Once you have identified the limiting reactant, you calculate how much of the other reactant it must have reacted with and subtract from the original amount.
My educated guess should be the 3rd one
The answer is (4) at the cathode, where reduction occurs. The Na+ gains one electron and become Na(l). So the reaction occurs at cathode and is reduction reaction.
Answer:
235/92U+10n→144/54Xe+90/38Sr+2/10n
Explanation:
- The nuclear reaction for the neutron-induced fission of u−235 to form xe−144 and sr−90 is represented by;
235/92U+10n→144/54Xe+90/38Sr+2/10n
- In nuclear fission reactions a heavy nuclide is split into two light nuclides and is coupled by the release of energy.
