Chemistry 103 lab 3 exam - Study guides, Class notes & Summaries
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NSG 6340 PREDICTOR STUDY GUIDE 2
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NSG 6340 PREDICTOR STUDY GUIDE 2 
1.	A 37-year-old female patient with a history of a single episode of depression and frequent complaints of PMS is being treated for hypothyroidism. Today she complains of poor concentration and fatigue. Initially, the NP should: 
a.	Question her further 
3.	Which of the following is an example of secondary prevention? 
a.	Annual influenza vaccination 
7.	A 35-year old female with a history of mitral valve prolapse is scheduled for routine dental cleaning. Accor...
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BS Mechanical Curriculum Rev 2024 DEPARTMENT OF MECHANICAL ENGINEERING PAKISTAN INSTITUTE OF ENGINEERING AND APPLIED SCIENCES (PIEAS) NILORE, ISLAMABAD
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Program Summary...............................................................................................................................................................4 
Semester Duration...............................................................................................................................................................4 
Eligibility........................................................................................................................................
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CHEM 103 Lab Exam 3 – Portage Learning
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CHEM 103 Lab Exam 3 – Portage Learning. A 1.1000 gram hydrate sample chosen from Na2CO3∙10H2O, AlCl3∙6H2O, 
MgCl2∙6H2O and BaCl2∙2H2O was heated and found to lose 0.5841 gram of H2O. (1) 
Show the calculation of the % H2O in the unknown hydrate sample. (2) Show the 
calculation of the % H2O in each of the hydrate compounds and identify the unknown 
hydrate from the list. 
Atomic weights: H = 1.008, O = 16.00. MWs: Na2CO3∙10H2O = 286.15, AlCl3∙6H2O = 
241.43, MgCl2∙6H2O = 203.301 ...
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Chem 103 portage learning Lab Exam 3
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Chem 103 portage learning Lab Exam 3
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CHEM 103 Module 2 Exam (General Chemistry) – Portage Learning
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CHEM 103 Module 2 Exam (General Chemistry) – Portage Learning. 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 14.01 = 14.01 
O = 1 x 16.00 = 16.00 
Br = 1 x 79.90 = 79.90 
Total Molecule Weight for C H NOBr= 199.02 
2 4 3 
7 5 
7 5 
1. 2Al + 3S + 12O = 342.17 
2. 7C + 5H + N + O + Br = 199.02 
Question 2 10 / 10 pts 
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CHEM 103 Module 2 Exam (Portage Learning)
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CHEM 103 Module 2 Exam (Portage Learning) 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 
M2: Exam - Requires Respondus LockDown Browser + Webcam: General Chemistry I w/Lab-2021- Kozminski 
3/18 
Total Molecular Weight for Al (SO ) = 342.17 
2. C H NOBr 
C = 7 x 12.01 = 84.07 
H = 5 x 1.008 ...
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CHEM 103 Module 2 Exam (General Chemistry) – Portage Learning
- Exam (elaborations) • 19 pages • 2023
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CHEM 103 Module 2 Exam (General Chemistry) – Portage Learning. 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 
M2: Exam - Requires Respondus LockDown Browser + Webcam: General Chemistry I w/Lab-2021- Kozminski 
3/18 
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 14.01 = 14.01 
O = 1 x 16.00 = 16.00 
Br = 1 x 79.90 = 79.90 
Total Molecule Weight for C H NOBr= 199.02
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FEMA IS-363: Introduction to Emergency Management for Higher Education Exam Questions and Answers 2024/2025 with complete solutions
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FEMA IS-363: Introduction to Emergency Management for Higher Education Exam Questions and Answers 2024/2025 with complete solutions 
 
FEMA IS-363: Introduction to Emergency Management for Higher Education Exam Questions and Answers 2024/2025 with complete solutions 
 
 
 
1.	Which type of exercise is most appropriate for an annual event to practice a coordinated response among all affected agencies and organizations? 
 
A.	workshop 
 
B.	full-scale exercise 
C.	drill 
 
D.	seminar 
 
2.	Which t...
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CHEM 103 Module 2 Exam (General Chemistry) – Portage Learning
- Exam (elaborations) • 19 pages • 2023
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CHEM 103 Module 2 Exam (General Chemistry) – Portage Learning. 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 
M2: Exam - Requires Respondus LockDown Browser + Webcam: General Chemistry I w/Lab-2021- Kozminski 
3/18 
Total Molecular Weight for Al (SO ) = 342.17 
2. C H NOBr 
C = 7 x 12.01 ...
-
CHEM 103 Module 2 Exam (Portage Learning)
- Exam (elaborations) • 19 pages • 2023
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Available in package deal
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- $19.49
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CHEM 103 Module 2 Exam (Portage Learning) 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 
M2: Exam - Requires Respondus LockDown Browser + Webcam: General Chemistry I w/Lab-2021- Kozminski 
3/18 
Total Molecular Weight for Al (SO ) = 342.17 
2. C H NOBr 
C = 7 x 12.01 = 84.07 
H = 5 x 1.008 ...
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