78.9g
That depends a lot on the pressure - at higher altitudes (less pressure), the boiling point is lower. At standard pressure (1 atm.), the answer is 100 degree Celsius.
Yes, the vapour pressure of water at 10°C is 1.2 kPa and at 50°C is 12.3 kPa.
Well, I am not quite sure, but I am sure you will be able to calculate it from this:TemperatureDensityVapor PressureoCg/mLtorr150.999102612.79160.998946013.64170.998777914.54180.998598615.49190.998408216.49200.998207117.55210.997995518.66220.997773519.84230.997541521.09240.997299522.40250.997047923.78260.996786725.24270.996516226.77280.996236528.38290.995947830.08300.995650231.86
Yes. As long as the pressure is below atmospheric pressure.
60Kpa
The flashpoint of propylene is approximately -108 degrees Celsius (-162 degrees Fahrenheit). This is the minimum temperature at which propylene vapors can ignite when in contact with an open flame or spark. It is important to handle and store propylene with caution to prevent any fire hazards.
If this solution is a mixture you would use Henry's or Raoult's Law. If this is pure water then the answer is already in the question.
The answer is about 30.9 kJ/mol
98.6 degrees Fahrenheit = 37 degrees Celsius.
A fixed quantity of gas at a constant pressure exhibits a temperature of 27 degrees Celsius and occupies a volume of 10.0 L. Use Charles's law to calculate: the temperature of the gas in degrees Celsius in atmospheres if the volume is increased to 16.0 L
K = (C + 273.15)
Rigid container holds hydrogen gas at a pressure of 3.0 atmospheres and a temperature of 2 degrees Celsius. The pressure if the temperature is raised to 10 degrees Celsius will be 15 atmospheres based on the law of pressure for gas.
3 K is equivalent to -270,15 Celsius degrees.
The freezing point of water is zero degrees Celsius at standard pressure.
-15 C
That depends a lot on the pressure - at higher altitudes (less pressure), the boiling point is lower. At standard pressure (1 atm.), the answer is 100 degree Celsius.
Using the ideal gas law (using torr instead of atm), we calculate that there would be .046 moles of CO, or 2.7x1022 molecules of carbon monoxide.