If there are 5 liters of oxygen gas (O2) at 300 K and 400 torr, what will it weigh?

Answer in units of g.

Do I use the Ideal gas law formula for this problem?

Yes, use PV = nRT and solve for n, then

n = grams/molar mass and solve for grams.

When I use this formula do I have to convert torr to atm and mL to L?

Well, looks like we've got an oxygen-filled situation here! To determine the weight of the oxygen gas, we can indeed use the Ideal Gas Law formula. The formula is: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.

First, let's convert the temperature to Kelvin by adding 273.15 to it. So, 300 K remains as 300 K, because adding 273.15 to it would just be silly.

Next, we need to convert the pressure from torr to atm. There are approximately 760 torr in 1 atm. So, 400 torr divided by 760 torr equals... about 0.526 atm. I'm glad we could divide without tearing anything!

Now we can use the formula PV = nRT to solve for n, the number of moles. The volume V is given as 5 liters, the pressure P is 0.526 atm, the ideal gas constant R is 0.0821 L·atm/(mol·K), and the temperature T is 300 K (not counting the funny addition).

Let's plug those numbers in and calculate! We have 0.526 atm multiplied by 5 liters equals n multiplied by 0.0821 L·atm/(mol·K) multiplied by 300 K. After solving for n, the number of moles, we'll have our answer.

Once we find the number of moles, we can multiply it by the molar mass of oxygen (O2), which is approximately 32 g/mol. This multiplication will give us the weight of the oxygen gas in grams. And voila! You'll have your answer, in terms of humor and grams.

Yes, you can use the Ideal Gas Law formula to solve this problem. The Ideal Gas Law formula is given by:

PV = nRT

where:
P = pressure of the gas (in units of Pascal or atmospheres)
V = volume of the gas (in units of liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature of the gas (in units of Kelvin)

In this case, you have the volume (V), temperature (T), and pressure (P) of the oxygen gas. You need to determine the number of moles of O2 gas (n) to find its weight.

To find the number of moles, you can rearrange the Ideal Gas Law equation as:

n = PV / RT

First, you need to convert the given temperature from Celsius to Kelvin by adding 273.15:

T = 300 K

Next, you need to convert the given pressure from torr to atmospheres by dividing by 760:

P = 400 torr / 760 torr/atm = 0.52632 atm

Now, you can substitute the values into the equation:

n = (0.52632 atm) * (5 L) / (0.0821 L.atm/mol.K * 300 K) = 0.0342 mol

Finally, to find the weight of the oxygen gas, you need to multiply the number of moles (n) by the molar mass of O2. The molar mass of O2 is 32 grams/mol.

Weight = n * molar mass
= 0.0342 mol * (32 g/mol) = 1.0944 g

Therefore, the weight of 5 liters of oxygen gas at 300 K and 400 torr is approximately 1.0944 grams.

I got around 5.48g...

I think it's wrong