2022 YIJC H2 CHEM PRELIM P2 MS

©YIJC [Turn over YISHUN INNOVA JUNIOR COLLEGE JC 2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUGGESTED ANSWERS CG INDEX NO CHEMISTRY Paper 2 Structured Questions Candidates answer on the Question Paper. Additional Materials: Data Booklet 9729/02 29 August 2022 2 hours READ THESE INSTRUCTIONS FIRST This document consists of 28 printed pages. For Examiner’s Use 1 / 14 2 / 17 3 / 12 4 / 16 5 / 16 Penalty units significant figures Overall / 75 Write your name and class in the spaces at the top of this page. Write in dark blue or black pen on both sides of the paper. You may use an HB pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid/tape. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. The number of marks is given in brackets [ ] at the end of each question or part question.

©YIJC [Turn over 2 Answer all the questions in the spaces provided. 1 Gallium, chromium and iron are metals that are found in Period 4 of the Periodic Table. (a) (i) Naturally occurring gallium, Ga, is a mixture of two isotopes. Table 1.2 shows the relative percentage abundance of the two isotopes. Table 1.2 relative mass relative % abundance 68.9256 60.11 70.9247 39.89 Calculate the relative atomic mass of Ga to four significant figures. Show your working. [2] Relative atomic mass of Ga = (68.9256 × 60.11) + (70.9247 × 39.89) 100 = 69.72 (ii) Gallium reacts with nitrogen to form gallium nitride, GaN. Calculate the volume of nitrogen gas needed to react with 10 kg of gallium at s.t.p.. [2] Ga(s) + ½N2(g) → GaN(s) Amount of Ga = 10000  69.7 = 143.47 mol Amount of N2 needed = 0.5 × 143.47 = 71.736 mol Volume of N2 needed = 71.736 × 22.7 = 1630 dm3 (b) Chromium(III) hydroxide, Cr(OH)3, is sparingly soluble in water. Cr(OH)3(s) ⇌ Cr3+(aq) + 3OH−(aq) (i) Write an expression for the solubility product, Ksp, of chromium(III) hydroxide, stating its units. [2] Ksp = [Cr3+][OH−]3, units: mol4 dm−12

©YIJC [Turn over 3 (ii) Calculate the concentration of OH −(aq) in a saturated solution of chromium( III) hydroxide, given the value of Ksp is 6.7 × 10−31. [2] Let the solubility of Cr(OH)3 be x mol dm−3 [Cr3+] = x mol dm−3; [OH−] = 3x mol dm−3 Ksp = [Cr3+][OH−]3 = (x)(3x)3 6.7 × 10−31 = 27x4 x = 1.2550 × 10−8 mol dm−3 [OH−] = 3 × (1.2550 × 10−8) = 3.77 × 10−8 mol dm−3 (iii) Describe and explain how the solubility of chromium(III) hydroxide is affected by adding Cr2(SO4)3(aq). [1] When Cr2(SO4)3(aq) is added, it increases the concentration of Cr3+. This cause the position of equilibrium to shift to the left to decrease [Cr3+], hence solubility of Cr(OH)3 decreases. (c) Iron(II) oxide, FeO, is used to form Fe3O4 as shown. 4FeO(s) → Fe(s) + Fe3O4(s) Each formula unit of Fe3O4 contains one Fe2+ and two Fe3+ ions. (i) Using oxidation number, show how the reaction can be described as a disproportionation reaction. [1] The oxidation number of Fe is simultaneously decreased from +2 in FeO to 0 in Fe and +2 in FeO to +3 in Fe3O4. Fe3O4(l) can be electrolysed using inert electrodes to form Fe. (ii) Write the half-equation for the reaction that occurs at the anode during the electrolysis of Fe3O4(l). [1] 2O2−(l) → O2(g) + 4e−

©YIJC [Turn over 4 (iii) Calculate the maximum mass of iron metal formed when Fe 3O4(l) is electrolysed for 6 hours using a current of 50 A. Assume that one Fe2+ and two Fe3+ ions are discharged at the same rate. [3] Q = I × t = 50 × (6 × 60 × 60) = 1.08 × 106 C = neF ne = (1.08 × 106)  (96500) = 11.19 mol Fe2+(l) + 2Fe3+(l) + 8e− → 3Fe(s) Amount of Fe formed = 3/8 × 11.19 = 4.20 mol Mass of Fe formed = 4.20 × 55.8 = 234 g [Total: 14] 2 (a) An organic compound A, C5HyO, undergoes complete combustion to produce CO2 and H2O. The value of y in the molecular formula of A can be determined by exploding it with an excess oxygen and analysing the products of the combustion. (i) Balance the following equation, in terms of y, for the complete combustion of one mole of compound A at 300 oC and 1 atm. C5HyO(g) + O2(g) → CO2(g) + H2O(g) [1] C5HyO(g) + ( 18 + y 4 )O2(g) → 5CO2(g) + ( y 2)H2O(g) (ii) When 10 cm3 of gaseous compound A was mixed with an excess oxygen at 300 oC and 1 atm, there is an expansion of volume by 25 cm3. Determine the value of y. [2] Using Avogadro’s Law where mole ratio = volume ratio, C5HyO(g) + ( 18 + y 4 )O2(g) → 5CO2(g) + ( y 2)H2O(g) Mole 1 ( 18 + y 4 ) 5 ( y 2) Volume 10 ( 18 + y 4 ) × 10 50 ( y 2) × 10 Volume of gases used up = Volume of A + Volume of O2 reacted = 10 + ( 18 + y 4 ) × 10 = ( 110 + 5y 2 ) cm3 Volume of gases produced = Volume of CO2 + Volume of H2O = 50 + ( y 2) × 10 = (50 + 5y) cm3

©YIJC [Turn over 5 Expansion in volume = Volume of gases produced – Volume of gases used up 25 = (50 + 5y) − ( 110 + 5y 2 ) y = 12 (iii) Draw the arrangement of the hybridised orbitals of one of the carbon atoms in Compound A. [1] (b) Draw the displayed formula of all the constitutional (structural) isomers with the formula C5H12. [1] (c) Cyanamide, NH 2CN, is an organic compound used in agriculture and in the synthesis of pharmaceuticals. The carbon atom is bonded to both nitrogen atoms in the cyanamide molecule. Cyanamide can be produce by the hydrolysis of calcium cyanamide in the presence of carbon dioxide. CaCN2 + H2O + CO2 → CaCO3 + NH2CN (i) Draw a ‘dot-and-cross’ diagram of the cyanamide molecule. You should distinguish carefully between electrons originating from the central atom and those from the other atoms. [1]

©YIJC [Turn over 6 (ii) Table 2.1 gives the melting points of CaCN2 and NH2CN. Table 2.1 compound melting point / oC CaCN2 1340 NH2CN 44 Explain, in terms of structure and bonding, the difference in melting point between CaCN2 and NH2CN [2] • CaCN2 has a higher melting point than NH2CN