devilsadvocate1728

The original problem is to balance the following redox equation "in base":

CrO4= + HSnO2- ↔ CrO2- + HSnO3-

Problem is that when I try to put OH- on the left and H2O on the right, I get a negative coefficient for OH-. I have similar problens when I try to put the OH- on the right instead. What am I missing here?

Most multiply-charged anions have less-charged versions stable in more acidic solutions, including the chromate ion CrO4--. However, most such anions are partially acidified simply by adding one or more hydrogen atoms to them, such as HCO3-, HPO4-- and H2PO4-, HSO4-, and so on. The partially acidified chromate ion is different in this respect, where two of the basic Cr2O4-- ions combine to form a Cr2O7-- ion rather than two HCrO4- ions implied by the older name for the dichromate ion "bichromate." Who, when, and how was this anomalous behavior discovered and what chemical consequences does it have?

I am also looking for references for more detailed info. These must be provided for "best answer."

This expression occurred when I tried to simplify the expression for degrees of freedom in a 2-sample t-test

[ (s1^2/N1 + s2^2/N2)^2 ] / [ (s1^2/N1)^2 / (N1 - 1) + (s2^2/N2)^2 / (N2 - 1) ]

where s1 = s2. According to my source at http://www.itl.nist.gov/div898/handbook/eda/sectio...

the above expression should simplify to

N1 + N2 - 2

But getting from here to there has proved a lot more difficult than I thought it might be.

What am I missing?

For the purposes of this problem, it may be assumed that:

* the probability density is continuous for the individual numbers

* the rounding is symmetric

* the rounding increments are small enough that the probability density varies slowly enough that the probability density is essentially uniform over the limits of each increment.

Also necessary is a mathematical justification for your answer, the more mathematically rigorous the better.