Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Chapter 15, Problem 1KT
Interpretation Introduction
Interpretation:
The key term corresponding to the definition “the amount of mass in one unit volume of matter” is to be stated.
Concept introduction:
The density is defined as the ratio of the mass of a body to its volume. The SI unit for density is
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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
metal:
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 15 Solutions
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Ch. 15 - Prob. 1CECh. 15 - Prob. 2CECh. 15 - Prob. 3CECh. 15 - Prob. 4CECh. 15 - Prob. 5CECh. 15 - Prob. 6CECh. 15 - Prob. 7CECh. 15 - Prob. 8CECh. 15 - Prob. 9CECh. 15 - Prob. 1KT
Ch. 15 - Prob. 2KTCh. 15 - Prob. 3KTCh. 15 - Prob. 4KTCh. 15 - Prob. 5KTCh. 15 - Prob. 6KTCh. 15 - Prob. 7KTCh. 15 - Prob. 8KTCh. 15 - Prob. 9KTCh. 15 - Prob. 10KTCh. 15 - Prob. 11KTCh. 15 - Prob. 12KTCh. 15 - Prob. 13KTCh. 15 - Prob. 14KTCh. 15 - Prob. 15KTCh. 15 - Prob. 16KTCh. 15 - Prob. 17KTCh. 15 - Prob. 18KTCh. 15 - Prob. 19KTCh. 15 - Prob. 20KTCh. 15 - Prob. 21KTCh. 15 - Prob. 22KTCh. 15 - Prob. 1ECh. 15 - Prob. 2ECh. 15 - Prob. 3ECh. 15 - Prob. 4ECh. 15 - Prob. 5ECh. 15 - Prob. 6ECh. 15 - Prob. 7ECh. 15 - Prob. 8ECh. 15 - Prob. 9ECh. 15 - Prob. 10ECh. 15 - Prob. 11ECh. 15 - Prob. 12ECh. 15 - Prob. 13ECh. 15 - Prob. 14ECh. 15 - Prob. 15ECh. 15 - Prob. 16ECh. 15 - Prob. 17ECh. 15 - Prob. 18ECh. 15 - Prob. 19ECh. 15 - Prob. 20ECh. 15 - Prob. 21ECh. 15 - Prob. 22ECh. 15 - Prob. 23ECh. 15 - Prob. 24ECh. 15 - Prob. 25ECh. 15 - Prob. 26ECh. 15 - Prob. 27ECh. 15 - Prob. 28ECh. 15 - Prob. 29ECh. 15 - Prob. 30ECh. 15 - Prob. 31ECh. 15 - Prob. 32ECh. 15 - Prob. 33ECh. 15 - Prob. 34ECh. 15 - Prob. 35ECh. 15 - Prob. 36ECh. 15 - Prob. 37ECh. 15 - Prob. 38ECh. 15 - Prob. 39ECh. 15 - Prob. 40ECh. 15 - Prob. 41ECh. 15 - Prob. 42ECh. 15 - Prob. 43ECh. 15 - Prob. 44ECh. 15 - Prob. 45ECh. 15 - Prob. 46ECh. 15 - Prob. 47ECh. 15 - Prob. 48ECh. 15 - Prob. 49ECh. 15 - Prob. 50ECh. 15 - Prob. 51ECh. 15 - Prob. 52ECh. 15 - Prob. 53ECh. 15 - Prob. 54ECh. 15 - Prob. 55ECh. 15 - Prob. 56ECh. 15 - Prob. 57ECh. 15 - Prob. 58ECh. 15 - Prob. 59ECh. 15 - Prob. 60ECh. 15 - Prob. 1STCh. 15 - Prob. 2STCh. 15 - Prob. 3STCh. 15 - Prob. 4STCh. 15 - Prob. 5STCh. 15 - Prob. 6STCh. 15 - Prob. 7STCh. 15 - Prob. 8STCh. 15 - Prob. 9STCh. 15 - Prob. 10STCh. 15 - Prob. 11STCh. 15 - Prob. 12STCh. 15 - Prob. 13STCh. 15 - Prob. 14STCh. 15 - Prob. 15STCh. 15 - Prob. 16ST
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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