The rate equation is given as:
k = A e^(- Ea / RT)
Dividing state 1 and state 2:
k1/k2 = e^(- Ea / RT1) / e^(- Ea / RT2)
k1/k2 = e^[- Ea / RT1 - (- Ea / RT2)]
k1/k2 = e^[- Ea / RT1 + Ea / RT2)]
Taking the ln of both sides:
ln (k1/k2) = - Ea / RT1 + Ea / RT2
ln (k1/k2) = - Ea / R (1/T1 - 1/T2)
Since k2 = 4k1, therefore k1/k2 = ¼
ln (1/4) = [- (56,000 J/mol) / (8.314 J / mol K)] (1/273
K – 1/ T2)
2.058 x 10^-4 = 1/273 – 1/T2
T2 = 289.25 K
Answer:
624510100
Explanation:
Doing a conversion factor:
![0,0006245101[km]*\frac{1000[m]}{1 km} *\frac{1x10^{9} nanometer}{1 m} =624510100 [nanometer]](https://tex.z-dn.net/?f=0%2C0006245101%5Bkm%5D%2A%5Cfrac%7B1000%5Bm%5D%7D%7B1%20km%7D%20%2A%5Cfrac%7B1x10%5E%7B9%7D%20nanometer%7D%7B1%20m%7D%20%3D624510100%20%5Bnanometer%5D)
Potassium oxide has the antifluorite structure. The antifluorite structure have compounds with the stoichiometry X₂Y, where X is the cation and Y is the anion. In the antifluorite structure <span>positions of the </span>cations<span> and </span>anions<span> are reversed relative to their positions in calcium fluoride.</span>
Potassium ions coordinated to 4 oxide ions, <span>potassium ions are all in the tetrahedral holes.</span>
Answer:
That is a compound. If it was an element it would either just be Na or Cl.
