Surname Centre Number Candidate Number Other Names 2 GCE AS/A Level 2410U10-1 NEW AS S16-2410U10-1 CHEMISTRY Unit 1 The Language of Chemistry, Structure of Matter and Simple Reactions A.M. FRIDAY, 27 May 2016 1 hour 30 minutes For Examiner s use only Question Maximum Mark Mark Awarded Section A 1. to 7. 10 ADDITIONAL MATERIALS In addition to this examination paper, you will need a: calculator; Data Booklet supplied by WJEC. Section B 8. 16 9. 11 10. 17 11. 13 12. 13 2410U101 01 INSTRUCTIONS TO CANDIDATES Total 80 Use black ink or black ball-point pen. Do not use gel pen or correction fluid. Write your name, centre number and candidate number in the spaces at the top of this page. Section A Answer all questions in the spaces provided. Section B Answer all questions in the spaces provided. Candidates are advised to allocate their time appropriately between Section A (10 marks) and Section B (70 marks). INFORMATION FOR CANDIDATES The number of marks is given in brackets at the end of each question or part-question. The maximum mark for this paper is 80. Your answers must be relevant and must make full use of the information given to be awarded full marks for a question. The assessment of the quality of extended response (QER) will take place in Q.11(c). If you run out of space, use the additional page(s) at the back of the booklet, taking care to number the question(s) correctly. MAY162410U10101 SM*(S16-2410U10-1)
2 SECTION A Examiner only Answer all questions in the spaces provided. 1. Using the convention of arrows to represent electrons, complete the electronic structure of an arsenic atom, As. [1] [Ar] argon core 3d 4s 4p 2. Give the oxidation number of molybdenum in the ion MoO 4 2. [1]... 3. The half-life of the radioactive isotope oxygen-15 is 2 minutes. (a) Give the atomic number, mass number and symbol of the nuclide produced when 1 atom of 15 O decays by the emission of one positron. [1]... (b) If a quantity of 15 O decays by positron emission, determine what fraction of 15 O is left after 10 minutes. [1]... 4. Explain why the second molar ionisation energy of sodium is greater than its first molar ionisation energy. [1] 02 (2410U10-1)
3 5. The explosive ADN has the structural formula shown below. Examiner only O 2 N O 2 N N _ NH 4 + Give the empirical formula of ADN.... [1] 6. Sodium carbonate, Na 2 CO 3, is made by heating sodium hydrogencarbonate, NaHCO 3. 2NaHCO 3 M r 84 Na 2 CO 3 M r 106 + CO 2 + H 2 O Calculate the atom economy of this reaction. [1] Atom economy =... % 2410U101 03 7. Hydrogen and iodine react together to give hydrogen iodide. H 2 (g) + I 2 (g) a 2HI(g) (a) Write the expression for the equilibrium constant in terms of concentration, K c, for this reaction. [1] (b) The equilibrium constant, K c, for this reaction has a value of 46.0 at a certain temperature. Calculate the equilibrium concentration of hydrogen at this temperature. The equilibrium concentration of iodine is 1.20 mol dm 3 and the equilibrium concentration of hydrogen iodide is 15.00 mol dm 3. [2] Concentration of hydrogen =... mol dm 3 10 03 (2410U10-1) Turn over.
4 SECTION B Examiner only Answer all questions in the spaces provided. 8. (a) Iron(III) oxide, Fe 2 O 3, and carbon monoxide react to give iron and carbon dioxide. Give the equation for this reaction. Explain why it is described as a redox process. [2] (b) The iron obtained from the blast furnace contains some sulfur as an impurity. This is removed during Basic Oxygen Steelmaking (BOS), by adding magnesium, which removes the sulfur as magnesium sulfide, MgS. 350 tonnes of impure iron used for BOS contain 0.02 % of sulfur. Calculate the mass, in kilograms, of magnesium needed to remove all the sulfur. [2] Mass =... kg (c) Magnesium sulfide has the same crystal structure as sodium chloride. Use the diagram below to show the crystal structure of magnesium sulfide, clearly labelling the formula of each species present. [2] 04 (2410U10-1)
(d) 5 Magnesium sulfide reacts with water producing gaseous hydrogen sulfide and magnesium hydroxide. MgS + 2H 2 O Mg(OH) 2 + H 2 S Examiner only (i) State what would be seen during the reaction apart from gas bubbles. [1] (ii) A student added a small sample of magnesium sulfide of mass 0.224 g to some water in a fume cupboard. Calculate the maximum volume, in cm 3, of hydrogen sulfide produced at 25 C. [3] Volume =... cm 3 2410U101 05 (e) When barium sulfide is added to water a similar reaction occurs to that described in part (d) with gaseous hydrogen sulfide and barium hydroxide being the products. State with a reason, how the observation for this reaction would be different from that seen with magnesium sulfide. [2] (f) Barium hydroxide can also be produced by adding barium oxide to water. Give the equation for this reaction and estimate the ph of the product. [2] 05 (2410U10-1) Turn over.
6 (g) Explain why it is unlikely that stable compounds containing Ba 3+ ions can exist. [2] Examiner only 16 06 (2410U10-1)
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8 9. (a) Ethanedioic acid (oxalic acid) has the formula (COOH) 2. Examiner only It can be made in the laboratory by oxidising sugar with a powerful oxidising agent (shown as [O] in the equation below). (i) Balance the equation. [1] C 12 H 22 O 11 +... [O]... (COOH) 2 +... H 2 O (ii) After purification, the oxalic acid is produced as white crystals of a hydrate, (COOH) 2. xh 2 O. On heating this hydrate, water is lost, leaving anhydrous oxalic acid. (COOH) 2. xh 2 O(s) (COOH) 2 (s) + xh 2 O(g) In an experiment 5.67 g of the hydrate were heated to constant mass, giving 4.05 g of the anhydrous acid. Calculate the value of x in (COOH) 2. xh 2 O. You must show your working. [3] x =... (b) Another method of producing oxalic acid in the laboratory is to heat potassium methanoate, HCOOK. This produces potassium oxalate, which is then acidified to give the acid. 2HCOOK (COOK) 2 + H 2 Some students carried out this experiment using the apparatus below. They collected the hydrogen gas produced over water. heat 08 (2410U10-1)
9 A few students obtained a smaller volume of hydrogen than expected. Examiner only Suggest two practical reasons why this may have occurred, apart from leaks. [2] 1... 2... (c) Gareth was asked to analyse a mixture of potassium methanoate and potassium oxalate. He weighed out 4.69 g of the mixture and carried out the method below. The mixture was completely dissolved in distilled water. Calcium chloride solution was added to this solution (only calcium oxalate was precipitated). The mixture was filtered and the precipitated calcium oxalate washed and dried. (i) Suggest how Gareth would know when enough calcium chloride solution had been added to react with all of the potassium oxalate present. [2] 2410U101 09 (ii) In his experiment Gareth obtained 2.49 g of pure dry calcium oxalate (M r 128). calcium chloride (aq) (COOK) 2 (COO) 2 Ca + 2KCl M r 166 M r 128 Calculate the percentage of potassium oxalate in the mixture, giving your answer to the appropriate number of significant figures. [3] Percentage of potassium oxalate =... % 11 09 (2410U10-1) Turn over.
10 10. (a) The mass spectrum of a silicon sample showed the presence of the three isotopes 28 Si, 29 Si and 30 Si. The percentage of 28 Si present was 92.2 % and the mass spectrum showed that the percentage of 29 Si present was twice that of 30 Si. Calculate the relative atomic mass of this sample of silicon. [3] You must show your working. Examiner only Relative atomic mass =... (b) Rhiannon studied the properties of silicon and found that its structure is similar to that of diamond. In her report she stated that the Si Si Si bond angle is 109.5 silicon is a poor conductor of electricity each silicon atom is bonded to four other silicon atoms. (i) State the name of the shape that has this bond angle. [1] (ii) Explain why solid silicon is a very poor electrical conductor. [1] (iii) Explain why the bonding between each atom is covalent. [1] 10 (2410U10-1)
(c) 11 Hydrofluoric acid, HF, is supplied as a solution that contains 50 % by mass of hydrogen fluoride. The density of this solution is 1.17 g cm 3. Calculate the concentration of this solution in mol dm 3. [2] Examiner only Concentration =... mol dm 3 (d) (i) Hydrofluoric acid will dissolve silica, SiO 2, to produce hexafluorosilicic acid. This acid contains the SiF 6 2 ion. Use the information from the table below to draw the shape of the SiF 6 2 ion, showing the F Si F bond angle. Give a reason for your answer. [3] Ion Number of bonding electron pairs Number of lone electron pairs on the central silicon atom SiF 6 2 6 0 11 (2410U10-1) Turn over.
12 (ii) Hexafluorosilicic acid, H 2 SiF 6, (M r 144) can be added to drinking water to promote good dental health. When added to water all the fluorine in the acid is available as fluoride ions. Examiner only The water in an area of Derbyshire contains 0.15 mg dm 3 of fluoride ions. Calculate how much hexafluorosilicic acid, in mg, should be added to each dm 3 of water to increase the fluoride level to 0.76 mg dm 3. [3] Mass of hexafluorosilicic acid =... mg (e) Solid magnesium hexafluorosilicate hydrolyses rapidly when added to water. One equation for this reaction is as follows. MgSiF 6 (s) + 2H 2 O(l) Mg 2+ (aq) + 6F (aq) + SiO 2 (s) + 4H + (aq) In an experiment 2.60 g of magnesium hexafluorosilicate were added to 1.00 dm 3 of water. Calculate the ph of the resulting mixture. [3] ph =... 17 12 (2410U10-1)
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14 11. (a) 2-Aminoethanol is a colourless liquid. Examiner only H H H N C C O H H H H 2-aminoethanol It is very soluble in water owing to its ability to hydrogen bond with water molecules. Complete the diagram above to show how 2-aminoethanol is able to hydrogen bond to water molecules. [4] (b) The boiling temperatures of 1,2-diaminoethane and 2-aminoethanol are shown in the table. Compound Formula Boiling temperature / C 1,2-diaminoethane H 2 N CH 2 CH 2 NH 2 117 2-aminoethanol H 2 N CH 2 CH 2 OH 170 Use these figures to comment on the strength of the intermolecular forces between the molecules in each compound suggesting reasons for your answer. [3] 14 (2410U10-1)
(c) 15 Ammonium sulfate can be prepared in the laboratory by neutralising aqueous sulfuric acid with ammonia solution in an acid-base titration. 2NH 3 (aq) + H 2 SO 4 (aq) (NH 4 ) 2 SO 4 (aq) Outline a method to obtain pure dry crystals of ammonium sulfate by this procedure. [6 QER] Examiner only 13 15 (2410U10-1) Turn over.
16 12. (a) Some students were given a mixture made up of about 50 % each of sodium chloride and sodium iodide and asked to design tests to show the presence and concentration of iodide ions in this mixture. Examiner only (i) One student decided to try electrolysis. He dissolved some of the mixture in distilled water to give a concentrated solution. He then passed electricity through the solution, using inert electrodes. The negative ions moved to the anode where they lost electrons. State what was seen at the anode to confirm the presence of iodide ions. Explain your answer, including a half-equation. [3] (ii) Another student decided to use a chemical test. She dissolved some of the mixture in distilled water and then added a little aqueous silver nitrate. After noting what was seen, she added some aqueous ammonia to the mixture and shook it. State and explain what occurred during this test. [3] 16 (2410U10-1)
17 (iii) In a further test the concentration of the iodide ions present was found by a titration with potassium iodate(v) dissolved in a strong acid. 11.24 g of the mixture of sodium chloride and sodium iodide were dissolved in distilled water and made up to 250 cm 3. The iodide ions present in 25.0 cm 3 of this solution required 18.00 cm 3 of the potassium iodate(v) solution of concentration 0.100 mol dm 3, to completely react. Examiner only Calculate the number of moles of iodide ions (present as sodium iodide, M r 150) in 25.0 cm 3 of the solution and hence the exact percentage of sodium iodide in the mixture. [3] (1 mol of potassium iodate(v) reacts with 2 mol of sodium iodide) Percentage of sodium iodide =... % (b) Hydrogen chloride is a colourless gas that absorbs at 278 nm in the ultraviolet region of the electromagnetic spectrum. Use this information and the data sheet to show that the energy of the H Cl bond is 431 kj mol 1. [2] 17 (2410U10-1) Turn over.
(c) 18 The table shows bond energy values and absorption maxima for the hydrogen halides. Examiner only H X Bond energy / kj mol 1 Absorption maximum / nm H F 562 212 H Cl 431 278 H Br 366 326 H I 299 400 H At The accepted range for wavelengths for the visible region of the electromagnetic spectrum is 400-700 nm. Suggest and explain why it is probable that H At would be a coloured gas. [2] 13 END OF PAPER 18 (2410U10-1)
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20 Question number Additional page, if required. Write the question number(s) in the left-hand margin. Examiner only 20 (2410U10-1)
GCE AS MARKING SCHEME SUMMER 2016 CHEMISTRY - NEW AS UNIT 1 2410U10-1
INTRODUCTION This marking scheme was used by WJEC for the 2016 examination. It was finalised after detailed discussion at examiners' conferences by all the examiners involved in the assessment. The conference was held shortly after the paper was taken so that reference could be made to the full range of candidates' responses, with photocopied scripts forming the basis of discussion. The aim of the conference was to ensure that the marking scheme was interpreted and applied in the same way by all examiners. It is hoped that this information will be of assistance to centres but it is recognised at the same time that, without the benefit of participation in the examiners' conference, teachers may have different views on certain matters of detail or interpretation. WJEC regrets that it cannot enter into any discussion or correspondence about this marking scheme. 1
GCE CHEMISTRY SUMMER 2016 MARK SCHEME AS UNIT 1 THE LANGUAGE OF CHEMISTRY, STRUCTURE OF MATTER AND SIMPLE REACTIONS MARK SCHEME GENERAL INSTRUCTIONS Recording of marks Examiners must mark in red ink. One tick must equate to one mark, apart from extended response questions where a level of response mark scheme is applied. Question totals should be written in the box at the end of the question. Question totals should be entered onto the grid on the front cover and these should be added to give the script total for each candidate. Extended response questions A level of response mark scheme is applied. The complete response should be read in order to establish the most appropriate band. Award the higher mark if there is a good match with content and communication criteria. Award the lower mark if either content or communication barely meets the criteria. Marking rules All work should be seen to have been marked. Marking schemes will indicate when explicit working is deemed to be a necessary part of a correct answer. Crossed out responses not replaced should be marked. Marking abbreviations The following may be used in marking schemes or in the marking of scripts to indicate reasons for the marks awarded. cao = correct answer only ecf = error carried forward bod = benefit of doubt Credit should be awarded for correct and relevant alternative responses which are not recorded in the mark scheme. 2
1. Section A Question Marking details Marks available AO1 AO2 AO3 Total Maths Prac 1 1 2. (+)6 1 1 3. (a) 1 1 N (b) 1 1 4. Any of following The second electron is being removed from a full shell that is nearer to the nucleus The second electron is being removed from a positive ion Greater effective nuclear charge on second electron Less shielding since second electron in a new shell 1 1 5. HNO 1 1 6. 63.1 1 1 1 7. (a) K c = [HI] 2 [H 2 ][I 2 ] 1 1 (b) [H 2 (g)] = [HI(g)] 2 (1) [I 2 (g)] K c 1 = 15.00 2 / 1.20 46.0 = 4.08 (mol dm 3) (1) 1 2 1 ecf possible from part (a) Section A total 9 1 0 10 2 0 3
Section B Question Marking details 8. (a) Fe 2 O 3 + 3CO 2Fe + 3CO 2 (1) Any of the following for (1) During this process the iron(iii) ions (in Fe 2 O 3 ) gain electrons (to produce iron); reduction is a process of electron gain The oxidation number of iron is reduced from +3 to 0; a reduction in (positive) oxidation number is reduction Carbon monoxide loses electrons; oxidation is a process of electron loss The oxidation number of carbon is increased from +2 to +4; an increase in (positive) oxidation number is oxidation Fe 2 O 3 loses oxygen and CO gains oxygen Marks available AO1 AO2 AO3 Total Maths Prac 2 2 (b) 350 tonnes of which 0.02 % is sulfur Mass of sulfur = 350 0.02 = 0.07 tonnes (1) 100 Mg + S MgS 0.07 tonnes sulfur needs 24.3 0.07 = 0.0530 tonnes 32.1 = 53.0 (kg) (1) ecf possible (c) Cubic structure shows alternating different ions (1) 2 2 1 Ions labelled as Mg 2+ and S 2 (1) 2 2 4
Marks available Question Marking details AO1 AO2 AO3 Total Maths Prac 8. (d) (i) White solid / precipitate (of magnesium hydroxide) 1 1 1 (ii) Number of mol of MgS = 0.224 = 0.00397 56.4 0.00397 mol of H 2 S also produced (1) 1 Volume of H 2 S = 0.00397 24.0 = 0.095(3) dm 3 (1) 2 = 95.(3) cm 3 (1) ecf possible Accept alternative method using pv = nrt (e) A colourless solution (1) 1 3 1 2 1 The solubility of the group 2 hydroxides increases down the group (1) 1 (f) BaO + H 2 O Ba(OH) 2 (1) 2 2 ph 7 (1) (g) Any two of following for (1) each Barium is in group 2 and has two outer electrons Too much energy is needed to remove a third electron This necessitates removing an electron from a shell nearer to the nucleus (to produce a Ba 3+ ion) 1 1 2 Question 8 total 9 7 0 16 2 2 5
Marks available Question Marking details AO1 AO2 AO3 Total Maths Prac 9. (a) (i) C 12 H 22 O 11 + 18 [O] 6 (COOH) 2 + 5 H 2 O 1 1 (ii) Mass of the anhydrous acid = 4.05g Moles of the anhydrous acid = 4.05 / 90 = 0.045 (1) Mass of the water lost = 1.62g Moles of water = 1.62/18.02 = 0.090 (1) Mole ratio acid : water is 1 :2 Value of x is 2 (1) 3 3 2 If no working is shown award (1) for the correct answer (b) Sample of potassium methanoate not pure / not all HCOOK reacted (1) Inadequate heating / not heated for long enough / not heated at a high enough temperature (1) 2 2 2 (c) (i) Allow to settle / test the filtrate (1) Add a few drops of calcium chloride solution and see if a precipitate forms / cloudiness (1) 2 2 2 (ii) Moles of calcium oxalate = 2.49/128 = 0.0195 (1) Number of moles of potassium oxalate is also 0.0195 Mass of potassium oxalate is 0.0195 166 = 3.24 g (1) % of potassium oxalate in mixture = 3.24 100 /4.69 1 = 69.1 (to 3 sig. figs.) (1) 3 3 (accept values from 68.9 to 69.1 depending on use of significant figures during the calculation) ecf possible Question 9 total 1 6 4 11 3 4 6
Question Marking details 10. (a) Total percentage of 29 Si and 30 Si is 100 92.2 = 7.8 % of 29 Si = 2 7.8 = 5.2 and % of 30 Si = 1 7.8 = 2.6 (1) 3 3 A r = (28 92.2) + (29 5.2) + (30 2.6) (1) 100 Marks available AO1 AO2 AO3 Total Maths Prac 1 1 1 A r = 2582 + 151 + 78 = 2811 = 28.1 (1) 100 100 Answer only no mark ecf possible 2 3 (b) (i) Tetrahedral 1 1 (ii) There are no free electrons or ions to carry the charge 1 1 (iii) Any of the following There are no electronegativity differences in the Si Si bond All the bonding electrons are shared equally between the four Si atoms Si cannot lose or gain 4 electrons 1 1 7
Question Marking details 10. (c) M r HF is 20.01 Solution contains 500g HF in 1000g solution Marks available AO1 AO2 AO3 Total Maths Prac (using V = m/d) 1000 g of the solution has a volume of 855 cm 3 or Number of moles of HF in 1000g / 855 cm 3 solution is 500 = 24.98 (1) 20.01 855 cm 3 contain 24.98 mol Concentration of HF = 29.2 mol dm 3 (1) 2 2 2 ecf possible (d) (i) (There are 6 bonding pairs of electrons and no lone pairs ) position of minimum repulsion taken up (1) 1 Drawing shows clear octahedral shape (1) Bond angle is 90 o equatorial / equatorial or 90 o equatorial / vertical (accept 180 o if vertical bonds only considered) (1) 2 3 8
Question Marking details 10. (ii) 1 mol of H 2 SiF 6 (144g) gives 6 mol of F ions (6 19g) = 114g (1) Marks available AO1 AO2 AO3 Total Maths Prac 114 mg of fluoride ions from 144 mg H 2 SiF 6 The increase in fluoride ion concentration needed is 0.76 0.15 = 0.61 mg dm 3 (1) Amount of H 2 SiF 6 needed is 144 0.61 = 0.77 mg (1) 114 3 3 2 Accept alternative method Award (3) for cao (e) MgSiF 6 M r 166 Moles of MgSiF 6 = 2.6/166 = 1.566 10 2 (1) [H + ] is 4 1.566 10 2 = 0.06265 mol dm 3 (1) ph = log 10 [H + ] = 1.20 (1) 3 3 2 ecf possible Question 10 total 3 9 5 17 8 0 9
Question Marking details 11. (a) Diagram should show: Polarisation of N H or O H bonds in 2-aminoethanol (1) Lone pairs of electrons used in hydrogen bonding to nitrogen or oxygen atoms (1) Marks available AO1 AO2 AO3 Total Maths Prac 2 Polarisation of water molecules (1) Hydrogen bonds indicated between 2-aminoethanol and water molecules using relevant nitrogen / oxygen and hydrogen atoms (1) 2 4 (b) If no water then 3 marks maximum The forces of attraction between molecules of 2-aminoethanol are stronger than the attractive forces between molecules of 1,2-diaminoethane (as the former has a higher boiling temperature) (1) This suggests that intermolecular hydrogen bonding between / involving O and H is stronger than the hydrogen bonding between N and H (1) 1 This is (probably) due to a greater electronegativity difference between O and H than between N and H / O more electronegative than N / size considerations (1) 2 3 10
Question Marking details 11. (c) Indicative content 1 aqueous sulfuric acid in burette (accept aqueous ammonia in burette) 2 measure volume of ammonia into flask 3 use of an indicator (not universal indicator) 4 titrate with aqueous sulfuric acid until colour of indicator just changes 5 read burette and repeat without indicator / use of decolorising charcoal and filter 6 concentrate neutralised solution 7 cool (concentrated) solution / leave to evaporate over time 8 filter and dry crystals Marks available AO1 AO2 AO3 Total Maths Prac 3 3 6 6 5-6 marks The method provided leads to pure dry crystals of ammonium sulfate. The candidate constructs a relevant and logically structured account including all key elements of the indicative content. Scientific conventions and vocabulary are used accurately throughout. 3-4 marks The method outlined leads to the production of a solution of ammonium sulfate. The candidate constructs a logically structured account including the main elements of the indicative content. scientific conventions and vocabulary are generally sound. The use of 1-2 marks The method provided leads to the production of a solution that contains ammonium sulfate. The candidate has given an outline method of the production of ammonium sulfate but a number of key points are missing. There is some evidence of the correct use of scientific conventions and vocabulary. 0 marks The candidate does not make any attempt or give an answer worthy of credit. Question 11 total 5 6 2 13 0 6 11
Question Marking details 12. (a) (i) The iodide ions (moved to the anode and) were oxidised / lost electrons forming iodine (1) Marks available AO1 AO2 AO3 Total Maths Prac 1 2I (aq) I 2 (aq) + 2e (1) (Aqueous) iodine was produced giving a yellow/brown coloration (around the anode) (1) 1 1 3 1 (ii) Both iodide and chloride ions formed precipitate yellow and white observed (1) On adding ammonia and shaking the white precipitate / silver chloride dissolves (1) Leaving silver iodide as a pale yellow solid / silver iodide does not dissolve (1) 3 3 3 (iii) Number of moles of potassium iodate(v) = 0.100 18.00 1000 = 1.8 10 3 (1) 1 Number of moles of NaI present in 25.0 cm 3 of the solution of the mixture = 1.8 10 3 2 = 3.6 10 3 Number of moles of sodium iodide in 250 cm 3 = 0.036 (1) Mass of sodium iodide present in 250 cm 3 of the solution of the mixture = 0.036 150 = 5.40 g % sodium iodide in the mixture = 5.40 100 = 48 (1) 11.24 3 3 ecf possible 12
Question Marking details 12. (b) f = c/ = 3 10 8 /278 10 9 = 1.079 10 15 (Hz) (1) Marks available AO1 AO2 AO3 Total Maths Prac E = hf E = 6.63 10 34 1.079 10 15 J = 7.154 10 19 J (per molecule) per mole 7.154 10 19 6.02 10 23 J mol 1 = 4.307 10 5 J mol 1 (1) 2 2 2 = 431 kj mol 1 (c) As the group is descended the bond energies decrease and the wavelengths increase / astatine is below iodine in the Periodic Table / max > 400nm (1) coloured gas linked with the visible region (1) 2 2 Question 12 total 1 7 5 13 3 4 13
SUMMARY OF MARKS ALLOCATED TO ASSESSMENT OBJECTIVES Question AO1 AO2 AO3 Total Maths Prac 1. to 7. 9 1 0 10 2 0 8. 9 7 0 16 2 2 9. 1 6 4 11 3 4 10. 3 9 5 17 8 0 11. 5 6 2 13 0 6 12. 1 7 5 13 3 4 Totals 28 36 16 80 18 16 WJEC GCE Chemistry AS Unit 1 MS/Summer 2016 14