3a. The acid and base react to form a salt solution of ammonium propionate. CaCO s + 2 HC H O aq CO g + H O l + Ca ( aq ) + 2 C H O ( aq)
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1 Chapter 5 Answers Practice Examples 1a M Cl 1b. (a) M F - ; (b).1 kg CaF a. (a) Al ( aq ) OH ( aq ) Al ( OH) ( s) (b) No reaction occurs. (c) Pb ( aq ) I ( aq ) PbI ( s) b. (a) Al ( aq ) PO ( aq ) AlPO ( s) (b) Ba ( aq ) SO ( aq ) BaSO ( s) (c) Pb ( aq ) CO ( aq ) PbCO ( s) a. The acid and base react to form a salt solution of ammonium propionate. NH aq HC H O aq NH aq C H O aq ( ) ( ) ( ) ( ) 5 5 b. ( ) ( ) ( ) ( ) CaCO s HC H O aq CO g H O l Ca ( aq ) C H O a. (a) It is not an oxidation reduction reaction; (b) This is an oxidation reduction reaction. b. Vanadium is oxidized, manganese is reduced. { } { } 5a. ( ) ( ) Oxidation : Al s Al aq e Reduction: ( ) ( ) H aq e H g ( ) ( ) ( ) ( ) Net equation : Al s 6 H aq Al aq H g 5b. Oxidation : Br ( aq ) Br ( l ) e Reduction: Cl ( g ) e Cl Net equation : Br ( aq ) Cl ( g ) Br ( l ) Cl 6a. MnO ( aq ) 8 H ( aq ) 5 Fe ( aq ) Mn ( aq ) H O(l) 5 Fe 6b. ( ) ( ) ( ) ( ) ( ) UO aq Cr O aq 8 H aq UO aq Cr aq H O(l) 7 7a. ( ) ( ) ( ) S s OH aq OCl aq SO ( aq ) H O(l) Cl Copyright 011 Pearson Canada Inc. 1
2 7b. ( ) ( ) ( ) MnO aq SO aq H O(l) MnO s SO ( aq ) OH 8a. Since the oxidation state of H is 0 in H (g) and is 1 in both NH (g) and H O(g), hydrogen is oxidized. A substance that is oxidized is called a reducing agent. The oxidation state of the element N decreases during this reaction, meaning that NO (g) is reduced. The substance that is reduced is called the oxidizing agent. 8b. Au has been oxidized and, thus, Au(s) (oxidization state = 0), is the reducing agent. O has been reduced and thus, O (g) (oxidation state = 0) is the oxidizing agent. 9a M 9b M 10a. 65.% Fe 10b M Integrative Example A. 9.89% B.1.% Exercises 1a. Weak electrolyte 1b. Strong electrolyte 1c. Strong electrolyte 1d. Nonelectrolyte. 1e. Strong electrolyte. HCl is practically 100% dissociated into ions. The apparatus should light up brightly. A solution of both HCl and HC H O will yield similar results. 5a. Barium bromide: strong electrolyte 5b. Propionic acid: weak electrolyte 5c. Ammonia: weak electrolyte 7a. 0.8 M K 7b. 0. M NO 7c M Al Copyright 011 Pearson Canada Inc.
3 7d M Na M OH 11a. 11b. 11c M Ca M K. 10 M Zn 1. The solution containing 8.1 mg K per ml gives the largest K of the three solutions mg MgI M 19a. Pb ( aq ) Br ( aq ) PbBr ( s) 19b. No reaction occurs (all are spectator ions). 19c. Fe ( aq ) OH Fe( OH) ( s) 1a. No reaction occurs. 1b. ( ) Cu aq CO CuCO ( s) 1c. ( ) Cu aq PO ( aq ) Cu ( PO ) ( s) a. Add KSO ( aq ); BaSO ( s ) will form and MgSO will not precipitate. b. H O ( l ); Na CO ( ) s dissolves, but MgCO (s) will not dissolve (appreciably). c. Add KCl(aq); AgCl(s) will form, while Cu(NO ) (s) will dissolve. 5a. ( ) ( ) ( ) ( ) Sr NO aq K SO aq : Sr aq SO ( aq ) SrSO ( s) 5b. ( ) ( ) ( ) Mg NO aq NaOH aq : Mg ( aq ) OH ( aq ) Mg ( OH) ( s) 5c. ( ) ( ) BaCl aq K SO aq : Ba (aq) SO (aq) BaSO (s) 7a. OH ( aq ) HC H O ( aq ) H O( l ) C H O 7b. No reaction occurs. This is the physical mixing of two acids. 7c. FeS( s ) H ( aq ) H S( g ) Fe Copyright 011 Pearson Canada Inc.
4 7d. HCO ( aq ) H ( aq ) "H CO ( aq )" H O ( l ) CO ( g) 7e. Mg ( s ) H ( aq ) Mg ( aq ) H ( g) 9. As a salt: ( ) NaHSO aq Na ( aq ) HSO As an acid: HSO ( aq ) OH ( aq ) H O( l ) SO 1. Use (b), NH (aq). NH affords the OH - ions necessary to form Mg(OH) (s). a. The O.S. of H is 1, that of O is, that of C is, and that of Mg is on each side of this equation. This is not a redox equation. b. The O.S. of Cl is 0 on the left and 1 on the right side of this equation. The O.S. of Br is 1 on the left and 0 on the right side of this equation. This is a redox reaction. c. The O.S. of Ag is 0 on the left and 1 on the right side of this equation. The O.S. of N is 5 on the left and on the right side of this equation. This is a redox reaction. d. On both sides of the equation the O.S. of O is, that of Ag is 1, and that of Cr is 6. Thus, this is not a redox equation. 5a. ( ) ( ) ( ) SO aq 6 H aq e S O aq H O(l) 5b. NO ( aq ) 10 H ( aq ) 8 e N O( g ) 5 H O( l) 5c. ( ) ( ) ( ) ( ) Al s OH aq Al OH aq e 7a. 10 I ( aq ) MnO ( aq ) 16 H 5 I ( s ) Mn ( aq ) 8 H O( l) 7b. N H ( l ) BrO N ( g ) Br ( aq ) 6 H O( l) 7c. Fe ( aq ) VO ( aq ) 6 H ( aq ) Fe ( aq ) VO ( aq ) H O( l) 7d. ( ) ( ) ( ) UO aq NO aq H aq UO ( aq ) NO( g ) H O( l) 9a. ( ) ( ) ( ) ( ) ( ) MnO s ClO aq OH aq MnO aq Cl aq H O(l) 9b. ( ) ( ) ( ) ( ) ( ) ( ) Fe OH s OCl aq OH aq FeO aq Cl aq 5 H O(l) 9c. ( ) ( ) ( ) 6 ClO (aq) 6 OH aq 5ClO aq Cl aq H O 9d. Ag(s) CrO - H O(l) Ag(aq) Cr(OH) (s) 5 OH - Copyright 011 Pearson Canada Inc.
5 1a. ( ) ( ) ( ) ( ) Cl g 6 OH aq 5 Cl aq ClO aq H O(l) 1b. S O ( aq ) H O(l) HSO ( aq ) S O a. ( ) ( ) 5 NO aq MnO aq 6 H 5 NO ( aq ) Mn ( aq ) H O( l) b. Mn (aq) MnO - (aq) OH - (aq) 5 MnO (s) H O (l) c. ( ) ( ) ( ) ( ) Cr O aq 8 H aq C H OH Cr aq 7 H O l CH CHO 7 5 5a. CH (g) NO(g) N (g) CO (g) H O(g) 5b. 16 H S(g) 8 SO (g) S 8 (s) 16 H O(g) 5c. 10 NH (g) Cl O(g) 6 NH Cl(s) N (g) H O(l) 7a. SO is the reducing agent; MnO is the oxidizing agent. 7b. H ( g) is the reducing agent; NO ( ) g is the oxidizing agent. 7c. Fe ( CN) 6 is the reducing agent; HO( ) ml NaOH ( aq ) soln ml KOH solution M NaOH M NaOH 57. Acidic 59. ml base 61. Answer is (d) M % Fe g Na CO Integrative and Advanced Exercises 71. Ca (aq) HPO (aq) Ca (PO ) (s) H (aq) ppm Mg aq is the oxidizing agent. Copyright 011 Pearson Canada Inc. 5
6 L 80a. FeS (s) 15 O (g) H O(l) Fe (aq) 8 SO (aq) H (aq) 80b.. g CaCO 8..6 g Cl g ClO (g) 88a. 5.9 g 88b L 89. % Mg(OH) = 1.6; %Al(OH) = % 9a. CaO(s) H O(l) Ca (aq) OH (aq) 9b. 0.0 kg Feature Problems 9. x = % MnO H PO (aq) OH (aq) PO (aq) H O(l) HPO (aq) OH (aq) PO (aq) H O(l) 5 Ca (aq) PO (aq) OH (aq) Ca 5 (PO ) OH(s) 97. Before the breath test: M; After the breath test 10 mol/l. Self-Assessment Exercises 10. The answer is (b). 10. The answer is (d). 10. The answer is (c) The answer is (a) I Pb PbI (s) CO H H O (l) CO (g) 108a. Zn PO Zn (PO ) (s) Copyright 011 Pearson Canada Inc. 6
7 108b. Cu OH Cu(OH) (s) 108c. Ni CO NiCO (s) 109a. Species oxidized: N in NO 109b. Species reduced: O 109c. Oxidizing agent: O 109d. Reducing agent: NO 109e. Gains electrons: O 109f. Loses electrons: NO 110. The answer is (b) The answer is (d). 11. The answer is (a). 11a. False 11b. True 11c. False 11d. False 11e. True 11a. No 11b. Yes 11c. Yes 11d. No Copyright 011 Pearson Canada Inc. 7
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