Transcript Relationships

Dimensional
Analysis
DHS Chemistry
ferrer
Setting Up Ratios
100 centimeters = 1 meter can be written as
100 cm
OR
1 m___
1m
100 cm
8 slices = 1 pizza can be written as
1 pizza
OR
8 slices___
8 slices
1 pizza
The Format
# x
1
#
#
x
#
#
x
#
#
=
OR
#
1
#
#
# =
#
Note: Unless you are using 1 as a spot filler,
all numbers should ALWAYS include units
Understanding order of operations &
typing in calculator keys correctly
Solve.
10
13
2
Method 1: 10 X 13 X 6 =
2 x4 =
6 =
4
780
8
then divide 97. 5
Method 2: 10 X 13 X 6 / 2 / 4 = 97.5
130 780 390 97.5
Method 3: 10 X 13 / 2 X 6 / 4 = 97.5
130 65 390 97.5
Try It
15 2
1 4
3
5
1.2
14
4=
5
0.308
Canceling out units
Just as numbers are multiplied and divided,
units are too. Cancel out all units that are
located in both the numerator and
denominator to reveal the left over units.
cm
in
cm
feet
in
yard = yard
feet
Try It
week day
hr
1 week day
min sec = sec
hr min
Putting the numbers &
units together
3 weeks 7 day
24 hr 60 min 60 sec =
1
1 week 1 day
1 hr 1 min
=1814400 sec
Rules?
I. One-Step
Conversions
I. One- Step Conversions
EX 1: Determine the number of slices in 282.3 pizzas
Step 1) what are you given? 282.3 pizzas
Step 2) What are you trying to get?
Number of slices
Step 3) List your relationships
1 pizza = 8 slices
(remember from pg 1 of the notes)
Step 4) Set up your problem
282.3 pizzas
1
8 slices =2258.4 slices
1 pizza
EX 2: Determine the number of dozen eggs in
3.8 x 103 eggs.
Step 3) List your relationships
12 eggs = 1 dozen
Step 4) Set up your problem
These should match
Step 1) what are you given? 3.8 X 103 eggs
Step 2) What are you trying to get? Dozen eggs
3.8 X 103 eggs 1 dozen =317 Dozen of
1
12 eggs
eggs
Practice
1. Determine the number of feet in 2821 inches.
Relationship
1 foot = 12 inches
want
Start with what’s given
2821 in
1
1 foot
12 in
= 235 feet
given
Practice
2. Determine the number of g in 0.03455 kg
Relationship
1000g = 1kg
want
Start with what’s given
0.03455 kg
1
1000 g = 34.6 g
1 kg
given
given
want
Relationships
1 mi = 1609.3 meter
1 mi = 1.6093 km
Start with what’s given
37 mi
1
99 mi
1
1.6093 km = 59.5 km
1 mi
1.6093 km = 159 km
1 mi
Solving the same
problem, but in a
different way
given
want
Relationships
1 mi = 1609.3 meter
1 km = 1000 m
Start with what’s given
37 mi
1
1609.3 m
1 mi
1 km = 59.5 km
1000 m
For every problem, start
with:
Given:
Want:
Relationships:
II. Multi-Step
Conversions
Ex. 1 Convert 0.115 km to cm
Given: 0.115 km
Want: ____ cm
Relationships: 100 cm = 1 m
1 km = 1000 m
Ex. 1 Convert 0.115 km to cm
Relationships
1km = 1000m
given
want
100cm = 1m
Start with what’s given
0.115 km
1
1000 m
1 km
100 cm
1m
= 1.15 x 104 cm
Ex. 2 Convert 323 mL to cups
Given: 323 mL
Want: ____ cups
Relationships: 1000 mL = 1 L
1 L = 1.06 quarts
1 quart = 4 cups
Ex. 2 Convert 323 mL to cups
Relationships
given
want
1000 mL = 1L
1L = 1.06 quarts
1 quart = 4 cups
Start with what’s given
323 mL
1
1L
1000 mL
1.06 qt.
1L
4 cups
1 qt.
= 1.37 cups
Practice 1: Convert 7.005 ft to mm
Given: 7.005 ft
Want: ____ mm
Relationships: 1 foot = 12 inches
1 inch = 2.54 cm
100cm = 1 m
1m = 1000mm
Practice 1: Convert 7.005 ft to mm
Relationships
1 foot = 12 inches
given
want
2.54cm = 1 in
100 cm = 1 m
1 m = 1000 mm
Start with what’s given
7.005 ft 12 in
1
1 ft
2.54 cm 1 m 1000 mm
1 in
100 cm 1 m
= 2135 mm
Practice 2: calculate the number of seconds in 2 years
Given: 2 years
Want: ____ seconds
Relationships: 1 year = 365.25 days
24hrs = 1 day
60 min = 1 hr
1 min = 60 sec
Practice 2: calculate the number of seconds in 2 years
Relationships
1 year = 365.25 days
want
given
1 day = 24 hours
1 hr = 60 min
1 min = 60 sec
Start with what’s given
2 years
1
365.25 days
1 year
24 hrs 60 min 60 sec
1 day
1 hr
1 min
= 6.31 X 10 7 seconds
What does that
price mean?
$1.99 = 1lb
III. Combination
Unit Conversions
Writing Clean Fractions
Take the following quantities and turn it into
a clean fraction as if you were starting a
dimensional analysis problem.
36.0 m/s
36.0 m OR
s
722 mi/hr
722 mi
hr
s___
36.0m
OR
hr__
722mi
Writing clean fractions from
the denominator
Remember from algebra, that a fraction in the
denominator can be multiplied by its reciprocal.
Take the following quantities and turn it into a
clean fraction as if you were starting a dimensional
analysis problem.
52m
46mi
727m/s
60mi/h
to
to
52m
1
s
= 0.0715 s
727m
46 mi hr
= 0.767 hr
1
60mi
EX 1: Convert 0.083 km/hr to m/s
Given: 0.083 km/hr
Want: ____ m/s
Relationships: 1 km = 1000m
1 hr = 60 min
1 min = 60 sec
EX 1: Convert 0.083 km/h to m/s
Relationships
1km = 1000m
given
want
1h = 60 min
1 min = 60 s
Start with what’s given
0.083km
1hr
1000m 1 hr
1 min
1 km 60 min 60 s
= 2.31 X 10-2 m/s
EX 2: Convert 2.85 g/mL to lb/gal
Given: 2.85 g/mL
Want: ____ lbs/gal
Relationships: 454g = 1lbs
1 gal = 4 quarts
1.06 qts = 1L
1 L = 1000mL
EX 2: Convert 2.85 g/mL to lb/gal
Relationships
454 g = 1lb
given
want
1000mL = 1 L
1.06qt = 1 L
4 qt = 1 gal
Start with what’s given
2.85 g
1mL
4 qt
1lb
1000mL 1 L
454 g 1 L
1.06 qt 1 gal
= 23.7 lb/gal
1. Convert 3.56 cm/s to ft/hr
Given: 3.56 cm/s
Want: ____ ft/hr
Relationships: 1 ft = 12 in
2.54 cm = 1 in
1 hr = 60 min
1 min = 60 sec
1. Convert 3.56 cm/s to ft/h
Relationships
2.54 cm = 1 in
given
want
12 in = 1 ft
60 s = 1 min
1 hr = 60 min
Start with what’s given
1 ft
60 s 60 min
3.56 cm 1 in
1s
2.54 cm 12 in 1 min 1 hr
= 420 ft/hr
2. Convert $25.00/feet to cents/cm
Given: $25.00/ft
Want: ____ cents/cm
Relationships: $1.00 = 100 cents
1 ft = 12 inches
2.54 cm = 1 in
2. Convert $25.00/feet to cents/cm
Relationships
$1 = 100 cents
given
want
2.54 cm = 1 in
12 in = 1 ft
Start with what’s given
$ 25.00 100 cents
1 ft
$1
1 ft
12 in
1 in
2.54 cm
= 82.0 cents/cm
Solving Word
Problems using
Dimensional
Analysis
If you are given multiple numbers in a
problem, only one number will be
your starting point. The other
numbers are relationships that you
will use in your problem. If you are
given multiple numbers and one of
them involves a “/” (e.g. m/s), then
always use the “/” as a relationship
and start with the other number.
If it helps, change
any combination
unit into a
relationship
ex. 0.05mL/s
0.05mL = 1 s
EX: A faucet is dripping at a rate of 0.05 mL/s.
How many liters of water will be lost in 1
day?
Given: 0.05mL/s
Want: ____ Liters
Relationships:
1 day
0.05mL = 1 sec
1 day = 24 hr
1000mL = 1 L
1 hr = 60 min
60 sec = 1 min
Relationship
EX: A faucet is dripping at a rate of 0.05 mL/s.
How many liters of water will be lost in 1 day?
want
given
Relationships
0.05 mL = 1 s
60 min = 1 hr
1000mL = 1 L
24 hr = 1 day
60s = 1 min
Start with what’s given
1 day 24 hr 60 min 60 s 0.05 mL 1 L
1000 mL
1 1 day 1 hr 1 min 1 s
= 4.32 L
Finally, note dimensional
analysis can be used
anytime you can say
something is equal to
something else. It does not
have to involve standard
conversion factors.
EX 1: Motion pictures are shown at a speed of
24 frames, or individual pictures, each second. If
a standard frame is 1.9 cm long, how long will
the strand of film be (in ft) for a 2 ½ hr movie?
Given: 2.5 hours
Want: ____ feet
Relationships:
24 frames = 1 sec
1 frame = 1.9cm
1 hr = 60 min
1 min = 60 sec
1 in = 2.54cm
1 ft = 12 in
EX 1: Motion pictures are shown at a speed of
24 frames, or individual pictures, each second. If a
Relationship
standard frame is 1.9 cm long, how long will the
relationship
want
strand of film be (in feet) for a 2 ½ hour movie?
want
Relationships
24 frames = 1 s
given
1 frame = 1.9cm
1 hr = 60 min
60 s = 1 min
1in = 2.54 cm
1 ft = 12 in
Start with what’s given
2.5 hr 60 min 60 s 24 frames 1.9 cm 1in 1 ft
1 1 hr 1 min 1 s
1 frame 2.54 12 in
= 13464.6 ft
cm
1. The Toyota Prius gets 60 mi/gal of gas. If each
trip to school is 3.5 km, how many trips can I
make with 10 gallons of gas?
Given: 10 gallons
Want: ____ trips
Relationships:
60 mi = 1 gal
1000m = 1km
1 trip = 3.5km
1 mile = 1609.3 m
1. The Toyota Prius gets 60 mi/gal of gas. If
Relationship
each trip to school is 3.5 km, how many trips
relationship
can I make with 10 gallons of gas?
want
given
Relationships
60 miles = 1 gal
1 mi = 1609.3m
Start with what’s given
10gal 60 mi
1
1 gal
1 trip = 3.5 km
1000m = 1km
1609.3 m 1 km
1 trip
1 mi
1000m 3.5 km
= 276 trips
2. Every second 2500 g of sulfuric acid flows
out of a pipe. How many kg of sulfuric acid
will flow in 1 day?
Given: 1 day
Want: ____ kg
Relationships:
1 sec = 2500g
60 s = 1 min
1kg = 1000g
60 min = 1 hr
24 hrs = 1 day
2. Every second 2500 g of sulfuric acid flows out
Relationship
of a pipe. How many kg of sulfuric acid will
flow in 1 day?
want
given
Relationships
1 s = 2500g
1000g = 1kg
Start with what’s given
60s = 1 min
60 min = 1 hr
24 hr = 1 day
1 day 24 hr 60 min 60 s 2500g 1 kg
1 s 1000g
1
1 day 1 hr 1 min
= 216000 kg
Tips for studying
• Try doing the review without using the
conversion sheet
• Quiz yourself with the conversions
• Do extra practice problems (your textbook
may have some)