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.
ScanSolve AI Answer
You're on a roll — 29) 3.22g of hydrated Sodium Sulphate, Na_2SO_4 · X H_2O were heated to a constant mass of 1.42g, determine the value of X in the formula. (Na = 23, S = 32, O = 16, H = 1). Step 1: Calculate the mass of anhydrous sodium sulphate and water. Mass of hydrated salt = 3.22 g Mass of anhydrous salt (Na_2SO_4) = 1.42 g Mass of water lost = Mass of hydrated salt - Mass of anhydrous salt Mass of water = 3.22 g - 1.42 g = 1.80 g Step 2: Calculate the molar mass of Na_2SO_4 and H_2O. Molar mass of Na_2SO_4 = (2 × 23) + 32 + (4 × 16) = 46 + 32 + 64 = 142 g/mol Molar mass of H_2O = (2 × 1) + 16 = 18 g/mol Step 3: Calculate the moles of anhydrous sodium sulphate and water. Moles of Na_2SO_4 = MassMolar mass = 1.42 g142 g/mol = 0.01 mol Moles of H_2O = MassMolar mass = 1.80 g18 g/mol = 0.1 mol Step 4: Determine the mole ratio to find X. Ratio of moles H_2O : moles Na_2SO_4 0.1 mol0.01 mol = 10 Therefore, X = 10. 30) 20 cm^3 of 2 M Sulphuric (IV) acid reacted completely with 3.2 g of WOH (O=16, H=1) Calculate the R.A.M of W in the formula WOH. Step 1: Write the balanced chemical equation. Sulphuric (IV) acid is H_2SO_3. Assuming WOH is a monobasic hydroxide, the reaction is: 2WOH(aq) + H_2SO_3(aq) → W_2SO_3(aq) + 2H_2O(l) Step 2: Calculate the moles of H_2SO_3. Volume of H_2SO_3 = 20 cm^3 = 0.020 L Molarity of H_2SO_3 = 2 M Moles of H_2SO_3 = Molarity × Volume = 2 mol/L × 0.020 L = 0.04 mol Step 3: Calculate the moles of WOH using the mole ratio from the balanced equation. From the equation, 1 mole of H_2SO_3 reacts with 2 moles of WOH. Moles of WOH = 2 × Moles of H_2SO_3 = 2 × 0.04 mol = 0.08 mol Step 4: Calculate the molar mass of WOH. Mass of WOH = 3.2 g Molar mass of WOH = MassMoles = 3.2 g0.08 mol = 40 g/mol Step 5: Calculate the R.A.M of W. Molar mass of OH = 16 (O) + 1 (H) = 17 g/mol R.A.M of W = Molar mass of WOH - Molar mass of OH = 40 g/mol - 17 g/mol = 23 g/mol 1) The figure below shows part of non-luminous flame. (a) (i) Describe an experiment that would confirm that region labeled A is not suitable for heating. Hold a thin copper wire or a wooden splinter* horizontally across the non-luminous flame, ensuring that one part of the wire/splinter is in region A (the inner blue cone) and another part is in the outer blue cone (the hottest part of the flame). Observe which part of the wire glows red first or which part of the splinter chars first. The part in the outer blue cone will glow red or char much faster than the part in region A, confirming that region A is cooler and thus not suitable for heating. Got more? Send 'em!