To find the volume occupied by 11.0 grams of a gas at STP (standard temperature and pressure), we can use the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
At STP, the pressure is 1 atm, the temperature is 273 K, and the ideal gas constant is 8.31 J/mol*K. We can therefore use the following equation to find the volume occupied by 11.0 grams of the gas:
V = (nRT)/P
To find the number of moles of gas, we can use the following formula:
n = m/M
Where m is the mass of the gas in grams and M is the molecular mass of the gas in grams/mol.
Substituting the given values into this formula, we find that the number of moles of gas is:
n = 11.0 g / 44.0 g/mol
= 0.25 mol
We can now substitute this value into the ideal gas law equation to find the volume occupied by the gas at STP:
V = (0.25 mol)(8.31 J/mol*K)(273 K)/(1 atm)
= 6.53 liters
Rounding this value to the nearest tenth of a liter, we find that the volume occupied by 11.0 grams of the gas at STP is approximately 6.5 liters