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Dr. Jim Bloxton CHEM 1021 Fall 2007 OPTIONAL PRACTICE FINAL EXAM ANSWER KEY CHAPTERS 1-9 Note: This is a partial practice Final Exam that I gave in Chem 1021 during the first summer session of 2006 for Chapters 1-9. This is optional practice material for chapters 1- 9. These are sample questions that may or may not be used on a final exam. Other questions that are not found on this sample final exam will be on the actual final exam. Learn the basic principles and then be prepared to apply the principles to problems that you have not seen before. This includes all material that was covered or contained in the Chem 1021 lecture text, lecture notes, lecture handouts, study guides, recitation notes, and recitation handouts. It is very important to review all practice and actual lecture quizzes, midterms, study guides and homework materials. Several other practice final exams could have been generated from these materials that would look very different from this practice final exam. Note: It is very important to remember that Chemistry is a cumulative topic and that the material that was covered in previous chapters needs to be reviewed as this material will continue to reappear in concepts that are needed to solve new problems. Note: The number and distribution of questions among the chapters on the actual Chem 1021 final exam may vary from the number and distribution of questions among the chapters on this practice Chem 1021 final exam. All work for all problems must be shown. Include proper units for each calculation. 1. ATP is a molecule that is used to store energy in the body. What type of energy is found in an ATP molecule? Potential energy 2. The density of methanol at 20 o C is 0.791 g/mL. What is the volume in L of a 391. cg sample? Show all calculations. First convert cg to g. There are 100 g in a cg. 391. cg x 1 g ---------------------------- = 3.91 g 100 cg Calculate volume in mL by using volume = mass ÷ density 3.91 g x 1 mL ---------------------------- = 4.94 mL 0.791 g Convert mL into L as follows: 4.94 mL x 1 L ----------------------------- = 0.00494 L 1000 mL
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3. What is hyperthermia? What is a treatment for hyperthermia? Hyperthermia is when body temperature is higher than normal. The use of a cold compress is a good treatment for hyperthermia since the cold compress will help absorb the excess heat that is present in an individual with hyperthermia. 4. The specific heat of water is 1.00 cal/g .o C. How many kcal are needed to raise the temperature of 40.3 kg of water from 98.6 o C to 103.2. o C? Show all calculations. Calculate ∆T as follows: ∆T = T final - T initial = 103.2 o C - 98.6 o C = 4.6 o C = ∆T Convert 40.3 kg to g as follows: 40.3 kg x 1000 g ---------------------------- = 40300 g 1 kg Calculate amount of heat in calories as follows: Amount of heat = SH x m x ∆T = 1.00 cal/g .o C x 40300 g x 4.6 o C = 185380 cal Convert 185380 cal to kcal as follows: 185380 cal x 1 kcal -------------------------------- = 185.380 kcal
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This note was uploaded on 03/29/2011 for the course CHEMISTRY 1021 taught by Professor Jimbloxton during the Fall '09 term at Temple.

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