We have to calculate the percentage of mercury removed from water after treatment.

The **percentage of mercury **that was **removed** from water can be calculated by this formula:

$\overline{){\mathbf{\%}}{\mathbf{}}{\mathbf{mercury}}{\mathbf{}}{\mathbf{removed}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{mercury}\mathbf{}\mathbf{removed}}{\mathbf{initial}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{mercury}}{\mathbf{\times}}{\mathbf{100}}{\mathbf{}}{\mathbf{\%}}}$

In 2009, a team from Northwestern University and Western Washington University reported the preparation of a new "spongy" material composed of nickel, molybdenum, and sulfur that excels at removing mercury from water. The density of this new material is 0.20 g/cm^{3}, and its surface area is 1242 m^{2} per gram of material.

A 10.0-mL sample of contaminated water had 7.748 mg of mercury in it. After treatment with 10.0 mg of the new spongy material, 0.001 mg of mercury remained in the contaminated water. What percentage of the mercury was removed from the water?

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