I am sttruggling with this question please help If a person exhales 33.0 g of CO2 in an hour, what volume (in L) does this amount occupy at 1.06 atm and 36.9 °C? Be sure your answer has the correct
number of significant figures.
Note: Reference the Fundamental constants table for additional information.
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm (1.06 atm in this case)
V = volume in liters (what we are trying to find)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (36.9 °C = 310.1 K)
First, we need to convert the mass of CO2 to moles. The molar mass of CO2 is approximately 44.01 g/mol. So,
33.0 g CO2 * (1 mol CO2 / 44.01 g CO2) = 0.75 mol CO2
Now, we can rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
V = (0.75 mol * 0.0821 L·atm/mol·K * 310.1 K) / 1.06 atm
V ≈ 18.06 L
Therefore, 33.0 grams of CO2 will occupy approximately 18.06 liters at 1.06 atm and 36.9 °C.