What volume will a mixture of 5.00mol of He and 0.800 mol of CO2 occupy at STP?
Total mols = ntotal = n for CO2 + n for He.
Then use PV = nRT and remember T must be in kelvin.
To find the volume of the mixture of gases at STP (Standard Temperature and Pressure), we can use the ideal gas law equation: PV = nRT.
First, let's determine the total number of moles of gas in the mixture. We have 5.00 mol of helium (He) and 0.800 mol of carbon dioxide (CO2). Therefore, the total number of moles (n) is:
n = 5.00 mol (He) + 0.800 mol (CO2) = 5.80 mol
At STP, the temperature (T) is 273 K and the pressure (P) is 1 atm. The ideal gas constant (R) is 0.0821 L·atm/(mol·K).
Now, we can rearrange the ideal gas law equation to solve for the volume (V):
V = (nRT) / P
Substituting the given values, we get:
V = (5.80 mol)(0.0821 L·atm/(mol·K))(273 K) / 1 atm
By calculating this expression, we find:
V ≈ 130.418 L
Therefore, the volume of the mixture of 5.00 mol of He and 0.800 mol of CO2 would be approximately 130.418 liters at STP.