Chem 103 module 2 - Study guides, Class notes & Summaries
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Docu PORTAGE LEARNING CHEM 103 MODULE 4EXAM 2023 WITH QUESTIONS AND CORRECT ANSWERS 100% PASS,PERIODIC TABLE IN THE LAST PAGE 2024ment title]
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Docu PORTAGE LEARNING CHEM 103 MODULE 4EXAM 2023 WITH QUESTIONS AND CORRECT ANSWERS 100% PASS,PERIODIC TABLE IN THE LAST PAGE 2024ment title]MODULE 1 EXAM 
 
Convert 845.3 to exponential form and explain your answer. 
1.	Convert 3.21 x 10-5 to ordinary form and explain your answer. 
 
 
1.	Convert 845.3 = larger than 1 = positive exponent, move decimal 2 places 
= 8.453 x 102 
2.	Convert 3.21 x 10-5 = negative exponent = smaller than 1, move decimal 5 places = 0.0000321 
 
Question 2	...
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CHEM 103 MODULE 2 EXAM .
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CHEM 103 MODULE 2 EXAM.Question 2 Click this link to access the Periodic Table. This may be helpful throughout the exam. 
Show the calculation of the number of moles in the given amount of the following substances. Report your answerto 3 significant figures. 
1. 13.0 grams of (NH4)2CO3 
2. 16.0 grams of C8H6NO4Br 
1. Moles = grams / molecular weight = 13.0 / 96.09 = 0.135 mole 
2. Moles = grams / molecular weight = 16.0 / 260.04 = 0.0615 mole 
Question 3 Click this link to access the Periodic Ta...
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CHEM 103 Module 1 to 6 Exam Questions and Answers (Portage Learning)
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CHEM 103 Module 1 to 6 Exam Questions and Answers (Portage Learning) Convert 845.3 to exponential form and explain your answer. 
2. Convert 3.21 x 10-5 to ordinary form and explain your answer. 
1.Convert 845.3 = larger than 1 = positive exponent, move decimal 2 places 
= 8.453 x 102 
2.Convert 3.21 x 10-5 = negative exponent = smaller than 1, move decimal 5 
places = 0. 
Question 2 
Click this link to access the Periodic Table. This may be helpful throughout 
the exam. 
Using the following info...
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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 Questions and Answers | Portage Learning (2023-2024)
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1.18 moles of C9H8NO4Cl 
Grams = 1.18 moles X 229.614 = 270.944 grams --> 270.9 grams of 
C H NO Cl 
C = 9 x 12.01 = 108.09 
H = 8 x 1.008 = 8.064 
N = 1 x 14.01 = 14.01 
O = 4 x 16.00 = 64 
Cl = 1 x 35.45 = 35.45 
Molecular Weight of C H NO Cl = 229.614 
3 4 2 
9 8 4 
9 8 4 
1. Grams = Moles x molecular weight = 1.05 x 310.18 = 325.7 
grams 
2. Grams = Moles x molecular weight = 1.18 x 229.61 = 270.9 
grams 
Question 4 10 / 10 pts 
You may find the following resources helpful: 
Scientific Ca...
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CHEM 103 Module 1 to 6 Exam Questions and Answers (Portage Learning)
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CHEM 103 Module 1 to 6 Exam Questions and Answers (Portage Learning) Show the calculation of the mass of a 18.6 ml sample of freon with 
density of 1.49 g/ml 
2. Show the calculation of the density of crude oil if 26.3 g occupies 30.5 
ml. 
1. M = D x V = 1.49 x 18.6 = 27.7 g 
2. D = M / V = 26.3 / 30.5 = 0.862 g/ml 
Question 5 
Click this link to access the Periodic Table. This may be helpful throughout 
the exam. 
1. 3.0600 contains ? significant figures. 
2. 0.0151 contains ? significant figu...
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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 
You may find the following resources helpful: 
Scientific Calculator (
Periodic Table 
(
Equation Table 
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downloa...
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CHEM 103 Module 2 exam with correct answers.
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CHEM 103 Module 2 exam with correct answers.CHEM 103 Module 2 exam with correct answers.
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CHEM 103 Module 2 Exam (Portage Learning)
- Exam (elaborations) • 19 pages • 2023
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Available in package deal
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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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Portage Learning CHEM 103 MODULE 2 EXAM 2022 with Questions and RAtionales
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Show the calculation of the molecular weight for the following 
compounds, reporting your answer to 2 places after the decimal. 
1. Al2(CO3)3 
2. C8H6NO4Cl 
1. 2Al + 3C + 9O = 233.99 
2. 8C + 6H + N + 4O + Cl = 215.59
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