Physical Sciences Grade 11 P2 Gauteng June 2019 Memo (Afr)

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION / GAUTENGSE DEPARTEMENT VAN ONDERWYS PROVINCIAL EXAMINATION / PROVINSIALE EKSAMEN 2019 GRADE / GRAAD 11 PHYSICAL SCIENCES FISIESE WETENSKAPPE PAPER 2 / VRAESTEL 2 MARKING GUIDELINES / NASIENRIGLYNE 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 (PAPER 2) / FISIESE WETENSKAPPE (VRAESTEL 2) MARKING GUIDELINES / NASIENRIGLYNE QUESTION 1: MULTIPLE-CHOICE QUESTIONS VRAAG 1: MEERVOUDIGEKEUSE-VRAE 1.1 A  (2) 1.2 B  (2) 1.3 B  (2) 1.4 A  (2) 1.5 A  (2) 1.6 D  (2) 1.7 A  (2) 1.8 C  (2) 1.9 D  (2) 1.10 B  (2) [20] 2

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 QUESTION 2 / VRAAG 2: 2.1 The bond energy of a compound is the energy needed to break one mole of its molecules into separate atoms.  OR The energy needed to break the bonds in a molecule.  (2) Die energie benodig om een mol van ʼn substans op te breek in atome.  OF Die energie benodig om ʼn binding te breek.  2.2 -426 kJ.mol-1  (1) 2.3 Bond length is the average distance between the nuclei of two bonded atoms. Bindingslengte is die afstand tussen die kerne van die twee atome wat gebind is.  (2) 2.4 74 x 10-12 m  (1) 2.5 At point 3 the force of attraction between the nucleus of the one atom and the electrons of the other atom balances out with the force of repulsion between the two nuclei, therefore making it more stable  (Potential energy is at the lowest). At point 4 the repulsion force between the nuclei is much stronger  than the attraction force between the nucleus of the one atom and the electrons of the other atom.  (Increased potential energy). (4) By punt 3 is die aantrekkingskrag tussen die kern van die een atoom en die elektrone van die ander atoom in balans met die afstotingskragte tussen die twee kerne.  Dus is die binding meer stabiel.  (Potensiële energie is op sy laagste). By punt 4 is die afstotingskragte tussen die twee kerne van die twee atome baie groter as die aantrekkingskragte tussen die kern van die een atoom en die elektrone van die ander atoom.  (Potensiële energie is verhoog). [10] 3

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 QUESTION 3 / VRAAG 3: 3.1 Electronegativity is a measure of the tendency of an atom in a molecule to attract bonding electrons.  OR A measure of an atom’s attractive force on bonding electrons to form a molecule.  (2) Elektronegatiwiteit is ʼn aanduiding van die geneigdheid van ʼn atoom om die bindingselektrone van ʼn molekule aan te trek.  OF ʼn Aanduiding van die atoom se aantrekkingskrag op die verbindingselektrone van ʼn molekule.  3.2 Aℓ EN = 1,5  S EN = 2,5  S has a higher EN and will thus form the negative ion   S het die hoogste EN en sal dus die negatiewe ioon vorm. (2) 3.3.1 O EN = 3,5  Mg EN = 1,2  EN = 2,3  ionic bond  ioonbinding (2) 3.3.2 Cℓ EN = 3,0  H EN = 2.1  EN = 0,9  polar covalent  polêr kovalent (2) 3.3.3 P EN = 2,1  H EN = 2.1  EN = 0  non-polar covalent / pure covalent   nie-polêr kovalent / suiwer kovalent (2) 4

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 3.4 Valence electrons are the electrons in the highest energy level of an atom.  OR The electrons in the outer most orbital of the atom. (2) Valenselektrone is die elektrone wat in die hoogste energievlak van die atoom voorkom.  OF Die elektrone in die buitenste orbitaal van die atoom. 3.5.1 (2)  3.5.2 (2)  3.5.3 (2)  3.5.4  OR / OF  (2) 3.6 THREE / 3  (2) DRIE 3.7.1 Tetrahedral  (2) Tetraëdries 3.7.2 Trigonal planar  (2) Trigonaal planêr 3.7.3 Bent / Angular  (2) Hoekig / gebuig 5

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 3.8.1 Dative covalent bond / Coordinate bond  (2) Datief kovalente binding / koördinatiewe binding 3.8.2    (3) [33] QUESTION 4 / VRAAG 4: 4.1.1 The temperature at which the vapour pressure of a substance equals atmospheric pressure.  (2) Kookpunt is die temperatuur waneer die dampdruk van ʼn vloeistof gelyk is aan die atmosferiese druk. 4.1.2 There are strong hydrogen bonds between the water molecules.  There are weaker dipole – dipole forces between CH3Cℓ molecules.  Hydrogen bonds are stronger than dipole – dipole forces Therefore more energy is needed to overcome these forces and the boiling point is higher.  (3) Daar is sterk waterstofbindings tussen die molekules van die water  en swak dipool-dipool kragte tussen die CH3Cℓ molekules . Dus is meer energie nodig om die waterstofbindings tussen die watermolekules te breek.  6

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 4.1.3 Benzene and chloroform have weak London forces between their molecules.  Benzene has a larger surface area therefore stronger intermolecular forces and thus a higher boiling point.  OR Chloroform has a smaller surface area therefore weaker London forces and a lower boiling pont. (2) Daar is swak Londonkragte ( geïnduseerde dipoolkragte) tussen die molekules van die chloroform en die bensien  maar die molekulêre massa van die bensien is baie hoër as die van die chloroform, dus is bensien se kookpunt hoër. OF Chloroform het ʼn kleiner oppervlakarea en dus het dit swakker Londonkragte en ʼn laer kookpunt. 4.2.1 MgCl2  (1) 4.2.2 Cl2  and / en CO2  (2) 4.2.3 H2O  and / en HF  (2) 4.2.4 HCl  (1) 4.2.5 MgCl2  (1) 4.3.1 Non-polar  (2) Nie-polêr 4.3.2 Polar  (2) Polêr 4.3.3 Non-polar  (2) Nie-polêr 7

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 4.4.1 Vapour pressure is the pressure exerted by a vapour at equilibrium with its liquid (2) in a closed system.  Dampdruk is die druk van gasmolekules van ʼn stof wat in kontak is met sy vloeistof, in ewewig, in ʼn geslote sisteem.  4.4.2 PH3 has a higher vapour pressure.  PH3 has London forces between its molecules.  NH3 has hydrogen bonds between its molecules.  Hydrogen bonds are stronger than London forces, thus less energy is required to weaken the London forces in PH3  (4) PH3 het ʼn hoër dampdruk.  PH3 het Londonkragte tussen die molekules.  NH3 het waterstofbindings tussen die molekules.  Waterstofbindings is sterker as Londonkragte, dus word minder energie benodig om die swak Londonkragte in PH3 te breek. [26] 8

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 QUESTION 5 / VRAAG 5: 5.1 Oil is a non-polar molecule and water is a polar molecule.  Oil is less dense than water.  (2) Olie is ʼn nie-polêre molekule waar water polêr is.  Olie is minder dig as water.  5.2 Density is the number of molecules per unit volume.  (2) Digtheid: die aantal molekules per volume eenheid.  5.3.1 London forces / dispersion forces.  (2) Londonkragte / dispersiekragte  5.3.2 Electrostatic (Ionic bond)  (2) Ioonbindings (Elektrostatiese aantrekkingskragte) 5.3.3 Water  (2) 5.3.4 Water is polar and has hydrogen bonds (dipole – dipole forces) between molecules.  KMnO2 is an ionic solid.  Water and KMnO4 therefore have intermolecular forces of similar strength , so KMnO4 can dissociate in the water, turning it purple.  (4) Water is polêr en het waterstofbindings (sterk dipool-dipoolkragte) tussen die molekules.  KMnO2 is ʼn ioniese vastestof.  Water en KMnO4 het dus dieselfde sterkte intermolekulêre kragte , dus kan KMnO4 in water dissosieer om dit pers te kleur.  [14] 9

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 QUESTION 6 / VRAAG 6: 6.1.1 ANY THREE (3) OF THE FOLLOWING:   Particles are all identical.  Particles are in continual motion in all directions.  Particles do not contribute to the volume of the gas.  There are no forces between the particles, or the particles and the wall of the container, except during collisions.  Collision is perfectly elastic, with no loss of total energy of the molecules.  Obey gas laws in all temperatures and pressures. (3) ENIGE DRIE (3) VAN DIE VOLGENDE:   Alle molekules / deeltjies is identies.  Deeltjies is konstant in beweging in alle rigtings.  Deeltjies dra nie by tot die volume van die gas nie.  Daar bestaan geen kragte tussen die deeltjies onderling, of tussen die deeltjies en die wande van die houer nie behalwe gedurende botsings.  Botsings tussen deeltjies is volkome elasties met geen verlies aan die totale energie van die molekules nie.  Gehoorsaam gaswette in alle temperature en druk. 6.1.2 High Temperatures  Low pressures  (2) Hoë Temperature Lae druk 10

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 6.2.1 Boyle’s Law apparatus  (1) Apparaat vir Boyle se Wet 6.2.2 The pressure of certain mass of an enclosed gas is inversely proportional to the volume it occupies at a constant temperature.  (2) Die druk van ʼn sekere massa van ʼn ingeslote gas is omgekeerd eweredig aan die volume, by ʼn konstante temperatuur. 6.2.3.1 Pressure  (1) Druk 6.2.3.2 Volume  (1) Volume 6.2.3.3 Mass  Temperature  (2) Massa Temperatuur 6.2.4 120 kPa  (2) 6.2.5 The shape would be the same  but it would be higher up  on the graph (Higher volume at higher pressure). (2) Die vorm van die grafiek sal dieselfde bly, maar dit sal hoër op beweeg / hoër op die grafiek lê (Hoër volume teen ʼn hoër druk).  6.3.1   V2 = 4,84 dm3 / ℓ  (5) 6.3.2 t = T – 273 = 282 – 273  = 9o C  (2) 6.3.3 His lungs will expand / might burst.  As the pressure decreases and the temperature increases the volume will increase, as V ∝ 1/P and V ∝ T.  (3) Sy longe sal uitsit en mag dalk selfs bars  aangesien die volume van ʼn ingeslote gas verhoog soos die druk verminder. Die volume sal dus drasties toeneem aangesien V ∝ 1/P en V ∝ T.  [26] 11

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 QUESTION 7 / VRAAG 7: 7.1.1 6H2O  + Al2S3  3H2S  + 2Al(OH)3  (3) 7.1.2 M(Al(OH)3) = 27 + 3(16 +1) = 78 g.mol-1  = 0,1282 … mol  = 0,0641 … mol m = nM = (0,0641 …)[1(27)+3(32)]  = 7,88 g (4) 7.1.3 (3) 7.2.1 n(HCl) = cV  = (0,4)(0,2)  = 0,08 mol   = 0,2 mol HCl : CaCO3 2 : 1 0,08 : 0,04 OR / OF 0,4 : 0,2  Therefore CaCO3 is in excess   CaCO3 is in oormaat (6) 12

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 7.2.2 n(CaCl2) = 0,08 x ½  = 0,04 mol  m = nM = (0,04)(40 + 2(35,5) = 4,44 g  (3) 7.2.3 actual reël (2) % yield  x 100 / % opbrengs  x 100 theoretical teoreties  = 90,09%  [21] 13

Downloaded from hlayiso.com Physical Sciences / Fisiese Wetenskappe MEMORANDUM (Paper / Vraestel 2) GRADE / GRAAD 11 Taxonomy Grid Recall Comprehension Analysis Evaluation Q no: Mark Q no: Mark Q no: Mark Q no: Mark 1.1. 2 1.2. & 4 1.4. 2 1.3. 1.5. – 6 1.8 2 1.7 1.9 & 4 1.10 2.1. 2 2.2. 1 2.5. 4 2.3. 2 2.4. 1 3.1. 2 3.3. 6 3.2. 2 3.4. 2 3.5. 8 3.6. 2 3.8.1. 2 3.7. 6 3.8.2. 3 4.1.1. 2 4.1.2. 3 4.1.3. 2 4.4.1. 2 4.3. 6 4.2. 7 4.4.2 4 5.2. 2 5.1. 2 5.3.4. 4 5.3. 6 6.1.1. 3 6.1.2. 2 6.2.4. 2 6.3.3. 3 6.2.1. 1 6.2.3. 4 6.2.5. 2 6.2.2. 2 6.3.2. 2 6.3.1 5 7.1. 2 7.2. 5 7.3. 3 7.4.2. 3 7.4.1. 6 7.4.3. 2 Total mark 24 63 47 16 Total %/100% P1&2: 15% P1:35%/P2:40% P1:40%/P2:35% P1&2: 10% Correct application of Bloom’s / Barrett’s Taxonomy: Level 1: Recall of information (what? which? when? list ; label; name; define; give; describe) Level 2: Understanding and using information (summarize; classify ; apply rules; discuss) Applying information (distinguish; specify; compare; design ;explain; investigate ;interpret; calculate; give your input) Level 3: Analysis of information (classify; explain; identify; interpret; compare; give reasons; prove; give causes and effects) Level 4: Synthesize information (summarize; construct; argue; create; relate; design; formulate) Evaluate information (judge; assess; evaluate; choose; support; compare; estimate) 14