Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The estimated average concentration of \(\mathrm{NO}_{2}\) in air in the United States in 2006 was \(0.016\) ppm. (a) Calculate the partial pressure of the \(\mathrm{NO}_{2}\) in a sample of this air when the atmospheric pressure is 755 torr \((99.1 \mathrm{kPa}) .\) (b) How many molecules of \(\mathrm{NO}_{2}\) are present under these conditions at \(20^{\circ} \mathrm{C}\) in a room that measures \(15 \times 14 \times 8 \mathrm{ft}\) ?

Short Answer

Expert verified
The partial pressure of NO2 in the air is \( 1.208\times10^{-2} \) torr. There are \( 1.904\times 10^{20} \) NO2 molecules in the room measuring \( 15 \times 14 \times 8 \mathrm{ft} \) at \( 20^{\circ} \mathrm{C} \) and 755 torr atmospheric pressure.

Step by step solution

01

(a) Calculate the partial pressure of NO2

To calculate the partial pressure of NO2, we can use the given concentration in ppm and the atmospheric pressure. The concentration in ppm can be understood as a ratio of the number of NO2 molecules to the total number of air molecules. Therefore, we can find the partial pressure by multiplying the atmospheric pressure by the concentration (as a fraction): Partial Pressure of NO2 = Atmospheric Pressure × (Concentration of NO2 / 10^6) Given concentration of NO2 = 0.016 ppm Atmospheric Pressure = 755 torr (99.1 kPa) Partial Pressure of NO2 = 755 torr × (0.016 / 10^6) = \( 1.208\times10^{-2} \) torr
02

(b) Find the number of NO2 molecules in the room

To find the number of NO2 molecules in the room, first, we'll use the Ideal Gas Law to determine the number of moles of NO2 present. Then, we will multiply the number of moles by Avogadro's number to obtain the number of molecules. Ideal Gas Law: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin. Given room dimensions, Volume (V) = 15ft x 14ft x 8ft = 1680 ft³ = 47.60 m³ (converting ft³ to m³ by multiplying by 0.0283168) Technically room temperature is 20°C, so, T = 293.15 K (converting to Kelvin by adding 273.15) We already have the partial pressure of NO2 (P) in torr, so we first convert it to kPa by multiplying with the conversion factor (1 torr = 0.133322 kPa): Partial Pressure of NO2 = \( 1.208\times10^{-2} \) torr × 0.133322 = \( 1.613\times10^{-3} \) kPa Now, we can solve for the number of moles (n) using the Ideal Gas Law: n = PV / RT n = \((1.613\times10^{-3} \mathrm{kPa})(47.60 \mathrm{m^3}) / ((8.314 \mathrm{J/(mol.K)})(293.15 \mathrm{K})\) = 3.16 × 10^-4 mol Lastly, we'll multiply the moles by Avogadro's number to determine the number of NO2 molecules: Number of NO2 molecules = \(3.16 \times 10^{-4} \mathrm{mol} \) × \( 6.022 \times 10^{23} \mathrm{molecules/mol} \) = \( 1.904\times 10^{20} \) NO2 molecules

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Air pollution in the Mexico City metropolitan area is among the worst in the world. The concentration of ozone in Mexico City has been measured at 441 ppb \((0.441 \mathrm{ppm})\). Mexico City sits at an altitude of 7400 feet, which means its atmospheric pressure is only \(0.67\) atm. Calculate the partial pressure of ozone at 441 ppb if the atmospheric pressure is 067 atm.

Alcohol-based fuels for automobiles lead to the production of formaldehyde \(\left(\mathrm{CH}_{2} \mathrm{O}\right)\) in exhaust gases. Formaldehyde undergoes photodissociation, which contributes to photochemical smog: $$ \mathrm{CH}_{2} \mathrm{O}+h v \longrightarrow \mathrm{CHO}+\mathrm{H} $$ The maximum wavelength of light that can cause this reaction is \(335 \mathrm{~nm}\). (a) In what part of the electromagnetic spectrum is light with this wavelength found? (b) What is the maximum strength of a bond, in \(\mathrm{kJ} / \mathrm{mol}\), that can be broken by absorption of a photon of \(335-\mathrm{nm}\) light? (c) Compare your answer from part (b) to the appropriate value from Table \(8.4\). What do you conclude about the \(\mathrm{C}-\mathrm{H}\) bond energy in formaldehyde? (d) Write out the formaldehyde photodissociation reaction, showing Lewis-dot structures

A first-stage recovery of magnesium from seawater is precipitation of \(\mathrm{Mg}(\mathrm{OH})_{2}\) with \(\mathrm{CaO}\) : \(\mathrm{Mg}^{2+}(a q)+\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(t) \longrightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{Ca}^{2+}(a q)\) What mass of \(\mathrm{CaO}\), in grams, is needed to precipitate \(1000 \mathrm{lb}\) of \(\mathrm{Mg}(\mathrm{OH})_{2} ?\)

The concentration of \(\mathrm{Ca}^{2+}\) in a particular water supply is \(5.7 \times 10^{-3} \mathrm{M}\). The concentration of bicarbonate ion, \(\mathrm{HCO}_{3}^{-}\), in the same water is \(1.7 \times 10^{-3} \mathrm{M}\). What masses of \(\mathrm{Ca}(\mathrm{OH})_{2}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) must be added to \(5.0 \times 10^{7} \mathrm{~L}\) of this water to reduce the level of \(\mathrm{Ca}^{2+}\) to \(20 \%\) of its original level?

The Henry's law constant for \(\mathrm{CO}_{2}\) in water at \(25^{\circ} \mathrm{C}\) is \(3.1 \times 10^{-2} \mathrm{M} \mathrm{atm}^{-1} .\) (a) What is the solubility of \(\mathrm{CO}_{2}\) in water at this temperature if the solution is in contact with air at normal atmospheric pressure? (b) Assume that all of this \(\mathrm{CO}_{2}\) is in the form of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) produced by the reaction between \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) : $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)-\longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$ What is the \(\mathrm{pH}\) of this solution?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free