This is another calculation. there are 0.123 moles inn this volume.
The answer is 0,125 moles.
0.125 moles
Air is a mixture.
You can use the equationPV=nRT. So there are 0.12231 moles inthat volume.
You solve this using the ideal gas law: PV = nRT and solve for n (moles)n = PV/RT = (2.62 atm)(47.8 L)/(0.0821 Latm/Kmol)(775K) n = 1.97 moles (3 significant figures)
0.125 moles
The answer is 0,125 moles.
0.125 moles
Air is a mixture.
You can use the equationPV=nRT. So there are 0.12231 moles inthat volume.
3 (L) / 22.4 (L/mol) = 0.13 mol of any gas at STPapex- 0.125 moles
You can use the ideal gas law, PV = nRT, to solve for the number of moles, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Rearrange the formula to solve for n, n = (PV) / (RT). Convert the temperature from Celsius to Kelvin (27°C + 273 = 300 K), and plug in the values to find the number of moles of the gas in the container.
To calculate the number of moles of nitrogen gas in the container, you can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Convert the pressure to atm, the volume to L, and the temperature to Kelvin. Then plug in the values and solve for n.
You solve this using the ideal gas law: PV = nRT and solve for n (moles)n = PV/RT = (2.62 atm)(47.8 L)/(0.0821 Latm/Kmol)(775K) n = 1.97 moles (3 significant figures)
3 (L) / 22.4 (L/mol) = 0.13 mol of any gas at STPapex- 0.125 moles
Using the ideal gas law (PV = nRT), we can calculate the number of moles of air in the container. Rearranging the formula to solve for n (moles), we get n = (PV) / (RT). Plugging in the values (P = 1 atm, V = 3 L, R = 0.0821 Latm/molK, T = 293 K) and solving for n gives us approximately 0.12 moles of air.
1 mole of gas occupies 22.4 liters at STP. 564/22.4 = 25.18 moles (2 decimal places)