Physical Sciences Grade 11 P2 November Mock Exam 2 2026 Memo

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. Copyright reserved Please turn over

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. Copyright reserved Please turn over

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] Copyright reserved Please turn over

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) Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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 Copyright reserved