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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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 Fall '06
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