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$\mathrm{Br}^{-}$is converted into $\mathrm{Br}_2$ by using
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The correct answer is:
$\mathrm{Cl}_2$
$\mathrm{Cl}_2+2 \mathrm{Br}^{-} \rightarrow 2 \mathrm{Cl}^{-}+\mathrm{Br}_2$
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