YOUR NAME (last name, first name)...answers... FALL 2018 MIDTERM. (Total number of pages = 10) (Total points = 110) (Total time = 110 minutes)

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YOUR NAME (last name, first name)...answers... FALL 2011 MIDTERM. (Total number of pages = 10) (Total points = 110) (Total time = 110 minutes)

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CHEM 14A Instructor: Dr. Laurence Lavelle YOUR NAME (last name, first name)...answers... STUDENT ID#... FALL 2018 MIDTERM (Total number of pages = 10) (Total points = 110) (Total time = 110 minutes) **Carefully remove the last two pages: Constants and Formulas, and Periodic Table.** YOUR DISCUSSION SECTION... YOUR TA s NAME.. WRITE IN PEN Do not use white-out. Show all your work. Check units and significant figures. Box your final answer. This is a closed book exam: Only a pen and simple scientific calculator are allowed. No other material allowed. Good Luck 1

Do not write on this page. QUESTION SCORE 1 2 3 4 5 6 7 8 TOTAL (max 110) 2

Q1. Compounds are being developed that can store hydrogen in vehicles. One reaction being studied for hydrogen storage is: Li3N(s) + H2(g) LiNH2(s) + LiH(s) (Concept question) A. Why do we balance chemical equations? () Conservation of mass in chemical reactions. Therefore the number (or moles) of atoms must be the same for the reactants and products. (Concept question) B. Write the balanced chemical equation for the above reaction. (4pt) Li3N(s) + 2 H2(g) LiNH2(s) + 2 LiH(s) C. How many moles of H2 are needed to react with 1.5 mg of Li3N? (4pt) Moles of H2 needed to react with 1.5 mg Li3N: = 1.5 10-3 1 mol Li g Li3N 3N 2 mol H2 = 8.6 10-5 mol H2 34.83 g Li3N 1 mol Li3N D. Calculate the mass of Li3N that will produce 0.650 mol LiH. (4pt) Mass of Li3N to produce 0.650 mol LiH: 1 mol Li = 0.650 mol LiH 3N 34.83 g Li3N = 11.3 g Li3N 2 mol LiH 1 mol Li3N -1sf -1units 3

Q2. Gallic acid can be used to prevent dysentery and is known to contain carbon, hydrogen, and oxygen. A chemist wanting to determine the molecular formula of gallic acid burned 1.000 g of the compound in an elemental analyzer. The products of the combustion were 1.811 g of CO2 and 0.3172 g of H2O. A. Determine the empirical formula of the compound. (1) 1.811 g CO 2 12.01 g C 1 2 44.01 g mol 1 mol CO2 = 0.041 15 mol CO = 0.4942 g C 0.3172 g H2O 2.0158 g H 1 2 18.02 g mol 1 mol H2O = 0.017 62 mol H O = 0.03552 g H 1.000 g (0.4942 + 0.03552)g = 0.4703 g O 0.4703 g O = 0.029 39 mol O 1 16.00 g mol Dividing each amount of moles by 0.02939 gives a ratio of C:H:O of 1.4:1.2:1 Multiplying through by 5 gives whole numbers as required for an empirical formula: CHO 7 6 5 3pt B. The molar mass was found to be 170.12 g.mol 1. What is the molecular formula of gallic acid? () Since the molar mass of this empirical formula matches the given value of about 170 g/mol, the molecular formula is the same as the empirical formula. CHO 7 6 5 4

Q3. When ultraviolet radiation is directed at an atom or a molecule, electrons can be ejected from the valence shell, and their kinetic energies are measured. Because the energy of the incoming ultraviolet photon is known and the kinetic energy of the outgoing electron is measured, the ionization energy, I, can be deduced from the fact that the total energy is conserved. (Concept question) A. Explain, comment, and show that the speed, V, of the ejected electron and the frequency, ν, of the incoming radiation are related by E = h ν = I + ½ mev 2. (5pt) The energy of the photon entering, Etotal, must be equal to the energy to eject the electron, Eejection, plus the energy that ends up as kinetic energy, Ekinetic to give: Etotal = Eejection + Ekinetic Etotal for the photon is h ν Eejection corresponds to the ionization energy, I. Ekinetic = ½mev 2 where m is the mass of the object and v is its velocity. B. Calculate the ionization energy of a rubidium atom, given that radiation of wavelength 58.4 nm produces electrons with a speed of 2450 km.s -1 ; recall that 1 J = 1 kg.m 2.s -2. (1) E total = hν = hcλ 1 = (6.62608 10 34 J s) (2.99792 10 8 m s 1 ) (58.4 10 9 m) 1 = 3.401 10 18 J = E ejection + E kinetic for equ for converting nm for answer E kinetic = 1 2 mv2 = 1 2 (9.10939 10 28 g) (2450 km s 1 ) 2 = 1 2 (9.10939 10 31 kg) (2.450 10 6 m s 1 ) 2 = 2.734 10 18 kg m 2 s 2 = 2.734 10 18 J for equ for converting km.s -1 for answer 3.401 10 E 18 ejection J = E ejection = 6.67 10 + 2.734 10 19 J 18 J 3pt for answer -1units 5

Q4A. Calculate the velocity of an electron with a wavelength of 2.00 10-10 m. (5pt) λ = h/p = h/(m v) v = h/(me λ) = 6.62608 10-34 J s / (9.1 10-31 kg 2.00 10-10 m) = 3.64 10 6 m.s -1 - units Q4B. (Concept questions, 3pt each) (9pt) Which of the following statements regarding electromagnetic radiation is true? E A. Electromagnetic radiation with a wavelength of 400 nm travels faster than that with a wavelength of 600 nm. B. The frequency of electromagnetic radiation determines how fast it travels. C. Electromagnetic radiation with a wavelength of 400 nm has a frequency that is smaller than that with a wavelength of 600 nm. D. Electromagnetic radiation with a wavelength of 600 nm travels faster than that with a wavelength of 400 nm. E. Electromagnetic radiation with a wavelength of 600 nm has a frequency that is smaller than that with a wavelength of 400 nm. D Which of the following experiments most directly supports de oglie's hypothesis of the wave nature of matter? A. black-body radiation B. the photoelectric effect C. α-particle scattering by a metal foil D. electron diffraction by a crystal E. the emission spectrum of the hydrogen atom Consider a hydrogen atom in an excited state with the quantum number n = 9. Which of the following statements is incorrect? C A. This hydrogen atom is higher in energy than the a hydrogen atom in which n = 8. B. If the quantum number l = 3, the magnetic quantum number ml can have seven different values. C. The lowest value that the magnetic quantum number ml can take is -9. D. The orbital angular momentum quantum number l can have any of the values 0, 1, 2, 3, 4, 5, 6, 7, 8. 6

Unless stated otherwise, in the following concept questions give the abbreviated (short) electron configuration. Q5A. Which member of each pair has the lower first ionization energy? Circle your answer. (3pt) (i) Ca or Mg (ii) Mg or Na (iii) Al or Na (i) Ca (ii) Na (iii) Na Q5B. Give the full ground-state electron configuration for bromine. (4pt) bromine, 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 5 Q5C. Write the ground-state electron configuration for Cr + and give the first two quantum numbers for the last electron. (5pt) [Ar]3d 5 3pt n = 3 l = 2 7

Q6A. Microwave ovens heat food by exciting water to a higher vibration mode. A typical microwave operates at a frequency of 2.45 GHz (1 GHz = 10 9 Hz). What is the energy and wavelength of this radiation? (6pt) E = hν = (6.626 x 10-34 Js) * (2.45 x 10 9 s -1 ) = 1.62 x 10-24 J λ = c/ν = (3.00 x 10 8 m/s)/(2.45 x 10 9 s -1 ) = 0.122 m - units -1sf (Concept questions) Q6B. What ion with a +1 charge is predicted to have the following ground-state electron configuration? [Ar]3d 10 4s 2 4p 3 (3pt) Se + Q6C. In class we discussed in detail the relationship between the quantum numbers n, l, ml, and ms. For an electron in a d-orbital what values can ml have? () -2, -1, 0, 1, 2 Q6D. Which specific subshell will accommodate an electron that has the following set of quantum numbers? (3pt) n = 4, l = 2, ml = +1, ms = +1/2 C A. 2f B. 4s C. 4d D. This is not a possible combination of quantum numbers. 8

Chemistry Community is about to reach 7 million page views this quarter. Many student questions and student-student discussion has centered on figuring out the structures and bonding in molecules as this is a major topic in Chem 14A. Q7A. Draw the most stable Lewis structure for each of the following compounds: P5 (4pt) P O3 (4pt) O O O - Q7B. The following Lewis structure was drawn for a period 3 element. Identify the element and explain your reasoning in detail. O (6pt) Cl E F structure has 24 electrons 7 from Cl, and 7 from F, and 6 from O E needs to have 4 valence electrons Identity is Si 3pt 9

In class there have been many excellent student questions. Q8A. Nitrite is important in biochemistry as a source of the potent vasodilator nitric oxide. Draw the most stable Lewis for nitrite, NO2. Predict the N-O bond lengths in NO2 given the N-O and N=O bond lengths of 140 and 120 pm, respectively. (6pt) 4pt Both ~130 pm Due to delocalized electrons (resonance). Q8B. A rich source of inorganic nitrate in the human body comes from diets rich in leafy green foods, such as spinach and arugula. Draw the most stable Lewis for nitrate, NO3. Why are all the bond lengths the same? (6pt) 4pt resonance 10