Physical Sciences Grade 12 P2 KwaZulu Natal Mock 2020 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES: CHEMISTRY (P2) PREPARATORY EXAMINATION MARKING GUIDELINES SEPTEMBER 2020 Time: 3 hours Marks: 150 NB. This marking guideline consists of 10 pages.

Downloaded from hlayiso.com 2 Physical Science P2 NSC Preparatory Examination 2020 QUESTION 1 1.1 D ✓✓ (2) 1.2 C ✓✓ (2) 1.3 D ✓✓ (2) 1.4 C✓✓ (2) 1.5 A✓✓ (2) 1.6 C✓✓ (2) 1.7 A✓✓ (2) 1.8 B✓✓ (2) 1.9 D✓✓ (2) 1.10 D✓✓ (2) [20] QUESTION 2 ✓ ✓ 2.1 3,3-diethylhexane (2) 2.2 A and D ✓ (1) 2.3 B✓ (1) 2.4 H CI I I  H C=C H whole structure ✓ chloroethene ✓ (3) 2.5 formyl group ✓ (1) 2.6 C ✓ (1) 2.7 propan-2-ol✓/2-propanol (1) [10] Copyright Reserved Please turn over

Downloaded from hlayiso.com 3 Physical Science P2 NSC Preparatory Examination 2020 QUESTION 3 3.1 Temperature at which the vapour pressure of a substance is equal to the (2) atmospheric pressure. ✓ ✓ 3.2 How does the type of functional group affect the boiling point of an organic compound? (2) OR What is the relationship between the functional groups of different organic compounds and their boiling points? Dependent and independent variable correctly identified i.e. boiling ✓ point and functional group Relationship between dependent and independent variables given ✓ in the form of a question that cannot be answered by YES or NO. 3.3 Molecular mass✓ (1) 3.4 Compound A has London/induced dipole forces ✓ whereas compound B has London forces and hydrogen bonds. ✓ Hydrogen bonds are stronger that London forces. ✓ (4) Therefore more energy is required to overcome the intermolecular forces in compound B. ✓ 3.5 Higher than. ✓ - The lower the boiling point, the higher the vapour pressure. ✓ (2) 3.6 Lower than. ✓ - Compound C has only one site for hydrogen bonding while compound B has (3) two sites for hydrogen bonding. ✓therefore more energy required to separate molecules of B✓ [14] Copyright Reserved Please turn over

Downloaded from hlayiso.com 4 Physical Science P2 NSC Preparatory Examination 2020 QUESTION 4 ✓ ✓ 4.1 2-methylbut-1-ene (2) 4.2.1 hydrohalogenation ✓ (1) 4.2.2 ✓ H CI H H I I I I ✓ whole structure H–C–C–C–C– H (4) I I I ✓ ✓ H H H 2-chloro-2-methylbutane 4.3.1 Pt / Pd /Ni /platinum/palladium/nickel ✓ H–C–H (1) ✓ ✓I 4.3.2 2-methylbutane (2) H 4.4.1 Dilute H2SO4  / dilute H3PO4 ✓ Mild heat ✓ (2) excess water H2SO4 OR H3PO4  4.4.2 Hydration ✓ (1) 4.4.3 Tertiary. ✓ The -OH/hydroxyl group is joined to the carbon that is joined to 3 other (2) carbon atoms. ✓ 4.4.4 H ✓ H H I I I H–C–C=C–C– H I I ✓ whole structure H H H–C–H ✓ I 2-methylbut-2-ene✓ (4) H [19] Copyright Reserved Please turn over

Downloaded from hlayiso.com 5 Physical Science P2 NSC Preparatory Examination 2020 QUESTION 5 5.1.1 Heptanoic acid ✓✓ / hexanoic acid (2) 5.1.2 ✓ ✓ H H H O H H H H H H I I I II I I I I I I ✓ H–C– C–C–O– C– C–C–C–C–C–C–H I I I I I I I I I (2) H H H H H H H H H (3) 5.2.1 The breaking of organic molecules into smaller more useable units. ✓✓ (2) 5.2.2 thermal✓ (1) 5.2.3 pentene / pent-1-ene / pent-2-ene ✓✓ (2) ✓ ✓ ✓bal 5.2.4 2C8H18 + 25O2 → 16CO2 + 18H2O (3) [13] Copyright Reserved Please turn over

Downloaded from hlayiso.com 6 Physical Science P2 NSC Preparatory Examination 2020 QUESTION 6 6.1 Change in concentration / mass / moles/ amount/ volume of reactants (or products) per unit time. ✓✓ (2) 6.2 3,0 dm3 ✓ (1) 6.3 Rate of Reaction = ΔV Δt = 3,0 - 0 ✓ 20 - 0 ✓ = 0,15 dm3⋅s−1 ✓ (3) 6.4 Increases ✓ - A higher concentration means that there is a greater number of particles per unit volume. ✓ This leads to an increase in the number of collisions per unit time✓. This leads to an increase in the number of effective collisions per unit time. ✓ (4) 6.5 21 12 Marking guideline 3 • Exp1 above Exp2✓ but end at same point for volume ✓ • Exp3-shows steeper gradient✓ but end volume is lower than Exp 1 and 2 ✓ (4) [14] Copyright Reserved Please turn over

Downloaded from hlayiso.com 7 Physical Science P2 NSC Preparatory Examination 2020 QUESTION 7 7.1 x axis- Kinetic Energy (of molecules)✓ y axis- % of molecules/no of particles✓ (2) 7.2 T2✓ (1) 7.3 It indicates the percentage of molecules that have more energy than the activation energy at a specific temperature. / It is the percentage of molecules that are capable of effective collisions. ✓ (1) 7.4 The area under the curves remains the same. ✓ The number of molecules in the reaction mixture stays the same/does not change. ✓ (2) 7.5 There are no molecules/particles with ZERO energy. ✓ (1) [7] QUESTION 8 8.1 When the equilibrium in a closed system is disturbed, the system will reinstate a new equilibrium by favouring the reaction that will oppose the disturbance.✓✓ (2) 8.2 Brown✓ (1) 8.3 Endothermic. ✓ An increase in temperature will favour the endothermic reaction in an equilibrium reaction. ✓ Since an increase in temperature resulted in an increase in the value of Kc, ✓ It can be concluded that the forward reaction is favoured✓ (4) 8.4 Add a catalyst✓ Increase the pressure. ✓ Increase concentration of reactant (any two) (2) Copyright Reserved Please turn over

Downloaded from hlayiso.com 8 Physical Science P2 NSC Preparatory Examination 2020 8.5 N2O4 NO2 Ratio 1 2 Initial mass 84,64g 0 84,64 Initial mole 92 ✓ 0 = 0,92 mol Change in mole 0,19✓ 0,38 Moles at equilibrium 0,73mol 0,38mol Equilibrium 0,73 0,38 concentration 2 2 (mol.dm-3) = 0,365 = 0,19✓ Kc = [NO2]2 ✓ [N2O4] = (0,19)2 ✓ 0,365 = 0,1 ✓ (9) Therefore the temperature is 300K✓ 8.6 Remains the same. ✓ Only change in temperature affects Kc. ✓✓ (3) [21] QUESTION 9 9.1 A standard solution is one whose concentration is precisely known. ✓✓ (2) 9.2 (C x V) dilute = (C x V) conc Cdilute = (C x V) conc Vdilute = 0,63 x 0,05 ✓ 1✓ = 0,0315 mol.dm-3✓ (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com 9 Physical Science P2 NSC Preparatory Examination 2020 9.3 Positive marking from Q9.2 nNaOH = C x V = 0,0315 x 0,04 ✓ = 1,26 x 10-3mol ✓ nNaOH: n C2H2O4 2 : 1 nC2H2O4 = 6,3 x 10-4mol ✓ mC2H2O4 = n x M ✓ = 6,3 x 10-4 x 90 ✓ = 0,0567g % purity = 0,0567 ✓ x100 0,25 1 = 22,68 % ✓ (7) 9.4 option 1 option 2 Kw = [H3O+][OH-] = 1x10-14 ✓ pOH = -log[OH-] ✓ 1x10-14 = [H3O+] (0,20) = -log (0,2) [H3O+] = 5 x10-14 ✓ = 0,70 ✓ pH = -log [H3O+] ✓ pH + pOH = 14 ✓ = -log (5 x10-14) pH + 0,70 = 14 = 13,3 ✓ pH = 13,3 ✓ (4) 9.5 Is the point in a titration where the indicator changes colour. ✓✓ (2) 9.6 CaVa = na CbVb nb • Sub on LHS✓✓ • Sub no moles on RHS✓ Ca (15) ✓ = 1✓(ratio) • Answer with unit✓ (0,2)(30,25) ✓ 1 Ca = 0,40 mol.dm-3 ✓ (4) Copyright Reserved Please turn over

Downloaded from hlayiso.com 10 Physical Science P2 NSC Preparatory Examination 2020 9.7 Option 1 Positive marking from Q9.6 M (CH3COOH) = 2(12) + 4(1) + 2(16) = 60g.mol-1 ✓ • Molar mass of CH3COOH. ✓ n (CH3COOH) diluted = C x V • Substitution 0,24 x 0,25 ✓ = 0,40 x 0,25 ✓ • Mass of 25g in vinegar✓ = 0,1 mol • Substitution of percentage✓ • Final answer✓ m (CH3COOH) dilute = n x M = 0,1 x 60 = 6,0g % CH3COOH in vinegar = 6,0 x 100 ✓ 25  1 = 24,0% ✓ Option 2 Positive marking from Q9.6 • Molar mass of CH3COOH. ✓ M (CH3COOH) = 2(12) + 4(1) + 2(16) • Substitution 0,4 x 60 ✓ = 60g.mol-1 ✓ • Mass of 25g in vinegar✓ • Substitution of percentage✓ m (CH3COOH) diluted = C x V • Final answer✓ = 0,40 x 60 x 0,25 ✓ • = 6.0 g % CH3COOH in vinegar = 6,0 x 100 ✓ 25  1 = 24,0% ✓ (5) 9.8.1 reaction of a salt with water ✓✓ (2) ✓(LHS) ✓ (RHS) 9.8.2 CH3COO- + H2O  CH3COOH + OH- (3) pH will increase✓ [32] TOTAL MARKS: 150 Copyright Reserved Please turn over