# Physical Sciences Grade 12 P2 2026 Memo (Eng)

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

- Page: https://www.hlayiso.com/library/grade-12-physical-sciences-2026-nsc-november-prediction-exam-memorandum-paper-2/71f47107-7458-4f47-9acd-0839f91b6815
- Updated: 2026-10-29
- Subject: Physical Sciences
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Prediction Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 12

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## About this document

Physical Sciences · Grade 12 · NSC November Prediction Exam · 2026 · English. Memorandum, 12 pages. Read online or download the PDF.

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### Page 1

hlayiso mock exams
                             Practice Series:
                             Grade 12 NSC Preparation
                             WWW.HLAYISO.COM

                                  NATIONAL
                             SENIOR CERTIFICATE

                                     GRADE 12

                      PHYSICAL SCIENCES: CHEMISTRY (P2)
                          PREDICTION EXAMINATION 2026
                               MARKING GUIDELINES

MARKS: 150

                     These marking guidelines consist of 12 pages.

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### Page 2

Physical Sciences/P2                         2                           Hlayiso/Prediction 2026
                                  NSC – Marking Guidelines

GENERAL MARKING GUIDELINES

  1. Calculations: Marks are awarded for the correct formula, correct substitution and the
     correct final answer WITH the correct unit, as indicated by the ticks (✓). A formula is not
     awarded a mark if it is not followed by a substitution. If an incorrect formula is used, no
     further marks are awarded for that calculation.
  2. Units: The final-answer mark is only awarded if the correct unit is given. Penalise an
     omitted or incorrect unit only ONCE per question.
  3. Rounding: Accept final answers rounded to a minimum of TWO decimal places. Accept
     small differences caused by rounding of intermediate values.
  4. Positive marking: Where an answer to a previous question is used in a later calculation,
     the later calculation is marked positively (consistent accuracy).
  5. Negative marking: If the choice (e.g. INCREASES or DECREASES; EXOTHERMIC or
     ENDOTHERMIC) is incorrect, no marks are awarded for the explanation that follows.
  6. Definitions: Where key words are underlined, deduct 1 mark for EACH key word or
     phrase that is omitted or used in an incorrect context.
  7. Equations: Ignore phases. Ignore double arrows (⇌) unless stated otherwise.
     Unbalanced equation: maximum 2/3. Incorrect reactants or products: 0 for that part.
  8. Organic names and formulae: Names must follow IUPAC conventions (hyphens,
     commas, alphabetical order of substituents). Condensed or semi-structural formulae
     are NOT accepted where STRUCTURAL formulae are asked: maximum 1/2 (or as
     indicated).

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### Page 3

Physical Sciences/P2                    3               Hlayiso/Prediction 2026
                             NSC – Marking Guidelines

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

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Physical Sciences/P2                            4                             Hlayiso/Prediction 2026
                                     NSC – Marking Guidelines

QUESTION 2
2.1.1   A✓                                                                                        (1)
2.1.2   C✓                                                                                        (1)
2.1.3   E and F ✓      (both required)                                                            (1)
2.2.1   2,4-dimethylhex-3-ene
        Marking criteria

             • Correct stem and position of the double bond, i.e. hex-3-ene ✓
             • Substituents (dimethyl) correctly identified ✓
             • IUPAC name completely correct, including numbering, hyphens and commas
               ✓

        Accept: 2,4-dimethyl-3-hexene.
                                                                                                  (3)
2.2.2   3-bromo-2,5-dimethylhexane
        Marking criteria

             • Correct stem, i.e. hexane ✓
             • All substituents (bromo and dimethyl) correctly identified ✓
             • IUPAC name completely correct, including numbering, hyphens, commas and
               alphabetical order ✓
                                                                                                  (3)
2.2.3   3-methylbutan-2-ol
        Marking: methylbutanol (correct stem and functional group) ✓; whole name completely
        correct (positions 3 and 2) ✓. Accept: 3-methyl-2-butanol.                                (2)
2.3     ANY ONE:
           H H H           H
        H    C   C     C   C   H
             H         H   O   H
                 C
             H    H
               H
        3-methylbutan-1-ol
        OR 2-methylbutan-2-ol OR 2-methylbutan-1-ol (drawn as structural formulae)
        Marking criteria

             • Functional group (–OH) on a DIFFERENT carbon atom of the same
               (branched) carbon chain ✓
             • Whole structure correct ✓

        Condensed or semi-structural formula: max 1/2. A straight-chain alcohol (e.g.
        pentan-2-ol) is a chain isomer: 0/2.
                                                                                                  (2)
2.4.1   Esterification / condensation ✓                                                           (1)

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### Page 5

Physical Sciences/P2                          5                         Hlayiso/Prediction 2026
                                   NSC – Marking Guidelines

2.4.2   OPTION 1
        Consider 100 g of Y:
                m 64,62
        n(C) = =           = 5,385 mol ✓
                M     12
                10,77                        100 – 64,62 – 10,77 24,61
        n(H) =        = 10,77 mol; n(O) =                            =    = 1,538 mol ✓
                  1                                     16             16
        C : H : O = 5,385 : 10,77 : 1,538 = 3,5 : 7 : 1 = 7 : 14 : 2 ✓
        Empirical formula C7 H14 O2 : M = 7(12) + 14(1) + 2(16) = 130 g·mol–1 = M(Y)
        ∴ Molecular formula: C7 H14 O2 ✓
        OPTION 2
        m(C) = 0,6462 × 130 = 84 g ⇒ n(C) = 12  84 = 7 ✓

        m(H) = 0,1077 × 130 = 14 g ⇒ n(H) = 14 ✓
        m(O) = 130 – 84 – 14 = 32 g ⇒ n(O) = 32 16 = 2 ✓
        Molecular formula: C7 H14 O2 ✓                                                      (4)
2.4.3       H   H      H   H   O       H   H
        H   C   C      C   C   C   O   C   C   H
            H H H H               H H
        (Ethyl pentanoate: compound D is pentanoic acid, so alcohol X is ethanol.)
        Marking criteria

            • Functional group (ester group –COO–) correct ✓
            • Whole structure correct ✓

        Condensed formula (CH3 (CH2 )3 COOCH2 CH3 ): max 1/2. Positive marking from
        QUESTION 2.4.2 for an ester consistent with the molecular formula obtained and
        made from pentanoic acid.
                                                                                           (2)
                                                                                          [20]

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### Page 6

Physical Sciences/P2                          6                      Hlayiso/Prediction 2026
                                   NSC – Marking Guidelines

QUESTION 3
3.1     The pressure exerted by a vapour at equilibrium with its liquid in a closed
        system. ✓✓                                                                       (2)
3.2     The compounds have the same molar mass (molecular mass), i.e. 74 g·mol–1 .
        ✓                                                                                (1)
3.3     D / propanoic acid ✓
        It has the lowest vapour pressure. ✓                                             (2)
3.4     Compound A (methyl ethanoate, an ester) has (London forces and)
        dipole-dipole forces between its molecules. ✓
        Compound D (propanoic acid) has (London forces, dipole-dipole forces and)
        hydrogen bonds between its molecules. ✓
        The intermolecular forces in A are weaker than those in D / hydrogen bonds
        are stronger than dipole-dipole forces. ✓
        Less energy is needed to overcome the intermolecular forces in A, so more
        molecules of A evaporate / A has the higher vapour pressure. ✓
        OR: the reverse argument for D.                                                  (4)
3.5.1   CHAIN ✓                                                                          (1)
3.5.2   Compound B is more branched / more compact / more spherical than C. ✓
        B has a smaller surface (contact) area over which intermolecular forces act,
        so its intermolecular (London) forces are weaker. ✓
        Less energy is needed to overcome the intermolecular forces in B, so more
        molecules evaporate / higher vapour pressure. ✓
        OR: the reverse argument for C (longer chain, larger surface area, stronger
        intermolecular forces).                                                         (3)
                                                                                       [13]

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### Page 7

Physical Sciences/P2                              7                         Hlayiso/Prediction 2026
                                       NSC – Marking Guidelines

QUESTION 4
4.1.1   Substitution ✓                                                                          (1)
4.1.2   Addition ✓                                                                              (1)
4.2.1   Hex-2-ene ✓✓
        Marking: hexene ✓; position 2 of the double bond ✓. Accept: 2-hexene.                   (2)
4.2.2           H    H     H   H   H        (hex-1-ene)
        H
            C   C    C     C   C   C    H
        H
                     H     H   H   H
        Marking criteria

            • Functional group (C=C double bond on carbon 1) ✓
            • Whole structure correct ✓

        Condensed formula (CH2 CHCH2 CH2 CH2 CH3 ): max 1/2.
                                                                                                (2)
4.3     Secondary ✓
        The carbon atom bonded to the halogen (Br) is bonded to TWO other carbon
        atoms. ✓                                                                                (2)
4.4         H Br H H H H
        H C C C C C C H + KOH
            H H H H H H
             →   H H H H H H
               H C C C C C C H + KBr + H2 O
                       H           H H H
        Marking criteria

            • Organic reactant (2-bromohexane) as structural formula ✓
            • Inorganic reactant (KOH) ✓
            • Organic product (hex-2-ene) as structural formula ✓
            • Inorganic products (KBr and H2 O) ✓
            • Balancing ✓

        Accept NaOH and NaBr. Ignore conditions written on the arrow.
        Accept 2-bromohexane → hex-2-ene + HBr (KOH/ethanol on arrow): organic
        reactant ✓, organic product ✓, balancing ✓, inorganic species ✓ (max 4/5).
        Hex-1-ene as organic product: max 4/5. Condensed or molecular formulae for the
        organic compounds: max 3/5.
                                                                                                (5)
4.5.1   Dilute potassium hydroxide / KOH(aq) / NaOH(aq) / water / H2 O ✓                        (1)
4.5.2   Mild heat / heat gently / dilute (aqueous) solution of the base ✓
        Accept: heat. Do not accept: concentrated base, ethanol as solvent, or strong heat.     (1)
4.6     2,3-dibromohexane ✓✓           (dibromohexane ✓; positions 2,3 ✓)                      (2)
                                                                                              [17]

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### Page 8

Physical Sciences/P2                            8                          Hlayiso/Prediction 2026
                                     NSC – Marking Guidelines

QUESTION 5
5.1     Change in concentration of reactants or products per unit time. ✓✓
        OR Change in amount/number of moles/volume/mass of reactants or products
        per (unit) time. ✓✓
        OR Rate of change in concentration/amount/volume/mass. ✓✓                              (2)
5.2.1   Concentration of the acid / H2 SO4 ✓                                                   (1)
5.2.2   ANY ONE: Temperature / mass of Mg / state of division (surface area) of Mg /
        volume of acid ✓                                                                       (1)
5.3.1   The concentration (amount) of the reactants decreases / the amount (surface
        area) of Mg decreases (fewer effective collisions per unit time). ✓                    (1)
                V     0,26
5.3.2   n(H2 ) =    =       ✓ = 1,08 × 10–2 mol
               Vm      24
        n(H2 SO4 )used = n(H2 ) = 1,08 × 10–2 mol    (ratio 1 : 1) ✓
               ∆n      1,08 × 10–2
        Rate =    =             ✓
               ∆t        60
        Rate = 1,81 × 10–4 mol·s–1 ✓
        Accept answers from 1,80 × 10–4 to 1,81 × 10–4 mol·s–1 . Using 260 cm3 and 24 000
        cm3 ·mol–1 is equivalent.                                                              (4)
                 m 0,36
5.3.3   n(Mg) =     =      = 0,015 mol ✓
                 M     24
        n(H2 ) = n(Mg) = 0,015 mol (Mg is the limiting reactant; ratio 1 : 1) ✓
        V = nVm = (0,015)(24) = 0,36 dm3 (360 cm3 ) ✓                                          (3)
5.4     EQUAL TO ✓                                                                             (1)
5.5     Powdered Mg has a larger surface area / more Mg particles are exposed to
        the acid. ✓
        More particles collide with the correct orientation and sufficient energy / more
        collisions occur. ✓
        More effective collisions per unit time / higher frequency of effective collisions.
        ✓                                                                                      (3)
5.6     Q✓
        At the higher temperature (35 ◦ C) the average kinetic energy of the particles
        is higher, so the curve is broader/flatter and shifted to higher kinetic energies /
        more particles have kinetic energy equal to or greater than Ea . ✓                     (2)
                                                                                              [18]

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### Page 9

Physical Sciences/P2                             9                               Hlayiso/Prediction 2026
                                      NSC – Marking Guidelines

QUESTION 6
6.1     When the equilibrium in a closed system is disturbed, the system will re-instate
        a new equilibrium by favouring the reaction that will oppose the disturbance.
        ✓✓                                                                                           (2)
6.2
        Marking criteria
                             m and substitution for n(Cℓ ) (0,12 mol) ✓
               • Formula n = M                          2
               • USING ratio PCℓ5 : PCℓ3 : Cℓ2 = 1 : 1 : 1 ✓
               • Divide equilibrium amounts by 2 dm3 ✓
               • Correct Kc expression (formulae in square brackets) ✓
               • Substitution of 0,04 and equilibrium concentrations into the Kc expression ✓
               • Equilibrium amount of PCℓ5 (0,18 mol) ✓
               • Initial amount of PCℓ5 = equilibrium amount + change (0,30 mol) ✓
               • Final answer: 62,55 g ✓
                    m 8,52
        n(Cℓ2 ) =     =    = 0,12 mol
                    M   71
                                                     PCℓ5      PCℓ3       Cℓ2
         Initial amount (mol)                          x        0          0
         Change in amount (mol)                      –0,12     +0,12   +0,12    ratio
         Equilibrium amount (mol)                   x – 0,12   0,12       0,12
                                                    x – 0,12
         Equilibrium concentration (mol·dm–3 )                 0,06     0,06    ÷2
                                                        2
               [PCℓ3 ][Cℓ2 ]                 (0,06)(0,06)
        Kc =                    ⇒   0,04 =
                 [PCℓ5 ]                        [PCℓ5 ]
        [PCℓ5 ] = 0,09 mol·dm–3 ⇒ n(PCℓ5 )eq = (0,09)(2) = 0,18 mol
        n(PCℓ5 )initial = 0,18 + 0,12 = 0,30 mol
        m = nM = (0,30)(208,5) = 62,55 g
        Accept 62,40 g – 62,55 g (rounding of M).
        No Kc expression, correct substitution: max 7/8. Wrong Kc expression: max 4/8 (no
        marks for substitution, equilibrium amount, initial amount and answer).                      (8)
6.3.1   PCℓ5 (g) was added / the concentration of PCℓ5 was increased. ✓                              (1)
6.3.2   A catalyst was added. ✓                                                                      (1)
6.4.1   INCREASED ✓                                                                                  (1)
6.4.2   The rate of the forward reaction increased more than the rate of the reverse
        reaction / the forward reaction is favoured. ✓
        An increase in temperature favours the endothermic reaction. ✓
        The forward reaction is endothermic (∆H > 0). ✓                                              (3)
6.4.3   GREATER THAN ✓                                                                               (1)

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### Page 10

Physical Sciences/P2                        10                          Hlayiso/Prediction 2026
                                  NSC – Marking Guidelines

6.5     INCREASES ✓
        An increase in pressure (decrease in volume) favours the reaction that
        produces the smaller number of moles of gas, i.e. the reverse reaction (1 mol
        gas versus 2 mol gas). ✓                                                            (2)
                                                                                           [19]

QUESTION 7
7.1.1   An acid that ionises incompletely (partially) in water. ✓✓                           (2)
7.1.2   HOCℓ ✓
        HOCℓ has a smaller Ka value, so it ionises to a smaller extent and its [H3 O+ ]
        is lower. ✓                                                                          (2)
7.1.3   BASIC ✓                                                                              (1)
7.1.4   OCℓ– (aq) + H2 O(ℓ) ⇌ HOCℓ(aq) + OH– (aq)
        Marking: reactants ✓ products ✓ balancing ✓. Ignore phases; accept →.
        Accept: NaOCℓ + H2 O ⇌ HOCℓ + Na+ + OH– (3/3).
        Also accept: NaOCℓ + H2 O ⇌ HOCℓ + NaOH (3/3).                                       (3)
7.2.1   Phenolphthalein ✓
        Oxalic acid is a weak acid and NaOH a strong base, so the pH at
        the equivalence point (endpoint) is greater than 7 (within the range of
        phenolphthalein). ✓                                                                  (2)
7.2.2   OPTION 1
        n((COOH)2 ) = cV = (0,10)(0,025) = 2,5 × 10–3 mol ✓
        n(NaOH) = 2(2,5 × 10–3 ) = 5,0 × 10–3 mol (ratio 1 : 2) ✓
                  n 5,0 × 10–3
        c(NaOH) = =                ✓ = 0,25 mol·dm–3 ✓
                  V      0,020
        OPTION 2
        ca Va na         (0,10)(25) 1
             =    ✓                  = ✓✓        cb = 0,25 mol·dm–3 ✓
        cb Vb nb           cb (20)     2                                                     (4)

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### Page 11

Physical Sciences/P2                           11                             Hlayiso/Prediction 2026
                                     NSC – Marking Guidelines

7.2.3
        Marking criteria

             • n(NaOH)initial = cV with substitution ✓
             • n((COOH)2 ) = cV with substitution and USING ratio 1 : 2 ✓
             • n(NaOH)excess = initial – reacted ✓
             • Divide by total volume 0,025 dm3 ✓
             • Use of Kw (or pOH) with substitution ✓
             • Final answer: 12,85 ✓

        n(NaOH)initial = (0,25)(0,015) = 3,75 × 10–3 mol
        n((COOH)2 ) = (0,10)(0,010) = 1,0 × 10–3 mol ⇒ n(NaOH)reacted = 2,0 ×
        10–3 mol
        n(NaOH)excess = 3,75 × 10–3 – 2,0 × 10–3 = 1,75 × 10–3 mol
                 1,75 × 10–3
        [OH– ] =               = 0,07 mol·dm–3
                    0,025
                                                1 × 10–14
        [H3 O+ ][OH– ] = 1 × 10–14 ⇒ [H3 O+ ] =            = 1,43 × 10–13 mol·dm–3
                                                   0,07
                              –13
        pH = – log(1,43 × 10 ) = 12,85
        OR: pOH = – log(0,07) = 1,15; pH = 14 – 1,15 = 12,85.
        POSITIVE MARKING FROM QUESTION 7.2.2.                                                      (6)
                                                                                                  [20]

QUESTION 8
8.1     Temperature: 25 ◦ C (298 K) ✓
        Concentration of the electrolytes (solutions): 1 mol·dm–3 ✓                                (2)
8.2     Ag ✓
        The mass of the Ag electrode increases, so Ag+ ions are reduced (gain
        electrons) at this electrode / reduction takes place at Ag. ✓                              (2)
8.3     Eθcell = Eθcathode – Eθanode ✓
        1,24 = 0,80 – Eθanode ✓✓
        Eθanode = –0,44 V ✓
        M is iron / Fe ✓
        Any other formula using unconventional abbreviations, e.g. Ecell = EOA – ERA , followed
        by correct substitution: 4/5.                                                              (5)
8.4     Fe(s) + 2Ag+ (aq) → Fe2+ (aq) + 2Ag(s)
        Marking: reactants ✓ products ✓ balancing ✓. Ignore phases.
        Fe + 2Ag+ ⇌ Fe2+ + 2Ag: 2/3        Fe2+ + 2Ag → Fe + 2Ag+ : 0/3
        POSITIVE MARKING FROM QUESTION 8.3 (metal identified there).                               (3)
                                                                                                  [12]

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### Page 12

Physical Sciences/P2                          12                         Hlayiso/Prediction 2026
                                    NSC – Marking Guidelines

QUESTION 9
9.1     The chemical process in which electrical energy is converted to chemical
        energy. ✓✓
        OR The use of electrical energy to produce a chemical change (a
        non-spontaneous reaction). ✓✓                                                         (2)
9.2     2Br– (aq) → Br2 (ℓ) + 2e– ✓✓
        Br2 + 2e– ← 2Br– : 2/2    2Br– ⇌ Br2 + 2e– : 1/2   Br2 + 2e– → 2Br– : 0/2
        Ignore phases; accept Br2 (aq).                                                       (2)
9.3     Br– ions are a STRONGER reducing agent than H2 O. ✓
        Br– ions are oxidised to Br2 (at the anode). ✓
        H2 O is not oxidised to O2 . ✓                                                        (3)
                  m 1,27
9.4     n(Cu) =     =      = 0,02 mol ✓
                  M 63,5
        n(e– ) = 2n(Cu) = 0,04 mol (Cu2+ + 2e– → Cu) ✓
        N = nNA = (0,04)(6,02 × 1023 ) ✓
        N = 2,41 × 1022 electrons ✓                                                           (4)
                                                                                             [11]
                                                                                    TOTAL:   150

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