<p>Oops my bad. Yeah, equation 3 should be the net ionic and it’s colorless to blue. (I don’t know about whether or not clear counts though.) I think I got the charge wrong. How did you guys know what charge Cu oxidized to?</p>
<p>I thought it was to 2+, given the reactions in the voltage table</p>
<p>Here’s what I got for 5. (Btw you guys can correct me. I’m pretty sure I didn’t get all the answers right haha.)</p>
<ol>
<li>(a) all single bonds with lone pairs on the N’s
(b) No: the shapes around N are trigonal pyramidal
(c) C2H6 is nonpolar & only has LDF
N2H4 is polar with hydrogen bonding
(d) 2N2H4 + 2H2O –> 4NH3 + O2
O2 doesn’t stay dissolved in water and NH3 is a base.
(e) redox: N is oxidized, O is reduced
(f) positive: liquid + gas –> gas + gas; also gaseous moles increases
(g) false because breaking bonds is endothermic</li>
</ol>
<p>I only mentioned dispersion forces (I didn’t say “london”) and hydrogen bonding. i.e. I didn’t say polar. Is that a problem . . . ?</p>
<p>I think if something has hydrogen bonding, it has to be polar. I was redundant (but I wanted to be safe).</p>
<p>Here’s what I think I got for 3. Again, I don’t have a calculator, but hopefully you guys will understand what I mean.</p>
<ol>
<li>(a) 2<em>.285.8 kJ
(b) 5-ish mol * (285.8kJ/2 mol)
(c) the answer to a + 2</em>44.0 kJ (Hess’s law)
(d) 2H2 + O2 -> 2H2O
(e) 0.40 + 0.83 V
(f) (i) 0.93 mol H2 * (1 mol e-/2 mol H2) = ? mol e-
(ii) I = q/t
? mol e- * (96500C/1 mol e-) = q
t = 600. s (given)
(g) Fuel cells based on H2 only release H2O. Fuel cells based on hydrocarbons release CO2, which is a greenhouse gas.</li>
</ol>
<p>Btw is anyone else taking the AP lit test tomorrow? I don’t know why I’m doing this right now. Procrastination.</p>
<p>For 2 was NaHCO3 what was used to clean up the spill?</p>
<p>Disclaimer: I’m probably wrong for a lot of this. This is what I put on my exam though, at least as far as I can remember.</p>
<p>1a. pH = 1.000, that’s the proper amount of sig figs btw
1bi. Kb = [NH4+][OH-]/[NH3]
1bii. By an ICE table, Kb = x^2/0.100, x = [OH-] = 1.3E-3 M
1ci. Ka = Kw/Kb = 1.0E-14/1.8E-5 = 5.6E-10
1cii. Initial [NH4+] is 0.0500 M from the NH4CL solution + 0.007 from the OH- solution = 0.0507 M
Initial [NH3] = 0.0494 M
Initial [H3O+] = 0
By an ICE table, [H3O+] = [H+] = 5.5E-10 so pH = 9.26
1di. I answered really generically: HCl is a strong acid, so adding it to a weak base (NH3) buffer limits the buffering capability, making it not a good buffer anymore
1dii. Since HCl is a strong acid, it completely dissociates, and this reaction takes precedence over the proton donation of NH4+. Basic stoich gives you 0.0025 moles of Cl-, which combine with just as many moles of NH4+ to form NH4Cl. This leaves you with 0.0507*(2/3)(0.075) - 0.0025 = 3.5E-5 moles of NH4+ left / 0.075 L = 4.7E-4 M NH4+</p>
<p>2ai. MVo = MVf, so Vo = MVf/Mo = 6*50/16 = 18.8 mL
2aii. Don the safety goggles and rubber gloves. Fill one graduated cylinder with 19 mL of 16 M HNO3, fill another with 50-19 = 31.2 mL of water, add the acid to the water and stir. (I suck at labs so I’m probably wrong.)
2aiii. Volumetric flask here is significant to the thousandth place. Our next-closest measurement, in terms of accuracy, is the graduated cylinder, which is only significant to the tenths place. This added significance is unusable, so it doesn’t pay to use such a specific instrument for this dilution.
2aiv. Use the 5% NaHCO3. The Na will neutralize the NO3-, and the HCO3 will combine with H+ to form water and carbon dioxide gas. Distilled water will merely dilute the acid, and NaCl will just form HCl on your hand…
2b. Precipitate mass = 29.2598 - 28.7210 (crucible + precipitate - crucible) = 0.5388 g AgCl * (1 mol AgCl / 143.023 g) = 3.767E-3 mol AgCl
2c. 3.767E-3 mol AgCl * (1 mol Ag/1 mol AgCl) * (107.87 g Ag / 1 mol Ag) = 0.4064 g Ag / 0.6489 g alloy * 100% = 62.62% silver by mass</p>
<p>3a. By Hess’s Law, 2 mol H2O * -285.8 kJ / mol - 0 kJ (0 kJ for standard heat of formation for monoelemental molecules or whatever) = -571.6 kJ/mol
3b. 10.0 g H2 * (1 mol H2 / 2.016 g H2) * (-571.6 kJ / 2 mol H2) = -1418 kJ, so 1418 kJ of heat released
3c. -571.6 kJ/mol + 2*44 kJ/mol = -483.6 kJ/mol
3d. Top reaction is reduced, lower is oxidized, so O2 + 2H2 –> 2H2O
3e. E°cell = E°red - E°ox = 0.40 - -0.83 = 1.23 V
3fi. 0.93 mol H2 * (2 mol e- / 1 mol H2) = 1.9 mol e-
3fii. 1.9 mol e- * (96500 C / 1 mol e-) * (1 / 600 s) = 310 A (lolwut)
3g. Hydrocarbon cells release carbon dioxide, which contributes, as a greenhouse gas, to global warming; hydrogen cells only release water.</p>
<p>4ai. Mg(OH)2(s) + 2H+ —> Mg2+ + 2H2O
4aii. 0.10 mol Mg(OH)2 * (2 mol H+ / 1 mol Mg(OH)2) * (1000 mL / 2.00 mol H+) = 1.0E2 mL HBr solution
4bi. 4Cl- + Co2+ –> Co[Cl-]4 2+
4bii. Cl- is a Lewis base since it provides both the electrons for the coordinate covalent bonds.
4ci. NOTE: I’m 99% confident that using either copper(I) or copper(II) will be accepted as accurate, given that both are valid copper cations found on the half-reactions sheet. With that being said, I did copper(I) for simplicity.
Cu(s) + Ag+ –> Ag(s) + Cu+
4cii. Since copper cations tend to form blue solutions, the solution will likely turn blue. Also, the copper wire will decrease in mass, and a silver precipitate will form at the bottom of the solution.</p>
<p>5a. Single bonds for everything, with no extra electrons on the Hs and a lone pair on both Ns. All formal charges sum to zero so this structure is ideal.
5b. All six atoms do not lie in the same plane. According to VSEPR, adding in nonbonding domains to an atom with three bonding domains on it is likely to create a trigonal pyramidal form in which the central atom is on a higher plane than the bonds. We are likely to see a sort of elongated pyramidal shape in this molecule.
[Hydrazine</a> - Wikipedia, the free encyclopedia](<a href=“http://en.wikipedia.org/wiki/Hydrazine]Hydrazine”>Hydrazine - Wikipedia) - check the ball and stick model. My prediction was sort of right hahaha
5c. N2H4 can form hydrogen bonds with other molecules due to the polarity of the H-N bonds. Hydrogen bonds are especially strong IMFs. Contrastingly, ethane is a nonpolar molecule that undergoes London dispersion forces. While LDFs can be strong on long-chain nonpolar molecules, on a molecule as small as ethane, its IMFs will be significantly wearer than hydrazine’s, so hydrazine has a higher boiling point.
5d. N2H4 + H2O –> N2H5+ + OH-
5e. Redox: N is reduced from +2 to 0, O is oxidized from 0 to +2
5f. dS is positive because there are more moles of gas in the products than in the reactants
5g. The statement is false because breaking bonds is an endothermic action (it requires energy and dH° is positive). dH° is negative here because the formation of the NN triple bond and the H-O bonds must release more energy than required to break the N-N, O=O, and H-N bonds.</p>
<p>6a. The pressure in the flask is equal to 100 tore because the equilibrium vapor pressure (the pressure at which there is an equilibrium between evaporation and condensation) is 100 torr. That value is the pressure in the flask once equilibrium is reached.
6b. Kinetic molecular theory states that pressure and kinetic energy are directly proportional. As you increase the flask’s temperature, you increase its kinetic energy, so pressure increases too. Also, when you heat a gas, it expands, but because the beaker is rigid, pressure increases as a result.
6ci. The order is zero because the reaction rate is constant regardless of what the ethanol concentration is, as shown in the leftmost graph. Also, were the reaction 1st or 2nd order, the graph of ln[ethanol] vs. time or 1/[ethanol] vs. time, respectively, would be linear. Since it is not, we can also show by deduction that the reaction is zero order.
6cii. rate = k
6ciii. ethanol disappears at a rate of (.0100 - .0080)/(500) = 2E-3/5E2 = .4E-5 = 4E-4 mol/Ls, and since rate = k, k = 4E-6 mol/Ls (moles per liter seconds)
6d. Po/no = Pf/nf. At the start of the reaction, only 1 mol of ethanol exists, but at the end of the reaction, 2 mols of gas exist (1 mol ethanal and 1 mol H2 gas). Thus, Pf = Po<em>nf/no = 0.40</em>2/1 = 0.80 atm</p>
<ol>
<li>(a) (i) M1V1 = M2V2
(ii) Put on goggles & gloves, etc. (don’t feel like typing this out)
(iii) You don’t need to be that precise/ that many sig figs
(iv) NaHCO3 because it’s a weak base that’ll neutralize the strong acid
(b) (29.2598g - 28.7210g)/(molar mass of AgCl)
(I’m not sure what you were supposed to use for the mass of crucible + precipitate. But I’m pretty sure you weren’t supposed to use the first one because that was probably off due to temperature.)
(c) (answer to b * molar mass of Ag)/0.6489 * 100%</li>
</ol>
<p>Keasbey Nights, our answers look similar. : D</p>
<p>Do you guys think it’s okay that for the silver/copper redox reaction I used Cu+, not Cu2+? I saw it on the half reactions sheet so I figured it was legit but it seems like a lot of people used Cu2+…</p>
<p>I’m hoping it’s one of those things where they’ll take either one.</p>
<p>2aii) I think that’s right, but I would have said to place it in the beaker. But unfortunately, I super rushed the problem and just said take 18.8 mL of the HNO3 and dilute it to 50 mL, totally not accounting for the fact that they might mix together and more than 32.2 mL of water would be needed to dilute it to 50 mL. </p>
<p>After the past posts, I feel a lot better-- I screwed up a little in some areas and definitely didn’t do as well on the equations as I should have. Thanks guys ! :)</p>
<p>Unbelievable! Typical of the collegeboard, they have linked the 2010 test to the AP Chem 2011 (Form A) test!!!
First they write a horrible test.
Now they do not even link it properly so that the teachers can see just how pathetic the questions were!</p>
<p>chembites - what?</p>
<p>Keasbey, the only questions we got different answers for were 1dii, 5d (I’m pretty sure I was wrong), and 6b (half the same, half different). Niceee.</p>
<p>I picked NaHCO3 as well, but my justification was that it would produce a gas which would be easier to clean up.
Yeah, I had no idea what I was doing.</p>
<p>@Keasbey Nights</p>
<p>Looks good. Pretty much the same thing i got. Except on #2 I got 20ml cause i sig fig from 6M HNO3 and I got 61.33% or something like that for percent mass. I think we just used different sig figs and stuff. But all the rest is same as me.</p>
<p>haha i got pretty much the exact same thing as Keasby for all questions except 1c,d and 4,b good stuff</p>
<p>Hey does anybody know when the scores of the exams are released?</p>
<p>I believe mid to late July.</p>
<p>And CollegeBoard is just evil for posting the 2011 FRQs but not the answer key.
<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>@flamingmango: shouldn’t #2 still be 2 sig figs because 6M and 16M don’t count since they are exact numbers?</p>