Physical Sciences Grade 11 P2 Gauteng November 2018 Memo

Downloaded from hlayiso.com - testpapers.co.za NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 PHYSICAL SCIENCES: CHEMISTRY (P2) FISIESE WETENSKAPPE: CHEMIE (V2) NOVEMBER 2018 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 16 pages. Hierdie nasienriglyne bestaan uit 16 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 2 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 B  (2) 1.2 C  (2) 1.3 D  (2) 1.4 B  (2) 1.5 C  (2) 1.6 A  (2) 1.7 A  (2) 1.8 B  (2) 1.9 D  (2) 1.10 D  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 3 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1 Chemical bond is mutual attraction between two atoms resulting from the simultaneous attraction between their nuclei and (outer) electrons.  Chemiese binding is die wedersydse aantrekking tussen twee atome as gevolg van die gelyktydige aantrekking tussen hulle kerne en (buite)- elektrone. (2) 2.2.1 H: C :: :N:  (2) 2.2.2 : O :H : :  H (2) 2.3 Linear  (1) Lineêr 2.4 3 – 2,5 = 0,5  (1) 2.5 Polar  Polêr (1) 2.6 CN has a higher order bond/triple bond with more orbitals overlapping  than CH, which is a single bond.  Thus CN bond needs more energy to break. CN het 'n hoër orde/drievoudige binding met meer orbitale wat oorvleuel as die CN enkel binding. Dus benodig die CN-binding meer energie om te breek. (2) 2.7 CN has a longer bond length than CH  because the H atom is smaller than the N atom.  CN het 'n groter bindingslengte as CH omdat die H-atoom kleiner as die N-atoom is (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 4 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 2.8 Yes/Ja  (1) - 2.9 • HCN has polar molecules with dipole-dipole forces.  • H 2 O has polar molecules with hydrogen bonds (dipole-dipole forces).  • If the forces are of the same order/comparable the substances will dissolve.  OR • Both molecules are polar  • HCN has dipole-dipole forces and H 2 O has (dipole-dipole forces) hydrogen bonds  • Like dissolve like.  • HCN het polêre molekules met dipool-dipoolkragte. • H 2 O het polêre molekules met waterstofbindings. • Indien die intermolekulêre kragte van dieselfde orde is, sal stowwe oplos OF • Beide molekules is polêr  • HCN het dipool-dipool kragte en H 2 O het (dipool-dipool kragte) waterstofbindings  • Soort los op in soort  (3) [17] QUESTION 3/VRAAG 3 3.1 • NH 3 has hydrogen bonds between the molecules  • N 2 has London forces/induced dipole forces  • NH 3 has stronger intermolecular forces than N 2 and therefore a higher boiling point than N 2  (Accept: more energy requires to overcome stronger forces of NH 3 ) OR • N 2 has weaker intermolecular forces than NH 3 and therefore a lower boiling point than NH 3 (Accept: less energy requires to overcome weaker forces of H 2 ) • NH 3 het waterstofbindings tussen die molekules • N 2 het Londonkragte/geïnduseerde dipoolkragte • NH 3 het sterker intermolekulêre kragte as N 2 en daarom 'n hoër kookpunt as N 2 OF • N 2 het swakker intermolekulêre kragte as NH 3 en daarom 'n laer kookpunt as NH 3 (Aanvaar: NH 3 vereis meer energie om sterker kragte te oorkom) (3) 3.2 H2  (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 5 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 3.3 • H 2 and N 2 both have weak London forces/induced dipole forces  • N 2 is a larger molecule/has a greater molecular mass/has a larger surface area than H 2  • and therefore N 2 has stronger intermolecular forces.  OR • H 2 is a smaller molecule/has a smaller molecular mass/has a smaller surface area than N 2  • and therefore H 2 has weaker intermolecular forces.  • H 2 en N 2 het beide swak Londonkragte/geïnduseerde dipoolkragte • N 2 is 'n groter molekule/groter molekulêre massa/groter oppervlakarea as H 2 • en daarom het N 2 sterker intermolekulêre kragte. OF • H 2 is 'n kleiner molekule/het kleiner molekulêre massa/kleiner oppervlakarea as N 2 • en daarom swakker intermolekulêre kragte. (3) 3.4 H2  It has the weakest intermolecular forces/London forces  It has the lowest boiling point  OR It has the weakest intermolecular forces/London forces  Boiling point is inversely proportional to vapour pressure  Dit het die swakste intermolekulêre kragte Dit het die laagste kookpunt OF Dit het die swakste intermolekulêre kragte/Londonkragte Kookpunt is omgekeerd eweredig aan die dampdruk (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 6 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1 Boyle's law/Boyle se wet  (1) 4.2 Criteria for hypothesis/Riglyne vir hipotese The dependent and independent variables are stated correctly.  Die afhanklike en onafhanklike veranderlikes korrek genoem. State the relationship between the dependent and independent  variables. Stel die verwantskap tussen die afhanklike en onafhanklike veranderlike. Dependent variable/afhanklike veranderlike: volume Independent variable/onafhanklike veranderlike: pressure/druk Example:/Voorbeeld: If the pressure of and enclosed gas increases the volume will decreases at constant temperature. The pressure of an enclosed gas is inversely proportional to the volume it occupies if the temperature is kept constant. Die druk van 'n ingeslote gas is omgekeerd eweredig aan die volume wat dit beslaan indien die temperatuur konstant gehou word. Indien die druk van ‘n ingeslote gas toeneem, sal die volume afneem. (2) 4.3 Graph of volume versus pressure Grafiek van volume teenoor druk 200 180 160 140 Volume (dm3) 120 100 80 60 40 20 0 0 50 100 150 200 250 300 Pressure/Druk (kPa) Refer to the last page of the memo for the graph drawn to scale Verwys na die laaste bladsy van die memo vir die skaalgrafiek Criteria for marking the graph/Nasienkriteria vir grafiek Use of correct scale on both axis  (If learners used table values as scale values maximum 1/ 3 for line drawn) Korrekte skaal op die asse (Indien leerders tabelwaardes as skaalwaardes gebruik maksimum 1/ 3 vir lyn getrek) At least five (5) points plotted correctly  Ten minste vyf (5) punte korrek gestip Curve is drawn  Kurwe getrek (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 7 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 4.4 Any set of values can be used from the table : Enige stel waardes vanaf die tabel kan gebruik word: p1V 1 = p 2V 2  70(174) = (300)V 2  V 2 = 40,6 dm3  (Accept/Aanvaar 40,32 – 40,8 dm3) (3) 4.5 At high pressure a gas starts to deviate from ideal gas behaviour  because the volume of the molecules of a gas and the intermolecular forces start to influence the measured value, causing it to be greater than the theoretical value calculated/Forces of repulsion between the gas particles prevents them from moving closer  By hoë druk sal 'n gas begin afwyk van ideale gasgedrag want die volume van die gasdeeltjies en intermolekulêre kragte begin die waarde van die volume beïnvloed, wat veroorsaak dat die gemete waarde groter is as die berekende waarde/Afstotingskragte veroorsaak dat gasdeeltjies nie nader aan mekaar kan beweeg nie (2) 4.6 Low/Laag (1) - 4.7 Temperature is an indication of the average kinetic energy of the molecules of a gas. If the temperature of a gas decreases, the molecules move slower and closer together up to a point where the gas will start to condense  and not behave like an ideal gas. OR The intermolecular forces of attraction becomes significant  then the gas condenses.  Temperatuur is die aanduiding van die gemiddelde kinetiese energie van die molekules van 'n gas. Indien die temperatuur afneem sal die molekules stadiger en nader aan mekaar beweeg tot by die punt waar die gas sal begin kondenseer sodat dit nie meer soos 'n ideale gas optree nie. OF Die intermolekulêre kragte word beduidend en dit veroorsaak dat die gas kondenseer. (2) 4.8 pV = nRT  (70 000)(174 x 10-3)  = n(8,31)(293)  n = 5 moles  (4) [18] QUESTION 5/VRAAG 5 5.1 m n= M 160 n=  40 n = 4 mole pV = nRT  (120 000)V = (4)(8,31)(288)  V = 0,08 m3  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 8 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 5.2 POSITIVE MARKING FROM QUESTION 5.1 POSITIEWE NASIEN VANAF VRAAG 5.1 OPTION 1/OPSIE 1 p1V1 p 2 V2 =  T1 T2  120(0,08) 240 V2 =  288 308 V 2 = 0,043 m3  (427,78 dm3) OPTION 2/OPSIE 2 pV = nRT  (240 000)V  = (4)(8,31)(308)  V = 0,043 m3  (426,58 dm3) (4) [8] QUESTION 6/VRAAG 6 6.1.1 One mole is the amount of a substance having the same number of particles as there are atoms in 12 g carbon-12.  Een mol is die stofhoeveelheid wat dieselfde getal deeltjies het as wat daar atome in 12 g koolstof-12 is. [2 or 0] (2) 6.1.2 Na 2 CO 3 + 2HCℓ  → 2NaCℓ + H 2 O + CO 2  (2) 6.1.3 Mark allocation/Nasienriglyne: • Formula for/ Formule vir volume  • Substitution of 0,306 and 24,45 /Vervanging van 0,306 en 24,45 • Using ratio/Gebruik verhouding  • Formula for/Formule vir mass  • Substitutions of moles and 106  /Vervanging van mole en 106 • Answer with units/Antwoord met eenheid  V n=  Vm 0,306 n=  24,45 n = 0,0125 mol of CO 2 Na 2 CO 3 : CO 2 1 : 1  (use of the ratio/gebruik die verhouding) n = 0,0125 mol of Na 2 CO 3 m n=  M  m 0,0125 = 106  m = 1,33 g  (1,325 – 1,33 g) (7) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 9 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 6.1.4 POSITIVE MARKING FROM QUESTION 6.3 POSITIEWE NASIEN VANAF VRAAG 6.3 OPTION 1/OPSIE 1 1,5 - 1,33  % CaCO 3 unreacted = × 100 1,5 = 11,33%  (11,33% - 11,67% ) OPTION 2/OPSIE 2 1,33  % CaCO 3 reacted = 1,5 x 100 = 88,67% % unreacted = 100 – 88,67 = 11,33%  OPTION 3/OPSIE 3 m n= M 1,5 n= 106 n = 0,0142 mol initial mol – reacted mol = 0,0142 – 0,0125 = 0,0017 mol unreacted 0,0017  % CaCO 3 unreacted = x 100 = 11,97%  (2) 0,0142 6.2.1 Sulphuric acid  (1) Swawelsuur Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 10 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 6.2.2 Mark allocation/Punte toekenning: • Any one of the formulae/Enige een van formules  • Substitution of/Vervanging van 3 g (5 g – 2 g)  • Ratio/Verhouding 1:1  • Substitution of moles and volume in dm3 /Vervanging van mol en volume in dm3  • Answer with units/Antwoord met eenheid  OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 m m n= n= M  any one of M  any one of 3 (5 – 2) 3  = the two = the two 65 formula/Enige 65 formula/Enige = 0,0462 mol of Zn een van = 0,0462 mol of Zn een van formules formules Ratio Zn : H 2 SO 4 Ratio Zn : H 2 SO 4 1:1  1:1  n(H 2 SO 4 ) = 0,0462 mol n(H 2 SO 4 ) = 0,0462 mol n m = nM c= = (0,0462)(98) V 0,0462 = 4,5276 g of H 2 SO 4 =  0,05 m = 0,92 mol·dm-3  c= MV 4,5276 =  (98)(0,05) = 0,92 mol·dm-3  OPTION 3/OPSIE 3  any one of m n= the two M formula/Enige 5 = een van 65 = 0,0769 mol of Zn initial formules m n= M 2 = 65 = 0,0308 mol of Zn final n used = 0,0769  – 0,0308 = 0,0461 mol Ratio Zn : H 2 SO 4 1:1  n(H 2 SO 4 ) = 0,0461 mol n c= V 0,0461 =  0,05 (5) = 0,92 mol·dm-3  [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 11 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 7/VRAAG 7 7.1 The minimum energy needed for a reaction to take place.  Die minimum energie benodig vir die reaksie om plaas te vind. [2 or 0] (2) 7.2 More energy is released than absorbed OR energy of products is less than energy of reactants OR ΔH < 0 OR ∆H is negative  Meer energie word afgegee as opgeneem OF energie van die produkte is minder as die energie van reaktante OF ΔH < 0 OF ∆H is negatief (1) 7.3 ΔH = H products - H reactants = 183,3 – 400  = -216,7 kJ·mol-1  (3) 7.4 2670 Ep (kJ·mol-1) Ea 400 183,3 Course of reaction/Reaksieverloop Criteria for marking/Nasienkriteria Lower curved line/Laer kurwelyn  Lower activation energy indicated  Laer aktiveringsenergie aangedui (2) 7.5 One mole of any gas occupies the same volume at the same temperature and pressure.  Een mol van enige gas beslaan dieselfde volume by dieselfde temperatuur en druk. [2 or 0] (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 12 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 7.6 OPTION 1/OPSIE 1 4NH 3 5O 2 4NO 6H 2 O Initial vol 6 9  0 0 Change in vol 6 7,5 6 9 Ratio  Final vol 0 1,5 6 9  3 Total volume = 1,5 + 6 + 9 = 16,5 dm  OPTION 2/OPSIE 2 Mol ratio 4 : 5 : 4 : 6 from balanced equation Mol verhouding 4 : 5 : 4 : 6 vanuit gebalanseerde vergelyking Volume ratio reacting 6 : 7,5 : 6 : 9  Volume verhouding wat reageer 6 : 7,5 : 6 : 9 O 2 in excess/in oormaat Only/Slegs 7,5 dm3 of/van 9 dm3 reacts/reageer 9 – 7,5 = 1,5 dm3   Total volume at the end of reaction = 6 + 9 + 1,5 = 16,5 dm3  Totale volume aan die einde van die reaksie = 6 + 9 + 1,5 = 16,5 dm3 (4) 7.7 m n= M 1,59 n(H) =  1 = 1,59 mol 22,2 n(N) =  14 = 1,5857 mol 76,2 n(O) =  16 = 4,625 mol H : N : O 1,59 1,5857 4,625 : :  divide by smallest number/ 1,5857 1,5857 1,5857 deel deur kleinste getal 1 : 1 : 3 Empirical formula/Empiriese formule is HNO 3  (5) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 13 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 A hydroxide ion can act as proton acceptor.  'n Hidroksiedioon kan optree as protonontvanger. (2) 8.2 Dative covalent bond  Datiefkovalente binding (1) 8.3 HCℓ(aq) + H 2 O(ℓ)  → H 3 O+(aq) + Cℓ-(aq)  OR/OF HCℓ(aq) + NH 3 (g)  → NH 4 +(aq) + Cℓ-(aq)  (2) 8.4 Concentration is the amount of solute per litre of solution.  Konsentrasie is die hoeveelheid opgeloste stof per liter van 'n oplossing. OR/OF Concentration is the number of moles of a substance per dm3 of solution. Konsentrasie is die aantal mol van 'n stof per dm3 –oplossing. (2) 8.5 OPTION 1/OPSIE 1 OPTION 2/OPSIE 2 n c1V1 = c2V2  c= V  (0,75)(5)  = c 2 (1000)  n c 2 = 3,75 x 10-3 mol·dm-3  0,75 =  5 n = 3,75 mol OPTION 3/OPSIE 3  5 c n =  c= 1000 0,75 V  3,75 c = 3,75 x 10-3 mol·dm-3  =  1000 = 3,75 x 10-3 mol·dm-3  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 14 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 8.6 Mark allocation/Punte toekenning • Usage of formula(e) of c = n/V and/or n = m/M • Usage or calculation of number of moles (3,75 mol) of HNO 3 • Ratio/Verhouding 2:1 • Usage of 74 g.mol-1 in formula n = m/M • Answer/Antwoord • Correct conclusion/Korrekte gevolgtrekking OPTION 1/OPSIE 1 POSITIVE MARKING FROM 8.5 n  any one of POSITIEWE NASIEN VANAF 8.5 c= OPTION 2/OPSIE 2 V the two  any one of n n formula/Enige c = 0,75 = V the two 5 een van n -3 formula/Enige formules 3,75 x 10 = n = 3,75 mol  1000 een van n = 3,75 mol  formules Ratio HNO 3 : Ca(OH) 2 2:1  Ratio HNO 3 : Ca(OH) 2 n(Ca(OH) 2 ) = 1,875 mol 2:1  n(Ca(OH) 2 ) = 1,875 mol m n= m M n= m M 1,875 = m 74  1,875 = m = 138,75 g  74  m = 138,75 g  No, it is insufficient.  Nee, dit is nie genoeg nie No, it is insufficient.  Nee, dit is nie genoeg nie POSITIVE MARKING FROM 8.5 POSITIEWE NASIEN VANAF 8.5 OPTION 3/OPSIE 3 m  any one of n= the two M 120 formula/Enige n= een van 74  n = 1,62 mol formules n = cV = 3,75 x 10-3(1000) = 3,75 mol  HNO 3 : Ca(OH) 2 2:1 n(Ca(OH) 2 ) = ½(3,75) = 1,875 mol  1,875 > 1,62 mol No it is insufficient  Nee dis nie genoeg nie (6) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 15 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1 Reduction is a decrease in oxidation number  Reduksie is die afname in oksidasiegetalle (2) 9.2 Mn is +7 / Mn7+  (1) 9.3 H 2 S / S2-  (1) 9.4 The oxidation number of S increases  from -2 to 0  Die oksidasiegetal van S neem toe van -2 na 0 (2) 9.5 MnO −4 / Mn+7  (1) 9.6 H 2 S(g) → S + 2H+ + 2e-  (2) 9.7 H 2 S → S + 2H+ + 2e- MnO −4 + 8H+ + 5e- → 2Mn2+ + 4H 2 O  2 MnO −4 + 5H 2 S + 6H+ → 2Mn2+ + 5S + 8H 2 O  balancing  equation (3) [12] QUESTION 10/VRAAG10 10.1 • Dangerous for workers because they can be trapped underground.  • Sinkholes  (Any relevant answer) • Gevaarlik vir werkers want hulle kan ondergronds vasgekeer word • Sinkgate (Enige relevante antwoord) (2) 10.2 O2  (1) 10.3 Oxidation number of O decreases from 0 (in O 2 ) to -2 (in NaOH)  Die oksidasiegetal van O neem af van 0 (in O 2 ) na -2 (in NaOH) (1) 10.4 Fe3+ + 3e- → Fe  (2) 10.5 2(56) %Fe = × 100%  160 = 70 %  (2) 10.6 m(Fe) = (0,65)(2 500)  or/of 65 (2 500) 100 = 1 625 kg Fe extracted/ontgin  (Accept/Aanvaar 1,625 x 106 g) (2) [10] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Physical Sciences P2/Fisiese Wetenskappe/V2 16 DBE/November 2018 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne SUBMIT THIS SHEET WITH THE ANSWER BOOK/ LEWER SAAM MET DIE ANTWOORDEBOEK IN NAME/NAAM _____________________________________CLASS/KLAS________ QUESTION/VRAAG 4.3 Graph of volume versus pressure/Grafiek van volume teenoor druk 200 Volume (dm3) 150 100 50 0 0 50 100 150 200 250 300 Pressure/Druk (kPa) Copyright reserved/Kopiereg voorbehou