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AS Chemistry Sample Assessment Materials Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) First teaching from September 2015 First certification from 2016 Issue 1

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Page 1: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

AS Chemistry

Sample Assessment MaterialsPearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0)First teaching from September 2015First certifi cation from 2016 Issue 1

Page 2: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed
Page 3: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

PearsonEdexcel Level 3Advanced Subsidiary GCE in Chemistry (8CH0) Sample Assessment Materials

First certification 2016

Page 4: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Edexcel, BTEC and LCCI qualifications

Edexcel, BTEC and LCCI qualifications are awarded by Pearson, the UK’s largest awarding body offering academic and vocational qualifications that are globally recognised and benchmarked. For further information, please visit our qualification websites at www.edexcel.com, www.btec.co.uk or www.lcci.org.uk. Alternatively, you can get in touch with us using the details on our contact us page at qualifications.pearson.com/contactus

About Pearson

Pearson is the world's leading learning company, with 40,000 employees in more than 70 countries working to help people of all ages to make measurable progress in their lives through learning. We put the learner at the centre of everything we do, because wherever learning flourishes, so do people. Find out more about how we can help you and your learners at qualifications.pearson.com

References to third party material made in these sample assessment materials are made in good faith. Pearson does not endorse, approve or accept responsibility for the content of materials, which may be subject to change, or any opinions expressed therein. (Material may include textbooks, journals, magazines and other publications and websites.)

All information in this document is correct at the time of publication.

Original origami artwork: Mark Bolitho

Origami photography: Pearson Education Ltd/Naki Kouyioumtzis

ISBN 978 1 446 91449 6

All the material in this publication is copyright © Pearson Education Limited 2015

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Contents

1 Introduction 1

2 General marking guidance 3

3 Paper 1 Core Inorganic and Physical Chemistry 5

4 Paper 1 Mark Scheme 33

5 Paper 2 Core Organic and Physical Chemistry 53

6 Paper 2 Mark Scheme 77

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Introduction

The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed for use in schools and colleges. It is part of a suite of GCE qualifications offered by Pearson.

These sample assessment materials have been developed to support this qualification and will be used as the benchmark to develop the assessment students will take.

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

1

Introduction

The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed for use in schools and colleges. It is part of a suite of GCE qualifications offered by Pearson.

These sample assessment materials have been developed to support this qualification and will be used as the benchmark to develop the assessment students will take.

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

2

General marking guidance

● All candidates must receive the same treatment. Examiners must mark the last candidate in exactly the same way as they mark the first.

● Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than be penalised for omissions.

● Examiners should mark according to the mark scheme – not according to their perception of where the grade boundaries may lie.

● All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme.

● Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification/indicative content will not be exhaustive.

● When examiners are in doubt regarding the application of the mark scheme to a candidate’s response, a senior examiner must be consulted before a mark is given.

● Crossed-out work should be marked unless the candidate has replaced it with an alternative response.

 

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

3

General marking guidance

● All candidates must receive the same treatment. Examiners must mark the last candidate in exactly the same way as they mark the first.

● Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than be penalised for omissions.

● Examiners should mark according to the mark scheme – not according to their perception of where the grade boundaries may lie.

● All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate’s response is not worthy of credit according to the mark scheme.

● Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification/indicative content will not be exhaustive.

● When examiners are in doubt regarding the application of the mark scheme to a candidate’s response, a senior examiner must be consulted before a mark is given.

● Crossed-out work should be marked unless the candidate has replaced it with an alternative response.

 

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Centre Number Candidate Number

Write your name hereSurname Other names

Total Marks

Paper Reference

Turn over

S47551A©2015 Pearson Education Ltd.

1/1/1/1/1/1/1/

*s47551A0128*

ChemistryAdvanced SubsidiaryPaper 1: Core Inorganic and Physical Chemistry

Sample Assessment Materials for first teaching September 2015Time: 1 hour 30 minutes 8CH0/01You must have: Data BookletScientific calculator, ruler

Instructions

• Use black ink or ball-point pen.• Fill in the boxes at the top of this page with your name, centre number and candidate number.• Answer all questions.• Answer the questions in the spaces provided

– there may be more space than you need.

Information

• The total mark for this paper is 80. • The marks for each question are shown in brackets

– use this as a guide as to how much time to spend on each question.• You may use a scientific calculator.• For questions marked with an *, marks will be awarded for your ability to

structure your answer logically showing the points that you make are related or follow on from each other where appropriate.

• A Periodic Table is printed on the back cover of this paper.

Advice

• Read each question carefully before you start to answer it.• Try to answer every question.• Check your answers if you have time at the end.• Show all your working in calculations and include units where appropriate.

Pearson Edexcel Level 3 GCE

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Answer ALL questions.

Write your answers in the spaces provided.

Some questions must be answered with a cross in a box . If you change your mind about an answer, put a line through the box

and then mark your new answer with a cross .

1 Many elements in the Periodic Table have different isotopes.

(a) What is meant by the term isotopes, with reference to sub-atomic particles?(2)

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(b) A radioactive compound containing the phosphide ion, 32P3–, is used in the treatment of skin cancer.

What is the electronic configuration of the phosphide ion, 32P3−?(1)

A 1s22s22p63s2

B 1s22s22p63s23p3

C 1s22s22p63s23p6

D 1s22s22p63s23p33d3

(c) A sample of silicon contains 92.2% 28Si and 4.67% 29Si, the remainder being 30Si.

Calculate the relative atomic mass of silicon in this sample, giving your answer to an appropriate number of significant figures.

(2)

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(d) Silicon tetrachloride, SiCl4, is a covalent substance which is a liquid at room temperature.

Calculate the number of molecules in 5.67 g of SiCl4.(2)

(Total for Question 1 = 7 marks)

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2 The molecules carbon dioxide, water and chloric(I) acid can be represented by these structures.

O=C=O OH H

O

H

Cl

All the bonds in these molecules are polar because the elements have different electronegativities.

(a) In terms of atomic structure, give two reasons why oxygen is more electronegative than carbon.

(2)

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(b) Explain why carbon dioxide is the only non-polar molecule of the three.(2)

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(c) The diagram shows the structure of a water molecule and of a chloric(I) acid molecule.

O

H

ClOH H

Draw a diagram to show how a hydrogen bond can form between the two molecules.

(2)

(Total for Question 2 = 6 marks)

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3 Phosphorus(V) chloride, PCl5, is a pale yellow solid which sublimes when heated.

Phosphorus(V) chloride may be prepared in the laboratory from dry phosphorus(III) chloride and dry chlorine, using the apparatus shown.

anhydrous calcium chloride tube

phosphorus(V) chloride

phosphorus(III) chloride

dry chlorine

(a) Which statement gives a reason why phosphorus(V) chloride has a higher melting temperature than phosphorus(III) chloride?

(1)

A PCI5 contains phosphorus with a higher oxidation number

B PCI5 has stronger intermolecular forces

C PCI5 molecules have more covalent bonds

D PCI5 molecules have stronger covalent bonds

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(b) (i) Explain an essential safety precaution for this preparation.(2)

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(ii) The reaction is exothermic.

Explain how the apparatus could be altered to maximise the formation of the product.

(2)

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(iii) Write an equation for the reaction between phosphorus(III) chloride and chlorine. State symbols are not required.

(1)

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(c) 4.17 g of phosphorus(V) chloride is reacted with ethanedioic acid, H2C2O4, to form phosphorus oxychloride as shown by the following equation:

PCl5(s) + H2C2O4(s) → POCl3(l) + 2HCl(g) + CO2(g) + CO(g)

The phosphorus oxychloride is then hydrolysed by water as shown by the following equation:

POCl3(l) + 3H2O(l) → H3PO4(aq) + 3HCl(g)

Calculate the total volume, in dm3, of hydrogen chloride gas produced by both reactions.

[1 mol of any gas occupies 24.0 dm3 at room temperature and pressure.](3)

(Total for Question 3 = 9 marks)

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4 This question is about disproportionation reactions.

(a) Solutions containing chlorate(I) ions can be used as household bleaches and disinfectants.

These solutions decompose on heating as shown by the following equation:

3ClO–(aq) → ClO3–(aq) + 2Cl–(aq)

What is the oxidation number of chlorine in each of these three ions?(1)

ClO– ClO3– Cl–

A +1 +3 –1

B –1 +3 +1

C –1 +5 +1

D +1 +5 –1

(b) A coin, of mass 5.00 g, contains silver. The coin is dissolved in 500 cm3 of concentrated nitric acid to form silver nitrate solution, AgNO3(aq).

50.0 cm3 of this solution is reacted with excess sodium chloride solution to form a precipitate of silver chloride, AgCl.

Ag+(aq) + Cl−(aq) → AgCl(s)

After filtering and drying, the mass of the precipitate was 0.617 g.

Calculate the percentage by mass of silver in the coin. Give your answer to an appropriate number of significant figures.

(4)

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(c) Silver nitrate reacts with iodine as shown by the following equation:

6AgNO3(aq) + 3I2(aq) + 3H2O(l) → AgIO3(aq) + 5AgI(s) + 6HNO3(aq)

The reaction is classified as a disproportionation reaction.

(i) What is meant by the term disproportionation?(2)

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(ii) Deduce the element undergoing disproportionation in this reaction.(1)

A Ag

B H

C I

D O

(Total for Question 4 = 8 marks)

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5 Compounds of Group 1 and Group 2 elements show trends in their properties.

(a) Which block of the Periodic Table contains the Group 2 elements?(1)

A s

B p

C d

D f

(b) Some rocks contain the compound strontium carbonate, SrCO3.

(i) Which of the following could represent successive ionisation energies, in kJ mol–1, for the Group 2 element strontium (Sr)?

(1)

A 1350 2370 3560 5020

B 400 2650 3850 5110

C 550 1060 4120 5440

D 640 1180 1980 6000

(ii) Write an equation to show the decomposition of strontium carbonate on heating.

Include state symbols.(2)

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*(iii) A student compares the rate of thermal decomposition of some Group 1 and Group 2 carbonates using the apparatus shown.

solid

limewater

HEAT

The student tried to keep the temperature of heating approximately constant by using a Bunsen burner with the same settings. The time taken for the limewater to go cloudy is shown in the table.

Sample Formula of carbonate

Time taken for limewater to turn cloudy / s

A CaCO3 40

B MgCO3 20

C K2CO3 does not decompose

D Li2CO3 35

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Discuss how the student can use observations from the experiment, and knowledge of the cations present in the carbonates, to justify the relative rate of decomposition.

(6)

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(Total for Question 5 = 10 marks)

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6 Ammonium carbamate can be used to make urea, a fertilizer. Ammonium carbamate has the empirical formula CH6N2O2.

(a) Ammonium carbamate contains 15.38% carbon, 7.69% hydrogen and 35.90% nitrogen by mass. The remainder is oxygen.

Show that the empirical formula of ammonium carbamate is CH6N2O2.(3)

(b) Ammonium carbamate is an ionic compound of formula H2NCOONH4. It contains ammonium ions, NH4

+, and carbamate ions, H2NCOO–. When heated, ammonium carbamate decomposes to release ammonia and carbon dioxide.

(i) Write the balanced equation to show the decomposition of ammonium carbamate into ammonia and carbon dioxide. State symbols are not required.

(1)

(ii) What is the shape of the ammonia molecule, NH3?(1)

A pyramidal

B tetrahedral

C trigonal planar

D V-shaped

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(iii) Complete the dot-and-cross diagram for the carbamate ion, H2NCOO–.

Use dots (•) for the N electrons, crosses (×) for both H and C electrons, circles (o) for O electrons and the symbol Δ for the extra electron on the O–.

(2)

N C

O

O–

H

H

(iv) Deduce the shape of the carbamate ion around the carbon atom. Justify your answer.

(3)

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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(Total for Question 6 = 10 marks)

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7 This question is about Group 7 and redox chemistry.

(a) Explain the trend in the boiling temperatures of the elements on descending Group 7, from fluorine to iodine.

(4)

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(b) Which element in Group 7 has the highest first ionisation energy? (1)

A iodine

B bromine

C chlorine

D fluorine

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(c) When concentrated sulfuric acid, H2SO4, is added to solid sodium bromide, the acid reacts to form a mixture of products, including sulfur dioxide, SO2, and bromine, Br2.

The conversion of bromide ions to bromine is shown by the half-equation:

2Br− → Br2 + 2e−

In parts (c)(i) and (ii), state symbols are not required.

(i) Write a half-equation to show the formation of SO2 from H2SO4.(1)

(ii) Hence write an overall equation for the reaction of Br− ions with H2SO4.(1)

(iii) Deduce the role of Br− ions in the reaction in (c)(ii).(1)

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(d) A student compares the reaction of solid sodium chloride and solid sodium iodide with concentrated sulfuric acid by adding 1 cm3 of acid to separate samples of the solid halides in test tubes.

(i) The reaction between solid sodium chloride and concentrated sulfuric acid produces steamy fumes. Identify the product responsible for this observation.

(1)

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(ii) Solid sodium iodide also reacts with concentrated sulfuric acid. Steamy fumes are produced initially, but a second reaction then takes place.

Give an observation that you would expect the student to make in this second reaction, and write appropriate equations to show the overall reaction taking place.

State symbols are not required.(3)

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(iii) Explain, in terms of the redox reactions occurring, why the solid halides form different products with concentrated sulfuric acid.

(2)

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(Total for Question 7 = 14 marks)

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BLANK PAGE

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8 A sample of trichloroethanoic acid was supplied to a laboratory by a chemical manufacturer. A technician at the laboratory was asked to check whether the percentage purity by mass of the acid was 99.9% as claimed on the label.

The technician used a titration method to determine the purity of the acid. The technician followed this method:

· The technician placed an empty glass bottle on a balance · After zeroing the balance, the technician added a sample of trichloroethanoic

acid to the bottle · The technician recorded the balance reading, accurate to 1 d.p., as 6.2 g · The technician transferred the acid to a beaker and dissolved the acid in a small

volume of distilled water · The technician poured this solution into a 250 cm3 volumetric flask and made the

solution level up to the mark with distilled water · The technician filled a burette with the acid solution · Using a pipette, 25.0 cm3 of 0.130 mol dm–3 sodium hydroxide solution was

transferred to a conical flask · Several 25.0 cm3 samples of the sodium hydroxide solution were titrated with the acid

solution and the results were recorded.

Results

Titration numbers

1 2 3 4 5

Burette reading (final)/cm3 23.15 45.40 22.45 45.20 22.20

Burette reading (initial)/cm3 0.00 23.15 0.15 22.50 0.00

Titre/cm3

Concordant titres ()

Table 1

(a) Complete Table 1 and hence calculate the mean titre in cm3.(2)

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(b) Trichloroethanoic acid has the formula CCl3COOH.

The equation for the reaction with sodium hydroxide is:

CCl3COOH(aq) + NaOH(aq) → CCl3COONa(aq) + H2O(l)

(i) Calculate the concentration of trichloroethanoic acid in g dm–3. Give your answer to one decimal place.

(3)

Turn over

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(ii) Show, using a calculation, that the percentage purity of trichloroethanoic acid is less than that claimed by the manufacturer.

(2)

(c) The uncertainties for the measurements made in this experiment were as follows.

Measurement Uncertainty Percentage error (%)

Each mass reading / g ±0.05

Volumetric flask volume / cm3 ±0.5 0.200

Pipette volume / cm3 ±0.04

Each burette reading / cm3 ±0.05 0.449

(i) Complete the table.(1)

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(ii) The total percentage error for the experiment can be estimated by adding together the four percentage errors.

Explain whether the manufacturer’s claim that the acid is 99.9% pure is correct.

Use your answer to (b)(ii) and the total percentage error for this experiment. (2)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(d) Identify three issues with the technician’s method and for each issue identify an improvement.

(6)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(Total for Question 8 = 16 marks)

TOTAL FOR PAPER = 80 MARKS

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*s47551A02728*

BLANK PAGE

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

33

CH

EMIS

TRY

AS

PA

PER

1 M

AR

KS

CH

EME

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

1(a

) An

answ

er t

hat

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

• at

om

s w

ith

the

sam

e nu

mbe

r of

pro

tons

(1

)

• w

ith

diff

eren

t nu

mbe

rs o

f ne

utro

ns

(1)

Rej

ect

‘Ele

men

ts w

ith

the

sam

e…’

Igno

re r

efer

ence

s to

the

sam

e nu

mbe

r of

el

ectr

ons

Igno

re ‘a

tom

s of

the

sam

e el

emen

t th

at d

iffer

on

ly in

mas

s nu

mbe

r’

2

1(b

) C

1

1(c

) •

calc

ulat

ion

of %

30Si a

nd s

ubst

itut

ion

into

ex

pres

sion

sho

win

g su

m o

f ab

unda

nce

x m

ass

num

ber

÷ t

otal

abu

ndan

ce

(1)

• ev

alua

tion

of

corr

ect

answ

er t

o 3

s.f.

(

1)

Exam

ple

of c

alcu

lation

: (9

2.2

x 28

) +

(4.

67 x

29)

+

( 3

.13

x 30

)

10

0

(=

28.

1093

) =

28.

1

C

orre

ct a

nsw

er w

ith

no w

orki

ng t

o 3.

s.f

scor

es

2 m

arks

Ig

nore

any

uni

ts

2

1(d

)

• ca

lcul

atio

n of

num

ber

of m

oles

of

mol

ecul

es

pres

ent

(1)

• us

e of

Avo

gadr

o nu

mbe

r to

con

vert

to

num

ber

of

mol

ecul

es

(1)

Exam

ple

of c

alcu

lation

: nu

mbe

r of

mol

es o

f m

olec

ules

= 5

.67

÷ 1

70.1

=

0.0

3333

...

num

ber

of m

olec

ules

= 0

.033

33..

.x 6

.02

x 10

23

= 2

.01

x 10

22

Allo

w 2

x 1

022

Cor

rect

ans

wer

no

wor

king

sco

res

2 m

arks

2

(To

tal

for

Qu

esti

on

1 =

7 m

arks

)

Page 40: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Qu

esti

on

N

um

ber

A

nsw

er

A

dd

itio

nal

Gu

idan

ce

Mar

k

2(a

) An

answ

er t

hat

mak

es r

efer

ence

to

the

follo

win

g po

ints

: O

(at

om)

• ha

s m

ore

prot

ons

/ ha

s gr

eate

r nu

clea

r ch

arge

(1)

has

smal

ler

(ato

mic

) ra

dius

(th

an C

ato

m)

(1)

Igno

re r

efer

ence

s to

shi

eldi

ng

Allo

w j

ust

‘sm

alle

r’

Allo

w r

ever

se a

rgum

ent

for

carb

on

2

2(b

) An

expl

anat

ion

that

mak

es r

efer

ence

to:

(o

nly

carb

on d

ioxi

de is

non

-pol

ar)

beca

use

only

car

bon

diox

ide

is s

ymm

etri

cal /

line

ar

(1

)

O

R

bond

pol

aritie

s ar

e ve

ctor

s /

vect

or q

uant

itie

s

• an

d th

eref

ore

the

bond

pol

aritie

s ca

ncel

(1

)

2

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

34

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

35

Qu

esti

on

N

um

ber

A

nsw

er

A

dd

itio

nal

Gu

idan

ce

Mar

k

2(c

)

lone

pai

r of

ele

ctro

ns o

n O

of

one

mol

ecul

e

(1

)

• δ+

sym

bol o

n on

e re

leva

nt H

ato

m A

ND

δ–

sym

bol

on o

ne r

elev

ant

O a

tom

(1)

If

no

repr

esen

tation

of

a hy

drog

en b

ond

(by

dash

ed li

ne o

r si

mila

r),

then

onl

y on

e of

the

se m

arks

can

be

awar

ded

No

pena

lty

for

show

ing

both

pos

sibl

e hy

drog

en

bond

s

2

(To

tal

for

Qu

esti

on

2 =

6 m

arks

)

Page 42: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Q

ues

tio

n

Nu

mb

er

An

swer

Ad

dit

ion

al G

uid

ance

M

ark

3(a

) B

1

3

(b)(

i)

An

expl

anat

ion

that

mak

es r

efer

ence

to

the

follo

win

g po

ints

: •

use

a fu

me

cupb

oard

(1)

as c

hlor

ine

is t

oxic

/ p

oiso

nous

(1

)

2

3(b

)(ii

) An

expl

anat

ion

that

mak

es r

efer

ence

to

the

follo

win

g po

ints

: •

cool

the

rea

ctio

n ve

ssel

/ s

urro

und

the

flask

with

cold

wat

er

(1)

in o

rder

to

prev

ent

subl

imat

ion

(of

PCl 5)

/ to

pr

even

t th

e PC

l 5 t

urni

ng in

to a

gas

(1)

2

3(b

)(ii

i)

• PC

l 3 +

Cl 2 →

PC

l 5

(1

)

Igno

re s

tate

sym

bols

, ev

en if

inco

rrec

t

1

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

36

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

37

Qu

esti

on

N

um

ber

A

nsw

er

A

dd

itio

nal

Gu

idan

ce

Mar

k

3(c

) •

calc

ulat

ion

of

mol

es P

Cl 5

(=

mol

es P

OC

l 3)

(

1)

• m

oles

HC

l = m

oles

PC

l 5 x

5

(

1)

volu

me

HC

l = m

oles

HC

l x 2

4 dm

3

(1)

Allo

w e

cf f

or s

teps

in c

alcu

lation

, ig

nore

si

gnifi

cant

fig

ures

in f

inal

ans

wer

exc

ept

one

sign

ifica

nt f

igur

e C

orre

ct a

nsw

er w

ith

no w

orki

ng s

core

s 3

mar

ks

Exam

ple

of c

alcu

lation

M

oles

PC

l 5 =

4.1

7

= 0

.02(

00)

(mol

)

20

8.5

M

oles

HC

l = 5

x 0

.02(

00)

= 0

.1(0

0) (

mol

)

Vol

ume

HC

l = 0

.1 x

24

= 2

.4 (

dm3 )

3

(To

tal

for

Qu

esti

on

3 =

9 m

arks

)

Page 44: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Qu

esti

on

N

um

ber

A

nsw

er

A

dd

itio

nal

Gu

idan

ce

Mar

k

4(a

) D

1

4(b

) •

calc

ulat

ion

of

mol

es A

gCl

(1)

tota

l mol

es o

f Ag

in 5

00 c

m3

(w

here

mol

es A

g =

mol

es A

gCl)

(1)

• ca

lcul

atio

n of

tot

al m

ass

of A

g

(1)

eval

uation

of

corr

ect

answ

er t

o 3.

s.f

usin

g %

by

mas

s of

Ag

= M

ass

Ag

x 1

00%

(1)

5.

00

allo

w e

cf f

or s

teps

in c

alcu

lation

; in

clud

ing

for

final

ans

wer

dep

ende

nt o

n ro

undi

ng in

ste

ps o

f th

e ca

lcul

atio

n.

corr

ect

answ

er t

o 3.

s.f

with

no w

orki

ng s

core

s 4

mar

ks

Exam

ple

of c

alcu

lation

M

oles

AgC

l =

0.

617

=

0.0

0430

.. (

mol

)

14

3.4

To

tal m

oles

Ag

=

0.0

0430

x 5

00

= 0

.043

0...

50.

0 M

ass

Ag

= 0

.043

0 x

107.

9 =

4.6

425.

.. =

4.6

4 (g

) %

by

mas

s of

Ag

= 4

.642

5…

x 10

0%

5.0

0

=

92.

9 %

4

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

38

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

39

Qu

esti

on

N

um

ber

A

nsw

er

A

dd

itio

nal

Gu

idan

ce

Mar

k

4(c

)(i)

An

answ

er t

hat

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

• (a

rea

ctio

n in

whi

ch a

n) e

lem

ent

(in

a sp

ecie

s)

(1

) •

is s

imul

tane

ousl

y ox

idis

ed a

nd r

educ

ed

/ fo

r w

hich

the

oxi

dation

num

ber

both

incr

ease

s an

d de

crea

ses

(in

the

sam

e re

action

)

(

1)

Rej

ect

‘ato

m’

2

4(c

)(ii

) C

1

(To

tal

for

Qu

esti

on

4 =

8 m

arks

)

Page 46: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Q

ues

tio

n

Nu

mb

er

An

swer

Ad

dit

ion

al G

uid

ance

M

ark

5(a

) A

1

5

(b)(

i)

C

1

5(b

)(ii

) SrC

O3(

s) →

SrO

(s)

+

CO

2(g)

• sp

ecie

s

(1

)

• st

ate

sym

bols

(

1)

2

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

40

Page 47: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

41

Q

ues

tio

n

Nu

mb

er

An

swer

Ad

dit

ion

al G

uid

ance

M

ark

*5

(b)(

iii)

Th

is q

uest

ion

asse

sses

a s

tude

nt’s

abi

lity

to s

how

a

cohe

rent

and

logi

cally

str

uctu

red

answ

er w

ith

linka

ges

and

fully

-sus

tain

ed r

easo

ning

. M

arks

are

aw

arde

d fo

r in

dica

tive

con

tent

and

for

how

the

an

swer

is s

truc

ture

d an

d sh

ows

lines

of

reas

onin

g.

The

follo

win

g ta

ble

show

s ho

w t

he m

arks

sho

uld

be

awar

ded

for

indi

cative

con

tent

. N

umbe

r of

indi

cative

m

arki

ng p

oint

s se

en in

an

swer

Num

ber

of m

arks

aw

arde

d fo

r in

dica

tive

m

arki

ng p

oint

s 6

4 5–

4 3

3–2

2 1

1 0

0

Gui

danc

e on

how

the

mar

k sc

hem

e sh

ould

be

appl

ied:

Th

e m

ark

for

indi

cative

con

tent

sho

uld

be

adde

d to

the

mar

k fo

r lin

es o

f re

ason

ing.

Fo

r ex

ampl

e, a

n an

swer

with

five

indi

cative

m

arki

ng p

oint

s, w

hich

is p

artial

ly s

truc

ture

d w

ith

som

e lin

kage

s an

d lin

es o

f re

ason

ing,

sc

ores

4 m

arks

(3

mar

ks f

or in

dica

tive

co

nten

t an

d 1

mar

k fo

r pa

rtia

l str

uctu

re a

nd

som

e lin

kage

s an

d lin

es o

f re

ason

ing)

.

If t

here

are

no

linka

ges

betw

een

poin

ts,

the

sam

e fiv

e in

dica

tive

mar

king

poi

nts

wou

ld

yiel

d an

ove

rall

scor

e of

3 m

arks

(3

mar

ks f

or

indi

cative

con

tent

and

no

mar

ks f

or li

nkag

es).

6

Page 48: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Qu

esti

on

N

um

ber

A

nsw

er

A

dd

itio

nal

Gu

idan

ce

Mar

k

*5

(b)(

iii)

co

nt.

Th

e fo

llow

ing

tabl

e sh

ows

how

the

mar

ks s

houl

d be

aw

arde

d fo

r st

ruct

ure

and

lines

of

reas

onin

g.

Ind

icat

ive

con

ten

t:

• C

loud

ines

s /

milk

ines

s /

form

atio

n of

whi

te p

pte

due

to r

eact

ion

betw

een

limew

ater

and

car

bon

diox

ide

The

shor

ter

the

tim

e (f

or l

imew

ater

to

go c

loud

y),

the

fast

er t

he r

ate

of d

ecom

posi

tion

Rat

e of

de

com

posi

tion

de

pend

s on

m

etal

io

n si

ze

and

cha

rge

/ ch

arge

den

sity

B f

aste

r th

an A

as

Mg2+

(ra

dius

) sm

alle

r th

an C

a2+

• B

fas

ter

than

D a

s ch

arge

den

sity

of

Mg2+

gre

ater

th

an L

i+ /

hig

her

char

ge o

f M

g2+ h

as m

ore

effe

ct

than

sm

alle

r ra

dius

of

Li+

• C

doe

s no

t de

com

pose

as

K+ h

as (

rela

tive

ly)

larg

e ra

dius

and

sm

all c

harg

e

N

umbe

r of

mar

ks

awar

ded

for

stru

ctur

e of

ans

wer

and

su

stai

ned

line

of

reas

onin

g

Ans

wer

sho

ws

a co

here

nt a

nd

logi

cal s

truc

ture

with

linka

ges

and

fully

sus

tain

ed li

nes

of

reas

onin

g de

mon

stra

ted

thro

ugho

ut.

2

Ans

wer

is p

artial

ly s

truc

ture

d w

ith

som

e lin

kage

s an

d lin

es o

f re

ason

ing.

1

Ans

wer

has

no

linka

ges

betw

een

poin

ts a

nd is

uns

truc

ture

d.

0

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

42

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43

(To

tal

for

Qu

esti

on

5 =

10

mar

ks)

Qu

esti

on

N

um

ber

A

nsw

er

A

dd

itio

nal

Gu

idan

ce

Mar

k

6(a

)

• ca

lcul

atio

n of

% b

y m

ass

of o

xyge

n

(1)

• ev

alua

tion

of

num

ber

of m

oles

of

C,

H,

N a

nd O

(

1)

conf

irm

atio

n of

rat

io 1

: 6

: 2

: 2

(1)

Exam

ple

of c

alcu

lation

(%

by

mas

s of

) O

= 4

1.03

(%)

C

:

H

:

N

:

O

1

5.38

:

7

.69

:

35

.90

:

41.0

3 1

2.0

1.0

14

.0

16.0

1.2

8

:

7.6

9

:

2.5

6

:

2.

56

1

.28

:

7

.69

:

2

.56

:

2.5

6 1

.28

1

.28

1

.28

1.2

8

=

1

:

6

:

2

:

2

3

Page 50: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Qu

esti

on

N

um

ber

A

nsw

er

A

dd

itio

nal

Gu

idan

ce

Mar

k

6(b

)(i)

H2N

CO

ON

H4 →

2N

H3

+ C

O2

(1

)

Igno

re s

tate

sym

bols

, ev

en if

inco

rrec

t 1

6(b

)(ii

) A

1

6

(b)(

iii)

• al

l ele

ctro

n pa

irs

corr

ectly

show

n fo

r C

=O

an

d C

—O−

(1

)

corr

ect

elec

tron

pai

rs f

or C

—N

bon

d an

d t

he —

NH

2 gr

oup

and

the

lone

pai

r on

N

(1)

2

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

44

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45

Q

ues

tio

n

Nu

mb

er

An

swer

Ad

dit

ion

al G

uid

ance

M

ark

6(b

)(iv

) •

shap

e: t

rigo

nal p

lana

r

(1)

just

ifica

tion

:

C=

O t

reat

ed a

s a

sing

le b

ond

pair

of

elec

tron

s/(s

hape

of

ion)

bas

ed o

n th

ree

bond

pai

rs

of e

lect

rons

(ar

ound

cen

tral

C a

tom

)/(s

hape

of

ion

base

d on

) th

ree

area

s of

ele

ctro

n de

nsity

(aro

und

cent

ral C

ato

m)/

(sha

pe o

f io

n ba

sed

on)

thre

e vo

lum

es o

f el

ectr

on d

ensi

ty (

arou

nd c

entr

al C

ato

m)

(1

) el

ectr

on p

airs

/ele

ctro

n re

gion

s re

pel t

o po

sition

s of

m

axim

um s

epar

atio

n/m

inim

um r

epul

sion

(1)

Rej

ect

‘ato

ms

repe

l’/‘b

onds

rep

el’/

Just

‘e

lect

rons

rep

el’

3

(To

tal

for

Qu

esti

on

6 =

10

mar

ks)

Page 52: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Qu

esti

on

N

um

ber

A

nsw

er

A

dd

itio

nal

Gu

idan

ce

Mar

k

7(a

) An

expl

anat

ion

that

mak

es r

efer

ence

to:

• bo

iling

tem

pera

ture

s in

crea

se f

rom

flu

orin

e to

io

dine

(1

)

• (a

s) m

ore

elec

tron

s pe

r (X

2) m

olec

ule

from

flu

orin

e

to io

dine

(1)

• (t

here

fore

the

) st

reng

th o

f Lo

ndon

for

ces

incr

ease

s fr

om f

luor

ine

to io

dine

(1)

Pl

us o

ne

from

:

• (s

o) m

ore

ener

gy r

equi

red

to s

epar

ate

mol

ecul

es

(1)

(so)

mor

e en

ergy

req

uire

d to

bre

ak t

he

inte

rmol

ecul

ar f

orce

s

(1

)

Allo

w m

olec

ules

incr

ease

in s

ize

/ m

ass

from

flu

orin

e to

iodi

ne

Allo

w ‘m

ore

Lond

on f

orce

s’ f

rom

flu

orin

e to

io

dine

Allo

w ‘m

ore

heat

’ nee

ded

to s

epar

ate

mol

ecul

es

Allo

w m

ore

ener

gy r

equi

red

to o

verc

ome

the

in

term

olec

ular

att

ract

ions

Rej

ect

‘mor

e en

ergy

req

uire

d to

bre

ak c

oval

ent

bond

s’

Allo

w r

ever

se a

rgum

ent

4

7(b

) D

1

7

(c)(

i)

• H

2SO

4 +

2H

+ +

2e–

→ S

O2

+ 2

H2O

o

r

• SO

42– +

4H

+ +

2e– →

SO

2 +

2H

2O

(1

)

Allo

w m

ultipl

es

Igno

re s

tate

sym

bols

, ev

en if

inco

rrec

t

1

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

46

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

47

Q

ues

tio

n

Nu

mb

er

An

swer

Ad

dit

ion

al G

uid

ance

M

ark

7(c

)(ii

) •

H2S

O4

+ 2

H+ +

2Br– →

SO

2 +

2H

2O +

Br 2

o

r

• SO

42– +

4H

+ +

2Br– →

SO

2 +

2H

2O +

Br 2

(1)

No

ecf

from

(c)

(i)

Allo

w m

ultipl

es

Igno

re s

tate

sym

bols

, ev

en if

inco

rrec

t

1

7(c

)(ii

i)

• re

duci

ng a

gent

/ele

ctro

n do

nor/

redu

ces

sulfu

ric

acid

/red

uces

H2S

O4

(1

)

1

7(d

)(i)

hydr

ogen

chl

orid

e /

HC

l

(

1)

Ig

nore

sta

te s

ymbo

ls

1

7(d

)(ii

) O

bser

vation

: •

blac

k so

lid /

gre

y so

lid /

pur

ple

vapo

ur

OR

pun

gent

gas

O

R y

ello

w s

olid

O

R g

as s

mel

ling

of r

otte

n eg

gs

(

1)

Equa

tion

s:

• N

aI +

H2S

O4 →

NaH

SO

4 +

HI

(1

)

• 2H

I +

H2S

O4 →

I 2

+

SO

2 +

2H

2O

OR

6H

I +

H2S

O4 →

3I

2 +

S +

4H

2O

OR

8H

I +

H2S

O4 →

4I

2 +

H2S

+ 4

H2O

(1)

2nd e

quat

ion

mus

t m

atch

obs

erva

tion

mad

e

Allo

w p

urpl

e so

lid

Allo

w 2

NaI

+ H

2SO

4 →

Na 2

SO

4 +

2I

Allo

w c

ombi

nation

s of

bot

h eq

uation

s fo

r bo

th

mar

ks

e.g.

2N

aI +

3H

2SO

4 →

2N

aHSO

4 +

I2

+ S

O2

+

2H2O

3

Page 54: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Qu

esti

on

N

um

ber

A

nsw

er

A

dd

itio

nal

Gu

idan

ce

Mar

k

7(d

)(ii

i)

An

expl

anat

ion

that

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

• io

dide

ions

are

bet

ter

redu

cing

age

nts

(tha

n ch

lori

de io

ns)

(1)

beca

use

iodi

de io

ns lo

se e

lect

rons

mor

e re

adily

/

elec

tron

s in

iodi

de io

ns a

re le

ss s

tron

gly

held

by

the

nucl

eus

(1

)

Allo

w H

I is

a b

ette

r re

duci

ng a

gent

(th

an H

Cl)

Allo

w r

ever

se a

rgum

ent

2

(To

tal

for

Qu

esti

on

7 =

14

mar

ks)

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

48

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

49

Qu

esti

on

N

um

ber

A

nsw

er

A

dd

itio

nal

Gu

idan

ce

Mar

k

8(a

) •

corr

ect

calc

ulat

ion

of a

ll m

ean

titr

es (

23.1

5 an

d 22

.25

and

22.3

0 an

d 22

.70

and

22.2

0)

(1)

• co

ncor

dant

titre

s tick

ed (

2, 3

and

5)

and

cal

cula

tion

of

mea

n titr

e =

22.

25 (

cm3 )

(1

)

(i.e

. th

ose

that

agr

ee w

ithi

n ±

0.20

cm

3 )

2

8(b

)(i)

calc

ulat

ion

of n

umbe

r of

mol

es t

rich

loro

etha

noic

ac

id (

= n

umbe

r of

mol

es o

f N

aOH

)

(1)

rear

rang

emen

t an

d ev

alua

tion

of

tric

hlor

oeth

anoi

c

acid

con

cent

ration

in m

ol d

m-3

(

1)

eval

uation

of

Mr o

f tr

ichl

oroe

than

oic

acid

and

co

nver

sion

to

conc

entr

atio

n in

g d

m-3

, to

1 d

p

(

1)

Allo

w e

cf f

or s

teps

in c

alcu

lation

; in

clud

ing

for

final

ans

wer

dep

ende

nt o

n ro

undi

ng in

ste

ps o

f th

e ca

lcul

atio

n.

Cor

rect

ans

wer

with

no w

orki

ng t

o 1d

p sc

ores

3

mar

ks

Exam

ple

of c

alcu

lation

mol

es a

cid

= m

oles

NaO

H =

(0.

130

x 25

.0)

10

00

= 3

.25

x 10

–3 / 0

.003

25 (

mol

) co

ncen

trat

ion

of a

cid

= 3

.25

x 10

–3 x

100

0

22

.25

=

0.1

46…

mol

dm

-3

conc

entr

atio

n ac

id in

g d

m3 =

0.1

46…

x 1

63.5

=

23.

9 g

dm-3

3

Page 56: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Qu

esti

on

N

um

ber

A

nsw

er

A

dd

itio

nal

Gu

idan

ce

Mar

k

8(b

)(ii

) •

calc

ulat

ion

of n

umbe

r of

gra

ms

of

tric

hlor

oeth

anoi

c ac

id in

250

cm

3

(

1)

calc

ulat

ion

of %

pur

ity,

sho

win

g it is

< 9

9.9%

(1)

O

R •

conv

ersi

on o

f m

easu

red

mas

s in

to t

heor

etic

al

conc

entr

atio

n in

g d

m-3

(1)

• ca

lcul

atio

n of

% p

urity,

sho

win

g it is

< 9

9.9%

(1)

Exam

ple

of c

alcu

lation

: m

ass

acid

in 2

50 c

m3

= 2

3.9

x 2

50

= 5

.975

g

1000

pu

rity

= 5

.975

x

100

%

= 9

6.4

%,

whi

ch is

<99

.9%

6.

2

OR

th

eore

tica

l con

cent

ration

= 6

.2 x

100

0 =

24.

8 g

dm-3

2

50

puri

ty =

23.

9

x 10

0%

= 9

6.4

%,

whi

ch is

< 9

9.9%

2

4.8

Allo

w e

cf o

n va

lue

in (

b)(i

)

2

8(c

)(i)

• (e

ach

mas

s re

adin

g) =

1.6

1 %

an

d (e

ach

pipe

tte

read

ing)

= 0

.160

%

(1

)

Exam

ple

of c

alcu

lation

Ea

ch m

ass

read

ing:

) 2

x 0.

05 x

100

%

6.

2

=

1.6

1%

Each

pip

ette

vol

ume:

±

0.0

4 x

100%

25

.0

= 0

.160

%

1

8(c

)(ii

) An

expl

anat

ion

that

mak

es r

efer

ence

to:

• To

tal %

err

or =

2.4

2%

(1)

• cl

aim

is n

ot c

orre

ct b

ecau

se 9

6.4

± 2

.42%

is

still

low

er t

han

the

man

ufac

ture

r’s

valu

e of

99

.9%

(1

)

ecf

on v

alue

obt

aine

d in

(c)

(i)

2

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

50

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

51

Q

ues

tio

n

Nu

mb

er

An

swer

Ad

dit

ion

al G

uid

ance

M

ark

8(d

) M

axim

um t

hre

e m

arks

for

issu

e id

entifie

d M

axim

um t

hre

e m

arks

for

impr

ovem

ent

iden

tifie

d w

hich

m

ust

be li

nked

with

asso

ciat

ed is

sue

iden

tifie

d

• is

sue:

mas

s of

(so

lid)

acid

not

acc

urat

ely

wei

ghed

ou

t

(1

) •

impr

ovem

ent:

wei

gh m

ass

of a

cid

by

diff

eren

ce/r

inse

out

the

wei

ghin

g bo

ttle

/use

a

bala

nce

read

ing

to 2

d.p

./us

e a

mor

e pr

ecis

e ba

lanc

e

(1

)

• is

sue:

som

e ac

id w

ill b

e le

ft in

the

bea

ker/

som

e ac

id

will

not

be

tran

sfer

red

to t

he v

olum

etri

c fla

sk

(1)

impr

ovem

ent:

rin

se o

ut t

he b

eake

r (i

n w

hich

the

so

lid a

cid

was

dis

solv

ed)

and

add

the

was

hing

s to

th

e vo

lum

etri

c fla

sk

(1)

issu

e: in

suff

icie

nt m

ixin

g of

the

so

lution

/con

cent

ration

of

the

solu

tion

will

not

be

unifo

rm

(

1)

impr

ovem

ent:

inve

rt t

he v

olum

etri

c fla

sk (

seve

ral

tim

es)

(1)

issu

e: b

uret

te n

ot r

inse

d

(1

)

• im

prov

emen

t: b

uret

te s

houl

d be

rin

sed

with

acid

so

lution

bef

ore

use

(1

)

Rej

ect

use

of a

‘mor

e ac

cura

te’ b

alan

ce

Allo

w p

ipet

te n

ot r

inse

d w

ith

sodi

um h

ydro

xide

6

(To

tal

for

Qu

esti

on

8 =

16

mar

k)

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

5353

Centre Number Candidate Number

Write your name hereSurname Other names

Total Marks

Paper Reference

Turn over

S47552A©2015 Pearson Education Ltd.

1/1/1/1/1/1/1/

*s47552A0124*

ChemistryAdvanced SubsidiaryPaper 2: Core Organic and Physical Chemistry

Sample Assessment Materials for first teaching September 2015Time: 1 hour 30 minutes 8CH0/02You must have:Data Booklet Scientific calculator, ruler

Instructions

• Use black ink or ball-point pen.• Fill in the boxes at the top of this page with your name, centre number and candidate number.• Answer all questions.• Answer the questions in the spaces provided

– there may be more space than you need.

Information

• The total mark for this paper is 80. • The marks for each question are shown in brackets

– use this as a guide as to how much time to spend on each question.• You may use a scientific calculator.• For questions marked with an *, marks will be awarded for your ability to

structure your answer logically showing the points that you make are related or follow on from each other where appropriate.

Advice

• Read each question carefully before you start to answer it.• Try to answer every question.• Check your answers if you have time at the end.• Show all your working in calculations and include units where appropriate.

Pearson Edexcel Level 3 GCE

Page 60: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

3

*S47552A0324* Turn over

(d) Incomplete combustion of butane produces a mixture of products.

Which substance is not produced during the incomplete combustion of butane? (1)

A C

B CO

C H2

D H2O

(e) A student is given a bottle, containing an alkane, with an incomplete label.

The student realises that the alkane could be hexane or heptane and wants to identify it.

The student injects a sample of the liquid, with a mass of 0.235 g, into a sealed gas syringe. The syringe is placed in an oven at 425 K and the liquid vaporises. The volume of the gas produced is 83 cm3 at a pressure of 100 kPa (100 000 N m−2).

Identify the alkane. Justify your answer by the use of a calculation. (4)

(Total for Question 1 = 8 marks)

he ane

2

*S47552A0224*

Answer ALL questions.

Write your answers in the spaces provided.

Some questions must be answered with a cross in a box . If you change your mind about an answer, put a line through the box

and then mark your new answer with a cross .

1 Methylpropane is an alkane.

(a) Draw the displayed formula of methylpropane.(1)

(b) Give the empirical formula of methylpropane.(1)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(c) A partially balanced equation for the complete combustion of butane is:

C4H10 + xO2 → 4CO2 + 5H2O

The number of moles of oxygen, x, needed to balance this equation is(1)

A 4.5

B 6.5

C 9

D 13

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

5454

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553

*S47552A0324* Turn over

(d) Incomplete combustion of butane produces a mixture of products.

Which substance is not produced during the incomplete combustion of butane? (1)

A C

B CO

C H2

D H2O

(e) A student is given a bottle, containing an alkane, with an incomplete label.

The student realises that the alkane could be hexane or heptane and wants to identify it.

The student injects a sample of the liquid, with a mass of 0.235 g, into a sealed gas syringe. The syringe is placed in an oven at 425 K and the liquid vaporises. The volume of the gas produced is 83 cm3 at a pressure of 100 kPa (100 000 N m−2).

Identify the alkane. Justify your answer by the use of a calculation. (4)

(Total for Question 1 = 8 marks)

he ane

55

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5

*S47552A0524* Turn over

(c) Which physical property should be kept constant when measuring an enthalpy change?

(1)

A concentration

B pressure

C temperature

D volume

(d) A reaction occurs under standard conditions. Which of these represents a possible standard condition?

(1)

A 1 g cm–3

B 4.18 J g–1 K–1

C 24 dm3

D 298 K

(Total for Question 2 = 6 marks)

4

*S47552A0424*

2 The equations for three reactions are:

reaction 1 C(s) + O2(g) → CO2(g) ΔrHd = –394 kJ mol–1

reaction 2 CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔrHd = –890 kJ mol–1

reaction 3 2H2(g) + O2(g) → 2H2O(l) ΔrHd = –572 kJ mol–1

(a) Give two reasons why all three reactions are classified as examples of combustion.(2)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(b) Only reaction 1 represents a standard enthalpy change of formation.

Give reasons why reactions 2 and 3 do not.(2)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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*S47552A0524* Turn over

(c) Which physical property should be kept constant when measuring an enthalpy change?

(1)

A concentration

B pressure

C temperature

D volume

(d) A reaction occurs under standard conditions. Which of these represents a possible standard condition?

(1)

A 1 g cm–3

B 4.18 J g–1 K–1

C 24 dm3

D 298 K

(Total for Question 2 = 6 marks)

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(b) When this reaction is used in industry, the catalyst is an alloy of platinum and rhodium.

The diagram shows the reaction profile for the uncatalysed reaction.

enthalpy

4NH3 + 5O2

4NO + 6H2O

progress of reaction

(i) On the diagram, draw the reaction profile for the catalysed reaction.(1)

(ii) Label the diagram to show

· the enthalpy change, ΔrH

· the activation energy, Ea

for the catalysed reaction.(2)

(c) Write the expression for the equilibrium constant, Kc, for this reaction.(1)

(Total for Question 3 = 10 marks)

6

*S47552A0624*

3 Ammonia is used in the manufacture of nitric acid.

The equation for one step in this manufacturing process is:

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) ΔrH = –900 kJ mol–1

*(a) A manufacturer carries out this reaction at a temperature of 1200 K and a pressure of 10 atm. A scientist proposes that a temperature of 1000 K should be used at the same pressure.

Evaluate the effects of making this change on the rate and yield of this reaction.(6)

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(b) When this reaction is used in industry, the catalyst is an alloy of platinum and rhodium.

The diagram shows the reaction profile for the uncatalysed reaction.

enthalpy

4NH3 + 5O2

4NO + 6H2O

progress of reaction

(i) On the diagram, draw the reaction profile for the catalysed reaction.(1)

(ii) Label the diagram to show

· the enthalpy change, ΔrH

· the activation energy, Ea

for the catalysed reaction.(2)

(c) Write the expression for the equilibrium constant, Kc, for this reaction.(1)

(Total for Question 3 = 10 marks)

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(b) The table shows the bond enthalpies for the N≡N and H–H bonds.

Bond Bond enthalpy / kJ mol−1

N≡N + 944

H–H + 436

Use this data, together with the standard enthalpy change of reaction, ΔrHd , for

the decomposition of ammonia, to calculate a value, in kJ mol−1, for the mean bond enthalpy of the N–H bond in ammonia.

(3)

(Total for Question 4 = 7 marks)

8

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4 This question is about the thermal decomposition of ammonia.

This reaction is catalysed by platinum and is represented by the equation:

2NH3(g) → N2(g) + 3H2(g) ΔrHd = + 92 kJ mol−1

The diagram shows a sketch of the Maxwell-Boltzmann curve for the distribution of molecular energies for a fixed amount of ammonia gas at a given temperature.

Ea represents the activation energy of the uncatalysed reaction.

(a) (i) On the diagram, draw a vertical line to represent the activation energy of the catalysed reaction. Label this line Ea (with catalyst).

(1)

(ii) Use the diagram to explain why the use of a catalyst increases the rate of decomposition of ammonia.

(3)

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fraction of molecules with energy, E

energy, E

Ea (no catalyst)

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(b) The table shows the bond enthalpies for the N≡N and H–H bonds.

Bond Bond enthalpy / kJ mol−1

N≡N + 944

H–H + 436

Use this data, together with the standard enthalpy change of reaction, ΔrHd , for

the decomposition of ammonia, to calculate a value, in kJ mol−1, for the mean bond enthalpy of the N–H bond in ammonia.

(3)

(Total for Question 4 = 7 marks)

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(a) Calculate the standard enthalpy change, in kJ mol–1, for the reaction, giving your answer to an appropriate number of significant figures.

(5)

10

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5 Some reactions involving copper(II) sulfate are investigated. The diagram shows the apparatus used to measure the temperature change when

anhydrous copper(II) sulfate, CuSO4(s), is dissolved in water.

plastic cup

thermometer

water

The chemical change can be represented by this equation:

CuSO4(s) + aq → CuSO4(aq)

The method is:

· Pour some water into the plastic cup and record the steady temperature · Add some anhydrous copper(II) sulfate and stir until it has all dissolved · Record the maximum temperature reached in the reaction

These results are recorded.

Mass of water used 50.0 g

Initial temperature of water 18.2 °C

Final temperature of water 25.4 °C

Mass of anhydrous copper(II) sulfate used 4.70 g

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(a) Calculate the standard enthalpy change, in kJ mol–1, for the reaction, giving your answer to an appropriate number of significant figures.

(5)

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(d) The experimental value in part (c) of ∆rHd

2 = –41 kJ mol–1 for the reaction involving anhydrous copper(II) sulfate was different from a data book value.

The anhydrous copper(II) sulfate used in this experiment was pale blue, rather than white.

Explain how this observation partly accounts for the difference between the experimental value and the data book value.

(3)

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(Total for Question 5 = 13 marks)

12

*S47552A01224*

(b) The method used in this experiment leads to an inaccurate value for the temperature change.

Describe an alternative method to find a more accurate value for the temperature change.

(3)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(c) The experiment is repeated using the same method but with hydrated copper(II) sulfate, CuSO4.5H2O(s), as well as with anhydrous copper(II) sulfate.

The same method is used to calculate the standard enthalpy changes for these reactions:

CuSO4.5H2O(s) + aq → CuSO4(aq) ∆rHd

1 = +8.5 kJ mol–1

CuSO4(s) + aq → CuSO4(aq) ∆rHd

2 = –41 kJ mol–1

Calculate the standard enthalpy change for this reaction:

CuSO4(s) + aq → CuSO4.5H2O(s)(2)

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

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*S47552A01324* Turn over

(d) The experimental value in part (c) of ∆rHd

2 = –41 kJ mol–1 for the reaction involving anhydrous copper(II) sulfate was different from a data book value.

The anhydrous copper(II) sulfate used in this experiment was pale blue, rather than white.

Explain how this observation partly accounts for the difference between the experimental value and the data book value.

(3)

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(Total for Question 5 = 13 marks)

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*S47552A01524* Turn over

(iii) Explain the relative rates of hydrolysis of 1-chlorobutane and 1-iodobutane in this experiment.

(3)

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(b) The student saw, on a website, a comparison of the rates of hydrolysis of these bromoalkanes:

F CH3CH2CH2CH2Br G CH3CH2CHBrCH3 H (CH3)3CBr

(i) The correct order for the rates of hydrolysis of these bromoalkanes, showing the fastest first, is:

(1)

A F > G > H

B G > F > H

C H > F > G

D H > G > F

(ii) Identify, with a reason, which of F, G and H is a tertiary bromoalkane.(2)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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(iii) A student said that bromoalkane F had the name 4-bromobutane.

Give a reason why this name is incorrect.(1)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(Total for Question 6 = 10 marks)

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6 A student investigates the rates of hydrolysis of three different halogenoalkanes. The student uses this method.

· 5 cm3 of ethanol is put into each of three test tubes and five drops of a different halogenoalkane is added to each.

· The test tubes are placed into a water bath at 50°C.

· 5 cm3 of aqueous silver nitrate is put into each of three clean test tubes in the water bath.

· When the solutions have reached the temperature of the water bath, 5 cm3 of the silver nitrate solution is mixed with each test tube containing a halogenoalkane in ethanol. A stop clock is started when the solutions are mixed.

· The time taken for a precipitate to appear in each test tube is recorded in a table.

Halogenoalkane Time taken for precipitate to appear

1-chlorobutane 20 minutes and 50 seconds

1-bromobutane 9 minutes and 15 seconds

1-iodobutane 5 seconds

(a) (i) State the colour of the precipitate formed in the 1-iodobutane test tube.(1)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(ii) Calculate the relative rates of reaction for each of the three halogenoalkanes. (2)

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(iii) Explain the relative rates of hydrolysis of 1-chlorobutane and 1-iodobutane in this experiment.

(3)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(b) The student saw, on a website, a comparison of the rates of hydrolysis of these bromoalkanes:

F CH3CH2CH2CH2Br G CH3CH2CHBrCH3 H (CH3)3CBr

(i) The correct order for the rates of hydrolysis of these bromoalkanes, showing the fastest first, is:

(1)

A F > G > H

B G > F > H

C H > F > G

D H > G > F

(ii) Identify, with a reason, which of F, G and H is a tertiary bromoalkane.(2)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(iii) A student said that bromoalkane F had the name 4-bromobutane.

Give a reason why this name is incorrect.(1)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(Total for Question 6 = 10 marks)

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(c) But-1-ene reacts with hydrogen bromide to form two different products.

(i) Analysis of one of the products showed that it contains 34.9% carbon and 6.60% hydrogen by mass.

Calculate the empirical formula of this product.(3)

(ii) Give the name of the major product of this reaction.(1)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(d) The carbon-carbon double bond in but-2-ene can be represented by C C.

A disadvantage of this representation is that it shows the two covalent bonds to be the same.

However, there are two different types of covalent bond (σ and π) in but-2-ene.

Compare and contrast these two different types of bonds.(3)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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7 But-1-ene and but-2-ene are unsaturated hydrocarbons with the molecular formula C4H8.

Each one reacts with a few drops of bromine in an addition reaction.

(a) Give the colour change that occurs in these reactions.(1)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(b) (i) Name the type of mechanism for these reactions.(1)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(ii) Draw the mechanism for the reaction between but-2-ene and bromine.(4)

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(c) But-1-ene reacts with hydrogen bromide to form two different products.

(i) Analysis of one of the products showed that it contains 34.9% carbon and 6.60% hydrogen by mass.

Calculate the empirical formula of this product.(3)

(ii) Give the name of the major product of this reaction.(1)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(d) The carbon-carbon double bond in but-2-ene can be represented by C C.

A disadvantage of this representation is that it shows the two covalent bonds to be the same.

However, there are two different types of covalent bond (σ and π) in but-2-ene.

Compare and contrast these two different types of bonds.(3)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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7019

*S47552A01924* Turn over

8 This question is about some isomeric alcohols with molecular formula C5H12O.

(a) The table shows some information about three of these alcohols. Give the structural formula of each of these alcohols.

(3)

Alcohol Description Structural formula

P the straight-chain primary alcohol

Q the secondary alcohol with a branched carbon chain

Rthe alcohol not oxidised by potassium dichromate(VI) in

dilute sulfuric acid

(b) T is another different isomeric alcohol with the molecular formula C5H12O. T is a secondary alcohol.

The most prominent peak in the mass spectrum of T occurred at m/z = 45, with another peak at m/z = 43

Deduce the structure of T. Justify your answer by identifying the structural formula of each fragment ion.

(3)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

18

*S47552A01824*

(e) Draw the skeletal formula of trans-but-2-ene.(1)

(Total for Question 7 = 14 marks)

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7119

*S47552A01924* Turn over

8 This question is about some isomeric alcohols with molecular formula C5H12O.

(a) The table shows some information about three of these alcohols. Give the structural formula of each of these alcohols.

(3)

Alcohol Description Structural formula

P the straight-chain primary alcohol

Q the secondary alcohol with a branched carbon chain

Rthe alcohol not oxidised by potassium dichromate(VI) in

dilute sulfuric acid

(b) T is another different isomeric alcohol with the molecular formula C5H12O. T is a secondary alcohol.

The most prominent peak in the mass spectrum of T occurred at m/z = 45, with another peak at m/z = 43

Deduce the structure of T. Justify your answer by identifying the structural formula of each fragment ion.

(3)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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7221

*S47552A02124*

(c) (i) Deduce, using the table of absorptions from the data booklet, the type of compound responsible for each spectrum.

Include in your answer references to wavenumbers and their corresponding bonds.

(2)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(ii) One of the two compounds, Y and Z, of mass 2.74 g is completely burned in oxygen.

Carbon dioxide and water are the only two products.

The masses of carbon dioxide and water formed are 5.89 g and 2.44 g respectively.

Show by calculation that this data is consistent with a formula of C5H10O2 for this compound.

(4)

(Total for Question 8 = 12 marks)

TOTAL FOR PAPER = 80 MARKS

20

*S47552A02024*

Some of the alcohols were heated with potassium dichromate(VI) and sulfuric acid. The organic compounds were separated from the reaction mixtures and purified.

The infrared spectra of two of these organic compounds are shown.

100

50

04000 3000 2000 1500 1000 500

Wavenumber (cm–1)

Tran

smitt

ance

(%)

Compound Y

100

50

04000 3000 2000 1500 1000 500

Wavenumber (cm–1)

Tran

smitt

ance

(%)

Compound Z

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7321

*S47552A02124*

(c) (i) Deduce, using the table of absorptions from the data booklet, the type of compound responsible for each spectrum.

Include in your answer references to wavenumbers and their corresponding bonds.

(2)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(ii) One of the two compounds, Y and Z, of mass 2.74 g is completely burned in oxygen.

Carbon dioxide and water are the only two products.

The masses of carbon dioxide and water formed are 5.89 g and 2.44 g respectively.

Show by calculation that this data is consistent with a formula of C5H10O2 for this compound.

(4)

(Total for Question 8 = 12 marks)

TOTAL FOR PAPER = 80 MARKS

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22

*S47552A02224*

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*S47552A02424*

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77Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

CH

EMIS

TRY

AS

PA

PER

2 M

AR

KS

CH

EME

Q

ues

tio

n

Nu

mb

er

An

swer

A

dd

itio

nal

Gu

idan

ce

Mar

k

1(a

)

(

1)

Ever

y at

om a

nd e

very

bon

d m

ust

be

show

n 1

1(b

) C

2H5

(1)

1

1

(c)

B

1

1

(d)

C

1

1

(e)

• re

-arr

ange

men

t of

pV =

nR

T eq

uation

(

1)

conv

ersi

on o

f vo

lum

e an

d su

bstitu

tion

(

1)

eval

uation

of

num

ber

of m

oles

of

alka

ne

(1

)

• ca

lcul

atio

n of

Mr,

usin

g nu

mbe

r of

mol

es a

nd m

ass

of

alka

ne t

o id

entify

it

(1)

Cor

rect

ans

wer

, no

wor

king

sco

res

4 m

arks

Ex

ampl

e of

cal

cula

tion

n

= p

V

R

T =

100

000

x (

83 x

10-6

)

8.3

1 x

425

= 0

.002

35…

(m

oles

) M

r =

0.2

35 /

0.0

0235

… =

99.

99 o

r 10

0 i.e

. th

is is

hep

tane

4

(To

tal

for

Qu

esti

on

1 =

8 m

arks

)

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Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

78

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

2(a

) An

answ

er t

hat

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

• (a

ll eq

uation

s) s

how

oxy

gen/

O2

as a

rea

ctan

t/on

the

left

(1)

(all)

ΔrH

o va

lues

are

neg

ativ

e /(

all)

rea

ctio

ns a

re

exot

herm

ic

(1

)

2

2(b

) An

answ

er t

hat

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

• re

action

2 is

not

, be

caus

e th

ere

is m

ore

than

1 p

rodu

ct

(1

)

• re

action

3 is

not

, be

caus

e 2

mol

es o

f pr

oduc

t ar

e fo

rmed

(1)

Acc

ept

beca

use

met

hane

/CH

4 is

not

an

elem

ent

2

2(c

) B

1

2

(d)

D

1

(T

ota

l fo

r Q

ues

tio

n 2

= 6

mar

ks)

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79

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

*3

(a)

This

que

stio

n as

sess

es a

stu

dent

’s a

bilit

y to

sho

w a

coh

eren

t an

d lo

gica

lly s

truc

ture

d an

swer

with

linka

ges

and

fully

-su

stai

ned

reas

onin

g.

Mar

ks a

re a

war

ded

for

indi

cative

con

tent

and

for

how

the

an

swer

is s

truc

ture

d an

d sh

ows

lines

of

reas

onin

g.

The

follo

win

g ta

ble

show

s ho

w t

he m

arks

sho

uld

be a

war

ded

for

indi

cative

con

tent

. N

umbe

r of

in

dica

tive

m

arki

ng

poin

ts s

een

in a

nsw

er

Num

ber

of

mar

ks

awar

ded

for

indi

cative

m

arki

ng

poin

ts

6 4

5–4

3 3–

2 2

1 1

0 0

Gui

danc

e on

how

the

mar

k sc

hem

e sh

ould

be

app

lied:

Th

e m

ark

for

indi

cative

con

tent

sho

uld

be

adde

d to

the

mar

k fo

r lin

es o

f re

ason

ing.

Fo

r ex

ampl

e, a

n an

swer

with

five

indi

cative

mar

king

poi

nts

that

is p

artial

ly

stru

ctur

ed w

ith

som

e lin

kage

s an

d lin

es o

f re

ason

ing,

sco

res

4 m

arks

(3

mar

ks f

or

indi

cative

con

tent

and

1 m

ark

for

part

ial

stru

ctur

e an

d so

me

linka

ges

and

lines

of

reas

onin

g).

If

the

re a

re n

o lin

kage

s be

twee

n po

ints

, th

e sa

me

five

indi

cative

mar

king

poi

nts

wou

ld y

ield

an

over

all s

core

of

3 m

arks

(3

mar

ks f

or in

dica

tive

con

tent

and

no

mar

ks

for

linka

ges)

.

6

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80

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

*3

(a)

con

t.

The

follo

win

g ta

ble

show

s ho

w t

he m

arks

sho

uld

be a

war

ded

for

stru

ctur

e an

d lin

es o

f re

ason

ing.

N

umbe

r of

mar

ks

awar

ded

for

stru

ctur

e of

ans

wer

and

su

stai

ned

line

of

reas

onin

g

Ans

wer

sho

ws

a co

here

nt a

nd

logi

cal s

truc

ture

with

linka

ges

and

fully

sus

tain

ed li

nes

of

reas

onin

g de

mon

stra

ted

thro

ugho

ut

2

Ans

wer

is p

artial

ly s

truc

ture

d w

ith

som

e lin

kage

s an

d lin

es o

f re

ason

ing

1

Ans

wer

has

no

linka

ges

betw

een

poin

ts a

nd is

uns

truc

ture

d 0

Ind

icat

ive

con

ten

t:

tem

pera

ture

dec

reas

e lo

wer

s th

e ra

te o

f th

e re

action

beca

use

ther

e ar

e fe

wer

mol

ecul

es/p

articl

es w

ith

E ≥

Ea

• an

d th

eref

ore

ther

e ar

e fe

wer

suc

cess

ful c

ollis

ions

per

se

cond

tem

pera

ture

dec

reas

e in

crea

ses

the

yiel

d (o

f th

e pr

oduc

t)

• be

caus

e th

e (f

orw

ard)

rea

ctio

n is

exo

ther

mic

low

er r

ate

and

incr

ease

d yi

eld

are

oppo

sing

fac

tors

and

it

is n

ot p

ossi

ble

to t

ell w

hich

has

gre

ater

eff

ect

on o

vera

ll yi

eld

in a

giv

en t

ime.

Acc

ept

slow

s do

wn

the

reac

tion

Acc

ept

shi

fts

the

posi

tion

of

equi

libri

um

to t

he r

ight

/in

forw

ard

dire

ctio

n

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81

3(b

)(ii

)

• Δ

rH s

how

n as

the

app

roxi

mat

ely

vert

ical

dis

tanc

e be

twee

n re

acta

nts

and

prod

ucts

(1)

E a s

how

n as

the

app

roxi

mat

ely

vert

ical

dis

tanc

e be

twee

n re

acta

nts

and

peak

of

draw

n cu

rve

for

cata

lyse

d re

action

(1)

ecf

from

can

dida

te’s

cur

ve

2

3(c

) K

c =

[!"!]![!

!!(!)]!

[!!!!]![!

!(!)]!

(1)

Sta

te s

ymbo

ls n

ot e

ssen

tial

1

(To

tal

for

Qu

esti

on

3 =

10

mar

ks)

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

3(b

)(i)

• cu

rve

star

ting

at

reac

tant

s le

vel,

endi

ng a

t pr

oduc

ts le

vel,

and

peak

ing

low

er t

han

orig

inal

cur

ve

(1

)

1

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Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

4(a

)(i)

(

1)

Line

nee

ds t

o be

to

the

left

of

E a (

no

cata

lyst

) bu

t to

the

rig

ht o

f th

e hu

mp

in t

he

curv

e Sha

ding

not

req

uire

d bu

t ca

n be

incl

uded

an

d us

ed in

(a)

(ii)

1

4(a

)(ii

) An

expl

anat

ion

that

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

• th

e ar

ea u

nder

the

cur

ve t

o th

e ri

ght

of t

he a

ctiv

atio

n en

ergy

rep

rese

nts

the

frac

tion

of

mol

ecul

es t

hat

have

E ≥

Ea

(1

)

• th

e ar

ea un

der

the

curv

e to

th

e ri

ght

of

E a (w

ith

cata

lyst

) is

gre

ater

tha

n th

e ar

ea u

nder

the

cur

ve t

o th

e ri

ght

of E

a (n

o ca

taly

st)

(1)

• (t

here

fore

) th

ere

are

mor

e su

cces

sful

col

lisio

ns (

per

seco

nd p

er u

nit

vol.)

(1

)

Acc

ept

desc

ript

ion

that

invo

lves

sha

ded

area

s un

der

the

curv

e Acc

ept

mor

e m

olec

ules

hav

e en

ough

ene

rgy

to r

eact

3

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83

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

4(b

) •

calc

ulat

ion

of t

otal

bon

ds f

orm

ed

(1)

• re

-arr

ange

men

t of

ΔrH

exp

ress

ion

(1)

eval

uation

of

final

ans

wer

(1)

Cor

rect

ans

wer

, no

wor

king

sco

res

3 m

arks

Ex

ampl

e of

cal

cula

tion

Σ(b

onds

for

med

) =

E(N≡

N)

+ 3

E(H

–H)

= 9

44 +

(3

x 43

6)

= 2

252

(kJ

mol−

1 )

Δ

rH o

= Σbo

nds

brok

en –

Σ(b

onds

for

med

),

so Σ

(bon

ds f

orm

ed)

= Σ

bond

s br

oken

– Δ

rH o

6E(N

–H)

= 2

252

– (-

92)

E(N

–H)

= (

+)

391

(kJ

mol−

1 )

Acc

ept

390.

7

3

(To

tal

for

Qu

esti

on

4 =

7 m

arks

)

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Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

5(a

) •

calc

ulat

ion

of

tem

pera

ture

cha

nge

and

subs

titu

tion

into

Q

= m

cΔT

(1)

conv

ersi

on t

o kJ

(1)

calc

ulat

ion

of a

mou

nt o

f co

pper

(II)

sul

fate

(1

)

• us

e of

ΔH

=

–  ! !

(1)

corr

ect

answ

er w

ith

sign

and

to

3. s

.f

(

1)

Allo

w e

cf f

or s

teps

in c

alcu

lation

; in

clud

ing

for

final

ans

wer

dep

ende

nt o

n ro

undi

ng in

st

eps

of t

he c

alcu

lation

. C

orre

ct a

nsw

er w

ith

sign

, to

3.s

.f a

nd n

o w

orki

ng s

core

s 5

mar

ks

Mar

k co

nseq

uent

ially

on

inco

rrec

t te

mpe

ratu

re c

hang

e Ex

ampl

e of

cal

cula

tion

Q =

50

.0 x

4.1

8 x

7.2

(J)

=

1504

.8 (

J)

Q =

 !"#$.!

!"""

= 1

.504

8 (k

J)

mol

es o

f co

pper

(II)

sul

fate

= 4

.70

÷ 1

59.6

= 0

.029

449

mol

Δ

H

=

–  ! !

=

1.50

48 ÷

0.0

2944

9

= (

– 51

.099

) =

51.1

(kJ

mol

–1)

5

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

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85

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

5(b

) A d

escr

iption

tha

t m

akes

ref

eren

ce t

o th

e fo

llow

ing

poin

ts:

reco

rd t

empe

ratu

re a

t tim

ed in

terv

als

(aft

er a

ddin

g th

e so

lid)

(1)

plot

the

se t

empe

ratu

re v

alue

s (o

n a

grap

h)

(1

)

• ex

trap

olat

e th

e lin

e of

bes

t fit

to

the

tim

e of

add

ing

the

solid

(1

)

3

5(c

) •

use

of H

ess

cycl

e or

exp

ress

ion

e.g.

Δ

rH =

ΔrH

2 – Δ

rH1

or

ΔrH

= Δ

Hso

lutio

n(an

hydr

ous)

– Δ

Hso

lutio

n(hy

drat

ed)

(1)

• co

rrec

t fin

al a

nsw

er w

ith

sign

(–

49.5

kJ

mol

–1)

(1

)

Cor

rect

ans

wer

with

sign

and

no

wor

king

sc

ores

2 m

arks

2

5(d

) An

expl

anat

ion

that

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

• (t

he p

ale

blue

col

our

show

s th

at)

anhy

drou

s co

pper

(II

) su

lfate

is p

artial

ly h

ydra

ted

(1

)

• th

eref

ore

less

anh

ydro

us c

oppe

r (I

I) s

ulfa

te r

eact

s w

ith

wat

er /

som

e of

the

(bl

ue)

solid

rea

cts

endo

ther

mic

ally

(1)

ther

efor

e le

ss t

herm

al e

nerg

y tr

ansf

erre

d /

heat

ene

rgy

rele

ased

dur

ing

expe

rim

ent

o

r (e

xper

imen

tal v

alue

) ha

s a

smal

ler

nega

tive

val

ue

(1)

3

(To

tal

for

Qu

esti

on

5 =

13

mar

ks)

Page 92: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Q

ues

tio

n

Nu

mb

er

An

swer

A

dd

itio

nal

Gu

idan

ce

Mar

k

6(a

)(i)

(pal

e) y

ello

w

(1)

1

6(a

)(ii

)

• co

nver

sion

to

seco

nds

and

use

of

!!"#$  !"  !"#

(1)

divi

de b

y lo

wes

t to

get

who

le n

umbe

r ra

tio

(

1)

Exam

ple

of c

alcu

lation

!

!"#$

:

! !!!

: ! !

or 8

.00

x 10

−4

:

1.80

x 1

0−3

:

0.20

1 :

2.25

: 2

50

2

6(a

)(ii

i)

An

expl

anat

ion

that

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

• th

e C

–Cl b

ond

is s

hort

er t

han

the

C–I

bon

d /

the

Cl a

tom

is s

mal

ler

than

the

I a

tom

(1

)

• th

eref

ore,

the

C–C

l bon

d is

str

onge

r th

an t

he C

–I

bond

(1

)

• th

eref

ore,

the

C–C

l bon

d ne

eds

mor

e en

ergy

to

brea

k, a

nd s

o th

e re

action

is s

low

er

(1

)

Allo

w r

ever

se a

rgum

ent

thro

ugho

ut

3

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

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87

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

6(b

)(i)

D

1

6(b

)(ii

) •

H/

(CH

3)3

CBr

(1)

• be

caus

e C

of

C–B

r is

joi

ned

to 3

alk

yl

grou

ps/j

oine

d to

3 c

arbo

n at

oms

(1

)

2

6(b

)(ii

i)

• nu

mbe

r pr

efix

use

d sh

ould

be

as lo

w a

s po

ssib

le /

br

omin

e at

om is

on

the

first

car

bon

atom

of

the

chai

n

(1

)

1

(To

tal

for

Qu

esti

on

6 =

10

mar

ks)

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Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

7(a

) (r

ed-)

brow

n to

col

ourl

ess

(1)

Allo

w o

rang

e fo

r re

d-br

own

Do

not

allo

w c

lear

for

col

ourl

ess

1

7(b

)(i)

el

ectr

ophi

lic (

addi

tion

)

(

1)

1

7(b

)(ii

)

part

ial c

harg

es o

n Br

atom

s

(

1)

curl

y ar

row

fro

m C

=C

bon

d to

(δ+

) Br

atom

AN

D

curl

y ar

row

fro

m B

r–Br

bond

to

(δ–)

Br

atom

(1)

• st

ruct

ure

of c

arbo

cation

with

+ c

harg

e

(

1)

curl

y ar

row

fro

m B

r− t

o C

with

+ c

harg

e AN

D

stru

ctur

e of

1,

2-di

brom

obut

ane

prod

uct

(1

)

Acc

ept

cycl

ic b

rom

oniu

m io

n w

ith

+ c

harg

e sh

own

in c

entr

e of

rin

g

4

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

88

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89

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

7(c

)(i)

calc

ulat

ion

of %

by

mas

s of

Br

(

1)

eval

uation

of

num

ber

of m

oles

of

C,

H,

and

Br

(1)

• fin

al e

mpi

rica

l for

mul

a ba

sed

on w

hole

num

ber

ratio

(1)

Exam

ple

of c

alcu

lation

%

Br

= 1

00 –

34.

9 –

6.60

= 5

8.5%

C

!".!

!" =

2.9

1

H !.!" !

= 6

.60

Br

!".!

!".!

= 0

.732

2.

91

=

3.9

8

6.60

=

9.

02

0.7

32

=

1

0.

732

0.

732

0.

732

i.e.

C4H

9Br

3

7(c

)(ii

) •

2–br

omob

utan

e

(1

)

1

Page 96: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

7(d

) An

answ

er t

hat

incl

udes

at

leas

t on

e si

mila

rity

and

one

di

ffer

ence

to

gain

max

imum

mar

ks:

Sim

ilari

ties

: •

both

invo

lve

the

over

lap

of t

wo

(ato

mic

) or

bita

ls

(eac

h co

ntai

ning

a s

ingl

e el

ectr

on)

(1)

• bo

th in

volv

e an

ele

ctro

stat

ic a

ttra

ctio

n be

twee

n a

bond

ing

pair

of

elec

tron

s an

d tw

o nu

clei

(1

) D

iffer

ence

s:

• th

e σ

bond

rep

rese

nts

elec

tron

den

sity

/ele

ctro

ns

betw

een

the

two

carb

on a

tom

s/nu

clei

/end

-on

over

lapp

ing

of a

tom

ic o

rbital

s

(1

) •

the π

bond

rep

rese

nts

elec

tron

den

sity

/ele

ctro

ns

abov

e an

d be

low

the

σ b

ond/

sid

eway

s ov

erla

ppin

g of

orb

ital

s

(

1)

the 𝜎𝜎

bond

is f

orm

ed b

y th

e ov

erla

p of

a s

ingl

e lo

be f

rom

one

(at

omic

) or

bita

l with

a si

ngle

lobe

fr

om t

he a

noth

er

(ato

mic

) or

bita

l

(1

)

• th

e 𝜋𝜋

bond

is f

orm

ed b

y th

e ov

erla

p of

tw

o lo

bes

from

one

(at

omic

) or

bita

l with

two

lobe

s of

an

othe

r (a

tom

ic)

orbi

tal

(1)

Acc

ept

elec

tros

tatic

attr

action

bet

wee

n a

shar

ed

pair

of

elec

tron

s an

d tw

o nu

clei

Acc

ept

end-

to-e

nd/h

ead-

on/a

xial

ove

rlap

ping

3

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90

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91

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

7(e

)

(1

)

Do

not

acce

pt a

str

uctu

ral o

r di

spla

yed

form

ula.

1

(T

ota

l fo

r Q

ues

tio

n 7

= 1

4 m

arks

)

Page 98: AS Chemistry - Past Papers · Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry (8CH0) ... The Pearson Edexcel Level 3 Advanced Subsidiary GCE in Chemistry is designed

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

8(a

) •

P

CH

3CH

2CH

2CH

2CH

2OH

(1

)

• Q

(C

H3)

2CH

CH

(OH

)CH

3

(1)

R

(CH

3)2C

(OH

)CH

2CH

3

(1)

Acc

ept

disp

laye

d fo

rmul

ae

3

8(b

) •

m/z

= 4

5 is

due

to

CH

3CH

(OH

)+

(1)

• m

/z =

43

is d

ue t

o C

H3C

H2C

H2+

(1

)

• th

eref

ore

the

stru

ctur

e is

CH

3CH

2CH

2CH

(OH

)CH

3

(1

)

Pena

lise

the

lack

of

a pl

us s

ign

once

onl

y (C

H3)

2CH

CH

(OH

)CH

3 n

ot a

llow

ed a

s it is

Q

3

8(c

)(i)

An

answ

er t

hat

mak

es r

efer

ence

to

the

follo

win

g po

ints

:

• Y

is a

ket

one

beca

use

the

spec

trum

con

tain

s an

ab

sorp

tion

for

C=

O (

carb

onyl

gro

up)

1700

– 1

720

cm-1

(1

)

• Z

is a

car

boxy

lic a

cid

beca

use

the

spec

trum

con

tain

s ab

sorp

tion

s fo

r C

=O

172

5-17

00 c

m-1

AN

D O

–H (

acid

s) 3

300

– 25

00 c

m-1

(1)

Abs

ence

of

bond

s or

abs

ence

of

wav

enum

ber

rang

es m

axim

um 1

Allo

w v

alue

s w

ithi

n th

e ra

nges

2

Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015

92

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93

Qu

esti

on

N

um

ber

A

nsw

er

Ad

dit

ion

al G

uid

ance

M

ark

8(c

)(ii

) •

calc

ulat

ion

of m

ass

of c

arbo

n fr

om m

ass

of C

O2

(1)

• ca

lcul

atio

n of

mas

s of

hyd

roge

n fr

om m

ass

of H

2O

(1)

subt

ract

ion

to f

ind

mas

s of

O,

and

eval

uation

of

num

ber

of m

oles

of

C,

H a

nd O

(1

)

• co

nfir

m w

hole

num

ber

ratio

(1

)

Exam

ple

of c

alcu

lation

Mas

s of

car

bon

= (!" !!

x 5

.89)

= 1

.606

g

Mas

s of

hyd

roge

n =

(! !"

x 2

.44)

= 0

.271

g

Mas

s of

oxy

gen

= 2

.74

– 1.

606

– 0.

271

= 0

.863

g

C !.!"!

!" =

0.1

34

H !.!"#

! =

0.2

71

O !

.!"#

!" =

0.0

54

Rat

io =

2.4

8 :

5.02

: 1

= 5

: 1

0 :

2

4

(To

tal

for

Qu

esti

on

8 =

12

mar

ks)

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