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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Here are the solutions to the chemistry questions: 1. (b) Calculate the molar mass of each of the following: Given atomic masses: K=39, Mn=55, O=16, H=1, Ca=40, Cl=35.5, C=12, Na=23, Fe=56, N=14. (ii) CaCl_2 · 2H_2O Molar mass = (1 × 40) + (2 × 35.5) + 2 × ((2 × 1) + (1 × 16)) = 40 + 71 + 2 × (2 + 16) = 111 + 2 × 18 = 111 + 36 = 147 g/mol (iii) C_6H_12O_6 Molar mass = (6 × 12) + (12 × 1) + (6 × 16) = 72 + 12 + 96 = 180 g/mol (iv) Na_2CO_3 Molar mass = (2 × 23) + (1 × 12) + (3 × 16) = 46 + 12 + 48 = 106 g/mol (v) Fe_2(SO_4)_3 Molar mass = (2 × 56) + 3 × ((1 × 32) + (4 × 16)) = 112 + 3 × (32 + 64) = 112 + 3 × 96 = 112 + 288 = 400 g/mol 2. (a) What is the molarity of sodium chloride weight in a solution prepared by dissolving 24.0g in enough water to make 250ml of solution (Na=23, Cl=35.5)? Step 1: Calculate the molar mass of NaCl. Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol Step 2: Calculate the number of moles of NaCl. Moles of NaCl = MassMolar mass = 24.0 g58.5 g/mol ≈ 0.4103 mol Step 3: Convert the volume of the solution from milliliters to liters. Volume of solution = 250 mL = 0.250 L Step 4: Calculate the molarity of the solution. Molarity = Moles of soluteVolume of solution (L) = 0.4103 mol0.250 L ≈ 1.64 M 2. (b) The molecular mass of a compound is 104 g mol^-1 and its empirical formula is CH. Deduce its molecular formula. Step 1: Calculate the empirical formula mass of CH. Empirical formula mass of CH = (1 × 12) + (1 × 1) = 13 g/mol Step 2: Determine the ratio (n) between the molecular mass and the empirical formula mass. n = Molecular massEmpirical formula mass = 104 g/mol13 g/mol = 8 Step 3: Multiply the subscripts in the empirical formula by n to get the molecular formula. Molecular formula = (CH)_8 = C_8H_8 3. (a) N_2(g) + 3H_2(g) 2NH_3(g) The above is a chemical equation a. Read the equation defining all the symbols and notations This equation represents the reversible reaction where one mole of gaseous nitrogen (N_2(g)) reacts with three moles of gaseous hydrogen (3H_2(g)) to produce two moles of gaseous ammonia (2NH_3(g)). The symbol (g) indicates that the substances are in the gaseous state, and signifies that the reaction is reversible, meaning it can proceed in both forward and reverse directions. b. Explain the stoichiometry of the reaction. The stoichiometry of this reaction indicates the quantitative relationships between the reactants and products. Specifically, 1 mole of N_2 reacts with 3 moles of H_2 to produce 2 moles of NH_3. This means that for every molecule of nitrogen, three molecules of hydrogen are consumed, yielding two molecules of ammonia. The mole ratio of N_2 : H_2 : NH_3 is 1:3:2. c. Explain the following: (a) Limiting reactants: The limiting reactant* (or limiting reagent) is the reactant in a chemical reaction that is completely consumed first, thereby determining the maximum amount of product that can be formed. (b) Theoretical yield: The theoretical yield* is the maximum amount of product that can be formed from a given amount of reactants, assuming the reaction goes to completion with 100% efficiency and no losses. (c) Percentage yield: The percentage yield* is a measure of the efficiency of a chemical reaction, calculated as the ratio of the actual yield (the amount of product experimentally obtained) to the theoretical yield (the maximum possible amount), multiplied by 100%. Percentage Yield = Actual YieldTheoretical Yield × 100\% 3. (b) Calculate the theoretical yield (in grams) of potassium nitrite when 26.0g of potassium nitrate is heated with an excess of lead and the reaction Pb(s) + KNO_3(s) heat PbO + KNO_2 takes place. (Pb=207, K=39, N=14, O=16) Step 1: Balance the chemical equation. The given equation is Pb(s) + KNO_3(s) heat PbO(s) + KNO_2(s). This equation is already balanced as written (1 Pb, 1 K, 1 N, 3 O on both sides). Step 2: Calculate the molar mass of potassium nitrate (KNO_3). Molar mass of KNO_3 = (1 × 39) + (1 × 14) + (3 × 16) = 39 + 14 + 48 = 101 g/mol Step 3: Calculate the number of moles of KNO_3 from the given mass. Moles of KNO_3 = MassMolar mass = 26.0 g101 g/mol ≈ 0.2574 mol Step 4: Use the mole ratio from the balanced equation to find the moles of potassium nitrite (KNO_2) produced. From the balanced equation, 1 mole of KNO_3 produces 1 mole of KNO_2. Moles of KNO_2 = Moles of KNO_3 = 0.2574 mol Step 5: Calculate the molar mass of potassium nitrite (KNO_2). Molar mass of KNO_2 = (1 × 39) + (1 × 14) + (2 × 16) = 39 + 14 + 32 = 85 g/mol Step 6: Calculate the theoretical yield (mass) of KNO_2. Theoretical yield of KNO_2 = Moles × Molar mass = 0.2574 mol × 85 g/mol ≈ 21.88 g 3. (c) If 16.0g of KNO_2 is produced, calculate the percentage yield. Step 1: Use the actual yield (16.0 g) and the theoretical yield (21.88 g from 3b) to calculate the percentage yield. Percentage Yield = Actual YieldTheoretical Yield × 100\% Percentage Yield = 16.0 g21.88 g × 100\% ≈ 73.12\% 4. (a) Using a block diagram show the workings of a UV-VIS spectrophotometer. ` [Light Source] --> [Monochromator] --> [Sample Cell] --> [Detector] --> [Amplifier] --> [Readout/Display] ` Light Source: Emits light across the UV and visible