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October 2, 2014

October 2, 2014

Posted by **JJ** on Monday, September 10, 2012 at 10:37am.

log10(rate) =

Would I just plug 10.60 in for the rate?

- chemistry. am i doing this right? -
**Steve**, Monday, September 10, 2012 at 11:30amI'd say no.

rate is a speed

10.6 seconds is a time.

- chemistry. am i doing this right? -
**JJ**, Monday, September 10, 2012 at 11:32amhow do i find the rate then?

- chemistry. am i doing this right? -
**Steve**, Monday, September 10, 2012 at 2:14pmno idea - read the problem again. Something is changing, and it changed by some amount in 10.6 seconds.

For example, if it's some volume, and it changed by 10m^3 in 10.6 seconds, then the rate would be

10m^3/10.6s = 0.94m^3/s

**That**is a rate

- chemistry. am i doing this right? -
**DrBob222**, Monday, September 10, 2012 at 3:11pmI'm almost certain the thing changing is moles of one of the reactants or products.

- chemistry. -
**Mac**, Monday, September 10, 2012 at 7:46pmthe answer is -log(10.6)

dont forget the negative

no, it doesn't make any sense to me, but i just did that same problem and that was the right answer

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