The condition for this process to be possible is that the atomic number $Z$ and mass number $A$ need to be conserved. Before the reaction, the atomic number $Z_b$ is

$$
begin{align*}
Z_b=92
end{align*}
$$

After the reaction, the atomic number $Z_a$ is

$$
begin{align*}
Z_a=54+Z(x)
end{align*}
$$

where we denoted the atomic number of the unknown element by $Z(x)$. By equating the atomic numbers before and after the reaction, we get

$$
begin{align*}
92=54+Z(x)
end{align*}
$$

or

$$
begin{align*}
boxed{Z(x)}=92-54=boxed{38}
end{align*}
$$

The mass number before the reaction $A_b$ is

$$
begin{align*}
A_b=236
end{align*}
$$

The mass number after the reaction $A_a$ is

$$
begin{align*}
A_a=140+2+A(x)
end{align*}
$$

where we denoted the mass number of the unknown element by $A(x)$. By equating the mass number before and after the reaction, we get

$$
begin{align*}
236=142+A(x)
end{align*}
$$

or

$$
begin{align*}
boxed{A(x)}=236-142=boxed{94}
end{align*}
$$

Thus, we conclude that the unknown element is strontium Sr.

a) In Problem 25, we were able to observe that the system experienced decay, therefore the mass of the system $boxed{text{decreased}}$

b) Given that the reaction released E = $184.7$ MeV, we can compute the change in mass using the formula:
$E_u = 931.5$ MeV

$$
begin{align*}
E &= (Delta m) times E_u \
Delta m &= dfrac{E}{E_u} \
&= dfrac{184.7 MeV}{931.5 MeV} \
&= 0.19828 u
end{align*}
$$

Hence, the system experienced a decrease in mass by $boxed{0.19828 u}$

a) mass of the system decreased

b) $Delta m = 0.19828$ u

The condition for this process to be possible is that the atomic number $Z$ and mass number $A$ need to be conserved. Before the reaction, the atomic number $Z_b$ is

$$
begin{align*}
Z_b=92
end{align*}
$$

After the reaction, the atomic number $Z_a$ is

$$
begin{align*}
Z_a=57+Z(x)
end{align*}
$$

where we denoted the atomic number of the unknown element by $Z(x)$. By equating the atomic numbers before and after the reaction, we get

$$
begin{align*}
92=57+Z(x)
end{align*}
$$

or

$$
begin{align*}
boxed{Z(x)}=92-57=boxed{35}
end{align*}
$$

The mass number before the reaction $A_b$ is

$$
begin{align*}
A_b=236
end{align*}
$$

The mass number after the reaction $A_a$ is

$$
begin{align*}
A_a=148+3+A(x)
end{align*}
$$

where we denoted the mass number of the unknown element by $A(x)$. By equating the mass number before and after the reaction, we get

$$
begin{align*}
236=151+A(x)
end{align*}
$$

or

$$
begin{align*}
boxed{A(x)}=236-151=boxed{85}
end{align*}
$$

Thus, we conclude that the unknown element is bromine Br.