An example for aerosol, foam, emulsion, sol and gel has to be given. Concept Introduction: Based on the particle size of the components, solutions can be of three types – true solutions, colloidal solutions and suspension. Colloidal solutions have the particle size of the range 1 - 1 0 0 0 n m . a colloidal solution is made of dispersed phase and dispersion medium which corresponds to solute and solvent respectively. Depending upon the physical state of dispersed phase and dispersion medium, there are various types of colloids such as gel, foam, alloy, sol etc.
An example for aerosol, foam, emulsion, sol and gel has to be given. Concept Introduction: Based on the particle size of the components, solutions can be of three types – true solutions, colloidal solutions and suspension. Colloidal solutions have the particle size of the range 1 - 1 0 0 0 n m . a colloidal solution is made of dispersed phase and dispersion medium which corresponds to solute and solvent respectively. Depending upon the physical state of dispersed phase and dispersion medium, there are various types of colloids such as gel, foam, alloy, sol etc.
Solution Summary: The author explains that aerosol, foam, emulsion, sol and gel have been given according to their particle size and dispersion medium.
An example for aerosol, foam, emulsion, sol and gel has to be given.
Concept Introduction:
Based on the particle size of the components, solutions can be of three types – true solutions, colloidal solutions and suspension. Colloidal solutions have the particle size of the range 1-1000nm. a colloidal solution is made of dispersed phase and dispersion medium which corresponds to solute and solvent respectively. Depending upon the physical state of dispersed phase and dispersion medium, there are various types of colloids such as gel, foam, alloy, sol etc.
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…
Chapter 12 Solutions
OWLv2 with Student Solutions Manual eBook for Ebbing/Gammon's General Chemistry, 11th Edition, [Instant Access], 4 terms (24 months)