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Chemistry CHEMISTRY COURSEWORK AND EXAM PREPARATION VCE AND CAREERS EXPO 2012 Preparation for VCE Units 3&4 Chemistry • Venue: Caulfield Racecourse • Date: Saturday 5 May • Time: 12:00 pm – 12:45 pm • Lecturer: Mr Brian Ellett National Educational Advancement Programs (Neap) Pty Ltd ABN 49 910 906 643 96–106 Pelham Street Carlton Victoria 3053 Telephone (03) 8341 8341 Facsimile (03) 8341 8300 This publication is independently produced for students of VCE. Although material may have been reproduced with the permission of the VCAA, the publication is in no way connected to or endorsed by the VCAA. The notes, handouts and other documents issued at lectures have been specifically researched and produced by Neap. Reproduction of the whole or part of any document constitutes an infringement of copyright. None of the material may be used or passed on to other persons without the prior written consent of Neap. CHEM_EXPO_12.FM Copyright © Neap 2012 Contents Neap programs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Overview of Chemistry Units 3&4 . . . . . . . . . . . . . . . . . . . 3 1.1 1.2 1.3 1.4 1.5 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Assessment and levels of achievement. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Achieving A/A+ grades in VCE Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Data sheets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 5 5 6 7 Coursework and exam preparation . . . . . . . . . . . . . . . . . . 17 2.1 2.2 2.3 2.4 Unit 3 – Area of study 1: Chemical analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . Unit 3 – Area of study 2: Organic chemical pathways . . . . . . . . . . . . . . . . . . . Unit 4 – Chemistry at work. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Copyright © Neap CHEM_EXPO_12.FM 17 28 29 32 1 VCE Unit 3 Chemistry: Coursework & Exam Preparation Neap programs 11-Session Intensive Programs Semester 1 – March 14 – June 3 2012 11 × 2-hour small group sessions covering Unit 3 topics and preparation for the mid-year exam for Biology, Chemistry or Physics. Semester 2 – August 1 – October 10 2012 11 × 2-hour small group sessions covering Unit 4 topics and preparation for the end-of-year exam for Biology, Chemistry, Physics, Maths Methods (CAS) and Specialist Maths. 6-Session Intensive Programs Semester 1 – Commencing 18 April 2012 Small group sessions for students who need help in, or want to consolidate, specific topics in Maths Methods (CAS) and Specialist Maths. Final Revision for Mid-Year Exams May 29 and June 5 2012 4 or 5-hour lectures covering Unit 3 topics and final preparation for the mid-year exam for Accounting, Biology, Chemistry, Physics and Psychology. Winter School July 2–6 2012 6 hour lectures giving students a head start on Unit 4 topics and concepts. Subjects include Biology, Chemistry, English, Maths Methods, Psychology and Physics. September Holiday Programs for End-of-Year Exams September 24–30 2012 Lectures covering entire Unit 4 or Units 3&4 courses in preparation for the end of year exams. Subjects offered include: English, Accounting, Legal Studies, Biology, Chemistry, Physics, Psychology, Maths Methods (CAS) and Specialist Maths. Final Revision for End-of-Year Exams October 13, 14, 20 and 21 2012 A 4-hour lecture covering Unit 4 topics and final preparation for the end-of-year exam for English, Accounting, Biology, Chemistry, Physics, Maths Methods (CAS). Summer School Mid January 2013 Lectures giving students a head start on Year 12 topics and concepts. The value of Neap programs Neap lectures provide opportunities for students to revise all examinable work, incorporating exam-style questions with fully-worked solutions. Logically sequenced, comprehensive lecture/study notes and summaries are provided to accompany each program or lecture. Exam techniques and hints are included in the notes. 2 CHEM_EXPO_12.FM Copyright © Neap Overview of Chemistry Units 3&4 Overview of Chemistry Units 3&4 1.1 Introduction The key aims of this lecture are to • revise and consolidate key concepts from Chemistry Units 1 and 2 so that students have a strong knowledge base upon which to build for Year 12; • focus on key concepts and knowledge required for Year 12 so that students are well prepared for the challenges that they will meet. The Year 11 Chemistry course encompasses Units 1 and 2: The Big Ideas of Chemistry and Environmental Chemistry. Success at Year 12 is based upon a thorough understanding of the Year 11 course material, with the topics organic chemistry, stoichiometry, acids and bases and redox chemistry being particularly important. The first part of this program will be devoted to reviewing and revising some of these key concepts from Units 1 and 2. The major topics that will be investigated are 1. atomic structure and electron subshell configuration; 2. the mole concept, including empirical and molecular formulas; 3. stoichiometry and equation writing; 4. acids and bases, including pH calculations; 5. redox reactions. In broader terms, the ability to accurately write the chemical formulas of compounds and subsequently balanced chemical equations is also essential. An understanding of bonding theory as it relates to ionic, metallic and molecular compounds is also necessary. The following shows how the coursework studied at Year 11 is related to that covered in Year 12. Year 11 topic Coverage in Year 12 atomic structure reviewed electron configuration reviewed mole concept, empirical formulas etc. reviewed bonding theory (ionic, covalent, metallic) expected knowledge intermolecular bonding theory (H bonds etc.) expected knowledge organic chemistry significantly extended into Biochemistry (proteins, nucleic acids etc.) stoichiometry significant topic, with applications across numerous areas of the course acids and bases significantly extended into volumetric analysis, pH calculations and equilibrium considerations of weak acids the atmosphere relatively minor topic, although the gas laws are important redox chemistry significant topic, with extensive consideration of all aspects of redox reactions, with particular emphasis on galvanic and electrolytic cells and industrial processes Copyright © Neap CHEM_EXPO_12.FM 3 VCE Unit 3 Chemistry: Coursework & Exam Preparation The Year 12 Chemistry course upon which you are embarking encompasses Units 3 and 4: Chemical Pathways and Chemistry at Work. Each unit is divided into two Areas of study: Unit 3 investigates Chemical Analysis and Organic Chemical Pathways while Unit 4 considers aspects of Industrial Chemistry and Supplying and Using Energy. Unit 3 Areas of study Chemical analysis, which looks at the various analytical techniques available to the chemist. The food we eat, the chemicals and fertilisers used to grow this food, the fuels we use for transport and energy and the wide range of medications that we use to prolong and enhance our lives all require thorough chemical analysis to ensure that they will perform their intended function and will not be harmful to either us or the environment. The techniques that we will investigate include gravimetric and volumetric analyses which can be performed in the school laboratory as well as more complex modern instrumental techniques involving spectroscopy and chromatography. Organic chemical pathways, where we will investigate synthetic organic chemistry and some important aspects of biochemistry. Knowledge of the homologous series of the alkanes gained from Unit 1 studies will enable us to branch out into the different functional groups and their chemical properties. The reaction pathways by which one organic molecule can be synthesised from another will be considered in detail. The structure and function of proteins and their use as disease markers is an important aspect of our study of biochemistry, as is the structure and bonding of DNA and its applications in forensic science. Unit 4 Areas of study Industrial chemistry, which focuses on the factors that affect the rate and extent of chemical reactions. Students study energy profiles and how equilibrium laws are applied to homogeneous equilibria. Experiments are conducted to investigate the effect of temperature, concentration of reagents, pressure and catalysts on the position of equilibrium of a reaction, and apply Le Chatelier’s principle to explain the results. The industrial production of one significant chemical selected from ammonia, ethene, sulfuric acid or nitric acid will be studied in detail. Supplying and using energy focuses on our use of different energy resources. The extent of energy resources and the advantages and disadvantages of their use will be investigated. Students will use calorimeters to measure the energy of chemical reactions. The electrochemical series will be used extensively to assist in the prediction of redox reactions in aqueous solutions. Students will construct and operate simple galvanic and electrolytic cells and use the electrochemical series to predict and explain their results. Faraday’s laws will be used to solve problems involving quantitative calculations for electrolysis reactions. 4 CHEM_EXPO_12.FM Copyright © Neap Overview of Chemistry Units 3&4 1.2 Assessment and levels of achievement There are two outcomes which must be met for Unit 3 and two for Unit 4, with each outcome equally weighted in terms of assessment to ensure that they have been successfully met. School-assessed coursework in Unit 3 contributes 17% of the total study score and the mid-year examination in June contributes a further 33%. This split is repeated in the second half of the year, with a final examination in November which assesses the coursework taught in Unit 4. In Unit 3, the outcomes are to “evaluate the suitability of techniques and instruments used in chemical analysis” and to “identify and explain the role of functional groups in organic reactions and construct reaction pathways using organic molecules”. School-assessed coursework will consist of three tasks across the unit: an extended experimental investigation worth 50% of the marks available, a written report on one practical activity from the other Area of study (25%) and another task such as the analysis of data using structured questions (25%). The extended experimental investigation will involve • between three and five hours of practical work, conducted in pairs or small groups; • preparation of a risk assessment and management sheet; • each individual presenting his/her own report on the task in a format to be decided by your teacher. In Unit 4, the outcomes are to “analyse the factors that determine the optimum conditions used in the industrial production of a selected chemical” and to “analyse chemical and energy transformations occurring in chemical reactions”. As in Unit 3, school-assessed coursework in Unit 4 contributes 17% of the total study score and the final examination in November 33%. School-assessed coursework will consist of three tasks across the unit: a summary report including annotations of three related practical activities worth 50% of the marks available, a written report on one practical activity from the other Area of Study (25%) and another task such as the analysis of data using structured questions (25%). 1.3 Achieving A/A+ grades in VCE Chemistry Chemistry, like Mathematics and many other disciplines, is a subject where knowledge is accumulated and built upon over a number of years. To achieve at a high standard requires a thorough understanding of the fundamental chemical principles that underpin the course. It is not sufficient to ‘cram’ before the examinations or SAC tasks and fill your head with facts – this method will yield a B grade at best. It is the nature of the subject and how it is examined that exam questions require students to reason – to logically deduce a rational response to information provided in (sometimes) unfamiliar contexts. Some of the hallmarks of the successful student are listed below. • works consistently through the year, revising systematically and reviewing new concepts as they arise • asks for assistance when unsure, rather than simply abandoning a concept and hoping that it will disappear • practises question types consistently and comprehensively. These may be from the textbook, revision guides, old examinations and so on • always endeavours to contextualise knowledge, such that they can see how different ideas ‘fit together’ and make sense • perhaps most importantly of all, maintains a sensible life balance: family and friends time, sport, music, social life, relaxation, spiritual renewal and school work. Some sacrifices will be necessary, but the complete abandonment of all of the pleasures in life will leave you miserable and unproductive. Copyright © Neap CHEM_EXPO_12.FM 5 VCE Unit 3 Chemistry: Coursework & Exam Preparation 1.4 Resources Each of the major publishers has produced textbooks designed to meet the requirements of the recently accredited Chemistry course. Your teacher is likely to have booklisted one of these texts for your use. Hogendoorn B, Commons C, Jarrett S, McKenzie C, Moseley W, Porter M, Williamson M, 2007, Heinemann Chemistry 2, 4th edn, Heinemann, Melbourne. Jeffery F, 2007, Chemistry smartstudy revision & exam guide for VCE Unit 3, Neap, Melbourne. Jeffery F, 2007, Chemistry smartstudy revision & exam guide for VCE Unit 4, Neap, Melbourne Jeffery F, Ellett B, O’Shea P, and Ellis J, 2006, Chemistry Dimensions 1, Pearson Education Australia, Melbourne. Jeffery F, Ellett B and O’Shea P, 2007, Chemistry Dimensions 2, Pearson Education Australia, Melbourne. Lukins N, Elvins C, Lohmeyer P, Ross B, Sanders R and Wilson G, 2006, Heinemann Chemistry 1, 4th edn, Heinemann, Melbourne. Sharwood J and Gordon J, 2006, Chemistry VCE Units 1 and 2, Nelson, Melbourne. Sharwood J and Gordon J, 2007, Chemistry VCE Units 3 and 4, Nelson, Melbourne. Taylor N, Derbogosian M, Ng W, Stubbs A and Stokes R, 2006, Study on Chemistry 1, Wiley and Sons, Brisbane. Taylor N, Derbogosian M, Ng W, Stubbs A and Stokes R, 2007, Study on Chemistry 2, Wiley and Sons, Brisbane. On the VCAA website you will find a full list of all VCE subjects on offer, including course information, past examination papers and examiner reports (including solutions). This is a most useful website for all students and should be accessed regularly: www.vcaa.vic.edu.au. The website of the Chemistry Education Association is a valuable resource that can be accessed at www.cea.asn.au. 6 CHEM_EXPO_12.FM Copyright © Neap Copyright © Neap 4 Be 9.0 21 Sc 44.9 22 Ti 47.9 79 Au symbol of element relative atomic mass 197.0 atomic number 29 Cu 63.5 30 Zn 65.4 5 B 10.8 6 C 12.0 7 N 14.0 8 O 16.0 9 F 19.0 beryllium 12 Mg 24.3 magnesium 20 Ca 40.1 calcium 38 Sr 87.6 strontium 56 Ba 137.3 barium 88 Ra (226) radium lithium 11 Na 23.0 sodium 19 K 39.1 potassium 37 Rb 85.5 rubidium 55 Cs 132.9 casesium CHEM_EXPO_12.FM 87 Fr (223) francium actinium 89 Ac (227) lanthanum 57 La 138.9 yttrium 39 Y 88.9 scandium 60 Nd 144.2 seaborgium 106 Sg (263) tungsten 74 W 183.8 molybdenum 42 Mo 95.9 chromium 24 Cr 52.0 91 Pa 231.0 protactinium 90 Th 232.0 thorium uranium 92 U 238.0 praseodymium neodymium 59 Pr 140.9 58 Ce 140.1 cerium dubnium 105 Db (262) tantalum 73 Ta 180.9 niobium 41 Nb 92.9 vanadium rutherfordium 104 Rf (261) hafnium 72 Hf 178.5 zirconium 40 Zr 91.2 titanium 23 V 50.9 neptunium 93 Np 237.1 promethium 61 Pm (145) bohrium 107 Bh (264) rhenium 75 Re 186.2 technetium 43 Tc 98.1 manganese 25 Mn 54.9 plutonium 94 Pu (244) samarium 62 Sm 150.3 hassium 108 Hs (265) osmium 76 Os 190.2 ruthenium 44 Ru 101.1 iron 26 Fe 55.8 gold americium 95 Am (243) europium 63 Eu 152.0 meitnerium 109 Mt (268) iridium 77 Ir 192.2 rhodium 45 Rh 102.9 cobalt 27 Co 58.9 111 Rg (272) gold 79 Au 197.0 silver 47 Ag 107.9 copper curium 96 Cm (251) gadolinium 64 Gd 157.2 berkelium 97 Bk (247) terbium 65 Tb 158.9 darmstadtium roentgenium 110 Ds (271) platinum 78 Pt 195.1 palladium 46 Pd 106.4 nickel 28 Ni 58.7 name of element californium 98 Cf (251) dysprosium 66 Dy 162.5 112 Uub mercury 80 Hg 200.6 cadmium 48 Cd 112.4 zinc einsteinium 99 Es (252) holmium 67 Ho 164.9 thallium 81 Tl 204.4 indium 49 In 114.8 gallium 31 Ga 69.7 aluminium 13 Al 27.0 boron fermium 100 Fm (257) erbium 68 Er 167.3 114 Uuq lead 82 Pb 207.2 tin 50 Sn 118.7 germanium 32 Ge 72.6 silicon 14 Si 28.1 carbon mendelevium 101 Md (258) thulium 69 Tm 168.9 bismuth 83 Bi 209.0 antimony 51 Sb 121.8 arsenic 33 As 74.9 phosphorus 15 P 31.0 nitrogen nobelium 102 No (259) ytterbium 70 Yb 173.0 116 Uuh polonium 84 Po (209) tellurium 52 Te 127.6 selenium 34 Se 79.0 sulfur 16 S 32.1 oxygen lawrencium 103 Lr (260) lutetium 71 Lu 175.0 astatine 85 At (210) iodine 53 I 126.9 bromine 35 Br 79.9 chlorine 17 Cl 35.5 fluorine 118 Uuo radon 86 Rn (222) xenon 54 Xe 131.3 krypton 36 Kr 83.8 argon 18 Ar 39.9 neon 10 Ne 20.2 helium hydrogen 3 Li 6.9 2 He 4.0 1 H 1.0 Overview of Chemistry Units 3&4 1.5 Data sheets 1. Periodic Table of the elements 7 VCE Unit 3 Chemistry: Coursework & Exam Preparation 2. The electrochemical series E° in volt F2(g) + 2e – – +2.87 2F (aq) + – H2O2(aq) + 2H (aq) + 2e + – Au (aq) + e – Cl2(g) + 2e +1.68 – +1.36 2Cl (aq) + – + – 3+ – Ag (aq) + e 2H2O(l) – +1.09 Ag(s) +0.80 + – H2O2(aq) – – 2+ – 4+ – Cu (aq) + 2e – + – 2+ – 2+ – 2+ – 2+ – 2+ – Pb (aq) + 2e Sn (aq) + 2e Ni (aq) + 2e Co (aq) + 2e Fe (aq) + 2e 2+ + 2e – +0.15 – +0.14 – 2+ Mn (aq) + 2e H2(g) 0.00 Pb(s) –0.13 Sn(s) –0.14 Ni(s) –0.23 Co(s) –0.28 Fe(s) –0.44 3+ H2(g) + 2OH – – Al (aq) + 3e 2+ Mg (aq) + 2e + Na (aq) + e – – 2+ – + – + – K (aq) + e Mn(s) Al(s) Mg(s) Na(s) Ca (aq) + 2e 8 H2S(g) –0.76 Zn(s) 2H2O(l) + 2e Li (aq) + e +0.34 2+ + +0.40 Sn (aq) S(s) + 2H (aq) + 2e 2H (aq) + 2e 4OH (aq) Cu(s) Sn (aq) + 2e +0.68 +0.54 2I (aq) O2(g) + 2H2O(l) + 4e Zn +0.77 Fe (aq) – +1.23 2Br (aq) O2(g) + 2H (aq) + 2e I2(s) + 2e – 2+ Fe (aq) + e +1.77 Au(s) O2(g) + 4H (aq) + 4e Br2(l) + 2e 2H2O(l) Ca(s) – –0.83 –1.03 –1.67 –2.34 –2.71 –2.87 K(s) –2.93 Li(s) –3.02 CHEM_EXPO_12.FM Copyright © Neap Overview of Chemistry Units 3&4 3. Physical constants 23 Avogadro’s constant (NA) = 6.02 × 10 Charge on one electron = –1.60 × 10 –19 mol –1 C –1 Faraday constant (F) = 96 500 C mol –1 –1 Gas constant (R) = 8.31 J K mol Ionic product for water (KW) = 1.00 × 10 –14 2 mol L –2 at 298 K (self ionisation constant) Molar volume (Vm) of an ideal gas at 273 K, 101.3 kPa (STP) = 22.4 L mol –1 –1 Molar volume (Vm) of an ideal gas at 298 K, 101.3 kPa (SLC) = 24.5 L mol –1 Specific heat capacity (c) of water = 4.18 J g Density (d) of water at 25°C = 1.00 g mL K –1 –1 1 atm = 101.3 kPa = 760 mmHg 0°C = 273 K 4. SI prefixes, their symbols and values SI prefix Symbol Value giga G 10 mega M 10 kilo k 10 deci d 10 centi c 10 milli m 10 micro µ 10 nano n 10 pico p 10 Copyright © Neap 9 6 3 –1 –2 –3 –6 –9 –12 CHEM_EXPO_12.FM 9 VCE Unit 3 Chemistry: Coursework & Exam Preparation 1 5. H NMR data Typical proton shift values relative to TMS = 0. These can differ slightly in different solvents. Where more than one proton environment is shown in the formula, the shift refers to the ones in bold letters. Type of proton Chemical shift (ppm) 0.9 1.3 R–CH3 R–CH2–R RCH CH 1.7 CH3 R3–CH 2.0 O CH 3 O 2.0 or CH3 C C OR NHR CH 3 R C 2.1 O R–CH2–X (X = F, Cl. Br or I) R–CH2–OH 3–4 3.6 O R 3.2 C NHCH2R R–O–CH3 or R–O–CH2R 3.3 O O C CH3 4.1 O R 4.1 C OCH2R 1–6 (varies considerably under different conditions) 1–5 R–O–H R–NH2 RHC 4.6–6.0 CH2 OH 7.0 H 7.3 O R 8.1 C NHCH 2R O R 9–10 C H O R 11.5 C O 10 H CHEM_EXPO_12.FM Copyright © Neap Overview of Chemistry Units 3&4 6. 13 C NMR data Type of carbon R–CH2–OH Chemical shift (ppm) 8–25 20–45 40–60 36–45 15–80 35–70 50–90 RC CR 75–95 RC CR 110–150 R–CH3 R–CH2–R R3–CH R4–C R–CH2–X RC–NH2 160–185 RCOOH 7. Infrared absorption data Characteristic range for infrared absorption. –1 Bond Wave number (cm ) 700–800 750–1100 1000–1300 C–Cl C–C C–O C C 1610–1680 C O 1670–1750 O–H (acids) 2500–3300 2850–3300 3200–3550 3350–3500 C–H O–H (alcohols) N–H (primary Copyright © Neap amines) CHEM_EXPO_12.FM 11 VCE Unit 3 Chemistry: Coursework & Exam Preparation 8. 2-amino acids (α-amino acids) Name Symbol alanine Ala Structure CH3 H2N CH COOH NH arginine Arg H2N CH2 CH2 CH2 CH COOH NH C NH2 O asparagine Asn H 2N aspartic acid C NH 2 CH COOH CH2 COOH CH COOH CH2 SH CH COOH Asp H2N cysteine CH 2 Cys H2N O glutamine Gln CH 2 CH 2 CH COOH CH2 CH2 H2N CH COOH H2N CH2 COOH H 2N glutamic acid glycine C NH 2 COOH Glu Gly N histidine isoleucine leucine His CH COOH CH3 CH CH2 H2N CH COOH CH3 CH CH3 Leu CH2 CH COOH CH2 CH2 CH COOH CH 2 CH 2 CH COOH CH2 CH2 NH2 Lys S CH 3 Met H 2N 12 CH3 Ile H2N methionine H H 2N H2N lysine N CH 2 CHEM_EXPO_12.FM Copyright © Neap Overview of Chemistry Units 3&4 Name Symbol Structure CH2 phenylalanine Phe H2N CH H proline Pro serine Ser threonine COOH COOH N CH2 OH H2N CH COOH CH3 CH OH H2N CH COOH Thr H N tryptophan Trp CH2 H2N CH COOH OH CH2 tyrosine valine Copyright © Neap Tyr H2N CH COOH CH3 CH CH3 H2N CH COOH Val CHEM_EXPO_12.FM 13 VCE Unit 3 Chemistry: Coursework & Exam Preparation 9. Formulas of some fatty acids Name lauric Formula C11H23COOH myristic C13H27COOH palmitic C15H31COOH palmitoleic C15H29COOH stearic C17H35COOH oleic C17H33COOH linoleic C17H31COOH linolenic C17H29COOH arachidic C19H39COOH arachidonic C19H31COOH 10. Structural formulas of some important biomolecules CH2OH H H O H H O HOH2C OH H H HO OH CH2OH O C OH H C OH H C OH H OH H H H H OH sucrose glycerol O HOH2C OH H H H H OH H deoxyribose NH2 O N C N C HC HN CH C N adenine 14 C N H C NH2 N C O guanine N H O C C N CH C N NH2 C CH N H cytosine CHEM_EXPO_12.FM C HN CH O O C CH N H thymine CH3 O– P O– O– phosphate Copyright © Neap Overview of Chemistry Units 3&4 11. Acid–base indicators Name pH range Ka Colour change Acid Base thymol blue 1.2–2.8 red yellow 2 × 10 methyl orange 3.1–4.4 red yellow 2 × 10 bromophenol blue 3.0–4.6 yellow blue 6 × 10 methyl red 4.2–6.3 red yellow 8 × 10 bromothymol blue 6.0–7.6 yellow blue 1 × 10 phenol red 6.8–8.4 yellow red 1 × 10 phenolphthalein 8.30–10.0 colourless red 5 × 10 –2 –4 –5 –6 –7 –8 –10 12. Acidity constants, Ka, of some weak acids Ka Name Formula ammonium ion NH4 5.6 × 10 benzoic C6H5COOH 6.4 × 10 boric H3BO3 5.8 × 10 ethanoic CH3COOH 1.7 × 10 hydrocyanic HCN 6.3 × 10 hydrofluoric HF 7.6 × 10 hypobromous HOBr 2.4 × 10 hypochlorous HOCl 2.9 × 10 lactic HC3H5O3 1.4 × 10 methanoic HCOOH 1.8 × 10 nitrous HNO2 7.2 × 10 propanoic C2H5COOH 1.3 × 10 Copyright © Neap + CHEM_EXPO_12.FM –10 –5 –10 –5 –10 –4 –9 –8 –4 –4 –4 –5 15 VCE Unit 3 Chemistry: Coursework & Exam Preparation 13. Values of molar enthalpy of combustions of some common fuels at 298 K and 101.3 kPa 16 –1 ∆H c (kJ mol ) Substance Formula State hydrogen H2 g –286 carbon (graphite) s –394 methane C CH4 g –889 ethane C2 H6 g –1557 propane C3 H8 g –2217 butane C4H10 g –2874 pentane C5H12 l –3509 hexane C6H14 l –4158 octane C8H18 l –5464 ethene C2 H4 g –1409 methanol CH3OH l –725 ethanol C2H5OH l –1364 1-propanol CH3CH2CH2OH l –2016 2-propanol CH3CHOHCH3 l –2003 glucose C6H12O6 s –2816 CHEM_EXPO_12.FM Copyright © Neap Coursework and exam preparation Coursework and exam preparation 2.1 Unit 3 – Area of study 1: Chemical analysis Calculations EMPIRICAL AND MOLECULAR FORMULAS The empirical formula of a compound shows the simplest numerical ratio of each type of atom or ion present in the compound. The molecular formula of a compound shows the actual numbers of atoms present in a molecule of the compound. Question 1 A compound of carbon, hydrogen and oxygen is found to contain 52.2% by mass carbon and 13.0% by mass hydrogen. Find the empirical formula of the compound. C H O mass ratio ----------------------------atomic mass ratio simplest ratio MOLE CONCEPT The mass of one atom is too small for laboratory work. Instead, chemists deal with the mass of a fixed large number of atoms of an element. The standard number adopted is the number of atoms in 12 grams of carbon. Experimentally it has been shown that 12 grams of carbon-12 isotope contains approximately 6 × 10 This is called Avogadro’s number. This value is determined experimentally to be 23 NA = (6.02252 ± 0.00028) × 10 . 23 atoms. A mole of a substance is the amount of substance containing the same number of molecules, ions, atoms or 12 particles as there are atoms in 12 grams of C. The unit for mole is mol. 23 Because all other atomic masses are found by comparison to carbon-12, it logically follows that if 6 × 10 23 carbon-12 atoms weigh 12 grams, then 6 × 10 atoms of any other element would weigh the same as the relative atomic mass expressed in grams. The molar mass of any substance is the mass of one mole of the –1 substance expressed in grams. The symbol is M, the unit is g mol . KMnO4 Copyright © Neap RFM = 158 amu CHEM_EXPO_12.FM M = 158 g mol –1 17 VCE Unit 3 Chemistry: Coursework & Exam Preparation C6H12O6 RMM = 180 amu M = 180 g mol –1 To calculate the number of moles of a substance, we make use of the following formulas. m • n = ----M –1 where n = number of mole in mol, m = mass in grams and M = molar mass in g mol • • N n = ------NA 23 where N = number of particles and NA = Avogadro’s constant = 6.02 × 10 n = cV where c = concentration in mol L • • –1 (molarity) and V = volume in L V n = ------VM where V = volume in L and VM = molar volume in L (VM = 22.4 L at STP or 24.5 L at SLC) pV n = ------RT –1 –1 where p = pressure kPa, V = volume in L, R = gas constant (8.31 J mol K ) and T = temperature in K Question 2 Which of the following contains the smallest number of atoms? A. B. C. D. 18 45.0 L of SO2 at STP 24 6.0 × 10 molecules of NO2 1.5 mole of O2 128 g of SO2 CHEM_EXPO_12.FM Copyright © Neap Coursework and exam preparation pH CALCULATIONS The pH scale was devised to allow a quantitative measure of the relative acidity or basicity of particular solutions. It is important to recognise that pH is a mathematical scale based on logarithms, so that a difference of one unit on the pH scale correlates with a ten-fold change in relative acidity or basicity. + – It can be experimentally proven that the product of the concentrations of H3O and OH ions in any aqueous solution remains a constant value at a specified temperature. This product is known as the ionisation constant of water (KW) where + – –14 KW = [H3O ][OH ] = 10 + – –7 In neutral solutions, [H3O ] = [OH ] = 10 at 25°C M at 25°C. + – + –7 + – + –7 In acidic solutions [H3O ] > [OH ], i.e. [H3O ] > 10 In basic solutions, [H3O ] < [OH ], i.e. [H3O ] < 10 – –7 – –7 M and [OH ] < 10 M and [OH ] > 10 M at 25°C. M at 25°C. Note that the ionisation constant expression forms the basis of the pH scale. + + pH = –log10[H ] = –log10[H3O ] or + + –pH [H ] = [H3O ] = 10 , i.e. at 25°C pH <7 is acidic, pH = 7 is neutral and pH >7 is basic Question 3 Calculate the pH of each of the following solutions. a. 1.0 × 10 –4 M HCl _________________________________________________________________________________ _________________________________________________________________________________ _________________________________________________________________________________ b. 10 –6 M NaOH _________________________________________________________________________________ _________________________________________________________________________________ _________________________________________________________________________________ c. 5.0 × 10 –2 M H2SO4 _________________________________________________________________________________ _________________________________________________________________________________ _________________________________________________________________________________ _________________________________________________________________________________ EXAM TIP: A common multiple-choice question involves determining the pH of the resulting solution from a neutralisation reaction. Be careful to note which reactant is in excess, as this will determine whether the final solution is acidic or alkaline. Copyright © Neap CHEM_EXPO_12.FM 19 VCE Unit 3 Chemistry: Coursework & Exam Preparation Flowcharts and diagrams Now that the year is underway you will have begun to study a number of analytical techniques. To understand these techniques, make use of diagrams and flowcharts. Try to summarise your practical exercises using these flowcharts. Consider two important analytical techniques: gravimetric analysis and volumetric analysis. GRAVIMETRIC ANALYSISGravimetric analysis is the process of determining the composition of a sample by formation of a precipitate and the measurement of mass. Weigh a sample of the consumer item. Mix with water to dissolve the soluble ion. Filter the mixture. Add excess reagent to form a precipitate. glass rod Filter the precipitate and wash with water. Buchner funnel Dry the precipitate in an oven until no more water is lost. (weigh to constant mass) mixture containing precipitate to pump filter flask Use mass of precipitate to perform calculations. VOLUMETRIC ANALYSIS Volumetric analysis is the process of determining the concentration of one solution by reacting it with another solution of known concentration. A standard solution is a solution whose concentration is accurately known. 1. Place weighed sample in a volumetric flask. 20 2. Half fill with water. Shake to dissolve the sample. CHEM_EXPO_12.FM 3. Add water to the calibration line. Shake again. Copyright © Neap Coursework and exam preparation In volumetric analysis, amounts in solution are determined by performing a titration. 1. A burette is filled with standard solution. 2. A pipette delivers a known volume (aliquot) of the unknown solution into the flask. 3. An indicator is added. 4. When the end point is reached, record the volume used (titre). An indicator is a compound whose colour as an acid is different from its conjugate base. An example of an indicator is phenolphthalein. An indicator must be added to the reaction mixture to detect the equivalence point. The equivalence point for a titration is when the solutions have been added in the mole ratio shown by the reaction equation. Prepare standard solution (acid or base) in the burette and record the initial volume. Transfer an aliquot of the unknown solution into a conical flask by pipette. Add several drops of suitable indicator to the flask. Titrate the unknown solution with the standard solution until the end point is reached. Record the titre (volume) of standard solution added. Repeat the titration until three concordant titres are obtained. Perform calculations. Copyright © Neap CHEM_EXPO_12.FM 21 VCE Unit 3 Chemistry: Coursework & Exam Preparation Sample questions Question 4 2+ The concentration of lead ions (Pb ) in polluted water is determined by adding sodium sulfate solution (Na2SO4) to exactly 500.0 mL of the water. The equation for the reaction is Pb(NO3)2(aq) + Na2SO4(aq)→ 2NaNO3(aq) + PbSO4(s) a. The PbSO4 precipitate was collected and dried in an oven at 110°C and weighed periodically. The last five weighings are shown in the table below. Weighing number Mass (g) 1 0.9731 2 0.9653 3 0.9614 4 0.9613 5 0.9616 Why is more than one weighing of the precipitate required? ________________________________________________________________________________ ________________________________________________________________________________ b. Calculate the mass of lead in the precipitate. ________________________________________________________________________________ ________________________________________________________________________________ c. Calculate the molarity of Pb 2+ ions in the polluted water sample. ________________________________________________________________________________ ________________________________________________________________________________ d. If the polluted water contained another ion which forms a precipitate with sulfate ions, how would this 2+ affect the calculated result for the concentration of Pb ions? ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ 22 CHEM_EXPO_12.FM Copyright © Neap Coursework and exam preparation Question 5 Household bleach contains the strong oxidant sodium hypochlorite (NaOCl) as its active ingredient. The concentration of NaOCl is generally known as the ‘available chlorine’ and is often expressed as the mass of active ingredient per unit volume of solution, or w/v. A student wishes to verify the available chlorine content in a sample of commercially available bleach, quoted as ‘40 g/L available chlorine’. To achieve this aim, 25.00 mL of the bleach is pipetted into a 250.0 mL standard flask and made up to the mark with deionised water. Three 20.00 mL aliquots of this diluted solution are pipetted into separate conical flasks and about 10 mL of acidified potassium iodide solution is added. The solution in the flask immediately becomes dark brown, as the available chlorine oxidises the iodide ions to iodine, according to the equation – – + – OCl (aq) + 2I (aq) + 2H (aq) → I2(aq) + Cl (aq) + H2O(l) This solution is titrated with a standard solution of 0.513 M sodium thiosulfate solution for each flask. An average titre volume of 9.57 mL was recorded. The relevant equation is 2– 2– – I2(aq) + 2S2O3 (aq) → S4O6 (aq) + 2I (aq) a. Calculate the average number of mol of S2O3 2– in the flask. _________________________________________________________________________________ _________________________________________________________________________________ _________________________________________________________________________________ b. Hence calculate the average number of mol of I2 reduced by the S2O3 2– ions. _________________________________________________________________________________ _________________________________________________________________________________ – Deduce the number of mol of OCl in the aliquot of diluted solution. _________________________________________________________________________________ _________________________________________________________________________________ _________________________________________________________________________________ c. – Calculate the number of mol of OCl in the original sample of bleach. _________________________________________________________________________________ _________________________________________________________________________________ _________________________________________________________________________________ d. – Hence deduce the number of mol of chlorine atoms (as Cl ) in the original sample of bleach. _________________________________________________________________________________ _________________________________________________________________________________ _________________________________________________________________________________ Copyright © Neap CHEM_EXPO_12.FM 23 VCE Unit 3 Chemistry: Coursework & Exam Preparation e. Calculate the mass of available chlorine in the original sample, and hence determine the w/v ratio. ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ Spectroscopic techniques There is a range of spectroscopic techniques used by the modern analytical chemist. These techniques can provide both qualitative information (which chemical substances are present in a sample) and quantitative information (how much of the chemical of interest is actually present). Several of the spectroscopic techniques that we will consider involve analysing the energy released or absorbed by molecules, atoms or electrons under particular circumstances. Indeed, the term spectroscopy refers to the ‘analysis of the spectrum’. 10–12 10–9 10–6 10–3 1 1000 short wave infrared wavelength (m) NOT TO SCALE VHF UHF shortest wavelength (highest gamma X-rays energy) rays ultraviolet visible light medium wave microwaves longest wavelength (lowest energy) The section of the electromagnetic spectrum with which we are most familiar is that which corresponds to visible light: the rainbow of red, orange, yellow, green, blue, indigo and violet. Radio waves, infrared and ultraviolet frequencies are also of great importance to the analytical chemist. MASS SPECTROSCOPY Mass spectroscopy differs from the other spectroscopic techniques in that it does not use energy from the electromagnetic spectrum to interact with particles. In this case, atoms or fragments of larger molecules are ionised by being stripped of electrons, accelerated by an electric field and then deflected by a magnetic field acting at right angles to the direction of their motion. In this way, the relative masses of the particles can be determined and this information used to help to identify particular compounds. ion accelerating voltage + – beam of lighter particles magnetic field gas inlet 1. vapouriser produces a gasous sample 3. accelerator causes ions to move 2. ioniser produces positive ions 4. deflector separates ions 5. detector 6. recorder produces a mass spectrum beam of heavier particles While the mass spectrometer was an essential tool in determining the relative atomic masses of the elements by accurately determining the masses and relative abundances of their isotopes, its main use now is in identifying chemical compounds on the basis of their characteristic mass spectra. 24 CHEM_EXPO_12.FM Copyright © Neap Coursework and exam preparation Question 6 The mass spectrum of 2-butenoic acid is shown below. relative abundance 20 40 60 80 m/z (mass/charge ratio) a. Draw the structural formula of 2-butenoic acid. b. Explain why a significant peak occurs at 86. 100 _________________________________________________________________________________ _________________________________________________________________________________ c. Write structural formulas for the fragments that caused peaks at 41, 45 and 69. Copyright © Neap CHEM_EXPO_12.FM 25 VCE Unit 3 Chemistry: Coursework & Exam Preparation ATOMIC ABSORPTION SPECTROSCOPY Developed in the 1950s by the CSIRO scientist Alan Walsh, the atomic absorption spectrometer is perhaps the most widely used analytical instrument in modern laboratories. It is extremely accurate and highly versatile – capable of testing for the presence of up to 68 elements. AAS is used extensively to test for the trace amounts of metals in food, drink, blood samples and urine samples. The fundamental principles of the atomic absorption spectrometer are similar to those of several other spectrometers: colorimeters, IR spectrometers and UV–visible spectrometers. Light of a particular frequency which is characteristic of the element being measured is passed through an atomised sample of the substance to be analysed. Electrons of this element absorb some of this energy and are excited to higher energy levels. If, for example, a technician wished to test for the presence of mercury in a sample of fish, they would fit the instrument with a lamp which emits the characteristic frequencies of mercury. Thus, any mercury atoms in the sample would absorb light of these frequencies and be excited. The amount of light absorbed as the beam passes through the solution can be recorded and will be directly proportional to the concentration of mercury present. The lamp emits light of the required wavelength (plus some unwanted wavelengths). atomic vapour in flame The detector measures pulsed light. The light is chopped into pulses. A solution of sample is sprayed into a flame. The monochromator and slit select light of a particular wavelength. Before the AAS can be used to quantify the amount of an element present it must first be calibrated. In this procedure, samples of known concentration of the element are passed through the device and their absorbances recorded. The absorbance of the sample being tested can then be determined from the calibration curve. EXAM TIP: Instrumental analysis forms a major part of the new Chemistry course. It is likely that students will be required to use information obtained from a range of analytical instruments, and synthesise the data to (for example) identify an unknown molecule. 26 CHEM_EXPO_12.FM Copyright © Neap Coursework and exam preparation Sample questions SECTION A Question 7 If 150 mL of 0.10 M NaOH is mixed with 50 mL of 0.30 M HCl, then the pH of the resultant solution is A. 1 B. 2 C. 7 D. 13 Question 8 Which of the following processes is involved in analysis by UV–visible spectroscopy? A. the absorption of energy by hydrogen nuclei B. the absorption of energy by molecules, which changes their vibrational and rotational movements C. the absorption of energy as electrons move between energy levelsthe movement of positively charged particles by combined magnetic and electric fields SECTION B Question 9 Many commercial detergents contain phosphorus in the form of sodium polyphosphate. The amount of phosphorus in a sample of detergent can be determined colorimetrically. 0.500 g of a solid detergent sample is dissolved in water and the solution made up to 500 mL. 20.0 mL samples of this solution are mixed with 2.00 mL of a standard molybdate solution. A blue colour develops that gives an absorbance reading of 0.130 in a simple colorimeter. Five standard solutions of sodium polyphosphate were similarly treated with the standard molybdate solution, and their absorbances are shown below. The graph shows the measured absorbance as a function of the mass of phosphorus (P) per litre in the standard polyphosphate solutions. 0.3 absorbance 0.2 0.1 0 0 10 30 20 –1 concentration of P in mg L a. –1 What is the concentration (in mg L ) of phosphorus (as P) in the solution? 1 mark What is the percentage by mass of phosphorus (P) in the detergent? 2 marks b. Explain why it is necessary to make up a set of standard solutions of sodium pyrophosphate. 1 mark Total 4 marks Copyright © Neap CHEM_EXPO_12.FM 27 VCE Unit 3 Chemistry: Coursework & Exam Preparation 2.2 Unit 3 – Area of study 2: Organic chemical pathways Unit 3 Area of study 2, Organic chemical pathways, investigates the structure, bonding and reactions of selected organic chemicals. Students study homologous series, functional groups, biomolecules and selected reaction pathways. The roles of organic molecules in the generation of drugs and biofuels are investigated, along with applications of organic molecules such as DNA in forensic analysis. Functional group Name of compound –Cl chloro –OH (hydroxyl) alkanol –NH2 amine –COOH (carboxyl) carboxylic acid –COO R′ ester –CONH2 amide System used to name compounds prefix –chloro suffix –ol prefix hydroxy– suffix –ylamine prefix suffix –oic acid prefix alkyl group ( R′ ) suffix –oate suffix –amide Question 10 Name the following compounds. a. (CH3)2CHCHCH2 ________________________________________________________________________________ b. CH3(CH2)4COOH ________________________________________________________________________________ c. CH3COO(CH2)3CH3 ________________________________________________________________________________ d. CH3CH(OH)CH3 _________________________________________________________________________ 28 CHEM_EXPO_12.FM Copyright © Neap Coursework and exam preparation EXAM TIP: In the biochemistry section of the course, the molecules under consideration are often large and complex. Practice locating and naming functional groups in these molecules, and recall how they can bond with one another. 2.3 Unit 4 – Chemistry at work Enthalpy of chemical reactions The enthalpy of a chemical reaction is a measure of its heat energy. In the previous lecture we defined the terms exothermic and endothermic in terms of whether energy was lost to, or gained from the surroundings as a reaction proceeded. Heat change diagrams are most useful tools in describing such changes in enthalpy. EXOTHERMIC ENDOTHERMIC When we incorporate the concept of enthalpy with the mole ratio of a chemical reaction we generate a thermochemical equation. These equations enable us to calculate the actual amount of energy lost or gained by a given amount of reactant as a result of a particular reaction. For example, CH4(g) + 2O2(g) → 2CO2(g) + 2H2O(l) ΔH = –890 kJ mol –1 The equation shows us that one mole of methane undergoes complete combustion in oxygen gas to produce two moles of carbon dioxide and water, and in the process produces 890 kJ of energy. RULES • If you reverse the direction of a reaction then the sign of the Δ H value also changes: e.g. 2CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔH = +890 kJ mol • –1 H values are in stoichiometric ratio to the reactants and products as written in the equation e.g. 2CH4(g) + 4O2(g) → 4CO2(g) + 4H2O(l) ΔH = –1780 kJ mol –1 • If equations are added or multiplied by a factor then so are their ΔH values. • Symbols of state must be clearly stated as ΔH values change with change of state. Copyright © Neap CHEM_EXPO_12.FM 29 VCE Unit 3 Chemistry: Coursework & Exam Preparation Question 11 a. Calculate ∆H for the reaction 2S(s) + 3O2(g) → 2SO3(g), given that –1 S(s) + O2(g) → SO2(g) ∆H = –297 kJ mol –1 2SO2(g) + O2(g) → 2SO3(g) ∆H = –196 kJ mol ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ b. Calculate the amount of energy released when 10.0 g of carbon burns in excess oxygen according to the following equation. C(s) + O2(g) → CO2(g) ∆H = –393.4 kJ mol –1 ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ c. What mass of propanol must be burnt in excess oxygen to produce 500 kJ of energy, given the following equation. 2C3H7OH(l) + 9O2(g) → 6CO2(g) + 8H2O(l) ∆H = –196 kJ mol –1 ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ 30 CHEM_EXPO_12.FM Copyright © Neap Coursework and exam preparation Question 12 The chemical equilibrium C2H6(g) C2H4(g) + H2(g) is endothermic in the forward direction. In order to increase the fraction of ethane converted to ethene and hydrogen at equilibrium, a chemist should A. B. C. D. raise both the temperature and the pressure. raise the temperature and lower the pressure. lower the temperature and raise the pressure. lower both the temperature and the pressure. Question 13 The chemical reaction – + 2– Br2(l) + SO2(aq) + 2H2O(l) → 2Br (aq) + 4H (aq) + SO4 (aq) proceeds readily and rapidly in aqueous solution. It is proposed that a galvanic cell be made using this reaction as a cell reaction in the cell shown below. V a. Label this diagram in order to show a cell based on this cell reaction. Label the positive and negative electrodes and give the chemical composition of each half-cell. (You may need to use some of the data from the separate data sheet provided.) b. Give the half-cell reactions that occur as the cell discharges. _________________________________________________________________________________ _________________________________________________________________________________ c. What would be the necessary chemical properties of the electrode materials used? _________________________________________________________________________________ _________________________________________________________________________________ Copyright © Neap CHEM_EXPO_12.FM 31 VCE Unit 3 Chemistry: Coursework & Exam Preparation d. Given the maximum voltage of single cell is 0.89 V, explain how you would recharge a cell that had been discharged. ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ e. What reaction would occur in the cell when it was being recharged? ________________________________________________________________________________ ________________________________________________________________________________ ________________________________________________________________________________ 2.4 Solutions Question 1 C H O mass ratio ----------------------------atomic mass 52.2 ---------12.0 13.0 ---------1.0 34.8 ---------16.0 ratio 4.35 13.0 2.175 simplest ratio 2 6 1 The empirical formula is C2H6O. Question 2 C V 45.0 In A, n(SO2) = ------- = ---------- = 2.0 mol V m 22.4 N(atoms) = 3 × N(SO2) = 3 × n(SO2) × NA = 3.6 × 10 24 In B, N(atoms) = 3 × N(NO2) = 3 × 6.0 × 10 24 25 = 1.8 × 10 In C, N(atoms) = 2 × N(O2) = 2 × n(O2) × NA = 1.8 × 10 24 m 128 In D, n(SO2) = ----- = --------- = 2.0 mol M 64 N(atoms) = 3 × N(SO2) = 3 × n(SO2) × NA = 3.6 × 10 32 CHEM_EXPO_12.FM 24 Copyright © Neap Coursework and exam preparation Question 3 a. strong acid, ∴complete ionisation + ∴[H ] = 1.0 × 10 –4 + –4 pH = –log[H ] = –log(1.0 × 10 ) = 4 b. soluble hydroxide, ∴complete dissociation – –6 ∴[OH ] = 10 M – 14 – 14 10 10 –8 + - = ------------ = 10 [H ] = --------------– –6 [ OH ] 10 + –8 pH = –log[H ] = –log(10 ) = 8 c. strong, diprotic acid, ∴complete ionisation + ∴[H ] = 2.5 × 10 –2 = 10 + –1 –1 pH = –log[H ] = –log(10 ) = 1 Question 4 a. To ensure the mass of precipitate was constant, indicating that it was dry. b. m 0.9614 n(PbSO4) = ----- = ---------------M 303.3 0.9614 n(Pb) = n(PbSO4), m(Pb) = n × M = ---------------- × 207.2 = 0.6568 g 303.3 c. d. 0.9614 2+ n(Pb ) in 500.0 mL = ---------------303.3 n 0.9614 / 303.3 –3 2+ - = 6.339 × 10 M ∴c(Pb ) = --- = ---------------------------------– 3 V 500.0 × 10 More precipitate would form, so the calculated result would be larger. Question 5 2– –3 –3 a. n(S2O3 ) = c × V = 0.513 × 9.57 × 10 b. 1 2– –3 n(I2) = --- × n ( S 2 O 3 ) = 2.45 × 10 mol 2 c. From the first equation, n(OCl ) = n(I2) = 2.45 × 10 d. 250.0 – – n(OCl ) in bleach sample = n(OCl ) in 20.00 mL × ------------- = 0.0307 mol 20.00 e. n(Cl) = n(OCl ) = 0.0307 mol f. m(Cl) = n × M = 0.0307 × 35.5 = 1.089 g 1.089 g in 25.00 mL x g in 1000 mL –1 x = 43.6 g L = 4.91 × 10 – –3 mol mol – Copyright © Neap CHEM_EXPO_12.FM 33 VCE Unit 3 Chemistry: Coursework & Exam Preparation Question 6 a. O H H H C C C C H H OH b. The relative atomic mass of butenoic acid is 86 and so this is the parent peak for this spectrum. It is generated by the loss of one electron from the molecule. c. The fragments at 41 , 45 and 69 are generated by the fragments + + + H H H O H O C C C H H , and C OH H H C C C C H H respectively. Question 7 n(NaOH) = c × V = 0.10 × 0.150 = 0.0150 mol n(HCl) = c × V = 0.30 × 0.050 = 0.0150 mol ∴n(HCl) = n(NaOH) ∴neutral ∴pH = 7 Question 8 C The absorption of energy by hydrogen nuclei takes place in NMR spectroscopy, so A is incorrect. Infrared spectroscopy measures the energy taken up by molecules as they vibrate and rotate, so B is incorrect. Electrons do not absorb energy in UV–visible spectroscopy, so C is correct. Movement of positively charged particles and fragments of molecules by electric and magnetic fields is a feature of mass spectrometry, so D is incorrect. Question 9 a. [phosphorus] = 13 mg per litre (from graph) b. Note that [phosphorus] is the same in the 20.0 mL sample and the 500 mL flask = 13 mg per litre. If 13 mg phosphorus in 1000 mL then x mg in 500 mL ∴mass of phosphorus on the original sample = 6.5 mg –3 6.5 × 10 100 % mass = ------------------------ × --------- = 1.3% m/m 0.500 1 c. A set of standard solutions were required to generate a calibration curve so that the absorbance of the fertiliser solution could be converted to a concentration. Question 10 a. 3-methylbut-1-ene b. hexanoic acid c. butyl ethanoate d. propan-2-ol 34 CHEM_EXPO_12.FM Copyright © Neap Coursework and exam preparation Question 11 a. To obtain the required equation we must multiply the first equation by 2 and add the result to the second equation. –1 2 × –297 + (–196) = –790 kJ mol b. c. m 10.0 n(C) = ----- = ---------- = 0.833 mol M 12.0 energy released = 0.833 × 393.4 = 328 kJ From the equation, 2 mol of propanol releases 196 kJ of energy so x mol of propanol releases 500 kJ of energy. x 500 --- = --------2 196 2 × 500 x = ------------------ = 5.10 mol 196 Mass of propanol = n × M = 5.10 × 60 = 306 g Question 12 ∆H > 0 We want an increase in the forward reaction. ∴raise T and lower P Question 13 a. V cathode (+) reduction 2– SO2(g)/SO4 (aq) + and H (aq) – Br2(l)/Br (aq) – b. Br2(l) + 2e → 2Br (aq) 2– + SO2(g) + 2H2O(l) → SO4 (aq) + 4H (aq) + 2e c. The electrode materials would need to be inert (unreactive) and good conductors of electricity. d. To recharge the cell, an electrical supply of at least 0.89 V would need to be supplied to the cell. e. As the cell is recharged, the reaction proceeds backwards. – 2– + 2Br (aq) + SO4 (aq) + 4H (aq) → Br2(l) + SO2(g) + 2H2O(l) Copyright © Neap CHEM_EXPO_12.FM 35