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Ch. 3 Scientific Measurement

Ch. 3 Scientific Measurement. 3.1 Using and Expressing Measurements. Measurements have quantity and unit. Scientific Notation. Scientific notation: useful for very large or very small numbers 1 g of hydrogen contains 602,000,000,000,000,000,000,000 hydrogen atoms

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Ch. 3 Scientific Measurement

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  1. Ch. 3 Scientific Measurement

  2. 3.1 Using and Expressing Measurements • Measurements have quantity and unit

  3. Scientific Notation • Scientific notation: useful for very large or very small numbers • 1 g of hydrogen contains 602,000,000,000,000,000,000,000 hydrogen atoms (or 6.02 x 1023 hydrogen atoms) • 1 atom of gold has a mass of 0.000000000000000000000327 g (or 3.27 x 10-22 g)

  4. Scientific Notation • For adding and subtracting, the exponents must be the same before the coefficients can be added or subtracted • To multiply numbers written in scientific notation: multiply coefficients and add the exponents • (2 x 104) x (6 x 101) = (2 x 6) x 104+1 = 12 x 105 =1.2 x 106 • To divide: divide coefficients and subtract the denominator exponent from the numerator exponent

  5. Practice • Solve and express the answer in scientific notation • (4.0 x 103) x (2.0 x 10-5) • (8.0 x 10-2) x (7.0 x 10-5) • (8.0 x 103) ÷ (4.0 x 103) • (6.0 x 104) ÷ (3.0 x 10-3) • (7.1 x 10-2) + (5x 10-3) • (2.7 x 103) + (4.0 x 102) • (9.4 x 107) − (4.3 x 107) • (3.5 x 1013) − (4 x 1012)

  6. Practice • Solve and express the answer in scientific notation • (4.0 x 103) x (2.0 x 10-5)= 8.0 x 10-2 • (8.0 x 10-2) x (7.0 x 10-5)= 5.6 x 10-6 • (8.0 x 103) ÷ (4.0 x 103)= 2.0 x 100 • (6.0 x 104) ÷ (3.0 x 10-3)= 2.0 x 107 • (7.1 x 10-2) + (5x 10-3)= 7.6 x 10-2 • (2.7 x 103) + (4.0 x 102)= 3.1 x 103 • (9.4 x 107) − (4.3 x 107)= 5.1 x 107 • (3.5 x 1013) − (4 x 1012)= 3.1 x 1013

  7. Accuracy, Precision, and Error • Accuracy: close to true value • Precision: close to each other

  8. Accuracy, Precision, and Error • Error: difference between experimental value and the true(accepted) value • Experimental value – true value • Percent error: x 100%

  9. Significant Figures (Sig. Fig’s) • Use to represent how certain we can be of a measurement • Last digit is estimated • Limited by the precision of the measuring tool • If measuring to 1 mL certainty, you can’t know to 0.001 mL • If measuring to 100 mL certainty, can’t know to 1 mL • If calculating, can only know to the least certain measurement

  10. Significant Figures • Any non-zero digit is significant • Zeros between non-zero digits are significant • Zeros to the left of the first non-zero digit are not significant • All zeros to the right of both a decimal point and a non-zero number are significant • For numbers that do not contain decimal points, trailing zeros are not significant (unless counting whole objects or using an accepted conversion factor)

  11. Practice Sig. Fig’s • How many sig. fig’s are in each of the following? • 0.05730 m • 8765 m • 0.00073 m • 40.007 m • 98.473 L • 0.00076321 cg • 57.048 m • 12.170 mL • 0.00749830 x 104 mm • 5200 g • 5.200 x 104dL

  12. Practice Sig. Fig’s • How many sig. fig’s are in each of the following? • 0.05730 m 4 • 8765 m 4 • 0.00073 m 2 • 40.007 m 5 • 8.47 L 3 • 0.00076321 cg 5 • 7.048 m 4 • 12.170 mL 5 • 0.00749830 x 104 mm6 • 5200 g 2 • 5.200 x 104dL4

  13. Rounding • When adding or subtracting, round to fewestdecimal places in given measurements • When multiplying or dividing, round to fewest significant figures in given measurements • Avoid rounding errors! Wait until the end of a multi-step problem to round. Until then, use as many digits as you are given.

  14. Practice Sig. Fig’s with Rounding • Round each of the following answers correctly. • 8.7 g + 15.43 g + 19 g = 43.13 g • 4.32 cm x 1.7 cm = 7.344 cm2 • 853.2 L – 627.443 L = 225.757 L • 38.742 m2 ÷ 0.421 m = 92.02375 m • 5.40 m x 3.21 m x 1.871 m = 32.431914 m3 • 5.47 m3 + 11 m3 + 87.300 m3 = 103.770 m3

  15. Practice Sig. Fig’s with Rounding • Round each of the following answers correctly. • 8.7 g + 15.43 g + 19 g = 43.13g • 4.32 cm x 1.7 cm = 7.344cm2 • 853.2 L – 627.443 L = 225.757L = 225.8 L • 38.742 m2 ÷ 0.421 m = 92.02375m • 5.40 m x 3.21 m x 1.871 m = 32.431914m3 • 5.47 m3 + 11 m3 + 87.300 m3 = 103.770m3 = 104 m3

  16. 3.2 Units of Measurement • SI Base Units • Length: meter (m) • Mass: kilogram (kg) • Temperature: kelvin (K) • Time: second (s) • Amount of substance: mole (mol) • Luminous intensity: candela (cd) • Electric current: ampere (A)

  17. Common Metric Prefixes

  18. Metric Units of Length

  19. Units of Volume

  20. Units of Mass

  21. Temperature Scales Temperature: measure of the average kinetic energy of particle motion

  22. Density • Ratio of mass of an object to its volume • D=m/v • Intensive property

  23. 3.3 Solving Conversion Problems • Conversion Factors: ratio of equivalent measurements; changes numerical value but not actual quantity measured • 100 cm / 1 m • 12 inches / 1 foot • 6.022 × 1023 particles / 1 mol • Changes numerical value, but not actual quantity • Come in pairs • 1000 g / 1 kg and 1 kg / 1000 g

  24. What conversion factors are necessary for the following problems? • How many seconds are there in an 8-hour workday? • How many minutes are there in one week? • How many seconds are in a 40-hour work week?

  25. Dimensional Analysis • A way to analyze and solve problems using the units (dimensions) of the measurements • Useful for unit conversions • Especially useful for multistep problems • AKA factor-label method

  26. Solve the following problems using dimensional analysis. 1. How many seconds are in a 47-hour work week? 2. An experiment requires each student use a 3.1 cm length of magnesium ribbon. How many students can do the experiment if there is a 104 cm length of magnesium ribbon available? 3. An atom of gold has a mass of 3.271 x 10-22 g. How many atoms of gold are in 8.921 g of gold?

  27. Solve the following problems using dimensional analysis. 1. How many seconds are in a 47-hour work week? 47 hr x x = 169,200 s = 1.7 x 105 s 2. An experiment requires each student use a 3.1 cm length of magnesium ribbon. How many students can do the experiment if there is a 104 cm length of magnesium ribbon available? 104 cm x = 33.5 students  33 students 3. An atom of gold has a mass of 3.271 x 10-22 g. How many atoms of gold are in 8.921 g of gold? 8.921 g x = 2.727 x 1022 atoms

  28. Solve the following problems using dimensional analysis. 4. Convert 0.0453 km to meters. 5. Convert 52.9 mL to liters. 6. Convert 0.0649 g to centigrams. 7. Convert 152869.4 g to megagrams.

  29. Solve the following problems using dimensional analysis. 4. Convert 0.0453 km to meters. 0.0453 km x = 45.3 m 5. Convert 52.9 mL to liters. 52.9 mL x =0.0529 L 6. Convert 0.0649 g to centigrams. 0.0649 g x = 6.49 cg 7. Convert 152869.4 g to megagrams. 152869.4 g x = 0.1528694 Mg

  30. Dimensional Analysis Activity • Each group will get a segment of the Cardinals’ roster. Please write on a separate page, not on the roster. • Convert heights from feet and inches to meters (2.54 cm = 1 inch). • Convert weights in pounds to masses in kilograms (454 g = 1 lb).

  31. Solve the following problems using dimensional analysis. 8. Convert 291.4 cm3to liters. 9. Convert 53.6 g of boron to cm3of boron. The density of boron is 2.34 g/cm3. 10. Convert 45.38 cm3of boron to grams of boron.

  32. Solve the following problems using dimensional analysis. 8. Convert 291.4 cm3to liters. 291.4 cm3x x = 0.2914 L 9. Convert 53.6 g of boron to cm3of boron. The density of boron is 2.34 g/cm3. 53.6 g boron x =22.9 cm3 boron 10. Convert 45.38 cm3of boron to grams of boron. 45.38cm3 boron x = 106 g boron

  33. Solve the following problems using dimensional analysis. 11. What is the mass, in grams, of a sample of cough syrup that has a volume of 50.0 cm3? The density of cough syrup is 0.950 g/cm3. 12. The radius of a particular atom is 0.573 nm. Express this in cm. 13. The diameter of Earth is 1.3 x 104 km. What is this diameter expressed in mm? In Mm?

  34. 11. What is the mass, in grams, of a sample of cough syrup that has a volume of 50.0 cm3? The density of cough syrup is 0.950 g/cm3. 50.0 cm3 cough syrup x = 47.5 g coughsyr. 12. The radius of a particular atom is 0.573 nm. Express this in cm. 0.573 nm x x = 5.73 x 10-8 cm 13. The diameter of Earth is 1.3 x 104 km. What is this diameter expressed in mm? In Mm? 1.3 x 104km x x = 1.3 x 1010 mm 1.3 x 104 km x x = 1.3 x 101 Mm

  35. Solve the following problems using dimensional analysis. 14. Gold has a density of 19.3 g/cm3. What is the density in kilograms per cubic meter? 15. There are 7.0 × 106 red blood cells in 1.0 mm3 of blood. How many red blood cells are in 1.0 L of blood?

  36. Solve the following problems using dimensional analysis. 14. Gold has a density of 19.3 g/cm3. What is the density in kilograms per cubic meter? 19.3 x x = 1.93 x 104 kg/m3 15. There are 7.0 × 106 red blood cells in 1.0 mm3 of blood. How many red blood cells are in 1.0 L of blood? 1.0 L x x = 7.0 x cells

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