<p>i put F > F+ + e- (delta h= postive).....i think that is right too</p>
<p>I thought the energy was a given though. I knew you needed energy but it said show the equation when energy is added. Why would you need to write energy? lol</p>
<p>I had a question about the ap chem frq's. Could someone tell me what the lab based question was? Also, would you actually have to do the lab to know how to answer the question, or was the question simply based on the concepts learned in the lab?</p>
<p>The lab based question was based on hydrates, warming it up to see how many moles of water are lost, and, I guess, determining the formula for the hydrate (somebody else elaborate with a better explanation...).</p>
<p>I dunno. The reason why I got some questions right was because I had remembered some aspects of the lab, such as using filter paper, and all. But the lab question should be answerable, I think, even if you haven't done the lab. I mean, most of the question dealt with stoichiometry anyway, so yeah...</p>
<p>Unofficial, somewhat abbreviated answer key. (I didn't bother showing the work, but you need to show work for credit. Not all the explanations are included.)
Question 1</p>
<p>a) Kp = P(CO)^2 / P(CO2)
b) n = .105
c) i) P(CO) = 6.74 atm ii) K = 27.9
d) No change. Catalysts affect rate, not equilibrium constants.
e) Decrease. Q = 2 which is less than K, therefore reaction will shift right. (Carbon is not a limiting reagent.)</p>
<p>Question 2
a) mass remains constant between second and third heating.
b) i) Mole of water = .102 ii) MgCl2 . 6 H2O
c) Calculated mass of water will be too high.
d) Add excess silver nitrate to solution. Weigh filter paper. Filter solution. Dry filter paper and precipitate. Weigh again. Subtract to find mass of ppt.
e) Moles = .191
f) 61.9%</p>
<p>Question 3
a) 0.96 V
b) -360 kJ/mol rxn
c) Greater than zero. (moles of gas increase)
d) i) Second order ii) first order
e) Rate = k [NO]^2[O2]
f) 7100 L^2 / (mol^-2 sec^-1) (thats liters-squared on top, and mols-squared and seconds on the bottom)</p>
<p>Question 4
a) Al(3+) + 4 OH- --> Al(OH)4^1- Complex ion decreases
b) 4 HCl + O2 --> 2 H2O + 2 Cl2 O2 is in excess
c) K2O + H2O --> 2 K+ + 2 OH- Turns pink</p>
<p>Question 5
a) F --> F+ + e-
b) Both have valence electrons in 2p (similar distance to nucleus), but F has more protons making electron harder to remove.
c) 1st IE for Xenon is less than fluorine. Xe has more protons, but valence electrons are much further from nucleus (5th energy level rather than 2nd for F) so they're easier to remove
d) XeO3 has one set of lone pairs on central atom. XeF4 has two sets of lone pairs on central atom.
e) i) Trigonal pyramid ii) sp3d2
f) polar. Presence of lone pair means molecule is not symmetrical. Dipoles don’t cancel out.</p>
<p>Question 6
a) solubility depends on similar intermolecular forces in solute and solvent. Pyridine is polar, benzene is nonpolar. Therefore pyridine is soluble in polar solvents like water.
b) BP depends on strength of interparticle attractions. Ether has only LDF and dipole-dipole. Alcohol has LDF and h-bonding which is stronger.
c) SiO2 is network covalent, SO2 is molecular. Much harder to break covalent bonds in SiO2 than the intermolecular attractions in SO2.
d) HCl has dipole-dipole and LDF. Cl2 has only LDF, but LDF increases with increasing numbers of electrons. (more electrons = more polarizable).</p>
<p>Damn, I'm a retard. I just realized I said that Xe first IE would be higher than F's. I don't know what I was thinking but thats ridicuoulsy easy. I hope its only worth one point though.</p>
<p>darn it...i wish i didn't see the above post by gfaith...if 75-80% of my free response is decent, can that put me into a 4 range...assuming my mc is just average...say 35-40 raw</p>
<p>for the ionization of fluorine i think you have to put that atomic fluorine is in the gaseous state (thats part of the definition of ionization energy)</p>
<p>for the Cl2 HCl one would it be enough to say that Cl2 is heavier or has a larger molecular mass than HCl and thus has stronger LDFs?</p>
<p>medkid06 - regarding the gaseous state of fluorine - you're absolutely right that it's part of the definition. On the other hand, they generally don't require states of matter to be included in reactions. (For example, they are never required in question 4.) For clarification of that detail we'll have to wait for the official scoring guide.</p>
<p>Regarding mass of Cl2 versus mass of HCl - that will definitely NOT earn the points. Comparing molar masses is a quick and dirty way of judging polarizability, but if often leaves students with the mistaken impression that it's the mass that matters (like it's some sort of gravity effect). It's NOT, and CB is kind of sensitive to that misconception. The scoring guides for past questions that were similar specifically pointed out that explanations regarding mass do NOT earn points. It's the polarizability of the electron cloud that matters.</p>
<p>My thoughts on the FR questions:</p>
<ol>
<li><p>I thought this question was pretty straightforward. My answers match all of gfaith's. Catalysts affect the rate of the reaction and not the equilibrium position, right?</p></li>
<li><p>On part a, I said something like that as the hydrate is heated, the mass decreases, indicating that the waters of hydration are coming off. I can't remember if I said that the hydrate is heated a sufficient number of times because the mass stays relatively constant between the second and third heating, suggesting that all of the waters of hydration have been drivin off.
: ( Finding the formula of the hydrate was pretty straightforward. While checking my work during the no-calculator part, I realized on the last section of this question that the mass percentage I calculated was not for the original mixture. I sat up the problem and everything without my calculator, but I'm not sure if I will receive credit since I did not have a simplified, numerical answer. What do you guys think?</p></li>
<li><p>I thought this question was pretty straightforward. Hey gfaith, are you sure your answer on 3a is correct? I got .28 V.</p></li>
<li><p>On the reaction between sodium hydroxide and aluminum hydroxide, I predicted that the product would be Al(OH)6 with a charge or -3. Now, however, after looking back I think the product is Al(OH)4 with a charge of -1. I'm not too sure if Collegeboard will accept my answer or not. After all the charge is correct. What do you guys think. The other two reactions were pretty straightforward.</p></li>
<li><p>During the test, for some reason, I could not figure out the ionization equation for atomic fluorine -- so hopefully that will only be worth a point! Aditionally, I'm not too sure if my prediction and explination for whether Xenon or Fluorine has the higher first ionization energy is correct or not. On the last part I said: Because of the presence of the lone pair on Xe, the bond polarities are not completely symmetrical to cancel each other. Therefore, the molecule is polar. Is 'bond polarities' acceptable terminology?</p></li>
<li><p>Parts c and d of this question were a little weird, but I think I said a few things that could potentially get me partial credit. Part c I said that SiO2 is a covalent network solid, and SO2 is a molecular solid. I also said something like covalent network solids generally have higher melting points or something -- can't remember exactly. Part d, i used a polarizability argument.</p></li>
</ol>
<p>Post any comments if you would like!
Thanks!</p>
<p>I agree with 0.96V for the answer to 3a. Did you remember to reverse the first reaction?</p>
<p>i thought E cell was E(cathode)- E(anode)....so i put .28V too.....anyone want to check this?....everything else i put agrees with you too...</p>
<p>The equation is E=E(cathode)-E(anode)</p>
<p>so.....the answer is .28V.....cuz
E(total)= 0.62
E(cathode)=????
E(anode)=-.34</p>
<p>so......0.62=(?????)-(-.34)</p>
<p>????= 0.28 V</p>
<p>Wow. I'm pretty sure I got almost everything right except reaction number 2 and the question about ionization energies of fluorine vs. xenon (silly mistake, don't ask). That can't be right. I probably missed units, or significant digits, or something.</p>
<p>And to the above, E(anode) was .34, not -.34.</p>
<p>^^^....yeah....but you had to flip it around right....?</p>
<p>You don't flip the anode when using that equation. My teacher always taught it to me as find the anode, flip it, and add it to the reduction. So that's how I solved it.</p>
<p>Im pretty sure the answer is .96</p>
<p>lol....oh well....i really dont care....that would be the only thing i missed on that question if i missed it....</p>