Question

The spectrophotometer really measures the percent of light that is transmitted through the solution. The instrument then converts %T (transmittance) into absorbance by using the equation you determined in the prelab section. If the absorbance is 0.85, what is the percent light transmitted through the color sample at this collected wavelength?

Answer 1

the equation is
y=0.073487(conc)+0.024531

Answer 2

what is the y variable in the equation?

you can use: \(Abs = -log (\dfrac{\%T}{100})\)

you should double check that formula in your notes, i don't know if it's 100% correct

Answer 3

my bad wrong one
that is for something else

Answer 4

so the equation is actually: \(Abs=-log(T)\)

you just need to convert that to a percent after

Answer 5

\[-\log_{.85/100} \]

Answer 6

-log(85%)

Answer 7

is the answer 1.92941

Answer 8

i don't think that's right.

Answer 9

it shouldn't be over 100%

Answer 10

i g-0.07058 after redoing and changing it to a decimal

Answer 11

-log(1-0.85)

Answer 12

absorbance is on the other side of the equal sign

Answer 13

-log(0.85)

Answer 14

no absorbance is on the other side, you're plugging into where transmittance goes

Answer 15

I do not understand

Answer 16

\(\huge \color{red}{Abs}=-log(T)\)

Answer 17

can you show me what you mean

Answer 18

0.85=-log(T)

Answer 19

okay
a

Answer 20

so I just need to figure out how to find T right

Answer 21

then convert it to a percent

Answer 22

so do I multiple 0.85 by 100 to get t then

Answer 23

no, once you find T, multiply by 100%