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I II III I. Units of Measurement Chapter 2 – Scientific Measurement.

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1 I II III I. Units of Measurement Chapter 2 – Scientific Measurement

2 Number vs. Quantity  Quantity – number + unit UNITS MATTER!!

3 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

4 A. Accuracy vs. Precision

5 B. Percent Error  Indicates accuracy of a measurement your value given value

6 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. % error = 2.94 %

7 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 2.31 cm

8  Same object with different rulers: C. Significant Figures

9  Counting Sig Figs  Digits from 1-9 are always significant.  Zeros between two other sig figs are always significant  One or more additional zeros to the right of both the decimal place and another sig digit are significant  Count all numbers EXCEPT:  Leading zeros  Trailing zeros without a decimal point -- 2,

10 , C. Significant Figures Counting Sig Fig Examples , sig figs 3 sig figs 2 sig figs

11 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 ) = g 324 g 4 SF3 SF

12 C. Significant Figures  Calculating with Sig Figs (cont’d)  Add/Subtract – The # with the lowest decimal value determines the place of the last sig fig in the answer mL mL 7.85 mL 224 g g 354 g  7.9 mL  350 g 3.75 mL mL 7.85 mL 224 g g 354 g

13 C. Significant Figures  Calculating with Sig Figs (cont’d)  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

14 C. Significant Figures 5. (15.30 g) ÷ (6.4 mL) Practice Problems = g/mL  18.1 g g g g 4 SF2 SF  2.4 g/mL 2 SF

15 D. Scientific Notation  A way to express any number as a number between 1 and 10 (coefficient) multiplied by 10 raised to a power (exponent)  Number of carbon atoms in the Hope diamond  460,000,000,000,000,000,000,000  4.6 x  Mass of one carbon atom  g  2 x g coefficient exponent

16 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 – all of them! 65,000 kg  6.5 × 10 4 kg

17 D. Scientific Notation 7. 2,400,000  g kg  km  10 4 mm Practice Problems 2.4  10 6  g 2.56  kg km 62,000 mm

18 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 = = 670 g/mol= 6.7 × 10 2 g/mol Type on your calculator:

19 D. Scientific Notation 11. (4 x 10 2 cm) x (1 x 10 8 cm) 12. (2.1 x kg) x (3.3 x 10 2 kg) 13. (6.25 x 10 2 ) ÷ (5.5 x 10 8 ) 14. (8.15 x 10 4 ) ÷ (4.39 x 10 1 ) 15. (6.02 x ) ÷ (1.201 x 10 1 ) Practice Problems 4  cm  kg x x x 10 22

20 Homework  Complete Worksheet “Using Measurements – Chapter 2”: due tomorrow!

21 II. Unit Conversions CH. 2 – MEASUREMENT

22 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?

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

24 A. SI Prefix Conversions mega-M10 6 deci-d10 -1 centi-c10 -2 milli-m10 -3 PrefixSymbolFactor micro-  nano-n10 -9 pico-p kilo-k10 3 move left move right BASE UNIT

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

26 B. 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

27 C. Density Mass (g) Volume (cm 3 )

28 C. 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

29 C. 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

30 I II III III. Dimensional Analysis Conversion Factors Problems CH. 2 – MEASUREMENT

31 A. Vocabulary  Dimensional Analysis  A tool often used in science for converting units within a measurement system  Conversion Factor  A numerical factor by which a quantity expressed in one system of units may be converted to another system

32 B. Dimensional Analysis  The “Factor-Label” Method  Units, or “labels” are canceled, or “factored” out

33 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.

34 Fractions in which the numerator and denominator are EQUAL quantities expressed in different units Example: 1 in. = 2.54 cm Factors: 1 in. and 2.54 cm 2.54 cm 1 in. C. Conversion Factors

35 Write conversion factors that relate each of the following pairs of units: 1. Liters and mL 2. Hours and minutes 3. Meters and kilometers C. Conversion Factors Learning Check: 1 L 1000 mL 1 hr 60 min 1000 m 1 km 1000 mL 1 L =

36 Conversion factor cancel By using dimensional analysis / factor-label method, the UNITS ensure that you have the conversion right side up, and the UNITS are calculated as well as the numbers! How many minutes are in 2.5 hours? 2.5 hr 1 x xx x 60 min 1 hr = 150 min

37  You have $7.25 in your pocket in quarters. How many quarters do you have? X = 29 quarters D. Dimensional Analysis Practice 7.25 dollars 1 dollar 4 quarters

38 How many seconds are in 1.4 days? Plan: days hr min seconds 1.4 days x 24 hr x 60 min x 60 sec = 1 1 day1 hr 1 min E. Dimensional Analysis Practice sec sec 1.2 x10 5 sec

39 D. Dimensional Analysis Practice  How many milliliters are in 1.00 quart of milk? 1.00 qt 1 L qt = 946 mL qtmL 1000 mL 1 L 

40  You have 1.5 pounds of gold. Find its volume in cm 3 if the density of gold is 19.3 g/cm 3. lbcm lb 1 kg 2.2 lb = 35 cm g 1 kg 1 cm g D. Dimensional Analysis Practice

41 5) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off? 8.0 cm1 in 2.54 cm = 3.1 in cmin D. Dimensional Analysis Practice

42 6) Roswell football needs 550 cm for a 1st down. How many yards is this? 550 cm 1 in 2.54 cm = 6.0 yd cmyd 1 ft 12 in 1 yd 3 ft D. Dimensional Analysis Practice

43 7) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire? 1.3 m 100 cm 1 m = 86 pieces mpieces 1 piece 1.5 cm D. Dimensional Analysis Practice

44  How many liters of water would fill a container that measures 75.0 in 3 ? 75.0 in 3 (2.54 cm) 3 (1 in) 3 = 1.23 L in 3 L 1 L 1000 cm 3 D. Dimensional Analysis Practice


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