Wow lol got it! Thanks man!
So it should be -
C = A/(eb)
A = absorbance
e = mole absorabtion
b = path length
c = concentration
0.555/(5250)(1.00) which is -
1.06E-4 which is 0.0000106
Lets say I got it right, haha and a 100 on the prelab thanks for the common sense help lol...
So I have tried this problem so many times now lol. I have one more attempt at the problem for credit. So I first tried to do:
absorbance = eLc
0.555 = (5.25 * 1.00) *C which I got C = 5.25 *1.00/.555
But this was wrong so I attempted to do:
C = A/(eb)
A = absorbance
e = mole...
What is the concentration of chromate??
Homework Statement
A solution of K2CrO4 has an absorbance value 0.555 in a cell of 1.00cm path-length at a wavelength of 370nm. The value of the molar absorbtivity is 5.25x103M-1cm-1 for K2CrO4 at the wavelength. What is the concentration of this...
1. A solution exhibits 58.8 of %Trasmittance value. Convert this value to absorbance. Report with Three significant figures.
Not sure how to do this?? Do I subtract 100 percent from 58.8 I have no clue what to do lol..