0,125 moles
3 (L) / 22.4 (L/mol) = 0.13 mol of any gas at STPapex- 0.125 moles
To determine the number of moles of air in the container, we need to 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 ideal gas constant, and T is the temperature in Kelvin. Rearranging the equation to solve for n, we get n = PV / RT. Plugging in the values (P = 1 atm, V = 3 L, T = 293 K, R = 0.0821 L.atm/mol.K), we find n = (1 atm * 3 L) / (0.0821 L.atm/mol.K * 293 K) = 0.123 moles.
1 mole is 6.022 141 293 x 1023 molecules. (Avogadro's constant). If you have 3 moles of CO2, then you have 3 x 6.022 141 293 x 1023 molecules = 1.806642388 x 1024 molecules!
1 mole (mol) of gas occupies a volume of 22.4 L at a temperature of 20 deg C and pressure of 1 ATMIn 0.325 L there will be only (0.325 / 22.4) molAt 0.914 ATM there will be only 0.914 of the number of moles that would be present at 1 ATMAt 19 deg C (292 deg Absolute or Kelvin) there will be (293 / 292) the number of moles that would be present at the standard temperature of 20 deg CTherefore the number of moles= (0.325 / 22.4) x 0.914 x (293 / 292)= 0.0133 mol
145 degrees Celsius = 293 degrees Fahrenheit
The answer is 0,125 moles.
0.125 moles
0.125 moles
Air is a mixture.
This is another calculation. there are 0.123 moles inn this volume.
You can use the equationPV=nRT. So there are 0.12231 moles inthat volume.
To calculate the number of moles of air in a 3L container at 1 ATM pressure and 293 K, we can use the ideal gas law, PV = nRT. Rearranging for n gives us n = PV / RT. Using R = 0.0821 L·ATM/(K·mol), we find: n = (1 ATM) * (3 L) / (0.0821 L·ATM/(K·mol) * 293 K) ≈ 0.123 moles. So, there are approximately 0.123 moles of air in the container.
3 (L) / 22.4 (L/mol) = 0.13 mol of any gas at STPapex- 0.125 moles
To find the number of moles of air in a 3L container at 1 ATM pressure and 293 K, we can use the ideal gas law, PV = nRT. Here, P = 1 ATM, V = 3 L, R = 0.0821 L·ATM/(K·mol), and T = 293 K. Rearranging the formula to solve for n gives us n = PV / RT. Substituting the values, we get n = (1 ATM)(3 L) / (0.0821 L·ATM/(K·mol) * 293 K) ≈ 0.124 moles of air.
3 (L) / 22.4 (L/mol) = 0.13 mol of any gas at STPapex- 0.125 moles
The answer is 97,66 moles.
To determine the number of moles of air in the container, we need to 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 ideal gas constant, and T is the temperature in Kelvin. Rearranging the equation to solve for n, we get n = PV / RT. Plugging in the values (P = 1 atm, V = 3 L, T = 293 K, R = 0.0821 L.atm/mol.K), we find n = (1 atm * 3 L) / (0.0821 L.atm/mol.K * 293 K) = 0.123 moles.