EBK GET READY FOR ORGANIC CHEMISTRY
2nd Edition
ISBN: 9780321830555
Author: KARTY
Publisher: VST
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Question
Chapter 17, Problem 17.2YT
Interpretation Introduction
Interpretation:
The reaction in solved problem
The new asymmetric carbon is to be identified, and stereoisomers that would be produced are to be drawn.
Concept introduction:
The carbon attached to four different atoms or groups is an asymmetric carbon.
The nucleophile can attack on the trigonal planer carbonyl carbon (C=O) from both sides, that is, from back side and from in front of the plane to form a mixture of stereoisomers with a tetrahedral carbon.
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2H2S(g)+3O2(g)→2SO2(g)+2H2O(g)
A 1.2mol sample of H2S(g) is combined with excess O2(g), and the reaction goes to completion.
Question
Which of the following predicts the theoretical yield of SO2(g) from the reaction?
Responses
1.2 g
Answer A: 1.2 grams
A
41 g
Answer B: 41 grams
B
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Answer C: 77 grams
C
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Answer D: 154 grams
D
Chapter 17 Solutions
EBK GET READY FOR ORGANIC CHEMISTRY
Ch. 17 - Prob. 17.1PCh. 17 - Prob. 17.2PCh. 17 - Prob. 17.3PCh. 17 - Prob. 17.4PCh. 17 - Prob. 17.5PCh. 17 - Prob. 17.6PCh. 17 - Prob. 17.7PCh. 17 - Prob. 17.8PCh. 17 - Prob. 17.9PCh. 17 - Prob. 17.10P
Ch. 17 - Prob. 17.11PCh. 17 - Prob. 17.12PCh. 17 - Prob. 17.13PCh. 17 - Prob. 17.14PCh. 17 - Prob. 17.15PCh. 17 - Prob. 17.16PCh. 17 - Prob. 17.17PCh. 17 - Prob. 17.18PCh. 17 - Prob. 17.19PCh. 17 - Prob. 17.20PCh. 17 - Prob. 17.21PCh. 17 - Prob. 17.22PCh. 17 - Prob. 17.23PCh. 17 - Prob. 17.24PCh. 17 - Prob. 17.25PCh. 17 - Prob. 17.26PCh. 17 - Prob. 17.27PCh. 17 - Prob. 17.28PCh. 17 - Prob. 17.29PCh. 17 - Prob. 17.30PCh. 17 - Prob. 17.31PCh. 17 - Prob. 17.32PCh. 17 - Prob. 17.33PCh. 17 - Prob. 17.34PCh. 17 - Prob. 17.35PCh. 17 - Prob. 17.36PCh. 17 - Prob. 17.37PCh. 17 - Prob. 17.38PCh. 17 - Prob. 17.39PCh. 17 - Prob. 17.40PCh. 17 - Prob. 17.41PCh. 17 - Prob. 17.42PCh. 17 - Prob. 17.43PCh. 17 - Prob. 17.44PCh. 17 - Prob. 17.45PCh. 17 - Prob. 17.46PCh. 17 - Prob. 17.47PCh. 17 - Prob. 17.48PCh. 17 - Prob. 17.49PCh. 17 - Prob. 17.50PCh. 17 - Prob. 17.51PCh. 17 - Prob. 17.52PCh. 17 - Prob. 17.53PCh. 17 - Prob. 17.54PCh. 17 - Prob. 17.55PCh. 17 - Prob. 17.56PCh. 17 - Prob. 17.57PCh. 17 - Prob. 17.58PCh. 17 - Prob. 17.59PCh. 17 - Prob. 17.60PCh. 17 - Prob. 17.61PCh. 17 - Prob. 17.62PCh. 17 - Prob. 17.63PCh. 17 - Prob. 17.64PCh. 17 - Prob. 17.65PCh. 17 - Prob. 17.66PCh. 17 - Prob. 17.67PCh. 17 - Prob. 17.68PCh. 17 - Prob. 17.69PCh. 17 - Prob. 17.70PCh. 17 - Prob. 17.71PCh. 17 - Prob. 17.72PCh. 17 - Prob. 17.73PCh. 17 - Prob. 17.74PCh. 17 - Prob. 17.75PCh. 17 - Prob. 17.76PCh. 17 - Prob. 17.77PCh. 17 - Prob. 17.78PCh. 17 - Prob. 17.79PCh. 17 - Prob. 17.80PCh. 17 - Prob. 17.81PCh. 17 - Prob. 17.82PCh. 17 - Prob. 17.83PCh. 17 - Prob. 17.84PCh. 17 - Prob. 17.1YTCh. 17 - Prob. 17.2YTCh. 17 - Prob. 17.3YTCh. 17 - Prob. 17.4YTCh. 17 - Prob. 17.5YTCh. 17 - Prob. 17.6YTCh. 17 - Prob. 17.7YTCh. 17 - Prob. 17.8YTCh. 17 - Prob. 17.9YTCh. 17 - Prob. 17.10YTCh. 17 - Prob. 17.11YTCh. 17 - Prob. 17.12YTCh. 17 - Prob. 17.13YTCh. 17 - Prob. 17.14YT
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