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 \(0.20 \mathrm{~g} / \mathrm{cm}^{3},\) and its surface area is \(1242 \mathrm{~m}^{2}\) per gram of material. (a) Calculate the volume of a (b) Calculate the surface area for \(10.0-\mathrm{mg}\) sample of this material. a \(10.0-\mathrm{mg}\) sample of this material. \((\mathbf{c})\) A \(10.0-\mathrm{mL}\) sample of contaminated water had \(7.748 \mathrm{mg}\) of mercury in it. After treatment with \(10.0 \mathrm{mg}\) of the new spongy material, \(0.001 \mathrm{mg}\) of mercury remained in the contaminated water. What percentage of the (d) What is the final mass mercury was removed from the water? of the spongy material after the exposure to mercury?

Step by step solution

01

Calculate the volume of a 10.0-mg sample

First, we need to convert the mass of the sample from milligrams (mg) to grams (g): \(10.0 \mathrm{~mg} = 0.010 \mathrm{~g}\) We are given the density of the material: \(0.20 \mathrm{~g/cm^3}\) Now we can calculate the volume using the formula: \(V = \frac{mass}{density}\) \(V = \frac{0.010 \mathrm{~g}}{0.20 \mathrm{~g/cm^3}}\) \(V = 0.05 \mathrm{~cm^3}\)

02

Calculate the surface area of a 10.0-mg sample

We are given the surface area per gram of the material: \(1242 \mathrm{~m^2/g}\) Now we can calculate the surface area using the formula: \(A = m \frac{1242 \mathrm{~m^2}}{1 \mathrm{~g}}\) \(A = 0.010 \mathrm{~g} \times \frac{1242 \mathrm{~m^2}}{1 \mathrm{~g}}\) \(A = 12.42 \mathrm{~m^2}\)

03

Calculate the percentage of mercury removed from the water

We are given the initial and final amounts of mercury in the water sample: Initial amount = 7.748 mg Final amount = 0.001 mg Amount of mercury removed = Initial amount - Final amount Amount of mercury removed = 7.748 mg - 0.001 mg = 7.747 mg Now we can calculate the percentage of mercury removed: \(\% = \frac{7.747 \mathrm{~mg}}{7.748 \mathrm{~mg}} \times 100\% \) \(\% = 99.99 \% \)

04

Calculate the final mass of the spongy material after exposure to mercury

Initial mass of spongy material = 10.0 mg Amount of mercury removed = 7.747 mg Final mass of spongy material = Initial mass + Amount of mercury removed Final mass of spongy material = 10.0 mg + 7.747 mg = 17.747 mg

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