makeup exam 1
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makeup exam 1

Course Number: CH 301, Fall 2009

College/University: University of Texas

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Version 056 Make Up Exam 1 Laude (53755) This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 10/05/2008 11:00 PM 001 10.0 points Which of the following is a permitted combination of quantum numbers? 1 1. n = 3, = 1, m = 2, ms = 2 1 2. n = 2, = 1, m = 0, ms = 2 1 3. n = 3, = 1, m = 0, ms = correct 2 4. n = 6, = 5, m =...

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056 Version Make Up Exam 1 Laude (53755) This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 10/05/2008 11:00 PM 001 10.0 points Which of the following is a permitted combination of quantum numbers? 1 1. n = 3, = 1, m = 2, ms = 2 1 2. n = 2, = 1, m = 0, ms = 2 1 3. n = 3, = 1, m = 0, ms = correct 2 4. n = 6, = 5, m = 5, ms = 1 5. n = 0, = 0, m = 0, ms = Explanation: 1 2 wavelength 150 nm? 1. 9.94 1019 J 2. 1.33 1018 J correct 3. 8.21 1019 J 4. 1.33 1019 J 5. 1.07 1019 J 6. 7.52 1019 J Explanation: = 150 nm h = 6.626 1034 J s. For a photon c = , so E = h = 1 c = 3 108 m/s hc 1 for n = 3, = 1, m = 2, ms = 2 m = 2 is not an allowed value 1 n = 0, = 0, m = 0, ms = 2 n = 0 is not an allowed value n = 6, = 5, m = 5, ms = 1 ms = 1 is not an allowed value 1 n = 2, = 1, m = 0, ms = 2 = 1 is not an allowed value 002 10.0 points Which of the following is the correct Lewis structure of Magnesium Chloride (MgCl2 )? 1. Cl Mg Cl where c is the speed of light and h is Plancks constant. hc = 6.63 1034 J s E= 3 108 m/s 1 109 nm 150 nm 1m 18 = 1.33 10 J 004 10.0 points Which of the following compounds has only single bonds? I) SF6 II) BN III) F2 IV) SiH4 1. II, IV 2. I, III, IV correct 2. Mg2+ , 2 Cl 3. Cl 4. Cl 5. 2 Mg Mg Mg 2+ correct Cl Cl , 2 Cl 3. II, III 4. III only Explanation: 003 10.0 points What is the energy, in Joules, of a photon of 5. I, II, III Version 056 Make Up Exam 1 Laude (53755) 6. I, II Explanation: 005 10.0 points Which of the following statements concerning the Schrdinger equation and its solutions is o true? I) The use of Cartesian coordinates simplies the solution to the Schrodinger equation for the hydrogen atom. II) An electrons exact location can be calculated using the Schrdinger equation. o III) The Schrdinger equation for the helium o atom contains three potential energy terms. 1. II only 2. I, II, III 3. II, III 4. I, II 5. III only correct 6. I only 7. I, III Explanation: Solutions to the Schrodinger equation are wave functions, which when squared express the probable location of electrons; but, the exact location cannot be known. Attractive potential energy terms are found in all solutions for all atoms. Repulsive potential energy terms are found only in atoms that have more than one electron, i.e. everything beyond hydrogen. Polar coordinates are preferred for 3-D solutions because they simplify the math. 006 10.0 points How many blocks make up the standard periodic table? 1. 5 2. 4 correct 3. 18 4. 3 5. 7 2 Explanation: The four blocks are the s-block, the p-block, the d-block, and the f -block. 007 10.0 points Using charge density, rank the lattice energy of the following compounds from least to greatest: KI, CaSe, BeO, FrAt. 1. BeO < CaSe < KI < FrAt 2. CaSe < BeO < FrAt < KI 3. KI < BeO < FrAt < CaSe 4. FrAt < KI < CaSe < BeO correct 5. KI < FrAt < BeO < CaSe Explanation: Wherever larger charges are involved, the bonding will be greater between the atoms and thus the lattice energy will be larger. Bonds between smaller atoms are also stronger. 008 10.0 points What is the de Broglie wavelength of a bird being chased by Schrodingers cat, Albert? The bird has a mass of 500 g and is ying at 70 km/h. 1. 1.89314 1038 m 2. 6.81531 1038 m 3. 6.81531 1035 m correct 4. Not enough information is given. 5. 1.89314 1035 m Explanation: Version 056 Make Up Exam 1 Laude (53755) m = 500 g = 0.5 kg 1000 m 1h v = 70 km/h 1 km 3600 s = 19.4444 m/s h h = p mv 2 3 O I I) O P O O = 1. Neither is correct 2. Both are correct 3. II 4. I correct Explanation: 6.626 1034 kgsm = (0.5 kg) (19.4444 m/s) = 6.81531 1035 m 009 10.0 points What is the formal charge on the labeled atoms in the Lewis structure for phosphate ion (PO4 3 ) shown below? O Oa Pb O 1. -2; 1; -2 2. 0; 0; -1 correct 3. 0; 0; 0 4. -2; -2; -2 5. 0; 1; -1 6. 0; 1; -3 Explanation: 010 10.0 points Based on formal charge considerations, which of the following is a better Lewis structure for phosphate ion PO4 3 ? O I) O P O O Oc 011 10.0 points Consider a lled n = 3 shell. For a given conguration within that shell, how many electrons will be assigned m = 1? 1. 18 2. 16 3. 8 4. 4 correct 5. 6 Explanation: The quantum number n = 3 describes an s, p and d subshell ( = 0, 1 and 2). However, because m = 1, the s subshell is disallowed ( must be 1 for m = 1), leaving only the p and d subshells. Within each of those subshells, each m can be assigned to 2 electrons, giving a total of 4 electrons with m = 1. 012 10.0 points Rank the following diatomic molecules in terms of increasing bond length: N2 , O2 , F2 . 1. F2 < N2 < O2 2. F2 < O2 < N2 3. Not enough information. Version 056 Make Up Exam 1 Laude (53755) 4. N2 < O2 < F2 correct 5. N2 < F2 < O2 Explanation: 013 10.0 points For the Hydrogen atom, an electron moving between which two energy levels in the Balmer series would have to absorb light with a frequency ( ) of 4.58 1014 Hz? 1. n = 1 and n = 3 2. n = 1 and n = 4 3. n = 2 and n = 4 4. n = 2 and n = 3 correct 5. n = 1 and n = 2 6. n = 3 and n = 5 Explanation: A frequency of 4.58 1014 Hz corresponds to a transition from n = 2 to n = 3 using the Rydberg equation. The fact that the Balmer series is mentioned also indicates that n1 must be 2. 014 10.0 points Which of the following would be the correct electron conguration for unununium, Uuu, also known as Roentgenium, Rg? 1. [Rn] 7s1 5f 14 6d9 2. [Rn] 7s2 5f 14 6d9 3. [Rn] 7s2 5f 14 6d10 4. [Rn] 7s0 5f 14 6d10 5. [Rn] 7s1 5f 14 6d10 correct Explanation: The Aufbau order of electron lling is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f , 5d, 6p, etc. s orbitals can hold 2 electrons, p orbitals 6 4 electrons, and d orbitals 10 electrons. Rg has 111 electrons. Radon, Rn, accounts for 86 of those. Using the Aufbau rule, we use up the remaining electrons. Remember that elements do not exist in a d9 state, instead, they borrow an electron from the nearest energy p orbital to become d10 . 015 10.0 points An electron has a fairly well dened position with an uncertainty of only 108 m. What would the minimum uncertainty in the electrons velocity be? (A electron has a mass of 9.1 1031 kg) 1. 5.8 103 m s1 correct 2. 5.3 103 m s1 3. 6.1 103 m s1 4. 4.6 103 m s1 5. 3.5 103 m s1 Explanation: h 4 x(mv ) v h 4mx 6.626 1034 J s 4 (9.1 1031 kg) (108 m) 5.8 103 m s1 016 10.0 points Which of the following aect the ordering of ionization energies of N, C and O? I) Increasing ENC results in increasing IE. II) Increasing ENC results in decreasing IE. III) Having a half-lled subshell is energetically stabilizing. IV) Having a half-full subshell is energetically destabilizing. 1. I, III correct 2. IV only Version 056 Make Up Exam 1 Laude (53755) 3. I only 4. II, IV 5. III only Explanation: Ionization energy increases across a given row due to increasing ENC, and a half-lled subshell results in added stability (lower energy). 017 10.0 points Which of the following statements concerning particle in a box is/are true? I) As the principle quantum number (n) increases, the energy levels become closer in value. II) The wavelength () is inversely proportional to the principle quantum number (n). III) The energy of the particle cannot be equal to zero. 1. III only 2. I and III I 3. and II 4. I only 5. II and III correct 6. II only 7. I, II and III Explanation: The energy of the particle in the box can never be zero; this is a consequence of the fact that it can never be stationary, and as a massive particle in motion, thus has to have some energy. The length of the box (L) and the principle energy level (n) determine the wavelength () of the system, and because n can only have integer values, can only have certain discrete values. The energy of the particle (E) is inversely proportional to L, 5 and thus as L increases, the dierent energy levels are closer to total degeneracy. 018 10.0 points Which of the following statement(s) is/are true? I) The failure of classical mechanics to predict the absorptions/emission spectra of gases is called the ultraviolet catastrophe. II) Quantum mechanics accurately predicted the behavior of blackbody radiators. III) The emission spectra of gases are discrete rather than continuous. IV) Any frequency of light will eject an electron from a metal surface as long as the intensity is sucient. 1. I, II and IV 2. III and IV 3. II, III, and IV 4. I and III 5. II and III correct Explanation: Classical mechanics predicted that the power radiated by a blackbody radiator would be proportional to the square of the frequency at which it emitted radiation, and thus approach innity as the frequency increased. This was false, since at higher frequencies blackbody radiators emit less, not more power. This was termed the ultraviolet catastrophe. Classical mechanics also predicted that the energy (velocity) of electrons emitted from a metal surface is proportional to the intensity of light. In reality, the energy (velocity) is only dependent upon the frequency of light. Once the threshold frequency is reached, however, the number of emitted electrons is proportional to the intensity of light. Classical mechanics also fails in explaining the discrete lines in absorption/emission spectrum, which are due to discrete energy levels of electrons in atoms. Version 056 Make Up Exam 1 Laude (53755) 019 10.0 points What is the correct electronic conguration for a ground-state Antimony(V) ion (Sb5+ )? 1. [Kr] 5s1 4d9 2. [Kr] 5s0 3f 14 4d10 3. [Kr] 5s 4d 5p 2 5 3 6 guration of atoms because of the stability of either a full or half-full outermost d-orbital, so you may need to account for this by shuing an electron from the (n 1) s orbital. Use the total number of electrons in Mg (twelve), and the above rules to get the right answer. 021 10.0 points Rank the following species from smallest to largest atomic radius: K, Mg, Rb, Ca. 1. Mg < K < Ca < Rb 2. Mg < Ca < Rb < K 3. Mg < Ca < K < Rb correct 4. Rb < K < Ca < Mg 5. Mg < Rb < K < Ca Explanation: Atomic radii increase down and to the left. 022 10.0 points What is the number of electrons around the central atom in XeF4 ? 1. 8 2. 6 3. 10 4. 12 correct 5. 4 Explanation: 023 10.0 points Approximate the electronegativity dierence (EN) of the CF bond in methane (CH3 F). 1. 1.3 2. 2.0 3. 1.8 4. [Kr] 5s0 4d10 correct 5. [Kr] 5s 4d 2 8 Explanation: The Aufbau order of electron lling is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f , 5d, 6p, etc. s orbitals can hold 2 electrons, p orbitals 6 electrons, and d orbitals 10 electrons. Note some exceptions do occur in the electron conguration of ions of main group metals (such as Antimoy). When forming an ion from a main group metal, electrons are removed rst from the highest energy p orbital followed by the highest energy s orbital. Finally use noble gas shorthand to get the answer: [Kr] 5s0 4d10 . 020 10.0 points What is the correct electronic conguration of magnesium, Mg? 1. 1s2 2s2 2p6 3s2 correct 2. 1s2 2s2 2p2 3s2 3. 1s2 2s4 2p6 4. 1s2 2s10 5. 1s2 2s2 2p8 Explanation: The Aufbau order of electron lling is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f , 5d, 6p, etc. s orbitals can hold 2 electrons, p orbitals 6 electrons, and d orbitals 10 electrons. Note some exceptions do occur in the electron con- Version 056 Make Up Exam 1 Laude (53755) 4. 0.8 5. 1.5 correct Explanation: 024 10.0 points Which of the following correctly rationalizes the increase in atomic radii down and to the left on the periodic table, based on what we discussed in class? 1. The elements are simply larger due to more protons and neutrons. 2. Because the elements are easier to ionize, they have a larger electronegativity, and therefore their electron anity is not sucient to reduce atomic radii. 3. None of the above. Atomic radii increase up and to the right. 4. As you move to the left across a period, decreasing ENC means the outer electrons are less tightly held and can move further from the nucleus. As you move down a group, the electrons occupy orbitals that are further from the nucleus. correct 5. Larger elements have an increasing proportion of d and f orbitals, which are intrinsically larger than all of the s and p orbitals. 6. The periodic table was set up to group atoms by size to facilitate comparisons. Explanation: We used ENC to rationalize the overall periodic trends in class. The nuclei are larger due to more protons and neutrons, but not the atom as a whole most of an atoms size is the space taken up by the electrons, the nucleus is tiny in comparison. Orbital size increases with the principal quantum number n so that a 7s orbital would probably be larger than a 4f orbital the d and f orbitals are not automatically larger. The periodic table was organized by properties of the macroscopic el2. 7 ements not by the size of the atoms. The other answer choice is just a mindless spewing of chemistry vocabulary. 025 10.0 points Which of the following is the correct Lewis structure of hydroxylamine (NH2OH)? H 1. H N O H H N H H 3. N H H 4. N H O H O O correct H H Explanation: 026 10.0 points Which of the following elements commonly forms covalent compounds in which it has fewer than 8 valence electrons? I) In II) B III) N IV) Br 1. II, III, IV 2. III only 3. II only correct 4. I, IV Version 056 Make Up Exam 1 Laude (53755) 4. Neither is correct 5. II, III 6. I only Explanation: 027 10.0 points Which of the following would be radicals? I) NO II) NF3 III) CN IV) PO4 1. I, II, III 2. I, III correct 3. II, IV 4. II, III 5. II, III, IV 6. I, IV Explanation: 028 10.0 points Which of the following structures demonstrates resonance? H I) H C H H I I) H C H C C 8 Explanation: Resonance is when there are two equivalent ways of drawing a Lewis structure, so you draw both, as the true structure is somewhere in between. You cant just put resonance arrows between two identical structures. You cant just mirror the structures. 029 10.0 points Which of the following hypervalent compounds cannot exist even in theory? 1. OCl6 correct 2. I3 3. XeF4 4. XeF2 5. SBr6 Explanation: 030 10.0 points Consider the electron lling diagram below: 3p 3s 2p 2s 1s O O C H C H H H H O H H O C Which of the following does it violate? I) The aufbau principle II) The Pauli exclusion principle III) Hunds rule 1. I only C H O O 2. I and III 3. I, II and III 1. Both are correct 2. II only correct 4. I and II correct 3. I only 5. II and III Version 056 Make Up Exam 1 Laude (53755) 6. III only 7. II only Explanation: The overlled 2s subshell violates the Pauli exclusion principle and the unlled 2p subshell violates the aufbau principle. 9

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Emil Skoda Company Adjusted Trial balance For the month ended June 30, 2010 Adjusted Trial Balance Account Titles Cash Accounts Receivable Supplies Accounts Payable Unearned Revenue Emil Skoda, Capital Emil Skoda, Drawing Service Revenue Salaries Expense
Salem Intl. - MBA - ACC510
Marketing Strategy General Description The marketing strategy of Magnus Enterprises LLC is based on the mission of the company. The mission is to create a cost-effective clothing line alternative for young men aged 18-35. Through our method of distributio