Answer:
I think it's B.compressibility, seems the only one to make sense
Cant see it well please find better view
Answer:
a) 
b) 
Explanation:
Equation of reaction:

Initial pressure 3 1 0
Pressure change 2P 1P 2P
Total pressure = (3-2P) + (1-P) + (2P)
Total Pressure = 3.75 atm
(3-2P) + (1-P) + (2P) = 3.75
4 - P = 3.75
P = 4 - 3.75
P = 0.25 atm
Let us calculate the pressure of each of the components of the reaction:
Pressure of XO2 = 3 - 2P = 3 - 2(0.25)
Pressure of XO2 =2.5 atm
Pressure of O2 = 1 - P = 1 -0.25
Pressure of O2 = 0.75 atm
Pressure of XO3 = 2P = 2 * 0.25
Pressure of XO3 = 0.5 atm
From the reaction, equilibrium constant can be calculated using the formula:
![K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20%5Cfrac%7B%5BPXO_%7B3%7D%5D%20%5E%7B2%7D%20%7D%7B%5BPXO_%7B2%7D%5D%20%5E%7B2%7D%5BPO_%7B2%7D%5D%20%7D)

Standard free energy:

b) value of k−1 at 27 °C, i.e. 300K



Answer:
its is letter c is your answer
Explanation:
I hope this help
A. low blood glucose <span>is an example of a condition in which homeostasis is maintained using a negative feedback loop</span>