CH. 2 - MEASUREMENT

I

II

III C. Johannesson

I. Using Measurements

(p. 44 - 57)

CH. 2 - MEASUREMENT

C. Johannesson

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

C. Johannesson

B. Percent Error

Indicates accuracy of a measurement

100literature

literaturealexperimenterror %

your value

accepted value

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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 = 2.9 %

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

Count all numbers EXCEPT:

Leading zeros -- 0.0025

Trailing zeros without a decimal point -- 2,500

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4. 0.080

3. 5,280

2. 402

1. 23.50


Counting 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

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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) = 324.103g

324 g

4 SF 3 SF3 SF

C. Johannesson


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

+ 4.1 mL

7.85 mL

224 g

+ 130 g

354 g 7.9 mL 350 g

3.75 mL

+ 4.1 mL

7.85 mL

224 g

+ 130 g

354 g

C. Johannesson


Calculating with Sig Figs (con’t) Exact Numbers do not limit the # of sig

figs in the answer.Counting numbers: 12 studentsExact conversions: 1 m = 100 cm “1” in any conversion: 1 in = 2.54 cm

C. Johannesson


5. (15.30 g) ÷ (6.4 mL)

Practice Problems

= 2.390625 g/mL

18.1 g

6. 18.9 g

- 0.84 g

18.06 g

4 SF 2 SF

2.4 g/mL2 SF

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D. Scientific Notation

Converting into Sci. Notation:

Move decimal until there’s 1 digit to its left. Places moved = exponent.

Large # (>1) positive exponentSmall # (<1) negative exponent

Only include sig figs.

65,000 kg 6.5 × 104 kg

C. Johannesson


7. 2,400,000

g

8. 0.00256 kg

9. 7 10-5 km

10. 6.2 104

mm

Practice Problems

2.4 106 g

2.56 10-3 kg

0.00007 km

62,000 mm

C. Johannesson


Calculating with Sci. Notation

(5.44 × 107 g) ÷ (8.1 × 104 mol) =

5.44EXPEXP

EEEE÷÷

EXPEXP

EEEE ENTERENTER

EXEEXE7 8.1 4

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

Type on your calculator:

C. Johannesson

E. Proportions

Direct Proportion

Inverse Proportion

xy

xy

1

y

x

y

x

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II

III C. Johannesson

II. Units of Measurement

(p. 33 - 39)

CH. 2 - MEASUREMENT

C. Johannesson

A. Number vs. Quantity

Quantity - number + unit

UNITS MATTER!!

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B. SI Units

Quantity Base Unit Abbrev.

Length

Mass

Time

Temp

meter

kilogram

second

kelvin

m

kg

s

K

Amount mole mol

Symbol

l

m

t

T

n

C. Johannesson

B. SI Units

mega- 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

C. Johannesson

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/cm3)mass per volume

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D. DensityM

ass

(g)

Volume (cm3)

Δx

Δyslope D

V

M

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Problem-Solving Steps

1. Analyze

2. Plan

3. Compute

4. Evaluate

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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 g

V

MD

C. Johannesson

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/mL

V = ?

M = 25 g

WORK:

V = M D

V = 25 g

0.87 g/mL

V = 29 mLV

MD

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II

III C. Johannesson

III. Unit Conversions

(p. 40 - 42)

CH. 2 - MEASUREMENT

C. Johannesson

A. SI Prefix Conversions

1. Find the difference between the

exponents of the two prefixes.

2. Move the decimal that many places.

To the leftor right?

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=


NUMBERUNIT

NUMBER

UNIT

532 m = _______ km0.532

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mega- 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

mo

ve le

ft

mo

ve r

igh

t BASE UNIT --- 100

C. Johannesson


1) 20 cm = ______________ m

2) 0.032 L = ______________ mL

3) 45 m = ______________ nm

4) 805 dm = ______________ km

0.2

0.0805

45,000

32

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3

3

cm

gcm

B. Dimensional Analysis

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

“factored” out

g

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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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Lining up conversion factors:

1 in = 2.54 cm

2.54 cm 2.54 cm

1 in = 2.54 cm

1 in 1 in

= 1

1 =

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1. Dimensional Analysis

How many milliliters are in 1.00 quart of milk?

1.00 qt 1 L

1.057 qt= 946 mL

qt mL

1000 mL

1 L

C. Johannesson


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= 35 cm3

1000 g

1 kg

1 cm3

19.3 g

C. Johannesson


How many liters of water would fill a container that measures 75.0 in3?

75.0 in3 (2.54 cm)3

(1 in)3= 0.191 L

in3 L

1 L

1000 cm3

C. Johannesson


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

8.0 cm 1 in

2.54 cm= 3.1 in

cm in

C. Johannesson


5) Taft football needs 550 cm for a 1st down. How many yards is this?

550 cm 1 in

2.54 cm= 6.0 yd

cm yd

1 ft

12 in

1 yd

3 ft

CH. 2 - MEASUREMENT

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