Answer Key, Problem Set 2 (Written) (plus some solutions to Mastering Problems from this unit)

Transcription

1 Chemistry 121 Mines, Fall 2018 Answer Key, Problem Set 2 (Written) (plus some solutions to Mastering Problems from this unit) 1. NT1; 2. NT2; ; ; 5. NT3; 6. NT4; ; & 2.48 (ADD: correct each false statement by crossing out and adding a few words to make it read true; (ADD part (b): and between mass number and isotopic mass ); (ADD: Add a column for complete isotopic symbol (i.e., with subscript and superscripts and charge [if not zero]) and fill that in appropriately for each Nanoscopic Pictures, Atomic Theory, and Classification of Matter 1. NT1. (a) Characterize the physical state(s) represented by each type of matter in each box, and (b) state whether the matter in the box (as a whole) is an element, a compound, a mixture, and whether or not it is homogeneous or heterogeneous. (iii) (iv) (v) (vi) (i) (ii) (i) (ii) (a) gas Nanoscopic entities are far apart from each other (relative to their size), and fill up the whole box (evenly spread out in the available space). The orientations of the molecules are random, indicating they can move past one another and rotate freely. (b) mixture (of substances) Two different kinds of nanoscopic entity [whose ratio could be different in a different sample). Both substances in the box are elements (entities are made of only one kind of atom), but one is monatomic (made of individual ["free"] atoms) and one is diatomic molecular (made of molecules having exactly two atoms each). homogeneous Although there are two types of nanoscopic entities, they are "mixed up" randomly with one another. So all parts of the sample (as long as you don't sample a single entity!) have the same composition. (a) liquid Nanoscopic entities are close enough (relative to their size) to be considered "touching" (remember that they are moving and bouncing around!), and are not filling up the whole box (they are not evenly spread out in the available space they occupy the bottom 2/3 of the box only). The orientations of the molecules are random, indicating they can move past one another and rotate freely, hence a liquid (not a solid). Even though we cannot see the orientation changes of a single atom because of its spherical shape, the single atoms in the element are not arranged in an ordered array, so that substance is not a solid. (b) mixture (of substances) Two different kinds of nanoscopic entity [whose ratio could be different in a different sample). One substance in the box is an element (monatomic so only one kind of atom), and the other is a compound (two different kinds of atom in each [triatomic] molecule). heterogeneous There are two types of nanoscopic entities, and they are not "mixed up" randomly with one another different parts of the sample have different composition. (iii) (a) gas Nanoscopic entities evenly spread out in the available space. => gas. Orientations of the molecules are random, indicating they can move past one another and rotate freely. (b) (pure substance), compound Only one kind of entity (diatomic molecule) but each is composed of more than one kind of atom. Note that the ratio of the two different kinds of atoms in this sample could not be in a different ratio in a different sample of this matter. That is why this is a (pure) substance and not a mixture. homogeneous Since there is only one kind of entity, all parts of the sample have the same composition. PS2-1

2 (iv) (a) liquid Nanoscopic entities are not evenly spread out in the available space they occupy the "bottom" half of the box, but not the top area). => not a gas. Single atoms are not arranged in an ordered array, implying they can move past one another => not a solid. (b) mixture (of substances) Two different kinds of nanoscopic entity [whose ratio could be different in a different sample). Both substances in the box are elements (entities are made of only one kind of atom), and both are monatomic (entities are single atoms, not molecules). homogeneous Although there are two types of nanoscopic entities, they are "mixed up" randomly with one another. So all parts of the sample (as long as you don't sample a single entity!) have the same composition. (v) (a) one liquid (molecules, dark) and one solid (atoms, open spheres) In both cases, the nanoscopic entities are not filling up the whole box (they are not evenly spread out in the available space they occupy the "bottom" half of the box only). However, the sample of molecules is disordered, with random orientations, indicating a liquid, and the sample of atoms is arranged in an ordered array, indicating it is a solid (although in this particular depiction, the atoms are farther apart than I d normally put them when I make a solid [this problem came from another source and I kept their spacing]). (b) mixture (of substances) Two different kinds of nanoscopic entity [whose ratio could be different in a different sample). Both substances in the box are elements (entities are made of only one kind of atom), but one is monatomic (made of individual ["free"] atoms) and one is diatomic molecular (made of molecules having exactly two atoms each). heterogeneous There are two types of nanoscopic entities, and they are not "mixed up" randomly with one another different parts of the sample have different composition. (vi) (a) solid Nanoscopic entities are not evenly spread out in the available space => not a gas. The entities are arranged in an ordered array => a solid. (b) (pure substance), element Only one kind of entity (diatomic molecule) and each is composed of only one kind of atom. homogeneous Since there is only one kind of entity, all parts of the sample have the same composition. 2. NT2. For box (ii) in NT1, answer the following questions: (a) How many molecules are in the box? (b) How many free atoms are in the box? (c) How many total atoms are in the box? (d) How many compounds are in the box? (Note: a compound is NOT the same thing as a molecule!) (e) How many elements are in the box? (Note: an element is NOT the same thing as an atom!). ANSWERS: (a) 6 molecules (b) 9 free atoms (c) 9 + 3(6) = 27 total atoms (each molecule contains 3 atoms) (d) 1 compound (there are 6 molecules of a single compound [all the molecules are identical]) (e) 1 element (there are 9 free atoms of a single element [all the free atoms are identical]) Laws of Definite Proportion and Multiple Proportions Sulfur and fluorine (rest of problem not shown in this key). Answers: SF6: Strategy: 3.56 g F ; SF4: 1g S 2.37 g F, Ratio of these two is 3:2, a small whole number ratio, so 1g S the data are consistent with the Law of Multiple Proportions. 1) Set up each sample s mass ratio (i.e., grams of F per gram of S ) as described in the prior problems. In this problem, we will assume that the Law of Definite Proportion is followed for each compound (i.e., that the mass ratio for the single sample of each type of compound is the (constant) mass ratio for that compound [in the prior problems, there were two samples of the PS2-2

3 same compound in this problem, there are two samples, but they are not the same compound!]) 2) The Law of Multiple Proportions states that the mass ratios of two (or more) compounds formed by the same two elements will be in a ratio of small whole numbers. In other words, the ratio of the two mass ratios will be a small whole number one. So calculate the ratio of the two mass ratios determined in (1) by dividing one mass ratio by the other. Look to see if it can be expressed as a ratio of small whole numbers. Execution of Strategy: 1) Sample of SF6: 4.45 g F 3.56 g F ; Sample of SF4: 1.25 g S 1g S 4.43 g F 1.87 g S g F 3.56 g F 2) Ratio of mass ratios = 1g S g F Since 1.50 equals the fraction, 2 1g S 1g S which is a small whole number ratio (3:2), the data are consistent with the Law. Dalton s Atomic Theory Question not retyped in this key. (a) All carbon atoms are identical. (b) An oxygen atom combines with 1.5 hydrogen atoms to form a water molecule. (c) Two oxygen atoms combine with a carbon atom to form a carbon dioxide molecule. (d) The formation of a compound often involves the destruction of one or more atoms. Answers: (b) and (d) are inconsistent ((a) and (c) are consistent). Note, however, that (a) is inconsistent with current theory, as abundant experimental data were collected in the early 1900 s in support of the existence of isotopes. Reasoning: (b) implies that a half of an atom exists, where Dalton s theory defined an atom as a whole object that could not be divided into parts. (d) implies that atoms can be destroyed, and Dalton s theory postulated that atoms were indestructible. Furthermore, Dalton s theory explained compound formation by the joining (bonding) of atoms with one another to form distinct entities called molecules (which had different properties that the individual atoms from which they were composed) rather than destruction of atoms. 5. NT3. Describe how Dalton s Atomic Theory explains (a) The Law of Conservation of Mass (b) The Law of Definite Proportion (a) Shortest answer: During chemical reaction, atoms merely rearrange (re-"partner" with other atoms), so there are exactly the same number of each atom type after the reaction as before. Thus the total mass (= mass of all the atoms) must be the same. Slightly longer answer: Dalton s theory proposes that atoms have a fixed and unchanging mass, and that when chemical reaction occurs, atoms merely rearrange (regroup with other atoms) but are neither created nor destroyed. If these postulates are both true, then there is no mass created nor destroyed during chemical change, and thus there is no mass change. So his theory explains any observation of mass conservation during chemical change. Imagine a set of Lego blocks in which there are a fixed number of pieces, and each type block has a different mass than another. If you start with them all separate, and then end with them all pieced together, the total mass of all the blocks will not change, right? Or if you rearranged them into a bunch of 2-piece or 3-piece units, the total mass of the set of blocks would be the same. This is fairly good analogy to the idea that chemical change simply involves rearrangement of atoms. PS2-3

4 (b) The Law of Definite Proportion states that when any sample of a given compound is separated chemically into its elements (no matter where it came from or how big it is, etc.), the mass ratio of the elements formed is constant). The atomic theory explains this as follows. It assumes that 1) any sample of a given compound contains a fixed ratio of atom types (different elements atoms) and 2) that each atom of a specific element has the same mass as any other atom of that type (but different from those of other elements). If these are both true, then it will follow that the ratio of the masses of the elements formed from a given sample of a given compound will always be the same. If you don t follow this argument, think of it this way. If you filled up a bunch of goodie bags with 2 Hershey bars and 3 lollipops, then the ratio of the mass of chocolate to lollipop would always be the same no matter how many bags you collected, as long as the mass of each Hershey bar is the same and the mass of each lollipop is the same. When you have a bigger sample of a molecular compound, its like having more goodie bags the amount changes, but the ratio of the number of atoms stays the same, and so the ratio of the masses of those atoms must stay the same as well. If you still cannot see this, come by my office and I will mathematically prove to you that in any sample of a compound, the mass ratio of elements = the atom ratio x the mass ratio of the atoms Now, #1 above assumes the atom ratio is constant, and the assumption in #2 above means that the mass ratio of atoms is also constant. Hence the mass ratio of elements in the compound must also be a constant! NOTE: The mass ratio need only be constant it is not generally a whole number. On the other hand, the atom ratio will be a (usually small) whole number ratio (because an atom is a thing and you cannot have a fraction of one). Any mass unit, however (like an amu or a g or whatever) is not a physical object you can have a fraction of one. For example, in the compound with the formula ClF3, the ratio of F atoms to Cl atoms is 3:1, a small whole number ratio. But the mass ratio of F to Cl is : 1 (because the average mass of a Cl atom is amu and that of a F atom is amu. (3 x 19.00)/(35.45) = 1.608) : 1 is not a small whole number ratio. Rutherford s Gold Foil Experiment and Model of Atom 6. NT4. This problem asks you to think about Rutherford s experiment, the qualitative version of Coulomb s Law, and distinguishing observations (raw data) from explanations (hypothesis or theory). Rutherford came up with the following hypothesis to try to provide a model that was consistent with his observations in the lab: All of the positive charge in an atom and most of the mass in an atom are concentrated into a very tiny space inside the atom. Some students say that his reasoning for how this explained his observations is as follows: If all of the mass is concentrated into a small space, then only a small fraction of the alpha particles that were in the beam would hit the small nucleus and get deflected or bounce back. I don t like this explanation for two reasons: 1) there is no mention of the charge of the nucleus and its importance; and 2) it uses the word hit, which is (in my opinion) patently wrong (see simulation from Ppt04!). This all being said: (a) Describe at least two of Rutherford s key observations, and (b) (without using the word hit!) describe how Rutherford s proposed model of the atom explains these (two) observations. Think of not only the sign of the charge on the nucleus but the magnitude of the charge as well (and of that on the alpha particle, too), and think about how the force of repulsion would vary as an alpha particle gets closer and closer to a nucleus (Does a particle have to hit another in order to be deflected or pushed back?). Answer: (a) 1) the vast majority of very energetic positively charged particles went straight through the foil; 2) the path of some small, but finite fraction of these particles was altered: some were deflected at small angles, and an even smaller fraction were deflected right back toward the source. (b) First of all, what was Rutherford s model? 1) all positive charge is in a tiny space called the nucleus; 2) nearly all the mass of the atom is also concentrated in this same tiny space. How does this model explain the observations? Explanation (short version): One might say Only a small fraction of the alpha particles got close enough to the very tiny nucleus to feel the large repulsive force needed for them to be deflected. The force arises because the nucleus is very positively charged in Rutherford s model, and the force of repulsion between two positively charged particles is greater if the magnitude of the PS2-4

5 charges is greater. The large mass of the nucleus prevents it from being moved; the alpha particle s trajectory is changed instead. Explanation (detailed version). Some large force must act on the alpha particles to alter their trajectory. No such force must act upon the majority of the alpha particles since most go straight through without being affected. If all of the positive charge in a gold atom were concentrated into a very tiny space, a very large force of repulsion would be felt by an alpha particle if it got very close to the nucleus; those not passing that close would feel very little force. So that is how this model explained the observations; only a very small fraction of the alpha particles got very close to any atom s nucleus in the foil because the nuclei are so very tiny. Those that did get that close felt an extremely large force because of the Coulomb s Law relationship between force, distance, and magnitude of charge, and it was repulsive because alpha particles and the nucleus are both positively charged. Furthermore, since all of the positive charge in the atom is concentrated in this small space, the magnitude of the charge is as large as possible for each given atom (note that atoms of different elements have a different number of protons in the nucleus, so the specific charge of a nucleus depends on what the identity of the atom is). The reason that it was the alpha particles paths that got altered rather than the nuclei (in principle they could have been pushed out of the atom by the strong repulsive force) was because (in Rutherford s model) the nucleus is hypothesized to be much more massive than an alpha particle (nearly all of the mass of the atom is concentrated in the nucleus in Rutherford s model). One need not invoke the word collide or hit since all that is required is a FORCE (this force acts at a distance, just like the force between two magnets oriented so that they repel. The magnets do not need to touch or hit one another for one s path to be deflected by the other) Question not retyped in this key. (a) Since electrons are smaller than protons, and since a hydrogen atom contains only one proton and one electron, it must follow that the volume of a hydrogen atom is mostly due to the proton. (b) A nitrogen atom has seven protons in its nucleus and seven electrons outside of its nucleus. (c) A phosphorus atom had 15 protons in its nucleus and 150 electrons outside of its nucleus. (d) The majority of the mass of a fluorine atom is due to its nine electrons. Answers: (a), (c) and (d) are inconsistent (only (b) is consistent) Reasoning: (a) In Rutherford s model, the volume of any atom s nucleus is tiny compared to the volume of the atom itself most of the volume of an atom is said to be composed of empty space, around which the electrons are whizzing about. The key here is that the nucleus is stationary in this model whereas the electrons are in motion, thereby occupying more space because of their motion (not because their own volume is large it is not). Do not confuse the volume of the atom with the volume of an electron. Rutherford s model does not directly state how the volume of an electron compares to that of the nucleus. It only compares the volume of the nucleus to the volume of the atom. Also, this question provides a good opportunity to review the difference between mass and volume, and the relationship to density. The nucleus has high mass but very small volume, so it has a tremendously high density. (b) is consistent because an atom in Rutherford s model is electrically neutral. Since a proton was assumed to have a charge exactly equal in magnitude to that of an electron (but with opposite sign), an equal number of protons and electrons is required to have a net charge of zero for an atom. (c) As (b) above states, in order to be electrically neutral, #p s = #e - s in an atom. (d) According to Rutherford s model, most of the mass is due to the nucleus the electrons comprise only a tiny fraction of the mass of an atom. PS2-5

6 Subatomic Particles, Isotopes, s, etc & (ADD: correct each false statement to make it read true) Which statements about subatomic particles are true? (a) If an atom has an equal number of protons and electrons, it will be charge-neutral. (see #10(b) above) True (b) Electrons are attracted to protons. (according to Coulomb s Law) (c) Electrons are much lighter than neutrons. (~1800x less mass than in a neutron) False (d) Protons have twice the mass of neutrons Which statements about subatomic particles are false? True False about the same as (a) Protons and electrons have charges of the same magnitude but opposite sign. (b) Protons have about the same mass as neutrons. All at least one (c) Some atoms don t have any protons. (# of p s identifies the element. Zero isn t an option!) False electrons (d) Protons and neutrons have charges of the same magnitude but opposite signs. (Neutrons have no charge) (with part (b) added) (a) Explain [Describe] the difference between Z (the atomic number) and A (the mass number) Answer: Both of these are (whole) numbers, but Z is the number of protons in the nucleus of an atom, whereas A is the sum of the number of protons and neutrons in the nucleus of a given atom. NOTE: Z is defined for any element, but A is not defined for an element only for an isotope (of an element). (b) (added): Describe the difference between mass number and isotopic mass. Answers: (1) Mass number is a number of physical particles (it is not a mass), whereas isotopic mass is an amount of mass (of some particles, bound together in a an atom). Thus, A does not have units, but isotopic mass does (units of mass, usually amu ) (2) Specifically, mass number is the # of p s and n s in an atom, and isotopic mass is the mass of a given atom (3) Mass number is, by definition a whole number, whereas the isotopic mass of an atoms is a not a whole number (although it is often very close to a whole number, in units of amu) (4) Note that the isotopic mass is the mass of not only the p s and n s, but also the electrons, even though they don t contribute very much in terms of %, they do contribute some mass! Fill in the blanks to complete the table. [Smaller fonted, unbolded items with no annotations were given.] (not complete) Si 14 Z A #p #e - #n Charge = p e = 0 Complete Si S 2- S e = -2 e = 18 = A Z = S Cu 2+ Cu e = +2 e = Cu 2 29 Z = 15 P =p e = P PS2-6

7 ================== END OF WRITTEN SET ==================== Answers to some of the Mastering Problems (with possibly different numbers) are below Laws of Definite Proportion and Multiple Proportions (2.32). Two samples of sodium chloride were decomposed (rest of problem not retyped in this key) NOTE: Number values may vary a bit from what is found in the Mastering program; values here are from the text problem. Answer: Yes, because both samples have a chlorine : sodium mass ratio of (g Cl / g Na). Reasoning / Work: The Law of Definite Proportion(s) states that when any sample of a given compound is separated chemically (i.e., decomposed) into elements, the mass ratio of the elements produced is always the same (i.e., constant or definite ). To calculate the mass ratio, simply divide one element s mass by the other. E.g., to get the Cl : Na mass ratio in each sample, 10.7 g Cl 17.3 g Cl Sample 1: g Cl/g Na ; Sample 2: g Cl/g Na 6.98 g Na 11.2 g Na Since these ratios are (within experimental uncertainty) the same, these data are consistent with the Law. (2.34). Upon decomposition, one sample of magnesium (rest of question not retyped in this key) NOTE: Number values may vary a bit from what is found in the Mastering program; values here are from the text problem. Answer: 2060 g (or 2.06 x 10 3 g) F Reasoning / Work: Since the mass ratio of elements produced by any given compound is a constant (Law of Definite Proportion(s)), you can use the data from the first sample to set up the ratio, and then use that ratio like a conversion factor, along with the mass of magnesium of the other sample, to find the mass of fluorine in it. Since the problem asks for grams you will also need to convert from kg to g. (NOTE: You could also set up an actual proportion rather than use the ratio like a conversion factor to find the mass of fluorine in the second sample.) Dimensional Analysis Approach: 2.57 kg F 1000 g g F 1.32 kg Mg x x 1.65 kg Mg 1kg Proportion Approach: 2056 g F 2060 g F 2.57 kg F 1.65 kg Mg x kg F x kg F 1.32 kg Mg x 1.32 kg Mg 2.57 kg F 1.65 kg Mg kg F 1000 g x 2060 g F 1kg (2.132). Question not retyped in this key. a) Two different samples of water are found to have the same ratio of hydrogen to oxygen. b) When hydrogen and oxygen react to form water, the mass of water formed is exactly equal to the mass of hydrogen and oxygen that reacted. c) The mass ratio of oxygen to hydrogen in water is 8:1. The mass ratio of oxygen to hydrogen in hydrogen peroxide (a compound that only contains hydrogen and oxygen) is 16:1. d) The ratio of H atoms to C atoms in methane is 4 : 1, while the ratio of H atoms to C atoms in ethylene is 2 : 1. NOTE: This problem is modified slightly from what is found in your text. There was no part (d) in the text problem. Answer: Only (c) PS2-7

8 ghanswer Key, Problem Set 2 Reasoning: First of all, the first two statements each only deal with one kind of compound (water)! Thus, they cannot be an example of the Law of Multiple Proportions, which involves more than one kind of compound. As noted in Problem #5, the Law of Multiple Proportion states that the mass ratios of two (or more) compounds that are formed from the same two elements must be in a small whole number ratio. In (c), the two mass ratios (16:1 and 8:1) are in a 2:1 ratio: 16gO181ggHO2 2 1, which is a small whole number ratio, so the statement is an example of the Law Statement (d) correctly states that the atom ratios in the two compounds are in a small whole number ratio, but this is basically a statement of theory (Dalton s atomic theory) that would explain the Law of Multiple Proportions. Since the statement itself does not include any mass ratios, it cannot be an example of the Law of Multiple Proprotions. In short, the Law involves mass ratios. Theory involves atom ratios. NOTE: Statement (a) is an example of the Law of Definite Proportion(s) and Statement (b) is an example of the Law of Conservation of Mass. You are responsible for knowing all three of these laws, and statements that are examples of them. Subatomic Particles, Isotopes, s, etc. (2.51). Write isotopic symbols of the form X-A (E.g., C-13) for each isotope. NOTE: The specific isotopes here are from the text problem; they may vary from what is found in the Mastering program. (a) Answer: Ag-107 (Ag Z = p s; A = = 107) (b) Answer: Ag-109 (Ag Z = p s; A = = 109) (c) Answer: U-238 (U Z = p s; A = = 238) (d) Answer: H-2 (H Z = 1 1 p; A = = 2) (2.54). Determine the number of protons and neutrons in each isotope. NOTE: The specific isotopes here are from the text problem; they may vary from what is found in the Mastering program. (a) Answer: 19 p, 21 n (Z = p s; A = 40 = 19 + n n = = 21) (b) Answer: 88 p, 138 n (Z = p s; n = = 138) (c) Answer: 43 p, 138 n (Z = p s; n = = 56) (d) Answer: 15 p, 18 n (Z = p s; n = = 18) MP (2.58). Determine the number of protons and electrons in each ion. NOTE: The specific items here are from the text problem; they may vary from what is found in the Mastering program. (a) p = 13, e = 10 (Al Z = p s; 3+ means 3 fewer electrons than protons e = 13-10) (b) p = 34, e = 36 (Se Z = 34 = # p s; 2- means 2 more electrons than protons e = ) (c) p = 31, e = 28 (Ga Z = 31 = # p s; 3+ means 3 fewer electrons than protons e = 31-3) (d) p = 38, e = 36 (Sr Z = 38 = # p s; 2+ means 2 fewer electrons than protons e = 38-2) PS2-8

9 Analogous to Problem For Extra Practice Only (if you so choose): (not complete) Z A #p #e - #n Charge Complete Na Se Cr Answers For Extra Practice Problem above: (not complete) Na 11 Z A #p #e - #n Charge = p e = 0 Complete Na Se 2- Se e = -2 e = 36 = A Z = Se Cr 3+ Cr e = +3 e = Cr 3 Z = 19 K =p e = K Organization of the Periodic Table (2.64). Write the symbol for each element and classify it as a metal, nonmetal, or metalloid. (NOTE: I have also listed the number of the Group and Period for each element. You are responsible for this information on an exam) NOTE: The specific elements here are from the text problem; they may vary from what is found in the Mastering program. Metal, Nonmetal, or Metalloid Group Period (a) gold Au Metal 11 (or 1B) 6 (b) fluorine F Nonmetal 17 (or 7A) 2 (c) sodium Na Metal 1 (or 1A) 3 (d) tin Sn Metal 14 (or 4A) 5 Reasoning: (e) argon Ar Nonmetal 18 (or 8A) 3 (1) Use the List of elements and their symbols (or the Periodic Table in Mastering) to determine the symbol of each element. Remember that element symbols are always one or two letters each if there is only one letter, it is capitalized; if there are two letters, the first is capitalized and the second is lowercase. Be particularly careful with the elements whose symbols are not based off of the current name, but of an older Latin or other derived name. For example: PS2-9

10 Ir is iridium, not iron (Fe); Gd is gadolinium, not gold (Au); S is sulfur, not sodium (Na); Mn is manganese, not magnesium (Mg); Ti is titanium, not tin (Sn); Ni is nickel, not nitrogen (N); Co is cobalt, not copper (Cu); Ca is calcium, not cadmium (Cd) nor carbon (C); Ar is argon, not arsensic (As) Also, please learn that fluorine is F (not Fl!) and that the u comes before the o (is it not spelled like flour!) Ultimately, the specific elements that you need to learn for Exam 1b will be listed on a handout for PS3. You do not need to learn all of them! (2) Elements to the left of the staircase (or stepladder ) on the periodic table are metals (except for hydrogen). Elements to the right of the staircase are nonmetals. Many elements that are touching the staircase are metalloids. (3) Groups are columns on the periodic table. They are also called families. There are two usage systems I will never ask you to memorize either of these. Just use whatever periodic table you are provided with. The main thing is that you know the difference between the columns (groups, families) and the rows (periods). Also, it is important to know that the chemical properties of the elements in a group (not a period) are similar (see Problem #18 below). (4) Periods are rows on the periodic table. The chemical properties of elements in a period on the table are generally not similar they change from element to element. (2.69). Which pair of elements do you expect to be most similar? Why? Give brief reasoning for your choice above. (a) N and Ni (not in same group) (b) Mo and Sn (not in same group) (c) Na and Mg (not in same group) (d) Cl and F (e) Si and P (in same group similar chemical properties)) (not in same group [group is column, not row]) NOTE: The fact that elements are arranged in the Periodic Table such that elements in the same group have similar properties is not an explanation of why they have the same properties!! Explanation of why involves hypothesis or theory, not law (this problem is about the application of the Periodic Law, not any theory). That s why I changed the wording of this question in this key (and in Mastering I can edit questions there). You just need to know that elements in the same column (group) of the periodic table have similar properties to answer this question. The Periodic Law is descriptive, not explanatory! (Note: really, it is the chemical properties that are the most similar; other properties trend in a certain way down a family, as we shall see later) PS2-10