The short segment of nylon 4 has to be drawn Concept Introduction: Polyamides: Nylon is a polyamide and is formed from the polymerization of 6-aminohexanoyl chloride and hence it is called nylon-6 Nylon 66 is formed from adipic acid and 1,6-hexanediamine which have 6 carbons each. Polyamide belongs to the class of step-growth polymer .

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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…

Answer 1

Question

Chapter 27.11, Problem 16P

(a)

Interpretation Introduction

Interpretation:

The short segment of nylon 4 has to be drawn

Concept Introduction:

Polyamides:

Nylon is a polyamide and is formed from the polymerization of 6-aminohexanoyl chloride and hence it is called nylon-6

Nylon 66 is formed from adipic acid and 1,6-hexanediamine which have 6 carbons each.

Polyamide belongs to the class of step-growth polymer.

(b

)

Interpretation Introduction

Interpretation:

The short segment of nylon 44 has to be drawn

Concept Introduction:

Polyamides:

Nylon is a polyamide and is formed from the polymerization of 6-aminohexanoyl chloride and hence it is called nylon-6

Nylon 66 is formed from adipic acid and 1,6-hexanediamine which have 6 carbons each.

Polyamide belongs to the class of step-growth polymer.

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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…

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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…