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1 Chem1C Second Midterm Spring 2006 (Kahn) Your la TA s name Grading Ruric: Your full name Your PERM # Your signature Q1 (5) Q2 (5) Q3 (5) Q4 (5) Q5 (5) Q6 (5) Q7 (8/4) Q8 (8/4) Q9 (6/3) Q10 (8/) Total (60) Comments y the grader:

2 1) Radioactivity. Common household smoke detectors use Americium isotope 241 as a source for α-particles that, in the asence of smoke, ionize air. One α-particle has enough energy to ionize aout hundred thousand nitrogen and oxygen molecules, resulting in a small steady ion current flow through the detector. Smoke particles asor and deflect α-particles causing loss of the ion current, which will e sensed y the detector. Which of the nuclear reactions elow correctly descries the process that generates α-particles from americium-241? ) Am n Am e ~ ν e ) Am Cm e ~ ν e Am Np 3) ) Pu Am e ~ ν e ) 95 Am + 0n 47 Ag+ 53I+ 0γ He α-particle is the helium nucleus. See assigned prolems in the syllaus if you got this wrong. 2) Transition metal complexes. Consider the four complexes of Zn 2+ with four different ligands: aqua, ammine, chloro, and cyano. In each complex, four identical ligands are ound to the central metal atom, i.e. we have tetraaquazinc(ii) cation, tetraamminezinc(ii) cation, tetrachlorozincate(ii) anion, and tetracyanozincate(ii) anion. Identify one correct statement aout these complexes. 1) All four complexes are paramagnetic 2) Solutions of all four complexes appear visually similar 3) The tetraammine complex is octahedral ut the tetraaqua complex is tetrahedral ecause the ammine ligand is a stronger ligand than the aqua ligand 4) The oservation that the addition of ammonia to the tetrachlorozincate(ii) complex leads to the increased electrical conductivity of the solution can e explained y postulating that the energy splitting etween the e g and t 2g level is smaller in the tetraamminezinc(ii) complex than in the tetrachlorozincate(ii) complex 5) When the positive charge of tetraamminezinc(ii) cation is neutralized y the sulfate counterion, a highly corrosive solution known as the cleaning solution is otained Because Zn 2+ has d 10 configuration, all five d-oritals are filled and there are no electronic transitions etween d-levels. No electronic transitions means no visile color, all are colorless.

3 3) Transition Metals and Coordination Compounds. Identify an incorrect statement 1) Solule coordination complexes of copper(i) tend to have four ligands and have intense lue color 2) Potassium permanganate can readily oxidize many organic molecules and for safety, permanganate and glycerol should not e stored in the same chemical cainet. 3) Both Co 3+ and Ni 2+ can have linkage isomers with SCN - as a ligand 4) There is one and only one pair of mirror isomers possile for the [Pt(H 2 O) 2 (CN) 2 Br 2 ] complex 5) For the same ligand, a tetrahedral complex typically asors light more toward the red end of the spectrum as compared to the octahedral complex Copper(I) tends to have four ligands ut solutions of its coordination complexes are colorless ecause Cu + has d 10 electronic configuration. Many Cu(II) coordination complexes have intense lue color. 4) Properties of solutions. Identify a correct statement. 1) In order to calculate the mass percent composition of the solution from the mole fraction of the solute, the molecular weight of the solute and the density of the pure solvent must e known. 2) Soluility of most solids in water depends significantly on the atmospheric pressure 3) Henry s law allows to calculate the vapor pressure aove the open container with dissolved solid if we know the Henry s constant and the mole fraction of the solid in the solution 4) Soluility of most gases in liquids decreases as the temperature is increased 5) Ferrofluids are true solutions of paramagnetic [Fe(NH 3 ) 4 (H 2 O) 2 ] 2+ ions in water Try serving warm Champagne in your next party if you do not elieve it.

4 5) Miscellaneous definition. Identify an incorrect definition. 1) Geometric isomers are species with the same formulas and same onds ut with different positions of atoms (e.g. with respect to the central atom) and with different chemical properties 2) Reduction is a process y which an oxidized species gains electrons and reduced species loses electrons 3) Chelate is a ligand that can form only one ond with the metal atom 4) Ideal inary solution of two liquids is a solution that oeys the relationship P solution A 0 A B 0 B = χ P + χ P 5) Lysis is a process in which acteria urst open when placed in a solution that is hypotonic when compared to the interior of the acteria Chelate is a ligand that can form more than one ond with the metal atom. 6) Liquids, Solids, Gases and Solutions. Identify an incorrect statement. 1) Ice melts at a temperature where the liquid water and solid ice have identical vapor pressures 2) The superheating is a phenomenon in which the liquid achieves temperatures aove its normal oiling temperature at the normal pressure 3) The enthalpy change that occurs when one mole of liquid is converted into one mole of vapor at the oiling point is called the heat of fusion 4) The temperature of the oiling anana oil remains constant as long as the pressure aove the oiling liquid solution is constant 5) Dissolved nonvolatile molecules reduce the vapor pressure of the volatile solvent That would e called heat of vaporization

5 Prolem Solving 7) Maltose is a disaccharide where two glucose molecules are joined together that is formed from germinated grains y the action of enzymes. Fermentation of maltose-containing solutions yields eer. Calculate the molar mass of maltose if a 13.41% solution of maltose in water has a oiling point of C. The molar oiling point elevation constant for water is 0.51 C kg/mol. (8 pts) ΔT = C i = 1 (like glucose, maltose does not dissociate) K = 0.51 C kg/mol Assume 100 g of solution, ut this is aritrary Step1. Find molality: m = ΔT / (i K ) = mol per kg of water Step2. Figure out the composition of solution: g maltose g water Step3: Find moles of maltose: mol maltose 1000 g water X mol maltose g water X = moles in 100 g of solution Step 4: Find molecular weight: M w = 342 g/mol g maltose mol maltose M w g maltose mol maltose 8) Ethanol (C 2 H 6 O) has a normal oiling point of 78.4 C, density of g/cm 3, and the heat of vaporization of ethanol at the oiling point is 42.4 kj/mol. Mount Everest is the tallest mountain on the Earth. With its atmospheric pressure of 240 torr and the temperature during the winter of 36 C it is a dangerous place to e. Calculate the oiling point of ethanol at the top of Mount Everest. The normal atmospheric pressure at the sea level is 760 torr. (8 pts) P ΔH norm vap = 1 1 ln R PEverest TEverest T norm 760 ln 240 = T Everest T = Everest T Everest = K = 52.6 C

6 9) Imagine that you have just synthesized a novel compound that is right pink liquid at room temperature and can e pulled toward strong magnets. Propose and descrie an experiment, including analysis of your data, that allows determination of the heat of vaporization of this liquid (6 pts) To determine ΔH vap for any liquid regardless of its color or magnetic properties, a plot of ln P vap versus 1/T should e constructed. Thus, one needs to measure the vapor pressure of the liquid at different temperatures and make a plot where ln P is on the Y- axis and 1/T is on the X-axis (note: temperature should e in Kelvins, ut pressure can e in any units due to the way logarithms work). Once we have the plot, calculate the slope, which is equal to -ΔH vap /R. ln P slope = - ΔH R 1/T Partial credit if the use of Clausius-Clapeyron equation was correctly invoked 10) Using the phase diagram of water that you learned in this course, provide illustrated explanation for the following two phenomena: (8 pts) a) Increase in the oiling point of the solution in the presence of a nonvolatile solute (4 pts) The lack thick line separating liquid and vapor phase shows the vapor pressure of water as a function of temperature. Nonvolatile solutes lower the vapor pressure of volatile liquids (Raoult s law) as shown y the green thin line. The oiling occurs when the vapor pressure of the liquid equals the atmospheric pressure (1 atm). It can e seen that the vapor pressure curve in the presence of solute meets the 1 atm line at higher temperature. ) Melting of ice at suzero temperatures and high pressures (4 pts) The line separating the solid and liquid phase in water phase diagram has a negative slope. Water exists as solid ice at suzero temperatures and 1 atm, however increasing the atmospheric pressure at constant temperature (red arrow) will get the water into liquid regime. P.S. The slope is negative ecause water at near 0 C is denser than ice at these temperatures.

Your full name Your PERM #

Chem1C Second Midterm Spring 26 (Kahn) Your lab TA s name Grading Rubric: Your full name Your PERM # Your signature Q1 (5) Q2 (5) Q3 (5) Q4 (5) Q5 (5) Q6 (5) Q7 (8/4) Q8 (8/4) Q9 (6/3) Q1 (8/4) Total (6)