Chapter 2 – Analyzing Data
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Transcript of Chapter 2 – Analyzing Data
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Chapter 2 – Analyzing Data
2.1 Units and Measurement [and standard problem solving technique illustrated using density]
2.2 Scientific Notation and Dimensional Analysis
2.3 Uncertainty in Data
2.4 Representing Data
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Section 2.1 Units and Measurements
• Know the first five of the seven base units in the SI system (time, length, mass, temperature and amount of substance) including their units, symbols, and the basic physical objects or phenomena on which they are defined.
• Convert between the Kelvin and Celsius temperature scales.
• Distinguish between and give examples of base units and derived units.
Chemists use an internationally recognized system of units to communicate their findings.
Objectives
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Section 2.1 Units and Measurements
• Use the equation for density to solve for an unknown quantity using the problem solving process described on page 38 in example problem 2.1.
• Understand both the concepts and experimental procedure involved in the Mini Lab on page 39 (Determine Density) and be able to answer the analysis questions listed at the end of the lab.
• Know all the prefixes from giga to nano in table 2.2 including their symbols and associated powers of ten and be able to use them in a numerical problem.
Objectives (cont)
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Section 2.1 Units and Measurements
Key Concepts• SI measurement units allow scientists to report
data to other scientists.
• Adding prefixes to SI units extends the range of possible measurements.
• To convert to Kelvin temperature, add 273.15 to the Celsius temperature. K = °C + 273.15
• Volume and density have derived units. Density, which is a ratio of mass to volume, can be used to identify an unknown sample of matter.
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Units of Measurement
SI - Systemè Internationale d’Unités
7 base units
Base unit is based on an object or an event in the physical world
Base unit is independent of other units
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Quantity Unit Abbrev.
Time second s
Length meter m
Mass kilogram kg
Temperature kelvin K
Amount of substance mole mol
Current ampere A
Luminous Intensity candela cd
The 7 SI Base Units (see Table 2.1)
Only base unit with prefix
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Prefixes
To better describe range of possible measurements, base units (and other units) are modified by using prefixes
Correspond to a particular power of ten
See table 2.2 (following)
Memorize value and the symbol (including upper /lower case) of prefixes in 10-9 to 106 range
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SI Prefixes – Page 33
Capital letters Greek letter
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SI Prefixes – Expanded
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Base Unit - TimeThe Second
In 1967, the 13th General Conference on Weights and Measures first defined the SI unit of time as the duration of 9,192,631,770 cycles of microwave light absorbed or emitted by the hyperfine transition of cesium-133 atoms in their ground state undisturbed by external fields
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Length
The Meter
Distance light travels through a vacuum in 1/(299 792 458) of a second
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Masskilogram (kg)
Only base unit whose standard is a physical object
Defined by platinum-iridium metal cylinder kept in Sèvres, France
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Temperaturekelvin (K)
The kelvin is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water
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Temperature
Kelvin scale• Unit kelvin (small k), abbreviation K (but
no degree sign) • Absolute zero = zero K
Celsius scale• Unit °C (with degree sign)• T(C) = T(K) - 273.15
Kelvin scale important for certain formulas we will use in later chapters
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Temperatures – K vs CBoiling point water373.15 K 100.00 C
Freezing point water273.15 K 0.00 C
Absolute zero0.00 K -273.15 C
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The Mole and Avogadro’s Number
Abbreviation is mol
SI base unit for amount of substance
Defined as number of representative particles (carbon atoms) in exactly 12 g of pure carbon-12
Mole of anything contains 6.022 X 1023 representative particles
6.022 X 1023 = Avogadro’s number
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Picky Details - Units
Abbreviations are avoidedProper
• s or second• cm3 or cubic centimeter• m/s or meter per second
Improper• sec• cc• mps
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Picky Details - Units
Unit symbols are unaltered in the plural
Proper• l = 75 cm
Improper• l = 75 cms
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Derived Units
Defined by a combination of base units
Velocity v = l/t m/s
Volume V = l l l m3
liter (L) cubic decimeter dm3
Density d = mass / V kg/m3
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Derived Unit - Density
Density = mass volume
Most common unit is g/cm3
Will return to using density in a word problem later in presentation
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Chapter 2 – Analyzing Data
2.1 Units of Measurement [and standard problem solving technique illustrated using density]
2.2 Scientific Notation and Dimensional Analysis
2.3 Uncertainty in Data
2.4 Representing Data
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Section 2.2 Scientific Notation and Dimensional Analysis
• Express numbers correctly in standard scientific notation, convert them to and from numbers not expressed in scientific notation, and perform standard arithmetic operations using them.
Scientists often express numbers in scientific notation.
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Section 2.2 Scientific Notation and Dimensional Analysis
Key Concepts
• A number expressed in scientific notation is written as a coefficient between 1 and 10 multiplied by 10 raised to a power. To add or subtract numbers in scientific notation, the numbers must have the same exponent.
• To multiply or divide numbers in scientific notation, multiply or divide the coefficients and then add or subtract the exponents, respectively.
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Scientific Notation
Expresses a number as a multiple of two factors:
• 1 number 10 3.1 -7.9• 10 raised to a power 103 10-7
100 = 1 10n > 1 n positive integer0 < 10-n < 1 n positive integer
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Scientific Notation
3.1 104 = 31000
3.1 10-4 = 0.00031• Have moved decimal point 4 places in
both cases
Some exponents match one of the standard SI prefixes
• 4.27 10-6 s = 4.27 s (microseconds)
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Scientific NotationAddition/Subtraction
• Must be same power of ten
Multiplication/Division• Multiply first factors• Add exponents if multiplying • Subtract exponents if dividing
Put result back into standard form• z.yyy 10n z = any non-zero digit• One digit in front of decimal point
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Practice – Scientific NotationGeneral
Problems 11(a-h), 12(a-d) page 41Problems 76(a-d), 77(a-d) page 62
Addition & SubtractionProblems 13(a-d), 14(a-d) page 42Problems 78(a-j) page 62
Multiplication & DivisionProblems 15(a-d), 16(a-d) page 43Problems 79(a-f) page 62
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Chapter 2 – Analyzing Data
2.1 Units of Measurement [and standard problem solving technique illustrated using density]
2.2 Scientific Notation and Dimensional Analysis
2.3 Uncertainty in Data
2.4 Representing Data
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Section 2.3 Uncertainty in Data
• Define and compare accuracy and precision and correctly identify which term or terms apply to a given value based on a description of how the value was determined.
• Calculate the percent error associated with a given measurement.
• Determine the number of significant figures associated with a given number.
Measurements contain uncertainties that affect how a result is presented.
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Section 2.3 Uncertainty in Data
• Determine the appropriate number of significant figures to record when using an analog measuring device such as a ruler.
• Determine the number of significant figures associated with a result obtained from simple arithmetic operations (addition, subtraction, multiplication, division) on numbers.
• Round a number to a specified number of significant digits.
Objectives (cont)
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Section 2.3 Uncertainty in Data
• Show all work for a problem following the Problem-Solving Process described on page 38 in example problem 2.1 while utilizing dimensional analysis, significant figures, rounding and standard algebra skills.
Objectives (cont)
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Section 2.3 Uncertainty in Data
Key Concepts
• An accurate measurement is close to the accepted value. A set of precise measurements shows little variation.
• The measurement device determines the degree of precision possible.
• Error is the difference between the measured value and the accepted value. Percent error gives the percent deviation from the accepted value.
error = experimental value – accepted value
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Section 2.3 Uncertainty in Data (cont.)
Key Concepts
• The number of significant figures reflects the precision of reported data.
• Calculations should be rounded to the correct number of significant figures.
• The rules for determining the number of significant figures in a number produced in a mathematical operation are different for multiplication/division and addition/subtraction.
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Accuracy and Precision
Accuracy refers to agreement of particular value with true value (sometimes true value is difficult to determine; may require appropriate calibration)
Precision refers to degree of agreement among several measurements of same quantity
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Calibration[Formal definition – don’t need to know]The set of operations which establish, under specified conditions, the relationship between values indicated by a measuring instrument or measuring system, and the corresponding standard or known values derived from the standard.
Purpose is to determine and/or improve the accuracy of the device being calibrated
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Calibration - ExampleTo calibrate an electronic balance, might use following approach:a) Assign a value of zero to the reading of the
output of the balance when nothing is on the balance pan
b) Assign a value equal to the value of a calibration mass (e.g., 1000.00 g) to the reading of the output of the balance when the calibration mass is placed on the balance pan
c) Assume linear behavior of the reading of the balance between these two calibration points
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Calibration - ExampleFor calibration to be possible, need a calibration mass (e.g., a 100.00 g mass)
The actual (true) value of the calibration mass has to be determined by the organization/company supplying the mass – this ultimately requires tracing back to the primary standard of mass (making comparisons that are linked to the kilogram standard mass)
It is only through calibration that the accuracy of a measurement device (i.e., a balance) can be determined and improved
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Accuracy and PrecisionRefer to figure 2.10
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Accuracy and PrecisionRefer to figure 2.10
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precise and accurate
precise but not accurate
Precision and Accuracy
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Precision and Accuracy
random error
Not precise, individual measurement not generally accurate but average is
systematic error
May be precise (only marginally for this example); average not accurate
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Percent Error
% Error = 100 (actual – measured) actual(actual – measured) = errorCan ignore plus or minus signs for now – need only absolute value of errorActual = “true value”In some cases, true value unknown(Often compare a predicted value from some model with measured value)
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Practice
Accuracy and PrecisionProblems 46, 48, page 54Problem 87 page 63
Percent ErrorProblems 32-34 page 49Problems 49, 51 page 54Problems 93(a-d), 94(a-d) page 63
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Significant Figures2 different types of numbers:
Exact
Measured
Exact numbers have infinite precision
Measured - obtained from measuring device, have error and limited precision
Measured number written to reflect both its numerical value and the precision to which it was measured
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Significant Figures and Measurement Precision
For lab data, sig figs determined by precision of measurement device
For digital device, last number to right in display is limit to precision
For analog device, precision limited by the estimated digit obtained from “eyeballing” reading between markings
Examples to follow
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Balances and Precision
Electronic analytical 0.0001 g
(digital)
Student 0.1 g(digital)
Triple beam 0.01 g(analog)
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Significant Figures
Mass of object measured on student balance (precision ± 0.1g) is 23.6 g
This quantity contains 3 significant figures, i.e., three experimentally meaningful digits
If same measurement made with analytical balance (precision ± 0.0001g) , mass might be 23.5820 g (6 sig. fig.)
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2 measurements of mass of same object
Same quantity described at two different levels of precision or certainty
Significant Figures and Precision
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Significant Figures Determined by Measuring Device
32.33 C 32.3 C
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Significant Figures Determined by Measuring Device
0.1 mL graduationsCan estimate to 0.01 mL
Reading 1.70 mL3 significant figures
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Significant Figures Determined by Measuring Device
0.1 cm graduationsCan estimate to 0.01 cm
Reading 5.22 cm3 significant figures
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What Is Nail Length?
~6.33 cm
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Learning Check
Length of wooden stick?1) 4.5 cm 2) 4.54 cm 3) 4.547 cm
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Measurement of Volume
Graduated Cylinder
Volumetric Flask
Buret VolumetricPipet
Syringe
Most accurate but only useable for one volume
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20.16 mL ±0.01mL
Measurement of volume using buret
Read at bottom of liquid curve (called the meniscus)
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Significant Figures
Rules for recognizing which digits are significant – see the “Problem Solving Strategy” on top of page 51 and the following slide
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Recognizing Significant (Sig) Figures
1. Non-zero numbers always significant• 72.3 (3 sf) 4.737x10-8 (4 sf)
2. Zeros between non-zero numbers are significant• 60.5 (3 sf) 7.3002x10-4 (5 sf)
3. All final zeros to right of decimal point are significant• 6.20 ( 3 sf) 5.47000x109 (6 sf)
4. Zeros acting as placeholders not significant• 0.00253 (3 sf) 43200 (3 sf)
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Recognizing Significant (Sig) Figures
5. Counting numbers (integers) and defined constants or relationships have an infinite number of significant figures (all are exact numbers or relations)• 60 s = 1 minute • 1 foot = 12 inches• 6 molecules•
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Significant Digits Practice
45.8736
.000239
.00023900
4.8000 104
48000
3.982106
1.00040
6
3
5
5
2
4
6
All digits count
Leading 0’s don’t
Trailing 0’s do
Trailing 0’s count
0’s don’t count w/o decimal
All digits count
0’s between digits count as well as trailing in decimal form
Determine number of significant digits in each of following:
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How many significant figures in each of following measurements?
5.13
100.01
0.0401
0.0050
220,000
1.90 x 103
153.000
1.0050
3
5
3
2
2
3
6
5
?
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Learning Check
Classify each of following as an exact or measured number or relationship
1 yard = 3 feet
Diameter of red blood cell = 6 x 10-4 cm
There are 6 hats on shelf
Gold melts at 1064°C
?
Ans: exact, measured, exact, measured
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Practice
Significant FiguresProblems 35(a-d), 36(a-d), 37 p. 51Problems 47, 50 page 54Problems 85, 88 page 63
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Rounding RulesSee “Problem Solving Strategy” on bottom of page 52
d = last significant digit r = digit to right of d
1. r < 5 then d d2. r > 5 then d d + 1
3. r = 5 and digit after r ≠ 0 then d d + 1
4. r = 5 and digit after r = 0 or nothing then
d d + 1 if d odd; d d if d even (d always ends up even using this rule)
(You may see variations on rule 4 in other places)
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Rounding
4965.03
780,582
1999.5
4965 0 dropped, <5
780,600 8 dropped, >5
Note: you must include 0’s
2.000x103 5 dropped, = 5
If wrote as 2000 would have 1 SF
Round following to 4 significant figures:
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Rounding
1.5587
.0037421
1367
128,522
1.6683x106
1.56
.00374
1370
129,000
1.67x106
Round following to 3 significant figures:
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Practice
RoundingProblems 38(a-d), 39(a-d) page 53Problems 91(a-f) page 63
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Sig Figures and Multiplication/Division
Answer must have same number of significant figures as number with fewest number of significant figures
24 x 3.26 x 5.774 = 451.75776 450= 4.5 x 102
2 sig figs allowed
6.38 × 2.0 = 12.76 13 (2 sig figs)
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Operations with Significant Figures – Adding or Subtracting
When adding or subtracting, number of decimal places in result should equal smallest number of decimal places in any term in sum
135 cm + 3.25 cm = 138 cm
0 digits after dp
135 cm term limits answer to units decimal value
2 digits after dp
0 digits after dp
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Sig Figures and Addition/Subtraction
Answer must have same number of digits to right of decimal point as value with fewest number of digits to right of decimal point (value with lowest precision)
• Note: not concerned with # of sig figures in numbers
11.0 + 5.7732 + 2.01 = 18.7832 18.8
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Addition/Subtraction
25.5 32.72 320
+34.270 ‑ 0.0049 +12.5
59.770 32.7151 332.5
59.8 32.72 330
Focus on least significant digit (digit with least precision)
Last example – special caseLeast significant digit prior to decimal point (can use sci. notation to justify)
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Addition/Subtraction – Special Case
Use scientific notation to handle case where number has zeros in front of decimal and no digits of any kind afterZeros in front of decimal are placeholders to indicate power of ten320 + 12.5 = 3.2x102 + 0.125x102
Now following ordinary rule, only allowed one digit after decimal point3.325x102 3.3x102 = 330
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Addition/Subtraction: No Decimal Point
135000 m + 3250 m = ????
Convert to scientific notation
1.35x105 m + 3.25x103 m
Convert to common exponent (the largest)
1.35x105 m + 0.0325x105 m
Apply standard rule regarding digits after dp
1.35x105 m + 0.0325x105 m = 1.38x105 m
13500 m + 3250 m = 138000 m
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Addition/Subtraction – Special Case
82000 + 5.32 = [82005.32] = 82000Special case (focus on last precise digit): 82000 = 8.2x104 5.32 = 0.000532x104
8.2x104 + 0.000532x104 = 8.2x104
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Addition/Subtraction
6.8 + 11.934 = 18.734 18.7 (1 place after decimal)
Result has 3 SF even though 6.8 has only 2 SF
0.56 + 0.153 = [0.713] = 0.71
10.0 - 9.8742 = [0.1258] = 0.1
10 – 9.8742 = [0.1258] = 0 [1x101 – 0.98742x101 = 0.01258x101= 0]
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Addition/SubtractionLimiting term: one having largest value of the least significant digit
ProblemLeast
significant digit
Answer
234 + 34.65 = 268.65 1’s place 2691.642x106 + 23x106 =
24.642x106 1’s place 25x106
100 + 34.56 = 134.56 100’s place 100150 + 28.57 = 178.57
10’s place 180
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Rounding RulesIf doing a multistep calculation, round off after last step provided all steps are either multiplication/division or addition/subtraction (can’t be mixed)
Round after series of additions or subtractions before doing a multiplication or division
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Sig Figures and Mixed OperationsSome calculations involve both multiplication/division and addition/subtraction
Must round intermediate result prior to switching to new category of operation
8.52 + 4.1586 18.73 + 153.2 =
= 8.52 + 77.89 + 153.2 = [239.61] = 239.6
(8.52 + 4.1586) (18.73 + 153.2) =
= 12.68 171.9 = [2179.692] = 2.180x103
(2180 has 3 SF, not 4 as required)
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Sig Figures and Mixed Operations
Calculate 5.000Tc – 25C, where Tc = Celsius temperature, for Tk = 298.1 K
?
Tc = Tk – 273.15 = 298.1- 273.15 = [24.95]
Tc = 25.0 C 5.000Tc = 125C
125C – 25C = 1.00x102 C
Not 100! (need 3 SF)
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Mixed Operations - Percent Error
% Error = 100 (actual – measured) actualCalculating % error always involves a mixed operation – subtraction followed by division
Must round after subtraction prior to dividing
Actual =16.24 g Meas. = 15.8 g % error ?
Error = 16.24 g – 15.8 g = [0.44 g] = 0.4 g
% Error = 100 0.4 / 16.24 = [2.463] = 2%
Not: 100 0.44 / 16.24 = [2.709] = 2.7%
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Mixed Operations - Percent Error
Note: Results shown in example problem 2.5 on page 49 are incorrectAll answers in that example should have one significant figure (i.e., 3%, not 3.14%)
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Sig Figures and Mixed OperationsAccepted value for density of copper = 8.92 g/cm3. 3 experiments to measure its density resulted in values of 8.74, 9.01, and 8.83 g/cm3. Calculate % error of average value of experiments.Avg = (8.74+9.01+8.83)/3 = 8.86 g/cm3
% Error=100(actual–measured)/actual% Error = 100 0.06 g/cm3/8.92 g/cm3
% Error = 0.7% (1 SF)
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Sig Figures and Mixed OperationsResult of performing the following? 5.00 (22 1.85)?22 1.85 = 20.15 rounds to 2.0x101
Units place is significant2.0x101 5.00 = 1.0x102
Answer has 2 significant digits because subtraction operation generated a 2 significant digit intermediate result
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Practice
Rounding – Addition/SubtractionProblems 40(a-b), 41(a-b) page 53Problems 92(a,b,d) page 63
Rounding – Multiplication/DivisionProblems 42(a-d), 43(a-d), 44 p. 54Problems 92(c,e) page 63
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Chapter 2 – Analyzing Data
2.1 Units of Measurement [and standard problem solving technique illustrated using density]
2.2 Scientific Notation and Dimensional Analysis (Conversions)
2.3 Uncertainty in Data
2.4 Representing Data
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Section 2.2 Scientific Notation and Dimensional Analysis
• Use dimensional analysis (aka “the factor-label method” or conversion factor) in a numerical problem to convert a given quantity to different units.
Scientists often solve problems using dimensional analysis.
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Section 2.2 Scientific Notation and Dimensional Analysis
Key Concepts
• Dimensional analysis uses conversion factors to solve problems.
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Dimensional Analysis(Conversions / Factor-Label)
In dimensional analysis always ask 3 questions:
1. What data are we given?
2. What quantity do we need?
3. What conversion factors are available to take us from what we are given to what we need?
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Dimensional AnalysisAlso referred to as factor-label method
1. Start with given quantity with its units2. Multiply by conversion factors until
desired final units are obtained3. Make sure units cancel in converting
Example: convert 48 km to meters48 km 1000 m
1 km
= 48000 m = 4.8 104 m
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Dimensional Analysis
Convert quantity 2.3 x 10-8 cm to nanometers (nm)
Determine conversion factorsCentimeter (cm) Meter (m)
1 cm = 0.01 m = 1 x 10-2 m
Meter (m) Nanometer (nm)
1 x 10-9 m = 1 nm
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Setup equation so cm and m units cancel out leaving only nm
m
nm
cm
mcm8103.2
Fill-in values for conversion factors and solve equation
nmm
nm
cm
mcm 23.0
101
1
1
01.0103.2
98
Dimensional Analysis
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Convert quantity 14 m/s to miles per hour (mi/hr)
Determine conversion factorsMeter (m) Kilometer (km)
Kilometer(km) Mile(mi)
1 mile = 1.6093 km 1000m = 1 km
Seconds (s) Minutes (min)
Minutes (min) Hours (hr)
60 sec = 1 min 60 min = 1 hr
Dimensional Analysis
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hr
min
min
s
km
mi
m
km/14 sm
Setup equation so m, km, s, and min cancel out leaving only miles and hours
Dimensional Analysis
Fill-in values for conversion factors and solve equation
hrmi
sm
/31
hr1
min60
min1
s60
km6093.1
mi1
m1000
km1/14
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Multiple Conversion Factors
Convert 100 km/h to m/s
100 km 1000 m 1 h 1 min h km 60 min 60 s
= 27.8 m/s
= 30 m/s (to have 1 SF)
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Multiple Conversion Factors
Baseball thrown at 89.6 miles per hour
Speed in meters per second?
m/s
1 mile = 1.609 km; 60 s = 1 min; 60 min = 1 h
speed = 89.6mile
h
mile/h m/h
1.609kmmile 103 m
km
= 1.44x105mh 1 h
60 min 1 min
60 s= 40.0
ms
3 SF
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Units Raised to a Power
Conversion factor must also be raised to that power
Area of circle = 28 in2 Area in cm2?
1 in = 2.54 cm (1 in)2 = (2.54 cm)2
Area= 28 in2
Area = 1.8 x 102 cm2
in2 cm2
(2.54 cm)2
(1 in)2 = 28 in2 6.45 cm2
1 in2
2 SF
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Unit Conversion w Powers: mL to cm3
liter (L) defined = cubic decimeter (dm3)milliliter (mL) = 10-3 L
= 10-3 L 1 dm3/L= 10-3 dm3
= 10-3 dm3 (10-1 m/dm)3
= 10-3 dm3 10-3 m3/dm3
= 10-6 m3
= 10-6 m3 (1 cm/10-2 m)3
= 10-6 m3 1 cm3/10-6 m3
= 1 cm3
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Convert 31,820 mi2 to square meters (m2)
Determine conversion factorsMile (mi) kilometer (km)
1 mile = 1.6093 km
kilometer (km) meter (m)
1000 m = 1 km
Dimensional Analysis
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Setup equation so mi2 and km2 cancel out leaving only m2 (must have squares)
22
2820,31km
m
mi
kmmi
22
2
1
1000
1
6093.1820,31
km
m
mi
kmmi
Dimensional Analysis
Fill-in values for conversion factors and solve equation
2102
26
2
22 102407.8
1
101
1
5898.2820,31 m
km
m
mi
kmmi
8.2411010 m2
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Unit Conversion with Powers
Convert 531 lb/ft3 to units of g/cm3
1 in = 2.54 cm, 12 in = 1 ft, 1 kg = 2.2046 lb
531 lb/ft3 x (1 kg/2.2046 lb) x (103 g/kg)
= 2.409x105 g/ft3
2.409x105 g/ft3 x (ft/12 in)3 x (in/2.54 cm)3
2.409x105 g/ft3 x (ft3/1728 in3) x (in3/16.39 cm3)= 8.507 g/cm3
= 8.51 g/cm3 3 SF
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PracticeDimensional Analysis (Conversions)
Practice Problems 17(a-c), 18(a-b), 19(a-h), 20 page 45Practice Problems 21-23 page 46Problems 27, 28, 30 page 46Problems 80(a-f), 81- 84 page 63
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Chapter 2 – Analyzing Data
2.1 Units of Measurement [and standard problem solving technique illustrated using density]
2.2 Scientific Notation and Dimensional Analysis
2.3 Uncertainty in Data
2.4 Representing Data
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Section 2.1 Units of Measurement
• Show all work for a problem following the Problem-Solving Process described on page 38 in example problem 2.1 while utilizing dimensional analysis, significant figures, rounding and standard algebra skills.
Objectives
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Word Problems in Chemistry
Will need to solve variety of such problems
Exists standard approach, which will also be used in physics
Illustrated in text using problem involving density
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Word Problem Involving Density
Problem 2.1, page 38
Sample of aluminum placed in 25 mL graduated cylinder containing 10.5 mL water. Water level rises to 13.5 mL.
Mass of aluminum sample?density of Al = 2.7 g/cm3 (also see Table R-7, page 971 for data)
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Required Problem-Solving StepsRead problem; make sure you understand what is being askedAnalyze problem to find unknown (mass)State defining equation (d = m/V)Re-write equation to solve for unknown
• mass = volume density• m = V d (use of symbols preferred)
Show all intermediate work with unitsEvaluate final answer – check that units and significant figures are correct
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Word Problem 2.1Solution shown on following slide differs from that shown on page 38Symbols (m) are used in place of words (mass)Subscripts used to convey additional information about the symbol (VAl instead of “volume of sample”; VAl vs VW to distinguish volume of aluminum from measured volume of water)Prefer use of symbols in your workSubscripts optional
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Review of Problem 2.1 Al=Aluminum W=Water f=final i=initial
dAl = 2.7 g/cm3
VAl = DVw = Vwf – Vwi (water displacement)
VAl = 13.5 mL – 10.5 mL = 3.0 mL
dAl = mAl / VAl (main formula; must include)
mAl = dAl VAl (formula solved for unknown)
mAl = 2.7 g 3.0 mL mL
= 8.1 g of Al
You must show symbols, values, and units in all your work using the above method
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Practice
Working word problemsPractice Problems 1 – 3 page 38Problem 9, page 39Problems 66 – 69 page 62Problems 104, 105, 107 page 64
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Chapter 2 – Analyzing Data
2.1 Units of Measurement [and standard problem solving technique illustrated using density]
2.2 Scientific Notation and Dimensional Analysis
2.3 Uncertainty in Data
2.4 Representing Data
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Section 2.4 Representing Data
• Create graphs to reveal patterns in data.
• Interpret data presented in graphs.
• Identify dependent and independent variables
• Create a properly labeled line graph from supplied data (one with a reasonable number of labeled tic marks, axes with labels, a graph title, and data points and trend lines that are easily viewed and interpreted).
Graphs visually depict data, making it easier to see patterns and trends.
Objectives
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Section 2.4 Representing Data
• Determine the numerical value and units of the slope of a straight line that is presented in a line graph.
• Define and distinguish between the processes of interpolation and extrapolation and use them to obtain predicted values from a trend line.
Objectives (cont)
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Section 2.4 Representing Data
Key Concepts
• Circle graphs show parts of a whole. Bar graphs show how a factor varies with time, location, or temperature.
• Independent (x-axis) variables and dependent (y-axis) variables can be related in a linear or a nonlinear manner. The slope of a straight line is defined as rise/run, or ∆y/∆x.
• Because line graph data are considered continuous, you can interpolate between data points or extrapolate beyond them.
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Pie ChartGood for showing breakdown of quantities that add to 100%
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Bar GraphTrends of quantities versus a discrete variable (month, year, sample number, etc.)
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Line GraphTrends of quantities versus a continuous variable (mass, time, volume, temperature)
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Line GraphIndependent variable (variable deliberately changed by experimenter) on x axisDependent variable on y axisTerm “plot of A versus B” means that B = independent variable (x) A = dependent (y)
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Line GraphsGraph should have title, axes labels with units, tick marks, data markers, and legend (if more than one set of data is plotted on same graph)If least squares/best-fit line used, get slope from
Slope = y2-y1 = Dy x2-x1 Dx
or get directly from softwarePay attention to the units of the slope
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Line Graphs
Graph displaying both data points and best fit line
Negative slope(Dy < 0)
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Line Graph - ExcelIn MS Excel, making a “line” chart (graph) will create a graph with a discrete x axis – not appropriate for most scientific uses
In Excel, make a scatter chart – it will have the necessary continuous x axis
When plotting data, use point markers (circles, etc.) only – do not connect points
When plotting lines use lines (solid, dashed, etc.) but no point markers
If variable have units, make sure they appear in axis labels – “Length (m)”
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Interpolation / Extrapolation
If have trendline (does not have to be a straight line), points on trendline curve considered to be continuousInterpolation – reading value from a point on curve that falls between recorded data pointsExtrapolation - reading value from a point on curve that extends beyond recorded data points (potentially dangerous, especially if overdone)
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Interpolation / Extrapolation
Extrapolated point at
elevation = 700 m
Interpolated point at
elevation = 350 m
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Practice
Line graphsProblems 53, 54, 58 page 58Problem 111 page 64
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End of Chapter