To determine the percentage composition of a compound of your choice and then use these percentage composition to determine the molecular formulas of the same, after that find the empirical formula with the help of percentage composition data. Concept Introduction: Percentage by mass of any element in the compound is the amount of that element in the total amount of the compound. The chemical formula which represents the simplest whole number atoms ratio present in the compound is said to be empirical formula. The chemical formula which shows the actual number of every atoms present in the compound is known as molecular formula. Number of moles can be calculated as follows; Nuumber of moles = mass in g molar mass .
To determine the percentage composition of a compound of your choice and then use these percentage composition to determine the molecular formulas of the same, after that find the empirical formula with the help of percentage composition data. Concept Introduction: Percentage by mass of any element in the compound is the amount of that element in the total amount of the compound. The chemical formula which represents the simplest whole number atoms ratio present in the compound is said to be empirical formula. The chemical formula which shows the actual number of every atoms present in the compound is known as molecular formula. Number of moles can be calculated as follows; Nuumber of moles = mass in g molar mass .
Solution Summary: The author explains the molecular formula of cyclopropane, which shows the simplest whole number atoms ratio present in the compound.
To determine the percentage composition of a compound of your choice and then use these percentage composition to determine the molecular formulas of the same, after that find the empirical formula with the help of percentage composition data.
Concept Introduction:
Percentage by mass of any element in the compound is the amount of that element in the total amount of the compound.
The chemical formula which represents the simplest whole number atoms ratio present in the compound is said to be empirical formula.
The chemical formula which shows the actual number of every atoms present in the compound is known as molecular formula.
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
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…
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Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY