Answer:
2.40 M
Explanation:
The molarity of a solution tells you how many moles of solute you get per liter of solution.
Notice that the problem provides you with the volume of the solution expressed in milliliters,
mL
. Right from the start, you should remember that you must convert this volume to liters by using the conversion factor
1 L
=
10
3
mL
Now, in order to get the number of moles of solute, you must use its molar mass. Now, molar masses are listed in grams per mol,
g mol
−
1
, which means that you're going to have to convert the mass of the sample from milligrams to grams
1 g
=
10
3
mg
Sodium chloride,
NaCl
, has a molar mass of
58.44 g mol
−
1
, which means that your sample will contain
unit conversion
280.0
mg
⋅
1
g
10
3
mg
⋅
molar mass
1 mole NaCl
58.44
g
=
0.004791 moles NaCl
This means that the molarity of the solution will be
c
=
n
solute
V
solution
c
=
0.004791 moles
2.00
⋅
10
−
3
L
=
2.40 M
The answer is rounded to three sig figs, the number of sig figs you have for the volume of the solution.
Answer:
pKa of the histidine = 9.67
Explanation:
The relation between standard Gibbs energy and equilibrium constant is shown below as:
R is Gas constant having value = 0.008314 kJ / K mol
Given temperature, T = 293 K
Given,
So, Applying in the equation as:-
Thus,
![\frac{[His]}{[His+]}=e^{\frac{15}{-0.008314\times 293}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHis%5D%7D%7B%5BHis%2B%5D%7D%3De%5E%7B%5Cfrac%7B15%7D%7B-0.008314%5Ctimes%20293%7D)
![\frac{[His]}{[His+]}=0.00211](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHis%5D%7D%7B%5BHis%2B%5D%7D%3D0.00211)
Also, considering:-
![pH=pKa+log\frac{[His]}{[His+]}](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%5Cfrac%7B%5BHis%5D%7D%7B%5BHis%2B%5D%7D)
Given that:- pH = 7.0
So, 
<u>pKa of the histidine = 9.67</u>
Answer:
both
Explanation:
A homolytic fission is said to have occurred when the breakage of a bond between two atoms leaves each of the bonding atoms with equal number of electrons. Homolytic fission often results in the creation of radicals.
Since homolytic fission yields two species with equal number of electrons(usually odd number of electrons), the products of such process can not be charged. They can not be nucleophiles because nucleophiles need to possess two electrons which can be shared with another chemical specie.