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NWTC General Chemistry Ch 02
 

NWTC General Chemistry Ch 02

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NWTC General Chemistry Ch 02 by Steve Sinclair

NWTC General Chemistry Ch 02 by Steve Sinclair

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  • Qualitative observations are descriptions of what you observe.Example: The substance is a gray solid.Quantitative observations are measurements that include both a number and a unit.Example: The mass of the substance is 3.42 g
  • Figure 2.1 Measuring temperature (°C) with various degrees of precision. Thermometers a and b are precise to 0.1°C while thermometer c is precise to 0.01°C.
  • 1 – 32 – 43 – zero between 3 & 44 – zero after 4If zero does not need to be there ( 3.040 = 3.04), then zero is significant
  • Only place holders
  • 240,391 has 6 sig figs240,300 - 4 sig fig but rounded down240,490 – 5 sig fig240,000 – 2 sig fig240,400 – 4 sig fig rounded up
  • 1587 g - 120 g = 1467 rounded to 1470 (3 sig fig) not 1500 (2 sig fig).
  • 132.56 g - 14.1 g = 118.46 rounded to 118.5 (4 sig fig) not 118 (3 sig fig)
  • 1.626 – 0.101 = 1.525 do not need to round
  • Figure 2.2 Comparison of the metric and American Systems of length measurement: 2.54cm = 1 in.
  • Why 6.00 not just 6?
  • What is wrong with 0.0003? Only 1 sig fig, either 0.000300 or 3.00 x10 -4
  • UnitdeciCentiMilliMicroNano
  • You have the same mass on earth and moon, but you have different weights
  • Smaller unit so bigger number
  • Neither 5.1247 nor 5.1 (precision answer)
  • Figure 2.5 Comparison of Celsius, Kelvin and Fahrenheit temperature scales.That is no negative kelvin
  • Figure 2.6 (a) Comparison of the volumes of equal masses (10.0 g) of water, sulfur and gold.(b) Comparison of the masses of equal volumes (1.00 cm3) of water, sulfur and gold.Water is at 4°C; the two solids at 20°C.

NWTC General Chemistry Ch 02 NWTC General Chemistry Ch 02 Presentation Transcript

  • Chapter 2 Standards for MeasurementCareful andaccuratemeasurementsfor eachingredient areessential whenbaking orcooking as wellas in thechemistry Introduction to General, Organic, and Biochemistry 10elaboratory. John Wiley & Sons, Inc Morris Hein, Scott Pattison, and Susan Arena
  • Chapter Outline2.1 Scientific Notations 2.6 Dimensional Analysis2.2 Measurement and 2.7 Measuring Mass and Uncertainty Volume2.3 Significant Figures 2.8 Measurement of Temperature2.4 Significant Figures in Calculations 2.9 Density2.5 The Metric System Copyright 2012 John Wiley & Sons, Inc
  • Measurement
  • Observations• Qualitative• Quantitative Copyright 2012 John Wiley & Sons, Inc
  • Observations• Qualitative observations are descriptions of what you observe. – Example: The substance is a gray solid.• Quantitative observations are measurements that include both a number and a unit. – Example: The mass of the substance is 3.42 g. Copyright 2012 John Wiley & Sons, Inc
  • Scientific NotationScientific notation is writing a number as the product of a number between 1 and 10 multiplied by 10 raised to some power.• Used to express very large numbers or very small numbers as powers of 10.• Write 59,400,000 in scientific notation – Move the decimal point so that it is located after the first nonzero digit (5.94) – Indicate the power of 10 needed for the move. (107)• 5.94×107 Copyright 2012 John Wiley & Sons, Inc
  • Scientific Notation• Exponent is equal to the number of places the decimal point is moved.• Sign on exponent indicates the direction the decimal was moved – Moved right  negative exponent – Moved left  positive exponent• Write 0.000350 in scientific notation – Move the decimal point so that it is located after the first nonzero digit (3.50) – Indicate the power of 10 needed for the move. (10-4)• 3.50×10-4 Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!Write 806,300,000 in scientific notation.a. 8.063×10-8b. 8.063×108c. 8063×10-5d. 8.063×105 Copyright 2012 John Wiley & Sons, Inc
  • Measurement and UncertaintyThe last digit in any measurement is an estimate.uncert estimate a. 21.2°C +.1°C +.01°C certain b. 22.0°C c. 22.11°C Copyright 2012 John Wiley & Sons, Inc
  • Significant FiguresSignificant Figures include both the certain part of the measurement as well as the estimate.Rules for Counting Significant Figures1. All nonzero digits are significant  21.2 has 3 significant figures2. An exact number has an infinite number of significant figures.  Counted numbers: 35 pennies  Defined numbers: 12 inches in one foot Copyright 2012 John Wiley & Sons, Inc
  • Significant FiguresRules for Counting Significant Figures (continued)3. A zero is significant when it is• between nonzero digits  403 has 3 significant figures• at the end of a number that includes a decimal point  0.050 has 2 significant figures  22.0 has 3 significant figures  20. has 2 significant figures Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!How many significant figures are found in 3.040×106?a. 2b. 3c. 4d. 5e. 6 Copyright 2012 John Wiley & Sons, Inc
  • Significant FiguresRules for Counting Significant Figures (continued)4. A zero is not significant when it is• before the first nonzero digits  0.0043 has 2 significant figures• a trailing zero in a number without a decimal point  2400 has 2 significant figures  9010 has 3 significant figures Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!How many significant figures are found in 0.056 m?a. 5b. 4c. 3d. 2e. 1 Copyright 2012 John Wiley & Sons, Inc
  • Significant FiguresWhy does 0.056 m have only 2 significant figures?• Leading zeros are not significant.Lets say we measure the width of sheet of paper: 5.6 cm (the 5 was certain and the 6 was estimated)• This length in meters is 0.056 m (100 cm / m)• We use significant figures rules to be sure that the answer is as precise as the original measurement! Copyright 2012 John Wiley & Sons, Inc
  • Rounding NumbersCalculations often result in excess digits in the answer (digits that are not significant).1. Round down when the first digit after those you want to retain is 4 or less  4.739899 rounded to 2 significant figures is 4.72. Round up when the first digit after those you want to retain is 5 or more  0.055893 round to 3 significant figures is 0.0559 Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!Round 240,391 to 4 significant figures.a. 240,300b. 240,490c. 240,000d. 240,400 Copyright 2012 John Wiley & Sons, Inc
  • Significant Figures in Calculations The result of the calculation cannot be more precise than the least precise measurement. For example: Calculate the area of a floor that is 12.5 ft by 10. ft 12.5 ft × 10. ft = 125 ft210. ft But the 10. has only 2 significant figures, so the correct answer is 130 ft2. 12.5 ft Copyright 2012 John Wiley & Sons, Inc
  • 2.3 has two significant figures. (190.6)(2.3) = 438.38190.6 has four Answer givensignificant figures. by calculator. The answer should have two significant figures because 2.3 is the number with the fewest significant figures. Round off this Drop these three digit to four. digits. 438.38 The correct answer is 440 or 4.4 x 102
  • Significant Figures in Calculations Calculations involving Multiplication or Division The result has as many significant figures as the measurement with the fewest significant figures .9.00 m × 100 m = 900 m2 (100 has only 1 significant figure)9.00 m × 100. m= 900. m2 (both have 3 significant figures )9.0 m × 100. m = 9.0×102 m2 (9.0 has 2 significant figures ) Copyright 2012 John Wiley & Sons, Inc
  • Significant Figures in Calculations Calculations involving Addition and Subtraction The result has the same precision (same number of decimal places) as the least precise measurement (the number with the fewest decimal places).1587 g - 120 g = ? 120 g is the least precise measurement. The answer must be rounded to 1470 g. Key Idea: Match precision rather than significant figures! Copyright 2012 John Wiley & Sons, Inc
  • Significant Figures in Calculations Calculations involving Addition and Subtraction The result has the same precision (same number of decimal places) as the least precise measurement (the number with the fewest decimal places).132.56 g - 14.1 g = ? 14.1 g is the least precise measurement. The answer must be rounded to 118.5 g. Copyright 2012 John Wiley & Sons, Inc
  • Add 125.17, 129 and 52.2 Least precise number. 125.17 Answer given 129. by calculator. 52.2 306.37Round off to theCorrect answer. nearest unit. 306.37
  • Your Turn!A student determined the mass of a weigh paper to be 0.101 g. He added CaCl2 to the weigh paper until the balance read 1.626 g. How much CaCl2 did he weigh out?a. 1.525 gb. 0.101 gc. 1.626 gd. 1.727 g Copyright 2012 John Wiley & Sons, Inc
  • Metric SystemThe metric system or International System (SI) is adecimal system of units that uses factors of 10 to expresslarger or smaller numbers of these units. Copyright 2012 John Wiley & Sons, Inc
  • Metric System Copyright 2012 John Wiley & Sons, Inc
  • Units of LengthExamples of equivalent measurements of length:1 km = 1000 m 1 cm = 0.01 m 1 nm = 10-9 m 100 cm = 1 m 109 nm = 1 m Copyright 2012 John Wiley & Sons, Inc
  • How big is a cm and a mm? 2.54 cm = 1 in 25.4 mm = 1 inFigure 2.2 Comparison of the metric and American Systems of length measurement Copyright 2012 John Wiley & Sons, Inc
  • Dimensional Analysis: Converting One Unit to Another• Read. Identify the known and unknown.• Plan. Identify the principles or equations needed to solve the problem.• Set up. Use dimensional analysis to solve the problem, canceling all units except the unit needed in the answer.• Calculate the answer and round for significant figures.• Check answer – Does it make sense? Copyright 2012 John Wiley & Sons, Inc
  • Dimensional Analysis• Using units to solve problems• Apply one or more conversion factors to cancel units of given value and convert to units in the answer. u n it 1 con version factor = u n it 2• Example: Convert 72.0 inches to feet. 1 ft72.0 in 6.00 ft 12 in 72.0 in 1 ft = 6.00 ft 12 in Copyright 2012 John Wiley & Sons, Inc
  • Conversion FactorsWhat are the conversion factors between kilometers and meters? 1 km = 1000 m Divide both sides by 1000 m to get 1 km 1 one conversion factor. 1000 m Divide both sides by 1 km to get the 1000 m 1 other conversion factor. 1 kmUse the conversion factor that has the unit you want tocancel in the denominator and the unit you are solvingfor in the numerator. Copyright 2012 John Wiley & Sons, Inc
  • Dimensional Analysis u n it 1 con version factor = u n it 2Calculate the number of km in 80700 m.• Unit1 is 80700 m and unit2 is km• Solution map (outline of conversion path): m  km• The conversion factor is 1 km 1000 m 1 km 80700 m = 80.7 km 1000 m 80700 m 1 km = 80.7 km 1 1000 m Copyright 2012 John Wiley & Sons, Inc
  • Dimensional Analysis u n it 1 con version factor = u n it 2Calculate the number of inches in 25 m.• Solution map: m  cm  in 100 cm 1 in• Two conversion factors are needed: 1m 2.54 cm 100 cm 1 in 25 m = 984.3 cm 1m 2.54 cm Round to 980 cm since 25 m has 2 significant figures. Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!Which of these calculations is set up properly to convert 35 mm to cm? Another way: 0.001 m 1 cm 1m 100 cma. 35 m m x 1 mm x 0.01 m 35 m m x 1000 m m x 1m = 3.5 cm 1m 0.01 cmb. 35 m m x 0.001 m m x 1mc. 35 m m x 1000 m x 1 cm 1 mm 100 m Copyright 2012 John Wiley & Sons, Inc
  • Dimensional Analysis u n it 1 con version factor = u n it 2 The volume of a box is 300. cm3. What is that volume in m3? • Unit1 is 300. cm3 and unit2 is m3 • Solution map: (cm  m)3 1m • The conversion factor is needed 3 times: 100 cm 1m 1m 1m 3 -4 3 300. cm × 3.00×10 m 100 cm 100 cm 100 cm300. cm cm cm 1m 1m 1m = 0.0003 m m m 100 cm Copyrightcm John Wiley & Sons, Inc 100 2012 100 cm
  • Dimensional Analysis u n it 1 con version factor = u n it 2Convert 45.0 km/hr to m/s• Solution map: km m and hr  mins• The conversion factors needed are 1000 m 1 hr 1 m in 1 km 60 m in 6 0 sec km 1000 m 1 hr 1 m in m 45.0 × = 12.5 hr 1 km 60 m in 60 sec s Copyright 2012 John Wiley & Sons, Inc
  • Your Turn! The diameter of an atom was determined and a value of 2.35 × 10–8 cm was obtained. How many nanometers is this? cm to nm is a 10 -2 to 10 -9 change related to a meter a. 2.35×10-1 nm which is 10 -7 b. 2.35×10-19 nm Since answer is c. 2.35×10-15 nm smaller, subtract 8-7=1 So 10 -1 d. 2.35×101 nm0.0000000235 cm 1 m 1,000,000,000 nm = 0.235 nm1 100 cm 1 m Copyright 2012 John Wiley & Sons, Inc
  • Mass and Weight• Mass is the amount of matter in the object. – Measured using a balance. – Independent of the location of the object.• Weight is a measure of the effect of gravity on the object. – Measured using a scale which measures force against a spring. – Depends on the location of the object. Copyright 2012 John Wiley & Sons, Inc
  • Metric Units of MassExamples of equivalent measurements of mass:1 kg = 1000 g 1 mg = 0.001 g 1 μg = 10-6 g 1000 mg = 1 g 106 μg = 1 g Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!The mass of a sample of chromium was determined to be 87.4 g. How many milligrams is this? g to mg Is the answer going toa. 8.74×103 mg be bigger or smaller?b. 8.74×104 mgc. 8.74×10-3 mgd. 8.74×10-2 mg 87.4 g 1000 mg = 87400 mg 1 1 g Copyright 2012 John Wiley & Sons, Inc
  • Units of MassCommonly used metric to American relationships: 2.205 lb = 1 kg 1 lb = 453.6 gConvert 6.30×105 mg to lb. Solution map: mg  g  lb 5 1 g 1 lb 5.30 10 m g × = 1.17 lb 1000 m g 453.6 g Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!A baby has a mass of 11.3 lbs. What is the baby’s mass in kg? There are 2.205 lb in one kg.a. 11.3 kgb. 5.12 kgc. 24.9 kgd. 0.195 kg 11.3 lbs 1 kg = 5.1247 mg 1 2.205 lbs Copyright 2012 John Wiley & Sons, Inc
  • Setting StandardsThe kg is the base unit of mass in the SI systemThe kg is defined as the mass of a Pt-Ir cylinder stored in a vault in Paris.The m is the base unit of length1 m is the distance light travels in 1 s. 299, 792, 458 Copyright 2012 John Wiley & Sons, Inc
  • Volume Measurement1 Liter is defined as the volume of 1 dm3 of water at 4°C. 1 L = 1000 mL 1 L = 1000 cm3 1 mL = 1 cm3 1 L = 106 μL Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!A 5.00×104 L sample of saline is equivalent to how many mL of saline?a. 500. mLb. 5.00×103 mLc. 5.00×1013 mLd. 50.0 mLe. 5.00×107 mL 50000 L 1 L 1000 mL = 50 mL 1 1000000 L 1 L Copyright 2012 John Wiley & Sons, Inc
  • Units of VolumeUseful metric to American relationships: 1 L =1.057 qt 946.1 mL = 1 qtA can of coke contains 355 mL of soda.A marinade recipe calls for 2.0 qt ofcoke. How many cans will you need? 9 4 6 .1 m L 1 can 2.0 qt × = 5.3 cans 1 qt 355 m L Copyright 2012 John Wiley & Sons, Inc
  • Thermal Energy and Temperature• Thermal energy is a form of energy associated with the motion of small particles of matter.• Temperature is a measure of the intensity of the thermal energy (or how hot a system is).• Heat is the flow of energy from a region of higher temperature to a region of lower temperature. Copyright 2012 John Wiley & Sons, Inc
  • Temperature MeasurementK = °C + 273.15°F = 1.8 x °C + 32 °F - 32°C = 1.8 Copyright 2012 John Wiley & Sons, Inc
  • Temperature MeasurementThermometers are often filled with liquid mercury, which melts at 234 K. What is the melting point of Hg in °F? •First solve for the Centigrade temperature: 234 K = °C + 273.15 °C = 234 - 273.15 = -39°C •Next solve for the Fahrenheit temperature: °F = 1.8 x -39°C + 32 = -38°F Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!Normal body temperature is 98.6°F. What is that temperature in °C?a. 66.6°C °F = 1.8 x °C + 32b. 119.9°C 0C = (0F - 32) / 1.8c. 37.0°C 0C = (98.6 - 32) / 1.8d. 72.6°Ce. 80.8°C Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!On a day in the summer of 1992, the temperature fell from 98 °F to 75 °F in just three hours. The temperature drop expressed in Celsius degrees (C°) wasa. 13°C 0C = (0F - 32) / 1.8b. 9°C 0C = (98- 32) / 1.8 0C = (75 - 32) / 1.8c. 45°C 0C = 37 0C = 24d. 41°Ce. 75°C T = (37- 24) 0C Copyright 2012 John Wiley & Sons, Inc
  • m ass d en sity = Density volu m eDensity is a physical characteristic of a substance that can be used in its identification.• Density is temperature dependent. For example, water d4°C = 1.00 g/mL but d25°C = 0.997 g/mL.Which substance is the most dense? Water is at 4°C; the two solids at 20°C. Copyright 2012 John Wiley & Sons, Inc
  • Density m ass d = volu m e U n it sS olid s an d liq u id s: g g 3 or cm mL gG ases: L Copyright 2012 John Wiley & Sons, Inc
  • Density by H2O DisplacementIf an object is more dense than water, it will sink, displacing a volume of water equal to the volume of the object.A 34.0 g metal cylinder is dropped into a graduated cylinder. If thewater level increases from 22.3 mL to 25.3 mL, what is the densityof the cylinder? •First determine the volume of the solid: 3 25.3 mL – 22.3 mL 3.0 mL = 3.0 cm •Next determine the density of the solid: m ass 34.0 g g d = 3 = 11 3 volum e 3.0 cm cm Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!Use Table 2.5 (page 35) to determine the identity of a substance with a density of 11 g/cm3.a. silverb. leadc. mercuryd. gold Copyright 2012 John Wiley & Sons, Inc
  • Specific Gravity• Specific gravity (sp gr) of a substance is the ratio of the density of that substance to the density of a reference substance (usually water at 4°C). density of a liquid or solid sp gr = density of w ater (1.00 g/m L)• It has no units and tells us how many times as heavy a liquid or a solid is as compared to the reference material. Copyright 2012 John Wiley & Sons, Inc
  • Density CalculationsDetermine the mass of 35.0 mL of ethyl alcohol. The density of ethyl alcohol is 0.789 g/mL.Approach 1: Using the density formula•Solve the density equation for mass: m ass volum e d = volum e volum e•Substitute the data and calculate: g m ass = volum e d = 35.0 m L 0.789 = 27.6 g mL Copyright 2012 John Wiley & Sons, Inc
  • Density CalculationsDetermine the mass of 35.0 mL of ethyl alcohol. The density of ethyl alcohol is 0.789 g/mL.Approach 2: Using dimensional analysis Solution map: mL  g u n it 1 con version factor = u n it 2 .789 g 35.0 mL 27.6 g 1 mL Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!Osmium is the most dense element (22.5 g/cm3). What is the volume of 225 g of the metal?a. 10.0 cm3b. 10 cm3c. 5060 cm3d. 0.100 cm 3 225 g 1 cm3 = 10 cm3 1 22.5 g Copyright 2012 John Wiley & Sons, Inc
  • Your Turn!A 109.35 g sample of brass is added to a 100 mL graduated cylinder with 55.5 mL of water. If the resulting water level is 68.0 mL, what is the density of the brass?a. 1.97 g/cm3 volume = 68 – 55.5 mLb. 1.61 g/cm3 Density = mass / volumec. 12.5 g/cm3 Density = 109.35 g / 12.5 mL Density = 8.75 g / mLd. 8.75 g/cm 3 1 mL = 1 g/cm3 8.75 mL = 8.75 g/cm3 Copyright 2012 John Wiley & Sons, Inc
  • Questions• Review Questions – Do 1, 3, 5, 7 – Practice later 2-14 even• Paired Questions – Do 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 43, 47, 51, 55 – Practice later every other even (2, 6, etc) Copyright 2012 John Wiley & Sons, Inc 1-61