Final answer:
The flashbulb contains 0.00300 grams of O2(g).
Step-by-step explanation:
To calculate the mass of O₂(g) present in the flashbulb, we need to use the ideal gas law equation. The equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin: T(K) = T(°C) + 273.15 = 18.0°C + 273.15 = 291.15 K. Next, we convert the volume from mL to L: V(L) = V(mL)/1000 = 1.70 mL/1000 = 0.00170 L. Rearranging the ideal gas law equation to solve for the number of moles (n): n = PV/RT.
Plug in the values: n = (2.30 atm)(0.00170 L)/(0.0821 L·atm/(mol·K))(291.15 K) = 0.0000937 mol. Finally, use the molar mass of oxygen, O₂, which is 32.00 g/mol, to calculate the mass of O₂(g) in grams: mass = n x molar mass = (0.0000937 mol)(32.00 g/mol) = 0.00300 g.