SECTION 2-3. Objectives 1. Distinguish between accuracy and precision 2. Determine the number of...
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![Page 1: SECTION 2-3. Objectives 1. Distinguish between accuracy and precision 2. Determine the number of significant figures in measurements 3. Perform mathematical.](https://reader037.fdocuments.us/reader037/viewer/2022110319/56649c825503460f94939f3c/html5/thumbnails/1.jpg)
USING SCIENTIFIC MEASUREMENTS
SECTION 2-3
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Objectives
1. Distinguish between accuracy and precision
2. Determine the number of significant figures in measurements
3. Perform mathematical operations involving significant figures
4. Convert measurements into scientific notation
5. Distinguish between inversely and directly proportional relationships
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Accuracy and Precision Accuracy – the closeness of measurements to
the correct or accepted value of the quantity measured
Precision – the closeness of a set of measurements of the same quantity made in the same way.
Visual Concept – Click Here
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Percent Error Calculated by subtracting the experimental
value from the accepted value, dividing the difference by the accepted value, and then multiplying by 100
Percent error = valueaccepted - valueexperimental x 100
valueaccepted
Sample ProblemA student measures the mass and volume of a substance and calculates its density as 1.40 g/mL. The correct, or accepted, value of the density is 1.30 g/mL. What is the percentage error of the student’s measurement?
Percent error
= valueaccepte
d
- valueexperimental
valueaccepte
d
= 1.30 g/mL- 1.40
g/mL 1.30 g/mL
= - 7.7 %
x 100
x 100
Video Concept – Click Here
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Error in Measurement
1. Skill of the measurer2. Measuring instruments3. Conditions of measurement
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Significant Figures Consists of all digits known with
certainty, plus one final digit
6.75 cm
certain
estimated
Total of 3 significant figures, to two places after the decimal.
52.8 mL
**Read from bottom of meniscus.
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Rules for determining significant zeros
Visual Concept – Click Here
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Sample ProblemHow many significant figures are in each of the following measurements?a) 28.6 g
3 s.f.
b) 3440. cm 4 s.f.
c) 910 m 2 s.f.
d) 0.046 04 L 4 s.f.
e) 0.006 700 0 kg 5 s.f.
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Rounding
Greater than 5 inc. by 1 42.68 42.7
Less than 5 stay 17.32 17.3 5, followed by nonzero inc. by 1 2.7851 2.79 5, not followed by nonzero inc. by 1 4.635 4.64 Preceded by odd digit 5, not followed by nonzero stays 78.65 78.6
Preceded by Even digit
Visual Concept – Click Here
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Addition/subtraction with significant figures
The answer must have the same number of digits to the right of the decimal point as there are in the measurement having the fewest digits to the right of the decimal point.
35. 1+ 2.3456
So : 37.4
37.4456
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Multiplication/Division with significant figures The answer can have no more significant
figures than are in the measurement with the fewest number of significant figures.
3.05 g ÷ 8.470 mL =0.360094451 g/mL
3 s.f.
4 s.f. Should be 3 s.f.
So : 0.360 g/mL
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Sample Problems
Carry out the following calculations. Expresseach answer to the correct number of significantfigures.a) 5.44 m - 2.6103 m =
Answer: 2.84 m
b) 2.4 g/mL x 15.82 mL =Answer: 38 g
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Conversion Factors and Significant Figures There is no uncertainty exact
conversion factors. Significant figures are only for
measurements, not known values!
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Scientific Notation Numbers are written in the form M x
10n, where the factor M is a number greater than or equal to 1 but less than 10 and n is a whole number. 65 ooo km M is 6.5 Decimal moved 4 places to left
X 104
So: 6.5 x 104 kmWhy? Makes very small or large numbers more workable60 200 000 000 000 000 000 000
molecules6.02 x 1023 molecules Visual Concept – Click
Here
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Extremely small numbers – negative exponent Ex: 0.0000000000567 g 5.67 x 10-11 g M should be in significant figures
Mathematical Operations Using Scientific Notation1. Addition and subtraction —These
operations can be performed only if the values have the same exponent (n factor).
a) example: 4.2 × 104 kg + 7.9 × 103 kg
OR
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2. Multiplication —The M factors are multiplied, and the exponents are added algebraically.
a) example: (5.23 × 106 µm)(7.1 × 10−2 µm)= (5.23 × 7.1)(106 × 10−2)= 37.133 × 104 µm2 = 3.7 × 105 µm2
3. Division — The M factors are divided, and the exponent of the denominator is subtracted from that of the numerator.
a) example:
= 0.6716049383 × 103
= 6.7 102 g/mol
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Using Sample Problems Analyze
Read the problem carefully at least twice and to analyze the information in it.
Plan Develop a plan for solving the problem.
Compute Substitute the data and necessary conversion factors.
Evaluate1) Check to see that the units are correct.2) Make an estimate of the expected answer.3) Expressed in significant figures.
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Sample Problem
1. Analyzemass = 3.057 kgdensity = 2.70 g/cm3
volume = ?2. Plan
conversion factor : 1000 g = 1 kg
Calculate the volume of a sample of aluminum that has a mass of 3.057 kg. The density of aluminum is 2.70 g/cm3.
DV
=m
DV =
m
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3. Computemass = 3.057 kg x
Round using sig figs ------- 1.13 × 103 cm3
4. Evaluate cm3, correct units. correct number of sig figs is three
1 kg
1000 g = 3057
g
DV =
m=
3057 g2.70
g/cm3
= 1132.222 . . . cm3
Video Concept – Click Here
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Direct Proportions Two quantities if
divided by the other gives a constant value If one doubles so does
the other y = kx
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Inverse Proportions Two quantities who’s product is a
constant If one doubles, the other is cut in half xy = k
Visual Concept – Click Here