Chem120 week 8 final exam - Study guides, Class notes & Summaries
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CHEM120 Week 8 Final Exam (Version 4) -with 100% verified solutions-2023-2024
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Highergrades
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QUESTIONS AND ANSWERS
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CHEM120 Week 8 Final Exam (Version 4) -with 100% verified solutions-2023-2024
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yankym58
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CHEM120 Week 8 Final Exam (Version 4) -with 100% verified solutions-
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CHEM120 Week 8 Final Exam (Version 4) -with 100% verified solutions-2023-2024
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CHEM120 Week 8 Final Exam (Version 4) -with 100% verified solutions-
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CHEM120 Week 8 Final Exam (Version 4) -with 100% verified solutions-2023-2024
- Exam (elaborations) • 4 pages • 2023
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GENIUSA
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CHEM120 Week 8 Final Exam (Version 4) -with 100% verified solutions-
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CHEM 120 Week 8 Final Exam COMPLETE (100% CORRECT SOLUTIONS) | Already GRADED A.
- Exam (elaborations) • 22 pages • 2021
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ProfGoodlucK
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6. (TCO 6) A gas at a temperature of 95 degrees C occupies a volume of 165 mL. Assuming constant pressure, determine the volume at 25 degrees C. Show your work. (Points : 5) Using Charles’ Law, ( V1/T1) = (V2/T2). First, convert temperature to KELVIN (T1 = t1 +273) Thus, T1 = 95 + 273 = 368. We have V1 (165 mL) & T2 = (25 + 273) = 298. V2 = (V1*T2)/T1 = (165 mL*298)/368 = 133.6 mL. 0 7 Short 16 7. (TCO 6) A sample of helium gas occupies 1021 mL at 719 mmHg. For a gas sample at constant tempera...
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CHEM120 Week 8 Final Exam-with 100% verified solutions-2023-2024
- Exam (elaborations) • 20 pages • 2023
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yankym58
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CHEM120 Week 8 Final Exam-with 100% verified solutions-
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CHEM 120 FINAL-EXAM WEEK 8 STUDY-GUIDE
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jessicaacademics
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CHEM 120 FINAL-EXAM WEEK 8 STUDY-GUIDE
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CHEM 120 Week8 Final Exam Review
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jonathanclyne
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CHEM 120 Week8 Final Exam Review
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CHEM 120 Week 8 Final Exam (Solved Q & A) | Highly Rated Paper | Already Graded A+
- Exam (elaborations) • 22 pages • 2021
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ProfGoodlucK
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6. (TCO 6) A gas at a temperature of 95 degrees C occupies a volume of 165 mL. Assuming constant pressure, determine the volume at 25 degrees C. Show your work. (Points : 5) Using Charles’ Law, ( V1/T1) = (V2/T2). First, convert temperature to KELVIN (T1 = t1 +273) Thus, T1 = 95 + 273 = 368. We have V1 (165 mL) & T2 = (25 + 273) = 298. V2 = (V1*T2)/T1 = (165 mL*298)/368 = 133.6 mL. 0 7 Short 16 7. (TCO 6) A sample of helium gas occupies 1021 mL at 719 mmHg. For a gas sample at constant tempera...
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chem 120 week 8 final exam review
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chem 120 week 8 final exam review
