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Mizzou CHEM 1320 - Exam 2 Study Guide
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Chem 1320 1st EditionExam# 2 Study Guide Lectures: 12-22Lecture 12 (February 23)I. Daltons Law of Partial PressureThis law explains the pressure of individual gasses are additive. Some assumptions are that gases mix homogeneously and do not react. Since gases are so far apart on average, we can assume that they behave independently. Some consequences are that moles of gas are adaptive, independent. Also, if volume and temperature are fixed gases, pressures are adaptive.PT= P1+P2+P3……PT= n1RT/V + n2RT/V…II. Simple Model of a GasSome behaviors are that a gas appears to be independent of the identity of a gas, is a good assumption. Also, there are deviations from ideal behavior and we want to understand why. Some assumptions are that non- interacting particles have no attraction between them. Also, allof the particles are in constant random motion. Lastly, volume of a gas particles are small compared to the system. Some consequences are that magnitude of the pressure given by how often and how hard the molecules strike. Gas molecules have an average kinetic energy and that each molecule has a different energy. Also that there is a spread of individual energies of gas molecules in any sample of gas. Lastly, as temperature increases, average kinetic energy of gas molecules increases. Lecture 13(February 25)I. Average Molecular SpeedThe first thing you do is start with kinetic energy. Kinetic energy is the energy of motion and the model consumes no other energy. For single particles it is KE= 1/2mu2. For multiple particles it isKE= 1/2mu2. KE ɣ T  1/2mu2 ɣ T or ½mu2 = (constant) T Constant= 3R/2 for one particle (3R/2NA)R= 8.314J/(mol x K)U2=3RT/mNA = 3RT/M  u= (3RT/M)1/2II. Real Gases- Short comings of the ideal gas model1. Gas phase particles are not billiard balls and there are altercative forces between particles (P actual = P ideal- correction) 2. Gas phase particles have a finite volume (V ideal = V container- correction)Van der Waal’s Equation of StatePV = nRTZ= nRT/ PV (Z= compressibility factor) (P actual – constant)(V) = nRT(P actual – constant)(V- constant)= nRTEQUATION: (P-an2/v2)(v-bn)= nRT ***Lecture 14 (February 27)I. Nature of the ElectronThe electron has a negative charge and e= 1.602 x 10-19. The nature of light is referred to as electromagnetic radiation. The speed is 2.998 x 109m/s or 3x108m/s. Frequency is the number of oscillations of a wave per second (hertz).Speed= 120m/4s= 30m/s= 67mph. The wavelength tracks 120/4= 30m 120/6= 20m Frequency is 4/4s = 1.0s-1Wavelength x frequency (30m/s)(1/sec)=30m/s(20m)(1.55s-1)=30m/sEnergy is when frequency goes up volume goes downWhat is the frequency of light if it’s wavelength is 700m10-1 m= 1nmII. Max Plank’s Insight Classical physics (waves) could predict long wavelength behavior. T2 > T1Lecture 15 (March 2)I. Light Hitting MatterWhat happens when light hits matter? This is a question that comes with a complex answer. Light interacts with matter in discrete amounts of low energy. Infrared light (lowest energy) is when matter heats up atoms and molecules to begin to move. Visible light excites electrons andis where chemical reactions can occur. UV light (highest energy) like an x-ray makes chemical (covalent) bonds. II. Albert Einstein and Elevator DoorsEin= Eout (energy balance) hv= KE+w KE=1/2mv2hv< w w= work function- There is a minimum frequency required to emit electrons and the intensity is the number of electronsLecture 16 (March 4)I. The Borgie Hypothesis: light (particle behavior) Einstein= E=mc2Plank= E=hv/wavelengthmc2= hc/wavelength= m/h(wavelength) > cm= h/wavelength p= h/wavelength, P= momentum= mass x velocityII. John Balmer1/wavelength = 1.1 x 107m-1 (-1/z2- 1/h2) h= 3,4,5 specific to hydrogen III. Neils Bohr and the hydrogen atom“planetary” modelAssumptions:1. Only circular orbits (balance of forces)2. Electrons can move between orbitsResults:1. ΔE= Ef-Ei2. ΔE= R4(1/ni2-1/nf2) (+) = absorption & (-) = emission Lecture 17 (March 6)I. NewtonNewton came up with the wave theory of motion. The formula is F=ma. To figure out the problem you have to first determine the behavior exactly. Then you must find the initial values and know everything. The Schrodinger equation is the Hamiltonian x wavelength function= energy x wavelength function. To find this you need to first fine the probability. Then find the wavelength function squared  electron density/ location probability. The uncertainty principal is when you don’t know everything. About a system, ΔxΔp≥h/4∏; x=position, p= momentum (mass x velocity) Some consequences are the Bohr Model vs. The Quantum Mechanical Model. The Bohr Model means it has circular orbits and electron positions. The Quantum Mechanical Model is orbital and has electron energy states with definite energy. The location also leads to probability. II. The Hydrogen Atom (Figure 7.18 in Book) There are quantum numbers (electron book keeping) associated with this. The n stands for principal quantum number (1,2,3,……).The l stands for angular momentum quantum number (shape of the orbital). Lastly, the ml stands for magnetic quantum number (the orientation). Lecture 18 (March 9)I. Orbitals S-orbitals: n, l=0, ml= 0P- orbitals: l=1, ml= -1,0,1D-orbitals: l=2, ml= -2,-1,0,1,2There are three different shapes of orbitals starting with the smallest (1s) medium (2s) and largest (3s). II. Quantum numbersQuantum numbers are an extra electron label. The Stem and Gerlach Experiment was on electron spin. m8= electron spin qn. The permitted values are ½ & -1/2. The significance accounts for electrons in atoms that interacts with magnetic fields.III. Pauli Exclusion PrincipalThis principal explains the orbital occupation constraints. No two electrons can have the same four quantum numbers in an atom. A shell is all the electrons with the same principal qn. A subshell is all the electrons with the same n & l. An orbital is all the electrons with the same n, l, & ml. IV. The Aufban PrincipalThis principal is the order of filling orbitals to give the lowest energy configuration. This includes the ground state or ground electronic state. 1s2s 2p3s 3p 3d4s 4p 4d 4f5s 5p 5d 5f 5g6s 6p 6d 6f 6g 6h7s 7p 7d 7f 7g 7h…….8s 8p 8d 8f 8g 8h…….V. Electron ConfigurationsH: (Z=1) = 1s1 (core electron)He: (Z=2) = 1s2 (core electron)TIP: to max the spin put the up arrow in each box first and then go back and put the down arrow. Hunds Rule states


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Mizzou CHEM 1320 - Exam 2 Study Guide

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