<p>@Frizzle: I just did my first FRQ this morning. I’m guessing I’ll get maybe 15/75 on MC and 1/4 of the points on Form A and B.</p>
<p>lol godzillla what are u aiming for? a 15/75 is pretty low…</p>
<p>If your aiming for a 4 or above how many should you look to answer correctly on the MC and Free response?</p>
<p>still need help with my question:
XY2(aq) <–> X^2+ (aq) + 2 Y^-
a soluble compound XY2 dissociates in water according to the equation above. In a .050 molal solution of the compound, the XY2(aq) species is 40% dissociated. In the solution the number of moles of particles of solutes per 1.0 Kg of water is closest to? </p>
<p>0.050mol/kg * 1kg = .050mol
So you make .050mol*.4 of X and .050 *.4 * 2 of Y and you have 0.050mol * .6 XY2 left over. Sum those and you get 0.90mol</p>
<p>here is the new mcq curve thingie:
<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>scroll down for the new one~
to see where u should be for a 5/4/3</p>
<p>I’m aiming for a 3. I did 45 MC in like 10 mins and got 15/45 LOL. I’m so going to bomb this.</p>
<p>ty vadata!</p>
<p>lol g0dzilla</p>
<p>i am aiming for a 4!</p>
<p>i really need it also to prove i wasn’t sleeping in class the whole time (wat my teacher said~). she says if i can pull a 4 or better she will give me a final average of an A- for a 4 and an A for a 5.</p>
<p>shooting for the A-. if i get a 3 or worse i end up with a b+</p>
<p>my overall plan is to get 50 MCQ right and try and get 40-50 points out of part 2s. according to this i could possibly get a 4:P:
<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>Praying that I at least get a 3 loll</p>
<p>I have another question on <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; Question 2, part f)</p>
<p>How do we figure that out? I couldn’t follow the given solution of CB.</p>
<p>XY2(aq) <–> X 2+ (aq) + 2 Y -</p>
<p>i .05 0 0</p>
<p>c -.02 (40%) +.02 +.04 (.02 x 2)</p>
<p>e .03 .02 .04</p>
<p>.03 + .02 + .04 = .09</p>
<p>How did anyone get .11?!</p>
<p>I have the most trouble on problems involving atomic structure/lewis dot structures and gases. Math is pretty solid for me.</p>
<p>can you guys help me understand this ap mc problem?
The nonvolatile compound ethylene gylcol, C(2)H(6)O(2), forms nearly ideal solutions with water. What is the vapor pressure of a solution made from 1 mole of C(2)H(6)O(2) and 9 moles of H(2)O if the vapor pressure of pure water at the same temp is 25 mm Hg?
a) 2.5 mm Hg
b) 7.5 mm Hg
c) 12.5 mm Hg
d) 22.5 mm Hg
e) 22.5 mm Hg</p>
<p>the answer is e, btw</p>
<p>Options d and e are the same?</p>
<p>my bad, i meant 27.5 mm Hg for e), the answer is d) 22.5 mm Hg, not e)</p>
<p>so according to my general knowledge anions are reduced and cations are oxidized?
i was looking at that galvanic question someone posted earlier…</p>
<p>LOL EDIT: I am mad sleepy and i wrote it wrong.</p>
<p>ANION IS OXIDIZED AND CATION IS REDUCED MYBADDDDDDDDDD</p>
<p>Nevermind let me see if e is not the answer. Ptotal = Pwater + P(C(2)H(6)O(2)) might be used</p>
<p>Okay so how I saw it was like… vapor pressure of a solute + solvent is greater than a pure solvent… thus it had to be greater than 25 mm Hg. e) is the only answer. </p>
<p>Maybe I looked at it the wrong way, but the answer is correct.</p>
<p>This is a vapor pressure lowering problem.<br>
P=XP°
where X is mole fraction of the solute, and P° is the vapor pressure of the pure solvent.
There are 10 total moles, so mole fraction of water is 9/10.
P=XP°
=(9/10)(25)
=22.5</p>
<p>@tomato
the partial pressure formula - raoult’s is P(solution) = P(h20) x mole fration
To get mole fraction = mole H2O/ mole (H2O + C2H6O2) = 9/10
so its 25x 9/10</p>
<p>^f my life, i did that and did the math wrong originally and got 2.5, thanks for the help :)</p>