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I
II
III C. Johannesson
I. Using Measurements
(p. 44 - 57)
CH. 2 - MEASUREMENT
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C. Johannesson
A. Accuracy vs. Precision
Accuracy - how close a measurement is to the accepted value
Precision - how close a series of measurements are to each other
ACCURATE = CORRECT
PRECISE = CONSISTENT
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C. Johannesson
B. Percent Error
Indicates accuracy of a measurement
100literature
literaturealexperimenterror %
your value
accepted value
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C. Johannesson
B. Percent Error
A student determines the density of a substance to be 1.40 g/mL. Find the % error if the accepted value of the density is 1.36 g/mL.
100g/mL 1.36
g/mL 1.36g/mL 1.40error %
% error = 2.9 %
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C. Johannesson
C. Significant Figures
Indicate precision of a measurement.
Recording Sig Figs Sig figs in a measurement include the
known digits plus a final estimated digit
2.35 cm
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C. Johannesson
C. Significant Figures
Counting Sig Figs (Table 2-5, p.47)
Count all numbers EXCEPT:
Leading zeros -- 0.0025
Trailing zeros without a decimal point -- 2,500
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C. Johannesson
4. 0.080
3. 5,280
2. 402
1. 23.50
C. Significant Figures
Counting Sig Fig Examples
1. 23.50
2. 402
3. 5,280
4. 0.080
4 sig figs
3 sig figs
3 sig figs
2 sig figs
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C. Johannesson
C. Significant Figures
Calculating with Sig Figs Multiply/Divide - The # with the fewest
sig figs determines the # of sig figs in the answer.
(13.91g/cm3)(23.3cm3) = 324.103g
324 g
4 SF 3 SF3 SF
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C. Johannesson
C. Significant Figures
Calculating with Sig Figs (con’t) Add/Subtract - The # with the lowest
decimal value determines the place of the last sig fig in the answer.
3.75 mL
+ 4.1 mL
7.85 mL
224 g
+ 130 g
354 g 7.9 mL 350 g
3.75 mL
+ 4.1 mL
7.85 mL
224 g
+ 130 g
354 g
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C. Johannesson
C. Significant Figures
Calculating with Sig Figs (con’t) Exact Numbers do not limit the # of sig
figs in the answer.Counting numbers: 12 studentsExact conversions: 1 m = 100 cm “1” in any conversion: 1 in = 2.54 cm
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C. Johannesson
C. Significant Figures
5. (15.30 g) ÷ (6.4 mL)
Practice Problems
= 2.390625 g/mL
18.1 g
6. 18.9 g
- 0.84 g
18.06 g
4 SF 2 SF
2.4 g/mL2 SF
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C. Johannesson
D. Scientific Notation
Converting into Sci. Notation:
Move decimal until there’s 1 digit to its left. Places moved = exponent.
Large # (>1) positive exponentSmall # (<1) negative exponent
Only include sig figs.
65,000 kg 6.5 × 104 kg
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C. Johannesson
D. Scientific Notation
7. 2,400,000
g
8. 0.00256 kg
9. 7 10-5 km
10. 6.2 104
mm
Practice Problems
2.4 106 g
2.56 10-3 kg
0.00007 km
62,000 mm
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C. Johannesson
D. Scientific Notation
Calculating with Sci. Notation
(5.44 × 107 g) ÷ (8.1 × 104 mol) =
5.44EXPEXP
EEEE÷÷
EXPEXP
EEEE ENTERENTER
EXEEXE7 8.1 4
= 671.6049383 = 670 g/mol = 6.7 × 102 g/mol
Type on your calculator:
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C. Johannesson
E. Proportions
Direct Proportion
Inverse Proportion
xy
xy
1
y
x
y
x
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I
II
III C. Johannesson
II. Units of Measurement
(p. 33 - 39)
CH. 2 - MEASUREMENT
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C. Johannesson
A. Number vs. Quantity
Quantity - number + unit
UNITS MATTER!!
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C. Johannesson
B. SI Units
Quantity Base Unit Abbrev.
Length
Mass
Time
Temp
meter
kilogram
second
kelvin
m
kg
s
K
Amount mole mol
Symbol
l
m
t
T
n
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C. Johannesson
B. SI Units
mega- M 106
deci- d 10-1
centi- c 10-2
milli- m 10-3
Prefix Symbol Factor
micro- 10-6
nano- n 10-9
pico- p 10-12
kilo- k 103
BASE UNIT --- 100
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C. Johannesson
C. Derived Units
Combination of base units.
Volume (m3 or cm3) length length length
D = MV
1 cm3 = 1 mL1 dm3 = 1 L
Density (kg/m3 or g/cm3)mass per volume
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C. Johannesson
D. DensityM
ass
(g)
Volume (cm3)
Δx
Δyslope D
V
M
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C. Johannesson
Problem-Solving Steps
1. Analyze
2. Plan
3. Compute
4. Evaluate
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C. Johannesson
D. Density
An object has a volume of 825 cm3 and a density of 13.6 g/cm3. Find its mass.
GIVEN:
V = 825 cm3
D = 13.6 g/cm3
M = ?
WORK:
M = DV
M = (13.6 g/cm3)(825cm3)
M = 11,200 g
V
MD
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C. Johannesson
D. Density
A liquid has a density of 0.87 g/mL. What volume is occupied by 25 g of the liquid?
GIVEN:
D = 0.87 g/mL
V = ?
M = 25 g
WORK:
V = M D
V = 25 g
0.87 g/mL
V = 29 mLV
MD
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I
II
III C. Johannesson
III. Unit Conversions
(p. 40 - 42)
CH. 2 - MEASUREMENT
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C. Johannesson
A. SI Prefix Conversions
1. Find the difference between the
exponents of the two prefixes.
2. Move the decimal that many places.
To the leftor right?
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C. Johannesson
=
A. SI Prefix Conversions
NUMBERUNIT
NUMBER
UNIT
532 m = _______ km0.532
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C. Johannesson
A. SI Prefix Conversions
mega- M 106
deci- d 10-1
centi- c 10-2
milli- m 10-3
Prefix Symbol Factor
micro- 10-6
nano- n 10-9
pico- p 10-12
kilo- k 103
mo
ve le
ft
mo
ve r
igh
t BASE UNIT --- 100
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C. Johannesson
A. SI Prefix Conversions
1) 20 cm = ______________ m
2) 0.032 L = ______________ mL
3) 45 m = ______________ nm
4) 805 dm = ______________ km
0.2
0.0805
45,000
32
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C. Johannesson
3
3
cm
gcm
B. Dimensional Analysis
The “Factor-Label” Method Units, or “labels” are canceled, or
“factored” out
g
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C. Johannesson
B. Dimensional Analysis
Steps:
1. Identify starting & ending units.
2. Line up conversion factors so units cancel.
3. Multiply all top numbers & divide by each bottom number.
4. Check units & answer.
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C. Johannesson
B. Dimensional Analysis
Lining up conversion factors:
1 in = 2.54 cm
2.54 cm 2.54 cm
1 in = 2.54 cm
1 in 1 in
= 1
1 =
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C. Johannesson
1. Dimensional Analysis
How many milliliters are in 1.00 quart of milk?
1.00 qt 1 L
1.057 qt= 946 mL
qt mL
1000 mL
1 L
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C. Johannesson
2. Dimensional Analysis
You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3.
lb cm3
1.5 lb 1 kg
2.2 lb= 35 cm3
1000 g
1 kg
1 cm3
19.3 g
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C. Johannesson
3. Dimensional Analysis
How many liters of water would fill a container that measures 75.0 in3?
75.0 in3 (2.54 cm)3
(1 in)3= 0.191 L
in3 L
1 L
1000 cm3
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C. Johannesson
4. Dimensional Analysis
4) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off?
8.0 cm 1 in
2.54 cm= 3.1 in
cm in
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C. Johannesson
5. Dimensional Analysis
5) Taft football needs 550 cm for a 1st down. How many yards is this?
550 cm 1 in
2.54 cm= 6.0 yd
cm yd
1 ft
12 in
1 yd
3 ft