(3 pts) 1. Which of the following is a correct chemical symbol for a monatomic species containing 13 protons, 14 neutrons, and 10 electrons?
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1 (3 pts) 1. Which of the following is a correct chemical symbol for a monatomic species containing 13 protons, 14 neutrons, and 10 electrons? (A) Si (B) Ne (C) Al 3+ (D) Al 3 (E) O 2 (3 pts) 2. The amount of the radioactive 14 C isotope present is measured to determine the age of artifacts by carbon dating. The nucleus of a 14 C atom contains. (A) 6 protons and 6 electrons (B) 7 protons and 7 neutrons (C) 7 protons and 7 electrons (D) 8 protons and 6 neutrons (E) 6 protons and 8 neutrons (3 pts) 3. Given that the formula for the ionic compound lithium selenate is Li2SeO4 what is the charge on a selenate ion? (A) 2 (B) 1 (C) 0 (D) 1+ (E) 2+ (3 pts) 4. Element X has three naturally occurring isotopes. Based on the isotopic abundances shown below, what is the identity of element X? Isotope Mass Abundance (amu) 28 X % 29 X % 30 X % (A) Mg (B) Al (C) Si (D) P (E) S (4 pts) 5. How many of the following statements are true? (i) A substance that dissolves in water to produce OH ions is a base. (ii) A weak acid is any acidic solution with a concentration < 1 M. (iii) A solution of HCl(aq) contains H3O + ions, Cl ions, and HCl molecules. (iv) When dissolved in water, NH3 acts as a weak base. (v) The reaction of a strong acid and a strong base always has H2O as a product. (A) 0 (B) 2 (C) 2 (D) 3 (E) 4 C L.S. Brown
2 NAME: (5 pts) 6. The compound shown below is citronellol, which is used in perfumes and insect repellants. What is its molar mass, rounded to the nearest whole number? HO (A) 154 g/mol (B) 156 g/mol (C) 122g/mol (D) 139 g/mol (E) 158 g/mol (5 pts) 7. What is the sum of all of the stoichiometric coefficients when the following equation is correctly balanced? (Use the smallest possible whole number coefficients. If your equation includes any coefficients of one, be sure to count those.) Ca3(PO4)2 + SiO2 + C CaSiO3 + P4 + CO 2 Ca3(PO4)2 + 6 SiO C 6 CaSiO3 + P CO (A) 25 (B) 32 (C) 16 (D) 42 (E) 35 (4 pts) 8. How many moles of oxygen atoms are in 2 moles of Al2(SO4)3? (A) 24 (B) 12 (C) 4 (D) 8 (E) 16 (5 pts) 9. A student needs to prepare ml of M NiCl2 solution for an experiment. What mass of solid NiCl2 should be used? (A) 9.32 g (B) g (C) 64.8 g (D) 6.48 g (E) g mol g L = g 1 L 1 mol 2018 L.S. Brown C3
3 (5 pts) 10. A beaker containing ml of 1.00 M solution of NaCl(aq) is left in a laboratory fume hood. After several days the concentration has risen to 1.33 M. How much water must have evaporated from the original solution? (Evaporation will remove only H2O, not any of the NaCl.) (A) 20.0 ml (B) 25.0 ml (C) 75.0 ml (D) 80.0 ml (E) 33.3 ml Concentration increases by factor of 4/3, so volume decreases by factor of 3/4. (5 pts) 11. What volume of M HCl would be needed to completely neutralize ml of M Ba(OH)2 in an acid-base reaction? (A) 6.00 ml (B) 18.0 ml (C) 32.0 ml (D) 24.0 ml (E) 12.0 ml 2 HCl + Ba(OH)2 2 H2O + BaCl L mol Ba(OH) 2 1 L 2 mol HCl 1 L = L = 32 ml 1 mol Ba(OH) mol HCl (5 pts) 12. Ammonia is formed from the reaction of H2 and N2. 3 H2(g) + N2(g) 2 NH3(g) Which of the following reactant mixtures could produce the greatest amount of product? (Assume that the reaction goes to completion with 100% yield.) (A) 1 mol H2 and 5 mol N2 (C) 2 mol H2 and 4 mol N2 (B) 5 mol H2 and 1 mol N2 (D) 3 mol H2 and 3 mol N2 (E) 4 mol H2 and 2 mol N2 This is a limiting reactant question. Moles of NH3 produced would be 2/3, 2, 4/3, 2, and 8/3, repectively for A E. C L.S. Brown
4 NAME: Use the following information to answer questions 13 & 14. Pb(NO3)2, KI, and KNO3 are all strong electrolytes. When solutions of KI and Pb(NO3)2 are mixed, they react according to the following equation. Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq) (4 pts) 13. A molecular scale illustration representing the two reactant solutions is shown below. = Pb 2+ = NO3 = K + = I Which of the following pictures best represents the expected products when the two solutions shown above are mixed? (A) (B) (C) (D) (E) (4 pts) 14. Which species act only as spectator ions in this reaction? (A) K + and I (B) K + and Pb 2+ (C) I and NO3 (D) K + and NO3 (E) Pb 2+ and I 2018 L.S. Brown C5
5 (6 pts) 15. Aluminum (Al, Z = 13, r = 2.70 g/cm 3 ) and strontium (Sr, Z = 38, r = 2.64 g/cm 3 ) have nearly the same density. Suppose you had 2 cubes, one made of pure aluminum and the other made of pure strontium, with each cube containing 1 mole of its element. Which cube would be larger, and what would the edge length of the larger cube be? (A) Al, 2.15 cm (B) Sr, 3.21 cm (C) Al, 3.16 cm (D) Sr, 5.76 cm (E) Both cubes would be the same size, 1.39 cm Molar masses are g/mol for Al and g/mol for Sr. Because the strontium is so much heavier, it will be the big cube. Use density to get volume, and then cube root to get edge length of cube g g cm 3 = cm3 and (33. 19) 1 3 = cm (6 pts) 16. On February 2 the market price for rhodium metal was $1860 per troy ounce. How many rhodium (Rh, Z = 45) atoms could you buy for $1.00? (1 troy ounce = g) (A) (B) (C) (D) (E) $1 1 oz t g 1 mol $ oz t g Rh atoms = atoms 1 mol C L.S. Brown
6 NAME: (10 pts) 17. The yellow color of corn, mangos, and egg yolks is due in part to the presence of zeaxanthin. Elemental analysis shows that zeaxanthin contains 84.45% carbon, 9.92% hydrogen, and 5.63% oxygen by mass. Another test shows that the molar mass of the compound is between 400 and 600 g/mol. Find the molecular formula for zeaxanthin. We can use the percentages to find the empirical formula, then it should turn out that only one multiple of it will be in the given mass range. Start by taking a 100 g sample and converting the masses to moles, then divide by the smallest number of moles: g C = mol C = g/mol 9.92 g H = mol H = g/mol 5.63 g O = mol O = g/mol So our empirical formula is C20H28O That would have a molar mass of ~284 g/mol. So the only multiple that will fall between 400 and 600 is to double that, giving us the molecular formula of: C40H56O L.S. Brown C7
7 (10 pts) 18. Calcium carbonate (limestone, CaCO3) dissolves in hydrochloric acid, producing water and carbon dioxide as shown in the following equation. Calcium chloride is a soluble ionic compound, so it dissociates to give calcium ions and chloride ions. Suppose g of solid CaCO3 is added to 725 ml of M HCl. Assuming that the volume of the final solution is still 725 ml, what concentration of Ca 2+ ions should be present? CaCO3(s) + 2 HCl(aq) H2O(!) + CO2(g) + CaCl2(aq) We are given the amounts of each reactant, so we need to start by finding the limiting reactant. Convert given info to moles: g CaCO3 = mol (MM = 100 g/mol) (0.725 L)(0.225 mol/l) = mol HCl From those conclude that HCl is limiting. (various ways to do this) Then calculate from the HCl: mol HCl 1 mol CaCl 2 2 mol HCl 1 mol Ca2Y 1 mol CaCl 2 = mol Ca 2Y And that number of moles is in the 725 ml volume, so molarity is: mol Ca 2+ = M Ca 2Y L If you chose the CaCl2 as the LR, you would get M Ca 2+ C L.S. Brown
8 NAME: (10 pts) 19. KCl and KBr are both ionic solids. A mixture of KCl and KBr has a mass of g. When this mixture is heated in the presence of excess Cl2, all of the KBr is converted to KCl. If the total mass of KCl present after this reaction is g, what percentage (by mass) of the original mixture was KBr? (HINT: Be sure that you understand why the mass of the sample has decreased. It may help if you write an equation for the reaction that converted the KBr to KCl.) The mass decreases because the reaction replaces a heavier Br with a lighter Cl. There are a couple of ways we could work the problem, but no matter what we do we will need to use the change in mass, which is just Dm = g g = g Maybe the easiest thing to do is to then say that if we had replaced 1 mole of Br with Cl, the change in mass would just be the difference in the molar masses: mbr mcl = = g/mol We can use those two numbers to find the number of moles of Br actually replaced: 1 mol Br g = mol Br g That s also the moles of KBr initially present, so we can convert it to mass: g KBr mol KBr = g KBr 1 mol KBr g KBr = so the sample was 34.7% KBr g total An alternative approach would be to find the final mass if the original sample had been 100% KBr. 1 mol KBr 1 mol KCl g KCl g KBr g 1 mol KBr 1 mol KCl = g So if it was 100% KBr, then we would have Dm = g g = g Divide the actual Dm of g by that to get the fraction of KBr: g = 0.347, which matches our previous result g 2018 L.S. Brown C9
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