Physical Sciences Grade 12 Paper 2 Gauteng Mock 2025 Memo

P(EBYSS GAUTENG PROVINCE PREPARATORY EXAMINATION VOORBEREIDENDE EKSAMEN 2025 MARKING GUIDELINES NASIENRIGLYNE 10842 PHYSICAL SCIENCES: CHEMISTRY FISIESE WETENSKAPPE: CHEMIE (PAPER/VRAESTEL 2) 11 pages/bladsye

10842/25 QUESTION/VRAAG 1 11 D vv 12 DB vv 13 A vv 14 C vv 1.5 Accept all answers Vv 16 D vv 17 A vv 18 Bovv 19 C vv 110 D vv QUESTION/VRAAG 2 2.1.1 Ketone/Ketoon ¥ 2.1.2 Haloalkane VY or alkyl halide/Haloalkane of alkielhalied 2.2 1-bromo-3-chloro-4-methylpentane/ 1-bromo-3-chloro-4-metielpentaan VV Vv Marking criteria/Nasienriglyne: ° Correct stem (pentane) ¥/korrekte stamnaam (pentaan) e All substituents (chloro, bromo and methyl) were correctly identified. V/alle substituente (chloro, bromo en metiel) is korrek geidentifiseer e IUPAC name is completely correct including numbering, sequence, hyphens and commas. ¥//UPAC naam is heeltemal korrek insluitend nommering, volgorde, koppeltekens en kommas 2.3 AY 2.4.1 H H H O Marking criteria/Nasienriglyne: u—¢—o_—¢__y e Functional group -COH on the first r | | carbon ¥/Funksionele groep -COH op H H H vv die eerste koolstof e Correct whole structure V/Korrekte volledige struktuur 2.4.2 Marking criteria/Nasienriglyne: \ if e Functional group -COOC- “/Funksionele H—C—O—C—H groep -COOC- | e Correct whole structure /Korrekte H vv volledige struktuur 2 (1) (1) (3) (1) (2) (2)

2.43 2.5 2.6.1 2.6.2 2.7 2.8 10842/25 a Marking criteria/Nasienriglyne: H oO oH e Correct stem — 3 carbons ¥/Korrekte 4} ty stam- 3 koolstowwe HF | i e Functional group -OH and a methyl on H-C—H the second carbon V the O must be i Vv bonded on the C/Funksionele groep -OH en metielgroep op tweede koolstof, die O moet op die C gebind wees Unsaturated indicates a mulitple bond between two C atoms. “ This compound has an oxygen atom, hydrocarbons do not have an Oxygen. v Onversadig dui aan om meervoudige bindings tussen C atome. Hierdie verbinding het ‘n suurstofatoom, koolwaterstowwe moet nie ‘n suurstof hé nie. Ev Av CsHi2 “+ 802 > 5CO2 + 6H20 V v_balancing/balansering Empirical forrmula/Empiriese formule = CH20 Molar mass/Molére massa = 150 g.mol* M(CH20) = 30 g.mot? Molecular formula/Molekulére formule = ? gb0 8 30 Molecular formula/Molekulére formule = 5 x CH2O =CsH1005 ¥ (2) (2) (1) (1) (3) (2) [21]

10842/25 QUESTION/VRAAG 3 3.1 3.2 3.3 3.4 3.5.1 3.5.2 3.5.3 The temperature at which the vapour pressure of a substance equals atmospheric pressure. ¥ “/ (2 or 0) Die temperatuur waarby die dampdruk van die stof gelyk is aan atmosferiese druk. (2) Different homologous series ¥ OR organic compounds with different functional groups Verskillende homoloé reekse OF organiese verbindings met verskillende funksionele groepe (1) e Both B and C have dipole-dipole forces. ¥ A has London forces. ¥ e The intermolecular forces in C are stronger than those in B and the weakest inA. ¥ e Therefore, C will require more energy to overcome the forces between the molecules thanin BorA. ¥ Beide B en C het dipool-dipoolkragte. A het London kragte. Die intermolekulére kragte in C is sterker as dié in B en die swakste in A. Daarom sal C meer energie benodig om die kragte tussen die molekules te oorkom as in B of A. (4) Marking criteria/Nasienkniteria Do not accept if the reason is given as inversely proportional. Moet nie aanvaar indien die rede gegee word as omgekeerd eweredig nie. Dv The highest boiling point (will have the lowest vapour pressure). V OR D has the strongest intermolecular forces and will need the most energy to weaken the intermolecular forces. OR D has the highest number of particles present in the vapour phase. Die hoogste kookpunt (sal die laagste dampdruk hé). OF D het die sterkste intermolekulére kragte en sal dus die meeste energie benodig om die intermolekulére kragte te verswak. OF D het die hoogste getal deeltjies in die dampfase. (2) Liquid/Vioeistof ¥ (1) (e} I —C—H ¥ (1) 2-Butanone OR butan-2-one OR butanone v/ 2-butanoon OF butan-2-oon OF butanoon NO marks for but-2-one GEEN punte vir but-2-oon (1) [12] 4

10842/25 QUESTION/VRAAG 4 4.4.1 Esterification OR Condensation/Verestering/Esterifikasie/Kondensasie V 4.1.2 Addition/AddisieY Accept: bromination/halogenation Aanvaar: brominering/halogenering 4.1.3 Elimination/Eliminasie V Accept: dehydration Aanvaar: dehidrasie noe ety woooe acenei 99-9994 H H H H HH OR/OF H H H HHH vv Marking criteria/Nasienriglyne: e Correct stem — six carbons “/Korrekte stam — ses koolstowwe e Functional group -OH on the 2nd or 3rd carbon/funksionele groep-OH op die 2de of 3de koolstof ¥ 4.3 HH OH OH an P (a es H-G-G——0 0" + Ht —e—e v— HE 6G 60-0 + H,0 haan ¥ ok on had d HHO Y v butan-1-ol / 1-butanol propanoic acid butan-1-ol / 1-butanol propanoéuur do not accept butanol/moenie butanol aanvaar nie Marking criteria/Nasienriglyne: e Whole structure correct for butan-1-ol ¥ /Korrekte volledige struktuur vir butan-1-ol e Correct name reactant 1/Korrekte naam reaktant 1 Whole structure correct for propanoic acid “/Korrekte volledige struktuur vir propanoésuur Correct name reactant 2/Korrekte naam reaktant 2 e Whole structure correct for ester “/Korrekte volledige struktuur vir ester Correct formula for water/Korrekte formule vir water 4.4 To act as a catalyst/Dit dien as 'n katalisator ¥ 4.5 Hydroxyl (group) /Hidroksiel(groep) ¥ (1) (1) (2)

4.6.1 4.6.2 4.6.3 4.6.4 10842/25 Water and a strong acid (suphuric acid/phosphoric acid) ~ Water en in sterk suur (swawelsuur/fosforsuur) i —¢— H i He — H =x I-90-0-=Z Marking criteria/Nasienriglyne: e Functional group -OH on the 2nd carbon “/Funksionele groep -OH op die 2de koolstof e Correct whole structure ¥/Korrekte volledige struktuur Hydration/Hidrasie ¥ By using their different boiling points. “/deur gebruik te maak van die verskillende kookpunte. QUESTION/VRAAG 5 5.1 5.2.1 5.2.2 5.2.3 5.2.4 The energy absorbed or released in a chemical reaction. YY (2 or 0) Die energie geabsorbeer of vrygestel in 'n chemiese reaksie. The concentration (of the sodium thiosulfate/NazS203.) ~/Konsentrasie (van die natriumtiosulfaat/Na2S2O3) Average rate of reaction/Gemiddelde tempo van reaksie _ 0,080 ¥ “35 v = 0,0023 (moldm*.s')v (2,3x10°) The sulphur/sulfur ¥ that forms is a solid and covers the visibility of the cross/ Precipitaite is formed/murky solution Die swawel wat vorm is 'n vaste stof en bedek die sigbaarheid van die kruis/ neersiag vorm/deinserige oplossing/ondeursigtige oplossing e As the concentration increases, the number of particles per unit volume increases Vv e More effective collisions per unit time W e The rate of reaction increase “ * Soos die konsentrasie toeneem, neem die aantal deeltiies per volume- eenheid toe e Meer effektiewe botsings vind per tydseenheid plaas e Die reaksietempo neem toe (1) (2) (1) (1) [18] (2) (1) (3) (1) (3)

5.3.1 5.3.2 10842/25 The minimum energy needed for a reaction to take place. VY (2 OR 0) Die minimum energie benodig vir 'n reaksie om plaas te vind. Potential energy/Potensiéle energie Course of reaction/Verioop van reaksie Marking criteria/Nasienkriteria: e The shape is endothermic. “/Die vorm is endotermies e Axes labelled correctly AND Ea (Activation energy) or AH indicated vif time on x-axis no marks /Asse korrek benoem EN Ea (aktiveringsenergie) AH aangedui / indien tyd op x-as geen punte QUESTION/VRAAG 6 6.1.1 It is a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction. vv Dit is 'n dinamiese ewewig wanneer die tempo van die voorwaartse reaksie gelyk is aan die tempo van die terugwaartse reaksie. According to Le Chatelier’s principle: e Anincrease in temperature favours the endothermic reaction v e The reverse reaction will be favoured v e The colour of the solution changes to yellow v Volgens Le Chatelier se beginsel: e —_'n Toename in temperatuur bevoordeel die endotermiese reaksie e Die terugwaartse reaksie sal bevoordee! word e Die kleur van die oplossing word geel (2) (2) [14] (2) (3)

6.2.1 6.2.2 10842/25 Increases ¥ « Adding concentrated HCl increases the concentration of the common ion H* , the system will react to decrease the concentration of the H* ions by favouring the forward reactionY *® More dichromate ions form, the colour of the solution changes to orange. v Verhoog e Die byvoeging van gekonsentreerde HCI verhoog die konsentrasie van die gemeenskaplike H* ioon, die sisteem sal reageer om die konsentrasie van die H* ione te verlaag deur die voorwaartse reaksie te bevoordeel. e Meer dichromaatione vorm, die kleur van die oplossing word oranje = [0a] Cc [co]2 (1,21 x 107) V © (1 x 10-2)2 K, = 10 Since the Ke new = KC original therefore the system is in equilibrium. v Ke value remains the same Aangesien die Ke nuwe = KC corspronktixe iS die sisteem in ewewig/Kc waarde bly dieselfde Marking criteria (a) Substitution of concentration into Kc expression. ¥ (b) n at equilibrium ¥ (c) Calculating the change in mole of CO v (d) Using the mole ratio of 2CO:1COz. ¥ (e) Calculating the new initial n of CO2 ¥ (f) subtracting mol ~ (g) final answer ¥ Nasienkriteria: (a) Vervanging van konsentrasie in Ke uitdrukking (b) n by ewewig (c) bereken die verandering in mol van CO (d) gebruik die mol verhouding van 2CO:1COz. (e) bereken die nuwe aanvanklike n van COz2 (f) aftrek van mol (g)_finale Antwoord OPTION 1/OPSIE 1 _ [C0] °~ {coP _ & 10 = ual (a) x=0,1 [C02] = [CO] = 0,1 moledm? (3) (3) (7)

10842/25 2CO COz Ninital/ Neanvanklik 11x10 | 1,445x10-v(g ANchange/ANverandering 0,089 v, (c) 0,0445v (ratio) (d) Nequilibrium/Newewig 0,1 0,1 ¥(b) Cequilibrium/Cewewig x x Nadded/bygevoeg = Nnew initial/nuwe aanvanklik — Noriginal mole at eq/oorspronklike mol by ew. 1,445 x 107 —1,21x 10° v (f) 0,143 mol “ (g) OPTION 2/OPSIE2 co CO2 Ratio 2 1 ni 1,1x10 1,21x10°3 + x An 1,1x107 + 2y 1,21x10°3 +x -y nt 0,1 0,1 c=n/V 0,1 0,1 1,1x10?%+ 2y = 0,1 2y = 0,089 y = 0,0445 moi 1,21x103 +x -y =0,1 1,21x103 + x — 0,0445 = 0,1 x = 0,14329 mol added [18]

10842/25 QUESTION/VRAAG 7 7.4.1 7.2.1 W2i2 A concentrated sodium hydroxide solution contains a large amount of base particles as compared to the water particles Vv in Gekonsentreerde natriumhidroksiedoplossing bevat 'n groot hoeveelheid basisdeeltiies in verhouding met die volume waterdeeltjies. (1) Decreases. Vv Since hydroxide ions increase, V the hydronium decreases because Kw is constant Vv Neem af. Omdat die hidroksiedione toeneem moet die hidronium ione verminder aangesien Ky konstant is (3) OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 pH = —log[H30*] Wor [Hs0*] =10°4 pH + pOH = 14 12v = —log[H3 0*] 12+pOH =14v pOH =2 {H,0*] = 1x107! pOH = -log[OH] ¥ & Ky = [H30*][0H7] ar = agIe] = 1xl04v : 1x 10-14 [OH~] = 1x10-* mol.dm=3v [OW] = Ty [OH-] = 1x 10-2 mol.dm-#v (4) 20,05 + 20,15 + 20,10 Average/Gemiddeld = : = 20,10 cm® ¥ (0,0201 dm) (1) POSITIVE MARKING FROM7.2.1/POSITIEWE NASIEN VANAF 7.2.1 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 nici = cV CV, my = 0,0958 (0,0201)v Vo To = 1,92558 x10 mol (0,0958)(20,1) 1 (20) 1 since/aangesien NNaOH : NHC! C2 = 0,0963 mol-:dm*? =1:1v “+ 1 NaoH = 1,92558 x 10° n + CNaOH = 5 v -3 = 192558x10-3 0,02 = 0,0963 mol.dm*? Marking criteria: (a) Calculating mol of HCl (b) Use ratio (c) Formula mark for cNaoH = = (d) Substitution of correct values (4) 10

10842/25 Nasienkriteria: (a) Bereken mol HCI (b) Gebruiksverhouding (c) Formulepunt cnaon = = (d)_Vervanging van korrekte waardes 7.2.3 N NaOH used/gebruik = CV = 0,0963 (0,02445) v = 2,354 x 10° mol 7 : _ 2,354x 1073 Ratio MnaoH : N H3P04 “+ NH3P04 used/gebruik = ——-—— 3 ied = 7,848 x 10“ mol in 10 cm? of solution/oplossingY . N H3Po4 in 250 cm of solution/oplossing = 2° (7,848 x 104)v ~ 10, = 0,01962 mol ~ m=nM = 0,01962(98) ¥ = 1,92276 g * % purity/suiwerheid =~" x 100% v = 19,23 %Y (Range/Reeks 19,11 — 19,30) (7) [20] QUESTION/VRAAG 8 8.1 (i) Mg/magnesium rod/magnesiumstaaf ¥ (ii) Cu/copper rod/koperstaaf Vv (iii) MgSO4/magnesium sulphate solution/magnesiumsulfaatoplossing V (iv) CuSO4/copper(Il)sulphate solution/koper(I!)sulfaatoplossing V (4) 8.2 Cu2* + 2e — Cuvv -1 mark if double arrow used/-7 indien dubbelpyle gebruik is (2) 8.3 E° ceil/se! = E°cathode/katode - E°anode/anode Y (any one of the three as on data sheet) =+0,34v - (-2,36)v = 2,7VV (4) 8.4 The salt bridge completes the circuit V OR It maintains the electrical neutrality of the two solutions by providing a pathway for the ions to be exchanged. OR separates the two half cells Die soutbrug voltooi die stroombaan OF Dit handhaaf die elektriese neutraliteit van die twee oplossings deur 'n pad te verskaf vir die ione om uitgeruil te word. OF skei die twee halfselle (1) 8.5.1 DECREASE/VERLAAG vv (2) 8.5.2 REMAINS THE SAME/BLY DIESELFDE ¥ (1) [14] 11

QUESTION/VRAAG 9 9.1 9.2 9.3 9.4 9.5 10842/25 A substance whose aqueous solution contains ions OR a substance that dissolves in water to form a solution that conducts electricity. vv in Stof waarvan die oplossing in water ione bevat OF 'n Stof wat in water oplos om 'n oplossing te vorm wat elektrisiteit gelei. Copper(Il)sulfate or copper(Il)chloride solution or any other copper(II) solution /koper(I)sulfaat of koper(II)chloride of enige ander koper‘(II) oplossing ¥ Cu — Cu? + 2e vv -1 mark if double arrow used/-1 indien dubbelpyle gebruik is Silver/Ag is a weaker reducing agent v than copper, v and silver will not be oxidised at the anode. Vv OR Copper is a stronger reducing agent than silver and silver will not be oxidised at the anode. Silwer/Ag is 'n swakker reduseermiddel as koper, en daarom sal silwer sal nie by die anode geoksideer word nie. OF Koper is ‘n sterker reduseermiddel as silwer en daarom sal silwer nie by die anode geoksideer word nie. If Cu2* is used/As Cu?+ gebruik word = 0,02519 mol Since 1 Cu atom gives off 2e/Aangesien 1 Cu-atoom 2e- afgee “ Ne- = 0,02519 (2) v = 0,05039 mol N 1 NA N = (0,05039)(6,02 x 1023)” = 3,034 x 102? electrons/elektrone Vv 12 Marking criteria: Y formula for calculating n Y substitution of molar mass 63,5 v¥ mole ratio used Y substitution of Na in formula v final answer Nasienkriteria: Y formule vir die berekenin van n ¥ vervanging van molére massa van 63,5 ¥ mol verhouding gebruik Y vervanging van Na in formule Y finale antwoord TOTAL/TOTAAL: (2) (1) (2) (3) (5) [13] 150