
General, Organic, & Biological Chemistry
3rd Edition
ISBN: 9780073511245
Author: Janice Gorzynski Smith Dr.
Publisher: McGraw-Hill Education
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Question
Chapter 23, Problem 23.74P
Interpretation Introduction
Interpretation:
The action of HCN in the synthesis of ATP should be explained.
Concept Introduction:
Aerobic respiration occurs in two steps:
- Glycolysis
- Citric acid cycle
Glycolysis is the first step that forms pyruvate as given below:
In the presence of oxygen means aerobic respiration, this pyruvate enters the Krebs cycle and extracts energy in the form of electrons transfer. Electrons are transferred from the pyruvate to the receptors like
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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 23 Solutions
General, Organic, & Biological Chemistry
Ch. 23.2 - Prob. 23.1PCh. 23.3 - Prob. 23.2PCh. 23.3 - Prob. 23.3PCh. 23.3 - Prob. 23.4PCh. 23.3 - Prob. 23.5PCh. 23.4 - Prob. 23.6PCh. 23.4 - Prob. 23.7PCh. 23.4 - Prob. 23.8PCh. 23.4 - Prob. 23.9PCh. 23.4 - Prob. 23.10P
Ch. 23.5 - Prob. 23.11PCh. 23.5 - Prob. 23.12PCh. 23.5 - Prob. 23.13PCh. 23.6 - Prob. 23.14PCh. 23.6 - Prob. 23.15PCh. 23.6 - Prob. 23.16PCh. 23 - Prob. 23.17PCh. 23 - Prob. 23.18PCh. 23 - Prob. 23.19PCh. 23 - Prob. 23.20PCh. 23 - Prob. 23.21PCh. 23 - Prob. 23.22PCh. 23 - Prob. 23.23PCh. 23 - Prob. 23.24PCh. 23 - Prob. 23.25PCh. 23 - Prob. 23.26PCh. 23 - Prob. 23.27PCh. 23 - Prob. 23.28PCh. 23 - The phosphorylation of glucose with forms glucose...Ch. 23 - Prob. 23.30PCh. 23 - Prob. 23.31PCh. 23 - Prob. 23.32PCh. 23 - Prob. 23.33PCh. 23 - Prob. 23.34PCh. 23 - Prob. 23.35PCh. 23 - Prob. 23.36PCh. 23 - Prob. 23.37PCh. 23 - Classify each substance as an oxidizing agent, a...Ch. 23 - Prob. 23.39PCh. 23 - Prob. 23.40PCh. 23 - Prob. 23.41PCh. 23 - Prob. 23.42PCh. 23 - Prob. 23.43PCh. 23 - Prob. 23.44PCh. 23 - Prob. 23.45PCh. 23 - Prob. 23.46PCh. 23 - Prob. 23.47PCh. 23 - Prob. 23.48PCh. 23 - Prob. 23.49PCh. 23 - Prob. 23.50PCh. 23 - Prob. 23.51PCh. 23 - Prob. 23.52PCh. 23 - Prob. 23.53PCh. 23 - Prob. 23.54PCh. 23 - Prob. 23.55PCh. 23 - Prob. 23.56PCh. 23 - Prob. 23.57PCh. 23 - Prob. 23.58PCh. 23 - Prob. 23.59PCh. 23 - Prob. 23.60PCh. 23 - Prob. 23.61PCh. 23 - Prob. 23.62PCh. 23 - Prob. 23.63PCh. 23 - Prob. 23.64PCh. 23 - Prob. 23.65PCh. 23 - Prob. 23.66PCh. 23 - Prob. 23.67PCh. 23 - Prob. 23.68PCh. 23 - Prob. 23.69PCh. 23 - Prob. 23.70PCh. 23 - Prob. 23.71PCh. 23 - Prob. 23.72PCh. 23 - Prob. 23.73PCh. 23 - Prob. 23.74PCh. 23 - Prob. 23.75CPCh. 23 - Prob. 23.76CP
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