Cu is +2
S is S
O is -2. 4*-2 = -8
All compounds add to zero.
Cu + S + 4*-2 = 0
2 + S +(-8) = 0
2 + S + -8 = 0
Solve for S.
S = 0 - 2 + 8 = ?
The oxidation number of sulphur in CuSO4.5H2O is ......
12 answers
10
R7o3b6e2l5
0973625793
+6 because SO4 has no bracket
The oxidizing agent is oxizided by the reduced agent
Ansewor
What about Hydrogen
What about Hydrogen
CuSo4,5H2O
Cu=+2
O=-2*4 =-8
+2+x-8+2-2=0
X=+6
Cu=+2
O=-2*4 =-8
+2+x-8+2-2=0
X=+6
6
The oxidation number of sulphur in CuSO4.5H2O is +6.