1.what percent of light passes through the sample if its absorbance a=2? 2. for the best calibration curve, the transmittance values (%t) should fall within the range from 10% to 90%. what is the approximate range of corresponding absorbance values? 3. what is the absorbance of the solution if the percent transmittance for it is 10%?

1. The formula for absorbance is given as:

A = log (Io / I)

where A is absorbance, Io is initial intensity, and I is final light intensity

log (Io / I) = A

log (Io / I) = 2

Io / I = 100

Taking the reverse which is I / Io:

I / Io = 1 / 100

I / Io = 0.01

Therefore this means that only 0.01 fraction of light or 1% passes through the sample.

2. What is meant by transmittance values is actually the value of I / Io. So calculating for A:

at 10% transmittance = 0.10

A = log (Io / I)

A = log (1 / 0.10)

A = 1

at 90% transmittance = 0.90

A = log (Io / I)

A = log (1 / 0.90)

A = 0.046

So the absorbance should be from 0.046 to 1

3. at 10% transmittance = 0.10

A = log (Io / I)

A = log (1 / 0.10)

A = 1

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

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