I. Calculations with Significant Figures A. Rules for Counting Significant Figures

advertisement
I. Calculations with Significant Figures
A. Rules for Counting Significant Figures
B. Rules for Significant Figures in Calculations
1. For Multiplication and Division
a. Count the number of significant figures in each measurement
c. Round the result so it has the same number of significant figures as the
measurement with the smallest number of significant figures
4.5 cm
2 sig figs
x
0.200 cm
3 sig figs
=
0.90 cm2
2 sig figs
2. For Addition and Subtraction
a. Result is limited by the number with the smallest number of
significant decimal places
b. Find last significant figure in each measurement
c. Find which one is “left-most”
d. Round answer to the same decimal place
450 mL
+
precise to 10’s place
27.5 mL
=
480 mL
precise to 0.1’s place
precise to 10’s place
C.
Rules for Rounding Off
1. If the digit to be removed is:
a. < 5, the preceding digit stays the same (1.33 becomes 1.3)
b. ≥ 5, the preceding digit is increased by 1 (1.36 becomes 1.4)
2.
In a series of calculations, carry the extra digits to the final result and then
round off.
1.5 x 2.54 = 3.81 ÷ 2.33 = 1.6351931 Rounds to 1.6
3.
II.
Don’t forget to add place-holding zeros if necessary to keep value the
same!! 3.496 (4 figs) rounds to 3.50 (3 sig figs) not 3.5 (2 sig figs)
Dimensional Analysis
A.
Converting units requires canceling units algebraically
1. Conversion factors: state an equivalency as a fraction that equals 1
1 day
24 h
1
24 h
1 day
or
1 day
1440 min
1
1440 min
1 day
2. Use “unit factors” to change units Example: 7.00 in = ? cm
2.54 cm
7.00 in 
 17.8 cm
1 in
3. Unit Factors can be inverted if that’s what is needed for units.
1 in
64.8 cm 
 25.5 in
2.54 cm
4. Multiple conversions can be strung together for complicated changes.
Example: how many miles is 10.0 km?
 1000m  1.094 yd  1mi 
  6.22mi


10.0km


 km  m  1760 yd 
5.
Check for correct procedure by making sure units are correct.
Example: convert 15 km/L to miles/gallon.
 15km  1000m  1.094 yd  1mi  1L  4qt  35mi











gal
 L  km  m  1760 yd  1.06qt  1gal 
III. Temperature
A.
Defining the Temperature Scales
1. Fahrenheit Scale, °F (water freezes at 32°F, boils at 212°F)
2. Celsius Scale, °C (water freezes at 0°C, boils at 100°C)
3. Kelvin Scale, K (water freezes at 273 K, boils at 373 K)
4. Kelvin Scale has no negative numbers, has absolute 0 at 0 K
B. Temperature Conversions
1. The size of the Celsius and Kelvin degrees is the same
a. TK = TC + 273.15
b. TC = TK – 273.15
2. The Fahrenheit and Celsius scales have different size degrees and are off
by 32 oF at the freezing point of water
a. 180 oF = 100 oC from freezing to boiling of water
b. Fahrenheit degrees are only 5/9 as big as Celsius degrees
9
TF  TC  32
5
5
TC  TF  32
9
3. Sample Exercises
a. 98.6 oF = ?
b. -40 oC = ? (this is a good point to remember)
c. 77 K = ?
IV. Density
A.
Density is a property of matter representing the mass per unit volume d = m/V
1. Density can be used to identify unknowns or test for composition
2. Density : solids > liquids >>> gases
3. Solids = g/cm3
Liquids = g/ml
Gases = g/L
4. Example 25.00 ml of isopropanol = 19.625 g d = ?
d = 19.624g / 25.00 ml = 0.7850 g/ml
V.
Classification of Matter
A.
B.
Matter = anything that has mass and takes up space
Matter exists in three states
State
Solid
Liquid
Gas
Shape
Keeps
Shape
Takes
Shape of
Container
Takes
Shape of
Container
Solid
Volume
Compress
Flow
Keeps
Volume
Keeps
Volume
No
No
No
Yes
Takes
Volume of
Container
Yes
Yes
Liquid
Gas
C. Pure substances and Mixtures
1. Pure Substances
a. All samples have the same physical and chemical properties
b. Constant Composition  all samples have the same composition
c. Homogeneous
d. Separate into components based on chemical methods
e. Compound = pure substance that can be broken down into simpler
substances called elements
f. Elements = pure substances that can’t be broken down any more
2. Mixtures
a. Different samples may show different properties
b. Variable composition
c. Homogeneous or Heterogeneous
d. Separate into components based on physical methods
3. Homogeneous = uniform throughout, appears to be one thing
a. pure substances [water]
b. solutions (homogeneous mixtures) [salt dissolved in water]
4. Heterogeneous = non-uniform, contains regions with different properties
than other regions
D. Separation of Mixtures uses Physical Property: no change in what substance is
1. Distillation: based on boiling point differences
2. Filtration: based on differences in states of matter
E. A Chemical Property is displayed only in a change in what the substance is
Mercury
+
A Physical Change
Iodine ---------> Mercuric Iodide
A Chemical Change
Download