8. Relax and do well.

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1 EM Exam II John II. Gelder March 5, 2002 Name TA's Name Section INSTRUTIONS: 1. This examination consists of a total of 8 different pages. The last three pages include a periodic table, a table of vapor pressures for water, a solubility table and a table of thermodynamic values. All work should be done in this booklet. 2. PRINT your name, TA's name and your lab section number now in the space at the top of this sheet. DO NOT SEPARATE TESE PAGES. 3. Answer all questions that you can and whenever called for show your work clearly. Your method of solving problems should pattern the approach used in lecture. You do not have to show your work for the multiple choice or short answer questions. 4. No credit will be awarded if your work is not shown in 6 and Point values are shown next to the problem number. 6. Budget your time for each of the questions. Some problems may have a low point value yet be very challenging. If you do not recognize the solution to a question quickly, skip it, and return to the question after completing the easier problems. 7. Look through the exam before beginning; plan your work; then begin. 8. Relax and do well. Page 2 Page 3 Page 4 Page 5 Page 6 TOTAL SORES (22) (18) (24) (18) (18) (100)

2 EM 1515 EXAM II PAGE 2 (9) 1. Write the chemical formula(s) of the product(s) and balance the following reactions. Identify all products phases as either (g)as, (l)iquid, (s)olid or (aq)ueous. Soluble ionic compounds should be written in the form of their component ions. a) Fe(NO 3 ) 3 (aq) + KSN(aq) FeSN 2+ (aq) + 3NO 3 (aq) + K + (aq) b) uso 4 (aq) + 2NaO(aq) 2Na + (aq) + 2SO 4 2 (aq) + u(o)2 (s) c) Na 2 O 3 (s) + 2l(aq) 2Na + (aq) + O 2 (g) + 2 O(l) + 2l (aq) (4) 2a. Write the ionic and net ionic chemical equation for 1a), 1b) or 1c). Ionic equation Na 2 O 3 (s) (aq) + 2l (aq) 2Na + (aq) + O 2 (g) + 2 O(l) + 2l (aq) Net Ionic equation Na 2 O 3 (s) (aq) 2Na + (aq) + O 2 (g) + 2 O(l) (9) 3. Identify the interparticle attractive force(s) present in the solids of the following substances. If more than one interparticle force, indicate which is the most important. a) NF 3 dipole-dipole and dispersion forces Since all of the atoms are in the second period the most important force is dipole-dipole. b) 3 N 2 hydrogen-bonding and dispersion forces. Again since all of the atoms are in the second period hydrogen-bonding is the most important attractive force. c) KBr ionic bonding is the only attractive force that is important. Dispersion forces are prsent but they are VERY small compared to ionic forces.

3 EM 1515 EXAM II PAGE 3 (12) 4. Account for each of the following observations about pairs of substances. In your answers, use appropriate principles of intermolecular forces. In each part, your answer must include references to both substances. a) F has normal boiling point = 20 where as l has a normal boiling point of 114. F is a polar compound with hydrogen-bonding and dispersion forces. l is a polar compound with dipole-dipole and dispersion forces. Dispersion forces are the most important in l. The hydrogen-bonding forces in F are much stronger compared to the dispersion forces in l. b) l 4 has normal boiling point = 76.7 where as Br 4 has a normal boiling point of 189. Both compounds are nonpolar so the only attractive forces are dispersion forces. Br 4 has more electrons and is more polarizable compared to l 4. So Br 4 has the higher boiling point. (6) 5. Give the name or draw the complete Lewis structure (showing all - bonds) for each of the following compounds. 4-ethyl-trans-2-heptene 3,3-dimethyl-1-butyne 2,3-dimethyl-2-pentene

4 EM 1515 EXAM II PAGE 4 (12) 6. Barium, Ba, crystallizes in one of the cubic unit cell systems. The edge length of its unit cell is 502 pm and the density of the metal is 3.50 g cm -3. Determine the number of atoms in the cubic unit cell and identify the type of cubic cell. 502 pm m 1 pm 10 2 cm 1 m = 5.02 x 10 8 cm Volume = (5.02 x 10 8 cm) 3 = 1.26 x cm x cm 3 g 3.50 cm 3 = 4.43 x g 4.43 x g 1 mol 137 g x10 23 atoms 1 mol = 1.95 atoms So the unit cell must be body centered cubic (12) 7a. Some solutions processes are exothermic while others are endothermic. Provide an explanation for this difference. There are three steps to determine for a solution; Step 1: Step 2: Step 3: overcome the solute-solute attractive forces (endothermic) overcome the solvent- solvent attractive forces (endothermic) form new solute- solvent attractive forces (exothermic) The solution process is exothermic when Step 3 is more negative, than the sum of Steps 1 and 2. The solution process is endothermic if the sum of Step 1 and 2 is more positive than the absolute value of Step 3. b) A substance with the formula 2 4 O 2 is very soluble in water, but insoluble in S 2. Suggest a structure for this substance that supports the solubility information. Indicate the intermolecular attractive force that explains the solubility. ydrogen-bonding occurs between the water molecules and the acetic acid molecule

5 EM 1515 EXAM II PAGE 5 (36) 8. An aqueous solution of Na 3 PO 4 is prepared by mixing 16.4 g Na 3 PO 4 with 500 g of water. a) calculate the molality of the solution. (6) 16.4 g Na 3 PO 1 mol g = mol mol kg = molal b) calculate the ideal freezing point of the solution. (6) Na 3 PO 4 (aq) 3Na + (aq) + PO 4 3 (aq) i = 4 (ideal) T f = ik f m = m m = 1.49 T f = 1.49 c) the experimental freezing point was found to be Explain why the experimental and ideal freezing point are different. (6) T f = ik f m 1.32 = i 1.86 m m = 1.49 i = 3.55 The experimental freezing point is higher because there are fewer particles in the solution compared tot he ideal solution. There must be ion-pairing occuring to reduce the number of particles.

6 EM 1515 EXAM II PAGE 6 8. (ontinued) d) a new aqueous solution of sodium phosphate, Na 3 PO 4, was prepared with a density of 1.05 g cm -3. The molality of this solution was determined to be molal. i) calculate the weight percent of Na 3 PO 4 in this solution. (6) mol Na 3 PO 4 per 1 kilogram of 2 O mol 164 g 1 mol = 52.5 grams weight % = 52.5 grams Na 3 PO grams Na 3 PO g 2 O 100 = 4.99 % ii) calculate the molarity of the solution. (6) total mass of the solution is grams grams 1 ml solution 1.05 g = 1002 ml = L mol L = M iii) describe how to prepare 1200 g of a molal solution beginning with a molal Na 3 PO 4 solution and distilled water. (6) 1200 gram solution 52.5 grams Na 3 PO grams Na 3 PO g 2 O = 59.9 g Na 3 PO 4 need A molal solution contains mol 164 g 1 mol = 82.0 grams Na 3 PO 4 or 82.0 grams Na 3 PO grams Na 3 PO g 2 O 59.9 g Na 3 PO grams Na 3 PO g 2 O 82.0 grams Na 3 PO 4 = 790 grams of molal solution This says that 790 grams of a molal Na 3 PO 4 solution contains 59.9 g of Na 3 PO 4. So if we add 1200 grams 790 grams solution = 410 grams of 2 O to 790 grams of molal Na 3 PO 4 solution we would have 1200 grams of a molal Na 3 PO 4 solution.

7 EM 1515 EXAM II PAGE IA Li IIA 4 Be Na Mg IIIA IVA VA VIA VIIA B N O F 9 10 Ne IIIB IVB VB VIB VIIB VIII IB IIB K a Sc Ti V r Mn Fe o Ni u Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag d In Sn Sb Te I Xe (98) s Ba La Periodic Table of the Elements f Al Si P S l VIIIA 2 Ta W Re 75 Os Ir Pt Au g 80 Tl Pb Bi Po At Rn (209) (210) Fr Ra Ac (223) (261) (262) (263) e Ar (222) Lanthanides Actinides e Pr Nd Pm Sm Eu Gd Tb Dy o Er Tm Yb Lu (145) Th Pa U Np Pu Am m Bk f Es Fm Md No Lr (244) (243) (247) (247) (251) (252) (257) (258) (259) (260) Useful Information 1 pm = m R = L atm mol K = P solution = χ solvent P solvent J mol K density of 2 O = 1.00 g cm x 1023 T = ikm k f ( 2 O) = 1.86 m k b( 2 O) = m edge length (l) = 2r edge length (l) = 2 2 r edge length (l) = 4r 3 G = - T S

8 EM 1515 EXAM II PAGE 8 Temperature ( ) Vapor Pressure(mmg) Temperature ( ) Vapor Pressure(mmg) Solubility Table Ion Solubility Exceptions NO 3 soluble none lo 4 soluble none l soluble except Ag +, g 2+ 2, *Pb 2+ I soluble except Ag +, g 2+ 2, Pb 2+ SO 2 4 soluble except a 2+, Ba 2+, Sr 2+, g 2+, Pb 2+, Ag + O 3 2 insoluble except Group IA and N 4 + PO 3 4 insoluble + except Group IA and N 4 - O insoluble except Group IA, *a 2+, Ba 2+, Sr 2+ S 2 insoluble except Group IA, IIA and N + 4 Na + soluble none N + 4 soluble none K + soluble none *slightly soluble

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