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Physical Sciences Grade 12 Paper 2 EC June 2014 Memo

MemorandumPhysical SciencesGrade 1220147 pages

Grade 12 Physical Sciences Paper 2 memorandum from the EC June 2014, 7 pages. Read online or get the free PDF download, question paper linked when available.

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EC June 2014

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Downloaded from hlayiso.com Physical Science/P2 1 June 2014 Common Test NSC – Memorandum PHYSICAL SCIENCES P2 MEMORANDUM COMMON TEST JUNE 2014 NATIONAL SENIOR CERTIFICATE GRADE 12 N.B. This memorandum consists of 7 pages including this page. Copyright Reserved Please Turn Over
Downloaded from hlayiso.com Physical Science/P2 2 June 2014 Common Test NSC – Memorandum SECTION A QUESTION 1 1.1 A  (2) 1.2 C  (2) 1.3 B  (2) 1.4 D  (2) 1.5 D  (2) 1.6 D  (2) 1.7 B  (2) [14] SECTION B QUESTION 2 2.1 2.1.1 Carboxylic acid  (1) 2.1.2 Butanal  (1) 2.1.3 H H l l C = C  l l H H (2) 2.2 Tertiary  (1) 2.3 They have the same molecular formula but different functional groups.  (2) [7] QUESTION 3 Copyright Reserved Please Turn Over
Downloaded from hlayiso.com Physical Science/P2 3 June 2014 Common Test NSC – Memorandum 3.1 Hydrocarbons are organic compounds that consist of hydrogen and carbon only.  (1) 3.2 3.2.1 Boiling point increases as the surface area increases  OR Boiling point decreases as the chain become more branched (1) 3.2.2 The more branches the smaller the surface area,  the weaker the intermolecular (Van der Waals) forces.  Lesser energy needed to separate the chains.  (the lower the boiling point). (3) 3.3 A.  (1) [6] QUESTION 4 4.1.1 CH3 CH2 CH2 CH2 -O-H  (2) heat  4.1.2 CH3 CH2 CH2 CH2 -O-H CH3CH2CH=CH2 + H2O. H2SO4 (conc)  H2O H2SO4 (conc) Excess heat  CH3CH2CHCH3 O -H (4) 4.2.1 Alcohol  (1)  heat 4.2.2 CH3CH=CH2 + H2O H2SO4 (conc)  CH3CHCH3  (3) O -H   heat  4.2.3 CH3CH2CH2Cℓ + NaOH (aq) CH3CH2CH2O-H + HCℓ (3) [13] Copyright Reserved Please Turn Over
Downloaded from hlayiso.com Physical Science/P2 4 June 2014 Common Test NSC – Memorandum QUESTION 5 5.1 Exothermic  ΔH = energy of the products – energy of the reactants ΔH = –197 – 0 ΔH = –197 kJmol–1  (2) 5.2 1580 kJmol–1  (1) 5.3 1777 kJmol–1  (1) 5.4 1537 kJmol–1  (1) [5] QUESTION 6 6.1 Zn(s) + 2HCℓ(aq)  → ZnCℓ2 (aq) + H2(g)  bal  (3) 6.2 surface area  / state of division (1) 6.3 To make the investigation fair  / To have 1 independent variable (1) 6.4 Volume of H2(g) is dependant on the amount of zinc used.  (1) ΔV 6.5 Average reaction rate = Δt 48  30 =  40  20 = 0,9 cm3.s-1  (3) 6.6 Greater than  (1) 6.7 Zn is powder, the more finely divided (greater surface area) the greater the contact area.  This will result in more effective collision occur simultaneously per unit time.  (2) 6.8 6.8.1 Remains the same.  (1) 6.8.2 Steeper.  (1) 6.9 Copper is a catalyst. Increase the rate of reaction.  (2) [16] Copyright Reserved Please Turn Over
Downloaded from hlayiso.com Physical Science/P2 5 June 2014 Common Test NSC – Memorandum QUESTION 7 7.1 Homogenous,  reactants and products are in the same phase.  (2) 7.2 Catalyst  (1) 7.3 The equilibrium in graph C is reached faster than in graph A while the concentration of the products in graph A and C are equal and also the concentration of the reactants in both graphs A and C are equal.  (2) 7.4 Exothermic.  (1) 7.5 If the temperature is increased from 5000C to 6000C the concentration of the products decreases and the reactants concentration increases.  The reverse reaction is thus favoured. According to Le Chateliers’ principle endothermic reaction is favoured when temperature is increased.  Thus the forward reaction is exothermic. (2) 7.6 H2(g) + I2(g) ⇌ 2HI(g) n(H2) = 1 2 = 0,5 mol n(I2) = 127 254 = 0,5 mol Mole Ratio H2 I2 2HI Initial mol (0,5 0,5)  0 Change (used/ formed) (0,4 0,4 0,8)  Equilibrium mol (0,1 0,1)  0,8  Equilibrium Concentration 0,1 0,1 0,8 C= n/v 2 2 2 (0,05) (0,05) (0,4)  Kc = [ HI ]2 [ H2 ] . [ I2 ]  Kc = (0,4)2 (0,05) (0,05)  Kc = 64  (8) [16] Copyright Reserved Please Turn Over
Downloaded from hlayiso.com Physical Science/P2 6 June 2014 Common Test NSC – Memorandum QUESTION 8 8.1 8.1.1 C  (1) 8.1.2 A  (1) 8.1.3 E  (1) 8.1.4 B  (1) 8.2 H2SO4 + 2H2O → 2H3O+ + SO42– pH = –log [ H3O+]  pH = – log (2 x 0,1)  pH = 0,698  (3) 8.3 Na2CO3 → Na+ + CO32-  CO32- + H2O  → HCO3– + OH–  (3) [10] Copyright Reserved Please Turn Over
Downloaded from hlayiso.com Physical Science/P2 7 June 2014 Common Test NSC – Memorandum QUESTION 9 9.1 Dilute acid contains a small number of moles of acid in proportion to the volume of water.  (1) 9.2 Standard solution is the solution of known concentration.  (1) 9.3 c = m MV  0,2  = m 56 x 0,3  m = 3,36g  (4) 9.4 To change the colour at the end point of titration.  OR To indicate/show when equivalent number of moles of acid and base have reacted. (1) 9.5 Bromothymol blue,  because the equivalent point will be approximately pH = 7 since H2SO4 is a strong acid and KOH is a strong base.  (2) 9.6 Ca x Va na  Cb x Vb nb  Ca x 20,15  1  15,04 x 0,2 2  Ca  0,075 mol.dm 3 (4) [13] TOTAL SECTION B: 86 GRAND TOTAL: 100 Copyright Reserved Please Turn Over

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