# chemistry. am i doing this right?

1.) One of the trials in this week's experiment in chemical kinetics was determined to be 10.60 seconds. Evaluate log10(rate).

log10(rate) =

Would I just plug 10.60 in for the rate?

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1. I'd say no.

rate is a speed
10.6 seconds is a time.

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2. how do i find the rate then?

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3. 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

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4. I'm almost certain the thing changing is moles of one of the reactants or products.

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5. 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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