how many grams of hydrogen occupy 40 ml at 300.15 K and 710 torr?

To determine the number of grams of hydrogen that occupy 40 mL at 300.15 K and 710 torr, we can use the ideal gas law equation:

PV = nRT

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

First, we need to convert the given values to the appropriate units:

40 mL = 0.04 L (since 1 L = 1000 mL)
300.15 K (no conversion needed)
710 torr = 0.934 atm (since 1 atm = 760 torr)

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

n = (PV) / (RT)

Substituting the given values:

n = (0.934 atm * 0.04 L) / (0.0821 L·atm/mol·K * 300.15 K)

Simplifying:

n = 0.03736 / 24.64857
n = 0.001514 moles

Finally, to find the grams of hydrogen, we need to multiply the number of moles by the molar mass of hydrogen. The molar mass of hydrogen is approximately 2 g/mol.

grams = 0.001514 moles * 2 g/mol
grams = 0.00303 grams

Therefore, approximately 0.00303 grams of hydrogen would occupy 40 mL at 300.15 K and 710 torr.