Answer:
![1.75272\ \text{m}^3](https://tex.z-dn.net/?f=1.75272%5C%20%5Ctext%7Bm%7D%5E3)
Explanation:
The breakdown reaction of ozone is as follows
![CF_3Cl + UV \rightarrow CF_3 + Cl](https://tex.z-dn.net/?f=CF_3Cl%20%2B%20UV%20%5Crightarrow%20CF_3%20%2B%20Cl)
![Cl + O_3 \rightarrow ClO + O_2](https://tex.z-dn.net/?f=Cl%20%2B%20O_3%20%5Crightarrow%20ClO%20%2B%20O_2)
![O_3 + UV \rightarrow O_2 + O](https://tex.z-dn.net/?f=O_3%20%2B%20UV%20%5Crightarrow%20O_2%20%2B%20O)
![ClO + O \rightarrow Cl + O_2](https://tex.z-dn.net/?f=ClO%20%2B%20O%20%5Crightarrow%20Cl%20%2B%20O_2)
It can be seen that 2 moles of ozone is required in the complete cycle
So for 10 cycles, 20 moles of ozone is required
m = Mass of
= 15.5 g
M = Molar mass of
= 104.46 g/mol
P = Pressure = 24.5 mmHg
T = Temperature = 232 K
R = Gas constant = ![62.363\ \text{L mmHg/K mol}](https://tex.z-dn.net/?f=62.363%5C%20%5Ctext%7BL%20mmHg%2FK%20mol%7D)
Number of moles is given by
![n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{15.5}{104.46}\\\Rightarrow n=0.1484\ \text{moles}](https://tex.z-dn.net/?f=n%3D%5Cdfrac%7Bm%7D%7BM%7D%5C%5C%5CRightarrow%20n%3D%5Cdfrac%7B15.5%7D%7B104.46%7D%5C%5C%5CRightarrow%20n%3D0.1484%5C%20%5Ctext%7Bmoles%7D)
![20\ \text{moles} = 20\times 0.1484 = 2.968\ \text{moles}](https://tex.z-dn.net/?f=20%5C%20%5Ctext%7Bmoles%7D%20%3D%2020%5Ctimes%200.1484%20%3D%202.968%5C%20%5Ctext%7Bmoles%7D)
From ideal gas law we have
![PV=nRT\\\Rightarrow V=\dfrac{nRT}{P}\\\Rightarrow V=\dfrac{2.968\times 62.363\times 232}{24.5}\\\Rightarrow V=1752.72\ \text{L}=1.75272\ \text{m}^3](https://tex.z-dn.net/?f=PV%3DnRT%5C%5C%5CRightarrow%20V%3D%5Cdfrac%7BnRT%7D%7BP%7D%5C%5C%5CRightarrow%20V%3D%5Cdfrac%7B2.968%5Ctimes%2062.363%5Ctimes%20232%7D%7B24.5%7D%5C%5C%5CRightarrow%20V%3D1752.72%5C%20%5Ctext%7BL%7D%3D1.75272%5C%20%5Ctext%7Bm%7D%5E3)
For 20 cycles of the reaction the volume of the ozone is
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