<p>kimberlyy - scoring guidelines don’t go up for a few months. Sucks, yeah? (Although that’s partially because they don’t even make the scoring guidelines for calc until they take a preliminary look at some student responses during grading week.)</p>
<p>on #2 for the 3 different masses with the crucible… If you use a scale that’s that precise you will get different amounts of mass each time. The temperature wouldn’t have affected the reading. You’re supposed to take the average of all three masses and subtract the crucible mass from them.</p>
<p>no you’re not.</p>
<p>You are heating the sample until all the water vaporizes…</p>
<p>@roobear That makes sense, but in class and on webassign and all problems, etc ive always taken sig figs into account for the M a solution. Usually it would say 6.00M solution. I dunno why collegeboard just said 6M.</p>
<p>But either way you get ±1 sig fig so its all good</p>
<p>what if for the one that asked “is N2H4 on the same plane”</p>
<p>i said No, because the nitrogens are tetrahedral.</p>
<p>is that good enough? i didn’t know they were trigonal planer</p>
<p>@FlamingMango: So if you get the sigfigs off by one digit it’s ok? So if I wrote 19mL and you wrote 20mL, we’re both right? :D</p>
<p>^I made it clear that my original answer was 18.75 but because of sig figs, I rounded to 19.</p>
<p>For #6 part c, I put that the order with respect to ethanol was zero, but then I said that the rate = k[ethanol]…
Then for c)iii, I think I actually calculated k correctly using the graph, but my units were off–I put sec^-1 instead of Mol/L-s-- because I used the rate law I made in c)ii to figure it out.</p>
<p>My question is how would the consistency points work here?
Part c)ii screwed me over because I didn’t get the right rate equation, but I got the concepts for c)i and half of c)iii.
Do you think they would give me points for the getting the correct value of k for c)iii or not because it’s not consistent with my answer in c)ii.</p>
<p>i feel much better knowing that everyone else had the same problems I did. I mean after taking random practice tests anywhere from 1994 to 2010 i though i had a handle on what was going on. The MC was not super easy, but i feel like it was better than some of my practice tests. im hoping for only like 20-25 tops wrong <em>Knock on wood</em>. the FRQ… DANG. i though after doing all the practice tests i would be prepared… NOPE. they were the hardest i have ever done. But according to my teacher (Who is amazing and taught us very well and usually has students get 4’s and 5’s) they were the hardest problems in the past 15 years. At my school we are all expecting a very generous curve. Dont worry guys, the tests may have seemed impossible but i think we will be surprised when the curve helps our scores out a lot!! <em>again knock on ood</em> Best of luck to you guys. I hope you all get the scores you wanted. And good luck to anyone with more AP tests. lets hope they arent as bad as this one was. :(</p>
<p>They have the 2011 FR including form b - up on the website.
Form b looks a bit more harder imo</p>
<p>@theeboy3: I said something along the lines of “The lone pairs are going to push the Hydrogen bonds in many different direction”. Yeah not the best answer but it seems right… right?</p>
<p>Form b people are lucky - the entire section is easy except for question 6 on form b.</p>
<p>I though the MC were normal, with some easy, and some hard</p>
<p>The FRQ’s though… Part A was ok, but Part B killed my soul
Q1: a and b were doable but I messed up on c and while I was fixing it, time was over
Q2: pretty good
Q3: also pretty good
Q4: What is a coordinate covalent bond?
Q5: N2H4 what on earth?
Q6: my teacher said those graphs would never be on the FR…</p>
<p>
</p>
<p>It’s a good thing I clarified that with my teacher the morning of the test; we had called it “complex ions” for the entire year.</p>
<p>^^^^^^^I believe it is called a complex ion as well. Does anyone know if the curve will be greater due to the removal of subtracted points for guessing? My teacher showed us the scale from last year of the scores depending on the MC score. I just really need to know how many I could have missed on the MC to get a 4 or 5.</p>
<p>I knew that the order was zero but I was freaking out because I thought it was unlikely that for the rate law was just rate = k. In the end I did write down rate = k[ethanol]^0 though, hopefully that gets points.</p>
<p>
Well, it is correct. Any reasonable grader will give you credit for it.</p>
<p>Keasbey,
your answers are great (and I’m sure good enough for a five), though there is one minor point I would change in a few of them.</p>
<p>In 6b. for example, it is not enough to say that kinetic theory says this. you should explain why. what they wanted was:</p>
<ol>
<li>The number of particles in the gaseous state will increase, as a greater number of particles will overcome intermolecular forces.</li>
<li>The force of each collision will increase, as these particles have a higher kinetic energy.
Since pressure is force/area, factor 1 increases pressure, because it increases the number of collisions with the sides of the container, and factor 2 increases the force of each individual collision with the sides of the container, increasing pressure.</li>
</ol>
<p>My reasons for 6b, as best as I remember them: </p>
<ol>
<li><p>Velocity increases, so momentum increases. In order for the collisions to be elastic, the container needs to exert more force, which increases the pressure.</p></li>
<li><p>Some BS about how because velocity increases, KE increases, so the pressure’s greater since the particles are faster. :P</p></li>
</ol>
<ol>
<li><p>Greater temperature = the molecules in the liquid are more energetic on average, thus more of them exceed the minimum energy requirement to go from liquid to gaseous form. More gas molecules = more collisions with the container = more pressure.</p></li>
<li><p>Greater temperature = more kinetic energy = molecules move faster. This means they have more momentum and hit the container more frequently = more pressure.</p></li>
</ol>
<p>This should be good, I hope.</p>