2. Used for numbers that are really big or really small A number in exponential form consists of a coefficient multiplied by a power of 10 10,000 1,000,000 546,000 0.00001 0.00751 0.00000029 12,450 15,230,000 0.0884 1 x 10 4 1 x 10 6 5.46 x 10 5 1 X 10 -5 7.51 x 10 -3 2.9 x 10 -7 1.245 x 10 4 1.523 x 10 7 8.84 x 10 -2

3. If the coefficient does not fall between 1 and 10, it must be re-written correctly If you move the decimal to the right, subtract from the exponent If you move the decimal to the left, add to the exponent 100 x 10 3 0.0001 x 10 12 4,490 x 10 -7 0.0065 x 10 3 0.090 x 10 -5 0.0112 x 10 8 150 X 10 12 200 x 10 -2 1 X 10 5 1 X 10 8 4.49 x 10 -4 6.5 x 10 0 = 6.5 9 X 10 -7 1.12 x 10 6 1.5 x 10 14 2 X 10 0

4. calculations involving scientific notation Enter the coefficient, then EE or Exp Do not enter “x 10” (2.4 x 10 6 ) (3.1 x 10 3 ) = (9.5 x 10 -7 ) (5 x 10 -4 ) = 4.8 x 10 9 = 2.4 x 10 2 7.5 x 10 -5 + 4.2 x 10 -6 = 7.44 x 10 9 4.75 x 10 -10 2 X 10 7 7.92 x 10 -5

5. Two types of data Qualitative descriptiveEx: the burner flame is hot Quantitative numericalEx: the flame is 1000 ° C Measurements should be both accurate and precise how close the experimental value is to the accepted or true value calculating percent error % E = |accepted –experimental| accepted X 100

6. how close the measurements are to each other when the experiment is repeated Ex: a student does an experiment to determine the density of lead; she repeats the experiment two more times and gets these results: 10.1 g/cm 3 9.4 g/cm 3 and 8.5 g/cm 3 comment on her precision If the actual density of lead is 11.4 g/cm 3, calculate her percent error using her average density as the experimental value poor |11.4 – 9.3| % E = 11.4 100 =18.4 % Avg.= 9.3

7. Mass (weight) Quantities and their units gram (g) kilogram (kg) milligram (mg) Length (distance) meter (m) kilometer (km) millimeter (mm) Example units Volume liter (L) milliliter (ml) any unit of length that is cubed

8. 1 mL = cm 3 1m Vol. = L x W x H Vol. = 1 m 3 Temperature Celsius, ° C Kelvin, K Celsius, ° C Kelvin, K Heat Joules (J) Calorie (Cal)

9. number of particles mole 1 mole = 6.02 x 10 23 use dimensional analysis to convert between these units 1. 12 inches = ? cm 1 in. = 2.54 cm in. cm 12in. 1 2.54 30.48 cm

10. 2. 95 miles = ? km1 km = 0.62 mi 3. 400 lbs. = ? kg 1 kg = 2.2 lbs 4. 250 grams = ? oz.1 oz = 28 g 95mi km lbs kg g g oz 1 0.62 400 2.2 1 250 28 1 153 km 182 kg 8.9 oz

11. 5. 500 cm = ? in. 1 in. = 2.54 cm kilo (k) 1000 times larger 10 times smaller (1/10 ) centi (c) deci (d) 100 times smaller (1/100) 500 cm cm in 2.54 1 197 in

12. milli(m) 1000 times smaller (1/1000) micro (µ) 1 x 10 6 times smaller(1/10 6 ) nano (n) 1 x 10 9 times smaller(1/10 9 ) Ex: 5 m = ? cm 5m m cm 1 100 500 cm

13. 1.50 L = ? mL 1 X 10 3 dg = ? g 0.025 g 1500 mL Ex: 25 mg = ? g 25 mg g 1000 1mg 1.50 L mL 1 1000L 1x 10 3 dg g 10 1dg 100 g

14. 2.4 g = ? cg 3 X 10 12 µg = ? g 0.9 dm = m 2.4g g cg 1 100 240 cg 3 x 10 12 µg g 1 X 10 6 1 3 X 10 6 g 0.9 dm m 10 1 0.09 m

15. 7.7 x 10 5 µm = ? m 200 g = ? kg 7.7 x 10 5 µm µm m1 1 X 10 6 0.77 m 200 g g 0.2 kg kg1 1000 0.25 L = ? mL 0.25 L L ml 1000 1 250 mL

16. 3.5 x 10 -4 g = ? mg 800 µL = L 200 m = ? cm 200 m m1 cm100 20,000 cm 3.5 x 10 -4 g g1 mg 1000 0.35 mg 800 µL 1 x10 6 L1 0.0008 dL µL

17. all the numbers in a measurement that are known with certainty plus one that is estimated 6.3 6.4 6.35 uncertain 3 sig figs 6.3500 incorrect

18. Rules for determining which numbers in a measurement are significant figures 1.Any number in a measurement that is not zero is a significant figure 1.Any number in a measurement that is not zero is a significant figure Ex: 213.5 m has 412, 567 m has 5 2. Zeros between nonzero numbers are significant figures 2. Zeros between nonzero numbers are significant figures Ex: 205 g has310.0002 g has 6 3.Zeros to the left of a number are not significant figures Ex: 0.078 L has20.00005 has 1

19. 4. Zeros to the right of a number and to the right of the decimal are significant figures 4. Zeros to the right of a number and to the right of the decimal are significant figures Ex: 2.00 has30.00100 has3 5.Zeros at the end of a number are not significant figures unless the decimal point is shown 5.Zeros at the end of a number are not significant figures unless the decimal point is shown Ex: 1200 has 21200. has4 For numbers in exponential form, look only at the coefficient and not the exponent 4.50 x 10 8 has3

20. Identify the number of significant figures in each measurement: 1.______ 250 9.______ 13,979 2.______ 35,02910.______ 3.00 x 10 2 3.______ 0.007511.______ 0.6000 4. ______ 900012.______ 50. 5.______ 0.008013.______ 4500 6.______ 10.0014.______ 0.002 7._______ 3.6 x 10 5 15.______ 3.040 8._______ 15,000 2 5 2 1 2 4 2 2 5 3 4 2 2 1 4

21. Rounding off numbers Begin counting from the first significant figure on the left; if the number being left off is 5 or higher, round up. Ex: round 65.31890 to 3 sig figs:65.3 round 0.05981 to 3 sig figs0.0598 round 43,925 to 2 sig figs44,000 or 4.4 x 10 4 round 545,858 to 4 sig figs 545,900 or 5.459 x 10 5 round 9.9992 x 10 -4 to 2 sig figs1.0 x 10 -3

22. Round each number to 3 sig figs, then to 2 then 1 sig fig 1.6.77510 2.0.04031 3.18.298 4.0.0011299 5.892.153 6. 57,320 6.78 6.8 7 18.3 18 20 0.00113 0.0011 0.001 892 890 900 57,320 57,000 60,000 0.0403 0.040 0.04

23. When rounding off the answer after a calculation, the answer cannot be more accurate than your least accurate measurement Multiplication and Division Rule: The number of sig figs in the answer is determined by the number with the fewest sig figs

24. Ex: 1.33 x 5.7 = 0.153 = 0.08 (5.00 x 10 3 ) ( 7.2598 x 10 2 ) = 7.581 =7.6 2 1.9125 = 3.6299 x 10 6 = 3.63 x 10 6

25. Perform each calculation and round to the correct number of significant figures 1.520 x 367 = 2. 2.5 x 9.821 = 3. 0.02430 = 0.95880 4.4 x 10 -8 = 1.5 x 10 -2 190,840 = 190,000 or 1.9 x 10 5 24.5525 = 25 0.02534418 = 0.02534 2.6666 x 10 -6 =3 X 10 -6

26. Addition and Subtraction Rule: The number of decimal places in the answer is determined by the number with the fewest decimal places Ex: 10.25 + 11.1 = 515.3215 - 30.42 = 1425 - 820.95 = 21.35 = 484.9015 = 604.05 = 21.4 484.90 604

27. Perform each calculation and round off to the correct number of significant figures 1.20.5 + 8.263 = 2.0.88 + 3.104 = 3.0.005 + 0.0066 = 4. 2291.7 - 1512.015 = 28.763 = 3.984 = 0.0116 = 779.685 = 28.8 3.98 0.012 779.7

28. Ratio of an object’s mass to its volume Density of water = 1 g/mL or 1 g/cm 3 Which is more dense: the water in a tub or in a small cup ? both water samples have the same density

29. Sample problems Calculate the density of a liquid if 50 mL of the liquid weighs 46.25 grams. D = M V D = 46.25 g 50 mL 0.925 0.9 g/mL A block of metal has the dimensions: 2.55 cm x 2.55 cm x 4.80 cm. If the mass of the block is 234.61 g, what is the density? A block of metal has the dimensions: 2.55 cm x 2.55 cm x 4.80 cm. If the mass of the block is 234.61 g, what is the density? V = L x W x HV = (2.55 cm)(2.55 cm)(4.80 cm) V = 31.212 cm 3 D = 234.61 g 7.51666 7.52 g/cm 3 31.212 cm 3

30. A marble weighing 53.87 g is placed in a graduated cylinder containing 40.0 mL of water. If the water rises to 64.9 mL, what is the density of the marble? V= 64.9 – 40.0 =24.9 mL D = 53.87 g 24.9 mL 2.163452.16 g/mL 1.

31. Calculate the mass of a piece of aluminum having a volume of 8.45 cm 3. The density of aluminum is 2.7 g/cm 3 Calculate the mass of a piece of aluminum having a volume of 8.45 cm 3. The density of aluminum is 2.7 g/cm 3 (8.45 cm 3 ) D = M V M= D x V M =(2.7 g/cm 3 ) 22.81523 g What volume of mercury weighs 25.0 grams? Density of mercury = 13.6 g/mL What volume of mercury weighs 25.0 grams? Density of mercury = 13.6 g/mL D = M V M= D x V V = M D V = 25.0 g 13.6 g/mL V = 1.8382351.84 mL

32. 1.What mass of gold (density= 19.3 g/cm 3 ) has a volume of 12.80 cm 3 ? D = M V M= D x VM = (19.3 g/cm 3 )(12.80 cm 3 ) M= 247.04 =247 g 2.Calculate the volume of a cork if its mass is 2.79 grams and it has a density of 0.25 g/cm 3 D = M V M= D x V V = M D V = 2.79 g 0.25 g/cm 3 11.16 = 11 cm 3

33. 3 scales B.P. water F.P. water Anders Celsius Lord Kelvin

34. Celsius Scale:Water freezes at 0°C, boils at 100°C Kelvin Scale:Water freezes at 273 K and boils at 373 K Absolute Zero: Lowest temperature that can be reached; all molecular motion stops K = °C + 273 0 Kelvin Ex: 25 ° C = ? K 298 K 37 °C = ? K 310 K -50 °C = ? K 223 K 0 K = ?°C -273 °C 600 K = ? °C 327°C

35. Energy that flows from a region of higher temp to lower temp units: Joules (J), Calories (Cal) Specific Heat Capacity: C, A measure of how well something stores heat specific heat of water:1.00 Cal/g°C 1 Cal = 4.18 J or 4.18 J/g°C high compared to most substances; water heats up slowly and cools off slowly.

36. Heat Calculations: q = mCΔT heat mass temp change specific heat Problems How many Joules of heat energy are needed to raise the temperature of 50.0 grams of water from 24.5°C to 75.0°C ? specific heat of water = 4.18 J/g °C How many Joules of heat energy are needed to raise the temperature of 50.0 grams of water from 24.5°C to 75.0°C ? specific heat of water = 4.18 J/g °C q= mCΔT m ΔTΔT C q = (50.0 g)(75.0 – 24.5)(4.18 J/g°C) 50.5°Cq = 10,554.510, 600 J

37. A piece of gold weighing 28 grams cools from 125°C to 23°C. To do this it must lose 362 Joules of heat energy. Calculate the specific heat of gold. q= mCΔT mΔTmΔTmΔTmΔT C = q m ΔT C = 362 J 28 g125- 23 102°C C = 0.12675 0.13 J/g°C

38. What mass of graphite can be heated from 30°C to 80°C by the addition of 1500 Joules of heat? Specific heat of graphite = 0.709 J/g°C What mass of graphite can be heated from 30°C to 80°C by the addition of 1500 Joules of heat? Specific heat of graphite = 0.709 J/g°C q= mCΔT C Δ TCΔTCΔT m = q CΔT m = 1500 J (0.709 J/g°C)(80-30) 50°C 42.3131 40 g

39. 1.How many Joules of heat are needed to increase the temperature of 100.0 grams of iron metal by 80.0°C? specific heat of iron = 0.45 J/g°C q = (100.0 g)(0.45 J/g°C)(80.0°C) q= 3600 J 2. 25 grams of water absorb 150 calories of heat. What will be the temp change of the water? T= q mC 150 Cal______ (25 g)(1.00 Cal/g °C) 6°C q= mCΔT