Physical Sciences Grade 12 P2 KwaZulu Natal Mock 2022 Memo

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Downloaded from hlayiso.com Downloaded from Stanmorephysics.com KWAZULU-NATAL PROVINCE s } EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 12 MARKS: 150 This marking guideline consists of 13 pages. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com PhysicaLScience September 2022 Preparatory Examination dwiloaded From Stanmoraphysies.£om QUESTION 1 1.4 Cvv (2) 1.2 Byv (2) 1.3 Dvv (2) 1.4 Avy (2) 15 Bvv (2) 1.6 Dvv (2) 1.7 Dvv (2) 1.8 A v¥ (accept C) (2) 1.9 Cvy (2) 1.10 Dvv (2) [20] QUESTION 2 2.1 2.1.1 CnHen-2 v (1) 2.1.2 H H | | _— — C — C — H | H Marking criteria: e functional groupY e All the substituents( 2 methyl groups and 1 ethyl group) correct “ e Whole structure correct” oA (3) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com PhysicaLScience September 2022 Preparatory Examination dwiloaded From Stanmorephysias.com oe 2.2.1. Organic compounds having the same molecular _formulay, but different functional groupsY (underlined words must be in correct contexts) (2) 2.2.2 Pentanal/2-methylbutanal/3-methylbutanal/2,2-dimethylpropanal vv (2) (functional group —anal Vv Everything correct v’) (If wrong functional group 0/2) 22a 2.3.1 hydroxylv (1) 232 H | H —cC—H H H | | Ho oS ee | | H H H —cC —H H Marking criteria: e Only functional group correct: Max: Xx 2, e Whole structure correct: 3, (2) 2.4 2.4.1 — esterification v (1) 2.4.2 butylv propanoateY (2) [14] Copyright Reserved Please Turn Over

Physic cience dwiloaded From Stanmorephysias.com Downloaded from hlayiso.com QUESTION 3 3.1 3.2 3.3 3.4 3.5 3.6 ag Copyright Reserved The temperature at which the vapour pressure equals atmospheric (external) pressure. vv (2 or 0) YES. v P, Q and R are straight chain primary alcohols/only ONE independent variable. v Boiling point increases V with increase in chain length/molecular mass. ” Intermolecular forces/Van der Waals forces/London forces/dispersion forces increase (becomes stronger) with increase in chain lengthy More energy needed to overcome/break intermolecular forces as chain length increases. Vv REMAINS UNCHANGEDY Py Any One P has the lowest boiling point “ OR P has the weakest intermolecular forcesv LESS THANY Intermolecular forces between molecules of alcohols are hydrogen bonding (in addition to London forces/dispersion forces). v Intermolecular forces between molecules of CeéHi4 are only London forces/dispersion forces. v London forces/dispersion forces are weaker than hydrogen bonding / Intermolecular forces in CeH14 are weaker/ Intermolecular forces in Q are stronger. “ Less energy needed to overcome/break intermolecular forces in CeH:4/ more energy needed to overcome Intermolecular forces in Q.” September 2022 Preparatory Examination (2) (2) (2) (2) (1) (2) (5) [16] Please Turn Over

Downloaded from hlayiso.com PhysicaLScience September 2022 Preparatory Examination owiloaded From Stanmorephysies.com QUESTION 4 41 Hydrogenation. ~ (1) 4.2 Hydrohalogenation/hydrobromination ~ (1) 4.3 Iv (1) 4.4.1 Strong basesY (1) 4.4.2 Reaction IV, base is dilutew Reaction III base is concentrated (in ethanol). v (2) 45 ¥ v lbtlestecs aes ipaes H-—C=C—C—H+H-—Br — H—C—C—C-—H H H Br H (3) 4.6 hydrolysis v (1) 4.7 (3) 48 dehydration” (1) 4.9 ethene vvv (3) [17] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical-Science: September 2022 Preparatory Examination owiloaded From Stanmorephysias.com QUESTION 5 521, Change in concentration of reactants/products per unit time. Change in amount/number of moles/volume/mass” of reactants or products per unit time. “ Amount/number of moles/volume/mass/ of products formed or reactants used per unit time. “ (2) a2 AV average rate = — At _ (104-64) ¥ (60-30) v =133 v ( cm*.s* ) (3) 5.3 lv The gradient / m/ slope of graph | is less steep than Il. “ or Took a longer time for rection to reach completion. ~ (2) 5.4 Catalystv (1) 5.5 e Acatalyst provides an alternate pathway of lower activation energy.” e More particles will have sufficient energy for an effective collision/ more molecules have kinetic energy equal to or greater than the activation energy.” e Number of effective collisions per unit time increases/frequency of effective collisions increases.“ (3) 5.6 Marking criteria: e Ratio: n(Mg) initial equals n(H2) final produced in reaction Il. v Formula: n = ee V, m e Correct substitution ( ) in the above formula p To calculate n(Mg) used(reacted) in reaction lin 150 sv n(Mg)initial - n(Mg)used/reacted v Formula: m = nMv Correct substitution of 24 with n Mg in the above formula. ” Final answer = 0,018 g. “ Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physic OuWhioa ‘eaFrom S tanmereaphysies.com 2022 Preparatory Examination OPTION 1 n(Mg)initial = n(H2)produced in EXP Il aM Vie 0,12 ‘ = v v 24,04 Either Either v = 4,99 x 10? mol n(Mg)used in‘EXP | = n(H2)produced in EXP | V = ve _ 0,102 24,04 = 4,24 x 10° mol n(Mg) remaining = 4,99x10° - 4,24 x 103v = 0,75 x 10? mols m(Mg) = nMv = (0,75 x 10°°)(24) v = 0,018 gv OPTION 2 nHz still to be produced = nMg “ mY _ 0,42-0,102 (4 mark for subtraction) 24,04 =7,49x10™ mol m=nxMvY = 7,49 x 10% x24 v =0,018g “ (8) [19] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical-Science: September 2022 Preparatory Examination owiloaded from Stanmorephysias.com QUESTION 6 6.1 The rate of forward reaction equals the rate of reverse reaction. Vv“ Notes IF: Forward reaction equals reverse reaction. Xx (2) 6.2 Reactants and products are ALL in the same phase. Vv (1) 6.3 Marking criteria: e n(Cfe) equilibrium= 0,4” e Using the correct mol ratiov’ e Calculating the quantity(mol) at equilibrium of all three substances v e Divide number of moles at equilibrium by 5 dm3v e Ke expression” e Correct substitution of equilibrium concentrations into Ke expression ~ e Substitute n(Oz) initial and M(O2) into m = nMv e Final answer 9,60 g v OPTION 1 HCt O2 HzO Clo Ratio 4 1: 2 2 Initial quantity (mol) 1 x 0 0 Seon Change (mol) 0,8 —_0,2— 0,4 0,4 eames a4 Quantity at equilibrium (mol) 1-0,8 x-0,2 |0+0,4) 0,4v , | x-02 | Ss Equilibrium concentration (mol-dm-3) | 9:04 5 0,08 0,08 | Divide by 5 [HOPIC!.1° = ; itution. 7, c [HC¢]*[0,] No Ke expression, correct substitution. Ye 2 2 . 800 = 10:08)'(0.08)" G,oey v ; 0,04)*| == Wrong K- expression (ooay(*2} % x = 0,3 mols n(Oz) = nM = (0,3) (32) ¥ = 9,60 gv (range 9,595312 to 9,60) (8) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com PhysicaLScience dwiloaded From Stanmorephysias.com September 2022 Preparatory Examination OPTION 2 HCt O2 H20 Ck Ratio 4 Z 2 2 Initial quantity (mol) 1 0,3 0 (e) Change (mol) 08 | 02 | 04 0,4 Using ratio v Quantity at equilibrium (mol) “4 , 0,8 0,1 0 +04) 0,4v Equilibrium concentration (mol-dm?) | 9.04 [02] 0,08 0,08 | Divide by 5“ x. — HOF Ice) * [HC¢}*[0,] (0,08)'(0,08)" (0,04) [o,] Wrong K, expression A [Oz] = 0,02 mol-dm? v No K. expression, correct substitution. We n(Oz) = nM = (0,3) (32) v = 9,60 gv (range 9,595312 to 9,60) 6.4 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. vv Note: Underlined phrases must be in correct context 6.5.1 Remains the same. “ 6.5.2 Increases. Vv 6.6 An increase in pressure favours the reaction that produces a fewer number of moles. ” The forward reaction is favoured. ¥ 6.7.1 Negative 6.7.2 Option 1 e When the temperature increases the reverse reaction is favoured. v e An increase in temperature favours the endothermic reaction. ” e Forward reaction is exothermic. ~ Option 2 e Kc decreases with an increase in temperature. v e Reverse reaction is favoured/concentration of reactants increases/concentration of products decreases/yield decreases. “ + Increase in temperature favours the endothermic reaction. v (2) (1) (1) (2) (1) (3) [21] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com September 2022 Preparatory Examination PhysicaLScience Swiloaded From Stanmorephysies.com QUESTION 7 7.1.1 Asubstance that acts as an acid and as a base. vv (2) 7.1.2 HSO,/hydrogen sulphate ion” (1) 7.2.1. Marking quidelines: > © Formula: pH = -log[HsO*] Vv _e Substitution: 0,1” “se Final answer: 1¥” pH = - log [H3O*] “ = - log 0,1v = 1v (3) 7.2.2 EQUALTO7Y (1) 7.2.3 | OPTION 1/OPSIE 1 Marking guidelines: n(OH’) = n(H*) ” e Mol ratio: n(OH)) = n(H*) =cVVv e n= cV (entire eq) = (0,1)(0,03) v ¢ Substitution of 0,1 = 0,003 mol ¥ ¢ Substitution of 0,03. e Final answer: 0,003 mol. OPTION 2 Marking guidelines: c.xV. n e Mol ratio aay S = e Formula: n= cV Spree: Mp Substitution of (0,15)(0,02) O1x 30 1 v e Final answer: 0,03 mol cbx 20 1 cb = 0,15 mol-dm* n=cv ¥ = (0,15)(0,02) “ (4) = 0,003 molsv Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physic OuWhloa ‘eaFrom S tanmere. pliysies.com> °° Preparatory Examination 7.2.4 Marking criteria: Calculate number of moles of hydroxide ions in 250 cm?.v Calculate number of moles of hydroxide ions in Ba(OH)2. ¥ Calculate number of moles NaOH = 0,02 v Mol ratio: number of moles of Ba(OH)2 : number of moles of OH.” Formulae: n= cV ¥ Substitute in the above formula. “ Final answer: 0,175 mol.dm’. eeeeeeu30e OPTION 1 n(OH)in 250 cm? = oe) v 0,0375 mols n(OH)in Ba(OH)2 n(OH’)totat - n(OH)NaOH 0,0375 0,02 v 0,0175 mols n(Ba(OH)z) Y% n(OH) Y (0,0175) ¥ 0,00875 mols c (Ba(OH)2) = a v c (Ba(OH)2) = Oem v 0,05 = 0,175 mol.dm?. ¥ OPTION 2 , 4 V(NaOH) in 20 cm? = 5 x 20 = 16cm* VBa(OH)2in 20cm? = 20- 16 = 4cm§ n(OH’) from NaOH = cV (0,1)(46 x 10°) 1,6 x 10° mols ¥ n(OH) from Ba(OH)2 = 3.x 103 1,6 x 103v 1,4 x 10° mols n(Ba(OH)2) = %n(OH) = %(1,4 x 105)v = 0,7 x 10° mols c (Ba(OH)2) = y v c (Ba(OH)2) = COON 0,004 = 0,175 mol.dm®. ¥ (7) [18] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical-Science: September 2022 Preparatory Examination Swiloaded From Stanmorephysies.com QUESTION 8 1 Voltaic/galvanic cell.“ (1) 8.2 Temperature: 25 °C/298 K. v Concentration of electrolytes: 1 mol.dm*. ” (2) 8.3 spontaneous. v No external energy is required Accept: cell potential is positive / cell is a galvanic cell. ~ (2) 8.4.1 Pt(s)/ Fe**(aq)(1 mol.dm’s), Fe**(aq)(1 mol.dmr?) // X*(aq)(1 mol.dm’?)/Xx(s) (3) Accept: Pt/Fe**, Fe?*//X*/x (MINUS 1 MARK FOR ANY ERROR) 84.2 X*(aq) +e — X(s) %¥v Ignore phases Notes © XaxXte (%) Xt+eex (Y) =Xtte 0 4 Be 0, X2xt+e (%) Xt eo X (%) e Ignore if charge on electron omitted. If a charge of an ion is omitted eg. X +e > X Max: (¥) (2) 8.5 Ei = Er ecicaon = E iicton v Notes 0,03v =EP suction —(0,77) e Accept any other correct formula from the Fo -o80v data sheet. reduction “"™* e Any other formula using unconventional : abbreviations, e.g. E°cei = E°oa- E°ra X is Agv followed by correct substitutions Max: 3/4 (4) [14] Copyright Reserved Please Turn Over

Physic Downloaded from hlayiso.com cience September 2022 Preparatory Examination owiloaded From Stanmorephysies.com QUESTION 9 9.1 9.2 9.3 9.4 9.5 9.5.1 9.5.2 A substance that forms free (positive and negative) ions when melted or dissolved. Vv OR A solution that conducts electricity. vv OR A liquid/solution/dissolved substance that conducts electricity through the movement of ions. v ” (2) Electrode P ¥ (1) Cu?*(aq) + 2e° — Cu(s) VY Ignore phases Notes e Cue Cu + 2e (%) Cu** + 2e = Cu (¥) Cu?* + 2e — Cu (%) Cu2*+ 2e = Cu (%) e Ignore if charge on electron omitted. e If acharge of anion is omitted eg. Cu+2e<Cu Max: (%) (2) Increases. v Reduction takes place at electrode P. v (2) Zinc ions(Zn2*)V and Copper ions(Cu2*)” (2) OPTION 1 Cu?* ions is a stronger oxidising agent than Zn?* ions” Cu?* will be reduced to Cu.¥ ) ; OPTION 2 _ :, (2) Zn* ions are a weaker oxidising agent than Cu?* ions” Zn** will therefore not be reduced to Zn. “ [11] TOTAL: 150 Copyright Reserved