The normal boiling point of ethanol is 78.4 °C and 101.3 KPa. The heat of vaporization for ethanol is 42.32 kJ/mol. Determine th
e vapor pressure of ethanol at 95.0 °C. Use the Clasius Clapeyron Equation: Ln [P2/P1] = - [∆Hvap /R] * [1/T2 - 1/T1]
2 answers:
Answer:
6.1×10^4Pa or 61KPa
Explanation:
The Clausius-Clapeyron equation is used to estimate the vapour pressure at different temperature, once the enthalpy of vaporization and the vapor pressure at another temperature is given in the question. The detailed solution is shown in the image attached. The temperatures were converted to kelvin and the energy value was converted from kilojoule to joule since the value of the gas constant was given in unit of joule per mole per kelvin. The fact that lnx=2.303logx was also applied in the solution.
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Answer: 4.5 moles of can be made from complete reaction of 3.0 moles of Al.
Explanation:
The given reaction equation is as follows.
This shows that 2 moles of Al reacts with 6 moles of HCl. So, the amount of HCl required to react with 1 mole Al is three times the amount of HCl.
Therefore, 3 moles of Al will react with 9 moles of HCl to give 3 moles of and
moles of
.
The reaction equation now will be as follows.
The moles can also be written as 4.5 moles.
Thus, we can conclude that 4.5 moles of can be made from complete reaction of 3.0 moles of Al.