1. What are we doing today Overview of chapter Time to work on review in groups Jeopardy Accuracy and Precision Video

2. Steps in the Scientific Method 1.Identify Problem or an Unknown …then Research the Problem 1.Hypothesis »tentative explanation/prediction of experimental observations, a testable prediction 3. Experiment to Test Hypothesis »gathering new information to decide whether the hypothesis is valid 4. Analysis 5. Conclusion »if hypothesis is wrong – start over »if hypothesis is correct – finished

3. Types of Data/Observations Observations are also called data. Qualitative data Quantitative data clear liquid -- -- e.g., e.g., Descriptions, physical appearance Measurements, numerical data 55 L or 83 o C

4. The Fundamental SI Units (le Système International, SI) Quantity (Symbol)Base UnitAbbrev. Length Mass Time Temp meter kilogram second kelvin m kg s K Amountmolemol (l) (m) (t) (T) (n)

5. SI Prefixes Common to Chemistry PrefixUnit Abbr.Exponent Kilok10 3 Hectoh10 2 Decada10 1 Unit10 0 Decid10 -1 Centic10 -2 Millim10 -3 Kids Have Died Using Drugs Cocaine and Marijuana King Henry Died by Drinking Chocolate Milk

6. SI Prefix Conversions 1) 20 cm = ______________ m 2) 0.032 L = ______________ mL 3) 45  m = ______________ nm 4) 805 dm = ______________ km 0.2 0.0805 45,000 32

7. C. Johannesson 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

8. C. Johannesson 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

9. Percent Error  Indicates accuracy of a measurement your value accepted value

10. 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 = 2.9 %

11. C. Johannesson 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 2.3 cm

12. Sig Fig Practice #1 How many significant figures in each of the following? 1.0070 m  5 sig figs 17.10 kg  4 sig figs 100,890 L  5 sig figs 3.29 x 10 3 s  3 sig figs 0.0054 cm  2 sig figs 3,200,000  2 sig figs

13. Sig Fig Practice #2 3.24 m x 7.0 m CalculationCalculator says:Answer 22.68 m 2 23 m 2 100.0 g ÷ 23.7 cm 3 4.219409283 g/cm 3 4.22 g/cm 3 0.02 cm x 2.371 cm 0.04742 cm 2 0.05 cm 2 710 m ÷ 3.0 s 236.6666667 m/s240 m/s 1818.2 lb x 3.23 ft5872.786 lb·ft 5870 lb·ft 1.030 g ÷ 2.87 mL 2.9561 g/mL2.96 g/mL

14. Sig Fig Practice #3 3.24 m + 7.0 m CalculationCalculator says:Answer 10.24 m 10.2 m 100.0 g - 23.73 g 76.27 g 76.3 g 0.02 cm + 2.371 cm 2.391 cm 2.39 cm 713.1 L - 3.872 L 709.228 L709.2 L 1818.2 lb + 3.37 lb1821.57 lb 1821.6 lb 2.030 mL - 1.870 mL 0.16 mL 0.160 mL

15. 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

16. Scientific Notation 2,400,000  g  0.00256 kg  7  10 -5 km  6.2  10 4 mm  Practice Problems 2.4  10 6  g 2.56  10 -3 kg 0.00007 km 62,000 mm

17. Scientific Notation  Calculating with Sci. Notation (5.44 × 10 7 g) ÷ (8.1 × 10 4 mol) = 5.44 2 nd EE ÷ ÷ 2 nd EE ENTER 78.1 4 = 671.6049383= 670 g/mol= 6.7 × 10 2 g/mol Type on your calculator:

18. Proportions  Direct Proportion  Inverse Proportion y x y x  The quotient of two variables is a constant  As the value of one variable increases, the other must also increase  As the value of one variable decreases, the other must also decrease  The graph of a direct proportion is a straight line  The product of two variables is a constant  As the value of one variable increases, the other must decrease  As the value of one variable decreases, the other must increase  The graph of an inverse proportion is a hyperbola

19. Dimensional Analysis  Steps: 1.Always start with the given quantity. 2.Identify starting & ending units. 3.Line up conversion factors so units cancel. 4.Multiply all top numbers & divide by each bottom number. 5.Check units & answer.

20. Dimensional Analysis  How many milliliters are in 1.00 quart of milk? 1.00 qt 1 L 1.057 qt = 946 mL qtmL 1000 mL 1 L 

21. 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. lbcm 3 1.5 lb 1 kg 2.2 lb = 35 cm 3 1000 g 1 kg 1 cm 3 19.3 g