Midterm Study Guide
Midterm Study Guide ENVS 237
Popular in Foundations of Environmental Science II
Popular in Environmental Science
This 6 page Study Guide was uploaded by Corinn Notetaker on Saturday September 26, 2015. The Study Guide belongs to ENVS 237 at Loyola University Chicago taught by Ping Jing in Fall 2015. Since its upload, it has received 192 views. For similar materials see Foundations of Environmental Science II in Environmental Science at Loyola University Chicago.
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Date Created: 09/26/15
ENVS 237 Fall 2015 Midterm Study Guide Chapter 1 Air Pollution 0 Common air pollutants Carbon Monoxide CO Nitrogen Oxides NOX Sulfur Dioxide SOZ Particulate matter PM Ozone 0 Primary vs Secondary air pollutants I 2 Primary Pollutantsdirectly emitted into the atmosphere by natural processes amp human activities Ex Carbon Monoxide CO Nitrogen Oxides NOX Sulfur Dioxide SOZ Particulate Matter PM Secondary Pollutants not directly emitted forms after primary pollutants go through reactions Ex Ozone O3 0 Factors of air pollution exposure I 2 3 Concentration of pollutant Time span of exposure Rate of inhalation o Balanced chemical reactions I Law of Conservation of Mass the number of atoms in the equation s reactants must equal the number of atoms in its products Ex BC 2502 CSz 4C0 atoms are equal on both sides ofthe equation 0 Factors of air pollutant concentration I 2 Sources the amount that enters the atmosphere emissions or chemical production Sinks the amount that is removed from the atmosphere chemical destruction or deposition Volume area for pollutants to disperse into Fluxes transfer of pollutants through wind movement 3 Effectiveness of the Clean Air Act SHORT ANSWER QUESTION 40 400 7 02 VOC quotNOX Popuiation 1 35 3 A t 30 300 2 2 2 25 2 1 20 200 g E 393 Iquot 15 g 5 n 0 o 10 100 a E m 1900 1910 1920 1930 1940 1950 1960 1970 1930 1990 2000 2013 Year Interpret the given gure 0 Change in emissions over time 0 Calculate total change in emissions EX What was the change in 802 emissions between 1970 amp 1990 19901970 22 million ST31 million ST W 9 31 million ST 9 X Change 0 Calculate per capita change EX What was the per capita change in 802 emissions between 1970 amp 1990 42 per capita change percapita1990percapita1970 gt 22280 31225gt 079 137 gt 042 x 100 per capita 1970 31225 137 Chapter 2 Ozone 0 Periodic table basics 1201 1 is Carbon 6 carbon s atom1c mass 6 is carbon s atomic number the number 12011 of protons amp neutrons in the atom o Electromagnetic waves Wavelength amp Frequency are inversely related therefore a longer wavelength 2 a lower frequency and vice versa All electromagnetic waves consist of photons particles of light Formula to compute energy per photon EEnergy hPlancllt s constant E szrequency 2 0 Stratospheric ozone The good ozone layer lies at an altitude of 2025 km Chapman Mechanism i Ozone formation 02 gt O O 1 Oxygen absorbs solar radiation amp dissociates 2 O reacts with Ozto form ozone O3 20o gt 0 ii Ozone destruction 1 Ozone absorbs solar radiation amp O3 O2 O decomposes back into O2 amp O OR 03 0 02 O2 2 Ozone reacts with oxygen to create O2 amp O2 Cl s catalytic destruction of ozone i In this process ozone is destroyed without consuming a single Chlorine atom ii One Cl atom can destroy 100000 O3 molecules CFCs chloro uorocarbons i Desirable traits a Nonreactive non ammable b Odorless c Colorless ii Depleting stratospheric ozone a 100 year lifespan b Exposure to UV radiation causes CFCs to release atomic Chlorine causing the catalytic destruction process Formation of the ozone hole when where how why should we be concerned i Antarctica ii Springtime SeptemberNovember iii Polar Vortex of PSCs a Polar Stratospheric Clouds PSCs form in Antarctic winter b Chemical reactions occur on clouds ice crystalsthese reactions create ozonedestroying molecules c Catalytic ozone depletion occurs in the light of the spring sun around October 0 Groundlevel ozone Formation of groundlevel ozone N02 sunliiht NO 0 00 o Relation between Ozone NOX and VOCs 025 4 if C 937 0 15 024 I 034 05C VOC 02 I f 39 f l 39 1 l i quotT Analyze the gure o o o o I r I l y o Increasedecrease 1n em1ss1ons 1n r I I 9 0 93 x x g o ogOfw 03932 cze relat1on to ozone E ms 39 v f x K Xlt o Recognize Noxlimited and VOC P s A x x xx 5 C 0 a x 39 xxa til l1m1ted areas 1 5quotr quot f quot o Interpret pol1cymak1ng ma N39SP Q39 im lications how do N OX and J Jp No1 39Q H VOC levels effects ozone levels 39 139 T EE f il o l Her i x L I J c 02 04 05 00 10 12 1 15 16 20 VOC mm Chapter 3 Global Climate Change 0 Energy balance 3 Laws of Radiation SHORT ANSWER QUESTION 0 StefanBoltzmann Law EEnergy Intensity E C T4 o Stefan Boltzmann constant 567 X 10398 Wm39ZK394 TTemperature in Kelvin o Wien s Law Amaximum wavelength A 2 898T TTemperature in Kelvin EX Calculate the intensity of solar radiation Sun s surface temperature 5780 K EoT4 567x108 Wmayf 578W 567X108Mhn2112X105 635X107Wm392 EX Calculate the wavelength at the peak of Earth s radiation Earth s surface temperature 2887 K A2898T 2898 2887 100 Mm micrometers Distinguish between the solar radiation and Earth s radiation 0 Peak solar radiationvisible light range 0 Peak Earth radiationinfrared range Remember ALL bodies emit radiation 0 The Greenhouse Effect Greenhouse gases absorb some of Earth s outgoing infrared radiation This absorption warms the lower atmosphere The atmosphere s warmth is radiated onto Earth s surface The greenhouse effect restricts this radiation s ability to escape into space therefore contributing to global warming and climate change H20 is the most abundant greenhouse gas but because its lifespan is so short it does not significantly contribute to climate change The Enhanced Greenhouse Effect is the intensi cation of the Greenhouse Effect by humans Human activity increases the concentration of greenhouse gases in the air causing temperature to further increase 3 Molar Mass SHORT ANSWER QUESTION The mole o A standard unit of measurement used in chemical reactions 0 Avogadro s number The number of molecules in a mole of a substance represented as this constant 602x1023 Molar mass 0 An element s molar mass is calculated by adding the atomic masses of each element 0 Represents mass of Avogadro s number of whatever particles speci ed atomsmolecules etc 1 mole 2 an Avogadro s number of particles EX Find the mass of 7 26X1022 molecules of Methane CH4 Carbon 12011 is Carbon s Hydqogen 10078 is Hydrogen s C6 atomic mass H atomic mass 12011 10078 Step 1 CH4 molar mass gt Carbon s molar mass gt 12 gmol 1 C atom Hydrogen s molar mass 1 gmol 4 H atoms 16 gmol per 1 mole of CH4 Step 2 Number of moles in Methane 72 6x1022 molecules of CH4 602 x 1023 molecules per mole 437 x 1046mol Step 3 Mass 16gmol x 437 x 1046 I 699 X 1047 g I Final Answer mass of726 X 1022 molecules of CH4 Good Luck Corinn Rutkoski
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