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Chapter 10: Problem 40

(a) If the pressure exerted by ozone, \(\mathrm{O}_{3}\), in the stratosphere is \(3.0 \times 10^{-3}\) atm and the temperature is \(250 \mathrm{~K}\), how many ozone molecules are in a liter? (b) Carbon dioxide makes up approximately \(0.04 \%\) of Earth's atmosphere. If you collect a 2.0 - \(\mathrm{L}\) sample from the atmosphere at sea level \((1.00\) atm \()\) on a warm day \(\left(27^{\circ} \mathrm{C}\right),\) how many \(\mathrm{CO}_{2}\) molecules are in your sample?

Short Answer

Expert verified
(a) The number of ozone molecules in a 1 liter volume under the given conditions is \(8.79 \times 10^{19}\) O3 molecules. (b) The number of CO2 molecules in the 2.0 L sample collected from Earth's atmosphere is \(1.96 \times 10^{19}\) CO2 molecules.

Step by step solution

01

Use the Ideal Gas Law formula for part (a)

The Ideal Gas Law is given by the formula: \(PV=nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the amount of gas in moles, \(R\) is the universal gas constant, and \(T\) is temperature. Now, let's plug in the values given in the problem to find the number of moles of O3 in a 1-liter volume. Given, \(P = 3.0 \times 10^{-3} \mathrm{atm}\), \(V = 1.0 \mathrm{L}\), \(T = 250 \mathrm{K}\), and \(R = 0.0821 \frac{\mathrm{L.atm}}{\mathrm{mol.K}}\) Rearrange the formula to find "n": \( n = \frac{PV}{RT} \)
02

Calculate the number of moles for part (a)

Plug in the values in the formula: \( n_\mathrm{O_3} = \frac{(3.0 \times 10^{-3}\mathrm{atm})(1.0\mathrm{L})}{(0.0821\frac{\mathrm{L.atm}}{\mathrm{mol.K}})(250\mathrm{K})} \) Calculate the number of moles: \(n_\mathrm{O_3} = 1.46 \times 10^{-4} \: \mathrm{mol}\)
03

Calculate the number of ozone molecules for part (a)

We'll use Avogadro's number to convert moles to molecules. The Avogadro's number is approximately \(6.022 \times 10^{23}\) molecules per mole. Number of O3 molecules = \(n_\mathrm{O_3} \times N_\mathrm{A}\) where \(N_\mathrm{A}\) is Avogadro's number. Number of O3 molecules = \( (1.46 \times 10^{-4}\: \mathrm{mol}) \times (6.022 \times 10^{23} \: \mathrm{molecules/mol}) \) Calculate the number of O3 molecules: Number of O3 molecules = \(8.79 \times 10^{19}\) O3 molecules
04

Use the Ideal Gas Law formula for part (b)

To find the number of CO2 molecules in the 2.0 L sample, first calculate the number of moles of CO2, and then convert it to molecules. Given, CO2 in Earth's atmosphere = \(0.04\%\) = \(0.0004\), collected sample = \(2.0\mathrm{L}\), pressure = \(1.00\mathrm{atm}\), and temperature = \(27^{\circ}\mathrm{C}\) = \(300\mathrm{K}\). Calculate the pressure exerted by CO2 only: \(P_\mathrm{CO2} = (0.0004)(1.00\mathrm{atm}) = 4.0 \times 10^{-4} \mathrm{atm}\) Now, plug in the values in the Ideal Gas Law formula and rearrange it to find the number of moles of CO2: \(n_\mathrm{CO2}=\frac{P_\mathrm{CO2}V}{RT}\)
05

Calculate the number of moles for part (b)

Plug in the values in the formula: \( n_\mathrm{CO2} = \frac{(4.0\times10^{-4}\mathrm{atm})(2.0\mathrm{L})}{(0.0821\frac{\mathrm{L.atm}}{\mathrm{mol.K}})(300\mathrm{K})} \) Calculate the number of moles: \(n_\mathrm{CO2} = 3.25 \times 10^{-5} \mathrm{mol}\)
06

Calculate the number of carbon dioxide molecules for part (b)

Use Avogadro's number to convert moles to molecules: Number of CO2 molecules = \(n_\mathrm{CO2} \times N_\mathrm{A}\) Number of CO2 molecules = \( (3.25 \times 10^{-5}\: \mathrm{mol}) \times (6.022 \times 10^{23} \: \mathrm{molecules/mol}) \) Calculate the number of CO2 molecules: Number of CO2 molecules = \(1.96 \times 10^{19}\) CO2 molecules

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Most popular questions from this chapter

Magnesium can be used as a "getter" in evacuated enclosures to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of \(0.452 \mathrm{~L}\) has a partial pressure of \(\mathrm{O}_{2}\) of \(3.5 \times 10^{-6}\) torr at \(27{ }^{\circ} \mathrm{C}\), what mass of magnesium will react according to the following equation? $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) $$

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