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>>
View the other Key Equations and Concepts in this
chapter
Working with Equations
>> Parts of this equation/concept include:
Several equations were introduced in this chapter. Working with equations is generally a matter of substituting the appropriate numbers for the correct variables and solving algebraically. However, when working with chemistry equations, it is important to use the correct units. In solving problems, it often helps to keep the units with the numbers and be sure the units cancel.
One equation, not numbered in the text, is for density, where
density = mass / volume
Since mass and volume have difference units, these units will not
cancel. Typical density units are g/mL. However, any units of mass
and volume can be used. The important thing is that the units match.
>> Example 1
How much does 95.4 mL of mercury, which has a density of 13.6
g/mL, weigh in kilograms?
Solution:
| density |
= |
mass / volume |
| 13.6 g/mL |
= |
mass / 95.4 mL |
| 1297.44 g |
= |
mass |
1297.44 g |
|
|
= |
1.29744 kg |
| |
= |
1.30 kg |
Density is a measured value and does have significant figures.
Both the volume and the density have three significant figures.
The kg/g relationship is exact (infinite significant figures),
so the final answer has three significant figures.
>> back
to the Top of the Page
| B. Temperature Relationships |
>> Fahrenheit and
Celsius, Equation 1.10
The relationship between Fahrenheit and Celsius is best remembered
with water. Water freezes at 32°F and 0C. Water freezes at
212°F and 100°C. That makes a 180° between freezing
and boiling on the Fahrenheit scale and 100° on the Celsius
scale. The equation then requires that the two scales start in the
same place (F – 32) and account for the different size of a
degree (180/100 = 9/5), so the equation can be written as
([F – 32]/C) = (9/5)
or rearranged as in the textbook
C = 5/9 (F – 32) (Equation
1.10)
Since the freezing and boiling points of water define these temperature
scales, all numbers in the equation are exact. The number of significant
figures will be determined by the temperature that is being converted.
>> Example 2
What is the temperature in Celsius at 84°F?
Solution:
| F |
= |
84
|
| C |
= |
5/9 (F – 32) |
| |
= |
5/9 (84 – 32) |
| |
= |
5/9( 52) |
| |
= |
28.8889 |
| |
= |
29°C (two significant figures) |
>> Example 3
What is the temperature in Fahrenheit at 25.0°C?
Solution:
| C |
= |
25.0
|
| 25.0 |
= |
5/9 (F – 32) |
| 25.0(9/5) |
= |
F – 32 |
| 25.0(9/5) + 32 |
= |
77.0°F |
>> Celsius and Kelvin,
Equation 1.11
Because Kelvin and Celsius have the same-size degree spacing, the
relationship is very simple.
C + 273.15 = K (Equation
1.11)
Since this is an addition equation, significant figures are determined
by the number of decimal places. The value of 273.15 is not exact.
>> Example 4
What is the temperature in Kelvin at 25°C?
Solution:
| K |
= |
25 + 273.15 |
| |
= |
298.15 |
| |
= |
298 K |
>> back
to the Top of the Page
>> Wavelength and
Frequency, Equation 1.1
  = c (Equation
1.1)
where is wavelength,
is frequency, and c is the speed of light. The key to using
this equation is proper units. Since the speed of light is usually
cited in meters per second (2.998 x 108 m/s),
must be in units of meters, and in
units of s–1 or Hz. If these are not the given units,
you must convert as appropriate.
>> Example 5
What is the frequency (in Hz) of light with a wavelength of 650
nm?
Solution:
= 650 nm(1 m / 109 nm) = 6.50 x 10–7 m
|
= |
c |
|
= |
| c |
|
|
|
| |
= |
| 2.998 x 108 m/s |
|
| 6.50 x 10–7 m |
|
| |
= |
4.61 x 1014 s–1 |
| |
= |
4.61 x 1014 Hz |
>> Example 6
What is the wavelength of a 7.47-MHz wave?
Solution:
| |
= |
7.47 MHz |
|
|
|
|
= 7.47 x 106 s–1 |
|
= |
c |
|
= |
| c |
|
|
|
| |
= |
| 2.998 x 108 m/s |
|
| 7.47 x 106 s–1 |
|
| |
= |
40.1 m |
>> Wavelength, Frequency, and Energy; Equation 1.3
E = h = hc / (Equation
1.3)
where E = energy (J), h = Planck's constant = 6.626
x 10–34 J • s,
= frequency (s–1 or Hz), c = speed of light
= 2.988 x 108 m/s, and
= wavelength (m). As with the previous equation, the key is to be
sure that the units are correct, changing units if necessary
>> Example 7
What is the energy of a wave with a frequency of 969 MHz?
Solution:
| |
= |
969 MHz |
|
|
= 9.69 x 108 Hz |
| E |
= |
h |
| |
= |
(6.626 x 10–34 J • s)(9.69 x 108 s–1) |
| |
= |
6.42 x 10–25 J |
Both Planck's constant and frequency have limited significant
figures. Be sure your answer has the same number of significant
figures as the value that has the fewest.
>> Example 8
What is the energy of a wave with a wavelength of 335.1 cm?
Solution:
| |
= |
335.1 cm |
|
|
= 3.351 m |
| E |
= |
| hc |
|
|
|
| |
= |
| (6.626 x 10–34 J • s)(2.998 x 108 m/s) |
|
| 3.351 m |
|
| |
= |
5.958 x 10–26 J |
>> Example 9
What is the wavelength and frequency of a wave with an energy
of 2.99 x 10–19 J?
Solution:
| E |
= |
h |
|
= |
|
|
= |
| 2.99 x 10–19 J |
|
| 6.626 x 10–34 J • s |
|
|
= |
4.51 x 1014 s–1 |
| |
|
|
|
= |
| c |
|
|
|
|
= |
| 2.998 x 108 m/s |
|
| 4.51 x 1014 s–1 |
|
|
= |
6.65 x 10–7 m |
>> View
the other Key Equations and Concepts in this chapter
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