Among the given conditions, the condition which represents a basic solution should be identified. pOH = 12.53 pH = 3.64 pOH<[H + ] [OH − ] > 1.0 × 10 − 7 M . Concept Introduction: Water is an amphoteric in nature implies it can act as an acid as well as base. Ionization of water is given as: H 2 O ( l ) + H 2 O ( l ) ⇀ ↽ H + ( a q ) + OH − ( a q ) Here, one water molecule is act as an acid by accepting proton and another molecule of water act as a base by donating proton. At 25 ° C , the actual concentrations of the products are: [H + ][OH − ]=1 × 10 − 14 To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of [H + ] in aqueous solution is small, by using the p scale in the form of pH scale, it is better way to represent acidity of solution. Thus, pH = -log [H + ] Similarly, the relation between pOH and [OH − ] is given by: pOH=-log [OH − ] The solution is basic when its pH is equal to greater than 7.
Among the given conditions, the condition which represents a basic solution should be identified. pOH = 12.53 pH = 3.64 pOH<[H + ] [OH − ] > 1.0 × 10 − 7 M . Concept Introduction: Water is an amphoteric in nature implies it can act as an acid as well as base. Ionization of water is given as: H 2 O ( l ) + H 2 O ( l ) ⇀ ↽ H + ( a q ) + OH − ( a q ) Here, one water molecule is act as an acid by accepting proton and another molecule of water act as a base by donating proton. At 25 ° C , the actual concentrations of the products are: [H + ][OH − ]=1 × 10 − 14 To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of [H + ] in aqueous solution is small, by using the p scale in the form of pH scale, it is better way to represent acidity of solution. Thus, pH = -log [H + ] Similarly, the relation between pOH and [OH − ] is given by: pOH=-log [OH − ] The solution is basic when its pH is equal to greater than 7.
Solution Summary: The author explains that water is an amphoteric in nature, which implies it can act as an acid as well as base.
Among the given conditions, the condition which represents a basic solution should be identified.
pOH = 12.53
pH=3.64
pOH<[H+]
[OH−]>1.0×10−7 M.
Concept Introduction:
Water is an amphoteric in nature implies it can act as an acid as well as base.
Ionization of water is given as:
H2O(l)+H2O(l)⇀↽H+(aq)+OH−(aq)
Here, one water molecule is act as an acid by accepting proton and another molecule of water act as a base by donating proton.
At 25°C, the actual concentrations of the products are:
[H+][OH−]=1×10−14
To express the small number, p scale is used, which implies to take the log of a number. Since, the concentration of [H+] in aqueous solution is small, by using the p scale in the form of pH scale, it is better way to represent acidity of solution.
Thus, pH = -log [H+]
Similarly, the relation between pOH and [OH−] is given by:
pOH=-log [OH−]
The solution is basic when its pH is equal to greater than 7.
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 16 Solutions
Student Solutions Manual for Zumdahl/DeCoste's Introductory Chemistry: A Foundation, 9th
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