UPLOADED BY AHMAD JUNDI
CHAPTER 1 KEYS TO THE STUDY OF
CHEMISTRY
END–OF–CHAPTER PROBLEMS
1.1 Plan: If only the form of the particles has changed and not the composition of the particles, a physical change has
taken place; if particles of a different composition result, a chemical change has taken place.
Solution:
a) The result in C represents a chemical change as the substances in A (red spheres) and B (blue spheres) have
reacted to become a different substance (particles consisting of one red and one blue sphere) represented in C.
There are molecules in C composed of the atoms from A and B.
b) The result in D represents a chemical change as again the atoms in A and B have reacted to form molecules of
a new substance.
c) The change from C to D is a physical change. The substance is the same in both C and D (molecules
consisting of one red sphere and one blue sphere) but is in the gas phase in C and in the liquid phase in D.
d) The sample has the same chemical properties in both C and D since it is the same substance but has different
physical properties.
1.2 Plan: Apply the definitions of the states of matter to a container. Next, apply these definitions to the examples.
Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids have
a definite volume. The volume of the container does not affect the volume of a solid or liquid.
Solution:
a) The helium fills the volume of the entire balloon. The addition or removal of helium will change the volume of
a balloon. Helium is a gas.
b) At room temperature, the mercury does not completely fill the thermometer. The surface of the liquid mercury
indicates the temperature.
c) The soup completely fills the bottom of the bowl, and it has a definite surface. The soup is a liquid, though it is
possible that solid particles of food will be present.
1.3 Plan: Define the terms and apply these definitions to the examples.
Solution:
Physical property – A characteristic shown by a substance itself, without interacting with or changing into other
substances.
Chemical property – A characteristic of a substance that appears as it interacts with, or transforms into, other
substances.
a) The change in color (yellow–green and silvery to white), and the change in physical state (gas and metal to
crystals) are examples of physical properties. The change in the physical properties indicates that a chemical
change occurred. Thus, the interaction between chlorine gas and sodium metal producing sodium chloride is an
example of a chemical property.
b) The sand and the iron are still present. Neither sand nor iron became something else. Colors along with
magnetism are physical properties. No chemical changes took place, so there are no chemical properties to
observe.
1.4 Plan: Define the terms and apply these definitions to the examples.
Solution:
Physical change – A change in which the physical form (or state) of a substance, but not its composition, is
altered.
Chemical change – A change in which a substance is converted into a different substance with different
composition and properties.
a) The changes in the physical form are physical changes. The physical changes indicate that there is also a
chemical change. Magnesium chloride has been converted to magnesium and chlorine.
b) The changes in color and form are physical changes. The physical changes indicate that there is also a
chemical change. Iron has been converted to a different substance, rust.
1-1
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ument may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or p
, UPLOADED BY AHMAD JUNDI
1.5 Plan: Apply the definitions of chemical and physical changes to the examples.
Solution:
a) Not a chemical change, but a physical change — simply cooling returns the soup to its original form.
b) There is a chemical change — cooling the toast will not “un–toast” the bread.
c) Even though the wood is now in smaller pieces, it is still wood. There has been no change in composition, thus
this is a physical change, and not a chemical change.
d) This is a chemical change converting the wood (and air) into different substances with different compositions.
The wood cannot be “unburned.”
1.6 Plan: If there is a physical change, in which the composition of the substance has not been altered, the process
can be reversed by a change in temperature. If there is a chemical change, in which the composition of the
substance has been altered, the process cannot be reversed by changing the temperature.
Solution:
a) and c) can be reversed with temperature; the dew can evaporate and the ice cream can be refrozen.
b) and d) involve chemical changes and cannot be reversed by changing the temperature since a chemical change
has taken place.
1.7 Plan: A system has a higher potential energy before the energy is released (used).
Solution:
a) The exhaust is lower in energy than the fuel by an amount of energy equal to that released as the fuel burns.
The fuel has a higher potential energy.
b) Wood, like the fuel, is higher in energy by the amount released as the wood burns.
1.8 Plan: Kinetic energy is energy due to the motion of an object.
Solution:
a) The sled sliding down the hill has higher kinetic energy than the unmoving sled.
b) The water falling over the dam (moving) has more kinetic energy than the water held by the dam.
1.9 Observations are the first step in the scientific approach. The first observation is that the toast has not popped out
of the toaster. The next step is a hypothesis (tentative explanation) to explain the observation. The hypothesis is
that the spring mechanism is stuck. Next, there will be a test of the hypothesis. In this case, the test is an
additional observation — the bread is unchanged. This observation leads to a new hypothesis — the toaster is
unplugged. This hypothesis leads to additional tests — seeing if the toaster is plugged in, and if it works when
plugged into a different outlet. The final test on the toaster leads to a new hypothesis — there is a problem with the
power in the kitchen. This hypothesis leads to the final test concerning the light in the kitchen.
1.10 A quantitative observation is easier to characterize and reproduce. A qualitative observation may be subjective
and open to interpretation.
a) This is qualitative. When has the sun completely risen?
b) The astronaut’s mass may be measured; thus, this is quantitative.
c) This is qualitative. Measuring the fraction of the ice above or below the surface would make this a
quantitative measurement.
d) The depth is known (measured) so this is quantitative.
1.11 A well-designed experiment must have the following essential features:
1) There must be two variables that are expected to be related.
2) There must be a way to control all the variables, so that only one at a time may be changed.
3) The results must be reproducible.
1.12 A model begins as a simplified version of the observed phenomena, designed to account for the observed effects,
explain how they take place, and to make predictions of experiments yet to be done. The model is improved by
further experiments. It should be flexible enough to allow for modifications as additional experimental results are
gathered.
1-2
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ument may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or p
, UPLOADED BY AHMAD JUNDI
1.13 The unit you begin with (feet) must be in the denominator to cancel. The unit desired (inches) must be in the
numerator. The feet will cancel leaving inches. If the conversion is inverted the answer would be in units of feet
squared per inch.
1.14 Plan: Review the table of conversions in the chapter or inside the back cover of the book. Write the conversion
factor so that the unit initially given will cancel, leaving the desired unit.
Solution:
a) To convert from in2 to cm2, use
( 2.54 cm )2 ; to convert from cm2 to m2, use (1 m )2
(1 in )2 (100 cm )2
b) To convert from km2 to m2, use
(1000 m )2 ; to convert from m2 to cm2, use (100 cm )2
(1 km )2 (1 m )2
c) This problem requires two conversion factors: one for distance and one for time. It does not matter which
conversion is done first. Alternate methods may be used.
To convert distance, mi to m, use:
1.609 km 1000 m
= 1.609x10 m/mi
3
1 mi 1 km
To convert time, h to s, use:
1 h 1 min
= 1 h/3600 s
60 min 60 s
1.609 x 103 m 1 h 0.4469 m•h
Therefore, the complete conversion factor is
= .
1 mi 3600 s mi•s
Do the units cancel when you start with a measurement of mi/h?
1000 g
d) To convert from pounds (lb) to grams (g), use .
2.205 lb
(1 ft )3 (1 in )3
To convert volume from ft3 to cm3 use, = 3.531x10–5 ft3/cm3.
(12 in )3 ( 2.54 cm )3
1.15 Plan: Review the table of conversions in the chapter or inside the back cover of the book. Write the conversion
factor so that the unit initially given will cancel, leaving the desired unit.
Solution:
a) This problem requires two conversion factors: one for distance and one for time. It does not matter which
conversion is done first. Alternate methods may be used.
1 in
To convert distance, cm to in, use:
2.54 cm
1 min
To convert time, min to s, use:
60 s
3 3 (100 cm )3 ; to convert from cm3 to in3, use (1 in )3
b) To convert from m to cm , use
(1 m )3 ( 2.54 cm )3
c) This problem requires two conversion factors: one for distance and one for time. It does not matter which
conversion is done first. Alternate methods may be used.
1 km
To convert distance, m to km, use:
1000 m
1-3
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ument may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or p
, UPLOADED BY AHMAD JUNDI
To convert time, s2 to h2, use:
( 60 s )2 ( 60 min )2 2
= 3600 s
(1 min )2 (1 h )2 h2
d) This problem requires two conversion factors: one for volume and one for time. It does not matter which
conversion is done first. Alternate methods may be used.
4 qt 1L
To convert volume, gal to qt, use: ; to convert qt to L, use:
1 gal 1.057 qt
1h
To convert time, h to min, use:
60 min
1.16 Plan: Review the definitions of extensive and intensive properties.
Solution:
An extensive property depends on the amount of material present. An intensive property is the same regardless of
how much material is present.
a) Mass is an extensive property. Changing the amount of material will change the mass.
b) Density is an intensive property. Changing the amount of material changes both the mass and the volume, but
the ratio (density) remains fixed.
c) Volume is an extensive property. Changing the amount of material will change the size (volume).
d) The melting point is an intensive property. The melting point depends on the substance, not on the amount of
substance.
1.17 Plan: Review the definitions of mass and weight.
Solution:
Mass is the quantity of material present, while weight is the interaction of gravity on mass. An object has a
definite mass regardless of its location; its weight will vary with location. The lower gravitational attraction on
the Moon will make an object appear to have approximately one-sixth its Earth weight. The object has the same
mass on the Moon and on Earth.
mass
1.18 Plan: Density = . An increase in mass or a decrease in volume will increase the density. A decrease
volume
in density will result if the mass is decreased or the volume increased.
Solution:
a) Density increases. The mass of the chlorine gas is not changed, but its volume is smaller.
b) Density remains the same. Neither the mass nor the volume of the solid has changed.
c) Density decreases. Water is one of the few substances that expands on freezing. The mass is constant, but the
volume increases.
d) Density increases. Iron, like most materials, contracts on cooling; thus the volume decreases while the mass
does not change.
e) Density remains the same. The water does not alter either the mass or the volume of the diamond.
1.19 Plan: Review the definitions of heat and temperature. The two temperature values must be compared using one
temperature scale, either Celsius or Fahrenheit.
Solution:
Heat is the energy that flows between objects at different temperatures while temperature is the measure of
how hot or cold a substance is relative to another substance. Heat is an extensive property while temperature is
an intensive property. It takes more heat to boil a gallon of water than to boil a teaspoon of water. However,
both water samples boil at the same temperature.
Convert 65°C to °F: T (in °F) = 9 T (in °C) + 32 = 9 (65°C) + 32 = 149°F
5 5
A temperature of 65°C is 149°F. Heat will flow from the hot water (65°C or 149°F) to the cooler water (65°F).
The 65°C water contains more heat than the cooler water.
1-4
w-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale o
ument may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or p
CHAPTER 1 KEYS TO THE STUDY OF
CHEMISTRY
END–OF–CHAPTER PROBLEMS
1.1 Plan: If only the form of the particles has changed and not the composition of the particles, a physical change has
taken place; if particles of a different composition result, a chemical change has taken place.
Solution:
a) The result in C represents a chemical change as the substances in A (red spheres) and B (blue spheres) have
reacted to become a different substance (particles consisting of one red and one blue sphere) represented in C.
There are molecules in C composed of the atoms from A and B.
b) The result in D represents a chemical change as again the atoms in A and B have reacted to form molecules of
a new substance.
c) The change from C to D is a physical change. The substance is the same in both C and D (molecules
consisting of one red sphere and one blue sphere) but is in the gas phase in C and in the liquid phase in D.
d) The sample has the same chemical properties in both C and D since it is the same substance but has different
physical properties.
1.2 Plan: Apply the definitions of the states of matter to a container. Next, apply these definitions to the examples.
Gas molecules fill the entire container; the volume of a gas is the volume of the container. Solids and liquids have
a definite volume. The volume of the container does not affect the volume of a solid or liquid.
Solution:
a) The helium fills the volume of the entire balloon. The addition or removal of helium will change the volume of
a balloon. Helium is a gas.
b) At room temperature, the mercury does not completely fill the thermometer. The surface of the liquid mercury
indicates the temperature.
c) The soup completely fills the bottom of the bowl, and it has a definite surface. The soup is a liquid, though it is
possible that solid particles of food will be present.
1.3 Plan: Define the terms and apply these definitions to the examples.
Solution:
Physical property – A characteristic shown by a substance itself, without interacting with or changing into other
substances.
Chemical property – A characteristic of a substance that appears as it interacts with, or transforms into, other
substances.
a) The change in color (yellow–green and silvery to white), and the change in physical state (gas and metal to
crystals) are examples of physical properties. The change in the physical properties indicates that a chemical
change occurred. Thus, the interaction between chlorine gas and sodium metal producing sodium chloride is an
example of a chemical property.
b) The sand and the iron are still present. Neither sand nor iron became something else. Colors along with
magnetism are physical properties. No chemical changes took place, so there are no chemical properties to
observe.
1.4 Plan: Define the terms and apply these definitions to the examples.
Solution:
Physical change – A change in which the physical form (or state) of a substance, but not its composition, is
altered.
Chemical change – A change in which a substance is converted into a different substance with different
composition and properties.
a) The changes in the physical form are physical changes. The physical changes indicate that there is also a
chemical change. Magnesium chloride has been converted to magnesium and chlorine.
b) The changes in color and form are physical changes. The physical changes indicate that there is also a
chemical change. Iron has been converted to a different substance, rust.
1-1
w-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale o
ument may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or p
, UPLOADED BY AHMAD JUNDI
1.5 Plan: Apply the definitions of chemical and physical changes to the examples.
Solution:
a) Not a chemical change, but a physical change — simply cooling returns the soup to its original form.
b) There is a chemical change — cooling the toast will not “un–toast” the bread.
c) Even though the wood is now in smaller pieces, it is still wood. There has been no change in composition, thus
this is a physical change, and not a chemical change.
d) This is a chemical change converting the wood (and air) into different substances with different compositions.
The wood cannot be “unburned.”
1.6 Plan: If there is a physical change, in which the composition of the substance has not been altered, the process
can be reversed by a change in temperature. If there is a chemical change, in which the composition of the
substance has been altered, the process cannot be reversed by changing the temperature.
Solution:
a) and c) can be reversed with temperature; the dew can evaporate and the ice cream can be refrozen.
b) and d) involve chemical changes and cannot be reversed by changing the temperature since a chemical change
has taken place.
1.7 Plan: A system has a higher potential energy before the energy is released (used).
Solution:
a) The exhaust is lower in energy than the fuel by an amount of energy equal to that released as the fuel burns.
The fuel has a higher potential energy.
b) Wood, like the fuel, is higher in energy by the amount released as the wood burns.
1.8 Plan: Kinetic energy is energy due to the motion of an object.
Solution:
a) The sled sliding down the hill has higher kinetic energy than the unmoving sled.
b) The water falling over the dam (moving) has more kinetic energy than the water held by the dam.
1.9 Observations are the first step in the scientific approach. The first observation is that the toast has not popped out
of the toaster. The next step is a hypothesis (tentative explanation) to explain the observation. The hypothesis is
that the spring mechanism is stuck. Next, there will be a test of the hypothesis. In this case, the test is an
additional observation — the bread is unchanged. This observation leads to a new hypothesis — the toaster is
unplugged. This hypothesis leads to additional tests — seeing if the toaster is plugged in, and if it works when
plugged into a different outlet. The final test on the toaster leads to a new hypothesis — there is a problem with the
power in the kitchen. This hypothesis leads to the final test concerning the light in the kitchen.
1.10 A quantitative observation is easier to characterize and reproduce. A qualitative observation may be subjective
and open to interpretation.
a) This is qualitative. When has the sun completely risen?
b) The astronaut’s mass may be measured; thus, this is quantitative.
c) This is qualitative. Measuring the fraction of the ice above or below the surface would make this a
quantitative measurement.
d) The depth is known (measured) so this is quantitative.
1.11 A well-designed experiment must have the following essential features:
1) There must be two variables that are expected to be related.
2) There must be a way to control all the variables, so that only one at a time may be changed.
3) The results must be reproducible.
1.12 A model begins as a simplified version of the observed phenomena, designed to account for the observed effects,
explain how they take place, and to make predictions of experiments yet to be done. The model is improved by
further experiments. It should be flexible enough to allow for modifications as additional experimental results are
gathered.
1-2
w-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale o
ument may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or p
, UPLOADED BY AHMAD JUNDI
1.13 The unit you begin with (feet) must be in the denominator to cancel. The unit desired (inches) must be in the
numerator. The feet will cancel leaving inches. If the conversion is inverted the answer would be in units of feet
squared per inch.
1.14 Plan: Review the table of conversions in the chapter or inside the back cover of the book. Write the conversion
factor so that the unit initially given will cancel, leaving the desired unit.
Solution:
a) To convert from in2 to cm2, use
( 2.54 cm )2 ; to convert from cm2 to m2, use (1 m )2
(1 in )2 (100 cm )2
b) To convert from km2 to m2, use
(1000 m )2 ; to convert from m2 to cm2, use (100 cm )2
(1 km )2 (1 m )2
c) This problem requires two conversion factors: one for distance and one for time. It does not matter which
conversion is done first. Alternate methods may be used.
To convert distance, mi to m, use:
1.609 km 1000 m
= 1.609x10 m/mi
3
1 mi 1 km
To convert time, h to s, use:
1 h 1 min
= 1 h/3600 s
60 min 60 s
1.609 x 103 m 1 h 0.4469 m•h
Therefore, the complete conversion factor is
= .
1 mi 3600 s mi•s
Do the units cancel when you start with a measurement of mi/h?
1000 g
d) To convert from pounds (lb) to grams (g), use .
2.205 lb
(1 ft )3 (1 in )3
To convert volume from ft3 to cm3 use, = 3.531x10–5 ft3/cm3.
(12 in )3 ( 2.54 cm )3
1.15 Plan: Review the table of conversions in the chapter or inside the back cover of the book. Write the conversion
factor so that the unit initially given will cancel, leaving the desired unit.
Solution:
a) This problem requires two conversion factors: one for distance and one for time. It does not matter which
conversion is done first. Alternate methods may be used.
1 in
To convert distance, cm to in, use:
2.54 cm
1 min
To convert time, min to s, use:
60 s
3 3 (100 cm )3 ; to convert from cm3 to in3, use (1 in )3
b) To convert from m to cm , use
(1 m )3 ( 2.54 cm )3
c) This problem requires two conversion factors: one for distance and one for time. It does not matter which
conversion is done first. Alternate methods may be used.
1 km
To convert distance, m to km, use:
1000 m
1-3
w-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale o
ument may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or p
, UPLOADED BY AHMAD JUNDI
To convert time, s2 to h2, use:
( 60 s )2 ( 60 min )2 2
= 3600 s
(1 min )2 (1 h )2 h2
d) This problem requires two conversion factors: one for volume and one for time. It does not matter which
conversion is done first. Alternate methods may be used.
4 qt 1L
To convert volume, gal to qt, use: ; to convert qt to L, use:
1 gal 1.057 qt
1h
To convert time, h to min, use:
60 min
1.16 Plan: Review the definitions of extensive and intensive properties.
Solution:
An extensive property depends on the amount of material present. An intensive property is the same regardless of
how much material is present.
a) Mass is an extensive property. Changing the amount of material will change the mass.
b) Density is an intensive property. Changing the amount of material changes both the mass and the volume, but
the ratio (density) remains fixed.
c) Volume is an extensive property. Changing the amount of material will change the size (volume).
d) The melting point is an intensive property. The melting point depends on the substance, not on the amount of
substance.
1.17 Plan: Review the definitions of mass and weight.
Solution:
Mass is the quantity of material present, while weight is the interaction of gravity on mass. An object has a
definite mass regardless of its location; its weight will vary with location. The lower gravitational attraction on
the Moon will make an object appear to have approximately one-sixth its Earth weight. The object has the same
mass on the Moon and on Earth.
mass
1.18 Plan: Density = . An increase in mass or a decrease in volume will increase the density. A decrease
volume
in density will result if the mass is decreased or the volume increased.
Solution:
a) Density increases. The mass of the chlorine gas is not changed, but its volume is smaller.
b) Density remains the same. Neither the mass nor the volume of the solid has changed.
c) Density decreases. Water is one of the few substances that expands on freezing. The mass is constant, but the
volume increases.
d) Density increases. Iron, like most materials, contracts on cooling; thus the volume decreases while the mass
does not change.
e) Density remains the same. The water does not alter either the mass or the volume of the diamond.
1.19 Plan: Review the definitions of heat and temperature. The two temperature values must be compared using one
temperature scale, either Celsius or Fahrenheit.
Solution:
Heat is the energy that flows between objects at different temperatures while temperature is the measure of
how hot or cold a substance is relative to another substance. Heat is an extensive property while temperature is
an intensive property. It takes more heat to boil a gallon of water than to boil a teaspoon of water. However,
both water samples boil at the same temperature.
Convert 65°C to °F: T (in °F) = 9 T (in °C) + 32 = 9 (65°C) + 32 = 149°F
5 5
A temperature of 65°C is 149°F. Heat will flow from the hot water (65°C or 149°F) to the cooler water (65°F).
The 65°C water contains more heat than the cooler water.
1-4
w-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale o
ument may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or p
