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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Step 1: Draw the dot and cross structures. a) Ammonia molecule (NH_3) Nitrogen has 5 valence electrons, and each hydrogen has 1 valence electron. Nitrogen forms three single covalent bonds with three hydrogen atoms and has one lone pair of electrons. c H \\ . \\ H · × N × · H \\ . \\ . (Where 'x' represents electrons from Nitrogen and '.' represents electrons from Hydrogen. The two dots above N represent the lone pair.) b) Hydronium ion (H_3O^+) Oxygen has 6 valence electrons, and each hydrogen has 1 valence electron. The +1 charge means one electron is lost. Oxygen forms three single covalent bonds with three hydrogen atoms and has one lone pair of electrons. c H \\ . \\ H · × O × · H \\ . \\ . ^+ (Where 'x' represents electrons from Oxygen and '.' represents electrons from Hydrogen. The two dots above O represent the lone pair.) c) Ammonium hydroxide (NH_4OH) This is an ionic compound consisting of the ammonium ion (NH_4^+) and the hydroxide ion (OH^-). Ammonium ion (NH_4^+) Nitrogen has 5 valence electrons, and each hydrogen has 1 valence electron. The +1 charge means one electron is lost. Nitrogen forms four single covalent bonds with four hydrogen atoms and has no lone pairs. c H \\ . \\ H · × N × · H \\ . \\ H ^+ (Where 'x' represents electrons from Nitrogen and '.' represents electrons from Hydrogen.) Hydroxide ion (OH^-) Oxygen has 6 valence electrons, and hydrogen has 1 valence electron. The -1 charge means one electron is gained. Oxygen forms one single covalent bond with hydrogen and has three lone pairs of electrons. . \\ . \\ H · × O × \\ . \\ . ^- (Where 'x' represents electrons from Oxygen and '.' represents electrons from Hydrogen. The six dots around O represent the three lone pairs.) Step 2: Calculate the R.A.M of X. The reaction is between XCO_3 and hydrochloric acid (HCl). The balanced chemical equation is: XCO_3(s) + 2HCl(aq) → XCl_2(aq) + H_2O(l) + CO_2(g) Calculate the moles of HCl: Volume of HCl = 20 cm^3 = 0.020 L Concentration of HCl = 1.8 M Moles of HCl = Concentration × Volume Moles of HCl = 1.8 mol/L × 0.020 L = 0.036 mol From the balanced equation, the mole ratio of XCO_3 to HCl is 1:2. Moles of XCO_3 = Moles of HCl2 Moles of XCO_3 = 0.036 mol2 = 0.018 mol Calculate the molar mass of XCO_3: Mass of XCO_3 = 1.312 g Molar mass of XCO_3 = MassMoles Molar mass of XCO_3 = 1.312 g0.018 mol ≈ 72.89 g/mol Calculate the R.A.M of X: Molar mass of XCO_3 = R.A.M of X + R.A.M of C + (3 × R.A.M of O) Given C = 12.0, O = 16.0. 72.89 = R.A.M of X + 12.0 + (3 × 16.0) 72.89 = R.A.M of X + 12.0 + 48.0 72.89 = R.A.M of X + 60.0 R.A.M of X = 72.89 - 60.0 R.A.M of X = 12.89 The R.A.M of X is approximately 12.9. Step 3: Write ionic equations for the reactions. a) Potassium iodide and lead nitrate Pb^2+(aq) + 2I^-(aq) → PbI_2(s) b) Magnesium sulphate and sodium carbonate Mg^2+(aq) + CO_3^2-(aq) → MgCO_3(s) c) Calcium carbonate and dilute nitric (V) acid CaCO_3(s) + 2H^+(aq) → Ca^2+(aq) + H_2O(l) + CO_2(g) What's next? Send 'em! 📸