B <span>Divide the chemical equation into two half-reaction equations, identifying which half-reaction is oxidation and which is reduction
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Answer:
X(Cl-35) = 75.95% => Answer 'A'
Explanation:
34.9689·X(Cl-35) + 36.9695·X(Cl-37) = 35.45; X = fractional abundance
X(Cl-35) + X(Cl-37) = 1 ⇒ X(Cl-37) = 1 - X(Cl-25)
34.9689·X(Cl-35) + 36.9695(1 - X(Cl-35)) = 35.45
34.9689·X(Cl-35) + 36.9695 - 36.9695·X(Cl-35) = 35.45
Rearrange ...
36.9695·X(Cl-35) - 34.9689·X(Cl-35) = 36.9689 - 35.45
2.0006·X(Cl-35) = 1.5195
X(Cl-35) = 1.5195/2.0006 = 0.7595 fractional abundance
⇒ % abundance = 75.95%
