Chapter 2a - Bakersfield College

Download Report

Transcript Chapter 2a - Bakersfield College

Chapter 2a
Measurements and
Calculations
Chapter 2
Table of Contents
2.1
2.2
2.3
2.4
2.5
Scientific Notation
Units
Measurements of Length, Volume, and Mass
Uncertainty in Measurement
Significant Figures
Return to TOC
Section 2.1
Scientific Notation
Measurement
• Quantitative observation.
• Has 2 parts – number
and unit.
 Number tells
comparison.
 Unit tells scale.
Return to TOC
Section 2.1
Scientific Notation
• Technique used to express very large or very
small numbers.
• Expresses a number as a product of a
number between 1 and 10 and the
appropriate power of 10.
Return to TOC
Section 2.1
Scientific Notation
Using Scientific Notation
• Any number can be represented as the
product of a number between 1 and 10 and a
power of 10 (either positive or negative).
• The power of 10 depends on the number of
places the decimal point is moved and in
which direction.
Return to TOC
Section 2.1
Scientific Notation
Using Scientific Notation
• The number of places the decimal point is
moved determines the power of 10. The
direction of the move determines whether the
power of 10 is positive or negative.
Return to TOC
Section 2.1
Scientific Notation
Using Scientific Notation
• If the decimal point is moved to the left, the
power of 10 is positive.
345 = 3.45 × 102
very large number
• If the decimal point is moved to the right, the
power of 10 is negative.
0.0671 = 6.71 × 10–2 very small number
In Webassign homework use format:
345 = 3.45e02
0.0671 = 6.71e-02
Return to TOC
Section 2.1
Scientific Notation
Concept Check
Which of the following correctly expresses
7,882 in scientific notation?
a)
b)
c)
d)
7.882 × 104
788.2 × 103
7.882 × 103
7.882 × 10–3
Return to TOC
Section 2.1
Scientific Notation
Concept Check
Which of the following correctly expresses
0.0000496 in scientific notation?
a)
b)
c)
d)
4.96 × 10–5
4.96 × 10–6
4.96 × 10–7
496 × 107
Return to TOC
Section 2.1
Scientific Notation
Precision vs. Accuracy
good precision
poor accuracy
poor precision
good accuracy
good precision
good accuracy
Return to TOC
Section 2.1
Scientific Notation
Measurement Accuracy
How long is this line?
There is no such thing as a totally accurate measurement!
Return to TOC
Section 2.2
Units
Nature of Measurement
Measurement
•
•
Quantitative observation consisting of two parts.
 number
 scale (unit)
Examples
 20 grams
 6.63 × 10–34 joule·seconds
If a CHP asks you what do you have and you answer I have 3
kilos, you may go to jail. You should have said I have 3 kg of
doughnuts for my chemistry instructor.
Return to TOC
Section 2.1
lll
Scientific Notation
British
SI System
Measurement in Chemistry
Length
Mass
Volume
Time
meter
gram
Liter
second
Km=1000m
Kg=1000g
KL=1000L
1min=60sec
100cm=1m 1000mg=1 g 1000mL=1L 60min=1hr
http://www.kickstarter.com/projects/52746223/the1000mm=1m
state-of-the-unit-the-kilogram-documentary-fil
Foot
pound
gallon
12in=1ft
16oz=1 lb
4qt=1gal
3ft=1yd
2000 lb=1 ton 2pts=1qt
5280ft=1mile
second
(same)
Return to TOC
Section 2.1
Scientific Notation
Conversion between British and SI Units
2.54 cm = 1 in
454 g = 1 lb
1 (cm)3 = 1 cc = 1 ml = 1 gwater
1.06 qt = 1 L
Return to TOC
Section 2.2
Units
Prefixes Used in the SI System
•
Prefixes are used to change the size of the unit.
Return to TOC
Section 2.3
Measurements of Length, Volume, and Mass
Length
•
Fundamental SI unit of length is the meter.
Return to TOC
Section 2.3
Measurements of Length, Volume, and Mass
Volume
•
•
•
•
•
Measure of the amount
of 3-D space occupied
by a substance.
SI unit = cubic meter
(m3)
Commonly measure
solid volume in cm3.
1 mL = 1 cm3
1 L = 1 dm3
Return to TOC
Section 2.3
Measurements of Length, Volume, and Mass
Mass
•
•
•
•
Measure of the amount
of matter present in an
object.
SI unit = kilogram (kg)
1 kg = 2.2046 lbs
1 lb = 453.59 g
Return to TOC
Section 2.3
Measurements of Length, Volume, and Mass
Concept Check
Choose the statement(s) that contain improper
use(s) of commonly used units (doesn’t make
sense)?




A gallon of milk is equal to about 4 L of milk.
A 200-lb man has a mass of about 90 kg.
A basketball player has a height of 7 m tall.
A nickel is 6.5 cm thick.
Return to TOC
Copyright © Cengage Learning. All rights reserved
Section 2.4
Uncertainty in Measurement
•
•
•
A digit that must be estimated is called
uncertain.
A measurement always has some degree of
uncertainty.
Record the certain digits and the first uncertain
digit (the estimated number).
Return to TOC
Section 2.4
Uncertainty in Measurement
Measurement of Length Using a Ruler
• The length of the pin occurs at about 2.85 cm.
 Certain digits: 2.85
Estimate between smallest
division!
 Uncertain digit: 2.85
Return to TOC
Copyright © Cengage Learning. All rights reserved
Section 2.4
Uncertainty in Measurement
Significant Figures
• Numbers that measure or contribute to our accuracy.
• The more significant figures we have the more accurate
our measurement.
• Significant figures are determined by our measurement
device or technique.
Return to TOC
Copyright © Cengage Learning. All rights reserved
Section 2.4
Uncertainty in Measurement
Rules of Determining the Number of Significant Figures
1. All non-zero digits are significant.
234 = 3 sig figs
1.333 = 4 sig figs
1,234.2 = 5 sig figs
2. All zeros between non-zero digits are
significant.
203 = 3 sig figs
1.003 = 4 sig figs 1,030.2 = 5 sig figs
Return to TOC
Section 2.4
Uncertainty in Measurement
Rules of Determining the Number of Significant Figures
3. All zeros to the right of the decimal and to the right of
the last non-zero digit are significant.
2.30 = 3 sig figs
1.000 = 4 sig figs
3.4500 = 5 sig figs
4. All zeros to the left of the first non-zero digit are NOT
significant.
0.0200 = 3 sig figs
0.1220 = 4 sig figs
0.000000012210 = 5 sig figs
Return to TOC
Section 2.4
Uncertainty in Measurement
Rules of Determining the Number of Significant Figures
5.
Zeros to the right of the first non-zero digit and to the
left of the decimal may or may not be significant.
They must be written in scientific notation.
2300 = 2.3 x 103 or 2.30 x 103 or 2.300 x 103
2 sig figs
3 sig figs
4 sig figs
Return to TOC
Section 2.4
Uncertainty in Measurement
Rules of Determining the Number of Significant Figures
6. Some numbers have infinite significant figures or are
exact numbers.
233 people
14 cats (unless in biology lab)
7 cars on the highway
36 schools in town
Return to TOC
Section 2.4
Uncertainty in Measurement
How many significant figures are in each of the
following?
1) 23.34
4 significant figures
2) 21.003
5 significant figures
3) .0003030
4 significant figures
4) 210
2 or 3 significant figures
5) 200 students
infinite significant figures
6) 3000
1, 2, 3, or 4 significant figures
Return to TOC
Section 2.4
Uncertainty in Measurement
Chapter 2b
Measurements and
Calculations
Return to TOC
Section 2.4
Uncertainty in Measurement
2.5
Significant Figures
2.6
Problem Solving and Dimensional Analysis
2.7
Temperature Conversions: An Approach to
Problem Solving
2.8
Density
Return to TOC
Section 2.4
Uncertainty in Measurement
Using Significant Figures in Calculations
Addition and Subtraction
1.
2.
3.
4.
Take out of scientific notation.
Line up the decimals.
Add or subtract.
Round off to first full column or round off to the
same number of decimal places as the least
number of decimal places in the data.
23.345 +14.5 + 0.523 = ?
23.345
14.5
+ 0.523
38.368 = 38.4 or three significant figures
Return to TOC
Section 2.4
Uncertainty in Measurement
Using Significant Figures in Calculations
Multiplication and Division
1.
2.
Do the multiplication or division.
Round answer off to the same number of
significant figures as the least number in the
data.
(23.345)(14.5)(0.523) = ?
177.0368075
= 177 or three significant figures
Return to TOC
Section 2.5
Significant Figures
Rules for Rounding Off
1. If the digit to be removed is less than 5, the
preceding digit stays the same.

5.64 rounds to 5.6 (if final result to 2 sig figs)
Return to TOC
Section 2.5
Significant Figures
Rules for Rounding Off
1. If the digit to be removed is equal to or greater
than 5, the preceding digit is increased by 1.


5.64 rounds to 5.6 (if final result to 2 sig figs)
3.861 rounds to 3.9 (if final result to 2 sig figs)
Return to TOC
Section 2.5
Significant Figures
Rules for Rounding Off
2. In a series of calculations, do within the
parenthesis first and determine the significant
figures and use that answer to calculate and
find the significant figures after the
multiplication and/or division.
Return to TOC
Section 2.5
Significant Figures
Concept Check
You have water in each graduated
cylinder shown. You then add both
samples to a beaker (assume that
all of the liquid is transferred).
How would you write the number
describing the total volume?
3.08 mL
What limits the precision of the
total volume?
2.80
1st graduated cylinder
+ .280
2ndgraduated cylinder
3.080 or 3.08 ml
Return to TOC
Section 2.6
Problem Solving and Dimensional Analysis
Example #1
A golfer putted a golf ball 6.8 ft across a green. How
many inches does this represent?
•
To convert from one unit to another, use the
equivalence statement that relates the two units.
1 ft = 12 in
The two unit factors are:
1 ft
12 in
and
12 in
1 ft
Return to TOC
Section 2.6
Problem Solving and Dimensional Analysis
Example #1
A golfer putted a golf ball 6.8 ft across a green. How
many inches does this represent?
•
Choose the appropriate conversion factor by looking at
the direction of the required change (make sure the
unwanted units cancel).
12 in

6.8 ft 
1 ft
in
Return to TOC
Copyright © Cengage Learning. All rights reserved
Section 2.6
Problem Solving and Dimensional Analysis
Example #1
A golfer putted a golf ball 6.8 ft across a green. How
many inches does this represent?
•
Multiply the quantity to be converted by the conversion
factor to give the quantity with the desired units.
12 in
 82 in
6.8 ft 
1 ft
•
Correct sig figs? Does my answer make sense?
Return to TOC
Copyright © Cengage Learning. All rights reserved
Section 2.6
Problem Solving and Dimensional Analysis
Example #2
An iron sample has a mass of 4.50 lb. What is the
mass of this sample in grams?
(1 kg = 2.2046 lbs; 1 kg = 1000 g)
1 kg
1000 g

= 2.04  103 g
4.50 lbs 
2.2046 lbs
1 kg
454 g
OR 4.50 lbs x -------------1 lb
= 2043g = 2.04x103 g
Return to TOC
Section 2.6
Problem Solving and Dimensional Analysis
Concept Check
What data would you need to estimate the money you
would spend on gasoline to drive your car from New
York to Los Angeles? Provide estimates of values and a
sample calculation.
Sample Answer:
Distance between New York and Los Angeles: 2500 miles
Average gas mileage: 25 miles per gallon
Average cost of gasoline: $3.25 per gallon
2500 mi 
1 gal
$3.25

= $325
25 mi
1 gal
= $(3.3x102)
Return to TOC
Section 2.7
Temperature Conversions: An Approach to Problem Solving
Three Systems for Measuring Temperature
•
•
•
Fahrenheit
Celsius
Kelvin
Gabriel Fahrenheit
Lord Kelvin
Return to TOC
Section 2.7
Temperature Conversions: An Approach to Problem Solving
The Three Major Temperature Scales
F = 1.8C + 32
C = (F-32)/1.8
K = C + 273
What is 35oC in oF? 95 oF
What is 90oF in oC?
32oC
What is 100K in oC? -173oC
Return to TOC
Section 2.7
Temperature Conversions: An Approach to Problem Solving
Exercise
The normal body temperature for a dog is
approximately 102oF. What is this equivalent to
on the Kelvin temperature scale?
a) 373 K
b) 312 K
c) 289 K
d) 202 K
C = (F-32)/1.8 = (102-32)/1.80 = 38.9oC
K = C + 273 = 38.9 + 273 = 312 K
Return to TOC
Section 2.7
Temperature Conversions: An Approach to Problem Solving
Exercise
At what temperature does C = F?
Return to TOC
Copyright © Cengage Learning. All rights reserved
Section 2.7
Temperature Conversions: An Approach to Problem Solving
Solution
•
•
Since °C equals °F, they both should be the same
value (designated as variable x).
Use one of the conversion equations such as:
TC 
•
T F
 32 
1.80
Substitute in the value of x for both T°C and T°F. Solve
for x.
Return to TOC
Copyright © Cengage Learning. All rights reserved
Section 2.7
Temperature Conversions: An Approach to Problem Solving
Solution
TC 
T F
 32 
1.80
x
x
 32 
1.80
1.80x = x -32
0.80x = -32
x = -32/0.80
x   40
So –40°C = –40°F
Return to TOC
Copyright © Cengage Learning. All rights reserved
Section 2.8
Density
•
•
Mass of substance per unit volume of the
substance.
Common units are g/cm3 or g/mL.
mass
Density =
volume
Return to TOC
Copyright © Cengage Learning. All rights reserved
Section 2.8
Density
Measuring the Volume of a Solid Object by Water Displacement
Return to TOC
Section 2.8
Density
Example #1
A certain mineral has a mass of 17.8 g and a volume of
2.35 cm3. What is the density of this mineral?
mass
Density =
volume
17.8 g
Density =
2.35 cm3
3
Density = 7.57 g/cm
Return to TOC
Copyright © Cengage Learning. All rights reserved
Section 2.8
Density
Example #2
What is the mass of a 49.6 mL sample of a liquid, which
has a density of 0.85 g/mL?
mass
Density =
volume
x
0.85 g/mL =
49.6 mL
mass = x = 42 g
OR
 49.6 mL  0.85 g/mL 
g
= 42 mL
Return to TOC
Section 2.8
Density
Exercise
If an object has a mass of 243.8 g and occupies a
volume of 0.125 L, what is the density of this
object in g/cm3?
a) 0.513
b) 1.95
c) 30.5
d) 1950
 243.8 g  1L  1mL 



 = 1.95g/ cm
 0.125 L  1000mL  1cm 
3
3
Return to TOC
Section 2.8
Density
Using Density as a Conversion Factor
How many lbs of sugar is in 945 gallons of 60.0 Brix (% sugar) orange
concentrate if the density of the concentrate is 1.2854 g/mL?
1 L 1000 mL 1.2854 gT 60.0 gS 1 lbs
945 gal 4 qt
1 gal 1.06qt 1 L
1 mL
100 gT 454gS
= 6057.865514lbs
= 6.06 x 103 lbs sugar
lbs of what? Coffee? Cocaine?
Return to TOC
Section 2.8
Density
Using Density as a Conversion Factor Using the Formula
How many lbs of sugar is in 256 L of 60.0 Brix (% sugar)
orange concentrate if the density of the concentrate is
1.2854 g/mL?
M
D=
Solve for Mass
DV = M
V
(1.2854 g/mL)(256,000 mL) = 329062.4 gT = 3.29 x 105 gT
3.29 x 105 gT
1 lbT
60.0 lbsS
454 gT 100 lbsT
= 434.8017621 lbsS
= 4.35 x 102 lbsS
= 435 lbsS
Return to TOC
Section 2.8
Density
Concept Check
Copper has a density of 8.96 g/cm3. If 75.0 g of copper
is added to 50.0 mL of water in a graduated cylinder, to
what volume reading will the water level in the cylinder
rise?
 1cm   1mL 
a) 8.4 mL
= 8.37mL Cu
 75.0g 



b) 41.6 mL
 8.96g   1cm 
c) 58.4 mL
d) 83.7 mL
8.37 mL Cu + 50.0 mL water = 58.4 mL
3
3
Return to TOC