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Q1. How many protons are there in 6.0 g of nitrogen gas?Avogadro constant, L = 6.022 × 1023 mol–1?A1.3 × 1023B9.0 × 1023C1.8 × 1024D3.6 × 1024Q2. Two sealed flasks with the same volume are left side by side.Flask A contains 4.0 × 10?3 mol of methane.Flask B contains 340?mg of a different gas.Both gases are at the same temperature and pressure.Which gas could be in Flask B??ACH2Cl2BHBrCKrDPF3Q3. A 30 cm3 sample of nitrogen was reacted with a 60 cm3 sample of fluorine according to the equationWhat is the volume of the gas mixture after the reaction, at constant temperature and pressure??A20 cm3B30 cm3C40 cm3D50 cm3Q4. A sample of 2.0?mol?dm?3 acid has a volume of 100?cm3What volume of water, in cm3, should be added to this acid to dilute the sample to a concentration of 1.5?mol?dm?3??A25B33.3C50D66.7Q5. What is the empirical formula of an oxide of nitrogen that contains 26% nitrogen by mass?ANO2BN2O3CN2O5DN4O5Q6. A student devised an experiment to find the concentration of sulfuric acid in a sample of battery acid.?????????A measuring cylinder was used to transfer 10 cm3 of battery acid to a volumetric flask.?????????Distilled water was added to the volumetric flask until the volume reached 250 cm3?????????A 25.0 cm3 sample of diluted acid was transferred from the volumetric flask to a conical flask using a pipette.?????????A few drops of methyl orange indicator were added to the acid in the conical flask before titrating the acid with sodium hydroxide.?????????The titration was repeated five times but concordant results were not obtained. (Note: Methyl orange is red in acid and yellow in alkali.)Which suggestion would improve the chances of obtaining concordant titres??AInvert the volumetric flask several times after adding the distilled water. BWash the pipette with distilled water between each titration.CAdd extra drops of indicator to the sample when nearing the end point in each titration.DUse a more concentrated solution of sodium hydroxide in the burette.Q7. Ethanol can be made from glucose by fermentation.C6H12O6 → 2C2H5OH + 2CO2In an experiment, 268 g of ethanol (Mr = 46.0) were made from 1.44 kg of glucose (Mr?=?180.0).What is the percentage yield?A 18.6%B 36.4% C 51.1%D 72.8%Q8. A sample of pure Mg(NO3)2 was decomposed by heating as shown in the equation below.2Mg(NO3)2(s)????2MgO(s) + 4NO2(g) + O2(g)(a) ????A 3.74 × 10?2 g sample of Mg(NO3)2 was completely decomposed by heating.Calculate the total volume, in cm3, of gas produced at 60.0 °C and 100 kPa.Give your answer to the appropriate number of significant figures.The gas constant R = 8.31 J K?1 mol?1.????Total volume of gas = ___________ cm3(5)(b) ????The mass of MgO obtained in this experiment is slightly less than that expected from the mass of Mg(NO3)2 used.Suggest one practical reason for this._________________________________________________________________________________________________________________________________________________________________________________________________________(1)(Total 6 marks)Q9. (a)???? Calcium phosphate reacts with aqueous nitric acid to produce phosphoric acid and calcium nitrate as shown in the equation.Ca3(PO4)2????+????6HNO3????2H3PO4????+????3Ca(NO3)2(i)??????A 7.26 g sample of calcium phosphate reacted completely when added to an excess of aqueous nitric acid to form 38.0 cm3 of solution.Calculate the concentration, in mol dm–3, of phosphoric acid in this solution.Give your answer to 3 significant figures.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(5)(ii)?????Calculate the percentage atom economy for the formation of calcium nitrate in this reaction.Give your answer to 1 decimal place.__________________________________________________________________________________________________________________________________________________________________________________________(2)(b)???? Write an equation to show the reaction between calcium hydroxide and phosphoric acid to produce calcium phosphate and water.___________________________________________________________________(1)(c)???? Calcium dihydrogenphosphate can be represented by the formula Ca(H2PO4)x where x is an integer.A 9.76 g sample of calcium dihydrogenphosphate contains 0.17 g of hydrogen, 2.59 g of phosphorus and 5.33 g of oxygen.Calculate the empirical formula and hence the value of x.Show your working.___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(4)(Total 12 marks)Q10. Copper can be produced from rock that contains CuFeS2(a)?????Balance the equations for the two stages in this process......CuFeS2 + .....O2 + .....SiO2 ??? .....Cu2S + .....Cu2O + .....SO2 + .....FeSiO3.....Cu2S + .....Cu2O ??? .....Cu + .....SO2(2)(b)?????Suggest two reasons why the sulfur dioxide by-product of this process is removed from the exhaust gases.Reason 1 _________________________________________________________________________________________________________________________________________________________________________________________________Reason 2 _________________________________________________________________________________________________________________________________________________________________________________________________(2)(c)?????A passenger jet contains 4050 kg of copper wiring.A rock sample contains 1.25% CuFeS2 by mass.Calculate the mass, in tonnes, of rock needed to produce enough copper wire for a passenger jet.???????????(1 tonne = 1000 kg)Mass of rock ____________________ tonnes(4)(d)?????Copper can also be produced by the reaction of carbon with copper(II) oxide according to the equation2CuO + C ? 2Cu + CO2Calculate the percentage atom economy for the production of copper by this process.Give your answer to the appropriate number of significant figures.Percentage atom economy ____________________(2)(Total 10 marks)Q11. A sample of bromine was analysed in a time of flight (TOF) mass spectrometer and found to contain two isotopes, 79Br and 81BrAfter electron impact ionisation, all of the ions were accelerated to the same kinetic energy (KE) and then travelled through a flight tube that was 0.950 m long.(a)?????The 79Br+ ions took 6.69 × 10–4 s to travel through the flight tube.Calculate the mass, in kg, of one ion of 79Br+Calculate the time taken for the 81Br+ ions to travel through the same flight tube.The Avogadro constant, L = 6.022 × 1023 mol–1KE = ?mv2 ????? where m = mass (kg) and v = speed (m s–1)???????????????where d = distance (m) and t = time (s)Mass of one ion of 79Br+ ____________________ kgTime taken by 81Br+ ions ____________________ s(5)(b)?????Explain how ions are detected and relative abundance is measured in a TOF mass spectrometer.__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)(Total 7 marks)Q12. A student added 627 mg of hydrated sodium carbonate (Na2CO3.xH2O) to 200 cm3 of 0.250 mol dm–3 hydrochloric acid in a beaker and stirred the mixture.After the reaction was complete, the resulting solution was transferred to a volumetric flask, made up to 250 cm3 with deionised water and mixed thoroughly.Several 25.0 cm3 portions of the resulting solution were titrated with 0.150 mol dm–3 aqueous sodium hydroxide. The mean titre was 26.60 cm3 of aqueous sodium hydroxide.Calculate the value of x in Na2CO3.xH2OShow your working.Give your answer as an integer.Value of x ____________________ (Total 7 marks)Mark schemesQ1.CQ2.AQ3.DQ4.BQ5.CQ6.AQ7.BQ8.(a)???? Stage 1Mr for Mg(NO3)2 = 148.3Moles of Mg(NO3)2 = ?? = 2.522 × 10-4 molExtended response calculation1Stage 2Total moles of gas produced = 5/2 × moles of Mg(NO3)2= 5/2 × 2.522 × 10–4 = 6.305 × 10–4If ratio in stage 2 is incorrect, maximum marks for stage 3 is 21Stage 3PV = nRT so volume of gas V = nRT / P1V ?=??= 1.745 × 10–5 m31V = 1.745 × 10–5 × 1 × 106 = 17.45 cm3 = 17.5 (cm3)Answer must be to 3 significant figures (answer could be 17.4 cm3 dependent on intermediate values)1(b)???? Some of the solid is lost in weighing product / solid is blown away with the gas1[6]Q9.(a)???? (i)??????M1 - Mr calcium phosphate = 310(.3)If Mr wrong, lose M1 and M5.1M2 - Moles calcium phosphate =????????(= 0.0234)?0.0234 moles can score M1 and M2.If Mr incorrect, can score M2 for .?Allow M2 and / or M3 to 2 significant figures here but will lose M5 if answer not 1.23.1M3 - Moles phosphoric acid = 2 × 0.0234 = 0.0468Allow student’s M2 × 2. If not multiplied by 2 then lose M3 and M5.1M4 - Vol phosphoric acid = 0.038(0) dm3If not 0.038(0) dm3 then lose M4 and M5.1Conc phosphoric acid = M5 = 1.23 (mol dm?3)This answer only – unless arithmetic or transcription error that has been penalised by 1 mark.Allow no units but incorrect units loses M5.1(ii)????? × 100????OR???? × 100?1 mark for both Mr correctly placed.= 71.5%2(b) ????3Ca(OH)2 + 2H3PO4 Ca3(PO4)2 + 6H2OAllow multiples.1(c) ????If x = 2 with no working, allow M4 only.Ca = 1.67 g (M1).1Mark for dividing by correct Ar in Ca and P (M2).If M1 incorrect can only score M2.1Correct ratio (M3).1CaH4P2O8????OR????Ca(H2PO4)2????OR????x = 2Value of x or correct formula (M4).1AlternativeCa???? ???? ???? ????H2PO4Ca = 1.67 g (M1).Mark for dividing by correct Ar / Mr in Ca and H2PO4 (M2).If M1 incorrect can only score M2.Correct ratio (M3).CaH4P2O8????OR????Ca(H2PO4)2????OR????x = 2Value of x or correct formula (M4).[12]Q10.(a)?????4CuFeS2???+???9 O2???+???4SiO2???????Cu2S???+???Cu2O???+???7SO2???+???4FeSiO3Allow multiples1Cu2S???+???2Cu2O???????6Cu + SO21(b)?????ANY TWO????Prevents acid rain (which damages buidlings / ecology)????Toxic OR causes breathing problems????Reduces waste product OR makes use of the waste OR improves atom economy OR Reduces need for sulfur mining OR used to produce sulfuric acid OR any named products2(c)?????M1, M2, M3 are process marksM1?????Mol Cu = (= 63780) 1M2?????Mass CuFeS2 = (63780) × 183.5 (= 1.17 × 107g)1M3?????Mass ore = (1.17 × 107) × 1M4?????Mass ore = 936 tonnes (Allow 936 –937)1Alternative methodM1 % of Cu in CuFeS2 = (63.5/183.5) × 100 = 34.6%M2 % of Cu in the rock = (34.6/100) × 1.25 = 0.4325%M3 mass of rock = 4050 × 100/0.4325 = 936416kgM4 mass of rock in tonnes = 936 tonnesNotesM1 Ar Cu must be usedM2 Mr CuFeS2 to have been usedM3 Grossing up for the mass of rockM4 Final answer correct in tonnes(d)?????% atom economy = × 1001= 74.3% must be 3sf1[10]Q11.(a)?????= 79 / (1000 × 6.022 × 1023) = 1.31 × 10–25 kg1Then either follow method 1 (or method 2 below)Do not mix and match methodsMethod 1?= 0.950 / 6.69 × 10–4= 1420 ms–1In method 1, M2 can be awarded in M31KE = ? mv2= ? × 1.312 × 10–25 × (1420)2= 1.32 × 10–19 JMark consequential to their velocity and mass. Allow mass of 79 etc.1V81 = ?= √ 1.963 × 106= 1.40 × 103 ms–1(allow 1.398 × 103 - 1.402 × 103)Mark consequential to their velocity and mass. Allow mass of 81 etc.1?= 6.80 × 10–4 sMark consequential to their M4Accept 6.77 – 6.80 × 10–4 s1Method 2m1(d/t1)2 = m2 (d/t1)2orm1 / t12 = m2 / t221t22 = t12 (m2/m1)Ort22 = (6.69 × 10–4)2 × (81/79)1t22 = 4.59 × 10–7Mark consequential to their M31t = 6.77 × 10–4 sMark consequential to their M4Accept 6.77 – 6.80 × 10–4 s1(b)?????ion hits the detector / negative plate and gains an electron1Not positive plate(relative) abundance is proportional to (the size of) the current1[7]Q12.M1????HCl added = 0.050 mol andNaOH used in titration = 3.99 × 10–3 mol1M2????So moles that would be needed to neutralise total excessHCl = 3.99 × 10–3 × 10 = 3.99 × 10–2 molAlternative: divide moles HCl by 10 = 0.005 and 0.005 - 3.99 × 10–3 = 0.001011M3????Therefore the moles of HCl reacted with the Na2CO3.xH2O =0.050 - 3.99 × 10–2 = 0.0101 molAlternative: 0.00101 × 10 to produce 0.01011M4????So moles Na2CO3.xH2O reacted with the HCl = 0.0101 / 2 = 5.05 x10–3 mol1M5????Conversion of mg to g = 0.627 (g) or 627 × 10–3 (g)1M6????xH2O = 0.627/5.05 × 10–3 -106.0 = 18 (.16)Alternative: mass Na2CO3 that reacted with the HCl 5.05 × 10–3 x106.0 = 0.5353 g and mass H2O = 0.627- 0.5353 = 0.0917 g1M7????so x = 1Alternative: 0.0917 /18.0 = 5.094 × 10–3 so ratioNa2CO3 to H2O = 1:1.009 ie 1:1 so × = 11[7] ................
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