TJC H1 CHEM P2 (A) ANS

2015 TJC H1 Chemistry Preliminary Exam s CANDIDATE NAME ANSWERS CIVICS GROUP / CENTER NUMBER S INDEX NUMBER CHEMISTRY 8872/02 Paper 2 1 September 2015 2 hours Candidates answer section A on the Question Paper. Additional Materials: Answer Paper, Graph Paper Data Booklet READ THESE INSTRUCTIONS FIRST Write your Civics Group, centre number, index number and name on all the work you hand in. Write in dark blue or black pen. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer two questions on separate answer paper. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question. This document consists of 16 printed pages. For Examiner’s Use A1 / 10 A2 / 10 A3 / 10 A4 / 10 Section B / 40 Paper 1 / 30 Total PRELIMINARY EXAMINATIONS HIGHER 1

2 2015 TJC H1 Chemistry Preliminary Exam [Turn over For examiner’s use 1 (a) An acidified solution of KC lO2 oxidises Fe2+(aq) to Fe 3+(aq). When 0.150 g of KC lO2 reacted with 0.500 mol dm -3 Fe2+(aq) in the presence of H +(aq), 11.30 cm 3 of Fe 2+(aq) was needed for complete reaction. (i) Suggest the final oxidation state of chlorine after complete reaction. Show your workings clearly.  No of moles of Fe2+ = 𝟏𝟏.𝟑𝟎 𝟏𝟎𝟎𝟎 𝐱 𝟎. 𝟓𝟎𝟎 = 𝟎. 𝟎𝟎𝟓𝟔𝟓 𝐦𝐨𝐥 No of moles of KClO2 = 𝟎.𝟏𝟓𝟎 𝟏𝟎𝟔.𝟔 =0.00141 mol  Mole ratio of KClO2 ;Fe2+ = 0.00141:0.00565 = 1: 4 KClO2≡ 4Fe2+  KClO2 gained 4e, since in KC lO2 is C l is +3 oxidation state and final state of Cl is -1 (ii) State the full electronic configuration of Fe3+.  1s22s22p63s23p63d5 [4] (b) The research student was given the second ionisation energies of seven consecutive elements in the Periodic Table as shown below: (i) Define, with the aid of an equation, what is meant by second ionisation energy of element B.  Second ionisation energy of element B is the minimum energy required to completely remove one mole of valence electrons from one mole of ground state ions in the gaseous state to form doubly charged gaseous ions.  B+(g) → B2+(g) + e Second ionisation energy /kJ mol-1 B C D E F G H       

3 2015 TJC H1 Chemistry Preliminary Exam [Turn over For examiner’s use (ii) With reference to the graph above, deduce the Group of the Periodic Table to which B is likely to belong. Explain your answer.  Element H has a lower second ionisation energy than element G so the second electron is removed from outer quantum shell and element H is a group II element.  Element B is in Group IV. OR  Element G has the largest second ionisation energy so the second electron is removed from inner quantum shell and element G is a group I element.  Element B is in Group IV. [4] (c) When the student passed a beam of protons through an electric field, it deflected 12° towards the negative plate. Under identical conditions, the student passed a beam of doubly charged particles J through the electric field. The angle of deflection was found to be 1.5° towards the positive plate. Identify the ion J. since J is doubly charged and the deflection is towards the positive plate : J2- angle of deflection ∝ charge mass  mass of J2- = (12 x 2) / 1.5 = 16  J is O2- [2] [Total: 10]

4 2015 TJC H1 Chemistry Preliminary Exam [Turn over For examiner’s use 2 (a) Acid rain has harmful effects on aquatic animals, plants and infrastructure. It contains sulfuric acid formed from sulfur trioxide. Sulfur trioxide is produced upon atmospheric oxidation of sulfur dioxide from coal burning. (i) Write an equation for the formation of sulfuric acid from sulfur trioxide.  SO3 + H2O  H2SO4 Even before the existence of acid rain, unpolluted rain water was slightly acidic due to dissolved CO2. The solubility of pure carbon dioxide gas in water is 88 cm 3 per 100 cm3 of water under room conditions. (ii) Given that air contains 0.033% by volume of carbon dioxide, calculate the concentration in mol dm3 of carbon dioxide dissolved in unpolluted rain water. If air contains 0.033% by volume of carbon dioxide,  Volume of CO2 dissolved per 100 cm3 water = 𝟎.𝟎𝟑𝟑 𝟏𝟎𝟎 × 𝟖𝟖 = 0.0290 cm3  No. of moles of CO2 dissolved per 100 cm3 water = 𝟎.𝟎𝟐𝟗𝟎 𝟐𝟒𝟎𝟎𝟎 = 1.21  106 mol  Concentration of dissolved CO2 in water = 𝟏.𝟐𝟏𝟏𝟎−𝟔 𝟏𝟎𝟎 𝟏𝟎𝟎𝟎 = 1.21  105 mol dm3 Dissolved carbon dioxide forms carbonic acid in water, causing the pH of unpolluted rain water to be 5.63. (iii) Calculate the [H+] in unpolluted rain water. pH = -lg[H+] = 5.63  [H+] = 105.63 = 2.34  106 mol dm3 Some fishes and shellfish die at low pH values. Lakes with limestone -rich soil can maintain a relatively stable pH even when acid rain falls due to the HCO 3 /CO3 2 buffer system. (iv) With aid of an equation, explain how lakes with limestone -rich soil maintain a relatively stable pH in presence of acid rain.  CO3 2 + H+  HCO3  The base, CO3 2 (from limestone), reacts with H+ to form HCO3 . Negligible changes in pH as H+ ions are removed. [6] (b) Oxides and halides of Period 3 elements have many applications. A farmer intends to add a magnesium -containing compound to his farmland to correct magnesium deficiency and raise soil pH. (i) Suggest whether the farmer should add magnesium oxide or magnesium chloride to his farmland.  Magnesium oxide

5 2015 TJC H1 Chemistry Preliminary Exam [Turn over For examiner’s use (ii) Explain your answer in (b)(i) by describing the reactions of the compound suggested with water and acid. Write equations where appropriate and s tate the pH of the solution formed when the compound is dissolved in water.  MgO reacts with water to a very small extent (or almost insoluble in water) due to its high lattice energy, forming magnesium hydroxide, Mg(OH) 2, which is sparingly soluble. pH of solution formed is 9.  MgO(s) + H2O(l) Mg(OH)2(aq) MgO also raises pH by removing/reacting with acid to form salt and water.  MgO(s) + 2H+(aq)  Mg2+(aq) + H2O(l) [4] [Total: 10]

6 2015 TJC H1 Chemistry Preliminary Exam [Turn over For examiner’s use 3 (a) Levulinic acid is a keto -acid which is derived from degradation of cellulose and is a potential precursor to biofuels. (i) Arrange pentanoic acid, levulinic acid and pentan-1-ol in order of increasing pK a, explaining your answer. ● Levulinic acid < pentanoic acid < pentan-1-ol ● Levulinic acid and pentanoic acid are both carboxylic acids and are stronger acids than pentan-1-ol which is an alcohol as the carboxylate anions are stabilised by the delocalisation of the negative charge over the two oxygen atoms . Levulinic acid has an electron-withdrawing carbonyl group which further disperses the negative charge of the carboxylate ion, thereby stabilising it and hence it is the strongest acid. ● The negative charge on the alkoxide is localised on the oxygen and the alkyl group (CH 3CH2CH2CH2 -) is electron -releasing which intensifie