<p>I was quite confident a few weeks ago because I really knew Thermo, Kinetics, ACid/Base and whatever. That helps a lot with the FR…
BUT we just did a practice MC set in class…and BOY it’s hard. I even studied a bit on bonding and whatnot the night before, but there were so many questions where I didn’t even know where to start or what info I needed to answer it.</p>
<p>Ohhhhhh dear. I guess it’s time to study.
Definitely won’t be there on Monday…</p>
<p>No calculator on the MC.
Calculator allowed on the first 3-4 problems, aka Equilibrium and other calculation based stuff.
No calculator for the equations and lab based question.</p>
<p>^ but are we allowed to program equations into our calculator beforehand? To save time on stuff takes a while to work out, like the quadratic equation for equilibrium, or to avoid elementary mistakes in calculation?</p>
<p>Uhh, free response and MC are worth the same (50% and 50%). And no, you do not need a 107/150 for a 5; it is much less than that. The curve is around 60%-65% for a 5. So to be safe, probably a 97/150 for a 5.</p>
<p>Any predictions for the FRQ other than the usual equilibrium and predict the reaction?</p>
<p>I’m guessing bonding and periodic trends will be the non calculator ones (according to PR those two appear very often). As for the two calculator ones, maybe gasses and thermo?</p>
<p>Cicero, I don’t know where you get your information from, but it says on Collegeboard that the MC is worth 45% and the FR is worth 55%. The same is said in the Kaplan 2008-2009 book.</p>
<p>FR
Equilibrium
Kinetics, Acid Base, Thermo, or anything else
Same as above
Reactions
5)Lab question/or another descriptive chemistry problem
Usually bonding theory</p>
<p>MC is basically simple formulas with simple calculations. Do a lot of rounding :D</p>
<p>So weird! It must’ve been an old test where I read it off from then. The format I listed out for the FRQ is pretty dead straight though, from doing so many AP Central tests.</p>
<p>Princeton and Kaplan are great, you might as well just throw yourself off a cliff using Barron’s. The AP Chem test is already hard enough, and Barrons isn’t necessary in this case. </p>
<p>PR and Kaplan are both comprehensive, but they’ll only reinforce what you already know, not reteach you concepts like the ARCO/Peterson’s book. They’re good for practice though.</p>
<p>How likely is it that titration will show up on free response?
When was the last year it showed up on?</p>
<p>Let’s hope it does this year. If you have probs with titration, this is not meant to discourage you or anything, but seriously just review the topic… I just finished going through it. Not that bad. :)</p>
<p>To put this out there, I’m using 5 steps … and it’s great. It’s midway through PR and Barrons and I think that’s precisely what I need right now because it’s more than review, and less than teaching the concepts in complete detail.</p>
<p>A typical titration problem will include the volume and concentration of the weak acid/base and then will include consecutive volumes added of the strong base/acid and you’ll have to figure out the pH changes for each one.</p>
<p>Basically 4 things to keep in mind while tackling this: 1. If it asks you to calculate the pH before one starts adding ANY strong base/acid, all you have is a simple weak acid/base solution of known concentration. This is just a K(a)/K(b) problem. 2. As the strong base/acid is added, a mixture of weak acid/base and its conjugate is formed. This is a buffer problem. First do stoich, to figure out the moles of the conjugate formed and then calculate the concentration (keeping in mind the total volume of the solution - this will be the initial volume + the volume of the strong base/acid). After this you simply apply the “Hasselhoff” equation. 3. When the equivalence point is reached (this happens when the moles of strong base/acid is equal to the moles of the weak acid/base) - the only thing in the solution is the conjugate. This is simply a K(b)/K(a) problem. Just treat the solution as a weak base/acid. For K(b) be careful as you’ll be calculating [OH] and if the question asks you to calculate [H], use pH + pOH = 14. 4. After the equivalence point, you have an excess of strong base/acid and this will determine the pH. This is like the first step.</p>
<p>Hope this helps.
Now what’s the chance of this being a free response this year? @giant - unfortunately, you don’t.</p>