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RE: #Chemistry Challenge 5
Ah, that's how many Liters are required for 100 km, yet we're going 608 km. SO instead, 200 should be divided by 30.4 making the answer 1.50 x 10^-4 mol∙l^-1∙s^-1
Ah, that's how many Liters are required for 100 km, yet we're going 608 km. SO instead, 200 should be divided by 30.4 making the answer 1.50 x 10^-4 mol∙l^-1∙s^-1
Sounds really good now - the only thing I wonder about is why you are using the factor 0.5?
Don't you need to use it to account for the molar ratios of the balanced equation?
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). :)