<p>i said for 5e it’s decomp because it’s combustion , a type of decomp?</p>
<p>I kept it positive.</p>
<p>1d) It was a buffer system technically (it had a weak base and its conjugate), but it wasn’t an effective buffer, which is what it asked, because the concentration was tiny compared to that of the added HCl.</p>
<p>4c) ii. Silver solid forms, but the solution also turns blue because of the copper (II) ions.</p>
<p>Also, for 3g I doubt you’d get points off, but hydrogen is really explosive as well, so that isn’t the best reason.</p>
<p>I don’t know why people keep saying that the curve is set beforehand. If you think about it, that makes NO SENSE. That means that the curve they used last year would be the same as the one they used this year. However, last year’s exam was one of the easiest CB has given while this year’s was one of the hardest. The cutoff to get a 5 will be much lower on our test.</p>
<p>This is how I think the curve works. CB has a standard distribution of the % of each score. What I mean is that each year, CB has about 15% 5’s, 20% 4’s, and what ever the other percents are. So they first grade each of the exams, and then adjust the cutoff for each score to match the percents. Anyone else think this is true?</p>
<p>I agree @bioboy12</p>
<p>for the reaction 4c …</p>
<p>I simply put Cu + Ag+ –> Cu+ + Ag</p>
<p>…? how many points would i get?</p>
<p>This is what i read: On the AP exam, the top 20% of all students taking the exam earn 5’s, the next 20% earn 4’s, then 3’s, etc.</p>
<p>Well you balanced, so one (: and right reactants so one more (:</p>
<p>@echoyjeff222:
You forgot to balance properly. Cu becomes Cu^2+. So the Ag on both sides has a coefficient of 2.</p>
<p>So I was talking to my chem teacher and he said that this year College Board changed their policy about the sig figs. They pick one question where they check sig figs and if you have it wrong then you get points taken off. But on the other questions, the number of sig figs you put doesn’t really matter. :)</p>
<p>Well, no, because echoyjeff222 wrote the product as Cu+ instead, so the charges ARE balanced.</p>
<p>^ good to know!</p>
<p>Cu can be +1 or +2 if its cuprous or cupric…if you set it up to be cuprous, the charge is +1</p>
<p>so did i get all the points ? o.O or did i get one off for not saying that it’s 2+</p>
<p>I think that they would want you to go with the most common oxidation state of Cu which is 2+. However, they also should give points for 1+ because it can exist…</p>
<p>According to the reduction potential charts, though, it is more favorable for Cu to become Cu2+. But CB is usually pretty lenient with these sorts of things, so you might get full credit.</p>
<p>Note: These are not my answers, they are Keasbey Nights answers to the FRQ. I just thought I would post it in this thread as well :)</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 - Wikipedia, the free encyclopedia - 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>
<p>a few questions … </p>
<p>for 5e. can it be decomposition since it’s a combustion reaction??
6c. if i started off with 1st order and did 0.693/k = 1/2 life to find k value and such, i would only miss the first and then get the rest, right?</p>
<p>@echoyjeff222:</p>
<p>I don’t think it can be decomp. because the oxidation numbers do change, and I don’t know if a reaction can be both redox and decomp.</p>
<p>However, I do know the answer to the second part! You would lose points on the first part, but assuming your method was correct, you would not be penalized for future parts involving that value.</p>
<p>whew!
thanks!</p>
<p>Anyone else think that the Form B FRQs looked A LOT easier? They got a stoichiometry/gas question!!!</p>