Physical Sciences Grade 12 Western Cape November 2016 Memo

Downloaded from hlayiso.com METRO CENTRAL EDUCATION DISTRICT GRADE 12 PHYSICAL SCIENCES: PAPER 2 (CHEMISTRY) SEPTEMBER 2016 – MARKING GUIDELINE MARKS: 150 TIME: 3 hours MARKING GUIDELINE This question MEMO consists of 12 pages.

Downloaded from hlayiso.com PHYSICAL SCIENCES/P2 2 WCED Metro Central/Sept 2016 MEMORANDUM QUESTION 1 1.1 D ✓✓ (2) 1.2 D ✓✓ (2) 1.3 A ✓✓ (2) 1.4 C ✓✓ (2) 1.5 B ✓✓ (2) 1.6 D ✓✓ (2) 1.7 C ✓✓ (2) 1.8 C ✓✓ (2) 1.9 C ✓✓ (2) 1.10 D ✓✓ (2) [20] Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/P2 3 WCED Metro Central/Sept 2016 MEMORANDUM QUESTION 2 2.1.1 There are single bonds between C-atoms / There are no multiple bonds between C atoms in their hydrocarbon chains.  / No double or triple bonds between C atoms (1)    (substituents must be in alphabetical order) 2.1.2 3-chloro-3-ethylhexane (3) 2.1.3 Propanal ✓ (1) 2.1.4 carboxylic acids ✓ (1) 2.1.5 Methanol ✓ (1)  3C single bonds 2.1.6 carboxylic acid functional group  (2) 2.2.1 Propanoic acid. ✓ (1) 2.2.2 In 100 g there will be 9,81 g H , 58,85g C 31,34g O number of moles: H : = 9,81 mole ✓ C : = 4,904 moles ✓ O : = 1,959 moles ✓ : : 5 : 2,5 : 1 10 : 5 : 2  Molar mass/Molêre massa: 10(1) + 5(12) + 2(16) = 102 g·mol-1 ✓ n=1 (6) C5H10O2 ✓ [16] Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/P2 4 WCED Metro Central/Sept 2016 MEMORANDUM QUESTION 3 3.1 The temperature at which the vapour pressure of a substance✓ equals atmospheric pressure. ✓ (2) 3.2 2-methylpropan-2-ol ✓ (1) 3.3 London force / momentarily dipole forces/ dispersion forces ✓ (1) 3.4.1 In both pentane and 2-methylbutane there are weak London/ dispersion forces present. ✓ 2-methylbutane is more spherical / has a smaller surface area than pentane✓ and therefore there are fewer/less intermolecular forces between its molecules and the energy required to overcome the intermolecular forces in 2-methylbutane is less than the energy required to overcome the intermolecular forces in pentane.  therefore a lower boiling point  (4) 3.4.2 2-methylpropan-2-ol have stronger hydrogen bonding between molecules  while pentane has weaker London/dispersion forces between its molecules.  Therefore more energy is required to overcome the IMF in 2-methylpropan-2-ol than in pentane.  And the more energy required the higher the boiling point.  (4) 3.5 2-methylpropan-2-ol ✓ (1) 3.6 n= M[CO2] = 12 + 2(16) = 44 (a) CO2: n= = 0,773 mol ✓ (b) 0,773 mol CO2 is created by 3 mol C4H10✓ (c) Mass C4,H10: m = n x M = 0,193 x [4(12) + 10(1)] = 11,19 g✓ (d) Percentage purity = x 100✓ = 43,05 % ✓ [Accept: -- 43,09%] (5) [18] Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/P2 5 WCED Metro Central/Sept 2016 MEMORANDUM QUESTION 4 4.1 Cracking of alkanes (1) 4.2.1 Addition polymerization ✓ (1) 4.2.2 polyethene/ polythene/ polyethelene  (any one) (1) 4.2.3  (any one) CH2 CH2 n (1) ✓ 4.3 ✓ Mild heat ✓ Dilute H2SO4 (3) + 4.4 Hydration ✓ (1) 4.5 Ethanol ✓ (1) 4.6 Excess of a conc. strong acid (H2SO4) ✓ Mild heat ✓ (2) 4.7.1 Substitution  NOT: Halogenation/Bromination (1) 4.7.2 Add NaBr in presence of dilute H2SO4  Mild heat ✓ add   (2) Add HBr and mild heat 4.7.3 Sodium hydroxide  water  (1) [15] Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/P2 6 WCED Metro Central/Sept 2016 MEMORANDUM QUESTION 5 5.1 Amount of mole Zn available at 0 s: = 0,00025 mol ✓ (2,462 x 10-4 mol) Amount of mole Zn available at 12 s: = 0,00014 mol ✓ (1,385 x 10-4 mol) Ave Rate = - (no mark for formula) ✓ = - 9,167 x 10-6 mol∙s-1 ✓ (= - 8,975 x 10-6 mol·s-1) Ave Rate = 9,167 x 10-6 mol·s-1 (= 8,975 x 10-6 mol·s-1) n = m/MM = (0,009 – 0,016)/ 65  = - 1,077 x 10-4 mol  Average Rate = - n/t  = - [-1,077 x 10-4 / (12 – 0)]  = 8,974 x 10-6 mol·s-1.  (4) 5.2 The HCℓ is used up/depleted / HCℓ is the limiting reactant ✓ (1) 5.3 The rate decreases as time passes. ✓ (1) 5.4.1 ✓✓ 5.4.2 ✓✓ (4) Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/P2 7 WCED Metro Central/Sept 2016 MEMORANDUM 5.5 ✓✓ (2) 5.6 An increase in temperature increases the number of particles having minimum kinetic energy.  Therefore there are more collisions per second/ unit time/ frequency of collision increases More effective collisions per second/ unit time / frequency of effective collision increases  (2) Which increases rate of reaction [14] Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/P2 8 WCED Metro Central/Sept 2016 MEMORANDUM QUESTION 6 6.1 It is a dynamic equilibrium when the rate of the forward reaction equals the rate of the reverse reaction and the reactions occur simultaneously. [2 or 0] (2) 6.2 N2 3 H2 2 NH3 33,6 24 Initial moles /28 = 1,2  /2 = 12  0 Change in moles - 1 - 3 + 2 5,6 Equilibrium moles /28 = 0,2  9  2 0,2 9 2 Equilibrium conc (= n/V) /5 = 0,04 /5 = 1,8 /5 = 0,4 [N2] = 0,04 mol·dm-3  AND [H2] = 1,8 mol·dm-3  mass of N2 used during the reaction? 33,6 - 5,6 = 28 g Therefore 1 mol N2 was used during the reaction. ✓ 1 mol N2 react with three mole H2 Therefore 6 g H2 was used during the reaction✓ concentration N2 at equilibrium: ✓ = 0,04 mol∙dm-3✓ Amount H2 at equilibrium: = ✓ =1,8 mol∙dm-3✓ (6) 6.3 Kc = ✓ = (0,4)2/(0,04)(1,8)3 ✓ = 0,686 ✓ (3) 6.4 Increases ✓ (1) 6.5 Exothermic ✓ (1) 6.6 When the temperature increases, the Kc value decreases, which means the concentration of the reactants increased and the concentration of the products decreased. ✓ Therefore the reverse reaction was favoured. ✓ An increase of temperature favours the endothermic reaction, ✓ therefore the forward reaction must be exothermic. (3) [16] Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/P2 9 WCED Metro Central/Sept 2016 MEMORANDUM QUESTION 7 7.1 It dissociates completely in water  to produce a high concentration of OH- ions.  (2) 7.2 (a) n= = = 0,158 mol ✓ HO (b) Ba(OH)2 2 Ba2+ (aq) + 2 OH- (aq) Therefore 0,158 mol Ba(OH)2 produces 2 x 0,158 = 0,316 mol OH- ✓ (c) Concentration of hydroxide ions: c = = 0,316/2 = 0,158 mol∙dm-3 (d) Kw = [OH-][H+] ✓ OR: 10-14 = [0,158][H+] ✓ Calc: pOH = -log [OH-]  [H+] = 6,329 x 10-14 mol·dm-3 = - log (0,158)  = 0,801  (e) pH = -log [H+] ✓ = - log[6,329 x 10-14] ✓  pH = 14 - 0,801  = 13,19 ✓ = 13,20  (7) 7.3 Burette ✓ (1) 7.4 An acid is a proton (H+-ion) donor. ✓✓ (2 or 0) (2) 7.5 Ba(OH)2 + 2 HCℓ BaCℓ2 + 2H2O nb = 1 na = 2 cb = 0,079 ca = 2,5 Vb = 2 dm3 Va = ? 0,158 mol Ba(OH)2 will be neutralized by 0,316 mol HCℓ✓ c= ✓ 2,5 = ✓ V = 0,126 dm-3 or 0,13 dm-3 ✓. CaVa /CbVb = na/nb   (2,5)Vb /(0,079)(2) = 2/1  Vb = 0,126 dm3  (4) Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/P2 10 WCED Metro Central/Sept 2016 MEMORANDUM 7.6 bromothymol blue changes colours when the pH is around 7. This is also the end point for a reaction between a strong acid and a strong base / ✓ Phenolphthalein is an effective indicator for a reaction between a strong base (2) and a weak acid. 7.7 REMAINS YELLOW ✓ (1) [19] QUESTION 8 8.1 Aℓ ✓ (1) 8.2.1 Aℓ → Aℓ3+ + 3 e- ✓✓ (2) 8.2.2 Co3+ ✓ (1) 8.3 Decreases ✓ (1) 8.4    Phase must be indicated. Aℓ(s) Aℓ3+ (aq) Co 3+ 2+ (aq) Co (aq) Pt(s)✓ (1 mol·dm-3) (1 mol·dm-3)  1 mol·dm-3 not necessary accept Aℓ(s) Aℓ3+ (aq) Co3+ (aq) Co2+(aq) , Pt(s)✓ (3) -3 (1 mol·dm ) (1 mol·dm-3) 8.5 Eϴ = Ereduction - Eoxidation ✓ = 1,81 - (-0,76) ✓ (3) = 2,57 V ✓ [11] Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/P2 11 WCED Metro Central/Sept 2016 MEMORANDUM QUESTION 9 9.1 The chemical process in which electrical energy  is converted to chemical energy  OR The use of electrical energy  to produce a chemical change . (2) 9.2.1 Chlorine gas/ Cℓ2  (1) 9.2.2 Hydrogen / H2  (1) 9.3 H2O has a stronger oxidizing ability than Na+ / Na+ is a weaker oxidizing agent than H2O  (2) 9.4 (a) Amount of mole Cℓ2 that formed: n= = = 0,1 mol Cℓ2 formed. (b) 0,1 mol Cℓ2 is formed from 0,2 mol NaCℓ✓ Initial amount of NaCℓ available n = cV = 2,5 x 0,5 ✓ = 1,25 mol ✓ (c) Amount NaCℓ left in solution after electrolysis: 1,25 - 0,2 = 1,05 mol ✓ (4) [10] Please turn over

Downloaded from hlayiso.com PHYSICAL SCIENCES/P2 12 WCED Metro Central/Sept 2016 MEMORANDUM QUESTION 10 10.1 10.1.1 Oxygen / O2 ✓ (1) 10.1.2 Haber process ✓ (1) 10.1.3 H2SO4 ✓ (1) 10.1.4 The temperature at which the reaction takes place is approx. 450 oC and water is a vapour. Also the H2SO4 that will be formed is a vapour/mist and cannot be (2) collected easily. ✓ 10.2.1 Nitrogen ✓and phosphorous ✓ (2) 10.2.2 Mass of nutrient = 35/100 x 40 = 14 g  50% of the fertilizer consist of phosphorous: Mass of phosphorus = 0,5 x 14 = 7 g ✓ n= = 7/31 ✓ = 0,226 mol ✓ (4) [10] TOTAL 150 Please turn over