Transcript Slide 1

EXAMPLE 3.1
Classifying Matter
Classify each of the following as a pure substance or a mixture. If it is a pure substance, classify it as an element or
a compound; if it is a mixture, classify it as homogeneous or heterogeneous.
(a) a lead weight
(b) seawater
(c) distilled water
(d) Italian salad dressing
Solution:
Begin by examining the alphabetical listing of pure elements inside the back cover of this text. If the substance
appears in that table, it is a pure substance and an element. If it is not in the table but is a pure substance, then it is a
compound. If the substance is not a pure substance, then it is a mixture. Use your knowledge about the mixture to
determine whether it is homogeneous or heterogeneous.
(a) Lead is listed in the table of elements. It is a pure substance and an element.
(b) Seawater is composed of several substances, including salt and water; it is a mixture. It has a uniform
composition, so it is a homogeneous mixture.
(c) Distilled water is not listed in the table of elements, but it is a pure substance (water); therefore, it is a compound.
(d) Italian salad dressing contains a number of substances and is therefore a mixture. It usually separates into at least
two distinct regions with different composition and is therefore a heterogeneous mixture.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.1
Classifying Matter
Continued
SKILLBUILDER 3.1
Classifying Matter
Classify each of the following as a pure substance or a mixture. If it is a pure substance, classify it as an
element or a compound. If it is a mixture, classify it as homogeneous or heterogeneous.
(a) mercury in a thermometer
(b) exhaled air
(c) minestrone soup
(d) sugar
FOR MORE PRACTICE
Example 3.12; Problems 29, 30, 31, 32, 33, 34.
Note: The answers to all Skillbuilders appear at the end of the chapter.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.2
Physical and Chemical Properties
Determine whether each of the following is a physical or chemical property.
(a) the tendency of copper to turn green when exposed to air
(b) the tendency of automobile paint to dull over time
(c) the tendency of gasoline to evaporate quickly when spilled
(d) the low mass (for a given volume) of aluminum relative to other metals
Solution:
(a) Copper turns green because it reacts with gases in air to form compounds; this is a chemical property.
(b) Automobile paint dulls over time because it can fade (decompose) due to sunlight or it can react with oxygen in
air. In either case, this is a chemical property.
(c) Gasoline evaporates quickly because it has a low boiling point; this is a physical property.
(d) Aluminum’s low mass (for a given volume) relative to other metals is due to its low density; this is a physical
property.
SKILLBUILDER 3.2
Physical and Chemical Properties
Determine whether each of the following is a physical or chemical property.
(a) the explosiveness of hydrogen gas
(b) the bronze color of copper
(c) the shiny appearance of silver
(d) the ability of dry ice to sublime (change from solid directly to vapor)
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 3.13; Problems 35, 36, 37, 38.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.3
Physical and Chemical Changes
Determine whether each of the following is a physical or chemical change.
(a) the rusting of iron
(b) the evaporation of fingernail-polish remover (acetone) from the skin
(c) the burning of coal
(d) the fading of a carpet upon repeated exposure to sunlight
Solution:
(a) Iron rusts because it reacts with oxygen in air to form iron oxide; therefore, this is a chemical change.
(b) When fingernail-polish remover (acetone) evaporates, it changes from liquid to gas, but it remains acetone;
therefore, this is a physical change.
(c) Coal burns because it reacts with oxygen in air to form carbon dioxide; this is a chemical change.
(d) A carpet fades on repeated exposure to sunlight because the molecules that give the carpet its color are
decomposed by sunlight; this is a chemical change.
SKILLBUILDER 3.3
Physical and Chemical Changes
Determine whether each of the following is a physical or chemical change.
(a) copper metal forming a blue solution when it is dropped into colorless nitric acid
(b) a passing train flattening a penny placed on a railroad track
(c) ice melting into liquid water
(d) a match igniting a firework
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 3.14; Problems 39, 40, 41, 42.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
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EXAMPLE 3.4
Conservation of Mass
A chemist forms 16.6 g of potassium iodide by combining 3.9 g of potassium with 12.7 g of iodine. Show that these
results are consistent with the law of conservation of mass.
Solution:
The sum of the masses of the potassium and iodine is:
3.9 g + 12.7 g = 16.6 g
The sum of the masses of potassium and iodine equals the mass of the product,
potassium iodide. The results are consistent with the law of conservation
of mass.
SKILLBUILDER 3.4
Conservation of Mass
Suppose 12 g of natural gas combines with 48 g of oxygen in a flame. The chemical change produces 33 g
of carbon dioxide and how many grams of water?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 3.15; Problems 43, 44, 45, 46, 47, 48.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.5
Conversion of Energy Units
A candy bar contains 225 Cal of nutritional energy. How many joules does it contain?
Solve this problem using the procedure to solve
numerical problems from Chapter 2 (Section 2.10).
Here you are given energy in Calories and asked to
convert it to energy in joules.
Given: 225 Cal
Find: J
Conversion Factors: 1000 calories = 1 Cal
4.184 J = 1 cal
Draw a solution map. Begin with Cal, convert to cal,
and then convert to J.
Solution Map:
Follow the solution map to solve the problem and
round to the correct number of significant figures.
SKILLBUILDER 3.5
Solution:
Conversion of Energy Units
The complete combustion of a small wooden match produces approximately 512 cal of heat. How many
kilojoules are produced?
SKILLBUILDER PLUS
Convert 2.75 ×104 kJ to calories.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 3.16; Problems 49, 50, 51, 52, 53, 54, 55, 56.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
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EXAMPLE 3.6
Exothermic and Endothermic Processes
Identify each of the following changes as exothermic or endothermic.
(a) wood burning in a fire
(b) ice melting
Solution:
(a) When wood burns, it emits heat into the surroundings. Therefore, the process is exothermic.
(b) When ice melts, it absorbs heat from the surroundings. For example, when ice melts in a glass of water, it cools
the water as the melting ice absorbs heat from the water. Therefore, the process is endothermic.
SKILLBUILDER 3.6
Exothermic and Endothermic Processes
Identify each of the following changes as exothermic or endothermic.
(a) water freezing into ice
(b) natural gas burning
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 59, 60, 61, 62.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.7
Converting between Celsius and Kelvin Temperature Scales
Convert –25°C to kelvins.
Set up the problem in the normal way. You are given
a temperature in °C and asked to find K. Use the
equation that shows the relationship between the
given quantity (°C) and the find quantity (K).
Given: -25 °C
Find: K
Equation:
K = °C + 273
Build the solution map.
Solution Map:
Follow the solution map to solve the problem by
substituting the correct value for °C and computing
the answer to the correct number of significant
figures.
SKILLBUILDER 3.7
Solution:
K = °C + 273
K = –25 °C + 273 = 248 K
Converting between Celsius and Kelvin Temperature Scales
Convert 358 K to Celsius.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 3.17; Problems 63c, 64d.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.8
Converting between Fahrenheit and Celsius Temperature
Scales
Convert 55 °F to Celsius.
Set up the problem in the normal way. You are given
a temperature in °F and asked to find °C. Use the
equation that shows the relationship between the
given quantity (°C) and the find quantity (K).
Given: 55 °F
Find: °C
Equation:
Build the solution map.
Solution Map:
Substitute the given value into the equation and
compute the answer to the correct number of
significant figures.
Solution:
SKILLBUILDER 3.8
Converting between Fahrenheit and Celsius Temperature
Scales
Convert 139 °C to Fahrenheit.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Example 3.18; Problems 63a, 64a,c.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.9
Converting between Fahrenheit and Kelvin Temperature
Scales
Convert 310 K to Fahrenheit.
Set up the problem in the normal way. You are given
a temperature in K and asked to find °F. This
problem requires two equations: one relating K and
°C and the other relating °C and °F.
Given: 310 K
Find: °F
Equation:
Build the solution map, which requires two steps: one
to convert K to °C and one to convert °C to °F.
Solution Map:
The first equation must be solved for °C. Substitute
into this equation to convert K to °C.
The second equation must be solved for °F.
Substitute into this equation to convert °C to °F
and compute the answer to the correct number of
significant figures.
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.9
Converting between Fahrenheit and Kelvin Temperature
Scales
Continued
SKILLBUILDER 3.9
Converting between Fahrenheit and Kelvin Temperature
Scales
Convert –321 °F to kelvins.
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 63b, d, 64b.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.10
Relating Heat Energy to Temperature Changes
Gallium is a solid metal at room temperature but melts at 29.9 °C. If you hold gallium in your hand, it melts from
body heat. How much heat must 2.5 g of gallium absorb from your hand to raise its temperature from 25.0 °C to
29.9 °C? The specific heat capacity of gallium is 0.372 J/g °C.
You are given the mass of gallium and its initial and
final temperatures and asked to find the amount of
heat absorbed. The equation that relates the given and
find quantities is the specific heat capacity equation.
Given: 2.5 g gallium
Ti = 25.0 °C
Tf = 29.9 °C
C = 0.372 J/g °C
Find: q
Equation:
The solution map shows that the specific heat
capacity equation relates the given and find
quantities. Before solving the problem, you must
gather the necessary quantities—C, m, and ∆T—in
the correct units. Then substitute the correct variables
into the equation, canceling units, and compute the
answer to the right number of significant figures.
Solution Map:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Solution:
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.10
Relating Heat Energy to Temperature Changes
Continued
SKILLBUILDER 3.10
Relating Heat Energy to Temperature Changes
You find a copper penny (pre-1982) in the snow and pick it up. How much heat is absorbed by the penny as
it warms from the temperature of the snow, –5.0 °C, to the temperature of your body, 37.0 °C? Assume
the penny is pure copper and has a mass of 3.10 g. You can find the heat capacity of copper in Table 3.4 (p.
71).
SKILLBUILDER PLUS
The temperature of a lead fishing weight rises from 26 °C to 38 °C as it absorbs 11.3 J of heat. What is the mass
of the fishing weight in grams?
FOR MORE PRACTICE
Example 3.19; Problems 71, 72, 73, 74.
* This is the amount of heat required to raise the temperature to the melting point. Actually melting the gallium requires
additional heat.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.11
Relating Heat Capacity to Temperature Changes
A chemistry student finds a shiny rock that she suspects is gold. She weighs the rock on a balance and obtains
the mass, 14.3 g. She then finds that the temperature of the rock rises from 25 °C to 52 °C upon absorption of
174 J of heat. Find the heat capacity of the rock and determine whether the value is consistent with the heat
capacity of gold.
You are given the mass of the “gold” rock, the
amount of heat absorbed, and the initial and final
temperature. You are asked to find the heat capacity.
The equation that relates the given and find quantities
is the heat capacity equation
Given:
The solution map shows how the heat capacity
equation relates the given and find quantities.
Solution Map:
Before solving the problem, you must gather the
necessary quantities—m, q, and ΔT—in the correct
units.
Solution:
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Find: C
Equation:
14.3 g
174 J of heat absorbed
Ti = 25 °C
Tf = 52 °C
m = 14.3 g
q = 174 J
ΔT = 52 °C 25 °C = 27 °C
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.11
Relating Heat Capacity to Temperature Changes
Continued
Solve the equation for C and then substitute the
correct variables into the equation, canceling units,
and compute the answer to the right number of
significant figures.
By comparing the computed value of the heat capacity (0.45 J/g °C) with the heat capacity of gold from Table 3.4
(0.128 J/g °C), we conclude that the rock could not be pure gold.
SKILLBUILDER 3.11
Relating Heat Capacity to Temperature Changes
A 328-g sample of water absorbs 5.78 × 103 J of heat. Find the change in temperature for the water. If the
water is initially at 25.0 °C, what is its final temperature?
FOR MORE PRACTICE
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Problems 81, 82, 83, 84.
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.12
Classifying Matter
Classify each of the following as a pure substance or a mixture. If it is a pure substance, classify it as an element or
compound. If it is a mixture, classify it as homogeneous or heterogeneous.
(a) pure silver
(b) swimming-pool water
(c) dry ice (solid carbon dioxide)
(d) blueberry muffin
Solution:
(a) Pure element; silver appears in the element table.
(b) Homogeneous mixture; pool water contains at least water and chlorine, and it is uniform throughout.
(c) Compound; dry ice is a pure substance (carbon dioxide), but it is not listed in the table.
(d) Heterogeneous mixture; a blueberry muffin is a mixture of several things and has nonuniform composition.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.13
Physical and Chemical Properties
Determine whether each of the following is a physical or chemical property.
(a) the tendency for platinum jewelry to scratch easily
(b) the ability of sulfuric acid to burn the skin
(c) the ability of hydrogen peroxide to bleach hair
(d) the density of lead relative to other metals
Solution:
(a) Physical; scratched platinum is still platinum.
(b) Chemical; the acid chemically reacts with the skin to produce the burn.
(c) Chemical; the hydrogen peroxide chemically reacts with hair to bleach it.
(d) Physical; the heaviness can be felt without changing the lead into anything else.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.14
Physical and Chemical Changes
Determine whether each of the following is a physical or chemical change.
(a) the explosion of gunpowder in the barrel of a gun
(b) the melting of gold in a furnace
(c) the bubbling that occurs upon mixing baking soda and vinegar
(d) the bubbling that occurs when water boils
Solution:
(a) Chemical; the gunpowder reacts with oxygen during the explosion.
(b) Physical; the liquid gold is still gold.
(c) Chemical; the bubbling is a result of a chemical reaction between the two substances to form new substances, one
of which is carbon dioxide released as bubbles.
(d) Physical; the bubbling is due to liquid water turning into gaseous water, but it is still water.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.15
Conservation of Mass
An automobile runs for 10 minutes and burns 47 g of gasoline. The gasoline combined with oxygen from air and
formed 132 g of carbon dioxide and 34 g of water. How much oxygen was consumed in the process?
Solution:
The total mass after the chemical change is:
132 g + 34 g = 166 g
The total mass before the change must also be 166 g.
47 g + oxygen = 166 g
So, the mass of oxygen consumed is the total mass (166 g) minus the mass of gasoline (47 g).
grams of oxygen = 166 g – 47 g = 119 g
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.16
Conservation of Energy Units
Convert 1.7 × 103 kWh (the amount of energy used by the average U.S. citizen in one week) into calories.
Given: 1.7 × 103 kWh
Find: cal
Conversion Factors:
1 kWh = 3.60 × 106 J
1 cal = 4.18 J
Solution Map:
Solution:
The unit of the answer, cal, is correct. The magnitude of the answer makes sense since cal is a smaller unit than
kWh.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.17
Converting between Celsius and Kelvin Temperature Scales
Convert 257 K to Celsius.
Given: 257 K
Find: °C
Equation: K = °C + 273
Solution Map:
Solution:
K = °C + 273
°C = K – 273
°C = 257 – 273 = –16 °C
The answer has the correct unit, and its magnitude seems correct (see Figure 3.17).
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.18
Converting between Fahrenheit and Celsius Temperature
Scales
Convert 62.0 °C to Fahrenheit.
Given: 62.0 °C
Find: °F
Equation:
Solution Map:
Solution:
The answer has the correct unit, and its magnitude seems correct (see Figure 3.17).
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.19
Energy, Temperature Change, and Heat Capacity Calculations
What is the temperature change in 355 mL of water upon absorption of 34 kJ of heat?
Given: 355 mL water; 34 kJ of heat
Find: ΔT
Equation: q = m · C · ΔT
Solution Map:
The value for q must be converted from kJ to J:
The value for m must be converted from milliliters to grams; use the density of water, 1.0 g/mL, to convert milliliters
to grams.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.
EXAMPLE 3.19
Energy, Temperature Change, and Heat Capacity Calculations
Continued
Solution:
The answer has the correct units, and the magnitude seems correct. If the magnitude of the answer were a huge
number—3 × 106 for example—we would go back and look for a mistake. If water were to go above 100 °C, it
would boil, so such a large answer would be unlikely.
Introductory Chemistry, Third Edition
By Nivaldo J. Tro
Copyright ©2009 by Pearson Education, Inc.
Upper Saddle River, New Jersey 07458
All rights reserved.