Asked by Feather
Determining oxidation numbers for remaining atoms in HIO3, H3PO4, Ba(OH)2, and BaPO4
I have no idea if I am doing this right I don't even know how to do this
For HIO3 I added the charges as +1 and 3(-2) which i got -5 then I did it the similar way for the others and the answers I got is for H3PO4 is -3
Then for Ba(OH)2 I got 0 then for BaPO4 I got 22.
Please show me how to do this correctly....
I have no idea if I am doing this right I don't even know how to do this
For HIO3 I added the charges as +1 and 3(-2) which i got -5 then I did it the similar way for the others and the answers I got is for H3PO4 is -3
Then for Ba(OH)2 I got 0 then for BaPO4 I got 22.
Please show me how to do this correctly....
Answers
Answered by
Mahnoor
I'm explaining for HI03
First we assign oxidation numbers to Oxygen and Hydrogen then to any other atom.
So oxygen is -2*3 and H is +1.
So now let I's oxidation number be x.
+1+x+(3*-2)=0
(0 because the overall compound has 0 charge)
+1+x-6=0
x-5=0
x=5
First we assign oxidation numbers to Oxygen and Hydrogen then to any other atom.
So oxygen is -2*3 and H is +1.
So now let I's oxidation number be x.
+1+x+(3*-2)=0
(0 because the overall compound has 0 charge)
+1+x-6=0
x-5=0
x=5
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