93.0 ml of o2 gas is collected over water at 0.930 atm and 10.0 c. what would be the volume of this dry gas at stp

Let us assume that the oxygen behaves as an ideal gas such that we can use the ideal gas equation to solve for the number of moles of O2.
                       PV = nRT   ; n = PV/RT
Substituting the known values,
               n = (0.930 atm)(93/1000 L) / (0.0821 L.atm/mol.K)(10 + 273.15K)
                             n = 3.72 x 10^-3 mols
At STP, the volume of each mol of gas is equal to 22.4 L.
                          volume = (3.73 x 10^-3 mols) x (22.4 L/1 mol)
                           volume = 0.0833 L or 83.34 mL