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*24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3...

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ADVANCED General Certificate of Education 2017 Chemistry Assessment Unit A2 2 assessing Analytical, Transition Metals, Electrochemistry and Further Organic Chemistry [AC222] MONDAY 19 JUNE, MORNING TIME 2 hours. INSTRUCTIONS TO CANDIDATES Write your Centre Number and Candidate Number in the spaces provided at the top of this page. Answer all fifteen questions. Answer all ten questions in Section A. Record your answers by marking the appropriate letter on the answer sheet provided. Use only the spaces numbered 1 to 10. Keep in sequence when answering. Answer all five questions in Section B. You must answer the questions in the spaces provided. Do not write outside the boxed area on each page or on blank pages. Complete in black ink only. Do not write with a gel pen. INFORMATION FOR CANDIDATES The total mark for this paper is 120. Quality of written communication will be assessed in Question 13(b). In Section A all questions carry equal marks, i.e. two marks for each question. In Section B the figures printed down the right-hand side of pages indicate the marks awarded to each question or part question. A Periodic Table of the Elements, containing some data, is included in this question paper. *AC222* *AC222* *24AC22201* *24AC22201* Centre Number Candidate Number 10506
Transcript
Page 1: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

ADVANCEDGeneral Certificate of Education

2017

ChemistryAssessment Unit A2 2assessingAnalytical, Transition Metals, Electrochemistry and Further Organic Chemistry[AC222]MONDAY 19 JUNE, MORNING

TIME2 hours.

INSTRUCTIONS TO CANDIDATESWrite your Centre Number and Candidate Number in the spaces provided at the top of this page.Answer all fifteen questions.Answer all ten questions in Section A. Record your answers by marking the appropriate letter on the answer sheet provided. Use only the spaces numbered 1 to 10. Keep in sequence when answering.Answer all five questions in Section B. You must answer the questions in the spaces provided.Do not write outside the boxed area on each page or on blank pages. Complete in black ink only. Do not write with a gel pen.

INFORMATION FOR CANDIDATESThe total mark for this paper is 120.Quality of written communication will be assessed in Question 13(b).In Section A all questions carry equal marks, i.e. two marks for each question.In Section B the figures printed down the right-hand side of pages indicate the marks awarded to each question or part question.A Periodic Table of the Elements, containing some data, is included in this question paper.

*AC222*

*AC222*

*24AC22201*

*24AC22201*

Centre Number

Candidate Number

10506

Page 2: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22202*

*24AC22202*

10506

Section A

For each of the following questions only one of the lettered responses (A–D) is correct.

Select the correct response in each case and mark its code letter by connecting the dots as illustrated on the answer sheet.

1 Which one of the following states a condition at which a standard hydrogen electrode operates?

A A platinum/palladium electrode

B A temperature of 20 °C

C Hydrogen gas at a pressure of 1 atmosphere

D Sulfuric acid at a concentration of 1.0 M

2 Which one of the following compounds is the most soluble in water?

A CH3CH2OCOCH3

B CH3CO2CH3

C C2H5NH2

D C6H13COOH

Page 3: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22203*

*24AC22203*

10506[Turn over

3 Phosphorus burns in oxygen to form phosphorus pentoxide which dissolves in water to form phosphoric acid. Sodium hydroxide neutralises phosphoric acid.

P4 1 5O2 → P4O10

P4O10 1 6H2O → 4H3PO4

3NaOH 1 H3PO4 → Na3PO4 1 3H2O

Which one of the following is the volume of 0.30 M sodium hydroxide required to neutralise the phosphoric acid formed from 0.31 g of phosphorus?

A 7 cm3

B 100 cm3

C 200 cm3

D 500 cm3

4 Which one of the following industrial processes is catalysed by a mixture of metals?

A Hardening of unsaturated oils

B Manufacture of sulfuric acid

C Oxidation of ammonia

D Synthesis of ammonia

5 Addition of an excess of a concentrated solution of ammonia to chloroethanoic acid gives

A CH2ClCONH2.

B CH2ClCO2NH4.

C NH2CH2CO2NH4.

D NH2CH2CONH2.

Page 4: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22204*

*24AC22204*

10506

6 Which one of the following amino acids can not be found in proteins?

A NH2CH2COOH

B NH2CH2CH(CH3)COOH

C NH2CH(COOH)CH2OH

D NH2CO(CH2)2CH(NH2)COOH

7 Which species can act as an oxidising agent but not as a reducing agent?

A Cl22

B Cl2

C ClO22

D ClO42

8 Two standard electrode potentials are shown below.

Sn21 1 2e2 ? Sn 20.14 V Mn21 1 2e2 ? Mn 21.19 V

Which one of the following is the strongest oxidising agent?

A Mn

B Mn21

C Sn

D Sn21

Page 5: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22205*

*24AC22205*

10506[Turn over

9 [Ar] is the electronic configuration of argon. Which one of the following is the electronic configuration of an iron(III) ion in the ground state?

A [Ar] 3d5

B [Ar] 3d6

C [Ar] 4s23d3

D [Ar] 4s13d4

10 The volatile product formed when ammonium ethanoate is heated with phosphorus pentoxide for an extended period is

A ethanamide.

B ethanenitrile.

C phosphoric acid.

D water.

Page 6: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22206*

*24AC22206*

10506

Section B

Answer all five questions in the spaces provided

11 The following table shows the reaction of some transition metal ion solutions with a few drops of aqueous ammonia solution and then with excess aqueous ammonia solution. Results for the first ion have been completed. Fill in the remaining spaces.

transition metal ion solutions

addition of aqueous ammonia solution

addition of excess aqueous ammonia solution

Mn21 white precipitate no further reaction

Fe21

Fe31

Co21

Ni21

Cu21

[8]

Page 7: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22207*

*24AC22207*

10506[Turn over

12 Naphthalene is the largest single constituent of coal tar (6%).

naphthalene

It is insoluble in water but very soluble in benzene. Naphthalene resembles benzene in many of its reactions but forms addition and substitution products more readily and is more easily oxidised and reduced.

(a) The structure of naphthalene showing double bonds, which is drawn above, is known as a Kekulé structure. There are two possible Kekulé structures for naphthalene. Draw the other one.

[1]

(b) The structures of benzene and naphthalene are based on delocalised π electrons. This results in the shape of the benzene molecule being flat.

(i) Explain what is meant by the term delocalised.

[2]

(ii) Explain what is meant by the term π electrons.

[2]

(iii) How many π electrons does one molecule of naphthalene have?

[1]

Page 8: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22208*

*24AC22208*

10506

(c) Benzene is a liquid at room temperature with a boiling point of 80 °C. Naphthalene is a solid at room temperature with a melting point of 80 °C.

Explain why benzene is a liquid and naphthalene is a solid.

[2]

(d) Benzene is very slowly oxidised by acidified potassium manganate(VII) but naphthalene is readily oxidised forming phthalic acid.

COOH

COOH

phthalic acid

(i) Draw the structures of the other two isomers of phthalic acid, labelling them A and B.

[2]

(ii) Which one of these phthalic acids (A or B) is used to make polyethylene terephthalate?

[1]

(iii) Draw the structure of one repeating unit of polyethylene terephthalate.

[2]

Page 9: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22209*

*24AC22209*

10506[Turn over

(e) One of the most important uses of naphthalene compounds is the formation of azo dyes. Orange 1 was the first acidic azo dye on the market. It is formed by the coupling of diazotised sulfanilic acid with 1-naphthol.

OH

N NHO3S

Orange 1

(i) Draw the structures of the two reactants which couple to form Orange 1.

[2] (ii) Describe how a diazotisation reaction is carried out.

[3]

(iii) Explain, in terms of electronic structure, why this azo dye is coloured.

[3]

Page 10: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22210*

*24AC22210*

10506

(iv) Orange 1 contains the sulfonic acid group, 2SO3H. Write the equation for the reaction of a sulfonic acid group with sodium hydroxide.

[1]

(v) Explain why the solubility of Orange 1 is greatly increased as a sodium salt.

[1]

(f) Naphthalene is brominated by boiling it in a solvent with bromine and a suitable catalyst to give 1-bromonaphthalene.

Br

1-bromonaphthalene

(i) Suggest the name of a catalyst used in the bromination of benzene and naphthalene.

[1]

Page 11: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22211*

*24AC22211*

10506[Turn over

(ii) Draw a flow scheme to show the mechanism for the catalysed bromination of naphthalene.

[3]

(iii) Explain why benzene is not easily brominated but ethene is.

[2]

Page 12: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22212*

*24AC22212*

10506

13 The amount of chlorine in water is readily determined by liberating iodine from potassium iodide.

2KI 1 Cl2 → 2KCl 1 I2

The iodine produced can then be titrated with sodium thiosulfate.

I2 1 2Na2S2O3 → Na2S4O6 1 2NaI

(a) A spatula measure of potassium iodide, an excess, was added to 75 cm3 of water through which chlorine had been bubbled. A 25.0 cm3 sample of the water was titrated against 0.01 M sodium thiosulfate solution. The titre value was 22.5 cm3.

(i) Name the indicator used in the titration.

[1]

(ii) When is the indicator added?

[1]

(iii) State the colour change at the end point.

[2]

(iv) Calculate the volume of chlorine gas bubbled into the water. Assume the gas is bubbled through at 20 °C and one atmosphere pressure and that all of the gas dissolves.

[4]

Page 13: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22213*

*24AC22213*

10506[Turn over

(b) Titration methods are not accurate with chlorine concentrations lower than 0.03 parts per million. Orthotoluidine reacts with chlorine to produce a yellow colour which can be used in colorimetry to determine the concentration of chlorine.

Explain, giving full experimental detail, how you would use colorimetry to determine the concentration of chlorine in a sample of water.

[4]

Quality of written communication [2]

Page 14: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22214*

*24AC22214*

10506

14 There are four amino compounds that have the formula C3H9N. They are shown below with their trivial names and their boiling points.

CH3CH2CH2NH2 CH3CH2NHCH3

propylamine, 48 °C ethylmethylamine, 37 °C

(CH3)2CHNH2 (CH3)3N

isopropylamine, 32 °C trimethylamine, 4 °C

(a) Explain why these amines are isomers.

[2]

(b) (i) Explain why trimethylamine has a lower boiling point than propylamine.

[2]

(ii) Describe how a mixture of milligrams of these isomers could be separated.

[2]

Page 15: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22215*

*24AC22215*

10506[Turn over

(c) Propylamine is made by the reaction of 1-chloropropane with ammonia to form a salt.

(i) Write an equation for this reaction.

[2]

(ii) How would you liberate the amine from the salt?

[1]

(d) Explain which of the amines are primary, secondary or tertiary amines.

[3]

(e) Give the systematic name for isopropylamine.

[2]

(f) The amines, when placed in a mass spectrometer, produce fragment ions. A major fragment ion in one mass spectrum had an m/e ratio of 43. Identify this fragment ion and use it to explain which amine(s) might produce it.

[2]

Page 16: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22216*

*24AC22216*

10506

(g) The nmr spectrum shown below is for one of the amines.

6 5 4 3 2 1 0

A

B

C

D

(i) Identify the chemical which causes the signal at 0.0.

[1]

(ii) Explain why the signal at 0.0 is a singlet.

[2]

Page 17: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22217*

*24AC22217*

10506[Turn over

(iii) Explain how the integration curve can be used to identify the hydrogen attached to the nitrogen.

[2]

(iv) Explain the splitting pattern for peaks B and D.

[2]

(v) Identify the amine.

[1]

(vi) Explain how infrared spectrometry could be used to confirm the identity of the amine.

[2]

Page 18: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22218*

*24AC22218*

10506

15 Cyanide ions react with thiosulfate ions to produce thiocyanate ions and sulfite ions.

CN2 1 S2O322 → SCN2 1 SO3

22

(a) What is the oxidation number of sulfur in thiosulfate and sulfite ions?

[2]

(b) (i) Name the ion which is used to show the presence of thiocyanate ions.

[1]

(ii) What is the colour produced when using this ion?

[1]

(iii) Write the equation for the reaction.

[2]

(c) The reaction of cyanide ions with thiosulfate ions is catalysed by rhodanase.

HN SH HN S2SH 1 S2O3

2 1 SO32

O O

rhodanase

HN S2SH HN SH 1 CN 1 SCN

O O

STEP 1

STEP 2

Page 19: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22219*

*24AC22219*

10506[Turn over

(i) Explain how the flow scheme confirms that rhodanase acts as a catalyst.

[2]

(ii) Rhodanase is an enzyme. Explain what is meant by the term enzyme.

[1]

(iii) Explain why the ability of rhodanase to act as a catalyst decreases with increasing temperature.

[2]

(iv) In the rhodanase structure the –SH group is attached to a cysteine group which is an amino acid.

CH2SHCH(NH2)COOH

cysteine

Write equations for the reactions of cysteine with sodium carbonate and with nitrous acid.

sodium carbonate

[2]

nitrous acid

[2]

(v) The structure –S–S– occurs in proteins. State its name and explain the role it plays in protein structure.

[2]

Page 20: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22220*

*24AC22220*

10506

(d) Cyanide ions, thiosulfate ions and thiocyanate ions can all form complexes.

(i) What name is given to these ions which can form complexes?

[1]

(ii) Cyanide ions form the complex ion [Fe(CN)6]32. Draw the shape of this complex ion, and state, in terms of the cyanide ion, why this shape is formed.

[2]

(iii) Thiosulfate ions form the complex [Ag(S2O3)2]32 when they remove silver bromide from photographic emulsions. What is the charge on the silver ion in the complex?

[1]

(iv) Thiocyanate ions have two structures i.e. 2S C N or S C N2. What are the implications for the structure of thiocyanate complexes?

[2]

Page 21: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

*24AC22221*

*24AC22221*

10506

THIS IS THE END OF THE QUESTION PAPER

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DO NOT WRITE ON THIS PAGE

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*24AC22222*

*24AC22222*

10506

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DO NOT WRITE ON THIS PAGE

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*24AC22223*

*24AC22223*

10506

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217599

Permission to reproduce all copyright material has been applied for.In some cases, efforts to contact copyright holders may have been unsuccessful and CCEAwill be happy to rectify any omissions of acknowledgement in future if notified.

DO NOT WRITE ON THIS PAGE

For Examiner’suse only

QuestionNumber Marks

Section A1–10

Section B1112131415

TotalMarks

*24AC22224*

*24AC22224*

Page 25: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

ADVANCEDGeneral Certificate of Education

2017

ChemistryAssessment Unit A2 2

assessingAnalytical, Transition Metals, Electrochemistry

and Further Organic Chemistry

[AC222]

MONDAY 19 JUNE, MORNING

10506.01 F

MARKSCHEME

Page 26: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

10506.01 F 2 [Turn over

General Marking Instructions

Introduction Mark schemes are published to assist teachers and students in their preparation for examinations.Through the mark schemes teachers and students will be able to see what examiners are looking for inresponse to questions and exactly where the marks have been awarded. The publishing of the markschemes may help to show that examiners are not concerned about finding out what a student does notknow but rather with rewarding students for what they do know.

The Purpose of Mark SchemesExamination papers are set and revised by teams of examiners and revisers appointed by the Council.The teams of examiners and revisers include experienced teachers who are familiar with the level andstandards expected of students in schools and colleges.

The job of the examiners is to set the questions and the mark schemes; and the job of the revisers is toreview the questions and mark schemes commenting on a large range of issues about which they mustbe satisfied before the question papers and mark schemes are finalised.

The questions and the mark schemes are developed in association with each other so that the issuesof differentiation and positive achievement can be addressed right from the start. Mark schemes,therefore, are regarded as part of an integral process which begins with the setting of questions andends with the marking of the examination.

The main purpose of the mark scheme is to provide a uniform basis for the marking process so that allthe markers are following exactly the same instructions and making the same judgements in so far asthis is possible. Before marking begins a standardising meeting is held where all the markers arebriefed using the mark scheme and samples of the students’ work in the form of scripts. Considerationis also given at this stage to any comments on the operational papers received from teachers and theirorganisations. During this meeting, and up to and including the end of the marking, there is provision foramendments to be made to the mark scheme. What is published represents this final form of the markscheme.

It is important to recognise that in some cases there may well be other correct responses which areequally acceptable to those published: the mark scheme can only cover those responses whichemerged in the examination. There may also be instances where certain judgements may have tobe left to the experience of the examiner, for example, where there is no absolute correct response – allteachers will be familiar with making such judgements.

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10506.01 F 3

AVAILABLEMARKS

Section A

1 C

2 C

3 B

4 C

5 C

6 B

7 D

8 D

9 A

10 B

[2] for each correct answer [20] 20

Section A 20

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10506.01 F 4 [Turn over

AVAILABLEMARKS

Section B

11 Fe2+ green precipitate no further reaction/insoluble Fe3+ rust/brown precipitate no further reaction/insoluble Co2+ blue precipitate yellow solution Ni2+ green precipitate blue solution Cu2+ blue precipitate dark/deep blue solution

[–1] for two errors [8] 8

12 (a) [1]

(b) (i) electrons spread over several atoms [2] (ii) electrons in p orbitals [1] which overlap (sideways) [1] [2] (iii) 5 × 2 = 10 [1]

(c) naphthalene is a larger molecule/greater mass/more electrons [1] van der Waals forces are greater [1]

(d) (i) COOH

COOHHOOC

COOH

A B

[2] (ii) B [1]

(iii) OC CO O CH2CH2 O

[2]

(e) (i)

HO3S N N+

OH

[2] [1] [1]

(ii) sodium nitrite + (dil or conc) hydrochloric acid [1] add amine [1] , 10°C [1] add naphthol [3]

(iii) electrons exclusively delocalised [1] energy levels closer [1] light (energy) raises electrons to higher (energy) levels removes/absorbs (a) colour (from the light) [1]

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10506.01 F 5

AVAILABLEMARKS

(iv) –SO3H + NaOH –SO3Na + H2O [1]

(v) ions are surrounded by water molecules which disperse the ions [1]

(f) (i) iron/iron bromide/iron(III) bromide etc [1]

(ii)

+ Br2 + FeBr4–+

HBrFeBr3

[3]

(iii) relative stabilities of the electron systems, benzene delocalised [1] in ethene electrons are available [1] 29

Br

+ HBr + FeBr3

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10506.01 F 6 [Turn over

AVAILABLEMARKS

13 (a) (i) starch [1]

(ii) when the iodine colour has largely disappeared/straw coloured [1]

(iii) blue-black to colourless [2]

(iv) 22.5 cm3 of 0.01 M thiosulfate contains 22.5 × 10–2 × 10–3 = 2.25 × 10–4 mol mol Cl2 in 25cm3 = 0.5 × 2.25 × 10–4 = 1.125 × 10–4 mol mol Cl2 in 75cm3 = 3 × 1.125 × 10–4 = 3.375 × 10–4 mol volume of gas = 24 × 3.375 × 10–4 dm3 = 0.0081 dm3 = 8.1cm3 [4]

(b) select the appropriate filter for the colorimeter use different concentrations of chlorine with orthotoluidine to produce a calibration curve match the colour of sample with the calibration curve [4] Quality of written communication [2] 14

14 (a) they have the same (molecular) formula but a different structure [2]

(b) (i) trimethylamine cannot form H bonds/only van der Waals [1] propylamine has H bonds between the amino groups [1]

(ii) use GLC/TLC/paper chromatography remove the amines at different retention times [2]

(c) (i) CH3CH2CH2Cl + NH3 CH3CH2CH2NH3+Cl– [2]

(ii) Add sodium hydroxide solution [1]

(d) primary: propylamine and isopropylamine + explanation secondary: ethylmethylamine + explanation tertiary: trimethylamine + explanation [3]

(e) 2-aminopropane [2]

(f) C3H7+ = 43 either CH3CH2CH2

+ or (CH3)2CH+ propylamine or isopropylamine [2]

(g) (i) tetramethylsilane [1]

(ii) four methyl groups/hydrogens/protons all in the same chemical environment [2]

(iii) integration curve is the ratio of hydrogen atoms [1] nitrogen is the lowest peak only one hydrogen [1]

(iv) D triplet due to CH3 next to CH2 [1] B quartet due to CH2 next to CH3 [1]

(v) ethylmethylamine [1]

(vi) compare the spectrum with those of the amines [1] the identical ones confirm which amine [1] 26

Page 31: *24AC22201* · The iodine produced can then be titrated with sodium thiosulfate. I 2 1 2Na 2S 2O 3 → Na 2S 4O 6 1 2NaI (a) 3A spatula measure of potassium iodide, an excess, was

10506.01 F 7

AVAILABLEMARKS

15 (a) the oxidation number of S in thiosulfate is +2 [1] in sulphite it is +4 [1] [2]

(b) (i) iron(III) [1]

(ii) (blood) red [1]

(iii) [Fe(H2O)6]3+ + SCN– [Fe(SCN)(H2O)5]

2+ + H2O [2] (c) (i) rhodanase takes part in the reaction in step 1 [1] rhodanase reformed in step 2 [1]

(ii) biological catalyst [1]

(iii) the structure of the protein is denatured/changed/H bonds broken [1] active site destroyed [1]

(iv) 2CH2SHCH(NH2)COOH + Na2CO3 2CH2SHCH(NH2)COONa + CO2 + H2O [2] CH2SHCH(NH2)COOH + HNO2 CH2SHCHOHCOOH + N2 + H2O [2]

(v) disulfide bridge [1] holding chains together [1]

(d) (i) ligands [1]

(ii) octahedral diagram [1] the cyanide ion is small [1] [2]

(iii) 1+/Ag+ [1]

(iv) the thiocyanate ion can attach itself to the central metal atom/ion [1] by either the sulfur atom or the nitrogen atom [1] 23

Section B 100

Total 120


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