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U-M EARTH 125 - Mendelian Genetics & More Proof for Natural Selection
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These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute. Outline of Last Lecture I. Important Reminders II. Reaction Quotient III. Temperature & Solubility & Supersaturated Outline of Today’s Lecture I. Quiz Question II. Freezing point and Henry’s Law III. Solutions and solubility Today’s Lecture Quiz Question: 1. Will precipitate form given the Ksp of PbCl2 is 1.7 x 10^-5 and [Pb2+] = 10^-2 M and [Cl-] = 10^-2 M? No because K > Q where Q = [10^-2][10^-2]^2 = 10^-6 Freezing point and Henry’s Law: 1. Assume the partial pressure of CO2 inside the bottle above the liquid is 2 atm. The Henry’s Law constant for CO2 is 0.117 M * atm-1. Determine the molar concentration of the gas, CO2. Answer: Cgas= kH Pgas From the question we know: kH = 0.117 M*atm-1 and Pgas = 2 atm Cgas= (0.117 M*atm-1)(2 atm) Cgas= 0.234 M 2. Determine the freezing point depression of the soda, given that Kf = 1.86 °C/m Assume that all we have is CO2 in pure water ∆T = -iKfm i = 1 (CO2 is a molecular solute) Answer: Kf =1.86°C/m(fromtable) m = moles solute/kg solvent Cgas= 0.234 moles solute/L solution (#moles CO2)/(mass H2O) = 0.234 mol/0.989704 kg = 0.236 m The Freezing point depression is then: ∆T = -iKfm = -1*(1.86 °C/m)(0.236 m) = -0.44°C 3. Does your answer support your explanation for why one of the soda’s froze upon opening and the other did not? Explain. Answer: No. Based on the small amount of solute dissolved in the solvent there is not enough CO2 to keep the soda solution from freezing in the -8°C salt bath. CH 302 LaBrake, Cynthia 2013 Spring Week 4 Lecture 7 February 5, 2013Solutions and Solubility: 1. Describe what happens thermodynamically when sodium acetate dissolves in water (include S, H, G) Answer: G Since the sodium acetate does dissolve in water spontaneously, G is negative S Since sodium acetate was in a crystal lattice and is now dispersed in water molecules, there are more microstates in the solution. That is, S has increased and is positive. H Since we observed that the solution got cold as the sodium acetate dissolved, we know that this is an endothermic process (heat pulled from surroundings and into the system) and H is positive. 2. When one crystal of sodium acetate is added to a super saturated solution of sodium acetate what happens to the solution? Answer: The solution is so saturated that the single crystal begins a cascading effect of most of the sodium acetate crystallizing or crashing out of solution. 3. Do you expect that the test tube should feel warm or cool? Explain, thermodynamically, your choice of warm or cool? Answer: This solid is then hot, which we should expect. If the dissolving process was endothermic, the crystallization process (the reverse of dissolving) should be exothermic. As stable bonds are made in the crystal, heat is released. 4. The sodium acetate concentration in the supersaturated tube was 20 molar. The Ksp for NaCH3CO2 = 25. In this case is Q > Ksp or < Ksp at room temperature. Does this calculated value of Q fit with your observation? Answer: [NaC2H3O2] = 20 Q = [Na+][C2H3O2-] Q = [20][20] = 400 (each ion is 20 M because the ions each exist in a 1:1 ratio with the original salt). 400 > 25 Q > Ksp Yes, this does fit with the observation as the sodium acetate will fall out of solution with a single crystal to initiate the reaction. It is somewhat surprising that this super saturated solution can exist at


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U-M EARTH 125 - Mendelian Genetics & More Proof for Natural Selection

Type: Lecture Note
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