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OWLv2 with Student Solutions Manual eBook for Ebbing/Gammon's General Chemistry, 11th Edition, [Instant Access], 4 terms (24 months)
11th Edition
ISBN: 9781305864900
Author: Darrell Ebbing; Steven D. Gammon
Publisher: Cengage Learning US
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Chapter 21, Problem 21.33QP
Interpretation Introduction
Interpretation:
The term catenation has to be described and an example of a Carbon compound that exhibits catenation has to be given.
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need help please and thanks dont understand only need help with C-F
Learning Goal:
As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT.
The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7).
Part A - Difference in binding free eenergies
Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol.
The margin of error is 2%.
Part B - Compare difference in free energy to the thermal…
need help please and thanks dont understand only need help with C-F
Learning Goal:
As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT.
The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7).
Part A - Difference in binding free eenergies
Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol.
The margin of error is 2%.
Part B - Compare difference in free energy to the thermal…
Please correct answer and don't used hand raiting
Chapter 21 Solutions
OWLv2 with Student Solutions Manual eBook for Ebbing/Gammon's General Chemistry, 11th Edition, [Instant Access], 4 terms (24 months)
Ch. 21.9 - Considering the fact that N2 makes up about 80% of...Ch. 21.10 - Prob. 21.2CCCh. 21 - Prob. 21.1QPCh. 21 - Prob. 21.2QPCh. 21 - Prob. 21.3QPCh. 21 - Prob. 21.4QPCh. 21 - Prob. 21.5QPCh. 21 - Prob. 21.6QPCh. 21 - Prob. 21.7QPCh. 21 - Prob. 21.8QP
Ch. 21 - Prob. 21.9QPCh. 21 - Prob. 21.10QPCh. 21 - Prob. 21.11QPCh. 21 - Prob. 21.12QPCh. 21 - Prob. 21.13QPCh. 21 - Prob. 21.14QPCh. 21 - Prob. 21.15QPCh. 21 - Prob. 21.16QPCh. 21 - Prob. 21.17QPCh. 21 - Prob. 21.18QPCh. 21 - Prob. 21.19QPCh. 21 - Prob. 21.20QPCh. 21 - Prob. 21.21QPCh. 21 - Prob. 21.22QPCh. 21 - Prob. 21.23QPCh. 21 - Prob. 21.24QPCh. 21 - Prob. 21.25QPCh. 21 - Prob. 21.26QPCh. 21 - Prob. 21.27QPCh. 21 - Prob. 21.28QPCh. 21 - Prob. 21.29QPCh. 21 - Prob. 21.30QPCh. 21 - Prob. 21.31QPCh. 21 - Prob. 21.32QPCh. 21 - Prob. 21.33QPCh. 21 - Prob. 21.34QPCh. 21 - Prob. 21.35QPCh. 21 - Prob. 21.36QPCh. 21 - Prob. 21.37QPCh. 21 - Prob. 21.38QPCh. 21 - Prob. 21.39QPCh. 21 - Prob. 21.40QPCh. 21 - Prob. 21.41QPCh. 21 - Describe the steps in the Ostwald process for the...Ch. 21 - Prob. 21.43QPCh. 21 - Prob. 21.44QPCh. 21 - Prob. 21.45QPCh. 21 - Prob. 21.46QPCh. 21 - Prob. 21.47QPCh. 21 - Prob. 21.48QPCh. 21 - What is the most important commercial means of...Ch. 21 - Prob. 21.50QPCh. 21 - Prob. 21.51QPCh. 21 - Prob. 21.52QPCh. 21 - Prob. 21.53QPCh. 21 - Prob. 21.54QPCh. 21 - Prob. 21.55QPCh. 21 - Prob. 21.56QPCh. 21 - Prob. 21.57QPCh. 21 - Prob. 21.58QPCh. 21 - Prob. 21.59QPCh. 21 - Prob. 21.60QPCh. 21 - Prob. 21.61QPCh. 21 - A test tube contains a solution of one of the...Ch. 21 - Prob. 21.63QPCh. 21 - Prob. 21.64QPCh. 21 - Prob. 21.65QPCh. 21 - Prob. 21.66QPCh. 21 - Prob. 21.67QPCh. 21 - Prob. 21.68QPCh. 21 - Prob. 21.69QPCh. 21 - Prob. 21.70QPCh. 21 - Prob. 21.71QPCh. 21 - Prob. 21.72QPCh. 21 - Prob. 21.73QPCh. 21 - Prob. 21.74QPCh. 21 - Prob. 21.75QPCh. 21 - Prob. 21.76QPCh. 21 - Prob. 21.77QPCh. 21 - Prob. 21.78QPCh. 21 - Prob. 21.79QPCh. 21 - Prob. 21.80QPCh. 21 - Prob. 21.81QPCh. 21 - Prob. 21.82QPCh. 21 - Prob. 21.83QPCh. 21 - Prob. 21.84QPCh. 21 - Prob. 21.85QPCh. 21 - Prob. 21.86QPCh. 21 - Sketch a diagram showing the formation of energy...Ch. 21 - Sketch a diagram showing the formation of energy...Ch. 21 - Prob. 21.89QPCh. 21 - Prob. 21.90QPCh. 21 - Prob. 21.91QPCh. 21 - Prob. 21.92QPCh. 21 - Prob. 21.93QPCh. 21 - Prob. 21.94QPCh. 21 - Francium was discovered as a minor decay product...Ch. 21 - Prob. 21.96QPCh. 21 - Prob. 21.97QPCh. 21 - Prob. 21.98QPCh. 21 - Prob. 21.99QPCh. 21 - Prob. 21.100QPCh. 21 - Prob. 21.101QPCh. 21 - Prob. 21.102QPCh. 21 - Prob. 21.103QPCh. 21 - Prob. 21.104QPCh. 21 - Prob. 21.105QPCh. 21 - Prob. 21.106QPCh. 21 - Prob. 21.107QPCh. 21 - Prob. 21.108QPCh. 21 - Prob. 21.109QPCh. 21 - Prob. 21.110QPCh. 21 - Prob. 21.111QPCh. 21 - Prob. 21.112QPCh. 21 - Prob. 21.113QPCh. 21 - Prob. 21.114QPCh. 21 - Prob. 21.115QPCh. 21 - Prob. 21.116QPCh. 21 - Prob. 21.117QPCh. 21 - Prob. 21.118QPCh. 21 - Prob. 21.119QPCh. 21 - Prob. 21.120QPCh. 21 - Prob. 21.121QPCh. 21 - Prob. 21.122QPCh. 21 - Prob. 21.123QPCh. 21 - Prob. 21.124QPCh. 21 - Prob. 21.125QPCh. 21 - Prob. 21.126QPCh. 21 - Prob. 21.127QPCh. 21 - Prob. 21.128QPCh. 21 - Prob. 21.129QPCh. 21 - Prob. 21.130QPCh. 21 - Prob. 21.131QPCh. 21 - Prob. 21.132QPCh. 21 - Prob. 21.133QPCh. 21 - Prob. 21.134QPCh. 21 - Prob. 21.135QPCh. 21 - Prob. 21.136QPCh. 21 - Prob. 21.137QPCh. 21 - Prob. 21.138QPCh. 21 - Prob. 21.139QPCh. 21 - Prob. 21.140QPCh. 21 - Prob. 21.141QPCh. 21 - Prob. 21.142QPCh. 21 - Prob. 21.143QPCh. 21 - Phosphorous acid, H3PO3, is oxidized to phosphoric...Ch. 21 - Prob. 21.145QPCh. 21 - Prob. 21.146QPCh. 21 - Prob. 21.147QPCh. 21 - Prob. 21.148QPCh. 21 - What are the oxidation numbers of sulfur in each...Ch. 21 - What are the oxidation numbers of sulfur in each...Ch. 21 - Prob. 21.151QPCh. 21 - Prob. 21.152QPCh. 21 - Prob. 21.153QPCh. 21 - Prob. 21.154QPCh. 21 - Prob. 21.155QPCh. 21 - Prob. 21.156QPCh. 21 - Chlorine can be prepared by oxidizing chloride ion...Ch. 21 - Prob. 21.158QPCh. 21 - Prob. 21.159QPCh. 21 - Prob. 21.160QPCh. 21 - Prob. 21.161QPCh. 21 - Prob. 21.162QPCh. 21 - Prob. 21.163QPCh. 21 - Prob. 21.164QPCh. 21 - Prob. 21.165QPCh. 21 - Prob. 21.166QPCh. 21 - Prob. 21.167QPCh. 21 - Xenon trioxide, XeO3, is reduced to xenon in...Ch. 21 - Prob. 21.169QPCh. 21 - Prob. 21.170QPCh. 21 - Prob. 21.171QPCh. 21 - Prob. 21.172QPCh. 21 - Prob. 21.173QPCh. 21 - Prob. 21.174QPCh. 21 - Prob. 21.175QPCh. 21 - Prob. 21.176QPCh. 21 - Prob. 21.177QPCh. 21 - Prob. 21.178QPCh. 21 - Prob. 21.179QPCh. 21 - Prob. 21.180QPCh. 21 - Prob. 21.181QPCh. 21 - Prob. 21.182QPCh. 21 - Prob. 21.183QPCh. 21 - Prob. 21.184QPCh. 21 - Prob. 21.185QPCh. 21 - Prob. 21.186QPCh. 21 - Prob. 21.187QPCh. 21 - Sodium perchlorate, NaClO4, is produced by...Ch. 21 - The amount of sodium hypochlorite in a bleach...Ch. 21 - Prob. 21.190QPCh. 21 - Prob. 21.191QPCh. 21 - Prob. 21.192QPCh. 21 - Prob. 21.193QPCh. 21 - Prob. 21.194QPCh. 21 - Prob. 21.195QPCh. 21 - Prob. 21.196QPCh. 21 - Prob. 21.197QPCh. 21 - Prob. 21.198QPCh. 21 - Prob. 21.199QPCh. 21 - Prob. 21.200QPCh. 21 - Prob. 21.201QPCh. 21 - Prob. 21.202QPCh. 21 - Prob. 21.203QPCh. 21 - Prob. 21.204QP
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Similar questions
- need help please and thanks dont understand a-b Learning Goal: As discussed during the lecture, the enzyme HIV-1 reverse transcriptae (HIV-RT) plays a significant role for the HIV virus and is an important drug target. Assume a concentration [E] of 2.00 µM (i.e. 2.00 x 10-6 mol/l) for HIV-RT. Two potential drug molecules, D1 and D2, were identified, which form stable complexes with the HIV-RT. The dissociation constant of the complex ED1 formed by HIV-RT and the drug D1 is 1.00 nM (i.e. 1.00 x 10-9). The dissociation constant of the complex ED2 formed by HIV-RT and the drug D2 is 100 nM (i.e. 1.00 x 10-7). Part A - Difference in binding free eenergies Compute the difference in binding free energy (at a physiological temperature T=310 K) for the complexes. Provide the difference as a positive numerical expression with three significant figures in kJ/mol. The margin of error is 2%. Part B - Compare difference in free energy to the thermal energy Divide the…arrow_forwardPlease correct answer and don't used hand raitingarrow_forwardPlease correct answer and don't used hand raitingarrow_forward
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