Log rate law problem

[r12214001]

Homework Statement

calculation

Relevant Equations

log problem

my question is in fig.

 

[lurflurf]

log0.1=-2.303
because that is what it is when you calculate it
did you calculate it?

you do not need to calculate it thought as the log and antilog cancel
$$\mathrm{antilog}(-1.040+\log(0.010))=\mathrm{antilog}(-1.040)\cdot\mathrm{antilog}(\log(0.100))
=\mathrm{antilog}(-1.040)\cdot(0.100)$$

 

[Borek]

ln means natural logarithm, base e.

 

[r12214001]

TKS i get it

 

[Chestermiller]

$$\ln{0.1}=\ln{\left(\frac{1}{10}\right)}=\ln{1}-\ln{10}=0-2.303=-2.303$$