2010 AP Chemistry Free Response Discussion

<p>As far as the sig figs, they pick one frq every year to check sig figs on. If you mess up once on that specific frq as far as bad rouding, they deduct one point and then ignore incorrect sig figs.</p>

<p>3d) no, because it would require the simultaneous collision of 6 particles. That doesn’t happen</p>

<p>Thanks Xs, I compared Q to the answer to part B for some reason. You are correct, Q=1.04x^-8 which is greater than 5x10-13.</p>

<p>Like one of them I wasn’t sure on significant figures so I wrote the unrounded answer then explained why I rounded. But now I realize I shouldn’t have rounded the final answer. They’ll figure it out tho</p>

<p>@ post 138</p>

<p>ΔH decreases.
The dissolution of urea is endothermic. So the overall temperature change will be larger and more negative. So according to q=mcΔT, a larger and more negative ΔT will cause q to decrease. </p>

<p>To relate this to something more common, the melting of ice is an endothermic process. If you add ice to a 25 degree water solution, the temperature change will be larger an more negative. This is somewhat analogous to the dissolution of urea, which is also endothermic.</p>

<p>Also if you used the ΔG=ΔH-TΔS, the final temperature of the whole solution will be lower in magnitude.
ΔH = ΔG + TΔS so when you compare the 25 degree solution against the solution that is lower than 25, there is a decrease in ΔH.</p>

<p>ΔH = 13960 J/mol @ 298 K while ΔH = 12560 J/mol @ 278 K</p>

<p>what if I said that it was “unlikely” because it was the sum of multiple half reactions??? and can someone respond to post 138</p>

<p>@xs0itg0esx
yea i put increase due to an increase in delta T</p>

<p>@ Takayu
did you calculated a negative value for 2e?</p>

<p>u guys think i will get partial credit for my different response that i posted on post #146</p>

<p>Not 100% convinced on 1F or 2F but I agree with everything else that I didn’t comment on as far as Takayu’s post. #4 was easy this year.</p>

<p>@ user2134, i got a positive value for 2E. The accepted value of ΔH from the given chart is 14.0. So If you use the Error equation, E = (Observed - accepted)/accepted x100 with 12.46, you get -11 percent. But the ΔH value of 12.46 is positive</p>

<p>Regarding #146, that in all likelihood is a one point problem so my guess is no credit but not sure.</p>

<p>Regarding #138, I thought this was one of hardest questions on the exam. I agree with your statement, tho. To dissolve it needs a certain amount of energy. At 5 degrees the urea itself has less energy and therefore would need to absorb more energy to dissolve creating a larger delta H.</p>

<p>true… i mean the statement is not false and I chosed the correct answer of “unlikely”… but i dont think that is what they are exactly looking for</p>

<p>can anyone write a scale of how much each subsection is worth?</p>

<p>I think it’s this:</p>

<p>MC = 50%, FRQ = 50%</p>

<p>FRQ is split into:
Q1: 20%
2: 20%
3: 20%
4: 10%
5: 15%
6: 15%</p>

<p>@nicksyn: haha thanks but any idea on the letters? ie. 1a – 1 pt b – 2 pts</p>

<p>conceptual ones are usually 1 pt. calculation ones are usually more if they require some work. equilbrium is always 1 pt. 1a and 1b should be 1 pts.</p>

<p>I know the Q> Ksp part is worth three points from prior tests</p>

<p>Wow the FRQ really sucked for me. I thought the solution for #4 would be colorless - how’d we know that it would be yellow?</p>