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CH. 2 - MEASUREMENT

CH. 2 - MEASUREMENT. I. Using Measurements (p. 44 - 57). 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. your value. accepted value.

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CH. 2 - MEASUREMENT

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

  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 C. Johannesson

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

  4. % error = 2.9 % 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. C. Johannesson

  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 C. Johannesson

  6. 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 C. Johannesson

  7. C. Significant Figures Counting Sig Fig Examples 1. 23.50 1. 23.50 4 sig figs 3 sig figs 2. 402 2. 402 3. 5,280 3. 5,280 3 sig figs 2 sig figs 4. 0.080 4. 0.080 C. Johannesson

  8. 3 SF 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 4 SF 3 SF 324g C. Johannesson

  9. 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. 224 g + 130 g 354 g 224 g + 130 g 354 g 3.75 mL + 4.1 mL 7.85 mL 3.75 mL + 4.1 mL 7.85 mL  350 g  7.9 mL 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 C. Johannesson

  11. 5. (15.30 g) ÷ (6.4 mL)  2.4 g/mL 2 SF C. Significant Figures Practice Problems 4 SF 2 SF = 2.390625 g/mL 6. 18.9 g - 0.84 g  18.1 g 18.06 g C. Johannesson

  12. 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 C. Johannesson

  13. 7. 2,400,000 g 8. 0.00256 kg 9. 7  10-5 km 10. 6.2  104 mm D. Scientific Notation Practice Problems 2.4  106 g 2.56  10-3 kg 0.00007 km 62,000 mm C. Johannesson

  14. EXE EXP EXP ENTER EE EE D. Scientific Notation • Calculating with Sci. Notation (5.44 × 107 g) ÷ (8.1 × 104 mol) = Type on your calculator: 5.44 7 8.1 4 ÷ = 671.6049383 = 670 g/mol = 6.7 × 102 g/mol C. Johannesson

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

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

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

  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 C. Johannesson

  19. kilo- mega- M k 103 106 deci- BASE UNIT d --- 100 10-1 centi- c 10-2 milli- m 10-3 micro-  10-6 nano- n 10-9 pico- p 10-12 B. SI Units Prefix Symbol Factor C. Johannesson

  20. M V D = C. Derived Units • Combination of base units. • Volume (m3 or cm3) • length  length  length 1 cm3 = 1 mL 1 dm3 = 1 L • Density (kg/m3 or g/cm3) • mass per volume C. Johannesson

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

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

  23. 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 C. Johannesson

  24. WORK: V = M D V = 25 g 0.87 g/mL 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 V = 29 mL C. Johannesson

  25. CH. 2 - MEASUREMENT III. Unit Conversions (p. 40 - 42) C. Johannesson

  26. To the left or right? A. SI Prefix Conversions 1. Find the difference between the exponents of the two prefixes. 2. Move the decimal that many places. C. Johannesson

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

  28. mega- kilo- k M 106 103 BASE UNIT deci- d --- 100 10-1 centi- c 10-2 milli- m 10-3 micro-  10-6 nano- n 10-9 pico- p 10-12 A. SI Prefix Conversions Prefix Symbol Factor move left move right C. Johannesson

  29. 1) 20 cm = ______________ m 2) 0.032 L = ______________ mL 3) 45 m = ______________ nm 4) 805 dm = ______________ km A. SI Prefix Conversions 0.2 32 45,000 0.0805 C. Johannesson

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

  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. C. Johannesson

  32. B. Dimensional Analysis • Lining up conversion factors: = 1 1 in = 2.54 cm 2.54 cm 2.54 cm 1 = 1 in = 2.54 cm 1 in 1 in C. Johannesson

  33. qt mL  1. Dimensional Analysis • How many milliliters are in 1.00 quart of milk? 1 L 1.057 qt 1000 mL 1 L 1.00 qt = 946 mL C. Johannesson

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

  35. in3 L 3. Dimensional Analysis • How many liters of water would fill a container that measures 75.0 in3? 1 L 1000 cm3 (2.54 cm)3 (1 in)3 75.0 in3 = 0.191 L C. Johannesson

  36. cm in 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 C. Johannesson

  37. cm yd 5. Dimensional Analysis 5) Taft football needs 550 cm for a 1st down. How many yards is this? 1 ft 12 in 1 yd 3 ft 1 in 2.54 cm 550 cm = 6.0 yd C. Johannesson

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