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Question 3.89: This problem may be done as two dilution problems with the t......

This problem may be done as two dilution problems with the two final molarities added, or, as done here, it may be done by calculating, then adding the moles and dividing by the total volume.

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M\text{ KBr}=\frac{\text{Total Moles KBr}}{\text{Total Volume}} =\frac{\text{Moles KBr from Solution 1 + Moles KBr from Solution 2}}{\text{Volume Solution 1 + Volume Solution 2}} \\ M\text{ KBr}=\frac{\left\lgroup\frac{0.053\text{ mol KBr}}{1\text{ L}} \right\rgroup (0.200\text{ L})+\left\lgroup\frac{0.078\text{ mol KBr}}{1\text{ L}} \right\rgroup (0.550\text{ L})}{0.200\text{ L}+0.550\text{ L}} =0.071333=0.071 \ M\text{ KBr}

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