Transcript Measurement

CH. 2 - MEASUREMENT
Section 2-2
I
II
III
Units of Measurement
A. Number vs. Quantity
 Quantity – is something that has magnitude, size,
or amount;
number + unit
UNITS MATTER!!
B. SI Units
 SI units of measurement is the system
that all scientists around the world
have agreed upon as the single
measurement system.
 SI units are defined in terms of
standards of measurement.
 The standards are objects or natural
phenomena that are of constant
value, easy to preserve and
reproduce, and practical in size.
C. SI Base 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
D. SI Prefixes – pg 35
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-
p
10-12
E. SI Prefix Conversions
 A conversion factor is a ratio derived from the
equality between two different units that can
be used to convert from one unit to the other.
1. Find the difference between the exponents of
the two prefixes.
2. Move the decimal that many places.
To the left
or right?
E. SI Prefix Conversions
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-
p
10-12
E. SI Prefix Conversions
 Large unit being converted to a
smaller unit:
 Move the decimal place to the right the
specified number of places
 Small unit being converted to a larger
unit:
 Move the decimal place to the left the
specified number of places
E. 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-
p
10-12
E. SI Prefix Conversions
532 m
NUMBER
UNIT
0.532 km
= _______
=
NUMBER
UNIT
E. SI Prefix Conversions
1) 2100 g =
2.1
______________
kg
2) 5.5 L =
5,500
______________
mL
3) 45 km =
45,000
______________
m
4) 85 m =
85,000
______________
mm
F. Derived Units
 Combination of base units.
 Volume (m3 or cm3) – the
amount of space occupied by
an object
 length  length  length
 Density (kg/m3 or g/cm3) –
the ratio of mass to volume
 mass per volume
1 cm3 = 1 mL
1 dm3 = 1 L
M
D=
V
G. Density
Mass (g)
Δy M
D

slope 
Δx V
Volume (cm3)
Density is a characteristic
physical property of a substance
and does not depend on the size
of the sample; the ratio of mass
to volume is constant.
G. Density
 Density is
dependent on
temperature.
 An increase in
temperature
usually causes a
decrease in
density for most
substances.
H. Problem-Solving Steps
1. Analyze
2. Plan
3. Compute
4. Evaluate
I. Solving Density Problems
 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
I. Solving Density Problems
 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
V = 29 mL
J. Dimensional Analysis
 The “Factor-Label” Method
 Units, or “labels” are canceled, or
“factored” out
g
cm 

g
3
cm
3
J. 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.
J. Dimensional Analysis
 Your European hairdresser wants to cut
your hair 8.0 cm shorter. How many inches
will he be cutting off? (Hint: 2.54 cm = 1
inch)
8.0 cm
1 in
2.54 cm
= 3.2 in
J. Dimensional Analysis
 How many milliliters are in 1.00 quart of
milk? (Hint: 1.057 qt = 1 L)
qt
1.00 qt
mL
1L
1000 mL
1.057 qt
1L
= 946 mL