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by: Moses O'Conner


Moses O'Conner
GPA 3.87


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Class Notes
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This 7 page Class Notes was uploaded by Moses O'Conner on Friday October 2, 2015. The Class Notes belongs to CHE 1101 at Appalachian State University taught by Staff in Fall. Since its upload, it has received 18 views. For similar materials see /class/217703/che-1101-appalachian-state-university in Chemistry at Appalachian State University.

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Date Created: 10/02/15
Matthew Parker Jp41743cp appstate edu Emmi Prnperties Chapter Well mm must we hgputy huh 7 Eegmrmquot nee Shielding E en m E ec ve Nuckzr Chzxgz amade electxans B B n mumquot Inlnlnns Eeeynm new hes m was end We eme exeeemne The eme exeeemne m Wane xeemguee wee exeeemne eeemgm peeememe We exeeme and sheem eeeheme The inn1 leavmg m mm electxms seemg we pramns me Enter electxms dan39t shield each V1112 ngen39s electxms are reexmg a gem ENC Lhanare bexy xun39s have 7 puma mm m number Dfpmlans 15 increasing rpmquot Ammi Radius The me afm alum Le m ght 7 ENC an me Enter electmns xs memg m electmns dm39t mm as m 9392 aemzses Tap m mm 7 ENC cmstmL but sue uf declmn mud xs memg n 5 mean 9392 Innis Radius Matthew Parker 5739 7 quot 5 M S jp41743cpappstateedu Chapter 6 Electronic Structurg of Atgms Bring me my monocle I want to look rich 39 Equations Units Notes 01 c 1 m u gt squotI I c gt m s Ehv E J I hisaconstant I E is the energy of pie h J I S photon for light 0 S l I Combine with the rst one to get in terms of it h l m I h is a constant 1 h J I 7 is not physically mv s meaningful for something m kg 39 like a baseball I A is physically meaningful 0 V gt quotW for something like an electron I If asked to nd the wavelength of something that has mass use this equation he 1 1 1 m I Relates the energy AE hp 218 010 8J2 2 1 frequency and wavelength 1 n f n quot 0 3 of a photon emitted or camS absorbed 1 during the transition on an electron hJIS fromstatenitosomestate quot 11f EQJ I Moreorlessthesamething n gt 123 as the Rydberg equation but this form has both energy and frequency already included Problems With Classical Physics Classical Newtonian Physics had problems explaining what was going on with I Blackbody radiation things glow red orange yellow and white as they beat up I The photoelectric effect certain metals emitted electrons when hit with a certain wavelength of light I The hydrogen line spectrum the spectrum of excited hydrogen always contained four lines Planck got around the radiation problem by assuming energy is quantized the second equation above Einstein ran with Planck s idea to gure out was going on with the photoelectric e ect The hydrogen line spectrum was t perfectly with the Rydberg equation although no one really knew why until Bohr s Model If energy was quantized maybe electrons were too Bohr working from the idea that electrons were located in xed orbits with xed energy levels produced ultimately the fourth equation and explained the hydrogen line spectrum problem perfectly Matthew Parker jp41743cpappstateedu Problems with the Bohr Model Didn t really work all that great for anything more complicated than hydrogen Second problem was the planetary model of the nucleus and the electrons If light had particle properties per Planck and Einstein then did particles have wave properties Yes as per De Broglie and his equation the third one Bohr s model sort of irnplodes at this point If an electron has wave properties it can t be precisely located in clean planetarylike orbits Quantum Mechanics Bohr s model was modi ed by taking into account the wave properties of an electron Fellow named Schrodinger models wave properties of the electron with hideously complicated math The math becomes known as the Schrodinger equation The solution to the equation the wave function V contains three variables Solving for the variables produces the rst three quantum numbers and the wave function squared urz gives the probability of nding an electron in a piece of real estate at a given instant Pieces of real estate are called orbitals to differentiate from Bohr s orbits Quantum Numbers The rst three were produced from Schrodinger s modeling of hydrogen the only thing the Schrodinger equation can actually be solved for The fourth was added later The Pauli Exclusion Principle A set of four quantum numbers models any electron any orbital in any atom working like an address sort of All quantum numbers except n and m are derived from the quantum number that came before it I n the principle quantum number More or less an indicator of an orbital s distance from the nucleus and by extension its energy n is always a whole number greater than 1 l234 I I the angular momentum quantum number The shape sort of an orbital takes on O en designated with a letter instead ofa number 0s lp 2d 3f For any given 11 the possible 1 s are a range from 0 to nl I 111 the magnetic quantum number Describes the way the orbital is pointing sort of For any given I the possible 11185 are a range from l to 1 I m the electron spin quantum number Which direction an electron is spirming in sort 0t comrter clockwise or clockwise m is always either 12 or 12 Definitions Shell the region of space de ned by a given n all electrons with a n of 2 are in the same shell Subshell a further division of the shell de ned by a given n and 1 all 2p n2 and l1 are in the same subshell Orbital a region of space occupied probably by a pair of electrons electrons in a given orbital will have the same n l and m They differ only in spin Electron Con gurations I Know which blocks on the periodic table correspond to which subshell lling p 225 I n is a rough indicator of the subshell s or orbital s distance from the nucleus On a periodic table count down from the rst row of elements starting on 1 o For 5 and p orbitals this is going to be their n the s and p electrons in sulfur are both in n3 shells o For 1 orbitals the n is one less than the counting would indicate the rst d subshell that lls is not the 4d but the 3d Iron has 3d electrons not 4d 0 For f orbitals the n is two less than the counting would seem to indicate the lanthanides go after lanthanum and the actinides after actinium Uranium has 5f electrons I Next count from subshell to subshell starting with hydrogen until you reach the element you re going to s subshells hold 2 electrons d subshells hold 10 f subshells hold 14 p subshells hold 6 Orbital Diagrams Gives a more detailed view of how electrons are distributed and more importantly which electrons are paired or unpaired It s a lot simpler to do this if you have an electron con guration to look at Matthew Parker 2638340 Session Tuesda 120903 600PM Chapters 9 5 and 19 O times when I meet people who are retired I put my foot in their mouth andl say quotWelcome to Venice This will be your last meal Carbonate Cng39 has three resonance forms A Draw all three resonance forms B identify the rmlecular geometry C state whether the molecule is polar or nonpolar and D identify the hybridization on the carbon IH What is the hybridization on the nitrogen atom l N C There are two types of carbon hybrids in the H C HH H H molecule What are they CC H C b How many pairs of 1 bonding electrons are there C C H in this molecule H f c C C What is the angle between the carbon atoms a b HCN H c A system does 156k of work on is surroundings and absorbs 235k of heat Calculate the change in internal energy of the system When 955g of solid NaOH is dissolved in 100g of water in a coffee cup calorimeter the temperature rises from 236 C to 474 C Calculate the AH in kJmol for the reaction NaOHs gt Naaq OHaq The specific heat of the solution is 4184 JgK Chanters Five and Nineteen T39 I heard you tellin Dave to shoot me in the brain with a laser out behind the dumpster Sign Convention Signs matter In the case of work and heat a positive sign means it is ENTERING while an negative sign means it is LEAVING Formulas Most ofthe work in these two chapters is relatively simple math39 the hard part is knowing which formula to use Formula Important Things When to use E q w The 1 sf law of Relating internal energy heat T 39 and work Always valid w P V Expansion work39 a gas Use when a gas it expanding and expanding against a pressure P you need to know something in Pascals does work w in about work Should be very Joules obvious as the problem will have to give you both an external The V is the change in pressure and a volume volume ofthe gas needs to be in m3 H qp ONLY AT CONSTANT but lots of things are at PRESSURE constant pressure Problems at constant pressure always explicitly say so q Temperature has to be in The word calorimeter really SPeCl CHeat m Kelvin Mass usually needs to gives things away be in rams 0 0 H an r nH products nH mm DO NOT FORGET MOLAR COEFFICEINT S This H0 isn t necessarily the same thing as standard molar enthalpy of formation Temperature has to be in Use when you want to know an 9 AS Kelvin Watch the sign on q39 if entropy change associated with a T the system is releasing heat given heat transfer at a given and you want to know the S temperature of the surroundings q needs to be positive H A sort of special case form of Phase transitions melting ASquot f the formula above Watch the freezing sublimation sign on H SO usquotproducts nS0eamms DO NOT FORGET MOLAR standard molar entropy COEFFICEINT S This S0 is not necessarily the same thing as S Matthew Parker 26378340 Sunday 5 0076 00 Tuesday 6 0077 00 Thursday 4 0075 00 Chapmr7Perindi vapenizs panama mathvesmmynzckquot Th Skidling E en m E ec ve Nuckzr Chzxgz amade eleckans nl Be 3 lnlnlnns quot QM n1 Np 9 Inlnmns Nag 24va Mum 0522 m m mm m m m mums Th2 mm mm m mm The mu leavmg the Nmagen39s electmns areulmg a gem ENC mm bexy xun39s ENC N WWWMWafg msmmamgwo 227th f a mm m Dr 5 n 5 luni Radius Matthew Parker jp41743cpappstateedu Chapter 11 Phase Changes Clients my asteroid This is obviously just another one of your stupid supervillain plots Phase Changes Gases and Liquids On the boundary between a liquid and a gas molecules are constantly shitting between states molecules are getting bumped up into the gas phase from the liquid evaporating or vaporizing and gas molecules are colliding with the liquid phase and sticking condensing Eventually the two rates equalize the rate of the molecules ying into the gas phase will equal the rate of the molecules condensing back into the gas phase There will be no net change in the number of molecules in either phase although individual molecules will still be moving back and forth a dynamic equilibrium between the two is established The vapor pressure of water from the last chapter is technically the dynamic equilibrium pressure that water will establish at a given temperature Heat of Vaporization and Boiling Point I AH the energy in kilojoules required to vaporize one mole of a liquid AHW does not vary with temperature it is unique to every substance It is a function of the intermolecular forces operating in the liquid the stronger the forces the more energy will be required to separate the molecules from each other and kick them into the gas phase An exponential relationship exists between the vapor pressure again technically the dynamic equilibrium pressure and the temperature of the gas and the liquids AHW In P AHmRT C which can be reworked to another form 111 PlP2 AHvapyR 11 21Txl 1n PrP2 AHVIpR TiT2T1T2 The signi cance of the second form is that it lacks the C and has two different pressures and temperatures allowing calculation of the vapor pressure at any given temperature or the temperature at any particular vapor pressure Since boiling occurs when the vapor pressure of a liquid equals the external pressure set one of the P s to equal 1 atm if that s the external pressure the previous form allows calculations of the boiling point Again higher intermolecular forces will result in a higher boiling point Critical Temperature and Pressure The preceding section concerned itself with moving from the liquid phase to the gas phase this one the opposite moving from the gas phase to the liquid Gases can be condensed by one of two methods D Decrease the temperature El Increase the pressure If you were paying attention in the last chapter you should remember that decreasing the temperature and increasing the pressure both decrease the volume of a gas The gas molecules come closer together they latch onto each other with whatever intermolecular forces they got going and they condense This only works to a certain point Beyond a certain temperature critical temperature a gas cannot be made to condense however much you increase the pressure Intermolecular attractions can only do so much if the molecules have too much kinetic energy no dice Really a lot easier to visualize this with a phase diagram Critical temperature exists hand in hand with a critical pressure the minimum amount of pressure you have to apply to get the gas to liquefy at its critical tempemture the two of them come together to form the critical point on a phase diagram Once again gases with low intermolecular forces will have low critical temperatures


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