Physical Sciences Grade 11 P2 North West November 2024 Memo

PROVINCIAL ASSESSMENT PROVINSIALE ASSESSERING GRADE/GRAAD 11 PHYSICAL SCIENCES P2 FISIESE WETENSKAPPE V2 MARKING GUIDELINES/NASIENRIGLYNE NOVEMBER 2024 MARKS/PUNTE: 150 These marking guidelines consists of 10 pages. Hierdie nasienriglyne bestaan uit 10 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 2 NW/November 2024 Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 1/VRAAG 1 1.1 D  (2) 1.2 D  (2) 1.3 B  (2) 1.4 B  (2) 1.5 A  (2) 1.6 C  (2) 1.7 B  (2) 1.8 A  (2) 1.9 A  (2) 1.10 D  (2) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 3 NW/November 2024 Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 2/VRAAG 2 2.1 Electronegativity as a measure of the tendency of an atom in a molecule to attract bonding electrons.  Elektronegatiwiteit is ‘n maatstaf van die neiging van ‘n atoom in ‘n molekuul om bindingselektrone aan te trek.  (2) 2.2.1 ∆E = 4 – 1 = 3; Ionic bond/ioniese binding. (2) 2.2.2 ∆E = 3 – 3 = 0; non polar/nie polêr. (2) 2.3.1 Valence electrons are the electrons in the highest energy level of an atom in which there are electrons.  Valenselektrone is die elektrone in die hoogste energievlak van ‘n atoom waarin daar elektrone is.  (2) 2.3.2 4 (1) 2.3.3  (2) 2.3.4  (2) 2.4.1 Linear/Liniêr (1) 2.4.2 Tetrahedral/Tetrahedries (1) 2.5.1 Bond length is the average distance between the nuclei of two bonded atoms. Bindingslengte is die gemiddelde afstand tussen die kerne van twee gebonde atome.  (2) 2.5.2 C2H4 have double bonds between carbon atoms  while C2H6 have single bonds between carbon atoms. Bond length decreases with the number of bonds between the atoms. C2H4 het dubbelbindings tussen koolstofatome terwyl C2H6 enkelbindings tussen koolstofatome het. Bindingslengte neem af met die aantal bindings tussen die atome.  (3) 2.5.3 The shorter the bond length the higher the bond energy. OR The longer the bond length the smaller the bond energy. Hoe korter die bindingslengte hoe hoër is die bindingsenergie. OF Hoe langer die bindingslengte hoe kleiner die bindingsenergie. (2) 2.5.4 SMALLER THAN/KLEINER AS (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 4 NW/November 2024 Grade/Graad 11 – Marking Guidelines/Nasienriglyne 2.6 Both CHCℓ3 and CCℓ4 molecules have polar bonds and both have a tetrahedral shape. CHCℓ3 molecule is asymmetrical/have uneven distribution of electrons while CCℓ4 molecule is symmetrical/ have even distribution of electrons. This difference in the charge distribution makes CHCℓ3 to be a polar molecule and CCℓ4 to be a non-polar molecule. Beide CHCℓ3 en CCℓ4 molekules het polêre bindings en beide het ‘n tetraëdriese vorm. CHCℓ3 molekule is asimmetries/het oneweredige verspreiding van elektrone terwyl CCℓ4 molekule simmetries is/ het eweredige verspreiding van elektrone. Die verskil in die lading verspreiding veroorsaak dat CHCℓ3 ‘n meer polêre molekule en CCℓ4 ‘n nie-polêre molekule is. (4) [27] QUESTION 3/VRAAG 3 3.1 Melting point is temperature at which the solid and liquid phases of a substance are at equilibrium. Smeltpunt is die temperatuur waarby die vaste- en vloeistoffases van ‘n stof in ewewig is. (2) 3.2.1 The stronger the intermolecular forces the higher the melting point.  (2) Hoe sterker die intermolekulêre kragte, hoe hoër die smeltpunt. 3.2.2 The molecular mass of HBr is larger than that of HCℓ. The intermolecular (London) forces between molecules of HBr are stronger than the intermolecular force between the molecules of HCℓ. More energy is required to overcome the intermolecular forces between the molecules of HBr than of HCℓ. Die molekulêre massa van HBr is groter as die van HCℓ. Die intermolekulêre (London) kragte tussen molekules van HBr is sterker as die intermolekulêre kragte tussen die molekules van HCℓ. Meer energie benodig om die intermolekulêre kragte tussen molekules van HBr te oorkom as van HCℓ. (3) 3.3 HI  (1) 3.4 CCℓ4 molecules have London forces only while H2O has (London forces) and hydrogen bond. Intermolecular forces are not of comparable strength. (2) CCℓ4 molekules het slegs Londonkragte terwyl H2O waterstofbindings het. Intermolekulêre kragte is nie van vergelykbare sterkte. [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 5 NW/November 2024 Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 4/VRAAG 4 4.1 Boyle’s law: the pressure of an enclosed gas is inversely proportional to the volume it occupies at constant temperature. Boyle se wet: die druk van ‘n ingeslote gas is omgekeerd eweredig aan die volume wat dit by konstante temperatuur beslaan.  (2) 4.2.1 Pressure/Druk  (1) 4.2.2 Mass (of gas) and temperature.  Massa (gas)  en temperatuur.  (2) 4.2 p1V1 = p2V2 128,5 × 55 = 201,93×V2 V2 = 35 cm3 (3) 4.3.1 • 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. • Collisions are perfectly elastic with no loss of total energy of the molecules. • All molecules are identical. • Collisions of particles on the surface cause pressure. (Any two) • Deeltjies in voortdurende beweging in alle rigtings. • Deeltjies dra nie by tot die volume van die gas nie. • Daar is geen kragte tussen die deeltjies, of die deeltjies en die kant van die houer, behalwe tydens botsings. • Botsings tussen die molekules is volkome elasties met geen verlies van kinetiese energie van die molekules. • Al die molekules is identies. • Botsings van deeltjies teen die oppervlak veroorsaak druk. (Enige twee) (2) 4.3.2 High temperature and low pressure. Hoë temperatuur  en lae druk. (2) 4.4 As the temperature increases the average kinetic energy of the molecules increases. pressure of the molecules on each other and on the sides of the container increases as there will be more collisions. As die temperatuur verhoog, verhoog die gemiddelde kinetiese energie van die molekules. druk uitgeoefen deur die molekules op mekaar en die kante van die houer verhoog en daar is meer botsings.  (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 6 NW/November 2024 Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 5/VRAAG 5 5.1 EXOTHERMIC. Energy is released OR ∆H < 0 OR more energy is released than absorbed. EKSOTERMIES. Energie word vrygestel OF ∆H < 0 OF meer energie word vrygestel as geabsorbeer. (2) 5.2 ∆H = HP – HR = -86-25 = -111 k∙J∙mol-1 (2) 5.3.1 Decreases/Afneem (1) 5.3.2 Remain the same/Bly dieselfde. (1) 5.4.1 Activated complex is the unstable transition state from reactants to products. (2) Geaktiveerde kompleks is die onstabiele oorgangstoestand van reaktanse na produkte. 5.4.2 154 k∙J∙mol-1 (2) 5.5 V SO3 : V SO2 2 : 2 V SO3 = V SO2 = 50 cm3 (2) [12] QUESTION 6/VRAAG 6 6.1 Empirical formula is the simplest whole-number ratio of atoms in a compound. Empiriese formule is die eenvoudigste heelgetal verhouding in ‘n verbinding. (2) 6.2 nC:nH:nN m n= M 74.02 8.71 17.27 : :  12 1 14 6.17 : 8.71 : 1.23 5 : 7 : 1 Empirical formula/Empiriese formule: C5H7N (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 7 NW/November 2024 Grade/Graad 11 – Marking Guidelines/Nasienriglyne 6.3 m n= M 40,57 0,25 =  M M = 162,28 g  Empirical formula mass/Empiriese formule massa = 5 ×12 + 7 × 1 + 14 = 81 Ratio of the empirical formula mass: molecular formula mass Verhouding van empiriese formule massa : molekulêre formule massa 81 : 162,28 1 : 2 ∴ Molecular formula/molekulêre formule: C10H14N2  (5) [12] QUESTION 7/VRAAG 7 7.1 Limiting reagent is a substance that is completely used up in a chemical reaction. Beperkte reaktans is ‘n stof wat volledig opgebruik word in ‘n chemiese reaksie.  (2) 7.2 Calcium carbonate/Kalsiumkarbonaat/CaCO3 (1) 7.3.1 m nCO2 = M 4,4 =  44 = 0,1 mol Any one/Enige een V nCO2 = Vm VCO 2 0,1 =  22,4 VCO2 = 2,24 dm3  (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 8 NW/November 2024 Grade/Graad 11 – Marking Guidelines/Nasienriglyne 7.3.2 n(CaCO3) : n(CO2) 1:1 n(CaCO3) reacted/gereageer = n(CO2) formed/gevorm = 0,1 mol mCaCO 3 n(CaCO3 ) =  M mCaCO 3 0,1 =  100 m(CaCO3 ) = 10 g  mass of CaCO3 reacted % purity = x 100% mass of IMPURE CaCO3 10 = x 100%  12 = 83 ,33%  (6) 7.3.3 n c=  V n 2=  0,15 n = 0,15 mol n(HCℓ) reacted/gereageer : n(CO2) formed/gevorm 2:1 1 n(HCℓ) reacted/gereageer = x 0,1 = 0.05 mol 2 n(HCℓ) in excess/oormaat = n(HCℓ) initial/begin - n(HCℓ) reacted/gereageer = 0,15 – 0,05  = 0,1 mol  (5) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 9 NW/November 2024 Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 8/VRAAG 8 8.1 Base is a proton/H+ ion acceptor.  Basis is ‘n proton/ H+- ioon ontvanger.  (2) 8.2 pH = - log[H3O+]  = - log(0,1) = 1 (3) 8.3 Cℓ-. (1) 8.4 It is a substance that can act as either acid or base. (2) Dit is ‘n stof wat as ‘n suur of ‘n basis kan optree.  8.5 HSO-4 + H2O ⇌ H3O+ + SO42-  (3) 8.6.1 m c=  MV 7,9 =  56(0,25) = 0,56 mol∙dm-3 (3) 8.6.2 ca Va na =  cb Vb nb ca x40 1 =  0,56 x 25 2 ca = 0,175 mol∙dm-3  (4) 8.6.3 c1V1 = c2V2  c1 × 10 = 0,175 × 500 c1 = 8,75 mol∙dm-3 (4) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Physical Sciences P2/Fisiese Wetenskappe V2 10 NW/November 2024 Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION 9/VRAAG 9 9.1 Reduction is a decrease in oxidation number. Reduksie is ‘n afname in oksidasiegetal.  (2) 9.2.1 +2  (1) 9.2.2 +3 (1) 9.3 Mn2+/ MnO . (3) Oxidation number of Mn2+ decreases. Oksidasiegetal van Mn2+ neem af.  9.4.1 Cu ⟶ Cu2+ + 2e-  (2) 9.4.2 NO3- + 4H+ + 3e- ⟶ NO + 2H2O  (2) 9.4.3 Cu ⟶ Cu2+ + 2e- × 3  NO3- + 4H+ + 3e- ⟶ NO + 2H2O × 2  3Cu + 2NO3- + 8H+ ⟶ 3 Cu2+ + 2NO + 4H2O (4) [15] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou