Physical Sciences Grade 12 P2 KwaZulu Natal Mock 2018 Memo

Downloaded from hlayiso.com Physical Sciences / P2 1 NSC-Memorandum Education ducation KwaZulu-Natal Department of Education PHYSICAL SCIENCES P2 (CHEMISTRY) PREPARATORY EXAMINATION SEPTEMBER 2018 MEMORANDUM NATIONAL SENIOR CERTIFICATE GRADE 12 MARKS : 150 This memorandum consists of 9 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 2018 NSC-Memorandum QUESTION 1 1.1 B  (2) 1.2 C (2) 1.3 C (2) 1.4 A (2) 1.5 B (2) 1.6 B (2) 1.7 D (2) 1.8 A (2) 1.9 A (2) 1.10 C (2) [20] QUESTION 2 2.1.1 E (1) 2.1.2 B (1) 2.1.3 D (1) 2.1.4 F (1) 2.1.5 G (1) 2.2.1 2,4,4-trimethylpent-2-ene (2) 2.2.2 CnH2n  (1) 2.3.1 ethanol (1) 2.3.2 sulphuric acid (1) [10] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 3 Preparatory Examination September 2018 NSC-Memorandum QUESTION 3 3.1.1 a series of organic compounds that can be described by the same general formula in which one member differs from the next with a CH2 group.  (2) 3.1.2 the temperature at which the vapour pressure equals atmospheric/external pressure.  (2 or 0) (2) 3.2 C As the boiling point increases the vapour pressure decreases.  C has the highest boiling point.  (3) 3.3 B (1) 3.4.1 118,50 ºC (1) 3.4.2In addition to London forces and dipole-dipole forces, C has two sites for hydrogen bonding between the molecules resulting in the strongest intermolecular forces occurring between molecules of C. The intermolecular forces between molecules of C require the most amount of energy to overcome.  C will therefore have the highest boiling point.  (4) [13] QUESTION 4 4.1.1 Addition/hydrohalogenation (1) 4.1.2 Substitution/hydrolysis (1) 4.2  Whole structure correct: 2/2  Only functional group correct H H H H 1/2 │ │ │ │  More than one functional group H ― C ― C ― C ― C ― H  0/2 │ │ │ │ H Br H H  2-bromobutane (3) 4.3 Secondary The carbon to which the ―O―H is bonded to, is bonded to TWO other carbon atoms.  (3) 4.4 Dehydration (1) 4.5 (Gentle) heat Aqueous/dilute strong base (accept NaOH(dilute) or KOH(dilute)  (2) 4.6.1 Compounds with the same molecular formula,  but different positions of the side chain/substituents/functional groups on the parent chain. (2) 4.6.2 Elimination (1) 4.6.3 H H H H H H H H │ │ │ │ │ │ │ │ H─C─C─C─C─H H ─ C ─ C ═ C ─ C ─ H  + H─Br │ │ │ │ │ │ H Br H H H H 1 mark for each reactant and product (3) [17] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 4 Preparatory Examination September 2018 NSC-Memorandum QUESTION 5 5.1.1 ANY ONE  The change in concentration of reactants/products per unit time.   Rate of change in concentration of reactants or products.   Change in amount/number of moles/volume/mass of reactants/products  per (unit) time.   Amount/number of moles/volume/mass of products formed OR reactants used  per (unit) time.  (2) 5.1.2 60 - 61(s)  (1) 5.1.3 n(CO2) = n(CaCO3)  Marking criteria m  Use mol ratio: n(CO2) = n(CaCO3) = 1:1  = (n  ) 86  40 m M  Substitute in n   100 M 86  40  Substitute 0,46 mols and = ( ) 100 V  Substitute 22,4 dm3 in n  = 0,46 mols Vm V  Final answer: V = 10,304 dm3 n Vm V 0,46  = 22,4  V = 10,304 dm3 (5) 5.1.4 40 g (1) 5.1.5 INCREASES (1) 5.1.6 See attached graph. (3)  Curve starts at 86 g and ends at 40g  The completion time is above 60 or 61s  The curve above the original 5.2.1 Collision theory  (1) 5.2.2The shaded areas in the distribution curves represent the number of molecules with sufficient kinetic energy to overcome the activation energy . An increase in the temperature of the system results in a greater number of particles with sufficient kinetic energy to overcome the activation energy of the reaction . This results in more effective collisions per unit time OR a higher chance of an effective collision occurring, resulting in a higher reaction rate. (3) [17] QUESTION 6 6.1 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) 6.2.1 REMAINS THE SAME (1) 6.2.2 INCREASES (1) 6.2.3 REMAINS THE SAME  (1) 6.2.4 INCREASES (1) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 5 Preparatory Examination September 2018 NSC-Memorandum Apply negative marking from 6.2.4 6.3 According to Le Chatelier’s Principle a decrease in temperature favours the exothermic reaction A decrease in temperature increases the equilibrium constant.Therefore the forward reaction is favoured (3) 6.4 Marking criteria:  Indicating that the number of mols of H2 decreases by an unknown amout  Correct mol ratio  Calculating in terms of x the quantity(mol) at equilibrium of all three substances  n  Substitute V = 4 dm3 in c  to determine concentration at equilibrium of H2/I2 V and HI.  Kc expression  Substitution of concentrations in Kc expression   Substitution of 49 for Kc m  Equation: n   M  Substituting in the above equation  Final answer: 399,36 g No Kc expression, correct substitution: Max. 9 10 Wrong Kc expression : Max. 6 10 H2 I2 HI Initial quantity(mol) 2 2 0 Change(mol) -x -x +2x Ratio  Quantity at 2-x 2-x 2x  equilibrium(mol) Equilibrium 2x 2x x Divide by 4 concentration(mol.dm-3) 4 4 2 x ( )2 [HI ] 2 2 Kc = =  = 49  [H 2 ][I 2 ] 2x 2x ( )( ) 4 4 x = 1,56 mol m(HI) = nM = (2)(1,56)(128)  = 399,36 g (10) [19] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 6 Preparatory Examination September 2018 NSC-Memorandum QUESTION 7 7.1 It dissociates/ionises completely in water to form a high concentration of OH- ions.  (1) 7.2 It contains a small amount (number of moles) of base in proportion to the volume of water (2) 7.3  Formula pH = - log [H3O+] / pOH = - log [OH-]   Substitute 13,45 for pH/0,55 for pOH  c(OH-) = 0,282 mol.dm-3  Using ratio of 1: 2 to calculate c((Ba(OH)2)   Formula m = cVM  Substituting into the above formula  Answer Option 1: Option 2: pH = - log [H3O+]  pOH = - log [OH-]  + 13,45 = - log [H3O ] 0,55 = - log [OH-] [H3O ] + = 3,54 x 10-14 mol.dm-3  [ OH-] = 0,282 mol.dm-3 [H3O+][OH-] = 1 x 10-14 c((Ba(OH)2) = 0,141 mol.dm-3 - -3 c(OH ) = 0,282 mol.dm  m = cVM c((Ba(OH)2) = 0,141 mol.dm-3 = (0,141)(0,25)(171)  m = cVM = 6,03 g = (0,141)(0,25)(171)  = 6,03 g (7) 7.4 Positive marking from question 7.3: concentration of Ba(OH)2 Marking guidelines n c  Va n a  Formulae: c  /n = cV/ a   V c b  Vb nb  Substitution of: 0,141 x 60 /0,141 x 0,06   Use mol ratio: na : nb = 2 : 1   Final answer: 33,84 cm3 / 0,03384 dm3  Option 1: Option 2: n(HCℓ) = 2n((Ba(OH)2) cA VA nA = 2cV cB VB = nB  = 2(0,141)(0,06)  0,5 VA 2 = 0.01692 mols 0,141 0,06 = 1  c(HCℓ) = n/V VA = 0,03384 dm3 0,5 = 0,01692/V 3 Accept VB = 60 cm V = 0,03384 dm3/33,84cm3  VA = 33,84 cm3 (4) [14] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 7 Preparatory Examination September 2018 NSC-Memorandum QUESTION 8 8.1 GALVANIC,  converts chemical energy to electrical energy or no dc power supply. (2) 8.2 Temperature of 25 ºC/298K Pressure 101,3 kPa Concentration of electrolyte of 1mol.dm-3 (3) 8.3 Chlorine (molecule)  (2) 8.4 OPTION 1 Notes  Accept any other correct formula from the EΘcell = EΘcathode - EΘanode  data sheet. = 1,36  - (-1,66)   Any other formula using unconventional abbreviations, e.g. E°cell = E°OA - E°RA = 3,02 V  followed by correct substitutions: E°sel = E°OM - E°RM Max/: 3 4 OPTION 2 Cℓ2(g) + 2e-  2Cℓ-(aq) 1,36 (V)   Aℓ(s) Aℓ 3+(aq) + 3e- -1,66 (V)  Aℓ(s) + Cℓ2(g) Aℓ 3+(aq) + 2Cℓ- (aq) 3,02 (V)  (4) 8.5 3Cℓ2(g) + 2Aℓ(s)  6Cℓ-(aq) + 2Aℓ 3+(aq) Notes  Reactants  Products  Balancing   Ignore phases.  Marking rule 6.3.10  Marking rule 3.9.  Marking rule 3.4: One mark is forfeited when the charge of an ion is omitted per equation (not for the charge on the electron) (3) 8.6.1 REMAINS THE SAME (1) 8.6.2 DECREASES (1) [16] QUESTION 9 9.1 A solution that conducts electricity through the movement of ions.  (1) 9.2 Cu2+ (1) 9.3.1 Decreases (1) - 9.3.2 Cu → Cu2+ + 2 e Notes  Cu2+ + 2e- ← Cu (2 ) Cu2+ + 2e- ⇌ Cu (0 ) 2 2 2+ Cu ⇌ Cu + 2e - ( 1 ) 2+ Cu + 2e → Cu ( 0 ) 2 2  Ignore if charge on electron is omitted.  If a charge of an ion is omitted e.g. Cu → Cu2 + 2e- is Cu → Cu2 + 2e- Max.: 1 2 (2) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 8 Preparatory Examination September 2018 NSC-Memorandum 9.3.4 Marking criteria  Calculate number of mols of cations: 2,259 x 1024 = n(6,023 x 1023) m  Formula: n  M m  Substitute calculated number of moles of cations and 63,5 in n   M  Final answer 238,125 g ne = nNA 2,259 x 1024 = n(6,023 x 1023)  n = 3,75 mols m = nM = (3,75)(63,5)  = 238,125 g.  (4) [9] QUESTION 10 10.1.1 Haber (1) 10.1.2 Catalytic oxidation of ammonia (1) 10.1.3 Nitrogen dioxide (1) 10.1.4 Ammonium nitrate (1) 10.2.1 Sulphuric acid/H2SO4 (1) 10.2.2 H2SO4 + 2NH3  (NH4)2SO4 Notes:  Reactants  Products  Balancing   Marking rule 6.3.10. (3) 10.3 %N = 14/20 x 36 = 25,2% Mass of N = 25,2/100 x m 12,60= 25,2/100 x m m = 50 kg  (4) 10.4 Fertiliser A Fertilizer A has a high percentage of Phosphorus compared to fertilizer B.  (3) [15] TOTAL MARKS: 150 Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences / P2 9 Preparatory Examination September 2018 NSC-Memorandum QUESTION 5.1.6 100 ● 80 ● lumps Mass of CaCO3 (g) ● 60 ● ● ● ● ● ● 40 20 0 0 10 20 30 40 50 60 70 Time (s) Copyright Reserved Please Turn Over