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I II III I. Using Measurements CH. 2 - MEASUREMENT

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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. Significant Figures Indicate precision of a measurement. Recording Sig Figs Sig figs in a measurement include the known digits plus a final estimated digit 2.32 cm

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C. Significant Figures Counting Sig Figs Count all numbers EXCEPT: Leading zeros -- 0.0025 Trailing zeros without a decimal point -- 2,500 USA??

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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 2 sig figs

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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/cm 3 )(23.3cm 3 ) = 324.103g 324 g 4 SF3 SF

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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 7.9 mL 3.75 mL + 4.1 mL 7.85 mL

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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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C. Significant Figures 5. (15.30 g) ÷ (6.4 mL) Practice Problems = 2.390625 g/mL 18.1 g 6.18.9g 7.- 0.84 g 18.06 g 4 SF2 SF 2.4 g/mL 2 SF

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D. Scientific Notation 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. 65,000 kg 6.5 × 10 4 kg

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D. Scientific Notation 7. 2,400,000 g 8. 0.00256 kg 9.7 10 -5 km 10.6.2 10 4 mm Practice Problems 2.4 10 6 g 2.56 10 -3 kg 0.00007 km 62,000 mm

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D. Scientific Notation Calculating with Sci. Notation (5.44 × 10 7 g) ÷ (8.1 × 10 4 mol) = 5.44 EXP EE ÷ ÷ EXP EE ENTER EXE 78.1 4 = 671.6049383= 670 g/mol= 6.7 × 10 2 g/mol Type on your calculator:

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E. SI Units QuantityBase UnitAbbrev. Length Mass Time Temp meter kilogram second kelvin m kg s K Amountmolemol Symbol l m t T n

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F. Derived Units Combination of base units. Volume (m 3 or cm 3 ) length length length D = MVMV 1 cm 3 = 1 mL 1 dm 3 = 1 L Density (kg/m 3 or g/cm 3 ) mass per volume

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

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Density An object has a volume of 825 cm 3 and a density of 13.6 g/cm 3. Find its mass. GIVEN: V = 825 cm 3 D = 13.6 g/cm 3 M = ? WORK : M = DV M = (13.6 g/cm 3 )(825cm 3 ) M = 11,200 g

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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 mL

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SI Prefix Conversions mega-M10 6 deci-d10 -1 centi-c10 -2 milli-m10 -3 PrefixSymbolFactor micro- 10 -6 nano-n10 -9 pico-p10 -12 kilo-k10 3 move left move right BASE UNIT---10 0

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SI Unit Conversions King Henry Died__drinking chocolate milk K H D __ d C M

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= SI Prefix Conversions NUMBER UNIT NUMBER UNIT 532 m = _______ km 0.532

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SI Prefix Conversions 1) 20 cm = ______________ m 2) 0.032 L = ______________ mL 3) 45 m = ____ mm 4) 805 dm = ______________ km 0.2 0.0805 45,000 32

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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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Dimensional Analysis Lining up conversion factors: 1 in = 2.54 cm 2.54 cm 1 in = 2.54 cm 1 in 1 in = 1 1 =

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Dimensional Analysis How many milliliters are in 1.00 quart of milk? (1L = 1.057 qt) 1.00 qt 1 L 1.057 qt = 946 mL qtmL 1000 mL 1 L

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Dimensional Analysis You have 1.5 pounds of gold. Find its volume in cm 3 if the density of gold is 19.3 g/cm 3. (1 kg = 2.2 lbs) lbcm 3 1.5 lb 1 kg 2.2 lb = 35 cm 3 1000 g 1 kg 1 cm 3 19.3 g

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Dimensional Analysis 5) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off? (1 in=2.54cm) 8.0 cm1 in 2.54 cm = 3.2 in cmin

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