Physical Sciences Grade 10 P2 November Mock Exam 2 2026 Memo

Mock hlayiso mock exams Practice Series: Grade 10 Exam Preparation WWW.HLAYISO.COM NATIONAL SENIOR CERTIFICATE GRADE 10 PHYSICAL SCIENCES: CHEMISTRY (P2) NOVEMBER EXAMINATION 2026 (MOCK) MARKING GUIDELINES MARKS: 100 These marking guidelines consist of 9 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 ✓ (the balancing mark requires correct phases where phases are asked for), unless otherwise indicated. 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. If a LETTER is asked for, accept the correct name/formula only if it is unambiguous. Copyright reserved Please turn over

Physical Sciences/P2 3 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 1 1.1 C ✓✓ (2) 1.2 A ✓✓ (2) 1.3 B ✓✓ (2) 1.4 D ✓✓ (2) 1.5 C ✓✓ (2) 1.6 C ✓✓ (2) 1.7 A ✓✓ (2) 1.8 B ✓✓ (2) [16] Notes for teachers: 1.2 – P has the lowest boiling point (hydrogen), so the least energy is needed to overcome the forces between its particles. 1.6 – total moles of atoms: A 1; B 0,5 × 4 = 2; C 0,4 × 8 = 3,2; D 1,5 × 2 = 3. 1.7 – V(NH3 ) = 23 × 30 = 20 cm3 (volume ratio = mole ratio). 1.8 – n = cV = 0,25 × 0,2 = 0,05 mol; m = nM = 0,05 × 40 = 2 g. QUESTION 2 2.1 A pure substance consisting of one type of atom (that cannot be broken down into simpler substances by chemical means). ✓✓ (2) 2.2.1 C✓ (1) 2.2.2 D✓ (1) 2.2.3 B✓ (1) 2.3 Steel (metal) is a GOOD conductor of heat, so heat is transferred quickly to the food. ✓ Plastic/wood is a POOR conductor of heat (thermal insulator), so the handle stays cool enough to hold. ✓ Accept for steel: high melting point / strong / does not rust / malleable (does not melt on the stove). Accept for plastic/wood: poor conductor of heat (thermal insulator) ONLY, since that is the reason the handle is safe to hold. (2) 2.4.1 Cu(OH)2 ✓✓ (Cu2+ and OH– ✓; correct ratio with brackets ✓) (2) 2.4.2 Potassium permanganate ✓ (1) [10] Copyright reserved Please turn over

Physical Sciences/P2 4 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 3 3.1 The temperature at which a liquid changes to a solid (at a given pressure). ✓✓ Accept: the temperature at which the solid and liquid phases of a substance are in equilibrium. (2) 3.2 69 ◦ C ✓ (1) 3.3 Energy is released as the particles move closer together and forces of attraction between the particles form / become stronger (potential energy decreases). ✓ The average KINETIC energy of the particles remains constant. ✓ Temperature is a measure of the average kinetic energy of the particles, so the temperature remains constant. ✓ (3) 3.4 Solid ✓ (1) 3.5 Beaker 2 ✓ At the higher temperature the water particles (and permanganate particles) have a HIGHER average kinetic energy. ✓ The particles move faster, so the permanganate particles spread out (diffuse) through the water faster. ✓ (3) [10] Copyright reserved Please turn over

Physical Sciences/P2 5 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 4 4.1.1 The mass of a particle on a scale where an atom of carbon-12 has a mass of 12. ✓✓ (2) 4.1.2 36 ✓ (65 – 29) (1) 4.1.3 Let the percentage of copper-63 be x; then copper-65 is (100 – x). ✓ 63x + 65(100 – x) Ar = ✓ 100 63x + 65(100 – x) 63,5 = ✓ 100 6 350 = 6 500 – 2x x = 75% ✓ Accept any correct method, e.g. using fractions x and (1 – x): 63x + 65(1 – x) = 63,5, x = 0,75. (4) 4.2 2p 2s 1s Marking criteria 1s and 2s orbitals each with a pair of electrons with opposite spins ✓ THREE 2p orbitals, each with ONE electron (Hund’s rule), arrows in the same direction ✓ (2) 4.3.1 18 ✓ (1) 4.3.2 16 ✓ (1) [11] Copyright reserved Please turn over

Physical Sciences/P2 6 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 5 5.1 The first ionisation energy INCREASES (generally) from sodium to argon / across the period. ✓ (1) 5.2 From Na to Ar the number of protons / nuclear charge INCREASES. ✓ The valence electrons are in the SAME energy level (3), so the atomic radius DECREASES. ✓ The attraction between the nucleus and the valence electrons INCREASES, so MORE energy is needed to remove an electron. ✓ (3) 5.3 Mg: 1s2 2s2 2p6 3s2 ; Aℓ: 1s2 2s2 2p6 3s2 3p1 ✓ The electron removed from aluminium is in the 3p sub-level, which has a HIGHER energy than the 3s sub-level (it is further from the nucleus). ✓ The 3p electron is therefore LESS strongly attracted by the nucleus, so LESS energy is needed to remove it. ✓ Accept for the 2nd mark: the 3p electron is in a higher-energy sub-level / is further from the nucleus. Reference to shielding by the 3s electrons may be credited but is NOT required. Both configurations must be correct for the first mark. Accept [Ne] 3s2 and [Ne] 3s2 3p1 . (3) 5.4 Sodium ✓ (1) [8] Copyright reserved Please turn over

Physical Sciences/P2 7 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 6 6.1 The transfer of electrons (from a metal to a non-metal) to form oppositely charged ions that attract each other (electrostatically). ✓✓ (2) 6.2.1 H× N ×H × H Marking criteria Three N–H bonding pairs ✓ ONE lone pair on N ✓ (2) 6.2.2 2– Na + × O + × Na [Na]+ × O× [Na]+ OR: 2[Na]+ [O]2– with 8 electrons shown around O. Marking criteria Lewis diagrams of TWO Na atoms (1 valence electron each) and O (6 valence electrons) ✓ Transfer of electrons shown: TWO Na+ ions with charges ✓ O2– ion with 8 electrons, brackets and charge ✓ (3) 6.3 (a) Metallic ✓ (b) Covalent ✓ (c) Ionic ✓ (3) 6.4 Mg3 N2 ✓ Ionic ✓ (2) [12] Copyright reserved Please turn over

Physical Sciences/P2 8 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 7 7.1.1 Decomposition ✓ (1) 7.1.2 Chemical ✓ NEW substances (CuO and CO2 ) with different properties are formed / bonds are broken and new bonds are formed. ✓ Accept: a new black substance and a gas are formed. A colour change ALONE is not accepted. (2) 7.2.1 C3 H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2 O(g) Marking criteria Reactants ✓ Products ✓ Balancing and phases ✓ Accept H2 O(ℓ). Phases are required for the third mark. (3) 7.2.2 Mass of reactants = 1(44) + 5(32) = 44 + 160 = 204 g ✓ Mass of products = 3(44) + 4(18) = 132 + 72 = 204 g ✓ Mass of reactants = mass of products, therefore mass is conserved. ✓ POSITIVE MARKING FROM 7.2.1 if the equation used is balanced. (3) 7.3 C H O Mass in 100 g 62,1 g 10,3 g 27,6 g m 62,1 10,3 27,6 n= = 5,18 = 10,3 = 1,73 M 12 1 16 Ratio 3 6 1 Dividing by the moles in 100 g (any one correct) ✓ all three n correct ✓ simplest ratio 3 : 6 : 1 ✓ Empirical formula: C3 H6 O (M = 58 g·mol–1 ) 116 =2✓ 58 Molecular formula: C6 H12 O2 ✓ Accept the ratio method directly from masses. POSITIVE MARKING from an incorrect empirical formula. (5) [14] Copyright reserved Please turn over

Physical Sciences/P2 9 Hlayiso/Mock 2026 NSC – Marking Guidelines QUESTION 8 8.1.1 M(NaHCO3 ) = 23 + 1 + 12 + 3(16) = 84 g·mol–1 m 8,4 n= ✓ = ✓ M 84 n = 0,1 mol ✓ (3) 8.1.2 n(Na2 CO3 ) = 12 n(NaHCO3 ) = 0,05 mol (mol ratio 2 : 1) ✓ m = nM = 0,05 × 106 ✓ m = 5,3 g ✓ POSITIVE MARKING FROM 8.1.1. (3) 8.1.3 n(CO2 ) = 21 n(NaHCO3 ) = 0,05 mol ✓ V V n= 0,05 = ✓ Vm 22,4 V = 1,12 dm3 ✓ POSITIVE MARKING FROM 8.1.1. (3) 8.1.4 n(H2 O) = 21 n(NaHCO3 ) = 12 (0,1) = 0,05 mol ✓ m = nM = 0,05 × 18 ✓ m = 0,9 g ✓ POSITIVE MARKING FROM 8.1.1. (3) n n 8.2.1 c= 0,1 = ✓ n = 0,025 mol ✓ V 0,25 m 7,15 M= = = 286 g·mol–1 ✓ n 0,025 106 + 18x = 286 ✓ x = 10 ✓ m 7,15 OR: c = : 0,1 = ✓✓; M = 286 g·mol–1 ✓; 106 + 18x = 286 ✓; x = 10 ✓ MV M(0,25) (5) 10(18) 8.2.2 % H2 O = × 100 ✓ 286 = 62,94% ✓ POSITIVE MARKING FROM 8.2.1. (2) [19] TOTAL: 100 Copyright reserved