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 10.0 -mg sample of this material. (b) Calculate the surface area for a 10.0 \(\mathrm{mg}\) sample of this material. (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 mercury was removed from the water? (d) What is the final mass of the spongy material after the exposure to mercury?

Step by step solution

01

(a) Calculate the volume of a 10.0 mg sample of the material

We are given the density of the material, which is defined as \(\rho = \frac{m}{V}\), where \(\rho\) is the density, m is the mass, and V is the volume. We want to find the volume of a 10.0 mg sample of the material. Rearranging the formula, we get \(V = \frac{m}{\rho}\). Given: Density: \(\rho = 0.20 \mathrm{~g}/\mathrm{cm}^{3}\) Mass: \(m = 10.0 \mathrm{~mg}\) (we must convert this to grams) Converting the mass to grams: \(m = 10.0 \mathrm{~mg} \times \frac{1 \mathrm{~g}}{1000 \mathrm{~mg}} = 0.010 \mathrm{~g}\) Now we can plug the values into the formula to find the volume: \(V = \frac{0.010 \mathrm{~g}}{0.20 \mathrm{~g}/\mathrm{cm}^{3}} = 0.050 \mathrm{~cm}^{3}\) Hence, the volume of a 10.0 mg sample of this material is 0.050 cm³.

02

(b) Calculate the surface area for a 10.0 mg sample

We are given the surface area per gram of material, which is 1242 m²/g. To find the surface area for a 10.0 mg sample, we need to multiply the surface area per gram by the mass of the sample in grams. We have the mass from part (a): \(m = 0.010 \mathrm{~g}\) Now, we can find the surface area of the 10.0 mg sample: Surface Area = (Surface Area per Gram) x (Mass) Surface Area = \(1242 \mathrm{~m}^{2}/\mathrm{g} \times 0.010 \mathrm{~g} = 12.42 \mathrm{~m}^{2}\) Hence, the surface area for a 10.0 mg sample of this material is 12.42 m².

03

(c) Calculate the percentage of mercury removed from contaminated water

Initially, the contaminated water sample has 7.748 mg of mercury in it. After treatment with the spongy material, only 0.001 mg of mercury remains in the water. To find the percentage of mercury removed, we need to compute the difference between the initial and final amount of mercury, then divide by the initial amount of mercury, and multiply by 100%. Percentage Removed = \( \frac{\text{Initial Amount} - \text{Final Amount}}{\text{Initial Amount}} \times 100\%\) Percentage Removed = \( \frac{7.748 \mathrm{~mg} - 0.001 \mathrm{~mg}}{7.748 \mathrm{~mg}} \times 100\% \) Percentage Removed = 99.99% Hence, 99.99% of the mercury was removed from the contaminated water.

04

(d) Calculate the final mass of the spongy material after exposure to mercury

Initially, the spongy material has a mass of 10.0 mg. The amount of mercury removed from the water sample was 7.748 mg - 0.001 mg = 7.747 mg. To find the final mass of the spongy material after exposure to mercury, we simply add the mass of the removed mercury to the initial mass of the spongy material. Final Mass of Spongy Material = Initial Mass + Mass of Removed Mercury Final Mass of Spongy Material = \(10.0 \mathrm{~mg} + 7.747 \mathrm{~mg} = 17.747 \mathrm{~mg}\) Hence, the final mass of the spongy material after exposure to mercury is 17.747 mg.

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