a volume of 122mL of argon, Ar, is collected at 500 degrees C and 758 torr. what does this sample weigh?

I answered this below something like half an hour ago.

To determine the weight of a gaseous sample, we can use the ideal gas law equation, which states:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)

First, let's convert the given values:

The volume of the sample is given as 122 mL. We need to convert it to liters by dividing it by 1000:
V = 122 mL / 1000 mL/L = 0.122 L

The temperature is given as 500 degrees C. We need to convert it to Kelvin by adding 273.15:
T = 500 + 273.15 = 773.15 K

The pressure is given as 758 torr. We need to convert it to atm by dividing it by 760 (since 1 atm = 760 torr):
P = 758 torr / 760 torr/atm = 0.9974 atm

Now, we can rearrange the ideal gas law equation to solve for n (moles of gas):

n = (PV) / (RT)

Substituting the given values:

n = (0.9974 atm * 0.122 L) / [(0.0821 L·atm/(mol·K)) * 773.15 K]

n = 0.122 L * 0.9974 atm / (0.0821 L·atm/(mol·K) * 773.15 K)

n = 0.0121 mol

Finally, to find the weight of the sample (molar mass of argon is 39.95 g/mol), we use the formula:

Weight = n * molar mass

Weight = 0.0121 mol * 39.95 g/mol

Weight = 0.482 g

Therefore, the sample weighs 0.482 grams.