N for n 3 b n f 2on for n 4 4 for f 0 on for

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Unformatted text preview: ) = for O b ) N 2F 4: 2(O.N. for: N) + 4(–O.N. for N)0+ 1(+1 for H) + 2(–2 for O.N. 0 N = +2 .N. for N = +3 Solution: + c) NH4OH: (O.N. for N) + 4(+1 for H) = +1 + O.N. for N = –3 a) NH2 : 4.69 (O.N. for N) – 3(+1)for H) +) 1(–2 for O) = 0 O.N. for N = –1 ) –2 3 +4 3 +4 d) HNO24.139 a)(O.N. forbN)(11(+11for H) d=2(–2 for O)Al2O3(s) + AlCl3(s) + 6 H2O(g) + 3 NO(g) ( a) NH4ClO s) + c Al(s) =0 O.N. for N = +3 b) N F : : 2O.N. for N) 4++ 4(– for F)+ 0 O.N. for N = +2 24 4.69 4.70 4.69 c) NH +: (O.N. foronsult 4(+1 for H) =ammonium perchlorate;O.N. for reducing agent = aluminum N) + +1 N = –3 a) +4 4 b) 4.70 c) Plan: CO)xidizing agent = the rules for assigning oxidation numbers. –1 +4 d –2 Table 4.3 for Solution: d) HNO2: (O.N. for N) + 1(+1 for H) + 2(–2 for O) = 0 O.N. for N = +3 b) Moles .theis = for assigning oxidation its O.N. Plan: Consult Table a) AsH3of gascombined with a nonmetal, sonumbers.is +1 (Rule 3). The O.N. for As is –3. 4.3 for H– rules a) +4 b) –1 c) b) H2AsO4–.2The O.N. of H in this compound is +1, or +2 for 2 H’s. Oxygen’s O.N. is –2, with +4 d) Solution: total O.N. of –8 a) AsH3. H is combined with1276.70 = 1.28 x 103 mol gas (Rule 3). The O.N. for As is –3. its = a nonmetal, so to O.N. is +1 4 times –2),rules for assigning an O.N. numbers. (+5) + so As needs + + 4.70 Plan: Consult Table (4.3 for thethis compound haveoxidationof 2 5: +2Oxygen’s(–8) = –1. b) H2AsO4–. The O.N. AsClvolumes:an O.N. of is1, total of –3, so As must have an O.N. of –2, with c) c) of H3in has Initial . Cl – +1, or +2 for H’s. O.N. is +3. S of –8 total O.N. olution: a) As = –3a nonmetal,sso +5 O.N. is)+1s (Rule 3). The O.N. for As is –3. b) A = its c A = +3 a) times H is combined withNH an O.N. (4 AsH3. –2), so As needs to have ClO4 = of +5: +2 + (+5) + (–8) = –1. V = 25.6410 L (unrounded) 4 b) H2AsO4–. The O.N. of H in this compound As must have an H’s. of +3. c) AsCl3. Cl has an O.N. of –1, total of –3, so is +1, or +2 for 2 O.N. Oxygen’s O.N. is –2, with total O.N. )of –8 –3 a As = b) As = +5Al = c) As = +3 V (4 times –2), so As needs to have an O.N. of +5: +2 + (+5) + (–8) = –1. c) AsCl . Cl has an O.N. of –1, total of –3, so As must have an O.N. of +3....
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