After-test review and Properties of Molecules
After-test review and Properties of Molecules CHEM-1070-30
Popular in General Chemistry I
Popular in Chemistry
This 3 page Class Notes was uploaded by Nina Kalkus on Monday October 26, 2015. The Class Notes belongs to CHEM-1070-30 at Tulane University taught by Schmehl, Russell in Fall 2015. Since its upload, it has received 114 views. For similar materials see General Chemistry I in Chemistry at Tulane University.
Reviews for After-test review and Properties of Molecules
Report this Material
What is Karma?
Karma is the currency of StudySoup.
Date Created: 10/26/15
Chem Notes 102015 Aftertest review Neutralize Take a basic or acidic solution and get it back to where there are no excess HA or OHA and water is produced Test Example NaOH H2804 gt l Nast4 Partially correct H20 Na H280439 112 g ofalloy gNa gK Titration 025 M H280400684 L20H1 H2804 0034 moles of OH39 0034 moles OH39 moles NaOHN a moles KOHK 0034 moles OH39 g Na1 mole23 g 112ggNa1 mole391 g Solve for grams sodium divide by total to nd percent weight Test Example ice water MgSO4 s gt solution with MgSO4 aq Heat released by MgSO4 when dissolved melt the ice at 0 degree centigrade warm icewater to 6 degrees centigrade warm all the water to nal temperature 91 kJmole50 g1204 gmole1000 JkJ 37800 J heat released by MgSO4 gained by solution 37800 J 6300 Jmole55 g 18 gmole 418 JgdegreeC55 g6 degrees C 418 JgdegreeC100gchange in temperature 37800 J 18400 J 1380 J 418 J change in temperature change in temperature 431 degrees C Tf 431 6 491 degrees C Properties of Molecules Light Electromagnetic Radiation wave nature Electric wave vector and electric eld wave vector perpendicular to each other wavelength 9 frequency v cycles per second s 1 Hertz Hz C speed of light v the shorter the spacing between the peaks the higher the frequency is going to be xray ultra violet visible light infrared microwave radiowave 1011 1008 107 10061004 1002 10001002 gt wavelength 10020 10018 1016 10015 10013 10012 10010 1008 1006gt Hz white light through a prism will come out at a different angle angle of refraction depends on wavelength visible lightgtshortest 9tviolet indigo blue green yellow orange red longest L What does this have to do with atoms Pass a current through a tube with H2 Pass emitted light through a prism wavelength mgt Rydberg found that for these series of lines 1 9 R 1n12 1n22 n1 and n2 integers n2gtn1 For hydrogen R 10974quot 10397 m391 Chem Notes 102215 Chapter 8 visible light 400 nm gt 800 nm 4107 gt 8107 C v Emission lines of hydrogen lamp As described by Rydberg 10 109603 cm391 1n12 1n22 Since C V 10F vC V 3288 1015 NW 1n22 with n12 All v will be between 052 103915 and 082 103915 gap between any two levels is decreasing as the levels get higher smaller and smaller frequency gap calculations for when n1 1 gt which of these two is visible light emission and what does the other represent Tungsten lightbulb individual atoms heat gt oscillatory motion more with more heats think about it as two balls attached by a spring energy rises either as we compress or eXpand looks like a quadratic formula gt Simple Harmonic Oscillator Classical Physics Theory intensity at any given wavelength as a function of wavelength intensity of light increases as wavelength decreases Reality more of a bell shaped curve intensity has a peak at middle wavelength Tungsten Light Black body emitters Planck only certain states that exist with any reasonable probability transitions between those states occur only when discreet amounts of energy are put in Energy is quantized E njust an integer gt hproportionality constant vcyclessec How do we rationalize the Classical physics theory Boltzman probability is proportional to the exponentialE processKbT Kb Boltzman proportionality constant JK exponentialE processKbT e39E messKb probability is decreasing with wavelength which is why classical physics theory doesn39t work Vacuum chamber 5 Light source 7 Na Metal Power 801er6 V volt meter UV light electrons would come off Na Metal and current could be measured Infrared electrons would not come off and current could not be measured Start having to apply voltage to stop the current at a certain light frequency V0 minimum frequency required to apply stopping voltage Photoelectric effect Einstein 1905 Energy of Light minimum energy to ionize the metal excess energy kinetic energy of ejected electron hv eV0 l2mu2 l Joule voltcoulomb egt coulomb charge u velocity n1 n2gt quantized states