Foundation Program Chemistry TRANSITION RESOURCE BOOK Sample Exams UNSW Foundation Studies UNSW Global Pty Limited UNSW Sydney NSW 2052 Australia Copyright 2019 All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner NOTES Some changes have occurred over the years to the Chemistry syllabus: ‘Ionisation Energy’ and the ‘Structure of the Atom’ were moved from Term 2 to Term 1, therefore the MID Program samples do not have these sorts of questions in them. Look in the sample FINAL papers for typical exam style questions on these topics. Due to the limited time in Term 1, Electrolysis will be assessed in the Final Exam (not in the MID Program exam). Student Number: __________________ Family Name: __________________ Other Names: __________________ Class Group: __________________ Transition Program Chemistry Trial Mid–Program Examination A Time Allowed: 1.5 hour Reading Time: 5 minutes TrialMIDProgram (Paper A) Question Possible Mark 1 6 2 6 3 6 4 6 5 6 6 10 7 10 8 10 TOTAL 60 F O U N D A T I O N S T U D I E S UNSW Foundation Studies UNSW Global Pty Limited UNSW Sydney NSW 2052 Australia Copyright 2019 All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner. DIRECTIONS TO CANDIDATES 1. Check that your copy of the examination paper is complete. This examination contains ten questions: • Questions 1 to 5 @ 6 marks each • Questions 6 to 8 @ 10 marks each The total marks available are 60. 2. Ensure that you have in the room: • a black or blue pen for writing answers • an electronic calculator • no unauthorised material 3. Five (5) minutes reading time is allowed. During reading time do not write at all. 4. Ensure that your name and examination number are clearly printed on the front page on the blank answer booklet. 5. Answer all questions in the answer booklet. 6. Hand in the answer booklet to the supervisor at the end of the examination. 1. There are seven elements whose names begin with the letter “b”. From this set of 7 elements write down the name of: (a) the largest atomic radius (b) an element which exists as diatomic molecules (c) an element which belongs to the halogen family (d) an element which is classified as an actinide (e) an element which has some metallic and some non-metallic properties (f) the most chemically reactive metallic element. (The name of an element may be used more than once or not at all). 2. Many household cleaners contain ammonia dissolved in water. In a consumer test, 25.0 mL of one brand of household cleaner were titrated against 0.822 M hydrochloric acid solution. The graph below shows the variation in pH during the titration. (a) Using a net ionic equation briefly explain why the ammonia solution has a high pH at the beginning of the titration. (b) Write down the name of one indicator which would be suitable for this titration and briefly describe how its colour changes during the titration. (c) Calculate the molar concentration of ammonia in the household cleaner. Volume hydrochloric acid added (mL) 0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 2.00 4.00 6.00 8.00 10.00 12.00 14.00 pH 3. Five different samples of the insoluble compound marshite provided the following analyses: Sample Number 1 2 3 4 5 Mass of marshite analysed (g) 2.40 3.30 5.10 6.33 8.43 Mass of copper obtained (g) 0.80 1.10 1.70 2.11 2.81 Mass of iodine obtained (g) 1.60 2.20 3.40 4.22 5.62 (a) Evaluate the ratio: (moles of copper atoms) ÷ (moles of iodine atoms) formed by analysis of marshite (b) Write down the chemical formula and systematic name of marshite (c) Marshite is precipitated when copper (II) sulfate solution is mixed with potassium iodide solution. Solid iodine is also produced. Write a net ionic equation to represent this reaction. 4. (a) Carbon monoxide gas can be made by blowing carbon dioxide gas over hot carbon. Write a thermochemical equation for this reaction. (b) Calculate the standard enthalpy of combustion (∆Hc in kJ/mol) of carbon monoxide. (c) In a steelworks carbon monoxide reacts with iron (III) oxide according to: 3CO (g) + Fe2O3 (s) → 2Fe (s) + 3CO2 (g) ∆H = − 22 kJ Use these data to calculate the standard enthalpy of formation (∆Hf in kJ/mol) of iron (III) oxide. 5. A student was given four test tubes marked A, B, C and D. Each contained one of the liquids: silver nitrate solution, sodium hydroxide solution, sodium carbonate solution and sucrose (table sugar) solution, though not necessarily in that order. The student added 5 mL of 1 M hydrochloric acid solution to each test tube and obtained the following results. Solution Change When Hydrochloric Acid Added A Bubbles of gas produced B No visible change but temperature increased C White precipitate formed D No change was detected Identify the contents and write a net ionic equation for the reaction in: (a) test tube A (b) test tube B (c) test tube C 6. The solubility of barium hydroxide (in grams per litre of solution) at various temperatures is tabulated below. Temperature (oC) 0 20 60 80 Solubility (g/L) 16.7 38.9 209.4 1014.0 (a) What mass of barium hydroxide would be dissolved in 250 mL of a saturated solution at 60 °C? (b) Calculate the molar concentration of the solution in (a) (c) If the solution in (a) was cooled to 20 oC calculate the mass of crystals which would be produced. (d) Calculate the molar concentration of hydroxide ions present in the solution in (c) at 20 oC after all the crystals were removed. (e) Calculate the pH of the solution in (c) at 20 oC after the crystals were removed 7. The graph below shows the electrical conductivity of the resulting solution formed when 80.0 mL of 0.100 M barium chloride solution is added slowly to 400 mL of a 0.0100 M solution of silver sulfate. V1 Electrical conductivity of the mixture Vo ⇑ Volume of barium chloride solution 0 (a) Determine the number of moles of silver sulfate required to make 400 mL of a 0.0100 M solution of silver sulfate. (b) Write a net ionic equation for any chemical reaction that occurred when the barium chloride solution is added to the silver sulfate solution. (c) Briefly explain why the solution conducts electricity at volume V0 but not significantly at volume V1. assessed in FINAL not in MID- PROGRAM EXAM (d) What volume of barium chloride solution has been added between volume V0 and volume V1 ? (e) Briefly explain why the electrical conductivity increases after volume V1. assessed in FINAL not in MID- PROGRAM EXAM 8. The density of formaldehyde vapour is 1.226 g/L at 101.3 kPa and 25.0oC. (a) How many moles of formaldehyde are present in 1.00 L of the vapour under these conditions? (b) What is the mass, in grams, of one mole of formaldehyde? (c) If formaldehyde is a compound of carbon, hydrogen and oxygen what is its chemical formula? (d) The toxic amount of formaldehyde in one litre of air is 6.0 µg. How many molecules are present in 6.0 µg of formaldehyde? (e) Formaldehyde undergoes the following reaction at 300 oC: Formaldehyde gas + Ammonia gas → urotropine gas + steam (6 litres) (4 litres) (1 litre) (6 litres) If all volumes were measured at the same temperature and pressure write down the chemical formula of urotropine. DATA SHEET Avogadro's constant, NA 6.02 × 1023 mol–1 Universal Gas Constant, R 8.314 J K–1 mol–1 Specific heat of water 4.180 J g–1 K–1 Density of water 1.00 g mL–1 Molar volume of ideal gas: at 101.3 kPa and 0oC 22.41 L mol–1 at 101.3 kPa and 25oC 24.46 L mol–1 Unit / Symbol Equivalences: ∆Hf = ∆fH M = mol L–1 Standard enthalpy of formation (∆Hf) : CO(g) –111 kJ mol–1 CO2(g) –393 kJ mol–1 TrialMIDProgramTransition A (Solutions) 1 (a) barium (b) bromine (c) bromine (d) berkelium (e) boron (f) barium 2 (a) Ammonia has a high pH at the beginning of the titration because it is a weak base and produces hydroxide ion in solution NH3(aq) + H2O(l) ⇔ NH4+(aq) + OH–(aq) (b) The titration involves a strong acid (HCl) and a weak base (NH3) and the titration curve shows that the pH at the equivalence point is approximately pH 5. The best indicator is methyl red with a pH colour change interval of pH 4.4 to 6.2. The colour change is from base to acid so will be yellow to orange (to red past end-point). (c) NH3(aq) + HCl(aq)) ⇔ NH4 Cl(aq) Moles HCl = 0.822 × 0.0125 = 0.01027(5) mole Moles NH3 = moles HCl = 0.010275 = [NH3] × 0.02500 [NH3] = 0.411 molar 3. (a) The table shows that the mass of copper to mass of iodine is always 1.00 to 2.00 hence any the calculation can be based on any pair of data points however the best answer will be obtained by using the largest values (as this minimizes errors): Moles Cu = 2.81/63.55 =0.0442(1) mole Moles I = 5.62/126.9 = 0.0442(8) Moles Cu/Moles I = 0.04428/0.04421 = 1.00(1) (b) CuI, Copper(I) iodide (c) 2Cu2+(aq) + 4I–(aq) 2CuI(s) + I2(s) 4. (a) C(s) + CO2(g) 2CO(g) ∆H = +171 kJ/mol (b) CO(g) + ½ O2(g) CO2(g) ∆H = ( –393) – (–111 + ½ × 0) = –282 kJ/mol (c) 3CO(g) + Fe2O3(s) 3CO2(g) + 2Fe(s) ∆H =(3 × –393 + 2× 0) – (3 × –111 + ∆Hf(Fe2O3 (s)) = –22 kJ/mol ∆Hf(Fe2O3 (s)) = –824 kJ/mol 5. (a) Bubbles of gas form when hydrochloric acid are added to A therefore the only possibility listed is that A must be a solution (liquid) that contains sodium carbonate: 2H+(aq) + CO32–(aq) CO2(g) + H2O(l) (b) The only combination that does not produce a precipitate or a gas yet does react with hydrochloric acid is NaOH. The ionic equation for a strong acid/strong base combination is: H+(aq) + OH–(aq) H2O(l) (c) A white precipitate is produced when silver nitrate and hydrochloric acid react together: Ag+(aq) + Cl–(aq) AgCl(s) 6. (a) Mass of Ba(OH)2 in 250 mL at 60oC = 0.250 × 209.4 = 52.4 g (b) Moles Ba(OH)2 (in 1 litre) = 209.4/(137.3 + 2 × 16.00 + 2 × 1.008) = 1.221 moles [Ba(OH)2] = 1.221 mol/L (c) Mass of Ba(OH)2 in 250 mL at 20oC = 250 × 38.9 = 9.7 g Mass of solid crystals on cooling = 52.4 – 9.7 = 42.7 g (d) Moles Ba(OH)2 (in 1 litre) = 38.9/(137.3 + 2 × 16.00 + 2 × 1.008) = 0.227(1) moles [Ba(OH)2] = 0.227 mol/L Therefore [OH–] = 2 × 0.227 = 0.454 mol/L (e) pOH = 0.34 ∴ pH = 13.66 7. (a) Moles Ag2SO4 = [Ag2SO4] × volume = 0.0100 × 0400 = 0.00400 moles (b) Since the solutions are reacting the reactants must be aqueous. The electrical conductivity drops to zero indicating that no soluble ionic compounds are present at the equivalence point. 2Ag+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq) 2AgCl(s) + BaSO4(s) (c) The solution at Vo is a solution of silver sulfate. Silver sulfate is a strong electrolyte and is completely dissociated into silver ions and sulfate ions and these ions are free to move through the solution and carry electric charge (conduct electricity) through solution. At the equivalence point, V1, all the reactants have reacted and there are no mobile ions in solution (they have precipitated out of solution) and hence are not free to move through solution so cannot the mixture at V1, cannot conduct electricity. (d) Moles Ag2SO4 = 0.00400 moles Moles BaCl2 = 0.00400 = 0.100 × Volume therefore Volume = 0.0400 L = 40.0 mL (e) As previously stated, there are no mobile ions in solution at the equivalence point, V1, and thus the mixture at V1, cannot conduct electricity. Adding more barium chloride solution to this mixture increases the amount of barium chloride in excess and the concentration of ions (barium ions and chloride ions) increases in solution (as there is no longer any silver sulfate to react, it being the limiting reagent after V1). The barium ions and chloride ions are free to move through the solution and thus the solution can conduct electricity. 8. (a) Either n = PV/RT = (101.3 × 1.00)/(8.3141 × 298) = 0.0408(8) mole = 0.0409 moles or n = volume/molar volume = 1.00/24.46 = 0.409 mole (b) n = mass/molar mass therefore molar mass = 1.226/0.0409 = 29.99 = 30.0 The mass of one mole of formaldehyde is 30.0 gram (c) CxHyOz Hence 12 x + y + 16 z = 30.0 The largest value for z is 1 hence formaldehyde contains only 1 oxygen. If formaldehyde contains 1 oxygen the largest value for x is 1 and hence formaldehyde must only contain 1 carbon. This suggests that formaldehyde must contain 2 H atoms. Therefore formaldehyde is CH2O. (d) n = mass/molar mass = 6.0 × 10–6/30.0 = 2.0 × 10–7 Number of molecules = moles × Avogadro’s number = 2.0 × 10–7 × 6.02 × 1023 Number of molecules = 1.2 × 1017 (e) 6CH2O + 4 NH3 1 Urotropine + 6 H2O Hence urotropine = C6H12N4 Student Number: __________________ Family Name: __________________ Other Names: __________________ Class Group: __________________ Transition Program Chemistry Trial Mid–Program Examination B Time Allowed: 1.5 hours Reading Time: 5 minutes Question Possible Mark 1 6 2 6 3 6 4 6 5 6 6 10 7 10 8 10 TOTAL 60 F O U N D A T I O N S T U D I E S UNSW Foundation Studies UNSW Global Pty Limited UNSW Sydney NSW 2052 Australia Copyright 2019 All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner. DIRECTIONS TO CANDIDATES 1. Check that your copy of the examination paper is complete. This examination contains ten questions: • Questions 1 to 5 @ 6 marks each • Questions 6 to 8 @ 10 marks each The total marks available are 60. 2. Ensure that you have in the room: • a black or blue pen for writing answers • an electronic calculator • no unauthorised material 5. Five (5) minutes reading time is allowed. During reading time do not write at all. 6. Ensure that your name and examination number are clearly printed on the front page on the blank answer booklet. 5. Answer all questions in the answer booklet. 6. Hand in the answer booklet to the supervisor at the end of the examination. 1. 1.00 g samples of two metals (A and B), each with a valency of +2, were treated with 50.0 mL of an excess of hydrochloric acid; 445 mL of gas at RTP were produced by the reaction of metal A with hydrochloric acid. There was no reaction when metal B was treated with hydrochloric acid. (a) Write an equation for the reaction between metal A and hydrochloric acid. (b) Calculate the atomic mass, in u or amu, for metal A. (c) Suggest a possible identity for metal B. Briefly justify your choice of identity for metal B. 2. A student set out to measure the concentration of a certain ammonia solution. This was done by titrating 25.0 mL of the ammonia solution against 0.106 M sulfuric acid solution to produce ammonium sulfate solution. (a) The sulfuric acid was added using the instrument shown in the diagram below. Write down the name of the instrument and the reading (in mL) shown. (b) Name an indicator that would be suitable for detecting the equivalence point of this titration and briefly describe how its colour would change during the titration. (c) Determine the concentration of the ammonia solution used in this titration. 3. During the electrolysis of silver nitrate solution the following reaction takes place: 4AgNO3(aq) + 2 H2O(l) → 4 Ag(s) + 4 HNO3(aq) + O2(g) (a) Water and solid silver nitrate are non-conductors of electricity. Briefly explain how a silver nitrate solution is able to conduct an electric current whilst both water and solid silver nitrate are not able to conduct electricity. Electrolysis assessed in FINAL not in MID- PROGRAM EXAM (b) At which electrode (negative or positive) does the silver form? Briefly explain your choice. Electrolysis assessed in FINAL not in MID- PROGRAM EXAM (c) Briefly explain why the pH decreases during the electrolysis of silver nitrate solution. 4. The following table shows the data obtained when several 100 mL samples of silver nitrate solution were treated with different volumes of 0.0525 M calcium chloride solution. The silver chloride that was produced by this reaction was filtered, dried and then weighed. Volume of calcium chloride solution added (mL) Volume of silver nitrate solution (mL) Mass of silver chloride precipitate (g) 20.0 100.0 0.301 40.0 100.0 0.602 60.0 100.0 0.903 80.0 100.0 1.204 100.0 100.0 1.355 120.0 100.0 1.355 140.0 100.0 1.355 160.0 100.0 1.355 (a) Write a nett ionic equation for the reaction that occurs. (b) Calculate the minimum volume of the calcium chloride solution required to completely react with 100 mL of the silver nitrate solution. (c) Calculate the concentration of the silver nitrate solution. 5. Consider the representation of the periodic table shown below: A D E J L M G R Q Using only the letters on shown in the table above: (a) Write the letter for the metallic element with valency 3. (b) Write the letter for the element with the largest molar volume. (c) Write the letters for three elements which you would expect to have similar properties (d) Write an equation for the reaction (if any) of element “E” with water. (e) Give the letter for the most reactive metal. (f) Which of these elements would exist, at 298 K and 100 kPa, as diatomic molecules? 6. A known volume of ammonia gas, (44.8 mL at STP), was passed repeatedly over hot iron until all the ammonia had decomposed. The iron was unchanged. On cooling the mixture to the original conditions, the volume of gas was found to have doubled. This gas mixture was then passed over heated copper(II) oxide (to remove hydrogen) and again cooled to the original conditions. (a) Calculate the number of moles of ammonia used in this experiment. (b) Why did the gas volume double? Explain using an equation. (c) Write an equation for the reaction with copper(II) oxide and the hydrogen. (d) What mass of copper(II) oxide reacted in the reaction in (c). (e) What volume of gas (at STP) remains after the reaction in (c). 7. The salt deposits at Strassfurt, Germany, contain a mixture of potassium chloride and magnesium chloride. The aqueous solubility curves of these two compounds are shown below: (a) Calculate the molar solubility of potassium chloride at 25°C. (b) 100 mL of a hot aqueous solution contains 50.0 gram of potassium chloride and 50.0 gram of magnesium chloride dissolved in it. Calculate the concentration of all the chloride ions in this solution. (c) Calculate the mass of each compound which would crystallise out of the solution if the solution in (b) was cooled to 20°C. Potassium chloride Temperature (ºC) 0 20 40 60 80 100 So lu bi lit y (g /1 00 m L so lu tio n) 20 40 60 80 Magnesium chloride (d) Draw a neat, labelled sketch of the apparatus suitable for separating the crystals from the solution in (c). (e) Potassium chloride dissolves in ether but magnesium chloride does not. Briefly describe, using a flow chart if necessary, a method for obtaining a pure sample of potassium chloride from a mixture of the two solids. 8. The following diagram represents the enthalpy relationships between hydrogen, oxygen, water and hydrogen peroxide. Use the data to answer the following questions. (a) Write a thermodynamic equation to represent the enthalpy of formation (∆Hf) of liquid hydrogen peroxide. (b) Write a thermochemical equation to represent the enthalpy of solution (∆Hsol) of hydrogen peroxide. E N T H A L P kJ/mol H2(g) + O2(g) H2O2(g) H2O2(l) H2O(g) H2O2(aq) ) H2O(l) H2O(s) 0 -133 -188 -191 -242 -285 -291 (c) What is the maximum amount of heat energy released when 1.50 mole of H2O2(l) is dissolved in water? (d) Calculate the enthalpy change (∆H) for the reaction: 2H2O2(aq) → O2(g) + 2H2O(l) (e) A 0.100 M hydrogen peroxide solution was prepared. The hydrogen peroxide decomposed according to the equation in part (c). What temperature change would be observed when 0.500 L of this 0.100 M hydrogen peroxide solution decomposes? Assume the density of the solution is 1.00 g/mL and the heat capacity of the solution is 4.18 J g–1 K–1. UNSW Foundation Studies UNSW Foundation Studies UNSW Sydney NSW 2052 Australia T: 61 2 9385 5396 | F: 61 2 9662 2651 | E:
[email protected] | W: www.ufs.unsw.edu.au UNSW Foundation Studies is an education group of UNSW Global Pty Limited, a not-for-profit provider of education, training and consulting services and a wholly owned enterprise of the University of New South Wales UNSW Foundation Studies programs are delivered under UNSW CRICOS Provider Code: 00098G UNSW Global Pty Limited ABN 62 086 418 582 DATA SHEET Avogadro's constant, NA 6.02 × 10 23 mol–1 Universal Gas Constant, R 8.314 J K–1 mol–1 Specific heat of water 4.180 J g–1 K–1 Density of water 1.00 g mL–1 Molar volume of ideal gas: at 101.3 kPa and 0oC 22.41 L mol–1 at 101.3 kPa and 25oC 24.46 L mol–1 Unit / Symbol Equivalences: ∆Hf = ∆fH M = mol L–1 TRIAL MIDPROGRAM CHEMISTRY B EXAM: ANSWERS: 1 (a) A + 2HCl ACl2 + H2 (b) moles H2 = 0.445/24.46 = 0.01882 mole moles A = moles H2 = 0.01882 mole Atomic mass A = 1.00/0.01882 = 55.0 amu (c) Metal B does not react with hydrochloric acid so it is possibly copper 2 (a) Burette: 29.60 mL (the second decimal place must be estimated for full marks) (b) Methyl orange red: yellow → orange → red. (the intermediate colour must be indicated for full marks) (c) 2NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq) [NH3] × 0.0250 = 2 × 0.106× 0.02960 ∴ [NH3] = 0.251 Molar. 3 (a) Pure water does not contain ions (at a sufficient concentration) to allow electrical conductivity to occur. Pure (solid) silver nitrate contains ions but these ions (Ag+(s) and NO3–(s)) are trapped in their lattice positions and not freely moving (they only vibrate) so are not free to move to carry electrical charge. Silver nitrate is a strong electrolyte and dissociates completely when dissolved in water (the ionic lattice is destroyed) and the aquated ions (Ag+(aq) and NO3–(aq))are free to move between the electrodes and to conduct electricity through the solution (NOTE: Some answers really do require a high level of detail to obtain full marks - most students do not perform well on this sort of extended discussion question as they leave out too much detail). (b) Ag+(aq) migrate to the cathode which is the negative electrode in electrolysis cells because opposite electrical charges attract (positive attracted to negative) (c) pH = –log10[H+] Nitric acid is produced during this reaction and nitric acid is a strong acid: HNO3(aq) H+(aq) + NO3–(aq). The pH decreases because the concentration of H+ increases in solution and since pH (NOTE: For full marks to be awarded students need to define pH and to identify why it decreases by identifying the compound that causes hydrogen ion concentration to increase.) 4. (a) 2AgNO3(aq) + CaCl2(aq) 2AgCl(s) + Ca(NO3)2(aq) Ag+(aq) + Cl–(aq) AgCl(s) (b) Mole mass AgCl = 107.9 + 35.45 = 143.35 Moles AgCl = 1.355/143.35 = 0.009452 mole Moles CaCl2 = ½ moles AgNO3 = 0.004726 mole volume CaCl2 = 0.04726/0.0525 = 89.99 mL = 90.0 mL (c) Moles AgNO3 = Moles AgCl 0.009452 mole [AgNO3] = 0.009452/0.100 = 0.0945 molar 5. (a) M (b) D (c) A, J, R (d) 2E + 2H2O 2EOH + H2 (e) Q (f) G 6. (a) mole mass NH3 = 0.0448/22.41 = 0.00200 mole (b) 2NH3(g) N2(g) + 3H2(g) 2 mole of gas decomposes to give 4 moles of gas therefore 2 litres of gas would decompose to give 4 litre of gas (Avogadro’s law: Volume gas ∝ moles gas) (c) CuO(s) + H2(g) Cu(s) + H2O(l) (d) moles CuO = Moles of H2. Moles of H2 = 3/2 × moles NH3 = 3/2 × 0.00200 = 0.00300 mole Moles CuO = 0.00300 therefore mass CuO = 0.00300 × (63.55 + 16.00) = 0.239 gram (e) At STP, water is not a gas therefore the only gas remaining (after the hydrogen is removed) is nitrogen. The moles of nitrogen is half the moles of ammonia hence the volume of nitrogen is half the volume of ammonia: Vol nitrogen = ½ × 44.8 = 22.4. mL 7. (a) mole mass KCl = 39.10 + 35.45 = 74.55 Mass KCl (dissolved at 25°C) = 36 gram Moles KCl = 36/74.55 = 0.48 mole [KCl] = moles/volume = 0.48/0.100 = 4.8 molar (b) [KCl] = (50/74.55) ÷ 0.100 = 6.7 molar [MgCl2] = (50/95.21) ÷ 0.100 = 5.25 = 5.3 molar [chloride ion] = 6.7 + 2 × 5.2 = 17.2 molar (c) At 20°C the solubility of KCl is approximately 33 g (from graph) so 17 g of KCl precipitates from a solution that contained 50 g of KCl. At 20°C the solubility of MgCl2 is approximately 54 g (from graph) so no MgCl2 will precipitate from a solution that only contains 50 g. (d) Draw a filter funnel and a beaker (don’t forget the filter paper) (e) Add ether – KCl dissolves. Filter the mixture (the liquid is a mixture (solution of KCl and ether) the solid is pure MgCl2. Allow the liquid to evaporate as this will leave pure KCl as a white solid. 8. (a) O2(g) + H2(g) H2O2(l) ∆Hf = –188 kJ/mol (b) H2O2(l) H2O2(aq) ∆Hsol = –191 – (–188) = –3 kJ/mol (c) heat = 3 × 1.50 = 4.5 kJ (d) 2H2O2(aq) O2(g) 2H2O(l) ∆H = {0 + 2 × (–285)} – {2 × –191} = – 188 kJ/mol (e) Moles H2O2 = 0.500 × 0.100 = 0.0500 mole Heat released = ½ × 0.050 × 188 = 4.70 kJ releases Heat = m×SH×∆T ∴ ∆T = 4700/(500 × 4.18) = 2.25 oC Student Number: __________________ Family Name: __________________ Other Names: __________________ Class Group: __________________ Transition Program Chemistry Trial Mid–Program Examination C Time Allowed: 1.5 hours Reading Time: 5 minutes TrialMIDProgram (Paper C) Question Possible Mark 1 6 2 6 3 6 4 6 5 6 6 10 7 10 8 10 TOTAL 60 F O U N D A T I O N S T U D I E S UNSW Foundation Studies UNSW Global Pty Limited UNSW Sydney NSW 2052 Australia Copyright 2019 All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner. DIRECTIONS TO CANDIDATES 1. Check that your copy of the examination paper is complete. This examination contains ten questions: • Questions 1 to 5 @ 6 marks each • Questions 6 to 8 @ 10 marks each The total marks available are 60. 2. Ensure that you have in the room: • a black or blue pen for writing answers • an electronic calculator • no unauthorised material 7. Five (5) minutes reading time is allowed. During reading time do not write at all. 8. Ensure that your name and examination number are clearly printed on the front page on the blank answer booklet. 5. Answer all questions in the answer booklet. 6. Hand in the answer booklet to the supervisor at the end of the examination. BLANK PAGE 1. Sodium hydrogencarbonate is manufactured by the Solvay Process. This consists of four steps: (i) Calcium carbonate is heated until it changes into calcium oxide and carbon dioxide. (ii) Calcium oxide reacts with ammonium chloride forming calcium chloride, ammonia and water. (iii) Ammonia, water and carbon dioxide combine to produce ammonium hydrogencarbonate. (iv) Ammonium hydrogencarbonate and sodium chloride react together to form sodium hydrogencarbonate and ammonium chloride. (a) (i) Write a balanced chemical equation for the reaction described in step (i) …………………………………………………………………………………………. (ii) Write a balanced chemical equation for the reaction described in step (ii) …………………………………………………………………………………………. (b) Determine the limiting reagent when 50.0kg of calcium oxide and 50.0 kg of ammonium chloride are reacted as described in step (ii). …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (c) Determine the maximum mass of ammonia that can be produced if 10.0 moles of calcium oxide is reacted with 10.0 moles of ammonium chloride as described in reaction (ii) …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. 2. Two glass flasks have volumes and temperatures as shown. Flask X Flask Y Flask X contains 1.50 moles of carbon dioxide gas. Flask Y contains 22.0 grams of dinitrogen monoxide. Evaluate the following ratios: (a) (mass of gas in X) ÷ (mass of gas in Y) …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (b) (number of gas molecules in X) ÷ (number of gas molecules in Y) …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (c) (pressure of gas in X) ÷ (pressure of gas in Y) …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. 6.0 litres 50 oC 2.0 litres 50 oC 3. When heated, aluminium and carbon react together producing aluminium carbide. Several samples of this compound were analysed and the results are shown on the following graph. (a) Calculate the greatest mass of aluminium carbide which could be obtained by heating 15.0g aluminium with 15.0g carbon. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (b) Using the data on the graph determine the simplest (empirical) formula for aluminium carbide. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (c) When water is added to aluminium carbide it forms aluminium hydroxide and methane. Write a balanced chemical equation for this reaction. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. 0 2.0 4.0 6.0 8.0 10.0 Mass of aluminium in aluminium carbide (g) 4.0 3.0 2.0 1.0 0 Mass of carbon in aluminium carbide (g) 4. The minerals nitre (sodium nitrate) and saltpetre (potassium nitrate) are sometimes found mixed together. (a) Sodium nitrate is soluble in alcohol but potassium nitrate is not. In a series of numbered steps briefly explain how you would obtain a pure solid sample of sodium nitrate from a mixture of the two compounds. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (b) The following graph shows their solubilities in water (in grams of solid present in 100 mL of saturated solution) at various temperatures. Calculate the molar concentration of a saturated potassium nitrate solution at 20oC. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (c) 100.0 mL of a hot aqueous solution contains 50.0 grams of sodium nitrate and 50.0 grams of potassium nitrate dissolved in it. If this solution was cooled to 20 oC calculate the mass of each compound which would crystallise out of the solution. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. 5. The following diagram shows some energy relationships (in kilojoules per mole) involving several compounds of barium. Use these data to calculate: (a) the enthalpy of formation (∆Hf, in kJ/mole) for barium oxide. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (b) the enthalpy of solution (∆Hsol, in kJ/mole) for barium hydroxide. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (c) the enthalpy change (∆H, in kJ) for the reaction of barium hydride with water forming solid barium hydroxide and hydrogen gas. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. 6. (1) (2) 25.0 mL HCl vs KOH 25.0mL HF vs KOH mL KOH added mL KOH added The titration curves were obtained when 0.100 M solutions of hydrochloric acid and hydrofluoric acid were titrated separately with two different samples of the same potassium hydroxide solution. (a) Briefly explain why the initial pH of the HF solution was higher than that of the HCl solution. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (b) Write a net ionic equation to represent the reaction which occurs in titration (1). …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (c) Suggest a suitable indicator for titration (2) and briefly describe its colour change during the reaction. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (d) Calculate the molar concentration of the potassium hydroxide solution. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (e) Briefly explain why the same volume of potassium hydroxide was needed to cause neutralisation in both titrations. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. 7. Argon was discovered in 1894. J. Rayleigh and W. Ramsay burned magnesium in air producing a solid mixture of magnesium oxide and magnesium nitride (Mg3N2). Almost 1% of the air was left unchanged. They concluded that this unreactive gas was a new element which they named argon. (a) Write a balanced equation for the reaction which produced magnesium nitride. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (b) Briefly describe how they could prove that the unreactive gas was a new element, and not an impurity or some strange new compound of nitrogen and oxygen. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (c) Rayleigh and Ramsay found that argon was monatomic with a density of 1.717 g/L at S.T.P. Using their data calculate the atomic weight of argon in a.m.u. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (d) The modern value for the atomic weight of argon is 40.0 a.m.u. Using this value calculate the mass of one argon atom in grams. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (e) Briefly explain why argon (atomic weight = 40.0 a.m.u.) is placed before potassium (atomic weight = 39.1 a.m.u.) in the periodic table. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. 8. Addition of 1.00 M hydrochloric acid solution to magnesium produces a colourless gas. (a) Write a net ionic equation to represent this reaction. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (b) Calculate the smallest volume of 1.00 M hydrochloric acid solution which would react completely with 5.00 grams of solid magnesium. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (c) Would the volume of gas produced at 25 oC and 100 kPa be smaller, larger or the same if 5.00 grams of aluminium were used instead of 5.00 grams of magnesium? Briefly justify your answer. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (d) A colourless gas is also produced if magnesium carbonate is used instead of magnesium. Write a net ionic equation to represent this reaction. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. (e) Would the volume of gas produced at 25 oC and 100 kPa be smaller, larger or the same if 5.00 grams of magnesium were reacted with excess 1.00 M HCl or if 5.00 g of magnesium were reacted with excess 0.500 M HF? Briefly justify your answer. …………………………………………………………………………………………. …………………………………………………………………………………………. …………………………………………………………………………………………. UNSW Foundation Studies UNSW Foundation Studies UNSW Sydney NSW 2052 Australia T: 61 2 9385 5396 | F: 61 2 9662 2651 | E:
[email protected] | W: www.ufs.unsw.edu.au UNSW Foundation Studies is an education group of UNSW Global Pty Limited, a not-for-profit provider of education, training and consulting services and a wholly owned enterprise of the University of New South Wales UNSW Foundation Studies programs are delivered under UNSW CRICOS Provider Code: 00098G UNSW Global Pty Limited ABN 62 086 418 582 DATA SHEET Avogadro's constant, NA 6.02 × 1023 mol–1 Universal Gas Constant, R 8.314 J K–1 mol–1 Specific heat of water 4.180 J g–1 K–1 Molar volume of ideal gas: at 101.3 kPa and 0 oC (S.T.P.) 22.41 L mol–1 at 101.3 kPa and 25 oC (R.T.P.) 24.46 L mol–1 TRIAL MIDPROGRAM CHEMISTRY EXAM PAPER C: ANSWERS: 1. (a) (i) CaCO3 → CaO + CO2 (ii) CaO + 2NH4Cl → CaCl2 + 2NH3 + H2O (b) Moles CaCO3 = 50,000/(40.1 + 16.0) = 891 mol Moles NH4Cl = 50,000/(14.0 + 4.0 + 35.5) = 935 mol Therefore CaO is in excess and ammonium chloride is limiting (as 891 moles of CaO would require 2 × 891 = 1782 mole of NH4Cl for complete reaction and there is only 935 mole of CaO) (c) 10.0 moles of CaO would require 20.0 moles NH4Cl for complete reaction hence NH4Cl is limiting in this case. Mole NH3 = Moles NH4Cl mass NH3/17.0 = 10.0 mass NH3 = 170 g (3 significant figures) 2. (a) mass X : mass Y = (1.50 x 44.0) : 22.0 = 3.00 (b) nx : ny = (mol in X) : (mol in Y) = (1.50) : (22.0 / 44.0) = 3.00 (c) P = nRT/V So Px : Py = (1.5 RT/ 6.0) : (0.5 RT/ 2.0) = 1.0 (NOTE: To score full marks in this type of question it is necessary to establish the relationship between moles and pressure) 3. (a) From the graph 6.0 g Al combines with 2.0 g C So 15.0 g Al combines with 5.0 g C to form 20.0 g aluminium carbide (b) (mol C atoms) : (mol Al atoms) = (2.0 / 12.0) : (6.0 / 27.0) = 0.75 = ¾ So empirical formula is Al4C3 (c) Al4C3 + 12H2O → 4Al(OH)3 + 3CH4 4. (a) (i) Add alcohol to the mixture to dissolve the sodium nitrate. (ii) Filter the mixture from step (i) (iii) Crystallise solid sodium nitrate from the filtrate of step (ii). (b) Mass of dissolved potassium nitrate in 100 mL is approximately 31 g hence 1 litre is 310 grams (a range of values around 30 to 35 g accepted). So molar concentration = 310 / 101.1 = 3.1 M (c) 19 grams (approximately) of potassium nitrate; zero sodium nitrate. 5. (a) ∆Hf = –554 kJ/mol (b) ∆Hsol = – 48 kJ/mol (c) BaH2 (s) + 2H2O (l) → Ba(OH)2 (s) + 2H2 (g) ∆H = (sum of ∆Hf of products) – (sum of ∆Hf of reactants) = (–945 + 0) – (–179 + 2 × –285) = –196 kJ 6. (a) Only a small percentage of HF molecules dissociate into H+(aq) and F– (aq) ions so that the H+(aq) concentration in 0.10 M HF solution is very much less than 0.10 M; therefore its pH is greater than 1. However essentially all the HCl molecules dissociate so that the H+(aq) concentration in the HCl solution is 0.10 M and its pH is 1. (b) HCl is a strong acid and KOH is a strong base hence spectator ions of Cl– and K+ can be cancelled: The required ionic equation is: H+ (aq) + OH– (aq) → H2O (l) (c) Phenolphthalein is suitable; it is colourless at the start of the titration and changes to faint pink at the equivalence point. (d) Moles of KOH = moles of acid = (25.0 × 10-3 × 0.10) = 2.5 × 10-3 Concentration of KOH = mol / L = (2.5 × 10-3) / (20.0 × 10-3) = 0.125 M (e) 1 mole of KOH reacts with 1 mole of acid whether the acid is strong (HCl) or weak (HF). Because the quantities of the two acids are equal so are the quantities of KOH required to neutralise them. 7. (a) N2 + 3 Mg → Mg3N2 (b) The emission spectrum of the gas would be analysed and compared with those of known elements. (c) Molar mass = 22.41 x 1.717 = 38.53 grams. So their calculated value for the atomic weight = 38.53 u (which was obviously inaccurate). (d) 6.02 × 1023 atoms = 40.0 g Therefore 1 atom = 40.0 ÷ 6.02 × 1023 = 6.64 × 10-23 g (e) Argon is placed in the same periodic group (noble gases) as the elements it resembles. Potassium has similar properties to the alkali metals and is grouped with them. 8. (a) Mg(s) + 2 H+(aq) → H2(g) + Mg2+(aq) (b) Moles Mg = 5.00/24.3 = 0.206 mol ∴ Moles H+ = 2 × 0.206 = 0.412 mol ∴ Volume 1.00 M HCl = 0.412 L = 412 mL (c) Al(s) + 3 H+(aq) → 3/2 H2(g) + Al3+(aq) Moles Al = 5.00/27.0 = 0.185 ∴ Moles H+ = 3 × 0.185 = 0.555 mol The calculations are needed to justify that Al will react with the greater volume o acid (as more moles of acid will react with 5.00 g of Al than with 5.00 g of Mg) (d) MgCO3(s) + 2 H+(aq) → CO2(g) + Mg2+(aq) + H2O(l) (e) Magnesium is in excess in each case hence the same volume of gas (hydrogen) is produced in each case (as yields are always based on limiting reagents). Transition Program Chemistry Sample Final Examination A Time Allowed: 2.5 hours Reading Time: 5 minutes Paper A Question Mark 1 6 marks 2 6 marks 3 6 marks 4 6 marks 5 6 marks 6 6 marks 7 6 marks 8 6 marks 9 6 marks 10 6 marks 11 10 marks 12 10 marks 13 10 marks 14 10 marks 15 10 marks 16 10 marks Total Student Number: _____________________ First Name: _____________________ Family Name: _____________________ Class Group _____________________ UNSW Foundation Studies UNSW Global Pty Limited UNSW Sydney NSW 2052 Australia Copyright 2017 All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner. DIRECTIONS TO CANDIDATES 1. Check that your copy of the examination paper is complete. This examination contains ten questions: • Questions 1 to 10 @ 6 marks each • Questions 11 to 16 @ 10 marks each The total marks available are 120. 2. Ensure that you have in the room: • a black or blue pen for writing answers • an electronic calculator • no unauthorised material 3. Five (5) minutes reading time is allowed. During reading time do not write at all. 4. Ensure that your name and examination number are clearly printed on the front page on the blank answer booklet. 5. Answer all questions in the answer booklet. 6. Hand in the answer booklet to the supervisor at the end of the examination. 1. A pure sample of the blue mineral chalcanthite was analysed according to the following scheme. (a) Write a net ionic equation to represent the reaction in Step 3. (b) Write a net ionic equation to represent the reaction in Step 4. (c) Using the data above determine the chemical formula of chalcanthite. 2. The following table shows boiling points of fluorine and the hydrogen halides. Substance Fluorine Hydrogen fluoride Hydrogen chloride Hydrogen bromide Hydrogen iodide Boiling point (K) 85 292 188 206 238 With reference to the intermolecular forces involved, briefly explain: (a) the trend in boiling points among hydrogen chloride, hydrogen bromide and hydrogen iodide. (b) the difference in the boiling points of hydrogen fluoride and hydrogen chloride. (c) the difference in the boiling points of fluorine and hydrogen chloride. 3. The list below contains three pairs of isomers. Draw the structural formula of each compound listed and briefly describe a quick, easily visible chemical test you could perform in the laboratory to distinguish between each member of the pair. (Balanced equations are unnecessary) (a) butanal and butanone (b) butanoic acid and methyl propanoate (c) 1-butene and cyclobutane 4. The diagram below shows details of a fuel cell developed by N.A.S.A. in 1979. (a) Write an ionic equation to represent the overall reaction occurring in this cell. (b) Which electrode (left or right on the diagram) is the cathode and what is its polarity (negative or positive)? (c) Calculate the E.M.F. generated by this cell under standard conditions. 5. Flask A contains 10.0 grams of argon gas at 50 oC. Flask B contains 5.00 grams of propyne (C3H4) gas at 50 oC. The tap between them is closed. (a) Evaluate the ratio: (number of gas particles in A) / (number of gas particles in B). (b) Evaluate the ratio: (average speed of gas particles in A)/ (average speed of gas particles in B). (c) The tap was opened and the two gases mixed. Predict the reading on the pressure gauge at 50 oC. 6. In 1911, H. Geiger and E. Marsden fired a beam of alpha particles from a radioactive material at a very thin sheet of gold. They observed that almost all the alpha particles went straight through the gold but a few were deflected through large angles. (a) Briefly explain why most of the alpha particles went straight through the gold atoms. (b) Briefly explain what caused a few of the alpha particles to be scattered. (c) In 1920, J. Chadwick repeated this experiment using an equivalent thickness of silver in place of the gold. He obtained similar results but found that the alpha particles were scattered less by the silver than by gold. Briefly explain this observation. Fluorescent screen Beam of alpha particles Source Thin sheet of gold foil 7. Briefly outline, with relevant net ionic equations, how you would convert: (a) solid nickel carbonate to a solution of nickel chloride. (b) nickel chloride solution to solid nickel hydroxide. (c) solid nickel hydroxide to a solution of nickel nitrate. (If convenient, you may use a flow chart to answer this question) 8. The pH of blood is maintained at 7.4 by several chemical reactions in the body. One of these is the reversible reaction between sodium dihydrogenphosphate (NaH2PO4) and sodium hydrogenphosphate (Na2HPO4) : H2PO4–(aq) ⇌ H+(aq) + HPO42–(aq) (a) Calculate the hydrogen ion concentration in a solution whose pH is 7.4. (b) What term describes a mixture of chemicals which maintains a solution at a particular pH ? (c) Briefly explain how this mixture is able to absorb small numbers of H+(aq) or OH– (aq) ions without changing the pH. 9. (a) Draw a Lewis (electron dot) formula to represent an ammonia molecule and predict its shape using the V.S.E.P.R. theory. (b) Draw a Lewis (electron dot) formula to represent a boron trifluoride molecule and predict its shape using the V.S.E.P.R. theory. (c) When mixed, ammonia and boron trifluoride undergo an addition reaction to form a compound of formula NH3BF3. Draw a structural formula to represent this compound and name the type of bond which joins the ammonia and boron trifluoride molecules together. 10. Cyanic acid (HOCN) is a weak acid with dissociation constant (Ka) = 1.2 × 10–4 (a) Write a net ionic equation to represent the dissociation of cyanic acid in aqueous solution. (b) Write down the formula for the conjugate base of cyanic acid. (c) Calculate the pH of a 0.10 M cyanic acid solution. (d) Calculate the percentage of cyanic acid molecules which are dissociated in a 0.100 M solution of the acid. (e) Pure cyanic acid changes into its isomer: H – O – C≡N(g) O = C = N – H (g) Use average bond energies to predict the enthalpy change (in kJ) for this reaction. 11. A saturated solution of magnesium hydroxide has a pH of 10.50 and is known as “milk of magnesia”. (a) Calculate the hydroxide ion concentration in milk of magnesia. (b) Write down an expression for the solubility product constant (KSP) of magnesium hydroxide and determine its value. (c) Calculate the aqueous solubility (in grams/litre) of magnesium hydroxide. (d) When 1M ammonia solution is added to milk of magnesia a white precipitate forms. Briefly explain the formation of this precipitate using appropriate ionic equations. (e) The precipitate in (d) dissolves when mixed with a 1M ammonium chloride solution. Using Le Chatelier’s Principle briefly explain why this occurs. 12. (a) The element rubidium exists naturally as a mixture of two isotopes: rubidium–85 and rubidium-87. Calculate the percentage of each isotope in a sample of the metal. (b) Rubidium reacts with bromine according to: 2Rb + Br2 ⇌ 2RbBr (m. pt. = 39oC) (m. pt. = –7oC) (m. pt. = 699oC) Briefly explain why the melting point of rubidium bromide is much higher than that of rubidium or bromine. (c) Sizes of the particles involved in the reaction in (b) are tabulated below: Element Atomic radius (nm) Ionic radius (nm) Rubidium 0.248 0.147 Bromine 0.114 0.195 Briefly explain why the radius of rubidium decreases and the radius of bromine increases during this reaction. (d) The first ionisation energy of bromine is 11400 kJ/mol. Writer a thermochemical equation to represent the first ionisation energy of bromine. 13. The amount of sulfur dioxide in polluted air can be measured by reacting it with potassium permanganate solution which undergoes the following half reaction: MnO4–(aq) + 8H+(aq) + 5e– Mn2+(aq) + 4H2O(l) (a) Sulfur dioxide is converted into sulfuric acid. Write an ionic half equation to represent this conversion and state whether it is oxidation or reduction. (b) Briefly explain why no special indicator is needed to detect the end point of the reaction between sulfur dioxide and potassium permanganate solution. (c) In one analysis, 25.0 mL of 0.0100 M potassium permanganate solution reacted completely with all the sulfur dioxide present in 10.0 litres of polluted air at R.T.P. Calculate the number of moles of sulfur dioxide involved in this reaction. (d) Calculate the concentration (in mg/L) of sulfur dioxide in the polluted air. (e) If barium chloride solution was added to the mixture produced in (c) calculate the mass of barium sulfate which would be precipitated. 14. Methylcyclopentane and cylcohexane form an equilibrium mixture according to the following equation: Methylcyclopentane ⇌ Cyclohexane At 25oC the above reaction forms an equilibrium mixture which contains 87.5% cyclohexane and 12.5% methylcyclopentane. (a) Rewrite the equation showing correct structural (graphic) formulas for both compounds. (b) Evaluate the equilibrium constant (Kc) for this reaction at 25 oC. (c) 4.00 moles of methylcyclopentane were added to a 1.00 litre container which was then sealed. The above reaction occurred and was allowed to reach equilibrium at 25oC. Calculate the number of moles of methylcyclopentane which were present in the container at equilibrium. (d) At 75oC the value of the equilibrium constant (Kc) for this reaction is 3.0. If the mixture in (c) was allowed to reach equilibrium at 75 oC how many moles of methylcyclopentane would be present ? (e) Is the reaction exothermic or endothermic ? Briefly justify your answer. 15. The diagram below shows some energy relationships (in kilojoules) among potassium, chlorine and potassium chloride. (a) Briefly explain the difference between “K+ (l) + Cl– (l)” and “K+ (aq) + Cl– (aq)”. (b) Using this chart determine the value (in kJ/mole) of: the enthalpy of atomisation (∆Ha ) of potassium the bond energy (ΒΕ) of chlorine the enthalpy of formation (∆Hf ) of potassium chloride the enthalpy of fusion (∆Hm ) of potassium chloride the enthalpy of solution (∆Hsol ) of potassium chloride the ionisation energy (I.E.) of potassium (c) Determine the value for the electron affinity of chlorine (marked “X” on the diagram). 16. The flow chart below summarises several reactions involving ethanal (acetaldehyde): (a) Draw a Lewis (electron dot) diagram to represent an ethanal molecule. (b) Briefly explain why ethanal is very soluble in water. (c) The toxic concentration of ethanal in 1 litre of air is 0.360 mg. Calculate the number of molecules present in 0.360 mg of ethanal. (d) Addition of water to ethene forms Compound X which, when heated with copper metal, changes into ethanal and hydrogen. Write a balanced equation for the conversion of X to ethanal and use oxidation numbers to prove that this is a redox reaction. (e) Ketene reacts with hydrogen to form ethanal according to : ketene gas + hydrogen gas ethanal gas (1.0 L at R.T.P.) (1.0 L at R.T.P.) (1.0 L at R.T.P.) Draw a structural formula to represent a ketene molecule and name the class to which this organic reaction belongs. DATA SHEET Avogadro's constant, NA 6.02 × 1023 mol–1 Faraday's constant, F 96486 C mol–1 Universal Gas Constant, R 8.314 J K–1 mol–1 Specific heat of water 4.180 J g–1 K–1 Molar volume of ideal gas: at 101.3 kPa and 0oC (S.T.P.) 22.41 L mol–1 at 101.3 kPa and 25oC (R.T.P.) 24.46 L mol–1 Standard enthalpies of formation (∆Hf ): CO (g) –111 kJ mol–1 CH4 (g) – 75 kJ mol–1 CH3CHO (g) –166 kJ mol–1 Average bond energies: O –H 463 kJ mol–1 C – O 358 kJ mol–1 C ≡ N 891 kJ mol–1 C = O 745 kJ mol–1 C = N 615 kJ mol–1 N – H 391 kJ mol–1 Standard reduction potentials: Cr3+(aq), Cr2+(aq) – 0.41 V Fe3+(aq), Fe2+(aq) + 0.77 V Answers to Transition Final Exam - Sample A 1. (a) Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq) (b) Ba2+ (aq) + SO42– (aq) BaSO4 (s) (c) mol Cu : mol BaSO4 : mol H2O = (0.64/63.5) : (2.33/233.4) : (0.90/18.0) = 0.010 : 0.0100 : 0.050 So formula of chalcanthite is CuSO4·5H2O (Check: 0.0100 mole CuSO4.5H2O does weigh 2.50 g) 2. (a) HCl is the most polar of these three molecules. HBr and HI are less polar.The boiling points for this series of compounds suggests dispersion forces are more important than dipole-dipole forces. The boiling points increase because the strength of the dispersion forces between the molecules becomes progressively larger as the number of electrons in each increases. (b) Hydrogen fluoride and hydrogen chloride are both polar molecules, however hydrogen fluoride is much more polar than hydrogen chloride (difference in electronegativity. Hydrogen fluoride has hydrogen bonds between its molecules which result in a higher boiling point than in hydrogen chloride which does not form hydrogen bonds. (c) Fluorine is non polar covalent molecule, hence only dispersion forces act between molecules. Hydrogen chloride is a polar covalent molecule thus both dispersion and dipole/dipole forces act between its molecules. Both fluorine and hydrogen chloride have the same number of electrons so their dispersion forces are of similar strength however hydrogen chloride also has dipole forces between its molecules so has a higher boiling point. 3. (a) Butanal (CH3–CH2–CH2–CHO) would change the colour of potassium dichromate solution from orange to green; butanone (CH3–CH2–CO–CH3) would not. (b) Butanoic acid (CH3–CH2–CH2–COOH) would give bubbles of carbon dioxide when added to a solution of sodium carbonate. methyl propanoate (CH3–O–CO–CH2–CH3) would not form bubbles when added to a solution of sodium carbonate. (c) 1–butene (CH2=CH-CH2-CH3) would immediately change bromine solution from brown to colourless; cyclobutane { (CH2)4} would not. 4. (a) Fe3+(aq) + Cr2+ (aq) Cr3+ (aq) + Fe2+ (aq) (b) The cathode is on the left hand side and is positive. (c) E°Cell = E°cathode - E°anode = +0.77 – (–0.41) = +1.18 V 5. (a) mol in A : mol in B = (10.0 / 40.0) : (5.00 / 40.0) = 2.0 (b) Temperature is a measure of average kinetic energy. The temperatures of each gas are equal therefore the average molecular kinetic energies are the same; since the masses of argon and propyne particles are both 40.0 amu the ratio of their average speeds is 1.0. (c) P = nRT/V = ((10.0 + 5.00)/ 40.0) × 8.314 × 323 / 10.0 = 101 kPa 6. (a) Most of the volume of a gold atom is empty space so that the alpha particles met no obstacles in travelling through it. (b) At the centre of an atom is a small, positively charged nucleus which repels any alpha particles (also positively charged) which come near it. (c) The positive charge on a silver atom’s nucleus is +47 and this is less repulsive to alpha particles than a gold nucleus whose charge is +79. 7. (a) Add hydrochloric acid to solid nickel carbonate: NiCO3 (s) + 2H+ (aq) Ni2+ (aq) + H2O (l) + CO2 (g) (b) Add sodium hydroxide to nickel chloride solution then filter: Ni2+ (aq) + 2OH– (aq) Ni(OH)2 (s) (c) Add nitric acid to nickel hydroxide: 2H+(aq) + Ni(OH)2 (s) Ni2+(aq) + 2H2O (l) 8. (a) [H+] = 10–pH = 10 –7.4 = 4.0 × 10–8 M (b) Buffer (c) Added H+ (aq) ions react with HPO42– (aq) ions so they are removed and cannot alter the pH. Added OH– (aq) ions react with H2PO4– (aq) ions to form HPO42– (aq) so they are removed and cannot alter the pH. Added acid is removed by reaction with HPO42–: HPO42– (aq) + H+ (aq) H2PO4– (aq) Added base is removed by reaction with H2PO4–: H2PO4– (aq) + OH– (aq) HPO42– (aq) Alternative answer and reasoning for part (c): The equilibrium mixture that forms with these ions is represented by the following equation: H2PO4– (aq) ⇌ HPO42– (aq) + H+ (aq) Le Chatelier’s principle predicts that the system will shift to minimise any disturbance therefore adding hydrogen ion causes the system to shift to the left to minimise the increase in [H+] thus minimising the change in pH. Similiarly, adding hydroxide ion causes a decrease in [H+]. Le Chatelier’s Principle predicts the system will shift to the right to minimise the decrease in [H+] thus minimising the change in pH. 9. (a) The Lewis diagram for ammonia: The nitrogen in ammonia is directly attached to three H atoms and to a lone pair giving an AB3E system (VSEPR theory) hence ammonia is pyramidal with a HNH bond angle of approximately 109 – 2 = 107o. (b) The Lewis diagram for boron trifluoride: The boron in boron trifluoride is only directly attached to three fluorine atoms giving an AB3 system (VSEPR theory) hence boron trifluoride is trigonal (flat triangular). (c) The link between the two molecules is a dative covalent bond. 10. (a) HOCN (aq) ⇌ H+ (aq) + OCN– (aq) (b) OCN– (c) A RICE table is not necessary but may help you to get the right answer (and help the marker to give you marks even if your final answer is incorrect) Reaction HOCN (aq) ⇌ H+ (aq) + OCN– (aq) Initial 0.10 0 0 Change – x + x + x Equilibrium 0.10 – x ≈ 0.10 + x + x If x << 0.10 than 0.10 – x ≈ 0.10 Ka = [H+][ OCN–]/[HOCN] = x2/0.10 = 1.2 × 10–4 ∴ x = [H+] = [ H+] = 3.5 × 10–3 ∴ pH = –log10[1.2 × 10–4] = 2.5 (d) Dissociation = (3.5 × 10–3 / 0.10 ) × 100 = 3.5 % (e) H – O – C≡N(g) O = C = N – H (g) ∆H = BEreactants – BEproducts ∆H = (463 + 358 + 891) – (745 + 615 + 391) ∆H = –39 kJ/mol 11. (a) pOH = 3.5 ∴ [OH–] = 10–3.5 = 3.16 × 10–4 M (b) [Mg2+] = ½ [OH–] Ksp = [Mg2+] × [OH–]2 = (3.16 × 10–4/2) × (3.16 × 10–4)2 = 1.6 × 10–11 (c) Molar solubility = (3.16 × 10–4 / 2) = 1.6 × 10–4 M Mass solubility = 1.6 × 10–4 × 58.3 = 9.3 × 10–3 g/L (d) Ammonia is a weak base and forms hydroxide ions in solution: NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq) System 1 Milk of magnesia is a saturated solution of magnesium hydroxide. Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH–(aq) System 2 Le Chatelier’s principle predicts an increase in [OH–] causes system 2 to shift to the left to minimise the increase in [OH–] and a precipitate of Mg(OH)2 forms. (e) Addition of NH4+(aq) from NH4Cl drives the system 1 in part (d) above to the left reducing the [OH–]. Le Chatelier’s principle predicts that system 2 will shift to the right to minimise the decrease in [OH–] hence Mg(OH)2 dissolves. 12. (a) In 100 Rb atoms there are y 85Rb atoms and (100 – y) 87Rb atoms. Their total mass is 85y + 87(100 – y) = 8550 ∴ y = 75 Hence 75% 85Rb and 25% 87Rb if Ar = 85.5 or 76.5% 85Rb and 23.5% 87Rb if Ar = 85.47 (b) Solid rubidium bromide is held together by ionic bonds between the Rb+ and Br– ions. These are stronger than the metallic bonds in solid rubidium or the dispersion forces between bromine molecules so RbBr has the highest melting point. (c) The size of a particle is determined by the distance of its outermost electron from the nucleus. When Rb forms Rb+ it loses its single electron in the outer (5th) shell and so shrinks in size. When Br forms Br – it gains an electron which repels all the other electrons in the outer shell thus causing the particle to expand in size. (d) Br(g) Br+(g) e–(g) ∆H = 11400 kJ/mol 13. (a) SO2 (aq) + 2H2O (l) ⇌ SO42– (aq) + 4H+ (aq) + 2e–; oxidation. (b) During the reaction MnO4– ions (purple) change to Mn2+ ions which are colourless. The colour change that indicates the equivalence point is the appearance of a faint pink colour indicating an excess of MnO4– ions. (c) 2MnO4–(aq) + 5SO2(aq) + 2H2O(l) 2 Mn2+(aq) + 5SO42–(aq) + 4H+(aq) Mol MnO4– = 0.010 x 0.0250 = 2.5 × 10–4 Mol SO2 = 5/2 × 2.50 × 10–4 = 6.25 × 10–4 (d) Mass of SO2 = 64.1 × 6.25 × 10–4 = 0.0400 grams [SO2] = 0.0400 × 1000 / 10.0 = 4.00 mg / L (e) Mol BaSO4 = mol SO2 = 6.25 × 10–4 ∴ mass = 233.4 × 6.25 × 10–4 = 0.146 g 14. (a) methylcyclopentane ⇌ cyclohexane (b) Kc = [cyclohexane] / [ methylcyclopentane ] = 87.5 / 12.5 = 7.0 (c) A RICE table is not necessary, but may be used: Reaction Methylcyclopentane ⇌ Cyclohexane Initial 4.0 0.00 Change – x + x Equilibrium 4.0 – x x Kc = x/(4.0 – x) = 7.0 ∴ x = 3.5 M [Methylcyclopentane] = 4.00 – 3.5 = 0.50 M ∴Moles Methylpentane =0.50 mol (as volume is 1.00 litre). (d) The same approach as used in part (c), with the Kc = 3.0: Kc = x/(4.0 – x) = 3.0 ∴ x = 3.0 M [Methylcyclopentane] = 4.00 – 3.0 = 1.0 M ∴Moles Methylpentane =1.0 mol (as volume is 1.00 litre). (e) Temperature is the only variable that changes the value of the equilibrium constant. In increasing the temperature from 25oC to 75oC the equilibrium constant decreases from 3.5 to 3.0 indicating that Le Chatelier shift has been towards reactants which is consistent with heat being a product and the reaction, as written, being exothermic. 15. (a) “K+(l) + Cl–(l)” refers to liquid (molten) KCl which is a pure compound (KCl) that has been heated to a temperature above its melting point. “K+(aq) + Cl–(aq)” refers to an aqueous solution of KCl which is a mixture containing KCl and water. (b) i. ∆Hatomisation (KCl) = +90 kJ / mol ii. Bond Energy(Cl–Cl) = +242 kJ / mol iii. ∆Hformation (KCl(s)) = – 436 kJ / mol iv. ∆Hmelting (KCl) = +27 kJ / mol v. ∆Hsolution (KCl(s)) = +19 kJ / mol vi. First ionisation energy (K) = + 425 kJ / mol (c) (X + 710) = (121 + 425 + 90 + 436) ∴ X = – 362 kJ (exothermic) 16. (a) (b) Both water and ethanal are small polar covalent molecules. Ethanal can form strong dipole/dipole bonds (called hydrogen bonds) with water molecules therefore ethanal is soluble in water. (c) Moles ethanal = 0.360 × 10–3/44.0 = 8.18 × 10–6 mol Number of molecules = 8.18 × 10–6 × 6.02 × 1023 = 4.93 × 1018 molecules (d) X is ethanol: C2H6O C2H4O + H2 Oxidation numbers: “C” changes from –2 to –1; “H” changes from +1 to 0 or C2H6O + ½ O2 C2H4O + H2O) Oxidation numbers: “C” changes from –2 to –1; “O” changes from 0 to –2 (e) 1 × Ketene + H2 C2H4O The molecular formula for ketene must be C2H2O: Ketene is CH2=C=O; the reaction is an addition reaction. Transition Program Chemistry Sample Final Examination B Time Allowed: 2.5 hours Reading Time: 5 minutes Paper B Question Mark 1 6 marks 2 6 marks 3 6 marks 4 6 marks 5 6 marks 6 6 marks 7 6 marks 8 6 marks 9 6 marks 10 6 marks 11 10 marks 12 10 marks 13 10 marks 14 10 marks 15 10 marks 16 10 marks Total Student Number: _____________________ First Name: _____________________ Family Name: _____________________ Class Group _____________________ UNSW Foundation Studies UNSW Global Pty Limited UNSW Sydney NSW 2052 Australia Copyright 2019 All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner. DIRECTIONS TO CANDIDATES 1. Check that your copy of the examination paper is complete. This examination contains ten questions: • Questions 1 to 10 @ 6 marks each • Questions 11 to 16 @ 10 marks each The total marks available are 120. 2. Ensure that you have in the room: • a black or blue pen for writing answers • an electronic calculator • no unauthorised material 3. Five (5) minutes reading time is allowed. During reading time do not write at all. 4. Ensure that your name and examination number are clearly printed on the front page on the blank answer booklet. 5. Answer all questions in the answer booklet. 6. Hand in the answer booklet to the supervisor at the end of the examination. 1. The compound “salt of Saturn” reacts with hydrochloric acid forming acetic acid and a precipitate of lead chloride. (a) Write a net ionic equation to represent the precipitation reaction. (b) The acetic acid solution required 28.6 mL of 0.700 M sodium hydroxide solution for complete neutralisation. Calculate the number of moles of acetic acid which were produced. (c) Write down the chemical formula of the compound “salt of Saturn”. 2. Three organic compounds (X, Y and Z) have the following properties: Compound Molecular weight (u.) Effect on blue litmus Effect on KMnO4 solution X 60.0 Change to red No change Y 60.0 No change Changes to colourless Z 60.0 No change No change (a) Compound X reacts with magnesium metal to produce a gas. Write a balanced chemical equation to represent this reaction. (b) When compound Y reacts with KMnO4 solution one of the products is propanone. Draw a structural formula to represent compound Y. (c) Compound Z has a sweet smell. Name compound Z and the class of organic compounds to which it belongs. 3. The electrical conductivity of several different liquids was tested in the equipment shown below: The following results are obtained: Liquid Conductivity Pure acetic acid None Acetic acid dissolved in water Poor Pure liquid ammonia None Ammonia dissolved in water Poor Mixture of acetic acid and ammonia dissolved in water Good Using appropriate ionic equations briefly explain why: (a) acetic acid conducted a small current when dissolved in water. (b) ammonia conducted a small current when dissolved in water. (c) a mixture of acetic acid and ammonia in water conducted a large current. 4. The pH of a saturated solution of silver hydroxide (AgOH) is 10.15. (a) Calculate the molar concentration of hydroxide ions in a saturated silver hydroxide solution. (b) Evaluate the solubility product (Ksp) of silver hydroxide. (c) Silver hydroxide forms when solid silver oxide (Ag2O) dissolves in water. Calculate the maximum mass of solid silver oxide which could dissolve in 1.0 litre of water. 5. Propene and cyclopropane both react with hydrogen to form propane. When heated propane decomposes into a mixture of ethene and methane. (a) Using structural formulas write a balanced chemical equation for the conversion of cyclopropane to propane. (b) Using enthalpy of formation data write a thermochemical equation for the conversion of propene to propane under standard conditions. (c) Using bond energy data write a thermochemical equation for the conversion of propane into ethene and methane under standard conditions. Cyclopropane PropanePropene Ethene and methane Add hydrogen gas (using a catalyst) Heat Add hydrogen gas (using a catalyst) 6. The following table lists physical properties of four different substances (A-D). Substance Melting Point (oC) Boiling Point (oC) Electrical Conductivity at R.T.P. A -205 -190 None B 727 1050 None C 1490 2900 Good D -104 8 None The substances, not necessarily in order, are: • cobalt (Co) • carbon monoxide (CO) • cobalt chloride (CoCl2) • phosgene (COCl2). (a) Identify cobalt chloride and briefly justify your selection. (b) Identify phosgene and briefly justify your selection. (c) Draw a Lewis (electron dot) formula to represent a phosgene molecule and use the VSEPR Theory to predict its shape. 7. For a stomach X-ray a patient is fed a “barium meal” which is actually a suspension of solid barium sulfate. (a) Calculate the molar concentration of barium ions in a saturated aqueous solution of barium sulfate. (b) It is important to keep the concentration of barium ions as low as possible because they are poisonous to humans. A “barium meal” contains a mixture of barium sulfate and aqueous potassium sulfate. Briefly explain, using Le Chatelier’s Principle or otherwise, why the concentration of barium ions in this solution is less than that in (a). (c) The solubility of barium carbonate is similar to that of barium sulfate. However barium carbonate is never used in a barium meal because the stomach fluids are very acidic. Briefly explain, using an appropriate ionic equation, why barium carbonate cannot be used as a barium meal. 8. Consider the following flowchart: (a) Addition of acidified potassium permanganate solution to 2-butene forms a single organic product (Compound X). Name Compound X and draw its structural formula. (b) Addition of hydrogen chloride gas to 2-butene forms a single organic product (Compound Y). Name Compound Y and draw its structural formula. (c) Ozonolysis of 2-butene forms a single organic product (Compound Z). Name Compound Z and draw its structural formula. 9. One mole of a hydrocarbon (compound A) weighs 56.0 grams. A reacts with water to produce two isomeric compounds (B and C). Compound B reacts with potassium permanganate solution forming an organic acid (D). Compound C does not react with potassium permanganate solution. (a) Calculate the density (in grams per litre) of gaseous A at 120 kPa and 87 oC. (b) Write down the name of Compound D and draw a structural formula to represent its molecule. (c) Write down the name of Compound A and also the name of the class of hydrocarbons to which it belongs. KMnO4 + H2SO4 O3 then Zn + H2O HCl 2-butene Compound X Compound Y Compound Z 10. The graph below shows the boiling points (in K) of the noble gases and hydrogen halides. Name the type of interparticle forces and identify the strongest interparticle force which is responsible for: (a) keeping xenon atoms clustered together in the liquid state. (b) causing hydrogen iodide to boil at a higher temperature than xenon. (c) causing hydrogen fluoride to boil at a higher temperature than hydrogen iodide. 2 3 4 5 PERIOD 100 200 300 Xe Kr Ar Ne HF HCl HBr HI BOILING POINT 11. The table below lists some properties of elements whose names begin with the letter “b”. Element A. No. A. W. m.p.(oC) b.p.(oC) Isotopes (natural) Barium 56 137.34 714 1640 130,132,135,136,137,138 Berkelium 97 (247) 1050 unknown Beryllium 4 9.012 1280 2770 9 Bismuth 83 208.98 271 1560 209 Boron 5 10.81 2300 2550 10, 11 Bromine 35 79.91 -7 58 79, 81 (a) Estimate the percentage of boron atoms in nature which have a mass number of 11. (b) Draw a Lewis (electron dot) diagram to represent a boron bromide molecule and briefly describe its shape. (c) Write down the number of protons, electrons and neutrons present in a beryllium Be2+ ion. (d) Write down the name of the non-radioactive element from this list which has the lowest first ionisation energy. (e) Briefly explain why the melting and boiling points of barium are much lower than those of beryllium. 12. Potassium formate (methanoate) can be produced by heating carbon monoxide with sodium hydroxide: CO + KOH → KHCO2 (a) Draw a Lewis (electron dot) diagram to represent the carbon monoxide molecule. (b) Draw a Lewis (electron dot) diagram to represent the hydroxide ion (OH–) in potassium hydroxide. (c) The diagram represents a formate (methanoate) ion. Use the VSEPR theory to predict the size of the angle marked “X” and briefly justify your answer. (d) Potassium formate (methanoate) dissolves in water producing an alkaline solution. Write a net ionic equation to represent the hydrolysis reaction that occurs. (e) Show that the reaction in (d) is a Bronsted-Lowry reaction and draw a circle around each base present in the equation. 13. The diagram shows a Daniell cell in which the two electrolytes are separated by a porous container through which ions can diffuse. (a) Name the metal which forms the anode of this cell. (b) Write a balanced half equation to represent the process which occurs at the cathode of this cell. (c) Calculate the E.M.F. generated by this cell under standard conditions. (d) If this cell pushes 0.100 moles of electrons through an external circuit calculate the change in mass of the copper can and indicate whether it is an increase or decrease. (e) As the cell operates Zn2+(aq) ions diffuse out of the pot into the copper(II) sulfate solution. Briefly explain why this occurs. 14. The diagram below shows some energy relationships among sodium, fluorine and sodium fluoride. All enthalpies are in kilojoules. (a) Briefly explain the difference between “Na+(l) + F–(l)” and “Na+(aq) + F– (aq)” (b) Using this chart determine the value (in kJ/mole) of : • the enthalpy of atomisation of sodium • the bond energy of fluorine • the ionisation energy of sodium • the enthalpy of vaporisation of fluorine • the enthalpy of formation of sodium fluoride • the enthalpy of solution of sodium fluoride (c) Determine the value for the electron affinity of fluorine (marked “X” on the diagram.). 15. Lead ions react with metallic tin according to Sn (s) + Pb2+ (aq) ⇌ Pb (s) + Sn2+ (aq) ∆H = -10 kJ 100.0 mL of 0.100 M lead (II) nitrate solution were added to 2.00 g of solid tin at 25 oC. At equilibrium the mass of solid tin present in the container was 1.11 g. (a) Calculate the molar concentration of Sn2+(aq) ions in the solution at equilibrium. (b) Write down a mathematical expression relating the equilibrium constant (Kc) to the ion concentrations and evaluate Kc at 25 oC. The concentration of Sn2+(aq) was monitored over a period of time when three separate disturbances (X, Y and Z) were made to the system. The results are shown graphically below: (c) At which point (X, Y or Z) was the mixture cooled ? (d) At which point (X, Y or Z) was water added to the mixture ? (e) At which point (X, Y or Z) was some NaCl solution added to the mixture ? (Briefly justify your answers for (c), (d) and (e)). 16. (a) Flask A contains 5.00 g of hydrazine gas (N2H4) in 6.00 litres. Flask B contains 5.00 g of oxygen in 4.00 litres. Both flasks are at 127 oC. Evaluate the ratio: (number of hydrazine molecules in A) ÷ (number of oxygen molecules in B). (b) When the tap between the two flasks is opened the gases react to form nitrogen and steam. Write a balanced chemical equation to represent this reaction. (c) Using oxidation numbers show that the reaction between hydrazine and oxygen is a redox reaction. (d) Calculate the pressure gauge reading at 127 oC when the reaction is complete. (e) Calculate the partial pressure of the nitrogen gas in the mixture in (i). DATA SHEET Avogadro's constant, NA 6.02 × 1023 mol–1 Universal Gas Constant, R 8.314 J K–1 mol–1 Specific heat of water 4.180 J g–1 K–1 Molar volume of ideal gas: at 101.3 kPa and 0 oC (S.T.P.) 22.41 L mol–1 at 101.3 kPa and 25 oC (R.T.P.) 24.46 L mol–1 Standard enthalpies of formation (∆Hf): propene (gas) + 20 kJ mol–1 propane (gas) – 104 kJ mol-1 Average bond energies: C – C 348 kJ mol–1 C = C 614 kJ mol–1 Standard reduction potentials: Zn2+(aq), Zn (s) – 0.76 V Cu2+(aq), Cu (s) + 0.35 V Solubility product constant: BaSO4 (s) 1.0 × 10-10 UNSW Foundation Studies UNSW Foundation Studies UNSW Sydney NSW 2052 Australia T: 61 2 9385 5396 | F: 61 2 9662 2651 | E:
[email protected] | W: www.ufs.unsw.edu.au UNSW Foundation Studies is an education group of UNSW Global Pty Limited, a not-for-profit provider of education, training and consulting services and a wholly owned enterprise of the University of New South Wales UNSW Foundation programs are delivered under UNSW CRICOS Provider Code: 00098G UNSW Global Pty Limited ABN 62 086 418 582 Answers to Final Exam - Sample B 1. (a) Pb2+(aq) + 2Cl– (aq) → PbCl2 (s) (b) Mol of acetic acid = mol of NaOH = 0.0286 x 0.700 = 0.0200 (c) Mol PbCl2 = 2.78 / 278.1 = 0.0100 = mol Pb in the salt ∴ Pb(CH3COO)2 (Check: 0.0100 mol Pb(CH3COO)2= 3.25 g) 2. (a) X is acetic acid: 2CH3COOH + Mg → Mg(CH3COO)2 + H2 (b) Y is 2-propanol: CH3 - CH(OH) - CH3 (c) Z is methyl formate (methanoate) and is an ester. 3. (a) Acetic acid is a weak acid and dissociated to form a few ions in aqueous solution and these ions are free to move through the solution (cations to the cathode and anions to the anode) thus conducting electric charge (current) through the solution: CH3COOH (aq) → CH3COO–(aq) + H+ (aq) Alternative approach to part (a): (a) Acetic acid is a weak acid and hydrolyses to form a few ions in aqueous solution and these ions are free to move through the solution (cations to the cathode and anions to the anode) thus conducting electric charge (current) through the solution: CH3COOH (aq) + H2O(l) → CH3COO–(aq) + H3O+ (aq) (b) Ammonia is a weak base that undergoes hydrolysis in water to produce some (<100% hydrolysis) hydroxide ions and ammonium ions in solution. Ammonium ions are free to move to the cathode, hydroxide ions to the anode, and the movement of ions between the electrodes is the electric current: NH3 (aq) + H2O → NH4+ (aq) + OH– (aq) (c) Ammonia (a weak base) and acetic acid (a weak acid) react together forming ammonium acetate (an ionic compound) which is completely dissociated in solution and hence produces a large concentration of anions and cations which can freely move between the electrodes and so can conduct a large current. CH3COOH (aq) + NH3 (aq) → NH4+ (aq) + CH3COO– (aq) 4. (a) pOH = 3.85 ∴ [OH– (aq)] = 10–3.85 = 1.4 × 10–4 M (b) Ksp = [Ag+ (aq)] × [OH– (aq)] = (1.4 × 10-4)2 = 2.0 × 10-8 (c) Ag2O + H2O → 2AgOH Mol Ag2O dissolved = ½ × mol AgOH Mass Ag2O dissolved = ½ × 1.4 × 10-4 × 231.8 = 0.016 g 5. (a) (b) CH3CH=CH2(g) + H2(g) CH3CH2CH3(g) ∆Hreaction = Σ(∆Hf of products) – Σ(∆Hf of reactants) = (–104) – (+20) = –124 kJ/mol The thermochemical equation: C3H6 (g) + H2 (g) C3H8 (g) ∆H = –124 kJ/mol (c) CH3CH2CH3(g) CH2=CH2(g) + CH4(g) ∆Hreaction = Σ(Bond energy of reactants) – Σ(Bond energy of products) = (2 × 348 + 8 C–H) – (614 + 8 C–H) = +82 kJ/mol The thermochemical equation: CH3CH2CH3(g) CH2=CH2(g) + CH4(g) ∆H = –124 kJ/mol NOTE: Structural, or condensed structural, formulae must be used in the answers to parts (a), (b) and (c) of this question. 6. (a) The best approach to answering this sort of question is to classify the possible compounds as belonging to a particular lattice type (see below) and then to match the identify substance A to D based on their physical properties.. Cobalt is a metallic substance which will conduct in the solid state therefor Cobalt is substance C Carbon monoxide is a polar covalent molecule. Covalent molecules tend to boil at temperatures much less than 727 oC thus CO is likely to be either A or D. CO is a gas at room temperatures and does not become a liquid on a cold day hence D is not CO as its melting point is too high. CO is substance A. Cobalt chloride is an ionic compound and ionic compounds tend to have higher melting points than polar covalent compounds (as + to – attractions tend to be stronger than δ+ to δ– attractions).CoCl2 is substance B. (b) Phosgene is the remaining substance. Phosgene is substance D. (c) The Lewis diagram for phosgene is: This shows carbon is bonded to three other atoms, giving an AB3 system that VSEPR theory predicts will adopt a Trigonal (triangular planar) shape with bond angles of 120o about the central carbon atom. 7. (a) BaSO4(s) ⇌ Ba2+(aq) + SO42–(aq) [Ba2+] = [SO42–] and [Ba2+] × [SO42–] = 1.0 × 10-10 ∴ [Ba2+] = 1.0 × 10-5 M (b) BaSO4(s) ⇌ Ba2+(aq) + SO42–(aq) Adding potassium sulfate causes the sulfate ion concentration to increase. Le Chatelier’s principle predicts the system will shift to minimise this increase in sulfate ion concentration by moving towards the left hand side (precipitating solid) as it reaches a new equilibrium position. Alternative approach to part (b): BaSO4(s) ⇌Ba2+(aq) + SO42–(aq) [Ba2+] × [SO42-] = 1.0 × 10-10 : Adding potassium sulfate increase [SO42-] (a common ion effect). The concentration of Ba2+ decreases to ensure that equilibrium is maintained. (c) The stomach contains acidic substances and these acidic materials will react with barium carbonate: BaCO3 (s) + 2H+ (aq) Ba2+ (aq) + H2O (l) + CO2 (g) The soluble would Ba2+ (aq) ions are poisonous hence barium carbonate cannot be used as a barium meal. 8. (a) ethanoic acid: CH3COOH. (b) 2–chlorobuane: CH3CH2ClCH2CH3 (c) ethanal: CH3CHO 9. (a) Moles of A (in 1.00 litre) = PV/RT = (120 × 1.00)/(8.314 360) = 0.0400 mol Mass of A (in 1.00 litre) = 0.0400 × 56.0 = 2.24 g so density = 2.24 g/L (several alternative approaches can be used, all are correct as long as the logic is correctly followed and applied) (b) A must be C4H8; B is a primary alcohol, C is a tertiary alcohol and D is methylpropanoic acid. CH3 - CH( CH3)-COOH (c) A is methylpropene and is an alkene. 10. (a) Xenon atoms are non–polar therefor the only forces that operate between Xe atoms are dispersion forces. (b) HI is not very polar but a weak dipole/dipole force operates between its molecules as well as the dispersion force. Xenon is non–polar therefore only a dispersion force operates between xenon atoms. Hydrogen iodide is the larger species (diatomic whereas Xenon is monatomic) hydrogen iodide has the larger dispersion force, an additional (very) weak dipole/dipole force giving it the larger total intermolecular force and higher boiling point. (c) Hydrogen fluoride is a small, very polar molecule. The dispersion force acting between HF molecules is small (as it is a small molecule). The dipole/dipole forces acting between HF molecules are very strong and are called “hydrogen bonds”. HI is a large non-polar molecule and its dispersion force is larger than the dispersion force in HF, but smaller than HF’s H–bond. 11. (a) In 100 B atoms there are y 10B atoms and (100 - y) 11B atoms Their total mass = 10y + 11(100 - y) = 1081 amu ∴ y = 19 Therefore 81% of boron atoms are 11B. (b) The Lewis diagram for Boron bromide BBr3 is: The Lewis diagram shows that the central boron atom in boron tribromide is attached to only three other atoms (no lone pairs) and thus is anAB3 system. VSEPR theory predicts that this arrangement will be trigonal (flat triangular) with bond angles of 120o. (c) 4 protons, 2 electrons and 5 neutrons. (d) Barium (e) Both barium and beryllium are group 2 metals. Metallic bonding is the result of the attraction between the metal cation and their “sea of shared electrons”. Both beryllium and barium form +2 cations and so the number of electrons in the sea of electrons is the same. Beryllium is a smaller atom than barium and thus the strength of attraction between a beryllium ion and the electrons in the sea of electrons is larger and thus beryllium has the higher boiling point. : Br : .. .. : Br : .. .. : Br : .. ..B Beryllium ions are smaller hence its nucleus is closer to its sea of electrons and is more strongly attracted to its sea of electrons hence beryllium’s metallic bond is stronger and beryllium has the higher boiling point. 12. (a) :C:::O: (b) (c) Carbon in the formate ion is connected to three other atoms hence is surrounded by three groups of bonding electrons which will repel each other to a trigonal planar arrangement with a bond angle of 120o. (d) A Brønsted/Lowry reaction is on in which a hydrogen ion is transferred from an acid to a base (to form the corresponding conjugate acid and base. HCO2–(aq) + H2O(l) ⇌ HCO2H(aq) + HO–(aq) (e) A Brønsted/Lowry reaction is on in which a hydrogen ion is transferred from an acid to a base (to form the corresponding conjugate acid and base. HCO2–(aq) and HO–(aq) are the two bases in this reaction. 13. (a) Zinc is the more active metal and will be more easily oxidised than copper. Oxidation occurs at the anode hence zinc is the anode. (b) Cu2+ (aq) + 2e– → Cu (s) (c) E°cell = E°cathode – E°anode = (+0.35) – (–0.76) = +1.11 V (d) Moles Cu = ½ × moles of electrons = 0.100 / 2 = 0.0500 mol ∴ Mass of Cu = 0.0500 × 63.5 = 3.18 g. The copper can increases in mass. (e) Zn2+ ions are cations that will migrate to the cathode to replace the Cu2+ ions that are being removed from solution as solid copper plates onto the They are needed to maintain electrical neutrality in the can where copper ions are continually changing into copper atoms. 14. (a) “Na+(l) + F– (l)” refers to liquid (molten) NaF and is a pure compound. “Na+(aq) + F– (aq)” refers to a solution of NaF dissolved in water and is a mixture. (b) ∆Hatomisation Na = +109 kJ/mol Bond Energy F–F = +160 kJ/mol ∆H1st Ionisation energy Na = +502 kJ/mol ∆Hvaporisation fluorine = +10 kJ/mol ∆Hformation NaF(s) = -574 kJ/mol ∆Hsolution NaF(s) = +10 kJ/mol (c) X + 920 = 574 + 109 +502 +80 ∴ X = –345 kJ/mol (exothermic) 15. (a) Mass of Sn reacting = 2.00 – 1.11 = 0.89 g Moles of Sn reacting = 0.89/118.7 of tin = 0.007497 = 0.00750 mol [Sn2+] = 0.00750 / 0.100 = 0.0750 M (b) A RICE table is not needed but is included for completeness. Reaction Sn(s) + Pb2+(aq) ⇌ Pb(s) + Sn2+(aq) Initial – 0.10 – 0 Change – –0.075 – +0.075 Equilibrium – 0.025 – +0.075 Solids are not included in calculations involving equilibrium constants and are not shown in this RICE table Kc = [Sn2+] / [Pb2+] = 0.075 / 0.025 = 3.0 (c) Le Chatelier’s principle predicts that system will shift to minimise the decrease in heat that occurs when the system is cooled the system. This is an exothermic reaction hence heat can be considered to be a product. Decreasing the heat content favours the forward direction which will result in an increase in [Sn2+] therefore the temperature was decreased at time Z. (d) Adding water will cause a sudden decrease in the concentration of [Sn2+] therefore water was added at time X. (e) A precipitate of PbCl2 (s) forms when NaCl is added to the equilibrium mixture. Le Chatelier’s principle predicts the system will shift to minimise the decrease in [Pb2+] by shifting to the left thus reducing [Sn2+ (aq)] as occurs at time Y. 16. (a) Moles N2H4 = 5.00/(2 × 14.07 + 4 × 1.008) 0.1554 mole = 0.155 mol Moles O2 = 5.00/(32.0) = 0.1563 = 0.156 mole (Moles N2H4/Moles O2) = 0.1554/0.1563) = 0.9946 = 0.995 (b) N2H4 + O2 → N2 + 2H2O (c) Oxidation number of “N” changes from -2 to 0; oxidation number of “O” changes from 0 to -2 (d) Mol N2H4 = mol O2 = 5.0 / 32.0 = 0.156 so mol N2 formed = 0.156 and mol H2O formed = 0.312. So total mol gas after reaction = 0.468 P = nRT/V = 0.468 x 8.314 x 400 / 10.0 = 156 kPa (e) One third of the product molecules are N2 so their pressure = 156 / 3 = 52 kPa Transition Program Chemistry Sample Final Examination C Time Allowed: 2.5 hours Reading Time: 5 minutes Paper C Question Mark 1 6 marks 2 6 marks 3 6 marks 4 6 marks 5 6 marks 6 6 marks 7 6 marks 8 6 marks 9 6 marks 10 6 marks 11 10 marks 12 10 marks 13 10 marks 14 10 marks 15 10 marks 16 10 marks Total Student Number: _____________________ First Name: _____________________ Family Name: _____________________ Class Group _____________________ UNSW Foundation Studies UNSW Global Pty Limited UNSW Sydney NSW 2052 Australia Copyright 2017 All rights reserved. Except under the conditions described in the Copyright Act 1968 of Australia and subsequent amendments, this publication may not be reproduced, in part or whole, without the permission of the copyright owner DIRECTIONS TO CANDIDATES 1. Check that your copy of the examination paper is complete. This examination contains ten questions: • Questions 1 to 10 @ 6 marks each • Questions 11 to 16 @ 10 marks each The total marks available are 120. 2. Ensure that you have in the room: • a black or blue pen for writing answers • an electronic calculator • no unauthorised material 3. Five (5) minutes reading time is allowed. During reading time do not write at all. 4. Ensure that your name and examination number are clearly printed on the front page on the blank answer booklet. 5. Answer all questions in the answer booklet. 6. Hand in the answer booklet to the supervisor at the end of the examination. 1. “Goslarite” is a mineral with formula ZnSO4.yH2O. It can be analysed according to the following scheme: Goslarite (1.44 g) Barium sulfate (1.17 g) Dissolve in water Add excess barium chloride solution then filter Colourless solution Colourless solution Step 1. Step 2. (a) Write a net ionic equation to represent the reaction that occurs in Step 2. (b) Calculate the number of moles present in 1.17 grams of barium sulfate. (c) Calculate the value of “y” in ZnSO4.yH2O. 2. For the eleven elements with atomic numbers 10 to 20 inclusive write down the name of the element which: (a) has the highest melting point (b) has the lowest boiling point (c) has the lowest first ionisation energy (d) has the highest electronegativity (e) has an 5 electrons in its valence electron shell (f) has 14 neutrons and 13 protons in each of its nuclei. (An element’s name may be used more than once). 3. Briefly explain why a potassium (K+) ion is: (a) smaller than a potassium atom (b) larger than a calcium (Ca2+) ion (c) larger than a sodium (Na+) ion. 4. The electrical conductivities of hydrogen bromide and sodium bromide when solid, liquid and in aqueous solution are listed in the following table: Compound Solid Liquid Aqueous Solution Hydrogen bromide None None Conducts Sodium bromide None Conducts Conducts (a) Briefly explain, using a diagram if necessary, why sodium bromide conducts in the liquid state but not as a solid. (b) Briefly explain, using a diagram if necessary, why hydrogen bromide does not conduct when in either the liquid or solid state. (c) Briefly explain, using an appropriate equation, why hydrogen bromide conducts in aqueous solution. 5. When heated, solid ammonium sulfite changes into ammonia, steam and sulfur dioxide: (NH4)2SO3 2NH3 + H2O + SO2 (a) Draw a Lewis (electron dot) diagram to represent the ammonium ion (NH4+) and use the VSEPR theory to predict its shape. (b) Draw a Lewis (electron dot) diagram to represent the sulfite ion (SO32−) and use the VSEPR theory to predict its shape. (c) The diagram represents a sulfur dioxide molecule. Use the VSEPR theory to predict the size of the angle marked “X” and briefly justify your answer. O = S O X 6. Hydrochloric acid and hydrofluoric acid are neutralised by sodium hydroxide: HCl (aq) + NaOH (aq) → NaCl (aq) + H2O(l) ∆H = − 57.7 kJ HF (aq) + NaOH (aq) → NaF (aq) + H2O(l) ∆H = − 47.7 kJ (a) Write a net ionic equation to represent the neutralisation of hydrochloric acid by sodium hydroxide solution. (b) Using a net ionic equation, briefly explain why the enthalpy change for the neutralisation of hydrochloric acid is different to the enthalpy change for the neutralisation of hydrofluoric acid. (c) Calculate the standard enthalpy change (in kJ/mol of hydrofluoric acid) for the reaction: HF (aq) → H+ (aq) + F− (aq) 7. The electrodes of an electric cell or battery can be identified with pole-finding paper. This is made by soaking filter paper in sodium sulfate solution to which some phenolphthalein indicator has been added. When the two electrodes of a cell are joined with pole-finding paper an electrolysis reaction takes place and a deep red-violet colour soon appears around one of them. (a) Name the ion that is responsible for the phenolphthalein turning red-violet? (b) Write an ion-electron half-equation to represent the formation of this ion from water during the electrolysis. (c) At which electrode (positive or negative) of the cell does the pole-finding paper turn red-violet? Briefly justify your answer. 8. Mylanta© antacid tablets are used for the relief of indigestion. Mylanta ROLLTABS 12 ANTACID TABLETS Active Ingredients: Calcium Carbonate 317 mg and Magnesium Hydroxide 64 mg (a) A human stomach contains 0.063 M hydrochloric acid solution. Calculate the pH of 0.063 M hydrochloric acid solution. (b) Write a net ionic equation for the reaction of solid calcium carbonate with hydrochloric acid solution. (c) Calculate the volume of 0.063 M hydrochloric acid solution which would react completely with 64 milligrams of solid magnesium hydroxide. 9. An old process for manufacturing sodium hydroxide involves reacting a saturated solution of sodium carbonate with a saturated solution of calcium hydroxide. (a) Write an ionic equation for this reaction. (b) 1.00 litre of saturated calcium hydroxide (containing 1.85 grams of solute) was completely converted into 1.00 litre of sodium hydroxide solution by this process. Calculate the pH of this sodium hydroxide solution. (c) A saturated solution of calcium carbonate contains 9.34 mg/L of CaCO3. Calculate the Ksp for CaCO3. 10. The diagram shows two flasks (A and B), each of which has a volume of 4.00 litres and is fitted with a pressure gauge. Flask A contains 5.00 grams of nitrogen dioxide gas. Flask B contains 2.00 grams of methylhydrazine (CN2H6) gas. The temperature of both flasks is 127 oC. (a) Calculate the ratio: (number of molecules in Flask A) ÷ (number of molecules in Flask B) (b) Calculate the ratio: (pressure in Flask A) ÷ (pressure in Flask B) (c) When the tap is opened the gases mix and react together producing nitrogen, carbon dioxide and steam: 10 NO2 (g) + 4 CN2H6 (g) 9 N2 (g) + 4 CO2 (g) + 12 H2O (g) Calculate the reading on the right side pressure gauge at 127 oC when the reaction is complete. 11. Heating propane gas converts it into a mixture of methane and ethene: propane methane + ethene The standard enthalpy change for this reaction is + 81 kJ/mol of propane. (a) Rewrite this equation using structural formulas for the three compounds. (b) Using information on the Data Sheet, calculate the standard enthalpy of formation (∆Hf, in kJ/mol) of ethene. (c) Using information on the Data Sheet, calculate the bond energy (in kJ/mol) of carbon-carbon double bonds. (d) Ethene and oxygen can be made to react together to form a gaseous compound called “oxirane”: ethene + oxygen gas → oxirane (1.0 litre) (0.50 litre) (1.0 litre) If all gas volumes are measured at R.T.P. write a fully balanced equation for this reaction. (e) The toxic quantity of oxirane in one litre of air is 0.18 mg. How many molecules are there in 0.18 mg of oxirane? 12. The solubility of copper chromate (CuCrO4) in water is 1.9 × 10-3 mol/L. (a) Calculate the solubility of copper chromate in water in grams per litre. (b) Using Le Chatelier’s principle or otherwise, briefly explain why the solubility of copper chromate decreases if some solid copper chloride is added to the water. (c) Write an expression relating the solubility product (Ksp) of copper chromate to the concentration of its ions in a saturated aqueous solution. (d) Calculate the solubility product (Ksp) of copper chromate in water. (e) Predict whether a precipitate of copper chromate would form if 100 mL of 2.00 × 10-3 M Cu(NO3)2 solution was mixed with 100 mL of 4.00 × 10-3 M Na2CrO4 solution. Briefly justify your answer. 13. Pure silicon is obtained by the following reaction: SiCl4 (g) + 4Na (s) → Si (s) + 4NaCl (s) Some properties of these substances are tabulated below: Substance Melting point (oC) Electrical conductivity (as solid) Electrical conductivity (when melted) SiCl4 – 68 Non-conductor Non-conductor Na + 98 Conductor Conductor Si + 1410 Non-conductor Non-conductor NaCl + 801 Non-conductor Conductor (a) Briefly explain how sodium is able to conduct an electric current. (b) Briefly describe the particles, and the type of attractive forces that exist between them, in solid sodium chloride. (c) Briefly explain why silicon has a much higher melting point than the other three substances. (d) Draw a Lewis (electron dot) diagram to represent a SiCl4 molecule and describe its shape. (e) Briefly explain why SiCl4 has a much lower melting point than the other three substances. 14. A 0.100 M ammonia solution in water has a pH of 11.1 (a) Calculate the molar concentration of hydroxide ions in a solution whose pH is equal to 11.1 (b) Write a net ionic equation to represent the reaction of ammonia with water to produce hydroxide ions. (c) Briefly explain whether ammonia is acting as an acid or a base in this reaction. (d) If solutions of ammonia and magnesium sulfate are mixed together a precipitate is formed. Write a net ionic equation to represent the formation of a precipitate when the two solutions are mixed. (e) If some solid ammonium sulfate is added to the mixture in (d) it is seen that the precipitate dissolves. Using Le Chatelier’s Principle briefly explain why the precipitate dissolved. 15. A solution of potassium permanganate can be standardised by titration with solid oxalic (ethanedioic) acid: KMnO4 + H2C2O4 + H2SO4 → MnSO4 + K2SO4 + CO2 + H2O (unbalanced) 29.00 mL of a certain potassium permanganate solution (acidified with sulfuric acid) were required to react completely with 0.100 g of solid oxalic acid H2C2O4. (a) Using oxidation numbers, show that this reaction involves oxidation and reduction. (b) Briefly explain the meaning of the term “standardised”. (c) Briefly explain why an indicator was not needed for this titration. (d) The permanganate changed according to: MnO4– (aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (l) Write an ion-electron half equation to represent the change which occurred to the oxalic acid during the titration. (e) Calculate the molar concentration of the potassium permanganate solution used in the titration. Solid oxalic acid Potassium permanganate solution 16. (a) The formula of lactic acid is: (a) Write the correct (I.U.P.A.C.) name of lactic acid. (b) The acid dissociation constant (Ka) for lactic acid is 1.38 × 10–4. Calculate the pH of 0.100 M lactic acid solution. (c) Calculate the percentage of lactic acid molecules in a 0.100 M aqueous solution of the compound which have become ionised. (d) Potassium lactate CH3 −CH(OH) −COOK dissolves in water producing an alkaline solution. Write a net ionic equation to represent the hydrolysisa reaction that occurs and draw a circle around the conjugate base of lactic acid. (e) Calculate the pH of 0.0200 M potassium lactate solution. DATA SHEET Avogadro's constant, NA 6.02 × 1023 mol–1 Universal Gas Constant, R 8.314 J K−1 mol–1 Specific heat of water 4.180 J g−1 K–1 Molar volume of ideal gas: at 101.3 kPa and 0 oC 22.41 L mol−1 at 101.3 kPa and 25 oC 24.46 L mol-1 Standard enthalpies of formation (∆Hf ): methane (g) − 75 kJ mol−1 propane (g) − 104 kJ mol−1 Average bond energies: C — C + 346 kJ mol−1 Standard reduction potentials (Eo) in alkaline conditions: Zn(OH)2 (s), Zn (s) – 1.22 V NiO2 (s), Ni(OH)2 (s) + 0.49 V Unit / Symbol Equivalences: ∆Hf = ∆ f H Ksp = Ks M = mol L-1 Answers to Final Exam - Sample C 1. (a) Ba2+(aq) + SO42-(aq) → BaSO4(s); (b) 5.01 × 10-3 mol; (c) y = 7.00 2. (a) silicon; (b) neon; (c) potassium; (d) chlorine; (e) phosphorus; (f) aluminium. 3. (a) A K+ ion has 18 electrons hence 2,8,8 with 3 occupied electron shells. A K atom 19 electrons and the 19th electron is in the 4th shell which is further from the nucleus thus making the atom larger than the ion. (b) A Ca2+ ion has the same configuration as a K+ ion but its extra proton pulls these electrons closer to the nucleus thus making the ion smaller than K+. (c) A Na+ ion is 10 electrons and thus only 2 occupied electron shells and it is smaller. 4. (a) Sodium bromide is made of Na+ and Br¯ ions. These can move, and so form a current, in the liquid state but cannot move when NaBr is solid. (b) Solid and liquid HBr consist of uncharged molecules so that a current of moving charges is impossible. (c) Hydrogen bromide dissolves in water forming ions: HBr (g) → H+ (aq) + Br¯ (aq) These ions can move and thus produce a current. 5. (a) tetrahedral; (b) 2- pyramidal; (c) <120 o; VSEPR theory predicts 3 sets of electrons surround the sulfur atom will repel to give a triangular planar arrangement 6. (a) H+(aq) + OH–(aq) H2O(l) . (b) HCl(aq) is a strong acid and is fully dissociated in solution. HF(aq) is a weak acid and is mostly in the undissociated form in solution: HF(aq) + OH–(aq) H2O(l) + F–(aq). (c) HF(aq) H+(aq) + F–(aq) ∆H = –47.7 - (-57.7) = +10.0 kJ/mol 7. (a) Hydroxide ion (turns phenolphthalein red) (b) 2H2O (l) + 2e– → H2 (g) + 2OH¯ (aq) (c) Mg(OH)2 + 2 HCl MgCl2 + 2H2O Moles Mg(OH)2 0.064/(24.3 + 2 ×17.0) = 0.00110 mole Moles HCl = 0.00220 mole Vol HCl = 0.00200/0.63 =0.0348 L = 35 mL (to 2 significant figures) 8. (a) pH = –log10(0.063) = 1.20, (b) CaCO3 (s) + 2H+ (aq) → Ca2+ (aq) + H2O (l) + CO2 (g) (c) Moles HCl = 2×moles Mg(OH)2 = 0.064/(24.3 + 34.0) = 0.00220 mole Mole HCl = 0.00220/0.063 = 0.0348 L = 35 mL (to 2 significant figures) + H 9. (a) CO32- (aq) + Ca2+ (aq) → CaCO3(s); (b) 12.70 (c) Moles CaCO3 = 0.00934/(40.1 + 12.0 + 48.0) = 9.33 × 10–5 , Ksp = 8.71 × 10–9 10. (a) 2.50; (b) 2.50 ((PV = nRT, hence ratio of pressures is the same as ratio of moles if T and V are constant); (c) 113 kPa 11. (a) CH3 – CH2 – CH3 → CH4 + CH2 = CH2 (b) + 52 kJ/mol (c) + 611 kJ/mol (d) 2C2H4 + O2 → 2C2H4O (e) 2.5 × 1018 12. (a) 0.34 g/L (b) Addition of extra Cu2+ (aq) ions from dissolved CuCl2 causes the system to adjust so as to remove some of these i.e. more solid CuCrO4 forms i.e. the solubility decreases (c) Ksp = [Cu2+] × [CrO42-] (d) Ksp = 3.6 × 10-6 (e) ion product = 2.00 × 10-6 so no precipitation occurs. 13. (a) Sodium atoms each have one loosely-held valence electron which can move under an applied voltage and thus form a current. (b) Sodium and chloride ions attract each other with ionic bonds (the electrostatic force of attraction between cations and anions. (c) Silicon atoms attract each other with covalent bonds which are stronger than the other types of attractive force. (d) tetrahedral shape (e) SiCl4 is non-polar so has only dispersion forces acting between its molecules. These are much weaker than the other types of force so the molecules need less heat energy to separate them. 14. (a) Concentration of OH¯(aq) = 10-2.9 = 1.26 × 10-3 M (b) NH3 (aq) + H2O (l) → NH4+ (aq) + OH¯ (aq) (c) NH3 is acting as a base because its aqueous solution contains more OH¯ (aq) ions than does water therefore [OH–] > [H+] (d) Mg2+ (aq) + 2 OH¯ (aq) → Mg(OH)2 (s) (e) Le Chatelier’s principle predicts addition of NH4+ (aq) ions forces the reaction in (b) to the left thus reducing the concentration of OH¯(aq). This then forces the reaction in (d) to left causing Mg(OH)2 (s) to dissolve. Cl x Si x x Cl Cl Cl xx x x x x x x x x x x x x x x x x x x x x x x x 15. (a) KMnO4 + H2C2O4 + H2SO4 → MnSO4 + K2SO4 + CO2 + H2O (“Mn” is reduced from +VII to +II; “C” is oxidised from +III to +IV) (b) “standardised” = “accurately measured concentration”. (c) KMnO4 changes from violet to colourless when it is oxidised to Mn2+ At the endpoint, the colourless solution in the conical flask turns pale pink from a slight excess of MnO4– (d) H2C2O4 (aq) → 2CO2 (g) + 2H+ (aq) + 2e (e) Conctn. = 0.0153 M 16. (a) 2-hydroxypropanoic acid (b) [H+] = 3.71 × 10-3 M so pH = 2.43 (c) 3.71 % (d) C3H5O3– (aq) + H2O (l) ⇌ C3H6O3 (aq) + OH– (aq) (e) pH = 8.08 UNSW Foundation Studies UNSW Foundation Studies UNSW Sydney NSW 2052 Australia T: 61 2 9385 5396 | F: 61 2 9662 2651 | E:
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