Explanation:
The given data is as follows.
= 100 mm Hg or 
= 0.13157 atm
= 
= (1080 + 273) K = 1357 K
= 
= (1220 + 273) K = 1493 K
= 600 mm Hg or 
= 0.7895 atm
R = 8.314 J/K mol
According to Clasius-Clapeyron equation,
![log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]](https://tex.z-dn.net/?f=log%28%5Cfrac%7B0.7895%7D%7B0.13157%7D%29%20%3D%20%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B2.303%20%5Ctimes%208.314%20J%2Fmol%20K%7D%5B%5Cfrac%7B1%7D%7B1357%20K%7D%20-%20%5Cfrac%7B1%7D%7B1493%20K%7D%5D)
![log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]](https://tex.z-dn.net/?f=log%20%286%29%20%3D%20%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B19.147%7D%5B%5Cfrac%7B%281493%20-%201357%29%20K%7D%7B1493%20K%20%5Ctimes%201357%20K%7D%5D)
0.77815 = 
= 
J/mol
= 
= 221.9 kJ/mol
Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.
Answer:
The air molecules that are surrounding the metal will speed up, and the molecules in the metal will slow down.
Explanation:
There will be a transfer of thermal energy (heat) from the hot metal plate to the surrounding air. This transfer of energy equates to a transfer of kinetic energy in the molecules. As the plate loses heat, the molecules in the plate will lose kinetic energy and slow down. As the surrounding air gains heat, the molecules will gain kinetic energy and speed up.
Air & noise pollution
Quarry waste
Damage to biodiversity
1 Aluminium is oxidised Al - 3e = Al⁺³
2 Chlorine is reduced Cl⁺⁷ + 8e = Cl⁻¹
3 Nitrogen is oxidised 2N⁻³ - 6e = N₂
Answer:
the difference is tyat eruptions of less gassy and more gassy is that the less gassy doesnt retain as much gas as the more gassy one and thus the eruption of the less gassy is less damage to the more gassy
