Molar absorption coefficient of aqueous solution containing triphosphate derivative has to be calculated where absorbance is 1.011 and cell length is 1 .5 mm . Concept Introduction: Equation for absorbance, A can be given as, A = − log T A = ε [ J ] L w h e r e , I = I n t e n s i t y o f i n c i d e n t r a d i a t i o n I 0 = I n t e n s i t y o f t r a n s m i t t e d r a d i a t i o n N = N u m b e r o f s p e c i e s a b s o r b e d [ J ] = M o l a r c o n c e n t r a t i o n L = P a t h l e n g t h

Question

Want to see the full answer?

Answer 1

Question

Chapter 11, Problem 11A.3E

(a)

Interpretation Introduction

Interpretation:

Molar absorption coefficient of aqueous solution containing triphosphate derivative has to be calculated where absorbance is 1.011 and cell length is 1.5 mm.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

(b)

Interpretation Introduction

Interpretation:

Transmittance of aqueous solution containing twice the concentration has to be calculated in percentage.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely: (a) 0.200 M HCl

Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely: (a) 0.000259 M HClO4

What is the pH of a 1.0 L buffer made with 0.300 mol of HF (Ka = 6.8 × 10⁻⁴) and 0.200 mol of NaF to which 0.160 mol of NaOH were added?

Answer 2

Question

Chapter 11, Problem 11A.3E

(a)

Interpretation Introduction

Interpretation:

Molar absorption coefficient of aqueous solution containing triphosphate derivative has to be calculated where absorbance is 1.011 and cell length is 1.5 mm.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

(b)

Interpretation Introduction

Interpretation:

Transmittance of aqueous solution containing twice the concentration has to be calculated in percentage.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Question

Chapter 11, Problem 11A.3E

(a)

Interpretation Introduction

Interpretation:

Molar absorption coefficient of aqueous solution containing triphosphate derivative has to be calculated where absorbance is 1.011 and cell length is 1.5 mm.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

(b)

Interpretation Introduction

Interpretation:

Transmittance of aqueous solution containing twice the concentration has to be calculated in percentage.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Question

Chapter 11, Problem 11A.3E

(a)

Interpretation Introduction

Interpretation:

Molar absorption coefficient of aqueous solution containing triphosphate derivative has to be calculated where absorbance is 1.011 and cell length is 1.5 mm.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

(b)

Interpretation Introduction

Interpretation:

Transmittance of aqueous solution containing twice the concentration has to be calculated in percentage.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Chapter 11, Problem 11A.3E

(a)

Interpretation Introduction

Interpretation:

Molar absorption coefficient of aqueous solution containing triphosphate derivative has to be calculated where absorbance is 1.011 and cell length is 1.5 mm.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

(b)

Interpretation Introduction

Interpretation:

Transmittance of aqueous solution containing twice the concentration has to be calculated in percentage.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

Answer 6

Chapter 11, Problem 11A.3E

(a)

Interpretation Introduction

Interpretation:

Molar absorption coefficient of aqueous solution containing triphosphate derivative has to be calculated where absorbance is 1.011 and cell length is 1.5 mm.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

Answer 7

Chapter 11, Problem 11A.3E

(a)

Interpretation Introduction

Interpretation:

Molar absorption coefficient of aqueous solution containing triphosphate derivative has to be calculated where absorbance is 1.011 and cell length is 1.5 mm.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

Answer 8

(b)

Interpretation Introduction

Interpretation:

Transmittance of aqueous solution containing twice the concentration has to be calculated in percentage.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

Answer 9

(b)

Interpretation Introduction

Interpretation:

Transmittance of aqueous solution containing twice the concentration has to be calculated in percentage.

Concept Introduction:

Equation for absorbance, A can be given as,

A = −log TA = ε[J]Lwhere,I = Intensity of incident radiationI0 = Intensity of transmitted radiationN = Number of species absorbed[J] = Molar concentration L = Path length

Answer 10

Expert Solution & Answer

Answer 11

Expert Solution & Answer

Answer 12

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 13

Ch. 11 - Prob. 11A.1ST Ch. 11 - Prob. 11A.2ST Ch. 11 - Prob. 11A.3ST Ch. 11 - Prob. 11B.1ST Ch. 11 - Prob. 11B.2ST Ch. 11 - Prob. 11B.3ST Ch. 11 - Prob. 11B.4ST Ch. 11 - Prob. 11B.5ST Ch. 11 - Prob. 11C.1ST Ch. 11 - Prob. 11C.2ST

Concept explainers

Want to see the full answer?

Chapter 11 Solutions