1. I. Using Measurements (p. 44 - 57)
CH. 2 - MEASUREMENT
I. Using Measurements
(p )
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2. 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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3. B. Percent Error Indicates accuracy of a measurement your value
accepted value
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4. B. Percent Error % error = 2.9 %
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.
% error = 2.9 %
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5. 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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6. C. Significant Figures Counting Sig Figs (Table 2-5, p.47)
Count all numbers EXCEPT:
Leading zeros
Trailing zeros without a decimal point -- 2,500
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7. Counting Sig Fig Examples
C. Significant Figures
Counting Sig Fig Examples
4 sig figs
3 sig figs
3. 5,280
3. 5,280
3 sig figs
2 sig figs
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8. C. Significant Figures (13.91g/cm3)(23.3cm3) = 324.103g 324 g
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) = g
4 SF
3 SF
3 SF
324 g
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9. C. Significant Figures 3.75 mL + 4.1 mL 7.85 mL 3.75 mL + 4.1 mL
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 mL
7.85 mL
3.75 mL mL
7.85 mL
224 g
+ 130 g
354 g
224 g
+ 130 g
354 g
 7.9 mL
 350 g
C. Johannesson

10. C. Significant Figures Calculating with Sig Figs (con’t)
Exact Numbers do not limit the # of sig figs in the answer.
Counting numbers: 12 students
Exact conversions: 1 m = 100 cm
“1” in any conversion: 1 in = 2.54 cm
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11. C. Significant Figures Practice Problems 5. (15.30 g) ÷ (6.4 mL)
4 SF
2 SF
= g/mL
 2.4 g/mL
2 SF g g
 18.1 g
18.06 g
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12. D. Scientific Notation 65,000 kg  6.5 × 104 kg
Converting into Sci. Notation:
Move decimal until there’s 1 digit to its left. Places moved = exponent.
Large # (>1)  positive exponent Small # (<1)  negative exponent
Only include sig figs.
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13. D. Scientific Notation Practice Problems 7. 2,400,000 g 8. 0.00256 kg
9. 7  10-5 km
 104 mm
2.4  106 g
2.56  10-3 kg km
62,000 mm
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14. D. Scientific Notation Calculating with Sci. Notation
(5.44 × 107 g) ÷ (8.1 × 104 mol) =
Type on your calculator:
EXP EE
EXP EE
ENTER
EXE
5.44 7
8.1 ÷ 4 =
= 670 g/mol
= 6.7 × 102 g/mol
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15. E. Proportions Direct Proportion Inverse Proportion y x y x
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16. II. Units of Measurement (p. 33 - 39)
CH. 2 - MEASUREMENT
II. Units of Measurement
(p )
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17. A. Number vs. Quantity Quantity - number + unit UNITS MATTER!!
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18. B. SI Units Quantity Symbol Base Unit Abbrev. Length l meter m Mass m
kilogram kg
Time t
second s
Temp T
kelvin K
Amount n
mole
mol
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19. B. SI Units Prefix Symbol Factor mega- M 106 kilo- k 103 BASE UNIT ---
100
deci- d
10-1
centi- c
10-2
milli- m
10-3
micro- 
10-6
nano- n
10-9
pico- p
10-12
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20. M V D = C. Derived Units 1 cm3 = 1 mL 1 dm3 = 1 L
Combination of base units.
Volume (m3 or cm3)
length  length  length
1 cm3 = 1 mL
1 dm3 = 1 L
D = M V
Density (kg/m3 or g/cm3)
mass per volume
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21. D. Density
Mass (g)
Volume (cm3)
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22. Problem-Solving Steps
1. Analyze
2. Plan
3. Compute
4. Evaluate
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23. D. Density V = 825 cm3 M = DV D = 13.6 g/cm3 M = (13.6 g/cm3)(825cm3)
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
C. Johannesson

24. D. Density D = 0.87 g/mL V = M V = ? M = 25 g V = 25 g 0.87 g/mL
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 = g
0.87 g/mL
V = 29 mL
C. Johannesson

25. III. Unit Conversions (p. 40 - 42)
CH. 2 - MEASUREMENT
III. Unit Conversions
(p )
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26. A. SI Prefix Conversions
1. Find the difference between the exponents of the two prefixes.
2. Move the decimal that many places.
To the left
or right?
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27. A. SI Prefix Conversions=
532 m = _______ km
0.532
NUMBER
UNIT
NUMBER
UNIT
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28. A. SI Prefix Conversions
Symbol
Factor
mega- M
106
kilo- k
103
BASE UNIT
---
100
deci- d
10-1
move left
move right
centi- c
10-2
milli- m
10-3
micro- 
10-6
nano- n
10-9
pico- p
10-12
C. Johannesson

29. A. SI Prefix Conversions
0.2
1) 20 cm = ______________ m
2) L = ______________ mL
3) 45 m = ______________ nm
4) 805 dm = ______________ km 32
45,000
0.0805
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30. B. Dimensional Analysis
The “Factor-Label” Method
Units, or “labels” are canceled, or “factored” out
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31. 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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32. B. Dimensional Analysis
Lining up conversion factors:
= 1
1 in = 2.54 cm
2.54 cm cm
1 =
1 in = 2.54 cm
1 in in
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33. 1. Dimensional Analysis 1.00 qt 1 L 1.057 qt 1000 mL 1 L = 946 mL qt
How many milliliters are in 1.00 quart of milk? qt mL
1.00 qt
1 L
1.057 qt
1000 mL
1 L
= 946 mL 
C. Johannesson

34. 2. Dimensional Analysis 1.5 lb 1 kg 2.2 lb 1000 g 1 kg 1 cm3 19.3 g
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
1000 g
1 kg
1 cm3
19.3 g
= 35 cm3
C. Johannesson

35. 3. Dimensional Analysis 75.0 in3 (2.54 cm)3 (1 in)3 1 L 1000 cm3
How many liters of water would fill a container that measures 75.0 in3?
in3 L
75.0 in3
(2.54 cm)3
(1 in)3
1 L
1000 cm3
= L
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36. 4. Dimensional Analysis 8.0 cm 1 in 2.54 cm = 3.1 in cm in
4) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off? cm in
8.0 cm
1 in
2.54 cm
= 3.1 in
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37. 5. Dimensional Analysis 550 cm 1 in 2.54 cm 1 ft 12 in 1 yd 3 ft
5) Taft football needs 550 cm for a 1st down. How many yards is this? cm yd
550 cm
1 in
2.54 cm
1 ft
12 in
1 yd
3 ft
= 6.0 yd
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