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Chapter 22: Problem 30

A typical concentration of \(\mathrm{O}_{3}\) in the ozone layer is \(5 \times 10^{12} \mathrm{O}_{3}\) molecules \(\mathrm{cm}^{-3} .\) What is the partial pressure of \(\mathrm{O}_{3},\) expressed in millimeters of mercury, in that layer? Assume a temperature of \(220 \mathrm{K}\).

Short Answer

Expert verified
The partial pressure of \(O_{3}\) in the ozone layer, at a temperature of 220K, is \(1.1 \times 10^{-3}\) mmHg.

Step by step solution

01

Conversion to moles

First, we need to convert \(5 \times 10^{12} O_{3}\) molecules/cm\(^3\) to moles/cm\(^3\). Given that 1 mole of any substance has \(6.022 \times 10^{23}\) molecules (Avogadro's Number), this can be done using the following equation: Mole concentration = Molecule concentration / Avogadro's Number = \(5 \times 10^{12} O_{3}\) molecules/cm\(^3\) / \(6.022 \times 10^{23}\) molecules/mole = \(8.3 \times 10^{-12}\) mole/cm\(^3\)
02

Apply ideal gas law

The ideal gas law (PV = nRT) can then be used to calculate pressure. Rearranging for the pressure gives us P=nRT/V. Note that the gas constant R is in L atm/mol K and the given temperature T is in Kelvin, so the pressure P will be in atm. Subtituting the values we have P = \(8.3 \times 10^{-12}\) mole/cm\(^3\) * 0.0821 L atm/mol K * 220 K. But, 1 cm\(^3\) equals 1e-3 L, so the volume V needs to be adjusted accordingly. Therefore, P= \(1.5 \times 10^{-6}\) atm.
03

Conversion to mmHg

Lastly, the pressure needs to be converted to mmHg, which can be done using the conversion 1 atm = 760 mmHg. Thus, P = \(1.5 \times 10^{-6}\) atm * 760 mmHg/atm = 1.1 \times 10^{-3} mmHg.

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Most popular questions from this chapter

Freshly prepared solutions containing iodide ion are colorless, but over time they usually turn yellow. Describe a plausible chemical reaction (or reactions) to account for this observation.

Although relatively rare, all of the following compounds exist. Based on what you know about related compounds (for example, from the periodic table), propose a plausible name or formula for each compound: (a) silver(I) astatide; (b) \(\mathrm{Na}_{4} \mathrm{XeO}_{6} ;\) (c) magnesium polonide; (d) \(\mathrm{H}_{2} \mathrm{TeO}_{3} ;\) (e) potassium thioselenate; (f) \(\mathrm{KAtO}_{4}\).

Zn can reduce \(\mathrm{NO}_{3}^{-}\) to \(\mathrm{NH}_{3}(\mathrm{g})\) in basic solution. (The following equation is not balanced.) $$\begin{aligned}\mathrm{NO}_{3}^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s})+& \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \\\&\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{g})\end{aligned}$$ The \(\mathrm{NH}_{3}\) can be neutralized with an excess of \(\mathrm{HCl}(\mathrm{aq})\) Then, the unreacted HCl can be titrated with NaOH. In this way a quantitative determination of \(\mathrm{NO}_{3}^{-}\) can be achieved. A 25.00 mL sample of nitrate solution was treated with zinc in basic solution. The \(\mathrm{NH}_{3}(\mathrm{g})\) was passed into \(50.00 \mathrm{mL}\) of \(0.1500 \mathrm{M} \mathrm{HCl} .\) The excess \(\mathrm{HCl}\) required \(32.10 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for its titration. What was the \(\left[\mathrm{NO}_{3}\right]\) in the original sample?

Write equations to show how to prepare \(\mathrm{H}_{2}(\mathrm{g})\) from each of the following substances: \((a) \mathrm{H}_{2} \mathrm{O} ;\) (b) \(\mathrm{HI}(\mathrm{aq})\) (c) \(\mathrm{Mg}(\mathrm{s}) ;\) (d) \(\mathrm{CO}(\mathrm{g})\). Use other common laboratory reactants as necessary, that is, water, acids or bases, metals, and so on.

The heavier halogens (Cl, Br, and I) form compounds in which the central halogen atom, \(X\), is bonded directly to oxygen and to fluorine. Several examples are known, including those with formulas of the type \(\mathrm{FXO}_{2}, \mathrm{FXO}_{3, \text { and } \mathrm{F}_{3} \mathrm{XO} . \text { The structures of these }}\) molecules are all consistent with the VSEPR model. Draw Lewis structures and predict the geometries of (a) chloryl fluoride, \(\mathrm{FClO}_{2}\); (b) perchloryl fluoride, \(\mathrm{FClO}_{3} ;(\mathrm{c}) \mathrm{F}_{3} \mathrm{ClO}\).
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