<u>Answer:</u> The vapor pressure of mercury at 322°C is 0.521 atm
<u>Explanation:</u>
To calculate the final pressure, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm
= final pressure which is vapor pressure of mercury = ?
= Enthalpy of vaporization = 58.51 kJ/mol = 58510 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature which is normal boiling point = ![356.7^oC=[356.7+273]K=629.7K](https://tex.z-dn.net/?f=356.7%5EoC%3D%5B356.7%2B273%5DK%3D629.7K)
= final temperature = ![322^oC=[322+273]K=595K](https://tex.z-dn.net/?f=322%5EoC%3D%5B322%2B273%5DK%3D595K)
Putting values in above equation, we get:
![\ln(\frac{P_2}{1})=\frac{58510J/mol}{8.314J/mol.K}[\frac{1}{629.7}-\frac{1}{595}]\\\\\ln P_2=-0.6517atm\\\\P_2=e^{-0.6517}=0.521atm](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7B1%7D%29%3D%5Cfrac%7B58510J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B629.7%7D-%5Cfrac%7B1%7D%7B595%7D%5D%5C%5C%5C%5C%5Cln%20P_2%3D-0.6517atm%5C%5C%5C%5CP_2%3De%5E%7B-0.6517%7D%3D0.521atm)
Hence, the vapor pressure of mercury at 322°C is 0.521 atm
Answer:
B
Explanation:
its solute state is gas and dissolve in (solvent) liquid which is water
I believe this would be an exothermic reaction, As time increases, the temperature goes up meaning the heat from the system is going into the surroundings. Which indicates an exothermic reaction.
Answer:
The same 45.98 g of NaCl will be produced since 2 moles of Na will react with 2 moles of NaCl. Hence the stoichiometric coefficient will be in a ratio of 1:1
I would say that the answer has to be C
Since there is no change in mols on both sides of the equation the mass is constant