Chem 108 Study guides, Class notes & Summaries
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CHEM 108 PRACTICE FINAL EXAM ANSWERS WITH 100% CORRECT ANSWERS DOWNLOAD TO SCORE A+
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MCAT AAMC FL 1 CHem and Physics Questions and Answers 2023
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MCAT AAMC FL 1 CHem and Physics Questions and Answers 2023 
In the chromatography of the reaction mixture, water absorbed on cellulose functioned as the stationary phase. What was the principal factor determining the migration of individual components in the sample? 
 
A.Hydrogen bonding 
B.Solute concentration 
C.Stationary phase concentration 
D.Thickness of paper 
The correct answer is A. 
 
This is an Organic Chemistry question that falls under the content category "Separation and purificat...
![CHEM 108 Module 1 Exam Answers- Portage Learning](/docpics/61b7aaf659d99_1439940.jpg)
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CHEM 108 Module 1 Exam Answers- Portage Learning
- Exam (elaborations) • 11 pages • 2021
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CHEM 108 Module 1 Exam Answers- Portage Learning/CHEM 108 Module 1 Exam Answers- Portage Learning/CHEM 108 Module 1 Exam Answers- Portage Learning/CHEM 108 Module 1 Exam Answers- Portage Learning/CHEM 108 Module 1 Exam Answers- Portage Learning
CHEM 108 COMPREHENSIVE BUNDLE
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CHEM 108 LAB 7- Portage Learning
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CHEM 108 LAB 7- Portage Learning/CHEM 108 LAB 7- Portage Learning/CHEM 108 LAB 7- Portage Learning/CHEM 108 LAB 7- Portage Learning/CHEM 108 LAB 7- Portage Learning/CHEM 108 LAB 7- Portage Learning
![CHEM 108 Module 3 Exam Answers- Portage Learning](/docpics/61b7a6fba523b_1439900.jpg)
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CHEM 108 Module 3 Exam Answers- Portage Learning
- Exam (elaborations) • 10 pages • 2021
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CHEM 108 Module 5 Exam Answers- Portage Learning
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Exam Page 1 
In the reaction of gaseous N2O5 to yield NO2 gas and O2 gas as shown below the following data table 
is obtained: 
2 N2O5 (g) → 4 NO2 (g) + O2 (g) 
Data Table #1 
Time (sec) [N2O5] [O2] 
0 0.200 M 0 
300 0.182 M 0.009 M 
600 0.166 M 0.017 M 
900 0.152 M 0.024 M 
1200 0.140 M 0.030 M 
1800 0.122 M 0.039 M 
2400 0.112 M 0.044 M 
3000 0.108 M 0.046 M 
Complete the following three problems: 
Using the [O2] data from the table show the calculation of the average rate over the measured ...
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CHEM 108 Module 2 Exam (Questions and Answers)
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CHEM 108 Module 2 Exam (Questions and Answers)|CHEM 108 Module 2 Exam (Questions and Answers)|CHEM 108 Module 2 Exam (Questions and Answers)
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CHEM 108 Module 2 Exam Answers- Portage Learning
- Exam (elaborations) • 10 pages • 2021
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CHEM 108 Module 3 Exam Answers- Portage Learning
- Exam (elaborations) • 10 pages • 2022
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Module 3 exam 
Exam Page 1 
Show the calculation of the new temperature of a gas sample which has an original volume of 700 ml 
when collected at 730 mm and 35oC when the volume becomes 480 ml at 1.15 atm. 
700ml/1000= 0.7 liters= Vi 
730mm/700= 1.042 atm=Pi 
35 ºC + 273 = 308K = Ti 
480ml/ 1000 = 0.04 liters = Vf 
1.15 amt= Pf 
 (1.042) ×(0.7)= (1.15) × (0.04) 
 (308) Tf 
 Tf= 12.50 = 17.15K 
 0.729
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