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defon
3 years ago
8

1.What percent of light passes through the sample if its absorbance A=2?

Chemistry
1 answer:
Dennis_Churaev [7]3 years ago
3 0

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)

<span>A = 1</span>

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Part A
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117 mL of 0.210 M K₂S solution

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The question asks about the volume of 0.210 M K₂S (potassium sulfide) solution required to completely react with 175 mL of 0.140 M Co(NO₃)₂ (cobalt(II) nitrate).

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K₂S + Co(NO₃)₂ → CoS + 2 KNO₃

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Learn more about:

molar concentration

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#learnwithBrainly

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