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PRACTICE EXAM 1-C CHEMISTRY E-1ax Problems from this exam will be discussed in sections prior to the exam. LAST NAME: ________________________________________ FIRST NAME: ________________________________________ You should give yourself 2 hours to complete this exam NOTES: 1. This exam has 6 pages with 6 problems, plus the cover sheet, useful information, periodic table, and scrap paper. 2. Note the point values of each exam question, and consider where you can best spend your time. 3. Answer all questions on the exam sheets. Put your final answers in the boxes provided. If you must use the back, then please indicate that clearly! 4. Do not use red pen! 5. For full credit, show all calculations; this also helps us award partial credit. 6. All numerical answers must include correct units for full credit. 1. ____________ 2. ____________ 3. ____________ 4. ____________ 5. ____________ 6. ____________ TOTAL: ___________ / 100 1 1. Potassium is an element that is biologically necessary for the proper functioning of nerve and muscle cells in the human body. A shortage of potassium, called hypokalemia, can result in abnormal heart rhythms and in some cases can be life-threatening. a) The most abundant natural isotope of potassium is 39 19 K . Determine the number of protons (p), neutrons (n), and electrons (e) in one atom of this isotope of potassium. (3 pts) # of p = b) # of n = Most people are able to maintain normal potassium levels simply through a standard diet because many different foods are rich in potassium. For example, one average banana contains 422 mg of potassium. Determine the number of potassium atoms in one average banana. (6 pts) number of K atoms = c) atoms Potassium levels in the blood are measured in milliequivalents (mEq) of potassium per liter of blood, or “mEq/L”. A normal-range potassium level is around 4.3 mEq/L. Determine the mass of potassium contained in 1.00 pint of blood with a potassium level of 4.3 mEq/L. (7 pts) Some Useful Conversions: 1 Eq = 1000 mEq 1 Eq of potassium = 1 mole of potassium 1 liter = 2.11 pints mass of potassium = 1 # of e = grams ( ______ / 16 pts) 2 2. Compound X contains only the elements C, H, and O. a) When 6.00 grams of Compound X are completely combusted in excess oxygen, 8.80 g of CO2 and 3.60 g of H2O are collected. Determine the empirical formula of Compound X. (6 pts) empirical formula: b) When 6.00 grams of Compound X are titrated with NaOH, it is determined that Compound X is a monoprotic acid, and 44.4 mL of 1.50-molar NaOH are required to completely neutralize the sample. Determine the molecular formula of Compound X. (6 pts) molecular formula: c) 2 Write a complete, balanced equation for the combustion of liquid Compound X. (Please include state symbols such as (s), (aq), etc.) (4 pts) ( ______ / 16 pts) 3 3. A hydrated salt is an ionic substance that has a specific number of water molecules enclosed within its crystal structure. One example of a hydrated salt is CoCl2·6H2O. As indicated in this chemical formula, this hydrated salt consists of the ionic solid cobalt (II) chloride with 6 molecules of water trapped in the crystal structure. a) When solid CoCl2·6H2O is heated, all of the water is driven off as gaseous water, and pure solid cobalt (II) chloride (without any water of hydration) will remain. In the box below, write a complete balanced equation for the complete dehydration of solid CoCl2·6H2O upon heating. (Please include state symbols such as (s), (aq), etc.) (4 pts) b) You are given 3.196 grams of solid CoCl2·6H2O. How many moles of hydrated salt is this? (Hint: Don’t forget to account for the mass of water in determining the molar mass!) (4 pts) number of moles = c) You heat this hydrated salt to drive off the water, but unfortunately you do not heat it long enough, and the reaction does not go to completion. The final mass of solid substances after the reaction is 1.854 grams. (Note that this includes the dehydrated CoCl2 product as well as some hydrated salt that has not been completely dehydrated!) Determine the number of moles of water that were removed from the original hydrated salt. (4 pts) moles of H2O = d) Determine the percent yield for this reaction by using a ratio of the moles of water actually removed from the hydrated salt to the total moles of water originally present in the hydrated salt. (4 pts) percent yield = 3 ( ______ / 16 pts) 4 4. a) Write the chemical formula for each of the following species. (1 pt each) Barium Thiocyanate Copper (I) Cyanide Aluminum Sulfide Sulfur Trioxide b) Write an acceptable chemical name for each of the following. (1 pt each) FeCO3 NH3 NO c) Write the oxidation state of each atom in the box above the atom. (1 pt per box) H2PO4 d) – MgH2 For parts (i) and (ii) below, consider the following balanced chemical reaction and circle the best answer to each question: Si + 2 Cl2 ! SiCl4 3 moles of Si are mixed with 4 moles of Cl2 and this reaction proceeds to completion. i) How many moles of SiCl4 will be formed? (2 pts) a) 2 moles b) 3 moles c) 4 moles d) 5 moles e) 7 moles d) 2 moles e) 3 moles ii) How many moles of Si will remain unreacted? (2 pts) a) 0 moles e) 4 b) 1 mole c) 1.5 moles In the space below, write a balanced net-ionic equation for the complete neutralization of sulfuric acid with potassium hydroxide. (You do not need to include state symbols such as (s), (aq), etc.) (You may wish to write the complete reaction and do some work on scrap paper, but we will only grade the net-ionic reaction written in the box below.) (2 pts) ( ______ / 18 pts) 5 5. The following two solutions are mixed together: 450.0 mL of 0.200-molar silver nitrate, AgNO3 350.0 mL of 0.200-molar sodium phosphate, Na3PO4 A yellow precipitate of silver phosphate, Ag3PO4, is formed in a double displacement reaction. a) Write a complete, balanced equation, and then a net ionic equation for this chemical reaction. (Please include state symbols such as (s), (aq), etc.) (6 pts) Complete: Net Ionic: b) Calculate the mass of silver phosphate that would be formed, assuming the reaction goes to completion, and determine the molar concentration of all ions in the resulting solution. (Note: Assuming that the reaction goes to completion, some of the concentrations may be effectively zero.) (12 pts) mass of Ag3PO4 formed = [Ag+] = [Na+] = 5 [NO3–] = [PO43–] = ( ______ / 18 pts) 6 6. You are given a mixture of barium hydroxide, Ba(OH)2, and strontium hydroxide, Sr(OH)2. You dissolve this mixture in water and titrate it with hydrochloric acid. Complete neutralization of the mixture requires 96.0 mL of 1.50 M HCl. (Note that Sr(OH)2 is a strong base that reacts with HCl in the same manner as Ba(OH)2.) After performing the titration, you add excess sodium sulfate to the resulting solution, completely precipitating solid barium sulfate, BaSO4, and solid strontium sulfate, SrSO4. A total of 15.31 grams of precipitate is collected. Determine the number of moles of Ba(OH)2 and Sr(OH)2 that were in the original mixture. (16 pts) number of moles of Ba(OH)2 = moles number of moles of Sr(OH)2 = moles 6 ( ______ / 16 pts) Scrap Paper Nothing on this page will be graded unless you clearly indicate on a specific problem that additional work is located here. NAME:_____________________________ Useful Information This page will NOT be collected after the exam. We will NOT grade anything written on this page. Avogadro’s Number = 6.02 ! 1023 Density of water at 25°C = 1.00 g/mL Unit Conversions: 1 mL = 1 cm3 1 foot = 12 inches 1 inch = 2.54 cm 1 pound = 453.6 grams 27 Co 28 Ni 29 Cu 30 Zn Fr Cs 227.03 Ac 138.91 89 La Y 88.91 57 (261) Rf 178.49 104 Hf Zr 91.22 72 47.88 40 Actinide series Lanthanide series Ra Ba Sr 44.96 39 231.04 Pa 232.04 Th 140.91 91 140.12 90 59 Pr 58 Ce (263) 183.85 [106] W Mo 95.94 74 52.00 42 (262) Ha 180.95 105 Ta Nb 92.91 73 50.94 41 238.03 U 144.24 92 60 Nd (262) 186.21 [107] Re Tc (98) 75 54.94 43 237.05 Np (145) 93 61 Pm (265) 190.20 [108] Os Ru 101.07 76 55.85 44 (244) Pu 150.36 94 62 Sm (266) 192.22 [109] Ir Rh 102.91 77 58.93 45 (243) Am 151.96 95 63 Eu 195.08 Pt Pd 106.42 78 58.69 46 (247) Cm 157.25 96 64 Gd 196.97 Au Ag 107.87 79 63.55 47 (247) Bk 158.93 97 65 Tb 200.59 Hg Cd 112.41 80 65.39 48 (251) Cf 162.50 98 66 Dy 204.38 Tl In 114.82 81 69.72 49 Ga (252) Es 164.93 99 67 Ho 207.20 Pb Sn 118.71 82 72.61 50 Ge (257) Fm 167.26 100 68 Er 208.98 Bi Sb 121.76 83 74.92 51 As (258) Md 168.93 101 69 Tm (209) Po Te 127.60 84 78.96 52 Se (259) No 173.04 102 70 Yb (210) At I 126.91 85 79.90 53 Br 35.45 35 Cl 226.03 26 Fe S 32.07 34 19.00 17 (223) 25 Mn P 30.97 33 16.00 16 137.33 88 24 Cr Si 28.09 32 14.01 15 132.91 87 V 23 Al 26.98 31 12.01 14 87.62 56 Rb Ca Mg 10.81 13 (260) Lr 174.97 103 71 Lu (222) Rn Xe 131.29 86 83.80 54 Kr Ar 39.95 36 20.18 18 Ne 85.47 55 Ti 22 F 40.08 38 21 Sc O 39.10 37 K Na Li 24.31 20 N 9 22.99 19 C 8 9.012 12 B 7 6.941 11 Be 6 4.003 10 5 4 1.008 3 2 He H 1 PERIODIC TABLE OF THE ELEMENTS