SI Practice Exam 1 Key

SI Practice Exam 1 Key - ENGR 145: Chemistry of Materials...

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Unformatted text preview: ENGR 145: Chemistry of Materials SI Practice Exam I 1. In an effort to reduce the commercial power consumption of your dorm room (even if only very, very slightly), you want to engineer an electricity generating device that uses the photoelectric effect. After doing research on the subject, you decide that you will place a cesium electrode in the sunlight and wire the electrode to power your room. a) Knowing that Cs has a work function of (D=3.36 X 10-19 J and that sunlight has a wavelength of A=500 nm, what will be the average kinetic energy of the generated electrons? .. n; '- , a) ORV L:th “'1 I: h) M ar *— c/Q i; I F :(c.oicyxo'3”35)(exw'”l KE= (KW-33" “’0 (‘51!0 M5 ’ - t. I hlq : Soaxtb‘qm E: 31??)cf0 '4 J— “!23 O:GZX(0 j , firm “Wormws‘ W “Baum-203 b) With what velocity will each electron be emitted from the electrode? (HINT: mass of e“ = 9.11 x 10-3]L kg) r I 2 K ameli— UI; 2CuE)/m¢ 2cm) 6-— F—EE zCe.Zon‘°-°J) U”: 3:9‘le05M/S c) Next year, if you were to transfer to a school that got more sunlight than Case (the intensity of the sunlight was higher), would the velocity of the photoelectrons increase? Why or why not? No, Utlacrly .‘5 dgperwlcn-t an waveleafim (and! fir,,cW/)J not ‘Hu; Integral), 04 4M light, d). In the sunnier climate, would your cesium electrode be able to generate more electricity (emit a larger number of electrons)? Why or why not? Yes, The more Michel ‘l'lq; [likid flue Iarga mumlo-ef of e leclrons are: emiHMJ’ Answer the following questions about electron configuration and periodicity. a) Draw the electron configuration of the following elements: Mg, P, S Ma '_ [N e\ '5 3 1 P i {N ex ‘5 5" '5 Q 3 :5: UN”) ‘55?” 394 have b) Based on these electron configurations, rank the-toug'compounds from lowest to highest ionization energy. Explain your reasoning. M 4 s 4 © Gem.me lom1m\\cn exam” magmas :50 “a atmgg 4H,. Jungle. “Qwuer‘ {3 ‘hos an hail-r" Sm“ P she,“ ulnng ‘9va95 9341\nle amé “\CTC39‘9 A‘s Lbnl‘iflllb A c) The presence of the d and f orbitals greatly diminishes changes in ionization energy between subsequent atoms in the periodic table. Suggest why. a s- 5; c: dad u\s {a (‘bv tie. (pkg—Atlanal sxblokl \Lj MA g‘muk “‘3 I “cleans the. gnu-sax} m.th 5&0 ex: exec mo. nwlmr P U l\ 0mg decSCG-S\nfl {p03 mslcbmldn Leora-olden ng‘vxd can '9, _ (1) Copper has an electron configuration different from what you would expect. Draw its configuration and explain why copper adopts this configuration. 4s} girdle *" 9mm an éi Li‘s e,— Jro the 3d 0‘\°M\‘ SW)“ 0" mednon Qmwles Cv MAMA 0~ L’ll' 4r \wVV‘ iv“ 9:10.“, when muens es slnhA‘H- 3. The properties of refractory ceramics are extremely important to the metal casting industry. To get desirable properties, it is preferred that the ceramic be made of molecules that form strongly ionic bonds. a) Using your knowledge of periodic trends, which of the following elements would you expect make a form a more desirable refractory ceramic with oxygen: Mg, Ca, Be, or Sr? Explain why in terms of the elements’ electronegativities. The Fatwa-(C 41(an In! Skew-5 m4 alga-cw; as Parcels ahc‘rtagr Tl‘grezfmf 71¢ st :51”? Cane wfi'A {wags} A5“) Wt.” palm Walk ekn‘j 5C5)“ Cdfamn 13' (603(5),: b) Calculate the percent ionic character of magnesium oxide (formula: MgO). The electronegativity of Mg is 1.31 and the electronegativity of O is 8.44. O/O'IC: l—f:PC—,2S—(KMfl-xo>l] 0/016 = I~ aspf-.ZS‘(I.31~ “am—:31] %Ic: , (07%? 6/0 1C: 4. Answer the following questions about periodic trends. a) For each pair, select which has the higher melting point. i. 0014, @ ii. 0F2,@ b) For each pair, identify the more electronegative species. 1. Rb,@ ii. ©Se 5. Oxidation numbers were determined for the following compounds. Circle all of the moleCules that have properly assigned oxidation numbers +2 —2 '10 +2 2* 8602 O4] 6W4 ‘1“ gimme 2 Hi -'L . H, J). S1501, (SOq>2- 6. Answer the following questions about interatomic bonds a) The 0-0 bonds in benzene (Cefls) are all 140pm. This length is greater than a double bond (135pm), but less than a single bond (147). Explain how the electronic structure for benzene might account for this behavior. «ye—«Eye The. (2551\an 33m; c m (3% benzene. Cw‘k“ e; 'lo become, AB\bCa~\‘ sol ’<\\mvg\~oujr 5th; Tl “>11 Slim. A“ six be at»; algal l‘ 5 , \nltu maéw‘l't l0 ‘5W‘Q3l9— ‘- 500‘h\1 \'3 and: _ 0“th b) Acetylene (C2H2), ethylene (021—14), and ethane (CgHs) all contain two bonded carbon atoms. Explain why acetylene is used in gas welding instead of ethylene or ethane. Resell. zllc=CH1 ,,\-\Cvc.ll3 P“ “‘3 \‘M- \‘5 his welsh. \s an to 1. 4. it w. \x lcéhlj ft“¢-‘L\V o. Omd 9 r0 Bums o bubk *twefirlu m Hhmt- R “Al-g 422M: 5 \wtyfl3 (malt-we. 1C¢\?\°‘ ‘00 m3 Cb“ T\\aul\: lo (~69 Lcl 4.. $U‘Llo.\nfi.& CQM\(¢UQ_3\Qn _ “"3 9&0?th “3 move. lMfiabAupl Jc‘nm haul él coMXuglxbfi (hated :6. AMA. 7 . In one of the final steps of the steel-making process, a sheet of rolled steel is immersed in a bath of nitric acid (HNO3). Because of its chemical structure and arrangement, nitric acid has the ability to remove surface tarnish from the surface of the metal. Below are three different isomers (chemical arrangements) of HN03 drawn as Lewis Dot Structures with formal charge. Briefly explain which you expect to be the stable form of the acid and why. lsomer 1 Isomer 2 Isomer 3 I-1 a, . {+3.} -. ""‘iO.’ . . <3“- - e“? - «4 'J '_ _ I Jr:- t_ * : N;Q;H :Q :(_)_::O:N:"" IQ . . N I n O - H /”“-. H x"*». 8. The Bohr model of the atom depicts the atom as a positively charged nucleus surrounded by electrons that travel in circular orbits. The Bohr model is a quantum—physics based model which explains some, but not all, of the properties of the atom. Circle all of the following that are true of the Bohr model. The Bohr model was the first model to provide a quantum mechanical explanation for the movement of electrons in an atom. r7 cult: liar 3\v%\g e“ alomg b) The Bohr model accurately predicts emission spectra for all atoms. @ The Bohr model provides a means for calculating the gap between energy levels for single-electron atoms. (1) The Bohr model was the first to introduce the concept of wave-particle duality. L1 meet. an” \MemeA quwlrun phages. OW“ GAMMA plague}. w \lk n O u “AW-Wig \Hj exe\qh¢;y10i\ . 9. The following are statements about metallic bonding. Circle all statements that are true. r—j Hal-r4 “.72 ,3 :is owk ‘lwmc a) Metallic bonding is a type of ionic bond. of bad b) Metallic bonding occurs in elements where the electronegativities are both relatively small. c) Metallic bonding can be approximated as the positive ion cores of elements being held together by the columbic attraction of their electrons. d) The “sea of electrons” in a metallic bond is formed by all of an element’s electrons, regardless of electron energy state. (I \\ L—) ,3th Valfnde glee-lam; ler.bw-‘I-€ '10 Sf“ 10.Answer the following questions about intraatomic bonds. a) Methane (CH4) and ammonia (NI-13) both form molecular solids. Explain why methane is a gas at room temperature While ammonia is a liquid. H ,v amp—t 9”?“ H H TM @‘fi‘mt ‘3 c' \M“ PM“ G“ N allows \‘l 4° S}ch ll~leovrls lam-luau.“ NR3 molewle‘i. “mesa mlmmclmns 05‘9- Sifl‘b ream llaw like V04" at: \N Wk \MXQ’W \‘lum 93‘ (A v“ mad-bung. b) Hydrocarbons are organic compounds containing only carbon and hydrogen. Consider the alkanes, chains of carbon atoms joined together by single bonds. As the number of carbon atoms in an alkane increases, so does its melting point. Using your knowledge of Van der Waals forces, explain this behavior. A5 \X‘séwwt‘MoM (5le large}. 36kt GeGOrXUwE go? V.» dc L-Uuo.\$ \“XQWAWV‘S were}; 9%. Tm». aphid“) MA mlgmlwwns ‘0th 91" ‘Emcceskwelj leg-«59f M\e,c,u\gg can)”; M¢¢ QNWMA .10 éfimpl’ \WU‘EH-h‘g Alma. thH-fi funky 11. Place the following compounds on a bond type triangle below and identify the type of bonding present. 3) AgCI (ENAg: 1.93,EN01: 3,15) Covaknt b) CN‘ (ENC = 2.55, ENN = 3.04) CUVfildn‘l 0) ngS (ENHg = 2.00, ENS = 2.55;) am. 11d d) Dy203 (ENDy = 1.22, END = 3.44) ionic 8) GaAs (ENG... = 1.81, ENAS = 2,18) seam“ f) LieN (ENLi = 0.98, ENN = 3.04) 50"“ U3: ils. ‘ ii- . .i' I. IIIIIIIII IIIIIIIII IIIIII IIIII III-III ‘IIIIIIII III III IIII III :=- III III-III- IIEIIIII iiii' i. A IIIEIAII a. === .a IIIIIEIIII I II II I I I I E i I II §_EE IIIIIIIIIIII IEIIIIIEEIEIIIIE E III. IIIIIIIII IIIIIIIIEII I II E E- IIIIEIIIII III-q- I: -EII Ii I IIIIIIIIEEIIII E ‘III Iis II IIIII IIIII IIII IIII IIII III,“ I ::::.=:.as.. IIIIIIIIII I ‘ I III-III? i‘ a V V :::=: MIAMI!E I I n E I: III-IMIIIIIINIIH Illllfllllllllflfll IIIIMIIIIIII IV V IIIluIIIIIIIInnIHIHIII I ‘llfl IIInIII I I '1 II‘ :::agl||||iy%§§%I§"':§:=:==:=:. IIuI 1- Ii Eli=l=l====lli IMIIIIIIIIIHIIIIRI cm 0.3 9.91.0 1.1 1.2 1.3 1.4 1.5 1.5 1.1 1.3 H} 20 2.1 2.2 2.3 9.4 2.5 ms 9.7 2.8 2.91.; 3.2 3.3 M as 3.6 3.7 3.8 3.9 4.0 4.17.2 Afl EM g) In class, we discussed some of the limitations of the bond type triangle. Briefly state one of these limitations. ’ Aswo—s denial L2 any % 4J1 +L€ .Followuig ,, -— SJOICL:0Mi—rf nd‘lL iahén “trio Office-‘1‘ ‘— mokcuksi arfflJO-‘fikm 2’. aJ‘WKS- Canna'lLlLt ‘— OiUCQ hoi' ACCOuA‘I’ ‘FN" QCWJWy 'bWJ“r\3 IIIII [III-III IIIIE IIIIIIII IIEIII III-II I IIIEII IIII I I! a. 12.Answer the following questions about hybridization. a) Draw the Lewis dot structures for acetylene, ethylene, and ethane and identify the hybridization of each carbon atom. b. u u HICZZLC'JA mend-.14 “.5613” H g» ":P $99. 3:33 b) Explain why the hybridization changes for each compound. Hint: think about the number of bonds on each carbon atom. Rios-x5 uubvcacp the. “\n\w\ balaflbuyr-lgoq “$1530.13 3‘1‘: (1‘be Jaw. “Qt—lg Ski-8319. \bofilg - “a 0.3).\\\¢na.\ D mag he U3 Qé (m digentvoltfi orb dish) Cor lom 9kwcx or \u Abe. C059— 5 (3 orlaxlab‘ {30“ Tl- l’o“(\'\”‘3 m Aou‘ele '\' 5““?l’1 \c‘O‘V‘A’i. 0) Based on this explanation, predict the hybridization on oxygen in 002. 3W wean w new: WW men has a gull $ skell- $Q \5 no‘l eff—$519. becwfie ‘l'he deuce dcgmm'll E) 96 ace. 5A,“ Clout gem-g.) I maxim} Mae. 3°0th bona- wvveo’ittéflltv 33L babeéigalsbn Pu’ls lone (Dan-5. \n spe- or\° ‘5’ ltb—M—L ACME bwflyt 6A6 ‘k‘ {ire- ‘Lb Make Q "-11 “d . d) Some atoms for h rid hybrids with the d orbitals as well as s and p orbitals. These orbitals are known as sp3dn hybrids, where n is the number of d orbitals involved in the hybridization. SE; is such a compound. What is the hybridization of sulfur in this compound? W9) ‘32 3w swab; bowl: Muu‘e. 3U! \Msbwe— m\m\~uls . This Cam lee, &¢J\\K'L\J‘°A Lkwutbh S Q93 80' \5¥r\1\ 35AM)“. 13. GaP is a semiconductor that is common in light emitting diodes (LED). The basic principle to an LED is quite similar to a solid-state laser: electrons at the bottom of the conduction band recombine with empty energy states at the top of the valence band. The energy that is freed by this recombination is emitted as a photon of light. a) Given that GaP has a band gap of Eg = 2.26 eV = 3.62 X 10-19 J, what wavelength of light is produced by the LED? 13th kc E: '67 In: P. 9i"— /’=3 q ca.mcm'w"")(3x’0 ) a: gilflleQ-‘IQ l9: Squhml b) What color is the light? (Drama 0) If you wanted to make an LED that produced blue light, would the band gap of the semiconductor need to be larger or smaller than that of GaP? Why? glue {5.24 a larger [A Hum flrceu h . I:—- C 3) larSEr g W‘C’MS SMHEr E9 ...
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SI Practice Exam 1 Key - ENGR 145: Chemistry of Materials...

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