In one of the neutron-induced fission reactions of U-235 (atomic mass = 235.043922 amu), the products are Ba-140 and Kr-93 (a radioactive gas). What volume of Kr-93 (at 25.0 °C and 1.0 atm) is produced when 1.00 g of U-235 undergoes this fission reaction?

Verified step by step guidance

1. Calculate the number of moles of U-235 using its atomic mass: \( \text{moles of U-235} = \frac{\text{mass of U-235}}{\text{atomic mass of U-235}} \).
2. Use the stoichiometry of the fission reaction to determine the moles of Kr-93 produced. Assume a 1:1 molar ratio between U-235 and Kr-93 for simplicity.
3. Apply the ideal gas law to find the volume of Kr-93 produced: \( PV = nRT \), 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.
4. Convert the temperature from Celsius to Kelvin: \( T(K) = T(°C) + 273.15 \).
5. Solve the ideal gas law equation for \( V \) to find the volume of Kr-93: \( V = \frac{nRT}{P} \).