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Chem 101 Exam 2 Study Guide

by: randomchic12

Chem 101 Exam 2 Study Guide Chem 101

Marketplace > Louisiana Tech University > Chemistry > Chem 101 > Chem 101 Exam 2 Study Guide
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This study guide covers material that will be on exam 2. It explains in detail the material from quizzes on moodle, the study guide provided by the professor, and problems discussed in class. The s...
General Chemistry
Jessica Wasserman
Study Guide
Chemistry, atoms, wavelengths, frequency, Speed of light, Quantum Mechanics, Electromagnetic Radiation, Electromagnetic spectrum, Electromagnetic Waves, Electromagnetism, Electromagnetics, color, light, Wave Intensity, insensity, ultraviolet, photoelectric, quantization, Planck's equation, EM waves, spectra, radiation, Gamma radiation, Quantum Chemistry, rutherford, Bohr Model, Bohr, Bohr Equation, Schrodinger Equation, Schrodinger, uncertainty principle, Heisenberg uncertainty principle, quantum numbers, electron, Magnetism, Orbitals, atomic orbitals, atomic orbital shapes, Energy, periodic table, Periodic Properties of Elements, periodic, mendeleev, electron configuration, configuration, irregular electron configurations, electron configuration of ions, valence electrons, cation, anions, anomalies, ions, paramagnetic, diamagnetic, atomic radius, atomic, ionic radius, Ionization energy, ionization, electron affinity, metals, metal, non-metals, metallic, p
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This 10 page Study Guide was uploaded by randomchic12 on Saturday April 16, 2016. The Study Guide belongs to Chem 101 at Louisiana Tech University taught by Jessica Wasserman in Spring 2016. Since its upload, it has received 61 views. For similar materials see General Chemistry in Chemistry at Louisiana Tech University.


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Date Created: 04/16/16
1 Chemistry 101 Exam 2 Study Guide *side note: answers to example problems that are underlined, italicized, and bolded are answers  to the quiz questions on moodle. The problems on moodle are just explained in this study guide. Chapter 7: The Quantum­Mechanical Model of the Atom ­Electromagnetic Radiation: Wavelength & Frequency Relationship; Nature of Light *­v = c/ λ  [important equation to remember] ­frequency = v ­measured in cycles per second (1/s) or s  which is equivalent to Hz ­wavelength =  λ ­wavelengths are measured in nanometers (nm) or meters (m) so in order  to get the correct answer measurement conversions may be necessary  (depending on what the question is asking for: nm or m) 9 ­convert nanometers to meters using: 1 m = 10 nm  ­9 ­convert meters to nanometers using: 10  = 1 nm ­speed of light = c ­always a constant; equivalent to 3.00 x 10  m/s ­rearrange equation as needed to find the different variables ­Example problems: ­1. Calculate the wavelength (in nm) of the red light emitted by a neon  sign with a frequency of 4.74 x 10  Hz4 ­Answer: 633 nm ­the problem gives the frequency and the speed of light is  known since it’s constant so solve for the wavelength ­ λ = c/v ­ λ =  3.00 x 10    m/s   = 6.33 x 10 m           4.74 x 10  1/s  ­ λ = 6.33 x 10 m but the answer needs to be in nm so  convert meters to nanometers ­6.33 x 10  m    x     1 nm     = 633 nm          10  m 2 ­2. Calculate the frequency of the red light emitted by a neon sign with a  wavelength of 659.9 nm ­Answer:  4.55 x 10      s ­the problem gives the wavelength and the speed of light is already known since it’s a constant so solve for the frequency ­v = c/  λ ­just plug in what is known and solve for the unknown converting  the answer to the proper unit measurement it asks for ­3. Determine the velocity of a medicine ball (m = 10.0 kg) with a  wavelength of 1.33 x 10 m. ­35  ­Answer: 4.98 m/s ­The Electromagnetic Spectrum ­left side= longer wavelengths, lower frequency, & lower energy ­right side= shorter wavelengths, higher frequency, & higher energy 3 ­Example problems: ­place the following types of electromagnetic radiation in order of  increasing wavelength: ultraviolet light; gamma rays; radio waves ­Answer:  gamma rays    ultraviolet light   radio waves ­gamma rays are on the right side of the electromagnetic  spectrum and therefore have short wavelengths. The further left you go on the spectrum the longer the wavelengths are.  The wavelengths increase as you go left. ­the visible light color of blue has a higher frequency than green, red,  yellow and orange ­blue is closer to the right side of the spectrum where there’s a  higher frequency. Red, orange, yellow, and green are closer to the  left side of the spectrum where the frequency is lower.  ­the location of the visible range of light is between ultraviolet and  infrared ­by looking at the picture of the electromagnetic spectrum one can  determine visible light is between infrared and ultraviolet ­The Photoelectric Effect *E = hv (remember that v = c/  λ) so therefore *E = hc/  λ [important equations to remember] ­E = energy ­h = Planck’s constant ­34 ­6.626 x 10  J/s ­v = frequency ­c = speed of light ­(constant) 3.00 x 10  m/s ­ λ = wavelength ­Example problems: ­1. Calculate the energy of the green light emitted, per photon, by a  mercury lamp with a frequency of 5.49 x 10  Hz.4 4 ­19 ­Answer:  3.64 x 10      J ­frequency is given in the equation and Planck’s constant is already known so set up the equation to solve for energy. ­E = hv ­34 14 ­19 ­(6.626 x 10  J/s) (5.49 x 10  Hz) = 3.637674 x 10 ­19 ­E = 3.64 x 10  J ­2. Calculate the energy of the orange light emitted, per photon, by a neon  sign with a frequency of 4.89 x 10 Hz. 14  ­19 ­Answer:  3.24 x 10      J ­frequency is given in the equation and Planck’s constant is already known so set up the equation to solve for energy. ­E = hv ­(6.626 x 10  J/s) (4.89 x 10 Hz) = 3.24011 x 10 ­19 ­E = 3.24 x 10  J19 ­3. As a photon increases the wavelength gets shorter ­Also take note that all of these statements are true: ­The emission spectrum of a particular element is always the same and can be used to  identify the element. ­Part of the Bohr model proposed that electrons in the hydrogen atom are located in  “stationary states” or particular orbits around the nucleus. ­The uncertainty principle states that we can never know both the exact location and  speed of an electron. ­An orbital is the volume in which we are most likely to find an electron. ­Example problems worked in class on the board by the professor: ­deals with the photoelectric effect: ­1. What is the energy of a mol of photons with a wavelength of 628 nm? ­E = hc/  λ ­E =  (6.626 x 10       J/s) (3.00 x 10     m/s) (6.02 x     photons/mol) 628 x 10  m9 ­avogadro’s number: 6.02 x 10 23 5 ­used in this equation after plugging in what is known in order to  get the energy for a mol of photons since that’s what the problem  asked for ­the wavelength of 628 nm needs to be converted to meters so that is done  by multiplying 628 by 10  m ­9 ­13  ­Answer: 1.91 x 10 J/mol   Chapter 8: Periodic Properties of the Elements ­Pauli exclusion principle­ no two electrons can have the same four quantum numbers ­Ground state electron configuration for I:  [Kr] 5s         5 5 ­Element that has a ground state electron configuration of [Kr] 5s 4d : Tc 2 5 ­Unpaired electrons present in the ground state P atom: 3 ­An atom of S has 6 valence electrons ­Halogens possess 7 valence electrons ­Charge for the most stable ion of nitrogen: ­3 ­Place the following in order of increasing atomic radius: As, O, Br ­Answer: O < Br < As + ­ ­reaction that represents the first ionization of O:  O (g)    O     (g) + e ­ground state electron configuration for Br:  [Ar] 4s    3d 2     4 6 6 More Practice Problems Covering Material from Chapters 7 & 8 Answers are in red 1. Determine the frequency and energy of a photon with a wavelength of:               a. 714 nm                   v=4.202*10  Hz      E=2.78*10  J ­19 14 ­19 b. 444 nm                   v=6.757*10  Hz      E= 4.48*10  J 5 ­28 c. 582 m                     v=5.15*10  Hz         E= 3.42*10  J 9 ­25 d. 245cm                     v=1.224*10  Hz      E= 8.11*10  J 2. Dental lasers are based on neodymium yttrium aluminum garnet (Nd­YAG) crystal, with  an emission wavelength of 1066 nm.  What is the frequency of this laser?   a. 2.812 × 10  Hz 4 3. Which orbital is not possible?   a. 6s b. 4p z c. 2d xy d. 3p x e. 5d yz 4. The representation 5d indicates which values for n and l.   a. n = 5; l = 0 b.  n = 5; l = 1 7 c.  n = 5; l = 2 d.  n = 5; l = 3 5. Which of the following is a list of allowed quantum numbers for an electron?     a. n = 1, l = 1, m = 0, l  = ½ s b.  n = 2, l = 0, m = 1,l  = ½ s  c.  n = 3, l = 0, m = 0, m  = ½ l  s d.  n = 4, l = 3, m = 4,l   = ­½ s 6. List all possible quantum numbers for the following elements:   a. Carbon    i. n= 1, 2;  l=0, 1;  m= ­1, l, 1,  m = +1/2, s1/2 b. Nickel i. n= 1, 2, 3, 4;  l= 0, 1, 2, 3;  m= ­3, ­2, ­1, 0, 1, 2, 3;  m = +1/2,­1/2 l s c. Magnesium i. n= 1, 2, 3;  l= 0, 1, 2;  m= ­2, l1, 0, 1, 2;  m = +1/2, ­1s2 d. Tin i. n= 1, 2, 3, 4, 5;  l= 0, 1, 2, 3, 4; m= ­4, ­3l ­2, ­1, 0, 1, 2, 3, 4;   m = +1/2,  s ­1/2 7. Write the electron configuration for the following elements:  a. Oxygen           [Ne] 2s  2p 2 4 10 1 b. Silver              [Kr] 4d  5s 2 2 c. Silicon            [Ne] 3s  3p d. Potassium       [Ar] 4s 1 8. What is the correct electron configuration for sulfur?   a. 1s 2s 2p 3s 3p 1 5 2 2 6 2 4 b. 1s 2s 2p 3s 3p c. 1s 2s 2p 3s 3p 2 2 8 d. 3s 3p 4 9. Write the electron configuration for the following ions:   2+                          a. Mg [Ne] b. S 2­             [Ar]               c. Fe 2+           [Ar]  3d 6       ­                                  d. I [Xe] 10.  What is the correct condensed electronic configuration for Fe3+?  a. [Ar]4s 3d 6 2 3 b. [Ar]4s 3d 5 c. [Ar]3d d. [Ar]4s 3d 9  11. Which of two species below are isoelectronic?   2­ 4­ a. O , C 2­ ­ b. N , Cl c. Mg, Na d. Li , Be + 12. Which of these statements is false?   a. Atomic size decreases in going from left to right across a row (period). b. First ionization energy decreases in going down a column (group). c. Second ionization energies are always larger than first ionization energies. d. The first electron affinity is always a large negative value. e. An anion is always larger than its parent neutral atom. 13. Arrange the following in order of decreasing atomic radii   9 a. Fe, H, O, He, Na i. Na >> Fe >> O >> H >> He b. Br, Mg, S, C, F i. Mg >> Br >> S >> C >> F 14. Which is the largest atom below?   a. O b. N c. Al d. Sr 15. Arrange the following in order of decreasing ionic radii   + ­ 2­ + ­ a. Li , F, S , Na , Cl 2­  ­  ­ +  + i. S >> Cl >> F >> Na >> Li b. Mg , O , K , Br, Al ­ 3+ i. Br >> K  >> O  >> Mg  >> Al 2+ 3+ 16. Arrange the following in order of decreasing first ionization energy                    a. K, Ca, F, Al, Na i. F >> Ca >> Al >> Na >> K  17. Which atom below has the greatest ionization energy?   a. Li b. Ne c. Na d. Kr 18. Arrange the following in order of decreasing election affinity   a. F, I, O, S, Na i. F >> I >> S >> O >> Na 10 19. Which process illustrates electron affinity?   a. Li → Li+ + e­ b. O + e­ → O­ c. O + O → O2 d. Na + Cl → Na+ + Cl­


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