"If the first ionization energy of Li is 520 kJ/mol, the approximate IE₂ would be $x \times 10^3$ kJ/mol. Find $x$ (given IE₂ ≈ 7300 kJ/mol)."

Question

Accepted Answer

"IE₂ of Li ≈ 7300 kJ/mol $= 73 \times 10^2$ kJ/mol. The question asks for $x$ in $x \times 10^3$: actually $7300 = 7.3 \times 10^3$, so $x = 7.3$."

If the first ionization energy of Li is 520 kJ/mol, the approximate IE₂ would be $x \times 10^3$ kJ/mol. Find $x$ (given IE₂ ≈ 7300 kJ/mol).