CHE 211 ANSWER KEY -­-­ answers in bold red EXAM 1, 2011 Multiple Choice (2% each) and Problems (10% each, partial credit possible) grouped by chapter Please select the BEST answer to each multiple choice question and mark it clearly on your answer sheet. Problems should be answered in the space immediately after each. SHOW ALL WORK! Thank you. CHAPTER 1 (BASICS) 14 MC pts. + 10 problem pts. = 24 pts. total possible 1. Which of the following processes is a chemical change? a. skin cools as rubbing alcohol evaporates from it b. an ice cube melts as it sits on a hot stovetop c. flowing water moves pebbles in a steam d. a flashlight dims as the battery in it discharges 2. Which of the following molecular scenes depicts a mixture of two elements, one made up of molecules? A 3. A beaker sits on a bench, and falls to the floor and shatters when a student bumps it accidentally. When did the beaker have the highest potential energy? a. just before shattering b. just after shattering c. when on the bench d. all these have the same energy 4. How many milliliters are there in ten liters? a. 100 b. 1000 c. 100,000 d. 10,000 5. A piece of gold weighing 94.4g is placed into water in a graduated cylinder, displacing 4.89 mL. Its density is: a. 0.0518 mL/g b. 462 g-­‐mL c. 19.3 g/mL d. 1.00 g/mL 6. In the Sahara desert, an early morning temperature reading was 5.5°C. What would be the Fahrenheit value? a. 50. ° b. 82° c. 42° d. 35° 7. If you carried out this calculation, what is the properly-­‐stated answer? (17.55 + 12.133)(√385)/19.92 = a. 29.238 b. 29.24 c. 29 d. 29.2 PROBLEM Answer as requested. If calculations are required, then please show those also. Three new lab technicians are being tested to evaluate the quality of their work. They are given a set of five standard weights which are guaranteed to weigh 25.00 g ±0.01 g and asked to weigh each of them once, then report their data to their supervisor. The data sets are shown at the left. Which worker is most precise and which is most accurate? Explain what these criteria mean and how you determined your answers. Moe Larry Curly 25.06 25.14 24.98 25.07 24.89 24.97 24.95 24.91 24.99 24.99 25.08 24.97 25.04 25.02 24.96 Mean 25.02 25.01 24.97 StdDev 0.05 0.11 0.01 Curly is most precise (consistent in measurements) because his standard deviation is so small, but he is less accurate than Larry (who was not very precise at all – he had the worst std dev). Larry’s accuracy is based on being closest to the correct weight, since accuracy is how close you are to the correct answer. Moe actually had very good precision and accuracy values, but neither was better than the specific results of the other two. Which technician had the greatest random error? Which had the most problems with systematic error? Larry had the most random error – his individual values were all over the place even though his mean value turned out to be best. Curly evidently had a systematic error, since all his readings were off in the same direction (low). Perhaps he did not zero his balance or had a bad one. CHE 211 Exam 1 CHAPTER 2 (MATTER) Page 2 14 MC pts. + 10 problem pts. = 24 pts. total possible 8. Which of these would be a homogeneous mixture? a. sugar water b. distilled water 9. Which postulate of Dalton’s Atomic Theory is still accepted as originally stated? a. each element is made up of indivisible atoms b. atoms of one element cannot be turned into another c. all atoms of one element are unique and identical d. compounds are formed by combining atoms c. iron powder + sand d. methane gas 10. What did J.J. Thomson discover in his research? a. the electron b. the nucleus 11. Who discovered the structure of atoms consists of a high density positive nucleus and much empty space? a. Millikan b. Thomson c. Curie d. Rutherford 12. Which of these is in the nonmetal region of the Periodic Table? a. S b. Rh c. Pb d. Pr 13. What is the proper name for the compound Mn2S3 would be: a. manganese sulfate b. dimanganese trisulfide c. manganese (III) sulfide d. magnesium sulfate 14. What would be the molecular mass of nitrogen triiodide? a. 409 D b. 169 D c. 278 D c. the proton d. the neutron d. 395 D PROBLEM Briefly define FIVE of the following six terms as discussed in Chapter 2 of the textbook (with examples): Definitions from textbook glossary; concepts must be the same but wording may vary significantly Ionic bonding Bonding based upon the idea that oppositely-­charged ions, formed by the complete transfer of one or more electrons from one atom to another, attract each other. Usually metals lose electrons and the nonmetals gain them Covalent bonding Atoms bonding through the mutual sharing of electrons between them; the nuclei of the bonding atoms are mutually attracted to the electrons being shared. Anion A negatively-­charged ion Cation A positively-­charged ion. Polyatomic ion An ion (anion or cation) consisting of two or more covalently bonded atoms with a net charge overall Isotope Atoms of a given atomic number (one element) that have differing numbers of neutrons in their nuclei, and thus differing masses. CHE 211 Exam 1 CHAPTER 3 (STOICHIOMETRY) Page 3 16 MC pts. + 10 problem pts. = 26 pts. total possible 15. 75.0 g of Na2SO4 is _______________________moles. a. 0.630 b. 0.528 16. How many ATOMS would there be in the 75.0 g of Na2SO4 from the previous question? a. 4.5 x 1025 b. 3.2 x 1023 c. 9.6 x 1023 d. 2.2 x 1024 17. While there are different types of formulas used for covalent compounds, usually ionic compounds just use… a. structural formulas b. molecular formulas c. atomic formulas d. empirical formulas 18. When the hydrocarbon heptane, C7H16, is burned, how many moles of O2 are required per mole of heptane? a. 11 b. 7 c. 15 d. 14 19. A fairly common ore of iron is Siderite. Analysis shows this is 48.2% Fe and 10.4% C by weight (remainder O) What is the empirical formula of Siderite? a. FeCO3 b. Fe2(CO3)3 c. Fe5CO4 d. Fe2CO3 20. When the most common ore of iron (Hematite) is processed, this happens: Fe2O3 + 3C  2 Fe + 3 CO. If we mix 7500 kg of Hematite and 5000 kg of coke (a form of C made from coal) what is the limiting reactant? a. the coke (C) b. both are equally limiting c. Fe2O3 d. CO 21. If the reaction in question 20 is run in a blast furnace and we get 4890 kg of pure Fe, what is the % yield? a. 89 b. 65 c. 98 d. 93 22. One analysis method for iron ore containing FeO is to dissolve it in H2SO4 and titrate it with KMnO4 as in: 10 FeSO4 + 8 H2SO4 + 2 KMnO4  5 Fe2(SO4)3 + 8 H2O + 2 MnSO4 + K2SO4 If you start with 0.0100 mol FeO, which is all converted to FeSO4, how much 0.025 M KMnO4 will react with it? a. 400. mL b. 200.mL c. 12.5 mL d. 80.0 mL c. 1.06 d. 1.89 PROBLEM SHOW WORK, either directly by doing the calculations below, or as setups for calculator use. Combustion analysis of a biochemical substance found that 10.00 g of a carefully weighed sample, when it was burned, gave 14.60 g CO2 and 6.00 g H2O. The compound also contained O, but as usual that cannot be measured directly. Determine the empirical formula of the biochemical substance. Note tips below: Note this problem is a slightly revised, simpler version of the last one on PLTL 2 First, determine how many moles of C are in 14.60 g CO2 and how many moles H are in 6.00 g H2O: (14.60 g CO2)(1 mol/44.0 g CO2)(1 mol C/1 mol CO2) = 0.332 mol C (6.00 g H2)(1 mol/18.0 g H2)(2 mol H/1 mol H2) = 0.667 mol H Now, how many grams of C and how many grams of H are in those numbers of moles of C and H? 0.332 mol C x 12.01 g C/mol C = 3.99 g C 0.667 mol H x 1.008 g/mol H = 0.672 g H Subtract from 10.00 g to find how much of the biochemical was oxygen. How many moles of O is this? 10.00 g substance – (3.99 g C + 0.67 g H) = 5.34 g O 5.34 g O x (1 mol O/16.0 g O) = 0.334 mol O You now have determined how many moles of C, H, and O were in the biochemical. What is the simplest molar ratio of these? This is the empirical formula! Divide all three #moles by the smallest: 0.332 mol C/0.332 = 1.00 mol C 0.667 mol H/0.332 = 2.01 mol H 0.334 mol O/0.332 = 1.10 mol O This is obviously a 1:2:1 ratio of C:H:O, so the formula is CH2O Extra Credit (5 pts) If the molar mass of this substance is about 350, what is its molecular formula? (you must have found the empirical formula for this answer to count) The mass of one CH2O unit is 12 + 2 + 16 = 30, and 350/30 ≈ 12, so the formula is C12H24O12 CHE 211 Exam 1 CHAPTER 4 (REACTION TYPES) Page 4 16 MC pts. + 10 problem pts. = 26 pts. total possible 23. Which species are omitted from the NET ionic equation for AgNO3(aq) + KBr(aq)  AgBr(s) + KNO3(aq) a. nitrate and potassium b. silver and bromide c. silver and nitrate d. silver and potassium 24. Which is most likely to NOT be a water-­‐soluble ionic substance? a. Na3PO4 b. NH4ClO4 c. Pb(NO3)2 25. Which statement is NOT true of typical strong acids? a. they are ionic compounds in their pure form b. they tend to be nearly 100% ionized in water c. they are covalent compounds of hydrogen d. they contain oxyanions or are hydrohalic species 26. 25.00 mL of a NaOH solution required 12.36 mL of 0.250 M HCl to neutralize it, so the NaOH concentration is a. 0.1236 M b. 0.506 M c. 0.5057M d. 0.124 M 27. In 2 MnO2 + 2 H2SO4  2 MnSO4 + O2 + 2 H2O, what element gets oxidized? What element gets reduced? a. O ; Mn b. H ; O c. Mn ; S d. S ; H 28. NbOCl3 is covalent, having all the other atoms bonded to the Niobium atom. What is Nb’s oxidation state? a. +3 b. +5 c. +4 d. +6 29. Many reactions of elements are redox reactions; which of these involves oxidation of a nonmetal element? a. O2 + 2 Ca  2 CaO b. Sn + Cl2  SnCl2 c. Na2O + H2O  2 NaOH d. 2 NaBr + F2  2 NaF + Br2 30. The term “double displacement” describes a number of reactions. What is another name for this type change? a. redox b. metathesis c. precipitation d. acid-­‐base reaction d. Fe(OH)3 PROBLEM BALANCE TWO of the following redox reactions. SHOW OXIDATION STATE CHANGES that occur as part of your process of determining the proper balance. For EXTRA CREDIT, BALANCE ALL THREE REACTIONS. The first reaction is from PLTL 3; the others from the Student Study Guide of the textbook KMnO4 + H2C2O4 + HCl  MnCl2 + KCl + H2O + CO2 Oxidation state changes: Mn+7  Mn+2 so it gains five electrons TWO C in oxalate go from C+3 to C+4, so they lose two electrons This means the ratio of permanganate to oxalate must be 2:5 for 10 electrons 2 KMnO4 + 5 H2C2O4 + 6 HCl  2 MnCl2 + 2 KCl + 8 H2O + 10 CO2 OCl– + I– + H3O+  Cl– + I2 + H2O Oxidation state changes here are Cl+1  Cl–1 requiring two electrons and 2 I–1  I2, which will release two elecrons Thus the electrons are in balance as long as we have 2 I and one Cl involved OCl– + 2 I– + 2 H3O+  Cl– + I2 + 3 H2O Sn + NaOH  Na2SnO2 + H2 In this sort of unusual reaction, Sn°  Sn2+, releasing two electrons, and 2 H goes from H+1 to H°, requiring two electrons Again, electrons are in balance if we have two H react with one Sn… Sn + 2 NaOH  Na2SnO2 + H2