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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
PearsonEdexcel Level 3Advanced Subsidiary GCE in Chemistry (8CH0) Sample Assessment Materials
First certification 2016
Edexcel, BTEC and LCCI qualifications
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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.)
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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
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
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.
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.
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.
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.
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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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 carbonateTime 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)
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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)
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
(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
Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
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*s47551A02628*
BLANK PAGE
Pearson Edexcel Level 3 Advanced Subsidiary GCE in ChemistrySample Assessment Materials – Issue 1 – February 2015 © Pearson Education Limited 2015
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*s47551A02728*
BLANK PAGE
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
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 %
30 S
i and
sub
stitut
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
)
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
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
)
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
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
)
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
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
)
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
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
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
+ g
reat
er
than
Li+
/ h
ighe
r ch
arge
of
Mg2
+ 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
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