Transcript I. Using Measurements

CH. 2 - MEASUREMENT
I. Using Measurements
I
(p. 44 - 57)
II
III
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
experim ent
al  literature
% error
 100
literature
your value
accepted value
C. Johannesson
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.
% error
1.40 g/m L 1.36 g/m L
1.36 g/m L
% error = 2.9 %
C. Johannesson
 100
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
C. Johannesson
C. Significant Figures
 Counting Sig Figs (Table 2-5, p.47)
 Count all numbers EXCEPT:
 Leading
zeros -- 0.0025
 Trailing
zeros without
a decimal point -- 2,500
C. Johannesson
C. Significant Figures
Counting Sig Fig Examples
1. 23.50
4 sig figs
2. 402
3 sig figs
3. 5,280
3 sig figs
4. 0.080
2 sig figs
C. Johannesson
C. Significant Figures
 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
4 SF
3 SF
3 SF
324 g
C. Johannesson
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.
3.75 mL
+ 4.1 mL
7.85 mL  7.9 mL
C. Johannesson
224 g
+ 130 g
354 g  350 g
C. Significant Figures
 Calculating with Sig Figs (con’t)
 Exact Numbers do not limit the # of sig
figs in the answer.
 Counting
 Exact
 “1”
numbers: 12 students
conversions: 1 m = 100 cm
in any conversion: 1 in = 2.54 cm
C. Johannesson
C. Significant Figures
Practice Problems
5. (15.30 g) ÷ (6.4 mL)
4 SF
2 SF
= 2.390625 g/mL  2.4 g/mL
2 SF
6. 18.9 g
- 0.84 g
18.06 g  18.1 g
C. Johannesson
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
C. Johannesson
 Only include sig
figs.
D. Scientific Notation
Practice Problems
7. 2,400,000 g
2.4 
8. 0.00256 kg
2.56 
9. 7  10-5 km
0.00007 km
10. 6.2  104 mm
62,000 mm
C. Johannesson
6
10
g
-3
10
kg
D. Scientific Notation
 Calculating with Sci. Notation
(5.44 × 107 g) ÷ (8.1 × 104 mol) =
Type on your calculator:
5.44
EXP
EE
7
÷
8.1
EXP
EE
4
EXE
ENTER
= 671.6049383 = 670 g/mol = 6.7 × 102 g/mol
C. Johannesson
E. Proportions
 Direct Proportion
y
y x
x
 Inverse Proportion
1
y
x
y
C. Johannesson
x
CH. 2 - MEASUREMENT
II. Units of Measurement
I
(p. 33 - 39)
II
III
C. Johannesson
A. Number vs. Quantity
 Quantity - number + unit
C. Johannesson
UNITS MATTER!!
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
C. Johannesson
B. SI Units
Prefix
mega-
Symbol
M
Factor
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-
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p
10-12
C. Derived Units
 Combination of base units.
1 cm3 = 1 mL
length  length  length
1 dm3 = 1 L
 Volume (m3 or cm3)
 Density (kg/m3 or g/cm3)
mass per volume
C. Johannesson
M
D=
V
D. Density
Mass (g)
Δy M
D

slope 
Δx V
Volume (cm3)
C. Johannesson
Problem-Solving Steps
1. Analyze
2. Plan
3. Compute
4. Evaluate
C. Johannesson
D. Density
 An object has a volume of 825 cm3 and a
density of 13.6 g/cm3. Find its mass.
GIVEN:
WORK:
V = 825 cm3
D = 13.6 g/cm3
M=?
M = DV
M
D
V
M = (13.6 g/cm3)(825cm3)
M = 11,200 g
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:
WORK:
D = 0.87 g/mL
V=?
M = 25 g
V=M
D
M
D
V
V=
25 g
0.87 g/mL
VC. Johannesson
= 29 mL
CH. 2 - MEASUREMENT
III. Unit Conversions
I
(p. 40 - 42)
II
III
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 left
or right?
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A. SI Prefix Conversions
532 m
NUMBER
UNIT
0.532 km
= _______
=
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NUMBER
UNIT
A. SI Prefix Conversions
move right
move left
Prefix
mega-
Symbol
M
Factor
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-
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p
10-12
A. SI Prefix Conversions
1) 20 cm =
0.2
______________
m
32
2) 0.032 L = ______________
mL
3) 45 m =
45,000
______________
nm
0.0805
4) 805 dm = ______________
km
C. Johannesson
B. Dimensional Analysis
 The “Factor-Label” Method
 Units, or “labels” are canceled, or
“factored” out
g
cm 

g
3
cm
3
C. Johannesson
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.
C. Johannesson
B. Dimensional Analysis
 Lining up conversion factors:
1 in = 2.54 cm
=1
2.54 cm 2.54 cm
1 in = 2.54 cm
1=
1 in
1 in
C. Johannesson
1. Dimensional Analysis
 How many milliliters are in 1.00 quart of
milk?
qt
mL
1.00 qt

1L
1000 mL
1.057 qt
1L
C. Johannesson
= 946 mL
2. Dimensional Analysis
 You have 1.5 pounds of gold. Find its
volume in cm3 if the density of gold is
19.3 g/cm3.
cm3
lb
1.5 lb 1 kg 1000 g 1 cm3
2.2 lb
1 kg
19.3 g
C. Johannesson
= 35 cm3
3. 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
1L
1000 cm3
C. Johannesson
= 0.191 L
4. Dimensional Analysis
4) 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
= 3.1 in
C. Johannesson
5. Dimensional Analysis
5) Taft football needs 550 cm for a 1st
down. How many yards is this?
cm
550 cm
yd
1 in
1 ft 1 yd
2.54 cm 12 in 3 ft
C. Johannesson
= 6.0 yd