Physical Sciences Grade 12 Limpopo Mock 2020 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 PHYSICAL SCIENCES: CHEMISTRY (P2) FISIESE WETENSKAPPE: CHEMIE (V2) SEPTEMBER 2020 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 10 pages. Hierdie memorandum bestaan uit 10 bladsye. Copyright reserved

Downloaded from hlayiso.com Physical Sciences P2 / Fisiese Wetenskappe V2 2 LIMPOPO September 2020 NSC Marking Guideline / NSS Nasienriglyn QUESTION 1/VRAAG 1 1.1 C  (2) 1.2 D  (2) 1.3 B  (2) 1.4 A  (2) 1.5 B  (2) 1.6 A  (2) 1.7 C  (2) 1.8 C  (2) 1.9 A  (2) 1.10 B  (2) [20] QUESTION 2/VRAAG 2 2.1.1 Tertiary (halo-alkane)  the carbon attached to the halogen/Br is attached to three other carbons Tersiêre (haloalkaan) die koolstof waaraan die halogeen/Br verbind is,is aan drie ander koolstowwe verbind (2) 2.1.2 2-bromo-2-methylbutane 2-bromo-2-metielbutaan / 2-broom-2-metielbutaan Marking criteria/Nasienriglyne • Butane/butaan  • Both substituents correct : bromo and methyl /Altwee substituente korrek: bromo en metiel • Everything correct / Alles reg (Any error e.g. hyphens omitted and/or incorrect sequence: (3) Enige fout, bv. koppeltekens weggelaat en/of verkeerde volgorde: Max./Maks: 2 ) 3 2.1.3 2-methyl-2-butene/ 2-methyl but-2-ene / 2-metiel-2-buteen / 2-metielbut-2-een Marking criteria/Nasienriglyne • But-2-ene/2-butene/But-2-een/2-buteen  • 2-methyl/2-metiel Any error e.g. hyphens omitted and/or incorrect sequence: Enige fout, bv. koppeltekens weggelaat en/of verkeerde volgorde: Max./Maks: 1 (2) 2 2.2.1 (A series of) organic compounds which have the same general formula OR which differ from each other by a CH2 group/unit / ‘n Homoloë reeks is ‘n reeks organiese verbindings wat deur dieselfde algemene formule beskryf word OF waarvan die een lid van die volgende lid verskil met ‘n CH 2-groep. (2) 2.2.2 Esters  (1) 2.2.3 H H H O H H H Marking criteria/nasienriglyne I I I II I I I Functional group H—C—C—C— O—C—C— C—C—H Everything else correct I I I I I I funksionele groep H H H H H H alles verder korrek (2) 2.2.4 Propanol and butanoïc acid / Propanol en butanoësuur (2) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences P2 / Fisiese Wetenskappe V2 3 LIMPOPO September 2020 NSC Marking Guideline / NSS Nasienriglyn 2.2.5 Marking criteria/nasienriglyne O Functional group II Everything else correct CH3CH2CH2CH2CH2CH3 COH funksionele groep alles verder korrek (2) 2.3 2C4H10 + 13O2 → 8CO2 + 10H2O (Reactants Products  Balancing) (reaktante, produkte, balansering) (3) [19] QUESTION 3/VRAAG 3 3.1.1 Organic molecules with the same molecular formular but different structural formule / (2) Organiese molekule met dieselfde molekulêre formule, maar verskillende struktuurformules. 3.1.2 CnH2n+2  (1) 3.1.3 Chain(isomers) / ketting(isomere) (1) 3.2.1 The temperature at which the vapour pressure of a substance equals atmospheric/external pressure Die temperatuur waar die dampdruk van ‘n stof gelyk is aan die atmosferiese/eksterne druk (2) 3.2.2 number of branches /aantal vertakkings (1) 3.2.3 Number of C and H atoms,/ molecular mass  Aantal C en H atome,/ molekulêre massa (1) 3.2.4 From A to C (Structure) Branching decreases/molecules become less compact/surface area increases (over which intermolecular forces acts) (Intermolecular forces) Stronger/more intermolecular forces/Van Der Waals forces/London forces (Energy) More energy needed to overcome intermolecular forces/Van Der Waals forces/ /London forces Van A na C (Struktuur) Vertakkings verminder/molekule word minder kompak/oppervlakte (waaroor intermolekulêre kragte werk) word groter (Intermolekulêre kragte} Sterker of meer intermolekulêre kragte /Van Der Waalskragte / Londonkragte {Energie} Meer energie benodig om intermolekulêre kragte /Van Der Waalskragte / Londonkragte te oorkom (3) 3.2.5 A Lowest boiling point / laagste kookpunt (2) 3.3 Aldehydes/ aldehiede (1) 3.4 D/butan-1-ol has hydrogen bonding forces between the molecules E/butanal has dipole-dipole forces between the molecules Hydrogen bonds are stronger than dipole-dipole forces D/butan-1-ol het waterstofbindings tussen die molekule E/butanaal het dipool-dipoolkragte tussen die molekule Waterstofbindings is sterker as dipool-dipool kragte (3) [17] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences P2 / Fisiese Wetenskappe V2 4 LIMPOPO September 2020 NSC Marking Guideline / NSS Nasienriglyn QUESTION 4/VRAAG 4 4.1.1 Addition / hydration Addisie / hidrasie//hidratering (1) 4.1.2 Substitution /halogenation/ chloronation Substitusie /halogenasie//halogenering//chloronering (1) 4.1.3 Elimination/dehydration Eliminasie//dehidrasie//dehidratering (1) 4.1.4 Substitution Substitusie (1) 4.2.1 H2SO4 / H3PO4  (1) Marking criteria/nasienriglyne 4.2.2 H H H Functional group correct/ I I I Funksionele groep korrek  H—C—C—C—H Whole molecule correct/ Molekuul korrek I I I Note: Accept OH. Line (bond) must be from C to O H O-H H Aanvaar OH. Lyn (binding) moet vanaf C na O wees  (2) Marking criteria/nasienriglyne 4.2.3 2-propanol / propan-2-ol propanol (2) Everything correct / Alles reg 4.3.1 Hydrogenation /hidrogenasie/hidrogenering (1) 4.3.2 H H H H H H I I I  I I I H—C = C—C—H + H - H → H—C—C—C—H I  I I I  H H H H Accept/aanvaar H2 (3) 4.4.1 2-chloropropane / 2-chloropropaan (2) 4.4.2 Sodium hydroxide /potassium hydroxide  Natriumhidroksied/ kaliumhidroksied (1) 4.4.3 Dilute base OR adding of water /verdunde basis OF byvoeging van water (Mild) heat/(Matige) hitte (2) [18] QUESTION 5/VRAAG 5 5.1.1 Endothermic reaction /endotermiese reaksie (1) 5.1.2 Energy is absorbed  OR Energy is required for reaction to take place OR Energy is absorbed from the surroundings Energie is geabsorbeer OF Energie word benodig vir die reaksie om plaas te vind OF Energie word geabsorbeer uit die omgewing. (1) 5.2.1 NO/gas escapes OR it is not a closed system NO/ gas ontsnap OF dit is nie ʼn geslote sisteem nie (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences P2 / Fisiese Wetenskappe V2 5 LIMPOPO September 2020 NSC Marking Guideline / NSS Nasienriglyn 5.2.2 ∆m Rate/tempo = - ∆t 6,3-0 =- = 0,06 g∙s-1  0-105 (accept/aanvaar -0,06 g∙s-1) (3) 5.2.3 Reaction is completed/all Cu(reactant) is used up(NOT equilibrium) Reaksie is voltooi/al die Cu(reaktante) is opgebruik (NIE ewewig nie) (1) 5.2.4 Temperature increased/heat is given off /exothermic reaction Accept: HNO3 removes CuO from Cu surface/ cleans copper surface Temperatuur neem toe/ hitte word vrygestel/ eksotermiese reaksie Aanvaar HNO3 verwyder CuO vanaf Cu oppervlak/ maak Cu oppervlak skoon (1) 5.2.5 Concentration of HNO3 decreased/ reactants are being used up _ Konsentrasie van HNO3 neem af /reaktanse opgebruik (1) 5.2.6 The number of particles has decreased Thus fewer/less effective collisions occur per second Die aantal deeltjies neem af Minder effektiewe botsings vind per sekonde plaas (2) 5.2.7 m 6,3  NO: n = = = 0,21 mol (Accept / Aanvaar 6,2 - 6,4) M 30 nCu : nNO 0,21 1 : 4 ∴ = 0,052 mol (Using ratio / toepassing van verhouding) 4 Cu: m = nM = 0,052 x 63,5 = 3,30 g  (4) 5.2.8 Increase the concentration of HNO3 Increase the temperature of the solution Use Cu powder / smaller pieces of Cu/increase the surface area of Cu  Verhoog die konsentrasie van die HNO3 Verhoog die temperatuur van die oplossing Gebruik Cu-poeier kleiner Cu stukkies/ vergroot die reaksie oppervlak van Cu (3) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences P2 / Fisiese Wetenskappe V2 6 LIMPOPO September 2020 NSC Marking Guideline / NSS Nasienriglyn 5.3 Marking guidelines/Nasienriglyne • Substitution of/vervanging van: 0,25x0,1  • Use mol raio/ gebruik molverhouding: 1:2 ; 0,025:0,0125  m • Formula/formule: n =  M • Substitute/vervang: 32 0,0075 0,24 • Substitute/vervang: OR/OF  0,0125 0,4 • Final answer/finale antwoord: 60 % n = cV = 0,25x0,1 = 0,025 mol HCℓ : S 2 : 1 0,025 : 0,0125  Option 1/opsie 1 Option 2/opsie 2 m 0,24 m = nM n= = = 0,0075 mol M 32 = 0,0125 x 32 = 0,4 g 0,0075 0,24 % opbrengs = x 100 % opbrengs = x 100 0,0125 0,4 = 60 %  = 60 % (6) [19] QUESTION 6/VRAAG 6 6.1 When an external stress (change in pressure, temperature or concentration) is applied to a closed system in chemical equalibrium, the equilibrium point will change in such a way as to counteract the stress. Wanneer die ewewig in ʼn geslote sisteem versteur word (verandering in druk,temperatuur of konsentrasie) stel die sisteem ʼn nuwe ewewig in deur die reaksie wat die versteuring teëwerk te bevoordeel. (2) 6.2 Reaction producing fewer moles/ less volume of gas favoured The reverse reaction is favoured The amount of ozone will increase Die reaksie wat minder gas vorm/ kleinervolume gas vorm word bevoordeel Die terugwaatse reaksie word bevoordeel Dus sal die hoeveelheid osoon vermeerder (3) 6.3.1 Reverse /terugwaarts (1) 6.3.2 Exothermic  /eksotermies (1) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences P2 / Fisiese Wetenskappe V2 7 LIMPOPO September 2020 NSC Marking Guideline / NSS Nasienriglyn 6.3.3 Decreases/verlaag (1) 6.4 A catalyst is a chemical substance which increases the rate of a reactionwithout undergoing a permanent change itself // ʼn Katalisator is ʼn chemiese stof wat die tempo van ʼn chemiese reaksie verhoog sonder om self ʼn permanente verandering te ondergaan. OR/OF A catalyst increases the rate of a reaction by providing an alternative route with lower activation energy. // ʼn Katalisator verhoog die tempo van ‘n reaksie deur ‘n alternatiewe roete van (2) laer aktiveringsenergie te verskaf. 6.5 Amount of oxygen remains the same A catalyst speeds up the rate of the forward and reverse reactions equally Die hoeveelheid van suurstof bly dieselfde. ʼn Katalisator verhoog die tempo van die voorwaarte en terugwaartse reaksies ewe vee (2) 6.6 Mark allocation/Puntetoekenning • Substitution of 0,72 mol NO at equilibrium or 0,36 mol.dm 3 if using concentrations/ vervanging van 0,72 mol by ewewig of 0,36 mol∙dm -3 as konsentrasie gebruik word. • Change in NO (0,54/ 1,08)/verandering in NO(0,54/ 1,08)  • USING ratio/GEBRUIK verhouding: 1:1:1  • Divide or multiply by volume/Gedeel deur of vermenigvuldig met volume (2 dm3)  • Correct Kc expression (formulae in square brackets).  Korrekte Kc -uitdrukking (formules tussen vierkanthakies). • Substitution of reactant and product concentrations/ Vervanging van (7) reaktans- en produkkonsentrasies.  • Correct final answer/Korrekte finale antwoord: 20,25 Moles/mol: O3 NO O2 NO2 Initial moles 0,6X2=1,2 0,9X2=1,8 0,73X2=1,46 0,55X2=1,10 Aanvanklik mol Change 1,08 (-) 1,08  (+) 1,08 1,08 /verandering Ratio Equilibrium 0,12 0,36X2=0,72 2,54 2,18  Ewewig  (moles / mol) ÷2 Concentration C=n/v 0,36 1,27 1,09 Konsentrasie =0,12/2=0,06 [O2 ][NO2 ] (1,27)(1,09) Kc = = = 64,09  (64,0-64,2) [O2 ][NO] (0,06)(0,36) Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences P2 / Fisiese Wetenskappe V2 8 LIMPOPO September 2020 NSC Marking Guideline / NSS Nasienriglyn O3 NO O2 NO2 Initial moles 0,6 0,9 0,73 0,55 Aanvanklik mol Change 0,54 (-) 0,54  (+) 0,54 0,54 Verandering Concentration 0,06 0,36  1,27 1,09 ratio Konsentrasie Concentration/konsentrasie x2 [O2 ][NO2 ] (1,27)(1,09) Kc = = = 64,09  (64,0-64,2) (7) [O2 ][NO] (0,06)(0,36) 6.7.1 Increases/ neem toe (1) 6.7.2 Remains the same/ bly dieselfde (1) 6.7.3 Increases/ neem toe (1) 6.7.4 Decreases/ neem af (1) 6.7.5 Remains the same / bly dieselfde (1) [24] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences P2 / Fisiese Wetenskappe V2 9 LIMPOPO September 2020 NSC Marking Guideline / NSS Nasienriglyn QUESTION 7/VRAAG 7 7.1 Burette / Buret (1) 7.2 _ C Titration of a weak acid and strong base OR _ CH3COOH is a weak acid and NaOH a strong base Titrasie van ‘n swak suur en sterk basis OF CH3COOH is ‘n swak suur en NaOH “n sterk basis. (2) 7.3 Weak acids ionizes incompletely in water to form a low concentration of H3O+ ions. + Swak sure ioniseer onvolledig in water om 'n lae konsentrasie H3O -ione te vorm (2) 7.4 NaOH → Na+(aq) + OH-(aq) [OH-] = 0,11mol∙dm-3  (1 mark for 0,11/ 1 punt vir 0,11)) -14 + - Kw = 1x10 = [H3O ][OH ]  1x10-14 = [H3O+](0,11) [H3O+] = 9,09x10-14  pH = -log[H3O+]  = - log (9,09x10-14) = 13,04 OR [OH-] = 0,11 mol∙dm-3  pOH = -log[OH-]  = - log 0,11 = 0,06  pH = 14 – pOH = 14 – 0,06 = 13,04 (5) 7.5 n=cxV = 0,11 x 0,0285  = 0,0031 mol  (0,003135 ) (2) 7.6 Positive marking from QUESTION 7.5/ Positiewe nasien vanaf VRAAG 7.5 n acid/suur : n base/basis = 1:1 n acid/suur = 0,0031 mol m acid/suur in 25cm3 = nxM = 0,0031 x 60 = 0,186g  m acid/suur in 100 cm3 = 0,186 x 4 = 0,744g 0,744 % etanoic acid/ etanoёsuur =  x 100 = 9,9 % (9,8 – 10) (5) 7,5 [17] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences P2 / Fisiese Wetenskappe V2 10 LIMPOPO September 2020 NSC Marking Guideline / NSS Nasienriglyn QUESTION 8/VRAAG 8 8.1 Diprotic /Diproties (1) 8.2 pH = -log[H3O+] 1,6 = -log[H3O+]  [H3O+] = 0,025  [H2SO4] = 0,0125 mol∙dm-3  (3) 8.3.1 Reaction of a salt with water / Die reaksie van ‘n sout met water (2) 8.3.2 Acidic / Suur NH+4 (aq) + H2O → NH3 + H3O (aq) + [H3O ] increases  /[H3O ] neem toe + + (4) [10] TOTAL/ TOTAAL: 150 Copyright reserved Please turn over