CHM 151 Exam 4 Chapters 9 and Part of 8

Name: _____KEY________________________ Section: _________________ CHM 151 Exam 4: Chapters 9 and part of 8 You must show all work to receive credit. Clearly mark your final answer! 1. (10 pts) Matching: Indicate how the following changes made to a system (shown on the left) will affect the pressure of the system (letter options on the right). The system is initially filled with neon atoms. Assume variables not mentioned are held constant. Note: Answer choices may be used more than once! Changes to system: _C__ Volume is doubled _A__ Temperature is doubled _E__ Temperature is halved and volume is halved _B__ Number of atoms is tripled _E__ Neon atoms are replaced with xenon atoms Pressure changes: a. Pressure is doubled b. Pressure is tripled c. Pressure is decreased by 1/2 d. Pressure is decreased by 1/3 e. No change _D__2. (5 pts) Which of the following statements about gases is true? a. Gases are highly compressible. b. There are relatively large distances between atoms/molecules of a gas. c. A gas expands spontaneously to fill its container. d. All of these are true. _E__ 3. (5 pts) A sample of gas (24.2g) initially at 4.00 atm was compressed from 8.00 L to 2.00 L at constant temperature. After the compression, the gas pressure was _____ atm. a. 1.00 b. 2.00 c. 4.00 d. 8.00 e. 16.0 P1V1 = P2V2 4. (6 pts) How many moles of gas occupy a volume of 60.82 L at 31.1oC and 367 torr? V = 60.82 L P = 367 torr = 0.4829 atm T = 31.1 oC = 304.25 K n = PV / RT = 0.4829 atm * 60.82 L 0.08206 Latm/molK * 304.25 K n = 1.18 mol _C_ 5. (5 pts) An unknown gas occupies a volume of 725 mL at STP and has a mass of 1.036 g. This gas could be a. nitrogen, N2 b. carbon dioxide, CO2 c. oxygen, O2 MM = dRT / P d. hydrogen, H2 e. chlorine, Cl2 page 1 of 4 Name: _________________________________ Section: _________________ 6. (12 pts) The reaction of calcium hydride, CaH2, with water is often used in cases where the expansion of volume of an object needs to be delayed (e.g., life vests). This reaction produces hydrogen gas and aqueous calcium hydroxide. Complete and balance the equation below (with phases) and then calculate the mass of CaH2 needed to generate 53.5 L of gas if the pressure of the gas is 814 torr at 21.7oC? CaH2(s) + 2 H2O(l)  2 H2 (g) + Ca(OH)2 (aq) V = 53.5 L P = 814 torr = 1.071 atm T = 21.7 oC = 294.85 K n = 2.36816 mol H2 g n = PV / RT = 1.071 atm * 53.5 L 0.08206 Latm/molK * 294.85 K 2.36816 mol H2 * (1 mol CaH2 / 2 mol H2) * (42.096 g CaH2 / 1 mol) 49.8 g CaH2 are needed 7. (6 pts) A closed container holds CH4 and H2 gases. The total pressure in the container is measured to be 723 torr. The container has 0.651 moles of CH4 and 0.184 moles of H2. Calculate the mole fractions and the partial pressures of both gases in the container. XCH4: (0.651 / 0.835) = 0.780 PCH4 = (0.780)(723 torr) = 564 torr (or 0.742 atm) XH2: (0.184 / 0.835) = 0.220 PH2 = (0.220)(723 torr) = 159 torr (or 0.209 atm) _A_ 8. (5 pts) In a balloon with a small pinhole leak, which gas will escape the balloon the fastest? a. H2O (g) b. N2 (g) c. O2 (g) d. Ar (g) e. CO2 (g) 9. (6 pts) For each blank, circle the word that correctly completes the sentence. A chemical reaction that absorbs heat from the surroundings is said to be _endo____ (endothermic or exothermic), has a _pos____ (positive or negative) value of H at constant pressure, and feels _cold_ (cold or warm) to the touch. page 2 of 4 Name: _________________________________ Section: _________________ _C_ 10. (5 pts) Specific heat is the a. amount of energy required to melt 1.00 g of a substance b. amount of heat energy needed to change 1.00 mol of a substance by 1.00 oC c. amount of heat energy needed to change 1.00 g of a substance by 1.00 oC d. amount of a substance that is heated by 1.00 oC e. the number of Kelvins that 1.00 g of a substance is raised by heating it for 1.00 minute 11. (8 pts) Calculate the amount of heat (in kJ) released by an acid-base neutralization reaction if a total of 99.56 g of reactants are used. This reaction is measured to go from 36.0oC to 24.3oC and can be assumed to have the specific heat of water (4.184 J/g·oC). q = mass * sp.ht. * T = (99.56 g)(4.184 J/g·oC)(24.3oC – 36.0oC) = -4873.74 J = -4.87 kJ 12. (8 pts) Determine if the following reactions are endothermic or exothermic (circle your answer): a. Condensation of steam Endothermic Exothermic b. Freezing lemonade Endothermic Exothermic c. Sublimation of NH4Cl Endothermic Exothermic d. Decomposition of CuCO3 Endothermic Exothermic 13. (8 pts) How much total heat (in kJ) is transferred when 6.781 grams of oxygen combusts as shown in the equation below? ___ CH4 (g) + _2_ O2(g) → _2_ H2O(g) + ___ CO2(g) ∆H = -980.0 kJ 6.781 g O2 * (1 mol / 32.00 g) * (-980.0 kJ / 2 mol O2) = -103.8 kJ 14. (6 pts) Sulfur trioxide can be synthesized by combining sulfur dioxide and oxygen gas, as described by the equation: 2SO2 (g) + O2 (g)  2SO3 (g) Ho = ? Use the equations below to calculate Ho, in kJ, for the reaction above. SO2 (g)  S (s) + O2 (g) Ho = 296 kJ (x 2) 2 SO3 (g)  2 S (s) + 3 O2 (g) Ho = 789 kJ (reverse) 2 SO2 (g)  2 S (s) + 2 O2 (g) 2 S (s) + 3 O2 (g)  2 SO3 (g) 2 SO2 (g) + O2 (g)  2SO3 (g) Ho = 592 kJ Ho = -789 kJ Ho = -197 kJ page 3 of 4 Name: _________________________________ Section: _________________ 15. (5 pt) What grade do you honestly feel you deserve in this class (based on the amount of effort you have put in and amount of material you have learned)? _A+__________ Why? Because this is the best class ever and I love chemistry!  Extra Credit (6 points) – Show your work on the back of this page. A 30.01 g block of Al at 5.00°C (specific heat = 0.897 J/g·oC) is dropped into a calorimeter containing 1.00 L of water at 75.0°C (specific heat = 4.184 J/g·oC). What is the final temperature of the system after it reaches thermal equilibrium (assuming no heat is lost from the calorimeter)? qH2O < 0 (loses heat) mass: 1000 g (density = 1.00 g / mL) sp. ht.: 4.184 J/g·oC Tf: ? Ti: 75.0oC qAl > 0 (gains heat) mass: 30.01 g sp. ht.: 0.897 J/g·oC Tf: ? Ti: 5.00oC - qH2O = qAl - (1000 g)(4.184 J/g·oC)(Tf – 75.0oC) = (30.01 g)(0.897 J/g·oC)(Tf – 5.00oC) -4184J(Tf – 75.0oC) = 26.92 J/oC(Tf – 5.00oC) -4184J/oC * Tf + 313800 J = 26.92 J/oC * Tf – 134.6 J 313934.6 J = 4210.92 J/oC * Tf Tf = 313934.6 J / 4210.92 J/oC Tf = 74.6oC page 4 of 4