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:
second carbon atom from the end
end carbon atom
Explanation:
Carbohydrates are naturally occurring organic compounds containing carbon, hydrogen and oxygen. The general molecular formula of Carbohydrates is .
Carbohydrates can be classified based on structures,
Carbohydrates with the structure of alkanals (-CHO) are known as aldose while those of the structure of alkanones (C=O) are known as ketose.
In stereochemistry , D series is a kind of configurational arrangement where the hydroxyl group attaches itself to the right hand side.
Thus; in naturally occurring D series of ketoses, the carbonyl group is found on carbon number <u>second carbon atom from the end </u>whereas in aldoses, the carbonyl group is found on carbon number <u> end carbon atom.</u>