14A_Fall2005_Midterm_Answers

14A_Fall2005_Midterm_Answers - CHEM 14A YOUR NAME...

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Unformatted text preview: CHEM 14A YOUR NAME .fi.fl.§WéK.~S... Instructor: Dr. Laurence Lavelle STUDENT ID# ............................. .. FALL 2005 MIDTERM (Total number of pages = 11) (Total points = 120) (Totai time = 2 hours) “Carefully remove last page which is your 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 Constants and Formulas Planck constant, n = 6.62608 x 10‘34 J - s Avogadro constant, NA = 6.02214 x 1023 mol’1 Rydberg constant, R = 3.28984 x 10l5 Hz Gas constant, R = 8.314 J.K‘1.nnol'1 Mass of electron, me = 9.1 x 10'31 kg Speed of light, c = 2.99792 x 108 ms‘1 C2 : Second radiation constant = 0.0144 K-m 0°C=273.15K 1L=1dm3 c = A v E = h v h R h E, = — ’7' 2 = — n p A Ax > -h— E — ——7h2n2 p X — 4n n ‘ 8 m L hZ ETOTAL WU) = EK WOO + V(X) WOO = - Snz m _ [Au ]INIT1AL SALT pH _ pKA + LOG [AH]INITIAL ACID Solution to AX2 + BX + C = 0 is 1atm = 101.325 kPa E 2 pc E = %mv2 2 £1All—(2)2 + V(x) \Mx) 2A —B +/— ,/B2 —4AC 115:3.14 p=mV QlA. Balance the following chemical equations: (8pt) (a) AgN03(s) —> Ag(s) + N02(g) + 02(g) (b) P285(s) + PC15(s) ——> PSCl3(g) (c) BF3(g) + NaH —> B2H6(g) + NaF(s) (d) LnC2(S) + H200) —> Ln(OH)3(S) + C2H2(g) + H2(g) (a) 2 AgNo3 (s) —> 2 Ag(s)+2 N02(g) + 02(g) (b) 1355(5) + 3 1>c15 (s) —> 5 PSC13(g) (c) 2 B133 (g) + 6 NaH(s) —> B2H6(g)+ 6 NaF(s) (d) 2 anz (s) + 6 H200) —> 2 Ln(OH)3 (s) + 2 C2H2(g)+H2 (g) 0.5 pt for each correct coefficient Q2A. Determine the empirical formula for Saccharin, a sweetening agent, which has mass composition 45.89% C, 2.75% H, 7.65% N, 26.20% 0, and 17.50% S. (7pt) For 100 g of saccharin, 45. moles of C = ——-89—g—1 = 3.821 mol lpt 12.01 g - molfi moles of H = Jig—4— : 2.73 mol lpt 1.0079 g . mol moles of N = = 0.546 mol 1pt 14.01 g ~ mol moles of O = = 1.638 mol lpt 16.00 g - mol“ moles of s = ——17—'59g—_T = 0.546 mol lpt 32.06 g - mol Dividing by 0.546 mol gives a ratio of 7.00 C: 5.00 H 2 1.00 N: 3.00 O: 1.00 S. The empirical formula is C7H5NO3S. 2pt B. A chemist found that 4.69 g of sulfur combined with fluorine to produce 15.81 g of a gas. What is the empirical formula of the gas? (Spt) moles of s = —4—‘69—g—1 = 0.146 mol lpt 32.06 g - mol‘ M = 0585 mol moles of F = l 19.00 g - mol" lpt Dividing by 0.146 mol gives a ratio of 4 F: l S. The formula is SFA. 3pt Q3. The velocity of an electron that is emitted from a metallic surface by a photon is 3.6 x 103 km“. A. What is the wavelength of the ejected electron? (lOpt) Use the de Broglie relationship, xi 2 hp'1 = h(mv)'l. 2pt (3.6 ><103 km - s“) (1000 m - km") = 3.6 x106 m - s:1 2pt for conversion. /1 = h(mv)_1 _ 6.626 08 ><10’34 J - s 2 t (9.109 39x10‘3‘ kg) (3.6><106 m-s" p = 2.0 ><10‘10 In fig 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 x10“ J - s) (2.50 ><1016 s“) = 1.66 ><10’l7 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 nm) of the radiation that caused photoejection of the electron? (16pt) 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 X 103 km ~s'1. The energy involved is the kinetic energy of the electron, which equals %mv2. E =1.66><10"7 J +lmv2 hoton p 2 l =1.66><10"7 J +E(9.1 ><10-31 kg) (3.6x106 ms“ 2 =1.66><10’” J + 5.9x10‘18 J = 2.3><10”'7 J 2pt for each step. But we are asked for the wavelength of the photon, which we can get from E = hv and c 2 wt or E = hC/t‘l. 2pt 2.3x10‘” kg~m2-s'2 = (6.626 08X10‘34 kg -m2 -s‘1) x (2.997 92x108 m-s“)/t“ ,1 = 2,6x10' 7m = 8.6nm 2pt for each step. —1pt for sf —1pt for no units or incorrect units QSA. An astronomer discovers a new red star and finds that the maximum intensity occurs at )L = 715 nm. What is the temperature of the surface of the star? (5pt) From Wien’s law: Tam = 2.88X10‘3 K-m. (T)(715><10‘9 m) = 2.88><10'3 K-m T:4.03><103K -lpt for no units B. Write the ground—state electron configuration of a lead atom. (6pt) How many unpaired electrons does it have ? _ i Z [Xe]4f145d106326p2 LJLJ LJ U LJ 1 l l l I C. What is the subshell notation and the number of orbitals having the quantum numbers n=4,l=2? (2pt) C 4d and 5 { ) D. For dinitrogen monoxide, the arrangement of the atoms is N—N-O. In the Lewis structure with a single bond between NN and a triple bond between NO, the formal charges on N, N, and 0, respectively, are (3P0 ~2,+1,+1 Opt—€401» Q6A. Write all possible Lewis structures of sulfur dioxide. Which structure is most stable? (8pt) SULFUR DIOXIDE, 802 O O O O 1 -1 -1 1 O 06 on u 00 g. 00 .00 g. 00 00— 8—.(2 90: 8—5.): .O— 82.. IST LEWIS STRUCTURE MAKES BIGGER CONTRIBUTI N TO RESONANCE HYBRID. The structure with the expanded valence is favored. (Z 14/? B. Which statement is true? (one answer) 1 l (4pt) A) Atoms with high ionization energies and high electron affinities are highly electronegative. B) Atoms with high ionization energies and high electron affinities have low electronegativities. C) The electronegativity of an atom depends only on the value of the ionization energy of the atom. D) Atoms with low ionization energies and low electron affinities have high electronegativities. E) The electronegativity of an atom is defined as l/2(E1ectron Affinity) of the atom. A) Atoms with high ionization energies and high electron affinities are highly electronegative. C. Which of the following compounds contains the strongest bonds to hydrogen? (4pt) (Circle one answer.) SiH4 CH4 H28 H20 HF How many unpaired electrons does it have ? Sf [Ar 3d4 TitlLiM/t D. Write the ground-state electron configuration for Cr2+. ’17 fipt) E. Give the ground—state electron configuration for I -. (Spt) How many unpaired electrons does it have ? D / A17 [Kr]4d105s25p6 LJ L! L./ L} ///l Q7. A. Give the VSEPR formula, name the molecular shape, give the hybridization of the central atom, state if the molecule is polar or non—polar, and estimate the bond angle. PC13 /Mfwwt VSEPR formula molecular shape MAS”; hybridization of central atom polar or non-polar estimated bond angle PCl3is pgéamidal with Cl—P—Cl bond angles of slightly less than 1095". AX3E; polar; B. GeH4 is tetrahedral with H—Ge—H angles of 1095". Note that GeH4 has the same electronic structure as CH 4 because Ge lies in the same group as 3 p GCH4 C. AX4 non—polar; sp3 C. AXZ Sl02 linear SP non-polar 1800 10 VSEPR formula molecular shape hybridization of central atom polar or non-polar estimated bond angle VSEPR formula molecular shape hybridization of central atom polar or non—polar estimated bond angle (5pt) (Spt) Q8. Give the VSEPR formula, name the molecular shape, give the hybridization of the central atom, and estimate the bond angle. A 3H2 VSEPR formula as" molec lar sha e H/ \H u p hybridization of central atom estimated bond angle AX2E2 bent sp3 less than 109.50 B 13+ VSEPR formula , .. .. .. + molecular shape "I\I/I‘. hybridization of central atom estimated bond angle 1; should be bent with a bond angle of slightly less than 109.5°. AXZEZ; sp3 C. SeO32' VSEPR formula : O : 2- molecular shape l hybridization of central atom estimated bond angle W Sle' is/tpyramidal with O—Se—O angles of slightly less than 1095". AX3E; Sp3 11 (4pt) (4pt) (4pt) ...
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14A_Fall2005_Midterm_Answers - CHEM 14A YOUR NAME...

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