Precision and Accuracy Errors in Scientific Measurements Precision - Refers to reproducibility or how close the measurements are to each other. Accuracy - Refers to how close a measurement is to the ‘true’ value. Systematic Error - produces values that are either all higher or all lower than the actual value. Random Error - in the absence of systematic error, produces some values that are higher and some that are lower than the actual value.
Rules for Determining Which Digits Are Significant 1. Make sure that the measured quantity has a decimal point. 2. Start at the left of the number and move right until you reach the first nonzero digit. 3. Count that digit and every digit to its right as significant. 4. Zeros that end a number and lie either after or before the decimal point are significant; thus 1.030 mL has four significant figures, and 5300. L has four significant figures also.
5. Situation unclear if there is no decimal point. We will adopt convention that numbers such as 5300 L have 2 sig. figs. A terminal decimal point is often used to clarify the situation, 5300. L has 4 sig. figs. Scientific notation is clearer, 5.30 x 10 3 L has 3 sig figs. 6. Exact numbers have an infinite number of significant figures. The alphabet has 26 letters. There are 12 eggs in a dozen eggs.
Examples of Significant Digits in Numbers Problem - Sig digits P 3-1a 0.0050 L P 3-1b18.00 g P 3-1c1.089 x 10 –6 L P 3-1d83.0001 L P 3-1e 0.006002 g
P 3-1f875,000 oz P 3-1g30,000 kg P 3-1h5.0000 m 3 P 3-1i23001.00 lbs P 3-1j0.000108 g P 3-1k1,470,000 L
Rules for Significant Figures in answers 1. For multiplication and division. The number with the least certainty limits the certainty of the result. therefore, the answer contains the same number of significant figures as there are in the measurement with the fewest significant figures. Multiply the following numbers: 9.2 cm x 6.8 cm x 0.3744 cm = 2. For addition and subtraction. The answer has the same number of decimal places as there are in the measurement with the fewest decimal places. Example, adding two volumes (a) 83.5 mL + 23.28 mL = Example subtracting two volumes: (b) 865.9 mL - 2.8121393 mL =
Rules for Rounding Off Numbers (1) In a series of calculations*, carry the extra digits through to the final result, then round off. ** (2) If the digit to be removed a. a.is less than 5, the preceding digit stays the same. For example, 1.33 rounds to 1.3. b. b.is equal to or greater than 5, the preceding digit is increased by one. For example, 1.36 rounds to 1.4. (3) When rounding, use only the first number to the right of the last significant figure. Do not round off sequentially. For example, the number 4.348 when rounded to two significant figures is 4.3, not 4.4. Notes: * Your TI-93 calculator has the round function which you can use to get the correct result. Find round by pressing the math key and moving to NUM. Its use is round(num, no of decimal places desired), e.g. round(2.746,1) =2.7. ** Your book will show intermediate results rounded off. Don’t use these rounded results to get the final answer.
Rounding Off Numbers – Problems (3-4a) Round 5.379 to three significant figures Ans: (3-4b) Round 5.379 to two significant figures Ans: We used the rule: If the digit removed is greater than or equal to 5, the preceding number increases by 1. (3-4a) Round 0.2413 to three significant figures Ans: (3-4b) Round 0.2413 to two significant figures Ans: We used the rule: If the digit removed is less than 5, the preceding number is unchanged
P 3-5: A small rectangular slab of lithium has the dimensions 20.9 mm by 11.1 mm by 11.9 mm. Its mass is 1.49 x 10 3 mg. What is the density of lithium in g/cm 3 ? Step 1: task: Find the density of lithium in g/cm 3. Step 2: given information: The mass is 1.49 x 10 3 mg and the dimensions are 20.9 by 11.1 by 11.9 mm.
Step 3: strategy: Use density = mass/volume, so the volume needs to be calculated. For a rectangular slab, use volume = length x width x height. Need to convert units. Step 4: set up the problem: Are we ready to calculate?
Problem 3-6: Volume by Displacement Problem: Calculate the density of an irregularly shaped metal object that has a mass of 567.85 g if, when it is placed into a 2.00 liter graduated cylinder containing 900.00 mL of water, the final volume of the water in the cylinder is 1277.56 mL ? Plan: Calculate the volume from the different volume readings, and calculate the density using the mass that was given. Solution: Volume = Density = mass volume
Definitions - Mass & Weight Mass - The quantity of matter an object contains kilogram - ( kg ) - the SI base unit of mass, is a platinum - iridium cylinder kept in Paris as a standard! Weight - depends upon an object’s mass and the strength of the gravitational field pulling on it, i.e. w = f = ma.
Problem 3-7: Computer Chips Future computers might use memory bits which require an area of a square with 0.250 m sides. (a) How many bits could be put on a 1.00 in x 1.00 in computer chip? (b) If each bit required that 25.0 % of its area to be coated with a gold film 10.0 nm thick,what mass of gold would be needed to make one chip? Approach: (a) (a)use A chip = (b) use = m/V
Temperature Scales and Interconversions Kelvin ( K ) - The “Absolute temperature scale” begins at absolute zero and only has positive values. Celsius ( o C ) - The temperature scale used by science, formally called centigrade and most commonly used scale around the world. Water freezes at 0 o C, and boils at 100 o C. Fahrenheit ( o F ) - Commonly used scale in America for our weather reports. Water freezes at 32 o F, and boils at 212 o F. T (in K) = T (in o C) + 273.15 T (in o C) = T (in K) - 273.15 T (in o F) = 9/5 T (in o C) + 32 T (in o C) = [ T (in o F) - 32 ] 5/9
Problem 3-8:Temperature Conversions (a) The boiling point of Liquid Nitrogen is -195.8 o C, what is the temperature in Kelvin and degrees Fahrenheit? T (in K) = T (in o C) + 273.15 T (in K) = T (in o F) = 9/5 T (in o C) + 32 T (in o F) = (b)The normal body temperature is 98.6 o F, what is it in Kelvin and degrees Celsius? T (in o C) = [ T (in o F) - 32] 5/9 T (in o C) = T (in K) = T (in o C) + 273.15 T (in K) =
Answers to Problems in Lecture #3 1. 1.(a)2; (b) 4; (c) 4; (d) 6; (e) 4; (f) 3; (g) 1; (h) 5; (i) 7 (j) 3; (k) 3 2. 2.23 cm 3 3(a) 106.8 mL 3(b) 863.1 mL 4.(a) 5.38; (b) 5.5; (c) 0.241; (d) 0.24 5.0.536 g/cm 3 6.1.5040 g / mL 7.31 g gold 8. (a) 77.4 K; -320.4 o F; (b) 37.0 o C; 310.2 K