Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Chapter 19, Problem 61E
Interpretation Introduction
Interpretation:
The reason as to why the boiling point of ethylamine is higher than that of propane is to be stated.
Concept introduction:
The derivatives of ammonia whose one or more hydrogen atom gets replaced by an alkyl or aryl group is known as
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Consider the "C NMR spectrum below.
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The spectrum belongs to which one of the following constitutional isomers of the compound C,H12? Select the single best answer.
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The structure of compound 1,3,5-trimethylbenzene (mesitylene) is given below.
How many signals would you expect to find in the 'H NMR spectrum of 1,3,5-trimethylbenzene (mesitylene)?
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1
How many signals do you expect in the 'H NMR spectrum for this molecule?
CI CI
Cl
Write the answer in the table below.
Also, in each of the drawing areas below is a copy of the molecule, with H atoms shown. In each copy, one of the H atoms is highlighted 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: Remember, a multiplet is considered one signal in the 'H NMR spectrum.
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.
No additional H atoms to highlight in
top molecule
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…
Chapter 19 Solutions
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Ch. 19 - Prob. 1CECh. 19 - Prob. 2CECh. 19 - Prob. 3CECh. 19 - Prob. 4CECh. 19 - Prob. 5CECh. 19 - Prob. 6CECh. 19 - Prob. 7CECh. 19 - Prob. 1KTCh. 19 - Prob. 2KTCh. 19 - Prob. 3KT
Ch. 19 - Prob. 4KTCh. 19 - Prob. 5KTCh. 19 - Prob. 6KTCh. 19 - Prob. 7KTCh. 19 - Prob. 8KTCh. 19 - Prob. 9KTCh. 19 - Prob. 10KTCh. 19 - Prob. 11KTCh. 19 - Prob. 12KTCh. 19 - Prob. 13KTCh. 19 - Prob. 14KTCh. 19 - Prob. 15KTCh. 19 - Prob. 16KTCh. 19 - Prob. 17KTCh. 19 - Prob. 18KTCh. 19 - Prob. 19KTCh. 19 - Prob. 20KTCh. 19 - Prob. 21KTCh. 19 - Prob. 22KTCh. 19 - Prob. 23KTCh. 19 - Prob. 24KTCh. 19 - Prob. 1ECh. 19 - Prob. 2ECh. 19 - Prob. 3ECh. 19 - Prob. 4ECh. 19 - Prob. 5ECh. 19 - Prob. 6ECh. 19 - Prob. 7ECh. 19 - Prob. 8ECh. 19 - Prob. 9ECh. 19 - Prob. 10ECh. 19 - Prob. 11ECh. 19 - Prob. 12ECh. 19 - Prob. 13ECh. 19 - Prob. 14ECh. 19 - Prob. 15ECh. 19 - Prob. 16ECh. 19 - Prob. 17ECh. 19 - Prob. 18ECh. 19 - Prob. 19ECh. 19 - Prob. 20ECh. 19 - Prob. 21ECh. 19 - Prob. 22ECh. 19 - Prob. 23ECh. 19 - Prob. 24ECh. 19 - Prob. 25ECh. 19 - Prob. 26ECh. 19 - Prob. 27ECh. 19 - Prob. 28ECh. 19 - Prob. 29ECh. 19 - Prob. 30ECh. 19 - Prob. 31ECh. 19 - Prob. 32ECh. 19 - Prob. 33ECh. 19 - Prob. 34ECh. 19 - Prob. 35ECh. 19 - Prob. 36ECh. 19 - Prob. 37ECh. 19 - Prob. 38ECh. 19 - Prob. 39ECh. 19 - Prob. 40ECh. 19 - Prob. 41ECh. 19 - Prob. 42ECh. 19 - Prob. 43ECh. 19 - Prob. 44ECh. 19 - Prob. 45ECh. 19 - Prob. 46ECh. 19 - Prob. 47ECh. 19 - Prob. 48ECh. 19 - Prob. 49ECh. 19 - Prob. 50ECh. 19 - Prob. 51ECh. 19 - Prob. 52ECh. 19 - Prob. 53ECh. 19 - Prob. 54ECh. 19 - Prob. 55ECh. 19 - Prob. 56ECh. 19 - Prob. 57ECh. 19 - Prob. 58ECh. 19 - Prob. 59ECh. 19 - Prob. 60ECh. 19 - Prob. 61ECh. 19 - Prob. 62ECh. 19 - Prob. 63ECh. 19 - Prob. 64ECh. 19 - Prob. 65ECh. 19 - Prob. 66ECh. 19 - Prob. 67ECh. 19 - Prob. 68ECh. 19 - Prob. 69ECh. 19 - Prob. 70ECh. 19 - Prob. 71ECh. 19 - Prob. 72ECh. 19 - Prob. 73ECh. 19 - Prob. 74ECh. 19 - Prob. 75ECh. 19 - Prob. 76ECh. 19 - Prob. 77ECh. 19 - Prob. 78ECh. 19 - Prob. 79ECh. 19 - Prob. 80ECh. 19 - Prob. 81ECh. 19 - Prob. 82ECh. 19 - Prob. 83ECh. 19 - Prob. 84ECh. 19 - Prob. 85ECh. 19 - Prob. 86ECh. 19 - Prob. 1STCh. 19 - Prob. 2STCh. 19 - Prob. 3STCh. 19 - Prob. 4STCh. 19 - Prob. 5STCh. 19 - Prob. 6STCh. 19 - Prob. 7STCh. 19 - Prob. 8STCh. 19 - Prob. 9STCh. 19 - Prob. 10STCh. 19 - Prob. 11STCh. 19 - Prob. 12STCh. 19 - Prob. 13STCh. 19 - Prob. 14STCh. 19 - Prob. 15STCh. 19 - Prob. 16STCh. 19 - Prob. 17STCh. 19 - Prob. 18ST
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- wrtie the balanced equation and find the E° when the following half- reactions are combined Zn2+(aq) + 2e---> Zn(s) E°= -0.763V Ag+(aq) + e---> Ag (s) E°=+0.799Varrow_forwardConsider this molecule: How many H atoms are in this molecule? How many different signals could be found in its 'H NMR spectrum? Note: A multiplet is considered one signal. ☐arrow_forwardStudy this 'H NMR spectrum, and then answer the questions about it in the table below. Check 1.0- 0.5- 0.0 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 0.0 What unit symbol should be written on the horizontal axis? What is the chemical shift & of the doublet? If there is no doublet, just check the box instead. Give your answer to 2 significant digits. What is the chemical shift of the signal immediately upfield of the doublet? If there is no doublet, or no signal upfield of it, check the box instead. What is the chemical shift & of the least deshielded proton? If you can't tell without more information, check the box instead. 血 8 = ☐ There is no doublet. 8 = ☐ No such signal. 8 = 0 Need more information.arrow_forward
- how many moles of H2O2 are required to react with 11g of N2H4 according to the following reaction? (atomic weights: N=14.01, H=1.008, O= 16.00) 7H2O2 + N2H4 -> 2HNO3 + 8H20arrow_forwardcalculate the number of moles of H2 produced from 0.78 moles of Ga and 1.92 moles HCL? 2Ga+6HCL->2GaCl3+3H2arrow_forwardan adult human breathes 0.50L of air at 1 atm with each breath. If a 50L air tank at 200 atm is available, how man y breaths will the tank providearrow_forward
- Using reaction free energy to predict equilibrium composition Consider the following equilibrium: 2NO2 (g) = N2O4(g) AGº = -5.4 kJ Now suppose a reaction vessel is filled with 4.53 atm of dinitrogen tetroxide (N2O4) at 279. °C. Answer the following questions about this system: Under these conditions, will the pressure of N2O4 tend to rise or fall? Is it possible to reverse this tendency by adding NO2? In other words, if you said the pressure of N2O4 will tend to rise, can that be changed to a tendency to fall by adding NO2? Similarly, if you said the pressure of N2O4 will tend to fall, can that be changed to a tendency to '2' rise by adding NO2? If you said the tendency can be reversed in the second question, calculate the minimum pressure of NO 2 needed to reverse it. Round your answer to 2 significant digits. 00 rise ☐ x10 fall yes no ☐ atm G Ar 1arrow_forwardWhy do we analyse salt?arrow_forwardCurved arrows are used to illustrate the flow of electrons. Using the provided starting and product structures, draw the curved electron-pushing arrows for the following reaction or mechanistic step(s). Be sure to account for all bond-breaking and bond-making steps. H H CH3OH, H+ H Select to Add Arrows H° 0:0 'H + Q HH ■ Select to Add Arrows CH3OH, H* H. H CH3OH, H+ HH ■ Select to Add Arrows i Please select a drawing or reagent from the question areaarrow_forward
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