Physical Sciences Grade 10 P2 National Exemplar 2015 Memo

Downloaded from hlayiso.com cs basic education ©) Department: Basic Education REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 10 " PHYSICAL SCIENCES: CHEMISTRY (P2) i FISIESE WETENSKAPPE: CHEMIE (V2) q MEMORANDUM NY nH q NOVEMBER 2015 { uy MARKS/PUNTE: 150 This memorandum consists of 10 pages. Hierdie memorandum bestaan uit 10 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 10/Graad 10 — Memorandum QUESTION 1/VRAAG 1 1.1 AVY 1.2 Byv 1.3 AVY 14 Cvv 1.5 Dvv 1.6 Dvv 1.7 Byv 1.8 Byv 1.9 Byv 1.10 Cvv QUESTION 2/VRAAG 2 2.1 2.1.1. ANY ONE: Copper/oxygen gas ¥ ENIGE EEN: Koper/suurstofgas 2.1.2 | ANY ONE: brass/salt solution/pure air ¥ ENIGE EEN: Geelkoper/soutoplossing/suiwer lug 2.1.3. oxygen (gas)/suurstof(gas) V 2.1.4 ANY ONE: magnesium oxide/table salt/sugar V ENIGE EEN: magnesiumoksied/tafelsout/suiker 2.1.5 Sandv 2.2 2.2.1 (a) Filtration/Filtrering/filtrasie V (b) Evaporation/Verdamping Vv (c) Sand v (d) Sugar solution/sugar and water v Suikeroplossing/suiker en water 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 10/Graad 10 — Memorandum 2.2.2 Physical (process)/Fisiese (proses) V No new substances are formed./Water only changes phase./The chemical composition of the substance is not altered. v Geen nuwe stowwe word gevorm nie./Water verander slegs van fase./Die chemiese samestelling van die stowwe verander nie. (2) [11] QUESTION 3/VRAAG 3 3.1 3.1.1 The temperature Vv at which the vapour pressure is equal to the external/atmospheric pressure. V Die temperatuur waarby die dampdruk gelyk is aan die eksterne/atmosferiese druk. (2) 3.1.2 (a) Dv (1) (b) BY (1) (c) Cv (1) 3.1.3 lv (1) 3.2 3.2.1 Liquid/Vloeistof v (1) 3.2.2 132°CV (1) 3.2.3 (a) Increases/Toeneem Vv (1) (b) Remains the same/Bly dieselfde V (1) 3.2.4 Heat energy is used to break forces between particles V resulting in a phase change V and not to change the speed at which particles move. Vv Warmte energie word gebruik om kragte te breek tussen deeltjies wat lei tot 'n faseverandering en nie tot verandering in die spoed waarteen deeltjies beweeg nie. OR/OF Heat energy is used to increase the potential energy Vv of the particles making them move further apart v resulting in a phase change. ¥ Warmte energie word gebruik om die potensiéle energie van die deeltjies te verhoog en veroorsaak dat hul verder_vanmekaar_beweeg en ‘n faseverandering tot gevolg het. (3) [13] 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 10/Graad 10 — Memorandum QUESTION 4/VRAAG 4 4A 4.1.1 4.1.2 4.2 4.2.1 4.2.2 The number of protons V in the nucleus of an atom. ” Die getal protone in die kern van 'n atoom. (2) Atoms of the same element having the same number of protons, V but different numbers of neutrons. ¥ Atome van dieselfde element met dieselfde getal protone maar verskillende getalle neutrone. OR/OR Atoms (of the same element) with the same atomic number Vv but different mass numbers. ¥ Atome (van dieselfde element) met dieselfde atoomgetal, maar verskillende massagetalle. (2) 1s*2s*2p°v 35"3p° v (2) OPTION 1/OPSIE 1: Marking criteria/Nasienriglyne %CL-35 = x and %CL-37 = y e x+y=100 OR/OF x = 100-y ORIOF y = 100 —x ¥ X FY = 100 Voce ce eeeteeeee ees 1 | Expression to calculate average 35 37 atomic mass of 100 atoms. v 35,5 = *(35)+ ¥87) | Uitdrukking om gemiddelde : 100 ieeeceeeetetetetee 2 atoommassa van 100 atome te 1in2: bereken. 35.5 = x(35) + (100 — x)37 e Final answer / Finale antwoord: 75 , 100 3550 = 35x + 3700 - 37x X= 75 V OPTION 2/OPSIE 2: %Ct-35 = x -. YCL-37 = 100 —- x ¥ ( x Jp5+(*05* r= 35,5 100 100 0,35x + 37 — 0,37x = 35,5 1,5 = 0,02x X=75¥ (3) 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 10/Graad 10 — Memorandum 43 43.1. 7v (1) 4.3.2 Covalent (bond)/Kovalente (binding) v (1) 4.3.3 on err VV 2 Sle Cl (2) ‘4 ef 44.4 054 ss Marking criteria/Nasienriglyne: e Main energy levels shown V ALI ALI AL Hoofenergieviakke aangedui n=3/ | 4 3px 3py 3p, (1, 2, 3 etc.). 3s « Orbitals labelled as s or p. Ignore if 4" level and 4s not shown. ” 1b} [4b 441 Orbitale genommer as s of p. n=2) 1411 2px 2p, 2p, Ignoreer as 4de viak en 4s nie | 2s getoon word nie. n=1 ¢ Nine electron pairs shown as 1b arrows in opposite directions in 1s blocks/circles. VW Nege elektronpare getoon as pyltjies in teenoorgestelde rigtings in blokke of sirkels. (3) 442 Ca*veacrv (2) QUESTION 5/VRAAG 5 5.1 5.1.1 The ionisation energy is the energy needed/absorbed to remove an electron, “ whilst electron affinity is energy released when an electron is taken in. Vv Die ionisasie energie is die energie benodig/geabsorbeer om 'n elektron te verwyder, terwy! elektronaffiniteit die energie is wat vrygestel word wanneer 'n elektron opgeneem word. (2) 5.1.2 Increase in effective nuclear charge from left to right in period. “ Toename in effektiewe kernlading van links na regs in 'n periode. OR/OF Decrease in atomic radius from left to right in a period. v Afname in atoomradius van links na regs in 'n periode. (1) 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 10/Graad 10 — Memorandum 5.1.3. Higher than/Hoér as Vv Losing a second electron will result in an unstable electron structure for sodium, V whilst losing a second electron will result in a stable/noble gas electron structure for magnesium. “ Verlies van 'n tweede elektron sal tot 'n onstabiele elektronstruktuur vir natrium lei, terwyl verlies van 'n tweede elektron tot 'n stabiele/edelgas elektronstruktuur vir magnesium sal lei. (3) 5.1.4 m n=— M -46, 23 =2mol cnergy neosouherdie benodig = 2 x 496 Vv = 992 kJ Vv (3) 5.1.5 Chlorine/Chioor Vv Highest electron affinity/Hoogste elektronaffiniteit V OR/OF Releases the most energy./Ste/ die meeste energie vry. (2) 5.2 5.2.1 (a) EV (1) (b) Cv (1) (c:) Av &DY (2) 522 aA Vaf.e08 J Nae + 3 Cle — >Na t | OR/OF + Nae —>Na +ev ee ee sClet+e —|: Gi] v ee ee - v. ee J— Na" +[sGrs] — Na'[t | 5.2.3 COp(s) ¥ + CO2(q) v (2) [21] 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 10/Graad 10 — Memorandum QUESTION 6/VRAAG 6 6.1 6.1.1 Metallic (bond)/Metaal (binding) Vv (1) 6.1.2 lonic (bond)/loniese (binding) V (1) 6.2 Chemical change/Chemiese verandering V A new substance is produced. v 'n Nuwe stof word gevorm. (2) 6.3 2Mg + O2 V> 2MgO Vv Bal. v Notes/Aantekeninge e Reactants Vv Products W Balancing ¥ Reaktanse V Produkte V Balansering V e Ignore/Ignoreer = e Marking rule 3.10/Nasienreé/ 3.10 (3) 64 2Mg +O2 — 2MgO Marking criteria/Nasienriglyne: 2(24) V + 2(16) Vv = 2(16 + 24) v e M(reactant 1) correctly calculated. ¥ 80 = 80 Vv M(reaktans 1) korrek bereken. e M(reactant 2) correctly calculated. Vv M(reaktans 2) korrek bereken. e M(product) correctly calculated. ¥ M(produk) korrek bereken. e M(reactants) = M(product) “ M(reaktanse) = M(produkte) (4) [11] QUESTION 7/VRAAG 7 7.1 Reaction in which a proton/hydrogen ion/H” is transferred from one reactant to another. vv Reaksie_waarin 'n proton/waterstofioon/H’ oorgedra word van een reaktans na 'n ander. (2) 7.2 18 cm? Vv (1) 7.3 7.3.1 Number of moles of solute Y per cubic decimetre/litre of solution. Vv Aantal mol opgeloste stof per kubieke desimeter/liter oplossing. (2) vo 73.2 vcaco, as” =40% v (3) 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 10/Graad 10 — Memorandum 7.3.3 POSITIVE MARKING FROM QUESTION 7.3.2 i.e. 0,1 g FROM GRAPH. POSITIEWE NASIEN VANAF VRAAG 7.3.2, m.a.w. 0,1 g VAN DIE GRAFIEK. n(CaCO,) = —__, n(HCt) = 2 x n(CaCO3) = 2 x 10™ mol ¥ Volume acid/volume suur. c=4y Vv -3 o1= 2x10 Vv V = 0,02 dm* v (5) [13] QUESTION 8/VRAAG 8 8.1 Solution in which the solvent is water. v Oplossing waarin die oplosmiddel water is. (1) 8.2 8.2.1 The process in which solid ionic crystals are broken up into ions “when dissolved in water. v Die proses _waarin vaste ioniese_kristalle opgebreek word in ione wanneer opgelos word in water. (2) 8.2.2 (NH4)2COa(s) ~—> 2NH} (aq) + CO? (aq) Y bal. ¥ (3) Notes/Aantekeninge e Reactants V Products ¥ Balancing Reaktanse V Produkte V Balansering V e Ignore/Ignoreer = and phases/en fases. e Marking rule 3.10/Nasienreé/ 3.10 8.3 8.3.1 (a) Conductivity/Ammeter reading ¥ Geleidingsvermoé/Ammeterlesing (1) (b) Type of compound/Concentration of ions ¥ Tipe verbinding/Konsentrasie van ione (1) 8.3.2 | Anincrease in ion concentration increases the conductivity. v '‘n Toename in ioonkonsentrasie verhoog die geleidingsvermoé. (1) 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 10/Graad 10 — Memorandum 83.3 NaCf>Na*+Cr v Forms two ions per mole of NaC?./Vorm twee ione per mol NaCt. W CaCt, > Ca + 2Crv Forms three ions per mole of CaCfz./Vorm drie ione per mo! CaClo. Vv OR/OF Concentration of ions in CaCt2(aq) is higher than in NaCt(aq). vv Konsentrasie van ione in CaCf2(aq) is hoér as in NaCé(aq). (4) 8.4 8.4.1 Redox/Redoks Vv (1) 8.4.2 Gas forming/Gasvorming Vv (1) 8.4.3 Precipitation/Presipitasie V (1) 8.5 Cu + 4HNO3 VY — Cu(NO3)2 + 2H20 + 2NO2 Vv Bal. v Notes/Aantekeninge e Reactants Vv Products W Balancing ¥ Reaktanse V Produkte V Balansering V e = Ignore/Ignoreer = ¢ Marking rule 3.10/Nasienreé/ 3.10 (3) QUESTION 9/VRAAG 9 9.1 9.1.1 One mole is the amount of substance V having the same number of particles as there are atoms in 12 g carbon-12. v Een mol is die hoeveelheid stof wat dieselfde aantal deeltiies bevat as wat daar atome is in 12 g koolstof-12. (2) 9.1.2 0,67 mol/2 mol ¥ 3 (1) 9.1.3 V(N2): V(H2): V(NH3) = 1:3:2 V(N2 reacted/reageer) = tV(He) 1 =e =8dm Volume Nz remains/oorbly = 10 - 8 V=2 dm? Volume of gas that remains/Volume gas wat oorbly = 2 + 16 Vv = 18 dm® gas ¥ (4) 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 10/Graad 10 — Memorandum 914 7n-™ M _ 80/7 0 Oo 2 4 = 40 mol¥ 9.1.5 V(N2)= + (40) x 22,4 v = 298,67 dm® v 9.2 m(O2) = 239 — 207 = 32 gv m Pb) =— n(Pb) = _ 207 / 207 =1mol n(0) =" -2y 16 =2mol n(Pb) : n(O) 1:2v - PbOs v QUESTION 10/VRAAG 10 10.1 All the water of the earth./Al die water van die aarde V 10.2 10.2.1. Evaporation/Verdamping V 10.2.2 _Condensation/Kondensasie V 10.2.3 Infiltration/percolation ¥ Infiltrering/Infiltrasie/perkolasie 10.2.4 Rain fall/Precipitation VW Reénval/Presipitasie/Neerslag 10.3 Increase in ground water levels./Toename in grondwater viakke. V 10.4 ANY TWO/ENIGE TWEE: Storage of water “/recreation areas v/job creation/irrigation of crops/ attraction of bird life/source of food Stoor van water/ontspanningsareas/werkskepping/besproeiing van gewasse/ aantrekking van voéllewe/bron van voedsel TOTAL/TOTAAL: Copyright reserved/Kopiereg voorbehou (1) (1) (1) (1) (1) (2) [8] 150