Physical Sciences Grade 12 Paper 2 Gauteng Mock 2016 Memo

Downloaded from hlayiso.com PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN 2016 MEMORANDUM PHYSICAL SCIENCES: CHEMISTRY P2 (10842) FISIESE WETENSKAPPE: CHEMIE V2 (10842) 13 pages

Downloaded from hlayiso.com 10842 / 16 GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION – 2016 PHYSICAL SCIENCES: CHEMISTRY / FISIESE WETENSKAPPE: CHEMIE (Second Paper / Tweede Vraestel) MEMORANDUM QUESTION / VRAAG 1 1.1 D  1.2 A  1.3 D  1.4 D  1.5 C  1.6 C  1.7 B  1.8 A  1.9 C  1.10 C  [20] QUESTION / VRAAG 2 2.1.1 Saturated / Versadig  (2) ANY ONE / ENIGE EEN: • Have ONLY single bonds. / Het slegs ENKELBINDINGE . • Have single bonds between C atoms./ Het enkelbindinge tussen koolstofatome. • Have no double OR triple bonds OR multiple bonds. / Het geen dubbel OF  meervoudige bindinge nie. • Contains the maximum number of H atoms bonded to C atoms./ Beskik oor die maksimum aantal H-atome wat aan koolstofatome kan bind.  2.1.2 2,3-dimethylbutane  2,3-dimetielbutaan (2) 2

Downloaded from hlayiso.com 10842 / 16 2.2 H Marking criteria / Nasienriglyne: H C H • Six C atoms in longest chain with functional group (carbonyl group) on C-2  H H O H • Two methyl substituents on C-4  H C C C C C H • Whole structure correct  H H H H C H • Ses C atome in langste ketting met funksionele groep (karboniel groep) op C-2  H C H • Twee metielsubstituente op C-4  H • Volledige struktuur korrek  (3) 2.3 2.3.1 C n H 2n  (1) 2.3.2 H H H H C C + H H → H C C H H H or H2 H H Marking criteria / Nasienriglyne:  reactants / reaktante  product / produk  balancing / balansering Rules / Reëls: Any additional reactant / product / incorrect balancing / arrow omitted Enige addisionele reaktante/produk /verkeerde balansering /pyltjie uitgelaat max 2/3 (3) 2.3.3 Hydrolysis / Hidrolise OR/OF Substitution / Vervanging  (1) 2.3.4 HBr  (1) 2.3.5 E (1) 2.3.6 E – concentrated / gekonsentreerd and/en G – diluted / verdun  OR / OF Base is more concentrated in reaction E than G. / Basis is meer gekonsentreerd in reaksie E as in G OR / OF Base in reaction E is dissolved in ethanol (no water added). / Basis in reaksie E is in etanol opgelos (geen water bygevoeg nie.) (1) 2.3.7 H 2 SO 4 /sulphuric acid / swaelsuur OR/OF H 3 PO 4 phosphoric acid / fosforsuur  (1) [16] 3

Downloaded from hlayiso.com 10842 / 16 QUESTION / VRAAG 3 3.1 3.1.1 Aldehyde / Aldehied  (1) 3.1.2 Ketone / Ketoon  (1) 3.2 3.2.1 When molecules have the same molecular formula  but different positions of the side chain, substituents or functional groups on the parent chain.  Molekule met dieselfde molekulêre formule, maar met verskillende posisies van die sykettings, substituente of funksionele groepe op die hoofketting. (2) 3.2.2 H Cl H Marking criteria / Nasienriglyne: • Structure correct/struktuur korrek 2 / 2 H C C C H • Any error/enige foute: 0 / 2 H H H (2) 3.3 3.3.1 Pentan-1-ol and butanoic acid  Pentan-1-ol en butanoësuur (2) 3.3.2 Esterification  or acid catalysed condensation. Esterifikasie of suurgekataliseerde kondensasie. (1) 3.4 3.4.1 Polyethene / Polieteen (1) 3.4.2  (1) 3.4.3 Addition polymer. / Addisie polimeer  (1) [12] QUESTION / VRAAG 4 4.1 4.1.1 Boiling point / Kookpunt  (1) 4.1.2 Type of organic compound / homologous series / functional group  Tipe organiese verbinding / homoloë reeks / funksionele groep (1) 4.1.3 Relative molecular mass / Relatiewe molekulêre massa  (1) 4.2 Lower than / Laer as  (1) Negative marking from Q 4.2 to Q 4.3 4

Downloaded from hlayiso.com 10842 / 16 4.3 Structure: Compound A / Pentane is less branched / no branches / less compact / less spherical / has larger surface area.  Intermolecular forces: Stronger / more intermolecular forces / induced dipole forces / London forces / dispersion forces.  Energy: More energy needed to overcome / break intermolecular forces.  OR Structure: Compound B / 2-methylbutane is more branched / more compact / more spherical / has a smaller surface area.  Intermolecular forces: Weaker / less intermolecular forces / induced dipole forces / London forces / dispersion forces.  Energy: Less energy needed to overcome / break intermolecular forces.  Struktuur: Verbinding A / Pentaan het geen sykettings / minder kompak / minder sferies / het groter oppervlakarea.  Intermolekulêre kragte: Sterker / meer intermolekulêre kragte / geïnduseerde dipool kragte / London kragte / dispersie kragte.  Energie: Meer energie nodig om intermolekulêre kragte te breek/oorkom  OF Struktuur: Verbinding B / 2-metielbutaan het (meer) sykettings / meer kompak / meer sferiese molekule / het kleiner oppervlakarea  Intermolekulêre kragte: swakker / minder intermolekulêre kragte / geïnduseerde dipool kragte / London kragte / dispersie kragte.  Energie: Minder energie nodig om intermolekulêre kragte te oorkom/breek  (3) 4.4 Hoër as Negatiewe nasien van VRAAG 4.4 na 4.5 (1) 4.5 • (D) propanoic acid has more sites for hydrogen bonding than C / forms dimers / is more polar than C.  • D has stronger / more intermolecular forces / dipole-dipole forces.  • D needs more energy to overcome / break the intermolecular forces.  (3) 5

Downloaded from hlayiso.com 10842 / 16 OR • (C) buthanol has less sites for hydrogen bonding than D / forms dimers / is more polar than D.  • C has weaker / less intermolecular forces / dipole-dipole forces.  • C needs less energy to overcome / break the intermolecular forces.  • (D) propanoësuur het meer gebiede vir waterstofbindinge as C / vorm dimere / is meer polêr as C.  • D het sterker / meer intermolekulêre kragte / dipool-dipool kragte.  • D benodig meer energie om intermolekulêre kragte te oorkom/breek  OF • (C) butanol het minder gebiede vir waterstofbinding as D / vorm dimere / is meer polêr as D.  • C het swakker / minder intermolekulêre kragte / dipool-dipool kragte.  • C benodig minder energie om intermolekulêre kragte te oorkom/breek  4.6 O O H H H H H  C H C C H C C O C O H H   H Methyl ethanoate OR H H ethyl methanoate Metieletanoaat etielmetanoaat (4) [12] 6

Downloaded from hlayiso.com 10842 / 16 QUESTION / VRAAG 5 5.1 Examples: Marking criteria: • How will a change in • Identify dependent and independent concentration influence the variables correct / Identifiseer reaction rate? afhanklike en onafhanklike • What is the relationship between veranderlikes korrek  concentration and reaction rate? • Ask a question about the relationship between the variables / Vra vraag oor • Hoe sal ’n verandering in verwantskap tussen afhanklike en die konsentrasie die reaksietempo onafhanklike veranderlikes  beïnvloed? • If question has a yes/no answer, no • Wat is die verwantskap tussen die marks / Indien die vraag ’n ja/nee konsentrasie en reaksietempo? antwoord het, geen punte. (2) 5.2 HNO 3 / Nitric acid / Salpetersuur.  The magnesium is used up / the magnesium is the limiting reagent.  Die magnesium word opgebruik / die magnesium is die beperkende reagens.  (2) 5.3 • Change in concentration of products / reactants  per (unit) time.  • Verandering in konsentrasie van produkte / reaktanse per (eenheids)tyd. OR / OF • Rate of change in concentration • Tempo van verandering in konsentrasie. OR / OF • Change in amount / number of moles / volume / mass of reactants / products  per (unit) time.  • Verandering in hoeveelheid / aantal / mol / volume / massa van reaktanse / produkte per (eenheids)tyd. OR / OF • Amount / number of moles / volume / mass of products formed or reactants used per (unit) time. • Die hoeveelheid / aantal / mol / volume / massa van produk gevorm of reaktanse gebruik per (eenheid) tyd. (2) 5.4 Δ n = 1,0 – 0,8  = 0,2 mol m n= M m 0,2 =  24 m = 4,8 g ∆m Average rate / gem. reaksietempo = ∆t 4,8  = 30 (- 0)  = 0,16 g·s-1  (5) 7

Downloaded from hlayiso.com 10842 / 16 5.5 Criteria for marking of graph / nasienriglyne: • Steeper slope below original graph. / Skuinser helling onder oorspronklike grafiek  • Intercepts x-axis earlier. / Sny x- as vroeër  (2) [13] QUESTION / VRAAG 6 6.1 6.1.1 Activation energy / Aktiveringsenergie  (1) 6.1.2 (a) An increase in the concentration of one or both of the reactants.  ʼn Verhoging in die konsentrasie van een of beide reaktanse.  (1) (b) Increase in temperature / Verhoging van temperatuur  • Increase in average kinetic energy of molecules. / More molecules have enough / sufficient kinetic energy. / Verhoging in gemiddelde kinetiese energie van molekules / Meer molekules het voldoende of genoeg kinetiese energie.  • More effective collisions per (unit) time. / Meer effektiewe botsings per (eenheids)tyd.  • Higher reaction rate. / Hoër reaksietempo  (4) 6.2 6.2.1 Marking criteria/Nasienriglyne: In terms of reducing agent/In terme van die reduseermiddel: • Cu is a weaker reducing agent  than H 2  and will not reduce H+ (to H2)  • Cu is ’n swakker reduseermiddel  as H 2  en sal nie die H+ (na H 2 ) reduseer nie. In terms of oxidising agent/In terme van die oksideermiddel: • H+ is a weaker oxidising agent  than Cu2+  and will not oxidise Cu (to Cu2+) • H+ is ’n swakker oksideermiddel  as Cu2+  en sal nie die Cu (to Cu2+) oksideer nie.  (NOTE: Compare the two reducing agents in the two half-reactions involved OR the two oxidising agents in the two half-reactions involved. NOTA: Vergelyk die twee reduseermiddels in die twee half reaksies OF die twee oksideermiddels in die twee halfreaksies.) OR / OF H+ (H 2 SO 4 ) is a weaker oxidizing agent than Cu (to Cu2+.) H+(H 2 SO 4 ) is swakker oksideermiddel as die Cu (na Cu2+) Note/Nota: No marks if referring to relative positions on the table./ Geen punte indien na die relatiewe posisies op die redokstabel verwys word nie. (3) 8

Downloaded from hlayiso.com 10842 / 16 6.2.2 (a) Greater than. / Groter as.  Surface area / state of division is larger in B. / Die oppervlak area (reaksie oppervlakte / toestand van verdeeldheid) in B is groter.  (2) (b) Smaller than / Kleiner as  The Cu acts as a catalyst for the reaction in test tube D. Die Cu tree as ʼn katalisator  op in proefbuis D. (2) [13] QUESTION / VRAAG 7 7.1 7.1.1 Exothermic / Eksotermies  (1) 7.1.2 Negative marking from QUESTION 7.1.1/ Negatiewe nasien van VRAAG 7.1.1 Marking criteria/Nasienriglyne: • K c decreases with increase in temperature. / K c verminder met verhoging in temperatuur.  • Reverse reaction is favoured. / [reactants] increase and [product] decreases. / Terugwaartse reaksie word bevoordeel / [reaktanse] verhoog en [produk] verlaag.  • An increase in temperature favours the endothermic reaction. / ’n Verhoging in temperatuur bevoordeel die endotermiese reaksie.  (3) 7.1.3 Remains the same / Bly dieselfde.  (1) 7.1.4 Only a change in temperature has an effect on the equilibrium constant / has an effect on K c . / Slegs ʼn verandering in temperatuur beinvloed waarde van K c  (1) 7.2 Marking criteria/Nasienriglyne: • n(O 2 )reacted/gereageer = n(O 2 )eq - n(O 2 )ini  • Use ratio/Gebruik verhouding 2 : 20 : 0  • n(SO 2 )eq = n(SO 2 )ini + n(SO 2 )change/verandering  • n(SO 3 )eq = n(SO 3 )ini - n(SO 3 )change/verandering • n eq divide by / gedeel deur 2  • K c expression/uitdrukking  • Substitute/vervang [ ] eq into Kc expression/uitdrukking  • Answer/antwoord: 125  9

Downloaded from hlayiso.com 10842 / 16 2SO 3 2SO 2 O2 Initial mol 24 0 0 Aanvanklike mol Change mol 20 20 10 Verandering in mol Use ratio  Equilibrium mol 24 – 2 (10) = 4  20 10  Mol by ewewig Concentration n 4 n 20 n 10 equilibrium c= = =2 c= = c= =  V 2 V 2 V 2 Ewewigskonsentrasie = 10 =5 Kc = [SO 2 ] [O 2 ] 2  [SO 3 ]2 = (10)2 (5)  (2)2 (7) = 125  [13] 10

Downloaded from hlayiso.com 10842 / 16 QUESTION / VRAAG 8 8.1 Completes the circuit / allows for the overall charge in the beakers to remain neutral / allows ions to flow between the beakers / electrical neutrality.  Voltooi die stroombaan / laat toe dat die totale lading in die bekers neutraal bly / laat ione vryelik tussen bekers vloei/ elektrise neutraliteit.  (1) 8.2 A galvanic (voltaic cell) converts chemical energy (change) into electrical energy,  whereas in an electrolytic cell, electrical energy is converted into chemical energy (change).  ʼn Galvaniese sel skakel chemiese energie in elektriese energie om, terwyl die elektrolitiese sel, elektriese energie in chemiese energie verander. (2) 8.3 Pt.  MnO 4 - is reduced to Mn 2+ / Reduction takes place.  MnO 4 - word na Mn 2+ gereduseer / Reduksie vind plaas.  (2) 8.4 Fe2+ (ions/ione)  (1) 8.5 5Fe2+(aq) + MnO 4 - (aq) + 8H+(aq) → 5Fe3+(aq) + Mn2+(aq) + 4H 2 O() (3) Notes / Aantekeninge • Reactants  Products  Balancing  Reaktanse  Produkte  Balansering  • Ignore / Ignoreer phases / fases • Marking rule 6.3.10 / Nasienreël 6.3.10 8.6 E cell = E cathode/katode – E anode  = (+1,51)  – (+0,77)  = +0,84 V  (4) [13] 11

Downloaded from hlayiso.com 10842 / 16 QUESTION / VRAAG 9 9.1 A solution / liquid / / dissolved substance that conducts electricity through the movement of ions. / ’n Oplossing / vloeistof /opgeloste stof wat elektrisiteit gelei deur die beweging van ione.  (2) 9.2 9.2.1 Silver nitrate / silver ethanoate / silver cyanide  Silvernitraat / silveretanoaat / silwersianied  (1) 9.2.2 Silver/silwer / Ag  (1) 9.2.3 Ag+ + e- → Ag  (2) 9.3 m Marking guidelines / Nasienriglyne: n=  M • Formula / ratio / Formule / verhouding  2 • Substitute/Vervang M (Ag) = 108  =  • Multiply by/Vermenigvuldig met N A  108 • Final answer = 1,11 x 1022 electrons = 1,85 x 10-2 mol Ag (accept 1,11 x 1022 – 1,12 x 1022) / Finale antwoord = 1,11 x 1022 elektrone  6,02 x 1023 x 1,85 x 10-2 (aanvaar 1,11 x 1022 – 1,12 x 1022) = 1,11 x 1022 electrons needed / elektrone nodig  (4) [10] 12

Downloaded from hlayiso.com 10842 / 16 QUESTION / VRAAG 10 10.1 pH = - log [H+] OR pH = - log [H 3 O+]  = - log (5,6 x 10-6)  = 5,25  (3) 10.2 10.2.1 n c=  V n = cV = (2,5)(0,5)  = 1,25 mol NaOH  (3) 10.2.2 POSITIVE MARKING FROM QUESTION 10.2.1 / POSITIEWE NASIEN VAN VRAAG 10.2.1 n acid = c a V a Marking guidelines/ Nasienriglyne: = (0,2)(0,095)  n • Substitution into/Vervang in c =  = 0,019 mol H 2 SO 4 V. • Using ratio/Gebruik verhouding 2:1  n(NaOH) = 2n(H 2 SO 4 ) • n(NaOH used/gebr ) = n(NaOH initial/aanvk ) - n(NaOH excess/oormt ) = 2 x 0,019  • n[Mg(NO 3 ) 2 ] = ½n(NaOH)  = 0,038 mol NaOH • Substitute/Vervang M[Mg(NO 3 ) 2 ] = 148 g·mol-1  • Final answer/Finale antwoord: 64,38 g  n(NaOH used/gebruik) = n(NaOH initial/aanvanklik) - n(NaOH excess/oormaat): 1,25 – 0,038  = 1,212 mol 1,212 n[Mg(NO 3 ) 2 ] =  2 = 0,606 mol m[Mg(NO 3 ) 2 ] = nM = 0,606 x 148  = 89,69 g  (6) [12] 13

Downloaded from hlayiso.com 10842 / 16 QUESTION / VRAAG 11 11.1 11.1.1 The proportion / ratio  of nitrogen, phosphorus and potassium (N.P.K).  Die verhouding  van stikstof, fosfor en kalium (N.P.K).  (2) 11.1.2 Percentage fertiliser / persentasie kunsmisstof  (1) 11.2 Ammonium nitrate/Ammoniumnitraat Marking guidelines: Nasienriglyne: (NH 4 NO 3 ):  • Molar masses / 28 Molare massa  %N = x 100 = 35,0 % N 80 • Multiply with 100 / Vermenigvuldig met (answers) 100  Urea/Ureum ((NH 2 ) 2 CO): • % N (35,0% and/en 28  46,7%)  %N = x 100 = 46,7 % N 60 • Identify urea / Urea – it has the highest percentage N / Identifiseer ureum  Ureum het die hoogste persentasie  (4) 11.3 11.3.1 Fractional distillation (of liquid air) / Fraksionele distillasie van (vloeibare lug).  (1) 11.3.2 N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (aq) Marking criteria / Nasienriglyne:  reactants / reaktanse  product / produk  balancing / balansering Rules/Riglyne: • Any additional reactant / product / incorrect balancing / arrow • omitted: max 2 / 3 • Enige addisionele reaegense of produkte / verkeerde balansering / pyle uitgelaat: maks 2 / 3 (3) 11.3.3 (NH 4 ) 2 SO 4  ammonium sulphate / ammoniumsulfaat  (2) [13] TOTAL / TOTAAL: 150 14