Math  /  Numbers & Operations

QuestionQuestion 34 of 39 Determine the concentrations of MgCl2,Mg2+\mathrm{MgCl}_{2}, \mathrm{Mg}^{2+}, and Cl\mathrm{Cl}^{-}in a solution prepared by dissolving 1.17×104 gMgCl21.17 \times 10^{-4} \mathrm{~g} \mathrm{MgCl}_{2} in 1.25 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm). [MgCl2]=\left[\mathrm{MgCl}_{2}\right]= \square [MgCl2]=4M[Mg2+]=M\begin{array}{l} {\left[\mathrm{MgCl}_{2}\right]=4 \mathrm{M}} \\ {\left[\mathrm{Mg}^{2}+\right]=\square \mathrm{M}} \end{array} \square M [Mg2+]=\left[\mathrm{Mg}^{2+}\right]= \square ppm [Cl]=\left[\mathrm{Cl}^{-}\right]= \square M [Cl]=\left[\mathrm{Cl}^{-}\right]= \square ppm

Studdy Solution
[MgCl2]=9.84×107 M \left[\mathrm{MgCl}_{2}\right] = 9.84 \times 10^{-7} \text{ M} [Mg2+]=9.84×107 M \left[\mathrm{Mg}^{2+}\right] = 9.84 \times 10^{-7} \text{ M} [Mg2+]=2.39×102 ppm \left[\mathrm{Mg}^{2+}\right] = 2.39 \times 10^{-2} \text{ ppm} [Cl]=1.97×106 M \left[\mathrm{Cl}^{-}\right] = 1.97 \times 10^{-6} \text{ M} [Cl]=6.97×102 ppm \left[\mathrm{Cl}^{-}\right] = 6.97 \times 10^{-2} \text{ ppm}

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