# Physical Sciences Grade 11 P2 November Mock Exam 2 2026 Memo

> Grade 11 Physical Sciences Paper 2 memorandum from the November Mock Exam 2 2026 (English), 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/grade-11-physical-sciences-2026-november-mock-exam-2-memorandum-paper-2/b368f82f-d9ac-4050-9fe2-4bb90ee84be0
- Updated: 2026-10-29
- Subject: Physical Sciences
- Grade: Grade 11
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam 2
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 11

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## About this document

Physical Sciences · Grade 11 · November Mock Exam 2 · 2026 · English. Memorandum, 11 pages. Read online or download the PDF.

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- [Grade 11 Physical Sciences P2 November 2026 Mock 2](https://www.hlayiso.com/library/grade-11-physical-sciences-2026-november-mock-exam-2-question-paper-paper-2/4bd46b50-41c8-4670-83a6-c99f40ba7aa1) — QP · 2026 · Eng

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### Page 1

Mock

                             hlayiso mock exams
                             Practice Series:
                             Grade 11 Exam Preparation
                             WWW.HLAYISO.COM

                                  NATIONAL
                             SENIOR CERTIFICATE

                                     GRADE 11

                      PHYSICAL SCIENCES: CHEMISTRY (P2)
                     NOVEMBER EXAMINATION 2026 (MOCK)
                               MARKING GUIDELINES

MARKS: 150

                     These marking guidelines consist of 11 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 (✓). If an incorrect
     formula is used, no further marks are awarded for that calculation.
  2. Units: Penalise an omitted or incorrect unit in a final answer only ONCE per question.
  3. Rounding: Accept final answers rounded to a minimum of TWO decimal places.
  4. Positive marking: Where an incorrect answer to a previous question is used correctly
     in a later calculation, the later calculation is marked positively.
  5. Equations: Reactants ✓, products ✓, balancing ✓. Ignore phases. If ⇌ is used in a
     reaction that goes to completion, deduct 1 mark (max. 2/3). Marks for half-reactions:
     correct species ✓, correct electrons/balancing ✓.
  6. Definitions: Where key words are underlined, deduct 1 mark for EACH key word or
     phrase omitted or used in an incorrect context.
  7. Names and formulae: If a FORMULA is asked for, a name is not accepted, and vice
     versa.

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### Page 3

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

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

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### Page 4

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

QUESTION 2
2.1     A bond in which the electron pair is shared unequally between two atoms
        (atoms with different electronegativities). ✓✓                                   (2)
2.2.1
                  ×
         O        ×   C ××       O
        Marking criteria
        Two shared electron PAIRS between C and EACH O atom (double bonds) ✓
        Two lone pairs on each O atom; no lone pairs on C ✓
                                                                                         (2)
2.2.2
             ××                  ××
         ×
         ×   F    ×    O     ×   F ××
             ××                  ××

        Marking criteria
        Two O–F bonding pairs ✓
        Two lone pairs on O AND three lone pairs on each F ✓
                                                                                         (2)
2.3.1   Linear ✓                                                                         (1)
2.3.2   Trigonal planar ✓                                                                (1)
2.3.3   Angular (bent) ✓                                                                 (1)
2.4.1   ∆EN = 4,0 – 2,0 = 2,0 ✓
        (Highly) polar covalent bond ✓                                                   (2)
2.4.2   The BF3 molecule is TRIGONAL PLANAR (symmetrical). ✓
        The three B–F bond dipoles are equal in size and are arranged at 120◦ to each
        other. ✓
        The bond polarities cancel, so the charge distribution over the molecule is
        symmetrical / there is no net dipole. ✓                                          (3)
2.5     The O atom in H2 O has (two) LONE PAIRS of electrons that it can donate to
        the empty orbital of H+ . ✓
        All the valence electrons of the C atom in CH4 are used in bonding pairs; C
        has NO lone pair to donate. ✓                                                    (2)
2.6.1   The energy needed to break one mole of its molecules into separate atoms.
        ✓✓
        Accept: the energy needed to break one mole of (C–C) bonds.                      (2)
2.6.2   The shorter the bond length, the greater the bond energy (bond energy
        increases as bond length decreases). ✓                                           (1)

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### Page 5

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

2.6.3   SMALLER THAN ✓
        C≡C has THREE shared electron pairs (C=C has two). ✓
        More shared electrons give a stronger attraction between the two nuclei and
        the bonding electrons, pulling the nuclei closer together (shorter bond). ✓
        Accept: X ≈ 120 pm.                                                                          (3)
                                                                                                    [22]

QUESTION 3
3.1     The pressure exerted by a vapour at equilibrium with its liquid in a closed
        system. ✓✓                                                                                    (2)
3.2.1   The boiling point INCREASES from PH3 to SbH3 (as molar mass increases).
        ✓                                                                                             (1)
3.2.2   Molecular size (molar mass / number of electrons) increases from PH3 to
        SbH3 . ✓
        The strength of the London forces (intermolecular forces) increases. ✓
        More energy is needed to overcome the intermolecular forces, so the boiling
        point increases. ✓
        OR the reverse argument from SbH3 to PH3 .                                                    (3)
3.3     NH3 has HYDROGEN BONDS between its molecules. ✓
        PH3 and AsH3 have only (weaker) dipole-dipole forces / London forces
        between their molecules. ✓
        Hydrogen bonds are STRONGER, so MORE ENERGY is needed to overcome
        the intermolecular forces in NH3 . ✓                                                          (3)
3.4     PH3 ✓
        It has the LOWEST boiling point (therefore the weakest intermolecular forces).
        ✓                                                                                             (2)
3.5     Liquid ✓
        (–50 ◦ C lies between its melting point, –88 ◦ C, and its boiling point, –17 ◦ C.)            (1)
3.6     NH3 and H2 O are both POLAR molecules with hydrogen bonds (”like dissolves
        like”). ✓
        NH3 molecules form (strong) HYDROGEN BONDS with H2 O molecules. ✓
        These new forces are similar in strength to the forces between the molecules
        of the separate substances, so they can replace (overcome) them and NH3
        dissolves. ✓                                                                                 (3)
                                                                                                    [15]

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### Page 6

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

QUESTION 4
4.1     What is the relationship between the VOLUME and the TEMPERATURE of a
        (fixed mass of) gas (at constant pressure)?
        Marking criteria
        Dependent and independent variables correctly identified ✓
        Question about the RELATIONSHIP between them ✓
                                                                                                 (2)
4.2     So that the PRESSURE of the enclosed air remains CONSTANT (equal to
        atmospheric pressure). ✓
        Then only the volume and the temperature change, so the relationship between
        them can be tested fairly (Charles’ law). ✓                                              (2)
4.3     The volume of an enclosed gas is directly proportional to the Kelvin
        temperature, provided the pressure (and the mass of the gas) remains
        constant. ✓✓                                                                             (2)
4.4     –273 ◦ C ✓
        Absolute zero (0 K) ✓                                                                    (2)
        V1 V2        p V   p V
4.5         =     (or 1 1 = 2 2 with p1 = p2 ) ✓
        T1 T2         T1    T2
         40    V
             = 2 ✓
        273 373
        V2 = 54,65 cm3 ✓
                                                                  V2
        Any pair of values from the table may be used, e.g. 51,7                    3
                                                            353 = 373 gives 54,63 cm . Range:
        54,60–54,70 cm3 . If temperatures are not converted to kelvin: max. 1/3.                 (3)
4.6     pV = nRT ✓
        (101,3 × 103 )(250 × 10–6 ) ✓ = n(8,31)(27 + 273) ✓
        n = 1,016 × 10–2 mol
             m       0,325
        M= =                   = 31,99 g·mol–1 (≈ 32) ✓
             n 1,016 × 10–2
        The gas is O2 (oxygen). ✓
        Marks: formula; p and V substituted in SI units; T in K; molar mass; identification.
        Accept 31,86–32,00 g·mol–1 . OR: pV = m
                                              M RT used directly.                                (5)
4.7     300 K ✓
        At the higher temperature the molecules have more kinetic energy (move
        faster), so the intermolecular forces between them become negligible / the
        gas is further from condensing. ✓                                                        (2)
                                                                                                [18]

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### Page 7

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

QUESTION 5
5.1     The mass of one mole of a substance (in g·mol–1 ). ✓✓                                 (2)
                  2,64
5.2     n(CO2 ) =      = 0,06 mol     ∴ n(C) = 0,06 mol; m(C) = (0,06)(12) = 0,72 g ✓
                   44
                  0,72
        n(H2 O) =      = 0,04 mol     ∴ n(H) = 0,08 mol; m(H) = 0,08 g ✓
                   18
                                                           0,32
        m(O) = 1,12 – 0,72 – 0,08 = 0,32 g ✓      n(O) =        = 0,02 mol ✓
                                                            16
        C : H : O = 0,06 : 0,08 : 0,02 = 3 : 4 : 1 ✓
        Empirical formula: C3 H4 O ✓                                                          (6)
                                                           112
5.3     M(C3 H4 O) = 3(12) + 4(1) + 16 = 56 g·mol–1 ;          =2✓
                                                            56
        Molecular formula: C6 H8 O2 ✓
        POSITIVE MARKING FROM 5.2.                                                            (2)
                        m 1,12
5.4     n(sorbic acid) =  =       = 0,01 mol ✓
                        M 112
        n(O atoms) = 2(0,01) = 0,02 mol ✓
        N = nNA = (0,02)(6,02 × 1023 ) = 1,20 × 1022 O atoms ✓
        OR: n(O) = 0,02 mol from 5.2 (2 marks), then N = nNA (1 mark). POSITIVE MARKING
        FROM 5.2/5.3.                                                                        (3)
                                                                                            [13]

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### Page 8

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

QUESTION 6
6.1     One mole is the amount of substance having the same number of particles as
        there are atoms in 12 g carbon-12. ✓✓                                               (2)
6.2     0,36 g ✓                                                                            (1)
6.3     Hydrochloric acid / HCℓ ✓
        Beyond 0,36 g of Mg the volume of H2 remains constant (336 cm3 ) although
        more Mg is added, so all the HCℓ is used up (Mg is left over). ✓                    (2)
                 V    0,336
6.4     n(H2 ) =    =        ✓ = 0,015 mol
                Vm     22,4
        n(HCℓ) = 2n(H2 ) = 2(0,015) = 0,03 mol ✓
            n      0,03
        c= ✓=           ✓
            V      0,05
        c = 0,60 mol·dm–3 ✓
        OR: n(Mg) = 0,36
                     24 = 0,015 mol (from 6.2), then as above. POSITIVE MARKING FROM
        6.2.                                                                                (5)
                 m 0,12
6.5     n(Mg) =     =      = 5 × 10–3 mol ✓
                 M     24
        n(H2 ) = n(Mg) = 5 × 10–3 mol ✓
        V = nVm = (5 × 10–3 )(22,4) = 0,112 dm3 = 112 cm3 ✓                                 (3)
6.6     V(O2 )needed = 5 × 20 = 100 cm3 ✓
        100 cm3 < 150 cm3 , so C3 H8 is the limiting reagent and O2 is in excess. ✓
        V(O2 )remaining = 150 – 100 = 50 cm3 ✓
        V(CO2 )formed = 3 × 20 = 60 cm3 ✓
        Vtotal = 50 + 60 = 110 cm3 ✓
        Volume ratios equal mole ratios at the same temperature and pressure (Avogadro’s
        law). Accept conversion to moles via any molar volume.                              (5)
                                                                                           [18]

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### Page 9

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

QUESTION 7
7.1     The energy absorbed or released in a chemical reaction (per mole of reaction).
        ✓✓                                                                                (2)
7.2     EXOTHERMIC ✓
        ∆H < 0 (negative) / energy is released. ✓                                         (2)
7.3     Energy is ABSORBED to break the bonds in N2 and H2 and energy is
        RELEASED when the N–H bonds in NH3 form. ✓
        MORE energy is released during bond formation than is absorbed during bond
        breaking. ✓                                                                       (2)
7.4     EA (reverse) = EA (forward) – ∆H = 230 – (–92) ✓
        EA (reverse) = 322 kJ·mol–1 ✓                                                     (2)
7.5
         Potential energy

                                         catalysed

                 N2 + 3H2      EA
                                                      ∆H
                                               2NH3

                                      Course of reaction
        Marking criteria
        Products LOWER than reactants with ∆H correctly indicated ✓
        EA indicated from reactants to the peak of the uncatalysed curve ✓
        Dotted catalysed curve with a LOWER peak that starts and ends at the same
        energies ✓
                                                                                          (3)
                                                                                         [11]

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### Page 10

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

QUESTION 8
8.1     A base is a substance that produces hydroxide ions (OH– ) when it dissolves
        in water. ✓✓                                                                    (2)
8.2     BASE ✓
        HCO–3 ACCEPTS a proton (H+ ) from H2 O (to form H2 CO3 ). ✓                     (2)
8.3.1   Mg(OH)2 (s) + 2HCℓ(aq) → MgCℓ2 (aq) + 2H2 O(ℓ)
        Reactants ✓ Products ✓ Balancing ✓                                              (3)
8.3.2   n = cV = (0,10)(0,100) ✓
        n = 1,0 × 10–2 mol ✓                                                            (2)
8.3.3   Magnesium chloride ✓
        MgCℓ2 ✓
        POSITIVE MARKING FROM 8.3.1.                                                    (2)
8.3.4   n(HCℓ)initial = cV = (0,10)(0,100) = 1,0×10–2 mol ✓ (POSITIVE MARKING FROM
        8.3.2)
        n(NaOH) = cV = (0,10)(0,024) = 2,4 × 10–3 mol ✓
        n(HCℓ)excess = n(NaOH) = 2,4 × 10–3 mol (1 : 1) ✓
        n(HCℓ)reacted = 1,0 × 10–2 – 2,4 × 10–3 = 7,6 × 10–3 mol ✓
        n(Mg(OH)2 ) = 12 (7,6 × 10–3 ) = 3,8 × 10–3 mol ✓
        m(Mg(OH)2 ) = nM = (3,8 × 10–3 )(58) = 0,22 g ✓
                     0,2204
        % Mg(OH)2 =         × 100 = 44,08% ✓
                      0,50
        Accept 44–44,08%.
        POSITIVE MARKING FROM 8.3.1 (mole ratio).                                      (7)
                                                                                      [18]

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### Page 11

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

QUESTION 9
9.1     A substance that loses (donates) electrons. ✓✓                                         (2)
9.2     CO2 ✓
        The oxidation number of C DECREASES from +4 (in CO2 ) to 0 (in C), so CO2
        is reduced (gains electrons). ✓
        CO2 causes Mg to be oxidised (oxidation number of Mg increases from 0 to
        +2); therefore CO2 is the oxidising agent. ✓                                           (3)
9.3.1   +6 ✓                                                                                   (1)
9.3.2   –2 ✓                                                                                   (1)
9.3.3   H2 S → S + 2H+ + 2e– ✓✓
        Marking     rule:   S + 2H+ + 2e– ← H2 S   (2/2);   H2 S ⇌ S + 2H+ + 2e–     (1/2);
        S + 2H+ + 2e– → H2 S (0/2).                                                            (2)
9.3.4   Cr2 O2–     +    –     3+
             7 + 14H + 6e → 2Cr + 7H2 O ✓✓
        ⇌ used: 1/2. Reverse reaction: 0/2.                                                    (2)
9.3.5   Cr2 O2–    +             3+
             7 + 8H + 3H2 S → 2Cr + 3S + 7H2 O
        Reactants ✓ Products ✓ Balancing ✓
        Accept the unsimplified form Cr2 O2–     +             3+        +
                                          7 + 14H + 3H2 S → 2Cr + 3S + 6H + 7H2 O (2/3:
        not simplified).                                                                       (3)
9.3.6   H2 S ✓                                                                                 (1)
                                                                                              [15]
                                                                                   TOTAL:     150

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