Answer:
1.34 grams of neon are present in the 2.5 L flask at a temperature of 300 K and a pressure of 0.8 atm.
Step-by-step explanation:
To determine the number of grams of neon present in a 2.5 L flask at a temperature of 300 K and a pressure of 0.8 atm, we can use the ideal gas law, which states that the pressure, volume, and temperature of a gas are directly proportional to the number of moles of the gas present.
We can rearrange the ideal gas law to solve for the number of moles of a gas present:
PV = nRT
Where:
P = the pressure of the gas
V = the volume of the gas
n = the number of moles of the gas
R = the universal gas constant
T = the temperature of the gas
In this case, we know the pressure, volume, and temperature of the gas, so we can solve for the number of moles of neon present:
n = PV / RT
Substituting in the values from the problem, we get:
n = (0.8 atm * 2.5 L) / (8.31 L * atm / mol * K * 300 K) = 0.067 mol
Since neon has a molar mass of 20.18 g/mol, we can convert the number of moles of neon to grams by multiplying the number of moles by the molar mass:
0.067 mol * 20.18 g/mol = 1.34 g
Therefore, there are approximately 1.34 grams of neon present in the 2.5 L flask at a temperature of 300 K and a pressure of 0.8 atm.