<p>There is an older version released from 1994 and the other one is from the latest sat II publication by college board.</p>
<p>Is Kaplan a good study book if I plan on getting 800?
I recently tried to read through Barron’s AP book… so many typos.</p>
<p>And also, will I be given equations like on the AP test? or will I have to memorize everything except the periodic table?</p>
<p>oh, and quick actual study question:</p>
<p>most formulas have freezing/melting point depression as dT = (K)(m)… why do they always neglect i (van hoff’s factor)???</p>
<p>this formula, excluding van hoffs was given to me in kaplans… now i have doubts to the legitimacy of this entire book. and im only halfway through ahah</p>
<p>Let’s keep this thread going!</p>
<p>
</p>
<p>If you want an 800, Barron’s is the way to go. It has its share of typos, but they’re easy to spot if you’re a half decent chem student and it really makes you pay attention.</p>
<p>
</p>
<p>Periodic table only is given. </p>
<p>
</p>
<p>The difference is in what m means. Kaplan’s formula is correct, but there needs to be a distinction</p>
<p>dT = (K)(m) where m = molality of the SOLUTION
dT = (K)(m)(i) where m = molality of the SOLUTE</p>
<p>In the first equation, m solution = i x m solute.</p>
<p>Anyone have any tips/facts that would be helpful?</p>
<p>Most questions about colligative properties (freezing point depression, etc.) aren’t based on real calculations. Usually, questions will ask you to determine which solution will have the lowest freezing point, and, in order to answer, all you really need to know is that ionic compounds tend to dissociate in water. For instance, NaCl fully dissociates in water, and so a 1M (molar, not molal) solution of this compound will really have a 2M concentration in ions. I took the Chem SAT in May, and the only question on this topic involved identifying the solution that would have the lowest freezing point. I don’t think you’ll ever be asked to calculate the actual freezing point.</p>
<p>Anyone helps me explain question 11 page 202 (Barron’s)? I think the answer is D rather than B.</p>
<p>It’s definitely B (heterogeneous equilibrium).</p>
<p>are you guys referring to Barron’s SAT or AP?</p>
<p>are you telling me barron’s sat has typos too?! lol…</p>
<p>oh and how does kaplan compare to the real sat?</p>
<p>rockermcr, can you help me explain it?</p>
<p>Heterogeneous equilibrium, otherwise known as phase equilibrium, occurs when there is a phase change that takes place over the course of the reaction. For instance, dissolving a soluble compound in water involves a phase chance. When the reaction of dissolution (by far the most common type of heterogeneous equilibrium, and doubtless the only you will encounter on the SAT) is at equilibrium, the solution is saturated with solute. For instance, let’s say you are dissolving BaCl2 crystals into water:</p>
<p>BaCl2 (s) –> BaCl2 (aq)</p>
<p>If you add just enough BaCl2 crystals to saturate the solution, ie when the maximum amount of BaCl2 at the given temperature is dissolved in water, the above reaction will go to completion. However, adding more solute to an already saturated solution will create the following reversible reaction:</p>
<p>BaCl2 (s) <–> BaCl2 (aq)</p>
<p>This will create an equilibrium, with certain molecules precipitating out of solution and others dissolving. However, there will* always* be the same amount dissolved, because for every new molecule that dissolves, one will precipitate.</p>
<p>Yeah let the “heterogeneous” part remind you that you’re working with different phases</p>
<p>A homogeneous equilibrium refers to one in which all constituents are in the same phase.</p>
<p>has anyone used the McGraw-Hill SAT II book for chem? How do the practice exams in that book compare to the real thing? easier? harder?</p>
<p>and so are Sparknotes and Barron’s a little harder than the real thing?</p>
<p>Just did the practice test from the CollegeBoard SAT II book. multiple choice wasn’t bad at all!! But man, those T/F/CE questions are tough. I missed more than half of those haha</p>
<p>I hope the actual test will be really close to the one in the CB book. It better not be anything like Barrons.</p>
<p>The Barron’s tests aren’t particularly difficult. They simply require the ability to do complex calculations in very little time and without a calculator. If a calculator were permitted on the tests, the Barron’s practice exams wouldn’t really be that difficult.</p>
<p>My scores on Barron’s vs. Sparknotes have been relatively close, and it’s said that Sparknotes is significantly harder than the real thing. Take it as you will.</p>
<p>rockermcr- can you explain why their dissociation has an effect on the freezing point and what effect it has? (lowers, right?)</p>
<p>Agreed, and I am using Barron’s, and am very satisfied with the work they’ve done. They seem to go into more detail, yet at the same time, making it easier to understand. I do agree the math is a little challenging without a calculator, but it’s not too bad. I like it a lot better than Kaplan’s, which is what I was using. I haven’t completely finished reviewing yet, but I got a 700 on the practice, so I’m pretty happy. The book also covered a few more topics that our teacher didn’t really go into detail upon that wasn’t in our textbook, like specific gravity and normality. The way in which they introduce topics seems very organized and makes a lot more sense. Great book! I feel prepared, haha.</p>
<p>^ Also dancer, yea, their dissociation lowers the freezing point. The reason why is that when the molecules go into solution, they interfere with the intermolecular bonds of the water molecules, causing the need of a lower temperature to slow down the solute molecules enough so that they don’t interfere with the bonds between the water molecules needed to form ice.</p>
<p>^ That is incorrect. </p>
<p>The cause of freezing-point depression is far, far beyond the scope of the SAT Chemistry exam. It is explained by the chemical potential of the solvent, and is related to the overall entropy of the solution. Don’t even bother looking any of that up.</p>