Chapter 2 Homework Packet 1. the First People to Attempt to Explain

Chapter 2 Homework Packet

1. The first people to attempt to explain why chemical changes occur were (see page 40 of text—early history) a) metallurgists b) physicians c) the Greeks d) physicists e) alchemists The Greeks were the first in recorded history to conceive of the idea of matter as being composed of elements. Around 450 B.C., Empedocles first suggested that various proportions of the elements, “earth, wind, fire and air,” constituted all matter. Aristotle subsequently expanded these ideas, suggesting that the four basic elements were characterized by varying degrees of the qualities, “hot, cold, moist and dry.” Changes (what we know as chemical changes) occurred by the addition of these qualities to one of the basic elements. Also in the 5th century B.C., Leucippus and his student Democritus proposed that matter was composed of tiny indivisible particles. They reasoned that if a bar of gold could be continuously divided, a particle would eventually be reached that, if divided further, would cease to exhibit the properties of gold. They called these theoretical particles atoms (Gr. atomos). Aristotle rejected this line of reasoning and the concept that matter was composed of these indivisible particles was abandoned until the scientists of the 1600-1800’s developed the atomic theory. Answer C 2. The Greeks proposed that matter consisted of four fundamental substances: (see page 40 of text—early history) a) atoms, fire, water, air b) atoms, metal, fire, air c) fire, earth, water, air d) fire, metal, water, air c) earth, metal, water, air See explanation to question 1. Answer C 3. The first chemist to perform truly quantitative experiments was a) Paracelsus b) Boyle c) Priestly d) Bauer e) Lavoisier Robert Boyle was one of the first truly quantitative scientists. He made observations of the relationship between pressure and volume of gases, leading to the understanding that pressure is inversely related volume, and providing one of the important clues to the development of atomic theory. He hypothesized that in order for the relationship he observed to exist, gases must be made of particles in motion. The motion of the particles crashing against the side of the container would be what caused the pressure of a volume of gas. If a volume of gas was compressed, the particles would be closer together, there would be more collisions of particles with the wall, and the pressure would increase. His ideas significantly influenced the idea

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 22 that not only are gases made of particles, but all matter is made of particles. Answer B

4. The scientist who discovered the law of conservation of mass and is also called the father of modern chemistry is a) Priestly b) Bauer c) Lavoisier d) Proust e) Boyle In the late 1700’s, Antoine Lavoisier perfected techniques of precise measurements of mass and for the first time, demonstrated that the amount of mass in reactants of a reaction consistently equaled the amount of mass in the resulting products. Because of the consistency with which this observation is made (that is, we always observe this) the absence of a change of mass during a normal chemical equation became known as the law of conservation of mass. Answer C

5. Which of the following pairs of compounds can be used to illustrate the law of multiple proportions? a) NH4 and NH4Cl b) ZnO2 and ZnCl2 c) H2O and HCl d) NO and NO2 e) CH4 and CO2 According to the law of multiple proportions, if two elements (for example, elements A and B) can form two different compounds, , if you determine the ratio of mass fractions (mass percents) of A to B in each compound, and then determine the ratio of these ratios (“the ratio of the ratios”) this ratio will be a ratio of small whole numbers. Because of the conditions of this natural law, the compounds involved must both be composed of the same two elements. In this case, the only compounds that meet this requirement are NO and NO2. Answer D

6. Which of the following pairs can be used to illustrate the law of multiple proportions? a) H2O and C12H22O11 b) H2SO4 and H2S c) SO and SO2 d) CO and CaCO3 e) KCl and KClO2 See explanation for question 5. The only compounds that meet the requirements of this law are SO and SO2. Answer C

7. According to the law of multiple 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.

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 23 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. e) None of these. The law of multiple proportions states that if elements A and B react to form two different compounds, the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small whole numbers. (A) False—this law states that the elements combine not in the same ratio but in the ratio of small whole numbers. (B) False—we know through observations that two elements commonly form more than one compound. (C) False—While it is true that the ratio of masses of elements in a compound is always the same, this is a statement of the law of definite proportions, not the law of multiple proportions. (D) False—While the total mass after a chemical change is the same as before the change, this is a statement of the law of conservation of mass, not the law of multiple proportions. (E) True—noe of the above relates to the law of multiple proportions. Answer E

8. A sample of chemical X is found to contain 5.0 grams of oxygen, 10.0 grams of carbon, and 20.0 grams of nitrogen. The law of definite proportion would predict that a 67 gram sample of chemical X should contain how many grams of carbon? Remember the strategy/show the work:

gC = (mass fraction of C)(sample mass) = (.2857)(67.0g) = 19 g C 10.0 g C mass fraction of C = = .2857 5.0 g O + 10.0 g C + 20.0 g N Think—I want the grams of C in a 67 gram sample of chemical X—do I know a formula that would provide me with this? Yes—I simply need to multiply sample mass by the mass fraction of C, and this will give me the number of grams of carbon in the sample. Write out this formula. Then, inspect the formula to realize that you already have the sample mass but you do not have the mass fraction. Then realize that you are given data for the masses of oxygen, carbon and nitrogen for a different sample of X, so you should be able to use this data to calculate the mass fraction of carbon. You quickly realize that a formula that would give you this would be to divide the mass of carbon, by the sum of the masses of all of the components. If you label the calculation, the calculation is simple enough that you would not need to write out the formula, just show the calculation. Then take this value and plug it back in to the original formula. There are 2 sig figs in the final result, limited by 5.0 g in the given values. Answer 19 g C

9. Consider the following two compounds: H2O and H2O2 . According to the law of multiple proportions, the ratio of hydrogen atoms per gram of oxygen in H2O to hydrogen atoms per gram of oxygen in H2O2 is a) 2:2 b) 4:1 c) 1:1 d) 2:1 e) 1:2

The fact that there is the same mass of H in each compound but twice the mass of O in H2O2, when the question is asking for the amount of H per gram of O, makes this question a little confusing. It would be much easier to find the ratio of O atoms per gram of H. But, think of it in the following way—how many hydrogen atoms are there in each compound for every one O atom? In H2O, for every 1 O atom, there are 2 H atoms. In H2O2, for every 1 O atom, there is only 1 H atom. So, for H2O compared to H2O2, for every gram of oxygen, the ratio of atoms of hydrogen is 2:1. Answer D

10. Which of the following statements from Dalton's atomic theory is no longer true, according to modern atomic theory? a) Atoms are not created or destroyed in chemical reactions. b) All atoms of a given element are identical. c) Atoms are indivisible in chemical reactions. d) Elements are made up of tiny particles called atoms. e) All of these statements are true according to modern atomic theory. At first glance it might appear that all of these statements are still true. However, at the time of the first statement of the atomic theory in the early 1800’s, investigators did not know about isotopes—they truly thought all atoms of a given element would be identical. However, with the discovery of neutrons in the mid 1900’s and the ability to accurately mass substances using mass spectrometry, we discovered that most atoms of each element are identical, and this is still true regarding the “chemical” properties of each atom. However, a small percentage of atoms of any element will have slightly different numbers of neutrons than the most common isotope of that element. Remember, this variation in numbers of neutrons does not affect the chemical properties of the atom, only the nuclear properties, which, in this class, we will not say much about. In any case, not all atoms of a given element are truly identical. Answer B

11. Avogadro's hypothesis states that: a) Each atom of oxygen is 16 times more massive than an atom of hydrogen. b) A given compound always contains exactly the same proportion of elements by mass. c) When two elements form a series of compounds, the ratios of masses that combine with 1 gram of the first element can always be reduced to small whole numbers. d) At the same temperature and pressure, equal volumes of different gases contain an equal number of particles. e) Mass is neither created nor destroyed in a chemical reaction. As a prequel, remember that three primary investigators developed laws that relate the temperature, pressure, volume, and number of particles in a sample of gas. The relationships determined by each can be remembered using the following table. All of these laws state the relationship between volume and some other aspect of a gas, number of particles, pressure and temperature. The three investigators were Avogadro, Boyle, and Charles. We can remember the feature of the gas related to volume and the investigator that discovered the relationship by lining up the first letters of the terms and the investigators alphabetically, as follows:

Relationship between Volume and Investigator

Number of particles (moles) Avogadro Pressure Boyle Temperature Charles

So, Avogadro discovered the relationship between volume of a gas and the number of particles in that volume, if pressure and temperature are constant. (A) False—our knowledge that oxygen is 16 times more massive than hydrogen is a result of Avogadro’s hypothesis, but not a statement of that law. (B) False—this is a statement of the law of definite proportions. (C) False—this is a statement of the law of multiple proportions. (D) True (as discussed above) (E) False—This is a statement of the law of conservation of mass. Answer D

12. The first scientist to show that atoms emit any negative particles was a) Ernest Rutherford b) William Thomson c) John Dalton d) J. J. Thomson e) Lord Kelvin In the late 1800’s JJ Thomson demonstrated that negatively charged particles that were smaller than atoms were emitted from a metal cathode and streamed toward a metal anode when a large potential difference was placed across them. In an un- or partially evacuated tube the particles caused the atmosphere inside the tube to glow, while in an evacuated tube the particles caused the material of the end of the tube to glow, and a shadow was cast on the end of the tube where the anode prevented particles from passing to the end of the tube. In studies using a magnetic field to deflect the stream of particles, Thomson showed that the particles were negatively charged, and had a charge to mass ratio much than that of a stream of atoms, and so, had much less mass. We now know these particles are electrons. Answer D 13. Many classic experiments have given us indirect evidence of the nature of the atom. Which of the experiments listed below did not give the results described? a) The Rutherford experiment proved the Thomson "plum-pudding" model of the atom to be essentially correct. b) The Rutherford experiment was useful in determining the nuclear charge on the atom. c) Millikan's oil-drop experiment showed that the charge on any particle was a simple multiple of the charge on the electron. d) The electric discharge tube proved that electrons have a negative charge. e) All of the above experiments gave the results described. (A) False—The Rutherford experiment disproved the plum pudding model of the atom. The plum pudding model stated that the atom was a particle composed of an amorphous positively charged material in which electrons were embedded in a pattern. The results of the Rutherford experiment suggested that the atom was largely empty space with a core of positively charged and neutral particles around

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 26 which the electrons were arranged in the empty space. (B) True—The Rutherford experiment was useful in helping to determine the positive charge of the nucleus. Rutherford knew the relative mass of the particles in the alpha particle stream. Based on the degree of deflection, he could determine using mathematical calculations what the charge and mass of the nucleus causing the deflection must be. (C) True—the oil-drop experiment determined the amount of charge that is present on a random assortment of oil-droplets. When the distribution of charges on the various oil droplets was reviewed it was found that the differences in charges were multiples of the number 1.60 x 10-19 C. It was assumed that is number was the charge of a single electron, and that droplets that had a multiple of this charge had an excess of the number of electrons this represented. (D) True—See discussion for question 12. (E) Not a true statement—A was a false statement. Answer A

14. The scientist whose alpha-particle scattering experiment led him to conclude that the nucleus of an atom contains a dense center of positive charge is a) J. J. Thomson b) Lord Kelvin c) Ernest Rutherford d) William Thomson e) John Dalton (A) False—JJ Thomson discovered the electron with his cathode ray tube investigations; (B) False—Lord Kelvin investigated energy changes and is credited with the development of an absolute temperature scale; (C) True—Rutherford is credited with the discovery of the nucleus with his alpha particle scattering discovery; (D) False—William Thomson is Lord Kelvin; (E) False—John Dalton is credited with the development of the atomic theory but had no knowledge of subatomic particles. Answer C

15. Alpha particles beamed at thin metal foil may a) pass directly through without changing direction b) be slightly diverted by attraction to electrons c) be reflected by direct contact with nuclei d) A and C e) A, B, and C Rutherford hypothesized that if the plum pudding model were true, and the atom was made up of amorphous material in which electron were embedded, because the electrons were so small he felt that the much larger alpha particles should pass right through the gold foil. He was surprised when a few were deflected, so he hypothesized that there was in fact, a small dense center of the atoms composed of positively charged particles that the alpha particles must be hitting. Still, most passed through without changing direction. So, at least A and C are correct. While we did not talk about it, it would actually be possible that if an alpha particle passed closely enough to an electron, as the alpha particle is a particle composed of two protons and two neutrons, the negatively charged electron would cause the positively charged alpha particle to deviate slightly. Answer E

16. Which one of the following statements about atomic structure is false? a) An atom is mostly empty space. b) Almost all of the mass of the atom is concentrated in the nucleus. c) The protons and neutrons in the nucleus are very tightly packed. d) The number of protons and neutrons is always the same in the neutral atom. e) All of the above statements (A-D) are true. (A) True—an atom is mostly empty space—the nucleus, containing the protons and neutrons, is contained in a tiny space at the center of the atom while the electrons are arranged in orbitals around the nucleus. However, the volume of space in which the electrons resides is vastly greater than that of the nucleus; (B) True-as the mass of electrons is tiny compared to that of protons and neutrons, as the nucleus contains the protons and neutrons, the majority of the mass is also concentrated their; (C) True—the protons and neutrons are held together by very strong forces and are compressed into a tiny space; (D) False—as the nucleus becomes larger as protons are added sequentially, the number of neutrons required to maintain the stability of the nucleus increases compared to the number of protons. So, while the number of neutrons approximates the number of protons in the early part of the periodic table, the number of neutron becomes greater and greater compared to the number of protons later in the table. (E) False—only A-C are true. Answer D

17. If the Thomson model of the atom had been correct, Rutherford would have observed: a) Alpha particles going through the foil with little or no deflection. b) Alpha particles greatly deflected by the metal foil. c) Alpha particles bouncing off the foil. d) Positive particles formed in the foil. e) None of the above observations is consistent with the Thomson model of the atom. See explanation for question 15. Answer A

If you need to review the basic concepts of atomic structure (protons, neutrons, electrons, atomic number (Z), mass number (A), and standard symbolization read section 2.5 in your text.

18. Bromine exists naturally as a mixture of bromine-79 and bromine-81 isotopes. An atom of bromine-79 contains a) 35 protons, 44 neutrons, 35 electrons b) 34 protons and 35 electrons, only c) 44 protons, 44 electrons, and 35 neutrons d) 35 protons, 79 neutrons, and 35 electrons e) 79 protons, 79 electrons, and 35 neutrons Remember that the atomic number is the number of protons (designated Z) while the mass number is the number of protons and neutrons (designated A). When we state the name of an element followed by a number, we are referring to the mass number. So, to say bromine-79 means we are referring to that isotope of bromine that has a

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 28 mass number of 79. To find the number of neutrons we would subtract the atomic number from the mass number. For bromine, which is atomic number 35, as there are 35 protons, this means there are 44 neutrons. Also, in a neutral atom, the number of electrons will be the same as the number of protons, so there will be 35 electrons. Answer A

19. Which of the following atomic symbols is incorrect? 14 a) 6 C 37 b) 17 Cl 32 c) 15 P 39 d) 19 K 14 e) 8 N (A) The symbol 14 C would indicate a carbon atom with 6 protons, which is correct- 6 carbon is element number 6. As the mass number is 14 this would indicate 8 neutrons which is possible for this element; (B) The symbol 37 C would indicate a 17 chlorine atom with 17 protons, which is correct-chlorine is element number 17. As the mass number is 37 this would indicate 20 neutrons which is possible for this element; (C) The symbol 32 P would indicate a phosphorus atom with 15 protons, 15 which is correct-phosphorus is element number 15. As the mass number is 32 this would indicate 17 neutrons which is possible for this element; (D) The symbol 39 K 19 would indicate a potassium atom with 19 protons, which is correct-potassium is element number 19. As the mass number is 39 this would indicate 20 neutrons which is possible for this element; (E) The symbol 14 N would indicate a nitrogen atom with 8 8 protons, which is incorrect-nitrogen is element number 7. Answer E

20. The element rhenium (Re) exists as two stable isotopes and 18 unstable isotopes. Rhenium-185 has in its nucleus a) 130 protons, 75 neutrons b) 75 protons, 110 neutrons c) not enough information d) 75 protons, 75 neutrons e) 75 protons, 130 neutrons Remember that a number following a statement of the name of an element is going to be the mass number—so, Rhenium-185 has a sum of protons and neutrons that equals 185. As the atomic number is 75, meaning 75 protons, a Rh atom contains 185-75 = 110 neutrons. Answer B

21. An isotope, X, of a particular element has an atomic number of 15 and a mass number of 31. Therefore: a) X is an isotope of phosphorus. b) X has 16 neutrons per atom. c) X has an atomic mass of 30.973. d) A and B. e) A, B, and C. Given an atomic number of 15, this means an element with 15 protons, which is phosphorus. The mass number is 31 which means there will be 31-15 = 16 neutrons. So, A and B are correct. However, while phosphorus as an element has an average atomic mass of 30.973, the specific isotope, X, does not. Therefore, C is incorrect. Answer D

40 2+ 22. 20 Ca has a) 20 protons, 20 neutrons, and 18 electrons b) 22 protons, 20 neutrons, and 20 electrons c) 20 protons, 22 neutrons, and 18 electrons d) 22 protons, 18 neutrons, and 18 electrons e) 20 protons, 20 neutrons, and 22 electrons In the notation show, the subscript before the symbol is the atomic number, so the element has 20 protons. The superscript before the number is the mass number, so the number of neutrons is 40-20=20. Because the charge on the Ca atom is 2+ it has lost 2 electrons, so instead of the normal number of 20, which the neutral calcium atom should have, it has 18 electrons. Answer A

23. Which of the following statements is (are) true? 14 14 a) 6 C and 7 N are isotopes of each other because their mass numbers are the same. 18 2− 20 b) 8 O has the same number of electrons as 10 Ne . c) A and B 18 19 d) 8 O and 9 F have the same number of neutrons. e) A and C (NOTE: Sorry about this question-I changed the order of the answers without realizing that some of the answers depended on the order of the answers. The way I have the question, both answers B and D are correct) (A) False-To say that two atoms are isotopes of each other would mean that they are the same element. Carbon and Nitrogen are not the same elements so they cannot be isotopes of each other. They would have to have the same atomic number, not the same mass number. (B) True-as oxygen has 8 protons, the neutral oxygen atom will also have 8 electrons. With a 2- charge, it has taken two more electrons and so has ten. The neutral neon atom has 10 protons according to its atomic number and so has 10 electrons. (C) False—A and B are not true; (D) True-Oxygen-18 has 8 protons according to its atomic number. As its mass number is 18, it has 18-8=10 neutrons. Fluorine-19 has 9 protons according to its atomic number. As its mass number is 19, it also has 19-9=10 neutrons. (E) A and C are not both true. Answer(s) B and D

24. A species with 12 protons and 10 electrons is a) Ne2+ b) Ti2+ c) Mg2+ d) Mg e) Ne2– Any atom with 12 protons has an atomic number of 12. The element with the atomic number of 12 is magnesium. So, your choices are limited to C and D. If the neutral magnesium atom has 12 protons it also has 12 electrons. Given the question, this means that it has lost 2 of its electrons and so has a 2+ charge. Therefore, the answer is Mg2+. Answer C

25. The numbers of protons, neutrons, and electrons in 39 K+ are: 19 a) 20 p, 19 n, 19 e b) 20 p, 19 n, 20 e c) 19 p, 20 n, 20 e d) 19 p, 20 n, 19 e e) 19 p, 20 n, 18 e Given the notation in the question, the subscript before the symbol is the atomic number, which is equal to the number of protons, so 19 protons. The superscript before the symbol is the mass number, and so, the sum of the number of protons and neutrons. Therefore, the number of neutrons is 39-19=20. Finally, in the neutral potassium atom there are 19 protons, so there should also be 19 electrons. However, this potassium atom is stated to have a 1+ charge, meaning it has lost an electron. Therefore, it has 18 electrons. Answer E

26. An ion is 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 the above are true. e) Two of the above are true. Ions can only be formed by adding or removing electrons. While removal or addition of a proton would cause an imbalance in charge, we cannot do this without destroying the beginning atom—we would actually change the element. Adding or subtracting neutrons would do nothing to the charge. Answer B

27. The formula of water, H2O, suggests: a) There is twice as much mass of hydrogen as oxygen in each molecule. b) There are two hydrogen atoms and one oxygen atom per water molecule. c) There is twice as much mass of oxygen as hydrogen in each molecule. d) There are two oxygen atoms and one hydrogen atom per water molecule. e) None of these.

The subscripts of chemical formulas indicate the ratios of the numbers of atoms of the different elements. Therefore, H2O indicates there are 2 H atoms for every 1 atom of oxygen. As atoms of different elements have different masses, the subscript tells you nothing about the masses of the elements or the ratio of masses. Answer B

28. All of the following are true except: a) Ions are formed by adding electrons to a neutral atom. b) Ions are formed by changing the number of protons in an atom's nucleus. c) Ions are formed by removing electrons from a neutral atom. d) An ion has a positive or negative charge. e) Metals tend to form positive ions. Remember that the formation of ions is all about losing or gaining electrons and the resulting charge. Ions are not formed by changing the number of protons—this would change the atomic number, and so, change the element. Additionally, it should be intuitive that metals lose electrons and so form positive ions and non-metals gain electrons and so form negative ions. Therefore, the only false statement in the above list is answer b. ANSWER B

29. Which of the following are incorrectly paired? a) K, alkali metal b) Ba, alkaline earth metal c) O, halogen d) Ne, noble gas e) Ni, transition metal You should know the names of the commonly named columns—column 1A is the alkali metals, column 2A is the alkaline earth metals, column 7A is the halogens, column 8A is the noble gases, all of the B columns are the transition metals. K is in column 1A so it is an alkali metal; Ba is in column 2A so it is an alkaline earth metal; O is in column 6A so it is not a halogen; neon is in column 8A so it is a noble gas; nickel is in column 8B so it is a transition metal. Answer C

30. Which of the following are incorrectly paired? a) Sr, alkaline earth metal b) Ta, transition metal c) F, halogen d) As, halogen e) V, transition metal You should know the names of the commonly named columns—column 1A is the alkali metals, column 2A is the alkaline earth metals, column 7A is the halogens, column 8A is the noble gases, all of the B columns are the transition metals. Sr is in column 2A so it is an alkaline earth metal; Ta is in column 5B so it is a transition metal; F is in column 7A so it is a halogen; As is in column 5A so it is not a halogen; V is in column 5B so it is a transition metal. Answer D

31. Which of the following are incorrectly paired? a) Phosphorus, Pr

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 32 b) Palladium, Pd c) Platinum, Pt d) Lead, Pb e) Potassium, K The periodic table you will have on the test will only have atomic symbols, not element names, so you will ultimately need to be familiar with pretty much every atomic symbol and its name (this does not mean you need to know its exact position on the periodic table or that you need to memorize the periodic table). This is a question that is simply checking your familiarity with the periodic table. In inspecting the periodic table for the symbols or names of the above elements you discover that the incorrectly paired element is Phosphorus Pr. Phosphorus has the atomic symbol P, while Pr is the symbol for Praseodymium. Answer A

32. Which of the following are incorrectly paired? a) Copper, Cu b) Carbon, C c) Cobalt, Co d) Calcium, Ca e) Cesium, Ce The periodic table you will have on the test will only have atomic symbols, not element names, so you will ultimately need to be familiar with pretty much every atomic symbol and its name (this does not mean you need to know its exact position on the periodic table or that you need to memorize the periodic table). This is a question that is simply checking your familiarity with the periodic table. In inspecting the periodic table for the symbols or names of the above elements you discover that the incorrectly paired element is Cesium Ce. Cesium has the atomic symbol Cs, while Ce is the symbol for Cerium. Answer E

33. Which of the following are incorrectly paired? a) Antimony, Sb b) Silicon, Si c) Silver, Ag d) Argon, Ar e) Astatine, As The periodic table you will have on the test will only have atomic symbols, not element names, so you will ultimately need to be familiar with pretty much every atomic symbol and its name (this does not mean you need to know its exact position on the periodic table or that you need to memorize the periodic table). This is a question that is simply checking your familiarity with the periodic table. In inspecting the periodic table for the symbols or names of the above elements you discover that the incorrectly paired element is Astatine As. Astatine has the atomic symbol At, while As is the symbol for Arsenic. Answer E

34. All of the following are characteristics of metals except: a) good conductors of heat b) malleable

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 33 c) ductile d) often lustrous e) tend to gain electrons in chemical reactions You should become familiar with all of the main properties of metals and non-metals. Further, we will ultimately become familiar with the features of the electronic structure of substances that are responsible for these properties. In the above list all of the features are characteristic of metals except for the gain of electrons in chemical reactions. Metals tend to lose electrons in chemical reactions. Answer E

35. All of the following are characteristics of nonmetals except: a) poor conductors of electricity b) often bond to each other by forming covalent bonds c) tend to form negative ions in chemical reactions with metals d) appear in the upper left-hand corner of the periodic table e) do not have a shiny (lustrous) appearance You should become familiar with all of the main properties of metals and non-metals. Further, we will ultimately become familiar with the features of the electronic structure of substances that are responsible for these properties. In the above list all of the features are characteristic of non-metals except for the appearing in the upper left-hand corner of the periodic table—this is where metals appear. Answer D

36. Which of the following has 61 neutrons, 47 protons, and 46 electrons? a) 108 Cd+ 47 b) 80 Pm 61 c) 108 Ag+ 47 d) 108 Pd– 46 e) 108 Ag 47 In the above notation the subscript in front of the atomic symbol is the Z number or the atomic number, which is equal to the number of protons. Therefore, we can eliminate (b) and (d) immediately. Also, element number 47 is Ag, so this immediately eliminates (a). The superscript in front of the symbol represents the mass number which is the sum of the neutrons and protons. Therefore, the mass number would be 61 + 47= 108. (c), and (e) both have mass numbers of 108 so this does not eliminate anything. As there are 47 protons in Ag, the number of electrons in the neutral silver atom is also 47. A given value of 46 electrons means that 1 electron has been given up so the charge must be +1. Therefore, c is the answer. Answer C

37. You are given a compound with the formula MCl2, in which M is a metal. You are told that the metal ion has 25 electrons. What is the identity of the metal? Given that the compound is MCl2, because you know Cl is in column 7, and so, its charge is -1, and there are 2 Cl’s, M must have a charge of +2. If the metal ion has 25

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 34 electrons in the +2 state then it must have lost 2 electrons, and so has 27 electrons in the neutral state. If it has 27 electrons in the neutral state then it must also have 27 protons, so its atomic number is 27. Therefore, the metal is cobalt. Answer Cobalt

38. Which of the following names is incorrect? a) aluminum(III) oxide b) diphosphorus pentoxide c) cobalt(II) chloride d) magnesium oxide e) All of the above names are correct. (a) Aluminum is a metal and oxygen is a non-metal so aluminum oxide would be an ionic compound. Aluminum is in column 3A so it would have a positive three charge. However, as it is neither a transition metal nor a post-transition metal, it can only have a +3 oxidation state so we never use a Roman numeral to designate its oxidation state. Therefore, this answer is incorrect. As is typical for the second component, we attach the –ide suffix for an element.

(b) phosphorus and oxygen are both non-metals so this compound is covalent. As with all covalent compounds, for the first component we use a number prefix if the number of atoms is two or greater; for the second component we always use a number prefix. As is typical for the second component, we attach the –ide suffix for an element. This answer would be a valid covalent compound name.

(c) Cobalt is a metal and chlorine is a non-metal so cobalt (II) chloride would be an ionic compound. Cobalt is a transition metal and so could potentially have multiple oxidation states—therefore, we designate it using a Roman numeral. As is typical for the second component, we attach the –ide suffix for an element. This answer would be a valid ionic compound name.

(d) Magnesium is a metal and oxygen is a non-metal so magnesium oxide would be an ionic compound. As magnesium is neither a transition metal nor a post-transition metal, and is in column 2A, it can only have a 2+ oxidation state and so we do not use a roman numeral to designate it—it is correctly designated. As is typical for the second component, we attach the –ide suffix for an element. This answer would be a valid ionic compound name.

(e) not all of the above names are correct so this is not a correct answer.

Answer A

39. Which of the following pairs is incorrect? a) iodine trichloride, ICl3 b) phosphorus pentoxide, P2O5 c) ammonia, NH3 d) sulfur hexafluoride, SF6

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 35 e) All of the above pairs are correct. (a) iodine and chlorine are both non-metals so this compound is covalent. As with all covalent compounds, for the first component we use a number prefix if the number of atoms is two or greater; for the second component we always use a number prefix. As is typical for the second component, we attach the –ide suffix for an element. As there is one iodine and there are three chlorines in the given formula, the name correctly describes the formula. This answer would be a valid covalent compound name.

(b) phosphorus and oxygen are both non-metals so this compound is covalent. As with all covalent compounds, for the first component we use a number prefix if the number of atoms is two or greater; for the second component we always use a number prefix. As is typical for the second component, we attach the –ide suffix for an element. As there is two phosphorus atoms and there are five oxygens in the given formula, while pentoxide is correct, the first component should be diphosphorus. This answer would not be a valid covalent compound name.

(c) nitrogen and hydrogen are both non-metals so this compound is covalent. As with all covalent compounds, for the first component we use a number prefix if the number of atoms is two or greater; for the second component we always use a number prefix. As is typical for the second component, we attach the –ide suffix for an element. Normally, as there is only one nitrogen and there are three hydrogens we would call this molecule nitrogen trihydride. However, this compound so commonly goes by the name ammonia that this has become a valid name for this compound.

(d) sulfur and fluorine are both non-metals so this compound is covalent. As with all covalent compounds, for the first component we use a number prefix if the number of atoms is two or greater; for the second component we always use a number prefix. As is typical for the second component, we attach the –ide suffix for an element. As there is one sulfur and there are six fluorines in the given formula, the name correctly describes the formula. This answer would be a valid covalent compound name.

(e) Not all of the above names are correct so this is not a correct answer.

Answer B

40. The correct name for LiCl is Li is the symbol for lithium, which is a metal, and Cl is the symbol for chlorine which is a non-metal, so LiCl is an ionic compound. In naming ionic compounds if both components are elements, the first name is the name of the element and the second name has the –ide suffix attached. Additionally, if the metal is either a transition metal or a post transition metal, because the oxidation state for the metal can vary, we use a roman numeral to indicate the number of the charge. In this case, as lithium is in column 2A and can only have one oxidation state, we don’t use a roman numeral. For ionic compounds there are no number prefixes. Therefore, the name of this compound is lithium chloride. Answer Lithium Chloride

41. How many oxygen atoms are there in one formula unit of Ca3(PO4)2? In this compound oxygen atoms are part of the phosphate polyatomic ion. As indicated by the subscript following the O symbol, there are four oxygen atoms in this polyatomic ion. As polyatomic ions contain multiple types of atoms, in order to avoid confusion, and to indicate a multiple number of polyatomic ions, parentheses are used to enclose the formula of the polyatomic ion if there are more than one, and a subscript indicating the number of polyatomic ions follows. When polyatomic ions are designated in this way, the number of atoms inside the parentheses is multiplied by the number of the subscript following the parentheses. So, there are two phosphate ions in the formula unit as indicated by the subscript 2 following the parentheses. This means there are a total of 8 oxygen atoms in this formula unit. Answer 8

42. How many oxygen atoms are there in 4 formula units of Al2(CO3)3? In this compound oxygen atoms are part of the carbonate polyatomic ion. As indicated by the subscript following the O symbol, there are three oxygen atoms in this polyatomic ion. As polyatomic ions contain multiple types of atoms, in order to avoid confusion, and to indicate a multiple number of polyatomic ions, parentheses are used to enclose the formula of the polyatomic ion if there are more than one, and a subscript indicating the number of polyatomic ions follows. When polyatomic ions are designated in this way, the number of atoms inside the parentheses is multiplied by the number of the subscript following the parentheses. So, there are three carbonate ions in the formula unit as indicated by the subscript 3 following the parentheses. This means there are a total of 9 oxygen atoms in this formula unit. For 4 formula units, the number of oxygen atoms would be 36 Answer 36

43. The correct name for FeO is Fe is the symbol for iron, which is a metal, and O is the symbol for oxygen which is a non-metal, so FeO is an ionic compound. In naming ionic compounds if both components are elements, the first name is the name of the element and the second name has the –ide suffix attached. Additionally, if the metal is either a transition metal or a post transition metal, because the oxidation state for the metal can vary, we use a roman numeral to indicate the number of the charge. In this case, because the charge of a single oxygen atom is -2, the charge of the iron atom must be +2, so we would use a roman numeral II. For ionic compounds there are no number prefixes. Therefore, the name of this compound is iron (II) oxide. Answer iron (II) oxide

44. The correct name for Ca2+ is The atomic symbol Ca stands for calcium. The 2+ indicates that it is a calcium ion. Because calcium can only have a 2+ state when it is an ion we do not need to designate any more specifically so we would just call this the calcium ion. Answer calcium ion

45. The correct name for V3+ is

The atomic symbol V stands for Vanadium. The 3+ indicates that it is a vanadium ion. However, because vanadium can have more than one oxidation state we need to designate the actual state by using a roman numeral. Therefore, we would call this the vanadium (III) ion. Answer vanadium (III) ion

46. The correct name for N3– is The atomic symbol N stands for nitrogen. The 3- indicates that it is a nitrogen ion. Because nitrogen can only have a 3+ state when it is an ion we do not need to designate any more specifically. Also, as this ion would be the second component of the name of an ionic compound, to name the ion we would attach the –ide suffix, so we would call this the nitride ion. Answer nitride ion

47. What is the subscript of barium in the formula of barium sulfate? First create the formula for barium sulfate. Barium is a 2A element with the symbol Ba—because it is in column 2A it will have a 2+ charge as an ion. Sulfate is a polyatomic ion (the clue is the –ate ending). You should know by now that the sulfate 2+ ion has a formula of SO4 and that its charge is 2-. Therefore, it will only take one Ba 2- ion to balance the charge of an SO4 ion. So, the formula is BaSO4. Because there is only one barium atom, there is no subscript. Answer no subscript

48. The formula for calcium bisulfate is Calcium is a metal and bisulfate is a polyatomic anion—the clue is the –ate suffix—so the compound is ionic. When we write the formula for an ionic compound the first component is simply the symbol for the element (unless it is the only positive + polyatomic ion NH4 , the ammonium ion). Then we leave a little space and write the formula of the polyatomic ion. In this case you would have to look it up—the - bisulfate ion is HSO4 .

(NOTE—the preferred terminology for this ion is the hydrogen sulfate ion. The term bisulfate comes from an old naming system. Consider these two compound: Na2SO4 and NaHSO4. In the second compound there is twice as much sulfate per sodium atom –that is, one sulfate per one sodium—as there is in the first compound—that is, one half of a sulfate per one sodium. So, the old chemists called the second compound “bisulfate” to indicate this. IUPAC would rather this be called the “hydrogen sulfate” - ion. This is similar for the HCO3 ion which is traditionally called the bicarbonate ion, but should be called the hydrogen carbonate ion.)

(NOTE—you could predict the formula of this ion by knowing the formula of the sulfate ion, SO4—as this ion has a -2 charge, and the hydrogen would add a 1+charge, the addition of a hydrogen ion to the sulfate ion in this manner would cancel one of - the negative charges giving the HSO4 formula and charge.)

Then we would remind ourselves of the charges of each component—Calcium is in column 2A so its charge would be 2+, while we have already said that the charge of the HSO4 ion is -1. Therefore, it would take two of these to balance the 2+ charge.

Remember that when we have more than one of a polyatomic ion we put parentheses around the polyatomic ion formula and then use a subscript to designate the number of polyatomic ion units. Therefore, the formula is Ca(HSO4)2. Answer Ca(HSO4)2

49. The formula for lithium dihydrogen phosphate is Lithium is a metal and dihydrogen phosphate is a polyatomic anion—the clue is the -ate suffix—so, the compound is ionic. When we write the formula for an ionic compound the first component is simply the symbol for the element (unless it is the + only positive polyatomic ion NH4 , the ammonium ion). Then we leave a little space and write the formula of the polyatomic ion. In this case you would have to look it up—the dihydrogen phosphate ion.

(NOTE—you could predict the formula of this ion by knowing the formula of the phosphate ion, PO4—as this ion has a -3 charge, and each hydrogen would add a 1+charge, the addition of two hydrogen ions to the phosphate ion in this manner - would cancel two of the negative charges giving the H2PO4 formula and charge.)

Then we would remind ourselves of the charges of each component—Lithium is in column 1A so its charge would be 1+, while we have already said that the charge of the H2PO4 ion is -1. Therefore, the compound would have a balanced charge with one of each component—the formula is LiH2PO4. Answer LiH2PO4

50. Which of the following is incorrectly named? a) Pb(NO3)2, lead(II) nitrate b) NH4ClO4, ammonium perchlorate 3– c) PO4 , phosphate ion d) Mg(OH)2, magnesium hydroxide e) NO3–, nitrite ion (a) Pb is a metal in column 4A and NO3 is the polyatomic nitrate anion so the compound is ionic. In naming ionic compounds, if the first component is an element, the name of the first component is the name of the element, modified by a roman numeral if the element is a transition metal or a post transition metal. If this is the case, the charge can be determined by inspecting the formula. In this case, as the formula has lead paired with two nitrate ions, each of which has a 1- charge, the charge on lead would be 2+, so the roman numeral (II) is appropriate. If the name of the second component is a polyatomic ion, the name of the polyatomic ion is used. Therefore, the name lead (II) nitrate is the name for this compound.

(b) NH4 is the only positively charged polyatomic ion, the ammonium ion, and ClO4 is the polyatomic perchlorate anion (you should know both of these names). So, this is an ionic compound. In naming ionic compounds, if the first component is the ammonium ion then the name of the first component is ammonium. If the name of the second component is a polyatomic ion, the name of the polyatomic ion is used. Therefore, the name ammonium perchlorate is the name for this compound.

(c) This is an oxoanion with phosphorus as the non-oxygen central atom. In naming oxoanions, the “main” form is assigned the –ate suffix, which is attached to the name of the non-oxygen central atom. The –ate form of these oxoanions should be learned, as the other forms (hypo-/-ite, -ite, per-/-ate), if they exist, can be determined from the –ate form. PO4 is the –ate form for the phosphorus oxoanion and it has a minus 3 charge. This is therefore the phosphate ion and is correctly named.

(d) Mg is a metal in column 2A, and OH- is the polyatomic hydroxide anion, so the compound is ionic. In naming ionic compounds, if the first component is an element, the name of the first component is the name of the element, modified by a roman numeral if the element is a transition metal or a post transition metal. This is not the case here. If the name of the second component is a polyatomic ion, the name of the polyatomic ion is used. Therefore, the name magnesium hydroxide is the name for this compound.

(e) There is no NO3- polyatomic ion. The nitrite ion is the oxoanion related to the - nitrate ion, NO3 . It has one less oxygen, as is the rule for the –ite form, and so has a - formula NO2 . Therefore, this pairing is incorrect.

Answer E

51. Which of the following is incorrectly named? 2– a) SO3 , sulfite ion 2– b) Cr2O7 , dichromate ion 3– c) PO4 , phosphate ion – d) ClO2 , chlorate ion e) CN–, cyanide ion (a) This is an oxoanion with sulfur as the non-oxygen central atom. In naming oxoanions, the “main” form is assigned the –ate suffix, which is attached to the name of the non-oxygen central atom. The –ate form of these oxoanions should be learned, as the other forms (hypo-/-ite, -ite, per-/-ate), if they exist, can be determined from the –ate form. SO4 is the –ate form for the sulfate oxoanion and it has a -2 charge. 2- SO3 has one less oxygen than the –ate form so it will be designated as the –ite form, 2- and will have the same charge as the –ate form. So, the SO3 ion is correctly named the sulfite ion.

(b) a couple of oxoanions also have forms where there are two of the non-oxygen atom—the chromate ion is one of these—when there are two of the non-oxygen atoms the prefix di- is added to the non-oxygen element name. For chromium, the 2- 2- –ate form is CrO4 and the Cr2 form is Cr2O7 and is called the dichromate ion. So, dichromate is the correct name for this ion.

the –ate form. PO4 is the –ate form for the phosphate oxoanion and it has a -3 charge. Therefore, phosphate is the correct name for this ion.

(d) This is an oxoanion with chlorine as the non-oxygen central atom. In naming oxoanions, the “main” form is assigned the –ate suffix, which is attached to the name of the non-oxygen central atom. The –ate form of these oxoanions should be learned, as the other forms (hypo-/-ite, -ite, per-/-ate), if they exist, can be determined from - the –ate form. ClO3 is the –ate form for the chlorate oxoanion and it has a -1 charge. - ClO2 has one less oxygen than the –ate form so it will be designated as the –ite form, - and will have the same charge as the –ate form. So, the ClO2 ion is not correctly named the chlorate ion, it should be the chlorite ion.

(e) A few polyatomic anions do not contain oxygen and also do not follow an easily identifiable pattern of structure an naming. These simply need to be memorized. The polyatomic ion containing a single atom each of C and N is called the cyanide ion and it has a -1 charge. So, this is correctly named.

Answer D

52. All of the following are in aqueous solution. Which is incorrectly named? a) H2SO4, sulfuric acid b) H2CO3, carbonic acid c) H3PO4, phosphoric acid d) HCN, cyanic acid e) HCl, hydrochloric acid Remember that we name oxoacids according to the form of the polyatomic ion to which they correspond.

If acids contain the –ate form of a polyatomic oxoanion they have the –ic suffix attached to the name. If acids contain the –ite form of a polyatomic oxoanion, they have the –ous suffix attached to the name. If acids contain the hypo- / -ite form of a polyatomic oxoanion, they have the hypo- prefix and –ous suffix attachted to the name. If acids contain the per- / -ate form of a polyatomic oxoanion, they have the per- prefix and –ic suffix attached to the name. If acids contain an –ide form of anion (for example, “chloride,” “sulfide,” “cyanide,” etc) they have the hydro- prefix and –ic suffix attached to the name. Organic acids will have formulas that contain anions that have carbon and hydrogen in them. The formula will also begin with the number of hydrogen atoms that ionize (separate from the hydrogens in the anion). The names of these will usually not be descriptive and so have no recognizable patterns. Therefore, these will need to be memorized. 2- (a) This acid contains the polyatomic ion SO4 which is the sulfate ion.

Therefore, it should have the –ic suffix, and sulfuric acid is the correct name. 2- (b) This acid contains the polyatomic ion CO3 which is the carbonate ion. Therefore, it should have the –ic suffix, and carbonic acid is the correct name. 3- (c) This acid contains the polyatomic ion PO4 which is the phosphate ion. Therefore, it should have the –ic suffix, and phosphoric acid is the correct name. (d) This acid contains the polyatomic ion CN-, which is the cyanide ion. Therefore, it should have the hydro- prefix and the –ic suffix—cyanic acid is not the correct answer—it should be hydrocyanic. (e) This acid contains the monoatomic ion Cl-, which is the chloride ion. Therefore, it should have the hydro- prefix and the –ic suffix, and hydrochloric acid is the correct name.

Answer D

53. All of the following are in aqueous solution. Which is incorrectly named? a) HC2H3O2, acetic acid b) HBr, bromic acid c) H2SO3, sulfurous acid d) HNO2, nitrous acid e) HClO3, chloric acid See discussion of acid names in answer to question 52.

(a) The anion of this acid contains carbon and hydrogen and so is an organic acid. - The C2H3O2 ion is the acetate anion. As it has the –ate suffix the acid name will have the –ic suffix, acetic acid. Therefore, the name is correct. (b) This acid contains the monoatomic ion Br-, which is the bromide ion. Therefore, it should have the hydro- prefix and the –ic suffix—bromic acid is not the correct answer—it should be hydrobromic. 2- (c) This acid contains the polyatomic ion SO3 which is the sulfite ion. Therefore, it should have the –ous suffix, and sulfurous acid is the correct name. - (d) This acid contains the polyatomic ion NO2 which is the nitrite ion. Therefore, it should have the –ous suffix, and nitrous acid is the correct name. - (e) This acid contains the polyatomic ion ClO3 which is the chlorate ion. Therefore, it should have the –ic suffix, and chloric acid is the correct name.

Answer B

54. Which of the following pairs is incorrect? a) NH4Br, ammonium bromide b) K2CO3, potassium carbonate c) BaPO4, barium phosphate d) CuCl, copper(I) chloride e) MnO2, manganese(IV) oxide (a) NH4 is the only positively charged polyatomic ion, the ammonium ion, and Br is a non-metal, so this is an ionic compound. In naming ionic compounds, if the first

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 42 component is the ammonium ion then the name of the first component is ammonium. If the second component is an element, the name of the second component is the name of the element with the –ide suffix attached. Therefore, the name ammonium bromide is the correct name for this compound.

2- (b) K is a metal in column 1A, and CO3 is the polyatomic carbonate anion, so the compound is ionic. In naming ionic compounds, if the first component is an element, the name of the first component is the name of the element, modified by a roman numeral if the element is a transition metal or a post transition metal. This is not the case here. If the name of the second component is a polyatomic ion, the name of the polyatomic ion is used. Therefore, the name potassium carbonate is the correct name for this compound.

3- (c) Ba is a metal in column 2A, and PO4 is the polyatomic phosphate anion, so the compound is ionic. In naming ionic compounds, if the first component is an element, the name of the first component is the name of the element, modified by a roman numeral if the element is a transition metal or a post transition metal. This is not the case here. If the name of the second component is a polyatomic ion, the name of the polyatomic ion is used. The problem here is that the charge for the phosphate ion is 3- while that of barium is 2+. Therefore, the formula should be Ba3(PO4)2. So, while the name is correct, the formula is not.

(d) Cu is a transition metal in column 1B, and Cl is a non-metal in column 7A so the compound is ionic. In naming ionic compounds, if the first component is an element, the name of the first component is the name of the element, modified by a roman numeral if the element is a transition metal or a post transition metal. As copper is a transition metal, the charge of the copper atom needs to be modified by a roman numeral. The correct charge can be determined by the given formula. As Cl is in column 7A, its charge if it takes an electron would be -1. As there is only one Cl in the formula, the charge of the single copper atom must be +1. Therefore, its roman numeral should be (I). If the second component is an element, the name of the second component is the element with the –ide suffix attached. Therefore, the name of copper (I) chloride is the correct name for this compound.

(e) Mn is a transition metal in column 5B, and O is a non-metal in column 6A so the compound is ionic. In naming ionic compounds, if the first component is an element, the name of the first component is the name of the element, modified by a roman numeral if the element is a transition metal or a post transition metal. As manganese is a transition metal, the charge of the manganese atom needs to be modified by a roman numeral. The correct charge can be determined by the given formula. As O is in column 6A, its charge if it takes two electrons would be -2. As there are two O’s in the formula, the charge of the single manganse atom must be +4. Therefore, its roman numeral should be (IV). If the second component is an element, the name of the second component is the element with the –ide suffix attached. Therefore, the name of manganese (IV) oxide is the correct name for this compound.

Answer C

55. Which of the following name(s) is(are) correct? 1. sulfide, S2– 2. ammonium chloride, NH4Cl 3. acetic acid, HC2H3O2 4. barium oxide, BaO a) all b) none c) 1, 2 d) 3, 4 e) 1, 3, 4 (1) S is a non-metal so when it form ions it will take electrons. As it is in column 6A, it will take 2 electrons so the designation S2- is correct. As the ion is monoatomic, and it would be the second component of an ionic compound, the –ide suffix would be attached to the element name. Therefore, the name sulfide would be the correct name.

(2) NH4 is the only positively charged polyatomic ion, the ammonium ion, and Cl is a non-metal, so this is an ionic compound. In naming ionic compounds, if the first component is the ammonium ion then the name of the first component is ammonium. If the second component is an element, the name of the second component is the name of the element with the –ide suffix attached. Therefore, the name ammonium chloride is the correct name for this compound.

(3) Organic acids will have formulas that contain anions that have carbon and hydrogen in them. The formula will also begin with the number of hydrogen atoms that ionize (separate from the hydrogens in the anion). The names of these will usually not be descriptive and so have no recognizable patterns. Therefore, these will need to be memorized. The anion of this acid contains carbon and hydrogen and - so is an organic acid. The C2H3O2 ion is the acetate anion. As it has the –ate suffix the acid name will have the –ic suffix, acetic acid. Therefore, the name is correct.

(4) Barium is a metal and oxygen is a non-metal so barium oxide would be an ionic compound. As barium is neither a transition metal nor a post-transition metal, and is in column 2A, it can only have a 2+ oxidation state and so we do not use a roman numeral to designate it—it is correctly designated. As oxygen is in column 6A and so would have a -2 charge, only 1 O would be needed to balance 1 Ba, so the subscripts are correct. As is typical for the second component, we attach the –ide suffix for an element. Therefore, barium oxide is the correct name for this formula.

All four are correct. Answer A

56. Which metals form cations with varying positive charges? a) Group 2 metals

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 44 b) Group 3 metals c) transition metals d) Group 1 metals e) metalloids Generally, group 1A and group 2A metals never form cations with varying charges. The post-transition group 3A metal can form cations with varying charges but usually don’t. Metalloids can form ions but also usually don’t form cations with varying charges. Only transition metals routinely form cations with varying charges. Answer C

57. Three samples of a solid substance composed of elements A and Z were prepared. The first contained 4.31 g A and 7.70 g Z. The second sample was 35.9% A and 64.1% Z. It was observed that 0.718 g A reacted with Z to form 2.00 g of the third sample. Show that these data illustrate the law of definite composition. The way to show that samples of a substance illustrate the law of definite composition is to show that the components of the substance demonstrate the same mass fractions (or mass percents).

Data for the first sample were provided as grams of each component from an analysis. Simply divide each number of grams by the total number of grams to get a mass fraction for each and then multiply times 100.

Data for the second sample were already provided as mass percents so there was no need to do anything for these other than list them.

Data for sample three provided a mass for A and a total mass. To get the mass% for A just divide the mass of A by the total mass. To get the mass% for B, subtract the mass for A from the total mass to get mass of B, then divide by the total mass. (6 points)

Sample1 4.31g 7.70g mass% A = x100 = 35.9% mass% B = x100 = 64.1% 4.31g + 7.70g 4.31g + 7.70g Sample 2 mass% A = 35.9% mass% B = 64.1% Sample 3 .718g 2.00g − .718g mass% A = x100 = 35.9% mass% B = x100 = 64.1% 2.00g 2.00g

58. Complete the following table. Remember that we use symbols for elements that provide us with information about the numbers of particles within atoms of the element.

In the first row you are given the symbol 206Pb. You know this is the symbol for lead, and that the superscript in front of the symbol is the mass number for a particular

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 45 isotope. From the symbol you can look on the periodic table and get the atomic number, which is also the number of protons. This is 82. Then, the mass number minus the number of protons provides you with the number of neutrons. 206-82 = 124. Then, because the symbol shows this is a neutral atom, the number of electrons will be the same as the number of protons—this is also 82. Then, as the net charge is the sum of positive and negative charges, the net charge on this atom is 0.

For the second row you are given that the number of protons is 31. This is the same as the atomic number for the atom and so is element Gallium, the symbol of which is Ga. You are also given the number of neutrons so you would want to put the mass number, 31+38=69, as a superscript in from of the symbol, 69Ga. You are also given that it has a 3+ charge so you would want to add this into the symbol as well as a superscript after the symbol, 69Ga3+. Finally, this atom has a 3+ charge this means it has lost three electrons. As the neutral atom would have the same number as the number of protons, this means this atom would have 31-3=28 electrons.

For the third row, you are given that the number of protons is 52, which is the same as the atomic number, so the element is Tellurium and its symbol is Te. You are also given the number of neutrons so you would want to put the mass number, 52+75=127, as a superscript in from of the symbol, 127Te. You are also given that its number of electrons is two greater than its number of protons so it has a -2 charge, so you would want to add this in the net charge column and as a superscript following the symbol, 127Te2-.

Finally, in the fourth row, you are given that the atom is a manganese 2+ ion. You can look on the periodic table to see that this is atomic number 25, and so has 25 protons. This means that the neutral atom also has 25 electrons. Given that the charge is 2+, this means that the atom has lost two electrons so the number of electrons it now has is 23. (10 points)

Symbol # Protons # Neutrons # Electrons Net Charge 206Pb 82 124 82 0 69 3+ Ga 31 38 28 3+

127Te2- 52 75 54 2- 2+ Mn 25 30 23 2+

59. Complete the following table. Remember that we use symbols for elements that provide us with information about the numbers of particles within atoms of the element.

In the first column we are given that an atom is Gallium by the atomic symbol Ga. The superscript in front of the symbol is the mass number meaning that the sum of the number of protons and neutrons is 69. Because we know the atom is gallium, we can look its atomic number up on the periodic table to find this is 31, which will be the same as the number of protons. The number of neutrons will then be the mass

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 46 number minus the atomic number, 69-31=38. As the number of protons is 31, when the atom is neutral it will also have 31 electrons. As its charge is 3+ as shown by the superscript following the symbol, it will have lost three electrons and its current number of electrons is 28.

In the second column, because the number of protons is 34, the atomic number is also 34 and this identifies the element as selenium, atomic symbol Se. As the number of neutrons is 46, the mass number, which is the sum of protons and neutrons, is 80. As the number of protons is 34, the number of electrons in a neutral atom is also 34. The net charge is given as 2- so the atom has gained 2 electrons. This means the atom has 36 electrons. The symbol should reflect the mass number as a superscript before the symbol and the charge as a superscript after the symbol. (9 points)

Symbol 69Ga3+ 80Se2- Number of protons 31 34 Number of neutrons 38 46 Number of electrons 28 36 Atomic number 31 34 Mass number 69 80 Net charge 3+ 2–

60. Arsenopyrite is a mineral containing As, Fe, and S. Classify each element as metal, nonmetal, or metalloid. As—arsenic: metalloid (3 points) Fe—iron: metal S—sulfur: non-metal

61. Write the symbol for each of the following elements.

a) silver Ag b) calcium Ca c) iodine I d) copper Cu e) phosphorus P (5 points)

62. Write the names of the following compounds: Remember the strategy Identify whether the first component is a metal or a non-metal: If a metal (or the ammonium ion), the compound is ionic If non-metal, it is covalent If ionic If NH4 is the first component, this is the ammonium ion and the first name is ammonium If the first component is an element, the first name is the name of the element If the element is a transition or post-transition element it

must be accompanied by a roman numeral to indicate the charge. The charge can be determined from the formula by knowing the number of negative charges of the anion If the element is not a transition or post-transition element it is not modified If the second component is an element the second name is the name of the element with –ide attached If the second component is a polyatomic ion the second name is the name of the polyatomic ion If covalent Both components will always be elements The first name is the name of the element This name will be modified by a number prefix if the number of atoms is more than one The second name is the name of the element with the –ide suffix attached This name will always be modified by a number prefix

(a) First component is a metal so ionic compound. First component is a transition metal so need element name modified by roman numeral. Charge of anion is -2 so the single Fe atom must also have a +2 charge, so iron (II). The second component is the sulfate polyatomic anion. Name: iron(II) sulfate. (b) First component is a metal so ionic compound. First component is a column 1A metal so element name does not need to be modified by a roman numeral. Second component is the acetate polyatomic anion. Name: sodium acetate. (c) First component is a metal so ionic compound. First component is a column 1A metal so element name does not need to be modified by a roman numeral. Second component is the nitrite polyatomic anion. Name: potassium nitrite (d) First component is a metal so ionic compound. First component is a column 2A metal so element name does not need to be modified by a roman numeral. Second component is the hydroxide polyatomic anion. Name calcium hydroxide (e) First component is a metal so ionic compound. First component is a transition metal so need element name modified by a roman numeral. Charge of anion is -2 so the single Ni atom must also have a -2 charge, so nickel (II). The second component is the carbonate polyatomic anion. Name: nickel (II) carbonate (2 points each)

63. Write the chemical formulas for the following compounds or ions. (a) You will need to know the chemical formulas for all of the polyatomic ions. You - need to include the charge. The formula for the nitrate ion is NO3 . (b) The name of the first component is a metal element and the name of the second component is a non-metal element so the compound is ionic. First, write down the symbols for each component with a little space between them (Al O). Then assign charges. The metal element is in column 3A and is not a post-transition metal so will always give up 3 electrons and have a charge of 3+. The non-metal element is in

⎛ Al O⎞ column 6A and so will always take two electrons and have a charge of 2-. Our ⎜ +3 −2 ⎟ ⎝ ⎠ trick is to simply switch the numbers of the charges to get the subscripts so the formula will be Al2O3.

(c) You will need to know the chemical formulas for all of the polyatomic ions. You + need to include the charge. The formula for the ammonium ion is NH4 .

(d) See discussion of acid names for question 52. This acid would be based on the - polyatomic perchlorate ion. This ion is ClO4 . As there is only a single negative + charge on this ion, only one H would be associated with it. The formula is HClO4.

(e) The name of the first component is a metal element and the name of the second component is a non-metal element so the compound is ionic. First, write down the symbols for each component with a little space between them (Cu Br). Then assign charges. We have already been told the charge of the copper atom with the roman numeral II—2+. The non-metal element is in column 7A and so will always take one ⎛ Cu Br⎞ electron and have a charge of 1-. Our trick is to simply switch the numbers ⎜ +2 −1 ⎟ ⎝ ⎠ of the charges to get the subscripts so the formula will be CuBr2. (2 points each)

64. How many atoms (total) are there in one formula unit of Ca3(PO4)2? 3 calcium atoms 2 phosphorus atoms 13 atoms 8 oxygen atoms

Name the following compounds: See discussion for question 62 to remind yourself of the strategy for naming. (2 points each)

65. Al2(SO4)3 First component is a metal so compound is ionic. First component is a column 3A, non-post transition metal so element name does not need to be modified by a roman numeral. The second component is the polyatomic sulfate anion. Name: aluminum sulfate

66. NH4NO3 First component is the polyatomic ammonium ion so compound is ionic and the first name is ammonium. The second component is the polyatomic nitrate anion. Name: ammonium nitrate

67. NaH First component is a metal so compound is ionic. First component is a column 1A metal so element name does not need to be modified by a roman numeral. The

Copyright © Cengage Learning. All rights reserved. Chapter Error! Unknown document property name.: Error! Unknown document property name. 49 second component is an element so the second name is the name of the element with the –ide suffix attached. Name: sodium hydride

68. K2Cr2O7 First component is a metal so compound is ionic. First component is a column 1A metal so element name does not need to be modified by a roman numeral. The second component is the polyatomic dichromate anion. Name: potassium dichromate

69. CCl4 First component is a non-metal so compound is covalent. First component has no subscript and so has no number prefix. First name is just the name of the element. Second component has a subscript of 4 and so has a number prefix of tetra-. Name of second component is the name of the element with the –ide suffix. Name: carbon tetrachloride

70. AgCl First component is a metal so compound is ionic. First component is a transition metal so need element name modified by a roman numeral. Second component is a column 7A non-metal element so it would take one electron and have a -1 charge. There is only one Cl, so charge on the Ag atom must be +1 to balance. The second name is the name of the element with the –ide suffix attached. Name: silver (I) chloride

71. CaSO4 First component is a metal so compound is ionic. First component is a column 1A metal so element does not need to be modified by a roman numeral. The second component is the polyatomic sulfate anion. Name: calcium sulfate

72. HNO2 Because this formula starts with a hydrogen and ends with a polyatomic ion, it is an acid. See the discussion for question 52 to remind yourself about acid names. The polyatomic ion this acid is based on is the nitrite ion—therefore, we use the –ous suffix. Name: nitrous acid

73. N2O3 First component is a non-metal so compound is covalent. First component has a subscript of 2 and so has a number prefix of di-. First name is just the name of the element with the prefix. Second component has a subscript of 3 and so has a number prefix of tetra-. Name of second component is the name of the element with the –ide suffix. Name: dinitrogen trioxide

74. SnI2 First component is a metal so compound is ionic. First component is a post-transition metal so need element name modified by a roman numeral. Second component is a column 7A non-metal element so it would take one electron and have a -1 charge.

There are two I’s, so charge on the Sn atom must be +2 to balance. The second name is the name of the element with the –ide suffix attached. Name: tin (II) iodide

Write the formula for: (2 points each) 75. sodium thiosulfate The name of the first component is a metal element and the name of the second component is a polyatomic anion so the compound is ionic. First, write down the symbols for each component with a little space between them (Na S2O3). Then assign charges. Because Na is in column 1A we know that its charge will be +1. The charge ⎛ Na S O ⎞ of the thiosulfate ion is -2. 2 3 Our trick is to simply switch the numbers of the ⎜ +1 −2 ⎟ ⎝ ⎠ charges to get the subscripts so the formula will be Na2S2O3.

76. iron(III) oxide The name of the first component is a metal element and the name of the second component is a non-metal element so the compound is ionic. First, write down the symbols for each component with a little space between them (Fe O). Then assign charges. We have already been told the charge of the iron atom with the roman numeral III—3+. The non-metal element is in column 6A and so will always take two ⎛ Fe O⎞ electrons and have a charge of 2-. Our trick is to simply switch the numbers ⎜ +3 −2 ⎟ ⎝ ⎠ of the charges to get the subscripts so the formula will be Fe2O3.

77. dichlorine heptoxide You are given a name with a non-metal for both components so the compound is covalent. First, write down the symbols for each element with a little space between them (Cl O). Then inspect the number prefixes. The number prefix for the first component is di-, so there will be 2 Cl’s. The number prefix for the second component is hept- so there will be 7 O’s. The formula will be Cl2O7.

78. cobalt(II) chloride You are given a name with a metal element for the first component (cobalt) and a non-metal for the second component so the compound is ionic. First, write down the symbols for each component with a little space between them (Co Cl). Then assign charges. You have already been told by the Roman numeral that the charge on cobalt is +2. Because chlorine is in column 7A it will take one electron so its charge ⎛ Co Cl⎞ will be -1. By inspection, you can see that it would take 2 Cl- ions to balance a ⎜ +2 −1 ⎟ ⎝ ⎠ single cobalt ion. Answer CoCl2

79. aluminum hydroxide You are given a name with a metal element for the first component (aluminum) and a polyatomic anion (hydroxide) for the second component, so the compound is ionic.

First, write down the symbols for each component with a little space between them (Al OH). Then assign charges. Al is in column 3A so its charge is +3 while the charge of the hydroxide is -1 (remember, for a polyatomic ion it is the whole ion—OH—that ⎛ Al OH⎞ has the -1 charge) . By inspection, you can see that it would take 3 OH’s at a ⎜ +3 −1 ⎟ ⎝ ⎠ -1 charge to balance a single Al ion at a +3 charge, so the formula is Al(OH)3 (remember that if there is more than a single polyatomic ion unit, parentheses are placed around the polyatomic ion and a subscript of the correct number is placed outside the parentheses). Answer Al(OH)3

80. sulfurous acid Remember that for acids which contain anions that are polyatomic oxo-ions the rules for naming correspond to the formula of the corresponding polyatomic ion. The –ous form of an acid corresponds to the –ite form of the polyatomic ion. Therefore, 2- “sulfurous” corresponds to sulfite—SO3 , which means there must be 2 H’s associated 2- with the SO3 anion. Answer H2SO3

81. nitric acid Remember that for acids which contain anions that are polyatomic oxo-ions the rules for naming correspond to the formula of the corresponding polyatomic ion. The –ic form of an acid corresponds to the –ate form of the polyatomic ion. Therefore, “nitric” - corresponds to nitrate—NO3 , which means there must be 1 H associated with the - NO3 anion. Anwer HNO3

82. phosphoric acid Remember that for acids which contain anions that are polyatomic oxo-ions the rules for naming correspond to the formula of the corresponding polyatomic ion. The –ic form of an acid corresponds to the –ate form of the polyatomic ion. Therefore, 3- “phsophoric” corresponds to phosphate—PO4 , which means there must be 3 H’s 3- associated with the PO4 anion. Anwer H3PO4

83. acetic acid This is an acid for which you will have to memorize the formula, at least to begin with. It is an organic acid (contains carbon) so the formula will begin with the hydrogen atom that “ionizes” as is the tradition for all organic acids—this is to help us remember that the organic compound is an acid. The remainder of the formula is “what’s left behind” when the acid ionizes—an organic anion of 1- charge. In this - case, the “acetate” anion is C2H3O2 . The complete molecule would be HC2H3O2. Another way to write a formula for an organic acid is to show the structure of the organic acid functional group (COOH) as being attached to the remainder of the molecule. If we remove a COOH group from the above acetic acid formula the remainder is CH3—so, this form of the molecule would be CH3COOH. Answer HC2H3O2 or CH3COOH

84. phosphorus trichloride

Both components are non-metals—therefore, the compound is covalent. The first component is phosphorus so write down P to begin the formula. The second element is chlorine so write down Cl as the second component. Remember that the first component will have a number prefix unless there is only one atom for that component—so, there is just atom of P. The second component will always have number prefix—in this case, tri—so, three atoms of Cl. Therefore, PCl3. Answer PCl3

Total 150 points