I'm not going to do all your work.
The formula is Pressure1 X Volume 1 X Temperaure 1(in Kelvins)= Pressure 2 X Volume 2 X temperaure 2
So 3.97L X 1479mm Hg X 231.15K = y volume X 760mm Hg (1 atmosphere) X 293.15 (20 degrees C)
I'll let you figure out the maths.
This depends on the temperature and the pressure. At standard temperature and pressure 1 mole will occupy 22.4 L, so multiply... 22.4 x 2.22 = 48.728 L at STP.
At standard temperature and pressure, 1 mole of any gas will occupy 22.4 liters. Set up a direct proportion of 22.4 liters/1 mole = 1 liter/x moles and solve for x. You get 0.045 moles.
At STP, 1 mole of gas occupies a volume of 22.4 liters. Thus, 4/5 moles of gas will occupy .8*22.4 liters.
To work this out there are two possible methods, the first uses knowledge about densities: At it boiling point, Liquid Nitrogen has a density of 807.0 grams per liter At Standard Temperature and Pressure, Nitrogen Gas has a density of 1.251 grams per liter Thus:- 25 liters of Liquid Nitrogen will weigh 25*807 = 20175 grams At STP 20175 grams of Nitrogen Gas will occupy 20175/1.251 = 16127.1 liters. The second method is to use the published expansion ratio for Liquid Nitrogen: Liquid Nitrogen has a liquid-to-gas expansion ratio of 1:694 at 20°C (68 °F) Thus:- 25*694=17350 liters. You will see the numbers are similar but slightly different, of the two I would be more confident in the density calculation method result. Although I suspect most people would use the expansion ratio method.
Boyle's Law: p*V = constant for 1. same amount of gas and 2. at same temperatureThis will do: V = (98.4*0.725)/142= 0.502 L
10 mg of 'standard' pure water, at standard temperature and pressure, occupy 0.01 mL of space.
This depends on the temperature and the pressure. At standard temperature and pressure 1 mole will occupy 22.4 L, so multiply... 22.4 x 2.22 = 48.728 L at STP.
Water's accepted density is 1.00 g/mL at standard temperature and pressure so depending on temperature the 1057 grams of water will occupy just about 1057 mL.
7.41 L
What you need to know to work this out is that:- Moles of gases at standard temperature pressure (With P and T constant) are proportional to the volume they occupy, divided by their specific gas constant.
According to the ideal gas law, all gases occupy about 22.4 liters per moleof space at standard temperature and pressure, so 22.4x2.56=57.34 liters.
... "are proportional to the volume they occupy, divided by their specific gas constant." With P and T constant... V1 / (n1 R1) = V2 / (n2 R2) = ...
Molar gas volume is the volume of ONE moel of gas. It only depends on the pressure and temperature, not on the kind of gas. Molar volume at standard temperature and standard pressure is always 22,4 Litres (for any gas)
The molar volume doesn't depend on the identity of the gas. One mole of any ideal gas at STP will occupy 22.4 liters.
At standard temperature and pressure, 1 mole of any gas will occupy 22.4 liters. Set up a direct proportion of 22.4 liters/1 mole = 1 liter/x moles and solve for x. You get 0.045 moles.
The volume of any gas is dependent on the pressure and temperature according to specific gas laws (Charles & Boyles) covered in any basic chemistry text. If my memory is correct the molecular weight of any gass will occupy 22.4 L of space at standard temperature and pressure.
The question, as stated, cannot be answered sensibly. Different gases have different densities and these are affected by their temperature and pressure. A ton of a light gas, such as hydrogen, will occupy roughly 11125 kilolitres at standard temperature and pressure. On the other hand a heavy elemental gas such as Radon will occupy roughly 103 kilolitres - only one hundredth as much.