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CH. 2 – MEASUREMENT AND CALCULATIONS PowerPoint Presentation
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CH. 2 – MEASUREMENT AND CALCULATIONS

CH. 2 – MEASUREMENT AND CALCULATIONS

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CH. 2 – MEASUREMENT AND CALCULATIONS

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  1. CH. 2 – MEASUREMENT AND CALCULATIONS I. Scientific Method

  2. The Scientific Method • A logical approach to solving problems or answering questions. • Starts with observation- noting and recording information and facts • hypothesis- educated guess or testable statement

  3. Steps in the Scientific Method 1. Observations (uses your senses) a) quantitative involves numbers = 95oF b) qualitative is word description = hot 2. Formulating hypotheses (ideas) 3. Performing experiments (the test) - gathers new information to help decide whether the hypothesis is valid

  4. Scientific Method • Controls- constants • Variables- changing conditions • Limit variables • We gather data and observations by doing the experiment • Modify hypothesis - repeat the cycle based on results

  5. Steps in the Scientific Method • Theorize (model) - explanation of some natural phenomenon • Many phenomena- construct a theory • Publish Results - Do other experts agree

  6. CH. 2 – MEASUREMENTs AND CALCULATIONS II. Units of Measurement

  7. Number vs. Quantity • Quantity - number + unit UNITS MATTER!!

  8. 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

  9. 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 SI Units Prefix Symbol Factor

  10. M V D = 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

  11. Density Mass (g) Volume (cm3)

  12. 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

  13. WORK: V = M D V = 25 g 0.87 g/mL 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

  14. CH. 2 – MEASUREMENTs AND CALCULATIONS III. Using Measurements

  15. 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

  16. your value accepted value Percent Error • Indicates accuracy of a measurement

  17. % error = 2.9 % 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.

  18. 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

  19. 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

  20. 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

  21. 3 SF 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

  22. 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

  23. 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

  24.  2.4 g/mL 2 SF Significant Figures Practice Problems 5. (15.30 g) ÷ (6.4 mL) 4 SF 2 SF = 2.390625 g/mL 6. 18.9 g - 0.84 g  18.1 g 18.06 g

  25. 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 exponentSmall # (<1)  negative exponent • Only include sig figs.

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

  27. EXP EXP ENTER EE EE 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

  28. y y x x Proportions • Direct Proportion • Inverse Proportion

  29. CH. 2 – MEASUREMENTS AND CALCULATIONS Unit Conversions

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

  31. 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.

  32. 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

  33. qt mL  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

  34. lb cm3 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

  35. cm in Dimensional Analysis 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 = 3.2 in

  36. cm yd Dimensional Analysis 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

  37. cm pieces Dimensional Analysis A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire? 1 piece 1.5 cm 100 cm 1 m 1.3 m = 86 pieces

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