Page 7 - KIV - Quimica 2 sec

Química 2° Secundaria

PROBLEMAS RESUELTOS
Hallar los estados de oxidación E.O.

x

1. H SO 2 3 1 S =
2
2 + x – 6 = 0 x = 4 +


2
2. K Cr O Cr =
1
x
2 2 7
2 + 2x – 14 = 0 x = +6

1

x
2
3.K Mn O Mn =
4
1 + x – 8 = 0 x = 7+


4. Na B O . 10 H O B =
1 x
2
2
4
2
7
+2 + 4x – 14 = 0 x = 3+



2
x
2
5. Ca (P O ) P =
2
4
3
+6 + 2x – 16 = 0 x = +5


6. NH N =
4
1
1
 x
(N H ) x + 4 = +1
4
x = -3

2
x

7. Cr O Cr =
4
x – 8 = -2  x = +6

to
4 Bimestre -148-