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7B.13 Numerator

Posted: Tue Mar 10, 2020 8:05 pm
by Emil Velasco 1H
For this problem, when it asks for one sixteenth or one fourth, we change the numerator (e.g. to 16/[A]o).
What's the math behind this?

Re: 7B.13 Numerator

Posted: Tue Mar 10, 2020 8:12 pm
by Alicia Lin 2F
This is because, for example, part a is asking for 1/16 of the original concentration. In the solution manual, they are setting [A]=(1/16)[A]0. Plugging this value into the 2nd order linear equation, you get 1/([A]0/16)=kt+(1/[A]0). The left side of this equation written in simpler terms is 16/[A]0. Use this same logic for the rest of the problem too.

Re: 7B.13 Numerator

Posted: Tue Mar 10, 2020 11:14 pm
by Sue Bin Park 2I
could we not just set [A] = 1/16 and [A]initial = 1?
EDIT: oh nvm we are given [A]initial. my bad