EKSIKA JOINT EVALUATION TEST CHEMISTRY MSCM PP2 1 i) ii) iii) iv) v) vi) vii) 2 D2O31 B/Potassium/k it has the biggest /largest atomic radius hence most reading loses electrons /least ionization energy. D has a larger/bigger atomic radius than G D has fewer number of protons ½ /lower 02 nuclear charge hence weaker nuclear attraction on its valence A1 On the grid 2.8.8.21 Period 4 group II Covalent bond. Element D has a higher ½ charge with smaller atomic radius that allows it to share electrons with G. F is more reactive than G.F has a smaller ½ atomic radius than G hence more ½ readily gains an electron. a) The enthalpy change that occurs when one mole of a substance is formed from its constituent element under temperature of 298k and a pressure of 760mmHg. b) i) -Enthalpy of combustion of hydrogen. -Enthalpy of formation of water. ii) C2H6(g)+ 72 O2 ½ Labeling of both axis correctly ½ ∆H=-1560KJ/mole ½ Energy KJ 2CO2(g)+3H 2O(l) ½ Reaction path/progress/course iii) 2CO2(g)+3H2O(l)→C2H6(g)+ 72 O2(g),∆H1=+1560 ½ 2C(s)+2O2(g)→2CO2(g)∆H2=-394x2=-788 ½ 3H2(g)+ 32 O2(g)→3H2O(l),∆H3=-286x3=-858 ½ 2C(s)+3H2(g)→C2H6(g) ∆H=-86KJ/mole ½ iv) i) © 2015 Eksika Joint Evaluation Test Heat change=mc∆+ 500 gx 4.2 Jxgk21.5 k 2mks 45,150 J 233/2 Chemistry Marking Scheme 1 MALTYTECH PRINTERS If 86,000 J 30 g of C 2 H 6 ii) c) 3 i) 45,150 J 45,150 Jx 30 g 86, 000 J 15.75 g of C 2 H 6 lattice energy/Reserve lattice energy1 ii) ∆H3=∆H1+∆H2 =+2489+-2659 ½ =-170KJ/mole ½ a) i) ii) iii) Al3+/Aluminum/Aluminum (iii) ion1 Amphoteric property/Amphoterism1 Al(OH)3(s)+3H+(aq)→Al3+(aq)+3H2O(l)1 b) i) Step 2=Excess Carbon(iv) Oxide1 Step 4=Dilute hydrochloric acid1 ii) Hydrated calcium sulphate is used in holdingbroken bones in position/plaster of Paris iii) Ca(HCO3)2(aq)+Ca(OH)2(aq)→2CaCO3(s)+2H2O(l)1 iv) To aqueous calcium chloride,add 1H2SO4/any soluble sulphate.Filter 1to get calcium sulphate as residue.Dry ½ the calcium sulphate between filter papers. c) Fe(s)+2HCl(g)→FeCl2(s)+H2(g) ½ 1800cm3 mole ratio 1800 No of moles of HCl 24, 000 0.075moles mole ratio of Fe : HCl 1 : 2 No of moles of Fe 0.075 2 0.0375moles 03 Mass of iron reacted 0.0375 x56 2.1g Mass of unreacted iron 5.0 2.1 2.9 g 4 a) i) Condensation polymerization1 ii) 1 © 2015 Eksika Joint Evaluation Test 233/2 Chemistry Marking Scheme 2 MALTYTECH PRINTERS b) i) M Q -Propyl hydrogen1 -2-Chloropropane1 ii) 2CH3CH2CH2OH(l)+9O2(g)→6CO2(g)+8H2O(l)1 iii) Ethanoic acid/CH3COOH1 1 iv) c) 5 v) Reagent -hydrogen gas1 Conditions -nickel catalyst and temperature tied 150-2500C do not accept a range of temperature1 vi) Put a spatula full of sodium hydrogen carbonate1/sodium carbonate to test tubes of propanoic acid and propanal-01 separately. In propanoic acid there is efference ½ while in propan-1-01 there is no inference i) 2-bromobutane1 ii) 2-chloropent-2-ene1 a) S2│S . it has tQ value of 0.00v/ it is the standard reference electrode b) Q ½ c) i) © 2015 Eksika Joint Evaluation Test 233/2 Chemistry Marking Scheme 3 MALTYTECH PRINTERS ii) 6 E.m.f = cell reduced – cell oxidated =+0.34-(-2.90) ½ =+3.24v ½ d) T(s)→T2+(aq)+2e-0.34 R2+(aq)+2e→R(s)-2.38 T(s)+R2+(aq)→T2+(aq)+R(s) tӨ=-2.72v1 There will be no ½ reaction between the solution of R2+ and the container of T.Overal e.m.f is negative ½ (-2.72v) a) i) X Y ii) Finely divided iron catalyst1 iii) Reducing property/Reduction1 iv) 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)1 v) Used as a nitrogenous fertilizer1 vi) Insert ½ a glowing splint.The glowing slint will be rekindled/relit/re ignited. ½ i) Dehydration1 ii) Solid B Gas C iii) -Used as fuel1 any -Used as a reducing agent of some metallic oxides iv) To absorb carbon(iv) oxide1 b) © 2015 Eksika Joint Evaluation Test -Nitrogen gas/N2(g) ½ -Hydrogen gas/H2(g) ½ -Copper ½ -Carbon(iv) oxide ½ 233/2 Chemistry Marking Scheme 4 MALTYTECH PRINTERS 7 i) Graph attached at the back 03 ii) Correct reading from the graph 27.5mm ½ + 0.2 Shown on the graph ½ 01 a ) PbNO 3 2aq 2 NI aq PbI 2 s NaNO iii) mole ratio of Pb NO3 : NaI 1 : 2 3 aq b) No.of moles NaI 0.002 x 2 0.004moles No of moles of Pb NO3 2 8 x 0.25 1000 0.002moles iv) Pb2+(aq)+2I-(aq)→PbI2(s)1 v) Sodium nitrate solution │NaNO3(aq)1 vi) Name a chemical reagent that can be added to the mixture to spread up the setting of the precipitate1 © 2015 Eksika Joint Evaluation Test 233/2 Chemistry Marking Scheme 5 MALTYTECH PRINTERS