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 — I need some additional information to solve parts (iii), (iv), and (v) of question 1. The image provided does not contain the concentration of the solution that reacted with solution X in the titration, nor the initial mass of X used to prepare the 250 cm^3 solution. Without this information, I cannot calculate the moles, molar mass, or concentration of solution X. --- PART B 2. In an experiment magnesium ribbon was heated in air. The product formed was found to be heavier than the original ribbon. Potassium manganate (VII) was on the other hand, heated in air and product formed was found to be lighter. Explain the differences on the observation made. When magnesium ribbon is heated in air, it reacts with oxygen from the air to form magnesium oxide. The mass of the oxygen adds to the mass of the magnesium, making the product heavier than the original ribbon. 2Mg(s) + O_2(g) → 2MgO(s) When potassium manganate (VII) is heated in air, it decomposes to form potassium manganate (VI), manganese (IV) oxide, and oxygen gas. The oxygen gas escapes, causing the solid products to be lighter than the original potassium manganate (VII). 2KMnO_4(s) → K_2MnO_4(s) + MnO_2(s) + O_2(g) 3. In a titration experiment 25cm^3 of a solution of Sodium Hydroxide containing 8g per litre was required for complete neutralization of 0.245g of a dibasic acid. Calculate the relative molecular mass of the acid (Na = 23.0, O = 16, H = 1). Step 1: Calculate the molar concentration of Sodium Hydroxide (NaOH). Molar mass of NaOH = 23.0 (Na) + 16.0 (O) + 1.0 (H) = 40.0 g/mol Concentration of NaOH in mol/L: Concentration = 8 g/L40.0 g/mol = 0.2 mol/L Step 2: Calculate the moles of NaOH used. Volume of NaOH = 25 cm^3 = 0.025 L Moles of NaOH = Concentration × Volume Moles of NaOH = 0.2 mol/L × 0.025 L = 0.005 mol Step 3: Determine the mole ratio from the balanced equation. A dibasic acid (H_2A) reacts with NaOH in a 1:2 mole ratio. H_2A (aq) + 2NaOH (aq) → Na_2A (aq) + 2H_2O (l) Step 4: Calculate the moles of the dibasic acid. Moles of H_2A = (1)/(2) × Moles of NaOH Moles of H_2A = (1)/(2) × 0.005 mol = 0.0025 mol Step 5: Calculate the relative molecular mass of the dibasic acid. Given mass of dibasic acid = 0.245 g Relative molecular mass = Mass of acidMoles of acid Relative molecular mass = 0.245 g0.0025 mol = 98 g/mol The relative molecular mass of the acid is 98. Send me the next one 📸