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chem_103_m2_exam

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Master Chemistry with Confidence: Chem 103 Module 2 Exam Prep! Are you ready to conquer your Chem 103 Module 2 exam and unlock the secrets of chemistry mastery? Look no further! Elevate your study experience with our comprehensive Chem 103 Module 2 Exam Prep guide – your ultimate companion for acing the exam and achieving academic excellence! Tailored specifically for students tackling Module 2 of Chemistry 103, this meticulously crafted resource is designed to streamline your study process and empower you with the knowledge and skills needed to succeed. Whether you're a chemistry enthusiast or navigating the complexities of chemical concepts for the first time, this guide will pave the way for your success. Here's why our Chem 103 Module 2 Exam Prep is a must-have: Thorough Content Coverage: Dive deep into the core concepts of Module 2 with comprehensive coverage of key topics, including chemical reactions, stoichiometry, gases, and more. You'll gain a solid understanding of foundational principles that will serve as the cornerstone of your chemistry knowledge. Practice Problems Galore: Reinforce your learning and sharpen your problem-solving skills with a wealth of practice problems carefully crafted to mirror the style and difficulty level of the actual exam. Each question is designed to challenge and engage you, ensuring that you're fully prepared for any curveballs the exam may throw your way. Clear Explanations and Solutions: Don't just memorize – truly understand! Benefit from clear and concise explanations accompanying each practice problem, guiding you through the solution step-by-step and helping you grasp even the most challenging concepts with ease. Visual Aids and Illustrations: Immerse yourself in the world of chemistry with vibrant illustrations, diagrams, and charts that bring complex concepts to life. Visual learners will appreciate the wealth of visual aids that simplify abstract ideas and make studying both enjoyable and effective. Accessible Anywhere, Anytime: Study on your terms with convenient access to our exam prep guide from any device – whether you're at home, in the library, or on the go. Our PDF format ensures that you can dive into your studies whenever inspiration strikes. Guaranteed Success: Join the ranks of countless students who have achieved academic excellence with the help of our Chem 103 Module 2 Exam Prep. With our comprehensive guide by your side, you'll approach the exam with confidence and emerge victorious.

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CHEM 103 M2: Exam Questions and Correct Answers

Attempt History
Attempt Time Score
LATEST Attempt 1 112 minutes 100 out of 100



.


Question 1 pts



You may find the following resources helpful:

Scientific Calculator (https://www.desmos.com/scientific)

Periodic Table
(https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_

Equation Table
(https://portagelearning.instructure.com/courses/948/files/271474/download?
download_frd=1)




Show the calculation of the molecular weight for the following
compounds, reporting your answer to 2 places after the decimal.



1. Al2(SO4)3



2. C7H5NOBr


Your Answer:

1. Al2(SO4)3

Al = 2 x 26.98 = 53.96

S = 3 x 32.07 = 96.21

O = 12 x 16.00 = 192
1/18

,Total Molecular Weight for Al2(SO4)3 = 342.17



2. C7H5NOBr

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 C7H5NOBr= 199.02




1. 2Al + 3S + 12O = 342.17



2. 7C + 5H + N + O + Br = 199.02




Question 2 pts



You may find the following resources helpful:

Scientific Calculator (https://www.desmos.com/scientific)

Periodic Table
(https://previous.nursingabc.com/upload/images/Help_file_picture/Periodic_

Equation Table
(https://portagelearning.instructure.com/courses/948/files/271474/download?
download_frd=1)




Show the calculation of the number of moles in the given amount of the
2/18

, following substances. Report your answerto 3 significant figures.



1. 13.0 grams of (NH4)2CO3


2. 16.0 grams of C8H6NO4Br


Your Answer:

Moles = grams / Molecular Weight

1. 13.0 grams of (NH4)2CO3

Moles = 13.0 g / 96.094 = .135 moles of (NH4)2CO3

N = 2 x 14.01 = 28.02

H = 8 x 1.008 = 8.064

C = 1 x 12.01 = 12.01

O = 3 x 16 = 48

Molecular weight is 96.094



2. 16.0 grams of C8H6NO4Br

Moles = 16.0 g / 260.038 = .0615 moles of C8H6NO4Br

C = 8 x 12.01 = 96.08

H = 6 x 1.008 = 6.048

N = 1 x 14.01 = 14.01

O = 4 x 16.00 = 64

Br = 1 x 79.90 = 79.90

Molecular weight of C8H6NO4Br = 260.038




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