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Dimensional Analysis

Dimensional Analysis. DHS Chemistry. Note:. From this point on, unless told otherwise, it is expected that all answers will be given with 3 sig figs . All numbers greater than 999 and less than 0.01 are to be reported in scientific notation. Dimensional Analysis.

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Dimensional Analysis

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  1. Dimensional Analysis DHS Chemistry

  2. Note: • From this point on, unless told otherwise, it is expected that all answers will be given with 3 sig figs. All numbers greater than 999 and less than 0.01 are to be reported in scientific notation

  3. Dimensional Analysis • Dimensional analysis is a powerful problem-solving method that is based on the fact that any number or expression can be multiplied by one without changing its value. It is the most important skill you will learn for the rest of your science career.

  4. Conversion factors = relationships • To use dimensional analysis you will use conversion factors and fraction cancellation. Anytime you can say two things are equal to each other (or the same as each other), you can make 2 conversion factors out of it --- each conversion factor is equal to 1.

  5. Setting Up Ratios • 100 centimeters = 1 meter can be written as • 100 cm OR 1 m___ • 1 m 100 cm • 8 slices = 1 pizza can be written as • 1 pizza OR 8 slices___ • 8 slices 1 pizza • 5 g/mL can be written as 5 g OR 1 mL • 1 mL 5 g

  6. The Format • # x # x # x # = • 1 # # # • OR • # # # = • 1 # # • Note: Unless you are using 1 as a spot filler, all numbers should ALWAYS include units

  7. Understanding order of operations & typing in calculator keys correctly • Solve. 10 13 6 = 2 4 • Method 1: 10 X 13 X 6 = 780 then divide 97. 5 2 x 4 = 8 • Method 2: 10 X 13 X 6 / 2 / 4 = 97.5 130 780 390 97.5 • Method 3: 10 X 13 / 2 X 6 / 4 = 97.5130 65 390 97.5

  8. Try It • 15 2 3 1.2 4 = • 1 4 5 14 5 • 0.309 3 sig figs

  9. Canceling out units • Just as numbers are multiplied and divided, units are too. Cancel out all units that are located in both the numerator and denominator to reveal the left over units. • cm in feet yard = cm in feet yard

  10. Try It • week day hr min sec = • 1 week day hr min sec

  11. Putting the numbers & units together • 3 weeks 7 day 24 hr 60 min 60 sec = • 1 1 week 1 day 1 hr 1 min =1814400 sec Rules?

  12. Using Dimensional Analysis • Using Dimensional Analysis • Start by writing GWR (given, want, relationships) • Identify the given. • Determine the units you want. • Choose the relationships that will allow you to convert from your given to the want.

  13. Using Dimensional Analysis • Set-up your problem: • Start with the given as a clean fraction. • Write your multiplication symbol and place your relationships carefully so everything above and below each other is equal to each other, and make sure your units cancel out diagonally. • Any units that did not cancel out are part of the units in your answer.

  14. I. One-Step Conversions

  15. I. One- Step ConversionsEX 1: Determine the number of slices in 282.3 pizzas • Step 1) what are you given? • Step 2) What unit(s) do you want? • Step 3) List your relationships • (remember from pg 1 of the notes) • Step 4) Set up your problem 282.3 pizzas Number of slices 1 pizza = 8 slices 282.3 pizzas 8 slices =2258.4 1 1 pizza 2260 slices

  16. EX 2: Determine the number of dozen eggs in 3.8 x 103 eggs. 3.8 X 103 eggs • Step 1) what are you given? • Step 2) What unit(s) do you want? • Step 3) List your relationships • Step 4) Set up your problem Dozen Eggs 12 eggs = 1 dozen These should match 1 dozen 3.8 X 103 eggs Dozen of eggs =317 1 12 eggs

  17. Practice • Determine the number of feet in 2821 inches. Relationship1 foot = 12 inches want given Start with what’s given 2821 in 1 foot feet = 235 1 12 in

  18. Practice • 2. Determine the number of g in 0.03455 kg Relationship1000g = 1kg want given Start with what’s given 0.03455 kg 1000 g g = 34.6 1 1 kg

  19. given want Relationships1 mi = 1609.3 meter 1 mi = 1.6093 km Start with what’s given 37 mi 1.6093 km = 59.5 km 1 mi 1 99 mi 1.6093 km = 159 km 1 mi 1

  20. Solving the same problem, but in a different way

  21. given want Relationships1 mi = 1609.3 meter 1 km = 1000 m Start with what’s given 37 mi 1609.3 m 1 km = 59.5 km 1 mi 1 1000 m

  22. For every problem, start with: • Given: • Want: • Relationships:

  23. II. Multi-Step Conversions

  24. II. Multi-step conversions • Of course, most problems are not simple 1-step conversions. Most problems will require multiple conversion factors. Note: in this class you MUST go through the root unit for any metric conversion. (root units: meter, gram, liter …) Tip: Cancel out all units until you get what you’re looking for.

  25. Given: Want: Relationships: 0.115 km ____ cm • Ex. 1 Convert 0.115 km to cm 100 cm = 1 m 1 km = 1000 m Start with what’s given 0.115 km 1000 m 100 cm 1 m 1 km 1 cm = 1.15 x 104

  26. Given: Want: Relationships: 323 mL ____ cups • Ex. 2 Convert 323 mL to cups 1000 mL = 1 L 1 L = 1.06 quarts 1 quart = 4 cups Start with what’s given 4 cups 1.06 qt. 323 mL 1 L 1 L 1000 mL 1 1 qt. cups = 1.37

  27. Given: Want: Relationships: 7.005 ft ____ mm • Practice 1: Convert 7.005 ft to mm 1 foot = 12 inches 1 inch = 2.54 cm 100cm = 1 m 1m = 1000mm Start with what’s given 1000 mm 12 in 2.54 cm 1 m 7.005 ft 1 1 ft 1 in 100 cm 1 m = 2135 mm

  28. Given: Want: Relationships: 2 years ____ seconds • Practice 2: calculate the number of seconds in 2 years 1 year = 365.25 days 24hrs = 1 day 60 min = 1 hr 1 min = 60 sec Start with what’s given 60 sec 24 hrs 60 min 2 years 365.25 days 1 1 year 1 day 1 hr 1 min = 6.31 X 10 7 seconds

  29. What does that price mean? $1.99 = 1lb

  30. Solving Word Problems using Dimensional Analysis

  31. If you are given multiple numbers in a problem, only one number will be your starting point. The other numbers are relationships that you will use in your problem. If you are given multiple numbers and one of them involves a “/” (e.g. m/s), then always use the “/” as a relationship and start with the other number.

  32. If it helps, change any combination unit into a relationshipex. 0.05mL/s0.05mL = 1 s

  33. Given: Want: Relationships: 0.05mL/s • EX: A faucet is dripping at a rate of 0.05 mL/s. How many liters of water will be lost in 1 day? 1 day ____ Liters 0.05mL = 1 sec 1000mL = 1 L 60 sec = 1 min 1 day = 24 hr 1 hr = 60 min

  34. Relationship want given • EX: A faucet is dripping at a rate of 0.05 mL/s. How many liters of water will be lost in 1 day? • Relationships0.05 mL = 1 s 60 min = 1 hr • 1000mL = 1 L 24 hr = 1 day • 60s = 1 min Start with what’s given 1 L 60 min 60 s 0.05 mL 1 day 24 hr 1 hr 1000 mL 1 1 day 1 min 1 s = 4.32 L

  35. Finally, note dimensional analysis can be used anytime you can say something is equal to something else. It does not have to involve standard conversion factors.

  36. EX 1: Motion pictures are shown at a speed of 24 frames, or individual pictures, each second. If a standard frame is 1.9 cm long, how long will the strand of film be (in ft) for a 2 ½ hr movie? Given: Want: Relationships: 2.5 hours ____ feet 24 frames = 1 sec 1 hr = 60 min 1 in = 2.54cm 1 frame = 1.9cm 1 min = 60 sec 1 ft = 12 in

  37. Relationship relationship want • EX 1: Motion pictures are shown at a speed of 24 frames, or individual pictures, each second. If a standard frame is 1.9 cm long, how long will the strand of film be (in feet) for a 2 ½ hour movie? want given • Relationships24 frames = 1 s1 frame = 1.9cm • 1 hr = 60 min 60 s = 1 min • 1in = 2.54 cm 1 ft = 12 in Start with what’s given 2.5 hr 60 s 24 frames 1.9 cm 1in 1 ft 60 min 1 min 12 in 1 1 hr 1 s 1 frame 2.54 cm = 13464.6 ft

  38. Given: Want: Relationships: 0.5g/mL • EX 2: The density of an unknown liquid is 0.5 g/mL. What is the mass (in kg) of 2.00 L? 2.00L ____ kg 1000g = 1kg 1L = 1000mL

  39. EX 2: The density of an unknown liquid is 0.5 g/mL. What is the mass (in kg) of 2.00 L? Relationship want • Relationships0.5 g = 1mL • 1000g = 1 kg • 1000mL = 1 L given 2.00L 1000mL 0.5g 1 kg 1 1 L 1mL 1000g =1.00 kg

  40. Given: Want: Relationships: 0.0899g/L • EX 3: What is the volume of 5 kg of hydrogen gas at room conditions? The density of hydrogen at room conditions is 0.0899 g/L. 5kg ____ Liters 1kg = 1000g

  41. EX 3: What is the volume of 5 kg of hydrogen gas at room conditions? The density of hydrogen at room conditions is 0.0899 g/L. want given Relationship • Relationships0.0899 g = 1L • 1000g = 1 kg 5 kg 1000g 1L 1 1 kg 0.0899g =55600 L

  42. Practice Box Answers • 1. The Toyota Prius gets 60 mi/gal of gas. If each trip to school is 3.5 km, how many trips can I make with 10 gallons of gas? 276 trips • 2. Every second 2500 g of sulfuric acid flows out of a pipe. How many kg of sulfuric acid will flow in 1 day? 216000 kg • 3. The density of aluminum is 2.7 g/cm3. What volume would a 5.0 kg block of aluminum have? 1.85 X 103 cm3 • 4.Oxygen has a density of 1.43 g/L at 0oC and 1 atm. What would be the mass of 5.0 x 105 L of oxygen at those conditions? 715000 g

  43. Given: Want: Relationships: • 1. The Toyota Prius gets 60 mi/gal of gas. If each trip to school is 3.5 km, how many trips can I make with 10 gallons of gas? 10 gallons ____ trips 60 mi = 1 gal 1 trip = 3.5km 1000m = 1km 1 mile = 1609.3 m

  44. Relationship relationship want • 1. The Toyota Prius gets 60 mi/gal of gas. If each trip to school is 3.5 km, how many trips can I make with 10 gallons of gas? given • Relationships60 miles = 1 gal1 trip = 3.5 km • 1 mi = 1609.3m 1000m = 1km Start with what’s given 10gal 1609.3 m 1 km 1 trip 60 mi 1 mi 1000m 1 1 gal 3.5 km = 276 trips

  45. 2. Every second 2500 g of sulfuric acid flows out of a pipe. How many kg of sulfuric acid will flow in 1 day? Given: Want: Relationships: 1 day ____ kg 1 sec = 2500g 1kg = 1000g 60 s = 1 min 60 min = 1 hr 24 hrs = 1 day

  46. Relationship want • 2. Every second 2500 g of sulfuric acid flows out of a pipe. How many kg of sulfuric acid will flow in 1 day? given • Relationships1 s = 2500g 60s = 1 min • 1000g = 1kg 60 min = 1 hr • 24 hr = 1 day Start with what’s given 1 day 60 min 60 s 2500g 1 kg 24 hr 1 hr 1 min 1 s 1 1 day 1000g = 216000 kg = 2.16 x 105 kg

  47. Given: Want: Relationships: 2.7g/cm3 • 3. The density of aluminum is 2.7 g/cm3. What volume would a 5.0 kg block of aluminum have? 5.0kg ____ cm3 1kg = 1000g

  48. 3. The density of aluminum is 2.7 g/cm3. What volume would a 5.0 kg block of aluminum have? Relationship want given • Relationships2.7 g = 1 cm3 • 1000g = 1 kg 5.0kg 1000g 1 cm3 1 1 kg 2.7g =1850 cm3

  49. Given: Want: Relationships: 1.43g/L • 4. Oxygen has a density of 1.43 g/L at 0oC and 1 atm. What would be the mass of 5.0 x 105 L of oxygen at those conditions? 5.0 x 105L ____ g

  50. 4. Oxygen has a density of 1.43 g/L at 0oC and 1 atm. What would be the mass of 5.0 x 105 L of oxygen at those conditions? Relationship want given • Relationships1.43 g = 1L 5.0 X 105L 1.43 g 1 1 L =715000 g

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