Answer: C REDUCTION
Explanation:
Guessed after knowing oxidation isn't the answer. Got right
The molecular weight of the substance.
Always use least amount of sig figs possible. So this 9.7 would be (answer): 2 sig figs
The ionization energy for a hydrogen atom in the n = 2 state is 328 kJ·mol⁻¹.
The <em>first ionization energy</em> of hydrogen is 1312.0 kJ·mol⁻¹.
Thus, H atoms in the <em>n</em> = 1 state have an energy of -1312.0 kJ·mol⁻¹ and an energy of 0 when <em>n</em> = ∞.
According to Bohr, Eₙ = k/<em>n</em>².
If <em>n</em> = 1, E₁= k/1² = k = -1312.0 kJ·mol⁻¹.
If <em>n</em> = 2, E₂ = k/2² = k/4 = (-1312.0 kJ·mol⁻¹)/4 = -328 kJ·mol⁻¹
∴ The ionization energy from <em>n</em> = 2 is 328 kJ·mol⁻¹
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NF3– 0.94– third
NCl3–0.12– second
NBr3–0.08– first
CF4–1.43– fourth
NBr3—NCl3—NF3—CF4
Lowest. Highest