Depends on what you want to separate
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.
I don't know but look on the internet or use a calculator
Answer:
5.2g copper (Cu) => 0.082 moles copper (2 sig.figs.)
Explanation:
mole conversions:
grams to moles => divide by formula wt.
moles to grams => multiply by formula wt.
gas volumes to moles => divide volume by 22.4Liters/mole (STP conditions only)
This problem:
mass to moles => divide by formula wt.
mass = 5.2g = 5.2g/63.5g/mole = 0.082 moles copper (2 sig.figs.)
Answer:
2.57 e-9
Explanation:
The formula is H3O=10^-Ph
10^-8.59=2.57 e-9