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Name: YOSSUARA PITTY D. DATE: 15/10/15 GRADE: 9th B I Quarter Chemistry Study Guide Lesson 5.1 : Light and Quantized Energy. (pages 136-145) VOCABULARY Electromagnetic radiation: is a form of energy that exhibits wavelike behavior as it travels through space. Wavelength: is the shortest distance between equivalent points on a continuous wave. Frequency: is the number of waves that pass a given point per second. Amplitude: the amplitude of a wave is the wave’s height from the origin to a crest or from the origin to a trough. Electromagnetic spectrum: the electromagnetic spectrum, also called the EM spectrum, includes all forms of electromagnetic radiation, with the only differences in the types of radiation being their frequencies and wavelengths. Quantum: a quantum is the minimum amount of energy that can be lost or gained by an atom. Plank's constant: Planck proposed that the energy emitted by hot objects was quantized. The energy of a quantum is given by the product of Planck’s constant and the frequency. E quantum = hν. Planck’s constant has a value of 6.626 x 10 -34 J·s, where J is the symbol for joule, the SI unit of energy. The equation shows that the energy of radiation increases as the radiation’s frequency, v, increases. Photoelectric effect: in the photoelectric effect, electrons, called photoelectrons, are emitted from a metal’s surface when light of a certain frequency, or higher than a certain frequency, shines on the surface. Photon: a photon is a massless particle that carries a quantum of energy. Atomic Emission Spectrum (AES): the atomic emission spectrum of an element is the set of frequencies of the electromagnetic waves emitted by atoms of the element.

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Page 1: Chem's Studyguide Iq

Name: YOSSUARA PITTY D. DATE: 15/10/15 GRADE: 9th B

I Quarter Chemistry Study Guide

Lesson 5.1: Light and Quantized Energy. (pages 136-145)

VOCABULARY

Electromagnetic radiation: is a form of energy that exhibits wavelike behavior as it travels through space.

Wavelength: is the shortest distance between equivalent points on a continuous wave.

Frequency: is the number of waves that pass a given point per second.

Amplitude: the amplitude of a wave is the wave’s height from the origin to a crest or from the origin to a trough.

Electromagnetic spectrum: the electromagnetic spectrum, also called the EM spectrum, includes all forms of electromagnetic radiation, with the only differences in the types of radiation being their frequencies and wavelengths.

Quantum: a quantum is the minimum amount of energy that can be lost or gained by an atom.

Plank's constant: Planck proposed that the energy emitted by hot objects was quantized. The energy of a quantum is given by the product of Planck’s constant and the frequency. Equantum = hν. Planck’s constant has a value of 6.626 x 10-34 J·s, where J is the symbol for joule, the SI unit of energy. The equation shows that the energy of radiation increases as the radiation’s frequency, v, increases.

Photoelectric effect: in the photoelectric effect, electrons, called photoelectrons, are emitted from a metal’s surface when light of a certain frequency, or higher than a certain frequency, shines on the surface.

Photon: a photon is a massless particle that carries a quantum of energy.

Atomic Emission Spectrum (AES): the atomic emission spectrum of an element is the set of frequencies of the electromagnetic waves emitted by atoms of the element.

EXERCISES (page 166)

Exercise 34 Define the following terms.

a.) Frequency: Frequency is the number of waves that pass a given point per second.

b.) Wavelength: Wavelength is the shortest distance between equivalent points on a continuous wave.

c.) Quantum: A quantum is the minimum amount of energy that can be lost or gained by an atom.

d.) Ground state: An atom’s ground state is its lowest allowable energy state.

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Exercise 35 Arrange the following types of electromagnetic radiation in order of increasing wavelength.

a.) ultraviolet light c.) radio wavesb.) microwaves d.) X rays

d.) X rays, a.) ultraviolet light, b.) microwaves, c.) radio waves

Exercise 37 What is the photoelectric effect?

A phenomenon in which a metal emits electrons when light of a sufficient frequency shines on it.

Exercise 45 Radiation Use Figure 5.20 to determine the following types of radiation.

a.) radiation with a frequency of 8.6 x 1011 s-1 infraredb.) radiation with a wavelength of 4.2 nm X rayc.) radiation with a frequency of 5.6 MHz AM radiod.) radiation that travels at a speed of 3.00 x 108 m/s any Electromagnetic Wave

Exercise 46 What is the wavelength of electromagnetic radiation with a frequency of 5.00 x 1012 Hz? What kind of electromagnetic radiation is this? (infrared radiation)

c = λν = c/ = (3.00 x 108 m/s) / (5.00 x 1012 s-1) = 6.00 x 10-5 m Hz = s-1

c = is the speed of light in a vacuum.λ = is the wavelength.ν = is the frequency.

Exercise 47 What is the frequency of electromagnetic radiation with a wavelength of 3.33 x 10 -8 m? What type of electromagnetic radiation is this? (UV radiation)

= c/ = (3.00 x 108 m/s) / (3.33 x 10-8 m) = 6.00 x 10-5 s-1 or Hz (Herzt)

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Exercise 48 What is the speed of an electromagnetic wave with a frequency of 1.33 x 1017 Hz and a wavelength of 2.25 nm?

1 nm = 1 x 10-9 m (2.25 nm) x (1 x 10-9 m/1 nm) = 2.25 x 10-9 m ; Hz = s-1

c = λν = (2.25 x 10-9 m)(1.33 x 1017s-1) =3.00 x 108 m/s

Exercise 49 What is the energy of a photon of red light that has a frequency of 4.48 x 1014 Hz?

Hz = s-1

Ephoton = h·ν = (6.626 x 10-34 J·s)(4.48 x 1014 s-1 ) = 2.97 x 10-19 J (Joules)

Ephoton represents energy.h is Planck’s constant = 6.626 x 10-34 J·sν represents frequency = 4.48 x 1014 Hz or s-1

Exercise 50 Mercury’s atomic emission spectrum is shown in Figure 5.21. Estimate the wavelength of the orange line. What is its frequency? What is the energy of a photon corresponding to the orange line emitted by the mercury atom?

Solution: wavelength = λ ≈ 615 nm ; 1 nm = 1 x 10-9 m (615 nm) x (1 x 10-9 m/1 nm) = 615 x 10-9 m ; Hz = s-1 ; c = 3.00 x 108 m/s

= c/ = (3.00 x 108 m/s) / (615 x 10-9 m) = 4.88 x 1014 s-1 or Hz (Herzt)

Ephoton = h·ν = (6.626 x 10-34 J·s) x (4.88 x 1014 s-1 ) = 3.23 x 10-19 J (Joules)

Exercise 51 What is the energy of an ultraviolet photon that has a wavelength of 1.18 x 10-8 m?

= c/ = (3.00 x 108 m/s) / (1.18 x 10-8 m) = 2.54 x 1016 s-1

Ephoton = h·ν = (6.626 x 10-34 J·s) x (2.54 x 1016 s-1) = 1.68 x 10-17 J (Joules)

Exercise 52 A photon has an energy of 2.93 x 10-25 J. What is its frequency? What type of electromagnetic radiation is the photon?

Ephoton = h·ν ν = Ephoton / h = (2.93 x 10-25 J) / (6.626 x 10-34 J·s) = 4.42 x 108 s-1 or Hertz

What type of electromagnetic radiation is the photon? Radio (FM wave or TV)

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EXERCISES (page 979, chapter 5: No.1, 2, 3, 4).

Exercise 1: What is the frequency of an electromagnetic wave that has a wavelength 4.55 x 10 -3 m? 1.00 x 10-12 m?

a) Answer:

c = λν

c = is the speed of light in a vacuum (3.00 x 108 m/s).λ = is the wavelength = 4.55 x 10-3 mν = is the frequency = ?ν = c / λ = (3.00 x 108 m/s) / (4.55 x 10-3 m) = 6.60 x 1010 s-1

b) Answer:

c = λν

c = is the speed of light in a vacuum (3.00 x 108 m/s).λ = is the wavelength = 1.00 x 10-12 mν = is the frequency = ?ν = c / λ = (3.00 x 108 m/s) / (1.00 x 10-12 m) = 3.00 x 1020 s-1

Exercise 2: Calculate the wavelength of a electromagnetic wave with a frequency of 8.68 x 1016 Hz; 5.0 x 1014 Hz; and 1.00 x 106 Hz.

Exercise 3: What is the energy of a quantum of visible light having a frequency of 5.45 x 1014 s-1?

Exercise 4: An X-ray has a frequency 1.28 x 1018 s-1. What is the energy of a quantum of the X-ray?

Lesson 4.4: Unstable Nuclei and Radioactive Decay. (pages 122-124)

VOCABULARY

Radioactivity: is a process by which substances spontaneously emitted radiation.

Radiation: the rays and particles emitted by radioactive material.

Nuclear reaction: a reaction that involves a change in an atom’s nucleus.

Radioactive decay: is the lose energy from the atom’s unstable nuclei by emitting radiation in a spontaneous process.

Alpha radiation: a type of radiation that was deflected toward a negatively charged plate. It is made up of alpha particles.

Alpha particle: contains two protons and two neutrons, and thus has 2+ charge, which explains why alpha particles are attracted to the negatively charged plate. An alpha particle is equivalent to a helium-4 nucleus and is represented by 4

2He or α.

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Nuclear equation: in a nuclear reaction, a new element is created as a result of the alpha decay of the unstable radium-226 nucleus. It shows the atomic numbers and mass numbers of the particle involved. The mass number is conserved in nuclear equations.

Beta radiation: the radiation that was deflected toward the positively charged plate. This radiation consists of fast-moving beta particle.

Beta particle: Each beta particle is an electron with a 1- charge. Beta particles are represented by the symbol β or e-. The negative charge of the beta particle explains why it is attracted to the positively charged plate.

Gamma ray: A gamma ray is a high-energy radiation that possesses no mass and is denoted by the symbol Ƴ. Because they are neutral, gamma rays are not deflected by electric or magnetic fields. They usually accompany alpha and beta radiation, and they account for most of the energy lost during radioactive decay. Because gamma rays are massless, the emission of gamma rays by themselves cannot result in the formation of a new atom.

EXERCISES (page 129)

Exercise 82 Define alpha particle, beta particle, and gamma ray.

Alpha particle: alpha particle is a helium atom with a 2+ charge.

Beta particle: beta particle is an electron

Gamma ray: gamma rays is high energy radiation

Exercise 85 Radioactive Emissions. What change in mass number occurs when a radioactive atom emits an alpha particle? A beta particle? A gamma particle?

α, mass number decreases by 4; β, no change in mass number; andƳ, no change in mass numberExercise 89 Boron-10 emits alpha particles and Cesium-137 emits beta particles. Write balanced nuclear reactions for each radioactive decay.

105B 6

3Li + 42He

137

55Cs 13756Ba + 0

-1e

Exercise 107 Radiation. Identify the two types of radiation shown in Figure 4.24. Explain your reasoning.

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The deflected beam is alpha radiation because it is deflected toward the negatively charged plate. The undeflected beam must be neutral Ƴ gamma radiation.

Lesson 4.3: How Atoms Differ. (pages 115-121)

VOCABULARY

Atomic number: is the number of protons in an atom. The number of protons in an atom identifies it as an atom of a particular element. The atomic number of an atom equals its number of protons and its number of electrons.

Isotope: atoms with the same number of protons but different numbers of neutrons.

Mass number: the mass number is the sum of the atomic number (or number of protons) and neutrons in the nucleus.

Atomic mass unit (amu): is defined as one-twelfth the mass of a carbon-12 atom.

Atomic mass: is the weighted average mass of the isotopes of an element. Because isotopes have different mass, the weighted average is not a whole number.

EXERCISES (page 129)

Exercise 71 Sulfur. Show that the atomic mass of the element sulfur is 32.065 amu.

Sulfur = S = (31.972amu)(0.9502) x (32.971amu)(0.0075) x (33.968 amu) (0.0421) x (35.967 amu) (0.0002) = 32.065 amu

Exercise 72 Fill in the blanks in Table 4.6.

Table 4.6 Chlorine and ZirconiumElement Cl Cl Zr ZrAtomic number 17 17 40 40Mass number 35 37 90 92Protons 17 17 40 40Neutrons 18 20 50 52Electrons 17 17 40 40

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Exercise 73 How many electrons, protons, and neutrons are contained in each atom?

a. 13255 Cs =

b. 5927 Co =

c. 16369 Tm =

d. 7030 Zn =

Symbol Electrons Protons Neutrons132

55 Cs 55 55 7759

27 Co 27 27 32163

69 Tm 69 69 9470

30 Zn 30 30 40

Exercise 74 How many electrons, protons, and neutrons are contained in each atom?

a. gallium-69=b. fluorine-23 =c. titanium-48 =d. tantalum-181 =

Symbol Electrons Protons NeutronsGa-69 31 31 38F-23 9 9 14Ti-48 22 22 26

Ta-181 73 73 108

Exercise 75 For each chemical symbol, determine the number of protons and electrons an atom of the element contains.

a. V = c. Ir =b. Mn = d. S =

Exercise 76 Gallium, which has an atomic mass of 69.723 amu, has two naturally occurring isotopes, Ga-69 and Ga-71. Which isotope occurs in greater abundance? Explain.

Ga-69 must be more abundant because the atomic mass of gallium is closer to the mass Ga-69 than the mass of Ga-71.

Exercise 77 Atomic Mass of Silver. Silver has two isotopes: 10747Ag, which has a mass of 106.905

amu and a percent abundance of 52.00%, and 10947 Ag, which has a mass of 108.905 amu and an

percent abundance of 48.00%. What is the atomic mass of silver?

Symbol Electrons ProtonsV 23 23

Mn 25 25Ir 77 77S 16 16

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(106.905 amu x 52.00) /100 = 55.59(108.905 amu x 0.4800) /100 = 52.27 107.86 amu

Exercise 78 Data for chromium’s four naturally occuring isotopes are provided in Table 4.7. Calculate chromium’s atomic mass.

Cr-50 = (4.35 x 49.946 amu) / 100 = 2.17Cr-52 = (83.79 x 51.941amu) /100 = 43.52Cr-53 = (9.50 x 52.941 amu) / 100 = 5.03Cr-54 = (2.36 x 53.939 amu) / 100 = 1.27 AAM = 51.99 amu

EXERCISES (page 130)

Exercise 96 Complete Table 4.8.

Table 4.8 Composition of Various IsotopesIsotope S-32 Ca-44 Zn-64 F-19 Na-23Atomic number 16 20 30 9 11Mass number 32 44 64 19 23Number of Protons 16 20 30 9 11Number of Neutrons 16 24 34 10 12Number of Electrons 16 20 30 9 11

Exercise 98 Is the charge of a nucleus positive, negative, or zero? The charge of an atom?

The nucleus is positively charged. The charge of an atom is neutral.

Exercise 102 Boron-10 and boron-11 are the naturally occurring isotopes of elemental boron. If boron has an atomic mass of 10.81 amu, which isotope occurs in greater abundance?

Boron-11 must occur in greater abundance because the atomic weight of boron-11 is much closer to the mass of Boron-11 than to the mass of Boron-10.

Exercise 104 Titanium. Use Table 4.9 to calculate the atomic mass of titanium.

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Ti-46 = (45.953 amu x 8.00) /100 = 3.68Ti-47 = (46.952 amu x 7.30) / 100 = 3.43Ti-48 = (47.948 amu x 73.80) / 100 = 35.39Ti-49 = (48.948 amu x 5.50) / 100 = 2.69Ti-50 = (49.945 amu x 5.40) / 100 = 2.70 AAM = 47.89 amu

EXERCISES (pages 978/979, section 3:1, 2, 3, 4,5, 6, 7 and 8)

Exercise 1: How many protons and electrons are in each of the following atoms?

a. galliumb. siliconc. cesiumd. calciume. molybdenumf. titanium

Exercise 2: What is the atomic number of each of the following elements?

a. an atom that contains 37 electronsb. an atom that contains 72 protonsc. an atom that contains 1 electrond. an atom that contains 85 protons

Exercise 3: Use the periodic table to write the name and symbol for each element identified in exercise 2?

Exercise 4: An isotope of copper contains 29 electrons, 29 protons, and 36 neutrons. What is the mass number of this isotope?

Exercise 5: An isotope of uranium contains 92 electrons and 144 neutrons? What is the mass number of this isotope?

Exercise 6: Use the periodic table to write the name and symbol for each of the following elements. Then, determine the number of electrons, protons and neutrons each contains.

a. yttrium-88b. arsenic-75c. xenon-129d. bromine-79e. gold-197

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f. helium-4

Exercise 7: An element has two naturally occurring isotopes: 14X and 15X. 14X has a mass of 14.00307 amu and a relative abundance of 99.63%. 15X has a mass of 15.00011 amu and a relative abundance of 0.37%. Identify the unknown element.

Exercise 8: Silver has two naturally occurring isotopes. Ag-107 has an abundance of 51.82% and a mass of 106.9 amu. Ag-109 has a relative abundance of 48.18% and a mass of 108.9 amu. Calculate the atomic mass of silver.

Lesson 4.2 Defining the atom. (pages 106 – 114)

VOCABULARY

Atom: is smallest particle of an element that retains the properties of the element.

Cathode ray: is a ray originating from the cathode and traveling to the anode.

Electron (e-): are negatively charged particles that are part of all forms of matter. An electron has a charge of 1-.

Nucleus: tiny dense region in the center of the atom that is positive charge and contains all of its mass. Proton (p+): is a subatomic particle carrying a charge equal to but opposite that of an electron; that is, a proton has a charge of 1+.

Neutron(n0): is a subatomic particle that has a mass nearly equal to that of a proton, but it carries no electric charge.

EXERCISES (page 128)

Exercise 44 Explain why atoms are electrically neutral.

Because, the number of positively charged protons equals the number of negatively charged electrons.Exercise 46 Which particles account for most of an atom’s mass?

The protons and neutrons

Exercise 48 Cathode-Ray Tubes. Which subatomic particle was discovered by researchers working with cathode-ray tubes?

The electron (e-)

QUESTIONS (page 130)

Question 91 Cathode-Ray Tube. Describe a cathode-ray tube and how it operates.

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A cathode ray tube has a metal electrode at each end and is filled with a gas at low pressure. One electrode is connected to the negative terminal of a battery (cathode), and the other is connected to the positive terminal (anode). When current flows, negatively charged particles called electrons are emitted from the cathode and travel through the tube to the anode.

Question 93 Gold Foil Experiment. How did the actual results of Rutherford’s gold foil experiment differ from the results he expected?

In short, Rutherford expected that the particles to be slightly deflected when they passed through a gold foil. Instead, he found that some were deflected at very large angles.

Lesson 3.4 Elements and Compounds. (pages 84- 90)

VOCABULARY

Element: an element is a substance that cannot be broken down into simpler substances by physical or chemical means.

Periodic table: the periodic table organizes the elements into a grid of horizontal rows called periods and vertical columns called groups or families.

Compound: a compound is made up of two or more different elements that are combined chemically. Most matter in the universe exists in the form of compounds.

Law of definite proportion: the law of definite proportions states that a compound is always composed of the same elements in the same proportion by mass, no matter how large or small the sample.

Percent by mass: the percent by mass is the ratio of the mass of each element to the total mass of the compound expressed as a percentage.

Law of multiple proportions: the law of multiple proportions states that when different compounds are formed by a combination of the same elements, different masses of one element combine with the same relative mass of the other element in a ratio of small whole numbers.

EXERCISES (page 90)

Exercise 28 Complete the table, and then analyze the data to determine if Compounds I and II are the same compound. If the compounds are different, use the law of multiple proportions to show the relationship between them.

Analysis Data of Two Iron CompoundsCompound Total

Mass(g)

Mass Fe(g)

Mass O(g)

Mass Percent Fe Mass percent O Mass Ratio(mass Fe/mass O)

I 75.00 52.46 22.54 %Fe=(52.46/75.00) x100=69.95%

%O=(22.54/75.00)x100=30.05%

69.95/30.05=2.3278 g Fe/1gO

Analysis Data of Two Iron CompoundsCompound Total Mass

(g)Mass Fe(g)

Mass O(g)

MassPercent Fe

MassPercent O

I 75.00 52.46 22.54II 56.00 43.53 12.47

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II 56.00 43.53 12.47 %Fe=(43.53/56.00) x100=77.73%

%O=(12.47/56.00) x100=22.27%

77.73/22.27=3.4903 g Fe/1g O

The compounds are different. Using the law of multiple proportions:

(mass ratio of Compound I) / (mass ratio of Compound II) = (2.32 g Fe/g O)/ (3.49 g Fe/g O) = 0.6647 aprox.=1.0x2= 2 (2:3)

(mass ratio of Compound II) / (mass ratio of Compound I) = (3.49 g Fe/g O)/ (2.32 g Fe/g O) = 1.50x2=3 (3:2)

EXERCISES (page 95)

Exercise 67 Name the elements contained in the following compounds.

a. sodium chloride (NaCl) sodium and chlorine c. ethanol ( C2H6O) carbon, hydrogen, and oxygenb. ammonia (NH3 ) nitrogen and hydrogen d. bromine (Br2 ) bromine

Exercise 72

a. What is the percent by mass of carbon in 44 g of carbon dioxide (CO2 )?

Percent by mass (%) = mass of element × 100 mass of compound

Percent by mass = (mass of carbon / mass of carbon dioxide) x 100 = (12 g / 44 g) x 100 = 27 %

b. What is the percent by mass of oxygen in 44 g of carbon dioxide (CO2 )? 2 oxygen atoms = 16 g x 2 = 32 g

Percent by mass = (mass of oxygen / mass of carbon dioxide) x 100 = (32 g / 44 g) x 100 = 73 %

or (100% - 27% = 73% ) if we use the result from part (a).Exercise 74 A 25.3-g sample of an unknown compound contains 0.8 g of oxygen. What is the percent by mass of oxygen in the compound?

Answer:

Mass of unknown compound= 25.3 gMass of oxygen in the sample= 0.8 g% O= ? % O= (Mass of oxygen in the sample/ Mass of unknown compound) x 100= = (0.8 g O / 25.3 g unknown compound) x100 = 3.16%O

Exercise 75 Magnesium combines with oxygen to form magnesium oxide. If 10.57 g of magnesium reacts completely with 6.96 g of oxygen, what is the percent by mass of oxygen in magnesium oxide?

Mass percentage oxygen = 6.96 g / (10.57 g + 6.96 g) = 39.7%

Lesson 3.3 Mixtures of Matter. (pages 80-83)

VOCABULARY

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Mixture: a mixture is a combination of two or more pure substances in which each pure substance retains its individual chemical properties.

Heterogeneous mixture: is a mixture that does not blend smoothly throughout and in which the individual substances remain distinct. The salad dressing mixture is an example of a heterogeneous mixture.

Homogeneous mixture: is a mixture that has constant composition throughout; it always has a single phase. If you cut two pieces out of a silver mercury amalgam, their compositions will be the same.

Solution: is a homogeneous mixture that has constant composition throughout; it always has a single phase.

Filtration: is a technique that uses a porous barrier to separate a solid from a liquid. Heterogeneous mixtures composed of solids and liquids are easily separated by filtration

Distillation: is a separation technique that is based on differences in the boiling points of the substances involved. Most homogeneous mixtures can be separated by distillation.

Crystallization: is a separation technique that results in the formation of pure solid particles of a substance from a solution containing the dissolved substance.

Sublimation: is the process during which a solid changes to vapor without melting, i.e. withoutgoing through the liquid phase.

Sublimation can be used to separate two solids present in a mixture when one of the solids sublimates but not the other.

Chromatography: is a technique that separates the components of a mixture (called the mobile phase) based on the ability of each component to travel or be drawn across the surface of another material (called the stationary phase).

EXERCISES (page 95)

Exercise 58 Describe a method that could be used to separate each mixture.

a. iron filings and sand It can be used a magnet to draw the iron filings from the sand.

b. sand and salt

It can be added water to the mixture to dissolve the salt. Then, we can filter the mixture to remove the sand, and then boil off the water so only the salt remains.

c. the components of ink

It should be used Paper chromatography used to separate the components of the ink. If enough ink is available, distillation may also be used, but is far more complicated than paper chromatography.

d. helium and oxygen gases

We can cool the gas mixture until it condenses, then distill the condensate.

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Exercise 61 Describe how a homogeneous mixture differs from a heterogeneous mixture.

Homogeneous mixtures contain a single phase. Contrary, Heterogeneous mixtures may have many phases.

Exercise 64 What is chromatography, and how does it work?

Chromatography is a technique used to separate components of a mixture.

Usually, the mobile phase is a gas or a liquid, and the stationary phase is a solid, such as chromatography paper. The separation occurs because the various components of the mixture spread through the paper at different rates. Components with the strongest attraction for the paper travel slower.

Lesson 3.2 Changes in Matter. (pages 76 – 79)

VOCABULARY

Physical change: a change which alters a substance without changing its composition.

Phase change: is a transition of matter from one state to another

Chemical change: is a process that involves one or more substances changing into new substances. It is commonly referred to as a chemical reaction. The new substances formed in the reaction have different compositions and different properties from the substances present before the reaction occurred.

Law of conservation of mass: the law of conservation of mass states that mass is neither created nor destroyed during a chemical reaction—it is conserved. In other words, the mass of the reactants equals the mass of the products.

Exercise 1: Create a table that describes the three common states of matter in terms of their shape, volume, and compressibility.

Shape Volume CompressibilitySolid Definite Definite IncompressibleLiquid Takes shape of

container and fills container to the extent of its own volume

Definiteincompressible

Gas Takes shape ofcontainer

Fillsvolume ofcontainer

Compressible

Exercise 2: Describe the characteristics that identify a sample of matter as a substance.

The sample of matter must have a uniform and unchanging composition to be a substance.

Exercise 3: Classify each of the following as a physical or a chemical property.

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a. Iron and oxygen form rust. Chemical Propertyb. Iron is more dense than aluminum. Physical Property c. Magnesium burns brightly when ignited. Chemical Propertyd. Oil and water do not mix. Physical Property e. Mercury melts at −39°C. Physical Property

Exercise 4: Organize. Create a chart that compares physical and chemical properties. Give two examples for each type of property.

PHYSICAL PROPERTIES CHEMICAL PROPERTIES

Physical properties can be observed without changing the composition of the substance or sample.

The ability of a substance to combine with or change into one or more other substances. Chemical properties change the composition of the sample or substance

EXAMPLES Mass Density Boiling Point Melting Point Conduction of electricity Shape Color Malleability Ductility Heat

EXAMPLES Fermentation Rusting Decomposition Explosion Oxidation Tarnishing Rotting Burning Corrotion

Exercise 5 Use the data in the table to answer the following questions.

How many grams of bromine reacted?

Answer: before reaction - after reaction amount of bromine that reacted = 100.0 g - 8.5 g = 91.5 g

How many grams of compound were formed? Answer: Before reaction – after reactionamount of compound formed = 100.0 g + 10.3 g = 110.3 g – 8.5 g = 101.8 g

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Exercise 6 From a laboratory process designed to separate water into hydrogen and oxygen gas, a student collected 10.0 g of hydrogen and 79.4 g of oxygen. How much water was originally involved in the process?

Answer:

Mass of the reactants = mass of the products

Mass of the reactants = mass of water in the process

Mass of the products = mass of hydrogen + mass of oxygen

Mass of water in the process = mass of hydrogen + mass of oxygen

Mass of water in the process = 10.0 g + 79.4 g = 89.4 g

Exercise 7 A student carefully placed 15.6 g of sodium in a reactor supplied with an excess quantity of chlorine gas. When the reaction was complete, the student obtained 39.7 g of sodium chloride.

Calculate how many grams of chlorine gas reacted.

Answer:

mass of the reactants = mass of the productsmass of sodium + mass of chlorine = mass of sodium chloridemass of sodium = 15.6 gmass of sodium chloride = 39.7 g

So, substituting and solving for mass of chlorine:mass of sodium + mass of chlorine = mass of sodium chloride 15.6 g + mass of chlorine = 39.7 g

mass of chlorine = 39.7 g - 15.6 g = 24.1 g used in the reaction.

How many grams of sodium reacted?

Because the sodium reacts with excess chlorine, all of the sodium (15.6 g) is used in the reaction.

Exercise 8 A 10.0-g sample of magnesium reacts with oxygen to form 16.6 g of magnesium oxide.How many grams of oxygen reacted?

Answer:

mass of the reactants = mass of the products

mass of magnesium + mass of oxygen = mass of magnesium oxidemass of magnesium= 10.0 gmass of magnesium oxide = 16.6 g

So, substituting and solving for mass of oxygen:

mass of magnesium + mass of oxygen = mass of magnesium oxide

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10.0 g + mass of oxygen = 16.6 g

mass of oxygen = 16.6 g - 10.0 g = 6.6 g

EXERCISES (page 79)

Exercise 10 Classify each example as a physical change or a chemical change.

a. crushing an aluminum can physical change

b. recycling used aluminum cans to make new aluminum can physical change

c. aluminum combining with oxygen to form aluminum oxide chemical change

Exercise 13 Calculate. Solve each of the following.

a. In the complete reaction of 22.99 g of sodium with 35.45 g of chlorine, whatmass of sodium chloride is formed?

Answer:

mass of sodium chloride = mass of sodium + mass of chloridemass of sodium chloride = 22.99 g + 35.45 g = 58.44 g

b. A 12.2-g sample of X reacts with a sample of Y to form 78.9 g of XY. Whatis the mass of Y that reacted?

Answer:

mass of X = 12.2 g mass of XY = 78.9 g

mass of X + mass of Y = mass XY

So, substituting and solving for mass of Y:

mass of Y = mass of XY - mass of X = 78.9 g - 12.2 g = 66.7 g

EXERCISES (page 94)

Exercise 43 Classify each as a physical change or a chemical change.

a. breaking a pencil in two physicalb. water freezing and forming icephysicalc. frying an eggchemicald. burning woodchemicale. leaves changing colors in the fallchemical

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Exercise 46 List four indicators that a chemical change has probably occurred.

Indicators of a chemical reaction:

a change in color a change in odor a change in temperature the production of a gas or a solid upon mixing.

Lesson 3.1 Properties of Matter. (pages 70–75)

VOCABULARY

State of matter: the different forms of matter (solid, liquid, and gas). All matter that exists naturally on Earth and it can be classified as one of these physical forms.

Solid: is a form of matter that has its own definite shape and volume. Wood, iron, paper, and sugar are all examples of solids.

Liquid: is a form of matter that flows, has constant volume, and takes the shape of its container. Common examples of liquids include water, blood, and mercury.

Gas: is a form of matter that not only flows to conform to the shape of its container but also fills the entire volume of its container.

Vapor: refers to the gaseous state of a substance that is a solid or a liquid at room temperature. For example, steam is a vapor because water exists as a liquid at room temperature.

Physical property: a physical property is a characteristic that can be observed or measured without changing the sample’s composition. Density, color, odor, hardness, melting point,and boiling point are common physical properties

Extensive property: an extensive property is dependent on the amount of substance present. For example, mass is an extensive property. Length and volume are also extensive properties.

Intensive property: an intensive property is independent of the amount of substance present. For example, the density of a substance (at constant temperature and pressure) is the same no matter how much substance is present.

Chemical property: is the ability of a substance to combine with or change into one or more other substances.

EXERCISES (page 75)

Exercise 3: Classify each of the following as a physical or a chemical property.

a. Iron and oxygen form rust.chemicalb. Iron is more dense than aluminum.physicalc. Magnesium burns brightly when ignited.chemicald. Oil and water do not mix.

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physicale. Mercury melts at −39°C.physicalEXERCISES (page 94)

Exercise 34: Identify each physical property as extensive or intensive.

a. melting point intensive

b. mass extensive

c. density intensive d. length extensive

Exercise 37: Classify each as either a solid, a liquid, or a gas at room temperature.

a. milk liquidb. air gasc. copper solid d. helium gase. diamond solid f. candle wax solid

Exercise 38: Classify each as a physical property or a chemical property.

a. Aluminum has a silvery color. physicalb. Gold has a density of 19 g/cm3. physicalc. Sodium ignites when dropped in water. chemical d. Water boils at 100°C. physicale. Silver tarnishes. chemicalf. Mercury is a liquid at room temperature. physical