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Study Guide for Exam 3 Chem 109

by: mkennedy24

Study Guide for Exam 3 Chem 109 Chem 109

Marketplace > University of Nebraska Lincoln > Chemistry > Chem 109 > Study Guide for Exam 3 Chem 109
General Chemistry
Eric Malina

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About this Document

This study guide includes examples for the equations that are given during the chapters for this exam as well as viable information that should be known as well.
General Chemistry
Eric Malina
Study Guide
Chemistry, General Chemistry
50 ?




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This 0 page Study Guide was uploaded by mkennedy24 on Tuesday March 15, 2016. The Study Guide belongs to Chem 109 at University of Nebraska Lincoln taught by Eric Malina in Spring 2016. Since its upload, it has received 96 views. For similar materials see General Chemistry in Chemistry at University of Nebraska Lincoln.

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Date Created: 03/15/16
Study Guide End of Chapter 558end to end of chapter 7 Chapter 5 Kinetic Molecular Theory 0 Objective Describe parts and apply where needed 0 Kinetic Molecular Theory o Particles themselves occupy no volume but still have a mass Argas l VAr VT 01 A Space taken up by moecues 0 is proportional to temperature in Kelvins At any given moment some particles are moving faster than others there is distribution of velocities Relationship to temperature is related to velocity a hotgreater average kinetic energy gas moves faster than a cold gasesser average kinetic energy 1 KEZEmv2 gt where m IS the mass and v2 IS the velocity No energv loss in a collision onv exchanges Elastic Collision Inelastic Collision 4 L L t gt lt o o States of Matter 0 Solid Particles still move Vibrations 0 Liquid Particles move together but not completely stuck together like solid 0 Gas Free motion but all moleculesparticles do not move at the same speed nalitiierilli39ll ul Glued lIJ E EEIE39 E EE legume FHquot mail Femmm egllar1i tla39rr Hamstit arrangerrmr H ar aragerrmr Haws QUEEH ll39i all Erect313 Failure around Ham criticz 51m emer39f L39f asa39tulu39s wilful mi 39l39quot strut Eiugaiurn quot i ll l I I ll Iii i1 ii 391 il illi Root Mean Square Velocity 7R8314 Product of NAm wherxg is avogadro s number and m is mass results in kgmol Example Calculate the root mean square velocity for molecular oxygen and nitrogen Which is faster at 25 C First gure out what the answer should be like Nitrogen should be faster because why It is lighter than oxygen Oxygen atomic mass 160g Nitrogen atomic mass 1401g Also DON T FORGET to convert temperature to Kelvins and Grams to Kilograms Since it says for molecular oxygen and nitrogen oxygen and nitrogen are diatomic so their masses in the equation need to be multiplied by 2 Oxygen1602 320 and Nitrogen140122802 This is a basic plug in the numbers problem so 38314 298K m0 4 4 um 0032kg 8 9 kgmol 02 38314 298K u mOZ39K 2515038k 8N2 ms 002802 kg mol Rate of Effusion 0 Because molecules move at different rates they will effuse from a container at different speeds A o Graham s Law of effusion allows us to compar effusion rates Th process which a gas escapes from a container into a vacuum through a Rate refers to amount per unit of time 0 Example C02 effuses at 1822 times faster than a certain unknown gas What is the molar mass of this unknown gas rateA MB rateCO2 Mnnknown 21822 gt1822 unknown By basic algebra we can square each side to get ride of the 18226221461i mol rate B M A rate square root M 1822622M444016 4401 g 2 Chapter 6 Thermochemistry 0 Ability to do work The blue ball has As the balls collide After exchanging energy kinetic energy while they exchange because energy is not the purple ball has quot created nor destroyed the bl b ll h d 0 Kinetic Energy versus Potential Energy WEE thae ijplznfae fSSCei ggy O Han nllrnlo hall n mnuo E l associated with temperature 0 Type of kinetic energy 0 Has to do with motion of atom molecules in substance 0 Due to the amount of energy stored built up Due to position or composition Raising an object or a set of objects against gravitational pull inrrpncpc nntpntinl pnprnv A I Associated with position of electrons to the nucleus 0 Type of potential energy 0 The atommolecules has stored energy unless it is released due to a reaction 0 First Law of Thermodynamics The Law Of conSEIVGLIUII Ul EIICIyy lb UGDILGIIy UCDLIIUIIIy the First Law of Thermodynamics in different words From Kinetic Molecular Theory we should begin to see the connection between temperature and molecular vibrations o Vibrations from faster water transfers energy to the slower ice I Faster because the water is armer q Slower because colder The colder the particle is the slowe cubes 0 Therefore the water looses energy because the vibrations transfer into the icecubes warming the icecubes up and making the water loose energy therefore cool down q system q Surroundings qlAl f V q Inn Since q is equal to Mc A T we can infer from the abovethat 0 Same thing as what is happening in lab with the metal and the water 0 Example A 500g piece of Aluminum at 500 C is placed in a beaker lled with 1000mL of water at 10 C Calculate the nal temperature of the system assume solution has a density of lgmL cAIo9o2 JgK Cwater4184 Energy ow is out of the aluminum into the water qAqH20 5000902Tf 5001004184Tf 100 gtTf1520C o Exothermic Fire is giving off heat 0 Endothermic Person is absorbing heat from re Chemical Reactions are also able to give off heat Think back to dissolving Mg in HCI acid Mg 2HC l MgC2 This was an exothermic reaction which gave off heat known as heat of reaction for the reaction above AHRXN 464 kJmol Keep in mind that AHRXN is for 1 mol Mg and 2 mol HCI The quotper molquot refers to molar equivalents outlines in the balanced equation 0 Example 100mL of 550 molar HCI is mixed with an equal amount of 550 molar NaOH The following reaction occurs HClaq NaOHaq l H20 The solution was originally at 225 C After the solution was at 263 C Calculate AHRXN for reaction Exothermic reaction is heated up szcATgt20418438gtq318 Because the energy was absorbed by the solution it had to be given off by the reaction qRXN2 318J But how much of the reaction caused this 1 L 550mol H 10 0mL 1000 mL 1 L 005500 moles Cl Looking at our equation our AHRXN should be for 1 mol HCI so 318 J k 1 H V H l 00550molHCl 578 8mol Cl 578mol C Example Calculate AHORXN N2H4 N204g I 2N20g 2H20g Reactants l elements Eements I products Alike terms are highlighted because those can be canceled t since there is the same N2H4AH 0 I N204 D I AH amount of each Side of the I D 2N20 2 AMA I D 2H20 2 AMA Shortcut AJORXNZAIF ZAHFreactants z products sum Values from above Products 2 mol816kJmol2 mol2418kjmol 3204kJmol Reactants 1mol506kjmol 1 mol111 kjmol617kjmol Products Reactants 3204 617 Chapter 7 Light 0 Speed of Light c o C 30108mS CZAD O 4 IFrequency Hz Wavelength m 0 Example What is the frequency of 650nm of light Convert 650 nm to meters by just adding x109 Pre xes Example Calculate the Wavelength in nm of the red light emitted by a barcode scanner that has a frequency of 462 If you are given the frequency you can easily gure out the wavelength using this equation c2m by rearranging the equation to t what we need we get A 1 Convert wavelength from meters to nanometers by using the conversion factor between the two 1nm 1039 gm Threshold Frequency Minimum frequency to eject electrons o No electrons are emitted from metal no matter how long the light shines on the metal 0 Lowfrequency long wavelength light does not eject electrons from metal regardless of its intensity or its duration o Highfrequency shortwavelength light does eject electrons even if its intensity is low 0 Albert Einstein quotLight energy must come in packetsquot o Photon Quantized packet of energy 0 Ezhv 4 Frequency I Planck s Constant 0 Example A nitrogen gas lazer With a wavelength of 337 nm contains 383 ml of energy How many photons does it contain H2 Given E 383 m pulse A2337nm Find Number of Photons 10 9m lnm 662610 34s3O108 C S 19 E 00n 25898510 J P v 337107m A2337nm 3371O7m 10 31 383mJ 238310 31 E 3 NumberofPhotonSZ Pulse 2 33983 10 J 6491015ph0t0ns EPW 58985 1 0 19 Section 74 The Wave Nature of Matter The de Broglie Wavelength the Uncertainty Principle and lndeterminacy de Broglie Relation A mv 0 Notice The velocity of a moving electron is related to its wavelength Knowing one is equivalent to knowing the other Heinsenberg s Uncertainty Principle 0 Conclusion Best we can do is probability 0 Quantum Numbers Uncertainty in position Uncertainty in velocity lt A constant n l principle number 0 Overall size and energy that we actually have Bohr put it as a distance away from the nucleus 0 n 123 00 positive integers for n values ll angular quantum number 0 Shape of orbital Subshell If n1 then the only possible number for is 0 according to equation for I n1 so 11O If n3 the only possible number for 1 is n1 l 312 012 0 O l s orbital 1 l p orbital 2 l d orbital 3 I f orbital M M sub l Magnetic Quantum Number Orientation in space orbital Integers of M 2 1 O 1 2 Ms M sub 5 l Spin Quantum Number not required by Schrodingers equation 0 Electron spin 0 Ms 12 12 0 Objective Know relationships between n l M Subshells in a principal level n is always equal to the nvalue Number of orbitals in a subshell is always equal to 21 Number of orbitals in an energy level is always n2 Energy level in atom and spectra Bohr s equation l Hydrogen atom o En 21810 Jn2 Example What is the wavelength of light emitted when electron move from n2 to n1 in a hydrogen atom AE 2181018J iZ i gt 2181018J 2 nF quotI O O O O 1 1 gtAE 16351018Jgt 66261034Js30108 S 16351018 gtx1216107meters A


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