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(For grading) Chem 107 - Hughbanks Exam 1 - Spring 2011 Name (Print) SOLUTIONS UIN # Section 503 Exam 1, Version # B Scores 1 2 3 4 5 6 7 8 9 10 11 Tot. /7 /7 /7 /7 /7 /7 /12 /10 /12 /12 /12 /100 On the last page of this exam, you’ve been given a periodic table and some physical constants. You’ll probably want to tear that page off the to use during the exam – you don’t need to turn it in with the rest of the exam. The exam contains 11 problems, with 6 numbered pages. You have the full 75 minutes to complete the exam. Please show ALL your work as clearly as possible – this will help us award you partial credit if appropriate. Even correct answers without supporting work may not receive credit. You may use an approved calculator for the exam, one without extensive programmable capabilities or the ability to store alphanumeric information. Print your name above, provide your UIN number, and sign the honor code statement below. On my honor as an Aggie, I will neither give nor receive unauthorized assistance on this exam. SIGNATURE: (1) (7 points) Which of the following is not a strong electrolyte? (a) KBr (b) CH3NH2 (c) H2SO4 (d) Ba(OH)2 (e) NaOH Ans. 1____ b_____ (2) (7 points) Consider the position of the elements rubidium (Rb), chlorine (Cl), and barium (Ba) in the periodic table. Which one of the following full sets of compounds is the most likely to be easily available (or exist at all)? (a) Ba2Cl, RbCl, RbCl2 (d) Cl2, Ba2Cl, RbCl (b) Cl2, BaCl, Rb2Cl (e) Cl2, BaCl2, RbCl (c) BaCl, RbCl, RbCl2 Ans. 2_____ e____ (3) (7 points) Calcium carbide, CaC2, is an important preliminary chemical for industries producing synthetic fabrics and plastics. CaC2 may be produced by heating calcium oxide with coke (a source of carbon). CaC2 is be reacted with nitrogen, N2 at 1000 – 1200 ˚C to produce another important chemical precursor called calcium cyanamide, CaCN2: CaC2(s) + N2(g) → CaCN2(s) + C(s) What is the amount of CaCN2 that can be produced from the reaction of excess nitrogen gas and 12.8 g of calcium carbide? Assume that the percent yield in this process is 85%. (a) 18.8 g Ans. 3_____ b____ (b) 13.6 g (c) 14.2 g (d) 16.0 g (e) 12.1 g (4) (7 points) Which of the following choices correctly completes this statement: A cation with a charge of +2 that has an atomic number of 12 and a mass number of 25, (a) has the same number of electrons as a Si atom. (b) has 10 electrons. (c) has the same number of protons as an Ne atom. (d) has 12 neutrons. (e) is a Si2+ ion. Ans. 4___ b ___ (5) (7 points) The following full sets of potential compounds are made up of common cations and anions. Which full sets are most likely to actually exist (all 3 compounds in a set must be sensible to qualify)? There may be more than one correct choice; enter ALL the correct letters, in alphabetical order. (a) NH4(SO4)2, ZnCO3, CaCN (b) NH4PO4, NH4Cl, KNO3 (d) NH4SO4, K2CO3, Ca(OH)2 (c) Zn3(PO4)2, MgSO4, KNO3 (e) (NH4)2SO4, K2CO3, Ca3(PO4)2 Ans. 5___ c, e___ (6) (7 points) Which of the following quantities are conserved in all chemical reactions? There may be more than one correct choice; enter ALL the correct letters, in alphabetical order. (a) Number of atoms of each element (b) Volume (c) Total number of atoms (d) Number of molecules (e) Mass Ans. 6__ a, c, e_ Questions 7-11 – show all work (58 pts) (7) (12 points) The combustion of heptane with oxygen is a well-known process. (a) (5 points) Write a balanced equation for the reaction of heptane, C7H16, with oxygen, O2, forming the products carbon dioxide, CO2, and water, H2O. C7H16 + 11 O2 → 7 CO2 + 8 H2O (b) (7 points) What mass of carbon dioxide will be produced by the complete reaction of 8.93 grams of heptane? (8.93 g C7H16)/(100.198 g/mol) = 0.08913 mol (× 7 mol CO2/mol C7H16) ⇒ 0.62386 mol CO2 produced (0.62386 mol CO2)(44.01 g/mol) = 27.45 g CO2 produced (8) (10 points) Boron has two naturally occurring isotopes, 10B and 11B. Use the data in the table given to calculate the missing isotopic molar mass of 10B. Give an answer with 4-digit precision. The molar mass of naturally occurring boron is 10.811 g/mol. Isotope 10 11 B B Isotopic Molar Mass (g/mol) 11.00931 Abundance(%) 19.78 80.22 Let x = isotopic molar mass of 10B (in g/mol) 0.1978x + 0.8022(11.00931) = 10.811 ⇒ x = + 0.8022(11.00931) = (1/0.1978)(10.811 – 8.83167) x = 10.01 g/mol (9) (12 points) The molecule shown below is one of many used in devices ranging from wristwatches to displays for laptop computers. These molecules form highly ordered “liquid crystals”. Electric fields are used to control the extent of ordering of the molecules, thus allowing segments of the liquid to be switched between transparent and opaque forms. (a) (7 points) Give the chemical formula for this liquid crystal molecule. N O H H H3C H C O C C C C H C H H C N C H H C C C C C CH2 H2C H CH2 CH3 C18H21NO (b) (5 points) How many carbon atoms are there in 1.0 g of this compound? Total molar mass = 267.358 g ⇒ (1.0 g)/(267.358 g/mol) = 3.74 × 10–3 mol in 1 g (18 C atoms/molecule)(3.74 × 10–3 mol)(6.022 × 1023/molecules/mol) = 4.054 × 1022 C atoms (10) (12 points) Cryolite, Na3AlF6, is used in the commercial production of aluminum metal from bauxite ore (Al2O3). Cryolite itself is produced by the following (unbalanced) reaction: . 6 NaOH + 1 Al2O3 + 12 HF → 2 Na3AlF6 + 9 H2 O (a) (5 points) Using the blanks provided, balance this equation with the smallest possible integer coefficients (Hint: One of the reactants has a coefficient of 1.) (b) (7 points) A mixture containing 800.0 kg of NaOH, 300.0 kg of Al2O3, and 600.0 kg of HF is heated to 950 ˚C and reacts to completion. What is the maximum mass of Na3AlF6 formed? ratio (800.0 kg NaOH)/(0.040 kg/mol) = 20000 mol NaOH 6.797 (300.0 kg Al2O3)/(0.10196 kg/mol) = 2942.3 mol Al2O3 1 (600.0 kg HF)/(0.02001 kg/mol) = 29985 mol HF 10.19 Compared to the quantity of Al2O3, there is an excess of NaOH and there is too little HF. Therefore, HF is the limiting reagent ⇒ only (2/12) × (29985 mol) Na3AlF6 can be formed. max. # moles of Na3AlF6 is 4997.5 ⇒ (4997.5 mol) × (0.20995 kg/mol) = 1049 kg (11) (12 points) One compound found in decaying flesh goes by the descriptive name cadaverene. Analysis of cadaverene shows it to contain 58.77% carbon, 13.81% hydrogen, and 27.42% nitrogen by mass. Reactions with strong acids show that cadaverene molecules each contain two basic –NH2 groups (it is a diamine) that can react with HCl solutions and pick up protons (H+) that convert these into two –NH3+ groups. Determine the molecular formula of cadaverene, assuming the –NH2 groups contain all the nitrogen. In a 100.0 g sample of cadaverene, we have 58.77g C, 13.81g H, and 27.42g N ratio 2.5 (58.77g C)/(12.01 g/mol C) = 4.893 mol C (13.81g H)/(1.008 g/mol H) = 13.70 mol H ⇒ 7.0 (27.42g N)/(14.01 g/mol N) = 1.957 mol N 1.0 ratio 5 ⇒ 14 2 Other data shows there are two nitrogen atoms, so formula is C5H14N2 (The actual structure is H2N NH2 .)