Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Chapter 9, Problem 39E
Interpretation Introduction

Interpretation:

The liters of nitrogen gas that reacts with 50.0L of hydrogen gas are to be calculated.

Concept introduction:

Chemical reactions are represented by chemical equations. In a chemical equation the reactants are represented on the left of the arrow while the products are represented on the right of the arrow. Stoichiometric coefficient is the number preceding each symbol in a reaction which determines the moles of the reactants and products in the reaction. The ratio of moles is termed as mole ratio. In stoichiometry problems, a known amount of product or reactant is given and the other product or reactant has to be calculated.

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1. Calculate the accurate monoisotopic mass (using all 1H, 12C, 14N, 160 and 35CI) for your product using the table in your lab manual. Don't include the Cl, since you should only have [M+H]*. Compare this to the value you see on the LC-MS printout. How much different are they? 2. There are four isotopic peaks for the [M+H]* ion at m/z 240, 241, 242 and 243. For one point of extra credit, explain what each of these is and why they are present. 3. There is a fragment ion at m/z 184. For one point of extra credit, identify this fragment and confirm by calculating the accurate monoisotopic mass. 4. The UV spectrum is also at the bottom of your printout. For one point of extra credit, look up the UV spectrum of bupropion on Google Images and compare to your spectrum. Do they match? Cite your source. 5. For most of you, there will be a second chromatographic peak whose m/z is 74 (to a round number). For one point of extra credit, see if you can identify this molecule as well and confirm by…
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Chapter 9 Solutions

Introductory Chemistry: Concepts and Critical Thinking (8th Edition)

Ch. 9 - Prob. 11CECh. 9 - Prob. 12CECh. 9 - Prob. 13CECh. 9 - Prob. 1KTCh. 9 - Prob. 2KTCh. 9 - Prob. 3KTCh. 9 - Prob. 4KTCh. 9 - Prob. 5KTCh. 9 - Prob. 6KTCh. 9 - Prob. 7KTCh. 9 - Prob. 8KTCh. 9 - Prob. 9KTCh. 9 - Prob. 10KTCh. 9 - Prob. 11KTCh. 9 - Prob. 12KTCh. 9 - Prob. 13KTCh. 9 - Prob. 14KTCh. 9 - Prob. 15KTCh. 9 - Prob. 1ECh. 9 - Prob. 2ECh. 9 - Prob. 3ECh. 9 - Prob. 4ECh. 9 - Prob. 5ECh. 9 - Prob. 6ECh. 9 - Prob. 7ECh. 9 - Prob. 8ECh. 9 - Prob. 9ECh. 9 - Prob. 10ECh. 9 - Prob. 11ECh. 9 - Prob. 12ECh. 9 - Prob. 13ECh. 9 - Prob. 14ECh. 9 - Prob. 15ECh. 9 - Prob. 16ECh. 9 - Prob. 17ECh. 9 - Prob. 18ECh. 9 - Prob. 19ECh. 9 - Prob. 20ECh. 9 - Prob. 21ECh. 9 - Prob. 22ECh. 9 - Prob. 23ECh. 9 - Prob. 24ECh. 9 - Prob. 25ECh. 9 - Prob. 26ECh. 9 - Prob. 27ECh. 9 - Prob. 28ECh. 9 - Prob. 29ECh. 9 - Prob. 30ECh. 9 - Prob. 31ECh. 9 - Prob. 32ECh. 9 - Prob. 33ECh. 9 - Prob. 34ECh. 9 - Prob. 35ECh. 9 - Prob. 36ECh. 9 - Prob. 37ECh. 9 - Prob. 38ECh. 9 - Prob. 39ECh. 9 - Prob. 40ECh. 9 - Prob. 41ECh. 9 - Prob. 42ECh. 9 - Prob. 43ECh. 9 - Prob. 44ECh. 9 - Prob. 45ECh. 9 - Prob. 46ECh. 9 - Prob. 47ECh. 9 - Prob. 48ECh. 9 - Prob. 49ECh. 9 - Prob. 50ECh. 9 - Prob. 51ECh. 9 - Prob. 52ECh. 9 - Prob. 53ECh. 9 - Prob. 54ECh. 9 - Prob. 55ECh. 9 - Prob. 56ECh. 9 - Prob. 57ECh. 9 - Prob. 58ECh. 9 - Prob. 59ECh. 9 - Prob. 60ECh. 9 - Prob. 61ECh. 9 - Prob. 62ECh. 9 - Prob. 63ECh. 9 - Prob. 64ECh. 9 - Prob. 65ECh. 9 - Prob. 66ECh. 9 - Prob. 67ECh. 9 - Prob. 68ECh. 9 - Prob. 69ECh. 9 - Prob. 70ECh. 9 - Prob. 71ECh. 9 - Prob. 72ECh. 9 - Prob. 73ECh. 9 - Prob. 74ECh. 9 - Prob. 75ECh. 9 - Prob. 76ECh. 9 - Prob. 77ECh. 9 - Prob. 78ECh. 9 - Prob. 79ECh. 9 - Prob. 80ECh. 9 - Prob. 81ECh. 9 - Prob. 82ECh. 9 - Prob. 83ECh. 9 - Prob. 84ECh. 9 - Prob. 85ECh. 9 - Prob. 86ECh. 9 - Prob. 87ECh. 9 - Prob. 88ECh. 9 - Prob. 89ECh. 9 - Prob. 90ECh. 9 - Prob. 1STCh. 9 - Prob. 2STCh. 9 - Prob. 3STCh. 9 - Prob. 4STCh. 9 - Prob. 5STCh. 9 - Prob. 6STCh. 9 - Prob. 7STCh. 9 - Prob. 8STCh. 9 - Prob. 9STCh. 9 - Prob. 10STCh. 9 - Prob. 11STCh. 9 - Prob. 12STCh. 9 - Prob. 13STCh. 9 - Prob. 14STCh. 9 - Prob. 15ST
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Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY