$$PV = nRT$$
where:
* P is the pressure in atmospheres (atm)
* V is the volume in liters (L)
* n is the number of moles of gas
* R is the ideal gas constant (0.08206 L⋅atm/mol⋅K)
* T is the temperature in Kelvin (K)
Rearranging the ideal gas law to solve for n, we get:
$$n = \frac{PV}{RT}$$
Substituting the given values into the equation, we get:
$$n = \frac{(2.99 \ atm)(1.55 \ L)}{(0.08206 \ L⋅atm/mol⋅K)(607 \ K)} = 0.0798 \ mol$$
To convert moles to grams, we need to multiply the number of moles by the molar mass of hydrogen gas. The molar mass of hydrogen gas is 2.016 g/mol. Therefore, the mass of hydrogen gas that occupies 1.55 L at 607 K and 2.99 ATM is:
$$mass = n × molar \ mass = (0.0798 \ mol)(2.016 \ g/mol) = 0.161 \ g$$
Therefore, 0.161 grams of hydrogen gas would occupy 1.55 L at 607 K and 2.99 ATM.
