Answers

2014-07-27T14:22:10+05:30
Let oxidation number of Cr be x
2+2x-14=0
2x-12=0
2x=12
x=6
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2014-07-29T16:34:42+05:30
Suppose O.N. of Cr is x than we can write 2+2x-14=0 2x=12 x=6
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