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A solution is made by mixing 200.0 ml of 0.300 M FeI3 (436.56 g / mol)...

A solution is made by mixing 200.0 ml of 0.300 M FeI3 (436.56 g / mol) with 130.0 ml of 0.800 M LiBr (86.8450 g/mol) with 150.0 ml of a 0.250M MgI2 (293.89 g/ mol) . What is the molar concentration of iodide ion in the new solution?

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Expert Solution

The required molar concentration of iodide ions in the new solution is 0.53125 M or 0.53125 mol/liter

queen_honey_blossom answered 2 years ago
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