Chemistry: Atoms First
Chemistry: Atoms First
3rd Edition
ISBN: 9781259638138
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 21, Problem 21.21QP
Interpretation Introduction

Interpretation: The number of ozone molecules present and the mass of ozone molecules should be determined.

Concept Introduction:

Atmosphere: The atmosphere is defined as air that is layer of gases which surrounds the earth due to gravity of earth.

The earth atmosphere consists of layers such as thermosphere, mesosphere, stratosphere and troposphere depending on the temperature and its composition.

Stratosphere:

It is found below the mesosphere layer in which the concentration of ozone and other gases are high. The increasing temperature for this layer with respect to increasing height is due to the presence of high concentration of ozone and other gases in it.

The increased temperature is due to the response of UV radiation from sun and hence ozone is formed due to this reaction and the use of ozone is that it prevents the UV radiation from the sun which is actually harmful.

Balanced Equation: A chemical reaction is considered as balanced equation if all the number of atoms, the charge and the mass of elements on both reactant and the product side are equal in the reaction.

Moles: It is the number of particles present in the given specific substance. The formula used to identify the moles present in given mass of a compound is as follows,

xmoles=GivenmassofthesampleMolecularmassofthesample

Molar mass: It is the total mass of the substance including all the number and the mass of the elements present in substance.

Expert Solution & Answer
Check Mark

Answer to Problem 21.21QP

The number of ozone molecules present in the stratosphere is.

The mass of ozone is 3.2×1012kg

To determine: The number of ozone molecules and the mass of the molecules present using given data’s.

Explanation of Solution

Determine the total volume.

Givendata: Radius = 6.371×106mThickness, h = 3.0×103mVolume of sphere = 4πr2h= 4×227×(6.371×106m)2(3.0×103m)=1.5×1012m3×1000L1m3=1.5×1015L

In order to find the number of molecules one needs to find the moles and to determine the moles present in the stratosphere we need to know the total volume present hence volume is obtained by substituting the given radius and thickness data in volume of sphere formula since earth considered as sphere.

Determine the molecules.

Moles of O3 present in given volumeMoles(1.5×1015L)×1 mol22.41L=6.7×1013molNumber of molecules,Molecules = 6.7×1013mol×6.022×1023molecules1 mol=4.0×1037molecules

First, the moles present is determined by dividing the volume of ozone present with the 22.41L since the gas at standard temperature and pressure condition it occupies that volume yields the value of 6.7×1013mol .

The next step is to find the number of molecules which is accomplished by multiplying the number of moles present with Avogadro’s number ( 6.022×1023molecules )

Determine the mass of the ozone molecules.

Mass of O3Mass(6.7×1013mol)×48g1 mol×103=3.2×1012kg

The mass of ozone molecules present is determined by using the given moles formula where the moles of ozone and the molar mass of ozone are known hence substituting it gives the mass of the ozone which is converted into kilograms by multiplying with 103 since 1 kg = 1000 g .

Conclusion

Conclusion

The number of molecules and the mass present is determined by using the moles formula.

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Chapter 21 Solutions

Chemistry: Atoms First

Ch. 21.5 - Prob. 21.5.1SRCh. 21.5 - Prob. 21.5.2SRCh. 21.8 - Prob. 21.3WECh. 21.8 - Prob. 3PPACh. 21.8 - Prob. 3PPBCh. 21.8 - Prob. 3PPCCh. 21.8 - Prob. 21.8.1SRCh. 21 - Prob. 21.1QPCh. 21 - Prob. 21.2QPCh. 21 - Prob. 21.3QPCh. 21 - Prob. 21.4QPCh. 21 - Prob. 21.5QPCh. 21 - Prob. 21.6QPCh. 21 - Prob. 21.7QPCh. 21 - Prob. 21.8QPCh. 21 - Prob. 21.9QPCh. 21 - Prob. 21.10QPCh. 21 - Prob. 21.11QPCh. 21 - Prob. 21.12QPCh. 21 - Prob. 21.13QPCh. 21 - Prob. 21.14QPCh. 21 - Prob. 21.15QPCh. 21 - Prob. 21.16QPCh. 21 - Prob. 21.17QPCh. 21 - Prob. 21.18QPCh. 21 - Prob. 21.19QPCh. 21 - Prob. 21.20QPCh. 21 - Prob. 21.21QPCh. 21 - Prob. 21.22QPCh. 21 - Prob. 21.23QPCh. 21 - Prob. 21.24QPCh. 21 - Prob. 21.25QPCh. 21 - Prob. 21.26QPCh. 21 - Prob. 21.27QPCh. 21 - Prob. 21.28QPCh. 21 - Prob. 21.29QPCh. 21 - Prob. 21.30QPCh. 21 - Prob. 21.31QPCh. 21 - Prob. 21.32QPCh. 21 - Prob. 21.33QPCh. 21 - Prob. 21.34QPCh. 21 - Prob. 21.35QPCh. 21 - Prob. 21.36QPCh. 21 - Prob. 21.37QPCh. 21 - Prob. 21.38QPCh. 21 - Prob. 21.39QPCh. 21 - Calcium oxide or quicklime (CaO) is used in...Ch. 21 - Prob. 21.41QPCh. 21 - Prob. 21.42QPCh. 21 - Prob. 21.43QPCh. 21 - Prob. 21.44QPCh. 21 - Prob. 21.45QPCh. 21 - Prob. 21.46QPCh. 21 - Prob. 21.47QPCh. 21 - Prob. 21.48QPCh. 21 - Prob. 21.49QPCh. 21 - Prob. 21.50QPCh. 21 - In which region of the atmosphere is ozone...Ch. 21 - The gas-phase decomposition of peroxyacetyl...Ch. 21 - Prob. 21.53QPCh. 21 - Prob. 21.54QPCh. 21 - Prob. 21.55QPCh. 21 - Prob. 21.56QPCh. 21 - Prob. 21.57QPCh. 21 - Prob. 21.58QPCh. 21 - A concentration of 8.00 102 ppm by volume of CO...Ch. 21 - Prob. 21.60QPCh. 21 - Briefly describe the harmful effects of the...Ch. 21 - Prob. 21.62QPCh. 21 - Prob. 21.63QPCh. 21 - Prob. 21.64QPCh. 21 - Prob. 21.65QPCh. 21 - Prob. 21.66QPCh. 21 - Prob. 21.67QPCh. 21 - A glass of water initially at pH 7.0 is exposed to...Ch. 21 - Prob. 21.69QPCh. 21 - Prob. 21.70QPCh. 21 - Describe the removal of SO2 by CaO (to form CaSO3)...Ch. 21 - Which of the following settings is the most...Ch. 21 - Prob. 21.73QPCh. 21 - Peroxyacetyl nitrate (PAN) undergoes thermal...Ch. 21 - Prob. 21.75QPCh. 21 - Prob. 21.76QPCh. 21 - The carbon dioxide level in the atmosphere today...Ch. 21 - A 14-m by 10-m by 3.0-m basement had a high radon...Ch. 21 - Prob. 21.79QPCh. 21 - A person was found dead of carbon monoxide...Ch. 21 - Prob. 21.81QPCh. 21 - As stated in the chapter, carbon monoxide has a...Ch. 21 - Prob. 21.83QPCh. 21 - Prob. 21.84QPCh. 21 - Prob. 21.85QPCh. 21 - Prob. 21.86QP
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