Question

Some notes for my self.

Answer 1

\[\nabla D(r)\nabla V _{e}(r)-D(r)K ^{2}\sinh[V _{e}(r)]+\frac{ 4 \pi Ze \rho(r) }{ k _{B} T}=0\]

Answer 2

\[K=\frac{ 8 \pi (Ze)^{2}I }{ D(r)k _{B}T }\]

Answer 3

\[I=\sum_{i=1}^{N}\frac{ Z _{i}^{2} c _{i} }{ 2 }\]

Answer 4

\[\large pH=pK _{a} + \log \left( \frac{ \left[ B \right] }{ \left[ A \right] } \right)\]
\[\large \left[ B \right]=\left[ S \right] ~ 10^{pH-pK _{a}}\]
@C9H8O4

Answer 5

Thanks :)

Answer 6

Unfortunately I can't evaluate the answer my self as I don't have a calculator :/

Answer 7

I figured it out, I was just unsure on the equation that I used when rearranging it, but now, the missing base of 1mol, enables the pka+log(base/acid) to be the same pH as given

Answer 8

Right, well, always best if you figure it out your self! :)

Answer 9

\[\large k _{r}(T)=A ~ \exp(\frac{ - B }{ T })\]
it did not correct it to B.

Answer 10

\[\large t _{1/2}(T)\frac{ \ln(2) }{ A*\exp(\frac{ B }{ T }) }\]

Answer 11

sorry my head is exploding.

Answer 12

Assuming T is always positive, it should look something like Except instead of 1, it would be some combination of the cosntants.

Answer 13

I think the limit would be \(\ln(2)/A\).

Answer 14

Looks just about what I expected, thanks man you are the best!

Answer 15

no problem