# Physical Sciences Grade 12 P2 NSC November Mock 2026 Memo

> Grade 12 Physical Sciences Paper 2 memorandum from the NSC November Mock Exam 2026 (English), 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-12-physical-sciences-2026-nsc-november-mock-exam-memorandum-paper-2/7d283a35-01f1-4cd3-afbc-376d737230ba
- Updated: 2026-10-29
- Subject: Physical Sciences
- Grade: Grade 12
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: NSC November Mock Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 10

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## About this document

Physical Sciences · Grade 12 · NSC November Mock Exam · 2026 · English. Memorandum, 10 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)
                             MOCK EXAMINATION 2026
                               MARKING GUIDELINES

  MARKS: 150

                     These marking guidelines consist of 10 pages.

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### Page 2

Physical Sciences/P2                         2                               Hlayiso/Mock 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.
  5. Negative marking: An incorrect answer that is based on a conceptual mistake cannot
     be correctly motivated. If the choice (e.g. INCREASES/DECREASES) is incorrect, no
     marks are awarded for the explanation.
  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 2/3 in
     equations).

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### Page 3

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

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

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### Page 4

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

QUESTION 2
2.1.1   C✓                                                                                    (1)
2.1.2   F✓                                                                                    (1)
2.1.3   A and D ✓      (both required)                                                        (1)
2.2.1   3-chloro-2,4-dimethylhexane
        Marking criteria

             • Correct stem, i.e. hexane ✓
             • All substituents (chloro and dimethyl) correctly identified ✓
             • IUPAC name completely correct, including numbering, hyphens, commas and
               alphabetical order ✓
                                                                                              (3)
2.2.2   2,3-dimethylpent-1-ene
        Marking criteria

             • Correct stem and position of double bond, i.e. pent-1-ene ✓
             • Substituents (dimethyl) correctly identified ✓
             • IUPAC name completely correct, including numbering, hyphens and commas
               ✓                                                                              (3)
2.2.3   Pentan-3-one ✓✓
        Marking: pentanone ✓; position 3 ✓. Accept: 3-pentanone (max 1/2).                    (2)
2.3.1   Esterification / condensation ✓                                                       (1)
2.3.2   (Concentrated) sulphuric acid / H2 SO4 ✓                                              (1)
2.3.3   Ethanol ✓                                                                             (1)
2.3.4        H   H     H   O         H     H
        H    C   C     C   C    O    C     C   H
             H   H     H             H     H
        Marking criteria

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

        Condensed formula (CH3 CH2 CH2 COOCH2 CH3 ): max 1/2.
                                                                                              (2)
2.3.5   Ethyl butanoate
        Marking: ethyl ✓ butanoate ✓                                                          (2)
2.4     Compounds with the same molecular formula, but different functional groups
        (belonging to different homologous series). ✓✓                                       (2)
                                                                                            [20]

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### Page 5

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

QUESTION 3
3.1     The temperature at which the vapour pressure of a substance equals
        atmospheric (external) pressure. ✓✓                                              (2)
3.2     The compounds have comparable (similar) molar masses. ✓                          (1)
3.3     A✓                                                                               (1)
3.4     Both C and D have hydrogen bonds (and London forces) between their
        molecules.
        D (propanoic acid) has TWO sites for hydrogen bonding / forms dimers / has
        more extensive hydrogen bonding than C. ✓
        The intermolecular forces in D are stronger than in C. ✓
        More energy is needed to overcome the intermolecular forces in D than in C.
        ✓                                                                                (3)
3.5     Between pentane (A) molecules there are London forces only. ✓
        Between butanal (B) molecules there are dipole-dipole forces (in addition to
        London forces). ✓
        The intermolecular forces in B are stronger than in A. ✓
        More energy is needed to overcome the intermolecular forces in B. ✓
        OR: the reverse argument for A.                                                  (4)
3.6     LOWER THAN ✓
        2,2-dimethylpropane is more branched (more compact/spherical), so it has a
        smaller surface (contact) area and weaker London forces between molecules.
        ✓                                                                               (2)
                                                                                       [13]

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### Page 6

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

QUESTION 4
4.1.1   Addition / hydrohalogenation / hydrobromination ✓                                      (1)
4.1.2   Substitution / hydrolysis ✓                                                            (1)
4.1.3   Elimination / dehydration ✓                                                            (1)
4.2         H   Br     H   H
        H   C    C     C   C   H
            H    H     H   H
        Marking: functional group (Br on C2) ✓; whole structure correct ✓. Condensed
        formula: max 1/2.                                                                      (2)
4.3.1   Dehydrohalogenation / dehydrobromination ✓                                             (1)
4.3.2   Concentrated (strong) base, e.g. NaOH or KOH, dissolved in ethanol ✓
        Strong heating (heat under reflux) ✓                                                   (2)
4.3.3   But-2-ene ✓✓
        Marking: butene ✓; position 2 ✓.                                                       (2)
4.4     H       H H H                               H O H H
            C C C C H + H O H → H C C C C H
        H            H H                            H H H H
        Marking criteria
        Reactants ✓      Product ✓    Balancing ✓
        Accept H2 O as molecular formula. Ignore the catalyst (H3 PO4 /H2 SO4 ) if written
        above the arrow. Condensed formulae: max 2/3.                                          (3)
4.5     C4 H8 + 6O2 → 4CO2 + 4H2 O
        Marking: reactants ✓ products ✓ balancing ✓                                            (3)
4.6     Propene ✓                                                                             (1)
                                                                                             [17]

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### Page 7

Physical Sciences/P2                          7                                  Hlayiso/Mock 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 HCℓ (the acid) ✓                                                       (1)
5.2.2   ANY ONE: Temperature / mass of CaCO3 / state of division (surface area)
        of CaCO3 / volume of HCℓ ✓                                                              (1)
5.3     Powdered CaCO3 (experiment 2) has a larger surface area / more particles
        are exposed to the acid. ✓
        More collisions take place (more particles collide with the correct orientation
        and sufficient energy). ✓
        More effective collisions per unit time / higher frequency of effective collisions.
        ✓                                                                                       (3)
5.4     At a higher temperature the (average) kinetic energy of the particles is higher.
        ✓
        More particles have kinetic energy equal to or greater than the activation
        energy. ✓
        More effective collisions per unit time / higher frequency of effective collisions.
        ✓                                                                                       (3)
5.5     n(CO2 ) = rate × ∆t = (1,5 × 10–4 )(60) = 9 × 10–3 mol ✓
        n(CaCO3 )reacted = n(CO2 ) = 9 × 10–3 mol (ratio 1 : 1) ✓
        m = nM = (9 × 10–3 )(100) = 0,90 g ✓
        mremaining = 2 – 0,90 = 1,10 g ✓                                                        (4)
5.6
         Volume of CO2
                        3

                               1

                                                    Time
        Marking criteria

            • Both graphs start at the origin and level off at the SAME final volume (CaCO3
              is the limiting reactant) ✓
            • Graph 3 has a steeper initial gradient than graph 1 ✓
            • Graph 3 levels off before graph 1 ✓
                                                                                               (3)
                                                                                              [17]

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### Page 8

Physical Sciences/P2                               8                                Hlayiso/Mock 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

               • USING ratio CO : H2 : CH3 OH = 1 : 2 : 1 ✓
               • Equilibrium amount of CO = initial – change (0,5 mol) ✓
               • Equilibrium amount of H2 = initial – change (0,6 mol) ✓
               • Equilibrium amount of CH3 OH = 0,3 mol ✓
               • Divide by 2 dm3 ✓
               • Correct Kc expression (formulae in square brackets) ✓
               • Substitution of equilibrium concentrations into Kc expression ✓
               • Final answer: 6,67 ✓

                                                      CO      H2       CH3 OH
         Initial amount (mol)                         0,8     1,2           0
         Change in amount (mol)                      –0,3    –0,6    +0,3 ratio ✓
         Equilibrium amount (mol)                    0,5 ✓   0,6 ✓      0,3 ✓
         Equilibrium concentration (mol·dm–3 )       0,25     0,3    0,15   ÷2✓

                [CH3 OH]                0,15
        Kc =              ✓     =                ✓    = 6,67 ✓
               [CO][H2 ]2           (0,25)(0,3)2
        No Kc expression, correct substitution: max 7/8. Wrong Kc expression: max 5/8 (no
        marks for substitution and answer).                                                        (8)
6.3.1   DECREASES ✓                                                                                (1)
6.3.2   The forward reaction is exothermic (the reverse reaction is endothermic). ✓
        An increase in temperature favours the endothermic (reverse) reaction. ✓
        The concentration of CH3 OH decreases and the concentrations of CO and H2
        increase (so Kc decreases). ✓                                                              (3)
6.4.1   INCREASES ✓                                                                                (1)
6.4.2   An increase in pressure favours the reaction that produces the smaller number
        of moles (volume) of gas. ✓
        The forward reaction produces fewer moles of gas (3 mol of reactant gas → 1
        mol of product gas). ✓
        The forward reaction is favoured. ✓                                                        (3)
6.5     REMAINS THE SAME ✓
        A catalyst increases the rates of the forward and reverse reactions equally (it
        does not change the position of the equilibrium). ✓                                       (2)
                                                                                                 [20]

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### Page 9

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

QUESTION 7
7.1     A base is a proton (H+ ion) acceptor. ✓✓                                             (2)
7.2.1   ACIDIC ✓                                                                             (1)
7.2.2   NH+4 (aq) + H2 O(ℓ) ⇌ NH3 (aq) + H3 O+ (aq)
        Marking: reactants ✓ products ✓ balancing ✓. Ignore phases; accept →.
        Accept: NH4 Cℓ + H2 O ⇌ NH3 + H3 O+ + Cℓ– (3/3).                                     (3)
7.3.1   OPTION 1
                     m 3,2
        n(NaOH) =       =     = 0,08 mol ✓
                     M 40
            n 0,08
        c= =           = 0,32 mol·dm–3 ✓
            V 0,25
        [OH– ] = 0,32 mol·dm–3
        [H3 O+ ][OH– ] = 1 × 10–14 ⇒ [H3 O+ ](0,32) = 1 × 10–14 ✓
        [H3 O+ ] = 3,13 × 10–14 mol·dm–3
        pH = – log(3,13 × 10–14 ) ✓ = 13,51 ✓
        OPTION 2
             m         3,2
        c=       =             ✓✓ = 0,32 mol·dm–3 pOH = – log(0,32) = 0,49 ✓
            MV (40)(0,25)
        pH = 14 – 0,49 ✓ = 13,51 ✓                                                           (5)
7.3.2
        Marking criteria

             • Formula n = cV / substitution for n(NaOH) ✓
             • Substitution for n(H2 SO4 )initial ✓
             • USING ratio n(H2 SO4 ) : n(NaOH) = 1 : 2 ✓
             • n(H2 SO4 )excess = initial – reacted ✓
             • n(H3 O+ ) = 2 × n(H2 SO4 )excess ✓
             • Divide by total volume 0,055 dm3 ✓
             • Formula pH = – log[H3 O+ ] with substitution ✓
             • Final answer: 1,74 ✓

        n(NaOH) = cV = (0,32)(0,025) = 8 × 10–3 mol ✓
        n(H2 SO4 )initial = (0,15)(0,030) = 4,5 × 10–3 mol ✓
        n(H2 SO4 )reacted = 12 (8 × 10–3 ) = 4 × 10–3 mol ✓
        n(H2 SO4 )excess = 4,5 × 10–3 – 4 × 10–3 = 5 × 10–4 mol ✓
        n(H3 O+ ) = 2(5 × 10–4 ) = 1 × 10–3 mol ✓
                  1 × 10–3
        [H3 O+ ] =         ✓ = 0,018 mol·dm–3
                   0,055
        pH = – log(0,0182) ✓ = 1,74 ✓
        POSITIVE MARKING FROM 7.3.1 (concentration of NaOH).                                (8)
                                                                                           [19]

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### Page 10

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

QUESTION 8
8.1     Temperature: 25 ◦ C (298 K) ✓
        Concentration of the electrolytes: 1 mol·dm–3 ✓                                            (2)
8.2     Ensures electrical neutrality in the half-cells / completes the circuit (allows
        ions to move between the half-cells). ✓                                                    (1)
8.3     Ni ✓                                                                                       (1)
8.4     Ni(s) + 2Ag+ (aq) → Ni2+ (aq) + 2Ag(s)
        Marking: reactants ✓ products ✓ balancing ✓.
        Ni + 2Ag+ ⇌ Ni2+ + 2Ag: 2/3       Ni2+ + 2Ag → Ni + 2Ag+ : 0/3      Ignore phases.         (3)
8.5     Eθcell = Eθcathode – Eθanode ✓
        Eθcell = 0,80 – (–0,27) ✓✓
        Eθcell = 1,07 V ✓
        Any other formula using unconventional abbreviations, e.g. Ecell = EOA – ERA , followed
        by correct substitution: 3/4.                                                              (4)
8.6     INCREASES ✓                                                                                (1)
                                                                                                  [12]

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     Cu2+ (aq) + 2e– → Cu(s) ✓✓
        Cu2+ + 2e– ⇌ Cu: 1/2       Cu → Cu2+ + 2e– : 0/2      Cu ← Cu2+ + 2e– : 2/2                (2)
9.3     Chlorine (gas) / Cℓ2 ✓                                                                     (1)
                 m 0,635
9.4     n(Cu) =     =       = 0,01 mol ✓
                 M    63,5
        n(Cℓ2 ) = n(Cu) = 0,01 mol (2 mol e– per mol of each) ✓
        V = nVm = (0,01)(22,4) ✓
        V = 0,22 dm3 (0,224 dm3 ) ✓                                                                (4)
9.5     REMAINS THE SAME ✓
        Cu is a stronger reducing agent than Cℓ– ions (and H2 O), so the copper anode
        is oxidised to Cu2+ ions. ✓
        Cu2+ ions go into solution at the anode at the same rate as Cu2+ ions are
        reduced to Cu at the cathode. ✓                                                            (3)
                                                                                                  [12]
                                                                                      TOTAL:      150

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