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Prentice Hall Chemistry (c) 2005

Prentice Hall Chemistry (c) 2005. Section Assessment Answers Chapter 9. By Daniel R. Barnes Init: < 11/20/2008. Say my name, say my name. 3. Group A metal ions have positive charges equal to their group numbers. 1A metals have +1 charges, 2A metals have +2 charges, etc.

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Prentice Hall Chemistry (c) 2005

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  1. Prentice Hall Chemistry(c) 2005 Section Assessment Answers Chapter 9 By Daniel R. Barnes Init: < 11/20/2008 Say my name, say my name

  2. 3. Group A metal ions have positive charges equal to their group numbers. 1A metals have +1 charges, 2A metals have +2 charges, etc.. Group A nonmetals have negative charges. The size of nonmetal charges can be calculated by subtracting 8 from their group number. 7A nonmetals have -1 charges (7 – 8 = -1), 6A nonmetals have -2 charges (6 – 8 = -2), etc.. 9.1 Section Assessment

  3. 4. Transition metal ions have positive charges that depend on how many electrons they lose. (Wow. That’s not news. ALL metals’ charges depend on how many electrons they lose.) EXAMPLE: Iron atoms lose either 2 or 3 electrons each, so iron ions are either 2+ or 3+. How do you know how many electrons a transition metal is going to lose? A transition metal’s group number (the one that ends with B) might tell you, but maybe not. 9.1 Section Assessment For instance, zinc is in group 2B, and it forms 2+ ions. That’s nice. However, gold is in group 1B, but it forms both 1+ and 3+ ions, or, at least ends up in the +1 or +3 oxidation state, anyway, with +3 being more common than +1. Then there’s cobalt, in group 8B, but does it become +8? Not a chance. +2 or +3 are the possible charges for cobalt.

  4. 5. Polyatomic ions almost always end in . . . . . . “ite” or “ate”. Look at Table 9.3 on page 257 if you don’t believe me. ClO- = hypochlorite CO32- = carbonate ClO2- = chlorite NO2- = nitrite ClO3- = chlorate ClO4- = perchlorate 9.1 Section Assessment NO3- = nitrate SO32- = sulfite NH4+ = ammonium SO42- = sulfate Don't worry. You won't have to know these on the tests.

  5. 6. Group 1A ions always have a +1 charge. Group 3A (aluminum) ions will have a +3 charge. Group 5A ions will tend to have a charge of -3. Why? Group 1A atoms have one valence electron each, and they tend to lose that single electron, resulting in a +1 charge. This sheds the entire outer shell, causing the shell beneath it, which is already full, to become the new outer shell. Atoms love having full outer shells. 9.1 Section Assessment Aluminum (group 3A) could either lose its 3 valence electrons or gain 5 (to get a full “octet” of 8), but losing 3 is less drastic than gaining 5, so it loses 3, resulting in a +3 charge. Group 5A atoms could either lose all 5 of their valence electrons or gain 3 to get an octet. It’s less drastic to gain 3, so that’s what they do. Gaining 3 e-’s gives you a -3 charge.

  6. 7. A polyatomic ion is made of two or more atoms, whereas a monatomic ion is just a single atom. A polyatomic ion is a molecule with a charge. 9.1 Section Assessment

  7. 8. PROMPT ION SYMBOL KIND of ION NAME of ION a. potassium K+ cation “potassium ion” b. oxygen O2- anion “oxide ion” c. tin (2e- lost) Sn2+ cation “tin (II) ion” 9.1 Section Assessment d. bromine Br- anion “bromide ion” e. beryllium Be2+ cation “beryllium ion” f. cobalt (3e- lost) Co3+ cation “cobalt (III) ion”

  8. 9. ION NAME SYMBOL or FORMULA a. ammonium ion NH4+ b. tin (II) ion Sn2+ 9.1 Section Assessment c. chromate ion CrO42- d. nitrate ion NO3-

  9. 14. When writing the name of a binary ionic compound, write the name of the cation (the + ion, usually a metal) first, followed by the name of the anion (the – ion, usually a nonmetal) For example: Salt is made of equal numbers of Na+ & Cl- ions. Salt is known as “sodium chloride”. 9.2 Section Assessment “Sodium” comes first because sodium (Na) is positive. “Chloride” comes last because it’s negative. Its name is also changed, because it’s the anion.

  10. 15. When writing the formula of a binary ionic compound, write the symbol of the cation first, followed by the symbol of the anion. After each symbol, write whatever subscripts you need to make the compound electrically neutral (balanced). There should be just as many +’s as –’s. For example: The formula for magnesium nitride is . . . . . . Mg3N2. Mg is from 2A, so Mg loses 2e- to become Mg2+ 9.2 Section Assessment N is from 5A, so N gains 3e- to become N3- Mg2+ N3- 2+ & 3- isn’t neutral yet. We need to add more ions until the combination is balanced. Mg2+ N3- Mg2+ Notice how magnesium’s charge became nitrogen’s subscript. CRISS CROSS! Nitrogen’s charge also became magnesium’s subscript.

  11. 16. When writing the name of a compound with polyatomic ions, as always, write the name of the cation first, followed by the name of the anion. When writing the formula for a compound with polyatomic ions, as always, write the symbol of the cation followed by the formula of the anion. As always, write subscripts where needed to make sure that the compound is balanced (neutral)(+’s = -’s). 9.2 Section Assessment

  12. 17. b. cesium sulfide: Cs2S 17. a. beryllium chloride: BeCl2 Cs+ S2- Be2+ Cl- Cs+ Cl- 17. c. sodium iodide: NaI 17. d. strontium oxide: SrO 9.2 Section Assessment Na+ I- Sr2+ O2-

  13. 18. a. Chromium (III) nitrite: Cr(NO2)3 Cr3+ NO2- NO2- NO2- 9.2 Section Assessment 18. b. sodium perchlorate: NaClO4 Na+ ClO4-

  14. 18. c. Magnesium hydrogen carbonate: Mg(HCO3)2 Mg2+ HCO3- HCO3- 9.2 Section Assessment 18. d. Calcium acetate: Ca(C2H3O2)2 Ca2+ C2H3O2- C2H3O2-

  15. 19. a. Mg2(SO4)3 = correct / incorrect? Mg2+ SO42- You only need one of each ion to get balanced. MgSO4 is the correct formula! 9.2 Section Assessment

  16. 19. b. Rb3As = correct / incorrect? Rb+ As3- Rb+ Rb+ 9.2 Section Assessment +1 does not balance out against +3. We need more +’s. Looks good. NO HATS ON CAMPUS!

  17. 19. c. BeCl3 = correct / incorrect? Be2+ Cl- Cl- We need more minuses to balance out the 2+ 9.2 Section Assessment BeCl2 is the correct formula!

  18. 19. d. NaF = correct / incorrect? Na+ F- The formula is just fine. 9.2 Section Assessment

  19. 20. Prefixes in the name of a molecular compound tell you how many atoms of an element there are in a molecule of that compound. The prefix that means “one atom” is . . . “mono” The prefix that means “two atoms” is . . . “di” 9.3 Section Assessment The prefix that means “three atoms” is . . . “tri” The prefix that means “four atoms” is . . . “tetra”

  20. 21. To write the formula for a binary molecular compound . . . . . . write the symbols of each element, with a subscript AFTER each element that says how many atoms of that element there are in a molecule of the compound. 22.a. NCl3 = nitrogen trichloride 22.b. BCl3 = boron trichloride 9.3 Section Assessment 22.c. NI3 = nitrogen triiodide 22.d. SO3 = sulfur trioxide 22.e. N2H4 = dinitrogen tetrahydride 22.f. N2O3 = dinitrogen trioxide

  21. 23.a. CS2 = carbon disulfide 23.b. carbon tetrabromide = CBr4 23.c. Cl2O7 = dichlorine heptoxide 23.d. Diphosphorus trioxide = P2O3 9.3 Section Assessment

  22. 24.a. phosphorus pentachloride = PCl5 24.b. iodine heptafluoride = IF7 24.c. chlorine trifluoride = ClF3 24.d. iodine dioxide = IO2 25. The proper name of SiCl4 is “silicon tetroxide”. 9.3 Section Assessment There is no “mono” before the “silicon” because if there’s only one atom of the first element, you just don’t say “mono”. If there’s only one atom of the second element, you do say “mono” in front of that one. Furthermore, since there are four chlorine atoms, not three, it’s it’s “tetrachloride”, not “trichloride”.

  23. 26. The rules for naming acids are the following: (1) When the anion ends in –ide, the acid name begins with the prefix hydro-. The stem of the anion has the suffix –ic and is followed by the word acid. (2) When the anion name ends in –ite, the acid name is the stem of the anion with the suffix –ous, followed by the word acid. (3) When the anion name ends in –ate, the acid name is the stem of the anion with the suffix –ic, followed by the word acid. 9.4 Section Assessment 27. The rules for writing the names of acids are used in reverse for writing the formulas of acids. 28. The name of a base is written by writing the name of the cation followed by the name of the anion.

  24. 29.a. HNO2 = nitric acid 29.b. HMnO4 = permanganic acid 29.c. HCN = hydrocyanic acid 29.d. H2S = hydrosulfuric acid (aka “hydrogen sulfide”) 30.a. LiOH = lithium hydroxide 9.4 Section Assessment 30.b. Pb(OH)2 = lead (II) hydroxide 30.c. Mg(OH)2 = magnesium hydroxide 30.d. Al(OH)3 = aluminum hydroxide

  25. 31.a. Ba(OH)2 = base (barium hydroxide) 31.b. HClO4 = acid (perchloric acid) 31.c. Fe(OH)3 = base (iron (III) hydroxide) 31.d. KOH = base (potassium hydroxide) 9.4 Section Assessment 32.a. carbonic acid = H2CO3 32.b. sulfurous acid = H2SO3 32.c. iron (III) hydroxide = Fe(OH)3 32.d. strontium hydroxide = Sr(OH)2

  26. 33. The element that generally appears in all acids is . . . hydrogen (H) The ion that generally appears in all bases is . . . hydroxide (OH) 9.4 Section Assessment

  27. 35. Two laws that describe how chemical compounds form are the law of definite proportions and the law of multiple proportions. 36. To use a flowchart, follow the arrows and answer the questions on the flowchart. 9.5 Section Assessment 37. Four guidelines used when writing the formulas of chemical compounds are:

  28. 37. Four guidelines used when writing the formulas of chemical compounds are: (1) An –ide ending generally indicates a binary compound. (2) An –ite or –ate ending means a polyatomic ion that inculdes oxygen in its formula. (3) Prefixes in a name generally indicate that a compound is molecular. 9.5 Section Assessment (4) A roman numeral after the name of a cation shows the ionic charge of the cation.

  29. 38. Compound A: 32.10 g Cu & 17.90 g Cl Compound B: 23.64 g Cu & 26.37 g Cl (64.2% Cu) Are these two compounds the same? Let’s calculate the %’s of each element in both compounds to find out. 32.10 g Cu 32.10 g Cu x 100% = x 100% 32.10 g Cu + 17.90 g Cl 50 g compound A 9.5 Section Assessment = 0.642 x 100% = 64.2% Cu in compound A

  30. 38. Compound A: 32.10 g Cu & 17.90 g Cl Compound B: 23.64 g Cu & 26.37 g Cl (64.2% Cu) (35.8% Cl) Are these two compounds the same? Let’s calculate the %’s of each element in both compounds to find out. 17.90 g Cl 17.90 g Cl x 100% = x 100% 32.10 g Cu + 17.90 g Cl 50 g compound A 9.5 Section Assessment = 0.358 x 100% = 35.8% Cl in compound A

  31. 38. Compound A: 32.10 g Cu & 17.90 g Cl Compound B: 23.64 g Cu & 26.37 g Cl (64.2% Cu) (35.8% Cl) (47.28% Cu) Are these two compounds the same? Let’s calculate the %’s of each element in both compounds to find out. 23.64 g Cu 23.64 g Cu x 100% = x 100% 23.64 g Cu + 26.37 g Cl 50 g compound B 9.5 Section Assessment = 0.4728 x 100% = 47.28% Cu in compound B

  32. 38. Compound A: 32.10 g Cu & 17.90 g Cl Compound B: 23.64 g Cu & 26.37 g Cl (64.2% Cu) (35.8% Cl) (47.28% Cu) (52.74% Cl) Are these two compounds the same? Let’s calculate the %’s of each element in both compounds to find out. 26.37 g Cl 26.37 g Cl x 100% = x 100% 23.64 g Cu + 26.37 g Cl 50 g compound B 9.5 Section Assessment = 0.5274 x 100% = 52.74% Cl in compound B

  33. 38. Compound A: 32.10 g Cu & 17.90 g Cl Compound B: 23.64 g Cu & 26.37 g Cl (64.2% Cu) (35.8% Cl) (47.28% Cu) (52.74% Cl) These two compounds are not the same. The elemental %’s in compound A are not the same as they are in compound B. The book asks you to calculate some ratio stuff and I’m not really sure what they mean, but I’m going to do some ratio stuff and see if I get the answer they give in the teacher’s edition. 9.5 Section Assessment 32.10 g Cu The Cu-to-Cl ratio in compound A is bigger than in compound B A: = 1.783 g Cu/g Cl 17.90 g Cl 23.64 g Cu B: If you divide 1.783 by 0.8965, you get . . . = 0.8965 g Cu/g Cl 26.37 g Cl 2.000 The ratio of the ratios (!?!) is 2:1.

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