The following practice examination contains 38 questions. The actual examination will also contain 38 questions valued at 3 points/question.

Transcription

1 Chem 121 Exam 3 Practice Winter 2018 The following practice examination contains 38 questions. The actual examination will also contain 38 questions valued at 3 points/question. Name: KEY G = H T S PV = nrt P 1 V 1 = P 2 V 2 P 1 /T 1 = P 2 /T 2 V 1 /T 1 = V 2 /T 2 K = o C R =.0821 L atm/mole K 1 atm = 760 mm Hg C g = kp g P T = P A + P B + + P N C g = kp g % w/v = (g solute/ml solution)100 M 1 V 1 = M 2 V 2 %C 1 V 1 = %C 2 V 2 1. Write the net ionic equation when solutions of Na 3 PO 4 and Na 2 CO 3 are mixed. If there is no reaction, write N.R. N.R. Notice the cation in both cases is sodium 2. Write the net ionic equation when solutions of MgSO 4 and KOH are mixed. If there is no reaction, write N.R. Mg +2 (aq) + 2OH - (aq) Mg(OH) 2 (s) 3. Write the net ionic equation when solutions of (NH 4 ) 2 CO 3 and Mn(NO 3 ) 4 are mixed. If there is no reaction, write N.R. Mn +4 (aq) + 2CO 3-2 (aq) Mn(CO 3 ) 2 (s) Notice the charge on the manganese is determined by the number of NO 3 - in Mn(NO 3 ) 4 and the realization chemical compounds are electrically neutral

2 4. Please give the products of the following acid-base reaction as shown (do not change lead coefficients). If there is no reaction, write N.R. H 3 PO 4 + 3NaOH Na 3 PO 4 + 3H 2 O 5. Please give the products of the following acid-base reaction as shown (do not change lead coefficients). If there is no reaction, write N.R. H 3 PO 4 + NaOH NaH 2 PO 4 + H 2 O Not enough equivalents of OH - to remove all hydrogens from phosphoric acid 6. Please give the products of the following acid-base reaction as shown (do not change lead coefficients). If there is no reaction, write N.R. HNO 3 + CH 3 CO 2 H NR Both are acids probably a good idea to review the common acids and bases we have discussed so far 7. Please balance the following acid-base reaction 1CaCO 3 + 2HCl 1CaCl 2 + 1CO 2 + 1H 2 O Swap 2H + for Ca +2. Note that H 2 CO 3 decomposes to CO 2 + H 2 O 8. Which of the following elemental species is the strongest reducing agent? a. F 2 b. O 2 c. N 2 d. Be e. Li Strong reducing agents give up electrons readily and are oxidized

3 9. Which of the following elemental species is the strongest oxidizing agent? a. KClO 4 b. KClO 3 c. KClO 2 d. KClO e. KCl We burned up a Flamin Hot Cheetos with KClO 3, where the Cl oxidation number = +5. KClO 4 has an oxidation number = +7. Perchlorate (ClO 4 - ) salts have been known to detonate and kill people 10. Which compound is oxidized and which is reduced in the following reaction? Go-gurt (l) + O 2 (g) CO 2 (g) + H 2 O (l) Oxidized: Go-gurt Reduced: Oxygen Bonus (3 EC): Use the half-reaction method to balance the redox reaction between elemental lithium and oxygen to form lithium oxide Oxidation: 4[Li 0 Li e - ] Reduction: O e - 2O -2 Overall: 4Li 0 + O 2 0 4Li O -2 (or 2Li 2 O) The number of electrons lost must equal the number of electrons gained 11. Given the reaction for the combustion of propane below, how many grams of propane are required to produce 66 g of CO 2? C 3 H 8 + 5O 2 3CO 2 + 4H 2 O a. 11 g b. 22 g c. 33 g d. 44 g a. 66 g 66 g CO 2 (mol CO 2 /44 g CO 2 )(mol C 3 H 8 /3 mol CO 2 )(44 g C 3 H 8 /mol C 3 H 8 )

4 Notice that to 2 sig. fig., carbon dioxide and propane have the same molecular weight. Since 1 mol propane generates 3 mol CO 2 (and they have the same MW) you need 1/3 the mass 12. How many grams of propane can be combusted with 40 g O 2? a. 11 g b. 22 g c. 33 g d. 44 g e. 66 g 40 g O 2 (O 2 /32 g O 2 )(mol C 3 H 8 /5 mol O 2 )(44 g C 3 H 8 /mol C 3 H 8 ) 13. If 11 g of propane are completely combusted in excess O 2, how many grams of CO 2 are produced? a. 11 g b. 22 g c. 33 g d. 44 g e. 66 g 11 g C 3 H 8 (mol C 3 H 8 /44 g C 3 H 8 )(3 mol CO 2 /mol C 3 H 8 )(44 g CO 2 /mol CO 2 ) 14. If 11 g of propane are combusted [in excess O 2 ] and 22 g of CO 2 are recovered, what is the percentage yield? a. 33% b. 50% c. 66% d. 100% e. 200% From question 13, theoretical yield = 33 g. Actual yield = 22 g % yield = (thr yield/actual yield)100 = (22 g/33 g) Reactions that proceed with an input of heat are termed a. Exothermic b. Endothermic c. Exergonic d. Endergonic

5 16. Which of the following is the principal means by which catalysts such as enzymes increase the rate of chemical reaction? a. By increasing temperature b. By increasing number of collisions c. By lowering the energy of activation (E a ) d. By tightly binding to substrate and product e. All of the above 17. Please draw the energy diagrams for the following exergonic reactions a. A slow reaction with a small free energy change use a solid line b. A fast reaction with a large free energy change use a dashed line Free Energy Reaction Progress Please refer to homework problem # Is the following reaction spontaneous at o C? Please show your work H 2 O (l) H 2 O (g) H = kcal/mol, S = kcal/mol K G = H T S = kcal/mol 373K( kcal/mol K) G = kcal/mol Spontaneous? G < 0, yes Well whaddya know - the transition of water from liquid to gas becomes spontaneous at the boiling point of water

6 Bonus (3 EC): Why is the value for S > 0 for the reaction shown above? Products are in the highly random gas phase 19. Please write the equilibrium constant for the following reaction. Mg(OH) 2 (s) Mg +2 (aq) + 2OH - (aq) K = [Mg +2 ][OH - ] 2 [Mg(OH) 2 ] First and foremost, think products over reactants. If it helps you, think Mg(OH) 2 (s) Mg +2 (aq) + OH - (aq) + OH - (aq) K = [ [Mg +2 ][OH - ][ OH - ] [Mg(OH) 2 ] = [Mg +2 ][OH - ] 2 [Mg(OH) 2 ] Bonus (3 EC): What will happen to the equilibrium concentration of Mg(OH) 2 (s) if an acid is present? a. Increase b. Decrease c. Increase or decrease, depending on the strength of the acid d. Remain Unchanged The acid will react with the hydroxide. By the principal of LeChâtelier, loss of product will cause the reaction to shift towards products to reestablish equilibrium. Mg(OH) 2 is a useful antacid as it provides low level concentrations of OH - to neutralize stomach acid, and ultimately dissolves as the system continually tries to reestablish equilibrium 20. Which of the following would you expect to have the highest boiling point? a. CH 3 CH 2 CH 2 CH 2 CH 3 b. CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 c. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 d. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 e. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 Greatest overall London forces

7 21. Which of the following would you expect to exert the highest vapor pressure? a. CH 3 CH 2 CH 2 CH 2 CH 3 b. CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 c. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 d. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 e. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 Lowest overall London forces, allowing more of the liquid to exist as a vapor in the gas phase 22. Which of the following would you expect to have the greatest solubility in water? a. CH 3 CH 2 CH 2 CH 2 CH 2 OH b. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 OH c. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 OH d. HOCH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 OH 4C:1O e. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CO 2 H 23. Which of the following would you expect to have the greatest solubility in a 0.1 M NaOH solution (ph = 13)? a. CH 3 CH 2 CH 2 CH 2 CH 2 OH b. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 OH c. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 OH d. HOCH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 OH e. CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CO 2 H Notice the CO 2 H motif, allowing (e) to act as an acid will react in the presence of NaOH generating the sodium salt 24. What is the final pressure in a closed piston system initially at 1.00 L and 1.00 atm that expands to a final volume of L? a x 10-2 atm b x 10-1 atm c x 10 1 atm d x 10 2 atm e x 10 3 atm P 1 V 1 = P 2 V 2 P 1 (V 1 /V 2 ) = P 2 = 1.00 atm(1.00 L/100.0 L) = P 2

8 25. What is the final pressure in a closed piston system initially at 1.00 L and 1.00 atm that is compressed to a final volume of 10.0 ml? a x 10-2 atm b x 10-1 atm c x 10 1 atm d x 10 2 atm e x 10 3 atm Again use P 1 V 1 = P 2 V 2 noting 10.0 ml = L P 1 (V 1 /V 2 ) = P 2 = 1.00 atm(1.00 L/ L) = P What is the final temperature in a rigid container, initially at 1.00 atm and 10.0 o C, which when heated exerts a pressure of 2.00 atm? a o C b o C c. 100 o C d. 293 o C e. 566 o C A rigid container means constant volume, so use P 1 /T 1 = P 2 /T 2 making certain you convert to absolute temperature. P 1 /T 1 = P 2 /T 2 T 2 = T 1 (P 2 /P 1 ) = 283 K(2.00 atm/1.00 atm) = 566 K = 293 o C 27. What is the final temperature in a 10.0 L balloon, initially 10.0 o C, which when heated occupies a volume of 20.0 L? a o C b o C c. 100 o C d. 293 o C e. 566 o C Constant pressure, so use V 1 /T 1 = V 2 /T 2 again making sure to convert to absolute temperature. Refer to the previous problem, noting the similarities 28. How many moles of air are in a human lung inflated to 3.0 L at 1.0 atm pressure and 37 o C? a mol b. 1.0 mol c. 3.0 mol d. 9.1 mol e. 76 mol Use PV = nrt n = PV/RT = (1.0 atm)(3.0 L)/( L atm/mole K)(310 K)

9 29. The Henry s law constant for O 2 dissolved in water is 1.0 x 10-3 mol/l 40 o C. How much O 2 is dissolved in a 1.0 x 10 3 L fish tank at 1.0 atm air pressure? Oxygen = 21 % total air pressure at sea level. The units on the Henry s law constant pretty much tell you everything you need to know 1.0 x 10-3 mol/l atm We need to multiply by L and atm to get to moles and then by the molar mass of oxygen to convert to mass (1.0 x 10-3 mol/l atm)(0.21 atm)(1000 L)(32 g/mol) Answer: 6.7 g A significant decrease in the solubility of oxygen as a function of 20 o C k = 1.0 x 10-3 mol/l atm, giving 8.7 g O 2 ( )/8.2 =.17 or a 17 % decrease in dissolved oxygen 30. Assuming no change in the percentage composition of air, what is the partial pressure of O 2 at the top of Mount Whitney, where P T = 440 mm Hg, given P O2 = 160 mm Hg when P T = 1.0 atm. Easiest to solve this homework problem by ratios x/440 mm Hg = 160 mm Hg/760 mm Hg x = 160 mm Hg(440/760) Answer: 93 mm Hg Temperature ( o C) Vapor Pressure (mm Hg) What is the relative humidity on a 30 o C day if the gas pressure due to water is 4.6 mm Hg? (4.6 mm Hg/31.8 mm Hg)100 Answer: 14 % 32. What is the dew point for the conditions listed in question 31? Answer: 0 o C

10 33. What is the molarity of a ml solution containing 4.00 g NaOH? a M b M c M d M e M 4.00 g NaOH(mol/40.0 g) mol/ L 34. How many grams of NaOH would be required to make ml of 2.00 M NaOH solution? a g b g c g d g e g L( 2.00 mol/l)(40.0 g/mol) 35. What is the w/v % of the solution in question 34? Show your work. 24 g/300.0 ml = 8.0 g/dl Answer: 8 % w/v 36. If mol NaOH were required for a reaction, what volume of 2.00 M NaOH would be required? a L b L c L d L e L mol(l/2.00 mol) = L

11 37. What volume of 6.00 M NaOH would be required to make 30.0 ml of 2.00 M NaOH? a ml b ml c ml d x 10 2 ml e x 10 2 ml Dilution! M 1 V 1 = M 2 V 2 (6.00 M)(V 1 ) = (2.00 M(30.0 ml) 38. Snow White was suffering from insomnia so her doctor prescribed an Ambien suspension. She took the prescription to her local Wicked Queen pharmacy, where the prescription instructed Dr. Grimhilde to make ml of 2.00% w/v Ambien suspension, equally divided into 6 red delicious apples. Since the evil queen had a 50.0 % concentrate available, how many ml of the concentrate will she need? a ml b ml c ml d ml e ml Dilution! %C 1 V 1 = %C 2 V 2 (50.0 %)(V 1 ) = (2.00 %(100.0 ml). Best question ever.