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Chemistry 161 Exam I October 2, 2013 Student Name (Print): _____TAVSS______________ The exam booklet has 25 questions for credit and one additional question to check the color of your exam booklet. Please answer all 26 questions on the OpScan sheet. There is no penalty for guessing. If you do not find the exact answer, bubble in the best choice. A proctor will check your picture ID, OpScan form, signature and calculator during the exam. The use of calculators with permanent memories (graphing calculators), cell phones, pagers, PDAs or other electronic devices other than a basic scientific calculator is expressly forbidden. The last page of the booklet contains a periodic table along with other useful data. The use of any other notes or information on this test or taking extra time to bubble in your answers will be considered a violation of the Academic Honesty provisions of the student code. Exam scores will be posted as soon as possible. On the Opscan Form: (Use a #2 pencil or darker) 1. Sign your name across the top of the form. 2. Code the following information ( blacken circles) Your Name (Last Name First) Your Ru ID Number [Start under Box A and continue to Box I ] Your Recitation Section Number in K & L [For sections 01-09, bubble 0 under box K ] Your Exam Form Number under box P Your EXAM FORM is ① Chem161f13e1v1 1

Your EXAM FORM is ① 1 Chem 161-2013 Exam I B&O Chapter 1 – Chemistry: Matter and Measurement Filed in: H&P Chapter 1 – Chemistry: Matter and Measurement Significant Figures/Precision/Accuracy A homogeneous mixture was prepared by mixing 0.15g A, 112.3g B and 24.55g C. The volume is found to be 85.80 cm 3 . What is the density of the mixture to correct number of significant digits? A. 1.5967 g/cm 3 B. 1.59674 g/cm 3 C. 1.6 g/cm 3 D. 1.60 g/cm 3 E . 1.597 g/cm 3 0.15g 112.3g 24.55g 137.00g The rule for addition-subtraction is to consider the smallest number of digits to the right of the decimal point. Since 112.3 has only one digit to the right of the decimal point, then the sum, 137.00, can only have one digit to the right of the decimal point, i.e., 137.0, which has four significant figures. Nevertheless, don’t round off in the middle, so the next process should use the full 137.00, but we should keep in mind that this number really has only four significant figures. D = g/cm 3 D = 137.00g/85.80cm 3 = 1.59674g/cm 3 Since 137.00 has four significant figures and 85.80 has four significant figures then the final answer must have only four significant figures. Rounding off gives 1.597g/cm 3 . E Chem161f13e1v1 2

2 Chem 161-2013 Exam I B&O Chapter 1 – Chemistry: Matter and Measurement Filed in: H&P Chapter 1 – Chemistry: Matter and Measurement Units/unit conversions/dimensional analysis Which of the following is not an SI base unit? A. kilogram B. mole C . liter D. Kelvin E. second The SI base units are: meter, kilogram, second, Kelvin, ampere, mol and candela. Liters are derived SI units. meter → cm → cm 3 → mL → L C Chem161f13e1v1 3

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3 Chem 161-2013 Exam I B&O Chapter 4 – Periodic Trends of the Elements Filed in: H&P Chapter 8 – Electron Configurations & Periodic Table Trends Periodic table trends Which of the following would have the highest 2nd ionization energy? A. Ba B. Ga C. Sr D . K E. Ca D Chem161f13e1v1 4

The first ionization energy is M → M + + 1e - . The second ionization energy is M + → M 2+ + 1e - Therefore, plot M + , and mentally consider the relative energy required to remove another electron to form M 2+ . 20 Ca + = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 56 Ba + = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 1 19 K + = 1s 2 2s 2 2p 6 3s 2 3p 6 38 Sr + = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 56 Ba + 31 Ga + = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 20 Ca + 19 K + 38 Sr + 31 Ga+ K + has its valence electrons in the third principal shell. The other four ions have their valence electrons in higher principal shells. This means that the K + ion has the highest ionization energy, in that the valence electrons are held relatively strongly by the protons in the nucleus because the valence shell is close to the nucleus. Now look at the effective nuclear charge of the ions. The effective nuclear charge on K + is far greater than the effective nuclear charge on any of the other ions, adding further strength to the conclusion that K + has the highest second ionization energy. This conclusion is different than what one would expect Chem161f13e1v1 5

regarding the generalization that atoms and ions in the lower left-hand corner of the periodic table have low ionization energies. Chem161f13e1v1 6

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4 Chem 161-2013 Exam I B&O Chapter 1 – Chemistry: Matter and Measurement Filed in: H&P Chapter 1 – Chemistry: Matter and Measurement Units/unit conversions/dimensional analysis What is the dollar value of a cube of gold measuring 2.00 inches on a side? Price of gold: $1366 per troy ounce 31.1 g = 1 troy ounce Density of gold: 19.3 g/cm 3 2.54 cm = 1 inch A . $1.11 x 10 5 B. $2.41 x 10 7 C. $2.98 x 10 4 D. $1.07 x 10 8 E. $5.96 x 10 6 Plan: in → in 3 → cm 3 → g → troy ounce → $ (2.0 in) 3 x (2.54cm/in) 3 x (19.3g/cm 3 ) x (1 troy ounce/31.1g) x ($1366/troy ounce) = $1.11 x 10 5 A Chem161f13e1v1 7

5 Chem 161-2013 Exam I B&O Chapter 2 – Atoms, Molecules and Ions Filed in: H&P Chapter 3A – Stoichiometry: Chemical Calculations Atoms, ions, moles and molecular wts. Which of the following are true statements? X. The mass of 1 mol of carbon-12 is exactly 12 g. Y. Carbon contains at least one natural isotope heavier than carbon-12. Z. The mass of one atom carbon-12 is 1.99 x 10 -23 g A. X only B. Y only C. Z only D. X and Z only E . X, Y, and Z X. True, by definition. Exactly 12 g (12.00000. . .) of C-12 has been set as the standard for defining one mole, i.e., Avogadro’s number of 6.022 x 10 23 atoms/mol. Y. True. Since the atomic mass of carbon is 12.01, and carbon contains a C-12 isotope, then carbon must contain an isotope greater than 12 in order for the average of the isotope masses to be equal to 12.01. Z. True. (12g/mol)/(6.022x10 23 atom/mol) = 1.99 x 10 -23 g/atom E 6 Chem 161-2013 Exam I B&O Chapter 3 – Quantum Theory and the Electronic Structure of Atoms Filed in: H&P Chapter 7 – Light and Atomic Structure Quantum nos./orbitals What is the maximum number of electrons that can be in orbitals having quantum numbers n = 3, m  =0? A . 6 B. 10 C. 4 D. 2 E. 0 QUANTUM NUMBERS n  m  m s 1 to ∞ shell 0 to n-1 subshl - l to + l orbitals +1/2, -1/2 spin A Chem161f13e1v1 8

1 s = 0 0 +1/2, -1/2 2 s = 0 0 +1/2, -1/2 p = 1 -1 0 +1 +1/2, -1/2 3 s = 0 0 +1/2, -1/2 p = 1 -1 0 +1 +1/2, -1/2 d = 2 -2 -1 0 +1 +2 +1/2, -1/2 n = 3 and m l = 0 can have six electrons (2 electrons for the s subshell, 2 electrons for the p subshell, and 2 electrons for the d subshell). 7 Chem 161-2013 Exam I B&O Chapter 3 – Quantum Theory and the Electronic Structure of Atoms Filed in: H&P Chapter 7 – Light and Atomic Structure Quantum mechanics/DeBroglie wave-particle duality/Planck’s equation/Photoelectric effect The threshold frequency to eject an electron from the surface of a metal is 1.95 x 10 14 s −1 . What is the kinetic energy of an electron ejected from the surface of the same metal if it is bombarded with a light with a wavelength of 358 nm? A. 5.55 x 10 −19 J B . 4.26 x 10 −19 J C. 1.29 x 10 −19 J D. 6.84 x 10 −19 J E. 8.13 x 10 −19 J B Chem161f13e1v1 9

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Energy total = Energy threshold + Energy kinetic Energy threshold = hυ = 6.626x10 -34 Js x 1.95x10 14 s -1 = 1.29207 x 10 -19 J Energy total = hc/λ = ((6.626x10 -34 Js) x (3x10 8 ms -1 ))/(358 x 10 -9 m) = 5.5525 x 10 -19 J Energy total = Energy threshold + Energy kinetic 5.5525 x 10 -19 J = (1.29207 x 10 -19 J) + Energy kinetic Energy kinetic = 4.26 x 10 -19 J 8 Chem 161-2013 Exam I B&O Chapter 3 – Quantum Theory and the Electronic Structure of Atoms Filed in: H&P Chapter 7 – Light and Atomic Structure Bohr hydrogen atom How much energy is required to excite an electron from energy level n = 2 to energy level n = 5 in the hydrogen atom? A . 4.58 x 10 -19 J B. 1.56 x 10 -18 J C. 1.94 x 10 -18 J D. 3.03 x 10 -19 J E. 1.82 x 10 -19 J Use the Bohr hydrogen atom equation: ∆E = -B((1/n f 2 ) – (1/n i 2 )) n f = 5; n i = 2 ∆E = -(2.179 x 10 -18 J) x ((1/5 2 ) – (1/2 2 )) = 4.58 x 10 -19 J A 9 Chem 161-2013 Exam I B&O Chapter 5 – Ionic and Covalent Compounds Filed in: H&P Chapter 2 - Atoms, Molecules and Ions Fundamental chemical laws Which has a larger lattice energy? NaCl or KI NaF or MgO MgS or ScN A. NaCl, NaF, MgS B. KI, MgO, ScN C . NaCl, MgO, ScN D. KI, NaF, MgS E. NaCl, NaF, ScN Coulomb’s law: F = (kQ 1 Q 2 )/D 2 C Chem161f13e1v1 10

NaCl vs. KI The charge on Na and K are each +1; the charge on Cl and I are each -1; therefore, Q 1 xQ 2 would be identical for both formulas, that is, -1. Na + is smaller than K + , because the valence electrons on K + are further from the nucleus than the valence electrons on Na + ; also Cl - is smaller than I - for the same reason. Therefore D is smaller for NaCl, meaning that the ions are closer together. Therefore, the lattice energy of NaCl is larger than for KI. NaF vs. MgO Na has a charge of +1; F has a charge of -1. Therefore, the multiple of these charges is -1. Mg has a charge of +2; O has a charge of -2. Therefore, the multiple of these charges is -4. Hence, based on the charges alone, one would expect MgO to have a larger lattice energy. Mg 2+ is smaller than Na + based on isoelectronic series trends; F - is smaller than O 2- based on isoelectronic series trends. Since the cation of MgO is smaller than the cation of NaF, but the anion of MgO is larger than the anion of NaF, one cannot make a decision of the relative bond distances without quantitative data. Let’s assume that the distances are a tradeoff, and therefore D for NaF is approximately equal to D for MgO. The differences in the charges, however, are strongly in favor of MgO having a larger lattice energy. Furthermore, D is of minor importance for determining lattice energy, as compared to Q 1 x Q 2 . MgS or ScN Mg has a charge of +2; S has a charge of -2. The multiple of these charges is -4. Nitride has a charge of -3. Although it isn’t clear what the charge on Sc is, by the fact that there is a mole ratio of 1:1 between Sc and N forces the charge on Sc to be +3. (Also, Sc is in transition metal group 3B, which are mainly +3 charges, but the student might not know that.) The multiple of these charges is -9. Based on charges alone, one would expect ScN to have a larger lattice energy than MgS. Now let’s talk about the relative lengths of the bonds, D. Mg 2+ is smaller than Sc 3+ (I think); S 2- is larger than N 3- (I think). It’s impossible for the students to know the relative distances in this comparison. However, the charge factor is much more important than the distance factor, so ScN has greater lattice energy than MgS. 10 Chem 161-2013 Exam I B&O Chapter 2 – Atoms, Molecules and Ions Filed in: H&P Chapter 2 - Atoms, Molecules and Ions Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes K has mass 40.9615 amu. What amount of neutrons is in a sample of K having a mass of 10.0 g? A. 10.9 moles B . 5.37 moles C. 31.0 mol D. 24.2 mol E. 18.6 mol The atomic mass of potassium = 40.9615g/mol. moles = g/MW = 10.0g/(40.9615g/mol) = 0.244 moles B Chem161f13e1v1 11

One mole of potassium has 19 protons and 22 neutrons. If there are 22 neutrons per mole, then how many neutrons are in 0.244 mole? 0.244 mol x (22 neutrons/mol) = 5.37 neutrons in 0.244 mol. 11 Chem 161-2013 Exam I B&O Chapter 4 – Periodic Trends of the Elements Filed in: H&P Chapter 8 – Electron Configurations & Periodic Table Trends Periodic table trends Which of the following is the best explanation for why the ionization energy of nitrogen is larger than the ionization energy of carbon? A. nitrogen has more electrons than carbon B. nitrogen is closer to a complete outer shell than carbon C. nitrogen has more singly occupied orbitals than carbon D . nitrogen has a higher effective nuclear charge than carbon E. nitrogen is a gas occupying 78% of the earth’s atmosphere. 6 C = 1s 2 2s 2 2p 2 7 N = 1s 2 2s 2 2p 3 4e- 2e- 2e- 5e- 6+ 7+ 4+ 5+ A. False. The fact that nitrogen has more electrons than carbon is irrelevant as compared to the effective nucler charge factor. If anything, more electrons would tend to repel each other making nitrogen have smaller ionization energy than carbon. B. False. True, but irrelevant. C. False. True, but irrelevant. D. True. The greater effective nuclear charge, which holds the valence electrons close to the nucleus, is a powerful force in inhibiting the loss of a valence electron. In addition, D Chem161f13e1v1 12

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removing an electron from a half-filled 2p subshell in N is more difficult than removing an electron from a partially-filled 2p subshell in C. E. False. True, but irrelevant. Chem161f13e1v1 13

12 Chem 161-2013 Exam I B&O Chapter 3 – Quantum Theory and the Electronic Structure of Atoms Filed in: H&P Chapter 7 – Light and Atomic Structure Quantum mechanics/DeBroglie wave-particle duality/Planck’s equation/Photoelectric effect How many moles of photons with a wavelength of 679 nm would need to strike a material in order for it to absorb 54.7 kJ of energy? A. 0.192 mole B . 0.310 mole C. 0.259 mole D. 0.671 mole E. 0.127 mole E photon = hc/λ E photon = ((6.626x10 -34 Js) x (3x10 8 ms -1 ))/(679x10 -9 m) = 2.9275 x 10 -19 J But there are 54700 J 54700J/(2.9275 x 10 -19 J/photon) = 1.8685 x 10 23 photons 1.8685 x 10 23 photons x (1 mol photons/(6.022 x 10 23 photons)) = 0.310 moles of photons B 13 Chem 161-2013 Exam I B&O Chapter 5 – Ionic and Covalent Compounds Filed in: H&P Chapter 2 - Atoms, Molecules and Ions Fundamental chemical laws Consider two ions of opposite charges separated by a distance d. Based on Coulomb’s Law, which of the following effects would double the electrostatic energy of the ionic bond? I. doubling the positive charge and halving the negative charge. II. doubling the distance and doubling the positive charge. III. halving the distance d. IV. doubling the negative charge. A. IV only B . III and IV C. I and III D. II and IV E. II and III Coulomb’s law: E = (kQ 1 Q 2 )/D Note: Important not to confuse E = (kQ 1 Q 2 )/D, which is Coulomb’s law for energy, with F = (kQ 1 Q 2 )/D 2 , which is Coulomb’s law for force. The relationship between energy and force is E = F x D. B Chem161f13e1v1 14

As reference points, let’s call the positive charge +1 and the negative charge -1; also, let’s call the distance 1. E = (k x 1 x 1)/(1) = 1k I. This makes Q 1 = 2 and Q 2 = ½; D = 1. E = (k x 2 x 1/2)/(1) = 1; therefore no effect on electrostatic energy. II. This makes Q 1 = 2 and Q 2 = 1. D = 2. E = (k x 2 x 1)/(2) =1; therefore, no effect on electrostatic energy.. III. This makes Q 1 = 1 and Q 2 = 1. D = ½. E = (k x 1 x 1)/(1/2) = 2; therefore, the electrostatic energy was doubled. IV. This makes Q 1 = 1 and Q 2 = 2. D = 1. E = (k x 1 x 2)/(1) = 2; therefore, the electrostatic energy was doubled. 14 Chem 161-2013 Exam I B&O Chapter 3 – Quantum Theory and the Electronic Structure of Atoms Filed in: H&P Chapter 7 – Light and Atomic Structure Light and matter In the hypothetical electromagnetic wave picture, the distance between two representative points is noted. Calculate the energy of this radiation. A. 7.36 x 10 -17 J B. 1.45 x 10 -17 J C. 4.23 x 10 -17 J D . 9.55 x 10 -17 J E. 7.22 x 10 -17 J The wavelength is the length of a repeating cycle, i.e., the length of a whole cycle. 1.04 nm is the length of half a cycle. Therefore, the wavelength would be 2.08 nm = 2.08 x 10 -9 m. E = hc/ λ E = ((6.626 x 10 -34 Js) x (3 x 10 8 m/s))/(2.08x10 -9 m) = 9.56 x 10 -17 J D 15 Chem 161-2013 Exam I B&O Chapter 3 – Quantum Theory and the Electronic Structure of Atoms Filed in: H&P Chapter 8 – Electron Configurations & Periodic Table Trends Aufbau Which of the following atoms or ions has exactly three singly occupied orbitals? X. Sc Y Mn 2+ Z. Co A. X only B. Y only C . Z only C Chem161f13e1v1 15

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D. X and Z only E. Y and Z only 21 Sc = [ 18 Ar]4s 2 3d 1 3d 1 4s 2 25 Mn 2+ = [ 18 Ar]3d 5 3d 5 27 Co = [ 18 Ar]4s 2 3d 7 3d 7 4s 2 Chem161f13e1v1 16

16 Chem 161-2013 Exam I B&O Chapter 4 – Periodic Trends of the Elements Filed in: H&P Chapter 8 – Electron Configurations & Periodic Table Trends Periodic table trends In each of the following pairs, which has the larger radius? W: C O X: P 3− As 3− Y: F – S 2− Z: S 2− Ca 2+ W X Y Z A. C As 3− F – Ca 2+ B. O P 3− F – S 2− C. C P 3− S 2− S 2− D. O As 3− S 2− Ca 2+ E . C As 3− S 2− S 2− E Chem161f13e1v1 17

Using the general rules for periodic trends, the larger radius is in the lower left-hand corner of the periodic table while the smaller radius is in the upper right-hand corner. Therefore, W: C is larger X: As 3- is larger Y: S 2- is larger Z: Ca 2+ would be considered to be larger at first glance, but based on isoelectronic series rules the S 2- should be larger. Consider concentric circle diagrams: 6 C = 1s 2 2s 2 2p 2 8 O = 1s 2 2s 2 2p 4 4e- 2e- 2e- 6e- 6+ 8+ 4+ 6+ C is larger due to smaller effective nuclear charge. 15 P 3- = 1s 2 2s 2 2p 6 3s 2 3p 6 33 As 3- = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 8e- 8e- 18e- 8e- 8e- 2e- 2e- 15+ 33+ 5+ 5+ As 3- is larger due to 4 vs. 3 valence shells. 16 S 2- = 1s 2 2s 2 2p 6 3s 2 3p 6 9 F - = 1s 2 2s 2 2p 5 8e- 8e- 8e- 2e- 2e- 9+ 16+ 7+ 6+ Chem161f13e1v1 18

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S 2- is larger due to 3 vs. 2 valence shells. 16 S 2- = 1s 2 2s 2 2p 6 3s 2 3p 6 20 Ca 2+ = 1s 2 2s 2 2p 6 3s 2 3p 6 8e- 8e- 8e- 8e- 2e- 2e- 16+ 20+ 6+ 10+ S 2- is larger due to greater effective nuclear charge. The same results are found using the graphic analysis: W: C is larger due to a smaller effective nuclear charge. A smaller effective nuclear charge has a smaller attraction between the positively charged nucleus and the valence electrons. X: As 3- is larger due to the valence shell being further from the nucleus. Further from the nucleus means less interaction between the positively charged nucleus and the valence electrons. Y: S 2- is larger due to the valence shell being further from the nucleus. Further from the nucleus means less interaction between the positively charged nucleus and the valence electrons. Z: S 2- is larger due to a smaller effective nuclear charge. A smaller effective nuclear charge has a smaller attraction between the positively charged nucleus and the valence electrons. 17 Chem 161-2013 Exam I B&O Chapter 2 – Atoms, Molecules and Ions Filed in: H&P Chapter 2 - Atoms, Molecules and Ions Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes By knowing the number of protons in a neutral atom, we also know: X. number of electrons in the atom Y. the symbol of the element Z. the number of neutrons in the atom A. X only B. Y only C . X and Y only D. Y and Z only E. X, Y, and Z C Chem161f13e1v1 19

X. The key word here is neutral atom. The number of protons in a neutral atom is also equal to the number of electrons in the atom. For example, neutral lithium has three protons and three electrons. Y. The number of protons in an atom is also equal to the atomic number, which identifies the symbol of the element. For example, three protons is atomic number 3. Atomic number 3 is lithium. Z. The number of protons in an atom is not related to the number of neutrons in that atom. For example, lithium has three protons, but can have three or four neutrons. That is, lithium can exist as isotopes Li-3 and Li-4. 18 Chem 161-2013 Exam I B&O Chapter 3 – Quantum Theory and the Electronic Structure of Atoms Filed in: H&P Chapter 8 – Electron Configurations & Periodic Table Trends Aufbau Which of the following ions does not have a noble gas electron configuration? A. K + B. As 3- C. Ba 2+ D . Ti 2+ E. Al 3+ 19 K = [ 18 Ar]4s 1 ; 19 K + = [ 18 Ar] which is the noble gas electron configuration of Ar 33 As = [ 18 Ar]4s 2 3d 10 4p 3 ; 33 As 3- = [ 18 Ar]4s 2 3d 10 4p 6 which is the noble gas electron configuration of 36 Kr. 56 Ba = [ 54 Xe]6s 2 ; 56 Ba 2+ = [ 54 Xe] which is the noble gas electron configuration of Xe 22 Ti = [ 18 Ar]4s 2 3d 2 ; 22 Ti 2+ = [ 18 Ar]4s 0 3d 2 = [ 18 Ar]3d 2 which is not a noble gas configuration. 13 Al = [ 10 Ne]3s 2 3p 1 ; 13 Al 3+ = [ 10 Ne]3s 0 3p 0 = 13 Al = [ 10 Ne] which is the noble gas configuration of Ne. D Chem161f13e1v1 20

19 Chem 161-2013 Exam I B&O Chapter 2 – Atoms, Molecules and Ions Filed in: H&P Chapter 3A – Stoichiometry: Chemical Calculations Atoms, ions, moles and molecular wts. What mass of carbon has the same number of atoms as in 6.00 g of helium? A. 6.00 g B. 8.00 g C. 12.0 g D . 18.0 g E. 24.0 g Plan: gHe → molHe → atomsHe → atomsC → molC → gC 6.00gHe/(4.003g/mol) x (6.022x10 23 atomHe/molHe) x (1atomC/1atomHe) x (1molC/6.022x10 23 atomC) x (12.01gC/molC) = 18.0gC Note that some of the above calculation can cancel out, so the calculation could simply be gHe → molHe → molC → gC. This makes sense because if the number of atoms of He and C are the same then the number of moles of He and C would be the same. D 20 Chem 161-2013 Exam I B&O Chapter 3 – Quantum Theory and the Electronic Structure of Atoms Filed in: H&P Chapter 7 – Light and Atomic Structure Quantum mechanics/DeBroglie wave-particle duality/Planck’s equation/Photoelectric effect an experiment showing the photoelectric effect, an incident beam of green light produced no ejected electrons from a metal. Which of the following is most likely to result in electrons ejected from that metal? Use blue light instead of green light, but keep the intensity (brightness) of the light the same. Use red light instead of green light, but keep the intensity (brightness) of the light the same. Increase the intensity (brightness) of the green light. A . X only B. Y only C. Z only D. X and Z only E. Y and Z only X. True. One atom of a metal emitting blue light has more energy than one atom of a metal emitting green light. Therefore, the blue light might be enough energy to eject an electron. Y. False. One atom of a metal emitting red light has less energy than one atom of a metal emitting green light. If green light doesn’t have enough energy to eject an electron, then certainly red light doesn’t have enough energy. Z. False. If one atom of green light doesn’t have enough energy to eject an electron, then increasing the green light intensity, that is, increasing the number of atoms emitting the green light, won’t provide the energy required to eject an electron. A Chem161f13e1v1 21

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Note: E = nhυ. “n” is the number of photons emitting light, which is another way of saying the “intensity” of the emitting light. According to Einstein, n must be equal to 1 to determine if an emission is strong enough to produce a photoelectric effect. (Analogy of a million ping-pong balls vs. one baseball to break a window.) Chem161f13e1v1 22

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21 Chem 161-2013 Exam I B&O Chapter 1 – Chemistry: Matter and Measurement Filed in: H&P Chapter 1 – Chemistry: Matter and Measurement Units/unit conversions/dimensional analysis What would be the volume of a block of iron that contains 7.93 x 10 24 atoms of iron? Density of iron = 7.874 g/cm 3 A. 735 cm 3 B. 281 cm 3 C. 1.67 cm 3 D. 54.7 cm 3 E . 93.4 cm 3 Plan: atoms → mol → mass → volume (7.93 x 10 24 atoms) x (1 mol/(6.022x10 23 atoms)) x (55.845g/mol) x (1cm 3 /7.874g) = 93.4 cm 3 E Chem161f13e1v1 23

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22 Chem 161-2013 Exam I B&O Chapter 2 – Atoms, Molecules and Ions Filed in: H&P Chapter 3A – Stoichiometry: Chemical Calculations Atoms, ions, moles and molecular wts. A certain element has just one natural isotope. If 6.4 ng of this element contains 3.04 x 10 13 atoms, identify the element. A. beryllium B. sodium C. gallium D. strontium E . iodine (3.04 x 10 13 atoms)/(6.022 x 10 23 atoms/mole) = 5.0482 x 10 -11 moles present moles = g/MW MW = g/mol = (6.4 ng x (10 -9 g/ng))/(5.0482 x 10 -11 mol) = 126.8 = atomic weight. Iodine is the closest, with atomic mass of 126.9g/mol. E Chem161f13e1v1 24

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23 Chem 161-2013 Exam I B&O Chapter 1 – Chemistry: Matter and Measurement Filed in: H&P Chapter 1 – Chemistry: Matter and Measurement Classification of matter Which of the following are physical changes? I. Solid carbon dioxide sublimes into gas at room temperature. II. Sugar dissolves in a cup of coffee. III. Iron rusts in open air after a period of time. IV. Metal sodium forms hydrogen gas when in contact with water. A. I, II and IV B. I, II, and III C . I and II D. II and IV E. III and IV A physical change in a change in the form of a substance, but not a change in the substance, itself. I. Change of state from a solid to a gas is a physical change. The carbon dioxide is still carbon dioxide. II. Changing state from a solid to aqueous is a physical change. The sugar is still sugar, but spread out in the aqueous solvent. III. Iron becomes rust, which is iron oxide. This is a new compound. It is a chemical change. IV. The Na becomes Na + while the H 2 O becomes H 2 . These are new compounds. Therefore, these are chemical changes. C 24 Chem 161-2013 Exam I B&O Chapter 1 – Chemistry: Matter and Measurement Filed in: H&P Chapter 1 – Chemistry: Matter and Measurement Classification of matter Which of the following are intensive properties? X. density Y. volume Z. boiling point A. X only B. Y only C. Z only D . X and Z only E. X, Y, and Z only An intensive property is one that doesn’t increase as the quantity increases. (An extensive property is a property that does increase as the quantity increases.) D Chem161f13e1v1 25

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X. True. As the quantity increases, e.g., from 10 mL to 20 mL, the density doesn’t increase; e.g., 10 mL of H 2 O has a density of 1.0g/mL; 20 mL of H 2 O has a density of 1.0g/mL. Y. False. As the quantity increases, e.g., from 10g to 20 g, the volume does increase; e.g., 10 g of H 2 O has a volume of 10 mL; 20g of H 2 O has a volume of 20 mL. Z. True. As the quantity increases, e.g., from 10 mL to 20 mL, the boiling point doesn’t increase; i.e., 10 mL of H 2 O has a boiling point of 100oC; 20 mL of H 2 O has a boiling point of 100 o C. 25 Chem 161-2013 Exam I B&O Chapter 2 – Atoms, Molecules and Ions Filed in: H&P Chapter 2 - Atoms, Molecules and Ions Nature of the atom (protons, neutrons, electrons, symbols, etc.) including isotopes Lithium forms compounds which are used in dry cells and storage batteries and in high-temperature lubricants. It has two naturally occurring isotopes, 6 Li (isotopic mass = 6.015121 amu) and 7 Li (isotopic mass = 7.016003 amu). Lithium has an atomic mass of 6.9409 amu. How many atoms of 6 Li are in a naturally occurring sample of lithium that contains a total of 4.00 x 10 20 atoms? A . 3.00 x 10 19 B. 4.60 x 10 19 C. 3.80 x 10 19 D. 5.00 x 10 19 E. 2.20 x 10 19 6 Li 7 Li Average 6.015121g/mol 7.016003g/mol 6.9409g/mol X 1-X (X x 6.015121) + ((1-X) x 7.016003) = 6.9409 X = 0.0750 = 7.50% That is, 7.50% of the total lithium is Li-6. 0.0750 x (4.00 x 10 20 atoms) = 3.00 x 10 19 atoms of Li-6 A Chem161f13e1v1 26

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