Chemistry Practice Exam 1 Discussion

[Stroodle]

n(OH-)=0.21
n(H+)=0.12

0.21-0.12=0.09 mol of OH-

C(OH-)=0.09/0.6=0.15

C(H+)=10^(-14)/0.15

-log(10^(-14)/0.15)=13.18

[chansthename]

Ahh, I didn't read the question (again)

I just found the pH of Ca(OH)2
wouldn't have got it anyway don't think.

[crayolé]

n(OH-)=0.21
n(H+)=0.12

0.21-0.12=0.09 mol of OH-

C(OH-)=0.09/0.6=0.15

C(H+)=10^(-14)/0.15

-log(10^(-14)/0.15)=13.18

Thanks :3

Sorry to turn this into a question thread but
An aqueous solution of potassium dichromate has a clear orange colour.
The answers say that during UV Vis, it absorbs blue light and allows orange to pass through, i thought it reflected the orange light? :/ thus measuring the amount of blue that passed through the solution (wasn't absorbed)

[vexx]

n(OH-)=0.21
n(H+)=0.12

0.21-0.12=0.09 mol of OH-

C(OH-)=0.09/0.6=0.15

C(H+)=10^(-14)/0.15

-log(10^(-14)/0.15)=13.18

Thanks :3

Sorry to turn this into a question thread but
An aqueous solution of potassium dichromate has a clear orange colour.
The answers say that during UV Vis, it absorbs blue light and allows orange to pass through, i thought it reflected the orange light? :/ thus measuring the amount of blue that passed through the solution (wasn't absorbed)

yeah, the solution of an orange colour will be:
-absorbing blue light
-reflecting orange light as it is not absorbing this, so if orange light is shined onto it, since it cannot absorb this light then it will just pass through.