quiz_3_soln

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online