V1/T1 = V2/T2 T1 = 278K, T2 = 333K
75 ml/278K = V2/333K
V2 = (75/278K)*(333K)
V2 = 89.8 ml
By decreasing the pressure with the volume kept constant.
50.0 grams of what gas? This is the ideal gas law. Pressure * Volume = moles gas * the R constant * temperature in Kelvin PV = nRT
decreases
747 mmHg
14kpa
468ml
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
By decreasing the pressure with the volume kept constant.
By decreasing the pressure with the volume kept constant.
The amount of any given gas that will dissolve in a liquid at a given temperature is directly proportional to the partial pressure of that gas.
98.6 degrees Fahrenheit = 37 degrees Celsius.
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.
Assuming constant pressure, the answer is:density = 0.789 x 20/15 = 1.052 kg/litre
78.9g
50.0 grams of what gas? This is the ideal gas law. Pressure * Volume = moles gas * the R constant * temperature in Kelvin PV = nRT
The answer is about 30.9 kJ/mol
Henry's law constant for Carbon Dioxide at 20 degrees Celsius is: 1,6*10^3 ATM