hem 121 Quiz 3 Practice Winter 2018 The following practice quiz contains 26 questions. The actual quiz will also contain 26 questions valued at 1 point/question Name KEY G = 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 + 273 R =.0821 L atm/mole K g = kp g P T = P A + P B + + P N g = kp g % w/v = (g solute/ml solution)100 M 1 V 1 = M 2 V 2 % 1 V 1 = % 2 V 2 1. Write the net ionic equation when solutions of Ba(N 3 ) 2 and Nal are mixed. If there is no reaction, write N.R. NR 2. Write the net ionic equation when solutions of Ba() 2 and ul 2 are mixed. If there is no reaction, write N.R. u +2 (aq) + 2 - (aq) u() 2 (s) 3. Write the net ionic equation when solutions of MgS 4 and ul 2 are mixed. If there is no reaction, write N.R. NR
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. l + N 3 N 4 l or N 4 + + l - 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. 2 3 + Li Li 3 + 2 Not enough hydroxide to pull off both + from carbonic 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. 3 P 4 + Al() 3 AlP 4 + 3 2 7. Please balance the following acid-base reaction. 1 3 P 4 + 3K 1K 3 P 4 + 3 2 8. Which compound is oxidized and which is reduced in the following reaction? Pyruvic acid + NAD Lactic Acid + NAD + xidized: NAD Reduced: Pyruvic acid Tough one oxidation as loss of (better, loss of electrons holding onto molecule) and reduction as gain of (better, gain of electrons allowing to be added) 9. As we oxidize food to 2, we reduce oxygen to water. An intermediate in the process is hydrogen peroxide, or dihydrogen dioxide. What is the oxidation state of oxygen in hydrogen peroxide? int: you may wish to draw the Lewis structure. -1 +1 +1-1 xidation State: -1 = +1 always, = -2 except when superseded by or as 2
10. Ethyl acetate reacts with hydrogen in the presence of a catalyst to yield ethyl alcohol. The balanced chemical equation for the reaction is 4 8 2 (l) + 2 2 (g) 2 2 6 (l) ow many grams of ethyl alcohol are produced by reaction of 12.0 g ethyl acetate with 2? a. 3.14 g b. 6.28 g c. 12.6 g d. 22.9 g e. 24.0 g (12.0 g 4 8 2 )(mol 4 8 2 /88.1 g 4 8 2 )(2 mol 2 6 /mol 4 8 2 )(46.1 g 2 6 /mol 2 6 ) 11. Referring to the preceding question, how many grams of 2 are needed to react with 12.0 g of ethyl acetate? a. 0.138 g b. 0.275 g c. 0.550 g d. 2.02 g e. 48.5 g (12.0 g 4 8 2 )(mol 4 8 2 /88.1 g 4 8 2 )(2 mol 2 /mol 4 8 2 )(2.02 g 2 /mol 2 ) 12. Referring again to the generation of ethanol from hydrogen and ethyl acetate, if 12.0 g ethyl acetate are reacted in excess 2 and 3.14 g ethyl alcohol are produced, what is the % yield? a. 100 % b. 50.0 % c. 25.0 % d. 13.7 % e. 13.1 % (3.14 g 2 6 /12.6 g 2 6 )100 12.6 g theoretical yield from question 10 13. Reactions that proceed with the evolution of heat energy are termed a. Exothermic b. Endothermic c. Exergonic d. Endergonic
14. Is the following reaction spontaneous at 25 o? Please show your work. 2 2 (g) + 2 (g) 2 2 (l) = -571.6 kj/mol, S = -0.3264 kj/mol K G = T S = -571.6 kj/mol 298K(-0.3264 kj/ mol K) G = T S = -571.6 kj/mol + 97.3 kj/ mol = -474.3 kj/mol Spontaneous? Very much so G << 0 Bonus (1 E): Is the reaction between hydrogen and oxygen spontaneous at all temperatures? If so, briefly explain. If not, at what temperature does the reaction change from spontaneous to non-spontaneous? No, the reaction is exothermic but proceeds with an opposing decrease in entropy. By setting G = 0 we can find the transition temperature: G = 0 = T S T = / S = -571.6 kj/mol -0.3264 kj/mol K = 1751 K 1751 K = 1478 o, close to the melting point of iron. Notice the negative signs cancel, leaving a positive value for the absolute temperature, as must be the case 15. What are the 3 principal determinants of reaction rate? a. ollisions with energy > E a or temperature b. Number of collisions or concentration of reactants c. rientation of collisions (major mechanism for enzyme catalysis) 16. onsider the important reaction series of reactions which allow us to control our p by regulating our breathing. Please write the equilibrium constant expression for the first reaction, the formation of 2 3 from 3 - and + 3 - (aq) + + (aq) 2 3 (aq) 2 (g) + 2 (l) K = [ 2 3 ] [ 3 ][ + ] Remember, molar concentration of products divided by the molar concentration of reactants. If there is more than one reactant or product they are multiplied together, and lead coefficients show up as exponents For aa + bb + mm + nn + K = [M]m [N] n [A] a [B] b
17. Based on the coupled chemical equations in question 16, what is the effect on the concentration of + in the body if 2 is removed by breathing? a. Increases b. Decreases c. Remains unchanged By the principal of Lehâtelier, removal of 2 causes 2 3 to shift towards [products] 2 and 2, which in turn causes more bicarbonate and + to react and form carbonic acid 18. Which of the following compounds can hydrogen bond with water? a. Nl 3 b. 3 2 c. l 2 d. 3 3 e. All of the above Water acts as both a hydrogen bond donor and hydrogen bond acceptor; essentially, this is why one can simply count oxygen & nitrogen and compare to the number of carbon when determining water solubility 19. Which of the following would you expect to have the highest boiling point? a. 3 2 2 2 2 2 3 b. 3 2 2 2 2 2 c. 2 2 2 2 2 d. 3 2 2 2 2 e. 3 2 2 2 2 Na Sodium indicates there must be an ionic bond and ion-ion interactions are very strong 20. Which of the following would you expect to have the greatest solubility in water? a. 3 2 2 2 2 2 3 b. 3 2 2 2 2 2 c. 2 2 2 2 2 d. 3 2 2 2 2 e. 3 2 2 2 2 Na Sodium indicates there must be an ionic bond and water loves to form ion-dipole interactions. If you see evidence of an ionic bond - game over where water solubility is concerned
Temperature ( o ) Vapor Pressure (mm g) 0 4.6 10 9.2 20 17.5 25 23.8 30 31.8 40 55.3 21. What is the relative humidity on a 40 o day if the gas pressure due to water is 9.2 mm g? 100 % R @ 40 o = 55.3 mm g 100(9.2 mm g/55.3 mm g) = 17 % Answer: 17 % 22. What is the dew point for the conditions indicated in question 21? The temperature where equilibrium/saturation is achieved (any temperature below would generate dew as equilibrium reestablishes) Answer: 10 o 23. What is the final pressure in a constant temperature system initially at 1.00 atm gas pressure and 1.00 L that is compressed to 1.00 ml? a. 1.00 x 10 3 atm b. 1.00 x 10 1 atm c. 1.00 x 10 0 atm d. 1.00 x 10-1 atm e. 1.00 x 10-3 atm P 1 V 1 = P 2 V 2 (1.00 atm)(1.00 L) = P 2 (1.00 x 10-3 L) 24. A closed 1.00 L Erlenmeyer flask at -73 o and 3.00 atm pressure is heated to 327 o. What is the final pressure? a. 0.670 atm b. 3.00 atm c. 4.48 atm d. 9.00 atm e. 13.4 atm Volume is a constant so use P 1 /T 1 = P 2 /T 2 remembering to convert to K 3.00 atm/200 K = P 2 /600 K
25. Now many moles of [ideal] gas are in a 44.8 L container if the temperature of the system is 819 o and the gas is exerting a pressure of 4.00 atm? a. 0.500 mol b. 1.00 mol c. 2.00 mol d. 4.00 mol e. 8.00 mol Use PV = nrt or n = PV/RT. 819 o = 1092 K. n = (4.00 atm)(44.8 L) (.0821 L atm/mole K)(1092 K) = 2.00 mol Alternatively, given 1.00 mol gas occupies 22.4 L at 1.00 atm and 273 K, double the volume double the moles of gas; 4x temperature, 4x pressure so number of moles of gas doesn t change 26. What is the partial pressure due to water on a day where the P T = 762.0 mm g, P N2 = 580.7 mm g, P 2 = 155.8 mm g, P Ar = 6.9 mm g and P 2 = 0.3 mm g. Show your work. P T = P N2 + P 2 + P Ar + P 2 + P 2 762.0 mm g = 580.7 mm g + 155.8 mm g + 6.9 mm g + 0.3 mm g + P 2 Answer: 18.3 mm g