Physical Sciences Grade 12 P2 Gauteng Mock 2023 Memo (Afr)

PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN 2023 MARKING GUIDELINES NASIENRIGLYNE PHYSICAL SCIENCES: CHEMISTRY (PAPER 2) (10842) FISIESE WETENSKAPPE: CHEMIE (VRAESTEL 2) (10842) 16 pages/bladsye

10842/23 QUESTION 1/VRAAG 1 1.1 D  (2) 1.2 D  (2) 1.3 B  (2) 1.4 C  (2) 1.5 A  (2) 1.6 C  (2) 1.7 A  (2) 1.8 B  (2) 1.9 D  (2) 1.10 B  (2) [20] QUESTION 2/VRAAG 2 2.1 2.1.1 TERTIARY alcohol  The carbon atom, to which the functional group/hydroxyl is bonded, is bonded (directly) to three other carbon atoms.  TERSIÊRE alkohol  Die koolstofatoom, waaraan die funksionele groep/hidroksiel gebind is, is direk gebind aan drie ander koolstofatome.  (2) 2.1.2 2-methyl  butan-2-ol  2-metiel  butan-2-ol  (2) 2.1.3 H H H  functional and methyl group H C C C C H  whole structure H H  funksionele en metielgroep H C H  algehele struktuur H If double bond on first C and methyl group – maximum one mark Indien dubbelbinding op eerste C en metielgroep-maksimum 1 punt (2) 2.2 2.2.1 Organic compounds with the same molecular formula,  but different functional groups  Organiese verbindings met dieselfde molekulêre formule, maar verskillende funksionele groepe  (2) 2

10842/23 2.2.2 Ketone  Ketoon  (1) 2.2.3 Pentanal  (2 or 0) Pentanaal  (2 of 0) (2) 2.2.4 E (1)  reactant/reaktant C5H12  O2 → CO2 + H2O 2.3 C5H12 + 8O2 → 5CO2 + 6H2O  balancing/balansering (3) (Balancing mark can only be awarded if all reactants and products are also correct) [15] (Balansering punt kan slegs toegeken word indien alle reaktante en produkte korrek is) Award marks for multiple balancing e.g. Ken punte toe vir veelvoude van balansering, vb 2:16:10:12 3

10842/23 QUESTION 3 3.1 Melting point is the temperature at which the solid and liquid phases of a substance are at equilibrium.  (2 or 0) Smeltpunt is die temperatuur waarby die vaste- en vloeistoffases van ʼn stof in ewewig is.  (2 of 0) (2) 3.2 Branching/Chain length/Surface area  Vertakking/Kettinglengte/Oppervlakarea  (1) 3.3 These molecules are structural isomers because they have the same molecular formula  but different structural formulae  Hierdie molekules is struktuurisomere omdat dit dieselfde molekulêre formule  maar verskillende struktuurformules het. (2) 3.4 London forces or dispersion forces or momentary dipole forces or induced dipole forces London kragte  of dispersie kragte of oombliklike dipoolkragte of geïnduseerde dipoolkragte (1) 3.5 MARKING CRITERIA: A Compare structures.  B Compare the strength of intermolecular forces.  C Compare the energy required to overcome intermolecular forces.  D State the difference in melting point.  NASIENRIGLYNE: A Vergelyk strukture.  B Vergelyk die sterkte van intermolekulêre kragte.  C Vergelyk die energie benodig om intermolekulêre kragte te oorkom.  D Stel die verskil in smeltpunt.  OPTION 1/OPSIE 1 (Symmetry/simmetrie) MOLECULE A MOLECULE B MOLEKUUL A MOLEKUUL B A  Less spherical/symmetrical  More spherical/symmetrical Minder series/simmetries Meer series/simmetries B  Weaker intermolecular forces  Stronger intermolecular forces Swakker intermolekulêre Sterker intermolekulêre kragte kragte C  Less energy needed to  More energy needed to overcome IMF overcome IMF Minder energie nodig om IMK Meer energie nodig om IMK te te oorkom oorkom D  Lower melting point (than B)  Higher melting point (than A) Laer smeltpunt (as B) Hoër smeltpunt (as A) (4) 4

10842/23 OPTION 2/OPSIE 2(Chainlength/Kettinglengte) MOLECULE A MOLECULE B MOLEKUUL A MOLEKUUL B  Less branched  More branched A Minder vertakkings Meer vertakkings  Larger surface area  Smaller surface area Groter oppervlaksarea Kleiner oppervlaksarea  Longer chain length  Shorter chain length Langer kettinglengte Korter kettinglengte B  Stronger intermolecular forces  Weaker intermolecular forces Sterker intermolekulêre kragte  Swakker intermolekulêre kragte C  More energy needed to  Less energy needed to overcome IMF overcome IMF Meer energie nodig om IMK te Minder energie nodig om IMK oorkom te oorkom D  Higher melting point (than B)  Lower melting point (than A) Hoër smeltpunt (as B) Laer smeltpunt (as A) 3.6 A 211 (mmHg)  B 235 (mmHg) C 319 (mmHg)  (2) 3.7 The boiling point of (A) is higher than the boiling point of (B).  Therefore the higher the boiling point the less the vapour pressure.  Die kookpunt van (A) is hoër as die kookpunt van (B). Hoe hoër die kookpunt, hoe laer is die dampdruk.  OR/OF (A) has lower vapour pressure than (B) . Therefore the lower the vapour pressure the higher the boiling point.  (A) het ‘n laer dampdruk as (B).  Hoe laer die dampdruk, hoe hoër is die kookpunt.  (3) [15] 5

10842/23 QUESTION 4/ VRAAG 4 4.1 4.1.1 Elimination/Dehydration  Eliminasie/Dehidrasie  (1) 4.1.2 Substitution/Hydrolysis  Substitusie/Hidrolise  (1) 4.2  Catalyst/Katalisator: H2SO4/H3PO4   Heat/Hitte  (2) 4.3 4.3.1 Water  (Must be NAME/Moet NAAM wees) (1) 4.3.2 H2SO4/H3PO4  (Must be CHEMICAL FORMULA/Moet CHEMIESE FORMULE wees) (1) 4.4 4.4.1 H H H H \ / l l C = C  + HBr  → H – C – C – Br  / \ l l H H H H  Structural formula of ethene  HBr  Structural formula of bromoethane  Struktuurformule van eteen  HBr  Struktuurformule van bromoetaan (3) 4.4.2 To avoid the formation of the hydroxyl group/an alcohol  Om die vorming van die hidroksielgroep/ʼn alkohol te voorkom  (1) 4.5 4.5.1 Esterification / Condensation Esterifikasie / Kondensasie (1) 4.5.2 (Concentrated) H2SO4  (Must be CHEMICAL FORMULA) (Gekonsentreerde) H2SO4  (Moet CHEMIESE FORMULE wees) (1) 4.5.3 H H O H H H H l l ll l l l l  Functional group H–C–C–O–C–C–C–C–C–H  Whole structure l l I l l l  Funksionele groep H H H H H H  Hele struktuur (2) 4.5.4 Ethyl pentanoate  Etiel  pentanoaat  (2) 6

10842/23 4.6 CH3CH2Br + NaOH → CH2CH2 + NaBr + H2O     Condensed structural formulae of organic reactant  Condensed structural formulae of organic product (Accept CH2=CH2)  NaBr and H2O as inorganic products (Both correct for one mark) Penalised by one mark if structural formulae are used  Gekondenseerde struktuurformule van organiese reaktans  Gekondenseerde struktuurformule van organiese produk (Aanvaar CH2=CH2)  NaBr en H2O as anorganiese produkte (Beide korrek vir een punt) Penaliseer met een punt indien struktuurformules gebruik is. (3) 4.7 4.7.1 Hexane  (or any branched alkane with formula C6H14 e.g. 2-methylpentane) Heksaan  (of enige vertakte alkaan met formule C6H14 vb 2-metielpentaan) (1) 4.7.2 In shorter chained alkenes and branched alkanes the surface area is less  / will have weaker IMF / less activation energy / more flammable / higher vapour pressure Korter ketting alkene en vertakte alkane het verminderde oppervlakarea  / swakker IMK/minder aktiveringsenergie/meer vlambaar/ hoër dampdruk (2) [22] 7

10842/23 QUESTION 5/ VRAAG 5 5.1  Temperature  Concentration of HNO3  (Any ONE)  Temperatuur  Konsentrasie van HNO3  (Enige EEN) (1) 5.2 Rate of reaction / volume of gas per unit time  Reaksietempo/volume gas per eenheid tyd (1) 5.3 ΔV Average rate/gem tempo = Δt 48  42  = 150  120  = 0,2 cm3∙s-1  (3) 5.4 LESS THAN /KLEINER AS  The gradient is less steep at 250 s than at 120 s, indicating a decreased reaction rate.  Die gradient is minder steil by 250 s as by 120 s, dit dui op ʼn verminderde reaksietempo.  OR  There will be fewer reactants left after 250 s resulting in fewer effective collisions per unit time. Daar sal minder reaktante oor wees na 250 s wat minder effektiewe botsings per eenheid tyd tot gevolg het. (2) 5.5 DECREASE/AFNEEM  (1) 8

10842/23 5.6 OPTION 1 OPTION 2  (A) Dividing volume of H2 (from  (A) Dividing volume of H2 (from graph) graph) by 24 dm3 by 24 dm3  (B) Using mole ratio for H2 to Mg  (B) Using mole ratio for H2 to Mg  (C) Multiplying n(Mg) with Mr(Mg) in  (C) Subtract mole Mg used from initial the correct formula mole of Mg  (D) Subtract mass Mg used from  (D) Multiplying n(Mg) with Mr(Mg) in initial mass of Mg correct formula  (E) Final answer  (E) Final answer n(H2) produced = V n(H2) produced = V Vm Vm A A = 0,06  = 0,06  24 24 = 2,5 x 10-3 mol = 2,5 x 10-3 mol From the balanced equation: From the balanced equation: n(Mg) : n(H2) n(Mg) : n(H2) B 1:1  B 1:1  n(Mg) = 2,5 x 10-3 mol n(Mg) = 2,5 x 10-3 mol Calculating mass of magnesium used: Calculating initial moles of magnesium: m(Mg) = nM m(Mg) = nM = (2,5 x 10-3)(24)  C 2 = n (24) = 0,06 g = 0,083 mol Calculating mass of magnesium left: Calculating moles of magnesium left: ∆m (Mg) = 2 – 0,06  D ∆n (Mg) = 0,083 – (2,5 x 10-3)  C = 1,94 g  E = 0,08 mol Calculating mass of magnesium left: m(Mg) = nM D = (0,08) (24)  = 1,94 g E (5) 9

10842/23 OPSIE 1 OPSIE 2  (A) Deel volume H2 (vanaf grafiek)  (A) Deel volume H2 (vanaf grafiek) deur deur 24 dm3 24 dm3  (B) Gebruik molverhouding H2 tot Mg  (B) Gebruik molverhouding H2 tot Mg  (C) Vermenigvuldig n(Mg) met Mr(Mg)  (C) Trek n(Mg) gebruik af van  (D) Trek massa Mg af van aanvanklike n(Mg) aanvanklike massa Mg  (D) Vermenigvuldig n(Mg) met Mr(Mg)  (E) Finale antwoord  (E) Finale antwoord n(H2) geproduseer = V n(H2) geproduseer = V Vm Vm A A = 0,06  = 0,06  24 24 = 2,5 x 10-3 mol = 2,5 x 10-3 mol Vanaf die gebalanseerde vergelyking: Vanaf die gebalanseerde vergelyking: n(Mg) : n(H2) n(Mg) : n(H2) B 1:1  B 1:1  n(Mg) = 2,5 x 10-3 mol n(Mg) = 2,5 x 10-3 mol Bereken die massa magnesium Bereken die aanvanklike mol magnesium: gebruik: m(Mg) = nM m(Mg) = nM 2 = n (24) = (2,5 x 10-3)(24)  C = 0,083 mol = 0,06 g Bereken die mol magnesium oor: Bereken die massa magnesium oor: ∆n (Mg) = 0,083 – (2,5 x 10-3)  C ∆m (Mg) = 2 – 0,06  = 0,08 mol = 1,94 g  E Bereken die massa magnesium oor: m(Mg) = nM D = (0,08) (24)  E = 1,94 g 5.7  Powders have increased/larger surface area/ more contact sites than solid tablets.   More effective collisions per unit time occurs.   Poeiers het ʼn verhoogde/groter oppervlaksarea as soliede tablette.   Meer effektiewe botsings per eenheid tyd  vind plaas. (2) [15] 10

10842/23 QUESTION 6/ VRAAG 6 6.1 When the equilibrium in a closed system is disturbed, the system will reinstate a new equilibrium by favouring the reaction that will oppose the disturbance. (part marks)  Wanneer die ewewig in ʼn geslote sisteem versteur word, stel die stelsel ʼn nuwe ewewig in deur die reaksie wat die versteuring teenwerk, te bevoordeel. (2) 6.2 6.2.1  Adding more CH4 will increase the concentration of CH4.   The forward reaction is favoured.   The yield of H2 will increase.   Toevoeging van CH4 sal die konsentrasie van CH4 laat toeneem.   Die voorwaartse reaksie word bevoordeel.   Die opbrengs van H2 sal toeneem.  (3) 6.2.2  An increase in pressure favours the reaction that produces the lower number of moles/number of molecules.   The reverse reaction is favoured.   The yield of H2 decreases.   ʼn Toename in druk bevoordeel die reaksie wat die minste mol/hoeveelheid molekules produseer.   Die terugwaartse reaksie word bevoordeel.   Die opbrengs van H2 verminder.  (3) 6.3 The reaction is in (dynamic/chemical) equilibrium / the rates of the forward and reverse reactions are equal. Die reaksie is in (dinamiese/chemiese) ewewig waar die tempo van die voorwaartse en terugwaartse reaksies dieselfde is.  (1) 6.4 MARKING CRITERIA: a) Mole at equilibrium 0,6  b) Use mole ratio 1 : 1 : 1 : 3  c) Subtract CH4 , H2O and add H2  d) Divide by 2 dm3  e) Correct Kc expression (formulae in square brackets)  f) Substitution of concentrations into correct Kc expression  g) Final answer (range 1,82 – 1,83)  Option with concentration can also be used. NASIENRIGLYNE: a. Mol by ewewig 0,6  b. Gebruik molverhouding 1 : 1 : 1 : 3  c. Aftrek van CH4 , H2O en optel van H2  d. Deel deur 2 dm3  e. Korrekte Kc uitdrukking (formules in vierkantige hakies)  f. Vervanging van konsentrasies in korrekte Kc uitdrukking  g. Finale antwoord 1,8225/1,823/1,82  11

10842/23 CH4 H2O CO H2 Initial amount (moles) 1,4 1,2 0 0 Aanvangshoeveelheid (mol) Change in amount (moles) 0,6 0,6 0,6 1,8 ratio Verandering in hoeveelheid (mol) b) Equilibrium amount (moles) 0,8 0,6 0,6 1,8 -/+ c) Ewewigshoeveelheid (mol) a) Concentration at equilibrium 0,4 0,3 0,3 0,9 ÷2 d) Konsentrasie by ewewig Kc = [CO] [H2]3 e)  No Kc expression, correct [CH4] [H2O] substitution: Max. 6/7  No square brackets: Max. 6/7 = (0,3) (0,9)3 f) (0,4)(0,3)  Wrong Kc expression: Max. 4/7 = 1,82 g)  Geen Kc uitdrukking, korrekte substitusie: Maks. 6/7  Geen blok-hakies: Maks. 6/7  Verkeerde Kc uitdrukking: Maks. 4 /7 (7) 6.5 DECREASE/AFNEEM  (1) [17] 12

10842/23 QUESTION 7/VRAAG 7 7.1 STRONG (acid)  Ionises completely (in water)  STERK (suur)  Ioniseer volledig in water  (2) 7.2 - HSO4 ; SO42-  AND/EN H2O ; H3O+  (2) 7.3 It is a 2 proton donor/ Dit is ‘n 2-protonskenker OR It donates 2 H+ / H3O+ ions/protons.  Dit skenk 2 H+ / H3O+ ione/protone OR It ionises to form 2 moles of H+ / H3O+ions Dit ioniseer en vorm 2 mol of H+ / H3O+ione (2) 7.4 - HSO4  (1) 7.5 7.5.1 c= n v MARKING CRITERIA: 0,09 = /0,6 a) a) Substitution into concentration b) Using mole ratio correctly -3 = 0,15 mol∙dm c) pH formula n(H2SO4) : n(H3O+) d) Substitution in pH formula 1:2 e) Final answer 0,15 : 0,3 b) pH = -log[H3O+] c) = - log(0,3) d) = 0,52 e) (5) 7.5.2 OPTION 1 OPTION 2 m n=  M m = (80% of 11,2) = 8,96 g 11,2 n= m  = 56 M = 0,2 mol 8,96 n= /56  Thus 0,2 x 80%  n = 0,16 mol of KOH  n = 0,16 mol of KOH  (4) 7.5.3 Positive marking from 7.5.2/Positiewe nasien vanaf 7.5.2 ACIDIC/SUUR  MARKING GUIDELINE: Mol ratio H2SO4 : KOH  Mole ratio 1:2  Indicating excess 0,08 : 0,16   Acidic (3) 13

10842/23 OR H2SO4 : KOH 0,09 : 0,18 (not enough/nie genoeg) (the 0,09 is given) ∴ There will be an excess of H2SO4  (0,01 mol of H2SO4 in excess). ∴ Daar sal ʼn oormaat H2SO4  wees (0,01 mol H2SO4 in oormaat). [19] 14

10842/23 QUESTION 8/ VRAAG 8 8.1 YES  emf is greater than zero /emf is positive  Accept reducing and oxidizing agents are used JA  EMK is groter as nul/is positief Aanvaar: Reduseer- en oksideermiddels is gebruik (2) 8.2 Mn  (1) 8.3 E°cell = E°cathode – E°anode  NOTES: 1,05 = X - (-1,18)   Accept any other correct formula X = - 0,13 (V)  from the data sheet Aanvaar enige ander korrekte X = Pb formule vanaf die inligtingsbladsy  Any other formula using unconventional abbreviations, e.g. followed by correct substitutions: 4 Ecell =EOM – ERM /5 (5) 8.4 8.4.1 Concentration/Konsentrasie is 1 mol∙dm-3  (2) Temperature/Temperatuur 298 K/25 °C  8.4.2 Mn(s) l Mn2+(aq) (1 mol∙dm-3) ll Pb2+(aq) (1 mol∙dm-3) l Pb(s) (2)   Accept /Aanvaar Mn l Mn2+ ll Pb2+ l Pb   8.4.3 Pb2+ + 2e- → Pb  (1/2 if double arrows) (2) [14] 15

10842/23 QUESTION 9/ VRAAG 9 9.1 Electrical to chemical (energy)  Elektriese na chemiese (energie)  (1) 9.2 B (1) 2+ - 9.3 Cu → Cu + 2e  (if double arrows maximum ½) (2) 9.4 REMAIN THE SAME /DIESELFDE BLY (1) 9.5 Zn is a stronger reducing agent  than Cu  therefore Cu2+ ions will be reduced to Cu  Zn is ‘n sterker reduseermiddel  as Cu  daarom sal Cu2+ ione reduseer tot Cu  OR/OF Zn2+ is a weaker oxidising agent and will not be reduced to deposit on the cathode Zn2+ is ‘n swakker oksideermiddel en sal nie reduseer om op die katode neer te slaan nie OR/OF Zn will be oxidised to Zn2+ Zn sal oksideer tot Zn2+ (3) 9.6 9.6.1 m n(Cu) =  M 15 =  63,5 = 0,24 mol  (0,236 mol) (3) 9.6.2 Positive marking from 9.6.1/Positiewe nasien vanaf 9.6.1 nCu : 2n e- n = 2 x 0,236  = 0,472 mol of e-  (2) [13] TOTAL/TOTAAL : 150 16