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RE: #Chemistry Challenge 5
No, if we had 6.579 mol/l isooctane at the beginning and 0 after 6.08 h, than ∆c in the timescale was - 6.579 mol/l.
So what is v now finally? :)
No, if we had 6.579 mol/l isooctane at the beginning and 0 after 6.08 h, than ∆c in the timescale was - 6.579 mol/l.
So what is v now finally? :)
Well, with this little help, we shall get v = - (delta c)/(delta t) = - (-6.579 mol/l)/21888 s = 3 x 10^(-4) mol/(l*s)
(sorry for the bad formatting)
Congrats @thepe! Finally you are the one who found the solution of part 4! Congrats also to @bluejay188 and @mcw who solved parts 1 to 3 and worked hard to find the solution of part 4 (and nearly solved it as well). :)