I'd say no.
rate is a speed
10.6 seconds is a time.
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?
5 answers
how do i find the rate then?
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
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
I'm almost certain the thing changing is moles of one of the reactants or products.
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
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
0 answers