The blank is - <span>Endoplasmic reticulum</span>
Answer:
1. <u>No, you cannot calculate the solubility of X in water at 26ºC.</u>
Explanation:
You cannot calculate the solubility of X in <em>water at 26 degrees Celsius </em>because you do not know whether the solution formed by dissolving the crystals in 3.00 liters of water is saturaed or not.
The only way to determine the solubility of the compound X is by dissolving the crystals in certain (measured) amount of water and making sure that some crystals remain undissolved, as a solid on the bottom of the beaker.
Next, you should filter the solution to remove the undissolved crystals. Then, weigh the solution, evaporate, wash, dry, and weigh the crystals.
Then you have the mass of the crystals dissolved and the mass of the solution which will let you calculate the mass of pure water, and then the solubility.
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>
Two parts are stage and coarse focus
