consider the nuclear equation below.\n$_{92}^{235}u\\longrightarrow x + _{2}^{4}he$\nwhat is the nuclide…

consider the nuclear equation below.\n$_{92}^{235}u\\longrightarrow x + _{2}^{4}he$\nwhat is the nuclide symbol of x?\n$_{94}^{231}pu$\n$_{90}^{235}th$\n$_{94}^{239}pu$\n$_{90}^{231}th$
Answer
Explanation:
Step1: Apply mass - number conservation
In a nuclear equation, the sum of mass - numbers on the left - hand side equals the sum of mass - numbers on the right - hand side. The mass - number of $^{235}{92}U$ is 235 and the mass - number of $^{4}{2}He$ is 4. Let the mass - number of $X$ be $A$. Then $235=A + 4$, so $A=235 - 4=231$.
Step2: Apply atomic - number conservation
The sum of atomic numbers on the left - hand side equals the sum of atomic numbers on the right - hand side. The atomic number of $^{235}{92}U$ is 92 and the atomic number of $^{4}{2}He$ is 2. Let the atomic number of $X$ be $Z$. Then $92=Z + 2$, so $Z=92 - 2 = 90$.
Step3: Identify the element
The element with atomic number $Z = 90$ is thorium, symbol Th. So the nuclide symbol of $X$ is $^{231}_{90}Th$.
Answer:
$^{231}_{90}Th$