Organic Chemistry
Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
Question
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Chapter 15, Problem 15.9P
Interpretation Introduction

(a)

Interpretation:

The concentration of isoprene in molL1 from the extinction coefficient and the observed absorbance at 222.5nm is to be calculated.

Concept introduction:

The Lambert-Beer law says that the material absorbance is dependent upon the factors such as thickness of the sample and the concentration of the solution. According to the law, the absorbance is directly proportional to the concentration as well as the thickness of the sample. It also tells the relation between the intensity of light and the absorption by the material.

Interpretation Introduction

(b)

Interpretation:

The extinction coefficient of isoprene at 235nm with reference to the concentration of isoprene solution is to be calculated.

Concept introduction:

The Lambert-Beer law says that the material absorbance is dependent upon the factors such as thickness of the sample and the concentration of the solution. According to the law, the absorbance is directly proportional to the concentration as well as the thickness of the sample. It also tells the relation between the intensity of light and the absorption by the material.

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The DPD colorimetric method is used to determine the amount of residual chlorine in the sample. In this method, the free chlorine in the sample oxidizes the colorless amine N,N-diethyl-p-phenylenediamine to a colored compound that absorbs strongly at 520 nm. The analysis of a set of calibration standards gave the following results:ppm Cl2 Absorbance0.00 0.00310.50 0.12321.00 0.32411.50 0.54382.00 0.76422.50 0.95513.00 1.1565A 10.00 mL water sample from a public water supply was diluted to 250 mL and analyzed for free chlorine residual, giving an absorbance of 0.3210.A. Find the equation of the line.B. What is the molar absorptivity of Cl2 (FW = 70.9) at 250 nm?C. Calculate the free chlorine residual of the sample as ppm Cl2.D. Calculate ppm BaCl2 (FW = 208.2)?

Chapter 15 Solutions

Organic Chemistry

Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Prob. 15.24PCh. 15 - Prob. 15.25PCh. 15 - Prob. 15.26PCh. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Prob. 15.41PCh. 15 - Prob. 15.42APCh. 15 - Prob. 15.43APCh. 15 - Prob. 15.44APCh. 15 - Prob. 15.45APCh. 15 - Prob. 15.46APCh. 15 - Prob. 15.47APCh. 15 - Prob. 15.48APCh. 15 - Prob. 15.49APCh. 15 - Prob. 15.50APCh. 15 - Prob. 15.51APCh. 15 - Prob. 15.52APCh. 15 - Prob. 15.53APCh. 15 - Prob. 15.54APCh. 15 - Prob. 15.55APCh. 15 - Prob. 15.56APCh. 15 - Prob. 15.57APCh. 15 - Prob. 15.58APCh. 15 - Prob. 15.59APCh. 15 - Prob. 15.60APCh. 15 - Prob. 15.61APCh. 15 - Prob. 15.62APCh. 15 - Prob. 15.63APCh. 15 - Prob. 15.64APCh. 15 - Prob. 15.65APCh. 15 - Prob. 15.66APCh. 15 - Prob. 15.67APCh. 15 - Prob. 15.68APCh. 15 - Prob. 15.69APCh. 15 - Prob. 15.70APCh. 15 - Prob. 15.71APCh. 15 - Prob. 15.72APCh. 15 - Prob. 15.73APCh. 15 - Prob. 15.74APCh. 15 - Prob. 15.75APCh. 15 - Prob. 15.76APCh. 15 - Prob. 15.77APCh. 15 - Prob. 15.78APCh. 15 - Prob. 15.79APCh. 15 - Prob. 15.80APCh. 15 - Prob. 15.81APCh. 15 - Prob. 15.82APCh. 15 - Prob. 15.83APCh. 15 - Prob. 15.84APCh. 15 - Prob. 15.85APCh. 15 - Prob. 15.86AP
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