what is the volume of 62.3 g of nitrogen gas at STP

What is the volume of 62.3 g of nitrogen gas at STP

What volume does 7.50 ´ 1020

molecules of O2 occupy at STP?

To find the volume of a gas at STP (Standard Temperature and Pressure), you can use the ideal gas law equation:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature

At STP, the pressure is 1 atm (atmosphere) and the temperature is 273.15 K (0 °C).

First, we need to find the number of moles of nitrogen gas. To do this, we'll use the molar mass of nitrogen, which is approximately 28.0134 g/mol.

Number of moles (n) = mass (m) / molar mass (M)
n = 62.3 g / 28.0134 g/mol
n ≈ 2.22 mol

Now we have the number of moles, so we can find the volume.

V = (nRT) / P
V = (2.22 mol * 0.0821 L·atm/mol·K * 273.15 K) / 1 atm

Calculating this equation, we find:

V ≈ 50.05 L

Therefore, the volume of 62.3 g of nitrogen gas at STP is approximately 50.05 liters.

1 mole of any gas at STP occupies 22.4 L. Nitrogen gas is N2 and 1 mole is 28 grams (look it up if you want a more accurate answer than 28).