CHEM 103 MODULE 2 (CHEM103)

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 (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 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 = 233.99 2. 8C + 6H + N + 4O + Cl = 215.59 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 th...

[Document subtitle]   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 = 233.99 2. 8C + 6H + N + 4O + Cl = 215.59 Question 2 ...