I. Using Measurements (p. 44 - 57) CH. 2 - MEASUREMENT I. Using Measurements (p. 44 - 57) C. Johannesson
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 C. Johannesson
B. Percent Error Indicates accuracy of a measurement your value accepted value C. Johannesson
B. Percent Error % error = 2.9 % 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 = 2.9 % C. Johannesson
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 2.35 cm C. Johannesson
C. 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 C. Johannesson
Counting Sig Fig Examples C. Significant Figures Counting Sig Fig Examples 1. 23.50 1. 23.50 4 sig figs 2. 402 2. 402 3 sig figs 3. 5,280 3. 5,280 3 sig figs 4. 0.080 4. 0.080 2 sig figs C. Johannesson
C. Significant Figures (13.91g/cm3)(23.3cm3) = 324.103g 324 g 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 3 SF 324 g C. Johannesson
C. Significant Figures 3.75 mL + 4.1 mL 7.85 mL 3.75 mL + 4.1 mL 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. 3.75 mL + 4.1 mL 7.85 mL 3.75 mL + 4.1 mL 7.85 mL 224 g + 130 g 354 g 224 g + 130 g 354 g 7.9 mL 350 g C. Johannesson
C. 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 C. Johannesson
C. Significant Figures Practice Problems 5. (15.30 g) ÷ (6.4 mL) 4 SF 2 SF = 2.390625 g/mL 2.4 g/mL 2 SF 6. 18.9 g - 0.84 g 18.1 g 18.06 g C. Johannesson
D. 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 exponent Small # (<1) negative exponent Only include sig figs. C. Johannesson
Mathematics and Chemistry Section Mathematics and Chemistry 1.1 Scientific Notation A number in the form a x 10n is written in scientific notation where 1 ≤ a < 10, and n is an integer. (An integer is a whole number, not a fraction, that can be positive, negative, or zero.) When moving the decimal point to the right, you reduce the exponent when using scientific notation. Right – REDUCE When moving the decimal point to the left, you make the exponent larger when using scientific notation. LEFT – LARGER Common powers of ten include 100 = 1, 101 = 10, 102 = 100, etc.
D. 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 C. Johannesson
D. Scientific Notation Calculating with Sci. Notation (5.44 × 107 g) ÷ (8.1 × 104 mol) = Type on your calculator: EXP EE EXP EE ENTER EXE 5.44 7 8.1 ÷ 4 = 671.6049383 = 670 g/mol = 6.7 × 102 g/mol C. Johannesson
E. Proportions Direct Proportion Inverse Proportion y x y x C. Johannesson
II. Units of Measurement (p. 33 - 39) CH. 2 - MEASUREMENT II. Units of Measurement (p. 33 - 39) C. Johannesson
Warm-Up Problems Sig. figs?: 0.030800, 8000, 250., .450 450kg = _____g .56mm = _____m If the density of gold is 19.3 g/mL. If you have 58 grams of gold, what is the volume? (d=m/v) (remember sig figs!) The moon is 250,000 miles away. How many feet is it from Earth?(1 mi =5280 ft) C. Johannesson
Warm-Up Problems Sig. figs?: 0.030800; 8000; 250. ; 0.450 450kg = _____g .56mm = _____m If the density of gold is 19.3 g/mL. If you have 58 grams of gold, what is the volume? (d=m/v) (remember sig figs!) The moon is 250,000. miles away. How many feet is it from Earth?(1 mi = 5280 ft) C. Johannesson
A. Number vs. Quantity Quantity - number + unit UNITS MATTER!! C. Johannesson
B. 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 C. Johannesson
B. SI Units Prefix Symbol Factor mega- M 106 kilo- k 103 BASE UNIT --- 100 deci- d 10-1 centi- c 10-2 milli- m 10-3 micro- 10-6 nano- n 10-9 pico- p 10-12 C. Johannesson
M V D = C. Derived Units 1 cm3 = 1 mL 1 dm3 = 1 L Combination of base units. Volume (m3 or cm3) length length length 1 cm3 = 1 mL 1 dm3 = 1 L D = M V Density (kg/m3 or g/cm3) mass per volume C. Johannesson
D. Density Mass (g) Volume (cm3) C. Johannesson
Problem-Solving Steps 1. Analyze 2. Plan 3. Compute 4. Evaluate C. Johannesson
D. Density V = 825 cm3 M = DV D = 13.6 g/cm3 M = (13.6 g/cm3)(825cm3) 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 C. Johannesson
D. Density D = 0.87 g/mL V = M V = ? M = 25 g V = 25 g 0.87 g/mL 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 C. Johannesson
III. Unit Conversions (p. 40 - 42) CH. 2 - MEASUREMENT III. Unit Conversions (p. 40 - 42) C. Johannesson
Warm-Up What is the formula for density? What units are used to express density? C. Johannesson
Warm-Up Convert the following to sci-notation 65000000g .0000765km 780mL .00000000008mol Calculate the percent error if you were supposed to get 2.6g but you actually got 3.5 g. (check notes for formula) C. Johannesson
Warm-Up Identify the following as homogenous or heterogenous mixtures: Orange juice Soda water Sand with different colored grains Raisin bran cereal Mixed up Kool-Aid Soil C. Johannesson
Warm-Up Convert the following to sci-notation 65000000g .0000765km 780mL .00000000008mol Calculate the percent error if you were supposed to get 2.6g but you actually got 3.5 g. (check notes for formula) C. Johannesson
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? C. Johannesson
A. SI Prefix Conversions = 532 m = _______ km 0.532 NUMBER UNIT NUMBER UNIT C. Johannesson
A. SI Prefix Conversions Symbol Factor mega- M 106 kilo- k 103 BASE UNIT --- 100 deci- d 10-1 move left move right centi- c 10-2 milli- m 10-3 micro- 10-6 nano- n 10-9 pico- p 10-12 C. Johannesson
A. 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 C. Johannesson
B. Dimensional Analysis The “Factor-Label” Method Units, or “labels” are canceled, or “factored” out C. Johannesson
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. C. Johannesson
B. 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 C. Johannesson
B. Dimensional Analysis How many milliliters are in 1.00 quart of milk? (1L=1.057qt) qt mL 1.00 qt 1 L 1.057 qt 1000 mL 1 L = 946 mL C. Johannesson
B. Dimensional Analysis You have 1.5 pounds of gold. Find its volume in cm3 if the density of gold is 19.3 g/cm3. (1kg=2.2lbs) lb cm3 1.5 lb 1 kg 2.2 lb 1000 g 1 kg 1 cm3 19.3 g = 35 cm3 C. Johannesson
B. Dimensional Analysis How many liters of water would fill a container that measures 75.0 in3? (2.54cm=1 in) in3 L 75.0 in3 (2.54 cm)3 (1 in)3 1 L 1000 cm3 = 1.23 L C. Johannesson
B. Dimensional Analysis 5) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off? (1 in = 2.54 cm) cm in 8.0 cm 1 in 2.54 cm = 3.2 in C. Johannesson
B. Dimensional Analysis 6) Taft football needs 550 cm for a 1st down. How many yards is this? (1 in = 2.54 cm) (1ft = 12in) (1yd= 3ft) cm yd 550 cm 1 in 2.54 cm 1 ft 12 in 1 yd 3 ft = 6.0 yd C. Johannesson
B. Dimensional Analysis 7) A piece of wire is 1.3 m long. How many 1.5-cm pieces can be cut from this wire? cm pieces 1.3 m 100 cm 1 m 1 piece 1.5 cm = 86 pieces C. Johannesson