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Unit 2. Measurement

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Unit 2. Measurement. Do Now. In your own words, what do you think is the difference between: Accuracy and Precision?. 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. - PowerPoint PPT Presentation

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Page 1: Unit 2.  Measurement

Unit 2. Measurement

Page 2: Unit 2.  Measurement

Do Now In your own words, what do you think is the

difference between:

Accuracy and Precision?

Page 3: Unit 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 = CORRECTPRECISE = CONSISTENT

Page 4: Unit 2.  Measurement

ACCURATE = CORRECTPRECISE = CONSISTENT

Page 5: Unit 2.  Measurement

B. Percent Error Indicates accuracy of a measurement

100literature

literaturealexperimenterror %

your value

accepted value

Page 6: Unit 2.  Measurement

B. 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.

100g/mL 1.36

g/mL 1.36g/mL 1.40error %

% error = 2.90 %

Page 7: Unit 2.  Measurement

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

Page 8: Unit 2.  Measurement

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

Page 9: Unit 2.  Measurement

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

Count all numbers EXCEPT:

Leading zeros -- 0.0025 (not significant)

Trailing zeros without a decimal point -- 2,500 (not Significant)

Zeros between numbers are significant

Page 10: Unit 2.  Measurement

4. 0.080

3. 5,280

2. 402

1. 23.50

C. Significant FiguresCounting Sig Fig Examples

1. 23.50

2. 402

3. 5,280

4. 0.080

Page 11: Unit 2.  Measurement

4. 0.080

3. 5,280

2. 402

1. 23.50

C. Significant FiguresCounting Sig Fig Examples

1. 23.50

2. 402

3. 5,280

4. 0.080

4 sig figs

3 sig figs

3 sig figs

2 sig figs

Page 12: Unit 2.  Measurement

C. Significant Figures Calculating with Sig Figs

Multiply/Divide –

The # with the fewest sig figs determines the # of sig figs in the answer.

Page 13: Unit 2.  Measurement

Multiplication and Division Rules Do the sum Round the answer to the least

significant figure in the problem

13.91g/cm3)(23.3cm3) = 324.103g 4SF 3SF 3SF

324g

Page 14: Unit 2.  Measurement

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.

Page 15: Unit 2.  Measurement

Addition and Subtraction Rules Stack the numbers so that the decimal

point is aligned Do the sum Figure out which number has least

decimal place (least precise/decimal area least far out)

Draw a line after the last number with the least decimal place

Round the digit by looking at the number that follows

Page 16: Unit 2.  Measurement

Example

3.75 mL + 4.1 mL 7.85 mL 7.9 mL

Page 17: Unit 2.  Measurement

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

Page 18: Unit 2.  Measurement

5. (15.30 g) ÷ (6.4 mL)

Practice Problems

= 2.390625 g/mL

18.1 g

6. 18.9 g- 0.84 g18.06 g

4 SF 2 SF

2.4 g/mL2 SF

Page 19: Unit 2.  Measurement

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 × 104 kg

Page 20: Unit 2.  Measurement

D. Scientific Notation

7. 2,400,000 g

8. 0.00256 kg

9. 7 10-5 km

10. 6.2 104 mm

Practice Problems

Page 21: Unit 2.  Measurement

D. Scientific Notation

7. 2,400,000 g

8. 0.00256 kg

9. 7 10-5 km

10. 6.2 104 mm

Practice Problems

2.4 106 g2.56 10-3 kg

0.00007 km62,000 mm

Page 22: Unit 2.  Measurement

D. Scientific Notation Calculating with Sci. Notation

(5.44 × 107 g) ÷ (8.1 × 104 mol) =

5.44EXPEE

÷EXPEE ENTER

EXE7 8.1 4

= 671.6049383 = 670 g/mol = 6.7 × 102 g/mol

Type on your calculator:

Page 23: Unit 2.  Measurement

E. Proportions Direct Proportion

Inverse Proportion

xy

xy 1

y

x

y

x

Page 24: Unit 2.  Measurement

Units of Measurement

Page 25: Unit 2.  Measurement

A. Number vs. Quantity Quantity - number + unit

UNITS MATTER!!

Page 26: Unit 2.  Measurement

B. SI UnitsQuantity Base Unit Abbrev.

Length

Mass

TimeTemp

meter

kilogram

secondkelvin

m

kg

sK

Amount mole mol

Symbol

l

m

tTn

Page 27: Unit 2.  Measurement

B. SI Unitsmega- M 106

deci- d 10-1

centi- c 10-2

milli- m 10-3

Prefix Symbol Factor

micro- 10-6

nano- n 10-9

pico- p 10-12

kilo- k 103

BASE UNIT --- 100

Page 28: Unit 2.  Measurement

C. Derived Units Combination of base units.

Volume (m3 or cm3) length length length

D = MV

1 cm3 = 1 mL1 dm3 = 1 L

Density (kg/m3 or g/mL or g/cm3)

mass per volume

Page 29: Unit 2.  Measurement

D. Density

Mas

s (g

)

Volume (cm3)

ΔxΔyslope D

VM

Page 30: Unit 2.  Measurement

Problem-Solving Steps

1. Analyze

2. Plan

3. Compute

4. Evaluate

Page 31: Unit 2.  Measurement

D. Density 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:

VMD

Page 32: Unit 2.  Measurement

D. Density 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 gVMD

Page 33: Unit 2.  Measurement

D. 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/mLV = ?M = 25 g

WORK:

VMD

Page 34: Unit 2.  Measurement

D. 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/mLV = ?M = 25 g

WORK:V = M D

V = 25 g 0.87 g/mLV = 29 mLV

MD

Page 35: Unit 2.  Measurement

III. Unit Conversions

Page 36: Unit 2.  Measurement

A. SI Prefix Conversions1. Find the difference between the exponents of the

two prefixes.

2. Move the decimal that many places.

To the leftor right?

Page 37: Unit 2.  Measurement

A. SI Prefix Conversionsmega- M 106

deci- d 10-1

centi- c 10-2

milli- m 10-3

Prefix Symbol Factor

micro- 10-6

nano- n 10-9

pico- p 10-12

kilo- k 103

mov

e le

ftm

ove

right

BASE UNIT --- 100

Page 38: Unit 2.  Measurement

A. SI Prefix Conversions1) 20 cm = ______________ m

2) 0.032 L = _____________ mL

3) 45 m = ______________ nm

4) 805 dm = ______________ km

Page 39: Unit 2.  Measurement

C. Johannesson

A. SI Prefix Conversions1) 20 cm = ______________ m

2) 0.032 L = ______________ mL

3) 45 m = ______________ nm

4) 805 dm = ______________ km

0.2

0.0805

45,000

32

Page 40: Unit 2.  Measurement

33

cmgcm

B. Dimensional Analysis The “Factor-Label” Method

Units, or “labels” are canceled, or “factored” out

g

Page 41: Unit 2.  Measurement

B. Dimensional Analysis Steps:

1. Identify starting & ending units.2. Line up conversion factors so units

cancel.3. Multiply all top numbers & divide by

bottom number.4. Check units & answer.

Page 42: Unit 2.  Measurement

B. Dimensional Analysis Lining up conversion factors:

ARE THESE THE SAME?1 in = 2.54 cm

2.54 cm 2.54 cm

1 in = 2.54 cm 1 in 1 in

= 1

1 =

Page 43: Unit 2.  Measurement

B. Dimensional Analysis

How many milliliters are in 1.00 quart of milk?

1.00 qt 1 L1.057 qt

= 946 mL

qt mL

1000 mL1 L

Page 44: Unit 2.  Measurement

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 kg2.2 lb

= 35 cm31000 g

1 kg1 cm3

19.3 g

Page 45: Unit 2.  Measurement

B. Dimensional Analysis How many liters of water would fill a container that

measures 75.0 in3?

75.0 in3 (2.54 cm)3

(1 in)3= 1.23 L

in3 L

1 L1000 cm3

Page 46: Unit 2.  Measurement

B. Dimensional Analysis

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

8.0 cm 1 in2.54 cm

= 3.2 in

cm in

Page 47: Unit 2.  Measurement

B. Dimensional Analysis6) Taft football needs 550 cm for a 1st down. How

many yards is this?

550 cm 1 in2.54 cm

= 6.0 yd

cm yd

1 ft12 in

1 yd3 ft

Page 48: Unit 2.  Measurement

B. Dimensional Analysis7) A piece of wire is 1.3 m long. How many 1.5-cm

pieces can be cut from this wire?

1.3 m 100 cm1 m

= 86 pieces

cm pieces

1 piece1.5 cm