Physical Sciences Grade 12 P2 NSC November 2025 Memo (Afr)

NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: CHEMISTRY (P2) FISIESE WETENSKAPPE: CHEMIE (V2) NOVEMBER 2025 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 22 pages. Hierdie nasienriglyne bestaan uit 22 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 2 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 C  (2) 1.2 B  (2) 1.3 A  (2) 1.4 D  (2) 1.5 A  (2) 1.6 A  (2) 1.7 A  (2) 1.8 B  (2) 1.9 C  (2) 1.10 D  (2) [20] QUESTION 2/VRAAG 2 2.1 2.1.1 D (1) 2.1.2 A and/en C  (1) 2.1.3 E and/en F  (1) 2.2 2.2.1 Marking criteria: • Correct stem i.e. butanone.  • Substituents (methyl) correctly identified.  • IUPAC name is completely correct including numbering, sequence, hyphens and commas.  Nasienkriteria: • Korrekte stam, d.i. butanoon. • Substituente (metiel) korrek geïdentifiseer. • IUPAC-naam heeltemal korrek insluitende nommering, volgorde, koppeltekens en kommas. 3-methylbutan-2-one /3-methylbutanone 3-metielbutan-2-oon/3-metielbutanoon ACCEPT/AANVAAR 3-methyl-2-butanone/ methylbutanone/3-metiel-2-butanoon/metielbutanoon (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 3 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 2.2.2 Marking criteria: • Correct stem i.e. heptane.  • Substituents (dichloro and dimethyl) correctly identified.  • IUPAC name is completely correct including numbering, sequence, hyphens and commas.  Nasienkriteria: • Korrekte stam, d.i. heptaan. • Substituente (dichloro en dimetiel) korrek geïdentifiseer. • IUPAC-naam heeltemal korrek insluitende nommering, volgorde, koppeltekens en kommas. 2,6-dichloro-2,5-dimethylheptane  2,6-dichloro-2,5-dimetielheptaan (3) 2.2.3 Marking criteria: • Correct stem i.e. hexene.  • Substituents (ethyl and methyl) correctly identified.  • IUPAC name is completely correct including numbering, sequence, hyphens and commas.  Nasienkriteria: • Korrekte stam, d.i. hekseen. • Substituente (etiel en metiel) korrek geïdentifiseer. • IUPAC-naam heeltemal korrek insluitende nommering, volgorde, koppeltekens en kommas. 3-ethyl-2-methylhex-2-ene /3-ethyl-2-methyl-2-hexene 3-etiel-2-metielheks-2-een/3-etiel-2-metiel-2-hekseen (3) 2.2.4 Marking criteria/Nasienkriteria: Marking criteria/Nasienkriteria: • Hydroxyl group on the 1st C-atom.  • Hydroxyl group on the 2nd C-atom.  Hidroksiel groep op 1ste C-atoom. Hidroksiel groep op 2de C-atoom. • Whole structure correct.  • Whole structure correct.  Hele struktuur korrek. Hele struktuur korrek.  H H H H H H H H  H O C C C C H H C C C C H H H H H H O H H H  IF/INDIEN • More than one functional group/wrong functional group: 0⁄ per molecule/molekule 2 Meer as een funksionele groep/foutiewe funksionele groep: • Condensed structural formulae used Gekondenseerde struktuurformules gebruik: Max/Maks. 2⁄4 • Bond between O and H not shown, accept. Binding tussen O en H nie gewys nie, aanvaar. (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 4 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 2.3.1 Combustion/oxidation/Verbranding/oksidasie  (1) 2.3.2 Marking criteria: Nasienkriteria: (a) Volume O2 and C3H8 used.  (a) Volume O2 en C3H8 gebruik.  (b) Volume CO2 and H2O produced (b) Volume CO2 en H2O gevorm (c) Volume O2 remaining  (c) Volume O2 wat oorbly  (d) Addition of the three volumes  (d) Optel van drie volumes  (e) Correct final answer 66 cm3 (e) Korrekte finale antwoord 66 cm3 or 0,066 dm3  of 0,066 dm3  OPTION 1/OPSIE 1: V(O2)used/gebruik = 5V(C3H8) = (5)(8)  (a) = 40 cm3 V(CO2)formed/gevrom = 3V(C3H8) = (3)(8) = 24 cm3 V(H2O)formed/gevrom = 4V(C3H8) = (4)(8)  (b) = 32 cm3 V(O2)unused = 50 – 40 = 10 cm3  (c) Volume gas in the container = 10 + 24 + 32  (d) Total volume gas in the container = 66 cm3  (e) OPTION 2/ OPSIE 2: C3H8 O2 CO2 H2O Initial volume (cm3) 8 50 0 0 Aanvanklike volume (cm3) Change volume (cm3)  (a) -8 -40 24 32  (b) Verandering in volume (cm ) 3 Final volume (cm3)  (c) 3 0 10 24 32 Finale volume (cm ) Total volume of gases in container 10 + 24 +32  (d) 3 (cm ) = 66 cm3  (e) Totale volume gas in houer (cm ) 3 ACCEPT/AANVAAR: If subtraction not shown, give mark for the 10 cm3/ Indien aftrekking nie getoon, gee punt vir 10 cm3. n(C3H8) : n(O2) : n(CO2) : n(H2O) 1 : 5 : 3 : 4  (a)  (b) V= 8 40 24 32  (c) Total volume = 10 + 24 + 32  (d) = 66 cm3  (e) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 5 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne ACCEPT/ AANVAAR: Moles calculated using 22,4 dm3 or any other molar gas volume. Mol bereken deur 22,4 dm3 of ander molêre gasvolume te gebruik. Marking criteria: Nasienkriteria: (a) V(O2) and V(C3H8)/ (a) V(O2) en V(C3H8)/ n(O2) and n(C3H8) used.  n(O2) en n(C3H8) gebruik.  (b) V(CO2) and V(H2O)/ (b) V(CO2) en V(H2O)/ n(CO2) and n(H2O) produced.  n(CO2) en n(H2O) gevorm. (c) V(O2)/ n(O2) remaining  (c) V(O2)/ n(O2) wat oorbly  (d) Addition of the three volumes/moles  (d) Optel van drie volumes/moles  (e) Correct final answer 66 cm3 (e) Korrekte finale antwoord 66 cm3 or 0,066 dm  3 of 0,066 dm3  V V n(C3H8) = n(C3H8) = Vm Vm 0,008 0,008 = = 22,4 22,4 = 3,571×10-4 mol = 3,571×10-4 mol n(O2)change = (5)3,571 x 10-4 n(O2)change = (5)3,571 x 10-4  (a) = 1,7855 x 10-3 mol = 1,7855 x 10-3 mol  (a) V V n(O2)change = n(O2)ini = Vm Vm V 0,05 1,7855 × 10-3 = = 22,4 22,4 V(O2)change = 0,04 dm3 = 2,232 x 10-3 mol V(O2)remaining = 0,05 – 0,04 n(O2)remaining = 0,01 dm3  (c) = 2,232 x 10-3 – 1,7855 x 10-3 n(CO2) = (3) 3,571×10-4 = 4,465 x 10-4 mol  (c) = 1,0713 x 10-3 mol n(CO2) = (3)3,571×10-4 V = 1,0713 x 10-3 mol n(CO2) = Vm V n(H2O) = (4)x 3,571×10-4  (b) 1,0713 x 10-3 = = 1,43 x 10-3 mol 22,4 V(CO2) = 0,024 dm3 nTotal n(H2O) = (4)3,571×10 -4 = 4,465 x 10-4 + 1,0713 x 10-3 +  (b) 1,43 x 10-3  (d) = 1,43 x 10-3 mol = 2,95 x 10-3 mol V n(H2O) = V Vm nTotal = V Vm 1,43 x 10-3 = V 22,4 2,95 x 10-3 = V(H2O) = 0,032 dm3 22,4 VTotal = 0,066 dm3  (e) VTotal = 0,01 + 0,024 + 0,032  (d) = 0,066 dm3  (e) (5) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 6 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 3/VRAAG 3 3.1 A series of organic compounds that can be described by the same general formula.  (1 OR 0) OR A series of organic compounds in which one member differs from the next by a CH2 group. 'n Reeks organiese verbindings wat deur dieselfde algemene formule beskryf kan word. (1 OF 0) OF 'n Reeks organiese verbindings waarin die een lid van die volgende verskil met 'n CH2-groep (1) 3.2 Ester  and carboxylic acid /Ester en karboksielsuur (2) 3.3 3.3.1 H H H O Marking criteria/Nasienkriteria: (a) Functional group correct.  H C C C C O H  Funksionele groep korrek.  (b) Whole structure correct.  H H H Hele struktuur korrek. (2) 3.3.2 Methyl propanoate/Propyl methanoate/Ethyl ethanoate  (2 OR/OF 0) Metielpropanoaat/Propielmetanoaat/Etieletanoaat (2) 3.4.1 Hydrogen bonds / Waterstofbindings  (1) 3.4.2 Dipole-dipole forces /Dipool-dipoolkragte  (1) 3.5 A The hydrogen bond is stronger than the dipole-dipole force.  OR The dipole-dipole force is weaker than the hydrogen bonds. OR Compound A has stronger intermolecular forces (than B). OR Compound B has weaker intermolecular forces (than A). Die waterstofbinding is sterker as die dipool-dipoolkrag. OF Die dipool-dipoolkragte is swakker as die waterstofbinding. OF Verbinding A het sterker intermolekulêre kragte (as B). OF Verbinding B het swakker intermolekulêre kragte (as A). (2) 3.6 Decreases/Afneem  (1) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 7 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1 4.1.1 2-bromo✓butane✓/2-bromobutaan (2) 4.1.2 NaBr/Sodium bromide/Natriumbromied ✓ (1) 4.1.3 Addition/Addisie ✓ Hydrohalogenation/Hydrobromination/Hydrohalogenering/Hidrobrominering ✓ (2) 4.1.4 (Concentrated) sulphuric acid/ H2SO4/Phosphoric acid/H3PO4/ (Gekonsentreerde) swaelsuur/Fosforsuur ✓ (1) 4.1.5 H H H H Marking criteria/Nasienkriteria:  (a) Correct functional group.  Funksionele groep korrek. H C C C C H (b) Whole structure correct.  Hele struktuur korrek. H  H IF/INDIEN • More than one functional group/wrong functional group: • Meer as een funksionele groep/foutiewe funksionele groep: 0⁄ 2 • Correct condensed formula: Korrekte gekondenseerde formule Max: 1⁄2 (2) 4.1.6 Concentrated strong base/Gekonsentreerde sterk basis  Concentrated/Gekonsentreerde NaOH/KOH/LiOH (1) 4.2 4.2.1 Marking criteria/Nasienkriteria If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark./Indien enige van die onderstreepte frases in die korrekte konteks uitgelaat is, trek 1 punt af. The underlined phrases must be in the correct context. / Die onderstreepte frases moet in die korrekte konteks wees. The chemical process/reaction in which longer chain hydrocarbon/alkane molecules/ are broken down to shorter (more useful) molecules.  Die chemiese proses/reaksie waarin langer kettingkoolwaterstof/alkaan- molekule afgebreek word in korter (meer bruikbare) molekules. (2) 4.2.2 Decolourisation/colour fades/becomes lighter in colour/colourless  Ontkleuring/kleur raak dowwer/word ligter van kleur/kleurloos (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 8 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 4.2.3 Marking criteria/Nasienkriteria: H  H (a) Correct functional group.  C C C H Funksionele groep korrek. H H H (b) Whole structure correct.  Hele struktuur korrek. (2) 4.2.4 X / C3H6 / Propene / Propeen  • X is unsaturated/has a double bond/is an alkene.  ANY ONE • X undergoes addition.  OR Alkenes are more reactive than alkanes/Unsaturated compounds react faster than saturated compounds. Addition reaction is faster than substitution Addition reaction does not need UV/light. • X is onversadig/besit ‘n dubbelbinding/is ‘n alkeen. ENIGE EEN • X ondergaan addisie. OF Alkene is meer reaktief as alkane./Onversadigde verbindings reageer vinniger as versadigde verbindings. Addisiereaksie is vinniger as substitusie. Addisiereaksie benodig nie UV/lig. (3) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 9 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 NOTE/LET WEL Give the mark for per unit time only if in context of reaction rate. Gee die punt vir per eenheid tyd slegs indien in konteks met reaksietempo. ANY ONE: • Change in concentration  of products/reactants per (unit) time.  • Change in amount/number of moles/volume/mass of products or reactants per (unit) time. • Amount/number of moles/volume/mass of products formed/reactants used per (unit) time. • Rate of change in concentration/amount/number of moles/volume/ mass. (2 or 0) ENIGE EEN: • Verandering in konsentrasie van produkte/reaktanse per (eenheid) tyd. • Verandering in hoeveelheid/getal mol/volume/massa van produkte of reaktanse per (eenheid) tyd. • Hoeveelheid/getal mol/volume/massa van produkte gevorm/reaktanse gebruik per (eenheid) tyd. • Tempo van verandering in konsentrasie/ hoeveelheid/aantal mol/ volume/massa. (2 of 0) (2) 5.2.1 10 (s)  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 10 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 5.2.2 Marking criteria Nasienkriteria: (a) Calculate the change: (a) Bereken die verandering: [O2]final – [O2]initial / [O2]finale – [O2]aanvank / n(O2)final – n(O2)initial / n(O2)finale – n(O2)aanvanklik / n(CO2)final – n(CO2)initial.  n(CO2)finale – n(CO2)aanvanklik  Accept: 0,265 to 0,27 for [O2]initial Aanvaar: 0,265 tot 0,27 vir [O2]aanvank (b) Substitute 10 s in rate formula.  (b) Vervang 10 s in tempoformule.  (c) Multiply rate / concentration of O2 by (f) Vermenigvuldig tempo/konsentrasie 3 dm3.  O2 met 3 dm3.  (d) USE mol ratio: (c) Gebruik molverhouding: n(O2) : n(CO2) = 1 : 2  n(O2) : n(CO2) = 1 : 2  (e) Final correct answer (d) Finale korrekte antwoord = 0,072 (mol∙s-1)  = 0,072 (mol∙s-1)  RANGE: 0,06 – 0,072 GEBIED: 0,06 – 0,072 OPTION 1/OPSIE 1: OPTION 2/OPSIE 2: Δc(O2 ) Rate/Tempo = - ∆c(O2) = 0,27 – 0,15 (a) Δt = 0,12 mol∙dm-3 0,15 - 0,27  (a) =-( ) 10 - 0 (b) ∆n(O2) = cV = 0,012 mol∙dm-3∙s-1 = 0,12(3) (c) = 0,36 mol Rate/Tempo (O2) in mol∙s-1 = cV ∆n(CO2) = 2n(O2) (d) = (0,012)(3) (c) = 2(0,36) = 0,036 mol∙s-1 = 0,72 mol Rate/Tempo (CO2) = 2 x rate (O2) ∆n Rate(CO2) = = 2 x 0,036 (d) ∆t = 0,072 (mol∙s-1)(e) 0,72 = 10 - 0 (b) = 0,072 (mol∙s-1) (e) OPTION 3/OPSIE 3: n(O2)at 0s = cV = 0,27 x 3 (c) = 0,81 mol n(O2)at 10s = cV = 0,15 x 3 = 0,45 mol ∆n Rate (O2) = - ∆t 0,45 - 0,81  (a) = -( )(b) 10 - 0 = 0,036 mol·s-1 Rate (CO2) = 2 x rate (O2) = 2 x 0,036 (d) -1 = 0,072 (mol·s ) (e) (5) (e) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 11 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 5.2.3 O2/Oxygen/Suurstof  (1) 5.2.4 Increases/Neem toe  Higher reaction rate./Concentration of reactants are higher.  Hoër reaksietempo./Konsentrasie van die reaktanse is hoër. ACCEPT/AANVAAR: pressure increased/druk verhoog. (2) 5.3.1 Absorption  Absorpsie (1) 5.3.2 Marking criteria/Nasienkriteria: If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark./Indien enige van die onderstreepte frases in die korrekte konteks uitgelaat is, trek 1 punt af. The underlined phrases must be in the correct context. / Die onderstreepte frases moet in die korrekte konteks wees. Unstable (high energy) transition state from reactants to products.  Onstabiele (hoë energie) oorgangs-toestand van reaktanse na produkte ACCEPT/AANVAAR: Unstable (high energy) transition state between reactants and products.  Onstabiele (hoë energie) oorgangs-toestand tussen reaktanse en produkte. (2) 5.3.3 • Catalyst provides an alternative path with lower activation energy / lowers the activation energy.  • More particles have sufficient (kinetic) energy / kinetic energy greater (or equal to) activation energy.  • More effective collisions per unit time/second.  OR Higher frequency of effective collisions. • Katalisator verskaf 'n alternatiewe roete met 'n laer aktiveringsenergie/ Verlaag die aktiveringsenergie. • Meer deeltjies het genoeg (kinetiese) energie/kinetiese energie groter (of gelyk aan) aktiveringsenergie. • Meer effektiewe botsings per eenheid tyd/sekonde. OF Hoër frekwensie van effektiewe botsings. (3) 5.3.4 Remains the same/Bly dieselfde  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 12 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 5.3.5  Marking criteria:  Curve Y • Both curves start at origin and have Number of particles / correct shape with peaks at same Ek.  • Peak of curve Y must be higher than curve X with peaks at same Ek.  Aantal deeltjies Curve X Nasienkriteria: • Beide kurwes begin by die oorsprong en het dieselfde vorm met maksimums by dieselfde Ek. • Maksimum van kurwe Y moet hoër wees as kurwe X met maksimums Kinetic energy/ Kinetiese energie by dieselfde Ek. IF/INDIEN: • Both curves not labelled./ 0⁄ Beide kurwes nie benoem 2 • Curves intersect at any other point, beside the origin. Kurwes kruis by enige ander punt as oorsprong. Max: 1⁄2. IF/INDIEN:  Curve X Number of particles / Aantal deeltjies Curve Y Max/Maks: 1⁄2 Kinetic energy/ Kinetiese energie (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 13 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 6/VRAAG 6 6.1 Marking criteria/Nasienkriteria: 6.1.1 If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark./Indien enige van die onderstreepte frases in die korrekte konteks uitgelaat is, trek 1 punt af. The underlined phrases must be in the correct context. / Die onderstreepte frases moet in die korrekte konteks wees. When the equilibrium in a closed system is disturbed, the system will re-instate a new equilibrium by favouring the reaction that will cancel/oppose the disturbance.  Wanneer die ewewig in 'n geslote sisteem versteur word, sal die sisteem 'n nuwe ewewig instel deur die reaksie te bevoordeel wat die versteuring kanselleer/teenwerk. (2) 6.1.2 (Mass) decreases./(Massa) Afneem.  (1) — — 6.1.3 • Decrease in amount of OH ions /concentration of OH - ions, favours the — reaction that increases the amount/concentration of OH ions.  OR — Acid/ HCℓ/H+ reacts with OH - ions. • The forward reaction is favoured  OR The amount/concentration of the products increases. — — • 'n Afname in hoeveelheid OH --ione /konsentrasie OH -ione bevoordeel — die reaksie wat die hoeveelheid/konsentrasie van OH -ione laat toeneem. OF — Suur /HCℓ/H+ reageer met OH ione. • Die voorwaartse reaksie is bevoordeel. OF Die hoeveelheid/konsentrasie van die produkte neem toe. (2) 6.2.1 Endothermic/Endotermies  (1) 6.2.2 • With an increase in the temperature the Kc value increases.  • The concentration of the products increases. OR Concentration of reactants decreases. OR The forward reaction is favoured.  • (According to Le Chatelier's principle) an increase in temperature favours the endothermic reaction.  • Met 'n toename in temperatuur neem die Kc-waarde toe. • Die konsentrasie van die produkte neem toe. OF Konsentrasie van die reaktante neem af. OF Voorwaartse reaksie is bevoordeel. • (Volgens Le Chatelier se beginsel) sal 'n toename in temperatuur die endotermiese reaksie bevoordeel. (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 14 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 6.2.3 CALCULATIONS USING MOLES BEREKENINGE WAT MOL GEBRUIK Marking criteria: 70 (a) Calculate number of moles NH4HS ( )  OR 1,37 moles 51 (b) USING RATIO: NH4HS : NH3 : H2S : = 1 : 1 : 1  (c) Calculate c(NH3) and c(H2S) at equilibrium (divide equilibrium moles by 3)  (d) Correct Kc expression  (e) Substitute Kc = 18 x 10-2  (f) n(NH4HS)eq = n(NH4HS)in - n(NH4HS)change OR m(NH4HS)eq = m(NH4HS)in - m(NH4HS)change  m (g) Substitute 51 in n =  M (h) CORRECT final answer: m = 5,61 g  Range: 4,96 – 5,74 g Nasienkriteria: 70 (a) Bereken aantal mol NH4HS ( )  OF 1,37 mol 51 (b) GEBRUIK VERHOUDING: NH4HS : NH3 : H2S : = 1 : 1 : 1  (c) Bereken c(NH3) en c(H2S) by ewewig (deel ewewig mol met 3)  (d) Korrekte Kc uitdrukking  (e) Vervang Kc = 18 x 10-2  (f) n(NH4HS)eq = n(NH4HS)in - n(NH4HS)change OF m(NH4HS)eq = m(NH4HS)in - m(NH4HS)change  m (g) Vervang 51 in n =  M (h) KORREKTE finale antwoord: m = 5,61 g  Gebied: 4,95 – 5,74 g OPTION 1/OPSIE 1: m n= M 70 (a) = 51 NH4HS(s) NH3(g) H2S(g) Initial amount (mol) Aanvanklike hoeveelheid (mol) 1,37 0 0 Change (mol) Verandering (mol) x x x (b) Equilibrium amount (mol) Ewewig hoeveelheid (mol) 1,37 - x x x Equilibrium concentration (mol∙dm-3) x x Ewewigskonsentrasie (mol∙dm-3) (c) 3 3 Kc = [NH3][H2S] (d) No Kc expression, correct substitution (e) x 2 Geen Kc- uitdrukking, korrekte substitusie: 18 x 10-2 = ( ) Max./Maks. 7⁄8 3 x = 1,27 Wrong Kc expression/ n(NH4HS)eq = 1,37 – 1,27 (f) = 0,1 mol Verkeerde Kc-uitdrukking: Max./Maks. 6⁄8 m(NH4HS)eq = nM = 0,1 x 51 (g) = 5,1 g (h) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 15 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 2/OPSIE 2: m n= Kc = [NH3][H2S] (d) M 70 (e) 18 x 10-2 = x2 = (a) x = 0,42 51 NH4HS(s) NH3(g) H2S(g) Initial amount (mol) Aanvanklike hoeveelheid (mol) 1,37 0 0 Change (mol) (b) Verandering (mol) 1,26 1,26 1,26 Equilibrium amount (mol) Ewewig hoeveelheid (mol) (f) 0,11 1,26 1,26 Equilibrium concentration (mol∙dm-3) (c) Ewewigskonsentrasie (mol∙dm-3) 0,42 0,42 m(NH4HS)eq = nM = 0,11 x 51 (g) = 5,61 g (h) No Kc expression, correct substitution Geen Kc- uitdrukking, korrekte substitusie: OR/OF Max./Maks. 7⁄8 m(NH4HS)change = nM = 1,26 x 51 (g) Wrong Kc expression/ = 64,26 g Verkeerde Kc-uitdrukking: Max./Maks. 6⁄8 m(NH4HS)eq = 70 – 64,26 (f) = 5,74 g (h) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 16 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne CALCULATIONS USING CONCENTRATION BEREKENINGE WAT KONSENTRASIE GEBRUIK NH4HS(s) NH3(g) H2S(g) Initial concentration (mol∙dm-3) 0 0 Aanvanklike konsentrasie (mol∙dm-3) Change (mol∙dm-3) Verandering (mol∙dm-3) (b) x x Equilibrium concentration (mol∙dm-3) Ewewigskonsentrasie (mol∙dm-3) x x (d) Kc = [NH3][H2S] No Kc expression, correct substitution (e) 18 x 10-2 = x2 Geen Kc- uitdrukking, korrekte substitusie: Max./Maks. 7⁄8 x = 0,424 mol∙dm-3 nNH3(change) = cV Wrong Kc expression/ = (0,424)(3) (c) Verkeerde Kc-uitdrukking: Max./Maks. 6⁄8 = 1,272 mol n(NH4HS)change = nNH3(change) = 1,272 mol (a) 70 n(NH4HS)eq = – 1,272 (f) 51 = 0,098 mol m(NH4HS)eq = nM = 0,098 x 51 (g) = 4,998 g (h) OR/OF m(NH4HS)change = nM = 1,272 x 51 (g) = 64,872 g m(NH4HS)eq = 70 – 64,872 (f) = 5,128 g (h) (8) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 17 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 ─ 7.1.1 H2PO4  ─ 2─ Ka of H2PO4 greater/higher than Ka of HPO4  ─ Accept: Ka of H2PO4 greater (2) 7.1.2 2─ HPO4  (1) 7.1.3 ─ 2─ H2PO4 OR/OF HPO4  (1) 7.1.4 Basic/Basies  (1) 7.1.5 Marking criteria/Nasienkriteria: • Reactants  Products  Balancing  Reaktanse  Produkte  Balansering  • Ignore/Ignoreer → and phases/en fases Marking rule 6.3.10/Nasienreël 6.3.10 2─ ─ HPO4 (aq) + H2O(ℓ) ⇌ H2PO4 (aq) + OH─ (aq) bal    OR/OF 2─ HPO4 (aq) + 2H2O(ℓ) ⇌ H3PO4 (aq) + 2OH─ (aq) ACCEPT/AANVAAR: Na2HPO4(aq) + H2O(ℓ) ⇌ NaH2PO4(aq) + NaOH(aq) − Na2 HPO4(aq) + H2O(ℓ) ⇌ H2PO4 (aq) + 2Na+(aq) + OH− (aq) Na2 HPO4(aq) + H2O(ℓ) ⇌ NaH2PO4 (aq) + NaOH(aq) Na2 HPO4(aq) + 2H2O(ℓ) ⇌ H3PO4(aq) + 2Na+(aq) + 2OH− (aq) Na2 HPO4(aq) + 2H2O(ℓ) ⇌ H3PO4(aq) + 2NaOH (aq) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 18 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 7.2.1 Marking criteria: Nasienkriteria: a) Any formula: pH = -log[H3O+] / a) Enige formule: pH = - log[H3O+] / pH = -log[H+] / [H3O+] = 10-pH/ pH = - log[H+] /[H3O+] = 10-pH / pOH = -log[OH─] / pOH = - log[OH-] / [H3O+][OH─] = 10-14 / [H3O+][OH─] = 10-14 / pH + pOH = 14  pH + pOH = 14  b) Substitute 12,62 in pH = -log[H3O+]/ b) Vervang 12,62 in pH = - log[H3O+]/ pH + pOH = 14  pH + pOH = 14  c) Substitute calculated [H3O+] in c) Vervang berekende [H3O+] in [H3O+][OH─] / [H3O+][OH─] / 1,38 in pOH = -log[OH─]  1,38 in pOH = -log[OH─]  d) Final answer: 0,04 mol∙dm-3  d) Finale antwoord: 0,04 mol∙dm-3  RANGE: 0,04 – 0,042 mol∙dm-3 GEBIED: 0,04 – 0,042 mol∙dm-3 OPTION 1/OPSIE 1 pH = - log[H3O+] Any one/Enige een (a) 12,62 (b) = - log[H3O+] OR/OF [H3O+] = 10-12,62 [H3O+] = 2,4 x 10-13 [H3O+][OH─] = 10-14 (c) (2,4 x 10-13)[OH─] = 1 x 10-14 [OH-] = 0,0417 mol∙dm-3 (d) (0,04) OPTION 2/OPSIE 2 pH + pOH = 14 (b) 12,62 + pOH = 14 Any one/Enige een (a) pOH = 1,38 pOH = -log[OH─] 1,38 (c) = -log[OH─] [OH-] = 0,042 mol∙dm-3 (d) (0,04) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 19 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne 7.2.2 POSITIVE MARKING FROM QUESTION 7.2.1/ POSITIEWE NASIEN VANAF VRAAG 7.2.1 Marking criteria Nasienkriteria: (a) Substitute: 0,2 mol·dm-3 AND (a) Vervang: 0,2 mol·dm-3 EN 0,015 dm3 in n = cV  0,015 dm3 in n = cV  (b) USING RATIO: (b) GEBRUIK VERHOUDING: n(OH-) : nHCℓ = 1 : 1 / n(Ba(OH)2) : n(OH-) : nHCℓ = 1 : 1 / n(Ba(OH)2) : n(HCℓ) = 1 : 2  n(HCℓ) = 1 : 2  (c) Substitute: c(OH─) AND (c) Vervang: c(OH─) EN 0,04 in n = cV  0,04 in n = cV  (d) Calculate n(OH─)ini (d) Bereken n(OH─)ini = n(OH─)reacted with HCℓ + n(OH─)fin  = n(OH─)reageer met HCℓ + n(OH─)finale (e) USING RATIO: (e) GEBRUIK VERHOUDING: n(Ba(OH)2) : n(OH-) = 1 : 2  n(Ba(OH)2) : n(OH-) = 1 : 2  (f) Multiply nBa(OH)2 in 25 cm3 by 4 (f) Vermenigvuldig nBa(OH)2 in OR 25 cm3 met 4 Divide by 0,025 dm3 AND multiply OF by 0,1 dm3  Deel deur 0,025 dm3 EN vermenigvuldig met 0,1 dm3 (g) Final correct answer: (g) Finale korrekte antwoord: 9,34 x 10-3 mol  9,34 x 10-3 mol  Range: 9,2 x 10-3 – 9,36 x 10-3 Gebied: 9,2 x 10-3 – 9,36 x 10-3 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 nHCℓ = cV nHCℓ = cV = (0,2)(0,015) (a) = (0,2)(0,015) (a) = 3 x 10-3 mol = 3 x 10-3 mol nOH─ = nHCℓ (b) 1 = 3 x 10-3 mol nBa(OH)2 = nHCℓ 2 1 nOH─final = cV = (3 x 10-3) (b) 2 = (0,0417)(0,04) (c) = 1,5 x 10-3 mol = 1,67 x 10-3 mol nOH─final = cV nOH─ ini = 3 x 10-3 + 1,67 x 10-3  (d) = (0,0417)(0,04) (c) = 4,67 x 10-3 mol = 1,67 x 10-3 mol 1 1 nBa(OH)2 = nOH─ ini nBa(OH)2 = nOH─ ini 2 2 1 1 = (1,67 x 10-3) (e) = (4,67 x 10-3) (e) 2 2 = 8,33 x 10-4 mol = 2,33 x 10-3 mol in 25 cm3 nBa(OH)2 ini= 1,5 x 10-3 + 8,33 x 10-4  (d) 3 In 100 cm = (4)(2,33 x 10 )-3  (f) = 2,33 x 10-3 mol in = 9,34 x 10 mol (g) -3 25 cm3 In 100 cm = (4)(2,33 x 10-3) (f) 3 = 9,34 x 10-3 mol (g) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 20 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne OPTION 3/OPSIE 3: (g) in all options/in alle opsies: nHCℓ = cV = (0,2)(0,015) (a) nBa(OH)2 in 25 cm3 = cV = 3 x 10-3 mol 2,33 x 10-3 mol = c(0,025) 1 cBa(OH)2 in 25 cm3 = 0,0936 mol·dm-3 nBa(OH)2 = nHCℓ 2 1 = (3 x 10-3) (b) nBa(OH)2 in 100 cm3 = cV 2 = 1,5 x 10-3 mol = (0,0936)(0,1) 1 = 0,0936 mol cBa(OH)2 = cOH─ final (e) 2 1 = (0,0417) 2 = 0,02085 mol·dm-3 nBa(OH)2 = cV = 0,02085 x 0,04 (c) = 8,34 x 10-4 mol nBa(OH)2 ini= 1,5 x 10-3 + 8,34 x 10-4  (d) = 2,34 x 10-3 mol in 25 cm3 In 100 cm3 = (4)(2,34 x 10-3) (f) = 9,36 x 10-3 mol (g) (8) [20] QUESTION 8/VRAAG 8 8.1 ANY ONE: • A substance whose (aqueous) solution contains ions.  (2 OR/OF 0) • Substance that dissolves in water to give a solution that conducts electricity (through movement of ions). OR • A substance that dissociates to form ions in water/in molten state. ENIGE EEN: • 'n Stof waarvan die (waterige) oplossing ione bevat. • 'n Stof wat in water oplos om 'n oplossing te vorm wat elektrisiteit gelei (deur die beweging van ione). OF • ‘n Stof wat dissosieer om ione in water te vorm/in gesmelte toestand (2) 8.2 Aℓ 3+/Aluminium ion/ioon  Aℓ is oxidised OR Aℓ is a stronger reducing agent.  Aℓ is geoksideer OF Aℓ is ’n sterker reduseermiddel. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 21 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne   8.3  Aℓ(s) | Aℓ3+(aq) || Zn2+(aq) | Zn(s) OR/OF Aℓ(s) | Aℓ 3+ (1 mol·dm-3) || Zn2+(aq) (1 mol·dm-3) | Zn(s) ACCEPT/AANVAAR Aℓ | Aℓ 3+ || Zn2+ | Zn (3) 8.4 Marking criteria Nasienkriteria: (a) Calculate n(Aℓ3+)  (a) Bereken n(Aℓ3+)  (b) USE mol ratio: (b) Gebruik molverhouding: nAℓ2(SO4)3 : nAℓ3+ = 1 : 2  nAℓ2(SO4)3 : nAℓ3+ = 1 : 2  (c) Multiply nAℓ2(SO4)3 by M = 342  (c) Vermenigvuldig nAℓ2(SO4)3 met (d) Final correct answer M = 342 = 42,75 g  (d) Finale korrekte antwoord = 42,75 g  OPTION 1/OPSIE 1: OPTION 2/OPSIE 2: n(Aℓ3+) = cV m c= = (1)(0,25) (a) VM = 0,25 mol (a) 1 = m (0,25) (342) (c) 1 nAℓ2(SO4)3 = n(Aℓ3+) 2 m = 85,5 g 1 = (0,25) (b) 2 Al2 (SO4 )3 → 2 mol·dm-3 Al3+ = 0,125 mol 1 m(Al2 (SO4 )3 ) = (85,5) (b) 2 m = 42,75 g (d) nAℓ2(SO4)3 = M m 0,125 = (c) 342 m = 42,75 g (d) (4) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 22 DBE/November 2025 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1 Electrolytic/Elektrolitiese ✓ (1) 9.2.1 Increases/Neem toe ✓ (1) 9.2.2 Decrease ✓ • More copper (II) ions/Cu2+ are reduced than formed./ Only copper (II) ions/Cu2+ are reduced. ✓ • Copper (II) ion/Cu2+ is a stronger oxidising agent than the zinc (II) ion/Zn2+ ✓ OR Zinc (II) ion/Zn2+ is a weaker oxidising agent than the copper (II) ion/Cu2+ Afneem • Meer koper(II)ione/Cu2+ word gereduseer as wat gevorm word./ Slegs koper(II)ione/Cu2+ word gereduseer. • Koper(II)ioon/Cu2+ is ʼn sterkter oksideermiddel as die sink(II)ioon/Zn2+ OF Sink(II)ioon/Zn2+ is ʼn swakker oksideermiddel as die koper(II)ioon/Cu2+ (3) 9.2.3 Marking criteria Nasienkriteria: (a) USING RATIO: (a) GEBRUIK VERHOUDING  n(Cu2+) : n(Zn2+) = 1 : 1  n(Cu2+) : n(Zn2+) = 1 : 1 m m (b) Formula: n =  (b) Formule: n =  M M m m (c) Substitute 65 AND n(Zn2+) in n =  (c) Vervang 65 EN n(Zn2+) in n =  M M (d) Subtraction of moles  (d) Aftrek van aantal mol  m m (e) Substitute 63,5 AND n(Cu) in n =  (e) Vervang 63,5 EN n(Cu) in n =  M M (f) Final correct answer: 9,6 g  (f) Finale korrekte antwoord: 9,6 g  (NO RANGE) (GEEN GEBIED) n(Zn2+) = n(Cu2+) = 0,05 mol (a) m(Zn) = nM (b) = (0,05)(65) (c) = 3,25 g Zn from R 0,15 – 0,05 (d) = 0,1 mol Cu m(Cu) = nM = 0,1(63,5) (e) = 6,35 g Cu from R Change in mass = 6,35 + 3,25 = 9,6 g (f) (decrease in mass) (6) [11] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou