Complete the following nuclear equations. (a) \(\frac{23}{11} \mathrm{Na}+? \longrightarrow_{11}^{24} \mathrm{Na}+_{1}^{1} \mathrm{H}\) (b) \(_{27}^{59} \mathrm{Co}+_{0}^{1} \mathrm{n} \longrightarrow_{25}^{56} \mathrm{Mn}+?\) (c) \(?+_{1}^{2} \mathrm{H} \longrightarrow_{94}^{240} \mathrm{Pu}+_{-1}^{0} \beta\) (d) \(^{246} \mathrm{Cm}+? \longrightarrow_{102}^{254} \mathrm{No}+5_{0}^{1} \mathrm{n}\) (e) \(^{238} \mathrm{U}+? \longrightarrow_{99}^{246} \mathrm{Es}+6 \frac{1}{0} \mathrm{n}\)

In any given nuclear reaction, the total atomic number (also known as the charge) and the total mass number (number of protons + neutrons) are both conserved. This means the sum of atomic numbers and mass numbers on one side of the equation should equal the sum on the other side.

(a) The unknown particle will have a mass number of \((24 - 23) = 1\) and an atomic number of \(11 - 11 = 0\), so it's a neutron, denoted as \(_{0}^{1}\mathrm{n}\). (b) The unknown particle will have a mass number of \((59 - 56) = 3\) and an atomic number of \(27 - 25 = 2\), so it's an alpha particle, denoted as \(_{2}^{4}\mathrm{He} \).(c) The unknown particle will have a mass number of \((240 - 2) = 238\) and an atomic number of \(94 - 1 = 93\), so it's a Neptunium atom, denoted as \(_{93}^{238}\mathrm{Np}\).(d) The unknown particle will have a mass number of \((246 - 254 + 5) = -3\) and an atomic number of \(96 - 102 = -6\), which isn't possible, indicating an error in the equation. (e) The unknown particle will have a mass number of \((238 - 246 + 6*1) = -2\) and an atomic number of \(92 - 99 = -7\), which isn't possible, indicating an error in the equation.

In this exercise, atomic and mass number conservation was used to identify unknown particles in nuclear equations. In equations (d) and (e), the conservation laws weren't followed, indicating errors in the equations.