This is a portion of a solved solution.

$\overline{){{\mathbf{M}}}_{{\mathbf{1}}}{{\mathbf{V}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{M}}}_{{\mathbf{2}}}{{\mathbf{V}}}_{{\mathbf{2}}}}$

$\frac{{\mathbf{M}}_{{\mathbf{1}}}{\mathbf{V}}_{{\mathbf{1}}}}{{\mathbf{M}}_{\mathbf{2}}}{\mathbf{=}}\frac{\overline{){\mathbf{M}}_{{\mathbf{2}}}}{\mathbf{V}}_{{\mathbf{2}}}}{\overline{){\mathbf{M}}_{\mathbf{2}}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\overline{){{\mathbf{V}}}_{{\mathbf{2}}}{\mathbf{=}}\frac{{\mathbf{M}}_{\mathbf{1}}{\mathbf{V}}_{\mathbf{1}}}{{\mathbf{M}}_{\mathbf{2}}}}$

In 2014, a major chemical leak at a facility in West Virginia released 7500 gallons of MCHM (4-methylcyclohexylmethanol, C_{8}H_{16}O) into the Elk River. The density of MCHM is 0.9074 g/mL.

How much farther down the river would the spill have to spread in order to achieve a "safe" MCHM concentration of 1.00 10^{-4} M? Assume the depth and width of the river are constant and the concentration of MCHM is uniform along the length of the spill.

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