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Chapter 22: Problem 17

The abundance of \(\mathrm{F}^{-}\) in seawater is \(1 \mathrm{g} \mathrm{F}^{-}\) per ton of seawater. Suppose that a commercially feasible method could be found to extract fluorine from seawater. (a) What mass of \(\mathrm{F}_{2}\) could be obtained from \(1 \mathrm{km}^{3}\) of seawater \(\left(d=1.03 \mathrm{g} \mathrm{cm}^{-3}\right) ?\) (b) Would the process resemble that for extracting bromine from seawater? Explain.

Short Answer

Expert verified
From 1 kilometer cubed of seawater, \(2.06 \times 10^6 kg\) of fluorine could be obtained. The process wouldn't resemble bromine extraction because the extraction principles are based on different chemical properties.

Step by step solution

01

Converting volume to mass

Calculate the mass of seawater in 1 kilometer cubed. 1 cubic kilometer of water contains \(1km^3 = 1km \times 1km \times 1km = 10^3m \times 10^3m \times 10^3m = 10^9m^3 \times (10^2cm/m)^3 = 10^{15} cm^3 \).The mass of this seawater would be \(Volume \times Density = 10^{15}cm^3 \times 1.03g/cm^3 = 1.03 \times 10^{15}g = 1.03 \times 10^9 \) tons.
02

Calculating fluorine amount

Calculate the amount of fluorine from this mass of seawater. Since every ton of seawater has 1 gram of F-, thus, the mass of F- would be \(1.03 \times 10^9 g \) (F- per ton of seawater). To find the mass of F2, which has a formula mass of approximately 38.00 g/mol remember F- indicates one fluorine atom and F2 indicates a diatomic molecule of fluorine (two fluorine atoms). Thus, the mass of F2 obtainable is \(1.03 \times 10^9 g \times \frac{38.00 g}{19.00 g} = 2.06 \times 10^9 g = 2.06 \times 10^6 kg\)
03

Comparing to bromine extraction

The process of extracting fluorine from seawater would be different than that of extracting bromine. Bromine extraction is primarily a result of its solubility and volatility properties, where it can be released from water through aeration. Fluorine, on the other hand, is present in seawater as a fluoride ion and is not volatile. Consequently, a different extraction process would be required.

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Most popular questions from this chapter

What is the acid anhydride of \((a) \mathrm{H}_{2} \mathrm{SO}_{4} ;(\mathrm{b}) \mathrm{H}_{2} \mathrm{SO}_{3}\) (c) \(\mathrm{HClO}_{4} ;\) (d) \(\mathrm{HIO}_{3}\) ?

Write a chemical equation for the hydrolysis of \(\mathrm{XeF}_{4}\) that yields \(\mathrm{XeO}_{3}, \mathrm{Xe}, \mathrm{O}_{2},\) and \(\mathrm{HF}\) as products.

Despite the fact that it has the higher molecular mass, XeO \(_{4}\) exists as a gas at \(298 \mathrm{K},\) whereas \(\mathrm{XeO}_{3}\) is a solid. Give a plausible explanation for this observation.

Use the following electrode potential diagram for basic solutions to classify each of the statements below as true or false. Assume standard conditions. $$\begin{aligned}\mathrm{SO}_{4}^{2-} \stackrel{-0.936 \mathrm{V}}{\longrightarrow} \mathrm{SO}_{3}^{2-} & \stackrel{-0.576 \mathrm{V}}{\longrightarrow} \\\& \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \stackrel{-0.74 \mathrm{V}}{\longrightarrow} \mathrm{S} \stackrel{-0.476 \mathrm{V}}{\longrightarrow} \mathrm{S}^{2-}\end{aligned}$$ (a) Sulfate \(\left(\mathrm{SO}_{4}^{2-}\right)\) is a stronger oxidant than thiosulfate \(\left(\mathrm{S}_{2} \mathrm{O}_{3}^{2}\right)\) in basic solution. (b) \(S^{2-}\) can be used as a reducing agent in basic solutions. (c) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) is stable with respect to disproportionation to \(\mathrm{SO}_{3}^{2-}\) and \(\mathrm{S}\) in basic solution.

\(\mathrm{O}_{3}(\mathrm{g})\) is a powerful oxidizing agent. Write equations to represent oxidation of \((a) I^{-}\) to \(I_{2}\) in acidic solution; (b) sulfur in the presence of moisture to sulfuric acid; (c) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) to \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}\) in basic solution. In each case \(\mathrm{O}_{3}(\mathrm{g})\) is reduced to \(\mathrm{O}_{2}(\mathrm{g})\).
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