High School Name: Individual Exam 1 Solutions: Chemical Structure and Properties

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1 Name (Last, First): ID Number: High School Name: Washington University Chemistry Tournament April 2, 2016 Individual Exam 1 Solutions: Chemical Structure and Properties Please write your full name, ID number, and high school at the top of every page. 45 minutes is allowed for this examination. This exam is 6 questions long, and has 14 numbered pages in total. Please check to make sure that your exam is complete, and report to the exam proctors if any pages are missing. Necessary equations and constants, as well as a periodic table, can be found at the end of the exam. Do not write on the scoring sheet on the last page of the exam. Scratch paper will not be permitted during this exam. If you run out of room on the front of the page, you can continue to work on the back of the page, provided that all answers for parts are clearly labelled. Only write work on the backs of pages that correspond to the question on the front of the page. Make sure to circle or box your final answer if appropriate. Correct answers with appropriate work will receive full credit. If work contains reasoning or justification that is partially correct, partial credit may be awarded. Explanations should fully answer the question and provide supporting evidence and logic. These should also be given in complete sentences. Correct answers without reasonable supporting work will not receive credit. If you cannot answer a question or the entire question, it is advised that you do not spend too much time on that question and proceed onto other parts of the exam. No electronics of any kind are allowed during the exam, with the exception of a non-programmable scientific calculator. Cell phones must be turned off, and watches must be removed. A clock will be projected in the exam room. The scoring policy can be found on the WUCT website. Any appeals must be made in writing in the appeals room, Lab Sciences 400. Cheating will not be tolerated on this exam. The cheating policy that has been listed on the WUCT website will be followed. Violators of this policy will be referred to the directors of the competition.

2 1. Understanding concepts surrounding dilution is crucial when working with solutions. Consider these following questions. a. A solution containing 5 mg of a restriction enzyme is dissolved into 1.00 ml of deionized water. 10% of the resulting solution is diluted to a volume of 1.50 ml. What is the total mass of enzyme in ml of that resulting solution? Assume that there is no volume change upon initial addition of enzyme. 5 mg 0.10 ml = 0.5 mg 1.00 ml 0.5 mg ml = mg 1.50 ml b. A lab that is studying bacteria obtains a sample of bacteria and names it Sample A. i. They dilute 10 µl of sample A to 1.0 ml using sterile water and spread it evenly onto a petri dish and observe that the next day there were 1900 colonies of bacteria on the plate. If 1.0 ml of sample A were spread on a petri dish, how many colonies of bacteria would you expect to grow? 1900 colonies 1000 µl sample A 10 µl sample A 1.0 ml sample A = colonies ii. The lab then takes 1 ml of sample A and dilutes it to 10 ml with sterile water and calls this Sample B. If 2.4 ml of sample B were spread on a petri dish, how many colonies of bacteria would you expect to grow? colonies 1.0 ml sample A 1.0 ml sample A 10 ml sample B 2.4 ml sample B = colonies iii. The lab takes a 1.0 ml aliquot of Sample B and dilutes with some volume of sterile water, calling the final solution Sample C. If 4.3 ml of sample C were spread on a petri dish, and 4085 colonies of bacteria grow, what was the dilution factor for this dilution? colonies ml sample B colonies 2.4 ml sample B = colonies ml sample B 4085 colonies 4.3 ml sample C = 950 colonies ml sample C 1 ml sample B 1 ml sample C 950 colonies = 20 ml sample C 1.0 ml of sample B is diluted to 20 ml sample C, so the dilution factor is 20. (This question continues on the next page) 2

3 c. Max is asked to dilute a 15 M HNO 3 solution to a 0.5 M HNO 3 solution. He adds water to the beaker of concentrated acid. The acid ended up splashing all over Max s lab space. Explain what Max did wrong on the lines provided below. Max added water to a highly concentrated strong acid solution. The heat released by the dissociation will cause the solution to boil and splash the acid solution out of the beaker. 3

4 2. A 4 L container contains 13.0 g of N2 gas and 20.0 g of O2 gas. a. The temperature of the gas mixture in the cylinder is decreased to 0. For the N 2 gas, calculate: i. The mole fraction 13.0 g N 2 1 mol N 2 28 g N 2 = mol N g O 2 1 mol O 2 = mol O 32 g O Χ N2 = = ii. The partial pressure (in atm) P N2 = nrt L atm (0.464 V = mol N 2 ) ( ) (273 K) mol K = 2.60 atm 4 L b. A small hole is made in the container. Does the ratio of O2 to N2 increase, decrease, or not change? Explain in 2 complete sentences or less on the lines provided below. Graham s Law states that the rate of effusion of a gas is inversely proportional to the square root of the mass of the gas molecules. N 2 molecules have a lower mass, so they will effuse at a higher rate. Thus, the ratio will increase. c. The hole is plugged and a new mixture of 34.0 g O 2 gas and g NO gas are added to the same container. The new mixture is sparked and reacts to completion to form NO 2 gas. i. Give the balanced equation for this reaction 2NO (g) + O 2(g) 2NO 2(g) ii. Calculate the total pressure, in atm, after the reaction proceeds and the final gas mixture reaches 298 K g O 2 1 mol O 2 1 mol NO = mol O 32 g O 2 ; g NO = mol NO 2 30 g NO If the reaction proceeds to completion, mol O 2, mol N 2 will be consumed, mol NO 2 will be produced. So total number of moles in the container will be ( mol NO) + (2.125 mol NO 2 ) = mol total. P total = nrt L atm (6.667 V = mol N 2 ) ( ) (298 K) mol K = atm 4 L 4

5 3. The table below details the color of light that corresponds to ranges of wavelengths of light. Additionally, if white light is shined onto a solution, and a species absorbs a given wavelength of light, the perceived color of solution for each wavelength is also given. Wavelength of Light (nm) Color of Light Perceived Color of Solution if Light is Absorbed Violet Yellow-Green Blue Yellow Green-Blue Orange-Red Green Purple Yellow-Green Violet Yellow Blue Orange Green-Blue Red Blue-Green Given this information, consider the following questions: a. Solutions of barium, lithium, and potassium ions are prepared. A platinum wire is dipped into each solution, and the wire is held over a Bunsen burner. The color of the Bunsen burner flame turned to green, red, and violet, respectively. List the three ions by increasing energy of emitted photons. Assume that the wavelength of the photon is the midpoint of the range for that color. E Barium = E green = hc ( m2 kg m = J E Lithium = E red = hc ( m2 kg m = J E Potassium = E violet = hc ( m2 kg m = J Thus, the order is: Lithium < Barium < Potassium. (This question continues on the next page.) 5

6 b. If certain transition metal ions are dissolved in water, the solution appears purple. When those ions are present in concentrated chloride ion solution, the solution appears blue. When the ions are present in a concentrated solution of ammonia, the color of the solution seems orange-red. Rank water, chloride, and ammonia from lowest to highest in terms of the energy of photon that they cause the ion to absorb. Assume that wavelength of a photon is the midpoint of the range for that color. Purple solution Green light absorbed, with wavelength of 530 nm. E green = hc ( m2 kg = J m Blue solution Yellow light absorbed, with wavelength of 590 nm. E yellow = hc ( m2 kg = J m Orange-Red solution Green-Blue light absorbed, with wavelength of 490 nm. E green blue = hc ( m2 kg = J m Thus, the order is: Chloride < Water < Ammonia c. An aqueous solution of [Fe(H 2 O) 6 ] 2+ ions has a peak absorbance at 620 nm. If oxidized to [Fe(H 2 O) 6 ] 3+, the magnitude of the energy split increases by J. Predict the wavelength of the peak absorbance of the [Fe(H 2 O) 6 ] 3+ ion, and give the color of the solution. E = hc ( m2 kg = J m ( J) + ( J) = J λ = hc = ( m2 kg )( m s E J = nm; yellow solution 6

7 4. Consider the following reaction: Cr (aq) + I 3(aq) + H 2 O 2(aq) CrO 4(aq) + IO 4(aq) ; -1 +6; -2 +7; -2 a. Above each element in the reaction, write the oxidation number of that element. Clearly indicate which number goes to which element. b. Which species are reduced and which are oxidized? Reduced: H 2 O 2 Oxidized: Cr 3+, I c. Give the balanced ionic reduction half-reactions involved in this redox reaction in acidic aqueous medium. Be careful, there may be more than one half-reaction. 2e + + 2H (aq) + H 2 O 2(aq) 2H 2 O (l) d. Give the balanced oxidation half-reactions involved in this redox reaction in acidic aqueous medium. Be careful, there may be more than one half-reaction. 4H 2 O (l) + Cr (aq) CrO 4(aq) + 8H (aq) + 3e 3 12H 2 O (l) + I 3(aq) 3IO 4(aq) H (aq) + 24e (This question continues on the next page.) 7

8 e. Give the overall balanced ionic redox reaction in basic aqueous medium. Your answer should not contain any electrons. 2(4H 2 O (l) + Cr (aq) CrO 4(aq) + 8H (aq) + 3e ) 3 2(12H 2 O (l) + I 3(aq) 27(2e + + 2H (aq) 3IO 4(aq) H (aq) + H 2 O 2(aq) 2H 2 O (l) ) + 24e ) Adding these reactions together and balancing to remove electrons, you get the following overall reaction in acidic aqueous medium: 2Cr (aq) + 2I 3(aq) + 27H 2 O 2(l) 2CrO 4(aq) + 6IO 4(aq) + 10H (aq) + 22H 2 O (l) Adding hydroxides to each side to adjust the medium to basic, 2Cr 3+ 3 (aq) + 2I 3(aq) + 27H 2 O 2 (l) + 10 OH 2 2CrO 4(aq) + 6IO 4(aq) + 32H 2 O (l) 8

9 5. Dr. Hofstadter is preparing pure solutions of dichloromethane (84.93 g/mol) and tetrahydrofuran (72.11 g/mol) for his physics experiment. He steps out of the laboratory to grab lunch, and unbeknownst to him, Dr. Cooper plays one of his infamous pranks by mixing the two volatile liquids together into a large, sealed container. When Dr. Hofstadter returns, he is furious and demands that Dr. Cooper file the paperwork to dispose of the mixture. The vapor pressure of dichloromethane at room temperature is torr, while the vapor pressure of tetrahydrofuran at room temperature is torr. The vapor pressure of the room-temperature mixture is torr. What is the mass composition of the mixture? Assume that these substances form an ideal solution. Let dichloromethane be denoted as DCM and tetrahydrofuran as THF. P DCM = Χ DCM torr; P THF = Χ THF torr Χ DCM = 1 Χ THF P DCM + P THF = torr (1 Χ THF ) (352.3 torr) + (Χ THF ) (151.4 torr) = torr torr (352.3 torr)(χ THF ) + (151.4 torr)(χ THF ) = torr Χ THF = 0.457; Χ DCM = Assume 1 mole of waste solution: mol THF, mol DCM mol THF mol DCM % THF by mass = % DCM by mass = g THF 1 mol THF g DCM 1 mol DCM = g THF = g DCM g (100%) = % g g g (100%) = 58.32% g g 9

10 6. Ions of transition metals often have empty d orbitals. This means that other compounds with lone pairs of electrons can bond with these transition-metal ions. The compounds that result are called coordination complexes. a. Consider the following reaction: Cr NH 3 [Cr(NH 3 ) 6 ] 3+. Does NH 3 act as a Lewis acid or base in the formation of the [Cr(NH 3 ) 6 ] 3+ complex? Explain in 3 complete sentences or less on the lines provided below. Cr 3+ has empty 3d orbitals to receive electron pairs into, while NH3 has a lone pair of electrons on the central nitrogen to donate. A Lewis base is defined as any compound that can donate an electron pair to another compound (Lewis acid) that can accept the electrons. Since NH3 has extra electrons to donate into the 3d orbitals of Cr 3+, it would be the Lewis base in this complex formation. b. The d orbitals of free atoms are degenerate. That is to say, all 5 d orbitals are all at the same energy level. Fill in the energy diagram below with the proper electrons for Cr +, and explain briefly why it is stable as a free ion on the lines provided below. The electronic configuration for Cr + is as follows: [Ar]3d 5. Each electron is in a separate 3d orbital, and have the same spin, following Hund s rule of maximum multiplicity. There are two explanations for this phenomenon. One is that half-filled or completely-filled subshells are intrinsically more stable, so the five electrons will form a half-filled d subshell. Another fuller explanation is that in neutral chromium, the 4s and 3d orbitals are quite close in energy. The energy it would take to pair electrons in the 4s orbital is greater than the energy to place the 6 th valence electron unpaired into a 3d orbital. The chromium cation loses the 4s orbital, leading to the given electron configuration. (This question continues on the next page.) 10

11 c. However, in the [Cr(NH 3 ) 6 ] 3+ complex, the d orbitals do not all have the same energy. Their relative energies are shown in the orbital energy diagram below. Following all electron filling rules, draw in the electrons for Cr 3+ and explain briefly on the lines provided below the diagram why this oxidation state of chromium is stable in this complex. This compound is neutral, meaning that the oxidation state of chromium is Cr 3+. This means the electron configuration is [Ar]3d 3. For the adjusted d-orbital energy levels, each of the three d electrons will be placed into each of the three lower-energy d-orbitals, and will all have the same spin. This oxidation state is particularly stable in this complex because its half-filled sub-shell increases symmetry and exchange energy, lowering energy. This configuration also minimizes pairing energy. (End of Exam) 11

Last Name: First Name: High School Name: Individual Exam 3 Solutions: Kinetics, Electrochemistry, and Thermodynamics

Last Name: First Name: High School Name: Washington University Chemistry Tournament April 2, 2016 Individual Exam 3 Solutions: Kinetics, Electrochemistry, and Thermodynamics Please write your full name