Answer:
Aluminum lodide - 680 degrees
fructose- 824 degrees
potassium bromide- 2,615 degrees
calcium bromide 3, 515 degrees
lowest boiling point is 680 and highest is 3,515.
hope this helps:)
Explanation:
I think there' will be a decrease in volume of the air.
Answer:

Explanation:
In this problem we only have information of the equilibrium, so we need to find a expression of the free energy in function of the constant of equilireium (Keq):

Being Keq:
![K_{eq}=\frac{[fructose][Pi]}{[Fructose-1-P]}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5Bfructose%5D%5BPi%5D%7D%7B%5BFructose-1-P%5D%7D)
Initial conditions:
![[Fructose-1-P]=0.2M](https://tex.z-dn.net/?f=%5BFructose-1-P%5D%3D0.2M)
![[Fructose]=0M](https://tex.z-dn.net/?f=%5BFructose%5D%3D0M)
![[Pi]=0M](https://tex.z-dn.net/?f=%5BPi%5D%3D0M)
Equilibrium conditions:
![[Fructose-1-P]=6.52*10^{-5}M](https://tex.z-dn.net/?f=%5BFructose-1-P%5D%3D6.52%2A10%5E%7B-5%7DM)
![[Fructose]=0.2M-6.52*10^{-5}M](https://tex.z-dn.net/?f=%5BFructose%5D%3D0.2M-6.52%2A10%5E%7B-5%7DM)
![[Pi]=0.2M-6.52*10^{-5}M](https://tex.z-dn.net/?f=%5BPi%5D%3D0.2M-6.52%2A10%5E%7B-5%7DM)


Free-energy for T=298K (standard):

