Practice Final Examinaiton CHEM 1515 Spring 2012 Chemistry 1515 Final Examination Fall 2009 Form A The examination is multiple choice Each question has one and only one correct answer Mark that one answer on your examination sheet provided with the examination booklet Be certain to use a number 2 pencil If you wish to change an answer you must completely erase the original mark Scoring Each question counts 5 points There are 34 total questions so the maximum possible score on the examination is 170 points There is no subtraction for incorrect answers If you mark more than one response to any question it will be counted completely wrong Your score on the examination 5 x number of questions answered correctly Step 1 Write you name in the space provided in the upper left hand corner of side 1 of the answer sheet Then use your 2 pencil to darken the circles corresponding to the letters of your name in the column below the letter Darken the open circle to indicate a blank Step 2 Write your Student ID number in the space provided on the lower left side of side 1 of the answer sheet Then use your 2 pencil to darken the circles corresponding to the numbers in your ID Step 3 Write 1515 in the space to the right of that for the ID number Then darken the circles corresponding to 1 5 1 5 in the columns below Step 4 Write your section number in the form 0x in the space where it asks for Section SEC number Then use your 2 pencil to darken the zero circle and the x circle underneath the digits 0 1 Step 5 Be certain to read the directions for marking the answers on Side 2 of the examination sheet Mark your responses exactly as directed 1 Step 6 Answer the examination questions and get them all correct 2 Data and Additional Information R gas constant 0 08206 L atm mol 1 K 1 8 314 J mol 1 K 1 1 987 cal mol 1 K 1 One Faraday 96 500 coulombs Kw for H2O at 298 15 K 10 14 C speed of light in a vacuum 2 997 x 108 m s 1 Clausius Clapeyron Equation for liquid vapor equilibrium Peq 760 torr exp Hvap T 1 Tb 1 R where Tb is the normal boiling temperature Quadratic formula to find the roots of AX2 BX C 0 X Temperature dependence of the equilibrium constant KC T KC TO exp Nernst Equation at 298 K E Eo ln where E is the cell voltage Integrated rate equation for a second order reaction involving reactants A and B ln ln A o B o k T t Arrhenius equation for reaction rates k T A exp 3 1 An investigator measures the equilibrium vapor pressure of a liquid at 300 K and at 310 K She finds that the equilibrium vapor pressure at 310 K is twice as large as it is at 300 K We therefore may determine that the molar heat of vaporization H vap lies in which of the following ranges A Hvap 45 000 J mol 1 B 40 000 J mol 1 Hvap 45 000 J mol 1 C 35 000 J mol 1 Hvap 40 000 J mol 1 D 25 000 J mol 1 Hvap 35 000 J mol 1 E 2 Hvap 25 000 J mol 1 The pressure temperature phase diagram for a pure substance is given below Pressure D C Liquid Solid B E vapor A Temperature At which point on the diagram A B C D or E is only one phase present A Point A B Point B C Point C D Point D E Point E 3 Two moles of solute A are dissolved in 750 grams of solvent B In order to compute the mole fraction XA A B C D E we need no additional information we need the molecular weight of A we need the molecular weight of B we need the molecular weight of both A and B we need the molecular weight of both A and B and the density of the solution 4 4 A solid solute A is to be dissolved in liquid solvent B If the solution formed is ideal the mole fraction of A in the saturated solution will A increase linearly with the temperature of the solution B increase as the molar heat of fusion of A increases C decrease exponentially as 1 T decreases D be independent of the temperature of the solution E do none of the above 5 Henry s law tells us that the solubility of a gas in an ideal solution will A be independent of the applied pressure of the gas B increase exponentially with the applied pressure of the gas C increase linearly with the applied pressure of the gas D decrease linearly with the applied pressure of the gas E do none of the above 6 Consider an ideal solution of liquids A and B at 298 K The pure component vapor pressures of A and B liquids are 400 mm Hg and 600 mm Hg respectively If the partial pressure of A g over the solution is 250 mm Hg we know that the mole fraction of A in the solution is A 0 25 B 1 0 C 0 4167 D 0 625 E 250 7 Consider the following gas phase equilibrium 2 NO2 g N2O4 57 2 kJ At equilibrium it is found that the pressure of NO 2 g is 0 861 atm and that of N 2O4 g is 5 atm Kp is therefore approximately A B C D E 6 74 atm 1 0 1484 atm 5 805 3 709 none of the above 5 8 The equilibrium 2 NO2 g N2O4 57 2 kJ will certainly shift toward N2O4 if A B we raise the temperature we add more NO2 g and raise the temperature C D E we decrease the pressure we increase the pressure and raise the temperature we lower the temperature and add more NO2 g 9 At a particular temperature the solubility product constant in water for BaSO 4 is 1 8 x 10 10 We therefore know that at saturation the number of moles per liter of BaSO 4 that have dissolved in the water will be about A 1 34 x 10 5 mol L 1 B 1 8 x 10 10 mol L 1 C 1 34 x 10 10 mol L 1 D 3 24 x 10 20 mol L 1 E 1 8 x 10 5 mol L 1 10 At 298 15 K the solubility products of BaSO4 and PbSO4 in water are 1 8 x 10 10 and 1 6 x 10 8 respectively Three beakers A B and C each containing one liter of water are placed on the table with the temperature of all beakers being 298 15 K 50 grams of BaSO 4 s are added to Beaker A 50 grams of PbSO4 s are added to Beaker B 50 grams of BaSO4 s and 50 …