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:
10.28 mol
Explanation:
S + 2O = SO2
(atm x L) ÷ (0.0821 x K)
(3.45 x 45.6) ÷ (0.0821 x 373)
=5.13726
Then round it to significant figures
=5.14
5.14 mol SO2 x (2 mol O ÷ 1 mol SO2)
=10.28 mol O
To find the mass you need to find the weight of a mol of the molecules by adding up the atomic mass.
N = 14.007 g/mol
H = 1.008 g/mol
S = 32.065 g/mol
O = 16 g/mol
2(14.007) + 8(1.008) + 32.065 + 4(16) = 132.143 g/mol
Now you know how much an entire mol weight you multiply it by how much you actually have
0.00456 * 132.143 = 0.603 g