Student Solutions Manual for Zumdahl/DeCoste's Introductory Chemistry: A Foundation, 9th
9th Edition
ISBN: 9781337399470
Author: ZUMDAHL, Steven S.; DeCoste, Donald J.
Publisher: Cengage Learning
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Chapter 19, Problem 100AP
Interpretation Introduction
Interpretation:
Interpret balanced
Concept Introduction:
The decaying of an atom takes place by three type of decay such as alpha decay, beta decay, gama decay.
In alpha decay, an alpha particle with mass 4m and electron 2e removed from nuclide while in beta decay two type of decay are possible beta (+) and beta (-). In gamma decay, nuclide will remain same to the original one.
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Learning Goal:
This question reviews the format for writing an element's written symbol. Recall that written symbols have a particular format. Written symbols use a form like this:
35 Cl
17
In this form the mass number, 35, is a stacked superscript. The atomic number, 17, is a stacked subscript. "CI" is the chemical symbol for the element chlorine. A general way to show this form is:
It is also correct to write symbols by leaving off the atomic number, as in the following form:
atomic number
mass number Symbol
35 Cl or
mass number Symbol
This is because if you write the element symbol, such as Cl, you know the atomic number is 17 from that symbol. Remember that the atomic number, or number of protons in the nucleus, is what defines the element. Thus, if 17 protons
are in the nucleus, the element can only be chlorine. Sometimes you will only see 35 C1, where the atomic number is not written.
Watch this video to review the format for written symbols.
In the following table each column…
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…
Chapter 19 Solutions
Student Solutions Manual for Zumdahl/DeCoste's Introductory Chemistry: A Foundation, 9th
Ch. 19.1 - Prob. 1CTCh. 19.1 - Prob. 19.1SCCh. 19.1 - Prob. 19.2SCCh. 19.3 - Exercise19.3Watches with numerals that glow in the...Ch. 19.8 - Prob. 1CTCh. 19 - Prob. 1ALQCh. 19 - Prob. 2ALQCh. 19 - Prob. 3ALQCh. 19 - Prob. 4ALQCh. 19 - Prob. 5ALQ
Ch. 19 - Prob. 6ALQCh. 19 - Prob. 7ALQCh. 19 - Prob. 8ALQCh. 19 - Prob. 9ALQCh. 19 - Prob. 10ALQCh. 19 - Prob. 1QAPCh. 19 - Prob. 2QAPCh. 19 - Prob. 3QAPCh. 19 - Prob. 4QAPCh. 19 - Prob. 5QAPCh. 19 - Prob. 6QAPCh. 19 - Prob. 7QAPCh. 19 - Prob. 8QAPCh. 19 - Prob. 9QAPCh. 19 - Prob. 10QAPCh. 19 - Prob. 11QAPCh. 19 - Prob. 12QAPCh. 19 - Prob. 13QAPCh. 19 - Prob. 14QAPCh. 19 - Prob. 15QAPCh. 19 - Prob. 16QAPCh. 19 - Prob. 17QAPCh. 19 - Prob. 18QAPCh. 19 - Prob. 19QAPCh. 19 - Prob. 20QAPCh. 19 - Prob. 21QAPCh. 19 - Prob. 22QAPCh. 19 - Prob. 23QAPCh. 19 - Prob. 24QAPCh. 19 - Prob. 25QAPCh. 19 - Prob. 26QAPCh. 19 - Prob. 27QAPCh. 19 - Prob. 28QAPCh. 19 - Prob. 29QAPCh. 19 - Prob. 30QAPCh. 19 - Prob. 31QAPCh. 19 - Prob. 32QAPCh. 19 - Prob. 33QAPCh. 19 - Prob. 34QAPCh. 19 - Prob. 35QAPCh. 19 - Prob. 36QAPCh. 19 - Prob. 37QAPCh. 19 - Prob. 38QAPCh. 19 - Prob. 39QAPCh. 19 - Prob. 40QAPCh. 19 - Prob. 41QAPCh. 19 - Prob. 42QAPCh. 19 - Prob. 43QAPCh. 19 - Prob. 44QAPCh. 19 - Prob. 45QAPCh. 19 - Prob. 46QAPCh. 19 - Prob. 47QAPCh. 19 - Prob. 48QAPCh. 19 - . How do the forces that hold an atomic nucleus...Ch. 19 - Prob. 50QAPCh. 19 - Prob. 51QAPCh. 19 - Prob. 52QAPCh. 19 - Prob. 53QAPCh. 19 - Prob. 54QAPCh. 19 - Prob. 55QAPCh. 19 - Prob. 56QAPCh. 19 - Prob. 57QAPCh. 19 - Prob. 58QAPCh. 19 - Prob. 59QAPCh. 19 - Prob. 60QAPCh. 19 - Prob. 61QAPCh. 19 - Prob. 62QAPCh. 19 - Prob. 63QAPCh. 19 - Prob. 64QAPCh. 19 - Prob. 65QAPCh. 19 - Prob. 66QAPCh. 19 - Prob. 67QAPCh. 19 - Prob. 68QAPCh. 19 - Prob. 69APCh. 19 - Prob. 70APCh. 19 - Prob. 71APCh. 19 - Prob. 72APCh. 19 - Prob. 73APCh. 19 - Prob. 74APCh. 19 - Prob. 75APCh. 19 - Prob. 76APCh. 19 - Prob. 77APCh. 19 - Prob. 78APCh. 19 - Prob. 79APCh. 19 - . The elements with atomic numbers of 93 or...Ch. 19 - Prob. 81APCh. 19 - Prob. 82APCh. 19 - Prob. 83APCh. 19 - Prob. 84APCh. 19 - Prob. 85APCh. 19 - Prob. 86APCh. 19 - Prob. 87APCh. 19 - Prob. 88APCh. 19 - Prob. 89APCh. 19 - Prob. 90APCh. 19 - Prob. 91APCh. 19 - Prob. 92APCh. 19 - Prob. 93APCh. 19 - Prob. 94APCh. 19 - . The element zinc in nature consists of five...Ch. 19 - . Aluminum exists in several isotopic forms,...Ch. 19 - Prob. 97APCh. 19 - Prob. 98APCh. 19 - Prob. 99APCh. 19 - Prob. 100APCh. 19 - Prob. 101APCh. 19 - Prob. 102APCh. 19 - Prob. 103APCh. 19 - Prob. 104CPCh. 19 - Prob. 105CPCh. 19 - Prob. 106CPCh. 19 - Prob. 107CP
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- Please correct answer and don't used hand raitingarrow_forwardneed 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_forward
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