<p> Supplemental Instruction: Chem177 Practice Exam 2 March 3, 2009 show all work and report answers in correct # of sig figs</p><p>1. Formic acid, HCHO2, is a weak electrolyte. What solute particles (only) are present in an aqueous solution of this compound? + – + - A. H and CHO2 B. H , CO, and OH + – C. HCHO2, H , and CHO2 D. HCHO2</p><p>2. An element that loses electrons is said to have been… A. Redoxed B. Oxidized C. Reduced D. Decomposed</p><p>3. What is an example of an extensive property? A. heat capacity B. density C. specific heat D. molarity</p><p>4. Which of the following solutions is the most acidic?</p><p>A. 0.5 M NH3 B. 0.3 M HNO3</p><p>C. 0.4 M HI D. 0.7 M CH3OH</p><p>5. A chemical reaction that absorbs heat from the surroundings is said to be ______and has a ______ΔH at constant pressure. A. endothermic; positive B. endothermic; negative C. exothermic; negative D. exothermic; positive E. exothermic; neutral</p><p>6. Enthalpy is symbolized by A. S B. q C. w D. H</p><p>7. Which of the following is not a state function? A. E B. q C. P D. All are state functions 7. A compound consists of 75.69% C, 8.80% H, and 15.51% O by mass. Determine the empirical formula as well as the molecular formula if the molar mass is determined to be 412 g/mol. *Assume 100 g 75.69g C x 1 mol/12.01 g = 6.30 mol C 8.80g H x 1 mol/1.01 g = 8.71 mol H 15.51g O x 1 mol/16.00 g = 0.97 mol O *divide all by 0.97 to equal 6.5 mol C, 9 mol H, 1 mol O, times by 2</p><p>Emp. formula = C13H18O2</p><p>Mol. formula= C26H36O4</p><p>8. Mg3N2(s) + 4H2SO4(aq) 3MgSO4(aq) + (NH4)2SO4(aq) A. Balance the equation.</p><p>B. Calculate the number of grams of MgSO4 theoretically formed when 4.6 g</p><p> of Mg3N2 is reacted with 125 mL of 2.5M solution of H2SO4.</p><p>*4.6 g x 1 mol/100.9g = 0.046 mol Mg3N2 *limiting (from equation)</p><p>*2.5 M = x moles/0.125 L x = 0.3125 mol H2SO4</p><p>0.046 mol Mg3N2 x 3 mol MgSO4/1 mol Mg3N2 x 120.3 g/1 mol = </p><p>16.60 g MgSO4 (16 with sig figs)</p><p>C. If the actual yield of MgSO4 is 12.36 g, what is the percent yield? 12.36 g / 16.60 g = 75 % yield</p><p>D. How many moles of excess reactant remain?</p><p>0.046 mol Mg3N2 x 4 mol H2SO4/ 1 mol Mg3N2 = 0.184 mol H2SO4 used in rxn 0.3125 – 0.184 = 0.1285 mol (0.13 with sig figs) 9. Complete the following reaction, then write the balanced net ionic reaction equation; show phases of all reactants and products.</p><p>2HC2H3O2(aq) + Ba(OH)2(aq) 2H20(l) + Ba(C2H3O2)2 (aq)</p><p>- HC2H3O2(aq) + OH- H20(l) + C2H3O2</p><p>Identify the precipitant (if present): ______(none)______Identify all spectator ions (if any present): ___Ba2+______</p><p>10. A stock solution of NaOH is 6.00M. How many mL of this solution is needed to make 500 mL of 0.75M NaOH solution?</p><p>0.75 M NaOH = x moles/ 0.500 L</p><p>0.375 moles NaOH/x L = 6 M</p><p> x = 62.5 mL NaOH ( 60 mL with sig figs)</p><p>11. Assign oxidation numbers and decide what is oxidized and what is reduced.</p><p>PbS(s) + 4H2O2(aq) PbSO4(s) + 4H2O(l)</p><p> oxidized: sulfur in PbS</p><p> reduced: oxygen in H2O2 12. 4Fe(s) + 3O2(g) 2Fe2O3(s) ∆H = -1648.4 kJ If 2.6 g of Fe is reacted in excess oxygen, how many kJ of heat is released?</p><p>2.6 g Fe x 1 mol/55.85 g x -1648.4 kJ/4 mol Fe = -19 kJ</p><p>13. If the specific heat of water is 4.18 J/g*K, how much heat (in J) is required to raise the temperature of 72.0 grams of water by 12⁰C?</p><p> q = 72.0 g (4.18 J/g*K) ( 12⁰C) q = 3600 J</p><p>14. 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) </p><p>A. Using the standard enthalpies of formation, calculate the ∆H of the reaction. (4(-393.5) + 2(-285.8)) - (2(226.7)) = -2599 kJ</p><p>B. Is work done in this reaction? __yep___ If so, calculate amount of work and the ∆E. w = - 1 atm ((4-7)(22.4L) = 67.16 L*atm w = 67.16 x 101.3 J/L*atm = 6797 J</p><p>∆E = -2599 + 6.79 kJ = -2593 kJ 15. Find the concentration of each ion present when 16 mL of 1.0M KNO3 is mixed</p><p> with 22 mL of 0.75M K2SO4.</p><p>1.0 M KNO3 = x mol/0.016 L = 0.016 mol KNO3</p><p>0.75 M K2SO4 = x mol/0.022 L = 0.0165 mol K2SO4</p><p>K+: 0.016 + 2(0.0165) = 0.049 mol/0.038 L = 1.29 M - NO3 : 0.016 mol/0.038 L = 0.42 M 2- SO4 : 0.0165 mol/0.038 L = 0.43 M </p>