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**I. Units of Measurement (p. 33 - 39)**

CH. 2 - MEASUREMENT I. Units of Measurement (p )

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A. Number vs. Quantity Quantity - number + unit UNITS MATTER!!

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**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

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**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

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**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

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D. Density Mass (g) Volume (cm3)

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**Problem-Solving Steps**

1. Analyze 2. Plan 3. Compute 4. Evaluate

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**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

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**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 = g 0.87 g/mL V = 29 mL

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**II. Using Measurements (p. 44 - 57)**

CH. 2 - MEASUREMENT II. Using Measurements (p )

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**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

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**B. Percent Error Indicates accuracy of a measurement your value**

accepted value

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**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 %

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**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

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**C. Significant Figures Counting Sig Figs (Table 2-5, p.47)**

Count all numbers EXCEPT: Leading zeros Trailing zeros without a decimal point -- 2,500

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**Counting Sig Fig Examples**

C. Significant Figures Counting Sig Fig Examples 4 sig figs 3 sig figs 3. 5,280 3. 5,280 3 sig figs 2 sig figs

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**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) = g 4 SF 3 SF 3 SF 324 g

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**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 mL 7.85 mL 3.75 mL mL 7.85 mL 224 g + 130 g 354 g 224 g + 130 g 354 g 7.9 mL 350 g

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**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

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**C. Significant Figures Practice Problems 5. (15.30 g) ÷ (6.4 mL)**

4 SF 2 SF = g/mL 2.4 g/mL 2 SF g g 18.1 g 18.06 g

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**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.

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**D. Scientific Notation Practice Problems 7. 2,400,000 g 8. 0.00256 kg**

9. 7 10-5 km 104 mm 2.4 106 g 2.56 10-3 kg km 62,000 mm

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**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 = = 670 g/mol = 6.7 × 102 g/mol

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**Chemistry Binder Organization**

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**III. Unit Conversions (p. 40 - 42)**

CH. 2 - MEASUREMENT III. Unit Conversions (p )

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**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

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**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

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D. Density Mass (g) Volume (cm3)

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**Problem-Solving Steps**

1. Analyze 2. Plan 3. Compute 4. Evaluate

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**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

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**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 = g 0.87 g/mL V = 29 mL

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Homework P. 54 Practice problems 1 & 2

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**Conversion Factors Problems**

Dimensional Analysis Conversion Factors Problems

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**B. Dimensional Analysis**

A tool often used in science for converting units within a measurement system Conversion Factor A numerical factor by which a quantity expressed in one system of units may be converted to another system

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**B. Dimensional Analysis**

The “Factor-Label” Method Units, or “labels” are canceled, or “factored” out

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**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.

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**B. Dimensional Analysis**

Lining up conversion factors: = 1 1 in = 2.54 cm 2.54 cm cm 1 = 1 in = 2.54 cm 1 in in

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**B. Dimensional Analysis**

How many milliliters are in 1.00 quart of milk? qt mL 1.00 qt 1 L 1.057 qt 1000 mL 1 L = 946 mL

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**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. lb cm3 1.5 lb 1 kg 2.2 lb 1000 g 1 kg 1 cm3 19.3 g = 35 cm3

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**B. Dimensional Analysis**

How many liters of water would fill a container that measures 75.0 in3? in3 L 75.0 in3 (2.54 cm)3 (1 in)3 1 L 1000 cm3 = 1.23 L

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**B. Dimensional Analysis**

5) Your European hairdresser wants to cut your hair 8.0 cm shorter. How many inches will he be cutting off? cm in 8.0 cm 1 in 2.54 cm = 3.2 in

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**B. Dimensional Analysis**

6) Taft football needs 550 cm for a 1st down. How many yards is this? cm yd 550 cm 1 in 2.54 cm 1 ft 12 in 1 yd 3 ft = 6.0 yd

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**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

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I II III Welcome to Chemistry 116!. Work to be turned in will only be accepted during the first 10 minutes of lab, or will be considered late Work.

I II III Welcome to Chemistry 116!. Work to be turned in will only be accepted during the first 10 minutes of lab, or will be considered late Work.

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