Chemistry and physics for nurse anesthesia _ a student-centered approach (2017).pdf

Chemistry and Physics for Nurse Anesthesia David Shubert, PhD, is a dean in the College of Arts and Sciences at Newman University, Wichita, Kansas, where until recently he designed and cotaught the combined chemistry–physics for nurse anesthesia course with Dr. Leyba. He was inducted into the nati...

Chemistry and Physics for Nurse Anesthesia David Shubert, PhD, is a dean in the College of Arts and Sciences at Newman University, Wichita, Kansas, where until recently he designed and cotaught the combined chemistry–physics for nurse anesthesia course with Dr. Leyba. He was inducted into the national Achievement Rewards for College Scientists (ARCS) Foundation prior to ­ joining Newman University in 1987. Although his area of expertise lies in organometallic chemistry, Dr. Shubert has taught intro- ductory, general, organic, analytical, and instrumental analytical chemistry, as well as bio- chemistry, earth and space science, general physical science, and c­hemistry/physics for nurse anesthesia. In 1995, Dr. Shubert was honored by his students and peers with the Teaching Excellence Award. He has worked as a chemical consultant with several local industries, includ- ing Vulcan Chemicals and BG Products. Working with colleagues from five independent col- leges, and with financial support from the National Science Foundation, the Camille and Henry Dreyfus Foundation, and the A. V. Davis Foundation, Dr. Shubert has offered numerous work- shops that provided training and access to modern instrumentation to high school chemistry teach- ers. Most recently, he was recognized by the Wichita Business Journal in 2014 as a Healthcare Hero. His research interests have evolved from organometallic chemistry to chemical education. John Leyba, PhD, is a professor of chemistry and head of the Department of Chemistry and Biochemistry, University of North Georgia, Dahlonega, Georgia. Previously, he was a faculty mem- ber at Newman University for 12 years, where he served as the chair of the Division of Science and Mathematics and where he cotaught the combined chemistry–physics for nurse anesthesia course with Dr. Shubert. Dr. Leyba has extensive industrial experience in the Department of Energy Complex. He was senior scientist, senior scientist A, and principal scientist with Westinghouse Savannah River Company. In addition, he was the radiochemistry group leader for Rust Federal Services Clemson Technical Center in Anderson, South Carolina. He was also a visiting assistant professor in the Chemistry Department and adjunct assistant professor in the Department of Environmental Engineering and Science at Clemson University. Dr. Leyba was also the Denver Area director of operations for Canberra Industries. His research interests involve the rapid chemical separation and detection of radioactive materials. Dr. Leyba has won several teaching awards including the Newman University 2005 Teaching Excellence Award and two Outstanding Didactic Instructor Awards from the Newman University Nurse Anesthesia Program. Sharon Niemann, DNAP, CRNA, is director of the Nurse Anesthesia Program, Newman University, Wichita, Kansas (since 2004). She has practiced as a certified registered nurse anesthetist since 1995 and has been employed as a cardiovascular nurse anesthetist with Anesthesia Consulting Service (Wichita) since certifying. She is a retired captain in the U.S. Army Reserve, having served overseas in spring 2001. Dr. Niemann serves on the Steering Committee, Mid-Continent Regional Center for Health Care Simulation and is a member of numerous professional organizations including the American Association of Nurse Anesthetists (AANA) and the Kansas Association of Nurse Anesthetists (KANA), where she has held numerous offices, including president and treasurer, and currently is the program chair. She was also an item writer for the Council on Certification of Nurse Anesthetists (2006–2009) and a member of Sigma Theta Tau International. Dr. Niemann has presented widely for KANA and other state and local organizations on topics such as conscious sedation, anesthesia for total joint replacement, and anesthesia for cardiovascular procedures. She teaches a variety of courses in the Newman University Nurse Anesthesia Program and is principal investigator and clinical advisor for multiple senior nurse anesthesia thesis projects. She received her Doctor of Nurse Anesthesia Practice degree at Texas Wesleyan University in 2016. Chemistry and Physics for Nurse Anesthesia A Student-Centered Approach Third Edition David Shubert, PhD John Leyba, PhD Sharon Niemann, DNAP, CRNA Copyright © 2017 Springer Publishing Company, LLC All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission of Springer Publishing Company, LLC, or authorization through payment of the appropriate fees to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400, fax 978-646-8600, [email protected] or on the Web at www.copyright.com. Springer Publishing Company, LLC 11 West 42nd Street New York, NY 10036 www.springerpub.com Acquisitions Editor: Margaret Zuccarini Compositor: diacriTech ISBN: 978-0-8261-0782-4 e-book ISBN: 978-0-8261-0783-1 Instructor’s Test Bank ISBN: 978-0-8261-0402-1 Instructor’s PowerPoints ISBN: 978-0-8261-0404-5 Online videos are available at http://www.springerpub.com/shubert-3rd-edition Instructor’s Materials: Qualified instructors may request supplements by e-mailing [email protected] 17 18 19 20 / 5 4 3 2 1 The author and the publisher of this Work have made every effort to use sources believed to be reliable to provide infor- mation that is accurate and compatible with the standards generally accepted at the time of publication. Because medi- cal science is continually advancing, our knowledge base continues to expand. Therefore, as new information becomes available, changes in procedures become necessary. We recommend that the reader always consult current research and specific institutional policies before performing any clinical procedure. The author and publisher shall not be liable for any special, consequential, or exemplary damages resulting, in whole or in part, from the readers’ use of, or reliance on, the information contained in this book. The publisher has no responsibility for the persistence or accuracy of URLs for external or third-party Internet websites referred to in this publication and does not guarantee that any content on such websites is, or will remain, accurate or appropriate. Library of Congress Cataloging-in-Publication Data Names: Shubert, David, author. | Leyba, John, author. | Niemann, Sharon, author. Title: Chemistry and physics for nurse anesthesia : a student-centered approach / David Shubert, John Leyba, Sharon Niemann. Description: Third edition. | New York, NY : Springer Publishing Company, LLC, | Includes bibliographical references and index. Identifiers: LCCN 2016047394| ISBN 9780826107824 | ISBN 9780826104021 (instructors test bank) | ISBN 9780826107831 (e-book) | ISBN 9780826104045 (instructors PowerPoints) Subjects: | MESH: Chemical Phenomena | Anesthetics—chemistry | Nurse Anesthetists | Physical Phenomena | Nurses’ Instruction Classification: LCC RT69 | NLM QD 31.3 | DDC 617.9/60231—dc23 LC record available at https://lccn.loc.gov/2016047394 Contact us to receive discount rates on bulk purchases. We can also customize our books to meet your needs. For more information please contact: [email protected] Printed in the United States of America by Bradford & Bigelow. Contents Videos ix Foreword G. Michael Caughlin, MD xi Preface xiii Share Chemistry and Physics for Nurse Anesthesia: A Student-Centered Approach, Third Edition 1. Measurement 1 Why This Matters: A Nurse Anesthetist’s View 1 A Review of Some Basic Mathematical Skills 1 Measurements and Significant Figures 12 Significant Figures in Calculations 14 Accuracy and Precision 15 SI (Metric System) 16 Absolute Zero and the Kelvin Scale 17 Conversion Factors 18 Density 21 Density and Specific Gravity 22 Summary 23 Review Questions for Measurement 25 2. A Review of Some Chemistry Basics 31 Why This Matters: A Nurse Anesthetist’s View 31 What Is Chemistry? 31 Atomic Structure and Dimension 34 Dalton’s Atomic Theory 37 The Periodic Table of the Elements 38 Some Common Elements 46 Chemical Nomenclature 49 Electrolytes 55 Stoichiometry 56 Summary 57 Review Questions for Chemistry Basics 59 Note: The video icons link to http://www.springerpub.com/shubert-3rd-edition, where corre- sponding videos can be found. v vi Contents 3. Basics of Physics Part 1 (Force and Pressure)   65 Why This Matters: A Nurse Anesthetist’s View   65 Newton’s Laws  66 Mass  66 Velocity  66 Acceleration  69 Force  72 Pressure  74 Atmospheric Pressure  77 Measuring Pressure  78 Oscillometry  82 Summary  83 Review Questions for Physics (Part 1)  85 4. Basics of Physics Part 2 (Work, Energy, and Power)  89 Why This Matters: A Nurse Anesthetist’s View  89 Work  90 State Functions  95 Energy  97 Thermodynamics  100 Specific Heat  104 Power  105 Summary  107 Review Questions for Physics (Part 2)  109 5. Fluids  113 Why This Matters: A Nurse Anesthetist’s View  113 Fluids: A Definition  114 Hydrostatics  114 Hydrodynamics: Moving Fluids  122 Viscosity  131 Summary  134 Review Questions for Fluids  135 6. The Gas Laws  137 Why This Matters: A Nurse Anesthetist’s View  137 The Empirical Gas Laws   137 The Ideal Gas Law  144 Dalton’s Law of Partial Pressures  146 Kinetic Molecular Theory of Gases  150 Graham’s Law of Effusion  152 Ideal Gases and Real Gases  153 Summary  155 Review Questions for Gases  158 7. States of Matter and Changes of State  165 Why This Matters: A Nurse Anesthetist’s View  165 Kinetic Molecular Theory of Matter  165 Chemical Bonding and the Octet Rule  167 Energy Changes and Changes of State  196 Summary  201 Review Questions for States of Matter and Changes of State  205  Contents vii 8. Solutions and Their Behavior  211 Why This Matters: A Nurse Anesthetist’s View  211 Some Basic Terminology  212 Solution Concentrations  212 Solubility  218 Energy Changes and the Solution Process  220 Factors Affecting Solubility  222 Colligative Properties of Solutions  224 Colloids  230 Summary  231 Review Questions for Solutions  234 9. Acids, Bases, and Buffers  241 Why This Matters: A Nurse Anesthetist’s View  241 Chemical Equilibria  242 Acids and Bases  246 Acid–Base Reactions  254 Measuring Acidity: The pH Function  255 Calculating the pH of Solutions  259 Other Acidic Species  269 Buffers  270 Summary  276 Review Questions for Acids and Bases  281 10. Electricity and Electrical Safety  289 Why This Matters: A Nurse Anesthetist’s View  289 Electricity and Electrical Charge  290 Ohm’s Law and Electrical Circuits  294 Semiconductors  302 Spectroscopy   304 Summary  315 Review Questions for Electricity  318 11. Classes of Organic Compounds  319 Why This Matters: A Nurse Anesthetist’s View  319 Functional Groups  319 Hydrocarbon Functional Groups  320 Alkenes  325 Organohalogen Compounds  329 Functional Groups Based on Water  330 Amines  334 Carbonyl Functional Groups  336 Summary  347 Review Question for Organic Compounds  349 12. Biochemistry  351 Why This Matters: A Nurse Anesthetist’s View  351 Biomolecules  351 Carbohydrates  352 Fischer Projections  353 Lipids  365 viii Contents Metabolism of Carbohydrates and Fatty Acids  370 Proteins and Amino Acids  375 Nucleic Acids  386 Summary  398 Review Questions for Biochemistry  401 13. Radiation and Radioactivity  405 Why This Matters: A Nurse Anesthetist’s View  405 Radiation  406 Electromagnetic Radiation  407 Radioactive Materials—An Introduction  409 Radioactivity  411 Radioactive Decay  411 Decay Rate  416 Half-Life  417 Ionizing Radiation Versus Nonionizing Radiation  418 Sources of Radioactive Materials  419 Radiation Exposure  420 Effects of Ionizing Radiation on Biological Systems  422 Medical Uses of Radionuclides (Nuclear Medicine)  423 Working With Radioactive Materials  425 Summary  427 Review Questions for Radiation and Radioactivity  430 14. Problem-Solving Skills and Answers to Review Questions  431 Using Your Calculator  431 Unit Conversions and the Use of Conversion Factors  432 A Simple Problem-Solving Methodology  435 Step-by-Step Solutions to End-of-Chapter Problems  439 Comprehensive List of Equations  537 Formulas and Constants  557 Glossary  559 Supplemental References  577 Index  581 Videos 1.1 Density and Specific Gravity 22 2.1 Conductivity and Electrolytes 55 3.1 Pressure in a Syringe 75 5.1 Equation of Continuity 123 5.2 Poiseuille’s Law 133 6.1 The Volume–Pressure Relationship: Boyle’s Law 138 6.2 The Volume–Temperature Relationship: Charles’s Law 139 6.3 Effusion, Diffusion, and Graham’s Law 152 7.1 VSEPR (Valence Shell Electron Pair Repulsion) Theory 173 7.2 Surfactants and Surface Tension 187 9.1 pH 255 10.1 Pulse Oximetry 305 10.2 Electrical Safety 307 12.1 Levels of Protein Structure 383 12.2 Denaturing a Protein 386 Note: Videos can be found at http://www.springerpub.com/shubert-3rd-edition. This link can also be found as a footnote on those pages in text where the video icon appears. ix Foreword Mastery of the basic sciences of chemistry and physics is a cornerstone to the beginning student’s development of a foundation of anesthesia principles. Chemistry helps anesthesia providers begin to understand pharmacology and cellular function, while physics directly deter- mines our physiological status and governs the principles of volatile anesthetic gases that we use on a daily basis. New clinicians’ understanding of these basic sciences is essential to learning theories and concepts that will govern their anesthesia practices. As an anesthesiologist with more than 30 years of involvement in both physician and nurse anesthesia education, I particularly appreciate the enhanced clinical focus in this edition of Chemistry and Physics for Nurse Anesthesia: A Student-Centered Approach, by David Shubert, PhD, John Leyba, PhD, and Sharon Niemann, DNAP, CRNA. Readers can immedi- ately see the relevance of chemistry and physics to both the theory and practice of anesthesia. The book’s authors provide a solid theoretical foundation to both qualitative and quantitative aspects of fluids, energy, and gases. Students have ample opportunity to test their understanding through the wide variety of end-of-chapter questions. The addition of a new concluding chapter enhances that experience, enabling students to obtain immediate feedback to help master the presented concepts. The Irish physicist William Thomson Kelvin stated, “There cannot be a greater mistake than that of looking superciliously upon practical applications of science. The life and soul of science is its practical application.” Indeed, the application of principles of chemistry and phys- ics influences our daily practice of anesthesia. Their thorough understanding is necessary to developing a logical, scientific approach to the difficult clinical situations we frequently encoun- ter in anesthesia. G. Michael Caughlin, MD Medical Director, Newman School of Nurse Anesthesia Assistant Clinical Professor of Anesthesiology University of Kansas School of Medicine, Wichita, Kansas xi Preface We have had the pleasure of teaching chemistry and physics for anesthesia for nearly 20 years to hundreds of nurse anesthesia students. The importance of the physical sciences in anesthesia goes without saying, and every one of our students knows this. We are continually impressed by the ability and motivation of our students, but we understand that they generally have only a minimal background in chemistry and almost no background in physics. They are bright and capable, but their math skills are often rusty. Our students are graduate students in nurse anesthesia, not graduate students in chemistry or in physics. So, it makes no sense to us to teach this course as though they were. Hence, this text attempts to bridge that gap. We cover mate- rial ranging from the most basic introductory chemistry concepts to topics treated in physical chemistry. This text provides a somewhat uneven treatment of a vast array of material that emphasizes the specific content in chemistry and physics that relates to anesthesia and offers illustrations that directly relate to anesthesia. The text has sufficient mathematical rigor to deal with the material, but no more than that. Our intent in writing this book is to facilitate a conversation about chemistry and physics concepts that are essential to understanding anesthesia, and especially the behavior of gases. Our writing is intentionally conversational and encouraging. We try to make the story as engag- ing as possible. Some might observe that our writing is unacceptably casual, but we believe it is more important for these students to master this material, rather than to learn how to sound like a stuffy college professor. In writing this third edition, we listened to our students, to your students, and to you, our fellow instructors. We have incorporated many excellent suggestions. Every chapter has an increased focus on clinical stories and applications. We recruited one of our respected colleagues, Sharon Niemann, to provide a clinician’s perspective on the material. The end-of-chapter problems have been expanded, edited, and reorganized, and detailed solutions are now provided in a new concluding chapter. We have included end-of-chapter summaries and an end-of-book glossary. Most exciting (to us, anyway) are the demonstration videos available, courtesy of Springer Publishing Company, which can be viewed at http://www.springerpub.com/shubert-3rd-edition. There are 15 videos illustrating concepts ranging from the various gas laws to fluid dynamics. xiii xiv Preface They might look like home videos (mostly because they are) but the science is real, and some of the demonstrations are original. In addition to this text, we have provided instructor’s material that includes a chapter-based test bank and PowerPoint slides. To receive these electronic files, faculty should please contact Springer Publishing Company at [email protected]. And (of course), we think we have eliminated every typo from the text. If (when) you find one, please let us know. Thank you for your support and ENJOY CHEMISTRY AND PHYSICS! David Shubert John Leyba Sharon Niemann  Share Chemistry and Physics for Nurse Anesthesia: A Student-Centered Approach, Third Edition C h a p t e r 1 Measurement Why This Matters: A Nurse Anesthetist’s View Why do we care, in this day of technology, if a certified registered nurse anesthetist can do math? Well, even today technology may fail, leaving us to use basic nursing skills of patient assessment and math to provide the best care for our patients. If we have basic math skills, we cannot only determine a drug dose or measure a base deficit, but we can also recognize whether the answer we obtain is reasonable. We may recognize that the answer “just doesn’t seem right” when the calculated dose is unexpectedly lower or higher than we anticipate. This may lead us to prevent a potential erroneous drug dose or calculate a fluid deficit with ­precision. A common example involves rechecking your calculations if you have to crack a second vial of medication to achieve the desired dose. Most medications are packaged in ­common doses (i.e., spinal bupivacaine comes as a 7.5% solution in a 2-mL vial). If your original ­calculation requires you to open a second vial to draw up 22 mg (2.9 mL) of bupivacaine, it is worthwhile to double check the dosage. If your patient is a basketball player who is 7 ft tall, it may be correct. If your patient is 5 ft, 4 in. tall, the original dosage may be catastrophic.  A REVIEW OF SOME BASIC MATHEMATICAL SKILLS Chemistry and physics are both very logical sciences, but both depend on math to translate concepts into application. Unfortunately, many students who struggle in these disciplines have more trouble with the mathematics than with the scientific concepts. Therefore, let us begin our exploration of chemistry and physics with a review of some basic math skills and concepts that you will use throughout this course. While this chapter reviews basic math skills, it cannot replace a basic understanding of college-level algebra. You will need a calculator for this course. Some of you may have fancy graphing cal- culators, but the capacity of these instruments far exceeds your mathematical needs for 1 2 Chapter 1 Measurement this course. Any calculator that is labeled as a “scientific calculator” will more than suffice. You should be able to buy an adequate calculator for less than $20. When dealing with a math problem, don’t reach for your calculator first. Whether or not the calculator gives you the correct answer depends on your ability to give it the correct information with which to work. Think about what operations you need to do first. See how far you can get with just a paper and pencil. You don’t have to do arithmetic computations, such as long division, by hand, but you should be able to perform the mathematical manipulations (e.g., solve the equation for x) without a calculator. In fact, it is typically best to solve the equation first for the variable of interest before plugging in numbers. As you become more adept at thinking your way through math problems, you are more likely to get your calculator to give you the right answer. Order of Operations There are four principal arithmetic operations: addition, subtraction, multiplication, and divi- sion. When an equation involves multiplication and/or division as well as addition and/or sub- traction, you need to do the multiplication and division operations before doing the addition and subtraction operations. For example, what does 12 plus 3 times 10 equal? x = 12 + 3 ⋅ 10 Since this equation mixes addition and multiplication, we need to do the multiplication first: x = 12 + 30 x = 42 Now, you evaluate this expression: 12 x= −3⋅2+4 4 Answer: x = 1 Parentheses and division bars (fraction bars) are both symbols of enclosure. Whenever there are symbols of enclosure, execute any arithmetic operations inside the symbol of enclosure first. The general rule is to work from the innermost symbol of enclosure to the outermost, multi- plying and dividing, then adding and subtracting. Consider this example, which involves both kinds of symbols of enclosure and order of operations. 12 + 3 ⋅ (4 + 2) x= 3.5 To solve this example, we need to first evaluate inside the parentheses, then multiply and finally add: 12 + 3 ⋅ (4 + 2) 12 + 3 ⋅ (6) 30 x= = = 3⋅5 3⋅5 3⋅5  A Review of Some Basic Mathematical Skills 3 Here we come to a very common error. Many students are tempted to divide 30 by 3 first, and then multiply by 5. But the division bar is a symbol of enclosure, so multiply 3 times 5 first and then divide 30 by 15. 30 x= =2 15 Let’s evaluate this expression: 10 + (6 + 4) ⋅ (5 − 3) x= 5 8 +9 4 Answer: x = 2 Negative numbers are common sources of confusion. Just remember that multiplying (or dividing) two positive numbers or two negative numbers gives a positive number. Multiplication or division involving a positive number and a negative number gives a negative number. Multiplying a number by negative 1 changes the sign of that number. Likewise, a negative num- ber can be expressed as a positive number times −1. Subtracting a negative number is the same as adding a positive number. Some examples are shown in the following: 4 ⋅ (−3) = −12 (−4) ⋅ (−3) = +12 −4x − (−3x) = −4x + 3x = −x Algebra: Solving Equations for an Unknown Quantity Addition and subtraction are inverse operators of each other, as are multiplication and division. Usually, a problem will involve solving an equation for some variable. You want to convert the equation into the form x = some number (or some expression). Therefore, when you want to iso- late a variable (i.e., “solve the equation”), you need to undo any operations that are tying up the variable by applying the inverse operation. That is, if the variable is multiplied by a number, you need to divide both sides of the equation by that number. If a variable is divided by some number, you need to multiply both sides of the equation by that number. This eliminates that number from the variable by converting it into a 1. Since 1 is the “multiplicative identity element,” you can ignore 1’s that are involved in multiplication or division. If some number is added to the variable, you need to subtract that number from both sides of the equation. If some number is subtracted from the variable, you need to add that number to both sides of the equation. This eliminates that number from the variable by converting it into a zero. Since zero is the “additive identity element,” you can ignore zeros that are involved in addition or subtraction. The critical thing to remember, though, is that whatever you do to one side of the equation, you have to do to the other side. Think of it as being fair. If you add a number to one side, you 4 Chapter 1 Measurement have to add the same number to the other side; otherwise, the equation would no longer be balanced. For example, if 12x = 180, what’s x? 12x = 180 x is multiplied by a factor of 12: Let’s divide both sides by 12: 12x 180 = 12 12 12 divided by 12 equals 1, and we can ignore it. 180 x= = 15 12 On your calculator, first enter the number on top (the numerator) and then divide by the number on the bottom (the denominator). Division operators are not commutative (i.e., a ÷ b ≠ b ÷ a), so the order in which you enter the numbers is critically important. On the other hand, multiplication and division are associative, so you can regroup fractions multiplied by a quantity. For example: 12 6 ⋅ 12 6 6⋅ = = 12 ⋅ = 24 3 3 3 Division signs (÷) are rarely used in the sciences. Usually, a division operation is signaled by a fractional notation. Just think of the horizontal bar as the division sign. For example, if x divided by 15 equals 180, what does x equal? x = 180 15 x is divided by a factor of 15: Let’s multiply both sides by 15: x 15 ⋅ = 180 ⋅ 15 15 15 divided by 15 equals 1, and on the left side of the equation it drops out. x = 180. 15 = 2,700 Sometimes the variable is in the denominator. Since anything in the denominator denotes the division operation, you can undo the division by multiplying, which is the inverse operation. For example: 15 = 180 x x is tied up in a division, so let’s multiply both sides by x: 15 x⋅ = 180 ⋅ x x  A Review of Some Basic Mathematical Skills 5 On the left side, x divided by x is 1, and we can ignore it. 15 = 180 ⋅ x Dividing both sides by 180 gives 15 = x = 0.083 180 For example, Charles’s Law is shown in the following and describes the effect of tempera- ture on the volume of a gas. Solve this equation for T2: V1 V2 = T1 T2 V2 ⋅ T1 Answer : T2 = V1 When you need to solve an equation that involves addition and/or subtraction as well as mul- tiplication and/or division, the order of operations reverses. First, you need to clear the addition and/or subtraction operations and then the multiplication and/or division. For example, the equation that relates Fahrenheit temperature to Celsius temperature is 9 °F = (°C) + 32 5 To solve this equation for °C, we need to remove the addition term (32), the multiplication factor (9), and the division factor (5). Remember that, since the 5 is below the division (or frac- tion) bar, it is included in the equation by a division operation. We have to start with clearing the addition operation: 9 °F − 32 = ⋅ °C + 32 − 32 5 This converts the 32 on the right into 0, which we can ignore. 9 °F − 32 = ⋅ °C 5 Now, the °C is multiplied by 9 and divided by 5, so we need to divide by 9 and multiply by 5. 5 9 5 (°F − 32) ⋅ = ⋅ °C ⋅ 9 5 9 Nine divided by 9 is 1, as is 5 divided by 5. This clears the right side of the equation, and we finally have: 5 (°F − 32) ⋅ = °C 9 First, notice the °F − 32 term contained in parentheses. When we multiply and divide by 5 and 9, respectively, these operations are applied to the entire left side of the equation, not just 6 Chapter 1 Measurement the °F term. Second, notice that you do not need to memorize two equations for ­converting between °F and °C. The first equation, °F = 9/5°C + 32, converts °C into °F. If you need to con- vert °F into °C, you don’t need to memorize °C = 5/9(°F − 32). You just need to solve the first equation for °C, and then plug in your numbers and evaluate the expression. This strategy will serve you well throughout your studies of both physics and chemistry. How can we solve this expression for x? (x + 5) =5 (x − 5) Answer: 7.5 Two other functions you’re likely to encounter are square roots and logarithms. You can review your college algebra book for the nuances of these functions. For our purposes, however, it is more important that you are able to use these functions to generate numerically correct answers. Your calculator should have an “x2” button and a “ x” button. These operators are inverses of each other. That is, if you enter 10 on your calculator and then press x2, you get 100. Now if you press x, you get 10 back. Each operator undoes the effect of the other. Let’s try another example. To solve this equation for x, we clear the square root operator by squaring both sides, and then solve for x. x + 3 = 10 ( x + 3)2 = 102 x + 3 = 100 x = 97 In extracting square roots, remember that you always obtain two answers: a positive root and a negative root. This is a mathematical reality, but in science typically only one of these answers has a physical reality. For example, starting from rest, the distance (x) an object travels, when undergoing constant acceleration (a) for a time interval (t), is given by x = 1/2 · at2. Let’s solve this equation for t when x = 6 and a = 3: 1 6= ⋅ 3t2 2 Dividing both sides by 3 and multiplying both sides by 2 gives t2 = 4. If we take the square root of both sides, t can be equal to either +2 or −2. However, the concept of negative time makes no physical sense, so we would report that answer in a physics class as t = 2 s. Logarithm functions come up fairly frequently in chemistry and physics, so you need to be able to handle them. Logarithms are mixed up exponents. You know 102 = 100. In this expres- sion, 10 is the base, 2 is the power, and 100 is the result. In other words, you need 2 factors of 10 to get to 100, 3 to reach 1,000, and so forth. A logarithm is another way to express this exponential expression. So, the log expression for 102 = 100 is: log10(100) = 2  A Review of Some Basic Mathematical Skills 7 Again, 10 is the base of the logarithm, and the 2 tells us that we need two factors of ten to reach 100. This becomes more challenging when we want the log of a number such as 110, which is not an integral power of 10. Fortunately, you do not need to calculate logs by hand. Your calculator should have a “log” button (for base 10, or common, logarithms) and a “10x” button (for antilog). Unless you are told otherwise, the operator log means base-10 logarithms. The log and antilog operators are inverses of each other. So, if you take the log of a number, and then the antilog of the result, you get the original number. For example, if we take the log of 110, we get 2.041. log(110) = 2.041 Then, if we take the antilog of 2.041, we get 110. Notice the antilog function can be repre- sented as either “antilog” or 10x: antilog (2.041) = 102.041 = 110 Depending on your calculator, you might have to enter the number first, and then press the operator key, or first press the operator key and then enter the number. You need to explore the operation of your calculator. One reason logs are so handy is the way they reduce multiplication and division to the sim- pler operations of addition and subtraction. The log of a product of two numbers is equal to the sum of the logs of those numbers: log(a ⋅ b) = log(a) + log(b) And the log of a ratio of two numbers is equal to the difference of the logs of the numbers:  a log   = log(a) − log(b)  b Perhaps most usefully, the log of a power is equal to the exponent in the power times the log of the base of the power: log(ab) = b ⋅ log(a) Use your calculator to verify for yourself that the following statements are true: log(10 ⋅ 5) = log(10) + log(5) = 1.699  10  log   = log(10) − log(5) = 0.3010  5 log(105 ) = 5 ⋅ log(10) = 5 There are other bases for logarithms. Your calculator should also be able to handle natural logarithms, which are based on the irrational number “e” rather than 10. The natural log key is “ln” and the natural antilog key is “ex.” 8 Chapter 1 Measurement Exponents Exponents are shorthand for the number of times a quantity is multiplied. We most commonly encounter exponents when using scientific notation, but units in physics are sometimes expo- nential, and this gives them a unique meaning. Imagine we measure the edge of a cube and find each edge to be 1 cm in length. What is the volume of the cube? The volume of a cube is given by length × width × height, so we start with: volume = 1 cm × 1 cm × 1 cm Let’s regroup the like terms (multiplication is associative, so it is okay to group the terms together in any order we like). volume = 1 × 1 × 1 × cm × cm × cm Well, 1 × 1 × 1 = 1, but what about the cm? Since there are three cm terms, the exponential expression of cm × cm × cm is cm3. Therefore, the volume of our cube is 1 cm3. Notice that cm connotes “length” but cm3 connotes volume, a different concept than length. There are some special cases of exponents. Any quantity (except zero) raised to the zero power equals one. 50 = 1 cm0 = 1 Any quantity (except zero) to the “−1” power is equal to the reciprocal of the quantity. Moving a quantity across a division bar changes the sign of the exponent. This is true for num- bers as well as variables or units. 1 1 2−1 = = 21 2 g = g ⋅ cm −3 cm3 When powers are multiplied, the exponents are added. 103 ⋅ 10−4 = 10−1 When powers are divided, the exponents are subtracted. cm3 = cm 2 cm1 Let’s evaluate this expression: 10 −3 ⋅ 10 6 10 −8 Answer: 1011  A Review of Some Basic Mathematical Skills 9 Scientific Notation Scientific notation uses exponents (powers of 10) for handling very large or very small numbers. A number in scientific notation consists of a number multiplied by a power of 10. The number is called the mantissa. In scientific notation, only one digit in the mantissa is to the left of the decimal place. The “order of magnitude” is expressed as a power of 10, and indicates how many places you had to move the decimal point so that only one digit remains to the left of the decimal point. 11,000,000 = 1.1 × 107 0.0000000045 = 4.5 × 10−9 If you move the decimal point to the left, the exponent is positive. If you move the decimal to the right, the exponent is negative. Memorized rules like this are easy to confuse. You might find it more logical to recognize that large numbers (greater than 1) have positive exponents when expressed in scientific notation. Small numbers (less than 1) have negative exponents. As an example, let’s express 1.02 × 10−3 and 4.26 × 104 in decimal form. Answer: 0.00102 and 42,600 Numbers in proper scientific notation have only one digit in front of the decimal place. Can you convert 12.5 × 10−3 into proper form? You have to move the decimal point one place to the left, but does the exponent increase or decrease? Is the correct answer 1.25 × 10−4 or 1.25 × 10−2? Rather than memorize a rule, think of it this way. You want your answer to be equal to the initial number. So, if you DECREASE 12.5 to 1.25, wouldn’t you need to INCREASE the expo- nent to cancel out this change? So, the correct answer is 1.25 × 10−2. As another example, let’s express 0.0034 × 10−25 in proper scientific notation. Answer: 3.4 × 10−28 You will probably use your calculator for most calculations. It is critical that you learn to use the scientific notation feature of your calculator properly. Calculators vary widely, but virtually all “scientific” calculators have either an “EE” key or an “EXP” key that is used for scientific notation. Unfortunately, these calculators also have a “10x” key, which is the “antilog” key and has nothing to do with scientific notation, so don’t use the 10x key when entering numbers in scientific notation. One common mistake is to include the keystrokes “× 10” when entering numbers in scientific notation. DO NOT TYPE “× 10”! The calculator knows this is part of the expression, and manually entering “× 10” will cause your answer to be off by a factor of 10. You should refer to the owner’s manual of your calculator for exact instructions. However, most calculators follow this protocol: 1. Enter the “mantissa” 2. Press the “EXP” key 3. Enter the exponent, changing the sign if necessary 10 Chapter 1 Measurement Let’s evaluate this expression: (4.8 × 10 −6 ) (1.2 × 10 3 )(2.0 × 10 −8 ) Answer: 2.0 × 10−1 Remember to follow the rules for symbols of enclosure. The division bar is a symbol of enclosure, so do the operations in the numerator and in the denominator before performing the division. By the way, did you notice you can easily solve this problem without a calculator? Try regrouping the mantissas separately from the powers of 10. 4.8 = 2 (1.2 ⋅ 2) and 10−6 10−6 10−6 −8 = 3+(−8) = −5 = 10−1 (10 ⋅ 10 ) 10 3 10 Therefore, the answer is 2.0 × 10−1. When representing a negative number using scientific notation, the negative sign goes in front of the mantissa. This negative sign has nothing to do with the sign in front of the exponent. For example, −0.0054 would be represented in scientific notation as −5.4 × 10−3. Graphing It is frequently very helpful to plot data. The data is often easier to analyze when it is presented so as to obtain a linear relationship. One of the most useful forms of graphing data is to use the “slope-intercept” equation for a straight line: y = mx + b The x variable appears on the horizontal axis and represents how you are manipulating the system. For this reason, it is called the independent variable, because it is not free to change as the system is manipulated. The y variable appears on the vertical axis and represents the measured response of a system. For this reason, it is called the dependent variable, because its value depends on the way in which you change the system. In an anesthesia setting, you would control the amount of anesthetic administered to a patient. This would be the independent variable. The amount of time a patient remains under anesthesia depends on the dose, which is, therefore, the dependent variable. If you were conducting a study and wanted to graph the rela- tionship between the dose and time of anesthesia, the dose should be graphed on the horizontal axis and the time on the vertical axis. It may not always be clear in every case which variable is independent and which is dependent. The slope of the line (m) relates how strongly, and in which sense, the dependent variable changes as you change the independent variable. If the line rises from left to right, the slope is  A Review of Some Basic Mathematical Skills 11 positive. If the line falls from left to right, the slope is negative. The slope can be calculated for a line by the equation: y2 − y1 m= x2 − x1 The intercept (b) mathematically describes the value of the function where it crosses the y-axis. In a physical sense, however, the intercept represents a correcting or offset value that relates the dependent variable to the independent variable. That is, when your controlled (independent) variable is zero, the experimental (dependent) variable may have an inherently nonzero value. Consider the simple example of relating temperatures on the Fahrenheit to the Celsius scale. Imagine a container of water at its freezing point. A Celsius thermometer registers 0°C, and a Fahrenheit thermometer registers 32°F. Similarly, when the water is boiling, the two thermom- eters register 100°C and 212°F, respectively. If you graph the two data pairs, and we consider the Fahrenheit temperature to be the dependent variable (although this assignment is somewhat arbitrary), our graph would look like Figure 1.1. 250 200 Degrees, Fahrenheit (100, 212) 150 100 50 (0, 32) 0 0 20 40 60 80 100 Degrees, Celsius Figure 1.1 Fahrenheit Versus Celsius Temperatures The line has a value of 32°F where it crosses the y-axis, so b = 32. The slope of the line is: y2 − y1 212 − 32 180 m= = = = 1.8 x2 − x1 100 − 0 100 So the equation of our line is: y = 1.8x + 32 But the y variable is °F and the x variable is °C, so we can also express this equation as: °F = 1.8°C + 32 Of course, this is the formula most of us memorized for converting between Fahrenheit and Celsius temperatures (notice that 1.8 is just the decimal equivalent of 9/5). One important 12 Chapter 1 Measurement lesson from this example is that formulas don’t just “fall from the sky.” They are experimentally (empirically) or theoretically derived. Second, it is a poor strategy to get through chemistry and physics by trying to just memorize formulas. You will have a much better understanding of the concepts if you focus on how the formulas are derived and the underlying reasons for the relationship between various quantities. MEASUREMENTS AND SIGNIFICANT FIGURES Science is based on observation and measurement. All measurements have an inherent error or uncertainty. Therefore, our scientific conclusions can’t be more reliable than the least reliable measurement. One tool to assess the reliability of a measurement is the number of digits that are reported. Significant figures are digits in a measured value that have a physical meaning and can be reproducibly determined. For example, what is the pressure reading in the pressure gauge in Figure 1.2? 4 3 5 2 6 1 7 psi Figure 1.2 Pressure Gauge/Significant Figures Could you report the pressure as 3.8945 psi? Well, the needle is clearly between 3 and 4; there’s no doubt about that. If we want another decimal place, we have to estimate. The next decimal place looks to be about 0.8, although some people might estimate it to be 0.7 or 0.9. This is where the uncertainty enters. So we report the pressure as 3.8 psi. Any more decimal places would be pure guesses with absolutely no reproducibility. Therefore, this measuring device is capable of delivering a maximum of two significant figures. As a matter of habit, you should always push measuring devices to their limits for delivering good data. That is, you should always estimate one decimal place past the smallest division on the scale. The concept of significant figures applies only to measured quantities, which always have some inherent uncertainty. Numbers not obtained using measuring devices are called exact  Measurements and Significant Figures 13 numbers. That is, they are easily countable, and there is absolutely no question as to the value. For example, most hands have four fingers and one thumb, and there is no question regarding the four or the one. Quantities that are defined are another example of exact numbers. For example, there are 12 in. in 1 ft and 2.54 cm in 1 in. The 12 and 2.54 are definitions; ­therefore, they are exact quantities. You can consider these kinds of measurements to have an infinite number of significant figures. Not all counted quantities are exact numbers. For example, counting very large numbers (e.g., the number of red blood cells under a microscope) will have some uncertainty, because the result is difficult to reproduce. In order to deal with measurements reported to you by other investigators, you need to rec- ognize how many significant figures a given measurement contains. Nonzero digits are always significant and never seem to confuse anyone. Zeros are more problematic. For example, if you measure your height as 1.7 m, does the reliability of this measurement change if you report it as 0.0017 km or 1,700 mm? Of course not. The zeros in these two numbers are only ­“placeholders” and simply separate the decimal place from the nonzero digits. The following statements summarize the rules regarding significant digits: Nonzero digits are always significant. Captive zeros (zeros between nonzero digits) are always significant. Leading zeros (zeros at the beginning of a number and located to the left of nonzero digits) are never significant. Trailing zeros (zeros at the end of a number and located to the right of nonzero digits) are only significant when the number contains a decimal point. A major exception to the aforementioned rules involves exact numbers and definitions (rela- tionships), which are considered to have an infinite number of significant digits. An example of an exact number is 12 eggs. An example definition involves the relationship between an inch and a centimeter. One inch is defined to be exactly 2.54 cm. One operational way to identify significant zeros is to convert the number into scientific notation. If the decimal point does not cross a zero when converting a number into scientific notation, the zero is always significant. Some additional examples are given in the following (significant figures are in bold). VALUE NUMBER OF SIGNIFICANT FIGURES 2.0 2 20.00 4 0.002 1 0.020 2 200 1 2.30 × 103 3 2,020 3 20,000 1 14 Chapter 1 Measurement  SIGNIFICANT FIGURES IN CALCULATIONS The result of a calculation cannot be more reliable than any of the measurements on which the calculation is based. The statistical way in which the inherent uncertainties in measurements are propagated through addition and subtraction operations is different from the way they are propagated in multiplication and division operations. A rigorous treatment of this difference is beyond the scope of our purposes, although certainly not beyond your capacity to understand it. You should examine a statistics text if you are interested in learning more. Fortunately, there are some simple rules for determining the number of significant figures to retain when perform- ing calculations involving measured quantities. When adding or subtracting, keep the smaller number of decimal places. When multiplying or dividing, keep the smaller number of significant figures. Apply these rules according to the algebraic hierarchy of calculations (consider symbols of enclosure first, then multiplication/division operations, and lastly addition/ subtraction operations). For example, how should we report the answer in the following problem to the proper num- ber of significant figures? 4.023 + 102.3 = ? When adding, the answer should have the same number of decimal places as the addend with the smallest number of decimal places. The second number has only one decimal place. Therefore, the answer can have only one decimal place and is correctly reported as 106.3. Let’s try a multiplication example. Evaluate the following expression and report the answer to the proper number of significant figures. 4.023 × 102.3 = ? Each number has four significant figures. Therefore, the correct answer is 411.6 (four signifi- cant figures). We’ll crank it up a notch and try a mixed operation example next. Evaluate the following example and report the answer to the proper number of significant figures. 4.023 − 3.954 =? 5.564 A division bar is a symbol of enclosure: Do the subtraction first, keeping three decimal places: 0.069 = 0.012 5.564 The numerator has two significant figures, while the denominator has four. Therefore, the answer should have only two significant figures.  Accuracy and Precision 15  ACCURACY AND PRECISION When reporting a measurement, a reasonable question is, How good is the measurement? There are two ways to answer this question: accuracy and precision. Accuracy is the agreement between experimental data and the “true” value, while precision is the agreement between replicate measurements. For example, imagine shooting two arrows at a target. If both arrows strike the bull’s-eye, your data are both accurate and precise. If both arrows land close together, but miss the bull’s-eye, the data will be precise but not accurate. If one arrow hits 5 in. to the right of the bull’s-eye, and the other arrow hits 5 in. to the left, then the average position of the arrows does hit the bull’s-eye. These data are accurate but not precise. Finally, if one arrow hits to the left of the bull’s-eye and the other arrow misses the target completely, the data are neither precise nor accurate. Which estimate of “goodness” is more important? Well, both of them are important. It is important to you and your patient that the reading on a pulse oximeter accurately represents the patient’s blood PO2. Likewise, you won’t have confidence in a pulse oximeter if it gives a different reading every time you look at it (assuming the patient’s condition has not changed). The accuracy of a measurement can be improved by making replicate measurements and taking the average. All measurements have some inherent uncertainty. But if the uncertainties are random, then half of the measurements should be too big and half too small. The errors will cancel each other out. Accuracy is assessed by calculating the percent error, which is given by the formula: measured value − “true” value %error = × 100% “true” value Notice the quotation marks around “true” value. A method is verified by testing it against known standards, and in this case, you do know the “true” value, because you prepared the sys- tem. However, in most cases, you will never know what the true value is. The patient certainly has some value for serum cholesterol level, but you don’t know what that is. The precision of a measurement is improved by careful lab technique and/or using instru- ments capable of yielding greater precision (signaled by showing more significant figures). Repeating a measurement doesn’t necessarily improve precision (except in that your technique might improve), but it does allow you to determine whether or not the measurement is, in fact, reproducible. In any case, a greater number of significant figures implies greater precision. Precision can be quantified by standard deviation. Unfortunately, we do not have time for a rigorous exploration of statistics, but you probably had a statistics course sometime during your undergraduate nursing program. So, the equation for calculating averages and standard deviations should be familiar. average = ∑ measurements number of measurements standard deviation = ∑ (measurements − average of measurements) 2 number of measurements − 1 16 Chapter 1 Measurement The smaller the ratio of the standard deviation to the average value, the better the precision. However, there is no absolute standard that defines good precision or bad precision. In general, you should become increasingly suspicious of measurements as the ratio of standard deviation to the average value increases. For an experienced analytical chemist, working with a well-known system, this ratio will typically be less than 0.1%, but this level of precision is unlikely in a clinical setting. For example, imagine measuring the (arterial) PO2 of a patient three times and obtaining these values: 105, 96.0, and 102 mmHg (in that order). What is the average value? Does the data seem precise? What is the standard deviation of the measurements? What is the percent error? First, notice that the values are not “trending.” That is, they seem to bounce around the aver- age value, and this should be reassuring. If each successive measurement is smaller or larger than the previous measurement, you should be concerned that there is a problem with the measurement system, instrument, or method. The average value is 101 mmHg and the stan- dard deviation is 4.6 mmHg. Since the standard deviation is much smaller than the average value (4.6/101 = 0.045 or 4.5%), these data are fairly precise. You can’t determine the percent error (and thereby assess the accuracy) because you don’t know the “true” value. If these mea- surements were obtained from an artificial system in which you knew the PO2 was equal to 100 mmHg (three significant figures), the percent error is 1%.  SI (METRIC SYSTEM) The metric system consists of a base unit and (sometimes) a prefix multiplier. Most scientists and health care providers use the metric system, and you are probably familiar with the com- mon base units and prefix multipliers. The base units describe the type of quantity measured: length, mass, or time. The SI system is sometimes called the MKS (meter, kilogram, second) system, because these are the standard units of length, mass, and time upon which derived quantities, such as energy, pressure, and force, are based. An older system is called the CGS (centimeter, gram, second) system. The derived CGS units are becoming extinct. Therefore, we focus on the MKS units. The prefix multipliers increase or decrease the size of the base unit, so that it more conve- niently describes the system being measured. The base units that we will be using are listed in Table 1.1. Chemists frequently express mass in units of grams, which is derived from the kilogram. There are three other SI base units (the mole, the candela, and the ampere). We will consider moles (amount of material) and amperes (electric current) in subsequent chapters. The candela is a unit of light intensity or luminosity and does not concern us at this point. Table 1.1 SI BASE UNITS BASE UNIT QUANTITY MEASURED ABBREVIATION Meter Length m Kilogram Mass kg Kelvin Temperature K Mole Amount of material mol  Absolute Zero and the Kelvin Scale 17 The prefix multipliers are given in Table 1.2. It is crucial that you learn all the multipliers, as well as their numerical meanings and abbreviations. Notice some abbreviations are capital let- ters, while others are lowercase letters. It is important that you not confuse these. The letter m is particularly overused as an abbreviation. Notice that the abbreviation for micro is the Greek letter mu (μ), which is equivalent to the English letter m. Table 1.2 SI PREFIX MULTIPLIERS NAME OF PREFIX MULTIPLIER NUMERICAL MEANING ABBREVIATION giga 1 × 109 G mega 1 × 106 M kilo 1 × 103 k deci 1 × 10−1 d centi 1 × 10−2 c milli 1 × 10−3 m micro 1 × 10−6 μ nano 1 × 10−9 n Table 1.3 gives some useful relationships between metric and English measurements for length, mass, and volume. There is a common error in this table. A pound is a unit of weight, not mass. So it is not correct to equate pounds (a weight unit) and kilograms (a mass unit). However, in a clinical setting, this distinction is almost never made. Weight is related to mass through Newton’s Second Law, a concept we explore in Chapter 3. Table 1.3 COMMON CONVERSION FACTORS LENGTH MASS VOLUME 1 in. = 2.54 cm (exactly) 1 lb = 0.454 kga 1 L = 1 dm3 1 mi = 1.609 km 1 lb = 16 oz 1 gal = 3.79 L 1 mi = 5,280 ft (exactly) 1 oz = 28.35 gb 1 mL = 1 cm3 1 grain = 64.80 mg 1 cc = 1 cm3 a This is correctly stated as 1 lb = 4.45 N. b This is commonly cited, even though it equates a mass quantity with a weight quantity.  ABSOLUTE ZERO AND THE KELVIN SCALE Much of physics and physical chemistry is based on the Kelvin temperature scale, which begins at absolute zero. Absolute zero is the coldest possible temperature. Therefore, there is no tem- perature lower than 0 K. Notice that it is not zero “degrees kelvin.” It is simply zero kelvins. On the Celsius temperature scale, 0 K is equivalent to −273.15°C. Since the temperature difference 18 Chapter 1 Measurement represented by 1 K is the same as the temperature difference represented by 1°C, converting between the two scales is easy: K = °C + 273.15 As an example, what is body temperature (98.6°F) on the Kelvin scale? We don’t have an equation to convert between K and °F. It would be a waste of time for you to memorize one, because we can simply use the two relationships we already know. First, let’s convert 98.6°F into degrees Celsius: °F = 1.8°C + 32 °C = (98.6°F − 32)/1.8 = 37.0°C Then we use the second relationship to convert °C into kelvins: K = °C + 273.15 K = 37.0°C + 273.15 = 310.2 K Now, you should be able to confirm that absolute zero is equivalent to −459.67°F. CONVERSION FACTORS The factor-label method, or using conversion factors, is arguably the single most important skill for physical science students to master. Most problems in chemistry and physics can be solved by allowing the units associated with the measurements to walk you through the calculations. If you manipulate the units so that the undesired units drop out of the calculation, your numerical answer is almost sure to be correct. The basic strategy is to identify a beginning quantity and desired final quantity. You multiply a conversion factor times the beginning quantity so that the units of the beginning quantity are mathematically replaced by the units of the desired quantity. Notice unit2 divided by unit1 equals 1 and we can ignore it.  unit2  beginning quantity unit1 ⋅  conversion factor = desired quantity unit 2  unit1  A conversion factor is a statement of equality between two measurements of the same object or property. As a beginning example, let’s explore converting between metric quantities. For example, a length of 1 m obviously has a length of 1 m: 1m=1m Of course, we are free to mathematically manipulate this equation to suit our needs. So, it is perfectly legitimate to divide both sides of the equation by 100, which gives: 1 × 10−2 m = 1 × 10−2 m  Conversion Factors 19 Recall the numerical meaning of “centi” is 1 × 10−2, so we can substitute the prefix multiplier “c” for its numerical equivalent. We don’t have to do this substitution to both sides because in trading “c” for its numerical equivalent, we haven’t changed the value on the right side. 1 × 10−2 m = 1 cm Notice that our conversion factor between metric quantities has an interesting form. One side gets the prefix multiplier letter (in this case, the “c”), and the other side gets the numerical equivalent of the multiplier (in this case, 1 × 10−2). Students sometimes get the number on the wrong side of the conversion factor. If you stick with the definition of the prefix multipliers given in Table 1.2, your conversion factors will always have a prefix multiplier on one side and a number on the other side. Seems fairer that way, doesn’t it? Of course, we could divide both sides of our conversion factor equation by 1 × 10−2, which would give: 1 m = 100 cm You might have learned this form in a previous class. Of course, either form of our conver- sion factor can be used. However, if you have ever had trouble converting between metric units, you might want to stick with the first version. Learn the prefix multipliers and their numerical meanings. Then, when you write a conversion factor expression, make sure one side gets the letter and the other side gets the number. Let’s explore a couple of other manipulations of our conversion factor. If we divide both sides by 1 cm, our conversion factor would look like: 1 × 10−2 m 1 cm = 1 cm 1 cm Notice that, in doing this, we have 1 cm divided by 1 cm. Any quantity (except zero) divided by itself equals 1. 1 × 10−2 m 1 cm = =1 1 cm 1 cm Notice further that the reciprocal of 1 is 1. Therefore, our conversion factor is equal to its own reciprocal. 1 × 10−2 m 1 cm = 1 cm 1 × 10−2 m This is important because 1 is the multiplicative identity element. This means we are free to multiply any beginning quantity times our conversion factor, and we will always get an equiva- lent quantity. Can you write a conversion factor expression for converting bytes into gigabytes? Answer: 1 Gbyte = 1 × 109 bytes 1 Gbyte or 1 × 109 bytes 1 × 109 bytes or 1 Gbyte 20 Chapter 1 Measurement The trick, of course, is knowing which form of the conversion factor we need to use. Fortunately, we can always let the units tell us how to solve the problem. Now, let’s try a concrete example: How many inches are equal to 25.0 cm? First, we need to recognize the beginning quantity is 25.0 cm. The beginning quantity is the one about which you have all the information (i.e., both the number and the unit). Next, we need a conversion factor that relates centimeters and inches. Conversion factors always involve (at least) two units. From Table 1.3, we see that 1 in. = 2.54 cm. We want the cm unit in the conversion factor in the denominator, so that it will divide out the cm unit in the beginning quantity. This will leave units of inches in the numerator, which are the units of the desired quantity. Now, if we do the arithmetic and divide 25.0 by 2.54, we obtain 9.84 as the numerical portion of the answer. Notice we have kept three significant figures, because 25.0 cm has three significant figures. More importantly, the units that remain are inches, which are what we wanted to end up with.  1 in.  25.0 cm  = 9.84 in.  2.54 cm  When using conversion factors, there are two key points to remember. First, this method works only if you are meticulous in including units with all numerical data. Second, if your final answer has the correct units, the answer is probably correct. If your final answer has the wrong units, the answer is almost certainly incorrect. Now let’s try an example that requires more than one conversion factor. A red blood cell is about 6.0 μm in diameter. How many nanometers is this? In this case, we don’t have a conver- sion factor that directly relates nanometers to micrometers. Fortunately, we can relate micro­ meters to meters, and we can relate meters to nanometers: 1 μm = 1 × 10−6 m 1 nm = 1 × 10−9 m If you are very proficient in using conversion factors, you can “stack them up” and do all the conversions in one step. However, we will generally proceed stepwise, so that we are less likely to make an error. First, we need to convert micrometers into meters:  1 × 10−6 m  −6 6.0 µ m   = 6.0 × 10 m  1µ m  Our second conversion factor changes meters into nanometers:  1 nm  6.0 × 10−6 m  = 6.0 × 103 nm  1 × 10−9 m  If a patient receives 8.0 L of oxygen per minute, how many milliliters per second is this? Notice we need to divide the units of L from the numerator and units of minutes from the denominator. You know that 60 s = 1 min and 1 mL = 1 × 10−3 L. Be careful to apply the conversion factor so that the unit we need to eliminate is on the opposite side of the division bar from the same unit in the conversion factor. Let’s change minutes into seconds first, and  Density 21 then change liters into milliliters. Of course, you can apply the conversion factors in any order you like. 8.0 L  1 min  0.13 L = min  60 s  s 0.13 L  1 m L  130 mL   = s  1 × 10−3 L  s DENSITY Density is the ratio of the mass of an object to its volume. Chemists usually abbreviate density with a “d” while physicists typically use the Greek letter ρ (rho). Since the course is chemistry and physics for nurse anesthesia, you will need to recognize both abbreviations. mass density = ρ = d = volume Notice that density will always have two units: a mass unit divided by a volume unit. For example, water has a density of 1.0 g/mL. Density can have a different combination of units, though. For example, water also has a density of 1,000 kg/m3 and 8.33 pounds/gallon. By the way, the density of water as 1.0 g/mL is a useful quantity to remember. Since the mass of 1 mL of water has a mass of 1 g, we can interchange grams and milliliters of water (as long as the temperature is close to 25°C). If 28.0 g of nitrogen gas occupies 22.4 L at 0°C and 1 atm of pressure, what is the density of N2 in grams per milliliter? By now, you should easily be able to convert 22.4 L into 22,400 mL, so we can plug the numerical values into the definition of density and find the density of nitro- gen under these conditions is 0.00125 g/mL. 28.0 g g d= = 1.25 × 10−3 22, 400 mL mL Since density always has two units and conversion factors have two units, guess what? Density is a conversion factor between mass and volume. Don’t memorize when you need to multiply by density or divide by density to achieve a required conversion. Let the units tell you what to do. What is the edge length of a wooden cube if the mass of the cube is 145 g? The density of this sample of wood is 0.79 g/cm3. Using density, we can convert the mass of the wooden cube into its volume.  1 cm3  145 g  = 184 cm3  0.79 g  Hold on a minute. The density is reported to only two significant figures, so we are entitled to retain only two significant figures in our final answer, but you see this intermediate answer with three significant figures. If you round off to the correct number of significant figures after each calculation, your final answer can be significantly different from the correct answer. This is 22 Chapter 1 Measurement called a rounding error. It is a good idea to carry a “guard digit” through a multistep calcula- tion, and only round the final answer to the correct number of significant figures. You can also prevent rounding errors by not clearing your calculator after each step and re-entering the number. Just leave each intermediate answer in the calculator’s memory and proceed with the next computation. To complete this problem, you need to know a little geometry. The volume of a cube is equal to the length of its edge cubed. So, if we take the cube root of the volume, we will obtain the length of the edge. volume of a cube = edge length3 edge length = 3 184 cm3 = 5.7 cm Now, it’s time to round the final answer to two significant figures. Notice the units also work out in this calculation. The cube root of cm3 is cm. DENSITY AND SPECIFIC GRAVITY Specific gravity is the ratio between an object’s density and the density of water. density of object Video 1.1 specific gravity = sg = density of water Specific gravity is, therefore, dimensionless and does not depend on the units. Specific grav- ity will always have the same numerical value, regardless of the units of density you choose to employ in determining the specific gravity. Any sample with a specific gravity greater than 1 is denser than water. Any sample that has a specific gravity less than 1 is less dense than water. There is often some confusion between specific gravity and density, because the density of water is 1.00 g/mL. This means the density expressed as grams per milliliter and specific gravity of a sample are numerically equal. For example, let’s calculate the specific gravity of mercury, which has a density of 13.6 g/mL. g 13.6 sg = mL g = 13.6 1.00 mL It appears that density and specific gravity are the same, but they aren’t. They are numerically equal when the units of density are expressed as grams per milliliter (g/mL). Specific gravity equals density only when the units of measure are grams per milliliter, because the density of water is 1 g/mL. To illustrate this difference, let’s start with different density units. One gallon of mercury has a mass of 113 lb (i.e., the density of mercury is 113 lb/gal). The density of water is 8.33 lb/gal. If we calculate the specific gravity using these values, we find the same value for the specific gravity. 113 lb gal sg = = 13.6 8.33 lb gal http://www.springerpub.com/shubert-3rd-edition-video-1.1  Summary 23 SUMMARY Laboratory and clinical analyses all require measurements, and all measurements have an inherent uncertainty. Scientific notation is a tool to deal with very large or very small numbers. A number expressed in scientific notation consists of a number having only one digit to the left of the decimal point, multiplied by some power of 10. Significant digits represent measurements that can be replicated or determined with a reasonable level of certainty. A digit is not significant if it represents a random guess in a measurement. Nonzero digits are always significant, captive zeros are always sig- nificant, leading zeros are never significant, and trailing zeros are only significant when the number contains a decimal point. Exact numbers and definitions (relationships) are considered to have an infinite number of significant digits. The zeros in boldface in the following are all significant. 1.0   0.010   100.0 These zeros in boldface are not significant: 0.01   10 When adding or subtracting measurements, the final answer should have the same number of decimal places as the smallest number of decimal places in the measurements. When multiplying or dividing measurements, the final answer should have the same number of significant digits as the measurement with the smallest number of significant digits. The conversion factor method is a tool to assist you in converting one unit of measurement into another. A conversion factor is an equality statement that relates the same quantity in two different units (e.g., 1 in. = 2.54 cm). Conversion factors are applied as a ratio so that the old units cancel out, and only the desired units remain. Accuracy represents agreement of a measurement with the true or accepted value, although the true or accepted value is typically not known in a clinical setting. Accuracy can be evaluated through a percent error calculation: measured value − true value %error = 100% true value Precision represents agreement among repeated measurements. A larger number of sig- nificant digits in a measurement implies greater precision. Precision can be evaluated with the standard deviation of the replicate measurements. The SI, or metric, system of measurements consists of base units, plus prefix multipliers. The SI unit of mass is the kilogram, the SI unit of length is the meter, and the SI unit of time is the second. Clinical data contains several derived units of 24 Chapter 1 Measurement measurement, such as volume, which is the cube of length. One liter is defined as 0.001 m3. The most commonly encountered prefix multipliers include kilo (103), milli (10−3); and centi (10−2); however, you should be familiar with all of the multipliers listed in Table 1.2. The SI unit of temperature is the kelvin. The Kelvin temperature scale begins at a­ bsolute zero, the coldest possible temperature. The change in temperature represented by 1 K is equivalent to a change in temperature of 1°C. The relationships between the three most common temperature scales (Kelvin, Celsius, and Fahrenheit) are: °F = 1.8°C + 32 K = °C + 273.15 Density is mass divided by volume. The SI unit of density is kg/m3, although clinical densities are more likely to be reported as g/cm3. Density, like all measurements that have units in terms of the ratio of two other units, can be used as a conversion factor between mass and volume. mass density = ρ = d = volume Specific gravity is defined as the density of an object divided by the density of water. The density of both substances must be expressed in the same units. Since the density of water is approximately 1.0 g/cm3, the specific gravity of an object is approximately equal to the density of the object, expressed as g/cm3. Specific gravity has no units and will have the same numerical value, regardless of the units of density used to ­determine it. Objects with a specific gravity less than one will float on water; objects with a ­specific gravity greater than one will sink. density of object specific gravity = sg = density of water  Review Questions for Measurement 25 REVIEW QUESTIONS FOR MEASUREMENT 1. Evaluate the following exponential expressions. You should not need to use your calculator. a. 103 · 10−4/10−8 b. (103)−1 c. (10450)/10−253 d. (cm2)(cm) 2. Put the following numbers into scientific notation. a. 23.4 b. 26.4 × 10−4 3. Use your calculator to simplify these expressions. a. 8.46 × 105/3.25 × 10−6 b. 8.02 × 10−2/6.02 × 1023 4. What is a significant figure? 5. What is an exact number? How many significant figures are in an exact number? 6. Differentiate between accuracy and precision and describe how to assess and improve each. 7. If lug nuts on your car should be tightened to a torque of 150 ft-lb, what is the corre- sponding metric torque in units of N-m? (1 lb = 4.45 N and 1 in. = 2.54 cm) 8. What is the volume of 1.0 cm3 in units of cubic inches? 9. How is density different from specific gravity? 10. The density of water is 62 lb/ft3. The specific gravity of gold is 19.3. What is the density of gold in lb/ft3? 11. The density of gold is 19.3 g/cm3. What is the mass in kilograms of a gold brick m ­ easuring 5.00 in. by 3.00 in. by 12.0 in.? 12. If PV = nRT, what does R = ? 13. The Clausius–Clapeyron equation states:  P  ∆H  1 1  ln  1  = −  P2  R  T2 T1  What does T1 equal? 14. Evaluate the value of k, which, though of no particular importance to you at this time, is the famous Boltzmann constant. (7.60 × 102 ) ⋅ (2.24 × 104 ) k= (273.15) ⋅ (6.02 × 1023 ) 26 Chapter 1 Measurement 15. Complete the following table of the metric prefix multipliers. MULTIPLIER ABBREVIATION MATHEMATICAL MEANING micro m 10−2 kilo n 106 16. This graph shows how the volume of a sample of gas responds to changing temperature. Estimate the volume of the gas at a temperature of 0°C. 50 40 30 Temperature, Celsius 20 10 0 -10 -20 -30 -40 -50 18 19 20 21 22 23 24 25 26 27 Volume, Liters 17. Consider the syringe pictured at the right. Each mark on the syringe represents 0.1 mL. What is the correct reading of the material in the syringe and the maxi- mum number of significant figures that can be reported? a. One significant figure, and the reading is 0.7 b. Two significant figures, and the reading is 0.7 c. Two significant figures, and the reading is 0.63 d. Three significant figures, and the reading is 0.635 18. Which of these statements is correct? a. The precision of a data set is improved by averaging multiple measurements. b. Standard deviation is a way of quantifying the precision of data.  Review Questions for Measurement 27 c. Accuracy refers to the reproducibility of a measurement. d. All of the above statements are correct. 19. Dr. Kyotay has ordered a new operating table from the ACME surgical supply. The table will support a total of 50 kg. What is the corresponding weight of the largest patient you can put on this table in pounds? a. 0.11 lb b. 110 lb c. 23 lb d. 230 lb 20. Which of these quantities represents the greatest mass? a. 104 ng b. 10−1 mg c. 10−8 kg d. All of these represent equivalent masses. 21. A patient has a temperature of 310 K. Should you be concerned? Or, rather, what is the patient’s temperature on the Fahrenheit scale? a. No, it’s 98.6° b. Not terribly, it’s 97.0° c. Not really, it’s 101° 22. A syringe has a stated capacity of 5 cc (or 5 mL). What is the volume of the syringe in microliters? a. 5,000 μL b. 0.005 μL c. 0.05 μL d. 5 × 106 μL 23. Ethyl alcohol has a density of 0.79 g/mL. What volume of ethyl alcohol is needed to treat methanol poisoning, if the physician orders 200 g of ethanol? a. 160 mL b. 250 mL c. 200 mL d. There is insufficient information to solve this question. 24. Water has a density of 8.3 lb/gal. What is the density of water in units of stones per bushel? A stone is 14 lb and there are 8 gal per bushel. a. 8.3 stones/bushel b. 15 stones/bushel c. 4.7 stones/bushel d. 0.074 stones/bushel 25. The reason specific gravity is often cited rather than density is that: a. Specific gravity has units that are consistent with the SI (metric) system. b. T  he numerical value of density depends on the mass and volume units chosen, whereas specific gravity is unit-independent. 28 Chapter 1 Measurement c. The specific gravity of bodily fluids is always close to 1. d. If the specific gravity of a substance is greater than 1, you know it will float on water. 26. Charles’s Law relates the volume of a gas to its temperature. The following table gives the temperature of a gas as a function of volume (when the pressure is 1 atm). Treat the volume as the dependent variable. Graph the data and determine the equation of the line that relates to these variables. VOLUME (L) TEMPERATURE (°C) 20.0 −29 30.0 93 40.0 215 50.0 336 27. Using conversion factors found in this chapter, convert each of these measurements into the indicated units. a. 45 cm into m b. 36 m into km c. 0.45 L into mL d. 0.97 ms into s e. 17 μg into g 28. Using conversion factors found in this chapter, convert each of these measurements into the indicated units. a. 26 Mg into mg b. 0.26 nm into mm c. 38 dL into cL d. 72 μm into nm e. 55 Mb into Gb 29. Using conversion factors found in this chapter, convert each of these measurements into the indicated units. a. 12 in. into cm b. 38 gal into L c. 55 lb into kg d. 85 km into mi e. 59 mg into grains 30. Using conversion factors found in this chapter, convert each of these measurements into the indicated units. a. 12.0 cu in. into cm3 b. 67 km2 into mi2 c. 9.8 m/s2 into ft/min2 d. 67 W/mi2 into W/km2 e. 86 lb/ft3 into g/cm3  Review Questions for Measurement 29 31. Use the conversion factors provided to perform each of these transformations. a. A urine sample has a concentration of 1.9 mEq sodium/mL. How many mEq of sodium are in 55 mL of urine? b. A car is traveling at 75 mi/hr. How far will the car go in 89 hr? c. A motorcycle has a mileage of 89 mi/gal. How many gallons of gasoline are required for a 45-mi trip? d. An IV has a drip rate of 52 mL/min. How long will it take to deliver 12 L of fluid? e. If the Kessel run represents a distance of 18 parsecs, and the Millennium Falcon can travel at a maximum velocity of 2.6 × 108 m/s (just less than the speed of light), how long will this journey take? 1 parsec = 3.26 light-years and 1 light-year = 9.5 × 1015 m. 32. The temperature of a sample is measured using both a Celsius thermometer and a Fahrenheit thermometer, and both thermometers register the same value. What is the temperature of the sample? Start with °F = 1.8°C + 32 and use the requirement in this example that °F = °C. 33. Can an object have a temperature measured on the Kelvin scale that is equal to the same temperature on the Fahrenheit scale? Explain. 34. A cube has an edge length of 1.5 cm and a mass of 15 g. What is the density of the cube in units of g/cm3? 35. A cylinder has a radius (R) of 5.0 in. and a height (H) of 12 in. If the cylinder has a mass of 1.5 lb, what is the density of the cylinder in lb/ft3? The volume of a cylinder is given by Vcylinder = πR2 H. 36. A sphere has a diameter of 34 m and a mass of 75 kg. What is the density of the sphere in units of kg/m3? The volume of a sphere is given by Vsphere = 4/3πR3, and the radius of a sphere is one-half of the diameter. C h a p t e r 2 A Review of Some Chemistry Basics Why This Matters: A Nurse Anesthe