(a) Interpretation: The percent transmittance of the solution for the absorbance of 0.042 is to be calculated. Concept introduction: The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution. The relation between the absorbance and transmittance is established by the Beer’s law. A = − log T

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 13, Problem 13.1QAP

Interpretation Introduction

(a)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.042$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.042$ .

Substitute the value in above equation.

$T%=(10−A)100T%=(10(−0.042))100=(0.908)100=90.8%$

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Interpretation Introduction

(b)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.917$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

Write the expression for the Beer’s law.

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.917$ .

Substitute $0.917$ for $A$ in equation (1).

$T%=(10−A)100T%=(10(−0.917))100=(0.121)100=12.1%$

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Interpretation Introduction

(c)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.245$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.245$ .

Substitute $0.245$ for $A$ in equation (1).

$T%=(10(−0.245))100=(0.569)100=56.9%$

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Interpretation Introduction

(d)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.379$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.379$ for $A$ in equation (1).

$T%=(10(−0.379))100=(0.418)100=41.8%$

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Interpretation Introduction

(e)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.435$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.435$ for $A$ in equation (1).

$T%=(10(−0.435))100=(0.367)100=36.7%$

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Interpretation Introduction

(f)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.513$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.513$ .

Substitute $0.513$ for $A$ in equation (1).

$T%=(10(−0.513))100=(0.307)100=30.7%$

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Draw product B, indicating what type of reaction occurs. F3C CF3 NH2 Me O .N. + B OMe

Benzimidazole E. State its formula. sState the differences in the formula with other benzimidazoles.

Draw product A, indicating what type of reaction occurs. F3C CN CF3 K2CO3, DMSO, H₂O2 A

Answer 2

Question

Chapter 13, Problem 13.1QAP

Interpretation Introduction

(a)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.042$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.042$ .

Substitute the value in above equation.

$T%=(10−A)100T%=(10(−0.042))100=(0.908)100=90.8%$

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Interpretation Introduction

(b)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.917$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

Write the expression for the Beer’s law.

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.917$ .

Substitute $0.917$ for $A$ in equation (1).

$T%=(10−A)100T%=(10(−0.917))100=(0.121)100=12.1%$

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Interpretation Introduction

(c)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.245$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.245$ .

Substitute $0.245$ for $A$ in equation (1).

$T%=(10(−0.245))100=(0.569)100=56.9%$

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Interpretation Introduction

(d)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.379$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.379$ for $A$ in equation (1).

$T%=(10(−0.379))100=(0.418)100=41.8%$

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Interpretation Introduction

(e)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.435$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.435$ for $A$ in equation (1).

$T%=(10(−0.435))100=(0.367)100=36.7%$

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Interpretation Introduction

(f)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.513$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.513$ .

Substitute $0.513$ for $A$ in equation (1).

$T%=(10(−0.513))100=(0.307)100=30.7%$

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 13, Problem 13.1QAP

Interpretation Introduction

(a)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.042$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.042$ .

Substitute the value in above equation.

$T%=(10−A)100T%=(10(−0.042))100=(0.908)100=90.8%$

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Interpretation Introduction

(b)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.917$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

Write the expression for the Beer’s law.

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.917$ .

Substitute $0.917$ for $A$ in equation (1).

$T%=(10−A)100T%=(10(−0.917))100=(0.121)100=12.1%$

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Interpretation Introduction

(c)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.245$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.245$ .

Substitute $0.245$ for $A$ in equation (1).

$T%=(10(−0.245))100=(0.569)100=56.9%$

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Interpretation Introduction

(d)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.379$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.379$ for $A$ in equation (1).

$T%=(10(−0.379))100=(0.418)100=41.8%$

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Interpretation Introduction

(e)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.435$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.435$ for $A$ in equation (1).

$T%=(10(−0.435))100=(0.367)100=36.7%$

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Interpretation Introduction

(f)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.513$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.513$ .

Substitute $0.513$ for $A$ in equation (1).

$T%=(10(−0.513))100=(0.307)100=30.7%$

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 13, Problem 13.1QAP

Interpretation Introduction

(a)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.042$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.042$ .

Substitute the value in above equation.

$T%=(10−A)100T%=(10(−0.042))100=(0.908)100=90.8%$

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Interpretation Introduction

(b)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.917$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

Write the expression for the Beer’s law.

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.917$ .

Substitute $0.917$ for $A$ in equation (1).

$T%=(10−A)100T%=(10(−0.917))100=(0.121)100=12.1%$

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Interpretation Introduction

(c)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.245$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.245$ .

Substitute $0.245$ for $A$ in equation (1).

$T%=(10(−0.245))100=(0.569)100=56.9%$

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Interpretation Introduction

(d)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.379$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.379$ for $A$ in equation (1).

$T%=(10(−0.379))100=(0.418)100=41.8%$

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Interpretation Introduction

(e)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.435$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.435$ for $A$ in equation (1).

$T%=(10(−0.435))100=(0.367)100=36.7%$

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Interpretation Introduction

(f)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.513$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.513$ .

Substitute $0.513$ for $A$ in equation (1).

$T%=(10(−0.513))100=(0.307)100=30.7%$

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 13, Problem 13.1QAP

Interpretation Introduction

(a)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.042$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.042$ .

Substitute the value in above equation.

$T%=(10−A)100T%=(10(−0.042))100=(0.908)100=90.8%$

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Interpretation Introduction

(b)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.917$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

Write the expression for the Beer’s law.

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.917$ .

Substitute $0.917$ for $A$ in equation (1).

$T%=(10−A)100T%=(10(−0.917))100=(0.121)100=12.1%$

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Interpretation Introduction

(c)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.245$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.245$ .

Substitute $0.245$ for $A$ in equation (1).

$T%=(10(−0.245))100=(0.569)100=56.9%$

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Interpretation Introduction

(d)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.379$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.379$ for $A$ in equation (1).

$T%=(10(−0.379))100=(0.418)100=41.8%$

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Interpretation Introduction

(e)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.435$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.435$ for $A$ in equation (1).

$T%=(10(−0.435))100=(0.367)100=36.7%$

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Interpretation Introduction

(f)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.513$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.513$ .

Substitute $0.513$ for $A$ in equation (1).

$T%=(10(−0.513))100=(0.307)100=30.7%$

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Answer 6

Chapter 13, Problem 13.1QAP

Interpretation Introduction

(a)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.042$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.042$ .

Substitute the value in above equation.

$T%=(10−A)100T%=(10(−0.042))100=(0.908)100=90.8%$

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Answer 7

Chapter 13, Problem 13.1QAP

Interpretation Introduction

(a)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.042$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Answer 8

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.042$ .

Substitute the value in above equation.

$T%=(10−A)100T%=(10(−0.042))100=(0.908)100=90.8%$

The percent transmittance for the absorbance of $0.042$ is $90.8%$ .

Answer 13

Interpretation Introduction

(b)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.917$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

Write the expression for the Beer’s law.

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.917$ .

Substitute $0.917$ for $A$ in equation (1).

$T%=(10−A)100T%=(10(−0.917))100=(0.121)100=12.1%$

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Answer 14

Interpretation Introduction

(b)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.917$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Answer 15

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

Write the expression for the Beer’s law.

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.917$ .

Substitute $0.917$ for $A$ in equation (1).

$T%=(10−A)100T%=(10(−0.917))100=(0.121)100=12.1%$

The percent transmittance for the absorbance of $0.917$ is $12.1%$ .

Answer 20

Interpretation Introduction

(c)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.245$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.245$ .

Substitute $0.245$ for $A$ in equation (1).

$T%=(10(−0.245))100=(0.569)100=56.9%$

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Answer 21

Interpretation Introduction

(c)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.245$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Answer 22

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Answer 23

Expert Solution

Answer 24

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ...... (1)

The value for $A$ is $0.245$ .

Substitute $0.245$ for $A$ in equation (1).

$T%=(10(−0.245))100=(0.569)100=56.9%$

The percent transmittance for the absorbance of $0.245$ is $56.9%$ .

Answer 27

Interpretation Introduction

(d)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.379$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.379$ for $A$ in equation (1).

$T%=(10(−0.379))100=(0.418)100=41.8%$

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Answer 28

Interpretation Introduction

(d)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.379$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Answer 29

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Answer 30

Expert Solution

Answer 31

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.379$ for $A$ in equation (1).

$T%=(10(−0.379))100=(0.418)100=41.8%$

The percent transmittance for the absorbance of $0.379$ is $41.8%$ .

Answer 34

Interpretation Introduction

(e)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.435$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.435$ for $A$ in equation (1).

$T%=(10(−0.435))100=(0.367)100=36.7%$

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Answer 35

Interpretation Introduction

(e)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.435$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Answer 36

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Answer 37

Expert Solution

Answer 38

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.379$ .

Substitute $0.435$ for $A$ in equation (1).

$T%=(10(−0.435))100=(0.367)100=36.7%$

The percent transmittance for the absorbance of $0.435$ is $36.7%$ .

Answer 41

Interpretation Introduction

(f)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.513$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.513$ .

Substitute $0.513$ for $A$ in equation (1).

$T%=(10(−0.513))100=(0.307)100=30.7%$

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Answer 42

Interpretation Introduction

(f)

Interpretation:

The percent transmittance of the solution for the absorbance of $0.513$ is to be calculated.

Concept introduction:

The percent transmittance of a solution is the ability of the solution to allow the monochromatic light to pass through it. The percent transmittance of a solution is the ratio of the intensity of monochromatic light incident on the solution to the intensity transmitted through the solution.

The relation between the absorbance and transmittance is established by the Beer’s law.

$A=−logT$

Answer 43

Expert Solution

Answer to Problem 13.1QAP

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Answer 44

Expert Solution

Answer 45

Expert Solution

Answer 46

Expert Solution

Answer 47

Explanation of Solution

The relation between the absorbance and transmittance is related by the Beer’s law.

The expression for the Beer’s law is:

$A=−logT$

Where,

$A$ is the absorbance of the solution.
$T$ is the transmittance of the solution.

Rearrange the above expression in order to obtain the percent transmittance

$10−A=10(log(T%/100))$

Simplify the expression for percent transmittance.

$T%=(10−A)100$ ......(1)

The value for $A$ is $0.513$ .

Substitute $0.513$ for $A$ in equation (1).

$T%=(10(−0.513))100=(0.307)100=30.7%$

The percent transmittance for the absorbance of $0.513$ is $30.7%$ .

Answer 48

Ch. 13 - Prob. 13.1QAP Ch. 13 - Prob. 13.2QAP Ch. 13 - Prob. 13.3QAP Ch. 13 - Prob. 13.4QAP Ch. 13 - Prob. 13.5QAP Ch. 13 - Prob. 13.6QAP Ch. 13 - Prob. 13.7QAP Ch. 13 - At 580 nm, which is the wavelength of its maximum...Ch. 13 - Prob. 13.9QAP Ch. 13 - Zinc(II) and the ligand L form a 1:1 complex that...

Answer to Problem 13.1QAP

Explanation of Solution

Answer to Problem 13.1QAP

Explanation of Solution

Answer to Problem 13.1QAP

Explanation of Solution

Answer to Problem 13.1QAP

Explanation of Solution

Answer to Problem 13.1QAP

Explanation of Solution

Answer to Problem 13.1QAP

Explanation of Solution

Want to see more full solutions like this?

Chapter 13 Solutions