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If you have an unknown quantity of gas at a pressure of 1.2 ATM a volume of 31 liters and a temperature of 87 c how many moles of gas do i have?
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∙ 11y ago
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∙ 11y ago
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One mole of oxygen gas at stp occupies 22.4 what?
It means there are 22.4 liters of an "ideal" gas at STP (standard temperature and pressure), implying that temperature = 273.15 K and pressure = 1 atm.
Assuming no change in temperature and pressure calculate the volume of O2 in liters required for the complete combustion of 14.9 L of butane C4H10?
Assuming no change in temperature and pressure, calculate the volume of O2 (in liters) required for the complete combustion of 14.9 L of butane (C4H10):
Fifty liters of gas is kept at a temperature of 200 K and under pressure of 15 ATM The temperature of the gas is increased to 400 K The pressure is decreased to 7.5 ATM What is the resulting vol?
200
At a certain temperature and pressure 0.02 mol of carbon dioxide has a volume of 3.1 L A 3.1 L sample of hydrogen at the same temperature and pressure?
contains the same number of molecules
A balloon is filled with helium. Its volume is 4.2L at 312.K. What will its volume in liters be at 285.K assuming no change in pressure?
Assuming pressure is constant, like you said, volume and temperature have a direct relationship. As temperature increases, volume increases; as temperature decreases, volume decreases. Setting up a algebraic direct proportion, you get approximately 3.84 liters for the balloon at 285 degrees K.
One mole of oxygen gas at stp occupies 22.4 what?
It means there are 22.4 liters of an "ideal" gas at STP (standard temperature and pressure), implying that temperature = 273.15 K and pressure = 1 atm.
What volume is occupied of 2.4 moles of chlorine?
54 liters at STP (standard temperature and pressure)
What volume is occupied of 0.48 moles of hydrogen?
0.48 liters at STP (standard temperature and pressure)
How do you Convert 20 liters gas to kilograms?
you need to know the density of the gas (and temperature and pressure)
Assuming no change in temperature and pressure calculate the volume of O2 in liters required for the complete combustion of 14.9 L of butane C4H10?
Assuming no change in temperature and pressure, calculate the volume of O2 (in liters) required for the complete combustion of 14.9 L of butane (C4H10):
What is the volume occupied by 0.25 mol of chlorine gas?
The volume of one mole of gas at a standard temperature and pressure is 22.4 liters. Multiply 22.4 liters by 0.25 moles to get a volume of 5.6 liters.
What determines the volume of gas?
Pressure and temperature. As pressure increases, volume decreases; as temperature increases, volume increases with it. At standard temperature and pressure (1 atm, 273 degrees Kelvin), one mole of a gas (6.022 x 1023 particles) has the volume of 22.4 liters.
What determines a gas's of volume?
Pressure and temperature. As pressure increases, volume decreases; as temperature increases, volume increases with it. At standard temperature and pressure (1 atm, 273 degrees Kelvin), one mole of a gas (6.022 x 1023 particles) has the volume of 22.4 liters.
Fifty liters of gas is kept at a temperature of 200 K and under pressure of 15 ATM The temperature of the gas is increased to 400 K The pressure is decreased to 7.5 ATM What is the resulting vol?
200
When a mole of gas occupies 22.4 Liters this is known as?
This is the molar volume of an ideal gas at a given temperature and pressure.
The volume of the balloon is 3 liters at a temperature of 300 degrees kelvin What will the volume of the balloon be if the temperature rises to 350?
3.5 litre if pressure is kept constant.
A sample of a gas has a volume of 2.0 liters at a pressure of 1.0 atmosphere When the volume increases to 4.0 liters at constant temperature the pressure will be?
For this you would use Boyle's Law, P1V1 = P2V2. The first pressure and volume variables are before the change, while the second set are after the change. In this case, the volume is being changed and the pressure has to be solved for. P1 = 1.00 ATM V1 = 2.0 L P2 = Unknown V2 = 4.00 L P1V1 = P2V2 1.00(2.0)=4.00P P= .5 ATM
