Physical Sciences Grade 12 KwaZulu Natal Mock 2016 Memo

Downloaded from hlayiso.com Physical Sciences / P2 1 Preparatory Examination September 2016 NSC-Memorandum Education ducation KwaZulu-Natal Department of Education REPUBLIC OF SOUTHAFRICA PHYSICAL SCIENCES P2 (CHEMISTRY) PREPARATORY NATIONAL GRAD MARKS : 150 TIME : 3hours This memorandum consists of 10 pages. The marking guidelines as per 2014 Examination Guidelines, pages 34-37 must be applied when marking this Paper. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 2 Preparatory Examination September 2016 NSC-Memorandum QUESTION 1 1.1 C  (2) 1.2 B (2) 1.3 D (2) 1.4 B (2) 1.5 A (2) 1.6 C (2) 1.7 B (2) 1.8 A (2) 1.9 C (2) 1.10 D (2) [20] QUESTION 2 2.1. H  (2) 2.2 A (1) 2.3 B  (2) 2.2.1 chloroethene accept 1-chloroethene  (2) 2.2.2 2-methylpentan-2-ol accept 2–methyl– 2-pentanol (2) 2.3 H H H H O H │ │ │ │ ║ │ H― C ― C ― C ― C ― C ― O │ │ │ H H H H―C―H  │ H Functional group correct  rest of structure correct  (2) [10] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 3 Preparatory Examination September 2016 NSC-Memorandum QUESTION 3 3.1.1 Boiling does not involve breaking the chemical (intra molecular) bonds between atoms. OR Boiling involves breaking the intermolecular forces between molecules. (1) 3.1.2 Pentane has a longer chain of carbons than methane, therefore a greater surface area. Hence more sites for London force attraction and stronger intermolecular forces. More energy is needed to break the stronger intermolecular forces. (3) 3.1.3 OPTION 1: OPTION 2: 2-methylbutane  (or methylbutane) 2,2-dimethylpropane  (or dimethylpropane) H H H H │ │ │ │ H H─C─C─C─C─H  │ │ │ │ H─C─H H H H H H H ─C─H │ │ │ H─C─C─C─H  H │ │ H H H─C─H │ H (3) 3.2 The pressure exerted by a vapour at equilibrium with its liquid in a closed system.  (2) 3.3.1 D  (2) 3.3.2 A  (2) 3.4 The carboxylic acid has the strongest intermolecular forces,  because there are two sites for hydrogen bonding in each carboxylic acid molecule. The stronger the intermolecular forces the lower the vapour pressure (at a given temperature).  The graph indicates that D has the lowest vapour pressure. (3) [16] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 4 Preparatory Examination September 2016 NSC-Memorandum QUESTION 4 4.1.1 methyl ethanoate  (2) 4.1.2 M (C3H6O2) = 74 n = m  M = 68,88   74 = 0,93 mol M (CH4O) = 32 1 : 1 (alcohol : acid) m = 0,93 x 32  = 29,78 g % purity : 29,78 x 100 = 59,57 %  50 (5) 4.2.1 A - hydration (or addition)  C - substitution  D - dehydration (or elimination)  E - combustion  (4) 4.2.2 (i) Dissolve C2H5Cl in ethanol.  (ii) Add concentrated NaOH solution.  (iii) Heat the mixture.  (3) 4.2.3 H2SO4 OR sulphuric acid  (1) 4.2.4 Polythene OR Polyethene  (1) 4.2.5 The chemical process in which longer chain hydrocarbon molecules are broken down to shorter molecules at high temperatures (and pressures). (2) 4.2.6 C4H10  (2) [20] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 5 Preparatory Examination September 2016 NSC-Memorandum QUESTION 5 5.1 The amount of product formed/reactant used up per unit time (per second). (2) 5.2 concentration of HCl (1) 5.3 Product CO2 is an insoluble gas and escapes from the beaker.  (2) 5.4 Calcium carbonate is used up and the reaction stops, no more CO2 formed.  (2) 5.5.1 graph B (1) 5.5.2 graph D  (1) 5.5.3 graph C (1) 5.6 graph C. Catalyst speeds up the rate of reaction, therefore gradient of graph will be steepest for same initial mass.  (4) 5.7 Remains the same.  (1) [15] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 6 Preparatory Examination September 2016 NSC-Memorandum QUESTION 6 6.1.1 Increases  (1) 6.1.2 When the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium by favouring the reaction that will oppose the disturbance.   (2) 6.1.3 Increase in temperature increases Kc  Increase in Kc indicates that the forward reaction has been favoured , Increase in temperature favours the endothermic reaction , Therefore the forward reaction is endothermic . (4) 6.1.4 Add a catalyst  Increase pressure OR decrease volume of container  (2) 6.2 A2 + 2B → 2AB Initial (mol) x 2 0  Used / formed -0,2 -0,4 0,4   RATIO Equilibrium (mol) x – 0,2 1,6 0,4  [equilibrium] x – 0,2 1,6 0,4  Addition for all three For conc. only  A2 + 2B → 2AB Initial conc.(mol.dm-3) x 2 0  RATIO Used / formed -0,2 -0,4 0,4   [equilibrium] x – 0,2 1,6 0,4  Addition for all three Kc = [AB]2  Marking criteria: [A2][B]2 • Initial mol/conc correctly indicated.  • Mol/conc of AB produced = 0,4. = (0,4)2  • Ratio applied correctly.  (x – 0,2) (1,6)2 • Equilibrium mol:  initial – used Initial + produced = 0,5 OR Equilibrium conc:  initial – used ∴ 0,5 [((x – 0,2) (1,6)2] = (0,4)2 Initial + produced • Equilibrium mol ÷ 1 = eq conc.  ∴ 1,28x - 0,256 = 0,16 • Correct Kc expression.  • Substitution of eq conc to Kc expression.  ∴ X = 0,325 mol  • Final answer.  Wrong Kc expression – (Max.: 4/8) Wrong values on the table but used in the calculation (positive marking) Copyright Reserved - (Max.: 3/8) no mark for answer. Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 7 Preparatory Examination September 2016 NSC-Memorandum (8) [17] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 8 Preparatory Examination September 2016 NSC-Memorandum QUESTION 7 7.1.1 Option 1: pH = - log [H3O+]  =1 ∴ [H3O+] = 10-1 = 0,1  ∴ [HCl] = 0,1 mol.dm-3  Option 2: pH = - log [H3O+]  =1 ∴ [HCl] = 0,1 mol.dm-3  (3) 7.1.2 increase  (1) 7.1.3 Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l) (3) Reactants  Products  Balancing  7.2 When an acid reacts with a base to produce a salt and water.  OR Chemically equivalent quantities of acid and base are reacted.  (2) 7.3 Option 1: 2NaOH + H2SO4 → Na2SO4 + 2H2O  n(NaOH) = c x V  = 0,1 x 0,012 = 0,0012 mol n(H2SO4) = ½ x n(NaOH)  = 0,0006 mol V(H2SO4) = n/C = 0,0006 / 0,05 = 0,012 dm3 V(H2SO4) that reacts with Ba(OH)2 = 54 – 12  = 42 cm3 (OR 0,054 - 0,012 = 0,042 dm3) Ba(OH)2 + H2SO4 → BaSO4 + 2H2O  n(H2SO4) that reacts with Ba(OH)2 = c x V = 0,05 x (42 x 10-3 mol) = 0,0021 mol n(Ba(OH)2) = 1 x 0,0021 = 0,0021 mol [Ba(OH)2] = n/v = 0,0021 x 0,048  = 0,04375 mol∙dm-3  Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 9 Preparatory Examination September 2016 NSC-Memorandum 7.3 Option 2: Option 3: 2NaOH + H2SO4 → Na2SO4 + 2H2O 2NaOH + H2SO4 → Na2SO4 + 2H2O n(NaOH) = c x V  = 0,1 x 0,012 na = Va x ca  nb Vb x cb = 0,0012 mol n(H2SO4) = ½ x n(NaOH)  1  = Va x 0,05  2 0,1 x 12 = 0,0006 mol n(H2SO4) tot = c x V Va = 12 cm3 = 0,05 x (54 x10-3) V(H2SO4) with Ba(OH)2 = 54 – 12  = 0,0027 mol = 42 cm3 Ba(OH)2 + H2SO4 → BaSO4 + 2H2O  n(H2SO4) with Ba(OH)2 na = Va x ca = 0,0027 - 0,0006  = 0,0021 mol nb Vb x cb Ba(OH)2 + H2SO4 → BaSO4 + 2H2O 1 = 0,05 x 42  n(Ba(OH)2) = 1 x 0,0021 1 48 x cb = 0,0021 mol [Ba(OH)2] = 0,04375 mol∙dm-3  [Ba(OH)2] = n / v = 0,0021 x 0,048  = 0,04375 mol∙dm-3  (8) [17] QUESTION 8 8.1.1 Pt or the electrode where oxygen reacts or where peroxide is formed (1) 8.1.2 Cu → Cu2+ + 2e-  Cu ⇌ Cu2+ + 2e- (1) (2) Cu ← Cu2+ + 2e- (0) 8.1.3 O2 + 2H+ + 2e-  → H2O2  (2) 8.1.4 EΘcell = EΘcathode - EΘanode  = 0,68  - 0,34  = 0,34 V  (4) 8.2 2Ce + 3Pd2+  → 2Ce3+ + 3Pd  (balance ) (3) [12] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 10 Preparatory Examination September 2016 NSC-Memorandum QUESTION 9 9.1 Electrical energy to chemical energy.  (1) 9.2 Decrease in mass/ decrease in size/ gets used up/ errosion (2) 9.3 Oxidation  (1) 9.4 Note: + - 9.4.1 Ag + e → Ag  Marking rule 3.3 (2) 9.4.2 Silver nitrate OR AgNO3 OR Silver ethanoate/silver acetate  CH3COOAg/ AgC2H3O2/ AgCH3CO2 (1) OR Ag+ ions 9.5 Rate of oxidation is equal to the rate of reduction.  ACCEPT: Reduction and oxidation take place simultaneously (2) [9] QUESTION 10 10.1.1 2SO2 + O2 ⇌ 2SO3  (ACCEPT: 2SO2 + O2 → 2SO3 ) (2) 10.1.2 Step 2  (1) 10.1.3 Vanadium pentoxide  (1) 10.2.1 Fertilisers must replenish nutrients used by growing of crops.  (1) 10.2.2 Any ONE: Excessive fertiliser seeps/washes into groundwater and contaminates Drinking Water OR Excessive fertiliser run-off can lead to eutrophication which depletes aquatic life which serves as a food source OR Excessive fertilisation can damage a crop which leads to a smaller harvest.  (1) 10.2.3 Eutrophication  (1) 10.2.4 (NH4)2SO4  (2) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 11 Preparatory Examination September 2016 NSC-Memorandum 10.3.1 Total percentage of active fertiliser. OR Combined percentage of N, P and K. (1) 10.3.2 OPTION 1 13 = 5  x 26  x x = 10  thus N = 3 N:P:K = 3: 2 : 5 (4) OPTION 2 (N + P + K) in the bag = 26/100 x 20 = 5,2 kg K in the bag = 13/100 x 5,2 = 2,6 kg 2,6/5,2 x 100  = 50% = 5 parts  P = 2 parts = 20 %  Therefore, N = 30% = 3 parts Unknown component = 3.  (4) [14] Copyright Reserved Please Turn Over