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Answer: Heat of vaporization is 41094 Joules
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
where,
= initial pressure at 429 K = 760 torr
= final pressure at 415 K = 515 torr
= enthalpy of vaporisation = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 429 K
= final temperature = 515 K
Now put all the given values in this formula, we get
![\log (\frac{515}{760}=\frac{\Delta H}{2.303\times 8.314J/mole.K}[\frac{1}{429K}-\frac{1}{415K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B515%7D%7B760%7D%3D%5Cfrac%7B%5CDelta%20H%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B429K%7D-%5Cfrac%7B1%7D%7B415K%7D%5D)
Thus the heat of vaporization is 41094 Joules
The variables in the ideal gas constant has V as the unit of liters and T has the unit of Kelvin. Thus, option C is correct.
The gas constant in an ideal gas equation has been the value of the energy absorbed by 1 mole of an ideal gas at standard temperature and pressure.
The value of R has been dependent on the units of volume, temperature and pressure of the ideal gas.
The given value of R has been 0.0821 L.atm/mol.K
The unit in gas constant has been L (Liter) for volume (V).
The unit of pressure (P) has been atm.
The unit of temperature (T) has been Kelvin (K).
Thus the gas law constant used by student has V has the unit of liters and T has the unit of Kelvin. Thus, option C is correct.
For more information about the gas constant, refer to the link:
brainly.com/question/24814070
Answer:
56.28 g
Explanation:
First change the grams of oxygen to moles.
(50.00 g)/(32.00 g/mol) = 1.5625 mol O₂
You have to use stoichiometry for the next part. Looking at the equation, you can see that for every 2 moles of H₂O, 1 mole of O₂ is produced. Convert from moles of O₂ to moles of H₂O using this relation.
(1.5625 mol O₂) × (2 mol H₂O/1 mol O₂) = 3.125 mol H₂O
Now convert moles of H₂O to grams.
(3.125 mol) × (18.01 g/mol) = 56.28125 g
Convert to significant figures.
56.28125 ≈ 56.28
Because it tilts on its axis
