Chem 121 - Study guides, Class notes & Summaries
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CHEM 121 Lab Report 10: Urinalysis
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CHEM 121 Lab Report 10: Urinalysis
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CHEM 121 Lab Report 2: Density Determination FALL 2023_Portage learning
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Experiment #: 2 
Title: Density Determination 
Purpose: The purpose of this experiment is to learn how to calculate the density of both liquids 
and solids in the chemistry lab by finding the volume and mass of different substances. 
Procedure: 
Density Determination of Water: 
1) A graduated cylinder is to be filled with water 
2) The temperature is taken using a digital thermometer of the water in the graduated 
cylinder 
3) A value is read and recorded in ºC for the temperature of the water ...
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CHEM 121 Lab Report 1: Equipment, Safety, and Mass/Volume Measurement
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CHEM 121 Lab Report 1: Equipment, Safety, and Mass/Volume Measurement
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CHEM 121 Lab Report 10: Urinalysis - Portage Learning
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CHEM 121 Lab Report 10: Urinalysis - Portage Learning
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CHEM 121 Lab Report 9: Inorganic Synthesis
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CHEM 121 Lab Report 9: Inorganic Synthesis
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CHEM 121 Lab Report 7(Acid Base Titrations) 2023 Portage Learning
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CHEM 121 Lab Report 7 Acid Base Titrations 2023
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CHEM 121 Lab Report 7: Acid Base Titrations
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CHEM 121 Lab Report 7: Acid Base Titrations
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PORTAGE LEARNING CHEM 121 MODULE 1 EXAM 2023|A GRADED|NEW!!!
- Exam (elaborations) • 26 pages • 2023
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1.	Convert 1005.3 to exponential form and explain your answer. 
 
2.	Convert 4.87 x 10-6 to ordinary form and explain your answer. Your Answer: 
 
 
1.) Convert 1005.3 to exponential form and explain your answer: 
 
All number values must be between 1-10. Because this value is greater than 1, the exponent remains positive. 
 
In order for this to happen, we have to move the decimal three places to the LEFT thus, the exponential form of 1005.3 is: 1.0053 X 103
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Chem 121 Sample Final Exam Questions and Answers 2024/2025 With Complete Solution; The University of British Columbia
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Chem 121 Sample Final Exam Questions and Answers 2024/2025 With Complete Solution; The University of British Columbia
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CHEM 121 Exam 2 SPRING 2024 Portage learning
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Show the calculation of the molecular weight for the following two compounds, reporting 
your answer to 2 places after the decimal. 
1. Ca3(PO4)2 
2. C9H8NO4Cl 
Your Answer: 
1. 3(40.08)+2(30.97+4(16))=310.18 g/mol Ca3(PO4)2 
2. 9(12.01)+8(1.008)+14.01+4(16)+35.45 = 229.61 g/mol C9H8NO4Cl 
1. 3Ca + 2P + 8O = 310.18 
2. 9C + 8H + N + 4O + Cl = 229.61 
Question 2 
9 / 10 pts 
You may find the following resources helpful: 
Scientific Calculator 
Links to an external site. 
Periodic Table 
Equation ...
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