Chem 103 module 1 to 6 Study guides, Class notes & Summaries
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Chem 103 Module 1 to 6 Exam with Verified answers (100 OUT OF 100) Portage learning (Latest Update)
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Chem 103 Module 1 to 6 Exam with Verified answers (100 OUT OF 100) Portage learning (Latest Update) 
 
 
MODULE 1 EXAM 
Question 1	 Click this link to access the Periodic Table. This may be helpful throughout the exam. 
1.	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 = smalle...
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CHEM 103: GENERAL ORGANIC CHEMISTRY Module 1 to 6 Exams, Final Exam and Lab Exam 1 to 6 with Verified answers Package Deal | Portage learning (Latest Update)
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CHEM 103: GENERAL ORGANIC CHEMISTRY Module 1 to 6 Exams, Final Exam and Lab Exam 1 to 6 with Verified answers Package Deal | Portage learning (Latest Update)
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Chem 103 Module 1 to 6 Exam answers Portage learning 2023 ( A+ GRADED 100% VERIFIED)
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Chem 103 Module 1 to 6 Exam answers Portage learning 2023 ( A+ GRADED 100% VERIFIED)
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Chem 103 Module 1 to 6 Exam answers Portage learning| Graded A+ Latest 2024
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Chem 103 Module 1 to 6 Exam answers Portage learning| Graded A+ Latest 2024 
 
Chem 103 Module 1 to 6 Exam answers Portage learning| Graded A+ Latest 2024 CHAME
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Chem 103 Module 1 to 6 Exam answers Portage learning ( A+ GRADED 100% VERIFIED)
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Chem 103 Module 1 to 6 Exam answers Portage learning ( A+ GRADED 100% VERIFIED)
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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 1 TO 6 EXAM LATEST 2022-2023 QUESTIONS & CORRECT ANSWERS PORTAGE LEARNING|COMPLETE AGRADE
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CHEM 103 MODULE 1 TO 6 EXAM LATEST 2022-2023 
QUESTIONS & CORRECT ANSWERS PORTAGE 
LEARNING|COMPLETE AGRADE
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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 1 to 6 Exam Questions and Answers (Portage Learning)
- Exam (elaborations) • 35 pages • 2023
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Available in package deal
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- $22.49
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CHEM 103 Module 1 to 6 Exam Questions and Answers (Portage Learning) Using the following information, do the conversions shown below, showing all 
work: 
1 ft = 12 inches 1 pound = 16 oz 1 gallon = 4 quarts 
1 mile = 5280 feet 1 ton = 2000 pounds 1 quart = 2 pints 
kilo (= 1000) milli (= 1/1000) centi (= 
1/100) deci (= 1/10) 
1. 24.6 grams = ? kg 
2. 6.3 ft = ? inches 
1. 24.6 grams x 1 kg / 1000 g = 0.0246 kg 
2. 6.3 ft x 12 in / 1 ft = 75.6 inches 
please always use the correct units in your ...
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Portage Learning / CHEM 103 / Module 1 to 6 Exam answers 2023|2024 UPDATED A+ RATED
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Click this link to access the Periodic Table. This may be helpful throughout 
the exam. 
1. 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. ...
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