Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Chapter 13, Problem 5E
Interpretation Introduction
Interpretation:
The partial pressure of nitrous oxide gas is to be calculated.
Concept introduction:
Henry’s law states that the partial pressure of a gas above liquid is directly proportional to the solubility of that gas in liquid at a constant temperature. For example,
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Relative Intensity
Part VI. consider the multi-step reaction below for compounds
A, B, and C.
These compounds were subjected to mass spectrometric analysis and
the following spectra for A, B, and C was obtained.
Draw the structure of B and C and match all three compounds
to the correct spectra.
Relative Intensity
Relative Intensity
100
HS-NJ-0547
80
60
31
20
S1
84
M+
absent
10
30
40
50
60
70
80
90
100
100-
MS2016-05353CM
80-
60
40
20
135 137
S2
164 166
0-m
25
50
75
100
125
150
m/z
60
100
MS-NJ-09-43
40
20
20
80
45
S3
25
50
75
100
125
150
175
m/z
Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following:
(a) match structures of isomers given their mass spectra below (spectra A and spectra B)
(b) Draw the fragments given the following prominent peaks from
each spectrum:
Spectra A m/2 =43 and 1/2-57
spectra B m/2 = 43
(c) why is 1/2=57 peak in spectrum A more intense compared
to the same peak in spectrum B.
Relative abundance
Relative abundance
100
A
50
29
29
0
10
-0
-0
100
B
50
720
30
41
43
57
71
4-0
40
50
60 70
m/z
43
57
8-0
m/z = 86
M
90 100
71
m/z = 86
M
-O
0
10 20 30
40 50
60
70
80
-88
m/z
90
100
Part IV. C6H5 CH2CH2OH is an aromatic compound which was subjected to Electron Ionization - mass
spectrometry (El-MS) analysis. Prominent m/2 values: m/2 = 104 and m/2 = 9) was obtained.
Draw the structures of these fragments.
Chapter 13 Solutions
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Ch. 13 - Prob. 1CECh. 13 - Prob. 2CECh. 13 - Prob. 3CECh. 13 - Prob. 4CECh. 13 - Prob. 5CECh. 13 - Prob. 6CECh. 13 - Prob. 7CECh. 13 - Prob. 8CECh. 13 - Prob. 9CECh. 13 - Prob. 10CE
Ch. 13 - Prob. 11CECh. 13 - Prob. 12CECh. 13 - Prob. 1KTCh. 13 - Prob. 2KTCh. 13 - Prob. 3KTCh. 13 - Prob. 4KTCh. 13 - Prob. 5KTCh. 13 - Prob. 6KTCh. 13 - Prob. 7KTCh. 13 - Prob. 8KTCh. 13 - Prob. 9KTCh. 13 - Prob. 10KTCh. 13 - Prob. 11KTCh. 13 - Prob. 12KTCh. 13 - Prob. 13KTCh. 13 - Prob. 14KTCh. 13 - Prob. 15KTCh. 13 - Prob. 16KTCh. 13 - Prob. 17KTCh. 13 - Prob. 18KTCh. 13 - Prob. 19KTCh. 13 - Prob. 20KTCh. 13 - Prob. 1ECh. 13 - Prob. 2ECh. 13 - Prob. 3ECh. 13 - Prob. 4ECh. 13 - Prob. 5ECh. 13 - Prob. 6ECh. 13 - Prob. 7ECh. 13 - Prob. 8ECh. 13 - Prob. 9ECh. 13 - Prob. 10ECh. 13 - Prob. 11ECh. 13 - Prob. 12ECh. 13 - Prob. 13ECh. 13 - Prob. 14ECh. 13 - Prob. 15ECh. 13 - Prob. 16ECh. 13 - Prob. 17ECh. 13 - Prob. 18ECh. 13 - Prob. 19ECh. 13 - Prob. 20ECh. 13 - Prob. 21ECh. 13 - Prob. 22ECh. 13 - Prob. 23ECh. 13 - Prob. 24ECh. 13 - Prob. 25ECh. 13 - Prob. 26ECh. 13 - Prob. 27ECh. 13 - Prob. 28ECh. 13 - Prob. 29ECh. 13 - Prob. 30ECh. 13 - Prob. 31ECh. 13 - Prob. 32ECh. 13 - Prob. 33ECh. 13 - Prob. 34ECh. 13 - Prob. 35ECh. 13 - Prob. 36ECh. 13 - Prob. 37ECh. 13 - Prob. 38ECh. 13 - Prob. 39ECh. 13 - Prob. 40ECh. 13 - Prob. 41ECh. 13 - Prob. 42ECh. 13 - Prob. 43ECh. 13 - Prob. 44ECh. 13 - Prob. 45ECh. 13 - Prob. 46ECh. 13 - Prob. 47ECh. 13 - Prob. 48ECh. 13 - Prob. 49ECh. 13 - Prob. 50ECh. 13 - Prob. 51ECh. 13 - Prob. 52ECh. 13 - Prob. 53ECh. 13 - Prob. 54ECh. 13 - Prob. 55ECh. 13 - Prob. 56ECh. 13 - Prob. 57ECh. 13 - Prob. 58ECh. 13 - Prob. 59ECh. 13 - Prob. 60ECh. 13 - Prob. 61ECh. 13 - Prob. 62ECh. 13 - Prob. 63ECh. 13 - Prob. 64ECh. 13 - Prob. 65ECh. 13 - Prob. 66ECh. 13 - Prob. 67ECh. 13 - Prob. 68ECh. 13 - Prob. 70ECh. 13 - Prob. 71ECh. 13 - Prob. 72ECh. 13 - Prob. 73ECh. 13 - Prob. 74ECh. 13 - Prob. 75ECh. 13 - Prob. 76ECh. 13 - Prob. 77ECh. 13 - Prob. 78ECh. 13 - Prob. 79ECh. 13 - Prob. 80ECh. 13 - Prob. 81ECh. 13 - Prob. 82ECh. 13 - Prob. 83ECh. 13 - Prob. 84ECh. 13 - Prob. 1STCh. 13 - Prob. 2STCh. 13 - Prob. 3STCh. 13 - Prob. 4STCh. 13 - Prob. 5STCh. 13 - Prob. 6STCh. 13 - Prob. 7STCh. 13 - Prob. 8STCh. 13 - Prob. 9STCh. 13 - Prob. 10STCh. 13 - Prob. 11STCh. 13 - Prob. 12STCh. 13 - Prob. 13STCh. 13 - Prob. 14STCh. 13 - Prob. 15STCh. 13 - Prob. 16ST
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- For each reaction shown below follow the curved arrows to complete each equationby showing the structure of the products. Identify the acid, the base, the conjugated acid andconjugated base. Consutl the pKa table and choose the direciton theequilibrium goes. However show the curved arrows. Please explain if possible.arrow_forwardA molecule shows peaks at 1379, 1327, 1249, 739 cm-1. Draw a diagram of the energy levels for such a molecule. Draw arrows for the possible transitions that could occur for the molecule. In the diagram imagine exciting an electron, what are its various options for getting back to the ground state? What process would promote radiation less decay? What do you expect for the lifetime of an electron in the T1 state? Why is phosphorescence emission weak in most substances? What could you do to a sample to enhance the likelihood that phosphorescence would occur over radiationless decay?arrow_forwardRank the indicated C—C bonds in increasing order of bond length. Explain as why to the difference.arrow_forward
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY