CHM 211 Test 1 Name: ____________________ Fall 2016 (Please print) Multiple Choice: (4 points each. Put answers in left margin as capital letters.) 1. Which of the following sets of measurements is most accurate for 5.00 g reference weight? A) 4.92 g, 5.00 g, 5.02 g C) 4.93 g, 5.01 g, 5.07 g E) 4.90 g, 4.95 g, 5.00 g B) 4.97 g, 4.98 g, 4.99 g D) 5.03 g, 5.03 g, 5.04 g 2. How many significant figures are in the answer to the following problem? (35 + 74) x 1226? A) 1 B) 2 C) 3 D) 4 E) 5 3. Which of the following is a pure substance? A) air B) concrete C) Jello® D) neon E) saltwater 4. A block has a temperature of 601 ºC. What is its temperature in Kelvins? A) 316 B) 328 C) 601 D) 874 F) 1110 5. Which of the following is an alkaline earth element? A) lithium C) oxygen E) neon B) beryllium D) fluorine 6. The correct number of protons, neutrons, and electrons in Se4+ (Se-79) is: A) 30p, 45n, 34e C) 34p, 45n, 34e E) 79p, 34n, 34e B) 34p, 45n, 30e D) 79p, 34n, 30e 7. Which of the following is least likely to represent a real compound? A) Al(C2H3O2)3 (Al(CH3CO2)3) C) K2SO4 E) NH4Cl2 B) CaBr2 D) Mg3(PO4)2 8. Which of the following symbols is used to indicate the addition of light to a chemical reaction? A) (g) B) h C) h D) (s) E) 9. Which of the following is an example of a decomposition reaction? A) N2 + 3 H2 2 NH3 D) C3H8 + 5 O2 3 CO2 + 4 H2O B) CaCO3 CaO + CO2 E) None of these is a decomposition reaction C) AgNO3 + NaCl AgCl + NaNO3 Discussion Questions: (You must show your work to receive credit.) 1. Define: (12 points) energy – the capacity to transfer heat or do work element – pure substances that consist of only one type of atom and cannot be broken into simpler substances stoichiometry – the quantity relationship between reacting chemical species 2. Magnesium is used in automobile wheels because it is “lighter” than steel. What is a more scientifically correct statement of this? (6 points) Magnesium is used in automobile wheels because it is less dense than steel. 3. Complete the following: (11 points) WO3 (s) + 3 H2 (g) W(s) + 3 H2O(g) When solutions of magnesium bromide and lithium carbonate are mixed, solid magnesium carbonate forms and lithium bromide remains in solution. MgBr2 (aq) + Li2CO3 (aq) MgCO3 (s) + 2 LiBr(aq) 4. For the following, give the name or formula where appropriate: (15 points) NiF2 – nickel(II) fluoride lithium sulfate – Li2SO4 N2O4 – dinitrogen tetroxide phosphorus trichloride – PCl3 HNO3 – nitric acid th 5. In early to mid-20 century warships, smoke screens were created by pouring TiCl4 into the ocean. It would react with the water to produce clouds of TiO2, which is now used as the base for house paint. If 454 g of TiCl4 are mixed with 454 g of water, how much TiO2 will be produced? How much of which reagent will be left over? (10 points) TiCl4 () + 2 H2O () TiO2 (s) + 4 HCl(aq) a) Determine the limiting reagent by first calculating the number of moles of each substance present. 1 molTiCl4 molTiCl4 = (454 gTiCl4) = 2.39 molTiCl4 189.7 gTiCl4 1 molH2O molH2O = (454 gH2O = 25.2 molH2O 18.01 g H2O Now determine how much water is required to use up all of the titanium(IV) chloride. 2 molH2O molH2O(required) = (2.39 molTiCl4) = 4.78 molH2O 1 molTiCl4 But you have 24.2 molH2O, which is more than enough so titanium(IV) chloride is limiting. 1 molTiCl4 1 molTiO2 79.87 gTiO2 mass|TiO2 = (454 gTiCl4) = 191 gTiO2 189.7 gTiCl4 1 molTiCL4 molTiO2 2 molH2O 18.02 g H2O b) massH2O(remaining) = 454 gH2O - (2.39 molTiCl4) = 368 gH2O 1 molTiCl4 1 molTiCl4 6. Milk of magnesia is one of the oldest antacids. The material is a slurry of a magnesium compound in water. The compound is 41.7% magnesium, 54.9% oxygen, and 3.5% hydrogen. The mineral weighs about 60 g per mole. What are its empirical and molecular formulas? (10 points) Assume 100 g of compound: 1 mol 1.72 molMg mol = (41.7 g ) Mg = 1.72 mol = 1 Mg Mg Mg 24.305 gMg 1.72 molMg 1 molO 3.43 molO molO = (54.9 gO) = 3.43 molO = 1.99 molO per molMg 15.9994 g O 1.72 molMg 1 molH 3.47 molH molH = (3.5 gH) = 3.47 molH = 2.02 H per molFe 1.00794 gH 1.72 molMg empirical formula = MgO2H2 24.0 gMg 1 molMg 16.0 g 2 mol 1.0 g 2 mol empirical weight = + O O + H H = 58 g/eu molMg eu molO eu molH eu 60 g eu empirical units = = 1.03 eu/molecule 1 eu/molecule molecule 58 g Thus the molecular formula is MgO2H2. .