Answer:
pKa of the histidine = 9.67
Explanation:
The relation between standard Gibbs energy and equilibrium constant is shown below as:
![\Delta{G^0} =-RT \ln \frac{[His]}{[His+]}](https://tex.z-dn.net/?f=%5CDelta%7BG%5E0%7D%20%3D-RT%20%5Cln%20%5Cfrac%7B%5BHis%5D%7D%7B%5BHis%2B%5D%7D)
R is Gas constant having value = 0.008314 kJ / K mol
Given temperature, T = 293 K
Given, 
So, Applying in the equation as:-
![15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}](https://tex.z-dn.net/?f=15%5C%20kJ%2Fmol%3D-0.008314%5C%20kJ%2FKmol%5Ctimes%20293%5C%20K%5Ctimes%20%5Cln%20%5Cfrac%7B%5BHis%5D%7D%7B%5BHis%2B%5D%7D)
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:
The molality of the glycerol solution is 2.960×10^-2 mol/kg
Explanation:
Number of moles of glycerol = Molarity × volume of solution = 2.950×10^-2 M × 1 L = 2.950×10^-2 moles
Mass of water = density × volume = 0.9982 g/mL × 998.7 mL = 996.90 g = 996.90/1000 = 0.9969 kg
Molality = number of moles of glycerol/mass of water in kg = 2.950×10^-2/0.9969 = 2.960×10^-2 mol/kg
Answer is 0.289nm.
Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.
wt % of Fe in Fe-V alloy = 85%
wt % of V in Fe-V alloy = 15%
We need to calculate edge length of the unit cell having bcc structure.
Using density formula,
For calculating edge length,
For calculating 
, we use the formula
Similarly for calculating 
, we use the formula
From the periodic table, masses of the two elements can be written
Specific density of both the elements are
Putting and 
formula's in edge length formula, we get
![a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}](https://tex.z-dn.net/?f=a%3D%5Cleft%20%5B%5Cfrac%7BZ%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B%28wt%5C%25%29_%7BFe%7D%7D%7BM_%7BFe%7D%7D%2B%5Cfrac%7B%28wt%5C%25%29_%7BFe%7D%7D%7BM_%7BFe%7D%7D%7D%20%20%5Cright%20%29%7D%7BN_A%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B%28wt%5C%25%29_V%7D%7B%5Crho_V%7D%2B%5Cfrac%7B%28wt%5C%25%29_V%7D%7B%5Crho_V%7D%7D%20%20%5Cright%20%29%7D%20%20%5Cright%20%5D%5E%7B1%2F3%7D)
![a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}](https://tex.z-dn.net/?f=a%3D%5Cleft%20%5B%5Cfrac%7B2atoms%2F%5Ctext%7Bunit%20cell%7D%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B85%5C%25%7D%7B55.85g%2Fmol%7D%2B%5Cfrac%7B15%5C%25%7D%7B50.941g%2Fmol%7D%7D%20%20%5Cright%20%29%7D%7B%286.023%5Ctimes10%5E%7B23%7Datoms%2Fmol%29%5Cleft%20%28%5Cfrac%7B100%7D%7B%5Cfrac%7B85%5C%25%7D%7B7.874g%2Fcm%5E3%7D%2B%5Cfrac%7B15%5C%25%7D%7B6.10g%2Fcm%5E3%7D%7D%20%20%5Cright%20%29%7D%20%20%5Cright%20%5D%5E%7B1%2F3%7D)
By calculating, we get
Socratic helps for you page
Answer:
Explanation:
Since salt is composed of two ions Na+ and Cl- it will attract to the opposite poles of the water and form a ion-dipole bond. This will cause the phenomenon freezing point depression where as the name states it will take a lower tempertature to freeze it. So instead of freezing at 0 degrees c it will freeze at a negative 2 or so on.
