Acceptable Answers Reject Mark

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Carmel Stone
5 years ago
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1 (a)(i) two marks Cl in Cl is 0 Goes to + in HClO Goes to in HCl () All three correct for two marks Any two correct for one mark Ignore correct oxidation numbers for other elements If three correct numbers given without saying what species they are in max for these two marks Third mark Only Cl + for oxidation number + Only Cl - for oxidation number - (treat each separately) For each incorrect oxidation number change for O and H, lose one mark. 3 Cl/Cl /the same element is both oxidized and reduced Allow same molecule/species/ type of atom is both oxidized and reduced if answer elsewhere has been in terms of chlorine Cl/Cl /the same element both increases and decreases in oxidation number Chlorine both loses and gains electrons () 0 to + described as reduction and/or 0 to - described as oxidation (for third mark) (a)(ii) Equilibrium moves to the left / moves in reverse direction / moves to increase concentration of reactants () Just reverse reaction is favoured To use up (some of) added HCl/ to react with added HCl/ to stop formation of HCl/ restores equilibrium by producing more chlorine and water () Second mark depends on first Just to counteract the change in the system To minimise effect of HCl Allow moves to decrease concentration of products/hcl for both marks

2 (b)(i) ClO + H + + e (- ) Cl + H O Equations without electrons ALLOW ClO + H + Cl + H O e (-) () I I + e ( ) ALLOW I e (-) I () Allow multiples Ignore state symbols even if incorrect (b)(ii) ClO + H + + I Cl + H O + I Equations including electrons Mark independently. No TE on (b)(i) (b)(iii) Moles thiosulfate = (4.0 x / 000) =. x 0-3 /. x 0-3 /0.00/ 0.00 (mol) ().0 x 0-3 (mol) x 0-3 / 0.00 Moles iodine = half moles of thiosulfate = 6.05 x 0-4 / 6. x 0-4 / / (mol) () 6.0 x 0-4 (mol) 6 x 0-4 (mol) Correct answer without working () (b)(iv) Moles ClO = 6.05 x 0 4 (mol) TE on (b)(ii) and (b)(iii): If ratio ClO :I = : answer is x answer to (b)(iii) If ratio ClO :I = : answer is half of answer to (b)(iii)

3 (b)(v) Concentration = (6.05 x 0 4 x 000/5) =.4 x 0 / 0.04/ 0.04/.4 x 0 (mol dm 3 ) Answers to significant figure TE. Answer to (b)(iv) x (b)(vi) (Minimum) amount of I to react with OCl - = x answer to (b)(iv) = x 6.05 x 0 4 =. x 0 3 (mol) () Allow TE for x answer to (b)(iv) Ignore s.f. KI is in excess if no calculation has been done. Moles of I (9.04 x 0 3 ) is more than this number of moles of ClO / I is in excess / KI is in excess / so that all the ClO can react () 9.04 x 0 3 mol I can react with 4.5 x 0-3 mol OCl - () Ignore s.f. TE from incorrect equation in (b)(ii) Moles OCl - (6.05 x 0 4 ) is less than this/ I is in excess / KI is in excess / so that all the ClO can react () (b)(vii) 0.30 x 00 / 4. (= ) =.4/. %

4 (b)(viii) Judgement (of colour change) at end point / adding starch too early in the titration / jet of burette not filled Some potassium iodide did not dissolve Errors must cause an increase in titre. Ignore Just Human error Just overshot endpoint Transfer errors / spillage Errors due to misreading burette / pipette Leaving funnel in burette Errors which affect both the students titre and an accurate titre using the same solutions e.g. impu solutions (c) (Cl radicals) break down ozone (layer)/ ozone depletion / ozone (layer) thinning Allow damage ozone (layer)/ react with ozone Global warming Causes acid rain Total = 7 marks

5 (a) (i) H O + CO H CO 3 (Allow atoms in H CO 3 in any order) Or H O + CO H + + HCO 3 Or H O + CO H + + CO 3 Or H 3 O + in place of H + IGNE STATE SYMBOLS EVEN IF INCRECT (a) (ii) H + + CO 3 H O + CO H CO 3 as a product LHS () RHS () H + + CO 3 HCO 3 H 3 O + + CO 3 3H O + CO LHS () RHS () Any other ions including spectator IGNE STATE SYMBOLS, EVEN IF INCRECT ions (e.g. Ca +, Cl ) in IGNE arrows the equation scores zero (b) (i) If collection over water is not somehow evident Conical flask and a delivery tube leaving the conical flask IGNE heat beneath conical flask () Inverted measuring cylinder with collection over water shown and cylinder above mouth of delivery tube () ALLOW collection over water to be shown/implied in the diagram without labels or other annotation

6 (b) (ii) Any method which is likely to bring the reactants into contact after the apparatus is sealed Method suggesting mixing the reactants and then putting bung in flask very quickly (b) (iii) ( =) /9.333 x 0 3 (mol) as answer Ignore SF except SF Ignore any incorrect units (b) (iv) CaCO 3 (s) + HCl(aq) CaCl (aq) + H O(l) + CO (g/aq) ALL FOUR state symbols must be correct for this mark (b) (v) (Mass of mol CaCO 3 = x 6) = 00 g ALLOW just 00 ALLOW any incorrect units ALLOW 00. g just 00. (Reason: this uses the Periodic Table value of A r = 40. for Ca) (b) (vi) (Mass of CaCO 3 = 00 x ) = (g) () IGNE sig figs including sf here NOTE: Moles of CaCO 3 consequential on answers to (b)(iii) and (b)(v) [NOTE: if A r = 40. used for Ca, then the answer = (g)] Percentage of CaCO 3 in the coral = 00 x /.3 = 8.6% () Final % answer is not given to 3 sf NOTE: If mass CaCO 3 used is 0.93, final answer is 8.3% [NOTE: if A r = 40. used for Ca, then the answers = (g) and 8.7%]

7 (b) (vii) (Different samples of) coral have different amounts of CaCO 3 /different proportions of CaCO 3 / different levels of CaCO 3 Answers that do not include any mention of CaCO 3 ALLOW calcium carbonate for CaCO 3 Only one sample of coral (was) used References to solubility of CO in water References to repeating the experiment at a different temperature

8 (a) (i) (COOH) CO + H + + e () MnO 4 + 8H + + 5e Mn + + 4H O () (a) (ii) 5(COOH) + MnO 4 + 6H + 0CO + Mn + + 8H O Equation with electrons left in ALLOW multiples ALLOW 5(COOH) + MnO 4 + 6H + 0CO + Mn + + 8H O + 0H + Ignore state symbols even if incorrect (a) (iii) Moles of MnO 4 =.30/000 x 0.00 =.3 x 0 4 TE for 5th mark if % is 5 (mol) () greater than 00% Moles of (COOH) in 0 cm 3 =.3 x 0 4 x 5/ =.85 x 0 4 (mol) () Moles of (COOH) in whole sample =.85 x 0 4 x 50 = 0.04(5) (mol) () Mass of acid = 0.04(5)x 90 =.7 g () % in leaves =.7/50 x 00 = 0.5 (%) () If ratio 5 : is not used, maximum (4) Rounding errors once in first 4 marks Final answers not quoted to dp e.g. if ratio :5 is used then percentage in leaves = 0.08% (a) (iv) ± 0.05 cm 3 () [(0.05 x ) /.3] x 00 = 0.88% ALLOW ±0.05 cm 3 [(0.05 x ) /.3] x 00 = 0.44% () () () ALLOW TE for second mark

9 (a) (v) Any two from: Only one titration carried out () Leaves may contain other substances that MnO 4 could oxidize/ react with () Errors in technique e.g. transfer errors Not all ethanedioic acid extracted from leaves () ALLOW temperature too low / below 60 C () Different amounts of acid from different leaves () (a) (vi) (Wearing gloves suggested as) ethanedioic acid is toxic / harmful References to weak acid rhubarb leaves are toxic /harmful () Rhubarb is toxic (Unnecessary because) it is (very) dilute / present in small amounts () ALLOW because is not absorbed through the skin Second mark is independent of the first (a) (vii) (Cloudiness due to) MnO (solid /precipitate) () Ignore colour of precipitate EITHER Suitable use of E θ (+0.34V) MnO 4 ions are a strong enough oxidizing agent to oxidize Cl ions () (b) (i) (s )s p 6 3s 3p 6 3d 5 (4s 0 ) 4s 3d 3 (b) (ii) Octahedral

Correct Answer Reject Mark. Mass of ethanoic acid = 0.04 x 60.1 = (2.404 g) Volume of ethanoic acid = =

Correct Answer Reject Mark. Mass of ethanoic acid = 0.04 x 60.1 = (2.404 g) Volume of ethanoic acid = = (a)(i) Mass of ethanoic acid = 0.04 x 60. = (.404 g) Volume of ethanoic acid =.404.049 =.97 =.3 (cm 3 ) Correct answer with no working () Ignore SF except only one 60.0 for molar mass which gives mass.4