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: 4.18925 kJ heat is needed to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.
Explanation:
Temperature of Solid
Melting temperature of Solid 
Temperature of liquid 
Specific heats of solid ethanol = 0.97 J/gK
Specific heats of liquid ethanol = 2.3 J/gK
Heat required to melt the the 25 g solid
at 159 K
= 159 K - 138 K = 21 K

Heat required to melt and raise the temperature of
upto 223 K
= 223 K - 159 K = 64 K

Total heat to convert solid ethanol to liquid ethanol at given temperature :
(1kJ=1000J)
Hence, 4.18925 kJ of heat will be required to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.
The element "X" is "O" (oxygen).
<h3>Calculation:</h3>
Given,
Chemical formula = Na₂CX₃
Formula mass = 106 amu
Molar mass of Na = 23 amu
Molar mass of C = 12 amu
To find,
Element X =?
We will equate the equation as follows,
2(23) + 12 + 3(y) = 106
46 + 12 + 3y =106
58 + 3y = 106
3y = 106 - 58
3y = 48
y = 48/3
y = 16
We know that Oxygen has molecular mass of 16. Therefore the element "X" is "O".
Learn more about molar mass here:
brainly.com/question/22997914
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We can use the dilution formula to find the volume of the diluted solution to be prepared
c1v1 = c2v2
Where c1 is concentration and v1 is volume of the concentrated solution
And c2 is concentration and v2 is volume of the diluted solution to be prepared
Substituting the values in the equation
15 M x 25 mL = 3 M x v2
v2 = 125 mL
The 25 mL concentrated solution should be diluted with distilled water upto 125 mL to make a 3 M solution