# Physical Sciences Grade 11 P2 November Mock 2026 Memo (Eng)

> Grade 11 Physical Sciences Paper 2 memorandum from the November Mock Exam 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-memorandum-paper-2/5844891c-70bd-4aa3-988d-88375f6034e5
- Updated: 2026-10-29
- Subject: Physical Sciences
- Grade: Grade 11
- Language: English
- Document type: Memorandum
- Year: 2026
- Exam period: November Mock Exam
- Paper: 2
- Publisher: Hlayiso
- Curriculum: CAPS
- Pages: 11

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## About this document

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

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### Page 1

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

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### Page 4

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

QUESTION 2
2.1     The sharing of electrons (at least one pair) between two atoms (non-metals)
        to form a molecule. ✓✓                                                              (2)
2.2.1
         H× O ×H
        Marking criteria
        Two O–H bonding pairs ✓
        Two lone pairs on O ✓
                                                                                            (2)
2.2.2
         H× N ×H
               ×
               H
        Marking criteria
        Three N–H bonding pairs ✓
        One lone pair on N ✓
                                                                                            (2)
2.3.1   Dative covalent bond (coordinate covalent bond) ✓                                   (1)
2.3.2   Tetrahedral ✓                                                                       (1)
2.4.1   A measure of the tendency of an atom in a molecule to attract bonding
        electrons. ✓✓                                                                       (2)
2.4.2   ∆EN = 3,5 – 2,1 = 1,4 ✓
        Polar covalent bond ✓                                                               (2)
2.4.3   The O–H bonds are polar (O is more electronegative than H). ✓
        The molecule has an angular (bent) shape ✓
        so it is asymmetrical: the charge distribution is uneven, giving a slightly
        negative O end (δ – ) and slightly positive H end (δ + ) (the bond polarities do
        not cancel). ✓                                                                      (3)
2.5.1   The average distance between the nuclei of two bonded atoms. ✓✓                     (2)
2.5.2   Bond length: 74 pm ✓
        Bond energy: 436 kJ·mol–1 ✓                                                         (2)
2.5.3   SMALLER THAN ✓
        Cℓ atoms are larger (greater atomic radius) than H atoms, so the bond is longer.
        ✓
        The longer the bond, the weaker the bond (the weaker the attraction between
        the nuclei and the shared electrons), so less energy is needed to break it. ✓       (3)
                                                                                           [22]

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### Page 5

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

QUESTION 3
3.1     Weak forces of attraction between molecules (or between ions and molecules).
        ✓✓                                                                                  (2)
3.2     The temperature at which the vapour pressure of a liquid equals the
        atmospheric (external) pressure. ✓✓                                                 (2)
3.3.1   Type of substance / type (strength) of intermolecular forces ✓                      (1)
3.3.2   Boiling point ✓                                                                     (1)
3.4.1   London forces (dispersion forces / induced dipole forces) ✓                         (1)
3.4.2   Hydrogen bonds ✓                                                                    (1)
3.5     A (CH4 ) is non-polar and has only weak London forces between its molecules.
        ✓
        C (H2 O) has strong hydrogen bonds between its molecules (in addition to
        London forces). ✓
        The intermolecular forces in C are much STRONGER than in A. ✓
        MORE ENERGY is needed to overcome the intermolecular forces in C, so C
        has a higher boiling point. ✓
        OR: Less energy is needed to overcome the weaker forces in A, so A has the lower
        boiling point.                                                                      (4)
3.6     A / CH4 ✓
        It has the lowest boiling point / weakest intermolecular forces. ✓                  (2)
3.7     Gas ✓                                                                               (1)
                                                                                           [15]

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### Page 6

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

QUESTION 4
4.1     The pressure of an enclosed gas is inversely proportional to the volume
        it occupies provided the temperature (and the mass of the gas) remains
        constant. ✓✓                                                                              (2)
4.2     Temperature ✓
        Mass of the gas / number of moles of gas ✓                                                (2)
4.3     p1 V1 = p2 V2 ✓
        (80)(30) = (150)X ✓
        X = 16 cm3 ✓
        Accept any other pair of values from the table, e.g. (200)(12) = (150)X. Also accept
        p1 V1 p2 V2
             =       with T1 = T2 .
         T1     T2                                                                                (3)
4.4     A straight line through the origin. ✓                                                     (1)
4.5     pV = nRT ✓
        (100 × 103 )(24 × 10–6 ) ✓ = n(8,31)(25 + 273) ✓
        n = 9,69 × 10–4 mol ✓
        Any correct pair of values from the table may be used, provided units are converted to
        Pa and m3 . Marks: formula; pressure and volume substitution in SI units; temperature
        substitution in K; answer.                                                                (4)
        p1 p2         p V  p V
4.6        =      (or 1 1 = 2 2 ) ✓
        T1 T2          T1   T2
        100    p
             = 2 ✓
        298 348
        p2 = 116,78 kPa ✓
        If temperatures are not converted to kelvin: max. 1/3.                                    (3)
4.7     At high pressure the gas molecules are forced very close together. ✓
        The volume of the molecules themselves is no longer negligible compared with
        the volume of the container. ✓
        The intermolecular forces between the molecules become significant (are no
        longer negligible). ✓                                                                     (3)
                                                                                                 [18]

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### Page 7

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

QUESTION 5
5.1.1   The simplest positive whole-number ratio of atoms in a compound. ✓✓                 (2)
5.1.2   Assume 100 g of compound:

                           C                     H              O
              m      40                6,7               53,3
         n=             = 3,33 mol ✓       = 6,7 mol ✓        = 3,33 mol ✓
              M      12                 1                 16
         Ratio             1                     2              1

        Simplest ratio C : H : O = 1 : 2 : 1 ✓
        Empirical formula: CH2 O ✓                                                          (5)
                                                     180
5.1.3   M(CH2 O) = 12 + 2(1) + 16 = 30 g·mol–1 ;         =6✓
                                                      30
        Molecular formula: C6 H12 O6 ✓
        POSITIVE MARKING FROM 5.1.2.                                                        (2)
5.2     m(H2 O) = 4,99 – 3,19 = 1,80 g ✓
                  m 1,80
        n(H2 O) = =         = 0,10 mol ✓
                  M    18
                     3,19
        n(CuSO4 ) =        = 0,02 mol ✓
                    159,5
           0,10
        x=       =5✓
           0,02                                                                            (4)
                                                                                          [13]

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### Page 8

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

QUESTION 6
6.1     The substance that is totally consumed (used up) when the reaction is
        complete (and therefore limits the amount of product formed). ✓✓                       (2)
                m 6,5
6.2     n(Zn) =    =      ✓ = 0,10 mol ✓
                M 65
        n(HCℓ) = cV = (1,2)(0,15) = 0,18 mol ✓
        0,10 mol Zn would need 2 × 0,10 = 0,20 mol HCℓ, but only 0,18 mol HCℓ is
        available. ✓
        HCℓ is the limiting reagent. ✓
        OR: 0,18 mol HCℓ needs only 0,09 mol Zn, which is less than the 0,10 mol available,
        so Zn is in excess and HCℓ is limiting.                                                (5)
6.3     n(H2 ) = 21 n(HCℓ) = 12 (0,18) = 0,09 mol ✓
        V = nVm = (0,09)(22,4) ✓
        V = 2,02 dm3 ✓
        POSITIVE MARKING FROM 6.2.                                                             (3)
6.4     n(Zn reacted) = 0,09 mol;    n(Zn in excess) = 0,10 – 0,09 = 0,01 mol ✓
        m = nM = (0,01)(65) ✓
        m = 0,65 g ✓
        POSITIVE MARKING FROM 6.2.                                                             (3)
                   V    2,8
6.5     n(CO2 ) =     =     = 0,125 mol ✓
                  Vm 22,4
        n(CaCO3 ) = n(CO2 ) = 0,125 mol ✓
        m(CaCO3 ) = nM = (0,125)(100) = 12,5 g ✓
                   12,5
        % purity =      × 100 ✓
                    15
        % purity = 83,33% ✓                                                                    (5)
                                                                                              [18]

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### Page 9

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

QUESTION 7
7.1     ENDOTHERMIC ✓
        The potential energy of the products is HIGHER than that of the reactants /
        ∆H > 0 / energy is absorbed. ✓                                                  (2)
7.2     The minimum energy needed for a reaction to take place. ✓✓                      (2)
7.3.1   EA = 320 – 60 = 260 kJ ✓                                                        (1)
7.3.2   ∆H = 150 – 60 = +90 kJ ✓                                                        (1)
7.4     An unstable transition state from reactants to products (with the highest
        potential energy). ✓✓                                                           (2)
                m 75
7.5     n(AB) =    =    = 2,5 mol ✓
                M 30
        2 mol AB : 90 kJ
                     2,5
        2,5 mol AB :     × 90 ✓
                      2
        = 112,5 kJ ABSORBED ✓
        POSITIVE MARKING FROM 7.1 and 7.3.2.                                           (3)
                                                                                      [11]

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### Page 10

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

QUESTION 8
8.1     An acid is a proton (H+ ion) donor. ✓✓                                           (2)
8.2.1   NH+4 and NH3 ✓✓
        H2 O and OH– ✓✓
        Mark each pair as a whole: 2 marks per correct pair.                             (4)
8.2.2   HCℓ(g) + H2 O(ℓ) → H3 O+ (aq) + Cℓ– (aq) Reactants ✓ Products ✓                  (2)
8.2.3   Ampholyte / amphiprotic substance ✓                                              (1)
8.3.1   A solution of precisely (exactly) known concentration. ✓✓                        (2)
             m
8.3.2   c=      ✓
            MV
               2,0
        c=             ✓
            (40)(0,25)
        c = 0,20 mol·dm–3 ✓
                 2 = 0,05 mol; c = n = 0,05 = 0,20 mol·dm–3 .
        OR: n = 40                 V   0,25                                              (3)
8.3.3   OPTION 1
        ca Va na
               =    ✓
        cb Vb nb
          ca (20)         1
                   ✓= ✓
        (0,20)(25)        2
        ca = 0,125 mol·dm–3 (0,13 mol·dm–3 ) ✓
        OPTION 2
        n(NaOH) = cV = (0,20)(0,025) = 5 × 10–3 mol ✓
        n(H2 SO4 ) = 12 (5 × 10–3 ) = 2,5 × 10–3 mol ✓
             n 2,5 × 10–3
        c=     =          ✓ = 0,125 mol·dm–3 ✓
             V    0,02
        POSITIVE MARKING FROM 8.3.2.                                                    (4)
                                                                                       [18]

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### Page 11

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

QUESTION 9
9.1     An increase in oxidation number. ✓✓                                                      (2)
9.2.1   +6 ✓                                                                                     (1)
9.2.2   +4 ✓                                                                                     (1)
9.3.1   Fe2+ → Fe3+ + e– ✓✓
        Marking rule: Fe3+ + e– ← Fe2+ (2/2); Fe2+ ⇌ Fe3+ + e– (1/2); Fe3+ + e– → Fe2+ (0/2).    (2)
9.3.2   MnO–4 + 8H+ + 5e– → Mn2+ + 4H2 O ✓✓
        ⇌ used: 1/2. Reverse reaction: 0/2.                                                      (2)
9.3.3   MnO–4 + 8H+ + 5Fe2+ → Mn2+ + 4H2 O + 5Fe3+
        Reactants ✓ Products ✓ Balancing ✓                                                       (3)
9.3.4   MnO–4 ✓                                                                                  (1)
9.4     The oxidation number of Zn INCREASES from 0 to +2. ✓
        Zn therefore loses electrons, i.e. Zn is oxidised. ✓
        Zn donates (gives) its electrons to Cu2+ and so causes Cu2+ to be reduced;
        therefore Zn is the reducing agent. ✓                                                    (3)
                                                                                                [15]
                                                                                    TOTAL:      150

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