<p>For the measuring out exactly 843 mL of water, you can’t just say “in a seperate beaker, get exactly 843mL and see if it fits exactly, overflows, or is too little” in the container at question. Why? Because when you pour water out of the beaker, not ALL of the water gets poured out, there is a little bit of water that still stays stuck to the other beaker due to water’s property of adhesion or cohesion (I can’t remember which). So what did u guys put for it?</p>
<p>I don’t think there is a way to easily get around that. Perhaps, weigh the flask with the water and calculate how much much water there is based on that. But I seriously doubt they will be that finicky.</p>
<p>The 2009 FRQ is online: <a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>^We’ve known this for… quite a while now.</p>
<p>Omg I couldn’t get question 1… for c and d what were we supposed to do??</p>
<p>All that you’re given is the initial concentration… you don’t know the equilibrium constant or the solubility…</p>
<p>1c:
i: should’ve been easy
ii:I got confused here too
iii: Use the reaction given and the Ka of HOcl to get this</p>
<p>1d:
i: ICE table + buffer
ii: Use the ratio HX/X- = [H+]/Ka or something like that.</p>
<p>Haha. Sorry, didn’t notice.</p>
<p>For #1</p>
<p>a) i) HOCl, its Ka is bigger
a) ii) Self explanatory
a) iii) Weaker, Cl is more electronegative and draws electrons away from H more, making it easier for H to pop off for HOCl and harder to pop off on HOI</p>
<p>b) HOCl + H2O <-> H3O+ + OCl-</p>
<p>c) i) Kb = [OH-][HOCl]/[OCl-]
c) ii) Kb = Kw/Ka = (1.0E-14)/(2.9E-8) = some number which I’m too lazy to calculate
c) iii) Plug in Kb in the equilibrium expression, [OCl-] = 1.2, [OH-]=[HOCl] = x, solve for x</p>
<p>d) i) pH = -log [H3O+], so [H3O+] = 10^-6.48
d) ii) pH = pKa + log [H-]/[HA], basically show that because log [A-]/[HA] must be less than zero, [A-]/[HA] < 1 and [HA] > [A-]</p>
<p>Can someone explain to me how Cl2 to had an order of 1/2?
I got that both were first order.
How did you guys determine the order of Cl2?
ALSO: The question which asked if Y had a faster rate than X, I said that the statement was false and stated that you must know reaction rates to determine that.</p>
<p>On the same problem(with reactions X,Y,Z), what was the answer to the question where it asked what would happen to the pressure as the reaction proceeded?
Wouldn’t it remain the same!
Possible reason: the vessel was sealed, therefore the volume is the same, and the temperature remained constant, and a solid has no affect on pressure.</p>
<p>On 1) d) ii), if i forgot to use calculations, but still got the answer correct, will that cost me a lot of points, or just one or two points?</p>
<p>Either they’ll make it so the right answer is a point and the justification another point, or the right answer + justification is one point. You’ll lose at most 2 points.</p>
<p>alright, so if i miss a total of like around ten points in the free response and a raw score of around 55-60 in the multiple choice, would that be good enough for a 5?</p>
<p>Easily a 5</p>
<p>Yeah, you pretty much need 66% of overall points for a 5.</p>
<p>thanks a lot, appreciate it.</p>
<p>this thread made me realize all of my stupid mistakes on the FRQ >.<</p>
<p>same haha. and can someone answer Kamikaze’s question about Cl’s order being 1/2? I didn’t get it either.</p>
<p>I miraculously put 1st order for the other one haha. I put “probably 1st order”</p>
<p>that counts right?</p>
<p>Step 2 is the rate determining step because it is the slowest
so
rate = k[CH4][Cl]
Cl2 is not in the equation so we look at Step 1 which has both Cl2 and Cl
and we see that K = [Cl]^2/[Cl2]
rearranging this -> [Cl] = sqrt(K x [Cl2])
then you plug [Cl] back into the rate equation for Step 2 and you get: rate = k sqrt(K)[CH4]sqrt([Cl2])
k x sqrt(K) basically can be multiplied together to form a new rate constant: k’
and sqrt [Cl2] is [Cl2]^.5 so the rate order is 1/2 in respect to Cl2</p>
<p>hope this helps</p>
<p>Do you get any points for a correct answer but a wrong explanation? o_o</p>
<p>We didn’t cover how to determine the rate of a reactant in a kinetics equation, but I guessed the order for CH4 correctly (through the wrong justification, however).</p>
<p>Some questions require both correct answer and explanation for a point. Sometimes, a question has two points possible, where corrcet anwer is worth one point and explanation is another point.</p>