ChemHL P2 M06 TZ0 M - c IB DIPLOMA PROGRAMME PROGRAMME DU...

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c IB DIPLOMA PROGRAMME PROGRAMME DU DIPLÔME DU BI PROGRAMA DEL DIPLOMA DEL BI M06/4/CHEMI/HP2/ENG/TZ0/XX/M+ 17 pages MARKSCHEME May 2006 CHEMISTRY Higher Level Paper 2
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– 2 – M06/4/CHEMI/HP2/ENG/TZ0/XX/M This markscheme is confidential and for the exclusive use of examiners in this examination session. It is the property of the International Baccalaureate and must not be reproduced or distributed to any other person without the authorization of IBCA.
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– 6 – M06/4/CHEMI/HP2/ENG/TZ0/XX/M SECTION A 1. (a) 61 2 2 2 2 C H 9O 6CO 6H O; + + [1] (b) (i) f products f reactants () HH H = ∑∆ −∑∆ 00 0 (6 394 6 242) ( 43); H −− 0 31 c 3773/ 3.8 10 (kJ mol ); H = × 0 [2] Accept 2 , 3 or 4 sf . Award [1] for / ( ) 3773 3.8 10 kJ mol ++ × . Allow ECF from (a) only if coefficients used . (ii) pr ( ) (6 189 6 214) (385 9 205); SS S = =× +× 0 11 c 188 (J K mol ); S = 0 [2] Accept only 3sf . Award [1] for –188 . Allow ECF from (a) only if coefficients used . (c) c ( G V = c H V c T) S V = 3800 (298 0.188); × 1 3900 kJ mol = . [2] Accept –3800 to –3900 . Accept 2, 3 or 4 sf . Allow ECF from (b) . Units needed for second mark . (d) spontaneous and G V negative; [1] Allow ECF from (c) . (e) 1 1 H × / 676; 2 1 H × / –394; 3 2 H × / – 484; 1 4 202 (kJ mol ); H = [4] Accept alternative methods . Correct answers score [4] . Award [3] for () ( / ) . 1 202 or 40 kJ kJ mol
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– 7 – M06/4/CHEMI/HP2/ENG/TZ0/XX/M 2. (a) rr (Tl) 203 0.2952 205 0.7048/ (Tl) 204.41; AA = (Br) 79 0.5069 81 0.4931/ (Br) 79.99; = r3 (TlBr ) 204.41 3 79.99 444.38/ 444.37; M =+ × = [3] Correct answer scores [3] . Ignore units of g or g mol –1 . Apply ECF to M r from A r values . (b) r M is an average value (because of the isotopes); each HBr molecule has its own value depending on which isotopes (of H or Br) it contains/ OWTTE ; [2] (c) 22 6261 02 6 1s 2s 2p 3s 3p 3d 4s 4p ; [1] Do not accept noble gas shortcut . No subscripts . (d) 2 Mg + ; [1] (e) 32 3 A l ,O ,Ne ,Na,F,N ; + + −− [2] Award [2] for any three, [1] for any two . 3. 3 23 n(Fe O ) 30 10 159.7 / n(Fe O ) 188 mol; ÷ = 3 n(C) 5.0 10 12.01/ n(C) 416 mol =×÷ = ; Fe O is the limiting reagent or implicit in calculation; n(Fe) 2 n(Fe O ) 2 188 376 mol; = m(Fe) 376 55.85 21 kg; = [5] Accept 2sf or 3sf, otherwise use –1(SF) . Correct final answers score [5] . Allow ECF . 4. (a) (i) (a species that) gains electrons (from another species) / causes electron loss; [1] (ii) changes by 3; reduced because its oxidation number decreased / 63 / + ++ + / it has gained electrons; [2] (b) (i) 686 666 CHO 2H 2e ; + [1] (ii) C H O 2Fe C H O 2H 2Fe ; + + [1]
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– 8 – M06/4/CHEMI/HP2/ENG/TZ0/XX/M 5. (a) same general formula; successive members differ by 2 CH ; Do not allow elements or just “they” .
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ChemHL P2 M06 TZ0 M - c IB DIPLOMA PROGRAMME PROGRAMME DU...

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