Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Chapter 3, Problem 61E
Interpretation Introduction
Interpretation:
The mass of carbon dioxide that is formed by the thermal decomposition of
Concept introduction:
The
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each pair of substrates below, choose the one that will react faster in a substitution reaction, assuming that:
1. the rate of substitution doesn't depend on nucleophile concentration and
2. the products are a roughly 50/50 mixture of enantiomers.
Substrate A
Substrate B
Faster Rate
X
Ś
CI
(Choose one)
(Choose one)
CI
Br
Explanation
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Br
(Choose one)
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NMR spectrum of ethyl acetate has signals whose chemical shifts are indicated below. Which hydrogen or set of hydrogens corresponds to the signal at
4.1 ppm? Select the single best answer.
The
H
O
HỌC—C—0—CH, CH,
2
A
ethyl acetate
H NMR: 1.3 ppm, 2.0 ppm, 4.1 ppm
Check
OA
B
OC
ch
B
C
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How many signals do you expect in the H NMR spectrum for this molecule?
Br Br
Write the answer below.
Also, in each of the drawing areas below is a copy of the molecule, with Hs shown. In each copy, one of the H atoms is colored red. Highlight in red all other H
atoms that would contribute to the same signal as the H already highlighted red
Note for advanced students: In this question, any multiplet is counted as one signal.
1
Number of signals in the 'H NMR spectrum.
For the molecule in the top drawing area, highlight in red any other H atoms that will contribute to
the same signal as the H atom already highlighted red.
If no other H atoms will contribute, check the box at right.
Check
For the molecule in the bottom drawing area, highlight in red any other H atoms that will contribute
to the same signal as the H atom already highlighted red.
If no other H atoms will contribute, check the box at right.
O
✓
No additional Hs to color in top
molecule
ง
No additional Hs to color in bottom…
Chapter 3 Solutions
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Ch. 3 - Prob. 1CECh. 3 - Prob. 2CECh. 3 - Prob. 3CECh. 3 - Prob. 4CECh. 3 - Prob. 5CECh. 3 - Prob. 6CECh. 3 - Prob. 7CECh. 3 - Prob. 8CECh. 3 - Prob. 9CECh. 3 - Prob. 10CE
Ch. 3 - Prob. 11CECh. 3 - Prob. 12CECh. 3 - Prob. 1KTCh. 3 - Prob. 2KTCh. 3 - Prob. 3KTCh. 3 - Prob. 4KTCh. 3 - Prob. 5KTCh. 3 - Prob. 6KTCh. 3 - Prob. 7KTCh. 3 - Prob. 8KTCh. 3 - Prob. 9KTCh. 3 - Prob. 10KTCh. 3 - Prob. 11KTCh. 3 - Prob. 12KTCh. 3 - Prob. 13KTCh. 3 - Prob. 14KTCh. 3 - Prob. 15KTCh. 3 - Prob. 16KTCh. 3 - Prob. 17KTCh. 3 - Prob. 18KTCh. 3 - Prob. 19KTCh. 3 - Prob. 20KTCh. 3 - Prob. 21KTCh. 3 - Prob. 22KTCh. 3 - Prob. 23KTCh. 3 - Prob. 24KTCh. 3 - Prob. 25KTCh. 3 - Prob. 26KTCh. 3 - Prob. 27KTCh. 3 - Prob. 28KTCh. 3 - Prob. 29KTCh. 3 - Prob. 30KTCh. 3 - Prob. 1ECh. 3 - Prob. 2ECh. 3 - Prob. 3ECh. 3 - Prob. 4ECh. 3 - Prob. 5ECh. 3 - Prob. 6ECh. 3 - Prob. 7ECh. 3 - Prob. 8ECh. 3 - Prob. 9ECh. 3 - Prob. 10ECh. 3 - Prob. 11ECh. 3 - Prob. 12ECh. 3 - Prob. 13ECh. 3 - Prob. 14ECh. 3 - Prob. 15ECh. 3 - Prob. 16ECh. 3 - Prob. 17ECh. 3 - Prob. 18ECh. 3 - Prob. 19ECh. 3 - Prob. 20ECh. 3 - Prob. 21ECh. 3 - Prob. 22ECh. 3 - Prob. 23ECh. 3 - Prob. 24ECh. 3 - Prob. 25ECh. 3 - Prob. 26ECh. 3 - Prob. 27ECh. 3 - Prob. 28ECh. 3 - Prob. 29ECh. 3 - Prob. 30ECh. 3 - Prob. 31ECh. 3 - Prob. 32ECh. 3 - Prob. 33ECh. 3 - Prob. 34ECh. 3 - Prob. 35ECh. 3 - Prob. 36ECh. 3 - Prob. 37ECh. 3 - Prob. 38ECh. 3 - Prob. 39ECh. 3 - Prob. 40ECh. 3 - Prob. 41ECh. 3 - Prob. 42ECh. 3 - Prob. 43ECh. 3 - Prob. 44ECh. 3 - Prob. 45ECh. 3 - Prob. 46ECh. 3 - Prob. 47ECh. 3 - Prob. 48ECh. 3 - Prob. 49ECh. 3 - Prob. 50ECh. 3 - Prob. 51ECh. 3 - Prob. 52ECh. 3 - Prob. 53ECh. 3 - Prob. 54ECh. 3 - Prob. 55ECh. 3 - Prob. 56ECh. 3 - Prob. 57ECh. 3 - Prob. 58ECh. 3 - Prob. 59ECh. 3 - Prob. 60ECh. 3 - Prob. 61ECh. 3 - Prob. 62ECh. 3 - Prob. 63ECh. 3 - Prob. 64ECh. 3 - Prob. 65ECh. 3 - Prob. 66ECh. 3 - Prob. 67ECh. 3 - Prob. 68ECh. 3 - Prob. 69ECh. 3 - Prob. 70ECh. 3 - Prob. 71ECh. 3 - Prob. 72ECh. 3 - Prob. 73ECh. 3 - Prob. 74ECh. 3 - Prob. 75ECh. 3 - Prob. 76ECh. 3 - Prob. 77ECh. 3 - Prob. 78ECh. 3 - Prob. 79ECh. 3 - Prob. 80ECh. 3 - Prob. 81ECh. 3 - Prob. 82ECh. 3 - Prob. 83ECh. 3 - Prob. 84ECh. 3 - Prob. 85ECh. 3 - Prob. 86ECh. 3 - Prob. 87ECh. 3 - Prob. 88ECh. 3 - Prob. 89ECh. 3 - Prob. 90ECh. 3 - Prob. 91ECh. 3 - Prob. 92ECh. 3 - Prob. 93ECh. 3 - Prob. 94ECh. 3 - Prob. 95ECh. 3 - Prob. 96ECh. 3 - Prob. 1STCh. 3 - Prob. 2STCh. 3 - Prob. 3STCh. 3 - Prob. 4STCh. 3 - Prob. 5STCh. 3 - Prob. 6STCh. 3 - Prob. 7STCh. 3 - Prob. 8STCh. 3 - Prob. 9STCh. 3 - Prob. 10STCh. 3 - Prob. 11STCh. 3 - Prob. 12STCh. 3 - Prob. 13STCh. 3 - Prob. 14STCh. 3 - Prob. 15STCh. 3 - Prob. 16STCh. 3 - Prob. 17STCh. 3 - Prob. 18STCh. 3 - Prob. 19STCh. 3 - Prob. 20ST
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- calculate the equilibrium concentration of H2 given that K= 0.017 at a constant temperature for this reaction. The inital concentration of HBr is 0.050 M.2HBr(g) ↔ H2(g) + Br2(g)a) 4.48 x 10-2 M b) 5.17 x 10-3 Mc) 1.03 x 10-2 Md) 1.70 x 10-2 Marrow_forwardtrue or falsegiven these two equilibria with their equilibrium constants:H2(g) + CI2(l) ↔ 2HCI(g) K= 0.006 CI2(l) ↔ CI2(g) K= 0.30The equilibrium contstant for the following reaction is 1.8H2(g) + CI2 ↔ 2HCI(g)arrow_forwardI2(g) + CI2(g) ↔ 2ICIK for this reaction is 81.9. Find the equilibrium concentration of I2 if the inital concentration of I2 and CI2 are 0.010 Marrow_forward
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