Answer:The molar heat of vaporization of molten lead is 178.03 kJ/mol.
Explanation:
The molar heats of fusion of lead
...(1)
The molar heats of sublimation of lead
...(2)
Adding (1) and (2)
![Pb(l)\rightarrow Pb(g),\Delta H_v=?](https://tex.z-dn.net/?f=Pb%28l%29%5Crightarrow%20Pb%28g%29%2C%5CDelta%20H_v%3D%3F)
![\Delta H_v=\Delta H_f+\Delta H_s=(-4.77) kJ/mol+182.8 kJ/mol=178.03 kJ/mol](https://tex.z-dn.net/?f=%5CDelta%20H_v%3D%5CDelta%20H_f%2B%5CDelta%20H_s%3D%28-4.77%29%20kJ%2Fmol%2B182.8%20kJ%2Fmol%3D178.03%20kJ%2Fmol)
The molar heat of vaporization of molten lead is 178.03 kJ/mol.
Temp must be Kelvin
38 C =
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311.15
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K
Volume at STP = 8.50 liters * (273.15 / 311.15) * (725 / 760) =
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7.1182746306
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Liters
The formula to use is:
Volume at STP = Present Volume * (273.15 / Present Temp °K) * (Present Pressure (Torr) / 760)
39 J
P=Fd/t i belive is the answer
Answer:
B. 3.0 g/ml
Explanation:
density formula: mass/volume
15/5=3