NYGH 2018-S3MYE-IP Chem Ans

Class Register Number Name NANYANG GIRLS' HIGH SCHOOL Mid-Year Examination 2018 Secondary Three IP Chemistry CHEMISTRY Mark Scheme 1 hour 30 minutes Monday 7 May 2018 0845 - 1015 READ THESE INSTRUCTIONS FIRST Write your class, register number and name in the spaces at the top of this page. Do not use paper clips, glue or correction fluid. There are 3 sections in this paper. Section A Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in the Multiple-Choice Answer Sheet provided. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Section B and C Answer all questions. The number of marks is given in brackets [ ] at the end of each question or part question. Candidates are reminded that all quantitative answers should include appropriate units. The use of an approved scientific calculator is expected, where appropriate. Candidates are advised to show all their working in a clear and orderly manner. Section A (20 marks) Section B (30 marks) Section C (10 marks) Total (60 marks) Setter: KXY, NHK, CT This document consists of 12 printed pages. NANYANG GIRLS' HIGH SCHOOL [Turn over

2018 Mid-Year Examination Secondary Three IP Chemistry 2 Nanyang Girls’ High School Setter: KXY, NHK, CT Choose the answer you consider correct and record your choice in the Multiple-Choice Answer Sheet provided. 1 D 2 B 3 D 4 B 5 C 6 B 7 B 8 C 9 D 10 B 11 D 12 A 13 C 14 B 15 D 16 D 17 D 18 C 19 B 20 B Section A Answer all questions.

2018 Mid-Year Examination Secondary Three IP Chemistry 3 Nanyang Girls’ High School Setter: KXY, NHK, CT [Turn Over] B1 Enzymes are biological catalysts and used in research laboratories and in industries. Enzymes called proteases can break down proteins by water to amino acids. The amino acids can be separated and identified by chromatography. The diagram below shows a typical chromatogram. (a) Given that the amino acids have the following Rf values: amino acid glutamic acid glycine alanine leucine Rf values 0.4 0.5 0.7 0.9 Identify the two amino acids on the chromatogram. [2] X is glutamic acid [1] and Y is alanine [1]. (b) A locating agent is used on the chromatogram. Explain why the chromatogram must be exposed to the locating agent. [1] Locating agent is required as amino acids are colourless/ so as to make the amino acids visible/ to show the positions of the samples/ to show the distance travelled by the samples. Section B Answer all questions. Write your answers in the spaces provided. solvent front starting line for samples sample Y sample X initial level of solvent

2018 Mid-Year Examination Secondary Three IP Chemistry 4 Nanyang Girls’ High School Setter: KXY, NHK, CT (c) Suggest another method to identify the amino acids on the chromatogram other than measuring Rf values. [1] The chromatogram may be compared with that of known amino acids/ reference samples/ standards. Total [ 4 ] B2 A student placed a crystal of copper(II) sulfate in a beaker of water. After one hour, the crystal had completely disappeared and a dense blue colour was observed in the water at the bottom of the beaker. After 48 hours, the blue colour had spread throughout the water. Using the kinetic particle theory, explain the observations. [2] Copper(II) sulfate crystals being soluble in water dissolves to release copper(II) ions and sulfate ions in water [1]. The ions randomly diffuse from a region of higher concentration starting from the bottom of the beaker to a region of lower concentration at the top of the beaker [1]. After 48 hours, the mixture is homogeneous with uniform concentration throughout the solution [1]. Any 2 of the 3 points will be awarded full marks. Total [ 2 ] copper(II) sulfate crystal after 1 hour after 48 hours water

2018 Mid-Year Examination Secondary Three IP Chemistry 5 Nanyang Girls’ High School Setter: KXY, NHK, CT [Turn Over] Total [ 5 ] B3 2.0 cm3 portions of aqueous sodium hydroxide were added to 4.0 cm3 aqueous cobalt(III) nitrate. Both solutions had a concentration of 1.00 mol dm–3. After each addition, the mixture was stirred, centrifuged and the height of the precipitate was measured. The results are shown on the following graph. (a) Write the ionic equation of the reaction between aqueous sodium hydroxide and aqueous cobalt(III) nitrate. [1] Co3+(aq) + 3OH–(aq) ž Co(OH)3(s) (b) On the same grid of the graph above, sketch the graph that would have been obtained if 4.0 cm3 of 1.00 mol dm–3 cobalt(II) nitrate had been used instead of cobalt(III) nitrate. Height of precipitate at 4 mm [1], turning point at 8 cm3 [1] [2] (c) The experiment is repeated with aluminium nitrate and aqueous sodium hydroxide of the same concentration and volume. Given that aluminium nitrate forms a precipitate that dissolves in excess aqueous sodium hydroxide to form a colourless solution, describe the shape of the graph obtained. [2] Starting from the origin, the height of the precipitate increases linearly as the volume of aqueous sodium hydroxide increases [1]. When more than 12 cm3 of aqueous sodium hydroxide is added, the height of the precipitate decreases and becomes zero [1]. volume of aqueous sodium hydroxide/ cm3 height of precipitate/ mm (b)

2018 Mid-Year Examination Secondary Three IP Chemistry 6 Nanyang Girls’ High School Setter: KXY, NHK, CT B4 Iron(II) sulfate is one of the cheapest industrial chemicals and is used to make inks and pigments of iron(III) oxide and prussian blue. Two experiments using iron(II) sulfate were conducted. (a) Experiment 1: Iron(II) sulfate crystallises as a green hydrated salt FeSO4• xH2O. To determine the extent of hydration, 4.16 g of the hydrated crystals were heated carefully until the mass of the anhydrous iron(II) sulfate remains constant at 2.28 g. (i) Calculate the number of moles in 2.28 g of anhydrous iron(II) sulfate. [1] Molar mass of anhydrous FeSO4 = 55.8 + 32.1 + 16.0 × 4 = 151.9 g mol–1 No. of moles of anhydrous FeSO𝟒 = 2.28151.9=0.0150 (3 s.f.) (ii) Calculate the number of moles of water lost. [2] Mass of H𝟐O lost = (4.16−2.28) g=1.88 g [𝟏] No. of moles of H𝟐O lost = 1.881.0× 2+16.0 =0.10444 5 s.f. =0.104 3 s.f. [𝟏] (iii) Calculate the number of moles of water of crystallisation in one mole of the hydrated salt. [1] In 0.0150 mol of FeSO𝟒, there is 0.10444 mol of H𝟐O No. of moles of water of crystallisation = 0.104440.0150 =6.96 =7 (b) Experiment 2: A nail containing iron reacted completely with dilute sulfuric acid to produce a solution of iron(II) sulfate. The solution was made up to 250 cm3 in a volumetric flask. In a titration, 25.0 cm3 of the iron(II) sulfate solution required 29.50 cm3 of 0.0200 mol dm–3 acidified potassium dichromate(VI), K2Cr2O7 for complete reaction. The reaction between acidifed potassium dichromate(VI) and iron(II) sulfate is as follows:

2018 Mid-Year Examination Secondary Three IP Chemistry 7 Nanyang Girls’ High School Setter: KXY, NHK, CT [Turn Over] *Deduct one mark from the overall paper for not leaving final answer in 3 significant figures a or for not using the 4 or 5 significant figure values for subsequent parts or for not writing units Cr2O72–(a