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Intro Chemistry - Study Guide Exam 2

by: Alexis Tate

Intro Chemistry - Study Guide Exam 2 13699

Marketplace > Appalachian State University > Chemistry > 13699 > Intro Chemistry Study Guide Exam 2
Alexis Tate

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This study guide compiles everything from chapter 7, chapter 3, and some of chapter 4. Also, notes from previous chapter were included.
Intro Chemistry I
Alexander Schwab
Study Guide
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This 12 page Study Guide was uploaded by Alexis Tate on Friday September 30, 2016. The Study Guide belongs to 13699 at Appalachian State University taught by Alexander Schwab in Fall 2016. Since its upload, it has received 44 views. For similar materials see Intro Chemistry I in Chemistry at Appalachian State University.


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Date Created: 09/30/16
Intro Chemistry: Exam II Study Guide Chapter 7  Know how to usethe equation for light:c =λ(v) 3 o c = speed of light(3x 10 m/s) o λ = wavelength (meters) o v =frequency (1/s or Hz)  *Wavelength and frequency are inversely proportional (as onegoes up the other goes down) o ExampleProblem:  Thewavelengthofasatellitesignalis656nm. What isits frequency? c =λ(v) 8 3.00 x 10 m/s= 65-9m(v) 656 nm= 656 x10 m 3.00 x 10 m/s 656 x10 m9 14 v = 4.57 x 10 Hz (1/s)  Equation for energy ofaphoton: E=h(v) o E= energy (joules) o h = Planck’s constant (6.626 x10 -34J x s) *no need to memorize this number o v =frequency (Hzor 1/s) o ExampleProblem:  A particlegivesoffanenergyatafrequencyof3.45 x10 Hz. How muchenergy didit giveoff? E= h(v) E= (6.626x 10 J(s))(3.45x10 Hz) E= 2.29 x 10 J1  Equation for electron (e)in a particular orbit:E=(-hcR )(1Hn ) =(-2.18x10 J)(1/n )8 2 -34 o h = Planck’s constant (6.626 x10 J x s) o c = speed of light(3x 10 m/s) o R =Rydberg’s constant (109678 cm )* no need to memorize this number H  Equation for the changein energy ofan electron that absorbsor releasesenergy:  ∆E= E final initial -  Keep in mind you areusing the energy equation that describes an e in a particular orbit o Example Problem:  Calculatethe ∆Ewhen e inH goes from n = 1 to n = 3.  n = 1is the initial  n = 3is the final 2 -18 2 E= (-hcR )H1/n )-18-2.18 x20 J)(1/n ) -18 E i(-2.18 x 10 J)(1/(1) )=-2.18 x 10 J E f(-hcR )H1/n ) =(-2.18 x10 J)(1/(3) )=-2.42 x 10 J -19 ∆E= E -f i ∆E= -2.42x10 J– (-2.18x10 J) -18 -18 ∆E=1.94x10 J *The answer being positive should make sensebecausetheelectron absorbed energy to getfrom the firstenergy level to the third energy level. Ifit was negative,that means the electron releasedenergy and jumped down anenergy level.  What is the energy ofthephoton absorbedduring this transition? Photon’s energy =l 1.94 x 10 Jl = 1.94x10 J -18 *This answer should ALWAYS bepositive becauseitdoes not have direction  Quantum Mechanics o Principle quantum number (n)  Any positiveinteger  All orbitals with the samevalue of n arein the sameshell (sameenergy level)  Energy levels on the periodic table run from 1 to 7 o Secondary Quantum Number (l)  Divides shells intosubshells  Values of l are limited by the valueof n  l is aninteger between 0and (n- 1)  This means l canNEVERequal n  l is always a positiveinteger  Values of l are denoted by a letter code Valueofl 0 1 2 3 4 5 Letter s p d f g h o Magnetic Quantum Number (m) l  Divides subshells intoindividual orbits  m lalues arelimited by l  m is aninteger between - l and + l l  So, when n = 3, l =2, m =l2, -1, 0,1, 2 o Spin Magnetic Quantum Number (m ) s  Spin is quantized "up" or "down"  m s+1/2 or -1/2 o Key Principles:  PauliExclusion Principle -no two e scan haveidenticalvaluesfor all4 quantumnumbers -  Hund'sRule-e s arespaced out asmuch as possiblebeforethey begin to pair up - -  Aufbau Principle - e sfillthelowest energy levelfirst;e s areusually in predicted spot 95% ofthetime o Example Problem:  What is the e configuration for *Forshorthandalwaysstart withthe Na? 2 2 6 1 noblegasrightabovetheatom's row. 1s , 2s ,2p ,3s ThisappliestonoblegasessoDONOT Shorthand:[Ne]3s 1 say [Ne],say[He]2s ,2p 6 *ALWAYS writeallofthe  How many unpaired e s? - "up" arrowsfirstbefore puttinginany "down" arrows.So forexample,in 2p thereshouldbe3"up" arrows beforeyoufillin3 1s 2s 2p 3s "down"arrows. - 1unpaired e o Whene sare unpaired,thewholeatomactsas a magnet * Electronsalwayshave  Attractedto amagnet oppositespinsinan orbitalindicatedbythe  Paramagnetic up and downarrows o Whene sare paired,e smagneticeffectscancelout  Notattracted to amagnet(repels)  Diamagnetic (Rough)PeriodicTable *Each rowhas a valueofn (Hydrogenwouldstartat n =1) and goesfrom1 to 7 l = 0 (s orbitals) l = 2 (d orbitals) l = 1 (p orbitals) l = 3 (forbitals) This is where the alkali and alkali earthmetals are (Groups 1A and 2A) This is where the nonmetals and metalloids are (Groups 3A through 8A) This is roughly where the transition metals are (Keep in mind you don’t start - 2 naming them in the e configuration until after 4s and then itjumps down to 3d) Theseare the two bottom rows (lanthanides and actinides)of the periodic table. (Keep in mind you don’t start naming them inthe e configuration 2 1 until after 6s ,5d and then it jumps down to 4f)  EffectiveNuclear charge - o Nucleus is shieldedby core e s o Z (neffear charge)  Z increasesfromleftto right onthe periodictableand *Zeffjustthe attraction eff of electronstothenucleus increasesas yougo up a column (protons)inanatom. Z iseff aFORCE  Trends in Atomic Radii o Atomic radii decreasesgoing up ina givengroup  Number of e s increases as yougo down a column o Atomic radii decreasesas you go left to right ina givenrow  Zeffcreases causing smallere cloud  Ionic (Ions) Radii o An anion will be largerthan its parent atom  Adding e s to an atom increases e repulsion (makes it bigger) o A cation will be smallerthan its parent atom  Removing e s from an atom decreases e repulsion (makes it smaller)  Ionization Energy (IE) o The minimum energy required to remove an e from anatom o I1- firstIE  Energy required to remove the first e - *IEis justdescribinghow  Na -->Na + e - much ENERGYit would - o I2- secondIE taketo removeane from  Energy required to remove the second e - an atom + 2+ -  Na -->Na +e  IEfor lightelements o I1< I2<I 3 o Core e s are very difficult to remove  1stIE trends o I1increasesfrom left to right,bottomto top  Larger Z effes it harder to remove e s -  Smaller atom alsomakes it harder to remove e s -  Fully filledsubshells arestablesoraises IE  1/2 filledsubshells arestablesoalsoraises IE  Electron Affinities - o Energy change associatedwithadding e s to gaseous atoms  If ∆ E >0, then the atom is unstableupon adding an e -  - If ∆ E <0, then the atom is stableor more stableupon adding an e o Rough trend: increasesfrom left to right,bottomto top (Rough)PeriodicTable Black arrows (increases):IEeff, ElectronAffinity Red arrows (decreases):atomicsize Chapter 3  Mole o 1 mole (mol) of atoms = 6.0221423 x10 atoms (Avogadro’s number) o Molar mass  What is the molar mass of sugar(C 6 12)?6 C = 6x (12.01g) Add thesetogether to H =12 x (1.0079g) O = 6 x(15.999g) getmolar mass (g/mol) 180.155 g/mol  What is the mass of 0.350 mol of sugar? - *Molar Mass inmost problems will be your conversion tool to get grams to moles or viceversa 0.350 mol 180.155 g = 63.1 g 1 mol  What is the mass of one sugarmolecule ingrams? -*Avogadro’s number inmost problems will be your conversion tool to getfrom moles to whatever you’re solving for. (i.e.molecules, atoms, etc.) -22 180.155 g 1 mol = 2.992 x 10 g mol 6.022 x 10 molecules * Differencebetween finding TOTAL number ofmoleculesand TOTAL number ofatoms- When calculating formolecules usuallyall you have to do is multiply Avogadro’s number by the number of moles of the compound. For atoms, you need to add up the total number of atoms in a compound and then multiply that total by Avogadro’s number. This phrasing may help * In anelement likeCl,there are Avogadro’s number of atoms (6.022 x 10 atoms of Cl make up the molecule Cl) * In one mole of a compound likesugar,there are Avogadro’s number of molecules 23 (6.022 x 10 molecules of sugarmake up one mole of sugar) o Example Problem:  Find the total number of molecules in2 moles for sugar. 23 = 1.2044 x 10 moleculesso, 2 mol 6.022 x 10 molecules 24 1 mol 1 x 10 moleculesb/c of1sig fig (2 mol)  Find the total number of atoms in 2 moles for sugar. - we alreadyknow the total number of molecules intwo moles of sugar - to getto atoms we need to add the total number of atoms inone molecule of sugar(6C + 6O + 12H =24 atoms) 6.022 x 10 molecules  24 atoms 1 mol = 1.44528 x 10 atomsso, 1.4 x 10 atomsb/c of2sig figs(24 mol)  Mass Percent Total mass ofelement atoms x 100 =% element Formula weight o Example:  What is the mass percentage of C in sugar? 6(12.011g) X 100 =40.002% 180.155g/mol  Empirical Formula o Example of Combustion Analysis  Let’s sayyou place 2.200 g of anorganic compound which should contain C, H,and O. CHanalysis tells us thereare 0.8247 g of C and 0.2768 g of H. What is the empirical formula for this compound? *For any question dealing with finding the empirical formula, always get eachindividual element in the compound to moles. If the problem gives you percents, just changethe percent signto grams (g).(Reasonwhy was explained innotes) 0.8247 g C 1 mol C = 0.06866 molC 12.011 g C 0.2768 g H 1 mol C = 0.2746 molH 1.0079 g H - All of the elements masses mustequal 2.200 g so, Mass ofO =2.200g –0.8247g C – 0.2768g H= 1.098]5g 1.0985g O 1 mol O = 0.06866 We’ll keep the 15.999g O (0.0687) whole valuefor molO accuracybut keep in mind sig figs. *After getting each element to moles,divide each answer by the element with the smallestnumber of moles. In this particular problem moles of O = moles of C so itdoes not matter which number you pick. 0.06866 mol C =1 C 0.06866 mol C 0.06866mol O = 1 O 0.06866 mol C 0.2746 mol H =3.999 H,soapproximately 4 0.06866 mol C EmpiricalFormula=CH O 4 *For any decimals always round to the nearest whole number unless itis in the middle between two numbers (~1.5)  This littlesaying may help you to remember the steps: Percent toMass Mass toMole Divideby thesmallest Round to thewhole  Mass ofProducts to Mass ofReactants o Example:  You drop 10.0 g of chromium into an enormous vat of HCl.Predict the amount of chromium(III) chloride that would be made. Molar Mass:Cr= 51.996 Cl = 3(35.45) 158.35 g/mol 10.0 g Cr 1 mol Cr 2 mol CrCl3 158.35g CrCl 3 51.996 g Cr 2 mol Cr 1 mol = 30.5 g CrCl3  Limiting Reagent o The limiting reagent determines how much of a substance you can make o Example:  You have 14.0 moles of H a2d 3.7 moles of O . Ho2many moles of H O 2 canyou make based off of this reaction? 2H + O  2H O 2 2 2 * How many moles ofO do y2u need to react with 14.0 moles of H ? 2 - Oneway to find the limiting reagent is to seewhich reactant cannot completely react with the other 14.0 mol H 2 1 mol O 2 Since you do not have = 7.0mol O 2 enough moles ofO to 2 2 mol H 2 react with 14.0 moles of H 2O 2s the limiting 3.7 mol O 2 2 mol H 2 = 7.4 molH O2 reagent 1 mol O 2  How many moles of H O c2uld you produce with 14.0 moles of H and 8.02moles of O 2 Which is the limiting reactant and which is the excess reactant? 2H 2 O 22H O 2 - Another way to find the limiting reagent is to convert both reactants to the desired product and seewhich produces a fewer amount 14.0 mol H 2 1 mol H 2 = 14.0 molH O2 2 mol H 2 8.0 mol O 2 2 mol H 2 = 16mol H O2 1 mol O 2 * Thelimiting reactant is H 2nd the excess reactantis O 2 o To find amount excess reagentleftover:  First, convert the limiting reagent’s amount (usuallymoles or grams)to the excess reagent’s amount (moles or grams)  Then, subtract that answer from the amount of excess givenin the problem  Example (Using the ammonia example from chapter 3 notes): o How many moles of excess N2is leftover from the reaction? 7.0 mol H2 1 mol N2 = 2.5mol N 2 3 mol H2 3.0 mol N2– 2.5 mol N 2 0.50 molof N le2t over  Reaction Yield o Actual Yield  Amount of desired product actuallyproduced by a chemical reaction (mass) o Theoretical Yield  Amount of desired product that is predicted to be produced by a chemical reaction o Percent Yield Percent yield = Actual Yield x 100 Theoretical Yield o Example:  Basedoffof lastexample’s answer, the predicted amount of silverthat would be produced was 1.59 g.The reaction actually produced 1.41 g of Ag (s). What was the theoretical yieldand percent yieldfor this reaction? 1.41 g x100 = 88.7% 1.59 g Chapter 4  Aqueous Solution o Homogenous mixture o Anything that can be dissolvedinwater is aqueous  Ions  Ionic substances  Some molecular substances (acids) o Solvent – thing that does the dissolving o Solute –thing that is dissolved  Solution Concentration o Describes givenamount of solute in solvent  Concentrated – means highconcentration  Dilute –means low concentration  Saturated – means concentration is as highas possible  Unsaturated – not saturated, means concentration is lower than a saturated solution  Supersaturated Solution • Are unstable • Too much stuff is dissolvedandcan leavethe solution atany moment • The precipitation canbe started by dust, vibration, or other things  Electrolytes o Substances that form hydrated ions in aqueous solutions (conduct electricity) o Write the chemical equations for the dissociationof the following:  Ammonium sulfate + - (NH 4 2O (4q)  2NH (aq)4+ SO (aq) 4  Nonelectrolytes o Compounds do not dissociateinto ions when dissolvedin water o Most molecular compounds  Methanol (CH OH3molecules aredispersed by H O (l) bu2 are not dissociateinto ions CH 3H(l) CH OH(a3)  Precipitation Reactions o Pb(NO ) (3 2reacts with KI (aq)and produces PbI (s): 2 o Pb(NO ) (3 2+ 2KI (aq)  PbI (s)+ 2KNO (aq) 3 + - o Keep in mind that KNO exis3s as twoseparate ions K and NO inthe solu3ion  Ionic Equations: o This is calledthemolecular equation: Calleda precipitate Pb(NO )3 2q) +2KI (aq)  PbI (s)2+2KNO (aq) 3 o This is the ionic equation: 2+ - + - + - Pb (aq) +2NO (aq3 +2K (aq) + 2I (aq)  PbI (s)+ 2K 2aq) + 2NO (aq) 3  Emphasizes that compounds existas ions ina solution  K and NO a3e spectator ions  Spectator ions canbe removed by yielding a net ionic equation  Pb (aq) +2I (aq)  PbI (s)2 *To write a net ionic equation all you need are the ions that make up the precipitate Extra Info  Know your polyatomic ions (Memorize)  Know how to name compounds o Ionic Compounds  Cation +anion  Second name (anion) ends in -ide  Na + Cl =NaCl  Sodium chloride  Only compounds with multiple charges geta roman numeral  Fe +O = Fe O 2 3  Iron(III) oxide  Fe +O = FeO  Iron(II) oxide o Molecular Compounds  Anion +anion  Useprefixes  PCl  Phosphorus monochloride (do not use the mono- prefix on the firstname)  C 2l 6  Dicarbon hexachloride o Polyatomic compounds  Cation/anion + polyatomic ion  NH +O = (NH ) O 4 4 2  Ammonium oxide  Na + SO 2-4=Na 2O 4  Sodium sulfate


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