Chemistry: A Molecular Approach, Books a la Carte Edition (4th Edition)
4th Edition
ISBN: 9780134113593
Author: Nivaldo J. Tro
Publisher: PEARSON
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Chapter 12, Problem 37E
Interpretation Introduction
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3. Consider the compounds below and determine if they are aromatic, antiaromatic, or
non-aromatic. In case of aromatic or anti-aromatic, please indicate number of I
electrons in the respective systems. (Hint: 1. Not all lone pair electrons were explicitly
drawn and you should be able to tell that the bonding electrons and lone pair electrons
should reside in which hybridized atomic orbital 2. You should consider ring strain-
flexibility and steric repulsion that facilitates adoption of aromaticity or avoidance of anti-
aromaticity)
H H
N
N:
NH2
N
Aromaticity
(Circle)
Aromatic Aromatic Aromatic Aromatic Aromatic
Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic
nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic
aromatic TT
electrons
Me
H
Me
Aromaticity
(Circle)
Aromatic Aromatic Aromatic
Aromatic Aromatic
Antiaromatic Antiaromatic Antiaromatic Antiaromatic Antiaromatic
nonaromatic nonaromatic nonaromatic nonaromatic nonaromatic
aromatic πT
electrons
H
HH…
A chemistry graduate student is studying the rate of this reaction:
2 HI (g) →H2(g) +12(g)
She fills a reaction vessel with HI and measures its concentration as the reaction proceeds:
time
(minutes)
[IH]
0
0.800M
1.0
0.301 M
2.0
0.185 M
3.0
0.134M
4.0
0.105 M
Use this data to answer the following questions.
Write the rate law for this reaction.
rate
= 0
Calculate the value of the rate constant k.
k =
Round your answer to 2 significant digits. Also be
sure your answer has the correct unit symbol.
None
Chapter 12 Solutions
Chemistry: A Molecular Approach, Books a la Carte Edition (4th Edition)
Ch. 12 - Prob. 1SAQCh. 12 - Prob. 2SAQCh. 12 - Prob. 3SAQCh. 12 - Prob. 4SAQCh. 12 - Prob. 5SAQCh. 12 - Prob. 6SAQCh. 12 - Prob. 7SAQCh. 12 - Q8. Which solid has the highest melting point?
a)...Ch. 12 - Prob. 9SAQCh. 12 - Prob. 10SAQ
Ch. 12 - Prob. 11SAQCh. 12 - Prob. 12SAQCh. 12 - Prob. 13SAQCh. 12 - Prob. 1ECh. 12 - Prob. 2ECh. 12 - Prob. 3ECh. 12 - Prob. 4ECh. 12 - Prob. 5ECh. 12 - Prob. 6ECh. 12 - Prob. 7ECh. 12 - Prob. 8ECh. 12 - Prob. 9ECh. 12 - Prob. 10ECh. 12 - Prob. 11ECh. 12 - Prob. 12ECh. 12 - Prob. 13ECh. 12 - Prob. 14ECh. 12 - Prob. 15ECh. 12 - Prob. 16ECh. 12 - Prob. 17ECh. 12 - Prob. 18ECh. 12 - Prob. 19ECh. 12 - Prob. 20ECh. 12 - Prob. 21ECh. 12 - Prob. 22ECh. 12 - Prob. 23ECh. 12 - Prob. 24ECh. 12 - Prob. 25ECh. 12 - Prob. 26ECh. 12 - Prob. 27ECh. 12 - Prob. 28ECh. 12 - Prob. 29ECh. 12 - Prob. 30ECh. 12 - Prob. 31ECh. 12 - Prob. 32ECh. 12 - Prob. 33ECh. 12 - Prob. 34ECh. 12 - Prob. 35ECh. 12 - Prob. 36ECh. 12 - 37. Polonium crystallizes with a simple cubic...Ch. 12 - Prob. 38ECh. 12 - Prob. 39ECh. 12 - Prob. 40ECh. 12 - Prob. 41ECh. 12 - Prob. 42ECh. 12 - Prob. 43ECh. 12 - Prob. 44ECh. 12 - Prob. 45ECh. 12 - Prob. 46ECh. 12 - Prob. 47ECh. 12 - Prob. 48ECh. 12 - Prob. 49ECh. 12 - Prob. 50ECh. 12 - Prob. 51ECh. 12 - Prob. 52ECh. 12 - Prob. 53ECh. 12 - Prob. 54ECh. 12 - Prob. 55ECh. 12 - Prob. 56ECh. 12 - Prob. 57ECh. 12 - Prob. 58ECh. 12 - Prob. 59ECh. 12 - Prob. 60ECh. 12 - Prob. 61ECh. 12 - Prob. 62ECh. 12 - Prob. 63ECh. 12 - Prob. 64ECh. 12 - Prob. 65ECh. 12 - Prob. 66ECh. 12 - Prob. 67ECh. 12 - Prob. 68ECh. 12 - Prob. 69ECh. 12 - Prob. 70ECh. 12 - 71. One kind of polyester is a condensation...Ch. 12 - 72. Nomex, a condensation copolymer used by...Ch. 12 - Prob. 73ECh. 12 - Prob. 74ECh. 12 - Prob. 75ECh. 12 - Prob. 76ECh. 12 - Prob. 77ECh. 12 - Prob. 78ECh. 12 - Prob. 79ECh. 12 - 80. Consider the body-centered cubic structure...Ch. 12 - Prob. 81ECh. 12 - Prob. 82ECh. 12 - Prob. 83ECh. 12 - Prob. 84ECh. 12 - Prob. 85ECh. 12 - Prob. 86ECh. 12 - Prob. 87ECh. 12 - Prob. 88ECh. 12 - Prob. 89ECh. 12 - Prob. 90ECh. 12 - Prob. 91ECh. 12 - Prob. 92ECh. 12 - Prob. 93ECh. 12 - Prob. 94ECh. 12 - Prob. 95ECh. 12 - Prob. 96ECh. 12 - Prob. 97ECh. 12 - Prob. 98ECh. 12 - Prob. 99ECh. 12 - Prob. 100ECh. 12 - Prob. 101ECh. 12 - Prob. 102ECh. 12 - Prob. 103ECh. 12 - Prob. 104ECh. 12 - Prob. 105DIA
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- in which spectral range of EMR, atomic and ionic lines of metal liesarrow_forwardQ2: Label the following molecules as chiral or achiral, and label each stereocenter as R or S. CI CH3 CH3 NH2 C CH3 CH3 Br CH3 X &p Bra 'CH 3 "CH3 X Br CH3 Me - N OMe O DuckDuckarrow_forward1. For the four structures provided, Please answer the following questions in the table below. a. Please draw π molecular orbital diagram (use the polygon-and-circle method if appropriate) and fill electrons in each molecular orbital b. Please indicate the number of π electrons c. Please indicate if each molecule provided is anti-aromatic, aromatic, or non- aromatic TT MO diagram Number of π e- Aromaticity Evaluation (X choose one) Non-aromatic Aromatic Anti-aromatic || ||| + IVarrow_forward
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