<p>but when you increase pressure in a system at equilibrium, equilibrium is has to be reestablished, thus the system will try to lower the pressure, but will never reach the original value. I am thinking Lechatlier principle but i could be wrong. </p>
<p>That’s what I thought too @Chimp22 … It couldn’t just even out perfectly </p>
<p>I’m pretty sure it doesn’t reach the original value when equilibrium is re-established.</p>
<p>@hypnotoad107 you have wrong number of sig figs in 6c.</p>
<p>If I got 50% on the frq and 75% on the mc! then what would my score be? Like an estimation, since no one really knows…</p>
<p>@terrytao it’s off by one so it still counts</p>
<p>@DAIMYO I agree. I understand that Kp = (CO2) but adding more molecules in a rigid container means that the overall pressure would increase simply because you have more molecules. Even after equilibrium those molecules don’t just disappear and make the pressure go back to normal. </p>
<p>well according to the expression, the reverse reaction would occur until the pressure is reestablished as normal.</p>
<p>I suppose it doesn’t state that the volume of solid CaCO3 is negligible, but I think that was intended to be assumed.</p>
<p>Hey guys, what was the ka value for question 2?</p>
<p>Hey so I took the exam but got form D, or the form that was not released on the college board website, does that mean it will never be released???</p>
<p>@magathar, it was about 1.29 x 10^-5. The pKa was around 4.89.</p>
<p>Oh wow, I nailed the FRQ. Only missed about 5 parts of questions.</p>
<p>The Kp = Pco2, so when more CO2 is added it will shift right until Kp = Pco2 again. That’s how all LeChatelier’s problems work. This problem just feels different because there is only one species involved in the K expression. To have a value other than 1.04 atm, the K value would have to change, and since the temperature didn’t change that wouldn’t be possible. So what happens is that when CO2 is injected, it will shift left and form CaCO3(s) until the pressure has once again been lowered to 1.04 atm.</p>
<p>The curve for chem is worrying me because of what happened to bio last year. Does anyone know if the terrible bio curve last year was due to the fact that more people did well on the new test or because lots of people just didn’t do that well? </p>
<p>This test is killing me… a score of 4 is 4 credits of chemistry and a score of 5 is 8 credits at my university. If only I didn’t make stupid mistakes cause of time I would be confident in a 5 but now I NEED TO KNOW THE GRADING SCALE.</p>
<p>@caffeinemolecule : For the AP bio exam last year, all the people I know generally found it very easy. I took the practice exam (from the secure teacher portal of ap central and administered by my teacher) and also found it to be common sense.</p>
<p>The people I know that took it are very smart (near the top of the class and taking AP Calc BC, Physics C, etc) and though they found it very easy they all got 4s. I think in general with the new exam style the questions are easier but it is very difficult to get in the 5 range…Just imagine that for AP Bio, out of 100 people (generally the hardest working students) taking the exam, you would need to beat 95 of them to get a 5. </p>
<p>@hypnotoad…I said that the reaction was first order because a graph of ln[cis-2-butene] over time would yield a straight line. Do you think I would get the point, since there was enough information in the problem to calculate moles of the gas, molarity of the gas and time of the reaction (half-life), and since they asked for a proof based on the available data in the table? </p>
<p>Was rate constant .00693 s^-1? </p>
<p>Also for the Ksp part of q1 is the answer just that it will precipitate because of the ridiculously low Ksp and that there is already [Ag+] that is much much greater than the given Ksp thus an unrealistically small mass of I- would have to be added for precipitation not to occur?</p>
<p>@dopecake… yea, that’s all correct.</p>