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4 Aqueous Reactions Solutions to Exercises 4.93 The two precipitates formed are due to AgCl(s) and SrSO₄(s). Since no precipitate forms on addition of hydroxide ion to the remaining solution, the other two possibilities, Ni²⁺ and Mn²⁺, are absent. 4.94 (a,b) Expt. 1 No reaction Expt. 2 (aq) + Ag₂CrO₄(s) red precipitate Expt. 3 Ca²⁺(aq) + CaCrO₄(s) yellow precipitate Expt. 4 (aq) + Ag₂C₂O₄(s) white precipitate Expt. 5 Ca²⁺(aq) + CaC₂O₄(s) white precipitate Expt. 6 Ag+ (aq) + Cl⁻(aq) AgCl(s) white precipitate 4.95 (a) (b) (c) (d) (e) CaCO₃(s) + [In (c), (d) and (e), one could also write the equation for formation of bicarbonate, e.g., Mg²⁺ + (aq).] 4.96 4NO(g) + 6H₂O(g). O=0 (a) redox reaction (b) N is oxidized, is reduced 2NO(g) + O₂(g) O=0 (a) redox reaction (b) N is oxidized, is reduced N = +4 N = +5 (a) redox reaction (b) N is oxidized (NO₂ N is reduced (NO₂ NO). A reaction where the same element is both oxidized and reduced is called disproportionation. 4.97 A metal on Table 4.5 can be oxidized by ions of the elements below it. Or, a metal on the table is able to displace the (metal) cations below it from their compounds. (a) Zn(s) + 2HNO₃(aq) + H₂(g) The substance that inflates the balloon is H₂(g). Of Zn, Cu and Hg, only Zn is above H on Table 4.5, so only Zn can displace H from (b) 65.39 1 mol Zn = 0.53525 = 0.535 mol Zn 100