SAJC_H1_CHEM_P1_Workout_solutions

1 2015 H1 Chemistry Preliminary Examination Paper 1 Workout Solutions 1 Trichloroisocyanuric acid is used c ommonly in swimmi ng pools as a disinfectant. The recommended concentration level of the acid is 1.50 mg per litre. Trichloroisocyanuric acid How many moles of chlorine atoms are present in a 2.50 x 106 litre Olympic-sized pool? Working: 3 atoms in 1 molecule of Trichloroisocyanuric acid No of moles of Trichloroisocyanuric acid in the pool = 14x335.5x316x3 12x3 10 x 1.50 x 10 x 2.50 36  =16.1 No. of moles of chlorine atoms = 3 x 16.1 = 48.3 (Answer: B) 2 Carbon sulfide, CS2 is a volatile flammable liquid used in the manufacture of cellophane. It combusts to form CO2 and SO2. What is the volume of gases produced after the complete combustion of 30 cm3 of CS2? Working: CS2 (g) + 3O2 (g)  CO2 (g) + 2SO2 (g) Volume of gas produced = 3 x 30 = 90 cm3 (Answer: C)

2 3 The active ingredient in antacids is calcium carbonate. To assess the percentage content of calcium carbonate in a tablet, a 0.50 g tablet was dissolved in 30 cm 3 of 0.150 mol dm-3 hydrochloric acid. The resultant solution required 21.30 cm3 of 0.100 mol dm-3 sodium hydroxide for complete reaction. What is the percentage mass of calcium carbonate in the antacid tablet? Working: NaOH + HCl  NaCl + H2O 2HCl + CaCO3  CaCl2 + H2O + CO2 No. of moles of NaOH = 3-10 x 2.13 0.1 x 1000 21.30  No. of moles of HCl reacted with CaCO3 = 33- 2.37x1010 x 2.13 - 0.150 x 1000 30.00  No. of moles of CaCO3 = -33 10 x 1.185 22.37x10  Mass of CaCO3 = 1.185 x 10-3 x (40 + 12 + 16x3) = 0.1185g % mass of CaCO3 = 23.7% 100 x 0.500 0.1185  (Answer: A) 4 720 cm 3 of H 2S gas is passed through 40 cm 3 of 0.500 mol dm -3 HNO3. Yellow precipitate of S was produced after the complete reaction as shown in the following half equation. Which of the following could be the nitrogen-containing product? (Assume that all volumes are measured at room temperature and pressure conditions) Working: No. of moles of H2S = 0.03 24000 720  No. of moles of HNO3 = 0.02 0.500x 1000 40  Mole ratio of H2S: HNO3 = 3: 2 1 mole of H2S lost 2 electrons  3 moles of H2S lost 6 electrons 2 moles of HNO3 gains 6 electrons  1 mole of HNO3 gain 3 electrons Original oxidation state of HNO3: (+1) + x + 3(-2) = 0 x = +5 Since HNO3 gains 3 electrons, new oxidation state of N = +2 (Answer: B)

3 Compound Oxidation State of N A N2 0 B NO +2 C NO2 +4 D NH3 -3 5 Barium carbonate is an insoluble salt composed of barium and carbonate ions . When the two ions are passed through an electric field, it was observed that the barium ions are deflected by an angle of −5.8o. What would be the angle of defection for the carbonate ions? Working: Angle of deflection = )mass charge( x k −5.8 = )137 2( x k  k = −397.3 Angle of deflection = )16x3 12 2-( x (-397.3  = 13.2o (Answer: C) 6 The following data shows the succ