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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 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 a 3L container at 1 ATM pressure and 293 k Use PVnrt?
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.
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 of 1 ATM pressure and 293?
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 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 a 3L container at 1 ATM pressure and 293 k Use PVnrt?
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.
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 L container at ATM pressure and 293 k?
To calculate the number of moles of air in a 1-liter container at 1 atm pressure and 293 K, we can use the ideal gas law, ( PV = nRT ). Here, ( P = 1 ) atm, ( V = 1 ) L, ( R = 0.0821 ) L·atm/(K·mol), and ( T = 293 ) K. Rearranging the formula gives ( n = \frac{PV}{RT} ), which results in ( n = \frac{(1 \text{ atm})(1 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)})(293 \text{ K})} \approx 0.041 moles ). Thus, there are approximately 0.041 moles of air in the container.
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 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 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 293 K Use PV nRT?
To find the number of moles of air in a 3 L container at 1 ATM pressure and 293 K, we can use the ideal gas law equation (PV = nRT). Here, (P = 1 , \text{ATM}), (V = 3 , \text{L}), (R = 0.0821 , \text{L} \cdot \text{ATM} / (\text{mol} \cdot \text{K})), and (T = 293 , \text{K}). Rearranging the equation to solve for (n), we get (n = \frac{PV}{RT} = \frac{(1)(3)}{(0.0821)(293)} \approx 0.123 , \text{moles}).
