Answer:
8
Explanation:
When you balance the entire equation, you should get:
C5H12 + 8O2 ---> 5CO2 + 6H2O
We first calculate the energy contained in one photon of this light using Planck's equation:
E = hc/λ
E = 6.63 x 10⁻³⁴ x 3 x 10⁸ / 590 x 10⁻⁹
E = 3.37 x 10⁻²² kJ/photon
Now, one mole of atoms will excite one mole of photons. This means that 6.02 x 10²³ photons will be excited
(3.37 x 10⁻²² kJ/photon) x (6.02 x 10²³ photons / mol)
The energy released will be 202.87 kJ/mol
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):


A chemical bond is the best classification.