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Review of Key Concepts on CH. 1 and CH.2

by: Sandy Ho

Review of Key Concepts on CH. 1 and CH.2 Chemistry 14A

Marketplace > University of California - Los Angeles > Chemistry > Chemistry 14A > Review of Key Concepts on CH 1 and CH 2
Sandy Ho
GPA 4.0
Chemistry 14A: Atomic and Molecular Structure, Equilibria, Acids, and Bases

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About this Document

includes notes on modules and lectures about CH. 1 and 2
Chemistry 14A: Atomic and Molecular Structure, Equilibria, Acids, and Bases
Class Notes
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This 7 page Class Notes was uploaded by Sandy Ho on Wednesday October 14, 2015. The Class Notes belongs to Chemistry 14A at University of California - Los Angeles taught by Lavelle in Summer 2015. Since its upload, it has received 95 views. For similar materials see Chemistry 14A: Atomic and Molecular Structure, Equilibria, Acids, and Bases in Chemistry at University of California - Los Angeles.


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Date Created: 10/14/15
Wave Properties of Electrons waves interact inphase constructive interference and outof phase destructive interference resulting in diffraction patterns 0 Ex xrays passing through a crystal constructive interference 0 peak overlaps peak trough overlaps trough 0 resulting interaction as higher peak and deeper trough 0 Ex the waves resulting from dropping one pebble into water destructive interference 0 peak overlaps with trough trough overlaps peak Q waves out of phase would cancel each other if the same size 0 resulting wave really small the waves resulting from dropping one pebble into water Vs the waves resulting from 2 pebbles dropped at the same time creating complicates ovenaps interacting particles like billard balls do not give rise to diffraction patterns only waveslight create diffraction patterns however electronsneutrons passing through a crystal show a diffraction pattern and therefore have wavelike properties quantum mechanics description of electrons o electrons have particlelike and wavelike properties 0 all particles have particlelike and wavelike properties 0 all matter has wavelike properties only noticed for objects like electrons with small mass and high velocity any moving particle with momentum p has wavelike properties with wavelength x 0 De Broglie equation k p mass x velocity h 663 x 103934 J o s in an atom an electron has wavelength of 103912 m Photoelectric Effect photoelectric experiment light ultraviolet radiation source hits metal surface and if electrons e are removed from metal their kinetic energy is measured E photon E energy remove e Eke E hv o h Planck s Constant 6626 x 10 34 J s 0 V frequency 0 E proportional to frequency hv work function 12 me e2 0 me mass 0 ve velocity 0 threshold energy energy of photon has to meet a certain value to eject electron unless E photon 2 E energy remove e then e not emitted even with high intensity light if light had only wave properties then increasing intensity should eject e instead it is made of photons which depended on wavelengths o C M 0 V frequency 0 k wavelength 0 c speed of light 3 x108 ms 1 to 1 ratio 1 photon 1 electron Mass Percentage Composition Empirical Formulas and Molecular Formulas 0 Examples include glucose benzene and vitamin C MOLECULAR COMPOUNDS are molecules with atoms held together by covalent bonds EMPIRICAL FORMULA shows relative number of atoms shows the molecules and te ratios within a compound Ex CHZO for glucose MOLECULAR FORMULA shows actual number of atoms ratio is the same as in the empirical formula simplyjust a multiple of the ratio shown in the empirical formula Ex C6H1206for glucose The first things to start off with You ll need to know the molecular formula in order to understand the properties of a molecule and to understand a chemical equa on Example of mass percentage composition 0 A sample of 10g and 5 g is carbon What is the mass percentage composition of carbon 50 C Example of calculating molecular formula C VITAMIN C Sample of 8 g Analyzed results show C 327 g H 0366 g 0 436 g 0 Mass of C 327 g8 gx 100 409 H 458 0 545 0 Molecular Formula represents whole numbers of atoms that means they must be integers no decimals to make up a molecule 0 Objective finding ratio converting masses to moles 0 Since Empirical formula is relative number of atoms easiest procedure is to imagine sample mass is 100 g Q In vitamin C Mass of C is 409 g Mass of H is 458 g Mass of O is 545 g 0 Divide these masses by the atomic mass per mole 0 MOL of c atoms409 g1201 gmol 341 MOL MOL of H Atoms454 MOL MOL of O atoms 341 MOL 0 Divide by smallest value 341 so the m for CH0 becomes 1 133 1 0 Want to make 133 the smallest whole number possible so multiply by 3 0 Empirical Formula for vitamin C is C3HO3 C To calculate molecular formula we need its molar mass From mass spectrometry the vitamin C is 176 14 g mol 0 Current molar mass for empirical formula8806 g mol 3C12 4H1 3 O16 Molar mass for empirical formula less than total molar mass Need to find a factor to see how much it must increase by I Note If molar mass for empirical formula is same as the total molar mass given then the empirical formula is the molecular formula C 17614 g mol8806 gmol12 Molecular is 2x molar mass of empirical 0 Molecular formula for vitamin C is C6H806 Heisenberg Uncertainty Principle light scattered from a baseball measures both momentum and position BUT light scattered from an electron will affect electron s momentum and position pathway will change 0 distance can only be approximated from between a certain distance to distance at atomic scale measurement process influences outcome 0 one variable velocity unknown so all the momentum is uncertain stated qualitatively there is a limit on the accuracy to which the momentum and position of a particle can be known simultaneously state quantitatively Heisenberg indeterminacy uncertainty equation 0 Ap gtlt Ax 24 indeterminacy in momentum x indeterminacy in position 24 p mv momentum mass x velocity h 663 x103934 Hz 0 O O Balancing Chemical Equations Chemical equations are used to represent All chemical reactions Law Of Conservation of Mass Lavoisier 1789 Total Mass Before Total Mass After Total Number of Atoms on Left Reactants Total Number of Atoms on Right Products Chemical Equations Must Always Be Balanced Use Stoichiometric Coefficients To Balance Equ Stoichiometric Greek For Element amp Measure Why do we balance chemical equations 0 A In a chemical reaction the number and identity of each atom does not change and therefore the total number of each atom type on the left of the chemical equation the reactants must equal balance the total number of each atom type on the right of the chemical equation the products Another way of saying this is that the total mass of the reactants must equal the total mass of the products Chemical equations are balanced to show that in a chemical reaction atoms are neither created nor destroyed Chemical equations must be balanced to represent conservation of mass in a chemical reaction 2 Na s 2 H20 a 2 Moles 2 Moles gt 2 Moles 1 Mole aq Aqueous Solution CaCo3s Cozg Greek letter delta indicates the reaction is heated Eg Combustion of Butane C4H10 C4Hlog 02g C02g H20g Balance C C4Hlog 4C02g H20g Balance H H4Hlog 4C02g Balance 0 C4H10g 132 02g a 4C02g 5H20g Multiplying both sides of equation by a number does not affect the balance of each atom type 2C4H10g 13 02g gt 8C02g 8C02g 10H20g Atomic Spectra quantum world certain valuesconditions must be metallowed for it to happen light can cause excitation of an electron in an atom analysis of light given off by excited atoms shows only photons of particular energy are given off 0 this tells us electrons in an atom have specific energies each element has its own unique spectral fingerprint by which it can be identified emissionabsorptionlineatomic spectroscopy different energy level photon of different energy increase the electron to another energy level from n1 to n2 higher energy level higher E electrons will only accept photons whose energy is enough to skip 1 or more level not all light accepted must match energy differences bw energy levels spectroscopic analysis of light given off by excited atoms shows only photons of particular energy v are given off helps analyze structure and identity of atom higher energy electrons can drop to lower energy level by giving off photons that make up the difference in energy Bohr frequency condition v Ah E o AE change in energy transition 0 when the frequency of the incoming light matches the energy diff for that energy transition the light will be absorbed Hatom electron in energy level n E 3 o can ONLY be used for hydrogen 0 use v ATE to change equation to v n 0 used to find energy diff between 2 energy levels Ex En2 En1 energy amount electron has to have to jump from level 1 to 2 o emissionlineatomic spectrum or H shows diff groups of spectral lines I visible region called Balmer series I ultraviolet region called Lyman series planck constant h 663 x 103934 J o s Rydberg s constant R 329 x 1015 Hz n energy level As n gt 00 E gt O electron at a high energy level but has 0 energyit will have to emit photons to get back to ground state doing so will make it less negative more than one electron requires a more complicated model because if one electron gets excited the other electrons will too 0000


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