(Updated 100426) 1 Sample Questions for Chem 002 Final WS10 1. MSDS (the rest listed on review): a. Proper attire – goggles, closed toe shoes, long pants or skirt or lab apron b. Acid Spill – neutralize with sodium bicarbonate c. Bunsen Burners – do not light if flammable biproducts (e.g., H2 gas) are present d. Phenolphthalein – has a laxative effect when ingested e. Types of radiation (listed below) are stopped by what type of material? alpha – paper or hand beta – aluminum (goes through paper, hand) gamma – lead (goes through paper, hand, aluminum) neutron – concrete (goes through paper, hand, aluminum, lead) 2. Radioactive Decay: a. Balance the following radioactive decay equations: 22286Rn → 21884Po + 42He 23490Th → 23491Pa + 0-1e + anti-υ b. Determine the specific decay constant, initial activity and half-life of a radioactive isotope. Given Time, minutes Counts/Min ln (Counts/Min) 0 2 14472 9.58 3 14328 9.57 4 14248 9.56 5 14095 9.55 6 13920 9.54 10 13359 9.50 1. Determine the specific decay constant, k, for this radioactive decay. k = -m m = (y2 – y1) / (x2 – x1) = (9.58 – 9.50) / (2 – 10) min = -0.01 k = 0.01 min-1 2. Determine the initial activity, Ao. y = mx + b ln A = -kt + lnAo 9.50 = -(0.01)(10) +lnAo 9.50 = -0.10 + lnAo 9.60 = lnAo Ao = elnAo = e9.60 = 14764 counts/min 3. Determine the half-life. t 1/2 = ln2 / k = 0.693 / 0.01 min-1 = 69.3 min(Updated 100426) 2 3. Heat of Neutralization: A reaction of 100mL of 1.35M HCl and 100mL of 1.76M NaOH is monitored and the following temperatures were recorded: starting temperature = 24.6 oC; and final temperature = 38.8 oC. Calculate the ΔH of this reaction. Given: Cp of solution (J/K) = 4.13 J/( g . K)*Volume of solution in mL (1 mL ≈ 1 g for aqueous soln) Cp of calorimeter (J/K) = 50 Q = (-total Cp* ΔT) ΔT = Tf - Ti ΔH = Q/n n = # of moles reacted a. Determine the change in temperature for the system. ΔT = Tf – Ti = 38.8 oC – 24.6 oC = 14.2 oC Since we calculated ΔT, 14.2 oC = 14.2 K b. Determine the Cp of the solution (J/K). Cp (soln) = 4.13 (J/g*K) x (1 g / 1 ml) x (200 ml) OR Cp (soln) = 4.13 (J/ml*K) x (200 ml) Cp (soln) = 826 J/K c. Determine the total Cp of the system. Cp (sys) = Cp (soln) + Cp (cal) = Cp (soln) = (826 +50) J/K = 876 J/K d. Determine the number of moles of the acid and the base. Which is the limiting reagent? (1.35 mole / L HCl) x (100 ml) x (1 L / 1000 ml) = 0.135 mole HCl (1.76 mole / L NaOH) x (100 ml) x (1 L / 1000 ml) = 0.176 mole NaOH Since HCl is the limiting reagent, the moles of solution is 0.135 mole. e. Determine the Heat Transfer, Q, for the reaction. Q = [ (-876 J/K) (14.2 K)] Q = -12,439.2 joule f. Determine the change in enthalpy, ΔH, for the reaction. ΔH = (-12,439.2 joule) / (0.135 mole soln) ΔH = -92,142 joule / mole(Updated 100426) 3 4. Heat of Fusion. An ice cube with mass 9.53 grams (presume Ti = 0 oC) is placed in a calorimeter containing 111.24 grams of distilled water at a temperature of 23.2 oC. After equilibration, the final temperature was 15.8 oC. Given: ΔHtotal = ΔHice + ΔHwater + ΔHcalorimeter + ΔHfus = 0 ΔHwater (J) = Cp,H2O *(mass)*ΔT ΔHice (J) = Cp,ice *(mass)*ΔT Hcalorimeter (J) = Cp,Cal *ΔT Cp,H2O = 4.18 J/( g.oC) Cp,Cal = 50 J/oC ΔT = Tf - Ti a. Determine the ΔHwater. ΔHwater (J) = Cp,H2O *(mass)*ΔT = 4.18 J/( g.oC) x (111.24 grams) x (15.8 oC – 23.2 oC) = -3441 J b. Determine the ΔHice. (Hint for Cp – The ice has melted.) Since Cp,ice = Cp,H2O ΔHice (J) = Cp,H2O *(mass)*ΔT = 4.18 J/( g.oC) x (9.53 grams) x (15.8 oC – 0.0 oC) = 629 J c. Determine the ΔHcalorimeter. Hcalorimeter (J) = Cp,Cal *ΔT = 50 J/oC x (15.8 oC – 23.2 oC) = -370 J d. Determine the ΔHfus for one gram of ice. (Hint: For a calorimeter (i.e., closed systems) ΔHtotal = 0) ΔHtotal = ΔHice + ΔHwater + ΔHcalorimeter + ΔHfus = 0 ΔHfus = – ( ΔHice + ΔHwater + ΔHcalorimeter ) = – (629 J – 3441 J – 370 J) = – 3182 J So ΔHfus for one gram of ice ΔHfus = – 3182 J / 9.53g = 334 J/g f. If heat transfers from the system (solute) to the surroundings (solvent), then ΔH is negative (ΔH < 0), and the reaction is defined as (endothermic / exothermic) and the temperature of the solvent will go (up / down). g. If heat transfers from the surroundings (solvent) to the system (solute), then ΔH is positive (ΔH > 0), and the reaction is defined as (endothermic / exothermic) and the temperature of the solvent will go (up / down). h. The heat of neutralization experiment was an (endothermic / exothermic) reaction . i. The heat of fusion experiment was an (endothermic / exothermic) reaction. j. This term means “the techniques that are used to measure enthalpy”: Calorimetry k. This term means “the energy needed to raise the temperature of an object 1o C”: Heat Capacity l. This term means “the energy needed to raise the temperature of one gram of a substance 1o C”: Specific Heat m. The heat capacity is an extrinsic property. Define intrinsic and extrinsic properties and give an example of each. Intrinsic properties are inherent properties usually physical. For example, when a piece of wood is cut, each piece still has the appearance of wood. Extrinsic properties are dependent upon the amount of an object present. For example, when a small piece of wood is burned it generates less heat, than when a large piece of wood is burned.(Updated 100426) 4 5. Antacids: You are given 1.12 M HCl and 1.48 M NaOH. The antacid you use contains 300 mg of CaCO3 and 100 mg of Al(OH)3. If the antacid dissolved in 35.0 ml of HCl and was then back titrated with 21.8 ml of NaOH, find the following: a. The original number mmoles of HCl used to dissolve the antacid and neutralize the base. (1.12 mmole / ml HCl) x (35.0 ml HCl) = 39.2 mmole HCl b. The number of mmoles of NaOH used to backtitrate the acid. (1.48 mmole / ml NaOH) x (21.8 ml NaOH) = 32.3 mmole NaOH c. The