What is the molar concentration of the br⁻ ions in 0.33 m cabr2(aq)?

CaBr2(aq) is an ionic compound which will have the releasing of 2 Br⁻ ions ions in water for every molecule of CaBr2 that dissolves.
CaBr2(s) --> Ca+(aq) + 2 Br⁻(aq)
            [Br⁻] = 0.33 mol CaBr2/1L × 2 mol Br⁻ / 1 mol CaBr2 = 0.66 M
The answer to this question is [Br⁻] = 0.66 M

A) the enthralling of formation for pure element in its standard state is always positivehope this : )

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