<p>so i did a practice exam today and got a 5…compared to barely a 3 a few weeks ago…cramming really does help!</p>
<p>I have a review session at school tomorrow for pretty much the whole day.</p>
<p>I still have to memorize all of the molecular shapes, names, and bond angles…lol</p>
<p>^^same here</p>
<p>Oh dam do we have to know all of the labs to?</p>
<p>Thanks a lot, Feed! That was very helpful.</p>
<p>Everyone should start asking questions, so knowledgeable people can help explain how to do problems. </p>
<p>anamai - how did you cram so well?! I got a 3 on my last practice exam, and I need to get a 5…</p>
<p>We only get equations for the free response. Also, I am SO hoping for titrations. And anything with Kinetics or Thermochemistry for the FRQ. Just because those are the big concepts you need to know. If it’s about hybridization or something, I’m SCREWED.</p>
<p>Hybridization isn’t hard. Basically count the number of effective pairs (double and triple bonds still count as one effective pair) around the central atom in a Lewis structure of the compound. Memorize the table for hybridization and you’ll be good to go.</p>
<p>I’m feeling pretty effed for labs, wonder what we can do about those? Also, for equilibrium, can anyone explain the difference between that and an acid base problem? For equilibrium, I always see problems that are like Ksp, Kc, Kb, Ka, Kp, etc. </p>
<p>Electrochem is fun too! D:</p>
<p>Yeah, anybody care to explain labs?</p>
<p>If you give me an example problem, I’d be happy to help :)</p>
<p>For significant figures on the free response, do we round off in each intermediate step?</p>
<p>Or are we supposed to carry all digits through to the final answer, then round?</p>
<p>^ The latter.
Rule of thumb for sig figs = Answer should be same as lowest sig figs for the numbers given in the question.</p>
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<p>I took AP Chem last year, and I always saved the rounding off until the very last step. In multi-step calculations, I found that once in a while, the solution would be off significantly enough to not get any credit on it.</p>
<p>I wouldn’t sweat it though. For problems that require rounding, they usually let you use calculators on those sections. For problems without calculators, the answer usually comes out to some nice number, like 6.8E-5.</p>
<p>Ka and Kb are Kacid and Kbase, and they multiply to 1<em>10^(-14)=Kw. For instance, you would use Ka for the reaction (excluding charges) HF+H2O–>H3O+F and Kb for the reaction F + H2O–>HF +OH multiply to 1</em>10^(-14). You use Ka when an acid is the reactant, Kb when the base is the reactant. </p>
<p>So, for acid base, you may have to use equilibrium equations, but they will be in the form of Ka and Kb. An example is if you are given .2 molar F, determine the pH. To do this, you first find the Kb (you are usually given the Ka, so just use 1e-14/Ka=Kb). Then, set up your rice table, and set Kb=[OH]<em>[HF]/[F]=</em>/[.2-x], where x is the change in concentrations due to reaching equilibrium. Solve for x, which equals the concetration of OH, and then take the -log to get pOH and then do 14-pOH to get pH. </p>
<p>You would use the above analysis, say, if you were asked to find the pH at the equivalence point of a weak acid (HF) titrated with a strong base (OH). The pH will be greater than 7.</p>
<p>For entirely strong acid and strong base problems, Ka and Kb should not be used. The acids and bases completely transform, thus equilibria equations are not relevant. You need to use equilibrium when dealing with a weak acid or weak base.</p>
<p>Quote:
I took AP Chem last year, and I always saved the rounding off until the very last step. In multi-step calculations, I found that once in a while, the solution would be off significantly enough to not get any credit on it. </p>
<p>It was far off if you rounded, or didn’t round? (during intermediate steps)</p>
<p>
It was far off when I did round during the intermediate steps. Like I said before though, some problems didn’t have that issue, but on the AP test, I wouldn’t risk the chance of having it happening.</p>
<p>My teacher pretty much drilled everything into my head, but I do have a question.</p>
<p>Would anyone care to explain the concept of freezing point depression and boiling point elevation?</p>
<p>The most I remember is that they are both influenced by i, or the number of particles (ions) in the solution. So like CaCl2 would have three particles, since there are three ions. Nonelectrolytes are either 1 particle or none at all, I can’t remember. This is usually found in the MC and descriptive chem section.</p>
<p>^I think I know what you are asking about (took ap chem last year). Freezing point depression and boiling point elevation occur when a solute is dissolved in the substance. This raises the boiling point (elevation) and lowers the freeqing point (depression). This is why salt is applied to ice to melt driveway ice and why antifreeze is applied to engine whatever to prevent freezing in cold conditions and also boiling in hot conditions. </p>
<p>The equations are the change in temperature of freezing= Kf<em>molality</em>i. Kf is a constant depending on the solvent (water = 1.86). i is the van hoff factor, which is basically how many dissolved ions there are (i = 1 for sugar, 2 for sodium chloride, 3 for magnesium chloride, etc.) Chane in temperature for boiling=Kb<em>molality</em>i, where Kb for water=.512 and everything else is the same as for freezing. Whatever you get for delta T, you either add to the boiling point (so delta T + 100) or subtract from the freezing point (so 0 - delta T). </p>
<p>Both are colligative properties. I don’t really know too well why this occurs.</p>