Chem 103 lab exam 3 Study guides, Class notes & Summaries
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![CHEM 103 Lab Exam 5 – Portage Learning](/docpics/641a2fef12691_2499146.jpg)
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CHEM 103 Lab Exam 5 – Portage Learning
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CHEM 103 Lab Exam 5 – Portage Learning. According to Boyle’s law, as the volume of a gas decreases, the 
pressure _________. 
3. (6 points) A sample of gas occupies a volume of 10.5 L when the pressure 
is 14.3 psi. Use Boyle’s Law to calculate the volume (L) when the pressure is 
decreased to 12.8psi. Show the calculation. 
Answer: 
1. A syringe was used, and the volume was read from the graduation 
markings.
![CHEM 103 Lab Exam 4 – Portage Learning](/docpics/641b8b0cb2a5d_2504213.jpg)
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CHEM 103 Lab Exam 4 – Portage Learning
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CHEM 103 Lab Exam 4 – Portage Learning. How much heat was gained by the tap water if the specific heat of water is 
4.18 J/K-g? 
2. (3 points) In one experiment in the video, HCl was mixed with another 
substance. What other substance was mixed with HCl? 
3. (4 points) One student conducts an acid-base reaction where he finds the 
following results. What is the enthalpy of the reaction (in KJ/mol)? 
Heat gained by solution 3875 J
![CHEM 103 Lab Exam 4 – Portage Learning](/docpics/641a00980d95c_2498545.jpg)
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CHEM 103 Lab Exam 4 – Portage Learning
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CHEM 103 Lab Exam 4 – Portage Learning. What does it mean that the heat of a reaction is an extensive 
property? 
2. (3 points) Does the following reaction give off heat or absorb heat? 
Explain briefly. 
C D H = +263Kcal/mol 
3. (4 points) In one experiment in the video, room temperature water was 
mixed with water at about 38 degrees. To what temperature (about) did the 
mixture of the two water samples come to after setting for a few minutes? 
Answer: 
1. The heat depends on the amount of s...
![CHEM 103 Lab Exam 5 – Portage Learning](/docpics/641a011382832_2498551.jpg)
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CHEM 103 Lab Exam 5 – Portage Learning
- Exam (elaborations) • 3 pages • 2023
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CHEM 103 Lab Exam 5 – Portage Learning. In the Boyle’s law experiment, how was the volume of gas at 
different pressures determined? 
2. (2 points) According to Boyle’s law, as the volume of a gas decreases, the 
pressure _________. 
3. (6 points) A sample of gas occupies a volume of 10.5 L when the pressure 
is 14.3 psi. Use Boyle’s Law to calculate the volume (L) when the pressure is 
decreased to 12.8psi. Show the calculation. 
Answer: 
1. A syringe was used, and the volume was read f...
![NURS 6531 FINAL EXAM STUDY GUIDE](/docpics/636649b47f589_2083395.jpg)
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NURS 6531 FINAL EXAM STUDY GUIDE
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NURS 6531 FINAL EXAM STUDY GUIDE 
 
 
NURS 6531 FINAL EXAM STUDY GUIDE 
 
PLEASE DO NOT SHARE IN THE GROUP AS A POST. EVERY MEMBER WHO HAS CONTRIBUTED HAS RECEIVED A COPY. THOSE WHO HAVE NOT CONTRIBUTED THEIR ASSIGNED PORTIONS HAVE BEEN EXCLUDED. 
 
Still waiting for: 
 
79. Severe depression 
 
80. GAD 
 
 
When I receive these topics, I will post them in the group and you can just copy them and paste them into the study guide. If these topics are not received by next weekend, I will t...
![CHEM 102 Winter 18 Final Exam (A) Questions and Answers,100% CORRECT](/docpics/64060035a8701_2441244.jpg)
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CHEM 102 Winter 18 Final Exam (A) Questions and Answers,100% CORRECT
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CHEM 102 Winter 18 Final Exam (A) Questions and Answers 
 
Potentially useful data: 
 
x 	1 atm = 760 torr = 760 mm Hg 
2 a 
[A]t = – kt + [A]o	ΔG° = ΔH° – T ΔS° 
ln[A]t = – kt + ln[A]o	ΔG = ΔG° + RT ln Q 
1/[A]t = kt +1/[A]o	ΔG° = – RT ln K 
First order half-life: t1/2 = ln2 / k	ΔG° = – nFE° 
 
pH pKa 
 
	log base 
acid 
 
E E R T ln Q 
n F 
 
R = 8.314 J mol-1 K-1	F = 96500 C/mol e- 
 
 
1.	Please choose the letter “A” as your a...
![Chem 103 Bundled Exams Latest Versions with complete solutions;100% guaranteed success](/docpics/5549876/6660c1fb72441_5549876_121_171.jpeg)
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Chem 103 Bundled Exams Latest Versions with complete solutions;100% guaranteed success
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Exam (elaborations)
Chem 103 Exam 1 review - UW Madison latest questions and answers all are correct graded A+

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CHEM 103 FINAL EXAM QUESTIONS & ANSWERS VERIFIED 100% CORRECT

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Chem 103 Lab Final latest questions and answers all are correct graded A+

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Exam (elaborations)
chem 103 multiple choice questions
![ATLS Case Scenario 2023 CDEM Self Study Modules Approach to Trauma](/docpics/63f5259ae333b_2396535.jpg)
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ATLS Case Scenario 2023 CDEM Self Study Modules Approach to Trauma
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ATLS Case Scenario CDEM Self 
Study Modules Approach to Trauma 
APPROVED BY: 
 
DEM Curriculum 
Home The Approach To... Specific Diseases DIEM Cases Downloads ML's Blog 
The Approach to Trauma 
WRITTEN BY: NICHOLAS E. KMAN, MD 
THE OHIO STATE UNIVERSITY 
EDITED BY: DAVID MANTHEY, MD 
WAKE FOREST MEDICAL SCHOOL 
Trauma is the leading cause of death in the first four decades of life in most developed countries. To 
this end, there are more than 5 million trauma-related deaths each year worldwide....
![ATLS Case Scenario CDEM Self Study Modules Approach to Trauma](/docpics/6436fa61acc27_2586867.jpg)
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ATLS Case Scenario CDEM Self Study Modules Approach to Trauma
- Exam (elaborations) • 14 pages • 2023
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ATLS Case Scenario CDEM Self 
Study Modules Approach to Trauma 
APPROVED BY: 
 
DEM Curriculum 
Home The Approach To... Specific Diseases DIEM Cases Downloads ML's Blog 
The Approach to Trauma 
WRITTEN BY: NICHOLAS E. KMAN, MD 
THE OHIO STATE UNIVERSITY 
EDITED BY: DAVID MANTHEY, MD 
WAKE FOREST MEDICAL SCHOOL 
Trauma is the leading cause of death in the first four decades of life in most developed countries. To 
this end, there are more than 5 million traumarelated 
deaths each year worldwide....
![CHEM 103 Exam 2 2021 Download to Score A](/docpics/622a5414cc8e6_1613475.jpg)
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CHEM 103 Exam 2 2021 Download to Score A
- Exam (elaborations) • 19 pages • 2022
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CHEM 103 Exam 2 2021. Show the calculation of the molecular weight for the following 
compounds, reporting your answer to 2 places after the decimal. 
1. Al (SO ) 
2. C H NOBr 
2 4 3 
7 5 
1. Al (SO ) 
Al = 2 x 26.98 = 53.96 
S = 3 x 32.07 = 96.21 
O = 12 x 16.00 = 192 
2 4 3 
6/29/2021 M2: Exam - Requires Respondus LockDown Browser + Webcam: General Chemistry I w/Lab-2021- Kozminski 

Total Molecular Weight for Al (SO ) = 342.17 
2. C H NOBr 
C = 7 x 12.01 = 84.07 
H = 5 x 1.008 = 5.04 
N = 1 x...
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