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Standards of Measurements Chapter 1.2. Accuracy and Precision Accuracy – how close a measured value is to the actual value Precision – how close the measured.

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Presentation on theme: "Standards of Measurements Chapter 1.2. Accuracy and Precision Accuracy – how close a measured value is to the actual value Precision – how close the measured."— Presentation transcript:

1 Standards of Measurements Chapter 1.2

2 Accuracy and Precision Accuracy – how close a measured value is to the actual value Precision – how close the measured values are to each other

3 Significant Figures All nonzero digits are significant.  1, 2, 3, 4, 5, 6, 7, 8, 9 Zeros within a number are always significant.  Both 4308 and 40.05 have four significant figures.

4 Significant Figures Zeros that set the decimal point are not significant.  470,000 has two significant figures. Trailing zeros that aren't needed to hold the decimal point are significant.  4.00 has three significant figures.

5 Significant Figures If the least precise measurement in a calculation has three significant figures, then the calculated answer can have at most three significant figures.  Mass = 34.73 grams  Volume = 4.42 cubic centimeters.  Rounding to three significant figures, the density is 7.86 grams per cubic centimeter.

6 Scientific Notation For large numbers, moving the decimal to the left will result in a positive number  346500 = 3.46 x 10 5 For small numbers, moving the decimal to the right will result in a negative number  0.000145 = 1.45 x 10 -4 For numbers less than 1 that are written in scientific notation, the exponent is negative.

7 Scientific Notation Before numbers in scientific notation can be added or subtracted, the exponents must be equal.  5.32 x 10 5 + 9.22 x 10 4  5.32 x 10 5 + 0.922 x 10 5  5.32 + 0.922 x 10 5  6.24 x 10 5

8 Scientific Notation When numbers in scientific notation are multiplied, only the number is multiplied. The exponents are added. (3.33 x 10 2 ) (2.71 x 10 4 ) (3.33) (2.71) x 10 2+4 9.02 x 10 6

9 Scientific Notation When numbers in scientific notation are divided, only the number is divided. The exponents are subtracted. 4.01 x 10 9 1.09 x 10 2 4.01 x 10 9-2 1.09 3.67 x 10 7

10 Scientific Notation A rectangular parking lot has a length of 1.1 × 10 3 meters and a width of 2.4 × 10 3 meters. What is the area of the parking lot? (1.1 x 10 3 m) (2.4 x 10 3 m) (1.1 x 2.4) (10 3+3 ) (m x m) 2.6 x 10 6 m 2

11 SI Units Kilo- (k) 1000 Milli- (m) Hecto- (h) Deka- (da) Base Unit Deci- (d) Centi- (c) 100 10 m, L, g 0.1 0.01 0.001 Mnemonic device: King Henry Died By Drinking Chocolate Milk

12 Metric System Meter (m) – The basic unit of length in the metric system Length – the distance from one point to another  A meter is slightly longer than a yard

13 Metric System Liter (L) – the basic unit of volume in the metric system  A liter is almost equal to a quart

14 Metric System Gram (g) – The basic unit of mass

15 Derived Units Combination of base units  Volume – length  width  height 1 cm 3 = 1 mL  Density – mass per unit volume (g/cm 3 ) D = MVMV D M V

16 Density 1) 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 x 825 cm 3 cm 3 1 M = 11,220 g D M V

17 Density 2) 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 D M V WORK : V = M D V = 25 g 0.87 g/mL V = 28.7 mL

18 Density 3) You have a sample with a mass of 620 g & a volume of 753 cm 3. Find the density. GIVEN: M = 620 g V = 753 cm 3 D = ? D M V WORK : D = M V D = 620 g 753 cm 3 D = 0.82 g/cm 3

19 Unit Factors

20 Slopes

21 Percentage Error

22 Temperature A degree Celsius is almost twice as large as a degree Fahrenheit. You can convert from one scale to the other by using one of the following formulas:

23 Temperature Convert 90 degrees Fahrenheit to Celsius  o C = 5/9 ( o F - 32)  o C = 5/9 (90 - 32)  o C = 0.55555555555555556 (58)  o C = 32.2

24 Temperature Convert 50 degrees Celsius to Fahrenheit  o F = 9/5 ( o C ) + 32  o F = 9/5 (50 ) + 32  o F = 1.8 (50) + 32  o F = 90 + 32  o F = 122

25 Temperature The SI base unit for temperature is the kelvin (K). A temperature of 0 K, or 0 kelvin, refers to the lowest possible temperature that can be reached. In degrees Celsius, this temperature is –273.15°C. To convert between kelvins and degrees Celsius, use the formula:

26 Temperature


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