# Physical Sciences Grade 12 Paper 2 Gauteng Mock 2014 Memo > Grade 12 Physical Sciences Paper 2 memorandum from the Gauteng Mock Exam 2014, 7 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/2014afr-gau-physical-science-p2-eng-memo/d9aeec80-7e75-4c10-8f2b-da3caef13236 - Updated: 2026-09-27 - Subject: Physical Sciences - Grade: Grade 12 - Document type: Memorandum - Year: 2014 - Exam period: Gauteng Mock Exam - Paper: 2 - Curriculum: CAPS - Pages: 7 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Physical Sciences](https://www.hlayiso.com/library/grade-12/physical-sciences) › [Physical Sciences Grade 12 Paper 2 Gauteng Mock 2014 Memo](https://www.hlayiso.com/library/2014afr-gau-physical-science-p2-eng-memo/d9aeec80-7e75-4c10-8f2b-da3caef13236) ## About this document Physical Sciences · Grade 12 · Gauteng Mock Exam · 2014. Memorandum, 7 pages. 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Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com PREPARATORY EXAMINATION 2014 MEMORANDUM SUBJECT: PHYSICAL SCIENCE P2 (10842) ### Page 2 Downloaded from hlayiso.com 10842/14 GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION PHYSICAL SCIENCE (Second Paper) MEMORANDUM QUESTION 1/VRAAG 1 1.1 B (2) 1.2 A (2) 1.3 C (2) 1.4 D (2) 1.5 D (2) 1.6 C (2) 1.7 C (2) 1.8 B (2) 1.9 C (2) 1.10 A (2) [20] QUESTION 2/VRAAG 2 2.1.1 Ketone (1) 2.1.2 D (1) 2.1.3 Chloroethane (1) 2.1.4 Ethanoic acid (1) 2 ### Page 3 Downloaded from hlayiso.com 10842/14 2.1.5 + H  + (4) 2.1.6 (2) 2.1.7 C3H8 + 5O2(reactants) 3CO2 + 4H2O (products)balancing (3) 2.1.8 Combustion/oxidation (1) 2.2.1 Chain isomers are compounds with the same molecular formula but the carbon atoms in the chain change position to form different arrangements/structural formulae. (2) 2.2.2 (4) 2.2.3 Butane (1) 2.2.4 Butane is less branched/ has a longer chain/ has a larger surface area.  There are stronger intermolecular forces/ Van der Waals forces between the molecules.  More energy needed to overcome intermolecular forces. (3) OR 2-methyl propane is more branched/ more spherical/ has a shorter chain/ has a smaller surface area. Weaker intermolecular forces/ Van der Waals forces. Less energy needed to overcome intermolecular forces. [24] 3 ### Page 4 Downloaded from hlayiso.com 10842/14 QUESTION 3/VRAAG 3 3.1.1 Addition/ Halogenation/ Bromination (1) 3.1.2 (2) 3.1.3 Yes (1) 3.2.1 Addition/ Hydration (1) 3.2.2. Propan – 2 – ol (2) 3.2.3 Alcohol/Alkanol (1) 3.3.1 Cracking is the breaking up of large organic molecules into smaller useful molecules. (2) 3.3.2 CH3 H3C C CH2 CH3 CH CH3 H3C (2)  3.4.1 H H │ │  -C-C- │ │ H H (2) 3.4.2 B (1) 3.4.3 Plastic bags, squeeze bottles, cling wrap, bullet-proof vests. (1) [16] QUESTION 4/VRAAG 4 4.1 temperature (1) 4.2 Rate of reaction (1) 4.3 Rate = gradient of graph= 60-0/30-0= 2 cm3·s-1 (3) – 4.4 Less than  (1) 4.5 The rate of change (or gradient) is less (1) 4.6 decrease  (1) 4.7 As the surface area increases, there are more particles available for reacting. Thus there will be more collisions per second, and thus a greater number of effective collisions per second and thus the reaction rate will increase. (3) 4 ### Page 5 Downloaded from hlayiso.com 10842/14 4.8 n(Mg) = m/M = 20/24 Alt. Method = 0,833 moles 24 g Mg reacts with 1 mol H2SO4  xg → 0,1 mol x = 2,4 g n (H 2SO4) =c x V =1 x0,1 mass left = 20-2,4 =0,1 moles = 17,6 g 0,1 moles of Mg will react with 0,1 moles of sulphuric acid nMg left = 0,83-0,1 moles= 0,733 moles thus mass of Mg = n x M = 0,733 x 24 = 17,6 g Option 1 Option 2  mass → mole  mole → mole  relationship  relationship  xg-0,1 mole  subtraction to obtain 2,4 g  subtraction  substitution  answer  answer (5) 4.9 A substance that increase the rate of a chemical reaction without undergoing (2) permanent change.  [18] QUESTION 5/VRAAG 5 5.1 X2 3Y2 ⇌ 2XY3 No of moles 1,65 2 0 Change 0,4 1,2 0,8 Ratio Moles at equm 1,25 0,8 0,8 (0,4 x2) X 2 Conc at equm (mol∙dm-3) 0,625 0,4 0,4 Kc = [XY3]2 [X2] [Y2 ]3  4 = [0,4]2  [X2][0,4]3 [X2] = [0,4]2 4[0,4]3 =0,625 mol∙dm-3  No. of mol X2 = 1,65 mol No. of mol Y2 = 2 mol (9) MARK ALLOCATION STARTING WITH Kc calculation and then working back to answer in table.  Calculation of Kc (5 marks)  Moles at equilibrium (1 marks)  Change (1 marks)  Initial Moles (2 marks) 5 ### Page 6 Downloaded from hlayiso.com 10842/14 5.2 endothermic (1) 5.3 As temp increases the Kc increases, [product] increases, indicating the forward reaction is favoured.  A temperature increase favours the endothermic reaction, thus the forward reaction is the endothermic reaction. (2) 5.4. 1 increase (1) 5.4.2 decrease (1) 5.5 X2 (1) 5.6 At a lower temperature the Kc is smaller indicating fewer products and more reactants.  (2) [17] QUESTION 6/VRAAG 6 6.1 It is a weak acid (1) 6.2 An acid that forms a large no of moles per volume of water (2) 6.3 H2PO4- + H2O ⇌ H3PO4 + OH-  (2) 6.4 H2O/OH- or H2PO4- /H3PO4 (1) 6.5.1 m = cMV = 58  x 0,2 x 0,5=5,8g (5) 6.5.2 0,1x2=0,2 mol·dm -3  (2) 6.5.3 Not safe positive marking 3 [OH-]=0.2 mol.dm  pOH = -log[OH-]= -log(0,2) =0,7 pH + pOH = 14 (5) pH=14-0,7=13,3 OR [H+][OH-] = 10-14 [H+](0,2) = 10-14 [H+] = 5 x 10-14 mol∙dm-3 pH = - log[H+] = - log (5 x 10-14) = 13,3 6.6 na = caVa nb cbVb 2/1 = (0,1)V  (0,2)25 V = 100 cm3 (4) [22] 6 ### Page 7 Downloaded from hlayiso.com 10842/14 QUESTION 7/VRAAG 7 7.1 The cell where a chemical compound is broken down with the aid of an electric (2) current.  7.2 Al  and O2-  OR aluminium ion and oxide ion 3+ (2) 7.3 2 O2- → O2 + 4e- (2) 7.4 Al3+ + 3e- → Al (2) 7.5 Eθcell/sel = Eθcathode/katode - Eθanode  = -1,66 – (- 0,56)  = - 1,1 V  (4) 2- 7.6 O ions in the melt are oxidised to produce O2 which then reacts with the graphite ( (1) carbon) rods to produce CO2 and so the rods get corroded away and have to be replaced regularly.  [13] QUESTION 8/VRAAG 8 8.1 The process in which electrons are lost by a substance. (2) 8.2 8.2.1 Aℓ (1) 8.2.2 Zn (1) 8.3 As reaction proceeds, the Ag+ ion concentration decreases while the Zn2+ ion concentration increases.According to Le Chatelier’s principle, the rate of reverse reaction will increase to reduce the increase in product ion concentration, thus favouring the reverse reaction and lowering the forward reaction leading to the decrease in the overall cell potential. (3) 8.4 Aℓ3+ + 3e- → Aℓ (2) 8.5 Chemical to electrical energy (1) [10] QUESTION 9/VRAAG 9 9.1 Vanadium pentoxide (1) 9.2 N2 (g) + H2 (g)  (reactants) ⇌ 2NH3 (g)  (products) balancing (3) 9.3 22% of 5 kg = 1,1 kg 7/11 x 1,1 = 0,7 kg = 700 g n=m/M = 700/14= 50 moles  (5) 9.4 Excess plants block sunlight from aquatic life and organisms die./Lack of oxygen. 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