Part 1: Multiple Choice (2 points each)Directions: Please circle the best answer for each of the following questions.20. Part 2: Short Answer21. Directions: Answer each of the following questions. Be sure to use complete sentences where appropriate. For full credit be sure to show all of your work.Grossmont College Chemistry 141 Exam 2 Fall 2015Name: ___________________________________Date: ________Instructor: Diana VanceExam 2Part 1: Multiple Choice (2 points each) Directions: Please circle the best answer for each of the following questions. 1. Which of the following is permitted in lab?a. Eatingb. Drinkingc. Smokingd. Close-toed shoese. Gum chewing2. For ∆Esys to always be +, what must be true? a. q = wb. +q > –wc. -q > +wd. –w > +q e. none of the above 3. Which of the following could be described as a polar covalent, partially ionized substance with poorconductivity? a. 0.1 M CH3CO2H (aq)b. Glacial acetic acid, CH3CO2H (l)c. 0.1 M HCld. 0.1 M NaCle. KClO3 (s)4. Under what conditions would chlorine gas be the least ideal? a. Low pressure and high temperatureb. Low pressure and low temperaturec. High pressure and high temperatured. High pressure and low temperaturee. Not enough information 5. Which statement is TRUE about kinetic molecular theory?a. A single particle does not move in a straight line.b. The size of the particle is large compared to the volume.c. The collisions of particles with one another is completely elastic.d. The average kinetic energy of a particle is not proportional to the temperature.e. none of the above 6.Page 1 of 7Grossmont College Chemistry 141 Exam 2 Fall 2015Name: ___________________________________Date: ________Instructor: Diana Vance7. Which one of the following statements is not correct?a. Solubility of a substance is the maximum amount that dissolves in a specified amount of solvent at a giventemperature.b. A saturated solution contains the maximum amount of solute dissolved in a solvent at equilibrium.c. A supersaturated solution contains more solute than predicted by the solubility of the substance.d. A supersaturated solution can be produced by slowly decreasing the temperature.e. A supersaturated solution can be produced by slowly adding the solvent.8. What is the oxidation number of P in H2PO4-? a. -5b. -6c. +5d. +6e. +49. One type of alcohol breathalyzer used by law enforcement involves the following reaction. The amount ofethanol present is determined by a change in color of the solution. What type of reaction is this?3CH3CH2OH (g) + 2K2Cr2O7 (aq) + 8H2SO4 (aq)  3CH3COOH (aq) + 2Cr2(SO4)3 (aq) + 2K2SO4 (aq) + 11H2O (l)a. Oxidation-reductionb. Acid-base neutralizationc. Decompositiond. Single replacemente. Double replacement 10. Which of the following is not a standard state?a. for a liquid, it is 25 °Cb. for a gas, it is 1 atmc. for a solution it is 1 Md. for a solid, it is 25 °Fe. for a liquid, it is 1 atm 11. Assume that you have a sample of gas in a cylinder with a moveable piston, as shown in diagram (1). The initialpressure, number of moles, and temperature of the gas are noted on the diagram. 12. Which diagram (2)-(4) most closely represents the result of doubling the pressure while keeping thetemperature and number of moles of gas constant? a. 2b. 3c. 4d. 3 & 4e. all of the above13.14.15.16.17.18.19.Page 2 of 7Grossmont College Chemistry 141 Exam 2 Fall 2015Name: ___________________________________Date: ________Instructor: Diana Vance20. Part 2: Short Answer 21. Directions: Answer each of the following questions. Be sure to use complete sentences where appropriate.For full credit be sure to show all of your work. 1. A 500 mg dietary supplement of L-lysine (an amino acid, 146.2 g/mol) required 68.4 mL of 22. 0.100 M NaOH to reach the end point. How many protons were removed for each L-lysine molecule in this titration (10 points)?23.68.4 mL NaOH soln×1 L1000 mL×0.100mol NaOH1 L NaOH soln=6.84 ×10−3mol NaOH24.500 mg L−lysine ×1 g1000 mg×1 mol L−lysine146.2 g L−lysine=3.42× 10−3mol L−lysine25.molar ratio=6.84 × 10−3mol NaOH3.42 ×10−3mol L−lysine=2 mol NaOH1 mol L−lysine26. Therefore, 2 moles of protons were removed from each L-lysine molecule. 2. A solution of hydrochloric acid is prepared by bubbling hydrogen chloride gas into water. Ifthe resulting solution has a pH of 0.9745 (10 points)a. what is the hydrogen ion concentration of the solution? 27.+¿⌉=10−pH=10−0.9745=0.1060 M⌈ H¿b. If 5.00 mL of this solution is diluted to 25.00 mL, what is the concentration of the newsolution? 28. M1 = 0.1060 M29. V1 = 5.00 mL30. V2 = 25.00 mL31. M2 = ? 32.33.M1V1=M2V2⇒ M2=M1V1V2=(0.1060 M)(5.00 mL)(25.00 mL)=0.0212 M3. Why is the sign of ∆H negative for an exothermic process (3 points)? 34.35. A change in enthalpy is the sum of the change of internal energy and the product of thesystem’s pressure and change in volume. 36.37.38.39.Page 3 of 7Grossmont College Chemistry 141 Exam 2 Fall 2015Name: ___________________________________Date: ________Instructor: Diana Vance40.4. Automobile airbags produce nitrogen gas based on the reaction (10 points): 41. 2 NaN3 (s) → 2 Na (l) + 3 N2 (g)a. If 2.25 g of sodium azide react to fill an air bag, how much P-V work will the nitrogengas do against an external pressure of 1.00 atm given that the density of nitrogen is 1.165g/L at 20 °C? 42.2.25 g Na N3×1 mol Na N365.011 g Na N3×3 mol N22mol Na N3×28.014 g N21 mol N2×1 L N21.165 g N2=1.25 L N2 43.w=−P ∆V =−(1.00 atm) (1.25 L)×101.325 J1 Latm=− 126 Jb. If the process releases 2.34 kJ of heat, what is ∆E for the system? 44.∆ E=q+w45.∆ E=−2.34 kJ +(−126 J ×1 kJ1000 J)=−2.34 kJ+(−0.126 kJ)=−2.47 kJ5. Exactly 10.0 mL of water at 25.0 °C is added to a hot iron skillet. All of the water is converted into steam at 100.0 °C (cwater = 4.184 J/g °C, ∆Hvaporization = 2.26 kJ/g, ∆Hfusion =333 kJ/g). The mass of the pan is 1.20 kg and the molar heat capacity of iron is 25.19 J/mol °C. What is the temperature change of the skillet (9 points)? 46.−qout=q¿47.−qskillet=qwater48.water 25.0 ℃−mFecFe∆TFe=mwatercwater∆ T¿100.0 ℃ ¿+mwater∆ Hvaporization49.water25.0 ℃∆ TFe=mwatercwater∆T¿100.0 ℃ +mwater∆ Hvaporization¿−mFecFe50.∆ TFe=(10.0 g)(4.184Jg ℃)(100.0 ℃ −25.0℃)×1 kJ1000 J+(10.0 g)(2.26kJg)−(1.20 kg)(25.19Jmol ℃)×1 mol Fe55.845 g Fe×1kJ1000 J×1000 g1kg51.∆ TFe=3.138