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 6CE
Interpretation Introduction
Interpretation:
Whether a hot cup of coffee or a warm cup of coffee dissolves more sugar is to be stated.
Concept introduction:
The term unsaturated is used for the solution in which more solute can be added. The term, saturated is used for the solution in which no more solute can be added. The term supersaturated is used for the solution in which solute is added more than its solubility.
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AG/F-2° V
3. Before proceeding with this problem you may want to glance at p. 466 of your textbook
where various oxo-phosphorus derivatives and their oxidation states are summarized.
Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14:
-0.93
+0.38
-0.50
-0.51 -0.06
H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3
Acidic solution
Basic solution
-0.28
-0.50
3--1.12
-1.57
-2.05 -0.89
PO HPO H₂PO₂ →P → PH3
-1.73
a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the
formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both
processes; comment.
(3 points)
0.5
PH
P
0.0
-0.5
-1.0-
-1.5-
-2.0
H.PO,
-2.3+
-3 -2
-1
1
2
3
2
H,PO,
b) Frost diagram for phosphorus under acidic
conditions is shown. Identify possible
disproportionation and comproportionation processes;
write out chemical equations describing them. (2 points)
H,PO
4
S
Oxidation stale, N
4. For the following complexes, draw the structures and give a d-electron count of the
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a) Tris(acetylacetonato)iron(III)
b) Hexabromoplatinate(2-)
c) Potassium diamminetetrabromocobaltate(III)
(6 points)
2. Calculate the overall formation constant for [Fe(CN)6]³, given that the overall formation
constant for [Fe(CN)6] 4 is ~1032, and that:
Fe3+ (aq) + e
= Fe²+ (aq)
E° = +0.77 V
[Fe(CN)6]³ (aq) + e¯ = [Fe(CN)6] (aq) E° = +0.36 V
(4 points)
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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- elow are experimentally determined van Deemter plots of column efficiency, H, vs. flow rate. H is a quantitative measurement of band broadening. The left plot is for a liquid chromatography application and the night is for gas chromatography. Compare and contrast these two plots in terms of the three band broadening mechanisms presented in this activity. How are they similar? How do they differ? Justify your answers.? 0.4 H (mm) 0.2 0.1- 0.3- 0 0.5 H (mm) 8.0 7.0 6.0 5.0 4.0- 3.0 T +++ 1.0 1.5 0 2.0 4.0 Flow Rate, u (cm/s) 6.0 8.0 Flow Rate, u (cm/s)arrow_forwardPredict the products of this organic reaction: + H ZH NaBH3CN H+ n. ? Click and drag to start drawing a structure. Xarrow_forwardWhat is the missing reactant R in this organic reaction? + R H3O+ + • Draw the structure of R in the drawing area below. • Be sure to use wedge and dash bonds if it's necessary to draw one particular enantiomer. Click and drag to start drawing a structure.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