Exam 4 fall 09 with answers

Exam 4 fall 09 with answers - CHM 2045 Final Exam (Form...

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CHM 2045 Final Exam (Form Code A) (Instructors: Gower, Harrison, Mitchell, Ucak) Fall 2009 Instructions: On your scantron sheet enter and bubble in your name , UF ID number (start on the leftmost space and leave any extra spaces blank), and Form Code (see above). You may retain your exam sheet (mark your answers on it and the scantron sheet) and any other sheets. Turn in only the scantron. Careful! Bubbling errors will not be negotiated . Possibly Useful Info: h = 6.626x10 -34 J∙s; c = 3 x 10 8 m/s; R = 0.0821 L•atm/mol•K; R = 8.314 J/ mol•K; specific heat capacity of water = 1.00 cal/g o C = 4.184 J/g o C; Rydberg constant = 1.0974 x 10 7 m -1 ; Rhc = 2.18 10 -18 J/atom 1. A solution of sucrose in water is 48.5% sucrose by mass and has a density of 1.118 g/mL. What mass of sucrose, in grams, is contained in 3.50 L of solution? 1) 5.5 x 10 2 g 2) 1.1 x 10 3 g 3) 1.9 x 10 3 g 4) 3.1 x 10 3 g 5) 3.9 x 10 3 g 2. The vitamin C (C 6 H 8 O 6 ) content in a tablet is analyzed by a reaction with bromine and then titration of the hydrobromic acid with standard base: C 6 H 8 O 6 + Br 2 C 6 H 6 O 6 + 2HBr HBr + NaOH NaBr + H 2 O One tablet was dissolved in water and reacted with bromine. The solution was then titrated using 21.60 mL of standard 0.1350 M NaOH. How much Vitamin C did the tablet contain in mg? 1) 323.7 mg 2) 632.5 mg 3) 513.6 mg 4) 449.2 mg 5) 256.8 mg 3. What volume of a 0.125 M oxalic acid (H 2 C 2 O 4 ) solution is required to titrate 147.9 mL of a 0.546 M NaOH solution? 1) 0.956 L 2) 0.116 L 3) 0.255 L 4) 0.323 L 5) 0.452 L 4. The effect of adding a catalyst to a reaction is to 1) increase the number of collisions between reactants. 2) increase the energy of the products. 3) stabilize the equilibrium constant of a reaction. 4) lower the activation energy of a reaction . 5) decrease the enthalpy change of a reaction. 5. Calculate the enthalpy change for the reaction NO( g ) + O( g ) → NO 2 ( g ) from the following data: NO(g) + O 3 ( g ) → NO 2 ( g ) + O 2 ( g ) Δ H = -198.9 kJ O 3 ( g ) → 1.5O 2 ( g ) Δ H = -142.3 kJ O 2 ( g ) → 2O( g ) Δ H = 495.0 kJ 1) -551.6 kJ 2) -304.1 kJ 3) 190.9 kJ 4) 153.8 kJ 5) 438.4 kJ 6. If 1.951 g of BaCl 2 x H 2 O yields 1.526 g of anhydrous BaSO 4 after treatment with sulfuric acid, calculate the value of x . (1) 1 (2) 2 (3) 3 (4) 4 (5) 5 7. Consider the following reaction: 2A
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Exam 4 fall 09 with answers - CHM 2045 Final Exam (Form...

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