Exam 1 study guide
Exam 1 study guide CHM 11500 - 002
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Popular in Chemistry
This 4 page Study Guide was uploaded by Zack Bales on Thursday February 4, 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 133 views. For similar materials see General Chemistry in Chemistry at Purdue University.
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Date Created: 02/04/16
CHM 115 Exam Study Guide 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 3 3 i. H 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+ 238 4 234Th 2. U α + ii. Beta Decay 1. β = -1e 234 234 2. Th β + Pa iii. Positron emission 1. β = 01e 38 0 38 2. K 1e + Ar iv. Electron Capture 1. 19Hg + 0-1 195Au v. Gamma 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
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