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How many moles of air are there in a 3 L container at 1 ATM pressure of 293 K?
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.
How many moles of air are there in a 3 l container at 1 ATM pressure and 2933 k?
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.
How many molecules are in 3 moles of carbon dioxide?
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!
How many moles are there in a sample of gas held in a 0.325 liter container at 0.914 ATM and 19 deg Celsius?
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
How many degrees Fahrenheit is 145 Celsius?
145 degrees Celsius = 293 degrees Fahrenheit
How Many Moles of air are there in a 3 L container of 1 ATM pressure and 293?
0.125 moles
How many moles are there in a 3 L container at 1 ATM pressure and 293 K?
The answer is 0,125 moles.
How many moles of air are there in a 3 L container a 1 ATM pressure and 293 K?
0.125 moles
How many moles of air are there in a 3 L container at 1 am pressure and 293 K?
Air is a mixture.
How many moles are there in a 3 L container at 1 ATM pressure and 293 k use PVnRT?
This is another calculation. there are 0.123 moles inn this volume.
How many moles of air there in a 3 L container at 1 ATM pressure and 293 K?
You can use the equationPV=nRT. So there are 0.12231 moles inthat volume.
How many moles of air are there in 3 L container at 1 ATM pressure and 293 K?
3 (L) / 22.4 (L/mol) = 0.13 mol of any gas at STPapex- 0.125 moles
How many moles of air are there in a 3L container at 1 ATM pressure and 293 K?
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.
How many moles of air are there in a 3 L container at 1 ATM pressure of 293 K?
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.
How many moles of ethane does it take to produce 293 moles of H2O?
The answer is 97,66 moles.
How many moles of air are there in a 3 l container at 1 ATM pressure and 2933 k?
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.
What is the pressure in the container of kPa when there is 1.09 mol of H2 is contained in a 2.00L container at 293K?
PV = nRT so --- P = nRT/V = 1.09(8.314)(293)/(2.00) = 1327.62109 kPa
