Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Chapter 16, Problem 1KT
Interpretation Introduction
Interpretation:
The key term corresponding to the definition “the principle that the rate of a
Concept introduction:
According to collision theory, molecules randomly collide with one another to form new bonds by breaking the old bonds. The effective collisions depend upon the frequency, energy, and the orientation of the colliding molecules.
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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 16 Solutions
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Ch. 16 - Prob. 1CECh. 16 - Prob. 2CECh. 16 - Prob. 3CECh. 16 - Prob. 4CECh. 16 - Prob. 5CECh. 16 - Prob. 6CECh. 16 - Prob. 7CECh. 16 - Prob. 8CECh. 16 - Prob. 9CECh. 16 - Prob. 10CE
Ch. 16 - Prob. 1KTCh. 16 - Prob. 2KTCh. 16 - Prob. 3KTCh. 16 - Prob. 4KTCh. 16 - Prob. 5KTCh. 16 - Prob. 6KTCh. 16 - Prob. 7KTCh. 16 - Prob. 8KTCh. 16 - Prob. 9KTCh. 16 - Prob. 10KTCh. 16 - Prob. 11KTCh. 16 - Prob. 12KTCh. 16 - Prob. 13KTCh. 16 - Prob. 14KTCh. 16 - Prob. 15KTCh. 16 - Prob. 16KTCh. 16 - Prob. 17KTCh. 16 - Prob. 18KTCh. 16 - Prob. 1ECh. 16 - Prob. 2ECh. 16 - Prob. 3ECh. 16 - Prob. 4ECh. 16 - Prob. 5ECh. 16 - Prob. 6ECh. 16 - Prob. 7ECh. 16 - Prob. 8ECh. 16 - Prob. 9ECh. 16 - Prob. 10ECh. 16 - Prob. 11ECh. 16 - Prob. 12ECh. 16 - Prob. 13ECh. 16 - Prob. 14ECh. 16 - Prob. 15ECh. 16 - Prob. 16ECh. 16 - Prob. 17ECh. 16 - Prob. 18ECh. 16 - Prob. 19ECh. 16 - Prob. 20ECh. 16 - Prob. 21ECh. 16 - Prob. 22ECh. 16 - Prob. 23ECh. 16 - Prob. 24ECh. 16 - Prob. 25ECh. 16 - Prob. 26ECh. 16 - Prob. 27ECh. 16 - Prob. 28ECh. 16 - Prob. 29ECh. 16 - Prob. 30ECh. 16 - Prob. 31ECh. 16 - Prob. 32ECh. 16 - Prob. 33ECh. 16 - Prob. 34ECh. 16 - Prob. 35ECh. 16 - Prob. 36ECh. 16 - Prob. 37ECh. 16 - Prob. 38ECh. 16 - Prob. 39ECh. 16 - Prob. 40ECh. 16 - Prob. 41ECh. 16 - Prob. 42ECh. 16 - Prob. 43ECh. 16 - Prob. 44ECh. 16 - Prob. 45ECh. 16 - Prob. 46ECh. 16 - Prob. 47ECh. 16 - Prob. 48ECh. 16 - Prob. 1STCh. 16 - Prob. 2STCh. 16 - Prob. 3STCh. 16 - Prob. 4STCh. 16 - Prob. 5STCh. 16 - Prob. 6STCh. 16 - Prob. 7STCh. 16 - Prob. 8STCh. 16 - Prob. 9STCh. 16 - Prob. 10STCh. 16 - Prob. 11STCh. 16 - Prob. 12STCh. 16 - Prob. 13STCh. 16 - Prob. 14STCh. 16 - Prob. 15ST
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- What is the missing reactant R in this organic reaction? N N H3O+ +R + • 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. fmarrow_forwardThe product on the right-hand side of this reaction can be prepared from two organic reactants, under the conditions shown above and below the arrow. Draw 1 and 2 below, in any arrangement you like. 1+2 NaBH3CN H+ N Click and drag to start drawing a structure. 5arrow_forwardAssign this HSQC Spectrum ( please editing clearly on the image)arrow_forward
- (a 4 shows scanning electron microscope (SEM) images of extruded actions of packing bed for two capillary columns of different diameters, al 750 (bottom image) and b) 30-μm-i.d. Both columns are packed with the same stationary phase, spherical particles with 1-um diameter. A) When the columns were prepared, the figure shows that the column with the larger diameter has more packing irregularities. Explain this observation. B) Predict what affect this should have on band broadening and discuss your prediction using the van Deemter terms. C) Does this figure support your explanations in application question 33? Explain why or why not and make any changes in your answers in light of this figure. Figure 4 SEM images of sections of packed columns for a) 750 and b) 30-um-i.d. capillary columns.³arrow_forwardfcrip = ↓ bandwidth Il temp 32. What impact (increase, decrease, or no change) does each of the following conditions have on the individual components of the van Deemter equation and consequently, band broadening? Increase temperature Longer column Using a gas mobile phase instead of liquid Smaller particle stationary phase Multiple Paths Diffusion Mass Transferarrow_forward34. Figure 3 shows Van Deemter plots for a solute molecule using different column inner diameters (i.d.). A) Predict whether decreasing the column inner diameters increase or decrease bandwidth. B) Predict which van Deemter equation coefficient (A, B, or C) has the greatest effect on increasing or decreasing bandwidth as a function of i.d. and justify your answer. Figure 3 Van Deemter plots for hydroquinone using different column inner diameters (i.d. in μm). The data was obtained from liquid chromatography experiments using fused-silica capillary columns packed with 1.0-μm particles. 35 20 H(um) 큰 20 15 90 0+ 1500 100 75 550 01 02 594 05 μ(cm/sec) 30 15 10arrow_forward
- 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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