<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
Permafrost soil (gelisol) is b) fragile.
Hello Mlbozer,
Hope your having a great day! Gasses are known for having molecules that 'run' in a way. They all are Spread Out, and bounce off each other. Its a big mess, and is more 'spread out' than even liquid, in fact its the most open form of the three! So your answer is false!
Thank you,
Darian D.
Answer:
148.2 g of H20
Explanation:
Equation of reaction: 4NH3 + 5O2 ---> 4NO + 6H20
From the equation above, 4 moles of ammonia reacts with 5 moles of oxygen gas to produce 6 moles water.
Molar mass of NH3 = 17 g/mol;
Molar mass of O2= 32 g/mol;
Molar mass of H2O = 18 g/mol
First, we determine the limiting reactant:
4*17 g of NH3 reacts with 5*32 g of O2
Mass Ratio = 68 : 160
Therefore, NH3 is the limiting reactant.
68 g of NH3 reacts to produce 6* 18 g of H20 = 108 g of H2O
93.3 g of NH3 will react to produce (93.3 * 108)/68 g of H20 = 148.2 g of H2O
Therefore, the maximum amount of H2O produced = 148.2 g