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# Gen Chemistry I CHM 1045

FIU

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This 27 page Class Notes was uploaded by Jesus Koepp on Monday October 12, 2015. The Class Notes belongs to CHM 1045 at Florida International University taught by Uma Swamy in Fall. Since its upload, it has received 19 views. For similar materials see /class/221825/chm-1045-florida-international-university in Chemistry at Florida International University.

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Date Created: 10/12/15

CHM 1045 Notes Chapter 1 Dr Palmer Graves Chemistry A Molecular Approach 1St Ed Nivaldo Tro Chapter 1 Matter Measurement and Problem Solving Palmer Graves Florida International University 2008 Prentice Hall 11 Identify the state and classify mixtures and compounds Readings 1 1 14 Problems 137 39 47 49 amp 50 Structure Determines Properties 0 the properties of matter are determined by the atoms and molecules that compose it carbon monoxide carbon dioxide 1 composed of one carbon atom l composed of one carbon atom and one oxygen atom and two oxygen atoms 2 colorless odorless gas 2 colorless odorless gas 3 burns with a blue ame 3 incombustible 4 binds to hemoglobin 4 does not bind to hemoglobin ygcn Oxygen atom atom Carbon Oxygen atom atom 1 Carbon Mum CHM 1045 Dr Palmer Notes Graves Atoms and Molecules I atoms are submicroscopic particles are the fundamental building blocks of all matter I molecules two or more atoms attached together 0 attachments are called bonds 0 attachments come in different strengths molecules come in different shapes and patterns Chemistry is the science that seeks to understand the behavior of matter by studying the behavior of atoms and molecules Chapter 1 Classifying Matter by Physical State 0 matter can be classi ed as solid liquid or gas based on the characteristics it exhibits Solids the particles in a solid are packed close together and are xed in position 7 though they may vibrate solids are incompressible pl q l solids retain their shape and volume solids don t easily ow Solid maucr CHM 1045 Notes Dr Palmer Graves Chapter 1 Crystalline Solids 0 crystalline solids particles are arranged in an orderly geometric pattern rwmllinc mel h particles are randomly distributed Without any long range pattern I 0 salt and diamonds l at3939 i Amorphous Sollds 0 amor hous solids 39 P on o 1 Plastic O Glass 7 charcoal i Charcoal r l mwv mum L1qu1ds particles are closely packed but they have some ability to move around liquids are incompressible liquids take the shape of their container and ow however they don t have enough freedom to escape and expand to ll the container l Liquidlnaller l CHM 1045 Notes Dr Palmer Graves Gases I particles have complete freedom from each other I the particles are constantly ying around bumping into each other and the container I in the gas state there is a lot of empty space between the particles Gaseous muncr C490 Chapter 1 Gases I gases are compressible I gases expand to ll and take the shape of their container I gas particles will ow Gas compressible Classi cation of Matter by Composition I pure substance composition is constant from one sample to another I made of a single type of atom or molecule all samples have the same characteristics I mixture composition will vary from one sample to another two or more types of atoms or molecules combined in variable proportions samples have the different characteristics CHM 1045 Notes Chapter 1 Dr Palmer Graves Classi cation of Matter by Composition a Iable 0 tie NO V r mpDSl n Pure Substances Yes Mixture l Classi cation of Pure Substances elements cannot be broken down into simpler substances by chemical reactions basic building blocks of matter composed of single type of atom compounds can be decomposed chemical combinations of elements composed of molecules that contain two or more different ldnds all molecules of a compound are identical7 so all samples of a compound behave the same way most natural pure substances are compounds Classi cation of Pure Substances l Pun Suhslanzesj Seuamble mm simple No substauws at r 7 l Elemem ompound V Helium PM water CHM 1045 Notes Chapter 1 Dr Palmer Graves Changes in Matter 0 physical changes alter the state or appearance of the matter Without altering the composition 0 chemical changes alter the composition of the matter during the chemical change the atoms that are present rearrange into new molecules all of the original atoms are still present Physical Changes in Matter Wave molecules hang31mm mm m gaseous 5m 1h nge Common Physical Changes Subliming of Dry Ice I processes that cause changes in the matter that do not change its composition state changes boiling condensing melting freezing subliming CHM 1045 Notes Chapter 1 Dr Palmer Graves Chemical Changes in Matter mm unlm Rusting iron is a a chemical change iquot 7 Iron atoms in the nail combine with oxygen atoms from O2 in the air Makes a new i J substance rust with a gt different composition Irnn midc i rm Common Chemical Changes processes that cause 39u 0 changes in the matter 0 G that change its 4 v b composition Q 393 J processes that release b 2 lots of energy 8 3y gt gt 4 0 burn1ng 7g 3 C3H8g 5 023 4 3 3023 4 H200 12 Express measured and calculated quantities in exponential form Correctly enter exponential notation in your calculator Readings Appendix LA Problems 1 95 amp 96 CHM 1045 Notes Chapter 1 Dr Palmer Graves Exponential Notation Appendix A 1 Express very large or very small numbers as a number between 1 and 10 times a power of 10 Example the number of atoms of carbon in a 12g sample is 600000000000000000000000 In exponential notation 6 x 1023 The weight of one carbon atom is z 000000000000000000000002g In exponential notation 2 x 103923g 13 Recognize SI units and give the numerical equivalent of the metric pre xes Readings 1 6 Problems 151 60 The Standard Units 0 Scientists have agreed on a set of international standard units for comparing all our measurements called the SI units 7 Syrt me International International System CHM 1045 Notes Dr Palmer Graves TABLE 11 SI Base Units Quantity Unit Symbol Length Meter m Mass Kilogram kg Time Second 5 Temperature Kelvin K Amount of substance Mole mo Electric current Ampere A Luminous intensity Candela cd Copyrighl 2005 Pearson Prentice Haw inc Chapter 1 Length 0 Measure of the two dimensional distance an object covers often need to measure lengths that are very long distances between stars or very short distances between atoms 0 SI unit meter About 39 37 inches longer than a yard Commonly use centimeters cm 1 m 100 cm 1cm001m10mm 1 inch 2 54 cm exactly quot Yurdstick 39 V 39 quot Mctcrstick 5mm zuJn pawn munM Hall Iquot Mass 0 Measure of the amount of matter present in an object 7 weight measures the gravitational pull on an object which depends on its mass SI unit kilogram kg 7 about 2 2 lbs MP Commonly measure mass in grams g or milligrams mg 7 1 kg 2 2046 pounds 1 lbs 453 59 g 7 1 kg 1000 g 103 g 7 lg 1000 mg 103 mg 71g0001kg10393 kg 7 1 mg 0001 g 10393g CHM 1045 Notes Dr Palmer Graves Mass I Measure of the amount of matter present in an object weight measures the gravitational pull on an object Which depends on its mass I SI unit 2 kilogram kg about 2 2 lbs I Commonly measure mass in grams g or milligrams mg Chapter 1 Time I measure of the duration of an event I SI units second s I 1 s is de ned as the period of time it takes for a speci c number of radiation events of a speci c transition from cesium 133 Temperature I measure of the average amount of ldnetic energy 7 higher temperature larger average kinetic energy Tro Chemistry A Molecular Approach in r 3mmquotvMmeMVth lv 10 CHM 1045 Notes Chapter 1 Dr Palmer Graves Temperature Scales Waler hails Fahrenheit Scale F 7 used in the U S Celsius Scale C 7 used in all other countries Kelvin Scale K 7 absolute scale WalEr lreezes I no negative numbers 7 directly proportional to average amount of kinetic x r v r r energy n r Absolute zero 7 0 K absolute zero FANIEK IHEH alarm Neler camngm r was V earson Prentice mm W Fahrenheit vs Celsius 0 a Celsius degree is 18 times larger than a Fahrenheit degree 0 the standard used for 0 on the Fahrenheit scale is a lower temperature than the standard used for 0 on the Celsius scale Kelvin vs Celsius 0 the size of a degree on the Kelvin scale is the same as on the Celsius scale though technically we don t call the divisions on the Kelvin scale degrees we called them kelvins so 1 kelvin is 18 times larger than 1 F the 0 standard on the Kelvin scale is a much lower temperature than on the Celsius scale ll CHM 1045 Notes Dr Palmer Graves Example 12 Convert 4000 C into K and F Chapter 1 Find the equation that relates the given quantity to the quantity you want to nd Given Find Equation 4000 C K K C 273 15 Since the equation is solved for the quantity you want to nd substitute and compute Find the equation that relates the given quantity to the quantity you want to nd Given Find Equation 40 00 C F Solve the equation for the quantity you want to nd Substitute and compute Related Units in the SI System standard unit by a power of 10 regardless of the standard un1t All units in the SI system are related to the The power of 10 is indicated by a pre x multiplier The pre x multipliers are always the same Report measurements with a unit that is close to the size of the quantity being measured TABLE 12 S Prefix Multipliers Prefix Symbol Multiplier cxa E l0110000010000000000 pcla P 100000000000000 tcra T l00000000000 giga 0 1000000000 mega M llO00 kiln k ll0 dcci d 01 centi c Mil milli m 0001 micro 0 0000001 ham 1 0000000001 pito p 0000000000001 lL mto f l0000000000000l arm a 000000000lI00l0l000l Copyright 2005 Pearson Prennce Hair Inc 12 CHM 1045 Notes Dr Palmer Graves Using Metric Pre xes 0 Replace pre x with the exponential form 0 m 10 3 0 253 mm m 0 Practice with the concept 15 pm 2 m 606 MHz 2 Hz 0 25 x 10 6 scopes Chapter 1 Volume Derived unit 4 L x W x h 7 llhm 41 Commonly measure solid volume in cubic I 39 7 centlmeters cm quotu39 u 1 m3 106 cm3 Mu 1 cm3 1061113 0 000001 m3 quot quotquotI39liigil IL39 gt V Commonly measure liquid or gas volume in milliliters mL 1 L is slightly larger than 1 quart 0 quotWW5 1mL10m3 Tro Charm stry A Molecular 38 Approach A101 tube animus Common Units and Their Equivalents Length 1 kilometer km 2 06214 mile mi 1 meter m 3937 inches in 1 meter m 1094 yards yd 1 foot ft 3048 centimeters cm 1 inch in 254 centimeters cm exactly 13 CHM 1045 Notes Chapter 1 Dr Palmer Graves Common Units and Their Equivalents Mass 1 kilogram km 2205 pounds lb 1 pound lb 45359 grams g 1 ounce oz 2835 grams g Volume 1 liter L 1000 milliliters mL 1 liter L 1000 cubic centimeters cm3 1 liter L 1057 quarts qt 1 Us gallon gal 3785 liters L 14 Calculate density from mass and volume and use density to calculate either mass or volume Readings 1 6 Problems 161 66 amp 117 Mass amp Volume 0 two main physical properties of matter 0 mass and volume are extensive properties 7 the value depends on the quantity of matter 7 extensive properties cannot be used to identify what type of matter something is if you are given a large gls containing 100 g ofa clear colorless liquid and asmall gls containing 25 g ofa clear colorless liquid 7 are both liquids the same stu 0 even though mass and volume are individual properties for a given type of matter they are related to each other CHM 1045 Notes Chapter 1 Dr Palmer Graves Volume vs Mass of Brass y838x Mass 9 nn 2 u an n 1 u an 1 u an an Volume cm3 Density Ratio of massV01ume is an intensive property value independent of the quantity of matter Solids gcm3 1 cm3 2 1 mL I 0 Liquids gmL lb 0 Gases gL Volume of a solid can be determined by water displacement Archimedes Principle Density solids gt liquids gtgtgt gases except ice is less dense than liquid water hquot TABLELA The Densin ufSume Common Substances at 20 C Substance Denle lgcm 0 For equal volumes denser Mm w object has larger mass M Mmquot W l 21 0 For equal masses denser object has smaller volume 0 Heating an object generally causes it to expand therefore the density changes with temperature Cnnvngm 12mm Pearson Pveane Hall in 15 CHM 1045 Notes Dr Palmer Graves Example 13 Decide if a ring with a mass of3 15 g that displaces 0233 cm3 of water is platinum Chapter 1 Find the equation that relates the given quantity to the quantity you want to nd Given Find Equation mass 3 15 g volume 0 233 cm3 density gcm3 Since the equation is solved for the quantity you want to nd and the units are correct substitute and compute Compare to accepted value of the intensive property platinum 15 Express calculated quantities in the proper number of signi cant gures Readings 17 Questions 177 82 What Is a Measurement 0 quantitative observation 0 every measurement has a number and a unit Meniscus 16 CHM 1045 Notes Chapter 1 Dr Palmer Graves A Measurement the unit tells you What standard you are comparing your object to 0 the number tells you The amount The uncertainty 0 every digit reported is certain except the last one which is an estimate Estimating the Last Digit 0 Analog instruments the last digit is an estimate of the space between the marks if possible 0 mentally diVide the space into 10 equal spaces then estimate how many spaces over the indicator mark is Estimation in Weighing Markings may 01 g I39Mimmcd reading 127 g Markings every 1 g Fsiimuled reading 12 g Cupyughl w 2005 Penman Prentice Hall in l7 CHM 1045 Notes Chapter 1 Dr Palmer Graves Uncertainty in Measured Numbers 0 uncertainty is due to limitations of 39 experimental design and random effects 0 limitations of the measurement 0 accuracy 7 how close a measurement comes to the actual value of the quantity 0 precision 7how quot a quot r Precision 0 random errors 7 errors that result from random uctuations 7 no speci c cause therefore cannot be corrected 0 every measurement has some random error 0 errors average out with a large number of measurements Accuracy systematic errors 7 limitations in the instruments technique or experimental design 7 reduced by using more accurate instruments better technique or better experimental design 0 we determine the accuracy of a measurement by comparison to a standard value 0 Who Cares CHM 1045 Notes Dr Palmer Graves MM g1 Accuracy vs Precision Immum implcilsu Inuczuvawqirczin39 Munu mau Studenx A Sludent c 5m n eapyngm oznus Pearsnn Prentice Hall in Chapter 1 Rules for Signi cant Figures 0 1 All digits 1 through 9 are signi cant 1562 has four signi cant digits 0 2 Leading zeros are not signi cant 000015 has only two sig gs 0 3 Con ned zeros are signi cant 3015 has four sig gs 0 4 Trailing zeros are signi cant 35000 has ve sig gs Example 15 Determining the Number of Signi cant Figures in a Number How many significant figures are in each of the following 0 04450 m 5 0003 km 10 dm 1 m 1 000 x 105 s 0 00002 mm 10000 m 19 CHM 1045 Notes Chapter 1 Dr Palmer Graves Rules for Sig Figs Multiplication amp Division Result has number of sig gs as least precise measurement Example 1563 55 284181818182 4sig gs 2sig gs 2 sig gs 28 Addition and Subtraction with Signi cant Figures when adding or subtiacting meauremenls with signi cant gures the result ha the same number of decimal places a the meaurement with the fewest number of decimal places 574 0823 26519214921 48 a 3965 0835 08 Rounding Off In a series of calculations mampd round at the nal ste p To round consider only the rst unknown If number to remove is less than ve drop If to remove is ve or more round up Examples round to 2 sig gs 254 2548 255 256 round to correct of sig gs 2654 55 7 35662 7 3482 4668 20 CHM 1045 Notes Chapter 1 Dr Palmer Graves Both MultiplicationDivision and Addition Subtraction with Signi cant Figures when doing different kinds of operations with measurements uu irate er evaluate the signi cant gures in the intermediate answer then do the remaining steps 489 x 557 7 23 16 Use Dimensional Analysis to convert quantities among different unit systems Readings 18 Problems 83 94 Units 0 Always write every number with its associated unit 0 Always include units in your calculations a you can do the same kind of operations on units as you can with numbers Iemxcmem2 I cm cm 1 7 using units as a guide to problem solving is called dimem39iomtl analyrix 21 CHM 1045 Notes Chapter 1 Dr Palmer Graves Problem Solving and Dimensional Analysis I Many problems in chemistry involve using relationships to convert one unit of measurement to another Conversion factors are relationships between two units Conversion factors generated from equivalence statements eg1mch2540mcang1ve I I Ii I III Dimensional Analysis 0 Goal Quantity given X conversion factor 0 Conversion Factors Factors equal to l Dimensional Analysis 0 Goal Quantity given X conversion factor 0 Conversion Factors Factors equal to l III l I h H111 22 CHM 1045 Notes Chapter 1 Dr Palmer Graves Problem Solving and Dimensional Analysis 0 Arrange conversion factors so given unit cancels Arrange conversion factor so given unit is on the bottom of the conversion factor 0 May string conversion factors Trial and error to nd something the given and desired units are related to H I Conceptual Plan 0 a conceptual plan visual outline that shows the strategic route required to solve a problem 0 for unit conversion focus on units and how to convert one to another 0 for problems that require equations the conceptual plan focuses on solving the equation to nd an unknown value Concept Plans and Conversion Factors 0 Convert inches into centimeters Find relationship equivalence 1 in 254 cm Write concept plan Change equivalence into conversion factors with starting units on the bottom 23 CHM 1045 Notes Chapter 1 Dr Palmer Graves Systematic Approach I Sort the inforrmtion from the problem 7 identify the given unit the goal unit any relationships 6P P sorneurnes rnay ant to work backwards each step involves a conversion factor or equauon I Apply the steps in the concept plan 7 check that units cancel properly 4 multiply terms across the top and divide by each bo om term I Check the answer 7 double check the setup to ensure the unit at the end is the one you wished to nd 4 check to see that the size of the number is reasonable since cenurneters are smaller than rnches convemng inches to centimeters should result in a larger number Example 17 Convert 176 yd to centimeters informaion nd solve the Dimensional Analysis I Multirstep conversions I Convert 183 ft to meters I ConceptPlan I 12in 1ft 1 in 254 cm amp 100 cm 1m 24 CHM 1045 Notes Dr Palmer Graves Practice Convert 300 mL to quarts 1 L 1057 qt Chapter 1 Concept Plans for Units Raised to Powers Convert cubic inches into cubic centimeters Find relationship equivalence 1 in 254 cm Write concept plan 3 Change equivalence into conversion factors With given unit on the bottom Example 19 Convert 570 L to cubic inches informaion nd in3 L 1mL1cm31mL10393L 1cm254m concept plan to solve the 25 CHM 1045 Notes Chapter 1 Dr Palmer Graves Practice 19 How many cubic centimeters are there in 21 1 yd3 Density as a Conversion Factor 0 can use density as a conversion factor between mass and volume 7 density ofHZO10 gmL 10gHZOlmLHZO 7 density obe 113 gcm3 113 g Pb 1 cm3 Pb How much does 40 cm3 of lead weigh 40 cm3 Pb x 11393 Pb 45 ng 1 cm3 Pb Example 110 What is the mass in kg of 173231 L informaion density 0 738 gm lmL0738g1mL10393L 1 concept plan to solve the 133x105kg 26 CHM 1045 Notes Dr Palmer Graves Problem Solving With Equations 0 When solving a problem involves using an equation the concept plan involves being given all the variables except the one you want to nd 0 Solve the equation for the variable you Wish to nd then substitute and compute Chapter 1 Using Density in Calculations Concept Plans a I llI 4 Example 112 Find the density of a metal cylinder with mass 83 g length 194 cm and radius 055 cm 0 Sort Given m 8 3 g information I 1 94 cm r 0 55 cm Find density gcm3 0 Strategize Concept Plan Relationships V 7L r21 d mV 0 Follow the Solution V 2 7t 0 55 cm2 1 94 cm concept plan V 1 436 cm3 to solve the problem 0 Sig gs and round 0 Check Check Units amp magnitude OK 27

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