Introductory Chemistry (6th Edition)
6th Edition
ISBN: 9780134302386
Author: Nivaldo J. Tro
Publisher: PEARSON
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Question
Chapter 17, Problem 27E
Interpretation Introduction
Interpretation:
The source of radon and its disadvantages are to be explained.
Concept introduction:
The radon is one of the radioactive gases. It is a product of the radioactive decay of Uranium.
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Students have asked these similar questions
Part I.
a) Elucidate
the structure of compound A using the following information.
• mass spectrum: m+ = 102, m/2=57
312=29
• IR spectrum:
1002.5
% TRANSMITTANCE
Ngg
50
40
30
20
90
80
70
60
MICRONS
5
8
9 10
12
13
14 15 16
19
1740 cm
M
10
0
4000
3600
3200
2800
2400
2000
1800
1600
13
• CNMR
'H
-NMR
Peak
8
ppm (H)
Integration
multiplicity
a
1.5 (3H)
triplet
b
1.3
1.5 (3H)
triplet
C
2.3
1 (2H)
quartet
d
4.1
1 (2H)
quartet
& ppm (c)
10
15
28
60
177 (C=0)
b) Elucidate the structure of compound B using the following information
13C/DEPT NMR 150.9 MHz
IIL
1400
WAVENUMBERS (CM-1)
DEPT-90
DEPT-135
85 80 75
70
65
60
55
50
45 40
35
30 25 20
ppm
1200
1000
800
600
400
•
Part II.
a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below:
• IR spectra
% TRANSMITTANCE
1002.5
90
80
70
60
50
40
30
20
10
0
4000
3600
3200
2800
2400
2000
1800
1600
• Information from 'HAMR
MICRONS
8 9 10
11
14 15 16
19
25
1400
WAVENUMBERS (CM-1)
1200
1000
800
600
400
peak
8 ppm
Integration
multiplicity
a
2.1
1.5 (3H)
Singlet
b
3.6
1 (2H)
singlet
с
3.8
1.5 (3H)
Singlet
d
6.8
1(2H)
doublet
7.1
1(2H)
doublet
Information from 13C-nmR
Normal carbon
29ppm
Dept 135
Dept -90
+
NO peak
NO peak
50 ppm
55 ppm
+
NO peak
114 ppm
t
126 ppm
No peak
NO peak
130 ppm
t
+
159 ppm
No peak
NO peak
207 ppm
по реак
NO peak
Could you redraw these and also explain how to solve them for me pleas
Chapter 17 Solutions
Introductory Chemistry (6th Edition)
Ch. 17 - Prob. 1SAQCh. 17 - Prob. 2SAQCh. 17 - Prob. 3SAQCh. 17 - Prob. 4SAQCh. 17 - Prob. 5SAQCh. 17 - Prob. 6SAQCh. 17 - Prob. 7SAQCh. 17 - Prob. 8SAQCh. 17 - Prob. 9SAQCh. 17 - Prob. 10SAQ
Ch. 17 - Prob. 1ECh. 17 - Prob. 2ECh. 17 - Prob. 3ECh. 17 - Prob. 4ECh. 17 - Prob. 5ECh. 17 - Prob. 6ECh. 17 - Prob. 7ECh. 17 - Prob. 8ECh. 17 - Prob. 9ECh. 17 - Prob. 10ECh. 17 - Prob. 11ECh. 17 - Prob. 12ECh. 17 - Prob. 13ECh. 17 - Prob. 14ECh. 17 - Prob. 15ECh. 17 - Prob. 16ECh. 17 - Prob. 17ECh. 17 - Prob. 18ECh. 17 - Prob. 19ECh. 17 - Prob. 20ECh. 17 - Prob. 21ECh. 17 - Prob. 22ECh. 17 - Prob. 23ECh. 17 - Prob. 24ECh. 17 - Prob. 25ECh. 17 - Prob. 26ECh. 17 - Prob. 27ECh. 17 - Prob. 28ECh. 17 - Prob. 29ECh. 17 - Prob. 30ECh. 17 - Prob. 31ECh. 17 - Prob. 32ECh. 17 - Prob. 33ECh. 17 - Prob. 34ECh. 17 - Prob. 35ECh. 17 - Prob. 36ECh. 17 - Prob. 37ECh. 17 - Prob. 38ECh. 17 - Prob. 39ECh. 17 - Prob. 40ECh. 17 - Prob. 41ECh. 17 - Prob. 42ECh. 17 - Prob. 43ECh. 17 - Prob. 44ECh. 17 - Prob. 45ECh. 17 - Prob. 46ECh. 17 - Prob. 47ECh. 17 - Prob. 48ECh. 17 - Prob. 49ECh. 17 - Prob. 50ECh. 17 - Draw the symbol for the isotope of lead that...Ch. 17 - Prob. 52ECh. 17 - Prob. 53ECh. 17 - Prob. 54ECh. 17 - Prob. 55ECh. 17 - Prob. 56ECh. 17 - Prob. 57ECh. 17 - Prob. 58ECh. 17 - Prob. 59ECh. 17 - Prob. 60ECh. 17 - Prob. 61ECh. 17 - Prob. 62ECh. 17 - Prob. 63ECh. 17 - Prob. 64ECh. 17 - Prob. 65ECh. 17 - Prob. 66ECh. 17 - Prob. 67ECh. 17 - Prob. 68ECh. 17 - Prob. 69ECh. 17 - Prob. 70ECh. 17 - Prob. 71ECh. 17 - Prob. 72ECh. 17 - Prob. 73ECh. 17 - Prob. 74ECh. 17 - Prob. 75ECh. 17 - Prob. 76ECh. 17 - Prob. 77ECh. 17 - Prob. 78ECh. 17 - Prob. 79ECh. 17 - Prob. 80ECh. 17 - Prob. 81ECh. 17 - Prob. 82ECh. 17 - Prob. 83ECh. 17 - Prob. 84ECh. 17 - Prob. 85ECh. 17 - Prob. 86ECh. 17 - Prob. 87ECh. 17 - Prob. 88ECh. 17 - Prob. 89ECh. 17 - Prob. 90ECh. 17 - Prob. 91ECh. 17 - 92. The fusion of deuterium and tritium produces J...Ch. 17 - Prob. 93ECh. 17 - Prob. 94ECh. 17 - Prob. 95ECh. 17 - Prob. 96ECh. 17 - Prob. 97ECh. 17 - Prob. 98ECh. 17 - Prob. 99ECh. 17 - Prob. 100ECh. 17 - Prob. 101ECh. 17 - Prob. 102ECh. 17 - Prob. 103QGWCh. 17 - Prob. 104QGWCh. 17 - Prob. 105QGWCh. 17 - Prob. 106QGWCh. 17 - Prob. 107DIA
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- In the image, the light blue sphere represents a mole of hydrogen atoms, the purple or teal spheres represent a mole of a conjugate base. A light blue sphere by itself is H+. Assuming there is 2.00 L of solution, answer the following: The Ka of the left & right solution is? The pH of the left & right solution is? The acid on the left & right is what kind of acid?arrow_forwardNonearrow_forwardNonearrow_forward
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