This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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You're on a roll — Here are the solutions to the remaining questions: 11. Write the observations and the equation of the reactions which occur when concentrated sulphuric (VI) acid reacts with each of the following: b) Copper (II) sulphate crystals Observation: The blue crystals turn white. Explanation: Concentrated sulphuric (VI) acid is a strong dehydrating agent and removes the water of crystallization from the hydrated copper (II) sulphate. Equation: CuSO_4 · 5H_2O(s) conc. H_2SO_4 CuSO_4(s) + 5H_2O(l) c) Copper metal followed by Potassium Manganate (VII) solution Reaction with Copper metal*: Observation: The copper metal dissolves, forming a blue solution. A colorless gas with a pungent smell is evolved. Equation: Cu(s) + 2H_2SO_4(conc) heat CuSO_4(aq) + SO_2(g) + 2H_2O(l) Reaction of the gas with Potassium Manganate (VII) solution*: Observation: The purple potassium manganate (VII) solution is decolorized (turns colorless). Equation: 5SO_2(g) + 2KMnO_4(aq) + 2H_2O(l) → K_2SO_4(aq) + 2MnSO_4(aq) + 2H_2SO_4(aq) d) Potassium Nitrate Observation: Colorless fumes (of nitric acid) are produced. Equation: KNO_3(s) + H_2SO_4(conc) heat HNO_3(g) + KHSO_4(s) 12. a) State Charles' Law. Charles' Law states that for a fixed mass of gas at a constant pressure, the volume is directly proportional to its absolute temperature. b) A gas occupies 450cm^3 at 27^. What volume would the gas occupy at 177^, if its pressure remains constant? Step 1: Convert temperatures to Kelvin. T_1 = 27^ + 273 = 300 K T_2 = 177^ + 273 = 450 K Step 2: Apply Charles' Law, (V_1)/(T_1) = (V_2)/(T_2). V_2 = (V_1 T_2)/(T_1) V_2 = 450 cm^3 × 450 K300 K V_2 = (202500)/(300) cm^3 V_2 = 675 cm^3 The new volume would be 675 cm^3. c) At a temperature of 57^, Nitrogen gas occupies a volume of 750cm^3. At what temperature, in degrees Celsius, will the gas occupy 100 cm^3? Step 1: Convert initial temperature to Kelvin. T_1 = 57^ + 273 = 330 K Step 2: Apply Charles' Law, (V_1)/(T_1) = (V_2)/(T_2). T_2 = (V_2 T_1)/(V_1) T_2 = 100 cm^3 × 330 K750 cm^3 T_2 = (33000)/(750) K T_2 = 44 K Step 3: Convert the final temperature back to degrees Celsius. T_2(in ^) = 44 K - 273 = -229^ The gas will occupy 100 cm^3 at -229^. 13. a) State Graham's Law of Diffusion. Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (or density) at constant temperature and pressure. b) It takes 20 seconds for 200cm^3 of Oxygen gas to diffuse across a porous plug. How long will it take an equal volume of Sulphur (IV) oxide to diffuse across the same plug? (S = 32.0, O = 16.0) Step 1: Calculate the molar masses of Oxygen (O_2) and Sulphur (IV) oxide (SO_2). M_O_2 = 2 × 16.0 = 32.0 g/mol M_SO_2 = 32.0 + (2 × 16.0) = 32.0 + 32.0 = 64.0 g/mol Step 2: Apply Graham's Law of Diffusion. For equal volumes, the ratio of times is directly proportional to the square root of the ratio of molar masses. Time_O_2Time_SO_2 = sqrt(M_O)_2M_SO_2 Step 3: Substitute the given values. 20 sTime_SO_2 = sqrt(32.0 g/mol)64.0 g/mol 20 sTime_SO_2 = sqrt(0.5) 20 sTime_SO_2 ≈ 0.7071 Step 4: Solve for Time_SO_2. Time_SO_2 = 20 s0.7071 Time_SO_2 ≈ 28.28 s It will take approximately 28.28 s for Sulphur (IV) oxide to diffuse. Send me the next one 📸