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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Hapa kuna majibu ya maswali yako: I) Step 1: Solid E reacts with dilute hydrochloric acid to produce Solution F and a colourless gas G. Solution F is later shown to contain copper(II) ions. If Solid E is copper(II) carbonate, it reacts with hydrochloric acid to form copper(II) chloride (Solution F), water, and carbon dioxide gas (colourless gas G). CuCO_3(s) + 2HCl(aq) → CuCl_2(aq) + H_2O(l) + CO_2(g) Therefore, solid E is copper(II) carbonate. Solid E: Copper(II) carbonate II) Step 2: From the reaction in Step 1, the colourless gas G produced is carbon dioxide. Gas G: Carbon dioxide III) Step 3: Solution H is the deep blue solution formed when the blue precipitate (copper(II) hydroxide) reacts with excess aqueous ammonia. This deep blue solution contains the tetraamminecopper(II) complex ion. Cu(OH)_2(s) + 4NH_3(aq) → [Cu(NH_3)_4]^2+(aq) + 2OH^-(aq) The formula of the compound (ion) in solution H is [Cu(NH_3)_4]^2+. Formula of the compound in solution H: [Cu(NH_3)_4]^2+ IV) Step 4: The reaction in Step II involves Solution F (containing Cu^2+(aq) ions) reacting with 3 drops of aqueous ammonia to form a blue precipitate, which is copper(II) hydroxide, Cu(OH)_2(s). Aqueous ammonia provides hydroxide ions. Ionic equation for the reaction in Step II: Cu^2+(aq) + 2NH_3(aq) + 2H_2O(l) → Cu(OH)_2(s) + 2NH_4^+(aq) That's 2 down. 3 left today — send the next one.