Recent content by Chaz32621

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...

Beer's law right??

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...

ahhh found it I actually found a table for it but I am still lost in how to do the math for it lol

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..