# I. Using Measurements (p )

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I. Using Measurements (p. 44 - 57)
CH. 2 - MEASUREMENT I. Using Measurements (p ) 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 C. Johannesson

B. Percent Error Indicates accuracy of a measurement your value
accepted value C. Johannesson

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 % 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 C. Johannesson

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 C. Johannesson

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 C. Johannesson

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 C. Johannesson

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

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 C. Johannesson

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 C. Johannesson

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. C. Johannesson

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 C. Johannesson

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 C. Johannesson

E. Proportions Direct Proportion Inverse Proportion y x y x
C. Johannesson

II. Units of Measurement (p. 33 - 39)
CH. 2 - MEASUREMENT II. Units of Measurement (p ) C. Johannesson

A. Number vs. Quantity Quantity - number + unit UNITS MATTER!!
C. Johannesson

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 C. Johannesson

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 C. Johannesson

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 C. Johannesson

D. Density Mass (g) Volume (cm3) C. Johannesson

Problem-Solving Steps
1. Analyze 2. Plan 3. Compute 4. Evaluate C. Johannesson

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

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

III. Unit Conversions (p. 40 - 42)
CH. 2 - MEASUREMENT III. Unit Conversions (p ) 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 left or right? C. Johannesson

A. SI Prefix Conversions
= 532 m = _______ km 0.532 NUMBER UNIT NUMBER UNIT C. Johannesson

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

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 C. Johannesson

B. Dimensional Analysis
The “Factor-Label” Method Units, or “labels” are canceled, or “factored” out 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. C. Johannesson

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 C. Johannesson

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

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

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 C. Johannesson

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 C. Johannesson

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 C. Johannesson

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