Physical Sciences Grade 12 P2 Eastern Cape Prelim 2024 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE KEREITI YA 12 LOETSE 2024 PHYSICAL SCIENCES P2 MARKING GUIDELINE/ FISIKALE SAENSESE P2 TATAISO YA HO TSHWAYA MATSHWAO: 150 This marking guideline consists of 21 pages./ Tataiso ena ya ho tshwaya e na le maqephe a 21. Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 2 PHYSICAL SCIENCES P2/FISIKALE SAENSESE P2 (EC/LOETSE 2024) QUESTION 1 / POTSO YA 1 1.1 C ✓✓ (2) 1.2 C ✓✓ (2) 1.3 B ✓✓ (2) 1.4 A ✓✓ (2) 1.5 B ✓✓ (2) 1.6 B ✓✓ (2) 1.7 C ✓✓ (2) 1.8 A ✓✓ (2) 1.9 B ✓✓ (2) 1.10 B ✓✓ (2) [20] Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/LOETSE 2024) PHYSICAL SCIENCES P1/FISIKALE SAENSESE P2 3 QUESTION 2 / POTSO YA 2 2.1 Marking criteria/ Mokgwa wa ho tshwaya If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Haeba e le nngwe ya mantswe/diratswana tse sehelletweng mola ka tlase di sihelletswe maemong a nepahetseng: -1 letshwao lentsweng/seratswaneng ka nngwe A bond or an atom or a group of atoms that determine(s) the physical and chemical properties of a group of organic compounds. ✓✓ Bond kapa athomo kapa lequlwana la diathomo le dethemaenang fisikale le chemikale phopophatisi tsa lequlwana la diokanike khompaonde (2) 2.2.1 E ✓ (1) 2.2.2 A ✓ (1) 2.2.3 C ✓ (1) 2.3 EQUAL TO. ✓ Compound E and ethyl methanoate are functional isomers / structural isomers / have the same molecular formula / same number and type of atoms / same number of C, H and O atoms ✓✓ E LEKANA LE Khompaonde ya E le ethyl methanoate ke difankeshenale isomer/diseteraketjharale isomer/di na le molecular fomulara/ nomoro e tshwanang kapa mofuta wa diathomo/nomoro tsa C, H le O tse tshwanang (3) 2.4.1 H Marking criteria/Mokgwa wa ho │ tshwaya H H― C ― H O H • Functional group correct ✓ ✓ Lequlwana la fankeshenale le │ │ II │ nepahetseng H ― C ― C ― C ― C ― H • Whole structure correct ✓ │ │ │ Seteraketjhara sohle se H H H nepahetse (2) 2.4.2 Propanoic acid / Propanoic asiti ✓✓ (2) 2.4.3 4-methylhex-2-yne Marking criteria/ Mokgwa wa ho 4-metielheks-2-yn tshwaya • Hexyne ✓ • Methyl ✓ • Whole name correct / Lebitso lohle (3) le nepahetse ✓ Marking criteria/ Mokgwa wa ho tshwaya 2.5 H H H • Functional group on 2nd carbon ✓ │ │ │ Lequlwana la fankeshenale le ho H ― C ― C ― C ― H 2nd carbon │ │✓ │ • Whole structure correct ✓ (2) Seteraketjhara sohle se nepahetse Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 4 PHYSICAL SCIENCES P2/FISIKALE SAENSESE P2 (EC/LOETSE 2024) H Cℓ H 2.6 C7H12 + 10 O2 ✓→ 7 CO2 + 6 H2O ✓ Bal ✓ NOTES / DINOTO • Reactant / reeketente ✓ Products/ diphorodakete ✓ Balancing / ho balansa ✓ (3) [20] QUESTION 3 / POTSO YA 3 3.1.1 Marking criteria/ Mokgwa wa ho tshwaya If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Haeba e le nngwe ya mantswe/diseratswana tse sehelletweng mola ka tlase di sihelletswe maemong a nepahetseng: -1 letshwao lentsweng/seratswaneng ka nngwe The temperature at which the vapour pressure of a substance / liquid equals the atmospheric pressure ✓✓ Themphereitjhara eo vapour pressure ya sabosetense/likhwiti e lekanang le atmospheric pressure (2) 3.1.2 All Primary alcohols / Ke diporayimari alekohole kaofela ✓ (1) 3.1.3 London forces / dispersion forces/ induced-dipole forces ✓ Di-London fose/ di-dispersion fose/ di-induced-dipole fose (1) 3.1.4 Pentan-1-ol / 1-pentanol ✓✓ (2) 3.2.1 Compound C / propanoic acid ✓ Khompaonte ya C / propanoic asiti (1) 3.2.2 Marking criteria/Mokgwa wa ho tshwaya • Type of intermolecular forces in compound A ✓ • Compound B and C have hydrogen bonds ✓ • Compare the number of sites for hydrogen bonding in B and C ✓ • Compare the strength of intermolecular forces ✓ • Compare energy required to overcome the intermolecular forces ✓ • Mofuta wa di-intermolecular fose ho khompaonde ya A • Khompande ya B le ya C di na le di-hydrogen bond • Bapisa nomoro ya di-site tsa hydrogen bonding ho B le ho C • Bapisa seterenthe sa di-intermolecular fose • Bapisa eneji e hlokehang ho hlola di-intermolecular fose Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/LOETSE 2024) PHYSICAL SCIENCES P1/FISIKALE SAENSESE P2 5 • Compound A/butanone has dipole-dipole forces ✓ (and London forces / dispersion forces/ induced-dipole forces) • Compound B/butan-1-ol and C/propanoic acid has hydrogen bonds ✓(and London forces / dispersion forces/ induced-dipole forces) • Compound B/butan-1-ol has one site for hydrogen bonding and C/propanoic acid has two sites for hydrogen bonding ✓ • Strength of the intermolecular forces increases from compound A / butanone to compound B / butan-1-ol to compound C / propanoic acid ✓ • More energy is needed to overcome the intermolecular forces in compound C / propanoic acid than compounds A/ butanone and B/butan-1-ol ✓ • Khompaonte ya A/butanone e na le di-dipole-dipole fose (le di-London fose/di-dispersion fose/ di-induced-dipole fose) • Khompaonde ya B/butan-1-ol le ya C/propanoic asisti e na le di-hydrogen bond (le di-London fose / di-dispersion fose/ di-induced-dipole fose) • Khompaonde ya B/butan-1-ol e na le site e le nngwe ya hydrogen bonding le C/propanoic asiti e na le di-site tse pedi tsa hydrogen bonding • Seterenthe sa di-intermolecular fose se a ata ho tloha ho khompaonde ya A / butanone ho ya ho khompaonde ya B / butan-1-ol ho ya ho khompaonde ya C / propanoic asiti • Eneji e ngata e hlokeha ho hlola di-intermolecular fose ho khompaonde ya C / propanoic asiti ho e na le khompaonde ya A/ butanone le B/butan- 1-ol ) (5) [12] Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 6 PHYSICAL SCIENCES P2/FISIKALE SAENSESE P2 (EC/LOETSE 2024) QUESTION 4 / POTSO YA 4 4.1.1 UNSATURATED ✓ It contains a double bond between its carbon-carbon atoms in the hydrocarbon chain ✓ UNSATURATED. E na le double bond pakeng tsa di-carbon-carbon athomo ho hydrocarbon chain (2) 4.1.2 Substitution / Hydrolysis of a haloalkane ✓ Sabostitjhushene/Haeterolisisi ya haloalkane (1) 4.1.3 Elimination / dehydrohalogenation of haloalkanes ✓ Elimineishene/dehydrohalogenation ya di-haloalkane (1) Marking criteria/Mokgwa wa ho 4.1.4 H H H H H tshwaya • Functional group correct ✓ H C C C C C H✓ Lequlwana la fankeshenale le nepahetse ✓ • Whole structure correct ✓ H Cℓ H H H (2) Seteraketjhara sohle se nepahetse 4.1.4 Pentan-2-ol / 2-pentanol ✓✓ (2) 4.1.6 Dilute strong base / NaOH ✓ and mild heat ✓ Beisi e matla e daelutuweng/NaOH le mild heat (2) 4.1.7 Positional isomers ✓✓ Di-positional isomer (2) Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/LOETSE 2024) PHYSICAL SCIENCES P1/FISIKALE SAENSESE P2 7 4.2 Marking criteria / Mokgwa wa ho tshwaya Reaction 1: Dehydration reaction and correct chemicals. ✓ Reactants with correct condensed structural formula ✓ Products with correct condensed structural formula ✓ Reekeshene ya 1: Dehydration reekeshene le dikhemikale tse nepahetseng. Direeketente tse nang le seteraketjharele fomulara se khondensuweng se nepahetse Diprodakete tse nang le seteraketjharale fomulara se khondensuweng se nepahetse Reaction 2: Addition reaction and correct chemicals used. ✓ Reactants with correct condensed structural formula ✓ Products with correct condensed structural formula ✓ Reekeshene ya 2: Addition reekeshene le dikhemikhale tse nepahetseng di sebedisitswe. Direeketenete tse nang le seteraketjharale fomulara se khondensuweng se nepahetse Diprodakete tse nang le seteraketjharale fomulara se khondensuweng se nepahetse Reaction 1 / Reekeshene ya 1 Conc H2SO4 CH3CH2CH2CH2OH ✓ Heat CH3CH2CHCH2 + H2O ✓ ✓(correct reaction / reekeshene e nepahetseng) Reaction 2 / Reekeshene ya 2 Pt CH3CH2CHCH2 + H2 ✓ CH3CH2CH2CH3 ✓ ✓(correct reaction / Heat reekeshene e nepahetseng) (6) [18] Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 8 PHYSICAL SCIENCES P2/FISIKALE SAENSESE P2 (EC/LOETSE 2024) QUESTION 5 / POTSO YA 5 5.1 Marking criteria/ Mokgwa wa ho tshwaya If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Haeba e le nngwe ya mantswe/diseratswana tse sehelletweng mola ka tlase di sihelletswe maemong a nepahetseng: -1 letshwao lentsweng/seratswaneng ka nngwe ANY ONE Change in concentration ✓of reactant or product per (unit) time. ✓ Change in amount/number of moles/volume/mass of products or reactants per (unit) time. ✓✓ Change in amount/number of moles/volume/mass of products formed or reactants used reactants per (unit) time. ✓✓ LE HA E LE EFE E LE NNGWE Ho tjhentjha ha konsentereshene ya reeketente kapa porotakete per (unit) time. Ho tjhentjha ha amaonte/number of moles/volume/mass ya diporotakete kapa direekenente per (unit) time. Ho tjhentjha ha amaonte/number of moles/volume/mass ya doporotakete tse entsweng kapa direeketente tse sebedisitsweng per (unit) time. OR/KAPA The rate of change in concentration / amount of moles / number of moles / volume / mass. ✓✓ (2 or 0). Reithi ya ho tjhentjha ha khosentereishene/ amaonte ya di-mole /number of moles / volume / mass. (2 or 0). (2) 5.2 Marking Criteria for investigative question / Mokgwa wa ho tshwaya investigative question The independent and dependent variables are stated ✓ Independent le dependent variables di hlalositswe Ask a question about the relationship between the independent and dependent variables ✓ Botsa potso ka kamano dipakeng tsa independent le dependent variables What is the effect of the increase / decrease / change of concentration on the reaction rate? ✓✓ Ke efe tshusumetso ya ho atisa/fokotsa/ho tjhentjha konsentereishene ho reaction rate? OR / KAPA Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/LOETSE 2024) PHYSICAL SCIENCES P1/FISIKALE SAENSESE P2 9 What is the relationship between the concentration and reaction rate? ✓✓ Ke efe kamano dipakeng tsa konsenteireshene le reaction rate? (2) 5.3.1 HIGHER THAN / KA HODIMO HO ✓ (1) Volume of hydrogen gas (cm3) / Marking criteria / Mokgwa wa ho 5.3.2 tshwaya • Gradient B of is higher than A  B A Gradient ya B e hodimo ho na le ya A • Final volume of curve A and B is the same and is horizontal / Volume ya hydrogen gas (cm3) Final volume ya mothinya wa A le B e a tshwana ho horizontal NOTE: A or B must be indicated Ignore the labels of the axes. ELELLWA: A le B di tlameha ho Time /Nako bontswa. Se natse di-leobole tsa di-axes (2) Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 10 PHYSICAL SCIENCES P2/FISIKALE SAENSESE P2 (EC/LOETSE 2024) 5.4 Marking criteria a) Subst. into rate equation ✓ V b) Subst. 486,62 and 25 000 into in n = ✓ Vm c) Use of mol ratio n (Zn) : n (H2) ✓ d) Subst. 0,02 and 65 into m = nM ✓ e) Final answer ✓ Mokwa wa ho tshwaya a) Sabst. ho ekweishene ya reithi V b) Sabst 486,62 le 25 000 ho n = Vm c) Tshebediso ya mol ratio n (Zn) : n (H2) d) Sabst 0,02 le 65 ho m = nM e) Karabo ya ho qetela ΔV rate = Δt V-0 8,39 = (𝑎) ✓ 58 V = 486,62 cm3 V n= Vm 486,62 n= (b) ✓ 24000 n(H2) = 0,020 mol n (Zn) = n (H2) = 0,020 (c) ✓ m (Zn) = nM m (Zn) = (0,020)(65) (d) ✓ m (Zn) = 1,30 g (e) ✓ (5) Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/LOETSE 2024) PHYSICAL SCIENCES P1/FISIKALE SAENSESE P2 11 5.5 INCREASES / NYOLOHA ✓ (1) 5.6 • At higher temperature the average kinetic energy of the particles is higher ✓ • More particles have enough / sufficient kinetic energy or more particles have kinetic energy equal or higher than the activation energy ✓ • More effective collisions per unit time/ Higher frequency of effective collisions ✓ • Ho themphereitjhara e hodimo average kinetic energy ya di-particle e a nyoloha • Di-particle tse ngata di na le enough / sufficient kinetic energy kapa pi- particle tse ngata di na le kinetic energy e lekanang kapa e hodimo ho na le ya activation energy • Di-effective collision tse ngata per unit time/ Frequency e hodimo ya di- effective collision (3) [16] Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 12 PHYSICAL SCIENCES P2/FISIKALE SAENSESE P2 (EC/LOETSE 2024) QUESTION 6 / POTSO YA 6 6.1.1 Marking criteria/ Mokgwa wa ho tshwaya If any of the underlined key words/phrases in the correct context are omitted: - 1 mark per word/phrase. Haeba e le nngwe ya mantswe/diseratswana tse sehelletweng mola ka tlase di sihelletswe maemong a nepahetseng: -1 letshwao lentsweng/seratswaneng ka nngwe When the equilibrium in a closed system is disturbed, the system will re- instate a new equilibrium by favouring the reaction that will oppose/cancel the disturbance.  Ha ekwiliboriamo ho closed system e phatsamisitswe, system we tla re- instata ekwiliboriamo e ntjha ka ho feiva reekeshene e tla opposa/khansela phatsamiso. (2) 6.1.2 Forward (reaction).  The equilibrium concentration of B is higher than A.  Fowade (reekeshene). Ekwiliboriamo konsentereishene ya B e hodimo hon a le ya A. (2) 6.1.3 INCREASES/ NYOLOHA  (1) 6.1.4 NO EFFECT/HA HO TSHUSUMETSO  (1) 6.1.5 EXOTHERMIC  (1) 6.1.6 • The concentration of A increased / [B] decreased  • (According to Le Chatelier’s principle) an increase in temperature favours the endothermic reaction.  • The reverse reaction was favoured / The equilibrium position shifted towards the left  • Konsentereishene ya A e nyolohile / [B] o theohile • (Ho ya ka Le Chatelier’s principle) ho nyoloha ha themphereitjhara ho feiva endothermic reaction. • Revese reekeshene e ile ya feivuwa / Equilibrium position e shifotetse ka ho le letshehadi (3) Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/LOETSE 2024) PHYSICAL SCIENCES P1/FISIKALE SAENSESE P2 13 6.2 OPTION 1: CONCENTRATION : Marking criteria a) Correct Kc expression with square brackets ✓ b) Substitution of 6,34 x 10-4 into correct kc expression ✓ c) Correct substitution of [Br2] into correct kc expression ✓ d) Use of the correct ratio [Br2] : [Br] ✓ e) Determine initial [Br2] ✓ f) Substitution into c = n/V ✓ g) Final answer ✓ KGETHO YA 1: KONSENTEREISHENE / Mokgwa wa ho tshwaya a) Kc expression e nepahetseng e nang le di-square bracket b) Sabosititjhushene ya 6,34 x 10-4 ho Kc expression e nepahetseng c) Sabosititjhushene e nepaheng ya [Br2] ho Kc expression e nepahetseng d) Tshebdiso ya ratio e nepahetseng [Br2] : [Br] e) Fumana [Br2] ya qalong • No Kc expression, correct f) Sabosititjhushene ho c = n/V substitution / Ha ho Kc expression, g) Karabo ya ho qetela sabosititjhusene e nepahetseng. Max / Maks 6/7 • Wrong Kc expression / Kc [Br]2 expression e fosahetse. Max. Maks Kc = (a) ✓ [Br2] 3/7 [Br]2 6,34 x 10-4(b) ✓ = (c) ✓ (2,074) [Br] = 0,03626 mol·dm-3 Br2 2 Br Initial conc 2,09213 ✓ (e) - Change in conc 0,03626 ✓ (d) -0,01813 Ho tjhentjha ha kons. Equilibrium conc. 2,074 0,03626 n V = c 1,05 V = (f) ✓ 2,09213 V = 0,50 dm3 (g) ✓ Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 14 PHYSICAL SCIENCES P2/FISIKALE SAENSESE P2 (EC/LOETSE 2024) OPTION 2: CONCENTRATION: Marking criteria a) Determine Δ [Br2] ✓ b) Use of the correct ratio [Br2] : [Br] ✓ c) Correct Kc expression with square brackets ✓ d) Substitution of 6,34 x 10-4 into correct Kc expression ✓ e) Correct substitution of [Br2] and [Br] into correct Kc expression ✓ f) Substitution into c = nV ✓ g) Final answer ✓ KGETHO YA 2: KONSENTEREISHENE / Mokgwa wa ho tshwaya a) Fumana Δ [Br2] b) Sebedisa ratio ya [Br2] : [Br] e nepahetseng c) Kc expression e nepahetseng e nang le di-square bracket d) Sabosititjhushene ya 6,34 x 10-4 e nepahetseng ho Kc-expression e) Sabosititjhushene ya [Br2] le [Br] tse nepahetseng ho Kc-expression f) Sabosititjhushene ho c = n/V g) Karabo ya ho qetela Δ[Br2] = x(0,865/100) = 0,0865x ✓ (a) Br2 2 Br Initial conc x - Change in conc -0,00865x 0,0173x ✓ (b) Ho tjhentjha ha knons Equilibrium conc. 2,074 0,0173x [Br]2 Kc = (c) ✓ [Br2] (0,0173x)2 6,34 x 10-4 (d) ✓ = (e) ✓ (2,074) x / [Br2]o = 2,096 mol·dm-3 n • No Kc expression, correct V = c substitution / Ha ho Kc expression, sabosititjhusene e nepahetseng. 1,05 Max / Maks 6/7 V = (f) ✓ 2,096 • Wrong Kc expression / Kc expression e fosahetse. Max. Maks V = 0,50 dm3 (g) ✓ 3/7 Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/LOETSE 2024) PHYSICAL SCIENCES P1/FISIKALE SAENSESE P2 15 OPTION 3: MOLE CALCULATION : Marking criteria a) Determine Δ mol Br2 ✓ b) Use of the correct ratio Br2 : Br ✓ c) Correct Kc expression with square brackets ✓ d) Substitution of 6,34 x 10-4 into correct Kc expression ✓ e) Correct substitution of [Br] and [Br] into correct Kc expression ✓ f) Substitution into c = nV ✓ g) Final answer ✓ KGETHO YA 3: KKAHETJHULEISHENE YA MOLE: Mokgwa wa ho tshwaya a) Fumana Δ mol ya Br2 b) Sebedisa ratio ya Br2 : Br e nepahetseng c) Kc expression e nepahetseng e nang le di-square bracket d) Sabosititjhushene ya 6,34 x 10-4 e nepahetseng ho Kc-expression e) Sabosititjhushene ya [Br2] le [Br] tse nepahetseng ho Kc-expression f) Sabosititjhushene ho c = n/V g) Karabo ya ho qetela Δn(Br2) = 1,05 x 0,865 / 100 = 0,0090825 mol ✓ (a) Br2 2 Br Initial mol 1,05 - Change in mol -0,0090825 0,0818165 ✓ (b) Ho tjhentjha ha mol Equilibrium mol. 2,074V 0,0818165 Equilibrium conc. 2,074 0,018165 / V Ekwiliboriamo kons [Br]2 Kc = (c) ✓ [Br2] • No Kc expression, correct substitution / Ha ho Kc expression, [Br]2 6,34 x 10-4 (d) ✓ = (e) ✓ sabosititjhusene e nepahetseng. (2,074) Max / Maks 6/7 • Wrong Kc expression / Kc [Br]o = 0,03626 mol·dm-3 expression e fosahetse. Max. Maks 3/7 n c = • V 0,018165 0,03626 = (f) ✓ V V = 0,50 dm3 (g) ✓ Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 16 PHYSICAL SCIENCES P2/FISIKALE SAENSESE P2 (EC/LOETSE 2024) OPTION 4: MOLE CALCULATION: Marking criteria a) Correct Kc expression with square brackets ✓ b) Substitution of 6,34 x 10-4 into correct Kc expression ✓ c) Correct substitution of [Br] and [Br] into correct Kc expression ✓ d) Determine change in mol Br ✓ e) Use of the correct ratio [Br2] : [Br] ✓ f) Substitution into ne = ni – Δn for Br2 ✓ g) Final answer ✓ KGETHO YA 4: KHALETJHULEISHENE YA MOL/Mokgwa wa ho tshwaya a) Kc expression e nepahetseng e nang le di-square bracket b) Sabosititjhushene ya 6,34 x 10-4 ho Kc expression e nepahetseng c) Sabosititjhushene e nepaheng ya [Br2] ho Kc expression e nepahetseng d) Fumana ho tjhentjha ha mol ho Br e) Tshebediso ya ratio e nepahetseng [Br2] : [Br] f) Sabosititjhushene ho ne = ni – Δn ho Br2 g) Karabo ya ho qetela • No Kc expression, correct [Br]2 substitution / Ha ho Kc expression, Kc = (a) ✓ sabosititjhusene e nepahetseng. [Br2] Max / Maks 6/7 [Br]2 • Wrong Kc expression / Kc 6,34 x 10-4 (b) ✓ = (c) ✓ expression e fosahetse. Max. Maks (2,074) 3/7 [Br] = 0,03626 mol·dm-3 • Br2 2 Br Initial mol 1,05 - Change in mol -0,01813V ✓ (e) 0,03626V ✓ (d) Ho tjhentjha ho mol Equilibrium mol. 2,074V 0,03626V Equilibrium conc. 2,074 0,03626 Ekwiliboriamo kons ne = ni – Δn 2,074V = 1,05 - 0,01813V ✓ (f) V = 0,5 dm3 ✓ (g) (7) [17] Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/LOETSE 2024) PHYSICAL SCIENCES P1/FISIKALE SAENSESE P2 17 QUESTION 7 / POTSO YA 7 7.1.1 Strong acids ionise completely in water ✓ to produce a high concentration of the hydronium ions / (H3O+) ✓ Di-strong acid ayonaisa metsing ka botlalo ✓ ho porodusa konsentereishene e hodimo ya di-hydronium ion / (H3O+) (2) 7.1.2 It donates two protons/ H+ / E doneitha di-proton tse pedi / H+ ✓ (1) 7.1.3 H2O ✓ (1) 7.1.4 HSO4 – ✓✓ (2) 7.1.5 Marking criteria/ Mokgwa wa ho tshwaya a) Ratio [H3O+] : [H2SO4] ✓ b) Formula / Fomulara pH = - log [H3O+] ✓ c) pH value substituted into formula /pH vaue e sabostitjhuwe ho fomulara✓ d) Final answer / Karabo ya ho qetela ✓ [H3O+] = 2(0,1) ✓ (a) = 0,2 mol·dm-3 pH = - log[H3O+] ✓ (b) pH = - log (0,2) ✓ (c) pH = 0,70 ✓ (d) (4) 7.2.1 OPTION 1/ KGETHO YA 1 OPTION 2 / KGETHO YA 2 m m c = ✓ n = MV M 1,2 1,2 c = ✓ n = (90)(50x10-3) 90 c = 0,27 mol· dm-3 ✓ n = 0,013 mol· dm-3 n c = ✓ V 0,013 c = ✓ 50x10-3 c = 0,26 mol· dm-3 ✓ (3) Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 18 PHYSICAL SCIENCES P2/FISIKALE SAENSESE P2 (EC/LOETSE 2024) 7.2.2 Positive marking from / Ho tshwaya Positive marking from / Ho ho phositifo ho tloha ho 7.2.1 tshwaya ho phositifo ho tloha Marking criteria / Tsela ya ho tshwaya ho 7.2.1 Marking criteria / Tsela ya ho a) Subst. values of H2C2O4 into n = cV / tshwaya Sabst divelu tsa H2C2O4 ho n = cV ✓ b) Using ratio / Sebedisa ratio H2C2O4: a) Subst. values of/Sabst. NaOH 1:2 ✓ Divelu tsa na / nb into/ho c) Subst of values of /Sebedisa divelu ca Va n = na ✓ tsa NaOH into/ ho c = n/V ✓ c V b b b d) Subst into values / Sabst ho divelu b) Subst. values of/ Sabst. NaOH into/ ho c1V1 = Divelu tsa caVa into/ho ca Va n c2V2 = na ✓ c V b b b e) Formula / Fomulara m = cMV ✓ c) Subst. values of/ Sabst. f) Subst into / Sabst ho m = cMV ✓ Divelu tsa Vb into/ho g) Final answer / Karabo ya ho qetela ca Va na ✓ = ✓ c V b b n b d) Subst into values/ Sabst ho n (H2C2O4) = cV divelu tsa NaOH into/ ho c1V1 = c2V2 ✓ = 0,27 x 25 x 10-3 (a) ✓ e) Formula / Fomulara m = cMV ✓ = 6,75 x 10-3 mol f) Subst into/Sabst ho m = cMV ✓ n (NaOH) = 2 x 6,75 x 10-3 (b) ✓ g) Final answer/Karabo ya ho qetela ✓ n (NaOH) = 0,0135 mol caVa na = n cbVb nb c(NaOH) = V (0,27)(25) (b) ✓ 1 = (a) ✓ 0,0135 cb(43,8) (c) ✓ 2 c(NaOH) = (c) ✓ 43,8 x 10-3 cb = 0,308 mol·dm-3 c(NaOH) = 0,308 mol·dm-3 c1V1 = c2V2 c1(25) = (0,308)(100) (d) ✓ c1 = 1,232 mol· dm-3 m = cMV (e) ✓ m = (1,232)(40)(2) (f) ✓ m = 98,56 g (g) ✓ (7) [20] Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/LOETSE 2024) PHYSICAL SCIENCES P1/FISIKALE SAENSESE P2 19 QUESTION 8 / POTSO YA 8 8.1 Chemical energy to electrical energy ✓✓ Chemical energy ho ya ho electrical energy (2) 8.2 Provide path for movement of ions / E porovaida tsela ya motsamaho wa di- ion Ensures electrical neutrality in the cell /E nnetefatsa electrical neutrality ho sele (Any one/ Le ha efe e le nngwe) ✓ (1) 8.3 Fe3+ + e- → Fe2+ ✓✓ (2) Marking criteria / Tsela ya ho tshwaya • Fe3+ + e - ⇌ Fe2+ ½ • Fe2+ ← Fe3+ + e- 2/2 • Fe2+ ⇌ Fe3+ + e - 0/2 • Fe2+ → Fe3+ + e - 0/2 Ignore if the charge omitted on electron / Se natse ha tjhatjhe ya eleketerone e sihelletswe 8.4 Eθcell = Eθcathode/reduction/oxidising agent –Eθanode/oxidation/reducing agent ✓ Marking criteria/Tsela ya ho tshwaya Eθcell = (0,77) ✓ – ( - 0,13) ✓ • Any other formula using unconventional abbreviation, e.g. / Eθcell = EθOA –Eθ RA followed by the correct Eθcell = 0,90 V ✓ substitution Max ¾ Le ha e le efe fomulara e sebedisang unconventional abbreviation, mohl. Eθsel = EθOM –Eθ RM e latelwe ke sabosititjhushene e (4) nepahetseng Maks ¾ 8.5.1 Increases / E a ata ✓ (1) 8.5.2 Remains the same / E dula e tshwana ✓ (1) 8.5.3 Increases / E a ata ✓ (1) 8.6 Zn is a stronger reducing agent than Pb ✓ More energy per unit charge is released for the reaction between Zn and Fe3+ / The reaction between Zn and Fe3+ is more strongly product-favoured / equilibrium position lies further to the right than Pb and Fe3+ ✓ Zn ke reducing agent e matla ho na le Pb ✓ Eneji e ngata e ya tswa per unit charge ho reekeshene e dipakeng tsa Zn le Fe3+ / Reekeshene e dipakeng tsa Zn le Fe3+ e product-favoured haholo ka matla/ equilibrium position e itshetlehile haholo ka ho le letona ho na le ya Pb le Fe3+ ✓ (2) [14] Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com 20 PHYSICAL SCIENCES P2/FISIKALE SAENSESE P2 (EC/LOETSE 2024) QUESTION 9 / POTSO YA 9 9.1 The chemical process in which electrical energy is converted to chemical energy (2 or 0) ✓✓ Chemical process eo electrical energy e fetohelang ho chemical energy (2 or 0) OR / KAPA The use of electrical energy to produce a chemical change (2 or 0) ✓✓ Ho sebediswa ha electrical energy ho porodusa chemical change (2 or 0) ✓✓ (2) 9.2 Battery/Leshala ✓ (1) 9.3 Hydrogen gas/ H2 ✓ (1) 9.4 2 Cℓ− (aq)+ 2e− → Cℓ2(g) ✓✓ Ignore phases / Se natse di-phase Marking criteria / Mokgwa wa ho tshwaya 1/2 • 2 Cℓ− (aq) + 2e− ⇌ Cℓ2(g) 2/2 • Cℓ2(g) ← 2 Cℓ− (aq)+ 2e− 0/2 • Cℓ2(g) ⇌ 2 Cℓ− (aq)+ 2e− 0/2 • Cℓ2(g) → 2 Cℓ− (aq)+ 2e− Ignore if the charge omitted on electron/ Se natse haeba tjhatjhe ya eleketerone e sihelletswe (2) 9.5 Pink / Pinki ✓ . OH− ✓ (2) 9.6.1 Zinc ✓ (1) Hosted by www.ecexams.co.za.

Downloaded from hlayiso.com (EC/LOETSE 2024) PHYSICAL SCIENCES P1/FISIKALE SAENSESE P2 21 9.6 2 Marking criteria / Mokgwa wa ho tshwaya a) Using ratio / Ho sebedisa ratio Cu: e- = 1:2 ✓ b) Subst. of values of / Subst. divelu tsa Cu into/ho m = nM ✓ c) Adding the mass of Cu reduced to initial mass / Ho eketsa mass ya Cu ka mora tlhwekiso d) Final answer / Karabo ya ho qetela ✓ n (Cu) = ½ x 1,38 x 10-2 (a) ✓ n (Cu) = 6,9 x 10-3 mol m = nM m = (6,9 x 10-3)(63,5) (b) ✓ m = 0,43815 g m (cathode) = 2 + 0,43815 (c) ✓ m (cathode) = 2,43815 g (d) ✓ (4) [13] MATSHWAO KAOFELA: 150 Hosted by www.ecexams.co.za.