Chem 103 module 2 exam - Study guides, Class notes & Summaries

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...

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...

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 ( downloa...

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...

CHEM 103 Module 2 exam with correct answers.CHEM 103 Module 2 exam with correct answers.

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

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 ...

CHEM 103 Module 2 exam with correct answersCHEM 103 Module 2 exam with correct answersCHEM 103 Module 2 exam with correct answersCHEM 103 Module 2 exam with correct answers

CHEM 103 MODULE 2 EXAM 2022 Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. 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 = 23

CHEM 103 Module 2 exam with correct answers Latest Updated Examination study Guide 2024