Physical Sciences Grade 12 Paper 2 North West Mock 2025 Memo

education Department: Education North West Provincial Government REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SRSSSPS SSS SRC RRR RRR STARTER RESET RRR PHYSICAL SCIENCES: CHEMISTRY (P2) FISIESE WETENSKAPPE: CHEMIE (V2) MARKING GUIDELINES/NAS/ENRIGLYNE SEPTEMBER 2025 MARKS/PUNTE: 150 These marking guidelines consist of 13 pages Hierdie nasienriglyne bestaan uit 13 bladsy Copyright reserved/Kopiereg voorbehou Please turn over/Blaim om asseblief

Physical Sciences/P2 NW/September 2025 QUESTION/VRAAG 1 1.D vv (2) 12 BO vv (2) 13 D vv (2) 14 A vv (2) 165 B vv (2) 16 A vv (2) 17° D vv (2) 18 C vv (2) 19 CC vw (2) 110 D vv (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaim om asseblief

Physical Sciences/P2 NW/September 2025 QUESTION/VRAAG 2 24. A bond or an atom or a group of atoms that determine(s) the physical and chemical properties of a group of organic compounds. v v (2 or 0) ‘n Chemiese binding of ‘n atom of ‘n groep atome wat die fisiese en chemiese eienskappe van ’n groep organiese verbindings bepaal. (2 of 0) (2) 2.2 i -¢€-0- G-— 4 | (1) 2.3 231 FY (1) 2.3.2 Butanal/Butanaal VV (2) 2:33) 2,3-dichloro-4,5-dimethylheptanev v ¥ 2,3-dichloro-4,5-dimetielheptaan Marking criteria/Nasienkriteria: e Correct stem i.e. heptane. ¥ e All substituents (dichloro and dimethyl) correctly identified.” e IUPAC name completely correct including numbering, sequence, hyphens and commas. ¥ Korrekte stam d.i heptaan ¥ e Alle sunstituente (dichioro en dimetiel) korrek geindentifiseer IUPAC naam heeltemal korrek insluitende nommering, volgorde, koppeitekens en kommas. (3) 2.3.4 H/O\H H pfu \r ot H— r oF c— ro 4 Marking criteria/Nasienkriteria H ¥v H Functional groupY H=C =H Whole strucure corectY Funksionele groep ¥ Hele struktuur korrek ¥ (2) | H 2.35 H nae H y | H=c —C. Cc— c-C-C-H toot I | H\WYFS/H H Marking criteria/Nasienkriteria Flourines on C3 ¥ Whole strucure correct ¥ Fluore op C3 ~ Hele struktuur korrek “ Copyright reserved/Kopiereg voorbehou Please turn over/Blaim om asseblief

Physical Sciences/P2 NW/September 2025 2.4.1 Organic molecules with the same molecular formula, “but different structural formulae ~ Organiese molekules met dieselfde molekulére formule, maar verskillende struktuurformules V (2) 2.4.2 Propan-1-ol/ 1-propanol Vv and Propan-2-ol/ 2-propanolY Propan-1-ol/ 1-propanol en Propan-2-ol/ 2-propanol ¥ (2) 2.5 Aldehyde¥ and ketone “OR carboxylic acid and ester Aldehied “en ketoon ~ OF karboksielsuur en ester (2) [19] QUESTION/VRAAG 3 3.1 The temperature at which the vapour pressure of a substance equals 3.2 3.2.1 3.2.2 3.3 atmospheric pressurev ¥ Die temperatuur waarby die dampdruk van ’n stof gelyk is aan die atmosferiese druk. Markingcriteria/Nasienkriteria If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark./ndien enige van die onderstreepte frases in die korrekte konteks uitgetaat is, trek 1 punt af. (2) Carboxylic acid/Karboksielsuurv (1) Hydrogen bond or dipole-dipole bond.“ Waterstofbinding of dipoo!l-dipool binding (1) From A to D boiling point increasesY e Structure: The chain length/number of carbon atoms/molar mass/ surface area/ contact area increases V from compound A to D. e Intermolecular forces: The strength of the London forces/induced dipole/Van der Waals/ intermolecular forces increase. Vv e Energy: More energy needed to overcome/break the intermolecular forces from A to D.v OR Boiling point decrese from D to AY e Structure: The_chain length/number of carbon atoms/molar mass/ surface area/ contact area decreases from compound D to A.” e Intermolecular forces: The strength of the intermolecular forces/London forces decrease from D toAv « Energy: Lesser energy needed to weaken/overcome/break the intermolecular forces from D to A.”

Physical Sciences/P2 NW/September 2025 © =©Struktuur: Die kettinglengte/aantal koolstowwe/molére massa/reaksieopperviak/kontakarea neem toe van verbinding A na D. « Intermolekulére kragte: Die sterkte van die Londonkragte/geinduseerde dipoolkragte/Van der Waalskragte neem toe. e Energie: Meer energie benodig om intermolekulére kragte te oorkom/breek van A na D. Van D na A neem die kookpunt af e = Struktuur: Die kettinglengte/ aantal koolstowwe/molére massa/reaksieopperviak/kontakarea neem af van verbinding D na A.. e Intermolekulére kragte: Die sterkte van die Londonkragte/geinduseerde dipoolkragte/Van der Waalskragte neem af. e =©Energie: Minder energie benodig om intermolekulére kragte te oorkom/breek van D (4) naA. 3.4 C/butanoic acid vit has the highesi/higher boiling pointY C/butanoésuur ¥ dit het die hoogste/hoér kookpunt ¥ (2) 3.5 Pentanoic acid/Pentanoésuur v ¥ (2) 3.6 e Compound D/Pentanoic acid/carboxylic acid has two sites for hydrogen bond v Esters has dipole-dipole forcesY Intermolecular forces in compound D/pentanoic acid are stronger than Intermolecular forces in esters V OR Intermolecular forces in ester are weaker than in compound D e Less energy needed to overcome/break the intermolecular forces in ester than in compound D/pentanoic acid.” OR More energy needed to overcome/break the intermolecular forces in D than in ester. e Verbinding D/Pentanoésuur/carboksielsuur het twee plekke vir waterstofbinding ~ Ester het dipool-dipoolkragte ~ Intermolekulérekragte in verbinding D/pentanoésuur is sterker as die intermolekulérekragte in esters ~ OF Intermolekulérekragte in ester is swakker as in vebinding D Minder energie benodig om die intermolekulére kragte te oorkom/breek in esters as in verbinding D/pentanoésuur.¥ OF Meer energie benodig om die intermolekulére kragte te oorkom/breek in verbinding D/pentanoésuur as in esters.v (4) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaim om asseblief

Physical Sciences/P2 NW/September 2025 QUESTION/VRAAG 4 4.1.1 4.1.2 4.2 4.3.4 4.3.2 44 4.5.1 4.5.2 Copyright reserved/Kopiereg voorbehou Substitution/Hydrolysis reaction. ¥ Substitusie/Hidrolise reaksie (1) NOTE/NOTA IPenalise 1 mark if a hyphen is missing/Penaliseer as daar ‘n koppelteken luitgelaat is. 4-chloropropane/1-chioropropaan v ¥ Primary alcohol “ Alcohol in which the carbon atom bonded to the functional group(-OH) is attached to one carbon atom. ¥ “ Prim6re alkohol Alkohol waar die koolstofatoom gebind aan die funksionele groep(-OH) gebind is aan een ander koolstofatoom. Dehydration reaction/Dehidrasie reaksie V CH3CH2CH20H ¥ > CH3CHCH2Y + H20 ¥ Propane/Propaan V¥ (2) (3) (1) (3) (2) INOTE/NOTA INo mark for dilute H2SO4/Geen punt vir verdunde H2SO4 Sulphuric acid/Hydrogen sulphate/H2SO4/Phosphoric acid/HsPO4 ¥ Swawelsuur/Waterstofsulfaat/H2S O4/Fosforsuur/ H3PO4 Marking criteria/Nasienkriteria: Correct structural formula for methanoic acid ¥ Correct structural formula for alcohol ¥ The whole structural formula for ester correct ¥ H20 ¥ Korrekte struktuurformule vir metanoésuur ¥ Korrekte struktuurformule vir alkohol ¥ Die hele struktuurformule vir die ester is korrek ¥ H20 ¥ IF/AINDIEN Condensed structural formula used/Gekondenseerde struktuur formule gebruik Max/Maks % Please turn over/Blaim om asseblief (1)

Physical Sciences/P2 NW/September 2025 H H H v fe) I | | ll v H-C- C-— CC —O-H + H-C —0O-H—> | { [ H H H HOT Ir + HoOv H-C -C —-C -O-C-H¥Y | | I H H H (4) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaim om asseblief

Physical Sciences/P2 NW/September 2025 QUESTION/VRAAG 5 a4 ANY ONE: Ei e Change in concentration Y of products/reactants per unit time.v Change in amount/number of mole/volume/mass “of products/reactants per unit time.v Amount/number of mole/volume/mass of products formed/reactants usedv er unit time.” Rate of change in concentration/amount/number of moles/volume/mass.v ¥ (2 or 0) NIGE EEN: Verandering in konsentrasie “van produkte/reaktanse per eenheidtyd. ~ Verandering in hoeveelheid/getal mol/volume/massa “van produkte/reaktanse per eenheidtyd. ~ Hoeveelheid/getal mol/volume/massa produkte gevorm/reaktanse gebruik~ per eenheidtyd. ~ Tempo van verandering in konsentrasie/hoeveelheid/getal (2) mol/volume/massa. “(2 or 0) NOTE/NOTA Give the mark for per unit time only if in correct context of reaction rate Gee die punt vir per eenheidtyd slegs indien in korrekte konteks met reaksietempo. 5.2 Surface area/State of divisionv Kontakopperviak/Toestand van verdeeldheid (1) 5.3. Experiment/Eksperiment Iv (1) 5.4 e The surface area increase/Experiment 2 has a larger surface area/_more particles exposed (to oxygen gas)” More particles will have enough kinetic energy” More effective collision per unit time/second/Higher frequency of effective collisions ¥ Die kontakoppervlak vergroot/Eksperiment 2 het ‘n groter kontakopperviak/meer deeltjies word blootgestel (aan suurstofgas) Meer deeltiies het genoeg/voldoende kinetiese energie Meer effektiewe botsings per eenheidtyd/sekonde/hoér frekwensie van effektiewe botsings (3) 5.5.1 Marking criteria/Nasienkriteria: ® Am (7,36-0) v & At (5,2-0) ¥ e _Answer/Antwoord 1,42 (g:s"') Vv Rate/Tempo =Am/At = (7,36-0) v/ (5,2-0) v = 1,42 (g-s1) v (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaim om asseblief

Physical Sciences/P2 5.5.2 5.6 NW/September 2025 Marking criteria: e Formula nav M e Substituting 94 g-mol' vin n= M USE mol ratio n{K) = 2n(K20) ¥ Substitution of 39 g-mol"’ and ‘ . m reacting mole in n “" v 6,11 « Substitution of —— (100)” 6,92 e Answer 88,25%~” Range: 88,25% - 89,05% Nasienkriteria: © Formule n= ~ M e Vervanging van 94 g-mol? vin n=— M e GEBRUIK molverhouding n(K) = 2n(K2O) ~ e = Vervanging van 39 g:mor 4 m en reagerende mol in n= oe e Vervanging van oa (100) “ e Antwoord 88,25% ~“ Range: 88,25% - 89,05% m n=—v M 7,36 eo n= 94 = 0,0783 mol \ n(K) = 2 x 0,783 ¥ = 0,1566 mol n(K) =m/M m(K) = 0,1566 x 39” =6,11g % purity/suiwerheid = c= (100) v = 88,25% ¥ Negative Y Exothermic reaction/energy of the product < energy of the reactant/ energy of the reactant > energy of the product “ Negatief ¥ Eksotermiese reaksie/energie van die produkte < energie van die reaktanse/ energie van die reaktanse > energie van die produkte ¥ ACCEPT/AANVAAR More energy is released than absorbed/Meer energie word vrygestel as wat geabsorbeer word. (6) Combustion reaction. It releases energy/Verbrandingsreaksie. Stel energie vry (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaim om asseblief [18]

Physical Sciences/P2 NW/September 2025 QUESTION/VRAAG 6 6.1 6.2 6.3 6.4 The dynamic equilibrium when) the rate of the forward reaction equals the rate of the reverse reaction.v v (2 or 0) OR The stage in a chemical reaction when the concentration of reactants and products remain constant (Die dinamiese ewewig wanneer) die tempo van die voorwaartse reaksie gelyk is aan die tempo van die terugwaartse reaksie. (2 of 0) OF Die stadium in ’n chemiese reaksie waar die konsentrasie van die reaktanse en produkte konstant bly. (2) Endothermic/Eksotermies VV e Decrease in temperature favours the the reaction that decreases temperature (exothermic reaction) ¥ e The reverse reaction is favouredY ¢ Concentration of reactant/NO2 increases ¥ OR Concentration of products/NO and Oz decreases e Afname in temperatuur bevoordeel die eksotermiese reaksie “ e Die terugwaartse reaksie word bevoordeel~ e Konsentrasie van reaktanse/NO2 neem toe ~ OF (3) Konsentrasie van produkte/NO en O2 neem af CALCULATIONS USING NUMBER OF MOLES/BEREKENING DEUR MOL TE GEBRUIK Marking criteria: a) Change in n(NOz) = equilibrium nNOz) - initial n(NOz) = x - 0,6 ¥ b) Using ratio: NO: N20: O2=2:2:1v4 c) Equitibrium n(NO) = intial n{NO) + change in n(NO) Equilibrium n(Oz) = intial n(Oz) + change in n(O2) v d) Dividingby2 v e) Correct Ke expression in square bracketsY f) Substitution of equilibrium constant’ g) Substitution of equilibrium concentrations into Kc expressionsY h) Final answer. 1,48 mol“ Nasienkriteria: a) Verandering in n(NO2) = ewewig nNO2) - aanvanklike n(NOz) = x -0,6” b) gebruik verhouding: NO : N20 ; O2 = 2: 2:1 Vv c) Ewewig n(NO) = aanvanklike n(NO) + verandering in n(NO) Ewewig n(O2) = aanvanklike n(O2) + verandering in n(O2) v7 d) Korrekte Ke uitdrukking in vierkantige hakkiesY e) Deeldeur2 v f) Vervanging van ewewigskonstante ¥ g) Vervanging van ewewigskonsentrasies in Ke uitdrukkingY i) _ Finale antwoord. 1,48 mol~” Copyright reserved/Kopiereg voorbehou Please turn over/Blaim om asseblief

Physical Sciences/P2 Mol 2NOz2 2NO O2 Initial mole x 0 0 Aanvanklike mol = Change in mole x-0,6V C_| x-0,6 | x-0,6/2> v Verandering in mol ——— Equilibrium mole 0,6 Cc x-0,6 (x-0,6/2) |v | by ewewig ———o ncentration 0,3. C¢ | (x-0,6)/2 x-0,6/2 + 2 DV Ke = [NO}x [Oa _” [NO2}2 = (x= 0,62)? x (x-0,6/4) “ x= 1,48 molv (0,3 0,48 % QUESTION/VRAAG 7 iaiA dete Copyright reserved/Kopiereg voorbehou Solution with known concentration vv Oplossing met bekende konsentrasie [Marking criteria/Nasienkriteria: Formula c = oY or n=n/M or c=n/V Dividing by/dee/ deur 0,5dm3 ¥ Substituting/vervanging van 40 g-mol' and 4gv Answer/AntwoordY OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 Ic = m/MV ¥ n=n/M c= 4/(40x0,5) v vw = 4/40 ¥ c= 0,2 mol+dm? ¥ =0.1 mol _ |Any one/Enige een ¥ c=n/V = 0.1/0.5 ¥ = 0.2 mol:dm? ¥ NW/September 2025 (8) [15] (2) (4) Marking criteria/Nasienkriteria: e Formula 22Va _ Ma v C.V, n, e Substitution on CaVaand Substitution on CeVp~/ Substitution on n= cV and n=cV e Ratio 1:1” e Final answer 0,17 mol: dm3’v” e Formule £2Va _Ma vw ChVp any Vervanging in CaVa & vervanging in CpVp” Vervanging van n=cV enn =cV Verhouding 1:1” Finale antwoord 0,17 mol :dnv3 Please turn over/Blaim om asseblief

Physical Sciences/P2 NW/September 2025 OPTION 4/OPSIE 1: OPTION 2/OPSIE 2: CV, _ Ny n (NaOH) = cV ¥ 8. 2G. CaX35 ly = 0,2 xX 0,03 02x30 2 ca = 0,17 mol«dm3 v = 0,006 mol Ratio: 1:1 n (HCl) = 0,006 mol ¥ c=2 v 4 = 2,006 ~ 0,035 = 0,17 mol-dm3 ¥ (4) 7.14 NOv It is a salt of strong acid and a strong base. v Dit is ‘n sout van ‘n sterk suur en 'n sterk basis 7.2.1 Neutral ¥ [H3O*] = [OH] = 1x107?’moldm* or pH=7 ¥ ACCEPTIAANVAAR [H3sO*] = [OH] 7.2.2 |OPTION 1/OPSIE 1: OPTION 2/OPSIE 2: pOH = -log [OH] ¥ [HsO*][OH}] = 1x10-4 pOH = -log (3x 10-7) v| |[HsO*][3x10°7] = 1x10-4 pOH=6,52 v¥ [H30*] = 3.33x10°% (moledm*) | “~~ Any one pH + pOH = 14 pH +6,52=14 v pH = -log [H30*] pH = 14-6,52 = -log [3.33x10*] v pH=7,48 ¥ =7.48 V QUESTION/VRAAG 8 8.1 e Concentration must be 1 mol-dm?v « Temperature must be 25°C/298KY e Konsentrasie moet 1 mol-dnr? wees~ e Temperatuur moet 25°C/298K wees” 8.2 Sodium chloride/NaCf/Potassium chloride/KC? or any suitable electrolyte” Natriumchloried/ NaCt/Kaliumchloried/ KC#/ of enige geskikte elektroliet ACCEPT/AANVAAR Any metal nitrates/Enige metal nitrate QR2 Flactrada wihara radiictinn tavanlarca/ / (2) (2) (5) [19] (2) (1)

Physical Sciences/P2 8.4 8.5 8.6 8.7 Copyright reserved/Kopiereg voorbehou NW/September 2025 Marking guidelines/Nasienkriteria: © Pb*%+2e= Pb (1/,) ° Pb < Pb*+2e (2/5) ¢ Pb = Pb%+2e (%/,) © Pb Pb%+2e- — (0/5) e Ignore charges omitted on electron//Ignoreer indien lading weggelaat is op elektron e If charge (+) omitted on Pb** / Indien (+) lading weggelaat is op Pb?* Max. (1/5) Example/Voorbeeld: Pb? + 2e° > Phv¥ Pb** + 2e° > Ph vv E®cei! = E® reduction — E®oxidationY E®eu = -0,13¥ — (-0,91) ~ = 0,78V¥ Notes/Aantekeninge: e Accept any other correct formula from the data sheet./Aanvaar enige ander korrekte formule vanaf gegewensblad e Any other formula using unconventional abbreviations, ¢.g E®csn = E° oa — E® ra followed by correct substitutions/Enige ander formule wat onkonvensionele afkortings gebruik bv. E®sei = E° om — E® rm gevolg deur korrekte vervangings: % arking criteria/Nasienkriteria: ° oxidation’ double lineY —_ reduction V e oksidasie ¥ dubbellyn“ —_ reduksie ¥ Cr(s)/ Cr2* (aq)(1 mol-dm3)VIl ¥ Pb2*(aq)(1 mol-dm3)/ Pb(s)” Increases/VerhoogY Cr is oxidised to Cr?*v Cr is geoksideer na Cr?* Please turn over/Blaim om asseblief (2) (4) (3) (2) [16]

Physical Sciences/P2 QUESTION/VRAAG 9 9.1 Cell in which electric energy is converted to chemical energy.v Y (2 or 0) Sel waarin elektriese energie omgeskakel is na chemiese energie .v V (2 or 0) 9.2 Copper/Koper (Cu)v 9.3 Xv Oxidation takes place at X/ X loses electrons V Oksiedasie vind by X plaas/X verloor elektrone 9.4 Copyright reserved/Kopiereg voorbehou NW/September 2025 Marking guidelines/Nasienkriteria: © Cu +2e = Cu (1/5) * Cus Cu*+2e (2/5) ¢ Cu = Cu%*+2e (9/,) * CuCu*+2e = (%,) e Ignore charges omitted on electron/fgnoreer indien lading weggelaat is op elektron e If charge (+) omitted on Cu?* Indien lading (+) lading weggelaat is op Cu2* Max. (1 />) Example/Voorbeeld: Cu? + 2e > Cuv Cu?* + 2e > Cuvv Please turn over/Blaim om asseblief (2) (1) (2) (2)

Physical Sciences/P2 9.5 Marking criteria/Nasienkriteria: « ReactantsY ProductsY Ignore double arrows Ignore phases Marking rule 6.3.10 {gnoreer dubbel pyle {gnoreer fases Nasienreél 6.3.10 BalancingY Reaktanse~ Produkte~ Balansering~ Cu**(aq) + 2C£(aq)” > Cu(s)“+ Cha balv ACCEPT/AANVAAR CuClz (aq) ¥ > Cu(s)¥+ Clo balv Copyright reserved/Kopiereg voorbehou NW/September 2025 (3) [10] TOTAL/ITOTAAL: 150