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4 Aqueous Reactions Solutions to Exercises 1.00 M KBr 0.0350 L = 0.0350 mol KBr; 0.600 M KBr 0.060 L = 0.0360 mol KBr 0.0350 mol KBr + 0.0360 mol KBr = 0.0710 mol KBr total 0.0710 mol KBr = 1.42 M KBr 0.0500 L soln 4.100 (a) 50 1 mL pg 1 10³ L mL 23.0 1 mol g Na Na = 10⁻⁹ = 2.2x10⁻⁹ M (b) 2.17 1L 10⁻⁹ soln mol Na X 1L X 6.02 X 1 mol Na Na atom = atom or ions/cm³ 4.101 (a) must replace the total positive (+) charge due to Ca²⁺ and Mg²⁺. Think of this as moles of charge rather than moles of particles. 0.020 mol Ca²⁺ X 1.5x10³ L 2 mol + change = 60 mol of + charge 1L water 1 mol Ca²⁺ 0.0040 mol X 1.5x10³ L 2 mol + charge = 12 mol of + charge 1L water 1 mol Mg²⁺ 72 moles of + charge must be replaced; 72 mol are needed. (b) 72 mol Na 1 1 mol mol NaCl X 58.44 1 mol NaCl NaCl = g = 4.2x10³ g Ag 4.102 + 0.02465 L NaOH soln 0.2500 mol NaOH X 1 mol 1 1L 2 mol NaOH 0.0500 =0.06163 M soln 4.103 (a) 10.45 g Sr(OH)₂ 1 mol Sr(OH)₂ X 0.05000 1 L = 1.7182 = 1.718 M Sr(OH)₂ (b) (c) Plan. mol Sr(OH)₂ = X mol ratio mol HNO₃ M HNO3. Solve. 1.7182 mol L Sr(OH)₂ x0.0239 L Sr(OH)₂ X 1 2 mol mol 0.0315 L 1 = 2.6073 = 2.61 M 4.104 mol OH- from NaOH(aq) + mol OH- from Zn(OH)₂(s) = mol from HBr mol = M HBr L HBr = 0.500 M HBr 0.350 L HBr = 0.175 mol H+ mol OH- from NaOH = M NaOH L NaOH = 0.500 M NaOH 0.0885 L NaOH = 0.04425 = 0.0443 mol OH- 102