INTRODUCTORY CHEMISTRY
8th Edition
ISBN: 2818000163285
Author: CORWIN
Publisher: PEARSON CO
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Chapter 13, Problem 39E
Interpretation Introduction
(a)
Interpretation:
The solution that contains
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.
Interpretation Introduction
(b)
Interpretation:
The solution that contains
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.
Interpretation Introduction
(c)
Interpretation:
The solution that contains
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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Students have asked these similar questions
For the decomposition reaction of N2O5(g): 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed:
N2O5 NO2 + NO3 (K1) | NO2 + NO3 → N2O5 (k-1) | NO2 + NO3 NO2 + O2 + NO (k2) | NO + N2O51 NO2 + NO2 + NO2 (K3)
→
Give the expression for the acceptable rate.
→
→
(A).
d[N205]
dt
==
2k,k₂[N₂O₂]
k₁+k₁₂
(B).
d[N2O5]
=-k₁[N₂O] + k₁[NO₂] [NO3] - k₂[NO₂]³
dt
(C).
d[N2O5]
=-k₁[N₂O] + k [NO] - k₂[NO] [NO]
d[N2O5]
(D).
=
dt
= -k₁[N2O5] - k¸[NO][N₂05]
dt
Do not apply the calculations, based on the approximation of the stationary state, to make them perform correctly. Basta discard
the 3 responses that you encounter that are obviously erroneous if you apply the formula to determine the speed of a reaction.
For the decomposition reaction of N2O5(g): 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed:
N2O5 NO2 + NO3 (K1) | NO2 + NO3 → N2O5 (k-1) | NO2 + NO3 NO2 + O2 + NO (k2) | NO + N2O51 NO2 + NO2 + NO2 (K3)
→
Give the expression for the acceptable rate.
→
→
(A).
d[N205]
dt
==
2k,k₂[N₂O₂]
k₁+k₁₂
(B).
d[N2O5]
=-k₁[N₂O] + k₁[NO₂] [NO3] - k₂[NO₂]³
dt
(C).
d[N2O5]
=-k₁[N₂O] + k [NO] - k₂[NO] [NO]
d[N2O5]
(D).
=
dt
= -k₁[N2O5] - k¸[NO][N₂05]
dt
Do not apply the calculations, based on the approximation of the stationary state, to make them perform correctly. Basta discard
the 3 responses that you encounter that are obviously erroneous if you apply the formula to determine the speed of a reaction.
R lactam or lactone considering as weak acid or weak base and why
Chapter 13 Solutions
INTRODUCTORY CHEMISTRY
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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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY