Physical Sciences Grade 11 P2 2015 Memo

Downloaded from hlayiso.com - NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 PHYSICAL SCIENCES: CHEMISTRY (P2) FISIESE WETENSKAPPE: CHEMIE (V2) NOVEMBER 2015 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 12 pages. Hierdie memorandum bestaan uit 12 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 2 DBE/November 2015 CAPS/KABV – Grade 11/Graad 11 – Memorandum QUESTION 1/VRAAG 1 1.1 D  (2) 1.2 C  (2) 1.3 D  (2) 1.4 B  (2) 1.5 D  (2) 1.6 B  (2) 1.7 B  (2) 1.8 A  (2) 1.9 A  (2) 1.10 D  (2) [20] QUESTION 2/VRAAG 2 2.1 The sharing of electrons  between (two) atoms (to form a molecule).  Die deling van elektrone tussen (twee) atome (om 'n molekuul te vorm). (2) 2.2 xx x x Marking criteria/Nasienriglyne: x O◦ H  x◦ • O atom shown with 8 electrons around it. H O-atoom getoon met 8 elektrone rondom dit. • Two electron pairs on O atom shared with two H atoms as shown. Twee elektronpare op O-atoom word gedeel met H-atome soos getoon. (2) 2.3 2.3.1 xx + H xx O◦x H  x◦ H (1) 2.3.2 • One atom/ion must have an empty valence shell / orbital.  Een atoom/ioon moet 'n leë valensskil/orbitaal hê. • The other atom must have a lone pair of electrons.  Die ander atoom moet 'n alleenpaar-elektrone hê. (2) 2.4 2.4.1 The tendency of an atom in a molecule  to attract bonding electrons closer to itself.  Die neiging van 'n atoom in 'n molekuul om bindingselektrone nader aan ditself aan te trek. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 3 DBE/November 2015 CAPS/KABV – Grade 11/Graad 11 – Memorandum 2.4.2 (a) ΔEN(between C and O/tussen C en O) =1  (1) (b) ΔEN(between H and O/tussen H en O) =1,4  (1) 2.4.3 • The bonds in both molecules are polar due to the difference in electronegativities  between C and O and H and O. Die bindings in beide molekule is polêr weens die verskil in elektronegatiwiteit tussen C en O en H en O. • The shape of the H 2 O molecule is angular  and therefore the molecule is polar  because one side of the molecule can be positive and the other side negative. Die vorm van die H 2 O-molekuul is hoekig en dus is die molekuul polêr omdat een kant van die molekuul positief en die anderkant negatief kan wees. • The shape of the CO 2 molecule is linear and thus it is non-polar  because the the charge distribution is symmetrical. Die vorm van die CO 2 -molekuul is liniêr en dus is die molekuul nie-polêr omdat die ladingsverspreiding simmetries is. (6) [17] QUESTION 3/VRAAG 3 3.1 3.1.1 The temperature at which the vapour pressure of a liquid equals external (or atmospheric) pressure.  Die temperatuur waarby die dampdruk van 'n vloeistof gelyk is aan die eksterne (of atmosferiese) druk. (2) 3.1.2 M = 6(12) + 14(1) = 86 g·mol-1  (1) 3.1.3 Gas  (1) 3.1.4 London/dispersion/induced dipole forces  London-/dispersie-/geïnduseerde-dipoolkragte (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 4 DBE/November 2015 CAPS/KABV – Grade 11/Graad 11 – Memorandum 3.1.5 • Molecular mass increases from A to E.  Molekulêre massa neem toe van A tot E. • Strength of intermolecular forces/London forces/dispersion forces/ induced dipole forces increases.  Sterkte van intermolekulêre kragte/Londonkragte/dispersiekragte/geïn- duseerde dipoolkragte neem toe. • More energy is needed to overcome/break the intermolecular forces.  Meer energie benodig om intermolekulêre kragte/Londonkragte/ dispersiekragte/geïnduseerde dipoolkragte te oorkom/breek. OR/OF • Molecular mass decreases from E to A.  Molekulêre massa neem af van E tot A. • Strength of intermolecular forces/London forces/dispersion forces/induced dipole forces decreases.  Sterkte van intermolekulêre kragte/Londonkragte /dispersiekragte/geïn- duseerde dipoolkragte neem af. • Less energy is needed to overcome the intermolecular forces.  Minder energie benodig om intermolekulêre kragte/Londonkragte/ dispersiekragte/geïnduseerde dipoolkragte te oorkom/breek. (3) 3.1.6 Higher than/Hoër as  (1) 3.2 3.2.1 H2S  (1) 3.2.2 Hydrogen bonding/Waterstofbinding  (1) 3.2.3 Hydrogen bonding between H 2 O molecules is stronger  than the London forces/dispersion forces/induced dipole forces or dipole-dipole forces between H 2 S molecules.  More energy is needed to overcome/break the intermolecular forces in water.  Waterstofbinding tussen H 2 O-molekule is sterker as die Londonkragte/ dispersiekragte/geïnduseerde dipoolkragte of dipool-dipoolkragte tussen H 2 S- molekule. Meer energie benodig om intermolekulêre kragte in water te oorkom/breek. (4) [15] QUESTION 4/VRAAG 4 4.1 Boyle’s law / Boyle se wet  (1) 4.2 Pressure Volume 1 Druk (kPa) (cm3) (cm-3) V 40 43 0,02 80 27 0,04 100 22 0,05  120 18 0,06 (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 5 DBE/November 2015 CAPS/KABV – Grade 11/Graad 11 – Memorandum 4.3 1 1 Graph of pressure versus /Grafiek van druk teenoor V V 120  100  80  60 p (kPa) 40  20 0 0,02 0,04 0,06 0,08 0,1 1 (cm-3) V Criteria for graph/Riglyne vir grafiek: Two points correctly plotted./Twee punte korrek gestip.  Four points correctly plotted./Vier punte korrek gestip.  Straight line of best fit drawn./ Reguitlyn van beste passing getrek.  The straight line will intercept origin if extended.  Reguitlyn gaan deur oorsprong wanneer dit verleng word. (4) 4.4 1 Notes/Aantekeninge: = 0,034  ∴ V = 29,41 cm3  IF/INDIEN: V Only answer given, allocate 2 marks./Slegs antwoord gegee, ken 2 punte toe. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 6 DBE/November 2015 CAPS/KABV – Grade 11/Graad 11 – Memorandum 4.5 4.5.1 pV = nRT  (100 x 103)  (22 x 10-6)  = n(8,31)(298)  ∴ n = 8,88 x 10-4 mol m n= M 2,49 × 10 −2 ∴ 8,89 x 10-4 =  M ∴M = 28,03 g·mol-1  (6) 4.5.2 N2  (1) 4.6 4.6.1 Volume (cm3)  Temperature/Temperatuur (K) (1) 4.6.2 At low temperatures, molecules move slower/with less kinetic energy.  Intermolecular forces become more effective.  The gas liquifies and the volume (open space in container) is larger than predicted for ideal gases.  By lae temperature beweeg molekule stadiger/met minder kinetiese energie. Intermolekulêre kragte word meer effektief. Die gas vervloei en die volume (oop ruimte in houer) is groter as voorspel vir ideale gasse. (3) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 7 DBE/November 2015 CAPS/KABV – Grade 11/Graad 11 – Memorandum QUESTION 5/VRAAG 5 5.1 CO 2  (1) 5.2 The amount of substance  having the same number of particles as there are atoms in 12 g carbon-12.  Die hoeveelheid stof wat dieselfde getal deeltjies het as wat daar atome is in 12 g koolstof-12. (2) 5.3 m n(NaHCO 3 ) =  M 3,36 =  84 = 0,04 mol  (3) 5.4 POSITIVE MARKING FROM QUESTION 5.3. POSITIEWE NASIEN VAN VRAAG 5.3. m n(H3C6H5O7 ) = M 1,8 =  192 = 0,01 mol (9,38 x 10-3 mol) n(NaHCO 3 needed/benodig) = 3n(H 3 C 6 H 5 O 7 ) = 3(0,01) mol  = 0,03 mol  n(NaHCO 3 ) < n(NaHCO 3 in antacid/in teensuurmiddel) H 3 C 6 H 5 O 7 /citric acid is the limiting reactant.  H 3 C 6 H 5 O 7 /sitroensuur is die beperkende reagens. (4) 5.5 POSITIVE MARKING FROM QUESTION 5.3 & 5.4. POSITIEWE NASIEN VAN VRAAG 5.3 & 5.4. n(NaHCO 3 in excess/oormaat) = 0,04 - 0,03  = 0,01 mol m(NaHCO 3 in excess/oormaat) = nM = (0,01)(84)  = 0,84 g  (3) 5.6 POSITIVE MARKING FROM QUESTION 5.3.1. POSITIEWE NASIEN VAN VRAAG 5.3.1. m Marking criteria/Nasienriglyne: n(CO 2 ) = • Using/Gebruik M(CO 2 ) = 44 g·mol-1 M  m • 3(CO 2 ) = n(NaHCO 3 ) ∴ 0,03 = • Final answer: 1,32 g 44  Finale antwoord: 1,32 g ∴ m(CO 2 ) = 1,32 g  (3) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 8 DBE/November 2015 CAPS/KABV – Grade 11/Graad 11 – Memorandum QUESTION 6/VRAAG 6 6.1 6.1.1 The amount of solute  per litre/cubic decimetre of solution.  Hoeveelheid opgeloste stof per liter/kubieke desimeter oplossing. (2) 6.1.2 n c=  V n 0,2 =  200 × 10 −3 = 0,04 mol  (3) 6.1.3 POSITIVE MARKING FROM QUESTION 6.1.2. POSITIEWE NASIEN VAN VRAAG 6.1.2. n(SO 2 ) = ½n(HCℓ) = ½(0,04)  = 0,02 mol V n= Vm ∴ V = (0,02)(22,4)  = 0,45 dm3  (3) 6.2 6.2.1 Marking criteria/Nasienriglyne: • Substitute/Vervang 44 g·mol-1.  • n(C) = n(CO 2 )  • Substitution/Vervang 12 g·mol-1.  • Final answer/Finale antwoord: 7,68 g  OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 m 12 n= %C in CO 2 = × 100 M 44 28,16 = 27,27% ∴ n(CO 2 ) = = 0,64 mol 44  m(C )in CO 2 = 27,27% of 28,16 g n(C) = n(CO 2 )  = 0,64 mol = 7,68 g  m(C) = nM = (0,64)(12)  = 7,68 g  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 9 DBE/November 2015 CAPS/KABV – Grade 11/Graad 11 – Memorandum 6.2.2 POSITIVE MARKING FROM QUESTION 6.3.1. POSITIEWE NASIEN VAN VRAAG 6.3.1. Marking criteria/Nasienriglyne: • Substitute M(H 2 O) to calculate n(H 2 O).  Vervang M(H 2 O) om n(H 2 O) te bereken. • n(H) = 2n(H 2 O).  • Substitution M(H) to calculate m(H).  Vervang M(H) om m(H) te bereken. • M(O) = m(menthol) – (m(C) + m(H))  • Substitution M(O) to calculate n(O).  Vervang M(O) om n(O)te bereken. • Ratio/Verhouding: mol C : mol H : mol O = 10 : 20 : 1  • Empirical formula/Empiriese formule: C 10 H 20 O  OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 n(H) and m(H) 2 %H in H 2 O = × 100 m 18 n= M = 11,11% 11,52 ∴ n(H 2 O) = m(H) in H 2 O = 11,11% of 11,52 g  18 = 0,64 mol = 1,28 g n(H) = 2n(H 2 O) m(O) = 9,984 – (7,68 + 1,28) = 2(0,64)  = 1,024 g  = 1,28 mol C : H : O m(H) = nM 7,68 1,28 1,024 : : = (1,28)(1)  12 1 16  = 1,28 g 0,64 : 1,28 : 0,064 n(O) and m(O) m(O) = 9,984 – (7,68 + 1,28)  10 : 20 : 1  = 1,024 g Empirical Formula: C 10 H 20 O  1,024 Empiriese formule: C 10 H 20 O n(O) =  16 = 0,064 mol Ratio: mol C : mol H : mol O 0,64 : 1,28 : 0,064 10 : 20 : 1  Empirical formula: C 10 H 20 O  Empiriese formule: C 10 H 20 O (7) 6.2.3 POSITIVE MARKING FROM QUESTION 6.2.2. POSITIEWE NASIEN VAN VRAAG 6.2.2. M(C 10 H 20 O) = 10(12) + 20 + 16 = 156 g·mol-1  Molecular formula/Molekulêre formule: C 10 H 20 O  (2) [21] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 10 DBE/November 2015 CAPS/KABV – Grade 11/Graad 11 – Memorandum QUESTION 7/VRAAG 7 7.1 Exothermic/Eksotermies  Reactants at higher energy than products./Products at lower energy than reactants./Energy is released./ΔH < 0.  Reaktanse by hoër energie as produkte./Produkte by laer energie as reakanse./Enegie is vrygestel./ ΔH < 0. (2) 7.2 7.2.1 A (1) 7.2.2 A–B (1) 7.2.3 B–C (1) 7.3 1 mol Ba(OH) 2 releases/stel vry: 116 kJ  0,18 mol Ba(OH) 2 releases/ stel vry: 0,18 x 116  = 20,88 kJ  (Accept answers in range/Aanvaar antwoorde in gebied: 20,3 – 20,88 kJ) (3) [8] QUESTION 8/VRAAG 8 8.1 An acid is a proton donor.  'n Suur is 'n protondonor/protonskenker. (2) 8.2 8.2.1 SO 24− (aq)  (1) 8.2.3 H 2 SO 4 (aq)  (1) 8.2.3 HSO −4 (aq)  (1) 8.3 8.3.1 Neutralisation / Neutralisasie  (1) 8.3.2 H 2 SO 4 (aq) + KOH(aq)  → K 2 SO 4 (aq) + 2H 2 O(ℓ)  Bal.  Notes/Aantekeninge: • Reactants  Products  Balancing:  Reaktanse Produkte Balansering • Ignore double arrows. / Ignoreer dubbelpyle. • Marking rule 6.3.10. / Nasienreël 6.3.10. (3) 8.3.3 Blue to yellow  Blou na geel (2) 8.3.4 Potassium sulphate / Kaliumsulfaat  (1) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 11 DBE/November 2015 CAPS/KABV – Grade 11/Graad 11 – Memorandum QUESTION 9/VRAAG 9 9.1 9.1.1 Oxidation is an increase in oxidation number. Oksidasie is 'n toename in oksidasiegetal. (2) 9.1.2 2Cr + 7O = -2 2Cr + (-14) = 2 Cr = +6  (1) 9.1.3 2H + 2O = 0 2 + 2O = 0 O = -1  (1) 9.2 9.2.1 A reducing agent loses/donates electrons.  'n Reduseermiddel verloor/skenk elektrone. (2) 9.2.2 Fe2+(aq)  (1) 9.2.3 Cℓ 2 (g)  (1) 9.2.4 Cℓ 2 (g) + 2e- → 2Cℓ- Marking guidelines/Nasienriglyne: • Cℓ 2 + 2e- ⇌ 2Cℓ- 1 2Cℓ- ⇌ Cℓ 2 + 2e- 0 2 2 - • 2Cℓ ← Cℓ 2 + 2e - 2 - 2Cℓ → Cℓ 2 + 2e - 0 2 2 (2) 9.2.5 Fe2+(aq) → Fe3+(aq) + e-  Marking guidelines/Nasienriglyne: • Fe2+ ⇌ Fe3+ + e- 1 Fe3+ + e- ⇌ Fe2+ 0 2 2 • Fe 3+ - 2+ + e ← Fe 2 Fe 3+ - + e → Fe 2+ 0 2 2 (2) 9.2.6 2Fe2+(aq) + Cℓ 2 (g) → 2Fe3+(aq) + 2Cℓ-  Notes/Aantekeninge: • Ignore double arrows./Ignoreer dubbelpyle. (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe V2 12 DBE/November 2015 CAPS/KABV – Grade 11/Graad 11 – Memorandum QUESTION 10/VRAAG 10 10.1 10.1.1 Remove impurities/Verwyder onsuiwerhede  (1) 10.1.2 Reducing agent/Formation of CO or CO 2  Reduseermiddel/Vorming van CO of CO 2 (1) 10.2 10.2.1 (Hot) air/(Warm) lug  (1) 10.2.2 Molten iron/pig iron/Fe(ℓ)  Gesmelte yster/ru-yster/Fe(ℓ) (1) 10.2.3 Slag/CaSiO 3 /metaalskuim  (1) 10.3 Carbon monoxide/carbon dioxide/nitrogen monoxide/nitrogen dioxide/nitrogen/ sulphur dioxide  Koolstofmonoksied/Koolstofdioksied/stikstofmonoksied/stikstofdioksied/swawel- dioksied (1) 10.4 10.4.1 Reducing agent/Reduseermiddel  (1) 10.4.2 Fe 2 O 3  (1) [8] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou