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Exam 1 Key - CHEM 25 “j Z Exam 1 Name Lehigh email Ii...

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Unformatted text preview: CHEM 25 “j Z , February 11, 2009 Exam 1 Name ' Lehigh email Ii) I’lease check the studio to which you are assigned: Mon PM Mon EVE Tues AM Tues PM Thurs AM Thurs PM ‘ INSTRUCTIONS: I. You must show your work to receive credit. 2. Use the correct number of significant figures in answers to all questions on the exam. 3. If necessary, please continue your solutions on the back of the preceding page (facing you). 4. There are a total of 7 problems on 9 pages. Count the pages before you begin. Multiple Choice (30 points.) 1. Clearly circle the BEST answer for each of the following questions (3 points each). A. If you had 6.3 g of each of the following substances, which would occupy the greatest volume? ' (1) ice (0120.917 g/mL) (2) beeswax (d=0.96 g/rnL) (3 water (d=0.997 g/mL) ocoa butter (dm0.91 g/mL) B. Germanium has five naturally occurring isotopes. Which is the most abundant isotope? (1) ”Go, 69.9240 amu (2) ”Ge, 71.9216 emu @73Ge, 72.9233 arnu (4) 74Ge, 73.9210 amu (5) 76Gc, 75.9214 amu C. Buffer solutions that maintain certain levels of pH or acidity are widely used in biochemical growth media. A commonly used buffer system has these two components: sodium dihydrogenphosphate and sodium mouohydrogenphosphate. Identify the pair showing the correct formulas . (1) NaHPO4 3133 N3(HPO4)2 (2) Na2H2P04 and NaHPO4 (3 NaPO4 and. NaHPO4 @NaH2P04 and NazflPO4 D. A tiny speck (2.0 X 10”"4 g) of Americium~24l is used in a smoke detector as an a particle source. 241Am has a molar mass of 241 .0567 g/mol. Assume the sample is isotopically pure. How many atoms are there in the sample? @50 X 1017 atoms (2) 1.2 x 1020 atoms (3) 2.9 x 1022 atoms (4) 6.0 x 1023 atoms E. A US. penny is composed of zinc (97.5%), plated with copper (2.5%), and weighs 2.500 g. Hydrochloric acid reacts with zinc but not with copper to form hydrogen gas (Hg) and zinc chloride (2.510;). A penny with a scored edge allows the HCl to react with the internal zinc metal, leaving a shell of unreacted copper. How many milligrams of hydrogen gas are formed by reacting a penny with excess HCl‘? (1) 0.765 mg (2) 30.0 mg 75.1 mg 4) 76.9 mg F. Pick the correct net ionic equation for the reaction that takes place between a solution of magnesium chloride and a solution of sodium hydroxide. (1 ) MgC12(aq) + Na0H(aq) —> Mg0H(s) + NaClz(aq) (2) MgClg(aq) + 2 NaOH(aq) we Mg(OH)2(s) + 2 NaCl(aq) @Mg: (aq) + 2 037219 —> Mg(OH)2(S) (4) Mg (aq) + 2 or (aq) + 2 Na (aq) + 2 omag) a Mg(OH)2(s) + 2 NaCl(s) G. Combustion analysis of an organic compound to detennine the percentages of carbon, hydrogen, and oxygen in the formula depends on which of the following assumption(s)? @The compound bums completely to form carbon dioxide and water. (2) All the oxygen in the products comes from the added oxygen gas (3) The mass of the resulting carbon dioxide 13 equal to the mass of the carbon' in the sample. (4) All of the above are correct. I H. If 50.0 mL of a 0.10 M solution of sodium chloride is mixed With-50.0 mL of 0.10 M magnesium chloride, What is the molar concentration of chloride in the resulting solution? (1) 0.050 M (2) 0.10 M (3 0.20M Q) 0.15 M I. If one regular Tums tablet contains 500 mg of Ca003(s), how many mi. of 1.0 M stomach acid (HCI) could it neutralize? $5.0 mi. 10. mL (3)50.1nL (4) 100. 1111. J. Nitrogen—fixing bacteria and plants are capable of converting atmospheric nitrogen to ammonia (NHg). This process is an example of (1) an acid base reaction. (2) a precipitation reaction. (3) an oxidation reaction. .a reduction reaction. Calculations 2.) (17 total points) Permanganate ion (M1104) is sometimes used as an antibacteriai and antiviral agent to purify water. Excess permanganate makes the water an unappetizing light brownish purple color. The excess permanganate can be removed from the water by bubbling 302 (g) through it until the color disappears. One of the products of the reaction is M11804 (5): q 2 H20 (1) + 5+sog (g) + 2 1x330; (aq) a» 2 {discus} + {sang-(am + 4 imaq) (a) (6 points) How many grams of 802(g) are required to treat 3.60 L (about 1 gallon) of water that is 0.227 M in M1104" ion? I answer (a): 13> l 3’ Work space: 3.(ooL x 0.771} mailing; 5M 50; x 54,07 350‘; W 2% P111001" . (met so; 3 l3l 330.29.. (b) (4 points) What is the meter concentration of sulfate ion in the final solution? (Assume the chemical reactions that take place do not change the volume of the solution.) - - lmriEO so _.. 1‘ {3:35on we)“ 3&0 , (game: 0,3LH “l Work Space: ~..— 6 [+013 SYMQSDL 3,. (a D L er€r\ (c) (4 points) Identify the species indicated from the reaction Oxidizing agent Reducing agent Element Element reduced oxidized (molecule, atom or ion molecule, atom or ion ((1) (3 points) Write the balanced molecular equation for the process that takes place when 802 (g) is bubbled into water: Answer (d): 3.) (10 total points) You are given a sample of a Group II metal carbonate. You react 1.437 g of the sample with excess HCI (aq) and determine that 0.750 g of a gas is given off. Answer the following questions: (i) (6 points) Detennine the identity of the metal in the metal carbonate and wn'te the formula for the compound in the space provided. Answer: Work space: 0,7608 0/01! M 1’? 0- (DWOI-l met COO): CO 440/]. 91 W MM *4ng WMPLW 1: €4,33a/mfla w(l2.0l%/mrE-+3(lb.00 élmall) :2 34.3 IOWOle 56m“? ................ W343é§1mmo H W: N5 (ii) (4 points) Write a balancg‘édm net Ionic quation that describes the reaction that occurs when the metal carbonate m (i) IS tr ’ l (aq). If you have not been able to establish the identity of the metal caxbonate m pent (i), you may use the symbol M for the metal and mite a general equation. 4.) (15 points) An analysis of an unknown liquid by mass spectrometry yielded the following ' mass spectrum: 1a; 107 t”: 3903 54,3 I... .‘§ M+'u138 Combustion analysis established the percent composition of the material as: C: 69.54% H: 7.30 % O I 93’1970 What is the molecular formula of the compound? Molecular formula: Cc? H i 0 OJ.» “"1 Work space: MSLijC/xhifii r. WmLf ia-Oij 6 PM 75’ I «#6 H Clfigo H9 17308 H ([11:03 H -" 7L-M2‘ll/mreH w 5 (Ha/fr! a \.L+cn5 "I" ._ LMMCM M I @ cm wide Wt“? 93 {lacs O X /W‘—e O lLiLi?5/VV€ O W l } mflf‘mvm I6 00 0 _ “‘” m‘ J l LN 73) MHEyMonqga- @231“: f; in" 7 12:6” 5.) (10 points) Fill in all of the blanks in the following table: Atomic Number (2) Number of Number of Mass Number (A) Neutrons fitectrens _ymbol 6.) (6 points) Synthetic chemists working for Merck come up with two alternative syntheses for a new anticancer drug to compete against taxol. Both procedures for making it require four steps. The yieid in each step for both methods are given on the table. Which procedure results in the greatest overali yield of the candidate drug? Mathematical work must be show to receive credit for your answer. _ Work space: ts) Many salts of lead (9b) are highly insoluble in water, which makes them useful as in paints. Unfortunately many of these lead pigments are quite toxic, so they must be in environments where there is no risk of them being ingested. You get a summer job in a laboratory where the toxicity of lead-based pigments is being evaluated, and you have been asked to prepare small batches of insoluble lead salts. You mix together 175 ml. of a solution that is 0.392 Min lead (II) nitrate with 245 mL of a second solution that is 0.375 M in sodium chloride (NaCl) and produce thereby a White solid known in the business as ‘sugar of lead.’ You need 10.0 g of ‘sugar of lead’ for the study. Do you get that much (at least 10.0 g) product from this reaction, or do you need to run it again to have enough material for your study? Answer: Yes — this synthesis yields “’0 Lg a 9 f . g of product VLO h’i ’Lfi which is enough for the study GR No -—— this synthesis yields only g of product and I need to run the reaction _ again. waofimfl +- QNnQMfl 4+ Pouch“) +':;LN&LNO:3(&_3) merit 245ml. (3.796221% 0315M on“ A IBEEMOIWMS)” m! Pbfig 509.! jpfgae 5: W03 3 P505.) [Moi Mm); ’W" a» 19“.! F i ‘ , p M PM . autism: 335;“: N Li X WLX 9.?(843W’6L ; 152.? to 71301;) 7 351% Mid [Mi Flatt; I @ OLE,“ LR (‘0,i?§b)(o {79?2333) I: .0 lei/33¢: ”4&6 {9.12ng 3 0.0681”; W206 Ho“ 1.. C0 geese) (o ; ea Se.) 3 e 043/19 We Ma (it .. L w ”a - wet 2U, {:01 We” QflrLE 9.1“; ‘9 )2 ._ -: 0 .e‘? {0.1 mw‘tCt— ...
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