Bundle: Chemistry for Today: General, Organic, and Biochemistry, Loose-Leaf Version, 9th + LMS Integrated OWLv2, 4 terms (24 months) Printed Access Card
9th Edition
ISBN: 9781337598255
Author: Spencer L. Seager
Publisher: Cengage Learning
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Chapter 17, Problem 17.43E
Interpretation Introduction
Interpretation:
The monosaccharide which is the component of all three of the disaccharides sucrose, maltose, and lactose is to be stated.
Concept Introduction:
Monosaccharides are the small units of simple sugars. Disaccharides are made up of two small units of monosaccharides. These small units are joined together by glycosidic linkage. Carbohydrates are naturally occurring organic compounds.
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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 17 Solutions
Bundle: Chemistry for Today: General, Organic, and Biochemistry, Loose-Leaf Version, 9th + LMS Integrated OWLv2, 4 terms (24 months) Printed Access Card
Ch. 17 - Prob. 17.1ECh. 17 - Describe whether each of the following substances...Ch. 17 - Prob. 17.3ECh. 17 - Prob. 17.4ECh. 17 - Prob. 17.5ECh. 17 - Why are carbon atoms 1 and 3 of glyceraldehyde not...Ch. 17 - Prob. 17.7ECh. 17 - Which of the following molecules can have...Ch. 17 - Which of the following molecules can have...Ch. 17 - Prob. 17.10E
Ch. 17 - Prob. 17.11ECh. 17 - Prob. 17.12ECh. 17 - Prob. 17.13ECh. 17 - Draw Fischer projections for both the D and L...Ch. 17 - Prob. 17.15ECh. 17 - Prob. 17.16ECh. 17 - Prob. 17.17ECh. 17 - Prob. 17.18ECh. 17 - Prob. 17.19ECh. 17 - Prob. 17.20ECh. 17 - Prob. 17.21ECh. 17 - Prob. 17.22ECh. 17 - Prob. 17.23ECh. 17 - Prob. 17.24ECh. 17 - Prob. 17.25ECh. 17 - Prob. 17.26ECh. 17 - Prob. 17.27ECh. 17 - Prob. 17.28ECh. 17 - Prob. 17.29ECh. 17 - Prob. 17.30ECh. 17 - Prob. 17.31ECh. 17 - Prob. 17.32ECh. 17 - Prob. 17.33ECh. 17 - Prob. 17.34ECh. 17 - Prob. 17.35ECh. 17 - Prob. 17.36ECh. 17 - Prob. 17.37ECh. 17 - Prob. 17.38ECh. 17 - Prob. 17.39ECh. 17 - Prob. 17.40ECh. 17 - Prob. 17.41ECh. 17 - Prob. 17.42ECh. 17 - Prob. 17.43ECh. 17 - Prob. 17.44ECh. 17 - Prob. 17.45ECh. 17 - Prob. 17.46ECh. 17 - Prob. 17.47ECh. 17 - Sucrose and honey are commonly used sweeteners....Ch. 17 - Prob. 17.49ECh. 17 - Prob. 17.50ECh. 17 - Prob. 17.51ECh. 17 - Prob. 17.52ECh. 17 - Prob. 17.53ECh. 17 - Prob. 17.54ECh. 17 - Prob. 17.55ECh. 17 - Prob. 17.56ECh. 17 - Prob. 17.57ECh. 17 - Prob. 17.58ECh. 17 - Prob. 17.59ECh. 17 - Prob. 17.60ECh. 17 - Prob. 17.61ECh. 17 - Prob. 17.62ECh. 17 - Prob. 17.63ECh. 17 - Prob. 17.64ECh. 17 - Prob. 17.65ECh. 17 - Prob. 17.66ECh. 17 - Prob. 17.67ECh. 17 - Prob. 17.68ECh. 17 - Prob. 17.69ECh. 17 - Prob. 17.70ECh. 17 - Prob. 17.71ECh. 17 - Prob. 17.72ECh. 17 - Prob. 17.73ECh. 17 - Glucose is a reducing sugar, which if boiled in...Ch. 17 - Prob. 17.75E
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