Chem120 week 5 exam - Study guides, Class notes & Summaries
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CHEM120 Week 8: Exam 3 (Units 5, 6, and 7) – With 100% Correct Answers- Download To Score An A+
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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 5 Exam 2 (Units 3 and 4)
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1.	A solution is prepared by dissolving 50 g of KCl (74.6 g/mol) to a volume of 1 L. What is the molarity of this solution? 
2.	You prepare a solution by dissolving 20 g of NaCl to a volume of 350 mL in water. What is the mass/volume % concentration of this solution? 
3.	When the following reaction goes in the forward direction, what is the base? 
4.	You are preparing a buffered solution. So far, your solution contains H2PO4- ions, what would you add to this solution to prepare your buffer sy...
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CHEM 120 Week 5 Exam 2 (Units 3 and 4)
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CHEM 120 Week 5 Exam 2 (Units 3 and 4)
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CHEM 120 Week 8 Exam 3 (Units 5, 6, and 7)
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CHEM 120 Week 8 Exam 3 (Units 5, 6, and 7)
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chem 120 final questions and answers to week 8
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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 (Sample Question, Resources)
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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 (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 8 Final Exam COMPLETE (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 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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CHEM120 FINAL EXAM -WEEK 8- PAGE 3- 100%
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CHEM120 FINAL EXAM -WEEK 8- PAGE 3- 100% 
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. 
Question: (TCO 7) (a, 5 pts) What is the mass/volume percent of a solution prepared by dissolving 43 g of NaOH in enough water to make ...
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