Physical Sciences Grade 12 P2 Gauteng Prelim 2024 Memo (Afr)

Downloaded from hlayiso.com PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN 2024 MARKING GUIDELINES NASIENRIGLYNE 10842 PHYSICAL SCIENCES: CHEMISTRY FISIESE WETENSKAPPE: CHEMIE (PAPER/VRAESTEL 2) 15 pages/bladsye

Downloaded from hlayiso.com 10842/24 QUESTION/VRAAG 1 1.1 C  (2) 1.2 A  (2) 1.3 B  (2) 1.4 C  (Accept A and D/Aanvaar A en D) (2) 1.5 D  (2) 1.6 A  (2) 1.7 B  (2) 1.8 B  (Accept A/Aanvaar A) (2) 1.9 C  (2) 1.10 B  (2) [20] 2

Downloaded from hlayiso.com 10842/24 QUESTION/VRAAG 2 2.1 Saturated  Only single bonds between C-atoms/No multiple bonds between C-atoms.  Versadig Slegs enkelbindings tussen C-atome/Geen meervoudige bindings tussen C-atome nie. (2) 2.2.1 E  (1) 2.2.2 D  (1) 2.2.3 B&E  (1) 2.3.1 Primary  The carbon that is bonded to the hydroxyl group (OH) is bonded to one carbon atom.  Primêre Die koolstof wat aan die hidroksielgroep (OH) gebind is, is gebind aan een ander koolstofatoom. (2) 2.3.2 MARKING CRITERIA/ NASIENKRITERIA  correct whole structure  correct IUPAC name Korrekte volledige struktuur Korrekte IUPAC naam butan-1-ol (2) 2.4.1   2,3-dibromo-5-methylheptane (2) 2,3-dibromo-5-metielheptaan 2.4.2 Formyl  group Formiel groep (1) 3

Downloaded from hlayiso.com 10842/24 2.5.1 100 – (40 + 53,3) = 6,7 C: H : O MARKING CRITERIA/ NASIENRIGLYNE 𝑚 40 6,7 53,3 𝑛= 𝑀 12 1 16 6,7 3,33 6,67 3,33 Divide by smallest number/ 3,33 3,33 3,33 deel deur kleinste getal 1 2 1 CH2O Molar mass of/Molêre massa van CH2O = 30 g∙mol-1 60 ÷ 30 = 2  ∴ C2H4O2  1 (NOTE: If only formula given, 4) 1 (4) (NOTA: As slegs formule gegee word, 4) 2.5.2 Ethanoic acid/Etanoësuur  Methyl methanoate/Metielmetanoaat  (2) [18] 4

Downloaded from hlayiso.com 10842/24 QUESTION/VRAAG 3 3.1 3.1.1 Organic molecules with the same molecular formula, but different structural formulae.  (2 or 0) Organiese molekule met dieselfde molekulêre formule, maar verskillende struktuurformules. (2) 3.1.2 OPTION 1:  B (Hexane) has a larger surface area/longer straight chain than A (2,3-dimethylbutane)   Stronger intermolecular forces/London forces/Van Der Waals forces   More energy required to overcome/weaken the intermolecular forces (IMF)  OR OPTION 2:  A (2,3-dimethylbutane) has a smaller surface area/more branched than B (hexane) (Accept: more spherical)  Weaker intermolecular forces/London forces/Van Der Waals forces  Less energy required to overcome/weaken the intermolecular forces (IMF) OPSIE 1:  B (Heksaan) het 'n groter oppervlakarea/langer reguitketting as A (2,3-dimetielbutaan)  Sterker intermolekulêre kragte/London kragte/Van der Waals kragte  Meer energie benodig om intermolekulêre kragte (IMK) te oorkom/verswak OF OPSIE 2:  A (2,3-dimetielbutaan) het 'n kleiner oppervlaksarea/meer vertakkings as B (heksaan) (Aanvaar: meer sferies)  Swakker intermolekulêre kragte/London kragte/Van der Waals kragte  Minder energie benodig om intermolekulêre kragte (IMK) te oorkom/verswak. (3) 5

Downloaded from hlayiso.com 10842/24 3.2 The pressure exerted by a vapour at equilibrium with its liquid in a closed system.  Die druk uitgeoefen deur 'n damp in ewewig met sy vloeistof in 'n geslote sisteem. Marking criteria If any one of the underlined key words phrases in the correct context (vapour pressure) is omitted, deduct 1 mark. Nasienkriteria Indien enige een van die onderstreepte woorde /frases in die korrekte konteks uitgelaat word, trek 1 punt af. (2) 3.3 B (Hexane)  Has weaker/less London (dispersion) forces than C (methyl propanoate) OR Has a lower boiling point than C (methyl propanoate)  B (Heksaan) Het swakker/minder London (dispersie) kragte as C (metielpropanoaat) OF Het 'n laer kookpunt as C (metielpropanoaat) (2) 3.4 3.4.1 Y (1) 3.4.2  C (methyl propanoate) has dipole-dipole forces and D (pentan-1-ol) has hydrogen bonding between molecules.   D has stronger intermolecular forces than C.   More energy required to overcome/weaken the intermolecular forces in D.   C (metielpropanoaat) het dipool-dipoool kragte en D (pentan-1-ol) het waterstofbinding tussen molekules.  D het sterker intermolekulêre kragte as C.  Meer energie benodig om intermolekulêre kragte te oorkom/verswak in D. (3) [13] 6

Downloaded from hlayiso.com 10842/24 QUESTION/VRAAG 4 4.1 4.1.1 Addition/Hydrohalogenation/Hydrobromination  Addisie/Hidrohalogenering/Hidrobrominering (1) 4.1.2 Elimination/Dehydrohalogenation/Dehydrobromination  Eliminasie/Dehidrohalogenering/Dehidrobrominering (1) 4.2 Pt/platinum  (Also accept/Aanvaar ook: Pd or/of Ni) (1) 4.3 4.3.1 Hydrolysis  Hidrolise (1) 4.3.2  Dilute strong base (NaOH(aq)/KOH(aq)/LiOH(aq)/sodium hydroxide(aq)/potassium hydroxide(aq)/lithium hydroxide(aq))  Heat  (NOTE: Do NOT accept temperature)  Verdunde sterk basis (NaOH(aq)/KOH(aq)/LiOH(aq)/natriumhidroksied(aq)/kaliumhidroksied (aq)/ litiumhidroksied(aq))  Hitte (NOTA: Moet NIE temperatuur aanvaar NIE) (2) 4.3.3 Propan-2-ol  (Accept/Aanvaar: 2-propanol) (2 or zero) (2) 4.4 + KBr (optional) (3) 4.5 4.5.1 Esterification/Condensation  Esterifikasie/Kondensasie (1) 4.5.2 Water  (NOTE: Do NOT accept chemical formula) (NOTA: Moet NIE die chemiese formule aanvaar nie) (1) 4.5.3 (Concentrated) sulphuric acid/ (Gekonsentreerde) swawelsuur  (NOTE: Do NOT accept chemical formula) (NOTA: Moet NIE die chemiese formule aanvaar nie) (1) 4.5.4 Hexanoic acid  (2 or zero) Heksanoësuur (2) 4.6  (NOTE: Indicate all four bonds around carbon) (NOTA: Dui al vier bindings rondom koolstof aan) (1) [17] 7

Downloaded from hlayiso.com 10842/24 QUESTION/VRAAG 5 5.1 5.1.1 OPTION 1: What is the relationship between the concentration and rate of reaction?  OR OPTION 2: What is the relationship between the state of division and rate of reaction?  OR OPTION 3: How is the rate of the reaction affected by the degree of purity? OPSIE 1: Wat is die verband tussen die konsentrasie en die reaksietempo? OF OPSIE 2: Wat is die verband tussen die toestand van verdeeldheid en die reaksietempo? OF OPSIE 3: Hoe word die reaksietempo beïnvloed deur die graad van suiwerheid? (2) 5.1.2 IF OPTION 1 IN 5.1.1: Temperature/state of division (surface area)/mass of impure CaCO3  IF OPTION 2 IN 5.1.1: Temperature/concentration/mass of impure CaCO3  IF OPTION 3 IN 5.1.1: Temperature/concentration/state of division (surface area)/mass of impure CaCO3  INDIEN OPSIE 1 IN 5.1.1: Temperatuur/toestand van verdeeldheid (oppervlakarea)/massa onsuiwer CaCO3 INDIEN OPSIE 2 IN 5.1.1: Temperatuur/konsentrasie/massa onsuiwer CaCO3 INDIEN OPSIE 3 IN 5.1.1: Temperatuur/konsentrasie/massa onsuiwer CaCO3 (1) 5.1.3 Rate of reaction/volume of gas per unit time  Reaksietempo/volume gas per eenheid tyd (1) 8

Downloaded from hlayiso.com 10842/24 5.2 Rate between time 0 s and 20 s is faster(higher)/rate between time 30 s and 50 s is slower(lower)  Tempo tussen 0 s en 20 s is vinniger(hoër)/tempo tussen 30 s en 50 s is stadiger(laer) (1) 5.3 Reaction stop/calcium carbonate is used up/reaction is complete.  (NOTE: Hydrochloric acid is in excess) Die reaksie stop/kalsiumkarbonaat is opgebruik/reaksie is voltooi. (LET WEL: Soutsuur is in oormaat) (1) 5.4 MARKING CRITERIA/NASIENKRITERIA (a) Any applicable formula Enige toepaslike formule (b) Dividing volume of CO2 (from graph) by 24 dm3 Deel volume CO2 (vanaf grafiek) deur 24 dm3 (c) Using mole ratio for CO2 to CaCO3 (1:1) Gebruik molverhouding vir CO2 tot CaCO3 (1:1) (d) Multiplying n(CaCO3) with 100 in the correct formula Vermenigvuldig n(CaCO3) met 100 in die korrekte formule (e) Final answer Finale antwoord 𝑉 𝑎 𝑛 = 𝑉𝑚  0,025 𝑏 =  24 = 1,042 x 10-3 mol From the balanced equation/van die gebalanseerde vergelyking: n(CO2) : n(CaCO3) m(CaCO3) = nM 𝑐 𝑑 1:1  = 1,042 x 10-3 x 100  n(CaCO3) = 1,042 x 10-3 mol 𝑒 = 0,104 g  (0,10g) (5) 5.5 Positive marking from QUESTION 5.4/ Positiewe nasien vanaf VRAAG 5.4 _pure mass sample/suiwer massa monster % purity/suiwerheid = impure mass sample/onsuiwer massa monster x 100 = 0,104 0,3 x 100  = 34,67 %  (33,33-34,67%) (2) 5.6 5.6.1 Steeper gradient/Steiler gradiënt  (1) 9

Downloaded from hlayiso.com 10842/24 5.6.2  Increase in concentration increases the number of particles per unit volume.   More effective collisions per unit time. /Frequency of effective collisions increases.   Rate of reaction increases.   Verhoging in konsentrasie verhoog die aantal deeltjies per eenheid volume.  Meer effektiewe botsings per eenheid tyd./Frekwensie van effektiewe botsings neem toe  Reaksietempo neem toe (3) [17] 10

Downloaded from hlayiso.com 10842/24 QUESTION/VRAAG 6 6.1 (Chemical equilibrium is a dynamic equilibrium) when the rate of the forward reaction equals the rate of the reverse reaction. (2 or 0)  (Chemiese ewewig is 'n dinamiese ewewig) wanneer die tempo van die voorwaartse reaksie gelyk is aan die tempo van die terugwaartse reaksie. (2) 6.2 6.2.1 More orange/Meer oranje  (1) 6.2.2  Nitric acid ionises completely in the solution thereby increasing the concentration of hydrogen ions   Forward reaction is favoured to decrease the concentration of hydrogen ions   Thus the concentration of (yellow) CrO42- ions decrease and the concentration of (orange) Cr2O72- ions increase   Salpetersuur ioniseer volledig in oplossing, dus verhoog die konsentrasie waterstofione.  Die voorwaartse reaksie word bevoordeel om die konsentrasie van die waterstofione te verminder.  Dus sal die konsentrasie van die (geel) CrO42- ione afneem en die konsentrasie van die (oranje) Cr2O72- ione toeneem. (3) 6.3 6.3.1 Exothermic/Eksotermies  (1) 6.3.2  An increase in temperature favours an endothermic reaction.   The mixture turns yellow, indicates that reverse reaction was favoured.   Thus the reverse reaction is endothermic and forward reaction is exothermic.  'n Toename in temperatuur bevoordeel 'n endotermiese reaksie.  Die mengsel word geel, dui daarop dat die terugwaartse reaksie bevoordeel was.  Dus is die terugwaartse reaksie endotermies en die voorwaartse reaksie is eksotermies. (3) 11

Downloaded from hlayiso.com 10842/24 6.4 MARKING CRITERIA/NASIENKRITERIA: (a) Calculate equilibrium moles Cr2O72- Bereken die ewewigsmol Cr2O72- (b) USE ratio 2CrO42-: 2H1+: 1Cr2O72- GEBRUIK die verhouding 2CrO42-: 2H1+: 1Cr2O72- (c) Calculate equilibrium moles of CrO42- and H1+ by subtraction. Bereken die ewewigsmol van CrO42- en H1+ deur aftrekking. (d) Divide equilibrium moles by 0,5 (mol·dm-3) Deel ewewigsmolle deur 0,5 (mol·dm-3) (e) Correct Kc expression (formulae in square brackets) Korrekte Kc uitdrukking (formules in blokhakies) (f) Substitution of Kc value into expression Vervang Kc waarde in Kc uitdrukking (g) Substitution of concentrations into correct Kc expression Vervang konsentrasies in korrekte Kc uitdrukking (h) Final answer: 10,25 mol Finale antwoord: 10,25 mol CrO42- H+ Cr2O72- Initial amount (moles) 11 x 0 Aanvanklike hoeveelheid (mol) Change in amount (moles) 9 9 4,5  b Verandering in hoeveelheid (mol) Equilibrium amount (moles) 2 c x-9 (9x0,5) Ewewigshoeveelheid (mol) =4,5  a Equilibrium concentration (mol·dm-3) 2÷0,5 (x-9)÷0,5 9 d Ewewigskonsentrasie (mol·dm-3) =4 =2x-18  No Kc expression, correct substitution Kc = [Cr2O72-] Max: ⅞ [CrO42-]2 [H1+]2 e  Wrong Kc expression Max: ⅝ (0,09) f = (9) (4) (2x -18)2 g 2  Geen Kc uitdrukking, korrekte (0,3)2 = (3)2 vervanging (4)2 (2x -18)2 Maks: ⅞ x = 10,25 (mol) h  Verkeerde Kc uitdrukking Maks: ⅝ (8) [18] 12

Downloaded from hlayiso.com 10842/24 QUESTION/VRAAG 7 7.1 7.1.1 pH = -log[H3O+]  or [H3O+] = 10-pH 4,2= -log[H3O+] [H3O+] = 10-4,2 [H3O+] = 6,31 x 10-5 mol·dm-3  (3) 7.1.2 WEAKER THAN  H2CO3 has smaller Ka value than H2SO4, indicating that it ionises incompletely/partially. OR  H2CO3 has a concentration of 0,02 mol·dm-3, but [H3O+] is calculated at 6,31 x 10-5 mol·dm-3. Therefore carbonic acid ionises incompletely/partially and is weaker than sulphuric acid that will ionise completely. OR  H2CO3 has a pH of 4,2. Sulphuric acid has a lower pH. (ACCEPT ANY ONE OF THE OPTIONS AS MOTIVATION)  SWAKKER AS  H2CO3 het 'n laer Ka waarde as H2SO4 wat aandui dat dit onvolledig ioniseer. OF  H2CO3 het 'n konsentrasie van 0,02 mol·dm-3, maar [H3O+] is bereken as 6,31 x 10-5 mol·dm-3. Dus het koolsuur/karbonaatsuur onvolledig ioniseer en is swakker as swawelsuur wat wel volledig ioniseer. OF  H2CO3 het ‘n pH van 4,2. Swawelsuur het 'n laer pH. (AANVAAR ENIGE EEN VAN DIE OPSIES AS MOTIVERING) (2) 7.2 7.2.1 Hydrolysis is the reaction of a salt with water.  Hidrolise is die reaksie van 'n sout met water. (1) 7.2.2 GREATER THAN 7/GROTER AS 7 (1) 7.2.3 CH3COO- + H2O → CH3COOH + OH-  (Accept/Aanvaar: CH3COONa + H2O → CH3COOH + NaOH)  Hydrolysis of acetate forms hydroxide ions OR  An increased concentration of hydroxide ions will make the solution more alkaline/increase pH OR  A weak acid (vinegar) reacting with a strong base (NaOH) results in an alkaline solution. (3) 13

Downloaded from hlayiso.com 10842/24  Hidrolise van asetaat vorm hidroksied-ione OF  'n Toename in die konsentrasie hidroksied-ione sal die oplossing meer alkalies maak/die pH verhoog OF  As swak suur (asyn) reageer met 'n sterk basis (NaOH) sal die oplossing alkalies wees 7.3 7.3.1 OPTION 1/OPSIE 1: OPTION 2/OPSIE 2: n(HCl) initially/aanvanklik = cV  n(HCl) initially/aanvanklik = cV  = (0,3)(0,025)  = (0,3)(0,025)  = 0,01 mol  = 0,0075 mol  (3) 7.3.2 OPTION 1/OPSIE 1: OPTION 2/OPSIE 2: Positive marking from 7.3.1 Positive marking from 7.3.1 n (NaOH) added/bygesit n (NaOH) added/bygesit = cV = cV = (0,1)(0,015)  = (0,1)(0,015)  = 0,0015 mol = 0,0015 mol n(NaOH): n(HCl) left /oor n(NaOH): n(HCl) left /oor 1:1  1:1  n(HCl) left /oor = 0,0015 mol n(HCl) left /oor = 0,0015 mol n(HCl) reacted/gereageer n(HCl) reacted/gereageer = n(HCl) initially/aanvanklik- n(HCl) left/oor = n(HCl) initially/aanvanklik- n(HCl) left/oor = (0,01) – (0,0015)  = (0,0075) – (0,0015)  = 0,0085 mol  = 0,006 mol  (4) 14

Downloaded from hlayiso.com 10842/24 7.3.3 OPTION 1/OPSIE 1: OPTION 2/OPSIE 2: Positive marking from 7.3.2 Positive marking from 7.3.2 n(Y2CO3): n(HCl) reacted with Y2CO3/ gereageer met Y2CO3 n(Y2CO3): n(HCl) reacted with Y2CO3/ 1:2 gereageer met Y2CO3 n(Y2CO3) = 0,0085÷2 1:2 = 0,00425 mol n(Y2CO3) = 0,006÷2 = 0,003 mol 𝑚 𝑛=𝑀 𝑚 0,5865 𝑛=𝑀 0,00425 =  𝑀 0,5865 = 138 g·mol-1 0,003 =  𝑀 = 195,5 g·mol-1 Mr (CO3) = 12+(3x16) = 60 g·mol-1 Mr (CO3) = 12+(3x16) = 60 g·mol-1 Mr (Y2) = 138 – 60  = 78 g·mol-1 Mr (Y2) = 195,5 – 60  = 135,5 g·mol-1 Ar (Y) = 78÷2 = 39 g·mol-1 Ar (Y) = 135,5÷2 = 67,75 g·mol-1 Element Y is POTASSIUM/K Element Y is KALIUM/K Element Y is ZINC/Zn (Ar=65) Element Y is SINK/Zn OR/OF Element Y is GALLIUM/Ga (Ar=70) OR/OF Element Y cannot be identified from periodic table/ Element Y kan nie vanaf die periodieke tabel identifiseer word nie (5) [22] 15

Downloaded from hlayiso.com 10842/24 QUESTION/VRAAG 8 8.1 8.1.1 NICKEL/NIKKEL  (1) 8.1.2 Cd → Cd2+ + 2e- Marking criteria/Nasienkriteria:  Reactants  Products  Reaktanse  Produkte   Ignore phases/Ignoreer fases  Ignore if charge omitted on electron./Ignoreer indien lading weggelaat op elektron.  Marking rule 6.3.10/Nasienreël 6.3.10: Cd2+ + 2e- ← Cd ( 2 2 ) Cd ⇌ Cd2+ + 2e- (1 ) 2 Cd2+ + 2e- ⇌ Cd ( 0 2 ) Cd ← Cd2+ + 2e- ( 0 ) 2 (2) 8.1.3 INCREASE  Increasing the Ni2+ ion concentration will favour the forward reaction TOENEEM Deur Ni2+ ioon konsentrasie te verhoog sal die voorwaartse reaksie bevoordeel word (2) 8.2 8.2.1 Cu(s) + 2Ag+(aq) → 2Ag(s) + Cu2+(aq)  bal  (3) 8.2.2 CATHODE/KATODE: (2Ag+ + 2e- → 2Ag) n(Ag+) reacted/gereageer = cV = (0,5)(0,2)  = 0,1 mol n(Cu) reacted/gereageer: n(Ag+) reacted/gereageer 1:2  n(Cu) reacted/gereageer = 0,1÷2 = 0,05 mol 16

Downloaded from hlayiso.com 10842/24 ANODE: (Cu → Cu2+ + 2e-) n(Cu) reacted/gereageer: n(Cu2+) released/vrygestel 1:1 2+ n(Cu ) released/vrygestel = 0,05 mol n(Cu2+) initially/aanvanklik = cV = (1)  x (0,2)  = 0,2 mol n(Cu2+) fina/finaal = n(Cu2+) initially/aanvanklik + n(Cu2+) released/vrygestel = (0,2) + (0,05)  = 0,25 mol c(Cu2+) final/finaal = n(Cu2+) final/finaal V 0,25 c = 0,2  = 1,25 mol·dm-3  (7) [15] 17

Downloaded from hlayiso.com 10842/24 QUESTION/VRAAG 9 9.1 The anode is the electrode where oxidation takes place.  Die anode is die elektrode waar oksidasie plaasvind. (2) 9.2 DO NOT MARK 9.2 X Concentrated sodium chloride/brine contains chlorine ions that will oxidise at electrode X/positive electrode OR Concentrated sodium chloride/brine enters the membrane cell on the anode side OR Chlorine ions are negative and will be attracted by a positive electrode (ANY ONE OF THE MOTIVATIONS FOR ONE MARK) X Gekonsentreerde natriumchloried/pekel bevat chloriedione wat sal oksideer by elektrode X/positiewe elektrode OF Gekonsentreerde natriumchloried/pekel dring die membraansel aan die anode kant binne OF Chloriedione is negatief en sal deur die positiewe elektrode aangetrek word (ENIGE EEN VAN DIE MOTIVERINGS VIR EEN PUNT) 9.3 9.3.1 Chlorine gas/Cℓ2  (ACCEPT: Hydrogen gas/H2) (2) Chloorgas/Cℓ2 (AANVAAR: Waterstof gas/H2) 9.3.2 DO NOT MARK 9.3.2 Water/H2O 9.4 2H2O + 2e- → H2 + 2OH- Marking criteria/Nasienkriteria:  Reactants  Products  Balancing  Reaktanse Produkte Balansering  Ignore phases/Ignoreer fases  Ignore if charge omitted on electron./Ignoreer indien lading weggelaat op elektron.  Marking rule 6.3.10/Nasienreël 6.3.10: 2H2O + 2e- ← H2 + 2OH- (1/3) H2 + 2OH- ⇌ 2H2O + 2e- (1/3) 2H2O + 2e- ⇌ H2 + 2OH- (2/3) H2 + 2OH- ← 2H2O + 2e- (3/3) (3) 9.5  H2O is a stronger oxidising agent than Na+ and is reduced (to H2) OR  + Na is a weaker oxidising agent than H2O and therefore H2O is reduced  H2O is 'n sterker oksideermiddel as Na+ en sal dus reduseer (tot H2) OF  Na is ‘n swakker oksideermiddel as H2O en sal dus reduseer H2O + (3) [10] TOTAL/TOTAAL: 150 18