products preceded by an arrow, may earn marks; reactants or products written without
an arrow earn none. Omitting the charge of an ion: –1 mark per equation. The error-carrying
principle does not apply to equations or half-reactions.
• Half-reactions (rule 6.3.3): the correct arrow is needed. For Ni −−→Ni2+ + 2e – (2
2): Ni2+ +
2e – ←−−Ni (2
2); Ni −−⇀
↽−−Ni2+ + 2e – (1
2); Ni ←−−Ni2+ + 2e – (0
2); Ni2+ + 2e – −−⇀
↽−−Ni (0
2).
• Organic chemistry (rules 6.3.2, 6.3.7–6.3.9): a structural formula must show ALL hydrogen
atoms and bonds; deduct marks for a condensed formula when a structural formula is asked. A
NAME given when a FORMULA is asked (or the reverse) forfeits the marks. IUPAC names need
the correct hyphens and commas (pent 1 ene: marks forfeited).
• Relative strengths (rule 6.3.4): explanations must compare the strengths of oxidising/reducing
agents; “X is above Y in Table 4” or “X is more reactive” earns no marks.
READING OF THE QUESTION PAPER
• Drawn up from photographs of the cover, pages 2–15 and the four data sheets (Table 1: Vm = 22,4
dm3·mol–1, NA = 6,02×1023 mol–1; periodic table; Tables 4A/4B). Q5.5 gives its own molar gas
volume, 24,4 dm3·mol–1 at 25 ◦C.
• Definitions are quoted from the DBE Physical Sciences Examination Guidelines, Grade 12
(2021); accepted alternatives follow recent DBE marking guidelines.
• Answers that depend on our reading of a diagram, or on how the DBE has marked a similar
item before, carry a short Hlayiso.com note. These notes are for learners and teachers, not for
marking.
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Physical Sciences/P2
3
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
QUESTION 1
1.1
A ✓✓
(2)
1.2
C ✓✓
(2)
1.3
C ✓✓
(2)
1.4
B ✓✓
(2)
1.5
D ✓✓
(2)
1.6
C ✓✓
(2)
1.7
B ✓✓
(2)
1.8
A ✓✓
(2)
1.9
D ✓✓
(2)
1.10
B ✓✓
(2)
[20]
Hlayiso.com notes (why each answer is correct):
1.1: Alkanes have only C–C and C–H bonds and are non-polar. Ketones (C=O), alcohols (O–H) and haloalkanes
(C–X) all contain polar bonds.
1.2: The alkyl group on the single-bonded O has four C atoms (butyl). The acid part, H–C(=O)–, has one C atom
(methanoate). The name is butyl methanoate.
1.3: The formyl group, –CHO, is the functional group of the aldehydes (2-methylheptanal). Heptan-2-one has a
carbonyl group, heptanoic acid a carboxyl group, and methyl pentanoate is an ester.
1.4: The products (R) lie below the reactants (Q), so the reaction is exothermic. (i) is false: P – Q is the activation
energy, not the activated complex. (ii) is true: ∆H = Eproducts – Ereactants = R – Q. (iii) is false: a catalyst lowers
the peak (the activation energy); it does not change Q or R.
1.5: At any pressure the yield is lower at 350 K than at 300 K, so a higher temperature favours the reverse
reaction and the forward reaction is exothermic. The yield rises with pressure, so the forward reaction produces
fewer moles of gas: q > r.
1.6: (i) is true: extra O2 favours the reverse reaction, which uses up H2O(g). (iii) is false: solid NH4NO3 is not
included in Kc. (ii) states the equilibrium condition (forward rate = reverse rate) in the wording the DBE used
in Nov 2015 Q6.1. That question compared the rates of formation of X3 and X2 for 3X2 ⇌2X3, and the memo
answer was “Equal to”. So C is the intended answer. Strictly, the 2 : 1 mole ratio means NH4NO3 forms twice as
fast (in mol per second) as O2. On that reading (ii) is false and the answer is A. Check the official memo, which
may accept A as well.
1.7: In the forward reaction H2O accepts a proton (base). In the reverse reaction SO2–
4 accepts a proton (base).
A is false: Ka is small, so HSO–
4 is weak. C is false: dilution does not change the strength of an acid. D is false:
removing water makes the solution more concentrated.
1.8: NH4Cℓis the salt of a strong acid and a weak base (acidic, pH < 7). NaCℓis neutral (pH = 7). Na2CO3 is
the salt of a weak acid and a strong base (basic, pH > 7).
1.9: In cell notation the anode is written on the left and the cathode on the right. Cr2+ is oxidised to Cr3+ at the
inert Pt electrode (anode), and Ag+ is reduced at the Ag electrode (cathode).
1.10: H2O (–0,83 V) is a stronger oxidising agent than Na+ (–2,71 V), so H2O, not Na+, is reduced at the cathode
(to H2 and OH–).
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Physical Sciences/P2
4
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
QUESTION 2
2.1.1
F and G ✓
(1)
2.1.2
D and E ✓
(1)
2.1.3
C ✓
Hlayiso.com note: F (5,5-dimethylhept-3-yne) and G (5-methyloct-3-yne) are both C9H16
alkynes with different chains. D (3-methylbutan-2-one, a ketone) and E (3-methylbutanal, an
aldehyde) are both C5H10O. C (C7H16) is the only alkane.
(1)
2.2
The alcohol is compound B, C6H14O. ANY TWO of:
H
C
H
H
C
O
H
C
H H H
C
H
H
C
H
H
C
H
H
H
2-methylpentan-2-ol ✓✓
H
C
H
H
C
H
H
C
O
H
C
H H H
C
H
H
C
H
H
H
3-methylpentan-3-ol ✓✓
H
C
H
H
C
O
H
C
H H H
C
C
H
H
H
H
C
H
H
H
2,3-dimethylbutan-2-ol ✓✓
Marking criteria
hlayiso.com
Per alcohol:
• Hydroxyl group on a tertiary C atom.
✓
• Whole structure correct (C6H14O). ✓
IF:
• More than one/wrong functional
group: 0
2 per molecule.
• Primary or secondary alcohol: 0
2 per
molecule. The same alcohol drawn
twice: mark once.
• Condensed structural formulae used:
Max 2
4.
• Bond between O and H not shown:
accept.
(4)
2.3.1
Marking criteria
hlayiso.com
• Correct stem, i.e. butane. ✓
• Substituents (trimethyl) correctly identified and numbered: IUPAC name
completely correct including numbering, sequence, hyphens and commas.
✓
2,2,3-trimethylbutane
(2)
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Physical Sciences/P2
5
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
2.3.2
Marking criteria
hlayiso.com
• Correct stem and functional group, i.e. butanal. ✓
• Substituent (methyl) on C3: IUPAC name completely correct including
numbering and hyphens. ✓
3-methylbutanal
NOTE: methylbutanal: 1
2 (2- and 3-methylbutanal are both possible).
(2)
2.3.3
Marking criteria
hlayiso.com
• Correct stem, i.e. heptyne. ✓
• Substituents (dimethyl) correctly identified. ✓
• IUPAC name completely correct including numbering, sequence, hyphens
and commas. ✓
5,5-dimethylhept-3-yne / 5,5-dimethyl-3-heptyne
Hlayiso.com note: The longest chain through the triple bond has 7 C atoms (the first branch
on the left is an ethyl group, CH2CH3). Numbering from the left gives the triple bond the lowest
number (3, not 4).
(3)
2.4
Marking criteria
hlayiso.com
(a) Substitute 22 and 44 g·mol–1 in n = m
M (for CO2). ✓
(b) Mass of H: 7,25–6 = 1,25 g (subtraction). ✓
(c) Moles of H: n(H) = 1,25 mol. ✓
(d) Mole ratio C : H = 0,5 : 1,25 = 2 : 5 (empirical formula C2H5). ✓
(e) Molecular formula C4H10 (CnH2n+2) and the name butane. ✓
OPTION 1
n(CO2) = m
M = 22
44 ✓= 0,5 mol
n(C) = 0,5 mol; m(C) = (0,5)(12) = 6 g
m(H) = 7,25–6 = 1,25 g ✓
n(H) = 1,25
1
= 1,25 mol ✓
C : H = 0,5 : 1,25 = 2 : 5 ✓(C2H5)
Alkane CnH2n+2: C4H10, butane ✓
OPTION 2
1 mol CnH2n+2 gives n mol CO2
n(CO2) = 22
44 ✓= 0,5 mol
n(alkane) = 0,5
n ✓
M = 14n+2: 7,25 = 0,5
n (14n+2) ✓✓
7,25n = 7n+1, n = 4
C4H10, butane ✓
Accept: 2-methylpropane/methylpropane (also C4H10).
C4H10 without a name: Max
4
5.
(5)
[19]
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Physical Sciences/P2
6
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
QUESTION 3
3.1
If any of the underlined key words/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
A ✓
Hlayiso.com note: A has the lowest vapour pressure, so it has the strongest intermolecular
forces. For straight-chain acids this means the longest chain.
(1)
3.3
London forces / dispersion forces / induced dipole forces ✓
(1)
3.4
LOWER THAN ✓
•
Carboxylic acids have (in addition to London forces and dipole-dipole forces)
hydrogen bonding between molecules. ✓
•
Esters have dipole-dipole forces (and London forces) between molecules.
✓
•
The intermolecular forces in the carboxylic acid are stronger, so more energy
is needed to overcome/break them (fewer molecules escape to the vapour
phase). ✓
OR: the intermolecular forces in the ester are weaker; less energy is needed to
overcome them.
Accept: the abbreviation IMF. IF HIGHER THAN or THE SAME AS: 0
4.
(4)
3.5.1
Both have a hydrogen atom bonded to a (highly electronegative) oxygen atom.
✓
Accept: both contain an –OH (hydroxyl) group / an O–H bond.
(1)
3.5.2
•
A (the carboxylic acid) has two sites for hydrogen bonding between
molecules / forms dimers, while the alcohol has one site for hydrogen
bonding. ✓
•
The intermolecular forces in A are stronger (than in the alcohol). ✓
(2)
[11]
QUESTION 4
4.1.1
If any of the underlined key words/phrases in the correct context is omitted, deduct 1
mark.
Compounds with the same molecular formula, but different positions of the
side chain/substituents/functional groups on the parent chain. ✓✓
(2)
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Physical Sciences/P2
7
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
4.1.2
H
C
H
H
C
C
H
H
H
C
H H H
C
H
O
H
C
H
H
C
H
H
H
Marking criteria
hlayiso.com
• Hydroxyl group on the third C atom
(of 2,2-dimethylpentane). ✓
• Whole structure correct. ✓
IF:
• More than one/wrong functional
group: 0
2.
• Condensed structural formula: Max
1
2.
Hlayiso.com note: A is 2,2-dimethylpentan-3-ol: reaction II (substitution) puts Br on the same
C atom, giving 3-bromo-2,2-dimethylpentane. Hydration of 4,4-dimethylpent-2-ene puts the
–OH on C2 or C3, so B is the positional isomer 4,4-dimethylpentan-2-ol.
(2)
4.1.3
Marking criteria
hlayiso.com
• Correct stem and position of the functional group, i.e. pentan-2-ol. ✓
• Substituents (4,4-dimethyl) correct: IUPAC name completely correct including
numbering, hyphens and commas. ✓
4,4-dimethylpentan-2-ol / 4,4-dimethyl-2-pentanol
(2)
4.1.4
Substitution ✓
Accept: bromination/halogenation.
(1)
4.1.5
Elimination / dehydrohalogenation / dehydrobromination ✓
(1)
4.1.6
KBr ✓and H2O ✓
Accept: NaBr/LiBr (with H2O), matching the strong base used (NaOH/KOH/LiOH).
IF: HBr given instead of the metal bromide: Max 1
2.
Hlayiso.com note: The paper does not name the base for reaction III. In dehydrohalogenation
(concentrated KOH/NaOH in ethanol, strong heat) the H and Br removed end up as the metal
bromide and water, as in DBE Nov 2024 4.6.2 (NaBr + H2O).
(2)
4.1.7
Alcohols are flammable/catch fire easily. ✓
Accept (ANY ONE): alcohol is volatile/evaporates too quickly; to heat evenly/gently;
a water bath gives a low temperature (DBE Nov 2018 2.1).
(1)
4.1.8
Catalyst ✓
Accept: speeds up the reaction / dehydrating agent (removes water).
(1)
4.1.9
Esterification / condensation ✓
(1)
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Physical Sciences/P2
8
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
4.1.10
C
O
O
C ✓
Accept:
C
O
O
or R–C(=O)–O–R′ drawn with the C=O shown. Condensed
(–COO–): 0
1.
Hlayiso.com note: Compound C is the ester 4,4-dimethylpentan-2-yl methanoate (methanoic
acid + B). The question asks only for the ester functional group.
(1)
4.2.1
Concentrated H2SO4 / (concentrated) sulphuric acid ✓
IF dilute H2SO4: 0
1.
(1)
4.2.2
Br2(ℓ) / bromine ✓
(1)
4.2.3
CH3CH(OH)CH2CH3 −−→CH3CH –– CHCH3 + H2O
(conc. H2SO4 and heat may be written on the arrow)
Marking criteria
hlayiso.com
• Functional group of the alcohol: –OH on the second C atom. ✓
• Whole condensed structural formula of butan-2-ol correct. ✓
• Functional group of the alkene: double bond between C2 and C3. ✓
• Whole condensed structural formula of but-2-ene correct. ✓
• H2O ✓
IF:
• Molecular formulae used: Max 1
5.
Full structural formulae used: deduct 1
mark.
• But-1-ene (CH2 –– CHCH2CH3) as the product: Max 3
5.
• H2SO4 written as a reactant or product (extra species): deduct 1 mark.
• Marking rule 6.3.10.
Accept: CH3CHOHCH2CH3 for the alcohol and CH3CHCHCH3 for the alkene (DBE
condensed style). Ignore conditions on the arrow.
Hlayiso.com note: X is butan-2-ol (hydrolysis of 2-bromobutane with dilute KOH). Y must
be but-2-ene, because adding Br2 in step 3 gives 2,3-dibromobutane. But-1-ene would give
1,2-dibromobutane, so it is not accepted for Y even though it is the minor dehydration product.
(5)
[21]
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Physical Sciences/P2
9
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
QUESTION 5
5.1
NOTE: Give the mark for “per unit time” only if in the 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
•
A higher concentration of HCℓmeans more (HCℓ) particles per unit volume.
✓
•
More effective collisions per unit time / higher frequency of effective
collisions. ✓
•
Higher reaction rate. ✓
OR (lower concentration): fewer particles per unit volume; fewer effective collisions
per unit time; lower reaction rate.
(3)
5.3.1
Q ✓
(1)
5.3.2
R ✓
Hlayiso.com note: Experiment 2 (pure powder) is fastest: P. Experiment 1 (impure powder)
has less Al2(CO3)3 in the 8 g sample: Q. Experiment 3 (impure granules) also has the smallest
surface area: R.
(1)
5.4
Experiment 3 has the smallest surface area / granules have a smaller
(exposed) surface area than a powder. ✓
Accept: larger particles; impure AND granules.
(1)
5.5
Marking criteria
hlayiso.com
(a) Substitute 5,1×10–3 and 430 s in rate = ∆V
∆t . ✓
(b) Substitute 24,4 dm3·mol–1 in n = V
Vm . ✓
(c) Use the mole ratio n(Aℓ2(CO3)3) : n(CO2) = 1 : 3. ✓
(d) Substitute 234 g·mol–1 in n = m
M. ✓
(e) Subtraction: 8–m(Aℓ2(CO3)3). ✓
(f) Final answer: 0,99 g (range: 0,98 – 0,99 g). ✓
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Physical Sciences/P2
10
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
OPTION 1
rate = ∆V
∆t : 5,1×10–3 =
V
430 ✓
V(CO2) = 2,193 dm3
n(CO2) = V
Vm
= 2,193
24,4 ✓= 0,0899
mol
n(Aℓ2(CO3)3) = 1
3(0,0899) ✓
= 0,02996 mol
m = nM = (0,02996)(234) ✓= 7,01 g
m(impurities) = 8–7,01 ✓
= 0,99 g ✓
OPTION 2
V(CO2) = (5,1×10–3)(430) ✓
= 2,193 dm3
234 g Aℓ2(CO3)3 gives
3(24,4) = 73,2 dm3 CO2 ✓✓
m = (234)(2,193)
73,2
✓= 7,01 g
m(impurities) = 8–7,01 ✓= 0,99 g ✓
M(Aℓ2(CO3)3) = 2(27)+3(12)+9(16) = 234 g·mol–1.
If 22,4 dm3·mol–1 is used: Max
5
6.
(6)
[14]
QUESTION 6
6.1
A closed system is isolated from its surroundings. ✓✓(2 or 0)
Accept: a system in which no matter/substances can enter or leave (it does not
exchange matter with its surroundings).
Hlayiso.com note: The Examination Guidelines (2021) wording is: “An open system
continuously interacts with its environment, while a closed system is isolated from its
surroundings.”
(2)
6.2
Dynamic equilibrium has been reached: the rate of the forward reaction equals
the rate of the reverse reaction. ✓
(1)
6.3.1
The volume of the syringe was decreased / the pressure was increased. ✓
(1)
6.3.2
N ✓✓
(2)
6.3.3
•
Kc = [CO2] (only the gas appears in the Kc expression). ✓
•
The temperature is constant, so Kc (and therefore the equilibrium [CO2])
remains the same. ✓
•
The increase in pressure favours the reverse reaction (fewer moles of gas),
so [CO2] decreases until it is again equal to its original equilibrium value. ✓
NOTE: L and M would both give a new equilibrium [CO2] different from the original,
i.e. a different Kc at the same temperature.
(3)
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Physical Sciences/P2
11
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
6.4
Marking criteria
hlayiso.com
(a) Substitute 4,3 g and 44 g·mol–1 in n = m
M. ✓
(b) Divide by 0,3 dm3 (c = n
V). ✓
(c) Kc = [CO2] = 0,33 (correct Kc expression and value). ✓
(d) Kc increases (0,33 > 0,25) when the temperature increases. ✓
(e) The forward reaction is favoured / [CO2] (product) increases. ✓
(f) An increase in temperature favours the endothermic reaction, so ∆H > 0. ✓
n(CO2) = m
M = 4,3
44 ✓= 0,0977 mol
c(CO2) = n
V = 0,0977
0,3
✓= 0,33 mol·dm–3
Kc = [CO2] = 0,33 ✓
∆H > 0:
•
Kc increased from 0,25 to 0,33 when the temperature was increased. ✓
•
The forward reaction is favoured (the concentration of the product
increases). ✓
•
An increase in temperature favours the endothermic reaction, so the forward
reaction is endothermic: ∆H > 0. ✓
IF: Kc expression with CaCO3 and/or CaO: no mark for (c).
∆H < 0: no marks for (d)
– (f).
(6)
6.5
REMAINS THE SAME ✓
Hlayiso.com note: CaCO3 is a solid: its concentration is constant and it does not appear in Kc
= [CO2]. At constant temperature and volume, [CO2] and the amount of CO2 do not change.
(1)
[16]
QUESTION 7
7.1.1
The reaction of a salt with water. ✓
Accept: the reaction of the cations or anions (of a salt) with water.
(1)
7.1.2
NH4+(aq) + H2O(ℓ) −−⇀
↽−−NH3(aq) + H3O+(aq)
✓✓bal ✓
Marking criteria
hlayiso.com
Reactants ✓
Products ✓
Balancing ✓
Ignore →and phases. Marking rule 6.3.10.
Accept: (NH4)2SO4(aq) + 2H2O(ℓ) −−⇀
↽−−2NH3(aq) + 2H3O+(aq) + SO42 – (aq).
(3)
7.2.1
Weak (base) ✓
(1)
7.2.2
n = cV ✓
n(HCℓ) = (0,17)(0,120) ✓
= 0,0204 mol (2,04×10–2 mol) ✓
IF 120 cm3 is not converted to dm3: Max 1
3.
(3)
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Physical Sciences/P2
12
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
7.2.3
POSITIVE MARKING FROM QUESTION 7.2.2
Marking criteria
hlayiso.com
(a) Substitute 0,2 and 0,0269 in n = cV. ✓
(b) Use the mole ratio n(HCℓ) : n(Ba(OH)2) = 2 : 1. ✓
(c) n(HCℓ)reacted with NH3 = n(HCℓ)initial –n(HCℓ)excess. ✓
(d) Use the mole ratio n(NH3) : n(HCℓ) = 1 : 1. ✓
(e) Use the mole ratio n((NH4)2SO4) : n(NH3) = 1 : 2. ✓
(f) Substitute 132 g·mol–1 in n = m
M. ✓
(g) Percentage: m
3 ×100. ✓
(h) Final answer: 21,21 % (range: 21,2 – 21,21 %). ✓
n(Ba(OH)2) = cV = (0,2)(0,0269) ✓= 5,38×10–3 mol
n(HCℓ)excess = 2n(Ba(OH)2) = 2(5,38×10–3) ✓= 1,076×10–2 mol
n(HCℓ)reacted = 0,0204–0,01076 ✓= 9,64×10–3 mol
n(NH3) = n(HCℓ)reacted = 9,64×10–3 mol ✓
n((NH4)2SO4) = 1
2n(NH3) = 1
2(9,64×10–3) ✓= 4,82×10–3 mol
m = nM = (4,82×10–3)(132) ✓= 0,636 g
%(NH4)2SO4 = 0,636
3
×100 ✓= 21,21% ✓
M((NH4)2SO4) = 2(14+4)+32+4(16) = 132 g·mol–1. NOTE: If the excess HCℓis not
subtracted (all 0,0204 mol used for NH3): Max 5
8.
(8)
7.3.1
YELLOW TO BLUE ✓
(1)
7.3.2
•
The OH– ions of the base react with/neutralise H3O+ ions, so [H3O+]
decreases. ✓
•
The forward reaction is favoured (to increase [H3O+]). ✓
•
[In–] increases / more (blue) In– ions form ([HIn] decreases). ✓
(3)
[20]
QUESTION 8
8.1
Ensures electrical neutrality in the cell (half-cells) / provides a path for the
movement of ions / restores the charge balance. ✓
“Completes the circuit” is given in the question: 0
1.
(1)
8.2
Cr2O72 – + 14H+ + 6Fe2+ −−→2Cr3+ + 6Fe3+ + 7H2O
✓✓bal ✓
Marking criteria
hlayiso.com
Reactants ✓
Products ✓
Balancing ✓
Ignore ⇌and phases. Marking rule 6.3.10.
Hlayiso.com note: Cr2O2–
7 (+1,33 V) is the stronger oxidising agent, so it is reduced in half-cell
B (cathode), and Fe2+ is oxidised in half-cell A (anode). Multiply Fe2+ −−→Fe3+ + e – by 6 to
balance the electrons.
(3)
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Physical Sciences/P2
13
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
8.3
OPTION 1
Eθ
cell = Eθ
cathode –Eθ
anode ✓
= 1,33 ✓– 0,77 ✓
= 0,56 V ✓
Notes: Accept any other correct formula from the data sheet. Any other formula using
unconventional abbreviations (e.g. Eθ
cell = Eθ
OA–Eθ
RA) followed by correct substitutions:
3
4.
OPTION 2
Cr2O72 – + 14H+ + 6e – −−→2Cr3+ + 7H2O
Eθ = +1,33 V ✓
6Fe2+ −−→6Fe3+ + 6e –
Eθ = –0,77 V ✓
Cr2O72 – + 14H+ + 6Fe2+ −−→2Cr3+ + 6Fe3+ + 7H2O
Eθ = +0,56 V ✓✓
(4)
8.4
FROM X TO Y ✓
Hlayiso.com note: X is connected to half-cell A, where Fe2+ is oxidised (anode). Electrons
flow in the external circuit from the anode to the cathode (half-cell B, connected to Y).
(1)
8.5
INCREASES ✓✓
Hlayiso.com note: H+ ions are used up in the reduction half-reaction at the cathode
(Cr2O72 – + 14H+ + 6e – −−→2Cr3+ + 7H2O), so [H+] decreases and the pH increases.
(2)
8.6.1
INCREASES ✓✓
Hlayiso.com note: Eθ
cell = 1,81–0,77 = 1,04 V, compared with 0,56 V before.
(2)
8.6.2
•
Co3+ is a stronger oxidising agent than Cr2O2–
7 . ✓
•
Co3+ has a greater tendency to be reduced (gain electrons from Fe2+),
so the difference in strength between the oxidising agent and the reducing
agent (Fe2+) is greater. ✓
OR: Cr2O2–
7 is a weaker oxidising agent than Co3+.
Position in Table 4 only: 0
2 (rule
6.3.4).
(2)
[15]
QUESTION 9
9.1
ANY ONE:
•
The chemical process in which electrical energy is converted to chemical
energy. ✓✓
•
The use of electrical energy to produce a chemical change.
•
Decomposition of an ionic compound by means of electrical energy.
•
The process during which an electric current passes through a solution/ionic
liquid/molten ionic compound.
(2 or 0)
(2)
9.2.1
+2 ✓
Accept: 2+ / II.
(1)
9.2.2
REDUCTION ✓
(1)
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Physical Sciences/P2
14
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
9.2.3
Ni −−→Ni2+ + 2e –
✓✓
Marking criteria (rule 6.3.3): Ni2+ + 2e – ←−−Ni 2
2;
Ni −−⇀
↽−−Ni2+ + 2e –
1
2;
Ni ←−−
Ni2+ + 2e –
0
2;
Ni2+ + 2e – −−⇀
↽−−Ni
0
2;
Ni2+ + 2e – −−→Ni
0
2. Ignore phases.
Hlayiso.com note: For electroplating, the plating metal (nickel) is the anode, R, connected
to the positive terminal. It dissolves and replaces the Ni2+ ions that are reduced on the coin
(cathode).
(2)
9.3
Marking criteria
hlayiso.com
(a) Formula n = m
M. ✓
(b) Substitute 2,95 g and 59 g·mol–1. ✓
(c) Use the mole ratio n(e–) : n(Ni) = 2 : 1. ✓
(d) Substitute 6,02×1023 in n = N
NA . ✓
(e) Final answer: 6,02×1022 (electrons). ✓
OPTION 1
n(Ni) = m
M ✓= 2,95
59 ✓= 0,05 mol
n(e–) = 2n(Ni) = 2(0,05) ✓= 0,1 mol
N = nNA = (0,1)(6,02×1023) ✓
= 6,02×1022 electrons ✓
OPTION 2
n(Ni) = m
M ✓= 2,95
59 ✓= 0,05 mol
N(Ni) = nNA = (0,05)(6,02×1023) ✓
= 3,01×1022 Ni atoms
N(e–) = 2(3,01×1022) ✓
= 6,02×1022 electrons ✓
(5)
9.4
Ni + Pb2+ −−→Ni2+ + Pb
✓✓bal ✓
Marking criteria
hlayiso.com
Reactants ✓
Products ✓
Balancing ✓
Ignore ⇌and phases. Marking rule 6.3.10.
Accept: Ni + Pb(NO3)2 −−→Ni(NO3)2 + Pb.
Hlayiso.com note: Ni (–0,27 V) is a stronger reducing agent than Pb (–0,13 V), so the nickel
plating reduces Pb2+ to Pb.
(3)
[14]
TOTAL:
150
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Physical Sciences/P2
15
NSC – Unofficial Marking Guidelines
Hlayiso.com/November 2026
HOW DID OUR PREDICTION DO?
The real November 2026 P2 compared with the Hlayiso 2026 Physical Sciences P2 Prediction
Q Topic (marks)
Real paper asked
Our closest prediction item(s)
Verdict Covered
1
MCQ (20)
Non-polar series; ester name; formyl group;
PE diagram; yield vs T and p; equilibrium
statements; HSO–
4; salt pH order; cell
notation; NaCℓelectrolysis
Functional group, PE diagram, catalyst,
pressure on equilibrium, conjugate pair,
OA/RA strength
/
10/20
2
Names, isomers
(19)
Isomer letters (3); two tertiary alcohols (4);
three names (7); alkane from combustion
data (5)
Letter table with isomers; three names;
isomer drawing; molecular formula by
calculation

14/19
3
Physical
properties (11)
Define VP (2); most C atoms (1); London
forces (1); acid vs ester VP (4); acid vs
alcohol (3)
Same definition; ester vs acid VP
explained with IMF types, almost word
for word

8/11
4
Organic
reactions (21)
Positional isomers (2); the two hydration
alcohols (4); substitution, elimination, KBr +
H2O (4); esterification set-up (4);
dehydration reagents, 5-mark equation (7)
HBr substitution, dehydrohalogenation,
conc. H2SO4, Br2, a 5-mark equation,
esterification. Water bath was “unlikely”

17/21
5
Rates (14)
Define rate (2); collision theory (3); curves,
surface area (3); impurity mass from
average rate (6)
Definition, collision theory, curve
identification, average rate and gas
stoichiometry

12/14
6
Equilibrium (16)
Closed system (2); constant [CO2] (1);
volume change, curve N (6); Kc from mass
→sign of ∆H (6); add solid (1)
Kc and temperature reasoning; volume
change. No RICE table; solid equilibria
were Tier 3
/
10/16
7
Acids, bases
(20)
Hydrolysis (4); weak base (1); n = cV (3);
back-titration, % purity (8); HIn colour, Le
Chatelier (4)
Salt hydrolysis equation;
excess-reagent calculation (pH, not %
purity); indicator choice
/
13/20
8
Galvanic cell
(15)
Salt bridge (1); net reaction (3); emf (4);
electron flow (1); pH (2); stronger OA (4)
Net reaction, emf, cathode and OA
reasoning. No salt-bridge or pH item

10/15
9
Electrolysis (14)
Define (2); Ni2+, reduction, anode (4);
electrons from 2,95 g Ni (5); Ni + Pb2+ (3)
Same definition; half-reaction;
electrons from mass deposited. We set
CuBr2, not electroplating

12/14
Marks of the real paper covered by our prediction
106/150
A learner who had practised our prediction paper would have met about 106 of the 150 marks (71%)
as familiar question types. The nine-question structure, topic order and even-year rotation calls held.
Two Tier-1 calls failed (no RICE-table Kc, no Maxwell-Boltzmann curve), and four items we rated
“unlikely” appeared.
What we got right
• Even-year calls: esterification, molecular formula
by calculation, salt hydrolysis, define vapour
pressure, emf and net reaction.
• Ester vs carboxylic acid vapour pressure with IMF
types (Q3.4).
• HBr substitution, dehydrohalogenation (KBr +
H2O), dehydration, Br2 addition, a 5-mark organic
equation.
• Reaction rate, collision theory, curves and rate
stoichiometry.
• Define electrolysis; electrons from the mass
deposited.
What we missed
• Tier 1: no RICE-table Kc (only Kc = [CO2]) and no
Maxwell-Boltzmann curve.
• “Unlikely” items that came: CaCO3 equilibrium,
water bath, HIn colour equilibrium, define positional
isomers.
• Electroplating in Q9; Ni + Pb2+.
• Back-titration for % purity; closed system;
hydrolysis definition; salt bridge; pH change.
• MCQs on ester names, salt pH order, cell notation.
Rating: 6/10.
The structure, topic order and even-year rotation calls were right (106/150
covered), and Q3, Q4, Q5 and Q9 were close. Two near-certain calls did not appear, and four
items we rated unlikely did.
Copyright reserved

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