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### Topic: Volume of a solution  (Read 4411 times)

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#### Tom

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##### Volume of a solution
« on: July 01, 2004, 04:35:30 PM »
The basic solution, MnO4 ¯ oxidizes NO2¯ to NO3¯ and is reduced to MnO2.  What volume of 0.10 M KmnO4 would be needed to oxidize 30ml of a 0.10 M NaNO2 solution?

My conclusion came out that it would require 40ml of solution, does that sound logical? Again the all the numbers are subscripted other than the 0.10M and 30ml.

Thanks

#### Mitch

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##### Re:Volume of a solution
« Reply #1 on: July 01, 2004, 05:54:31 PM »
I suggest starting by writing the balanced chemical equation.
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#### Tom

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##### Re:Volume of a solution
« Reply #2 on: July 02, 2004, 11:22:10 AM »
I used this formula

D=M divided by V

#### Mitch

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##### Re:Volume of a solution
« Reply #3 on: July 03, 2004, 03:56:45 PM »
I don't think that'll do it. You'll have to write the balanced chemical formula.
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