Posted by JJ on Monday, September 10, 2012 at 10:37am.
1.) One of the trials in this week's experiment in chemical kinetics was determined to be 10.60 seconds. Evaluate log10(rate).
Would I just plug 10.60 in for the rate?
- chemistry. am i doing this right? - Steve, Monday, September 10, 2012 at 11:30am
I'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 11:32am
how do i find the rate then?
- chemistry. am i doing this right? - Steve, Monday, September 10, 2012 at 2:14pm
no 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 3:11pm
I'm almost certain the thing changing is moles of one of the reactants or products.
- chemistry. - Mac, Monday, September 10, 2012 at 7:46pm
the 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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