<p>Not high at all. I know my local section nominees’ average last year was about a 90% which is 54.</p>
<p>Hey I have a few questions about one of the nat’l rounds, specifically 2006 Part I, i thought there were some errors or maybe i just suck?? anyways would like clarification on: </p>
<h1>23 and #30 on 2006 Nat’l Part I</h1>
<p>For 23, I thought it was B) because more gas molecules form, but it’s D??
and for 30, I thought it was D) but they say its B), but when I set up B as the rate-determining step I get rate = k [H2][NO]^2 / [H2O] ??</p>
<p>don’t talk about the test pleaseeeee not until march 31…</p>
<p>im talking about a released 2006 Nat’l Round Part I, anybody?? just go the chem olympiad website to find the questions I’m referring to, im too lazy to type it out</p>
<p>Lolol2 u. He was talking about the 2006 nationals</p>
<p>Lol Harvey is talking about a past exam ^</p>
<p>For 23, that is an error.
First of all, D isn’t even balanced. (oxygens unbalanced)</p>
<p>IF ΔE = ΔH + w and w = -PΔV
ΔE = ΔH - PΔV
PΔV=ΔnRT
Therefore
ΔE = ΔH - ΔnRT</p>
<p>Δn = -1 for D (by the way n = mol of gas).</p>
<p>That means ΔE = ΔH + RT and it’s exothermic, adding the +RT term only makes it less negative.</p>
<p>Answer is actually B because Δn= 1 (A also has Δn=1 but it is not exothermic).</p>
<p>You can ignore [H2O] because you can’t exactly have a concentration of water</p>
<p>^ okay, but the H2O is in gaseous phase? but ignore it always? i thought you just ignore in liquid phase</p>
<p>and thanks for explanation</p>
<p>Yes ignore it anyway</p>
<p>My teacher says you generally do not include water in rate laws regardless of phase</p>
<p>thats strange because Zumdahl says to include H2O if it’s in gas phase…but I guess in the problem they assume H2O in the reaction mechanism is liquid…</p>
<p>Do we have to memorize all reduction potentials (lol) and constants like mass of protons?</p>
<p>@Mansu007: the easy way to determine what you have to memorize is to look at previous tests.</p>
<p>What did you guys think about the test, was it easier or harder than normal?</p>
<p>Can someone explain to me this question from the 2012 exam?
. Determine the volume of 0.125 M NaOH required to
titrate to the equivalence point 25.0 mL of a 0.175 M
solution of a monoprotic weak acid that is 20% ionized.
(A) 7.00 mL (B) 17.9 mL
(C) 28.0 mL (D) 35.0 mL</p>
<p>I keep getting 7 mL and i’m not sure what i’m doing incorrectly. Only 20% of the moles from the monoprotic acid will be titrated, correct?</p>
<p>^ equivalence point doesn’t have to do with the ionization of the acid, is just the moles of what is being titrated</p>
<p>does anyone taking the test have a swagsack?</p>
<p>can someone help me with this? </p>
<p>Use the given standard reduction potentials to determine
the reduction potential for this half-reaction.
MnO4–(aq) + 3e– + 4H+ => MnO2(s) + 2H2O(l)</p>
<p>Reaction Eo
MnO4–(aq) + e–=> MnO42– (aq) +0.564 V
MnO42–(aq) + 2e– + 4H+ => MnO2(s) + 2H2O(l) +2.261 V</p>
<p>(A) 1.695 V (B) 2.825 V
(C) 3.389 V (D) 5.086 V</p>
<p>since the rxn’s are extremely similar in element compositions, you need to convert to delta G and solve it that way</p>
<p>You can just add the reactions together. So basically you add the reduction potentials</p>
<p>Yeah I also think that you just add the reduction potentials in this situation.</p>
<p>On another note… does anyone have any idea what the cutoff for honors on the national exam is? I read somewhere that usually it’s around 43 but I just wanna check.</p>
<p>all i know is that you need a large, and i mean large, swagsack to make the top 20</p>