CHM 115 Final Exam Study guide
CHM 115 Final Exam Study guide CHM 11500 - 002
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This 11 page Study Guide was uploaded by Zack Bales on Sunday May 1, 2016. The Study Guide belongs to CHM 11500 - 002 at Purdue University taught by Chittaranjan Das,Suzanne C Doucette in Fall 2015. Since its upload, it has received 209 views. For similar materials see General Chemistry in Chemistry at Purdue University.
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Date Created: 05/01/16
CHM 115 Exam Study Guide Exam 1 A. Basic Units a. kg, m, s, mol, K, A, Pa B. Atom a. Proton, neutron, electron b. Isotope i. Different number of neutrons in atom ii. Cation – positive ion iii. Anion – negative ion c. Atoms combine molecules d. Law of Mass conversion e. Law of Multiple Proportions C. Stoich a. 1 mole = 6.022e23 b. g/mol = amu c. Steps to stoich i. Balance equation ii. Convert units iii. Use the mole ratio from the equation iv. Convert to desired units d. Limiting Reagent i. When there isn’t enough of one substance to fully react with another ii. Balance Equation iii. Determine moles of reagent iv. Use stoich to determine LR v. Calc the amount of product using LR e. Solutions i. Solute dissolves in a solvent to form a solution ii. Molarity 1. M = moles/L -- moles is moles of solute, Liters is of solution iii. M1V1 = M2V2 1. Use this only for dilutions f. Ho many moles of Na atoms corresponds to 1.56e21 atoms of Na? i. (1.56e21 atoms)*(1 mole/6.022e23 atoms) = 2.59e-3 moles Na g. How many grams? i. (2.59e-3 mol)* (22.99g/1 mole) = 0.05 g Na ii. 22.99 is the atomic mass of Na D. Acid Base Reaction a. Acid H+ (proton) i. HCl b. Base OH- (hydroxide) i. NaOH c. Acid – Base reaction neutralization d. Strong Acids i. HCl, HBr, HI, HNO3, H2SO4, HClO4 e. Weak Acids i. HF, H3PO4, CH3COOH f. Strong Bases i. LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 g. Weak Bases i. NH3 h. Hx H+ (aq) + X- (aq) i. Acid reaction i. MOH M+(aq) + OH-(aq) i. Base reaction E. Polyatomic ions a. NH4+, H3O+, CH3OO-, CN-, OH-, HCO3-, more in review packet F. Dimensional Analysis a. 55.0 mph to cm/s i. (55.0 mi/hr)*(5280ft/1mi)*(12 in/1ft)*(2.54cm/1in)*(1hr/3600s) = 2460 cm/s 1. All units except what you want should cancel G. Sig Figs a. All nonzero digits are significant b. Zeros between nonzero figures are significant c. Leading zeros on the left of the decimal d. 12.6 to 2 sig figs = 13 e. 11.5 odd digit before dropped number round up to 12 f. 12.5 even digit before dropped number round down to 12 g. Aka round to nearest even number with appropriate sig figs h. 0.12545 4 sig figs .1254 i. Multiplication fewest sig figs i. 12.54*2.11 = 26.459426.4 j. Addition fewest decimal places i. 12.54+2.115 = 18.055 18.1 H. Energies a. Potential – stored energy b. Kinetic – proportional to stored energy i. K=(1/2)m(v^2) ii. E = K + U 1. U = Potential I. Pressure a. 1 Pa = 1 N/(m^2) b. 1 atm = 101325 Pa c. 1/760 atm = 1 torr = 1 mm Hg = 133.322 Pa J. I deal Gas Law a. PV = nRT b. P = pressure, V = volume, n = moles, R = 0.0206 (atm*L)/ (mol*K), T =temp in K c. K = 273.15 = degrees C K. Daltons Law of Partial Pressures a. Ptot = P1 + P2+ P3 + … b. P=(nRT)/V c. Ptot =( (n1 + n2 +n3+ …)RT) / V L. Nuclear Chemistry a. Nucleon, nucleus, nuclide b. Hydrogen Isotopes i. 3H He + β c. Chemical vs Nuclear Reactions i. Chemical 1. Never changes identity 2. Electrons 3. Small energy change ii. Nuclear 1. Changes identity of atom 2. Nucleus 3. Large energy change d. Decay i. Alpha 1. α = He 2+ 2. 23U α + 234Th ii. Beta Decay 1. β = -1e 2. 23Th β + 234Pa iii. Positron emission 1. β = 01e 2. 3K 0 e + 38Ar 1 iv. Electron Capture 1. 19Hg + 0-1 195Au v. Gamma 0 1. 0ϒ emitted usually accompanies another reaction e. Band of Stability i. Shows modes of decay ii. Above = beta decay iii. Below = positron emission or electron capture iv. Past band = alpha decay v. Use n/p ratio to find where the atom lays on the BoS f. Geiger counter detects radiation g. Half life i. Rate of decay = dN/dt = -kN ii. Integrated rate = ln(N/No) = kt iii. k = ln(2)/t 1. t = half life iv. If Pu-293 has a half life of 24000 yrs, starting with 1kg, after 96000 yrs we have 1/16kg left 1. 96000/24000 = 4 half lifes 2. ½ * ½ * ½ * ½ = 1/16 v. t = 4.5e9 find k 1. ln(2)/4.5e9 = 1.54e-10 h. Mass defect i. N.B.E = Nuclear Binding Energy 1. Energy that keeps nucleus together ii. NBE = delta E iii. Delta E = mc^2 1. C = 3e8 iv. 0.1123 amu * (1.66e-27kg/1 amu) * (3e8) = 1.68e -11 J v. Greater NBE = more stable atom i. Fusion i. Produces heavier nucleus from 2 lighter nuclei ii. Mass is less than 56 j. Fission i. Produces 2 lighter nuclei from heavier nucleus ii. Mass is greater than 56 M. Atomic Structure a. Short wave lengths have more energy b. High amplitudes are brighter c. Wavelength (nm) d. Frequency (s-1 = Hz) e. C=wavelength * frequency f. Longer wavelengths yield low frequencies g. Energy increases as wavelength shortens i. E = hv 1. H = 6.626e-34 2. v = frequency h. visible light ranges from 400 nm – 750 nm i. less than 400 nm = ultra violet, xray, and gamma rays ii. more than 750 nm = infrared, microwave, and radio waves i. Photoelectric effect i. Electrons in metals exist in different and specific energy states ii. Photons whose frequency matches or exceeds the energy state of the electron will be absorbed iii. If the photon energy (frequency) is less than the electron energy level, the photon is not absorbed iv. The electron moves to a higher energy state and is ejected from the surface of the metal v. The electrons are attracted to the positive anode of a battery, causing a flow of current j. Each element emits an electron at different wavelengths producing different colors of light k. Rydberg equation i. 1/wavelength = R(1/ni^2 – 1/nf^2) ii. n = orbital l. Bohr’s model i. 2d model of an atom electrons are in specific orbitals m. Quantum Mechanical Model i. H = ii. Probability distribution 1. Electrons have a probability of being found at distances from nucleus a. Further = less likely to be found b. Still possible to be found at outer orbitals Exam 2 Quantum Numbers o n – (1-7) o ℓ - (n-1) down to zero o m - (-ℓ to ℓ ) o m - +1/2 s NRG Orbitals o Filling orbitals Add electrons to lowest NRG first 2 e per orbital with opposite spins - Fill each with one e before filling with 2 Types of electrons o Outer – highest n o Valence – involved in bonds o Core – lower n levels Composition of Solutions o Ions are stabilized by solution o Non electrolytes retain structure when dissolved o Gases become less soluable as temp increases o Solids become more soluble as temp increases o Solute + solvent solution Spectroscopy o Molecules absorb light at different energies o Absorbance = εℓc Atomic Radii o Distance between atoms o Ionization Energy Energy needed to remove electron o Electron Affinity Change in energy accompanying additional electrons Larger EA = more energy Metallic o Main group – ionic oxides, basic o Main group nonmetals – covalent oxides, acidic o Amphoteric oxides – acid or base o Oxidation loses electron, reduction gain electrons Bonds o Cations are smaller than parent atoms o Anions are larger than parent atoms o Magnetic Properties Spinning electrons generate magnetic field Paramagnetic: one or more d orbital with unpaired electron, weak Diamagnetic: only one paired electron set not attracted by magnetic field o Chemical bond Ionic – metal + nonmetal Covalent – nonmetal + nonmetal Metallic – metal + metal Atoms are unpaired in ionic and metallic Enthalpy of Rxn ΔH o ΔH = sum ofΔHf products – sum of ΔHf reactants o ΔHoverall = ΔH1 + ΔH2 + ΔH3 + … o Lattice Energy -ΔHlat = - ΔH final - (ΔH1 + ΔH2 + …) o ΔHrxn = sum Bonds formed – sum bonds broken Exam 3 Lewis Structures o Show pairs of electrons in atoms and how they are shared o Octet rule: must have 8 elctrons Exceptions: hydrogen, Lithium (2e), Sulfate (6e) o Steps Count valence electrons Determine central atom Draw bonds Distribute remaining electrons o Formal charge = initial e – final e VSEPR o Valence Shell Electron Pair Repulsion Pairs want to be as far apart as possible Double and triple bonds count as one electron group Electron groups determine geometry 2 electron groups = linear 3 electron groups = trigonal planar or bent 4 electron groups = tetrahedral, trigonal pyramidal, or bent 5 electron groups = trigonal bipyramidal 6 electron groups = octahedral o Electronegativity increases up and right along periodic table o Large difference = more polar o Add up dipoles to get molecular dipole Intermolecular Forces o Ion > dipole> induced dipole o Ion dipole is weak alone but strong together o H-bond acceptor: N, O, F o H-bond donor: H-N, H-O, H-F Orbitals o S+P = sp o S+p+p = sp 2 o S+p+p+p = sp 3 3 o S+p+p+p+d = sp d o #s+#p+#d o Number of electrons groups = # bonds + # lone pairs 3 atoms + 1 lonne pair = 4 e groups -> sp 3 o Single bond = sigma (σ) bond o Double bond = sigma bond and pi (π) bond Hydrocarbons o Skeleton formula gives structure and bond angle but not atoms o Isomers: some molecular formula but different structure o Alkane = single bond o Alkene = double bond o Alkyne = triple bond o Alcohol C-OH o Functional groups Polymers o Degree of polymerization = # of monomers/polymers (R) n o Branching decreases density and melting point Less contact lower IMF lower MP Less molecules/volume lower density Higher MW Higher MP Linear Higher MP o Addition vs Condensation Addition: No loss of molecules and few monomer are reactive Condensation: lose small molecules and all monomers are reactive IR Spectroscopy o V = 1/(2πC) * √(f[m1+m2]/m1*m2) F = force constant M= mass of atom F increases V increases More bonds = more f E=hv=hc/λ Wave number = V/C Energy highlow Wavelenght short long UV Visible IR Microwave Post Exam 3 Proteins o Peptides: linear amide backbone o Amino acids o Ionic side chains form strongest interactions with water o Non polar has weakest/ avoid water o Primary: sequence of amino acids Backbone has H-bonding sites o Secondary: arrange into sheet or helix Alpha helix: H-bonds vertical to axis Beta sheet: H-bonds between strands o Tertiary: 3-D shape o Quaternary: arrangement of chains into protein More than one chain involved Lipids, Solubility, electrolytes o Saturated fatty acid –strong chain no double bonds o Trans fatty acid – double bonds, strong chain o Cis fatty acid – bent chain o C =xk H,cpx o Electrolyte = electricity + carrier o strong electrolytes are salts (ionic bonds) and strong acids and bases NaCl, NaOH, MgCl ,2AgNO 3 o weak electrolytes are weak acids and weak bases o nonelectrolytes are covalently bound molecules CCl4, CH3OH o Moles dissolved * van’t Hoff factor = total moles dissolved particles Colligative Properties o o Raoult’s Law = t o ΔT =b m b K=boiling/freezing point Solid State Chemistry o Single crystals have one large order, maximize bonding and IMF o Polycrystalline solids have multiple smaller orders o Amorphous solids have no order o Unit cells Simple cubic Body-centered cubic Face-centered cubic Corner atom = 1/8 Face atom = ½ Coordination numbers Simple cubic = 6 Bcc = 8 Fcc = 12 Unit cell = atoms per cell/unit cell volume A = edge length r = radius Simple cubic r=a/2 Fcc/bcc a +b =r 2 o Tetrahedral coord =4 o Octahedral coord = 6 o Cubic coord = 8 Advanced Structures o Conductors have no energy gap between filled and unfilled orbitals; Semiconductors have a small gap Insulators have a large gap. o Doping a semiconductor by adding a few atoms with either more or less valence electrons results in formation of : o an n-type semiconductor (more electrons) that conducts negative charges easily, or a p-type semiconductor (less electrons) that conducts positive charges easily.
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