I II III I. Using Measurements CH. 2 - MEASUREMENT

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. Significant Figures  Indicate precision of a measurement.  Recording Sig Figs  Sig figs in a measurement include the known digits plus a final estimated digit 2.32 cm

C. Significant Figures  Counting Sig Figs  Count all numbers EXCEPT:  Leading zeros  Trailing zeros without a decimal point -- 2,500  USA??

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

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

C. Significant Figures  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 mL mL 7.85 mL  7.9 mL 3.75 mL mL 7.85 mL

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. 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

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. 65,000 kg  6.5 × 10 4 kg

D. Scientific Notation 7. 2,400,000  g kg 9.7  km  10 4 mm Practice Problems 2.4  10 6  g 2.56  kg km 62,000 mm

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:

E. SI Units QuantityBase UnitAbbrev. Length Mass Time Temp meter kilogram second kelvin m kg s K Amountmolemol Symbol l m t T n

F. 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

Problem-Solving Steps 1. Analyze 2. Plan 3. Compute 4. Evaluate

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

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

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

SI Unit Conversions  King Henry Died__drinking chocolate milk  K H D __ d C M

= SI Prefix Conversions NUMBER UNIT NUMBER UNIT 532 m = _______ km 0.532

SI Prefix Conversions 1) 20 cm = ______________ m 2) L = ______________ mL 3) 45 m = ____ mm 4) 805 dm = ______________ km ,000 32

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.

Dimensional Analysis  Lining up conversion factors: 1 in = 2.54 cm 2.54 cm 1 in = 2.54 cm 1 in 1 in = 1 1 =

Dimensional Analysis  How many milliliters are in 1.00 quart of milk? (1L = qt) 1.00 qt 1 L qt = 946 mL qtmL 1000 mL 1 L 

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

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=2.54cm) 8.0 cm1 in 2.54 cm = 3.2 in cmin