Chem120 week 3 exam Study guides, Class notes & Summaries

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CHEM 120 Week 3 Exam 1 (Information)
  • CHEM 120 Week 3 Exam 1 (Information)

  • Exam (elaborations) • 3 pages • 2023
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  • CHEM120 Students: The first exam in your science course is approaching. The General Education Academic Support is here to help. What can you do ahead of Exam 1? 1. Attend a live exam review Thursday, May 12th 6 PM MT Exam 1 Review Join Meeting here: Click here to join 2. Watch the recording of an exam review Click here to watch recorded exam review 3. Schedule a 1:1 appointment with a tutor Click here to book 4. Review the topic videos from weeks 1 and 2 Click her to review ...
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CHEM 120 Week 3 Exam 1 (Units 1 and 2)
  • CHEM 120 Week 3 Exam 1 (Units 1 and 2)

  • Exam (elaborations) • 3 pages • 2023
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  • Answer the following questions (1 point each) to add up to 5 points back to your exam score. 1. Question: Choose any isotope of an element and answer the following a. Write the isotopic formula b. How many neutrons in the isotope? c. How many protons are in the isotope? d. How many electrons are in the isotope? e. What is the mass number for the isotope? 2. Question: Choose any molecular compound with 3 to 5 atoms in it. a. Draw the Lewis structure. b. Give step by step directions...
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CHEM 120 Week 8 Final Exam 2023(100% CORRECT SOLUTIONS) | Already GRADED A.
  • CHEM 120 Week 8 Final Exam 2023(100% CORRECT SOLUTIONS) | Already GRADED A.

  • Exam (elaborations) • 22 pages • 2021
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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 3 Exam 1 (Information)
  • CHEM 120 Week 3 Exam 1 (Information)

  • Exam (elaborations) • 13 pages • 2023
  • CHEM 120 Week 3 Exam 1 (Information)
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chem 120 final questions and answers to week 8
  • chem 120 final questions and answers to week 8

  • Exam (elaborations) • 4 pages • 2024
  • Week 8 : Final Exam Questions - Final Exam CHEM120 Final-Exam- 1. Question: (TCO 7) (a, 5 pts) Given that the molar mass of H3PO4 is 97.994 grams, determine the number of grams of H3PO4 needed to prepare 0.25L of a 0.2M H3PO4 solution. Show your work. (b, 5 pts) What volume, in Liters, of a 0.2 M H3PO4 solution can be prepared by diluting 50 mL of a 5M H3PO4 solution? Show your work. (Pts. : 10) A. Molarity = moles of solute/liters of solution moles of solute = 0.2 M*0.25 L = 0.05 m...
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CHEM 120 Week 3 Exam 1 (Information)
  • CHEM 120 Week 3 Exam 1 (Information)

  • Other • 3 pages • 2023
  • CHEM 120 Week 3 Exam 1 (Information)
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CHEM120 Week 8: Exam 3 (Units 5, 6, and 7) – With 100% Correct Answers- Download To Score An A+
  • CHEM120 Week 8: Exam 3 (Units 5, 6, and 7) – With 100% Correct Answers- Download To Score An A+

  • Exam (elaborations) • 4 pages • 2024
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  • CHEM120 Week 8: Exam 3 (Units 5, 6, and 7) – With 100% Correct Answers- Download To Score An A+ 1. Question: (TCO 7) (a, 5 pts) Given that the molar mass of H3PO4 is 97.994 grams, determine t he number of grams of H3PO4 needed to prepare 0.25L of a 0.2M H3PO4 solution. Show your work. (b, 5 pts) What volume, in Liters, of a 0.2 M H3PO4 solution can be prepared by diluting 50 mL of a 5M H3PO4 solution? Show your work. (Pts. : 10) A. Molarity = moles of solute/liters of solution moles of solute ...
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CHEM 120 Week 8 Final Exam (Solved Q & A) | Highly Rated Paper | Already Graded A+
  • CHEM 120 Week 8 Final Exam (Solved Q & A) | Highly Rated Paper | Already Graded A+

  • Exam (elaborations) • 22 pages • 2021
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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 Week8 Final Exam Review
  • CHEM 120 Week8 Final Exam Review

  • Exam (elaborations) • 7 pages • 2023
  • CHEM 120 Week8 Final Exam Review
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CHEM 120 FINAL EXAM (Version 3 ) Latest Complete Questions & Answers, Already Graded A; Chamberlain.
  • CHEM 120 FINAL EXAM (Version 3 ) Latest Complete Questions & Answers, Already Graded A; Chamberlain.

  • Study guide • 19 pages • 2020
  • CHEM 120 FINAL EXAM 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) 7. (TCO 6) A sample of helium gas occupies 1021 mL at 719 mmHg. For a gas sample at constant temperature, determine the volume of helium at 745 mmHg. Show your work. (Points : 5) 8. (TCO 12) If one strand of a DNA double helix has the sequence T T A G C G A C G C, what is the sequenc...
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