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

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 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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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] Copyright reserved Please turn over

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 Copyright reserved