key_final - FINAL EXAM Mar 19 2001 CHEMISTRY 20A Winter 01...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
1 FINAL EXAM CHEMISTRY 20A Winter 01 SOLUTIONS Mar. 19, 2001 . . Student Name Bobby N. Johnny S. Rebecca E. Circle Name of TA . . Student I.D. No. PLEASE WRITE YOUR NAME & Circle Name of your TA . Then PLEASE WRITE YOUR NAME and ADDRESS ON THE BACK OF THE EXAM in the appropriate address label. If you want the exam returned, then PLACE $0.76 POSTAGE in the appropriate place. READ DIRECTIONS on THIS PAGE, but DO NOT TURN PAGE UNTIL TOLD TO DO SO . DIRECTIONS . All work and answers should be written on the exam. Please write legibly and clearly. Wherever applicable, show your work, justify approximations, and explain what you are doing. It is usually useful, when possible, to work problem out algebraically first. Partial credit will be given; correct statements will count in your favor, incorrect ones will count against you, unintelligible or illegible ones will not count at all. Allocate your time between problems judiciously. This is a closed book exam, but you may make use of a single 8.5" by 11" sheet of handwritten notes (both sides). This exam consists of 16 pages. NOTE that each part of each question can be answered independently, whether or not you are able to answer an earlier part; however, the answers to later parts may depend upon statements included in the previous parts. LEAVE OUT ONE OF THE 41 POINT PROBLEMS. MARK CLEARLY BELOW WHICH ONE YOU HAVE LEFT OUT. Problem. Max Grade Grade 1 27 2 27 Class average on Final. 3 34 4 41 YOUR GRADE for THE COURSE: 5 41 6 41 7 41 8 41 . Total 252
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2 Some possibly useful data: Molecular weights can be obtained from periodic table h = 6.63x10 -34 J s |e| = 1.60 x 10 -19 C. c = 3.00x10 8 m s -1 m electron = 9.11x10 -31 kg N av = 6.02x10 23 particles mol -1 m proton • m neutron = 1.67x10 -27 kg 1 eV = 1.60x10 -19 J Some possibly useful formulas. E = h ν KE = (1/2)mv 2 λ m = h/mv c = λ ν F = ma = -dU/dr (minus slope of U) U elect = ± e 2 4 πε o 1 r . For convenient numerical use U elect = 1.3 eV ( nm r ) = 1,250 kJ mol -1 ( nm r ) nm = nanometer (length) E n = - Z 2 R H n 2 = - (1/2) Z 2 e 4 m e (4 πε o h) 2 1 n 2 = -13 eV Z 2 n 2 = - 1,250 kJ Z 2 n 2
Image of page 2
3 1. When 14.8 g of a compound containing only carbon, hydrogen and oxygen was burned in excess oxygen, 18.0 g of H 2 O and 35.2 g of CO 2 were formed. What is the empirical formula of the compound, i.e. , in the formula C k H m O n what are the simplest integral values for k,n,m? 18.0 g of H 2 O is 1 mol, i.e ., 2 mol of H. 35.2 g of CO 2 is 0.8 mol , i.e . 0.8 mol of C. 2 mol of H is 2.0 g. 0.8 mol of C is 9.6 g. Thus the mass of H and C in the compound is 2.0 + 9.6 g = 11.6 g. Thus the mass of oxygen in the compound is 14.8 g - 11.6 g = 3.2 g. 3.2 g of O is 0.2 mol. So in moles we have C .8 H 2 O .2 . Usually we prefer integral numbers, so multiply through by 5: C 4 H 10 O. 2. 60 g of carbon (C) are burned in a container with 56 g of oxygen (O 2 ). At the end all the oxygen is converted to CO and CO 2 , but although there is no O 2 left, 36 g of C remain unburned (unchanged). What mole fraction of the remaining CO/CO 2 gas mixture is CO? (Mole fraction is moles of CO divided by the total number of moles of CO and CO 2 .) This problem is straightforward but requires a bit of thought; be sure you understand the reaction.
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern