i ii iii c. johannesson i. using measurements (p. 44 - 57) ch. 2 - measurement
TRANSCRIPT
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
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.
100g/mL 1.36
g/mL 1.36g/mL 1.40error %
% error = 2.9 %
C. Johannesson
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
4. 0.080
3. 5,280
2. 402
1. 23.50
C. Significant Figures
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
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
324 g
4 SF 3 SF3 SF
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
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
C. Significant Figures
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
C. Significant Figures
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
C. Johannesson
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
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 g
2.56 10-3 kg
0.00007 km
62,000 mm
C. Johannesson
D. Scientific Notation
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
I
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!!
C. Johannesson
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
C. Johannesson
D. DensityM
ass
(g)
Volume (cm3)
Δx
Δyslope D
V
M
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:
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
I
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?
C. Johannesson
=
A. SI Prefix Conversions
NUMBERUNIT
NUMBER
UNIT
532 m = _______ km0.532
C. Johannesson
A. SI Prefix Conversions
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
A. SI Prefix Conversions
1) 20 cm = ______________ m
2) 0.032 L = ______________ mL
3) 45 m = ______________ nm
4) 805 dm = ______________ km
0.2
0.0805
45,000
32
C. Johannesson
3
3
cm
gcm
B. Dimensional Analysis
The “Factor-Label” Method Units, or “labels” are canceled, or
“factored” out
g
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
2.54 cm 2.54 cm
1 in = 2.54 cm
1 in 1 in
= 1
1 =
C. Johannesson
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
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.
lb cm3
1.5 lb 1 kg
2.2 lb= 35 cm3
1000 g
1 kg
1 cm3
19.3 g
C. Johannesson
3. 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= 0.191 L
in3 L
1 L
1000 cm3
C. Johannesson
4. Dimensional Analysis
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. Dimensional Analysis
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