How many moles of oxygen will occupy a volume of 2 5 l at STP?

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So, there will be 0.122 moles of oxygen gas.

How many moles are in 2l of oxygen gas at STP?

Assuming that the gas is at standard temperature and pressure (STP), one mole of any gas occupies 22.4 L . This means the number of moles of O2 is 222.4=0.089 mol .

How many moles of oxygen will occupy a volume of 2.5 liters at 1.2 atm and 25ºc?

1. How many moles of oxygen will occupy a volume of 2.5 liters at 1.2 atm and 25° C? PV = V (1.2 n. 0821) (298) = 123 mols 2.

How many moles of oxygen are in 5.5 L canister at STP?

There are 0.25mol of O2 in the 5.5-L canister.

What is the volume of one mole of oxygen at STP?

Finally, use the fact that one mole of oxygen would occupy a volume of 22.4 L at STP to figure out the volume of the oxygen in this question at STP.

What is the volume of 2 moles of gas at STP?

2: A mole of any gas occupies 22.4L at standard temperature and pressure (0oC and 1atm).

What is the volume in liters of 2.00 moles of hydrogen at STP?

According to this, one mole of hydrogen gas at STP will occupy 22.4 L of volume. Therefore, 2 moles of hydrogen gas will occupy a volume twice of 22.4 L i.e. 44.8 L. So, we can see from the above discussion that 2 moles of hydrogen gas occupies 44.8L volume.

How many moles of gas does it take to occupy 120 liters at a pressure of 2.3 atm and a temperature of 340 K?

n = PV (2.3atm/120 L) =9.89 mols P = 2.3 atm RT (0,0821 (afm) ( 340k) T = 340k n=?

0.7327 g/L.

What is the volume of 0.5 moles of gas at STP?

0.5 moles⋅22.4 L/mol=11.2 L , and so on.

What volume does 16.0 g of O2 occupy at STP?

1 mole of a gas occupy =22.4L volume at STP therefore, 16gO2 gas =0.5 mole of O2 gas will occupy =0.5×22.41=11.2L at STP.

What is the mass of 2 mole of NaOH?

Answer: The molar mass of the compound NaOH is 40 g/mol.

How do you calculate oxygen volume?

Calculate the volume (in cubic meters) of gaseous oxygen using the ideal gas law: multiply the amount of oxygen (in moles) by temperature and the molar gas constant followed by dividing the product by pressure.

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