Unformatted text preview: Calculations In Chemistry
A Note to Students
The goal of these lessons is to help you solve calculations in firstyear chemistry. This is
only one part of a course in chemistry, but it is often the most challenging.
Provisions: A spiral notebook is suggested as a place to write your work when solving the
problems in these lessons. You will also need
• two packs of 100 3x5 index cards (two or more colors are preferred) plus a small
assortment of rubber bands, and • a pack of large (3 to 6 inch long) sticky notes to use as cover sheets. Choosing a Calculator: As you solve problems, use the same calculator that you will be
allowed to use during tests, to learn and practice the rules for that calculator before tests.
Many courses will not allow the use of a graphing calculator or other calculators with
extensive memory during tests. If no type of calculator is specified for your course, any
inexpensive calculator with a 1/x or x1 , yx or ^ , log or 10x , and ln functions
will be sufficient for most calculations in firstyear chemistry.
Buy two identical calculators if possible. If one becomes broken or lost, you will have a
familiar backup if the bookstore is sold out later in the term.
When to Start: You will receive the maximum benefit from these lessons by completing
the lessons on a topic before it is addressed in lecture.
Where to Start: The order of these lessons may not always match the order in which topics
are covered in your course. If you are using these modules as part of a course, complete the
lessons in the order in which they are assigned by your instructor.
If you are using these lessons “on your own” to assist with a course,
• First, determine the topics that will be covered on your next graded problem set,
quiz, or test. • Find those topics in the Table of Contents. • Download the modules that include the topics. • Find the prerequisite lessons for the topic, listed at the beginning of the module or
lesson. Download and print those lessons. Do the prerequisites, then the topics
related to your next graded assignments. • Follow the instructions on “How to Use These Lessons” on page 1. If you begin these lessons after the start of your course, as time permits, review prior topics
starting with Module 1. You will need all of these introductory modules for later topics and for your final exam.
Check back for updates at www.ChemReview.Net . ©2011 www.ChemReview.Net v. u6 Page ii Table of Contents
*****
Volume 1
How to Use These Lessons ............................................................................................... 1
Module 1 – Scientific Notation........................................................................................ 2
Lesson 1A:
Lesson 1B:
Lesson 1C:
Lesson 1D: Moving the Decimal.............................................................................................. 3
Calculations Using Exponential Notation ......................................................... 9
Estimating Exponential Calculations................................................................ 16
Special Project The Atoms (Part 1) ................................................................. 23 Module 2 – The Metric System...................................................................................... 25
Lesson 2A:
Lesson 2B:
Lesson 2C:
Lesson 2D: Metric Fundamentals .......................................................................................... 25
Metric Prefix Formats ......................................................................................... 32
Cognitive Science  and Flashcards ................................................................. 37
Calculations With Units...................................................................................... 43 Module 3 – Significant Figures...................................................................................... 48
Lesson 3A:
Lesson 3B:
Lesson 3C:
Lesson 3D: Rules for Significant Figures .............................................................................. 48
Sig Figs  Special Cases...................................................................................... 53
Sig Fig Summary and Practice........................................................................... 56
The Atoms –Part 2 ............................................................................................... 59 Module 4 – Conversion Factors ..................................................................................... 60
Lesson 4A:
Lesson 4B:
Lesson 4C:
Lesson 4D:
Lesson 4E:
Lesson 4F:
Lesson 4G: Conversion Factor Basics.................................................................................... 60
Single Step Conversions ..................................................................................... 63
MultiStep Conversions...................................................................................... 67
English/Metric Conversions.............................................................................. 69
Ratio Unit Conversions....................................................................................... 72
The Atoms –Part 3 ............................................................................................... 77
Review Quiz For Modules 14 ............................................................................. 78 Module 5 – Word Problems............................................................................................ 80
Lesson 5A:
Lesson 5B:
Lesson 5C:
Lesson 5D:
Lesson 5E:
Lesson 5F:
Lesson 5G: Answer Units  Single Or Ratio? ...................................................................... 80
Mining The DATA .............................................................................................. 82
Solving For Single Units ..................................................................................... 85
Finding the Given ................................................................................................. 90
Some Chemistry Practice.................................................................................... 93
Area and Volume Conversions ......................................................................... 95
Densities of Solids: Solving Equations .......................................................... 101 Module 6 – Atoms, Ions, and Periodicity .................................................................. 108
Lesson 6A:
Lesson 6B:
Lesson 6C:
Lesson 6D:
Lesson 6E:
Lesson 6F: Atoms .................................................................................................................. 108
The Nucleus, Isotopes, and Atomic Mass ...................................................... 113
Atoms, Compounds, and Formulas................................................................ 121
The Periodic Table............................................................................................. 126
A Flashcard Review System............................................................................. 130
The Atoms –Part 4 ............................................................................................. 132 ©2011 www.ChemReview.Net v. u6 Page iii Module 7 – Writing Names and Formulas.................................................................133
Lesson 7A:
Lesson 7B:
Lesson 7C:
Lesson 7D:
Lesson 7E: Naming Elements and Covalent Compounds ...............................................133
Naming Ions .......................................................................................................138
Names and Formulas for Ionic Compounds..................................................149
Naming Acids.....................................................................................................161
Review Quiz For Modules 57............................................................................164 Module 8 – Grams and Moles ......................................................................................168
Lesson 8A:
Lesson 8B:
Lesson 8C:
Lesson 8D: The Mole..............................................................................................................168
Grams Per Mole (Molar Mass) .........................................................................169
Converting Between Grams and Moles ..........................................................172
Converting Particles, Moles, and Grams ........................................................176 Module 9 – Mole Applications.....................................................................................182
Lesson 9A:
Lesson 9B:
Lesson 9C:
Lesson 9D: Fractions and Percentages.................................................................................182
Empirical Formulas............................................................................................186
Empirical Formulas from Mass or % Mass.....................................................187
Mass Fraction, Mass Percent, Percent Composition......................................194 Module 10 – Balanced Equations and Stoichiometry..............................................201
Lesson 10A:
Lesson 10B:
Lesson 10C:
Lesson 10D:
Lesson 10E:
Lesson 10F:
Lesson 10G:
Lesson 10H:
Lesson 10I: Chemical Reactions and Equations .................................................................201
Balancing Equations ..........................................................................................204
Using Coefficients  Molecules to Molecules ................................................210
Mole to Mole Conversions ................................................................................212
Conversion Stoichiometry ................................................................................215
Percent Yield .......................................................................................................222
Finding the Limiting Reactant..........................................................................227
Final Mixture Amounts – and RICE Tables....................................................234
Review Quiz For Modules 810..........................................................................248 Module 11 – Molarity..................................................................................................... 251
Lesson 11A:
Lesson 11B:
Lesson 11C:
Lesson 11D:
Lesson 11E:
Lesson 11F:
Lesson 11G: Ratio Unit Review ..............................................................................................251
Word Problems with Ratio Answers...............................................................252
Molarity ...............................................................................................................258
Conversions and Careers ..................................................................................264
Units and Dimensions ......................................................................................268
Ratios versus Two Related Amounts .............................................................274
Solving Problems With Parts ...........................................................................279 Module 12 – Molarity Applications ............................................................................290
Lesson 12A:
Lesson 12B:
Lesson 12C:
Lesson 12D:
Lesson 12E:
Lesson 12F: Dilution ...............................................................................................................290
Ion Concentrations.............................................................................................300
Solution Stoichiometry ......................................................................................307
Solution Reactions and Limiting Reactants....................................................310
Reaction Stoichiometry For Ratio Units..........................................................315
Review Quiz For Modules 1112........................................................................323 Module 13 – Ionic Equations and Precipitates ......................................................... 326
Lesson 13A:
Lesson 13B:
Lesson 13C:
Lesson 13D: Predicting Solubility for Ionic Compounds ...................................................326
Total and Net Ionic Equations..........................................................................330
Predicting Precipitation.....................................................................................334
Precipitate and Gravimetric Calculations.......................................................341 ©2011 www.ChemReview.Net v. u6 Page iv Module 14 – AcidBase Neutralization ...................................................................... 349
Lesson 14A:
Lesson 14B:
Lesson 14C:
Lesson 14D:
Lesson 14E: Ions in AcidBase Neutralization .................................................................... 349
Balancing Hydroxide Neutralization ............................................................. 353
AcidHydroxide Neutralization Calculations ............................................... 361
Neutralization Calculations in Parts............................................................... 367
Carbonate Neutralization................................................................................. 374 Module 15 – Redox Reactions ...................................................................................... 382
Lesson 15A:
Lesson 15B:
Lesson 15C:
Lesson 15D:
Lesson 15E: Oxidation Numbers .......................................................................................... 382
Balancing Charge............................................................................................... 387
Oxidizing and Reducing Agents .................................................................... 389
Balancing Redox Using Oxidation Numbers ................................................ 393
Redox Stoichiometry........................................................................................ 398 Module 16 – HalfReaction Balancing........................................................................ 402
Lesson 16A:
Lesson 16B:
Lesson 16C:
Lesson 16D:
Lesson 16E: Constructing HalfReactions – The CAWHe! Method................................ 402
Balancing By Adding HalfReactions ............................................................. 408
Separating Redox Into HalfReactions ........................................................... 411
Balancing Redox With Spectators Present ..................................................... 415
Review Quiz For Modules 1316 ....................................................................... 421 Volume 2
Module 17 – Ideal Gases ............................................................................................... 425
Lesson 17A:
Lesson 17B:
Lesson 17C:
Lesson 17D:
Lesson 17E:
Lesson 17F:
Lesson 17G:
Lesson 17H: Gas Fundamentals............................................................................................. 425
Gases at STP ....................................................................................................... 429
Complex Unit Cancellation.............................................................................. 435
The Ideal Gas Law and Solving Equations .................................................... 440
Choosing Consistent Units............................................................................... 443
Density, Molar Mass, and Choosing Equations ............................................ 448
Using the Combined Equation ........................................................................ 455
Gas Law Summary and Practice ..................................................................... 461 Module 18 – Gas Labs, Gas Reactions........................................................................ 466
Lesson 18A:
Lesson 18B:
Lesson 18C:
Lesson 18D: Charles’ Law; Graphing Direct Proportions.................................................. 466
Boyle’s Law; Graphs of Inverse Proportions ................................................. 473
Avogadro’s Hypothesis; Gas Stoichiometry.................................................. 476
Dalton’s Law of Partial Pressures ................................................................... 485 Module 19 – Kinetic Molecular Theory ..................................................................... 493
Lesson 19A:
Lesson 19B:
Lesson 19C:
Lesson 19D:
Lesson 19E: Squares and Square Roots ................................................................................ 493
Kinetic Molecular Theory................................................................................. 501
Converting to SI Base Units ............................................................................. 504
KMT Calculations.............................................................................................. 509
Graham’s Law.................................................................................................... 519 Module 20 – Graphing .................................................................................................. 523
Lesson 20A:
Lesson 20B:
Lesson 20C:
Lesson 20D:
Lesson 20E:
Lesson 20F: Graphing Fundamentals................................................................................... 523
The Specific Equation for a Line...................................................................... 532
Graphing Experimental Data........................................................................... 542
Deriving Equations From Linear Data ........................................................... 551
Linear Equations Not Directly Proportional ................................................. 562
Graphing Inverse Proportions......................................................................... 567 ©2011 www.ChemReview.Net v. u6 Page v Module 21 – Phases Changes and Energy..................................................................577
Lesson 21A:
Lesson 21B:
Lesson 21C:
Lesson 21D:
Lesson 21E: Phases and Phase Changes ...............................................................................577
Specific Heat Capacity and Equations ............................................................590
Water, Energy, and Consistent Units ..............................................................598
Calculating Joules Using Unit Cancellation ...................................................603
Calorimetry .........................................................................................................609 Module 22 – Heats Of Reaction (ΔH) ......................................................................... 617
Lesson 22A:
Lesson 22B:
Lesson 22C:
Lesson 22D:
Lesson 22E: Energy, Heat, and Work ...................................................................................617
Exo And Endothermic Reactions ...................................................................625
Adding ΔH Equations (Hess’s Law) ...............................................................630
Heats of Formation and Element Formulas ...................................................636
Using Summation to Find ΔH ..........................................................................644 Module 23 – Light and Spectra ...................................................................................649
Lesson 23A:
Lesson 23B:
Lesson 23C:
Lesson 23D:
Lesson 23E:
Lesson 23F:
Lesson 23G: Waves ..................................................................................................................649
Waves and Consistent Units.............................................................................654
Planck's Constant ..............................................................................................659
DeBroglie’s Wavelength ...................................................................................663
The Hydrogen Atom Spectrum........................................................................668
The Wave Equation Model ..............................................................................674
Quantum Numbers ...........................................................................................676 Module 24 – Electron Configuration...........................................................................680
Lesson 24A:
Lesson 24B:
Lesson 24C:
Lesson 24D:
Lesson 24E: The MultiElectron Atom ..................................................................................680
Shorthand Electron Configurations.................................................................684
Abbreviated Electron Configurations .............................................................687
The Periodic Table and Electron Configuration ...........................................691
Electron Configurations: Exceptions and Ions .............................................696 Module 25 – Bonding ....................................................................................................701
Lesson 25A:
Lesson 25B:
Lesson 25C:
Lesson 25D:
Lesson 25E:
Lesson 25F:
Lesson 25G:
Lesson 25H: Covalent Bonds ..................................................................................................701
Molecular Shapes and Bond Angles................................................................706
Electronegativity ................................................................................................714
Molecular Polarity..............................................................................................717
Solubility .............................................................................................................724
Double and Triple Bonds ..................................................................................728
Ion Dot Diagrams...............................................................................................733
Orbital Models for Bonding..............................................................................735 Module 26 – Mixtures and Colligative Properties ................................................... 740
Lesson 26A:
Lesson 26B:
Lesson 26C:
Lesson 26D: Measures of Solution Composition .................................................................740
Concentration in Percent or PPM ....................................................................746
Colligative Properties and Gas Pressures.......................................................750
Colligative Properties of Solutions ..................................................................758 Module 27 – Kinetics: Rate Laws................................................................................768
Lesson 27A:
Lesson 27B:
Lesson 27C:
Lesson 27D:
Lesson 27E:
Lesson 27F: Kinetics Fundamentals .....................................................................................768
Rate Laws ............................................................................................................773
Integrated Rate Law Zero Order ...................................................................782
Base 10 Logarithms ...........................................................................................790
Natural Log Calculations .................................................................................799
Integrated Rate Law  First Order ..................................................................807 ©2011 www.ChemReview.Net v. u6 Page vi Lesson 27G:
Lesson 27H:
Lesson 27I: Reciprocal Math................................................................................................. 817
Integrated Rate Law  Second Order ............................................................. 822
HalfLife Calculations....................................................................................... 829 Volume 3
Module 28 – Equilibrium.............................................................................................. 837
Lesson 28A:
Lesson 28B:
Lesson 28C:
Lesson 28D:
Lesson 28E:
Lesson 28F:
Lesson 28G:
Lesson 28H:
Lesson 28I:
Lesson 28J: Le Châtelier’s Principle..................................................................................... 838
Powers and Roots of Exponential Notation................................................... 850
Equilibrium Constants...................................................................................... 860
K Values ............................................................................................................. 867
Kp Calculations .................................................................................................. 870
K and Rice Moles Tables .................................................................................... 876
K Calculations From Initial Concentrations .................................................. 883
Q: The Reaction Quotient ................................................................................. 889
Calculations Using K and Q ............................................................................. 892
Solving Quadratic Equations ........................................................................... 899 Module 29 – AcidBase Fundamentals....................................................................... 910
Lesson 29A:
Lesson 29B:
Lesson 29C:
Lesson 29D:
Lesson 29E:
Lesson 29F: AcidBase Math Review ................................................................................... 910
Kw Calculations: H+ and OH─....................................................................... 913
Strong Acid Solutions ....................................................................................... 917
The [OH─] in Strong Acid Solutions .............................................................. 922
Strong Base Solutions........................................................................................ 925
The pH System................................................................................................... 928 Module 30 – Weak Acids and Bases ........................................................................... 939
Lesson 30A:
Lesson 30B:
Lesson 30C:
Lesson 30D:
Lesson 30E:
Lesson 30F:
Lesson 30G: Ka Math and Approximation Equations ........................................................ 939
Weak Acids and Ka Expressions ..................................................................... 943
Ka Calculations .................................................................................................. 950
Percent Dissociation and Shortcuts................................................................. 959
Solving Ka Using the Quadratic Formula ...................................................... 963
Weak Bases and Kb Calculations..................................................................... 966
Polyprotic Acids ................................................................................................ 976 Module 31 – BrønstedLowry Definitions................................................................. 982
Lesson 31A:
Lesson 31B: BrønstedLowry Acids and Bases ................................................................... 982
Which Acids and Bases Will React?................................................................ 986 Module 32 – pH of Salts................................................................................................ 995
Lesson 32A:
Lesson 32B:
Lesson 32C:
Lesson 32D: The AcidBase Behavior of Salts...................................................................... 995
Will A Salt AcidBase React? ........................................................................ 1002
Calculating the pH of a Salt Solution .......................................................... 1006
Salts That Contain Amphoteric Ions............................................................. 1011 Module 33 – Buffers..................................................................................................... 1016
Lesson 33A:
Lesson 33B:
Lesson 33C:
Lesson 33D: AcidBase Common Ions, Buffers ................................................................. 1016
Buffer Example ................................................................................................ 1019
Buffer Components ......................................................................................... 1025
Methodical Buffer Calculations..................................................................... 1029 ©2011 www.ChemReview.Net v. u6 Page vii Lesson 33E:
Lesson 33F: Buffer Quick Steps ...........................................................................................1033
The HendersonHasselbalch Equation..........................................................1040 Module 34 – pH During Titration ............................................................................. 1047
Lesson 34A:
Lesson 34B:
Lesson 34C:
Lesson 34D:
Lesson 34E: pH In Mixtures .................................................................................................1047
pH After Neutralization..................................................................................1052
Distinguishing Types of AcidBase Calculations ........................................1062
pH During StrongStrong Titration ...............................................................1067
Titration pH: Weak by Strong .......................................................................1079 Module 35 – Solubility Equilibrium.........................................................................1091
Lesson 35A:
Lesson 35B:
Lesson 35C:
Lesson 35D:
Lesson 35E: Slightly Soluble Ionic Compounds ................................................................1091
Ksp Calculations ...............................................................................................1094
Solubility and Common Ions..........................................................................1102
pH and Solubility .............................................................................................1109
Quantitative Precipitation Prediction............................................................1113 Module 36 – Thermodynamics...................................................................................1125
Lesson 36A:
Lesson 36B:
Lesson 36C:
Lesson 36D:
Lesson 36E:
Lesson 36F:
Lesson 36G: Review: Energy and Heats of Reaction ........................................................1125
Entropy and Spontaneity ...............................................................................1130
Free Energy .......................................................................................................1138
Standard State Values......................................................................................1144
Adding ΔG° Equations....................................................................................1149
Free Energy at NonStandard Conditions ....................................................1152
Free Energy and K ............................................................................................1157 Module 37 – Electrochemistry .................................................................................... 1164
Lesson 37A:
Lesson 37B:
Lesson 37C:
Lesson 37D:
Lesson 37E:
Lesson 37F: Redox Fundamentals .......................................................................................1164
Charges and Electrical Work ..........................................................................1171
Standard Reduction Potentials.......................................................................1175
NonStandard Potentials: The Nernst Equation .........................................1178
Predicting Which Redox Reactions Go .........................................................1184
Calculating Cell Potential ...............................................................................1191 Module 38 – Electrochemical Cells ........................................................................... 1201
Lesson 38A:
Lesson 38B:
Lesson 38C:
Lesson 38D:
Lesson 38E: Cells and Batteries............................................................................................1201
Anodes and Cathodes .....................................................................................1211
Depleted Batteries and Concentration Cells.................................................1220
Electrolysis ........................................................................................................1228
Amperes and Electrochemical Calculations.................................................1231 Module 39 – Nuclear Chemistry ................................................................................1240
Lesson 39A:
Lesson 39B:
Lesson 39C:
Lesson 39D: The Nucleus  Review .....................................................................................1240
Radioactive Decay Reactions..........................................................................1245
Fission and Fusion ...........................................................................................1250
Radioactive HalfLife Calculations................................................................1253 ••••• ©2011 www.ChemReview.Net v. u6 Page viii Module 1 – Scientific Notation How to Use These Lessons
1. Read the lesson. Work the questions (Q). As you read, use this method.
• As you turn to two new facing pages in this book, cover the page on the right
with a sheet of paper. • As you start any new page, if you see 5 stars ( * * * * * ) on the page,
cover the text below the stars. As a cover sheet, use either overlapping sticky
notes
or a folded sheet of paper. • In your problem notebook, write your answer to the question (Q) above the
* * * * * . Then slide down the cover sheet to the next set of * * * * *
and check your answer. If you need a hint, read a part of the answer, then
recover the answer and try the problem again. 2. Memorize the rules, then do the Practice.
The goal in learning is to move rules and concepts into memory. To begin, when
working questions (Q) in a lesson, you may look back at the rules, but make an
effort to commit the rules to memory before starting the Practice problems.
Try every other problem of a Practice set on the first day and the remaining
problems in your next study session. This spacing will help you to remember new
material. On both days, try to work the Practice without looking back at the rules.
Answers to the Practice are at the end of each lesson. If you need a hint, read a
part of the answer and try again.
3. How many Practice problems should you do? It depends on your background.
These lessons are intended to
• refresh your memory on topics you once knew, and • fillin the gaps for topics that are less familiar. If you know a topic well, read the lesson for review, then do a few problems on
each Practice set. Be sure to do the last problem (usually the most challenging).
If a topic is unfamiliar, do more problems.
4. Work Practice problems at least 3 days a week. Chemistry is cumulative. To
solve problems, what you learn early you will need in memory later. To retain
what you learn, space your study of a topic over several days.
Science has found that your memory tends to retain what it uses repeatedly, but to
remember for only a few days what you do not practice over several days. If you
wait until a quiz deadline to study, what you learn may remain in memory for a
day or two, but on later tests and exams, it will tend to be forgotten.
Begin lessons on new topics early, preferably before the topic is covered in lecture.
5. Memorize what must be memorized. Use flashcards and other memory aids.
Chemistry is not easy, but you will achieve success if you work at a steady pace. ©2011 www.ChemReview.Net v. u6 Page 1 Module 1 – Scientific Notation If you have previously taken a course in chemistry, many topics in Modules 1 to 4 will
be review. Therefore: if you can pass the pretest for a lesson, skip the lesson. If you
need a bit of review to refresh your memory, do the last few problems of each Practice
set. On topics that are less familiar, complete more Practice. Module 1 – Scientific Notation
Timing: Module 1 should be done as soon as you are assigned problems that use
exponential notation. If possible, do these lessons before attempting problems in other
textbooks.
Additional Math Topics
Powers and roots of exponential notation are covered in Lesson 28B.
Complex units such as
are covered in Lesson 17C. atm ● L
(mole)( atm ● L )
mole ● K Those lessons may be done at any time after Module 1.
Calculators and Exponential Notation
To multiply 492 x 7.36, the calculator is a useful tool. However, when using exponential
notation, you will make fewer mistakes if you do as much exponential math as you can
without a calculator. These lessons will review the rules for doing exponential math “in
your head.”
The majority of problems in Module 1 will not require a calculator. Problems that require a
calculator will be clearly identified.
You are encouraged to try complex problems with the calculator after you have tried them
without. This should help you to decide when, and when not, to use a calculator. Notation Terminology
In science, we often deal with very large and very small numbers.
For example: A drop of water contains about 1,500,000,000,000,000,000,000 molecules.
An atom of neon has a radius of about 0.000 000 007 0 centimeters.
When values are expressed as “regular numbers,” such as 123 or 0.024 or the numbers
above, they are said to be in fixed decimal or fixed notation.
Very large and small numbers are more clearly expressed in exponential notation: writing
a number times 10 to a wholenumber power. For the measurements above, we can write
• A drop of water contains about 1.5 x 1021 molecules. • An atom of neon gas has a radius of about 7.0 x 10―9 centimeters. ©2011 www.ChemReview.Net v. u6 Page 2 Module 1 – Scientific Notation Values represented in exponential notation can be described as having three parts.
For example, in ─ 6.5 x 10―4,
• The ─ in front is the sign. • the 6.5 is termed the significand or decimal or digit or mantissa or coefficient. • The 10―4 is the exponential term: the base is 10 and the exponent (or power) is ―4. In these lessons we will refer to the two parts of exponential notation after the sign as the
significand and exponential term.
↓
─ 6.5 x 10―4
↑
↑
significand
exponential sign You should also learn (and use) any alternate terminology preferred in your course.
***** Lesson 1A: Moving the Decimal
Pretest: Do not use a calculator. If you get a perfect score on this pretest, skip to Lesson 1B.
Otherwise, complete Lesson 1A. Answers are at the end of each lesson.
1. Write these in scientific notation.
a. 9,400 x 103 = ___________________ b. 0.042 x 106 = _________________ c. ─ 0.0067 x 10―2 = _________________ d. ─ 77 = _________________ 2. Write these answers in fixed decimal notation.
a. 14/10,000 = b. 0.194 x 1000 = c. 470= ***** Powers of 10
Below are the numbers that correspond to powers of 10. Note the relationship between the
exponents and position of the decimal point in the fixed decimal numbers as you go down the
sequence.
106 = 1,000,000
103 = 1,000 = 10 x 10 x 10
102 = 100
1 0 1 = 10
100 = 1 (Anything to the zero power equals one.) 10―1 = 0.1
10―2 = 0.01 = 1/102 = 1/100
10―3 = 0.001 ©2011 www.ChemReview.Net v. u6 Page 3 Module 1 – Scientific Notation When converting from powers of 10 to fixed decimal numbers, use these steps.
1. To change a positive power of 10 to a fixed decimal number,
• write 1, then after 1 add the number of zeros equal to the exponent.
Example: 102 = 100 Another way to state this rule: ∪∪↑ • From 1, move the decimal to the right by the number of places in the exponent. 2. To convert a negative power of 10 to fixed notation,
• From 1, move the decimal to the left by the number of places equal to the exponent
of 10 after its negative sign.
Question: ***** 10―2 = (fill in the fixed decimal number.) (See How To Use These Lessons, Point 1, on page 1). Answer: Practice A: 10―2 = 0 .01 ↑∪∪ Write your answers, then check them at the end of this lesson. 1. Write these as regular numbers without an exponential term.
a. 104 = _______________
c. 107 = _______________ b. 10―4 = ______________
d. 10―5 = ______________ e. 100 = ________ Multiplying and Dividing By 10, 100, 1000
To multiply or divide by numbers that are positive wholenumber powers of 10, such as 100
or 10,000, use these rules.
1. When multiplying a number by a 10, 100, 1000, etc., move the decimal to the right by the
number of zeros in the 10, 100, or 1000.
Examples: 72 x 100 = 7,200 ─ 0.0624 x 1,000 = ─ 62.4 2. When dividing a number by 10, 100, or 1000, etc., move the decimal to the left by the
number of zeros in the 10, 100, or 1000.
Q: Answer as fixed decimal numbers: 34.6/1000 = 0.47/100 = *****
Answers: 34.6/1000 = 0.0346 0.47/100 = 0.0047 ↑∪∪∪ 3. When writing a number that has a value between ─1 and 1 (a number that “begins with
a decimal point”), always place a zero in front of the decimal point.
Example: Do not write .42 or ─ .74 ; do write 0.42 or ─ 0.74 During your written calculations, the zero in front helps in seeing your decimals. ©2011 www.ChemReview.Net v. u6 Page 4 Module 1 – Scientific Notation Practice B: Write your answers, then check them at the end of this lesson. 1. When dividing by 1,000 move the decimal to the _______________ by _____ places.
2. Write these answers as fixed decimal numbers.
a. 0.42 x 1000 = b. 63/100 = c. ─ 74.6/10,000 = Converting Exponential Notation to Numbers
To convert from exponential notation (such as ─ 4 x 103 ) to fixed decimal notation
(─ 4,000 ), use these rules.
1. The sign in front does not change. The sign is independent of the terms after the sign.
2. If, in the exponential notation, the significand is multiplied by a
• positive power of 10, move the decimal point in the significand to the right by the
same number of places as the value of the exponent;
Examples: • 2 x 102 = 2 00 ∪∪↑ ─ 0 .0033 x 103 =
∪∪∪↑ ─ 3.3 negative power of 10, move the decimal point in the significand to the left by the
same number of places as the number after the minus sign of the exponent.
Examples: 2 x 10―2 = 0 .02 ─ 7,653.8 x 10―3 = ─ 7 .6 5 3 8 ↑∪∪ Practice C: ↑∪∪∪ Convert these to fixed decimal notation. 1. 3 x 103 = _____________________ 2. 5.5 x 10―4 = _________________________ 3. 0.77 x 106 = __________________ 4. ─ 95 x 10―4 = ______________________ Changing Exponential to Scientific Notation
In chemistry, it is often required that numbers that are very large or very small be written
in scientific notation. One reason to use scientific notation is that it makes values easier to
compare: there are many equivalent ways to write a value in exponential notation, but
only one correct way to write the value in scientific notation.
Scientific notation is simply a special case of exponential notation in which the significand
is 1 or greater, but less than 10, and is multiplied by 10 to a wholenumber power.
Another way to say this: in scientific notation, the decimal point in the significand must be
after the first digit that is not a zero.
Example: ─ 0.057x 10―2 in scientific notation is written as ─ 5.7 x 10―4 .
The decimal must be moved to after the first number that is not a zero: the 5. ©2011 www.ChemReview.Net v. u6 Page 5 Module 1 – Scientific Notation To convert from exponential to scientific notation,
• move the decimal in the significand to after the first digit that is not a zero, then • adjust the exponent to keep the same numeric value.
When moving a decimal point,
1. The sign in front does not change.
2. If moving the decimal Y times to make the significand larger, make the power of 10
smaller by a count of Y.
Example. Converting exponential to scientific notation: 0.045 x 105 = 4 . 5 x 103
∪∪↑ The decimal must be after the 4. Move the decimal two times to the right. This
makes the significand 100 times larger. To keep the same numeric value, lower the
exponent by 2, making the 10x value 100 times smaller.
3. When moving the decimal Y times to make the significand smaller, make the power of
10 larger by a count of Y.
Q. Convert to scientific notation: ─ 8 , 544 x 10 ― 7 =
*****
Answer: ─ 8 , 544 x 10 ― 7 = ─ 8 . 544 x 10 ― 4
↑∪∪∪ You must move the decimal 3 places to the left. This makes the significand 1,000
times smaller. To keep the same numeric value, increase the exponent by 3,
making the 10x value 1,000 times larger.
Remember, 10─4 is 1,000 times larger than 10─7.
It helps to recite, every time you move a decimal, for the terms after the sign in front:
“If one gets smaller, the other gets larger. If one gets larger, the other gets smaller.” Practice D: Change these to scientific notation. Check answers at the end of the lesson. 1. 5,420 x 103 = 2. 0.0067 x 10― 4 = ___________________ 3. 0.020 x 103 = 4. ─ 870 x 10― 4 = ___________________ 5. 0.00492 x 10― 12 = 6. ─ 602 x 1021 = ____________________ ©2011 www.ChemReview.Net v. u6 Page 6 Module 1 – Scientific Notation Converting Numbers to Scientific Notation
To convert regular (fixed decimal) numbers to scientific notation, use these rules.
• Any number to the zero power equals one.
20 = 1. • 420 = 1. Exponential notation most often uses 100 = 1. Since any number can be multiplied by one without changing its value, any number
can be multiplied by 100 without changing its value.
Example: 42 = 42 x 1 = 42 x 100 in exponential notation
= 4.2 x 101 in scientific notation. To convert fixed notation to scientific notation, the steps are
1. Add x 100 after the number.
2. Apply the rules that convert exponential to scientific notation.
•
• Write the decimal after the first digit that is not a zero. •
Try: Do not change the sign in front.
Adjust the power of 10 to compensate for moving the decimal. Q. Using the steps above, convert these to scientific notation.
a. 943 b. ─ 0.00036 *****
Answers: 943 = 943 x 100 = 9.43 x 102 in scientific notation.
─ 0.00036 = ─ 0.00036 x 100 = ─ 3.6 x 10―4 in scientific notation. When converting to scientific notation, a positive fixed decimal number that is
• larger than one has a positive exponent (zero and above) in scientific notation; • between zero and one has a negative exponent in scientific notation; and • the number of places that the decimal in a number moves is the number after the sign
in its exponent. These same rules apply to numbers after a negative sign in front. The sign in front is
independent of the numbers after it.
Note how these three rules apply to the two answers above.
Note also that in both exponential and scientific notation, whether the sign in front is
positive or negative has no relation to the sign of the exponential term. The sign in front
shows whether a value is positive or negative. The exponential term indicates only the
position of the decimal point. ©2011 www.ChemReview.Net v. u6 Page 7 Module 1 – Scientific Notation Practice E
1. Which lettered parts in Problem 2 below must have exponentials that are negative
when written in scientific notation?
2. Change these to scientific notation.
a. 6,280 = b. 0.0093 = _________________________ c. 0.741 = _____________________ d. ─ 1,280,000 = _____________________ 3. Complete the problems in the pretest at the beginning of this lesson. Study Summary
In your problem notebook,
• write a list of rules in this lesson that were unfamiliar or you found helpful. • Condense your wording, number the points, and write and recite the rules until you
can write them from memory. Then complete the problems below.
The Role of Practice
Do as many Practice problems as you need to feel “quiz ready.”
• If the material in a lesson is relatively easy review, do the last problem on each
series of similar problems. • If the lesson is less easy, put a check by ( ) and then work every 2nd or 3rd
problem. If you miss one, do some similar problem in the set. • Save a few problems for your next study session  and quiz/test review. During Examples and Q problems, you may look back at the rules, but practice writing
and recalling new rules from memory before starting the Practice.
If you use the Practice to learn the rules, it will be difficult to find time for all of the
problems you will need to do. If you use Practice to apply rules that are in memory,
you will need to solve fewer problems to be “quiz ready.” Practice F: Check ( ) and do every other letter. If you miss one, do another letter for
that set. Save a few parts for your next study session.
1. Write these answers in fixed decimal notation.
a. 924/10,000 = b. 24.3 x 1000 = c. ─ 0.024/10 = 2. Convert to scientific notation.
a. 0.55 x 105 ©2011 www.ChemReview.Net b. 0.0092 x 100 v. u6 c. 940 x 10―6 d. 0.00032 x 101 Page 8 Module 1 – Scientific Notation 3. Write these numbers in scientific notation.
a. 7,700 b. 160,000,000 ANSWERS
Pretest: d. 0.00067 (To make answer pages easy to locate, use a sticky note.) 1a. 9.4 x 106 2a. 0.0014 c. 0.023 2b. 194 1b. 4.2 x 104 1c. ─ 6.7 x 10―5 1d. ─ 7.7 x 101 2 c. 1 Practice A: 1a. 10,000 b. 0.0001 c. 10,000,000 d. 0.00001. e. 1 Practice B: 1. When dividing by 1,000 , move the decimal to the left by 3 places.
2a. 0.42 x 1000 = 420 2b. 63/100 = 0.63 (must have zero in front) Practice C: 1. 3,000 2. 0.00055 3. 770,000 c. ─ 74.6/10,000 = ─ 0.00746 4. ─ 0.0095 Practice D: 1. 5.42 x 106 2. 6.7 x 10―7 3. 2.0 x 101 4. ─8.7 x 10―2 5. 4.92 x 10―15 6. ─6.02 x 1023
Practice E: 1. 2b and 2c
Practice F: 1a. 0.0924 2a. 6.28 x 103 2b. 9.3 x 10―3 1b. 24,300 2a. 5.5 x 104 2b. 9.2 x 10―1 3a. 7.7 x 103
***** 3b. 1.6 x 108 2c. 7.41 x 10―1 2d. ─ 1.28 x 106 1c. ─ 0.0024 2c. 9.4 x 10―4 2d. 3.2 x 10―3 3c. 2.3 x 10―2 3d. 6.7 x 10―4 Lesson 1B: Calculations Using Exponential Notation
Pretest: If you can answer all of these three questions correctly, you may skip to Lesson 1C.
Otherwise, complete Lesson 1B. Answers are at the end of this lesson.
Do not use a calculator. Convert final answers to scientific notation.
1. (2.0 x 10―4) (6.0 x 1023) = 2. 1023
=
(100)(3.0 x 10―8) 3. (─ 6.0 x 10―18) ─ (─ 2.89 x 10―16) =
***** Mental Arithmetic
In chemistry, you must be able to solve simple or estimated calculations without a
calculator to speed your work and as a check on your calculator answers. This mental
arithmetic is simplified by using exponential notation. In this lesson, we will review the
rules for doing exponential calculations “in your head.” ©2011 www.ChemReview.Net v. u6 Page 9 Module 1 – Scientific Notation Adding and Subtracting Exponential Notation
To add or subtract exponential notation without a calculator, the standard rules of
arithmetic can be applied – if all of the numbers have the same exponential term.
Rewriting numbers to have the same exponential term usually results in values that are
not in scientific notation. That’s OK. During calculations, the rule is to work in exponential
notation, to allow flexibility with decimal point positions, then to convert to scientific
notation at the final step.
To add or subtract numbers with exponential terms, you may convert all of the exponential
terms to any consistent power of 10. However, it usually simplifies the arithmetic if you
convert all values to the same exponential as the largest of the exponential terms being
added or subtracted.
The rule is
To add or subtract exponential notation by hand, make all of the exponents the same.
The steps are
To add or subtract exponential notation without a calculator,
1. Rewrite each number so that all of the significands are multiplied by the same power
of 10. Converting to the highest power of 10 being added or subtracted is suggested.
2. Write the significands and exponentials in columns: numbers under numbers
(lining up the decimal points), x under x, exponentials under exponentials.
3. Add or subtract the significands using standard arithmetic, then attach the common
power of 10 to the answer.
4. Convert the final answer to scientific notation.
Follow how the steps are applied in this
Example: ( 40.71 x 108 ) + ( 222 x 106 ) = ( 40.71 x 108 ) + ( 2.22 x 108 ) =
40.71 x 108
+ 2.22 x 108
42.93 x 108 = 4.293 x 109 Using the steps and the method shown in the example, try the following problem without a
calculator. In this problem, do not round numbers during or after the calculation.
Q. ( 32.464 x 101 ) ─ (16.2 x 10―1 ) = ? *****
A. ( 32.464 x 101 ) ─ (16.2 x 10―1 ) = ( 32.464 x 101 ) ─ (0.162 x 10+1 ) =
32.464 x 101
─ 0.162 x 101
32.302 x 101 ©2011 www.ChemReview.Net v. u6 (101 has a higher value than 10―1)
= 3.2302 x 102 Page 10 Module 1 – Scientific Notation Let’s do problem 1 again. This time, below, convert the exponential notation to regular
numbers, do the arithmetic, then convert the final answer to scientific notation.
32.464 x 101
─ 16.2 x 10―1 =
= *****
32.464 x 101 =
─ 1 6 .2
x 10―1 = 324.64
─ 1.62
323.02 = 3.2302 x 102 The answer is the same either way, as it must be. This “convert to regular numbers”
method is an option when the exponents are close to 0. However, for exponents such as
1023 or 10―17, it is easier to use the method above that includes the exponential, but adjusts
so that all of the exponentials are the same. Practice A: Try these without a calculator. On these, don’t round. Do convert final
answers to scientific notation. Do the odds first, then the evens if you need more practice.
1. 64.202 x 1023
+ 13.2
x 1021 2. (61 x 10―7) + (2.25 x 10―5) + (212.0 x 10―6) = 3. ( ― 54 x 10―20 ) + ( ― 2.18 x 10―18 ) = 4. ( ― 21.46 x 10―17 ) ― ( ― 3,250 x 10―19 ) = Multiplying and Dividing Powers of 10
The following boxed rules should be recited until they can be recalled from memory.
1. When you multiply exponentials, you add the exponents.
Examples: 2. 103 x 102 = 105 10―5 x 10―2 = 10―7 10―3 x 105 = 102 When you divide exponentials, you subtract the exponents.
Examples: 103/102 = 101 10―5/102 = 10―7 10―5/10―2 = 10―3 When subtracting, remember: Minus a minus is a plus. 106―(―3) = 106+3 = 109
3. When you take the reciprocal of an exponential, change the sign. ©2011 www.ChemReview.Net v. u6 Page 11 Module 1 – Scientific Notation This rule is often remembered as:
When you take an exponential term from the bottom to the top, change its sign.
1 = 10―3 ; 1/10―5 = 105
103
Why does this work? Rule 2:
1
= 100
103
103
Example: = 100―3 = 10―3 4. When fractions include several terms, it may help to simplify the top and bottom
separately, then divide.
Example: 10―3
= 10―3 =
103
105 x 10―2 10―6 Try the following problem.
Q. Without using a calculator, simplify the top, then the bottom, then divide.
10―3 x 10―4
105 x 10―8 = = *****
Answer: 10―3 x 10―4 =
105 x 10―8 10―7 = 10―7―(―3) = 10―7+3 = 10―4
10―3 Practice B: Write answers as 10 to a power. Do not use a calculator. Do the odds first,
then the evens if you need more practice.
1. 106 x 102 = 2. 10―5 x 10―6 = 3. 10―5 =
10―4 4. 10―3 =
105 5. 1
=
―4
10 6. 1/1023 = 7. 103 x 10―5 =
10―2 x 10―4 8. 105 x 1023 =
10―1 x 10―6 9. 100 x 10―2 =
1,000 x 106 10. 10―3 x 1023 =
10 x 1,000 ©2011 www.ChemReview.Net v. u6 Page 12 Module 1 – Scientific Notation Multiplying and Dividing in Exponential Notation
These are the rules we use most often.
1. When multiplying and dividing using exponential notation, handle the significands and
exponents separately.
Do number math using number rules, and exponential math using exponential rules.
Then combine the two parts.
Apply rule 1 to the following three problems.
a. Do not use a calculator: (2 x 103) (4 x 1023) =
*****
For numbers, use number rules. 2 times 4 is 8
For exponentials, use exponential rules. 103 x 1023 = 103+23 = 1026
Then combine the two parts: (2 x 103) (4 x 1023) = 8 x 1026
b. Do the significand math on a calculator but try the exponential math in your
head for (2.4 x 10―3) (3.5 x 1023) =
*****
Handle significands and exponents separately.
• Use a calculator for the numbers. • Do the exponentials in your head. 10―3 x 1023 = 1020 • Then combine. 2.4 x 3.5 = 8.4 (2.4 x 10―3) (3.5 x 1023) = (2.4 x 3.5) x (10―3 x 1023) = 8.4 x 1020
We will review how much to round answers in Module 3. Until then, round numbers
and significands in your answers to two digits unless otherwise noted.
c. Do significand math on a calculator but exponential math without a calculator.
6.5 x 1023 =
4.1 x 10―8
*****
Answer: 6.5 x 1023 = 6.5 x 1023 = 1.585 x [1023 ― (―8) ] = 1.6 x 1031
4.1
10―8
4.1 x 10―8 2. When dividing, if an exponential term does not have a significand, add a 1 x in front of
the exponential so that the numbernumber division is clear.
Apply Rule 2 to the following problem. Do not use a calculator.
=
10―14
―8
2.0 x 10
***** ©2011 www.ChemReview.Net v. u6 Page 13 Module 1 – Scientific Notation 10―14
=
―8
2.0 x 10 Answer: 1
2.0 x
x 10―14
10―8 = 0.50 x 10―6 = 5.0 x 10―7 Practice C
Study the two rules above, then apply them from memory to these problems. To have
room for careful work, solve these in your notebook.
Do the odds first, then the evens if you need more practice. Try these first without a
calculator, then check your mental arithmetic with a calculator if needed. Write final
answers in scientific notation, rounding significands to two digits.
1. (2.0 x 101) (6.0 x 1023) =
3. 5. 3.0 x 10―21
― 2.0 x 103 4. = 6.0 x 10―23
2.0 x 10―4 6. = 10―14
― 5.0 x 10―3 2. (5.0 x 10―3) (1.5 x 1015) = 1014
4.0 x 10―4 = = 7. Complete the three problems in the pretest at the beginning of this lesson. Study Summary
In your problem notebook, write a list of rules in Lesson 1B that were unfamiliar, need
reinforcement, or you found helpful. Then condense your list. Add this new list to your
numbered points from Lesson 1A. Write and recite the combined list until you can write all
of the points from memory. Then do the problems below. Practice D
Start by doing every other letter. If you get those right, go to the next number. If not, do a
few more of that number. Save one part of each question for your next study session.
1. Try these without a calculator. Convert final answers to scientific notation.
a. 3 x (6.0 x 1023) = b. 1/2 x (6.0 x 1023) = c. 0.70 x (6.0 x 1023) = d. 103 x (6.0 x 1023) = e. 10―2 x (6.0 x 1023) = f. (― 0.5 x 10―2)(6.0 x 1023) = g. 1
=
12
10 h. i. 3.0 x 1024
6.0 x 1023 = ©2011 www.ChemReview.Net j. v. u6 1/10―9 =
2.0 x 1018 =
6.0 x 1023 Page 14 Module 1 – Scientific Notation k. 1.0 x 10―14 =
4.0 x 10―5 l. 1010
=
―5
2.0 x 10 2. Use a calculator for the numbers, but not for the exponents.
a. 2.46 x 1019 =
6.0 x 1023 b. 10―14
0.0072 = 3. Do not use a calculator. Write answers as a power of 10.
a. 107 x 10―2 =
10 x 10―5 b. 10―23 x 10―5 =
10―5 x 100 4. Convert to scientific notation in the final answer. Do not round during these.
a. ( 74 x 105 ) + ( 4.09 x 107 ) =
b. (5.122 x 10―9 ) ― ( ― 12,914 x 10―12 ) = ANSWERS
1. 1.2 x 1020 Pretest. In scientific notation: 2. 3.3 x 1028 3. 2.83 x 10―16 Practice A: You may do the arithmetic in any way you choose that results in these final answers.
1. 64.202 x 1023 =
+ 13.2 x 1021 64.202 x 1023
+ 0.132 x 1023
64.334 x 1023 = 6.4334 x 1024 2. (61 x 10―7) + (2.25 x 10―5) + (212.0 x 10―6) = (0.61 x 10―5) + (2.25 x 10―5) + (21.20 x 10―5) =
0.61
2.25
+ 21.20
24.06
3. x
x
x
x 10―5
10―5
10―5
10―5 (10―5 is the highest value of the three exponentials)
= 2.406 x 10―4 (― 54 x 10―20 ) + ( ― 2.18 x 10―18 ) = (― 0.54 x 10―18 ) + ( ― 2.18 x 10―18 ) =
─ 0.54 x 10―18
─ 2.18 x 10―18
─ 2.72 x 10―18 4. ( 10―18 is higher in value than 10―20 ) (― 21.46 x 10―17 ) ― ( ― 3,250 x 10―19 ) =
( + 3,250 x 10―19 ) ― ( 21.46 x 10―17 ) = ( + 32.50 x 10―17 ) ― ( 21.46 x 10―17 ) =
─ ©2011 www.ChemReview.Net v. u6 32.50 x 10―17
21.46 x 10―17
11.04 x 10―17 = 1.104 x 10―16 Page 15 Module 1 – Scientific Notation Practice B
1. 108
9. 2. 10―11 3. 10―1 4. 10―8 5. 104 100 x 10―2 = 102 x 10―2 = 100 = 10―9
103 x 106
109
1,000 x 106 10. 6. 10―23 7. 104 8. 1035 10―3 x 1023 = 1020 = 1016
10 x 1,000
104 (For 9 and 10, you may use different steps, but you must arrive at the same answer.) Practice C
1. 1.2 x 1025 2. 7.5 x 1012 3. ― 1.5 x 10―24 4. 3.0 x 10―19 5. ― 2.0 x 10―12 6. 2.5 x 1017 Practice D
1a. 3 x (6.0 x 1023) = 18 x 1023 = 1.8 x 1024 1b. 1/2 x (6.0 x 1023) = 3.0 x 1023 1c. 0.70 x (6.0 x 1023) = 4.2 x 1023 1d. 103 x (6.0 x 1023) = 6.0 x 1026 1e. 10―2 x (6.0 x 1023) = 6.0 x 1021 1g. 1f. (― 0.5 x 10―2)(6.0 x 1023) = ― 3.0 x 1021 10―12 1
=
1012 1h. 1i. 3.0 x 1024 = 3.0 x 1024
6.0 x 1023
6.0 x 1023 1/10―9 = 109 = 0.50 x 101 = 5.0 x 100 ( = 5.0 ) 1j. 2.0 x 1018 = 0.33 x 10―5 = 3.3 x 10―6
6.0 x 1023
1 x 1010
2.0 x 10―5 1.0 x 10―14 = 0.25 x 10―9 = 2.5 x 10―10
4.0 x 10―5 1k. = 0.50 x 1015 = 5.0 x 1014 1l. 1010
2.0 x 10―5 2a. 2.46 x 1019 = 0.41 x 10―4 = 4.1 x 10―5
6.0 x 1023 2b. 10―14
0.0072 = = 1.0 x 10―14 =
7.2 x 10―3 1.0 x 10―14 = 0.14 x 10―11 = 1.4 x 10―12
7.2 x 10―3 3a. 107 x 10―2 = 105 = 109
10―4
101 x 10―5
4a. ( 74 x 105 ) + ( 4.09 x 107 ) = 3b. 10―23 x 10―5
10―5 x 102 4b. (5.122 x 10―9 ) ― ( ― 12,914 x 10―12 ) = = ( 0.74 x 107 ) + ( 4.09 x 107 ) =
0.74 x 107
+ 4.09 x 107
4.83 x 107 = 10―25 = (5.122 x 10―9 ) + ( 12.914 x 10―9 ) =
+ 5.122 x 10―9
12.914 x 10―9
18.036 x 10―9 = 1.8036 x 10―8 ***** ©2011 www.ChemReview.Net v. u6 Page 16 Module 1 – Scientific Notation Lesson 1C: Estimating Exponential Calculations
Pretest: If you can solve both problems correctly, skip this lesson. Convert your final
answers to scientific notation. Check your answers at the end of this lesson.
(10―9)(1015)
(4 x 10―4)(2 x 10―2) 1. Solve this problem
without a calculator.
2. For this problem,
use a calculator as needed. = (3.15 x 103)(4.0 x 10―24) =
(2.6 x 10―2)(5.5 x 10―5) ***** Choosing a Calculator
If you have not already done so, please read Choosing a Calculator under Notes to the Student
in the preface to these lessons. Complex Calculations
The prior lessons covered the fundamental rules for exponential notation. For longer
calculations, the rules are the same. The challenges are keeping track of the numbers and
using the calculator correctly. The steps below will help you to simplify complex
calculations, minimize dataentry mistakes, and quickly check your answers.
Let’s try the following calculation two ways.
(7.4 x 10―2)(6.02 x 1023) =
(2.6 x 103)(5.5 x 10―5)
Method 1. Do numbers and exponents separately.
Work the calculation above using the following steps.
a. Do the numbers on the calculator. Ignoring the exponentials, use the calculator to
multiply all of the significands on top. Write the result. Then multiply all the
significands on the bottom and write the result. Divide, write your answer rounded
to two digits, and then check below.
* * * * * (See How To Use These Lessons, Point 1, on page 1).
7.4 x 6.02
2.6 x 5.5 = 44.55 =
14.3 3.1 b. Then exponents. Starting from the original problem, look only at the powers of 10.
Try to solve the exponential math “in your head” without the calculator. Write the
answer for the top, then the bottom, then divide.
*****
10―2 x 1023 = 1021 = 1021―(―2) = 1023
10―2
103 x 10―5
c. Now combine the significand and exponential and write the final answer.
***** ©2011 www.ChemReview.Net v. u6 Page 17 Module 1 – Scientific Notation 3.1 x 1023
Note that by grouping the numbers and exponents separately, you did not need to
enter the exponents into your calculator. To multiply and divide powers of 10, you
simply add and subtract whole numbers.
Let’s try the calculation a second way.
Method 2. All on the calculator.
Enter all of the numbers and exponents into your calculator. (Your calculator manual,
which is usually available online, can help.) Write your final answer in scientific
notation. Round the significand to two digits.
*****
On most calculators, you will need to use an E or EE or EXP or ^ key, rather than
the times key, to enter a “10 to a power” term.
If you needed that hint, try again, and then check below.
* ****
Note how your calculator displays the exponential term in answers. The exponent
may be set apart at the right, sometimes with an E in front.
Your calculator answer, rounded, should be the same as with Method 1: 3.1 x 1023 . Which way was easier? “Numbers, then exponents,” or “all on the calculator?” How you
do the arithmetic is up to you, but “numbers, then exponents” is often quicker and easier. Checking Calculator Results
Whenever a complex calculation is done on a calculator, you must do the calculation a
second time, using different steps, to catch errors in calculator use. Calculator results can be
checked either by using a different key sequence or by estimating answers.
“Mental arithmetic estimation” is often the fastest way to check a calculator answer. To
learn this method, let’s use the calculation that was done in the first section of this lesson.
(7.4 x 10―2)(6.02 x 1023) =
(2.6 x 103)(5.5 x 10―5)
Apply the following steps to the problem above.
1. Estimate the numbers first. Ignoring the exponentials, round and then multiply all
of the top significands, and write the result. Round and multiply the bottom
significands. Write the result. Then write a rounded estimate of the answer when
you divide those two numbers, and then check below.
*****
Your rounding might be
7x6 = 7
3
3x6 ≈2 (the ≈ sign means approximately equals) If your mental arithmetic is good, you can estimate the number math on the paper
without a calculator. The estimate needs to be fast, but does not need to be exact. If ©2011 www.ChemReview.Net v. u6 Page 18 Module 1 – Scientific Notation needed, evaluate the rounded top and bottom numbers on the calculator, but try to
practice the arithmetic “in your head.”
2. Simplify the exponents. The exponents can be solved by adding and subtracting
whole numbers. Try the exponential math without the calculator.
*****
10―2 x 1023 = 1021 = 1021― (―2) = 1023
10―2
103 x 10―5
3. Combine the estimated number and exponential answers. Compare this estimated
answer to the answer found when you did this calculation in the section above
using a calculator. Are they close?
*****
The estimate is 2 x 1023. The answer with the calculator was 3.1 x 1023. Allowing
for rounding, the two results are close.
If your fast, rounded, “done in your head” answer is close to the calculator answer, it
is likely that the calculator answer is correct. If the two answers are far apart, check
your work.
Estimating Number Division
If you know your multiplication tables, and if you memorize these simple decimal
equivalents to help in estimating division, you may be able to do many numeric estimates
without a calculator.
1/2 = 0.50 1/3 = 0.33 1/4 = 0.25 1/5 = 0.20 2/3 = 0.67 3/4 = 0.75 1/8 = 0.125
The method used to get your final answer should be slow and careful. Your checking
method should use different steps or calculator keys, or, if time is a factor, should use
rounded numbers and quick mental arithmetic.
On timed tests, you may want to do the exact calculation first, and then go back at the end,
if time is available, and use rounded numbers as a check. When doing a calculation the
second time, try not to look back at the first answer until after you write the estimate. If you
look back, by the power of suggestion, you will often arrive at the first answer whether it is
correct or not.
For complex operations on a calculator, work each calculation a second time using
rounded numbers and/or different steps or keys. ©2011 www.ChemReview.Net v. u6 Page 19 Module 1 – Scientific Notation Practice
For problems 14, you will need to know the “fraction to decimal equivalent” conversions
in the box above. If you need practice, try this.
• On a sheet of paper, draw 5 columns and 7
rows. List the fractions down the middle
column. • Write the decimal equivalents of the fractions
at the far right. • Fold over those answers and repeat at the far left. Fold over those and repeat. 1/2
1/3
1/4
… To start, complete the even numbered problems. If you get those right, go to the next
lesson. If you need more practice, do the odds.
Then try these next four without a calculator. Convert final answers to scientific notation.
1. =
4 x 103
7)
(2.00)(3.0 x 10 2. 1
=
(4.0 x 109)(2.0 x 103) 3. (3 x 10―3)(8.0 x 10―5) =
(6.0 x 1011)(2.0 x 10―3) 4. (3 x 10―3)(3.0 x 10―2) =
(9.0 x 10―6)(2.0 x 101) For problems 58 below, in your notebook
• First write an estimate based on rounded numbers, then exponentials. Try to do this
estimate without using a calculator. • Then calculate a more precise answer. You may
o plug the entire calculation into the calculator, or o use the “numbers on calculator, exponents on paper” method, or o experiment with both approaches to see which is best for you. Convert both the estimate and the final answer to scientific notation. Round the significand
in the answer to two digits. Use the calculator that you will be allowed to use on quizzes
and tests.
(3.62 x 104)(6.3 x 10―10) =
(4.2 x 10―4)(9.8 x 10―5) 6. 7. (1.6 x 10―3)(4.49 x 10―5) =
(2.1 x 103)(8.2 x 106) 8. 5. ©2011 www.ChemReview.Net v. u6 10―2
=
(750)(2.8 x 10―15)
1
=
(4.9 x 10―2)(7.2 x 10―5) Page 20 Module 1 – Scientific Notation 9. For additional practice, do the two pretest problems at the beginning of this lesson. ANSWERS
Pretest: 1. 1.25 x 1011 2. 8.8 x 10―15 Practice: You may do the arithmetic using different steps than below, but you must get the same answer.
4 x 103― 7 =
6 1. 4 x 103
=
7)
(2.00)(3.0 x 10 2. 1
(4.0 x 109)(2.0 x 103) 3. ( 3 x 10―3 )(8.0 x 10―5) =
( 2 6.0 x 1011)(2.0 x 10―3) 4. = =
1
8 x 1012 2 x 10―4 = 0.67 x 10―4 = 6.7 x 10―5
3
1 x 10―12 = 0.125 x 10―12 = 1.25 x 10―13
8 8 x 10―3―5 = 2 x 10―8 = 2 x 10―8―8 = 2.0 x 10―16
4
1011―3
108 ( 3 x 10―3 )( 3.0 x 10―2 ) = 9 x 10―3―2 = 0.50 x 10―5
18
10―6+1
10―5
( 9.0 x 10―6 )( 2.0 x 101 ) = 0.50 = 5.0 x 10―1 5. First the estimate. The rounding for the numbers might be
4 x 6 = 0.6 For the exponents: 104 x 10―10 = 10―6 = 109 x 10―6 = 103
4 x 10
10―4 x 10―5
10―9
≈ 0.6 x 103 ≈ 6 x 102 (estimate) in scientific notation.
For the precise answer, doing numbers and exponents separately,
(3.62 x 104)(6.3 x 10―10) = 3.62 x 6.3 = 0.55
(4.2 x 10―4)(9.8 x 10―5)
4.2 x 9.8 The exponents are done as in the estimate above. = 0.55 x 103 = 5.5 x 102 (final) in scientific notation.
This is close to the estimate, a check that the more precise answer is correct.
6. Estimate: 1 ≈ 1 = 0.05 ;
7 x 3 20 10―2
= 10―2― (―13) = 1011
(102)(10―15) 0.05 x 1011 = 5 x 109 (estimate)
Numbers on calculator: = 0.048
1
7.5 x 2.8 FINAL: 0.048 x 1011 = 4.8 x 109 Exponents – same as in estimate. (close to the estimate) 7. You might estimate, for the numbers first,
1.6 x 4.49 = 2 x 4 = 0.5 For the exponents: 10―3 x 10―5 = 10―8 = 10―17
109
2.1 x 8.2
2x8
103 x 106
= 0.5 x 10―17 = 5 x 10―18 (estimate) ©2011 www.ChemReview.Net v. u6 Page 21 Module 1 – Scientific Notation More precisely, using numbers then exponents, with numbers on the calculator,
1.6 x 4.49 = 0.42
2.1 x 8.2 The exponents are done as in the estimate above. 0.42 x 10―17 = 4.2 x 10―18
8. Estimate: This is close to the estimate. Check! 1 ≈ 1 ≈ 0.03 ;
5x7
35 1
= 1/(10―7) = 107
(10―2)(10―5) 0.03 x 107 ≈ 3 x 105 (estimate)
Numbers on calculator = 1
= 0.028
4.9 x 7.2 Exponents – see estimate. FINAL: 0.028 x 107 = 2.8 x 105 (close to the estimate)
***** ©2011 www.ChemReview.Net v. u6 Page 22 Module 1 – Scientific Notation Lesson 1D: Special Project – The Atoms (Part 1)
At the center of chemistry are atoms: the building blocks of matter. There are 91 different
kinds of atoms found in the earth’s crust. When a substance is stable at room temperature
and pressure and contains only one kind of atom, the substance is said to be an element,
and the atoms are in their elemental form.
The Periodic Table helps in predicting the properties of the elements and atoms. In firstyear chemistry, about 40 of the atoms are frequently encountered. Quick, automatic
conversion between of the names and symbols of those atoms “in your head” will speed
and simplify solving problems.
To begin to learn those atoms, your assignment is:
• For the 12 atoms below, memorize the name, and symbol, and the position of the
atom in the table. • For each atom, given either its symbol or name, be able to write the other. • Be able to fill in an empty chart like the one below with these atom names and
symbols. Periodic Table
1A 2A 3A 4A 5A 6A 7A 1 8A
2 H He Hydrogen
3 Helium
4 Li
Lithium
11 5 6 7 8 9 10 Be B C N O F Ne Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon 12 Na Mg
Sodium Magnesium ***** ©2011 www.ChemReview.Net v. u6 Page 23 Module 2 – The Metric System SUMMARY – Scientific Notation
1. When writing a number between ─1 and 1, place a zero in front of the decimal point.
Do not write .42 or ─ .74 ; do write 0.42 or ─ 0.74
2. Exponential notation represents numeric values in three parts:
• a sign in front showing whether the value is positive or negative; • a number (the significand); • times a base taken to a power (the exponential term). 3. In scientific notation, the significand must be a number that is 1 or greater, but less than
10, and the exponential term must be 10 to a wholenumber power. This places the
decimal point in the significand after the first number which is not a zero.
4. When moving a decimal in exponential notation, the sign in front never changes.
5. To keep the same numeric value when moving the decimal of a number in base 10
exponential notation, if you
• move the decimal Y times to make the significand larger, make the exponent
smaller by a count of Y;
• move the decimal Y times to make the significand smaller, make the exponent
larger by a count of Y.
Recite and repeat to remember: When moving the decimal, for the numbers after the
sign in front,
“If one gets smaller, the other gets larger. If one gets larger, the other gets smaller.”
6. To add or subtract exponential notation by hand, all of the values must be converted to
have the same exponential term.
• Convert all of the values to have the same power of 10.
• List the significands and exponential in columns.
• Add or subtract the significands.
• Attach the common exponential term to the answer.
7. In multiplication and division using scientific or exponential notation, handle numbers
and exponential terms separately. Recite and repeat to remember:
• Do numbers by number rules and exponents by exponential rules.
• When you multiply exponentials, you add the exponents.
• When you divide exponentials, you subtract the exponents.
• When you take an exponential term to a power, you multiply the exponents.
• To take the reciprocal of an exponential, change the sign of the exponent. 8. In calculations using exponential notation, try the significands on the calculator but the
exponents on paper.
9. For complex operations on a calculator, do each calculation a second time using rounded
numbers and/or different steps or keys. ####
©2011 www.ChemReview.Net v. u6 Page 24 Module 2 – The Metric System Module 2 – The Metric System
Lesson 2A: Metric Fundamentals
Have you previously mastered the metric system? If you get a perfect score on the
following pretest, you may skip to Lesson 2B. If not, complete Lesson 2A. Pretest: Write answers to these, then check your answers at the end of Lesson 2A.
1. What is the mass, in kilograms, of 150 cc’s of liquid water?
2. How many cm3 are in a liter? How many dm3 are in a liter? 3. 2.5 pascals is how many millipascals?
4. 3,500 centigrams is how many kilograms?
***** The Importance of Units
The fastest and most effective way to solve problems in chemistry is to focus on the units
that measure quantities. In science, measurements and calculations are done using the
metric system.
All measurement systems begin with standards defining fundamental quantities that
include distance, mass, and time. Distance
The metric distance unit is the meter, abbreviated m. One meter is about 39.3 inches,
slightly longer than one yard. A meter stick is usually numbered in centimeters. 102030405060708090
Just as a dollar can be divided into 100 cents, and a century is 100 years, a meter is divided
into 100 centimeters. The centimeter, abbreviated cm, is 1/100th of a meter. As equalities,
we can write
1 centimeter ≡ 1/100th of a meter ≡ 10─2 meters and 1 meter ≡ 100 centimeters The symbol ≡ means “is defined as equal to” and/or “is exactly equal to.”
A centimeter can be divided into 10 millimeters (mm). Each millimeter is 1/1000th of a
meter. In relation to the meter,
1 millimeter ≡ 1/1,000th of a meter ≡ 10─3 meters and 1 meter ≡ 1,000 millimeters A meter stick can also be divided into 10 decimeters (dm). As equalities,
1 decimeter ≡ 1/10th of a meter ≡ 10─1 meters and 1 meter ≡ 10 decimeters One decimeter is also equal to 10 centimeters.
Long distances are usually measured in kilometers (km). ©2011 www.ChemReview.Net v. u6 1 kilometer ≡ 1,000 meters Page 25 Module 2 – The Metric System What do you need to remember from the above? You will need to be able to write from
memory the following two rules.
1. The “meterstick” equalities
1 METER ≡ 10 deciMETERS ≡ 100 centiMETERS ≡ 1,000 milliMETERS
2. The “one prefix” definitions
1 milliMETER ≡ 1 mm ≡ 10―3 METERS ( ≡ 1/1000th METER ≡ 0.001 METERS )
1 centiMETER ≡ 1 cm ≡ 10―2 METERS ( ≡ 1/100th METER ≡ 0.01 METERS ) 1 deciMETER ≡ 1 dm ≡ 10―1 METERS ( ≡ 1/10th METER ≡ 0.1 METERS ) 1 kiloMETER ≡ 1 km ≡ 103 METERS (≡ 1,000 METERS) To help in remembering the meterstick equalities, visualize a meter stick. Recall what the
numbers and marks on a meter stick mean. Use that image to help you to write the
equalities above.
To help in remembering the kilometer definition, visualize 1,000 meter sticks in a row.
That’s a distance of one kilometer. 1 kilometer ≡ 1,000 meter sticks.
Once you commit Rule 1 to memory, Rule 2 should be easy to write because it is
mathematically equivalent. Rule 1 uses the “1 meter =” format and Rule 2 uses the
“1prefix” format. Because both formats are used in calculations, we will need to be
familiar with both ways of defining the prefix relationships.
Rules 1 and 2 are especially important because of Rule
3. You may substitute any unit for METER in the equalities above.
Rule 3 means that the prefix relationships that are true for meters are true for any units of
measure. The three rules above allow us to write a wide range of equalities that we can
use to solve science calculations, such as
1 liter ≡ 1,000 milliliters 1 centigram ≡ 10―2 grams 1 kilocalorie ≡ 103 calories To use kilo , deci, centi or milli with any units, you simply need to be able to write or recall
from memory the metric equalities in Rules 1 and 2 above. Practice A: Write Rules 1 and 2 until you can do so from memory. Learn Rule 3.
Then complete these problems without looking back at the rules.
1. From memory, add exponential terms to these blanks.
a. 1 millimeter = ______meters b. 1 deciliter = _______ liter 2. From memory, add full metric prefixes to these blanks.
a. 1000 grams = 1 _________gram ©2011 www.ChemReview.Net v. u6 b. 10―2 liters = 1 ___________liter Page 26 Module 2 – The Metric System Volume
Volume is the amount of threedimensional space that a material or shape occupies.
Volume is termed a derived quantity, rather than a fundamental quantity, because it is
derived from distance. Any volume unit can be converted to a distance unit cubed.
A cube that is 1 centimeter wide by 1 cm high by 1 cm long has a volume of one cubic
centimeter (1 cm3). In biology and medicine, cm3 is often abbreviated as “cc.”
In chemistry, cubic centimeters are usually referred to as milliliters, abbreviated as mL.
One milliliter is defined as exactly one cubic centimeter. Based on this definition, since
• 1,000 milliMETERs ≡ 1 METER, and 1,000 millianythings ≡ 1 anything, • 1,000 milliLITERS is defined as 1 liter (1 L). The mL is a convenient measure for smaller volumes, while the liter (about 1.1 quarts) is
preferred when measuring larger volumes.
One liter is the same as one cubic decimeter ( 1 dm3 ). Note how these units are related.
• The volume of a cube that is 10 cm x 10 cm x 10 cm = 1,000 cm3 = 1,000 mL • Since 10 cm ≡ 1 dm, the volume of this same cube can be calculated as
1 dm x 1 dm x 1 dm ≡ 1 cubic decimeter ≡ 1 dm3 Based on the above, by definition, all of the following terms are equal.
1,000 cm3 ≡ 1,000 mL ≡ 1 L ≡ 1 dm3
What do you need to remember about volume? For now, just two more sets of equalities.
4. 1 milliliter (mL) ≡ 1 cm3 ≡ 1 cc
5. 1 liter ≡ 1,000 mL ≡ 1,000 cm3 ≡ 1 dm3 Mass
Mass measures the amount of matter in an object. If you have studied physics, you know
that mass and weight are not the same. In chemistry, however, unless stated otherwise, we
assume that mass is measured at the constant gravity of the earth’s surface. In that case,
mass and weight are directly proportional and can be measured with the same instruments.
The metric baseunit for mass is the gram. One gram (g) was originally defined as the mass
of one cubic centimeter of liquid water at 4° Celsius, the temperature at which water has its
highest density. The modern SI definition for one gram is a bit more complicated, but it is
still very close to the historic definition. We will often use that historic definition in
calculations involving liquid water if high precision is not required.
For a given mass of liquid water at 4° C, its volume increases by a small amount with
changes in temperature. The volume increases more if the water freezes or boils. However,
for most calculations for liquid water at any temperature, the following rule may be used.
6. 1 cm3 H2O (liquid) ≡ 1 mL H2O (l) ≈ 1.00 gram H2O(l) ©2011 www.ChemReview.Net v. u6 ( ≈ means approximately) Page 27 Module 2 – The Metric System Temperature
Metric temperature scales are defined by the properties of water. The unit of the
temperature scale is termed a degree Celsius (ºC).
0ºC = the freezing point of water.
100ºC = the boiling point of water at a pressure of one atmosphere.
Room temperature is generally between 20ºC (which is 68ºF) and 25ºC (77ºF). Time: The base unit for time in the metric system is the second.
Unit and Prefix Abbreviations
The following list of abbreviations for metric units should also be committed to memory.
Given the unit, you need to be able to write the abbreviation, and given the abbreviation,
you need to be able to write the unit.
Unlike other abbreviations, abbreviations for metric units do not have periods at the end.
Units: m = meter g = gram s = second L = liter = dm3 = cubic decimeter
The most frequently used prefixes: k = kilo cm3 = cubic centimeter = mL = “cc”
d = deci c = centi m = milli Practice B: Write Rules 3 to 6 until you can do so from memory. Learn the unit and
prefix abbreviations as well. Then complete the following problems without looking back
at the above.
1. Fill in the prefix abbreviations: 1 m = 10 ____m = 100 ____m = 1000 ____m 2. From memory, add metric prefix abbreviations to these blanks.
a. 103 g = 1 ____g b. 10―3 s = 1 ___s 3. From memory, add fixed decimal numbers to these blanks.
a. 1000 cm3 = _________ mL b. 100 cc H2O (l) = _______ grams H2O (l) 4. Add fixed decimal numbers: 1 liter ≡ ________ mL ≡ _______ cm3 ≡ ________ dm3 SI Units
The modern metric system (Le Système International d’Unités) is referred to as the SI system
and is based on what are termed the SI units. SI units are a subset of metric units that
chooses one preferred metric unit as the standard for measuring each physical quantity.
The SI standard unit for distance is the meter, for mass is the kilogram, and for time is the
second. Historically, the SI system is derived from what in physics was termed the mks
system because it measured in units of meters, kilograms, and seconds.
In physics, and in many chemistry calculations that are based on relationships derived from
physics, using SI units is essential to simplify calculations. ©2011 www.ChemReview.Net v. u6 Page 28 Module 2 – The Metric System However, for dealing with laboratoryscale quantities, chemistry often measures and
calculates in units that are not SI, but are metric. For example, in chemistry we generally
measure mass in grams instead of kilograms. In Modules 4 and 5, you will learn to convert
between SI and nonSI units. Learning the Metric Fundamentals
A strategy that can help in problemsolving is to start each homework assignment, quiz, or
test by writing recently memorized rules at the top of your paper. By writing the rules at
the beginning, you avoid having to
remember them under time pressure
Metric Basics
later in the test.
1. 1 METER
≡ 10 deciMETERS
We will use equalities to solve most
≡ 100 centiMETERS
problems. The 7 metric basics define
≡ 1000 milliMETERS
the equalities that we will use most
1,000 METERS ≡ 1 kiloMETER
often. A Note on Memorization
A goal of these lessons is to minimize
what you must memorize. However, it
is not possible to eliminate
memorization from science courses.
When there are facts which you must
memorize in order to solve problems,
these lessons will tell you. This is one
of those times.
Memorize the table of metric basics in
the box at the right. You will need to
write them automatically, from
memory, as part of most assignments in
chemistry. 2. 1 milliMETER ≡ 1 mm ≡ 10─3 METER
1 centiMETER ≡ 1 cm ≡ 10─2 METER
1 deciMETER ≡ 1 dm ≡ 10─1 METER
1 kiloMETER ≡ 1 km ≡ 103 METER
3. Any word can be substituted for METER
above.
4. 1 mL ≡ 1 cm3 ≡ 1 cc
5. 1 liter ≡ 1000 mL ≡ 1000 cm3 ≡ 1 dm3
6. 1 cm3 H2O(liquid) ≡ 1 mL H2O(l)
≈ 1.00 gram H2O(l)
7. meter ≡ m ; gram ≡ g ; second ≡ s Memorization Tips
When you memorize, it helps to use as many senses as you can.
• Say the rules out loud, over and over, as you would learn lines for a play. • Write the equations several times, in the same way and order each time. • Organize the rules into patterns, rhymes, or mnemonics. • Number the rules so you know which rule you forgot, and when to stop. • Picture real objects.
o Sketch a meter stick, then write the first two metric rules and compare to your
sketch. ©2011 www.ChemReview.Net v. u6 Page 29 Module 2 – The Metric System o
o Write METER in ALL CAPS for the first two rules as a reminder that you that
you can substitute ANYTHING for METER.
For volume, mentally picture a 1 cm x 1 cm x 1 cm = 1 cm3 cube.
Call it one mL. Fill it with water to make a mass of 1.00 grams.
├1 cm ┤ After repetition, you will recall new rules automatically. That’s the goal. Practice C: Study the 7 rules in the metric basics table above, then write the table on
paper from memory. Repeat until you can write all parts of the table from memory. Then
cement your knowledge by doing these problems. Check your answers below.
1. In your mind, picture a kilometer and a millimeter. Which is larger?
2. Which is larger, a kilojoule or a millijoule?
3. Name four units that can be used to measure volume in the metric system.
4. How many centimeters are on a meter stick?
5. How large is a kiloliter?
6. What is the mass of 15 milliliters of liquid water? 7. One liter of liquid water has what mass? 8. What is the volume of one gram of ice? 9. Fill in the portion of the Periodic Table below for the first 12 atoms. ©2011 www.ChemReview.Net v. u6 Page 30 Module 2 – The Metric System ANSWERS
Pretest: 1. 0.15 kg 2. 1,000 cm3, 1 dm3 3. 2,500 millipascals 4. 0.035 kg Practice A
1a. 1 millimeter = 10―3 meters 1b. 1 deciliter = 10―1 liter 2a. 1000 grams = 1kilo gram 2b. 10―2 liters = 1 c enti liter Practice B
1.
3a. 1000 cm3 = 1000 mL 4. 1 liter ≡ 1000 mL ≡ 1000 cm3 ≡ 2a. 103 g = 1 kg 1 m = 10 dm = 100 cm = 1000 mm 2b. 10―3 s = 1 ms 3b. 100 cc H2O (l) = 100 grams H2O (l) 1 dm3 Practice C
1. A kilometer 2. A kilojoule 3. Possible answers include cubic centimeters, milliliters, liters, cubic decimeters, cubic meters, and any
metric distance unit cubed.
4. 100 5. 1,000 liters 6. 15 grams 7. 1,000 grams or one kilogram 8. These lessons have not supplied the answer. Water expands when it freezes. So far, we only know the
answer for liquid water.
9. See Periodic Table.
***** ©2011 www.ChemReview.Net v. u6 Page 31 Module 2 – The Metric System Lesson 2B: Metric Prefixes
Pretest: If you have previously mastered use of the prefixes in the table below, try the
Practice B problems at the end of this lesson. If you get those right, you may skip this
lesson.
***** Prefix Additional Prefixes tera T x 1012 For measurements of very large or very small
quantities, prefixes larger than kilo and smaller
than milli may be used. The 13 prefixes
encountered most frequently are listed in the
table at the right. Note that giga G x 109 mega M x 106 kilo k x 103 hecto h x 102 deka da x 101 deci d x 10―1 centi c x 10―2 milli m x 10―3 micro μ (mu) or u x 10―6 nano n x 10―9 pico p x 10―12 • Outside the range between ─ 3 and 3,
metric prefixes are abbreviations of powers
of 10 that are divisible by 3. • When the full prefix name is written, the
first letter is not normally capitalized. • For prefixes above k (kilo), the abbreviation
for a prefix must be capitalized. • For the prefixes k and below, all letters of
the abbreviation must be lower case. Using Prefixes Abbreviation Means femtofx 10―15
A metric prefix is interchangeable with the
exponential term it represents. For example, during measurements and/or calculations,
• An exponential term can be substituted for its equivalent metric prefix.
Examples: 7.0 milliliters = 7.0 x 10―3 liters
5.6 kg = 5.6 x 103 g
43 nanometers = 43 nm = 43 x 10―9 m • A metric prefix can be substituted for its equivalent exponential term.
Examples: 3.5 x 10―12 meters = 3.5 picometers = 3.5 pm
7.2 x 106 watts = 7.2 megawatts In calculations, we will often need to convert between a prefix and its equivalent
exponential term. One way to do this is to apply the prefix definitions.
Q1. From memory, fill in these blanks with prefixes.
a. 103 grams = 1 _________gram b. 2 x 10―3 meters = 2 ________ meters Q2. From memory, fill in these blanks with prefix abbreviations.
a. 2.6 x 10―1 L = 2.6 ____L ©2011 www.ChemReview.Net v. u6 b. 6 x 10―2 g = 6 ____g Page 32 Module 2 – The Metric System Q3. Fill in these blanks with exponential terms (use the table above if needed).
a. 1 gigajoule = 1 x _______ joules b. 9 μm = 9 x ________ m *****
Answers
1a. 103 grams = 1 kilogram 1b. 2 x 10―3 meters = 2 millimeters 2a. 2.6 x 10―1 L = 2.6 dL. 2b. 6 x 10―2 g = 6 cg 3a. 1 gigajoule = 1 x 109 joules 3b. 9 μm = 9 x 10─6 m From the prefix definitions, even if you are not yet familiar with the quantity that a unit is
measuring, you can convert between its prefixunit value and its value using exponentials. Science Versus ComputerScience Prefixes
Computer science, which calculates based on powers of 2, uses slightly different definitions
for prefixes, such as kilo = 210 = 1,024 instead of 1,000.
However, in chemistry and all other sciences, for all base units, the prefix to powerof10
relationships in the metricprefix table are exact definitions. Learning the Additional Prefixes
To solve calculations, you will need to recall the rows in the table of 13 metric prefixes
quickly and automatically. To begin, practice writing the table from memory. To help,
look for patterns and use memory devices. Note
• tera = T = 10Twelve and nano (which connotes small) = 10─nine . Focusing on those two can help to “anchor” the prefixes near them in the table.
Then make a selfquiz: on a sheet of paper, draw a table 3 columns across and 14 rows
down. In the top row, write
Prefix Abbreviation Means Then fill in the table. Repeat writing the table until you can do so from memory, without
looking back. Once you can do so, try to do the problems below without looking back at
your table. Practice A: Use a sticky note to mark the answer page at the end of this lesson. 1. From memory, add exponential terms to these blanks.
a. 7 microseconds = 7 x ______ seconds b. 9 fg = 9 x ________ g c. 8 cm = 8 x ________ m d. 1 ng = 1 x ________ g 2. From memory, add full metric prefixes to these blanks.
a. 6 x 10─2 amps = 6 ____________ amps ©2011 www.ChemReview.Net v. u6 b. 45 x 109 watts = 45 ________watts Page 33 Module 2 – The Metric System 3. From memory, add metric prefix abbreviations to these blanks.
a. 1012 g = 1 ____g b. 10―12 s = 1 ___s c. 6 x 10─9 m = 6 ____ m d. 5 x 10─1 L = 5 ____ L e. 4 x 101 L = 4 ______ L f. 16 x 106 Hz =16 _________Hz 4. When writing prefix abbreviations by hand, write so that you can distinguish between
(add a prefix abbreviation) 5 x 10─3 g = 5 ____g and 5 x 106 g = 5 ____g 5. For which prefix abbreviations is the first letter always capitalized?
6. Write 0.30 gigameters/second without a prefix, in scientific notation. Converting Between Prefix Formats
To solve calculations in chemistry, we will often use conversion factors that are
constructed from metric prefix definitions. For those definitions, we have learned two
types of equalities.
• Our “meter stick” equalities are based on what one unit is equal to:
1 METER ≡ 10 deciMETERS ≡ 100 centiMETERS ≡ 1,000 milliMETERS • Our prefix definitions are based on what one prefix is equal to, such as nano = 10─9 . It is essential to be able to correctly write both forms of the metric definitions, because work
in science often uses both.
For example, to convert between milliliters and liters, we can use either
• 1 mL = 10─3 L , based on what 1 milli means, or • 1,000 mL = 1 L ; which is an easytovisualize definition of one liter. Those two equalities are equivalent. The second is simply the first with the numbers on
both sides multiplied by 1,000.
However, note that 1 mL = 10─3 L , but 1 L = 103 mL . The numbers in the equalities
change depending on whether the 1 is in front of the prefix or the unit. Which format
should we use? How do we avoid errors?
In these lessons, we will generally use the one prefix equalities to solve problems. After
learning the fundamental definitions for the 10 prefixes in the table, such as 1 milli =
10─3, using the definitions makes conversions easy to check.
Once those prefix, abbreviation, and meanings are in memory, we will then need to “watch
where the 1 is.”
If you need to write or check prefix equalities in the “one unit =” format, you can derive
them from the one prefix definitions, by writing the table if needed. ©2011 www.ChemReview.Net v. u6 Page 34 Module 2 – The Metric System For example, 1 gram = ______ micrograms?
• Since 1 microanything = 10─6 anythings, then • 1 microgram = 10─6 grams • To get a 1 in front of gram, we multiply both sides by 106, so • 1 gram = 106 micrograms ( = 106 μg = 1,000,000 micrograms ) The steps above can be summarized as the reciprocal rule for prefixes:
If 1 prefix = 10a , 1 unit = 10─a prefixunits
Another way to state the reciprocal rule for prefixes:
To change a prefix definition between the “1 prefix = “ format and the “1 unit = “
format, change the sign of the exponent.
If you need to check your logic, write the most familiar example:
Since 1 milliliter = 10─3 liter , then 1 liter = 103 milliliters = 1,000 mL
Fill in these blanks with exponential terms.
Q1. 1 nanogram = 1 x ______ grams , so 1 gram = 1 x ______ nanograms Q2. 1 dL = 1 x ______ liters , 1 L = 1 x ______ dL so *****
Answers
A1. 1 nanogram = 1 x 10─9 grams , so 1 gram = 1 x 109 nanograms A2. 1 dL = 1 x 10─1 liters , 1 L = 1 x 101 dL = 10 dL so To summarize:
• When using metric prefix definitions, be careful to note whether the 1 is in front of
the prefix or the unit. • To avoid confusing the signs of the exponential terms in prefix definitions,
memorize the table of 13 one prefix definitions. Then, if you need an equality with a
“1 unit = 10x prefixunit” format, reverse the sign of the prefix definition. Practice B: Write the table of the 13 metric prefixes until you can do so from memory,
then try to do these without consulting the table. 1. Fill in the blanks with exponential terms.
a. 1 terasecond = 1 x ______ seconds , so 1 second = 1 x ______ teraseconds b. 1 µg = 1 x ______ grams , 1 g = 1 x ______ µg ©2011 www.ChemReview.Net v. u6 so Page 35 Module 2 – The Metric System 2. Apply the reciprocal rule to add exponential terms to these one unit equalities.
a. 1 gram = _______ centigrams b. 1 meter = ___________ picometers c. 1 s = ___________ ms d. 1 s = ___________ Ms 3. Add exponential terms to these blanks. Watch where the 1 is!
a. 1 micromole = _____ moles b. 1 g = 1 x _______ Gg c. 1 hectogram = 1 x _______ grams d. _______ kg = 1 g e. _______ ns = 1 s f. 1 fL = ________ L ANSWERS
Practice A
1. a. 7 microseconds = 7 x 10─6 seconds
c. 8 cm = 8 x 10─2 m
2. a. b. 9 fg = 9 x 10─15 g
d. 1 ng = 1 x 10─9 g 6 x 10─2 amps = 6 centiamps b. 45 x 109 watts = 45 gigawatts a. 1012 g = 1 Tg b. 10―12 s = 1 ps c. 6 x 10─9 m = 6 nm d. 5 x 10─1 L = 5 dL e. 4 x 101 L = 4 daL 3. f. 16 x 106 Hz =16 MHz 4. 5 mg and 5 Mg 6. 3.0 x 108 meters/second 5. M, G, and T. Practice B
1. a. 1 terasecond = 1 x 1012 seconds , so
b. 1 µg = 1 x 10─6 grams ,
2. a. 1 gram = 102 centigrams so 1 second = 1 x 10─12 teraseconds
1 g = 1 x 106 µg ( For “ 1 unit = “, take reciprocal (reverse sign) of prefix meaning ) b. 1 meter = 1012 picometers
3. a. 1 micromole = 10─6 moles c. 1 s = 103 ms d. 1 s = 1 x 10─6 Ms b. 1 g = 1 x 10─9 Gg c. 1 hectogram = 1 x 102 grams d. 10─3 kg = 1 g e. 109 ns = 1s f. 1 fL = 10─15 L ***** ©2011 www.ChemReview.Net v. u6 Page 36 Module 2 – The Metric System Lesson 2C: Cognitive Science – and Flashcards
In this lesson, you will learn a system that will help you to automatically recall the
vocabulary needed to read science with comprehension and the facts needed to solve
calculations.
Cognitive science studies how the mind works and how it learns. The model that science
uses to describe learning includes the following fundamentals.
• The purpose of learning is to solve problems. You solve problems using
information from your immediate environment and your memory. The human brain contains different types of memory, including
• Working memory: the part of your brain where you solve problems. • Shortterm memory: information that you remember for only a few days. • Longterm memory: information that you can recall for many years. Working memory is limited, but human longterm memory has enormous capacity. The
goal of learning is to move new information from short into longterm memory so that it
can be recalled by working memory for years after initial study. If information is not
moved into longterm memory, useful longterm learning has not taken place.
Children learn speech naturally, but most other learning requires repeated thought about
the meaning of new information, plus practice at recalling new facts and using new skills
that is timed in ways that encourage the brain to move new learning from short to longterm
memory.
The following principles of cognitive science will be helpful to keep in mind during your
study of chemistry and other disciplines.
1. Learning is cumulative. Experts in a field learn new information quickly because they
already have in longterm memory a storehouse of knowledge about the context
surrounding new information. That storehouse must be developed over time, with
practice.
2. Learning is incremental (done in small pieces). Especially for an unfamiliar subject,
there is a limit to how much new information you can move into longterm memory in
a short amount of time. Knowledge is extended and refined gradually. In learning,
steady wins the race.
3. Your brain can do parallel processing. Though adding information to long term
memory is a gradual process, studies indicate that your brain can work on separately
remembering what something looks like, where you saw it, what it sounds like, how
you say it, how you write it, and what it means, all at the same time. The cues
associated with each separate type of memory can help to trigger the recall of
information needed to solve a problem, so it helps to use multiple strategies. When
learning new information: listen to it, see it, say it, write it, and try to connect it to other
information that helps you to remember what it means. ©2011 www.ChemReview.Net v. u6 Page 37 Module 2 – The Metric System 4. The working memory in your brain is limited. Working memory is where you think.
Try multiplying 556 by 23 in your head. Now try it with a pencil, a paper, and your
head. Because of limitations in working memory, manipulating multiple pieces of new
information “in your head” is difficult. Learning stepwise procedures (standard
algorithms) that write the results of middle steps is one way to reduce “cognitive load”
during problem solving.
5. “Automaticity in the fundamentals” is another learning strategy that can help to
overcome limitations in working memory. When you can recall facts quickly due to
repeated practice, more working memory is available for higher level thought.
You can do work that is automatic while you think (most of us can think while walking),
but it is difficult to think about more than one problem at once.
6. Concepts are crucial. Your brain works to construct a “conceptual framework” to
categorize knowledge being learned so that you can recall facts and procedures when
you need them. The brain tends to store information in longterm memory only if it is
in agreement with your framework of concepts. In addition, if you have a more
complete and accurate understanding of “the big picture,” your brain is better able to
judge which information should be selected to solve a problem.
Concepts do not replace the need to move key facts and procedures into your longterm
memory, but concepts speed initial learning, recall, and appropriate application of your
knowledge in longterm memory.
7. “You can always look it up” is a poor strategy for problemsolving. Your working
memory is quite limited in how much information it can manipulate that is not in your
longterm memory. The more information you must stop to look up, the less likely you
will be able to follow your train of thought to the end of a complex problem.
How can you promote the retention of needed fundamentals? It takes practice, but some
forms of practice are more effective than others. Attention to the following factors can
improve your retention of information in longterm memory.
1. Overlearning. Practice until you are perfect only once and you will tend to recall new
information for only a few days. To be able to recall new facts and skills for more than
a few days, repeated practice to perfection is necessary.
2. The spacing effect. To retain what you learn, 20 minutes of study spaced over 3 days is
more effective than one hour of study for one day.
Studies of “massed versus distributed practice” show that if the initial learning of facts
and vocabulary is practiced over 34 days, then revisited weekly for 23 weeks, then
monthly for 34 months, it can often be recalled for decades thereafter.
3. Effort. Experts in a field usually attribute their success to “hard work over an extended
period of time” rather than “talent.”
4. Core skills. The facts and processes you should practice most often are those needed
most often in the discipline. ©2011 www.ChemReview.Net v. u6 Page 38 Module 2 – The Metric System 5. Get a good night’s sleep. There is considerable evidence that while you sleep, your
brain reviews the experience of your day to decide what to store in longterm memory.
Sufficient sleep promotes retention of what you learn.
[For additional science that relates to learning, see Willingham, Daniel [2007] Cognition: The
Thinking Animal. Prentice Hall, and Bruer, John T. [1994] Schools for Thought. MIT Press.] ] Practice A
1. What is “overlearning?” 2. What is the “spacing effect?” 3. What are two learning strategies that can help to overcome inherent limitations on the
manipulation of new information in your working memory? Flashcards
What is more important in learning: Knowing the facts or the concepts? Cognitive studies
have found that the answer is: both. However, to “think as an expert,” you need a
storehouse of factual information in memory that you can apply to new and unique
problems.
In these lessons, we will use the following flashcard system to master fundamentals that
need to be recalled automatically in order to efficiently solve problems. Using this system,
you will make two types of flashcards:
• “Oneway cards” for questions that make sense in one direction; and • “Twoway” cards for facts that need to be recalled in both directions. If you have access to about 30 3 x 5 index cards, you can get started now. Plan to buy
tomorrow about 100200 3x5 index cards, lined or unlined. (A variety of colors is helpful
but not essential.) Complete these steps.
1. On 1215 of your 30 initial cards (of the same color if possible), cut a triangle off the topright corner, making cards like this:
These cards will be used for questions that go in one direction.
Keeping the notch at the top right will identify the front side.
2. Using the following table, cover the answers in the right column with a folded sheet or
index card. For each question in the left column, verbally answer, then slide the cover
sheet down to check your answer. Put a check beside questions that you answer
accurately and without hesitation. When done, write the questions and answers
without checks onto the notched cards.
Frontside of cards (with notch at top right): Back Side  Answers To convert to scientific notation, move the
decimal to… After the first number not a zero If you make the significand larger Make the exponent smaller 420 Any number to the zero power = 1 ©2011 www.ChemReview.Net v. u6 Page 39 Module 2 – The Metric System To add or subtract in exponential notation Make all exponents the same Simplify 1/10─x 10x To divide exponentials Subtract the exponents To bring an exponent from the bottom of a
fraction to the top Change its sign 1 cc ≡ 1 ___ ≡ 1 ___ 1 cc ≡ 1 cm3 ≡ 1 mL 0.0018 in scientific notation = 1.8 x 10─3 1 L ≡ ___ mL ≡ ___ dm3 1 L ≡ 1,000 mL ≡ 1 dm3 To multiply exponentials Add the exponents Simplify 1/10x 10─x 74 in scientific notation = 7.4 x 101 The historic definition of 1 gram The mass of 1 cm3 of liquid water at 4ºC. 8x7 56 42/6 7 Any multiplication or division up to 12’s that you cannot answer instantly? Add to
your list of onesided cards. If you need a calculator to do number math, parts of
chemistry such as “balancing an equation” will be frustrating. With flashcard practice,
you will quickly be able to remember what you need to know.
3. To make “twoway” cards, use the index cards as they are, without a notch cut.
For the following cards, first cover the right column, then put a check on the left if you
can answer the left column question quickly and correctly. Then cover the left column
and check the right side if you can answer the rightside automatically.
When done, if a row does not have two checks, make the flashcard.
Twoway cards (without a notch):
103 g or 1,000 g = 1 __g 1 kg = ____ g Boiling temperature of water 100 degrees Celsius  if 1 atm. pressure 1 nanometer = 1 x ___ meters 1 ____meter = 1 x 10─9 meters Freezing temperature of water 0 degrees Celsius 4.7 x10─3 = ______________(nbr) 0.0047 = 4.7 x10? 1 GHz =10? Hz 109 Hz = 1 __Hz 2/3 = 0.? 0.666… = ? / ? 1 pL = 10? L 10─12 L = 1 __L 1/80 = 0.? 0.0125 = 1 / ? 3/4 = 0.? 0.75 = ? / ? 1 dm3 = 1 ___ 1 L = 1 __ 1/8 = 0.? 0.125 = 1 / ? 1/4 = 0.? 0.25 = 1 / ? ©2011 www.ChemReview.Net v. u6 Page 40 Module 2 – The Metric System More twoway cards (without a notch) for the metricprefix definitions.
kilo = x 10? x 103 = ? Prefix d = x 10? x 10─1 = ? abbr. micro =? abbr. µ = ? pref. nano = x 10? x 10─9 = ? pref. m = x 10? x 10─3 = ? abbr. mega =? abbr. M = ? pref. giga = x 10? x 109 = ? Prefix T = x 10? x 1012 = ? abbr. deka =? abbr. da = ? pref. milli = x 10? x 10─3 = ? pref. k = x 10? x 103 = ? abbr. pico =? abbr. p = ? prefix deci = x 10? x 10─1 = ? pref. f = x 10? x 10─15 = ? abb deci =? abbr. d = ? prefix tera = x 10? x 1012 = ? pref. µ = x 10? x 10─6 = ? abbr. hecto =? abbr. h = ? prefix pico = x 10? x 10─12 = ? pref G = x 10? x 109 = ? abbr. tera =? abbr. T = ? prefix hecto = x 10? x 102 = ? Prefix da = x 10? x 101 = ? abbr. milli =? abbr. m = ? pref. deka = x 10? x 101 = ? Prefix p = x 10? x 10─12 = ? abb femto =? abbr. f = ? prefix femto = x 10? x 10─15 = ? pref c = x 10? x 10─2 = ? abbr. giga =? abbr. G = ? pref. mega = x 10? x 106 = ? Prefix h = x 10? x 102 = ? abbr. nano =? abbr. n = ? prefix micro = x 10? x 10─6 = ? pref. M = x 10? x 106 = ? abbr. centi =? abbr. c = ? prefix centi = x 10? x 10─2 = ? pref. n = x 10? x 10─9 = ? abbr. kilo =? abbrev. k = ? prefix Which cards you need will depend on your prior knowledge, but when in doubt, make
the card. On fundamentals, you need quick, confident, accurate recall  every time.
4. Practice with one type of card at a time.
• For frontsided cards, if you get a card right quickly, place it in the got it stack. If
you miss a card, say it. Close your eyes. Say it again. And again. If needed, write
it several times. Return that card to the bottom of the do deck. Practice until every
card is in the gotit deck. • For twosided cards, do the same steps as above in one direction, then the other. 5. Master the cards at least once, then apply them to the Practice on the topic of the new
cards. Treat Practice as a practice test.
6. For 3 days in a row, repeat those steps. Repeat again before working assigned
problems, before your next quiz, and before your next test that includes this material.
7. Make cards for new topics early: before the lectures on a topic if possible. Mastering
fundamentals first will help in understanding lecture.
8. Rubber band and carry new cards. Practice during “down times.”
9. After a few modules or topics, change card colors.
This system requires an initial investment of time, but in the long run it will save time and
improve achievement.
The above flashcards are examples. Add cards of your design and choosing as needed. ©2011 www.ChemReview.Net v. u6 Page 41 Module 2 – The Metric System Flashcards, Charts, or Lists?
What is the best strategy for learning new information? Use multiple strategies: numbered
lists, mnemonics, phrases that rhyme, flashcards, reciting, and writing what must be
remembered. Practiced repeatedly, spaced over time.
For complex information, automatic recall may be less important than being able to
methodically write out a chart for information that falls into patterns.
For the metric system, learning flashcards and the prefix chart and picturing the meterstick
relationships all help to fix these fundamentals in memory. Practice B: Run your set of flashcards until all cards are in the “gotit” pile. Then try
these problems. Make additional cards if needed. Run the cards again in a day or two.
1. Fill in the blanks.
Format: 1 prefix 1 base unit 1 μMETER = ______ METERS 1 METER = _______ μMETERS 1 gigawatt = ______ watts 1 watt = ________ gigawatts 1 nanoliter = ______ liter _______ nanoliters = 1 liter 2. Add exponential terms to these blanks. Watch where the 1 is!
a. 1 picocurie = ___________ curies b. 1 megawatt = ___________watts c. 1 dag = ____________ g d. 1 mole = ___________ millimoles e. 1 m = ___________ nm f. 1 kPa = ____________ Pa 3. Do these without a calculator.
a. 10─6/10─8 = b. 1/5 = ___.___ ___ c. 1/50 = ___.___ ___ ___ 4. For the following atoms, write the symbol.
a. Helium = _______ b. Hydrogen = ________ c. Sodium = __________ 5. For the following symbols, write the atom name.
a. N = _______________ ©2011 www.ChemReview.Net v. u6 b. Ne = ______________ c. B = _____________ Page 42 Module 2 – The Metric System ANSWERS
Practice A
1. Repeated practice to perfection. 2. Study over several days gives better retention than “cramming.”
3. Learning stepwise procedures (algorithms) and learning fundamentals until they can be recalled
automatically. Practice B
1. 1 μMETER = 10─6 METERS 1 METER = 106 μMETERS 1 gigawatt = 109 watts 1 watt = 10─9 gigawatts 1 nanoliter = 10─9 liters 109 nanoliters = 1 liter 2. a. 1 picocurie = 10─12 curies
d. 1 mole = 103 millimoles b. 1 megawatt = 106 watts
e. 1 m = 109 nm 3. a. 10─6/10─8 = 10─6 + 8 = 102
4a. Helium = He
5a. N = Nitrogen b. 1/5 = 0.20 4b. Hydrogen = H
5b. Ne = Neon c. 1 dag = 101 g
f. 1 kPa = 103 Pa c. 1/50 = 0.020 4c. Sodium = Na 5c. B = Boron ***** Lesson 2D: Calculations With Units
Pretest: If you can do the following two problems correctly, you may skip this lesson.
Answers are at the end of the lesson.
1. Find the volume of a sphere that is 4.0 cm in diameter. (Vsphere = 4/3π r3 ).
2. Multiply: 2.0 g • m •
s2 3.0 m • 6.0 x 102 s =
4.0 x 10─2 *****
(Try doing this lesson without a calculator except as noted.) Adding and Subtracting With Units
Many calculations in mathematics consist of numbers without units. In science, however,
calculations are nearly always based on measurements of physical quantities. A
measurement consists of a numeric value and its unit.
When doing calculations in science, it is essential to write the unit after the numbers in
measurements and calculations. Why?
• Units give physical meaning to a quantity. • Units are the best indicators of what steps are needed to solve problems, and • Units provide a check that you have done a calculation correctly. ©2011 www.ChemReview.Net v. u6 Page 43 Module 2 – The Metric System When solving calculations, the math must take into account both the numbers and their
units. Use the following three rules.
Rule 1. When adding or subtracting, the units must be the same in the quantities being
added and subtracted, and those same units must be added to the answer.
Rule 1 is logical. Apply it to these two examples.
A. 5 apples + 2 apples = _________
B. 5 apples + 2 oranges = __________
*****
A is easy. B cannot be added. It makes sense that you can add two numbers that refer to
apples, but you can’t add apples and oranges. By Rule 1, you can add numbers that have
the same units, but you cannot add numbers directly that do not have the same units.
Apply Rule 1 to this problem: 14.0 grams
─ 7.5 grams *****
14.0 grams
─ 7.5 grams
6.5 grams If the units are all the same, you can add or subtract numbers,
but you must add the common unit to the answer. Multiplying and Dividing With Units
The rule for multiplying and dividing with units is different, but logical.
Rule 2. When multiplying and dividing units, the units multiply and divide.
Complete this example of unit math:
cm x cm = ________ .
*****
Units obey the laws of algebra. Try: cm5 = _____________
cm x cm = cm2
cm2
*****
cm5 = can be solved as
cm • cm • cm • cm • cm = cm3
c m • cm
cm2
or by using the rules for exponential terms:
cm5 = cm5─2 = cm3
cm2 Both methods arrive at the same answer (as they must). Rule 3. When multiplying and dividing, group numbers, exponentials, and units
separately. Solve the three parts separately, then recombine the terms.
Apply Rule 3 to this problem: If a postage stamp has the
dimensions 2.0 cm x 4.0 cm, the surface area of one side of the stamp = ____________
***** ©2011 www.ChemReview.Net v. u6 Page 44 Module 2 – The Metric System Area of a rectangle = l x w =
= 2.0 cm x 4.0 cm = (2.0 x 4.0) x (cm x cm) = 8.0 cm2 = 8.0 square centimeters
By Rule 2, the units must obey the rules of multiplication and division. By Rule 3, the unit
math is done separately from the number math.
Units follow the familiar laws of multiplication, division, and powers, including “like units
cancel.”
Apply Rule 3 to these: a. 8.0 L6
2.0 L2
*****
a. 8.0 L6 = 2.0 L2 8.0 • L6 =
2.0 • L2 = __________ 4.0 L4 b. b. 9.0 m6 = _________
3.0 m6 9.0 m6 = 3.0 (with no unit.) 3.0 m6 In science, the unit math must be done as part of calculations. A calculated unit must be
included as part of calculated answers (except in rare cases, such as part b above, when all
of the units cancel).
On the following problem, apply separately the math rules for numbers, exponential terms,
and units.
12 x 10─3 m4 = ____________
3.0 x 102 m2
*****
12 x 10─3 m4 12 • 10─3 • m4
3.0 • 102 • m2 = 3.0 x 102 m2 = 4.0 x 10─5 m2 When calculating, you often need to use a calculator to do the number math, but both the
exponential and unit math nearly always should be done without a calculator.
In the problems above, the units were all the same. However, units that are different can
also be multiplied and divided by the usual laws of algebra. Complete this calculation:
4.0 g • m • 3.0 m •
6.0 s
=
9.0 x 10─4 m2
s2
*****
When multiplying and dividing, do the number, exponential, and unit math separately.
4.0 g • m • 3.0 m •
s2 6.0 s
9.0 x 10─4 m2 = • g • m • m • s = 8.0 x 104 g
s •s
s
m2
10─4 72 • 1
9.0 This answer unit can also be written as g • s─1 , but you will find it helpful to use the x/y
unit format until we work with mathematical equations later in the course. ©2011 www.ChemReview.Net v. u6 Page 45 Module 2 – The Metric System Practice: Do not use a calculator except as noted. If you need just a few reminders, do
Problems 11 and 14. If you need more practice, do more. After completing each problem,
check your answer below. If you miss a problem, review the rules to figure out why before
continuing.
1. 16 cm ─ 2 cm = 2. 12 cm • 2 cm = 3. (m4)(m) = 4. m4 / m = 5. 105 =
10─2 6.
8. 9. =
24 L5
3.0 L─4 11. 12 x 10─2 L • g • 2.0 m •
4.0 s3
s
6.0 x 10─5 L2 3.0 g / 9.0 g = 10. 7. 3.0 meters • 9.0 meters = s─5
s2 18 x 10─3 g • m5
3.0 x 101 m2 = = = 12. A rectangular box has dimensions of 2.0 cm x 4.0 cm x 6.0 cm. Without a calculator,
calculate its volume. 13. Do pretest problem 1 at the beginning of this lesson (use a calculator). 14. Do pretest problem 2 at the beginning of this lesson (without a calculator). ANSWERS Both the number and the unit must be written and correct. Pretest: See answers to Problems 13 and 14 below.
1. 14 cm
6. s─7 3. m(4+1) = m5 2. 24 cm2
7. 27 meters2 11. 16 x 103 g • m • s2
L 8. 0.33 (no unit)
12. 4. m(4─1) = m3
9. 8.0 L9 5. 107
10. 6.0 x 10─4 g • m3 Vrectangular solid = length times width times height = 48 cm3 13. Diameter = 4.0 cm, radius = 2.0 cm.
Vsphere = 4/3 π r3 = 4/3 π (2.0 cm)3 = 4/3 π (8.0 cm3) = (32/3) π cm3 = 34 cm3
2.0 g • m • 3.0 m • 6.0 x 102 s = (2.0)(3.0)(6.0) • 104 • g • m • m • s = 9.0 x 104 g • m2
s
s2
4.0 x 10─2
4.0
s2
*****
14. ©2011 www.ChemReview.Net v. u6 Page 46 Module 2 – The Metric System SUMMARY – The Metric System
1. 1 METER ≡ 10 deciMETERS
≡ 100 centiMETERS
≡ 1000 milliMETERS Prefix Abbreviation Means tera T x 1012 giga G x 109 mega M x 106 1 centiMETER ≡ 1 cm = 10─2 METER kilo k x 103 1 deciMETER ≡ 1 dm = 10─1 METER hecto h x 102 1 kiloMETER ≡ 1 km = 103 METER deka da x 101 deci d x 10―1 centi c x 10―2 milli m x 10―3 micro μ (mu) or u x 10―6 nano n x 10―9 pico p x 10―12 femto f x 10―15 1,000 METERS ≡ 1 kiloMETER
2. 1 milliMETER ≡ 1 mm = 10─3 METER 3. Any unit can be substituted for METER above.
4. 1 cm3 ≡ 1 mL ≡ 1 cc
5. 1 liter ≡ 1000 mL ≡ 1 dm3
6. 1 cm3 H2O(l) ≡ 1 mL H2O(l) = 1.00 g H2O (l)
7. meter = m ; gram = g ; second = s
8. If prefix = 10a , 1 unit = 10─a prefixunits
9. To change a prefix definition from a “1 prefixunit = “ format to a “1 base unit = “ format,
change the exponent sign.
10. Rules for units in calculations. a. When adding or subtracting, the units must be the same in the numbers being added
and subtracted, and those same units must be added to the answer.
b. When multiplying and dividing units, the units multiply and divide.
c. When multiplying and dividing, group the numbers, exponentials, and units
separately. Solve the three parts, then recombine the terms.
##### ©2011 www.ChemReview.Net v. u6 Page 47 Module 3 – Significant Figures Module 3 – Significant Figures
Pretest: If you think you know how to use significant figures correctly, take the following
pretest to be sure. Check your answers at bottom of this page. If you do all of the pretest
perfectly, skip Module 3.
1. How many significant figures are in each of these?
a. 0.002030 b. 670.0 c. 670 d. 2 (exactly) 2. Round these numbers as indicated.
a. 62.75 to the tenths place. b. 0.090852 to 3 sf. 3. Use a calculator, then express your answer as a number with proper significant figures
and units attached.
4.701 x 103 L2 • g • 0.0401 s─2 •
s2 23.060 s4 • (an exact 4) =
6.0 x 10─5 L 4. Solve without a calculator. Write your answer in scientific notation with proper units
(56 x 10―10 cm) ─ (49.6 x 10―11 cm) = and significant figures.
***** Lesson 3A: Rules for Significant Figures
Nearly all measurements have uncertainty. In science, we need to express
• how much uncertainty exists in measurements, and • the uncertainty in calculations based on measurements. The differentials studied in calculus provide one method to find a precise range of the
uncertainty in calculations based on measurements, but differentials can be timeconsuming.
An easier method for expressing uncertainty is significant figures, abbreviated in these
lessons as sf.
Other methods measure uncertainty more accurately, but significant figures provide a
approximation of uncertainty that, compared to other methods, is easy to use in
calculations. In firstyear chemistry, significant figures in nearly all cases will be the
method of choice to indicate an approximation of the uncertainty in measurements and
calculations. ** * * *
Pretest Answers: Your answers must match these exactly.
1a. 4 1b. 4 1c. 2 1d. Infinite sf 2a. 62.8 2b. 0.0909 3. 2.9 x 108 L • g 4. 5.1 x 10─9 cm
** * * * ©2011 www.ChemReview.Net v. u6 Page 48 Module 3 – Significant Figures Significant Figures: Fundamentals
Use these rules when recording measurements and rounding calculations. 1. To Record a Measurement
Write all the digits you are sure of, plus the first digit that you must estimate in the
measurement: the first doubtful digit (the first uncertain digit). Then stop.
When writing a measurement using significant figures, the last digit is the first
doubtful digit. Round measurements to the highest place with doubt.
Example:
If a scale reads mass to the thousandths place, but under the conditions of the
experiment the uncertainty in the measurement is + 0.02 grams, we can write
12.432 g + 0.02 g using plusminus notation to record uncertainty. However, in a calculation, if we need to multiply or divide by that measured
value, the math to include the + can be timeconsuming. So, to convert the
measurement to significant figures notation, we write
12.43 g
When using significant figures to indicate uncertainty, the last place written in a
measurement is the first place with doubt. The + showed that the highest place
with doubt is the hundredths place. To convert to significant figures, we round
the recorded digits back to that place, then remove the + .
We convert the measurement to significant figures notation because in
calculations, the math when using numbers like 12.43 follows familiar rules. 2. To Add and Subtract Using Significant Figures
a. First, add or subtract as you normally would.
b. Next, search the numbers for the doubtful digit in the highest place. The answer’s
doubtful digit must be in that place. Round the answer to that place.
Example: 2 3 .1
+ 16.01
+ 1.008
4 0 .1 1 8 = 40.1 This answer must be rounded to 40.1 because the tenths place has doubt.
The tenths is the highest place with doubt among the numbers added.
Recall that the tenths place is higher than the hundredths place, which is higher
than the thousandths place. ©2011 www.ChemReview.Net v. u6 Page 49 Module 3 – Significant Figures c. The logic: If you add a number with doubt in the tenths place to a number with
doubt in the hundredths place, the answer has doubt in the tenths place.
A doubtful digit is significant, but numbers after it are not.
In a measurement, if the number in a given place is doubtful, numbers after that
place are garbage. We allow one doubtful digit in answers, but no garbage.
d. Another way to state this rule: When adding or subtracting, round your answer
back to the last full column on the right. This will be the first column of numbers,
moving right to left ( ), with no blanks above.
The blank space after a doubtful digit indicates that we have no idea what that
number is, so we cannot get a significant number in the answer in that column.
e. When adding or subtracting using a calculator: Underline the highest place with
doubt in the numbers being added and subtracted. Round your calculator answer
to that place.
Using a calculator, apply the rule to: Q. 4 3 + 1.00 —2.008 = *****
A. 4 3 + 1 .00 —2.008 = 4 1 .992 on the calculator = 4 2 in significant figures
Among the numbers being added and subtracted, the highest doubt is in the one’s
place. In chemistry calculations, you must round your final answer to that place.
f. When adding and subtracting exponential notation (see Lesson 1B), first make the
exponential terms the same, then apply the rules above to the significand of the
answer.
This rule is in agreement with the general rule: when adding and subtracting,
round to the highest place with doubt.
Example: 2.8 x 10 ─ 8
― 1 34 x 10 ─ 11 =
― 2 .8
x 10─8
0 .134 x 10 ─ 8
2 .6 66 x 10 ─ 8 = 2 .7 x 10 ─ 8 Summary: When adding or subtracting, round your final answer back to
● the highest place with doubt, which is also
When adding or subtracting in columns, this is also
● the leftmost place with doubt, which is also
● the last full column on the right, which is also
● the last column to the right without a blank space. ©2011 www.ChemReview.Net v. u6 Page 50 Module 3 – Significant Figures Practice A: First memorize the rules above. Then do the problems. When finished,
check your answers at the end of the lesson. 1. Convert these from plusminus notation to significant figures notation.
a. 65.316 mL + 0.05 mL b. 5.2 cm + 0.1 cm c. 1.8642 km + 0.22 km d. 16.8 ºC + 1 ºC 2. Add and subtract, with our without calculator. Round your final answer to the proper
number of significant figures.
a. 23 . 1
+ 23.1
1 6 . 01 b. 2.016 + 32.18 + 64.5 = c. 1.976 x 10 ─ 13
― 7 .3
x 10 ─ 14 3 . Use a calculator. Round your final answer to the proper number of significant figures.
a. 2.016 + 32.18 + 64.5 = b. 16.00 — 4.034 —1.008 = 3. To Count Significant Figures
When multiplying and dividing, we need to count the number of significant figures in a
measurement. To count the number of sf, count the sure digits plus the doubtful digit.
The doubtful digit is significant.
This rule means that for numbers in a measurement that do not include zeros, the count
of sf in a measurement is simply the number of digits shown.
Examples: 123 meters has 3 sf. 14.27 grams has 4 sf. In exponential notation, to find the number of sf , look only at the significand. The
exponential term does not affect the number of significant figures.
Example: 2.99 x 108 meters/second has 3 sf. 4. To Multiply and Divide
This is the rule we will use most often.
a. First multiply or divide as you normally would.
b. Then count the number of sf in each of the numbers you are multiplying or dividing.
c. Your answer can have no more sf than the measurement with the least sf that you
multiplied or divided by. Round the answer back to that number of sf.
Example: 3.1865 cm x 8.8 cm = 28.041 = 28 cm2 (must round to 2 sf)
^5 sf
^2 sf
^2 sf ©2011 www.ChemReview.Net v. u6 Page 51 Module 3 – Significant Figures Summary: Multiplying and Dividing
If you multiply and/or divide a 10sf number and a 9sf number and a 2sf
number, you must round your answer to 2 sf. 5. When Moving the Decimal: do not change the number of sf in a significand.
Q. Convert 424.7 x 10─11 to scientific notation. A. 4.247 x 10─9 6. In Calculations With Steps or Parts
The rules for sf should be applied at the end of a calculation.
In problems that have several separate parts (1a, 1b, etc.), and earlier answers are used
for later parts, many instructors prefer that you carry one extra sf until the end of a
calculation, then round to proper sf at the final step. This method minimizes changes
in the final doubtful digit due to rounding in the steps. Practice B: First memorize the rules above. Then do the problems. When finished,
check your answers at the end of the lesson. 1. Multiply and divide using a calculator. Write the first six digits of the calculator result,
then write the final answer, with units, and with the proper number of sf.
a. 3.42 cm times 2.3 cm2 =
b. 74.3 L2 divided by 12.4 L =
2. Convert to scientific notation: a. 0.0060 x 10─15 3. a. 9.76573 x 1.3 = A = ANSWERS: b. 1,027 x 10─1 b. A/2.5 = Your answers must match these exactly. Practice A
65.32 mL The highest place with doubt is hundredths. 1a. 65.316 mL + 0.05 mL When converting to sf, write all the sure digits. At the first place with doubt, round and stop.
5.2 cm 1b. 5.2 cm + 0.1 cm 1.9 km The highest doubt is in tenth’s place. Round to back tenths. 1c. 1.8642 km + 0.22 km
1d. 16.8 ºC + 1 ºC
2. (a) 17 ºC 23.1
+ 23.1
16.01 Highest doubt is in tenth’s place. Round to tenths. Doubt in the one’s place. Round back to the highest place with doubt.
(b) 2.016 + 32.18 + 64.5
= 98.696 =
Must round to 98.7 62.21 Must round to 62.2 (c) 1.976 x 10─13
― 0.73 x 10─13
1.246 x 10─13 Must round to 1.25 x 10─13 3a. 2.016 + 32.18 + 64.5 = 98.696  Must round to 9 8.7
3b. 16.0 0 — 4.034 —1.008 = 10.95 8 Must round to 1 0.96 ©2011 www.ChemReview.Net v. u6 Page 52 Module 3 – Significant Figures Practice B
1a. 7.9 cm3 (2 sf) 2a. 6.0 x 10─18 1b. 5.99 L (3 sf) 2b. 1.027 x 102 3a. 12.7 If this answer were not used in part b, the proper answer would be 13 (2 sf), but since we need the
answer in part b, it is often preferred to carry an extra sf. 3b. 12.7/2.5 = 5.1
***** Lesson 3B: Special Cases
When using significant figures to express uncertainty, there are special rules for zeros, and
exact numbers, and rounding off a 5.
1. Rounding. If the number beyond the place you are rounding to is
a. Less than 5: Drop it (round down). Example: 1.342 rounded to tenths = 1.3
b. Greater than 5: Round up. Example: 1.748 = 1.75 (rounded to underlined place) c. A 5 followed by any nonzero digits: Round up. Example: 1.02502 = 1.03 2. Rounding a lone 5
A lone 5 is a 5 without following digits or a 5 followed by zeros.
To round off a lone 5, some instructors prefer the simple “round 5 up” rule. Others
prefer a slightly more precise “engineer’s rule” described as follows.
a. If the number in front of a lone 5 being rounded off is even, round down by
dropping the 5.
Example: 1.45 = 1.4 b. If the number in front of a lone 5 is odd, round it up.
Example: 1.3500 = 1.4 A 5 followed by one or more zeros is rounded in the same way as a “lone 5.”
Rounding a lone 5, the rule is “even in front of 5, leave it. Odd? Round up.”
Why not always round 5 up? On a number line, a 5 is exactly halfway between 0 and
10. If you always round 5 up in a large number of calculations, your average will be
slightly high. When sending a rover on a 300 million mile trajectory to Mars, if you
calculate slightly high, you may miss your target by thousands of miles.
The “even leave it, odd up” rule rounds a lone 5 down half the time and up half the
time. This keeps the average of rounding 5 in the middle, where it should be.
Rounding off a lone 5 or a 0.1500 is not a case that occurs often in calculations, but when
it does, use the rounding rule preferred by the instructor in your course. Practice A: Round to the underlined place. Check answers at the end of this lesson. 1. 0.00212
4. 23.25 2. 0.0994
5. 0.065500 ©2011 www.ChemReview.Net v. u6 3. 20.0561
6. 0.0750 7. 2.659 x 10─3 Page 53 Module 3 – Significant Figures 3. Zeros. When do zeros count as sf? There are four cases. a. Leading zeros (zeros in front of all other digits) are never significant.
Example: 0.0006 has one sf. . (Zeros in front never count.) b. Zeros embedded between other digits are always significant.
Example: 300.07 has 5 sf. (Zeros sandwiched by sf count.)
c. Zeros after all other digits as well as after the decimal point are significant.
Example: 565.0 has 4 sf. You would not need to include that zero if it were not
significant.
d. Zeros after all other digits but before the decimal are assumed to be not significant.
Example: 300 is assumed to have 1 sf, meaning “give or take at least 100.”
When a number is written as 300, or 250, it is not clear whether the zeros are
significant. Many science textbooks address this problem by using this rule:
• “500 meters” means one sf, but • “500. meters,” with an unneeded decimal point added after a zero, means 3 sf. These modules will use that convention on occasion as well.
However, the best way to avoid ambiguity in the number of significant figures is to
write numbers in scientific notation.
4 x 102 has one sf; 4.00 x 102 has 3 sf.
In exponential notation, only the significand contains the significant figures.
In scientific notation, all of the digits in the significand are significant .
*****
Why are zeros complicated? Zero has multiple uses in our numbering system.
In cases 3a and 3d above, the zeros are simply “indicating the place for the decimal.” In
that role, they are not significant as measurements. In the other two cases, the zeros
represent numeric values. When the zero represents “a number between a 9 and a 1 in a
measurement,” it is significant. 4. Exact numbers. Measurements with no uncertainty have an infinite number of sf. Exact
numbers do not add uncertainty to calculations.
• If you multiply a 3 sf number by an exact number, round your answer to 3 sf. This rule means that exact numbers are ignored when deciding the sf in a calculated
answer. In chemistry, we use this rule in situations including the following.
a. Numbers in definitions are exact.
Example: The relationship “1 km = 1000 meters,” is a definition of kilo and not a
measurement with uncertainty. Both the 1 and the 1000 are exact numbers.
Multiplying or dividing by those exact numbers will not limit the number of sf in
your answer. ©2011 www.ChemReview.Net v. u6 Page 54 Module 3 – Significant Figures b. The number 1 in nearly all cases is exact.
Example: The conversion “1 km = 0.62 miles” is a legitimate approximation, but it
is not a definition ( ≡ ) and is not exactly correct. The 1 is therefore assumed to be
exact, but the 0.62 has uncertainty and has 2 sf.
c. Whole numbers (such as 2 or 6), if they are a measure of exact quantities (such as 2
people or 6 molecules), are also exact numbers with infinite sf.
d. Coefficients and subscripts in chemical formulas and equations are exact.
Example: 2 H2 + 1 O2 2 H2O All of those numbers are exact. You will be reminded about these exactnumber cases as we encounter them. For now,
simply remember that exact numbers
• have infinite sf, and • do not limit the sf in an answer. Practice B
Write the number of sf in these.
1. 0.0075 2. 600.3 6. 2.06 x 10─9 3. 178.40 7. 0.060 x 103 4. 4640. 8. 0.02090 x 105 5. 800
9. 3 (exact) ANSWERS
Practice A
1. 0.00212 rounds to 0.0021 2. 0.0994 rounds to 0.10 3. 20.0561 rounds to 20.06 4. 23.25 rounds to 23.2. by the engineer’s “lone 5: even, leave it” rule, or 23.3 by the “round lone 5 up” rule.
5. 0.065500 rounds to 0.066 by both rules. Eng: the lone 5 to be rounded follows an odd 5. Round “odd up.”
6. 0.0750 rounds to 0.08 by both rules. Engineers: When rounding a lone 5, use “even leave it, odd up.”
7. 2.659 x 10─3 rounds to 2.7 x 10─3 By all rules: when rounding a 5 followed by nonzeros, round up.
Practice B
1. 0.0075 has 2 sf. (Zeros in front never count.) 2. 600.3 has 4 sf. (Sandwiched zeros count.) 3. 178.40 has 5 sf. (Zeros after the decimal and after all the numbers count.)
4. 4640. has 4 sf. (Zeros after the numbers but before a written decimal count.)
5. 800 has 1 sf. (Zeros after all numbers but before the decimal place usually don’t count.)
6. 2.06 x 10─9 has 3 sf. (The significand in front contains and determines the sf.)
7. 0.060 x 103 has 2 sf. (The significand contains the sf. Leading zeros never count.)
8. 0.02090 x 105 has 4 sf. (The significand contains the sf. Leading zeros never count. The rest here do.)
9. 3 (exact) Infinite sf. Exact numbers have no uncertainty and infinite sf.
*****
©2011 www.ChemReview.Net v. u6 Page 55 Module 3 – Significant Figures Lesson 3C: Summary and Practice
First, memorize the rules.
1. When expressing a measurement in significant figures, include the first doubtful
digit, then stop. Round measurements to the doubtful digit’s place. 2. When counting significant figures, include the doubtful digit. 3. When adding and subtracting sf,
a. find the measurement that has doubt in the highest place.
b. Round your answer to that place. 4. When multiplying and dividing,
a. find the number in the calculation that has the least number of sf.
b. Round your answer to that number of sf. 5. In exponential notation, the sf are in the significand. 6. When moving a decimal, keep the same number of sf in the significand. 7. When solving a problem with parts, carry an extra sf until the final step. 8. To round off a lone 5, use the rule preferred by your instructor. Either always
round up, or use “even in front of 5, leave it. Odd? Round up.” 9. For zeros,
a. Zeros in front of all other numbers are never significant.
b. Sandwiched zeros are always significant.
c. Zeros after the other numbers and after the decimal are significant.
d. Zeros after all numbers but before the decimal place are not significant, but if
an unneeded decimal point is shown after a zero, that zero is significant. 10. Exact numbers have infinite sf.
For reinforcement, make the flashcards you need using the method in Lesson 2C.
Frontside (with notch at top right):
Writing measurements in sf, stop where?
Counting the number of sf, which digits count?
Adding and subtracting, round to where?
Multiplying and dividing, round how?
In counting sf, zeros in front
Sandwiched zeros
Zeros after numbers and after decimal
Zeros after numbers but before decimal ©2011 www.ChemReview.Net v. u6 Back Side  Answers
At the first doubtful digit
All the sure, plus the doubtful digit
The column with doubt in highest place
= last full column
Least # of sf in calculation = # sf allowed
Never count
count
count
Probably don’t count Page 56 Module 3 – Significant Figures Zeros followed by unneeded decimal
Exact numbers have count
Infinite sf Run the flashcards until perfect, then start the problems below. Practice: Try every other problem on day 1. Try the rest on day 2 of your practice. 1. Write the number of sf in these.
a. 107.42 b. 10.04 f. 1590.0 c. 13.40 g. 320 x 109 h. 14 (exact) d. 0.00640 e. 0.043 x 10─4 i. 2500 j. 4200. 2. Round to the place indicated.
a. 5.15 cm (tenths place) b. 31.84 meters (3 sf) c. 0.819 mL (hundredths place) d. 0.0635 cm2 (2 sf) e. 0.04070 g (2 sf) f. 6.255 cm (tenths place) 3. Addition and Subtraction: Round answers to proper sf.
a. 1.008
+ 1.008
3 2.00 b. 17.65
― 9 .7 c. 39.1 + 124.0 + 14.05 = 4. Multiplication and Division: Write the first 6 digits given by your calculator. Then
write the answer with the proper number of sf and proper units.
a. 13.8612 cm x 2.02 cm = b. 4.4 meters x 8.312 meters2 = c. 2.03 cm2/1.2 cm = d. 0.5223 cm3/0.040 cm = 5. Use a calculator. Answer in scientific notation with proper sf.
a. (2.25 x 10─2)(6.0 x 1023) b. (6.022 x 1023) / (1.50 x 10─2) 6. Convert these from + to sf notation.
a. 2.0646 m + 0.050 m b. 5.04 nm + 0.12 nm c. 12.675 g + 0.20 g d. 24.81 ºC + 1.0 ºC Answer 7 and 8 in scientific notation, with proper units and sf:
4.00 s4
• (an exact 2) =
7. 5.60 x 10─2 L2 • g • 0.090 s─3 •
s
6.02 x 10─5 L3
8. Without a calculator:
(─ 50.0 x 10―14 g) ─ (─ 49.6 x 10―12 g) = ©2011 www.ChemReview.Net v. u6 Page 57 Module 3 – Significant Figures 9. For additional practice, solve the problems in the pretest at the beginning of this module. ANSWERS
1. a.
c.
d.
e.
f.
g.
h.
i.
j. 107.42 5 sf (Sandwiched zeros count.)
b. 10.04 4 (Sandwiched zeros count.)
13.40
4 (Zeros after numbers and after the decimal count.)
0.00640 3 (Zeros in front never count, but zeros both after #s and after the decimal count.)
0.043 x 10─4 2 (Zeros in front never count. The significand contains and determines the sf.)
1590.0 5 (The last 0 counts since after #s and after decimal. This sandwiches the first 0.)
2 (Zeros after numbers but before the decimal usually don’t count.)
320 x 109
14 (exact) Infinite (Exact numbers have infinite sf.)
2500 2 (Zeros at the end before the decimal usually don’t count.)
4200. 4 (The decimal at the end means the 0 before it counts, and first 0 is sandwiched.) 2. a. 5.15 cm (tenths place) 5.2 cm (Round up by both lone 5 rules. Eng: 1 is odd, round 5 up.)
b. 31.84 meters (3 sf) 31.8 meters (3rd digit is last digit: rounding off a 4, round down.)
c. 0.819 mL (hundredths place) 0.82 mL (9 rounds up.)
d. 0.0635 cm2 (2 sf) 0.064 cm2 (Leading zeros never count.. Round to 2nd sf, up by both rules.)
e 0.04070 g (2 sf) 0.041 grams (Zeros in front never count.)
f.
3. a. 6.255 cm (tenths place) 6.3 cm (Rounding a 5 followed by other digits, always round up.)
1.008
+ 1.008
32.00
34.016 = 34.02 b. 17.65
c.
 9.7
7.95 = 8.0 (5 up or 9=odd up) 39.1
+ 124.0
14.05
177.15 = 177.2 4. For help with unit math, see Lesson 2B. For help with exponential math, see Module 1.
a. 13.8612 cm x 2.02 cm = 27.9996 = 28.0 cm2 (3 sf)
b. 4.4 meters x 8.312 meters2 = 36.5728 = 37 meter3 (2 sf, 5 plus other digits, always round up)
c. 2.03 cm2/1. 2 cm = 1.69166 = 1.7 cm (2 sf) d. 0.5223 cm3/0.040 cm = 13.0575 = 13 cm2 (2 sf)
5. a. (2.25 x 10─2)(6.0 x 1023) = 13.5 x 1021 = 1.4 x 1022 in scientific notation (2 sf)
b. (6.022 x 1023) / (1.50 x 10─2) = 4.01 x 1025 (3 sf)
6. a. 2.0646 m + 0.050 m 2.06 m The highest doubt is in the hundredth’s place. Round to that place.
b. 5.04 nm + 0.12 nm 5.0 nm The highest doubt is in the tenth’s place. Round to that place. c. 12.675 g + 0.20 g 12.7 g The highest doubt is in the tenth’s place. Round to that place. d. 24.81 ºC + 1.0 ºC 25 ºC ©2011 www.ChemReview.Net v. u6 The highest doubt is in the one’s place. Round to that place. Page 58 Module 4 – Conversion Factors 7. = 5.60 • 0.090 • 4.00 • (an exact 2) • 10─2 • L2 • g • s─3 • s4 = 0.67 x 103 = 6.7 x 102 g
L
s • L3
6.02
10─5
The 0.090 limits the answer to 2 sf. Exact numbers do not affect sf. For unit cancellation, see Lesson 2D.
Group and handle numbers, exponentials, and units separately.
8. (─ 50.0 x 10―14 g) ─ (─ 49.6 x 10―12 g) + 49.6
x 10―12 g
─ 0.500 x 10―12 g
x 10―12 g
49.1 = = 4.91 x 10―11 g Numbers added or subtracted must have same exponents and units (see Lessons 1B, 2B). Adjusting to
the highest exponent in the series (─12 is higher than ─14) often helps with sf. In moving the decimal
point, do not change the number of sf. Apply the rules for sf rounding at the end of a calculation. *** * * Lesson 3D: The Atoms – Part 2
To continue to learn the atoms encountered most often, your assignment is:
• For the 20 atoms below, memorize the name, symbol, and position in this table. For
each atom, given its symbol or name, be able to write the other. • Be able to fill in an empty table of this shape with those names and symbols in their
proper places. (The “atomic numbers” shown above each symbol are optional.) Periodic Table
1A 2A 3A 4A 5A 6A 7A 1 8A
2 H He Hydrogen
3 Helium
4 Li
Lithium
11 5 Be 7 B Beryllium Carbon 13 Na Mg 14 Al Sodium Magnesium 8 C Boron 12 19 6 N
Nitrogen
15 Si Aluminum 9 O
Oxygen
16 P 10 F
Fluorine
17 S Silicon Phosphorus Sulfur Ne
Neon
18 Cl Ar Chlorine Argon 20 K Ca Potassium Calcium ##### ©2011 www.ChemReview.Net v. u6 Page 59 Module 4 – Conversion Factors Module 4 – Conversion Factors
Prerequisites: Module 4 requires knowledge of exponential math and metric
fundamentals in Lessons 1A, 1B, 2A, 2B, 3A and 3B. The other lessons in Modules 13 will
be helpful, but not essential, for Module 4. Pretests: If the use of conversion factors is easy review, try the last two problems in each
lesson. If you get those right, skip the lesson. If they are not easy, complete the lesson.
***** Lesson 4A: Conversion Factor Basics
Conversion factors can be used to change from one unit of measure to another, or to find
equivalent measurements of substances or processes. A conversion factor is a ratio (a
fraction) made from two measured quantities that are equal or equivalent in a problem. A
conversion factor is a fraction that equals one.
Conversion factors have a value of unity (1) because they are made from equalities. For any
fraction in which the top and bottom are equal, its value is one.
For example: 7 =1
7
Or, since 1 milliliter = 10─3 liters ; 10─3 L = 1
1 mL and 1 mL =1 10─3 L These last two fractions are typical conversion factors. Any fraction that equals one rightside up will also equal one upside down. Any conversion factor can be inverted (flipped
over) for use if necessary, and it will still equal one.
When converting between liters and milliliters, all of these are legal conversion factors:
1 mL
10─3 L 1000 mL
1L 103 mL
1L 3,000 mL
3L All are equal to one. All of those fractions are mathematically equivalent because they all represent the same
ratio. Upside down, each fraction is also a legitimate conversion factor, because the top and
bottom are equal, and its value is one.
In solving calculations, the conversions that are preferred if available are those that are made
from fundamental definitions, such as “milli = 10─3.” However, each of the four forms
above is legal to use in converting between milliliters and liters, and either of the first three
forms may be encountered during calculations solved in science textbooks.
If a series of terms are equal, any two of those terms can be used as a conversion factor.
Example: Since 1 meter = 10 decimeters = 100 centimeters = 1000 millimeters
Then each of the following (and others) is a legitimate conversion factor:
1000 mm
1m ©2011 www.ChemReview.Net 1 mm
10─3 m v. u6 1 0 2 cm
1m 100 cm
10 dm 10 cm
1 dm Page 60 Module 4 – Conversion Factors Let’s try an example of conversionfactor math. Try the following problem. Show your
work on this page or in your problem notebook, then check your answer below.
7.5 kilometers • 103 meters =
1 kilometer Multiply
*****
Answer ( * * * mean: cover below, write your answer, then check below.)
7.5 kilometers • 103 meters = (7.5 • 103) meters = 7.5 x 103 meters
1 kilometer
1 When these terms are multiplied, the “like units” on the top and bottom cancel, leaving
meters as the unit on top.
Since the conversion factor multiplies the given quantity by one, the answer equals the given
amount that we started with. This answer means that 7,500 meters is the same as 7.5 km.
Multiplying a given quantity by a conversion factor changes the units that measure the
quantity but does not change its amount. The result is what we started with, measured in
different units.
This process answers a question posed in many science problems: From the units we are
given, how can we obtain the units we want?
Our method of solving calculations will focus on finding equal or equivalent quantities.
Using those equalities, we will construct conversion factors to solve problems.
***** Summary
• Conversion factors are made from two measured quantities that are either defined
as equal or are equivalent or equal in the problem. • Conversion factors have a value of one, because the top and bottom terms are
equal or equivalent. • Any equality can be made into a conversion (a fraction or ratio) equal to one. • When the units are set up to cancel correctly, given numbers and units, multiplied
by conversions, will result in the WANTED numbers and units.
Units tell you where to write the numbers to solve a calculation correctly. To check the metric conversion factors encountered most often, use these rules.
Since conversions must be equal on the top and bottom, the equalities that define
the metric prefixes can be used to write and check conversions.
• 1 milliunit or 1 m(unit abbreviation) must be above or below 10─3 units ;
• 1 centi or 1 c must be above or below 10─2 ;
• 1 kilo or 1 k must be above or below 103 . ©2011 www.ChemReview.Net v. u6 Page 61 Module 4 – Conversion Factors Practice: Try every other lettered problem. Check your answers frequently. If you miss
one on a section, try a few more. Answers are on the next page.
1. Multiply the conversion factors. Cancel units that cancel, then group the numbers and
do the math. Write the answer number and unit in scientific notation.
a.
b. 10─2 gram •
1 centigram 225 centigrams • • 1.5 hours • 60 minutes
1 hour 1 kilogram = 103 grams 60 seconds =
1 minute 2. To be legal, the top and bottom of conversion factors must be equal. Label these
conversion factors as legal or illegal. e. c. 1.00 g H2O
1 mL H2O b. 1000 L
1 mL a. 1000 mL
1 liter f. 103 cm3
1L 1 mL
1 cc g. 103 kilowatts
1 watt d. h. 10─2 volt
1 centivolt 1 kilocalorie
103 calories 3. Place a 1 in front of the unit with a prefix, then complete the conversion factor.
a. grams
kilograms b. mole
nanomole c. picocurie
curie 4. Add numbers to make legal conversion factors, with at least one of the numbers in each
conversion factor being a 1.
a. b. centijoules
joules liters
cubic cm cm3 c.
mL 5. Finish these.
a. 27A • 2T • 4W
8A
3T b. 2.5 meters • =
= 1 cm
10─2 meter c. 95 km •
hour 0.625 miles
1 km d. 60 s • 1 kilometer =
27 meters •
seconds
1 min.
103 meters ©2011 www.ChemReview.Net v. u6 = Page 62 Module 4 – Conversion Factors ANSWERS
1 10─2 gram
1 centigram a. 225 centigrams • • 1 kilogram = 225 x 10─2 x 1 kg = 2.25 x 10─3 kg
103 grams
1 x 103 The answer means that 2.25 x 10─3 kg is equal to 225 cg.
b. 1.5 hours • 60 minutes • 60 seconds = 1.5 x 60 x 60 s = 5,400 s or 5.4 x 103 s
1 hour
1 minute
1
Recall that s is the abbreviation for seconds. This answer means that 1.5 hours is equal to 5,400 s.
1000 L
1 mL Legal
e. 1 mL_
1 cc c. 1.00 g H2O
1 mL H2O Illegal b. 2. a. 1000 mL
1 liter Legal IF liquid water f. 103 cm3
1L Legal g. 103 kilowatts
1 watt Legal 103 grams
1 kilogram 4. a. 1 centijoule or 100 centijoules
10─2 joules
1 joule b. Legal Legal 10─9 mole
1 nanomole
b. 10─2 volt___
1 centivolt h. 1 kilocalorie_
103 calories Illegal 3. a. d. c. 1 picocurie
10─12 curie
1 liter or
1000 cc. 10─3 liters
1 cubic cm c. 1 cm3
1 mL Either fixed decimal numbers (such as 100) or equivalent exponentials (102 ) may be used in conversions.
5. a. 27A • 2T • 4W
8A
3T = 27A • 2T • 4W
3T
8A = 27 • 2 • 4 • W = 9W
8• 3 = 2.5 x 102 cm = 250 cm b. 2.5 meters • c. 95 km • 0.625 miles = 95 • 0.625 mi = 59 miles
1
hr
hour
hour
1 km d. 27 meters •
seconds 1 cm
─2 meter
10 60 seconds
1 min • 1 kilometer = 27 • 60
103 meters
103 km = 1.6 km
min
min ***** Lesson 4B: Single Step Conversions
In the previous lesson, conversion factors were supplied. In this lesson, you will learn to
make your own conversion factors to solve problems. Let’s learn the method with a simple
example.
Q. How many years is 925 days?
In your notebook, write an answer to each step below. ©2011 www.ChemReview.Net v. u6 Page 63 Module 4 – Conversion Factors Steps for Solving with Conversion Factors
1. Begin by writing a question mark (?) and then the unit you are looking for in the
problem, the answer unit.
2. Next write an equal sign. It means, “OK, that part of the problem is done. From here
on, leave the answer unit alone.” You don’t cancel the answer unit, and you don’t
multiply by it.
3. After the = sign, write the number and unit you are given (the known quantity).
*****
At this point, in your notebook should be ? years = 925 days 4. Next, write a • and a line _______ for a conversion factor to multiply by.
5. A key step: write the unit of the given quantity in the denominator (on the bottom) of
the conversion factor. Leave room for a number in front.
Do not put the given number in the conversion factor  just the given unit.
? years = 925 days • ______________
days
This step puts the given unit where it must be to cancel and tells you one part of what
the next conversion must include.
6. Next, write the answer unit on the top of the conversion factor.
? years = 925 days • year
days 7. Add numbers that make the numerator and denominator of the conversion factor equal.
In a legal conversion factor, the top and bottom quantities must be equal or equivalent.
8. Cancel the units that you set up to cancel.
9. If the unit on the right side after cancellation is the answer unit, stop adding
conversions. Write an = sign. Multiply the given quantity by the conversion factor.
Write the number and the uncanceled unit. Done!
Finish the above steps, then check your answer below.
*****
? years = 925 days • 1 year
= 925 years = 2.53 years
365
365 days (SF: 1 is exact, 925 has 3 sf, 365 has 3 sf (1 yr. = 365.24 days is more precise), round to 3 sf.)
You may need to look back at the above steps, but you should not need to memorize them.
By doing the following problems, you will quickly learn what you need to know. Practice: After each numbered problem, check your answers at the end of this lesson.
Look back at the steps if needed.
For the problems in this practice section, write conversions in which one of the numbers (in
the numerator or the denominator) is a 1.
©2011 www.ChemReview.Net v. u6 Page 64 Module 4 – Conversion Factors If these are easy, do every third letter. If you miss a few, do a few more.
1. Add numbers to make these conversion factors legal, cancel the units that cancel,
multiply the given by the conversion, and write your answer.
a. ? days = 96 hours • day
=
24 hours b. ? mL = 3.50 liters • 1 mL
liter = 2. To start these, put the unit of the given quantity where it will cancel. Then finish the
conversion factor, do the math, and write your answer with its unit.
a. ? seconds = 0.25 minutes • sec. = 1
kilogram = b. ? kilograms = 250 grams •
10 3
c. ? days = 2.73 years • 365 d. ? years = 200. days • 1 =
= 3. You should not need to memorize the written rules for arranging conversion factors,
however, it is helpful to use this “single unit starting template.”
When solving for single units, begin with
? unit WANTED = number and UNIT given • _________________
UNIT given
The template emphasizes that your first conversion factor puts the given unit (but not
the given number) where it will cancel.
a. ? months = 5.0 years • b. ? liters = 350 mL • = = c. ? minutes = 5.5 hours ©2011 www.ChemReview.Net v. u6 Page 65 Module 4 – Conversion Factors 4. Use the starting template to find how many hours equal 390 minutes.
? 5. ? milligrams = 0.85 kg • gram • _____________ =
kg
gram ANSWERS
Some but not all unit cancellations are shown. For your answer to be correct, it must include its unit.
Your conversions may be in different formats, such as 1 meter = 100 cm or 1 cm = 10─2 meters, as long as
the answer is the same as below.
1. a. ? days = 96 hours • 1 day
24 hours b. ? mL = 3.50 liters • = 96 days = 4.0 days
24
= 3.50 • 103 mL = 3.50 x 103 mL 1 mL 10─3 liter (SF: 3.50 has 3 sf, prefix definitions are exact with infinite sf, answer is rounded to 3 sf)
2. Your conversions may be different (for example, you may use 1,000 mL = 1 L or 1 mL = 10─3 L), but
you must get the same answer.
60 sec. = 0.25 • 60 sec. = 15 s
1 minute
(SF: 0.25 has 2 sf, 1 min = 60 sec. is a definition with infinite sf, answer is rounded to 2 sf) a. ? seconds = 0.25 minutes • b. ? kilograms = 250 grams • 1 kilogram
103 grams = 250 kg = 0.25 kg
103 c. ? days = 2.73 years • 365 days
1 year = 2.73 • 365 days = 996 days d. ? years = 200. days • 1 year
365 days = 200 years = 0.548 years
365 = 60. months
12 months
1 year
(SF: 5.0 has 2 sf, 12 mo. = 1 yr. is a definition with infinite sf, round to 2 sf , the 60. decimal means 2 sf) 3. a. ? months = 5.0 years • b. ? liters = 350 mL • 10─3 liter
1 mL = 350 x 10─3 liters = 0.35 L (m = milli = 10─3. SF: 350 has 2 sf, prefix definitions are exact with infinite sf, round to 2 sf)
c. ? minutes = 5.5 hours • 60 minutes
1 hour ©2011 www.ChemReview.Net v. u6 = 330 minutes Page 66 Module 4 – Conversion Factors 4. ? hours = 390 minutes • 1 hour
=
60 minutes 6.5 hours 1 mg
= 0.85 x 106 mg = 8.5 x 105 mg
5. ? milligrams = 0.85 kg • 103 gram •
1 kg
10─3 gram
***** Lesson 4C: MultiStep Conversions
In Problem 5 at the end of the previous lesson, we did not know a direct conversion from
kilograms to milligrams. However, we knew a conversion from kilograms to grams, and
another from grams to milligrams.
In most problems, you will not know a single conversion from the given to wanted unit, but
there will be known conversions that you can chain together to solve the problem.
Try this twostep conversion, based on Problem 5 above. Answer in scientific notation.
Q. ? milliseconds = 0.25 minutes *****
A. ? milliseconds = 0.25 minutes • 60 s • 1 ms = 15 x 103 ms = 1.5 x 104 ms
1 min.
10─3 s The 0.25 has two sf, both conversions are exact definitions that do not affect the significant
figures in the answer, so the answer is written with two sf.
The rules are, when Solving With Multiple Conversions • If the unit on the right after you cancel units is not the answer unit, get rid of it.
Write it in the next conversion factor where it will cancel. • Finish the next conversion with a known conversion, one that either includes the
answer unit, or gets you closer to the answer unit. • In making conversions, set up units to cancel, but add numbers that make legal
conversions. ©2011 www.ChemReview.Net v. u6 Page 67 Module 4 – Conversion Factors Practice
These are in pairs. If Part A is easy, go to Part A of the next question. If you need help with
Part A, do Part B for more practice.
gram • ______________ = 1. a. ? gigagrams = 760 milligrams • b. ? cg = 4.2 kg • • _______________ =
g 2. a. ? years = 2.63 x 104 hours • • = hr • b. ? seconds = 1.00 days • • = g H2O(l) • _____________ = 3. a. ? µg H2O (l) = 1.5 cc H2O(l) • b. ? kg H2O(liquid) = 5.5 liter H2O(l)• • • = ANSWERS
For visibility, not all cancellations are shown, but cancellations should be marked on your paper.
Your conversions may be different (for example, you may use 1,000 mL = 1 L or 1 mL = 10─3 L ), but you
must get the same answer.
1a. ? gigagrams = 760 milligrams • 10─3 g •
1 mg 1 Gg_ = 760 x 10─12 Gg = 7.6 x 10─10 Gg
109 g b. ? cg = 4.2 kg • 103 g • 1 cg = 4.2 x 105 cg
1 kg
10─2 g
2a. ? years = 2.63 x 104 hours • 1 day
24 hr. • 1 yr = 2.63 x 104 = 3.00 years
365 days
24 • 365 b. ? seconds = 1.00 days • 24 hr • 60 min • 60 s
1 min
1 day
1 hr ©2011 www.ChemReview.Net v. u6 = 8.64 x 104 s Page 68 Module 4 – Conversion Factors 3a. ? µg H2O(l) = 1.5 cc H2O(l) • 1.00 g H2O(l) • 1 µg = 1.5 x 106 µg H2O(l)
1 cc H2O(l) b. ? kg H2O(l) = 5.5 liter H2O(l) • 10─6 g 1 mL • 1.00 g H2O(l) • 1 kg = 5.5 = 5.5 kg H2O(l)
1 mL H2O(l)
10─3 L
103 g
100 ***** Lesson 4D: English/Metric Conversions
Using Familiar Conversions
All of the conversions between units that we have used so far have had the number 1 on
either the top or the bottom, but a one is not required in a legal conversion.
Both “1 kilometer = 1,000 meters” and “3 kilometers = 3,000 meters” are true equalities,
and both equalities could be used to make legal conversion factors. In most cases, however,
conversions with a 1 are preferred.
Why? We want conversions to be familiar, so that we can write them automatically and
quickly check that they are correct. Definitions are usually based on one of one component,
such as “1 km = 103 meters.” Definitions are the most familiar equalities and are therefore
preferred in conversions.
However, some conversions may be familiar even if they do not include a 1. For example,
many cans of soft drinks are labeled “12.0 fluid ounces (355 mL).” This supplies an
equality for Englishtometric volume units: 12.0 fluid ounces = 355 mL. That is a legal
conversion and, because its numbers and units are seen often, it is a good conversion to use
because it is easy to remember and check. Bridge Conversions
Science calculations often involve a key bridge conversion between one unit system,
quantity, or substance, and another. To learn the rules for bridge conversions, we will use
the conversions between units in the metric and English measurement systems.
For example, a bridge conversion between metric and Englishsystem distance units is
2.54 centimeters ≡ 1 inch
In countries that use English units, this is now the exact definition of an inch. Using this
equality, we can convert between metric and English measurements of distance.
Any metricEnglish distance equality can be used to convert between distance
measurements in the two systems. Another metricEnglish conversion for distance that is
frequently used (but not exact) is 0.621 mile = 1 km . (When determining the significant
figures, for conversions based on equalities that are not exact definitions, assume that an
integer 1 is exact, but the other number is precise only to the number of sf shown.)
In problems that require bridge conversions, our strategy to will be to “head for the
bridge,” to begin by converting to one of the two units in the bridge conversion. ©2011 www.ChemReview.Net v. u6 Page 69 Module 4 – Conversion Factors When a problem needs a bridge conversion, use these steps.
1) First convert the given unit to the unit in the bridge conversion that is in the same
system as the given unit.
2) Next, multiply by the bridge conversion. The bridge conversion crosses over from
the given system to the WANTED system.
3) Multiply by other conversions in the WANTED system to get the answer unit
WANTED.
Conversions between the metric and English systems provide a way to practice the bridgeconversion methods that we will use in chemical reaction calculations. Add these English
distanceunit definitions to your list of memorized conversions.
12 inches ≡ 1 foot 3 feet ≡ 1 yard 5,280 feet ≡ 1 mile Also commit to memory this metrictoEnglish bridge conversion for distance.
2.54 cm ≡ 1 inch
Then cover the answer below and apply the steps and conversions above to this problem.
Q. ? feet = 1.00 meter *****
Answer
Since the wanted unit is English, and the given unit is metric, an English/metric bridge is
needed.
Step 1: Head for the bridge. Since the given unit (meters) is metric system, convert to
the metric unit used in the bridge conversion (2.54 cm = 1 inch)  centimeters.
? feet = 1.00 meter • 1 cm
10─2 m • _______
cm Note the start of the next conversion. Since cm is not the wanted answer unit, cm must be
put in the next conversion where it will cancel. If you start the “next unit to cancel”
conversion automatically after finishing the prior conversion, it helps to arrange and choose
the next conversion.
Adjust and complete your work if needed.
*****
Step 2: Complete the bridge that converts to the system of the answer: English units.
? feet = 1.00 meter •
* 1 cm
10─2 m **** • 1 inch • __________
2.54 cm
inch Step 3: Get rid of the unit you’ve got. Get to the unit you want.
? feet = 1.00 meter • ©2011 www.ChemReview.Net 1 cm • 1 inch • 1 foot
10─2 m 2.54 cm 12 inches v. u6 = 3.28 feet Page 70 Module 4 – Conversion Factors The answer tells us that 1.00 meter (the given quantity) is equal to 3.28 feet.
Some science problems take 10 or more conversions to solve. However, if you know that a
bridge conversion is needed, “heading for the bridge” breaks the problem into pieces,
which will simplify your navigation to the answer. Practice: Use the inchtocentimeter bridge conversion above. Start by doing every
other problem. Do more if you need more practice.
1. ? cm = 12.0 inches • __________ = 2. ? inches = 1.00 meters • __________ • __________ 3. For ? inches = 760. mm
a. To what unit do you aim to convert the given in the initial conversions? Why?
b. Solve: ? inches = 760. mm 4. ? mm = 0.500 yards
5. For ? km = 1.00 mile , to convert using 1 inch = 2.54 cm ,
a. To what unit to you aim to convert the given in the initial conversions? Why?
b. Solve: ? km = 1.00 mile 6. Use as a bridge for metric mass and English weight units, 1 kilogram = 2.2 lbs.
? grams = 7.7 lbs
7. Use the “soda can” volume conversion (12.0 fluid ounces = 355 mL).
? fl. oz. = 2.00 liters
8. For the following symbols, write the name of the atom.
a. C = _______________ ©2011 www.ChemReview.Net v. u6 b. Cl = ______________ c. Ca = _____________ Page 71 Module 4 – Conversion Factors ANSWERS
In these answers, some but not all of the unit cancellations are shown. The definition 1 cm = 10 mm may be
used for mm to cm conversions. Doing so will change the number of conversions but not the answer.
1. ? cm = 12.0 inches • 2.54 cm = 12.0 • 2.54 = 30.5 cm (check how many cm are on a 12 inch ruler)
1 inch
2. ? inches = 1.00 meters • 1 cm • 1 inch = 1 x 102 = 0.394 x 102 in. = 39.4 inches
2.54 cm
2.54
10─2 m 3a. Aim to convert the given unit (mm) to the one unit in the bridge conversion that is in the same system
(English or metric) as the given. Cm is the bridge unit that is in the same measurement system as mm.
3b. ? inches = 760. mm • 10─3 meter • 1 cm • 1 inch = 760 x 10─1 in. = 29.9 inches
2.54
1 mm
10─2 m 2.54 cm
SF: 760., with the decimal after the 0, means 3 sf. Metric definitions and 1 have infinite sf. The answer
must be rounded to 3 sf (see Module 3).
4. ? mm = 0.500 yd. • 3 ft. • 12 in. • 2.54 cm • 10─2 meter • 1 mm = 457 mm
1 yd. 1 ft.
1 inch
1 cm
10─3 m
5a. Aim to convert the given unit (miles) to the bridge unit in the same system (English or metric) as the given.
Inches is in the same system as miles.
5b. ? km = 1.00 mile • 5,280 ft. • 12 in. • 2.54 cm • 10─2 m • 1 km
1 inch
1 cm
1 mile
1 ft.
103 m = 1.61 km SF: Assume an integer 1 that is part of any equality or conversion is exact, with infinite sf.
6. ? grams = 7.7 lbs • 1 kg • 103 grams
2.2 lb
1 kg = 3.5 x 103 grams SF: 7.7 and 2.2 have 2 sf. A 1 and metricprefix definitions have infinite sf. Round the answer to 2 sf.
7. ? fluid ounces = 2.00 liters • 1 mL • 12.0 fl. oz. = 67.6 fl. oz.
355 mL
10─3 L
8a. C = Carbon b. Cl = Chlorine (Check this answer on
any 2liter soda bottle.) c. Ca = Calcium ***** Lesson 4E: Ratio Unit Conversions
Long Distance Cancellation
The order in which numbers are multiplied does not affect the result. For example,
1 x 2 x 3 has the same answer as 3 x 2 x 1.
The same is true when multiplying symbols or units. While some sequences may be easier
to set up or understand, from a mathematical perspective the order of multiplication does
not affect the answer. ©2011 www.ChemReview.Net v. u6 Page 72 Module 4 – Conversion Factors The following problem is an example of how units can cancel in separated as well as
adjacent conversions. Try
Q1: Multiply. Cancel units that cancel. Write the answer number and its unit.
12 meters • 60 sec. • 60 min. • 1 kilometer • 0.62 miles =
sec.
1 min
1 hour
1000 meters
1 kilometer
*****
12 meters • 60 sec. • 60 min. • 1 kilometer • 0.62 miles = 27 miles
1 min.
1 hour
1000 meters
1 kilometer
hr.
sec.
This answer means that a speed of 12 meters/sec is the same as 27 miles/hour. Ratio Units in the Answer
In these lessons, we will use the term single unit to describe a unit that has one kind of base
unit in the numerator but no denominator (which means the denominator is 1). Single
units measure amounts. Meters, grams, minutes, milliliters, and cm3 are all single units.
These base units may have prefixes or powers, but they must be, or be equivalent to, a unit
that measures a fundamental quantity.
We will use the term ratio unit to describe a fraction that has one base unit in the numerator
and one base unit in the denominator. If a problem asks you to find
meters per second or meters/second or meters
second or m • s─1 all of those terms are identical, and the problem is asking for a ratio unit. During conversion
calculations, all ratio units should be written in the fraction form with a top and bottom.
In Module 11, we will address in detail the different characteristics of single units and ratio
units. For now, the distinctions above will allow us to solve problems. Converting the Denominator
In solving for single units, we have used a starting template that includes canceling a given
single unit.
When solving for single units, begin with
? unit WANTED = # and UNIT given • __________________
UNIT given
When solving for ratio units, we may need to cancel a denominator (bottom) unit to start a
problem. To do so, we will loosen our starting rule to say this. ©2011 www.ChemReview.Net v. u6 Page 73 Module 4 – Conversion Factors When Solving With Conversion Factors
If a unit to the right of the equal sign, in or after the given, on the top or the bottom,
• matches a unit in the answer unit, in both what it is and where it is, circle that
unit on the right side and do not convert it further; • is not what you WANT, put it where it will cancel, and convert until it matches
what you WANT. After canceling units, if the unit or units to the right of the equal sign match the
answer unit, stop adding conversions, do the math, and write the answer.
Q2: Use the rule above to solve.
? cm = 0.50 cm • __________ =
min.
s
*****
Answer
? cm = 0.50 cm
min
s • 60 s
=
1 min. 30. cm
min. Start by comparing the wanted units to the given units.
Since you WANT cm on top, and are given cm on top, circle cm to say, “The top is
done. Leave the top alone.”
On the bottom, you have seconds, but you WANT minutes. Put seconds where it
will cancel. Convert to minutes on the bottom.
When the units on the right match the units you WANT on the left for the answer,
stop conversions and do the math. Practice A: Do Problem 2. Then do Problem 1 if you need more practice. 1. ? g = 355
dL g • __________ =
L 2. ? meters = 4.2 x 105 meters • __________ • __________ =
second
hour ©2011 www.ChemReview.Net v. u6 Page 74 Module 4 – Conversion Factors Converting Both Top and Bottom Units
Many problems require converting both numerator and denominator units. In the
following problem, an order to convert both units is specified. Write what must be placed
in the blanks to make legal conversions, cancel units, do the math, and write then check
your answer below.
Q3. ? meters = 740 cm •
s
min. • minutes = cm *****
Answer
? meters = 740 cm • 10─2 meters •
s
min
1 cm 1 min
60 s = 0.12 meters
s In the given on the right, cm is not the unit WANTED on top, so put it where it will
cancel, and convert to the unit you want on top.
Next, since minutes are on the bottom on the right, but seconds are WANTED, put
minutes where it will cancel. Convert to the seconds WANTED.
When chaining conversions, which unit you convert first — the top or bottom unit —
makes no difference. The order in which you multiply factors does not change the answer.
On the following problem, no order for the conversions is specified. Add legal conversions
in any order, solve, then check your answer below. Before doing the math, double check
each conversion, one at a time, to make sure it is legal.
= Q4. ? centigrams
liter 0.550 x 10─2 g
mL *****
Answer: Your conversions may be in a different order.
? centigrams = 0.550 x 10─2 g •
liter Practice B mL 1 cg •
10─2 g 1 mL = 5.50 x 102 cg 10─3 L L Do every other part, and more if you need more practice. 1. On these, an order of conversion is specified. Write what must be placed in the blanks
to make legal conversions, then solve.
a. ? miles
hour = b. ? meters =
s 80.7 feet •
sec.
250. feet •
min. ©2011 www.ChemReview.Net v. u6 mile • __________ • __________=
min.
min. • ________ • _________ • __________=
1 inch Page 75 Module 4 – Conversion Factors 2. Add conversions in any order and solve.
a. ? km =
hour 1.17 x 104 mm
sec b. ? ng
mL 47 x 102 mg
dm3 feet =
sec. 95 meters
minute c. ? = ANSWERS
Practice A
1. ? g = 355 g
dL
L • 10─1 L
1 dL = 35.5 g
dL 2. ? meters = 4.2 x 105 meters • 1 hour • 1 min
s
hour
60 min
60 s = 1.2 x 102 meters
s Practice B
1a. ? miles
hour = 80.7 feet •
sec. 1 mile • 60 sec. • 60 min. = 55.0 miles
5,280 feet 1 min.
1 hour
hour 1b. ? meters = 250. feet • 1 min. • 12 inches • 2.54 cm • 10─2 meter = 1.27 meters
sec.
min. 60 sec.
1 foot
1 inch
1 cm
s
(SF: 250. due to the decimal has 3 sf, all other conversions are definitions, answer is rounded to 3 sf)
2a. ? km = 1.17 x 104 mm • 10─3 m •
hour
sec
1 mm 1 km
103 m • 60 sec • 60 min = 42.1 km
1 min
1 hour
hr 2b. ? ng = 47 x 102 mg • 1 dm3 • 10─3 L • 10─3 g • 1 ng
mL
dm3
1L
1 mL
1 mg
10─9 g = 4.7 x 106 ng
mL Hint: an English/metric bridge conversion for distance units is needed. Head for the bridge: first convert
the given metric distance unit to the metric distance unit used in your known bridge conversion.
*****
? feet = 95 meters • 1 min • 1 cm • 1 inch • 1 foot = 5.2 feet
2c. sec. min 60 s 10─2 m 2.54 cm 12 in. s ***** ©2011 www.ChemReview.Net v. u6 Page 76 Module 4 – Conversion Factors Lesson 4F: The Atoms – Part 3
To continue to learn the most often encountered atoms, your assignment is:
• For the first 20 atoms, plus the first and last two columns in the periodic table,
memorize the name, symbol, and the position of the atom. For each atom, given either
its symbol or name, be able to write the other. • Be able to fill in a blank table with these names and symbols. ***** Periodic Table
1A 2A 3A 4A 5A 6A 7A 1 8A
2 H He Hydrogen
3 Helium
4 Li Lithium
11 5 Be B Beryllium Boron 12 13 Na Mg 7 C Carbon
14 Al Sodium Magnesium
19 6 Aluminum 8 N Nitrogen
15 Si 9 O Oxygen
16 P 10 F Fluorine
17 S Silicon Phosphorus Sulfur Neon 18 Cl Ar Chlorine Argon 20 K Ne 36 Ca Br Potassium Calcium Bromine Kr
Krypton
54 Rb Sr I Rubidium Strontium Iodine Xe
Xenon
86 Cs Ba
Cesium At Rn Barium Fr Radon Ra Francium Astatine Radium ©2011 www.ChemReview.Net v. u6 Page 77 Module 4 – Conversion Factors Lesson 4G: Review Quiz For Modules 14
Use a calculator and scratch paper, but no notes or tables. Write answers to calculations in
proper significant figures. Except as noted, convert your answers to scientific notation.
To answer multiple choice questions, it is suggested that you
• Solve as if the question is not multiple choice, • Then circle your answer among the choices provided. Set a 20minute limit, then check your answers after the Summary that follows.
1023 1. (From Lesson 1C): = (1.25 x 1010)(4 .0 x 10―6)
a. 2.0 x 1018 b. 5.0 x 1018 c. 0.20 x 1019 d. 2.0 x 1020 e. 5.0 x 10―19 2. (Lesson 1B): (─ 60.0 x 10―16) ─ (─ 4.29 x 10―14) =
a. 4.8 x 10―16 b. 3.69 x 10―14 c. 3.7 x 10―14 d. 4.8 x 10―16 e. 4.89 x 10―14 3. (Lesson 2A): 15 mL of liquid water has what mass in kg?
a. 1.5 x 10―3 kg b. 15 x 10―4 kg c. 1.5 x 10―4 kg d. 1.5 x 10―4 kg e. 1.5 x 10―2 kg 4. (Lesson 2D): 5.00 x 10─2 L3 • m • 2.00 m •
2.0 s3
• (an exact 2) =
─5 L2
s
8.00 x 10
a. 1.00 x 10―4 m2 • s2 • L b. 5.00 x 103 m2 • s2 • L d. 1.0 x 103 m • s2 • L c. 5.0 x 103 m2 • s2 • L e. 5.0 x 10―3 m2 • s2 • L 5. (Lesson 3B): State your answer in proper significant figures
but do not convert to scientific notation.
+
a. 255.00 b. 255.0 c. 255.008 d. 255.1 1 .008
238.00
16.00 e. 255.01 6. (Lesson 4D): If 1 kg = 2.20 lb., solve
? mg = 4.0 x 10─2 lb.
a. 8.8 x 10―7 mg
7. (4E): ? kg
mL b. 8.8 x 104 mg
= c. 1.8 x 10―7 mg d. 1.8 x 104 mg e. 8.8 x 101 mg 2.4 x 105 μg
dm3 a. 2.4 x 10―7 kg b. 2.4 x 105 kg c. 2.4 x 10―10 kg d. 2.4 x 10―5 kg
mL
mL
mL
mL
8. (Lesson 3D): For the following symbols, write the name of the atom.
a. K = __________________ b. S = ______________ e. 2.4 x 10―4 kg
mL c. Na = _____________ *****
©2011 www.ChemReview.Net v. u6 Page 78 Module 4 – Conversion Factors SUMMARY: Conversion Factors
1. Conversion factors are fractions or ratios made from two entities that are equal or
equivalent. Conversion factors have a value of one.
2. An equality can be written as a conversion or fraction or ratio that is equal to one.
3. When solving a problem, first write the unit WANTED, then an = sign.
4. Solving for single units, start conversion factors with
? unit WANTED = # and UNIT given • ________________
UNIT given
5. Finish each conversion factor with the answer unit or with a unit that takes you closer
to the answer unit.
6. In making conversions, set up units to cancel, but add numbers that make legal
conversions.
7. Chain your conversions so that the units cancel to get rid of the unit you’ve got and get
to the unit you WANT.
8. When the unit on the right is the unit of the answer on the left, stop conversion factors.
Complete the number math. Write the answer and its unit.
9. Units determine the placement of the numbers to get the right answer.
10. If you plan on a career in a sciencerelated field, add these to your flashcard collection.
Frontside (with notch at top right): Back Side  Answers 1 inch = ? cm
1 kg = ? pounds
12 fluid oz. = ? mL 2.54 cm
2.2 lb.
355 mL ***** ANSWERS – Module 14 Review Quiz
Only partial solutions are provided below.
1. a. 2.0 x 1018 1/5 x 1023―10+6 = 0.20 x 1019 = 2.0 x 1018 2. b. 3.69 x 10―14 (+ 4.29 x 10―14) ─ (0.600 x 10―14) = net doubt in hundredths place 3. e. 1.5 x 10―2 kg 1.00 g H2O = 1 mL H2O ; 1.00 kg H2O = 1 L H2O 4. c. 5.0 x 103 m2 • s2 • L (2 sf ) 5. e. 255.01 (Adding and subtracting, round to highest place with doubt.) 6. d. 1.8 x 104 mg 7. a. 2.4 x 10―7 kg/mL 8a. K = Potassium b. S = Sulfur L = dm3 c. Na = Sodium ##### ©2011 www.ChemReview.Net v. u6 Page 79 ...
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This note was uploaded on 11/16/2011 for the course CHM 1025 taught by Professor J during the Summer '09 term at Santa Fe College.
 Summer '09
 J
 Chemistry

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