Physical Sciences Grade 11 Paper 2 Gauteng June 2018 Memo

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION PROVINSIALE EKSAMEN JUNE / JUNIE 2018 GRADE / GRAAD 11 PHYSICAL SCIENCES FISIESE WETENSKAPPE PAPER / VRAESTEL 2 MEMORANDUM 14 pages / bladsye 1

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 GAUTENG DEPARTMENT OF EDUCATION GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION PROVINSIALE EKSAMEN PHYSICAL SCIENCES / FISIESE WETENSKAPPE (Paper / Vraestel 2) QUESTION 1: MULTIPLE-CHOICE QUESTIONS VRAAG 1: MEERVOUDIGEKEUSE-VRAE 1.1 D (2) 1.2 A (2) 1.3 A (2) 1.4 B (2) 1.5 C (2) 1.6 A (2) 1.7 C (2) 1.8 B (2) 1.9 D (2) 1.10 A (2) [20] 2

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 QUESTION 2 / VRAAG 2: 2.1 Intermolecular forces are the attraction forces between molecules. Intermolekulêre kragte is die aantrekkingskrag tussen molekules. (2) 2.2.1 Hydrogen bonds  Waterstofbindings  (1) 2.2.2 Dipole-dipole forces  Dipool-dipool kragte  (1) 2.2.3 The strength of the intermolecular forces increases as the molecular size increases.  HI has a bigger molecular mass than HCℓ thus HI has a higher melting point.  More energy is needed to break the stronger dipole- dipole forces in HI.  Die krag van die intermolekulêre kragte verhoog soos die molekulêre grootte verhoog.  HI het ʼn grooter molekulêre massa as die HCℓ dus het HI ʼn hoër kookpunt.  Meer energie is nodig om die sterker dipool – dipoolkragte in HI te breek.  (6) 2.2.4 Hydrogen iodide.  Waterstofjodied.  (2) 2.3.1 Boiling point is the temperature at which a liquid’s vapor pressure is equal to its atmospheric pressure.  Kookpunt is die temperatuur waarteen die dampdruk van ʼn vloeistof gelyk is aan die atmosferiese druk.  (2) 2.3.2 There are strong Hydrogen bonds between the molecules of the methanol  and weak London forces (induced dipole forces) between the molecules of the methane.  Thus more energy is needed to break the hydrogen bonds between the molecules of the methanol.  Daar is sterk waterstofbindings tussen die molekules van die metanol  en swak London kragte ( geïnduseerde dipool kragte) tussen die molekules van die metaan.  Dus word meer energie benodig om die waterstofbindings tussen die metanol molekules te breek.  (3) [17] 3

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 QUESTION 3 / VRAAG 3: 3.1 A way of representing atoms or molecules by showing valence electrons as dots / crosses surrounding the element symbol.  ʼn Manier om atome of molekules voor te stel deur die valensie-elektrone as kolletjies / kruisies om die simbool van die element te toon.  (2) 3.2.1 NaCℓ – Ionic bond  / Ioniese binding H2O & NH3 – polar covalent bond / polêre kovalente binding H2 – non-polar covalent bond  / nie-polêre kovalente binding (3) 3.2.2 NaCℓ – ionic / not a molecule/ ionies / nie ʼn molekule H2O – Angular / bent / hoekig / gebuig  NH3 – Pyramidal / piramidaal  H2 – linear / linieêr  (3) 3.2.3 H2O / water to form H3O+ / water om H3O+ te vorm  and / en NH3 / ammonium / amoniak to form NH4+ / NH3 / ammonium / ammoniak om NH4+ te vorm  (2) 3.2.4  (3) OR / OF  [13] 4

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 QUESTION 4 / VRAAG 4: 4.1 Strong Hydrogen bonds  /Sterk waterstofbindings (2) 4.2 Heat of vaporisation is the amount of heat required to make water evaporate.  Verdampingshitte is die hoeveelheid hitte benodig om water te laat verdamp.  (2) 4.3 POLAR  Because of 1. The EN difference, H2O EN = 3,5 – 2,1 = 1,4  polar covalent  O attract shared electron pair more than H  2. Because of the non-symmetrical shape and two lone pairs on the oxygen.  (4) POLÊR  omdat 1. H2O EN = 3,5 – 2,1 = 1.4  polêr kovalent  O trek die gedeelde elektronpaar meer as die waterstof.  2. As gevolg van die nie-simmetriese vorm en die twee paar ongepaarde elektrone op die suurstof.  4.4 Molecular dipoles occur due to the unequal sharing of electrons  between atoms in a molecule. Those atoms that are more electronegative pull the bonded electrons closer to themselves creating a positive and a negative side to the opposite sides of the molecule.  (2) ʼn Molekulêre dipool ontstaan wanneer daar ʼn oneweredige verdeling van elektrone tussen die atome  in die molekule ontstaan a.g.v. ʼn groot verskil in elektronegatiwiteit. Dit gee dan aanleiding tot die vorming van ʼn positiewe en negatiewe lading aan die teenoorgestelde kante van die molekule.  4.5 KCℓ  (1) 4.6 KCℓ – forms an ionic bond and will be able to dissociate in water to form K + and Cℓ - ions,  where I2 forms a non-polar bond and will not mix with the polar water molecules.  (2) KCℓ – vorm ʼn ioniese binding wat in staat is om te dissosieer in water om K+ en Cℓ - ione te vorm,  terwyl I2 ʼn nie-polêre molekuul is en dus nie met die polêre watermolekules sal meng nie.  4.7 Capillary action  – the adhesion forces between the molecules of different origin e.g. water and glass are bigger than the cohesion forces of the water molecules. (3) Kapillêre werking  – die adhesiekragte tussen die molekules van verskillende stowwe is groter as die adhesiekragte tussen die watermolekules.  [16] 5

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 QUESTION 5 / VRAAG 5: 5.1 Ideal gas: It is a hypothetical gas that will obey all the gas laws under all conditions of pressure and temperature.  (two marks or none) (2) Ideale gas: ʼn Hipotetiese gas wat al die gaswette nakom onder alle omstandighede van temperatuur en druk.  (twee of geen punte) 5.2 Hydrogen gas (H2) OR Helium Gas (He) (2) Waterstofgas (H2) OF Heliumgas (He) 5.3 Temperature  (2) Temperatuur  5.4.1 Dependent variable: Volume of the gas  Independent variable: Different temperatures of the water bath.  (2) Afhanklike veranderlike: Volume van die gas  Onafhanklike veranderlike: Verskillende temperature van die waterbaddens.  5.4.2 Charles’s law (2) Charles se wet  5.4.3 Type of gas Mass of gas Pressure of the gas (any two)  (2) Tipe gas Massa van die gas Druk van die gas (enige twee)  5.4.4 58 – 60 cm3  (2) 5.4.5 pV = nRT  p=? p x 7,5 x 10 -4  = 12 x 8,31 x 298  V = 750 ÷ (100)3 = 7,5 x 10 -4 m3 p= 12 x 8,31 x 298 T = 298 K 7,5 x 10 -4 R = 8,31 n = 12 mol p = 39622080 Pa / 3,96 x 107 Pa  (4) 6

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 5.5.1 LOWER THAN  MINDER AS  (1) 5.5.2 1 mole of gas at the same temperature and pressure has the same volume (pV =nRT  p  n)  if the pressure is ½ then the molar mass is less  (2) 1 mol van enige gas teen dieselfde temperatuur en druk het dieselfde volume (pV =nRT  p  n)   dus as die druk ½ is, is die molêre massa minder [21] 7

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 QUESTION 6 / VRAAG 6: 6.1.1 Zn(s) + 2 HCℓ (aq)  ZnCℓ2 (s) + H2 (g)  (2) 6.1.2 Zn(s) + 2 HCℓ (aq)  ZnCℓ2(s) + H2(g) Mole ratio: 1 2 1 1 Mol verhouding: Initial mole n =m/M = 50 / 36,5 Aanvangs = 45,5g / 65 = 1,37 mol  mol = 0,7 mol  Change in Mol ratio / mol Have enough. verhouding: Verandering Het genoeg 1,37 ÷2 =0,685 in mol Zn mol  limiting reactant  (4) Beperkende reagens 6.1.3 Zn(s) + 2 HCℓ (aq)  ZnCℓ2 (s) + H2 (g) Change in mol Used 0,685 If 1,37 mol used Verandering mol (mole gebruik in mol ratio used / mol verhouding gebruik)  Mol left or 0,7 – 0,685 formed at = 0,015 mol  end m =n x M Mol = 0,015 x 65 oorgelaat of = 0,975 g  (3) aan die Zn left / einde Zn oor gevorm 8

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 6.1.4 Zn(s) + 2 HCℓ (aq)  ZnCℓ2 (s) + H2 (g) Change in mol 0,137 0,685 mol Verandering (mole ratio used / in mol gebruik)  Mol left or n = V/Vm  formed at = 0,685 ÷ 22,4 end = 15,34 dm3  Mol oor / gevorm aan (3) die einde 6.2.1 CH4 + 2 O2  CO2 + 2 H2O Mole ratio: 1 2 1 2 Mol verhouding: Initial mole 3 excess / 0 0 Aanvangs oortollig mol Change in mol If all used / Ratio 1:1  / Verandering Indien alles verhouding 1:1 in mol gebruik  3 mole 3 Mol left or m =n xM formed at = 3 x (44)  end =132g  Mol oor / CO2 that can be gevorm aan produced die einde CO2 wat geproduseer kan word (4) 9

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 6.2.2 % Yield / opbrengs = actual yield / ware opbrengs  Max possible yield / maks moontlike opbrengs = 87 x 100  132 = 65,91%  (3) 6.3 C H O 39,9 g 6,7 g 53,4 g = 100g m = 39,9 6,7 53,4  row / ry M 12 1 16 ÷ smallest 3,3 6,7 3,3  row / ry kleinste 3,3 3,3 3,3 Mol ratio: 1 : 2 : 1  row / ry IPF = CH2O = M = 12 + 2 + 16 = 30 g.mol-1  Molecular Formula / Molekulêre formule = 60 ÷ 30 = 2 x CH2O C2H4O2  (6) [25] 10

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 QUESTION 7 / VRAAG 7: 7.1 n = m/M = 12 65 = 0,1846 mol  But in each mole of NaN3 there are 3 moles of N Maar in elke mol NaN3 is daar 3 mol N nN = 0,1846 x 3 = 0,554 mol  n = No NA  0,554 = No ÷ 6,02 x 10 23   No N atoms / atome = 3,45 x 10 23 atoms / atome  (5) 7.2 2NaN3 (s)  → 2Na(s) + 3N2(g)  (3) 7.3 2NaN3 (s) → 2Na(s) + 3N2(g) n = V VM  = 85 22,4 Use mole ratio = 3,795 mol  Gebruik mol vehouding 2,53 mol   Mass NaN3 needed to fill bag:  Massa NaN3 benodig om sak te vul: m=nxM = 2,53 x 65  = 164,45 g  (5) 7.4 The chemical formula that shows the simplest ratio between the atoms of a compound.  (2) Die chemiese formule wat die eenvoudigste verhouding tussen die atome in ʼn verbinding aantoon. 11

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 7.5 Na N % 35,39 64,61  = 100 m = 35,39 64,61 M 23 14 1,549 4,615 ÷ smallest 1,549 4,615 kleinste 1,549 1,549 Mol ratio: 1 : 3  row / ry  IPF = NaN3  9 Empiriese formule = NaN3 (3) 7.6.1 DECREASE  (2) VERMINDER 7.6.2 p  T Pressure of a gas is directly proportional to temperature / Druk van ʼn gas is direk verwant aan die temperatuur daarvan If the temperature decreases, the average kinetic energy of the gas molecules decreases  and the pressure decreases.  (3) As die temperatuur van die gas afneem sal die gemiddelde kinetiese energie van al die deeltjies ook afneem en dus sal die druk verlaag. [23] 12

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 QUESTION 8 / VRAAG 8: 8.1. Oxygen  (2) Suurstof 8.2 2KCℓO3   2 KCℓ  + 3 O2  balancing / balansering One for each correct formula / Een punt vir elke korrekte formule. (4) 8.3 n = m/M = 16,2 122,5 = 0,1322 mol  In each mole of KCℓO3 there is 1 mol of K  In elke mol KCℓO3 is daar 1 mol K n = No NA 0,1322 = No ÷ 6,02 x 10 23   No K ions/ ione = 7,97 x 10 22 ions/ ione  (4) 8.4 2KCℓO3  2 KCℓ + 3 O2 If all off 16,2 g KCℓO3 decomposed As 16,2 g KCℓO3 alles ontbind n = m/M = 16,2 use ratio  122,5 gebruiksverhouding = 0,132 mol KCℓO3 0,132 KCℓ m=nxM = 0,132 x 74,5 = 9,83g  % Purity = Actual yield x 100 %Suiwerheid = ware opbrengs x100 Max yield  maks moontlike opbrengs = 7,2 x 100 9,83  = 73,25 %  (5) [15] 13

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 Taxonomy Grid / Taksonomietabel Recall / Onthou Comprehension / Analysis / Evaluation / Verstaan Analiseer Evalueer Q no: / Mark / Q no: / Mark / Q no: / Mark / Q no: / Mark / Vr nr: Punt Vr nr: Punt Vr nr: Punt Vr nr: Punt 1.1 2 1.3 2 1.7 2 8.1 4 1.2 2 1.4 2 1.10 2 9.2 7 2.1 2 1.5 2 3.3 4 9.3 3 3.1 4 1.6 2 3.5 2 4.1 1 1.8 2 4.4 2 5.3.1 2 1.9 2 4.5 2 5.3.2 2 2.2 4 5.1 5 5.3.3 2 2.3 6 5.2 2 5.4 2 3.2 2 6.4 2 6.1 3 3.4 4 6.5 2 7.4.2 2 4.2 2 6.7 5 4.3 2 7.2 3 6.2 6 7.4.3 6 6.3 2 8.3 3 6.6 2 8.4 4 7.1 2 9.1 6 7.3 6 7.4.1 2 7.4.4 4 8.2 3 Total mark / 16% 24 39,33% 59 34,67% 52 9,3 % 14 Totale punte Total / Totaal % P1&2: 15% P1:35%/P2:40% P1:40%/P2:35% P1&2: 10% / 100% Correct application of Bloom’s / Barrett’s Taxonomy: / Korrekte toepassing van Bloom / Barrett se taksonomie: Level 1: Recall of information (what? which? when? list; label; name; define; give; describe) Vlak 1: Oproep van inligting (wat? watter? wanneer? lys; benoem; definieer; voorsien; beskryf) Level 2: Understanding and using information (summarise; classify; apply rules; discuss) Applying information (distinguish; specify; compare; design; explain; investigate; interpret; calculate; give your input) Vlak 2: Verstaan en gebruik inligting (som op klassifiseer); pas reëls toe; bespreek) Level 3: Analysis of information (classify; explain; identify; interpret; compare; give reasons; prove; give causes and effects) Vlak 3: Toepassing van inligting (onderskei; spesifiseer; vergelyk; ontwerp; verduidelik; ondersoek; interreteer; bereken; gee jou opinie) Level 4: Synthesize information (summarize; construct; argue; create; relate; design; formulate) Evaluate information (judge; assess; evaluate; choose; support; compare; estimate) Vlak 4: Sintetiseer inligting (som op; konstrueer; argumenteer; skep; formuleer) Evalueer inligting (beoordeel; assesseer; evalueer; kies; ondersteun; vergelyk; skat) 14