The enthalpy of vaporization of Substance X is 23.0 Round your answer to 2 significant digits. 0 atm 0 x10 X kJ and its normal boiling point is 10. °C. Calculate the vapor pressure of X at -18. °C. mol
The enthalpy of vaporization of Substance X is 23.0 Round your answer to 2 significant digits. 0 atm 0 x10 X kJ and its normal boiling point is 10. °C. Calculate the vapor pressure of X at -18. °C. mol
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![The enthalpy of vaporization of Substance X is 23.0
Round your answer to 2 significant digits.
0
atm
x10
×
kJ
and its normal boiling point is 10. °C. Calculate the vapor pressure of X at - 18. °C.
mol
Ś](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5a3df371-d475-482a-84ce-057c931f1fc6%2F09aa4922-59e0-4ee0-a30d-f477704409d7%2F6aoa7a_processed.png&w=3840&q=75)
Transcribed Image Text:The enthalpy of vaporization of Substance X is 23.0
Round your answer to 2 significant digits.
0
atm
x10
×
kJ
and its normal boiling point is 10. °C. Calculate the vapor pressure of X at - 18. °C.
mol
Ś
Expert Solution
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Step 1: Given information:
The enthalpy of vaporization, of substance X is given as
.
The relationship between kJ and J is given by an expression as follows:
Converting in J/mol.
The normal boiling point of the substance X is given as .
The temperature at which the vapor pressure of a liquid becomes equal to the atmospheric pressure is called the normal boiling point of that liquid.
We have to determine the vapor pressure of the substance X at .
The relationship between the temperature in degrees Celsius and the temperature in Kelvin is given by an expression as follows:
Converting in Kelvin.
Converting in Kelvin.
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