42 Review for Final

42 Review for Final - Final Exam Monday, May 18, 11:00 AM...

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1 1 Final Exam Monday, May 18, 11:00 AM to 1:30 PM Rooms TBA, see Bb Cover page w/ info will be posted on Bb Be sure to bring Stony Brook ID card 2 or more #2 pencils Calculator with extra batteries Do not bring cell phones, beepers, iPods, other electronic devices, textbook, and notes. Read and follow directions on the cover page. Be sure to print and bubble in your ID number correctly 2 The truth about test-prep tutoring! Some test-prep services make bold claims. Here is some data. CHE 132 students surveyed, Spring 2009 Regarding experiences in CHE 129 & 131, Fall 2008 GPA = average grade in CHE 129 or 131 of the students in each category 37 2.57 C+ Other 29 2.63 C+ J 35 2.82 B- S 202 2.82 B- Chem LC 272 2.85 B- None Number GPA Service
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2 3 Dr Dave’s End-o-the-Semester Help Sessions May 12 – 14: Tuesday, Wednesday, Thursday Space Reserved in Chemistry Learning Center for CHE 132 Study Groups Chemistry Room 321 Dr Dave will be there 4:00 – 6:00 PM These sessions are student driven. Bring problems you want help solving! Concepts that need clarification! Questions you want answered! And they are free! 4 #18 Calculate the value of K sp for CdS from the following data. (1) CdS(s) + 2e Cd(s) + S 2– (aq) E o = – 1.21 V (2) Cd 2+ (aq) + 2e Cd(s) E o = – 0.40 V Think: K sp reaction?, K G o E o , Use E o =(RT/nF)lnK Turn (2) into an oxidation half cell (3) Cd(s) Cd 2+ (aq) + 2e - E o = + 0.40 V Combine (1) and (3), write the K sp & cell reaction and the standard cell potential. CdS(s) Cd 2+ (aq) + S 2- (aq) E o = -0.81 V -0.81 V = (8.314Jmol -1 K -1 x298K)/(2x96,485C/mol)lnK sp lnK sp = -63.09 K sp =4.0x10 -28
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3 5 #20 Suppose an aluminum beer can weighs 40.0 g. For how long would a current of 100.0 amp need to be passed through a molten AlF 3 electrolysis cell to produce enough aluminum to replace a discarded beer can? (Al 27.0 g/mol, F 19.0 g/mol) Think: Electrons needed = Electrons supplied by current hr 1.19 s 4290 ti me x C/s 100 C 429,000 x x C 429,000 C/mol 96,485 mol 4.44 mol 4.44 electrons of mol 1.48 3 Need Al mol 1.48 g/mol) g/(27.0 40.0 Al of g 40.0 Need 6 #21 How much lead sulfide will dissolve in a liter of water at a pH of 7.5? You should know the following. Sequential reactions increase the solubility. Examples: formation of weak acids or complexes. PbS(s) Pb 2+ (aq) + S 2- (aq) K sp S 2- (aq) + H 2 O HS - + OH - K bS2- HS - + H 2 O H 2 S + OH - K bHS- Solubility given by [Pb 2+ ] [Pb 2+ ]=[S 2- ]+[HS - ]+[H 2 S]
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4 7 Chapter 6 Energy and Chemical Reactions Conservation of Energy, E E system = q + w + q = energy transferred into the system + w = work that is done on the system Enthalpy, H H system = q P = E system –w expansion w expansion = - P V = - P (V final –V initial ) 8 Hess’s Law If Reaction 3 = Reaction 1 + Reaction 2 Then H o 3 = H o 1 + H o 2 Works for any state function, not just enthalpy A state function only depends on the state or situation and not on the path for getting there.
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This note was uploaded on 02/13/2011 for the course CHE 132 taught by Professor Hanson during the Spring '08 term at SUNY Stony Brook.

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42 Review for Final - Final Exam Monday, May 18, 11:00 AM...

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