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?
I'd say no.
rate is a speed
10.6 seconds is a time.
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
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
To evaluate log10(rate), you would need the actual value for the rate. If the rate was given as 10.60, then yes, you would simply plug in 10.60 into the equation.
Therefore, log10(rate) = log10(10.60)