CHEM1310_FinalW10_PartB_Answers - THE UNIVERSITY OF...

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THE UNIVERSITY OF MANITOBA April 17, 2010 FINAL EXAMINATION PAGE NO: 1 of 5 DEPARTMENT & COURSE NO: CHEM 1310 TIME: 2 HOURS EXAMINATION: University I Chemistry: Intro. To Phys. Chem. EXAMINER: CHEM 1310 Committee Part B will be marked out of a raw score of 32. This score will be multiplied by 5/8 for a total of 20 marks (or 40% of the 50 marks allotted to this final examination). Part B should require approximately 80-90 min. to complete. DO ALL of QUESTIONS 1 THROUGH 6. NAME: ( Please Print Legibly ) SIGNATURE: STUDENT NUMBER: Examination Room: Seat Number: For Examiner’s use only: 1: 2: 3: ________________ 4: 5: 6: ________________ Total: /32 INSTRUCTIONS: No examination booklets are needed. The questions are to be answered on the pages of this examination as indicated. Write in INK, please. ALL NUMERICAL ANSWERS MUST BE JUSTIFIED WITH PROPER AND CLEAR CALCULATIONS. Blank pages, to the extent that they are not needed for your answers, may also be used for any rough work required to prepare your answer. NOTE THAT YOU MUST ANSWER ALL SIX QUESTIONS ON THIS PAPER. The allocated marks for each question are shown in brackets to the right of the page after the question. Some information that you may find useful in answering this examination will be found on the DATA PAGES attached to Part A of this examination.
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FINAL EXAM PART B CHEM 1310 University I Chemistry: Intro. to Phys. Chem. Page: 2 of 5 (if more space is required to answer these questions, continue on the back side of this page) 1) Photosynthesis in higher plants is a complex process in which glucose is synthesized from atmospheric CO 2 and water in the presence of light energy. 6 CO 2 (g) + 6 H 2 O(l) C 6 H 12 O 6 (s) + 6 O 2 (g) It is estimated that the net amount of carbon dioxide fixed by photosynthesis on the landmass of the Earth is 55 x 10 16 g yr -1 . Calculate the amount of energy stored by photosynthesis on land per year. (5 marks) The enthalpy change for the above reaction is Δ Hº = -1273.3 + 6 (0.0) – 6 (-393.5) – 6 (-285.8) = 2802.5 kJ Since 6 moles of carbon dioxide are consumed, the energy stored per mole of CO 2 is (2802.5)/6 = 467.1 kJ/mol of CO 2 The molar mass of is 12.011 + 2 (16.00) = 44.01 g/mol Therefore the number of moles of CO 2 fixed per year = (55 x 10
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CHEM1310_FinalW10_PartB_Answers - THE UNIVERSITY OF...

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