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14am1k - CHEM 14A YOUR NAME WWII/5K3 Instructor Dr Laurence...

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Unformatted text preview: CHEM 14A YOUR NAME WWII/5K3 Instructor: Dr. Laurence Lavelle STUDENT |D# ............................... FALL 2006 MIDTE RM (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 (Show all your work on this paper, check units and significant figures.) Good Luck Do not write on this page. QUESTION SCORE QlA. A solution is prepared by dissolving 0.500 g KCl, 0.500 g K28, and 0.500 g K3PO4 in 0.500 L water. What is the concentration in the final solution of (a) potassium ions; (b) sulfide (82') ions? Total moles of K4: 0.500gKCl + 0.500ngs 74.55g mar1 110.26g mar1 + 0.500gK3P04 3m01K+ 212.27 g -m0l_1 1mol K3P04 =6. 71 x 10‘3 moi + 9.07 X10; moi + 7.07 x (4’4) [2.29%33g mol If} Mol arity of K: 0.500L (b) molarity of sz‘ 0.500g K25 1molS_____2 110.26g moi~1 lmolK s 0. 500L -1pt for sf (W 4.'258x10 M 2mol K+ lmol K25 10'3 moi = 2.29 x 10‘2 moi 0M] 1=.]907x103M @7114) —1pt for no units or incorrect units B. Diazepam, a drug used to treat anxiety, is 67.49% C, 4.60% H, 12.45% Cl, 9.84% N, and 5.62% 0. What is the empirical formula of the compound? For 100 g of compound, Moles ofC= 67 49g =5619 mo] 12. 01g mol Moles of H = 4 ————6——0g :4 563 moi 1. 008g- mol Moles of C1 = fl— 35.45g -m0l' Moles of N = 9 —_8___4g 2.0 7024 mol 14. Olg- mol Moles ofO= —i——623 =03512 moi 16. 00 g mol (W W) 1]=0.3512 moi (Of) <44) W} (6130 (6190 Dividing each number by 0.3512 mol gives a ratio of 16.00 C: 13.00 H: 1.00 Cl : 2.00N: 1.00 O. The empirical formula is C16H13C1N20 [7% Q2. The electron in a hydrogen atom is excited to a 4d orbital. Calculate the energy of the photon released if the electron were then to move to each of the following orbitals: (a) Is; (b) 2p; (c) 25; (d) 45. (e) If the outermost electron in a potassium atom were excited to a 4d orbital and then fell to the same orbitals, describe qualitatively how the emission spectrum would differ from that of hydrogen (do not do any calculations). Explain your answer. (20pt) wt) (W hen _ 2.18 x 10*18J Given that En = — —2— = 2 7’1 7’1 and. AE = E final‘EinitiaF (a) for an electron to fall from the 4d to Is level, the energy of the photon is AE = El—E4 = ._ 1 1 —2.18 X10 18]):5 — —?1 = — 2.04 X 10'18 J (the negative sign means that energy is l 4 (W) M ’ 2/; released or emitted); + Z , 0 ¢ / J— (b) 4% of Ad 07“? X 0 (b) Similarly, an electron moving from the 4d to 2p level will emit a photon of 4.09 x 10’19 J. WWW/“744(4) would if it were moving to the 2p orbital;lthis is due to the fact that all orbitals having the same principal quantum number n are degenerate (i.e. they have the same energy). Z / (d) (fl/M MW {/2 M) (d) In a hydrogen atom, no photon would be emitted on moving between orbitals possessing the same 11 (due to degeneracy of the 4d and the 4s orbitals in hydrogen). 2/117 (6) /"t. (e) Since potassium has both more electrons and protons than hydrogen, the individual orbitals ———____. -—-——_————————- (C) ( (c) same as (b); [in electron moving from the 4d to 25 emits the same amount of energy as it 2 WT /Mwithin a given shell will have different energies (arising from the attractions and repulsions of electrons with the nucleus and other electrons in the atom). As a result we would expect to see emission lines for all the transitions. The emission spectrum for potassium would have more i lines than hydrogen. { 2M. Q3. The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 ><103 kmsl. A. What is the wavelength of the ejected electron? (8pt) Use the de Broglie relationship, 2 = hp"1 = h(mv)’1. 2pt me = (9.109 39 ><10'28 g) (1kg/1000 g) = 9.109 39 x10“ kg (3.6 x103 km - s“) (1000 m'km‘1)= 3.6 ><106 m - s‘1 lpt for each conversion. /l = h(mv)_1 _ 6.626 08 ><10'34 J - s 2 t (9.109 39><10‘31 kg) (3.6><106 m-s'1 p = 2.0 X 10—10 m 2pt B. No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50 X 1016 Hz. How much energy is required to remove the electron from the metal surface? (6pt) E = hv = (6.626 08 x1034 J . s) (2.50 ><1016 s'l) =1.66><10‘” J 2pt for each step -1pt for sf —1pt for no units or incorrect units Q4. The following question follows on from question 3. What is the wavelength (in mm) of the radiation that caused photoejection of the electron? (12pt) The photon needs to contain enough energy to eject the electron from the surface as well as to cause it to move at 3.6 ><103 km - s‘]. The energy involved is the kinetic energy of the electron, which equals émvz. Emon = 1.66 ><10‘17 J + % mv2 = 1.66 ><10‘17 J + 5(9109 39 ><10‘31 kg) (3.6 ><106 m - s’1)2 21.66><10‘l7 J + 5.9><10“8 J = 2.115210“ 1 lpt for each of the above statements or steps (6pt max). But we are asked for the wavelength of the photon, which we can get from] E = hv and c = 122. or E = hC/l—l. 2pt 2.3x10'17 kg-mZ-s‘2 = (6.626 08x10’3“ kg-mz vs—l) x (2997 92-108 m‘S—1)fl.-1 1: 3' 33x M‘qm 2pt for each step (6pt max). —1pt for no units or incorrect units QSA. Write the ground-state electron configuration of Cl. How many unpaired electron(s) does it have ? [Ne] 382 3p5 one unpaired electron (271‘) @W B. Write the ground—state electron configuration of Pb“. How many unpaired electrons does it have ? [Xe] 4f14 5d10 6s2 none (m) (m) C. What is the subshell notation and the number of orbitals having the quantum numbers n=3,l=2? 3d and 5 (2,1) (”of“) D. How many elements can have valence electrons with the quantum numbers n = 2, l = l ‘? 6 (W (4pt) MM) (4130 (3130 Q6A. The following Lewis structure was drawn for a Period 3 element. Identify the element. (8pt) :ZCiI—E—Ql: l :Cl 2 The structure has a total of 32 electrons. Of these, 21 are accounted for by the chlorines (3 Cl’s X 7 valence electrons each). (2pt) Of the 11 electrons remaining, 6 come from the oxygen. (Zpt) This leaves 5 electrons unaccounted for; these must come from E. (2pt) Therefore, E must be a member of the nitrogen family and since it is a third period element E must be phosphorous (P). (2pt) B. Two contributions to the resonance structure are shown below for each species. Determine the formal Charge on each atom and then, if possible, identify the Lewis structure of lower energy for each species. Circle your answer. (8pt) T ‘ H’T' T T' ’TT ‘ -___--__. .. .. I z'o—s o: - :o: :0: :o: : o: (w (w :6- 1 1 () :1. III . O a o—slo: .. I :6z'1 1 (b) i‘- |+2 0 0-8—0: | H 1:9: H lower energy Q7. Give the VSEPR formula, name the molecular shape, state if the molecule is polar or non-polar, and estimate the bond angle. A. PBr; (4W) VSEPR formula in {0 molecular shape / 1 \ Kf ; polar or non-polar :K/I gr -. estimated bond angle PBr3 is pyramidal with Br———P—~Br bond angles of slightly less than lO9.5°. AX3E; polar B. CCI4 /CK (4130 l 706. VSEPR formula C ‘9 molecular shape / polar or non-polar $6. a ' estimated bond angle C. C02 (4P0 VSEPR formula '0‘ __ C _._.. '0 molecular shape polar or non-polar estimated bond angle AX2 linear non—polar 180O Q8. Give the VSEPR formula, name the molecular shape, and estimate the bond angle. A. SCHZ H H \ / VSEPR formula ‘ Se , molecular shape estimated bond angle AXZEz bent less than 10950 B. XeOF4 . 0 . =F H P ,. \ / .. VSEPR formula X6 \ -- molecular shape . o / 0 0 F: ' : F " estlmated bond angle AX5E square pyramidal distorted octahedral (less than 90 or less than 180) C. 3032' " VSEPR formula /5 \ -. molecular shape ____ '0 H .03 estimated bond angle ____ 3032' is pyramidal with O—-—S——O angles of slightly less than 109.5°. AXBE; 10 (3130 (3pt) (3pt) ...
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