what volume would 10.5g of nitrogen gas, N2, occupy at 200.K and 2.02atm?

PV = nRT and solve for V.

n = grams/molar mass
Remember T must be in kelvin

To find the volume occupied by 10.5g of nitrogen gas (N2) at 200 K and 2.02 atm, we need to use the ideal gas law equation:

PV = nRT

where:
- P is the pressure (in atm)
- V is the volume (in liters)
- n is the number of moles
- R is the ideal gas constant (0.0821 L•atm/mol•K)
- T is the temperature (in Kelvin)

First, we need to calculate the number of moles of N2 using the mass of N2 and its molar mass. The molar mass of nitrogen gas (N2) is 28.02 g/mol [(14.01 g/mol for each nitrogen atom) x 2].

Number of moles (n) = mass (m) / molar mass (M)

n = 10.5 g / 28.02 g/mol
n ≈ 0.375 mol

Now we can substitute the given values into the ideal gas law equation:

PV = nRT

V = (nRT) / P

V = (0.375 mol) * (0.0821 L·atm/mol·K) * (200 K) / 2.02 atm

V ≈ 7.30 L

Therefore, 10.5g of nitrogen gas (N2) would occupy approximately 7.30 liters at 200 K and 2.02 atm.