JDHS AP Chemistry 2020-2021 Summer Assignment

AP Chemistry Summer Assignment 2020-2021

Welcome to AP Chemistry. I hope that you are looking forward to the challenge of taking a college-level chemistry course as much as I am looking forward to helping you to prepare for the AP Chemistry Exam administered on Friday May 7, 2021. In the interest of saving some valuable time, and to let us hit the ground running when school starts back up again in September, it is important that we make sure that you have a solid foundation in some basic topics that were covered in your first-year chemistry course. There is a lot of material that we need to cover next year, so it is important that we hit the ground running. It is also very important that you do a significant amount of independent study including weekends, vacations, and other breaks. If you or your family are planning on taking a trip during one of the extended breaks next year, I strongly recommend that trip to occur during winter break rather than spring break, if at all possible. With the exam being so early in May, it is important for us to complete the curriculum in early-to-mid-April to allow sufficient time to practice past AP exam questions.

Thus, there are some topics that you should remember from your first year of chemistry that you need to review this summer so that we all start with the same basic background. You should arrive in September with a mastery of the basics of significant digits, scientific notation, proper use of units, simple unit conversions, factor-label method (dimensional analysis) and the methods used to analyze laboratory data. You must know the names, symbols and charges of the common monatomic ions. A list of polyatomic ions that must be memorized is attached. Be able to write formulas with charges when given the name and vice-versa. You also need to know how to name acids (such as hydrochloric acid and sulfuric acid). I have also attached a guide to naming compounds that was used in my Honors Chemistry class. When doing calculations, you will be expected to correctly use the factor-label method (aka Dimensional Analysis) to show your work on all homework, tests, and labs and to use the proper units on all numbers. A list of SI unit prefixes is also attached, and these prefixes, along with the conversion factors they represent, must also be memorized. Be sure you can write answers to mathematical operations using the correct number of significant digits (also called significant figures). The College Board (for the most part) allows leeway of one significant digit in either direction, so I follow a similar policy.

Your summer assignment (Labeled as HW #1-1 and HW #2-1) is listed on page 3 of this packet and starts on page 4. Chapters 1 and 2 are found at the end of this PDF. Do the problems to the best of your ability. This material is meant to be a review. You may find that you remember most of this content, but do not assume you fully remember it. The full summer assignment is due on the first day of class, Tuesday September 8th. It will be checked and counted as a homework assignment. In the event that school is still being conducted remotely in September, I will announce instructions for submission at that time. We will review the full summer assignment on that Thursday and Friday. There will be a full period test on the contents of the summer assignment on or about Friday September 11th (final date TBA).

Summary of the first few days of school:

Tuesday September 8: Full summer assignment (HW #1-1 and #2-1) will be checked at start of class. In class, we will mainly discuss Chapter 1 and HW #1-1.

Wednesday September 9: In class, we will continue to discuss Chapters 1 and 2.

Thursday September 10: Last chance to discuss Chapters 1-2. We will start Chapter 3 in class.

Friday September 11: TEST: Chapters 1-2 (70 pts) Common Polyatomic Ions: You must have these memorized!

Ion Name + NH4 Ammonium - NO2 Nitrite - NO3 Nitrate OH- Hydroxide CN- Cyanide - MnO4 Permanganate - HCO3 Hydrogen Carbonate (Bicarbonate) ClO- Hypochlorite - ClO2 Chlorite - ClO3 Chlorate - ClO4 Perchlorate - C2H3O2 Acetate 2- CO3 Carbonate 2- SO3 Sulfite 2- SO4 Sulfate 2- O2 Peroxide 2- CrO4 Chromate 2- Cr2O7 Dichromate 3- PO4 Phosphate 3- PO3 Phosphite

Also memorize: Aluminum ion Al3+ Silver ion Ag1+ Zinc ion Zn2+ Cadmium ion Cd2+

Prefixes Used with SI Units Prefix Symbol Meaning Example Tera- T 1,000,000,000,000 or 1012 1 terameter (Tm) = 1´1012 m Giga- G 1,000,000,000 or 109 1 gigameter (Gm) = 1´109 m Mega- M 1,000,000 or 106 1 megameter (Mm) = 1´106 m Kilo- k 1,000 or 103 1 kilometer (km) = 1´103 m Deci- d 1/10 or 10–1 1 decimeter (dm) = 1´10–1 m Centi- c 1/100 or 10–2 1 centimeter (cm) = 1´10–2 m Milli- m 1/1000 or 10–3 1 millimeter (mm) = 1´10–3 m Micro- µ 1/1,000,000 or 10–6 1 micrometer (µm) = 1´10–6 m Nano- n 1/1,000,000,000 or 10–9 1 nanometer (nm) = 1´10–9 m Pico- p 1/1,000,000,000,000 or 10–12 1 picometer (pm) = 1´10–12 m

Summer Assignment

• This full assignment (HW # 1-1 and HW #2-1, Significant Figures worksheet, and Nomenclature worksheet) is due on the 1st day of school—Tuesday, September 8th. • This assignment will count as a homework assignment for the 1st quarter (worth 40 points) • I expect you to read the book to brush up on the material in Chapters 1 and 2. • You must show your work to get credit for the homework. For calculation questions, you must use factor label method (unit conversions or dimensional analysis). Thus, every number must have units. This factor label method is crucial in this course because it allows you to show your work for multistep calculations quickly, clearly and fully. College Board awards points for implied calculations (like converting to moles before converting to something else) where the calculation is built into a string of dimensional analysis even though a number was never explicitly calculated for that step. • Please write all final answers with the proper number of significant figures. For more help on significant figures, do the Significant Figures worksheet attached to this packet. • For more help with Naming Compounds, do the Nomenclature worksheet attached to this packet.

Chapter 1

READ CHAPTER 1

HW #1-1: DO PROBLEMS, pg. 32 #22, 23, 33, 35, 43, 47, 53, 55, 57, 73, 75, 81, 83, 85, 87, 91, 95, 97, 103, *123, + SIGNIFICANT DIGITS WORKSHEET

Important information to help solve these problems: 1 kg = 2.205 lbs 1 ton = 2000 lbs 1 mi = 1.609 km 1 in = 2.54 cm Speed of light = c = 3.00´108 m/s 1 mL = 1 cm3 V of a cylinder = πr2h

Chapter 2

READ CHAPTER 2

HW # 2-1: DO PROBLEMS pg. 65 #39, 57, 61, 63, 65, 67, 71, 73, 85, 99, 103, + NOMENCLATURE WORKSHEET

Chapter 1 Questions and Exercises:

Important information to help solve these problems: 1 kg = 2.205 lbs 1 ton = 2000 lbs 1 mi = 1.609 km 1 in = 2.54 cm Speed of light = c = 3.00´108 m/s 1 mL = 1 cm3 V of a cylinder = πr2h

22. What is the difference between random error and systematic error?

23. A measurement is a quantitative observation involving both a number and a unit. What is a qualitative observation? What are the SI units for mass, length, and volume? What is the assumed uncertainty in a number (unless stated otherwise)? The uncertainty of a measurement depends on the precision of the measuring device. Explain.

33. How many significant figures are there in each of the following values?

a. 6.07 ´ 10-15 b. 0.003840

c. 17.00 d. 8 ´ 108

e. 463.8052 f. 300

g. 301 h. 300.

35. Round each of the following numbers to the indicated number of significant digits, and write the answer in correct scientific notation:

a. 0.00034159 to 3 digits b. 103.351 ´ 102 to 4 digits

c. 17.9915 to 5 digits d. 3.365 ´ 105 to 3 digits

43. Perform each of the following conversions:

a. 8.43 cm to millimeters b. 2.41 ´ 102 cm to meters

c. 294.5 nm to centimeters d. 1.445 ´ 104 m to kilometers

47. Using the following exact conversion factors (not taken into consideration for significant figures) to perform the stated calculations: (5.5 yd = 1 rod, 40 rods = 1 furlong, 8 furlongs = 1 mile)

a. The Kentucky Derby race is 1.25 miles. How long is the race in rods, furlongs, meters, and kilometers?

b. A marathon race is 26 miles, 385 yards. What is this distance in rods, furlongs, meters, and kilometers?

53. Science fiction often uses nautical analogies to describe space travel. If the starship USS Enterprise is traveling at warp factor 1.71, what is its speed in knots and in miles per hour? (Warp 171 = 5.00 times the speed of light, speed of light (c) = 3.00 ´ 108 m/s, 1 knot = 2030 yd/hour)

55. You are driving 65 miles/hr and take your eyes off the road for “just a second.” What distance (in feet) do you travel in this time?

57. The dosage for an antibiotic is prescribed at 8.0 mg per kilogram of bodyweight, taken twice daily for two weeks. What total mass of antibiotic will be taken by a 180 lb person for the two-week period?

73. Diamonds are measured in carats, and 1 carat = 0.200 g. the density of diamond is 3.51 g/cm3.

a. What is the density of a 5.0-carat diamond?

b. What is the mass in carats of a diamond measuring 2.8 mL?

75. A sample containing 33.42 g of metal pellets is poured into a graduated cylinder initially containing 12.7 mL of water, causing the water level in the cylinder to rise to 21.6 mL. Calculate the density of the metal.

81. The density of osmium (the densest metal) is 22.57 g/cm3. If a 1.00-kg rectangular block of osmium has two dimensions of 4.00 ´ 4.00 cm, calculate the third dimension of the block.

83. Match each description below with the following microscopic pictures. More than one picture may fit each description. A picture may be used more than once or not used at all.

a. A gaseous compound b. A mixture of two gaseous elements

b. A solid element d. A mixture of a gaseous element and a gaseous compound

85. Classify each of the following as homogeneous or heterogeneous:

a. A door b. the air you breathe

c. a cup of black coffee d. the water you drink

e. salsa f. your lab partner

87. Classify each of the following as a mixture or a pure substance. Of the pure substances, which are elements and which are compounds?

a. blood b. the oceans

c. iron d. brass

e. uranium f. wine

g. leather h. table salt 91. Classify the following as physical or chemical changes:

a. Moth balls vaporize in a closet

b. Hydrofluoric acid attacks glass and is used to etch calibration marks on glass laboratory utensils

c. A French chef making a sauce with brandy is able to boil off the alcohol from the brandy, leaving behind just the brandy flavoring

d. Chemistry majors sometimes get holes in the cotton jeans they wear to lab because of acid spills

95. Lipitor, a pharmaceutical drug that has been shown to lower “bad” cholesterol levels while raising “good” cholesterol levels in patients taking the drug, had over $3 billion in sales in 2015. Assuming one 2.5-g pill contains 4.0% of the active ingredient by mass, what mass in kg of active ingredient is present in one bottle of 100 pills?

97. The contents of one 40. lb bag of topsoil will cover 10. Square feet of ground to a depth of 1.0 inch. What number of bags is needed to cover a plot that measures 200. by 300. m to a depth of 4.0 cm>

103. A person with high cholesterol has 250 mg of cholesterol per 100.0 mL of blood. If the total blood volume of the person is 5.4 L, what is the total mass (in grams) of cholesterol present in the person’s blood?

*123. Sterling silver is a solid solution of silver and copper. If a piece of a sterling silver necklace has a mass of 105.0 g and a volume of 10.12 mL, calculate the mass percent of copper in the piece of necklace. Assume that the volume of the silver plus the volume of copper present equals the total volume. (density of silver = 10.49 g/cm3, density of copper = 8.96 g/cm3)

Mass Percent of copper = (mass of copper/total mass) ´ 100%

Chapter 2 Questions and Exercises:

39. A sample of chloroform is found to contain 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen. If a second sample of chloroform is found to contain 30.0 g of carbon, what is the total mass of chloroform in the second sample?

57. a. Classify the following elements as metals or nonmetals:

Mg Si Rn Ti

Ge Eu Au B

Am Bi At Br

b. The distinction between metals and nonmetals is really not a clear one. Some elements, called metalloids, are intermediate in their properties. Which of these elements would you reclassify as metalloids? Which other elements in the periodic table would you expect to be metalloids?

# 61. Write the atomic symbol "! for each of the following isotopes:

a. Z = 8, number of neutrons = 9 b. The isotope of chlorine in which A = 37

c. Z = 27, A = 60 d. 26 protons, 31 neutrons

e. The isotope of I with a mass number of 131 f. Z = 3, number of neutrons =4

# 63. Write the atomic symbol "! for each of the following atoms:

a. 11 protons, 12 neutrons, 10 electrons

b. 9 protons, 10 neutrons, 11 electrons

c. 8 protons, 8 neutrons, 8 electrons

65. How many protons and neutrons are in the nucleus of each of the following atoms? In a neutral atom of each element, how many electrons are present?

a. 79Br b. 81Br

c. 239Pu d. 133Cs

e. 3H f. 56Fe

67. For each of the following ions, indicate the number of protons and electrons the ion contains

a. Ba2+ b. Zn2+

c. N3- d. Rb+

e. Co3+ f. Te2-

71. Complete the following table: Symbol Protons Neutrons Electrons Net Charge &'( %&$ 20 20 2+ 23 28 20 (% '%) 35 44 36 15 16 3-

73. Would you expect each of the following atoms to gain or lose electrons when forming ions? What ion is the most likely in each case?

a. Ra b. In

c. P d. Te

e. Br f. Rb 85. Elements in the same family often form oxyanions of the same general formula. The anions are named in similar fashion. What are the names of the oxyanions of selenium and tellurium:

2- 2- a. SeO4 b. SeO3

2- 2- c. TeO4 d. TeO3

99. The isotope of an unknown element, X has a mass number of 79. The most stable isotope has 36 electrons and forms a binary compound with sodium, having a formula of Na2X. Which of the following statements is(are) true? For the false statements, correct them:

a. The binary compound between X and fluorine will be a covalent compound.

b. The isotope of X contains 38 protons.

c. The isotope of X contains 41 neutrons

d. The identity of X is strontium, Sr.

103. An element’s most stable ion forms an iconic compound with bromine, having the formula XBr2. If the ion of element X has a mass number of 230 and has 86 electrons, what is the identity of the element, and how many neutrons does it have? WKS Name______Significant Digits Date______Period______(Significant Figures)

Part A: Determining the number of significant digits in a measurement: 1. Digits other than zero are always significant. Ex: 96 g 2 sig figs Ex: 61.4 g 3 sig figs Ex: 1.5345 5 sig figs

2. One or more final zeros used after the decimal point are always significant. Ex: 4.70 L 3 sig figs Ex: 576.980 L 6 sig figs When there is a decimal, trailing zeros Ex: 45.9000 L 6 sig figs ARE significant.

3. Zeros between other digits are always significant. Ex: 5.089 mL 4 sig figs Ex: 80017 mL 5 sig figs Ex: 480.020 mL 6 sig figs

4. For numbers less than 1, all zeros at the beginning are insignificant until one hits a digit other than 0. Ex: 0.0052 cm 2 sig figs Ex: 0.0306 cm 3 sig figs When there is a decimal, leading zeros Ex: 0.00400 cm 3 sig figs are NOT significant.

5. Unclear number of significant digits: When a measurement ends with one or more zeros which are not after the decimal point, the number of significant digits is unclear. In these cases, one could clarify the precision by writing the measurement in scientific notation. When there is NO Ex: 3870 m 3 sig figs (at least), but could be 4 sig figs decimal, trailing zeros Is it 3.87 x 10 3 or 3.870 x 103 ? are ambiguous. Thus,

one generally must Ex: 500 m 1 sig fig (at least), but could be 2 sig figs or 3 sig figs assume that they are Is it 5 x 102 or 5.0 x 102 or 5.00 x 102 ? NOT significant.

Part A Practice: Determine the number of significant digits for the following measurements. If the number of significant digits is unclear, write “unclear” and list all the possible number of significant digits it could have. For example: 4050 L unclear, 3 or 4 possible Also, convert each of the measurements to scientific notation.

# of sig figs Write number in Scientific Notation 1) 678 g 2) 4.098 mm 3) 0.0089 s 4) 0.07608 mL 5) 5098 cm 6) 57.0010 min 7) 63100 km 8) 0.000001 pm 9) 600.0200 g 10) 20 m

More PART A Practice: Round all of the following numbers to three significant digits.

11) 8.97893 ______12) 0.004524 ______13) 2.995 ______14) 354,680 ______15) 5.903 x 105 ______

Part B: Keeping proper number of significant digits/decimal places when doing calculations Multiplication and Division: The answer may contain only as many significant digits as the measurement with the LEAST number of significant digits. Ex: (2.30 m) (4.565 m) = 10.4995 (on calc) = 10.5 m2 3 sig figs 4 sig figs 3 sig figs

5.2 x 10-4 m (2 sig figs) Ex: = = 1.710526316 x 10-4 (on calc) = 1.7 x 10-4 m/s 3.04 s ( 3 sig figs) 2 sig

figs Addition and Subtraction: Do NOT count significant digits. Instead, just look at the number of digits after the decimal point. The answer may contain only as many decimal places as the measurement having the LEAST number of decimal places. Ex: 4.56 m Ex: 7678.0 g + 0.32 m – 3.907 g 4.88 m (only 2 decimal places) 7674.1 g (only 1 decimal place)

Practice: Make the following calculations and round answers to have the correct number of significant digits or decimal places. Please write answers in normal and scientific notation. Write UNITS!!!

16) (5.09 cm) (3.4 cm) =

0.000434 g 17) = 0.0045 mL

18) (4001 cm) (2.34 cm) (40.1 cm) =

19) (6.890 x 10 3 m) ( 5.00 x 105 m) =

4.98 x 10-5 m 20) = 4.500 x 102 s

21) 5.879 s + 89.98 s = (Normal notation only is fine.)

22) 0.0989 g − 0.0078 g = (Normal notation only is fine.)

35.5 g- 1.2 g 23) = 1.234 x 103 mL Notes – Naming & Formulas of Ionic Compounds

Ionic Compounds (Metal/Nonmetal)

• Ionic compound represented by Formula unit: – The simplest whole ratio of the ions present in an ionic compound. – Not the same as a “molecule”: can’t have individual formula units alone

• Binary compound contains only two kinds of elements – Each ion is a monatomic ion–has only one atom in it § Main group so charge (oxidation number) comes from column: • Metal ion oxidation numbers 1+, 2+, 3+ according to distance from noble gases • Nonmetal ion oxidation numbers 1–, 2–, 3– according to distance from noble gases

• Ternary & Quaternary Ionic Compounds – Contain one or two polyatomic ions: Group of covalently bonded atoms that has a charge: § Most polyatomic ions called “oxyanions”: contain central atom bonded to at least one oxygen atom • Central atom give the root of ion name (i.e. N ® nitrate, Cl ® chlorate) • Name (ending/prefix) depends on number of oxygen atoms 2– O O O S O O N

2– O – O – – ¨ e.g. SO4 : ; NO3 : ; OH : [O–H] – – ¨ acetate has two formulas: C2H3O2 preferred, CH3COO more traditional

• Writing names from the formulas – Name cation first from periodic table or chart – Name anion from periodic table (change ending to –ide) or chart – Examples: KBr: potassium bromide; CaI2: calcium iodide; Al2Te3: aluminum telluride; Li2SO3: lithium sulfite; NH4I: ammonium bromide; NaHCO3: sodium hydrogen carbonate

• Determining formulas from names – Determine symbol/formula and charge of cation – Determine symbol/formula and charge of anion (remember –ide if an element from PT, –ate and –ite if a polyatomic ion) – “Criss-cross” charges to equalize + and – charges (LCM), Reduce subscripts to smallest whole numbers § If more than one polyatomic ion, put formula in parentheses and the subscript outside to the right • NEVER change the formula (draw a circle around it to remind yourself) § Remove charges from final answer 3+ 2– 2+ 2– – Examples: aluminum selenide: Al Se ® Al2Se3; calcium sulfide: Ca S ® Ca2S2 ® CaS; sodium carbonate: + 2– 2+ – Na CO3 ® Na2CO3; calcium bromate: Ca BrO3 ® Ca(BrO3)2

• Transition Metals (d-block and p-block metals) – Have the ability to form multiple ions (except for Al3+, Cd2+, Zn2+, and Ag+) – Cannot determine charge directly from the formula § Must use the anion it is paired with to deduce the charge – charges of anions are consistent § Examples: Iron can form 3+ or 2+ charge à FeO (O is 2-, so Fe is 2+) or Fe2O3 (O is 2-, so Fe is 3+) – When writing the name, the charge of the transition metal cation must be specified using a roman numeral 2+ 3+ § Examples: FeO (made with Fe ) à Iron (II) Oxide and Fe2O3 (made with Fe ) à Iron (III) Oxide

Molecular Compounds (Nonmetal/Nonmetal)

• Naming Binary molecular Compounds (two non-metals, not starting w/H) – Name first element without changes, then second element with –ide ending – Put prefix to indicate the number of atoms if more than one Mono = 1 Hexa = 6 Di = 2 Hepta = 7 Tri = 3 Octa = 8 Tetra = 4 Nona = 9 Penta = 5 Deca = 10 § Use mono- for elements in the second position only – Examples: PBr3 (phosphorous tribromide), NO (nitrogen monoxide), CO2 (carbon dioxide)

• Determining Formula of Binary molecular compounds § Write symbols from name § Add subscripts to match prefixes § Examples: carbon tetraiodide (CI4); dinitrogen pentoxide (N2O5); sulfur hexafluoride (SF6)

• Naming Acids (compounds starting with H) – Hydrogen exists as H+ (a proton) and pairs with anions to form an acid – Binary Acids (H+ and a monatomic anion) § Add enough H+ to balance charge of anion § Name: Hydro______ic Acid § Examples: HCl – Hydrochloric Acid; H3N – Hydronitric Acid; H2S – Hydrosulfuric Acid – Polyatomic Ions § Add enough H+ to balance charge of the polyatomic anion § Name: “Ate-ic Ite-ous” -- Replacing suffix of the polyatomic ion • If the polyatomic ion ends with -ate, the acid name ends with -ic ¨ Examples: HNO3 (Nitrate) – Nitric Acid; H2SO4 (Sulfate) – Sulfuric Acid • If the polyatomic ion ends with -ite, the acid name ends with -ous ¨ Examples: HNO2 (Nitrite) – Nitrous Acid; H2SO3 (Sulfite) – Sulfurous Acid

WKS #1 Name______Nomenclature Period______Date______

Write the correct formula for the following compounds listed below.

1. manganese dioxide 21. acetic acid

2. sulfur dioxide 22. copper (II) nitrite

3. hydroarsenic acid 23. nitrogen dioxide

4. chlorous acid 24. phosphorus trichloride

5. silver chloride 25. sodium phosphate

6. copper (II) hydroxide 26. potassium carbonate

7. ammonium sulfide 27. hydroselenic acid

8. nickel (II) bromide 28. lead (IV) chloride

9. iron (III) oxide 29. tin (II) bromide

10. bromic acid 30. ammonium hydroxide

11. ammonium hydrogen sulfate 31. periodic acid

12. mercury (II) sulfate 32. iron (II) hydroxide

13. iron (III) dichromate 33. carbon dioxide

14. magnesium phosphate 34. dinitrogen pentoxide

15. cadmium bicarbonate 35. silver oxide

16. zinc chlorite 36. aluminum nitride

17. chromic acid 37. manganese (II) nitrite

18. diphosphorus pentoxide 38. ammonium carbonate

19. aluminum phosphate 39. aluminum oxide

20. sulfurous acid 40. antimony pentasulfide Write the correct name for the following compounds listed below.

41. CO 63. RaBr2

42. MgBr2 64. NaMnO4

43. SnCl2 65. PbI2

44. N2O 66. CaS

45. NH4F 67. Bi2Te3

46. AsCl5 68. KClO4

47. KHCO3 69. HgBr2

48. K2O 70. Li2O2

49. BaS2O3 71. P3N5

50. ZnO 72. CuSO3

51. NaClO 73. FePO4

52. SrS 74. PbCr2O7

53. Al(BrO3)3 75. CuNO3

54. PF3 76. K2SO4

55. Pd(CN)2 77. AgC2H3O2

56. Zn(HSO4)2 78. TeI4

57. Mg(ClO2)2 79. Cd3(PO4)2

58. Ca(MnO4)2 80. Ag2S

59. H2S 81. H3PO4

60. Be(NO3)2 82. Pb(HCO3)4

61. NiCrO4 83. ZnF2

62. HClO 84. HIO4

Chapter 1

Chemical Foundations

§ (1.1) Chemistry: An overview § (1.2) The scientific method § (1.3) Units of measurement § (1.4) Uncertainty in measurement § (1.5) Significant figures and calculations § (1.6) Learning to solve problems systematically § (1.7) Dimensional analysis § (1.8) Temperature § (1.9) Density § (1.10) Classification of matter

Chemistry - Introduction § Humans have believed that matter is composed of atoms § Recent development - Individual atoms can be viewed by using a scanning tunneling microscope (STM) § STM - Uses an electron current from a tiny needle to probe the surface of a substance

Scanning Tunneling Microscope (STM) Images § Depict bridges (electrons) that interconnect atoms

Chemistry - Complexity and Challenge § Nature of atoms is complex § Components do not behave like objects in the macroscopic world § Macroscopic world - One that is experienced by humans § Challenge § To understand the connection between the macroscopic world and the microscopic world of atoms and molecules

Critical Thinking § The scanning tunneling microscope allows us to see atoms § What if you were sent back in time before the invention of the scanning tunneling microscope? § What evidence could you give to support the theory that all matter is made of atoms and molecules?

Configuration of Atoms § Properties of a substance can be determined by the way in which atoms are organized in that substance § Example - Water is composed two atoms, hydrogen and oxygen § Two hydrogen atoms and one oxygen atom are bound together to form the water molecule

Configuration of Atoms (continued 1)

§ When an electric current is passed through water, it decomposes to hydrogen and oxygen § Both chemical elements exist naturally as diatomic (two- atom) molecules

Configuration of Atoms (continued 2)

§ Decomposition of water to its component elements can be represented in the following manner:

Fundamental Concepts of Chemistry § Matter is composed of various types of atoms § By reorganizing the way the atoms are attached to each other, one substance changes to another

Science § Framework for gaining and organizing knowledge § Procedure for processing and understanding certain types of information § Scientific method § Lies at the center of scientific inquiry § Varies based on the nature of a specific problem and the particular investigator involved

Figure 1.3 - Fundamental Steps of the Scientific Method

Steps in the Scientific Method 1. Make observations § Qualitative observations do not involve numbers § Quantitative observations (measurements) involve both a number and a unit 2. Formulate a hypothesis § Hypothesis: Possible explanation for an observation

Steps in the Scientific Method (continued) 3. Perform experiments to test the hypothesis § Gather new information that would enable the scientist to ascertain the validity of the hypothesis § Experiments always produce new observations that bring the process back to the beginning again

Scientific Models § Theory (model): Set of tested hypotheses that gives an overall explanation of a natural phenomenon § Explanation of why nature behaves in a certain way § Constantly refined or replaced as more information becomes available § Explains observed natural behavior in terms of human experiences

Scientific Models (continued) § Observations § Events that are witnessed and can be recorded § Natural law: A summary of observed (measurable) behavior § Example - Law of conservation of mass, which states that the total mass of materials is unaffected by a chemical change in those materials

Law versus Theory

Law Theory

• Statement on • Human inventions generally observed • Attempts to explain behavior why something • Summarizes what happens happens

Science: Drawbacks § Focusing on a theory may limit one’s ability to see alternative explanations § Scientists are humans, and humans have prejudices § Science is affected by profit motives, budgets, fads, wars, and religious beliefs

Critical Thinking § What if everyone in the government used the scientific method to analyze and solve society’s problems, and politics were never involved in the solutions? § How would this be different from the present situation, and would it be better or worse?

Join In (1) § Which of the following is an example of a quantitative observation? a. Solution A is a darker red color than solution B b. The grass is green c. Substance A has a greater mass than substance B d. The temperature of the water is 45C

Measurement § Consists of two essential parts, a number and a scale (unit) § Standard systems of units § English system - Used in the United States § Metric system § SI system (International System) § Based on the metric system and units derived from the metric system

Table 1.1 - Fundamental SI Units

Table 1.2 - Prefixes Used in the SI System

Table 1.3 - Some Examples of Commonly Used Units

Figure 1.5 - Measurement of Volume

Mass and Weight § Mass: Measure of the resistance of an object to a change in its state of motion § Measured by the force necessary to give an object a certain acceleration § Weight: Force exerted by gravity on an object § Varies with the strength of the gravitational field

Critical Thinking § What if you were not allowed to use units for one day? § How would this affect your life for that day?

Certain and Uncertain Digits § Certain digits § Numbers that remain the same regardless of who measures them § Uncertain digits § Digits that must be estimated and therefore vary § While reporting a measurement, record all certain digits plus the first uncertain digit

Measurement of Volume Using a Buret § Volume is read at the bottom of the liquid curve, which is called the meniscus § Meniscus of the liquid occurs at about 20.15 mL § Certain digits - 20.1 § Uncertain digit - 20.15

Uncertainty § A measurement always has some degree of uncertainty § Uncertainty of a measurement depends on the precision of the measuring device § Significant figures: Numbers in which the certain digits and the first uncertain digit are recorded § Uncertainty in the last number is always assumed to be 1 unless otherwise indicated

Example 1.1 - Uncertainty in Measurement § In analyzing a sample of polluted water, a chemist measured out a 25.00-mL water sample with a pipet § At another point in the analysis, the chemist used a graduated cylinder to measure 25 mL of a solution § What is the difference between the measurements 25.00 mL and 25 mL?

Example 1.1 - Solution § Even though the two volume measurements appear to be equal, they convey different information § The quantity 25 mL means that the volume is between 24 mL and 26 mL, whereas the quantity 25.00 mL means that the volume is between 24.99 mL and 25.01 mL § The pipet measures volume with much greater precision than does the graduated cylinder

Precision and Accuracy § Accuracy: Agreement of a particular value with the true value § Precision: Degree of agreement among several measurements of the same quantity § Represents the reproducibility of a given type of measurement

Types of Errors § Random error (intermediate error) § Measurement has an equal probability of being low or high § Occurs in estimating the value of the last digit of a measurement § Systematic error (determinate error) § Occurs in the same direction each time § Either always high or always low

An Illustration of the Types of Errors

Large random errors Small random errors Small random errors and a large and no systematic systematic error error

Quantitative Work § Precision is often used as an indication of accuracy § Assumption - Average of a series of precise measurements is accurate or close to the true value § Valid if systematic errors are absent

Example 1.2 - Precision and Accuracy § To check the accuracy of a graduated cylinder, a student filled the cylinder to the 25-mL mark using water delivered from a buret and then read the volume delivered

Example 1.2 - Precision and Accuracy (continued) § Following are the results of five trials:

§ Is the graduated cylinder accurate?

Example 1.2 - Solution § The results of the trials show good precision (for a graduated cylinder) § The student has good technique § Note that the average value measured using the buret is significantly different from 25 mL § Thus, this graduated cylinder is not very accurate § It produces a systematic error (in this case, the indicated result is low for each measurement)

Join In (2) § The glassware shown below is called a buret § The buret is filled to the zero mark (at the top) with a solution, and the solution is transferred to a beaker § What volume of transferred solution should be reported? a. 20 mL b. 22 mL c. 22.0 mL d. 22.00 mL e. 25 mL

Join In (3) § The boiling point of a liquid was measured in the lab, with the following results:

Trial Boiling point 1 22.0C 0.1 2 22.1C 0.1 3 21.9C 0.1

§ The actual boiling point of the liquid is 28.7C

Join In (3) (continued) § The results of the determination of the boiling point are: a. accurate and precise b. precise but inaccurate c. accurate but imprecise d. inaccurate and imprecise

Join In (4) § _____ reflects the reproducibility of a given type of measurement a. Accuracy b. Precision c. Certainty d. Systematic error e. Random error

Join In (5) § _____ is the agreement of a particular value with the true value a. Accuracy b. Precision c. Certainty d. Systematic error e. Random error

Rules for Counting Significant Figures 1. Nonzero integers § Always count as significant figures 2. Zeros - There are three classes of zeros § Leading zeros § Captive zeros § Trailing zeros

Rules for Counting Significant Figures (continued) 3. Exact numbers § Determined by counting and not by using a measuring device § Assumed to have an infinite number of significant figures § Can arise from definitions § Example - 2 in 2πr

Copyright ©2018 Cengage Learning. All Rights Reserved. 48 Section 1.5 Significant Figures and Calculations Counting Significant Figures (Sig. Figs.): Classes of Zeroes § Leading zeros § Zeros that precede all the nonzero digits § Do not count as sig. figs. § Example - 0.0025 has only two sig. figs. § Captive zeros § Zeros between nonzero digits § Always count as sig. figs. § Example - 1.008 has four sig. figs.

Copyright ©2018 Cengage Learning. All Rights Reserved. 49 Section 1.5 Significant Figures and Calculations Counting Significant Figures (Sig. Figs.): Classes of Zeroes (continued) § Trailing zeros § Zeros at the right end of the number § Significant only if the number contains a decimal point § Examples -100 has only one sig fig whereas 1.00102 has three sig. figs.

Exponential Notation § Advantages § Number of significant figures can be easily indicated § Fewer zeros are required to write a very large or very small number § Example § 0.000060 is much more conveniently represented as 6.0 10–5 § Number has two significant figures

Interactive Example 1.3 - Significant Figures § Give the number of significant figures for each of the following results: a. A student’s extraction procedure on tea yields 0.0105 g of caffeine b. A chemist records a mass of 0.050080 g in an analysis c. In an experiment a span of time is determined to be 8.05010–3 s

Interactive Example 1.3 - Solution a. The number contains three significant figures § The zeros to the left of the 1 are leading zeros and are not significant, but the remaining zero (a captive zero) is significant b. The number contains five significant figures § The leading zeros (to the left of the 5) are not significant § The captive zeros between the 5 and the 8 are significant Copyright ©2018 Cengage Learning. All Rights Reserved. Section 1.5 Significant Figures and Calculations

Interactive Example 1.3 - Solution (continued)

§ The trailing zero to the right of the 8 is significant because the number contains a decimal point c. This number has four significant figures § Both zeros are significant

Copyright ©2018 Cengage Learning. All Rights Reserved. Section 1.5 Significant Figures and Calculations Rules for Significant Figures in Mathematical Operations § Multiplication or division § Number of significant figures in the result is the same as the number in the least precise measurement used in the calculation 4.56 ´ 1.4 = 6.38¾¾¾¾®corrected 6.4

Limiting term has two Two significant figures significant figures § Product should have only two significant figures

Copyright ©2018 Cengage Learning. All Rights Reserved. 55 Section 1.5 Significant Figures and Calculations Rules for Significant Figures in Mathematical Operations (continued) § Addition or subtraction § Result has the same number of decimal places as the least precise measurement used in the calculation § Example 12.11 18.0 Limiting term has one decimal place 1.013 31.123¾¾¾¾®corrected 31.1

One decimal place

Rules for Rounding § In a series of calculations, round off only after carrying the extra digits through to the final result § If the digit to be removed is: § Less than 5 - Preceding digit stays the same § Example - 1.33 rounds to 1.3 § Greater than or equal to 5 - Preceding digit is increased by 1 § Example - 1.36 to 1.4 § Do not round sequentially Copyright ©2018 Cengage Learning. All Rights Reserved. Section 1.5 Significant Figures and Calculations Interactive Example 1.4 - Significant Figures in Mathematical Operations § Carry out the following mathematical operations, and give each result with the correct number of significant figures a. 1.05 10–3 6.135 b. 21 – 13.8

Copyright ©2018 Cengage Learning. All Rights Reserved. Section 1.5 Significant Figures and Calculations Interactive Example 1.4 - Significant Figures in Mathematical Operations (continued) c. As part of a lab assignment to determine the value of the gas constant (R), a student measured the pressure (P), volume (V), and temperature (T) for a sample of gas, where PV R = T § The following values were obtained: § P = 2.560 § T = 275.15 § V = 8.8 § Calculate R to the correct number of significant figures Copyright ©2018 Cengage Learning. All Rights Reserved. Section 1.5 Significant Figures and Calculations

Interactive Example 1.4 - Solution a. The result is 1.7110–4, which has three significant figures because the term with the least precision (1.0510–3) has three significant figures b. The result is 7 with no decimal point because the number with the least number of decimal places (21) has none

Interactive Example 1.4 - Solution (continued 1) c. PV (2.560)( 8.8) R == T 275.15 § The correct procedure for obtaining the final result can be represented as follows: (2.560)( 8.8) 22.528 = = 0.0818753 275.15 275.15 = 0.082 = 8.2×10-2 = R

Interactive Example 1.4 - Solution (continued 2)

§ The final result must be rounded to two significant figures because 8.8 (the least precise measurement) has two significant figures § To show the effects of rounding at intermediate steps, carry out the calculation as follows: Rounded to two significant figures

(2.560)( 8.8) 22.528 23 = = 275.15 275.15 275.15

Interactive Example 1.4 - Solution (continued 3)

§ Now we proceed with the next calculation 23 = 0.0835908 275.15 § Rounded to two significant figures, this result is 0.084 = 8.4 10–2 § Note that intermediate rounding gives a significantly different result than that obtained by rounding only at the end

Interactive Example 1.4 - Solution (continued 4)

§ Again, we must reemphasize that in your calculations you should round only at the end § Rounding is carried out at intermediate steps in this text (to always show the correct number of significant figures) § The final answer given in the text may differ slightly from the one you obtain (rounding only at the end)

Join In (6) § Express 3140 in scientific notation a. 3.14 103 b. 3.14 10–3 c. 3.140 103 d. 3.140 10–3

Join In (7) § After performing a calculation in the lab, the display on your calculator reads 0.023060070 § If the number in the answer is to have five significant figures, what result should you report? a. 0.0230 b. 0.00231 c. 0.023060 d. 0.2367 e. 0.02306

Join In (8) § How many significant figures are in the number 0.03040? a. 1 b. 2 c. 3 d. 4 e. 5

Join In (9) § The beakers below have different precisions

Join In (9) (continued) § You pour the water from these three beakers into one container § What is the volume in this container reported to the correct number of significant figures? a. 78.817 mL b. 78.82 mL c. 78.8 mL d. 79 mL

Learning to Solve Problems Systematically § Questions to ask while approaching a problem § Where am I going? § What do I know? § How do I get there?

Dimensional Analysis (Unit Factor Method) § Helps convert a given result from one system of units to another

Copyright ©2018 Cengage Learning. All Rights Reserved. 71 Section 1.7 Dimensional Analysis Problem-Solving Strategy - Converting from One Unit to Another § Use the equivalence statement that relates the two units § Derive the appropriate unit factor by looking at the direction of the required change to cancel the unwanted units § Multiply the quantity to be converted by the unit factor to give the quantity with the desired units

Interactive Example 1.6 - Unit Conversions II § You want to order a bicycle with a 25.5-in frame, but the sizes in the catalog are given only in centimeters § What size should you order?

Interactive Example 1.6 - Solution § Where are we going? § To convert from inches to centimeters § What do we know? § The size needed is 25.5 in § How do we get there? § Since we want to convert from inches to centimeters, we need the equivalence statement 2.54 cm = 1 in

Interactive Example 1.6 - Solution (continued)

§ The correct unit factor in this case is 2.54 cm 1 in 2.54 cm 25.5 in × = 64.8 cm 1 in

Interactive Example 1.7 - Unit Conversions III § A student has entered a 10.0-km run § How long is the run in miles?

Interactive Example 1.7 - Solution § Where are we going? § To convert from kilometers to miles § What do we know? § The run is 10.00 km long § How do we get there? § This conversion can be accomplished in several different ways

Interactive Example 1.7 - Solution (continued 1)

§ Since we have the equivalence statement 1 m = 1.094 yd, we will proceed by a path that uses this fact § Before we start any calculations, let us consider our strategy § We have kilometers, which we want to change to miles § We can do this by the following route:

Interactive Example 1.7 - Solution (continued 2)

§ To proceed in this way, we need the following equivalence statements: 1 km = 1000 m 1 m = 1.094 yd 1760 yd = 1 mi § To make sure the process is clear, we will proceed step by step

Interactive Example 1.7 - Solution (continued 3) § Kilometers to meters 1000 m 10.0 km × = 1.00 ×104 m 1 km § Meters to yards 1.094 yd 1.00 ×104 m × = 1.094 ×104 yd 1 m

§ Note that we should have only three significant figures in the result

Interactive Example 1.7 - Solution (continued 4)

§ Since this is an intermediate result, we will carry the extra digit § Remember, round off only the final result § Yards to miles 1 mi 1.094 ×104 yd × = 6.216 mi 1760 yd § Note in this case that 1 mi equals exactly 1760 yd by designation

Interactive Example 1.7 - Solution (continued 5)

§ Thus 1760 is an exact number § Since the distance was originally given as 10.0 km, the result can have only three significant figures and should be rounded to 6.22 mi 10.0 km = 6.22 mi § Alternatively, we can combine the steps: 1000 m 1.094 yd 1 mi 10.0 km × × × = 6.22 mi 1 km 1 m 1760 yd

Interactive Example 1.9 - Unit Conversions V § A Japanese car is advertised as having a gas mileage of 15 km/L § Convert this rating to miles per gallon

Interactive Example 1.9 - Solution § Where are we going? § To convert gas mileage from 15 kilometers per liter to miles per gallon § What do we know? § The gas mileage is 15 km/L

Interactive Example 1.9 - Solution (continued) § How do we get there? § We use the following unit factors to make the required conversion: Result obtained by rounding only at the end of the calculation

15 km 1000 m 1.094 yd 1 mi 1 L 4 qt × × × × × = 35 mi/gal L 1 km 1 m 1760 yd 1.06 qt 1 gal

Systems for Measuring Temperature § Celsius scale and Kelvin scale are used in the physical sciences § Size of the temperature unit (the degree) is the same § Temperature in Celsius units is designated C, and temperature in Kelvin scale is symbolized by the letter K § Fahrenheit scale is used in the engineering sciences

Kelvin and Celsius Scales § Differ in their zero points § Conversion between the scales requires an adjustment for the different zero points

Temperature (Kelvin) = temperature (Celsius) + 273.15

Temperature (Celsius) = temperature (Kelvin) - 273.15

Fahrenheit and Celsius Scales § Degree sizes and the zero points are different § Conversion between these scales considers two adjustments § One for degree size § One for the zero point

Fahrenheit and Celsius Scales: Difference in Degree Size § 212F = 100C and 32F = 0C

212 – 32 = 180 Fahrenheit degrees = 100 – 0 = 100 Celsius degrees

§ Thus, 180 on the Fahrenheit scale is equivalent to 100 on the Celsius scale, and the unit factor is

180° F 9° F or 100° C 5° C

Fahrenheit and Celsius Scales: Different Zero Points § Converting from Fahrenheit to Celsius

5°C (TT- 32°F) = FC9°F

§ TF - Temperature on the Fahrenheit scale

§ TC - Temperature on the Celsius scale

9°F TT = × + 32°F FC5°C

§ TF - Temperature on the Fahrenheit scale

§ TC - Temperature on the Celsius scale

Example 1.12 - Temperature Conversions II § One interesting feature of the Celsius and Fahrenheit scales is that –40C and –40F represent the same temperature § Verify that this is true

Example 1.12 - Solution § Where are we going? § To show that –40C = –40F § What do we know? § The relationship between the Celsius and Fahrenheit scales § How do we get there? § The difference between 32F and –40F is 72F § The difference between 0C and –40C is 40C

Example 1.12 - Solution (continued)

§ The ratio of these is 72°F 8 × 9°F 9°F = = 40°C 8 × 5°C 5°C

§ Thus –40C is equivalent to –40F

Relationship between the Fahrenheit and Celsius Scales § 40on both the Fahrenheit and Celsius scales represents the same temperature Number of Fahrenheit degreesT --( 40) 9° F = F = Number of Celsius degreesTC --( 40) 5° C

T + 40 9F° F = TC + 40 5° C

Interactive Example 1.13 - Temperature Conversions III § Liquid nitrogen, which is often used as a coolant for low- temperature experiments, has a boiling point of 77 K § What is this temperature on the Fahrenheit scale?

Interactive Example 1.13 - Solution § Where are we going? § To convert 77 K to the Fahrenheit scale § What do we know? § The relationship between the Kelvin and Fahrenheit scales § How do we get there? § We will first convert 77 K to the Celsius scale

TC = TK – 273.15 = 77 – 273.15 = – 196C

Interactive Example 1.13 - Solution (continued)

§ Now convert to the Fahrenheit scale T + 40 9F° F = TC + 40 5° C TT + 40 + 40 9F° FF== -196° C + 40- 156° C 5° C

9F° T + 40 =(- 156° C) =- 281° F F 5C°

TF =- 281° F- 40= - 321° F

Exercise § Convert the following Celsius temperatures to Kelvin and to Fahrenheit degrees a. Temperature of someone with a fever, 39.2C 312.4 K; 102.6F b. Cold wintery day, –25C 248 K; –13F

Exercise (continued) § Convert the following Celsius temperatures to Kelvin and to Fahrenheit degrees c. Lowest possible temperature, –273C 0 K; –459F d. Melting-point temperature of sodium chloride, 801C 1074 K; 1473.8F

Density

mass Density = volume § Property of matter that is used as an identification tag for substances § Density of a liquid can be determined easily by weighing an accurately known volume of liquid

Interactive Example 1.14 - Determining Density § A chemist, trying to identify an unknown liquid, finds that 25.00 cm3 of the substance has a mass of 19.625 g at 20C

Interactive Example 1.14 - Determining Density (continued)

§ The following are the names and densities of the compounds that might be the liquid:

§ Which of these compounds is the most likely to be the unknown liquid?

Interactive Example 1.14 - Solution § Where are we going? § To calculate the density of the unknown liquid § What do we know? § The mass of a given volume of the liquid § How do we get there? § To identify the unknown substance, we must determine its density

Interactive Example 1.14 - Solution (continued)

§ Density can be determined by using its definition mass 19.625 g Density = = = 0.7850 g/cm3 volume 25.00 cm3

§ This density corresponds exactly to that of isopropyl alcohol, which therefore most likely is the unknown liquid § However, note that the density of ethanol is also very close § To be sure that the compound is isopropyl alcohol, we should run several more density experiments

Join In (10) § A 25 g cylinder of iron (d = 7.87g/mL) and a 1.0 gram pellet of copper (d = 8.96 g/mL) are placed in 500 mL of water (d = 0.9982 g/mL) § Predict whether each will float or sink in water a. Iron will float, and copper will sink b. Iron will sink, and copper will float c. Iron and copper will sink d. Iron and copper will float e. More information is needed

Matter § Anything that occupies space and has mass § Has many levels of organization and is complex § Exists in three states § Solid § Liquid § Gas

Properties of a Solid § Rigid § Fixed volume and shape § Slightly compressible

Solid: The water molecules are locked into rigid positions and are close together

Properties of a Liquid § Definite volume § No specific shape § Assumes the shape of its container § Slightly compressible

Liquid: The water molecules are still close together but can move around to some extent

Properties of a Gas § No fixed volume or shape § Takes on the shape and volume of its container § Highly compressible § Relatively easy to decrease the volume of a gas

Gas: The water molecules are far apart and move randomly

Mixtures § Have variable composition § Classification § Homogeneous mixture: Has visibly indistinguishable parts and is often called a solution § Heterogeneous mixture: Has visibly distinguishable parts § Can be separated into pure substances, which have constant compositions, by physical methods

Physical Change § Change in the form of a substance § No change in the chemical composition of the substance § Example § Boiling or freezing of water § Used to separate a mixture into pure compounds § Will not break compounds into elements

Methods for Separating Components in a Mixture

Distillation Filtration

Chromatography

Distillation § Depends on the differences in the volatility of the components § One-stage distillation process involves heating the mixture in a distillation device § Most volatile component vaporizes at the lowest temperature § Vapor is passed through a condenser, where it condenses back into its liquid state

One-Stage Distillation - Drawback § When a mixture contains several volatile components, the one-step distillation does not give a pure substance in the receiving flask § More elaborate methods are required

Figure 1.12 - Simple Laboratory Distillation Apparatus

Filtration § Used when a mixture consists of a solid and a liquid § Mixture is poured onto a mesh, such as filter paper, which passes the liquid and leaves the solid behind

Chromatography § General name applied to a series of methods that use a system with two states (phases) of matter § Mobile phase - Liquid or gas § Stationary phase - Solid § Separation occurs because the components of the mixture have different affinities for the two phases § They move through the system at different rates

Chromatography (continued)

§ Component with a high affinity for the mobile phase will quickly go through the chromatographic system as compared to one with a high affinity for the solid phase § Paper chromatography: Uses a strip of porous paper for the stationary phase

Figure 1.13 - Paper Chromatography of Ink

A dot of the mixture to be separated is placed The paper acts as a wick to at one end of a sheet of porous paper draw up the liquid Copyright ©2018 Cengage Learning. All Rights Reserved. Section 1.10 Classification of Matter

Pure Substances § Either compounds or free elements § Compound: Substance with a constant composition that can be broken down into its elements via chemical processes § Given substance becomes a new substance or substances with different properties and different composition § Element: Substance that cannot be broken down into simpler substances by physical or chemical means

Figure 1.14 - The Organization of Matter

Join In (11) § A solution is also a: a. heterogeneous mixture b. homogeneous mixture c. compound d. distilled mixture e. pure mixture

Join In (12) § Which of the following statements is false? a. Solutions are always homogeneous mixtures b. Atoms that make up a solid are mostly open space c. Elements can exist as atoms or molecules d. Compounds can exist as elements or molecules

Atoms, Molecules, and Ions

§ (2.1) The early history of chemistry § (2.2) Fundamental chemical laws § (2.3) Dalton’s atomic theory § (2.4) Early experiments to characterize the atom § (2.5) The modern view of atomic structure: An introduction § (2.6) Molecules and ions § (2.7) An introduction to the periodic table § (2.8) Naming simple compounds

Early History of Chemistry § Applications of chemistry before 1000 B.C. § Use of embalming fluids § Processing of natural ores to produce metals for ornaments and weapons § The Greeks (400 B.C.) § Proposed that matter was composed of four fundamental substances (earth, fire, air, and water)

Early History of Chemistry (continued)

§ Considered the question of whether matter is infinitely divisible or is composed of small, indivisible particles § Alchemists dominated the field of chemistry for the next 2000 years § Helped discover several elements § Learned to prepare mineral acids

Modern Chemistry § Foundations were laid in the 16th century by: § Georg Bauer, who developed systematic metallurgy § Paracelsus, who discovered the medicinal application of minerals § Robert Boyle (1627–1691) § Performed quantitative experiments to measure the relationship between the pressure and volume of air

Modern Chemistry (continued 1)

§ Developed the first experimental definition of an element § A substance is an element unless it can be broken down into two or more simpler substances § Held on to certain alchemists’ views § Metals were not true elements, and eventually, a method to change one metal to another will be found

Modern Chemistry (continued 2) § 17th and 18th century § Interest in the phenomenon of combustion rose § Joseph Priestley discovered oxygen gas § Georg Stahl § Suggested that a substance called phlogiston flowed out of burning material § Postulated that substances that burn in closed containers eventually stop burning since the air in the container is saturated with phlogiston

Modern Chemistry (continued 3)

§ Oxygen vigorously supported combustion and was thus supposed to be low in phlogiston § Was originally called dephlogisticated air

Antoine Lavoisier § Explained the true nature of combustion § Regarded measurement as the essential operation of chemistry § Verified the law of conservation of mass § Law of conservation of mass: Mass is neither created nor destroyed in a chemical reaction § Conducted quantitative experiments that showed that combustion involved oxygen, not phlogiston Copyright © Cengage Learning. All rights reserved 9 Copyright ©2018 Cengage Learning. All Rights Reserved. Section 2.2 Fundamental Chemical Laws

Antoine Lavoisier (continued) § Discovered that life was supported by a process that also involved oxygen and was similar in many ways to combustion

Joseph Proust § Proposed the principle of the constant composition of compounds or Proust’s law or the law of definite proportion § Law of definite proportion: A given compound always contains exactly the same proportion of elements by mass

John Dalton's Law of Multiple Proportions § When two elements form a series of compounds, the ratios of the masses of the second element that combine with 1 g of the first element can always be reduced to small whole numbers

12 Copyright ©2018 Cengage Learning. All Rights Reserved. Section 2.2 Fundamental Chemical Laws Example 2.1 - Illustrating the Law of Multiple Proportions § The following data were collected for several compounds of nitrogen and oxygen:

Example 2.1 - Solution § For the law of multiple proportions to hold, the ratios of the masses of nitrogen combining with 1 g of oxygen in each pair of compounds should be small whole numbers § Therefore, compute the ratios as follows:

A1.7502 = = B0.87501

Example 2.1 - Solution (continued)

B 0.8750 2 = = C 0.4375 1 A1.7504 = = C0.43751

§ These results support the law of multiple proportions

Join In (1) § According to the law of definite proportions: a. If the same two elements form two different compounds, they do so in the same ratio b. It is not possible for the same two elements to form more than one compound c. The ratio of the masses of the elements in a compound is always the same d. The total mass after a chemical change is the same as before the change Copyright ©2018 Cengage Learning. All Rights Reserved. Section 2.3 Dalton’s Atomic Theory

Dalton’s Atomic Theory § Each element is made up of tiny particles called atoms § Atoms of a given element are identical § Atoms of different elements are different in some fundamental way or ways

Dalton’s Atomic Theory (continued) § Chemical compounds are formed when atoms of different elements combine with each other § A given compound always has the same relative numbers and types of atoms § Chemical reactions involve reorganization of the atoms § Atoms themselves are not changed in a chemical reaction

Table of Atomic Masses § Dalton prepared the first table of atomic masses (atomic weights) based on the assumption that nature would be as simple as possible § Many masses were proved to be wrong because of Dalton’s incorrect assumptions about the formulas of certain compounds

Joseph Gay-Lussac § Measured the volumes of gases that reacted with each other under the same temperature and pressure conditions

Avogadro’s Hypothesis § At the same temperature and pressure, equal volumes of different gases contain the same number of particles § Makes sense if the distances between the particles in a gas are very great compared with the sizes of the particles § Volume of a gas is determined by the number of molecules present, not by the size of the individual particles

The spheres represent atoms in the molecules

Join In (2) § How many of the following did Dalton not discuss in his atomic theory? I. Isotopes a. I II. Ions b. II III. Protons c. III IV. Neutrons d. IV V. Electrons e. V

J. J. Thomson § Studied electric discharges in partially evacuated tubes called cathode-ray tubes

J. J. Thomson (continued 1) § Postulated that the cathode ray was a stream of negatively charged particles (electrons) § Cathode ray was produced at the negative electrode when high voltage was applied to the tube § Repelled by the negative pole of an applied electric field

J. J. Thomson (continued 2) § Determined the charge-to-mass ratio of an electron e =- 1.76 × 108 C/g m

§ e - Charge on the electron (in coulombs) § m - Electron mass (in grams)

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 26 Section 2.4 Early Experiments to Characterize the Atom J. J. Thomson - Development of the Plum Pudding Model § Primary goal of the cathode-ray tube experiments was to understand the structure of an atom § Assumptions § All atoms must contain electrons § Electrons can be produced from electrodes made of various metals § Atoms must contain some amount of positive charge § Atoms were known to be electrically neutral

J. J. Thomson’s Plum Pudding Model § Atoms consist of a diffuse cloud of positive charge with the negative electrons embedded randomly in it

Robert Millikan § Performed experiments involving charged oil drops, which helped determine the magnitude of electron charge § Used this value and the charge-to-mass ratio to calculate the mass of an electron as 9.11 10–31 kg

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 29 Section 2.4 Early Experiments to Characterize the Atom Figure 2.10 (a) - Schematic Representation of the Apparatus Used to Determine the Charge on the Electron

Henri Becquerel § Discovered radioactivity by observing the spontaneous emission of radiation by uranium § Observed that a mineral containing uranium produces its image on a photographic plate in the absence of light

Types of Radioactive Emission § Gamma rays (γ) § High-energy light § Beta particles (b) § High-speed electrons § Alpha particles (α) § Particles with 2+ charge § Mass is 7300 times that of the electron

Rutherford’s Experiment § Carried out to test the accuracy of Thomson’s plum pudding model § Involved directing α particles at a thin sheet of metal foil § Expectation § α particles will pass through the foil with minor deflections in their paths

Rutherford’s Experiment - Results § Most α particles passed through the foil § Atom is mostly open space § Many particles were deflected at large angles § Including those that had a close encounter with the massive positive center of the atom § Some particles were reflected § Including those that made a direct hit on the massive positive center

Rutherford’s Experiment - Conclusions § Result of the experiment can be explained in terms of a nuclear atom § Has a dense center of positive charge called the nucleus with electrons moving around the nucleus at a distance that is large relative to the nuclear radius

Figure 2.13 - Rutherford’s Experiment

Critical Thinking § You have learned about three different models of the atom: Dalton’s model, Thomson’s model, and Rutherford’s model § What if Dalton was correct? What would Rutherford have expected from his experiments with gold foil? § What if Thomson was correct? What would Rutherford have expected from his experiments with gold foil?

Copyright ©2018 Cengage Learning. All Rights Reserved. Section 2.5 The Modern View of Atomic Structure: An Introduction Atomic Structure § Nucleus is assumed to contain: § Protons: Have a positive charge that is equal in magnitude to the electron’s negative charge § Neutrons: Have virtually the same mass as a proton but no charge § Atoms of different elements, which have different numbers of protons and electrons, exhibit different chemical behavior

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 39 Section 2.5 The Modern View of Atomic Structure: An Introduction Features of the Nucleus § Small in size compared to the overall size of the atom § High density

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 40 Section 2.5 The Modern View of Atomic Structure: An Introduction Table 2.1 - The Mass and Charge of the Electron, Proton, and Neutron

Copyright ©2018 Cengage Learning. All Rights Reserved. Section 2.5 The Modern View of Atomic Structure: An Introduction Isotopes § Atoms with the same number of protons but different numbers of neutrons § Depict almost identical chemical properties § In nature, most elements contain mixtures of isotopes

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 42 Section 2.5 The Modern View of Atomic Structure: An Introduction Figure 2.15 - Two Isotopes of Sodium

Copyright ©2018 Cengage Learning. All Rights Reserved. Section 2.5 The Modern View of Atomic Structure: An Introduction Identifying Isotopes § Atomic number (Z): Number of protons § Written as a subscript § Mass number (A): Total number of protons and neutrons § Written as a superscript

Mass number 23 Element symbol 22 Na Atomic number

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 44 Section 2.5 The Modern View of Atomic Structure: An Introduction Critical Thinking § The average diameter of an atom is 210–10 m § What if the average diameter of an atom were 1 cm? § How tall would you be?

Copyright ©2018 Cengage Learning. All Rights Reserved. Section 2.5 The Modern View of Atomic Structure: An Introduction Interactive Example 2.2 - Writing the Symbols for Atoms § Write the symbol for the atom that has an atomic number of 9 and a mass number of 19 § How many electrons and how many neutrons does this atom have?

Copyright ©2018 Cengage Learning. All Rights Reserved. Section 2.5 The Modern View of Atomic Structure: An Introduction Interactive Example 2.2 - Solution § The atomic number 9 means the atom has 9 protons § This element is called fluorine, symbolized by F § The atom is represented as follows: 19 9 F § The atom is called fluorine nineteen

Interactive Example 2.2 - Solution (continued) § Since the atom has 9 protons, it also must have 9 electrons to achieve electrical neutrality § The mass number gives the total number of protons and neutrons, which means that this atom has 10 neutrons

Copyright ©2018 Cengage Learning. All Rights Reserved. Section 2.5 The Modern View of Atomic Structure: An Introduction Exercise § How many protons and neutrons are in the nucleus of each of the following atoms? § In a neutral atom of each element, how many electrons are present? 1. 79Br 35 p, 44 n, 35 e 2. 81Br 35 p, 46 n, 35 e 3. 239Pu 94 p, 145 n, 94 e 4. 133Cs 55 p, 78 n, 55 e

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 49 Section 2.5 The Modern View of Atomic Structure: An Introduction Join In (3) § The element rhenium (Re) exists as 2 stable isotopes and 18 unstable isotopes § The nucleus of rhenium-185 contains: a. 75 protons and 75 neutrons b. 75 protons and 130 neutrons c. 130 protons and 75 neutrons d. 75 protons and 110 neutrons

Copyright ©2018 Cengage Learning. All Rights Reserved. Section 2.5 The Modern View of Atomic Structure: An Introduction Join In (4) § Which of the following statements are true? I. The number of protons is the same for all neutral atoms of an element II. The number of electrons is the same for all neutral atoms of an element III. The number of neutrons is the same for all neutral atoms of an element

Join In (4) (continued)

a. I, II, and III are true b. Only I and II are true c. Only II and III are true d. Only I and III are true e. I, II, and III are false

Chemical Bonds § Forces that hold atoms together in compounds § Covalent bond: Formed by sharing electrons § Resulting collection of atoms is called a molecule, which can be represented in the following ways: § Chemical formula § Structural formula § Space-filling model § Ball-and-stick model

Chemical Formula § Uses symbols of elements to indicate types of atoms present in the molecule § Subscripts indicate the relative number of atoms § Example

§ Formula for carbon dioxide is CO2

§ Implies that each molecule of CO2 contains one atom of carbon and two atoms of oxygen

Other Methods for Representing a Molecule § Structural formula § Depicts individual bonds in a molecule § May or may not indicate the actual shape of the molecule § Space-filling model § Illustrates the relative sizes of atoms and their relative orientation in the molecule § Ball-and-stick model

Figure 2.16 - Molecular Representations of Methane

Structural formula Space-filling model Ball-and-stick model

Ion § Atom or group of atoms with a net positive or negative charge § Cation: Positive ion formed by losing electrons § Anion: Negative ion formed by gaining electrons § Ionic bonding: Force of attraction between oppositely charged ions

Ion Formation - Example § Sodium chloride § Formed when a neutral Cl and Na react § Electron is transferred from a Na atom to a Cl atom

Cation formation: Na¾¾® Na+ + e- Anion formation: Cl + e--¾¾® Cl

Ionic Solids § Solids containing oppositely charged ions § Can consist of: § Simple ions § Example - Sodium chloride (NaCl) § Polyatomic ions: Contain many atoms

§ Example - Ammonium nitrate (NH4NO3) contains ammonium + – ions (NH4 ) and nitrate ions (NO3 )

Exercise § Would you expect each of the following atoms to gain or lose electrons when forming ions? § What ion is the most likely in each case?

a. Ra Loses 2 e– to form Ra2+ b. In Loses 3 e– to form In3+ c. P Gains 3 e– to form P3– d. Te Gains 2 e– to form Te2–

Join In (5) § How is an ion formed? a. By either adding or subtracting protons from the atom b. By either adding or subtracting electrons from the atom c. By either adding or subtracting neutrons from the atom d. All of these are true e. Two of these are true Copyright ©2018 Cengage Learning. All Rights Reserved. Section 2.6 Molecules and Ions

Join In (6) § An element’s most stable ion forms an ionic compound with chlorine and has the formula XCl2 § If the ion has 36 electrons, what is the element from which the ion comes? a. Kr b. Se c. Sr d. Rb e. None of these

The Periodic Table § Provides information about all known elements § Letters in boxes are symbols of elements § Number above every symbol is the element’s atomic number

Atomic number Element symbol (lead)

Figure 2.19 - The Periodic Table

Metals and Nonmetals

Metals Nonmetals

• Efficient conductors of • Lack the physical heat and electricity, properties that malleable, and ductile characterize metals • Have a lustrous • Tend to gain electrons in appearance reactions with metals to • Tend to lose electrons to form negative ions form positive ions • Often bond to each other by forming covalent bonds

Structure of the Periodic Table: Groups or Families § Elements in the vertical columns with similar chemical properties § Alkali metals § Members of Group 1A § Very active elements that readily form ions with a 1+ charge when they react with nonmetals

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 66 Section 2.7 An Introduction to the Periodic Table Structure of the Periodic Table: Groups or Families (continued 1) § Alkaline earth metals § Members of Group 2A § Form ions with a 2+ charge when they react with nonmetals

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 67 Section 2.7 An Introduction to the Periodic Table Structure of the Periodic Table: Groups or Families (continued 2) § Halogens: Members of Group 7A § Form diatomic molecules § React with metals to form salts containing ions with a 1– charge (exception - Astatine) § Noble gases: Members of Group 8A § Exist under normal conditions as monatomic gases § Have little chemical reactivity

Structure of the Periodic Table: Periods § Horizontal rows of elements § Horizontal row 1 is called the first period, row 2 is called the second period, and so on

Join In (7) § _____ form ions with a 2+ charge when they react with nonmetals a. Alkali metals b. Alkaline earth metals c. Halogens d. Noble gases e. None of these

Binary Compounds § Composed of two elements § Include covalent and ionic compounds § Binary ionic compounds: Contain a cation, which is written first in the formula, and an anion

Naming Binary Ionic Compounds (Type I) § Cation is always named first and the anion second § Monatomic cation takes its name from the name of the parent element § Monatomic anion is named by taking the root of the element name and adding -ide

Table 2.3 - Common Monatomic Cations and Anions

b. AlCl3 c. LiH

Interactive Example 2.3 - Solution a. CsF is cesium fluoride b. AlCl3 is aluminum chloride c. LiH is lithium hydride

§ Notice that, in each case, the cation is named first and then the anion is named

Binary Ionic Compounds (Type II) § Nomenclature for metals that form more than one type of cation § Charge of the metal cation is indicated by a Roman numeral § Common metals that don’t require Roman numerals § Group 1A elements, which form only 1+ ions § Group 2A elements, which form only 2+ ions § Aluminum, which forms only Al3+ § Silver and zinc (Ag+ and Zn2+)

Binary Ionic Compounds (Type II) (continued) § Alternative nomenclature § Ion with the higher charge has a name ending in -ic, and the one with the lower charge has a name ending in -ous

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 77 Section 2.8 Naming Simple Compounds Interactive Example 2.5 - Naming Type II Binary Compounds 1. Give the systematic name for each of the following compounds:

a. CuCl b. HgO c. Fe2O3 2. Given the following systematic names, write the formula for each compound: a. Manganese(IV) oxide b. Lead(II) chloride

Interactive Example 2.5 - Solution § All of these compounds include a metal that can form more than one type of cation § Thus we must first determine the charge on each cation § This can be done by recognizing that a compound must be electrically neutral; that is, the positive and negative charges must exactly balance

Interactive Example 2.5 - Solution (1)

1. Formula Name Comments a. CuCl Copper(I) chloride Because the anion is Cl–, the cation must be Cu+ (for charge balance), which requires a Roman numeral I b. HgO Mercury(II) oxide Because the anion is O2–, the cation must be Hg2+ [mercury(II)] 2– c. Fe2O3 Iron(III) oxide The three O ions carry a total charge of 6–, so two Fe3+ ions [iron(III)] are needed to give a 6+ charge

Interactive Example 2.5 - Solution (2)

2. Name Formula Comments 2– a. Manganese(IV) MnO2 Two O ions (total charge 4–) are oxide required by the Mn4+ ion [manganese(IV)] – b. Lead(II) chloride PbCl2 Two Cl ions are required by the Pb2+ ion [lead(II)] for charge balance

Critical Thinking § We can use the periodic table to tell us something about the stable ions formed by many atoms § For example, the atoms in column 1 always form 1+ ions and the transition metals, however, can form more than one type of stable ion § What if each transition metal ion had only one possible charge? How would the naming of compounds be different?

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 82 Section 2.8 Naming Simple Compounds Figure 2.20 - Flowchart for Naming Binary Ionic Compounds

Interactive Example 2.6 - Naming Binary Compounds 1. Give the systematic name for each of the following compounds:

a. CoBr2

b. CaCl2 2. Given the following systematic names, write the formula for each compound: a. Chromium(III) chloride b. Gallium iodide

Interactive Example 2.6 - Solution (1)

1. Formula Name Comments

a. CoBr2 Cobalt(II) bromide Cobalt is a transition metal; the compound name must have a Roman numeral The two Br– ions must be balanced by a Co2+ ion

b. CaCl2 Calcium chloride Calcium, an alkaline earth metal, forms only the Ca2+ ion A Roman numeral is not necessary

Interactive Example 2.6 - Solution (2)

2. Name Formula Comments

a. Chromium(III) chloride CrCl3 Chromium(III) indicates that Cr3+ is present, so 3 Cl– ions are needed for charge balance

b. Gallium iodide GaI3 Gallium always forms 3+ ions, so 3 I– ions are required for charge balance

Figure 2.21 - Common Cations and Anions

Polyatomic Ions § Assigned special names that must be memorized for naming compounds § Oxyanions: Anions that contain an atom of a given element and different numbers of O atoms § When there are two members in the series: § Name of the member with the smaller number of O atoms ends with -ite § Name of the member with the larger number of O atoms ends with -ate

Polyatomic Ions (continued)

§ When more than two oxyanions make up a series: § Use the prefix hypo- (less than) to name members of the series with the fewest O atoms § Use the prefix per- (more than) to name members of the series with the most O atoms

Table 2.5 - Common Polyatomic Ions

Copyright ©2018 Cengage Learning. All Rights Reserved. Section 2.8 Naming Simple Compounds Interactive Example 2.7 - Naming Compounds Containing Polyatomic Ions 1. Give the systematic name for each of the following compounds:

a. Na2SO4

b. Mn(OH)2 2. Given the following systematic names, write the formula for each compound: a. Sodium hydrogen carbonate b. Sodium selenate

Interactive Example 2.7 - Solution (1)

1. Formula Name Comments

a. Na2SO4 Sodium sulfate

b. Mn(OH)2 Manganese(II) hydroxide Transition metal—name must contain a Roman numeral The Mn2+ ion balances three OH– ions

Interactive Example 2.7 - Solution (2)

2. Name Formula Comments

a. Sodium hydrogen NaHCO3 Often called sodium bicarbonate carbonate

b. Sodium selenate Na2SeO4 Atoms in the same group, like sulfur and selenium, often form similar ions that are named similarly 2– 2– Thus SeO4 is selenate, like SO4 (sulfate)

Binary Covalent Compounds (Type III) § Formed between two nonmetals § Naming binary covalent compounds § First element in the formula is named first, using the full element name § Second element is named as if it were an anion § Prefixes are used to denote the numbers of atoms present § Prefix mono- is never used for naming the first element Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 94 Section 2.8 Naming Simple Compounds Table 2.6 - Prefixes Used to Indicate Number in Chemical Names

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 95 Section 2.8 Naming Simple Compounds Interactive Example 2.8 - Naming Type III Binary Compounds 1. Name each of the following compounds:

a. PCl5

b. PCl3

c. SO2 2. From the following systematic names, write the formula for each compound: a. Sulfur hexafluoride b. Sulfur trioxide

Interactive Example 2.8 - Solution (1)

1. Formula Name

a. PCl5 Phosphorus pentachloride

b. PCl3 Phosphorus trichloride

c. SO2 Sulfur dioxide

Interactive Example 2.8 - Solution (2)

2. Name Formula

a. Sulfur hexafluoride SF6

b. Sulfur trioxide SO3

Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 99 Section 2.8 Naming Simple Compounds Figure 2.23 - Overall Strategy for Naming Chemical Compounds

Acids § Molecules in which one or more H+ ions are attached to an anion § Nomenclature rules depend on whether the anion contains oxygen § If the anion ends in -ide, the acid is named with the prefix hydro- and the suffix -ic § When the anion contains oxygen, based on the name of the anion, the acidic name is formed from the root name of the anion with a suffix of -ic or -ous Copyright ©2018 Cengage Learning. All Rights Reserved. Copyright © Cengage Learning. All rights reserved 101 Section 2.8 Naming Simple Compounds

Naming Acids When the Anion Contains Oxygen § If the anion name ends in -ate, the suffix -ic is added to the root name § If the anion has an -ite ending, the -ite is replaced by -ous

Table 2.8 - Names of Some Oxygen-Containing Acids

Critical Thinking § In this chapter, you have learned a systematic way to name chemical compounds § What if all compounds had only common names? § What problems would this cause?

Figure 2.24 - A Flowchart for Naming Acids

Exercise § Name each of the following compounds: a. CuI Copper(I) iodide

b. S4N4 Tetrasulfur tetranitride

c. NaHCO Sodium hydrogen carbonate or sodium 3 bicarbonate

d. BaCrO4 Barium chromate

Join In (8) § Which is not the correct chemical formula for the compound named?

a. Potassium phosphate, K3PO4 b. Iron(II) oxide, FeO

c. Calcium carbonate, CaCO3 d. Sodium sulfide, NaS

e. Lithium nitrate, LiNO3

Join In (9) § Which of the following is named incorrectly?

a. Li2O, lithium oxide

b. FePO4, iron(III) phosphate c. HF, hydrogen fluoride

d. BaCl2, barium dichloride

e. Mg3N2, magnesium nitride

Join In (10) § Which is the correct formula for copper(I) sulfide? a. CuS

b. Cu2S

c. CuS2

d. Cu2S2 e. None of these

Join In (11) § Which of the following is the correct chemical formula for iron(III) oxide? a. FeO

b. Fe3O

c. FeO3

d. Fe2O3

e. Fe3O2

Join In (12) § What is the correct name for the compound with the formula Mg3(PO4)2? a. Trimagnesium diphosphate b. Magnesium(II) phosphate c. Magnesium phosphate d. Magnesium(II) diphosphate e. Magnesium(III) diphosphate

Join In (13) § What is the correct name for the acid with the formula HFO? a. Fluoric acid b. Hydrofluoric acid c. Hydrofluorous acid d. Hypofluorous acid e. Perfluoric acid