<p>I got my results (score) but not the correct answer/ my test booklet back yet.</p>
<p>Any thoughts on this one?</p>
<p>Calculate the hydronium ion concentration in 50.0 mL of
0.10 M NaH2AsO4.
(K1 = 6.0 </p>
<p>Did you mistype D? I’m getting 2.5x10^-2. Just do an ICE dissociation two times and you’ll get the answer, not a particularly hard problem.</p>
<p>Since its a salt of a polyprotic acid, you use: 1/2 * (pKa1 + pKa2) = pH</p>
<p>the H2AsO4- has amphiprotic properties, we must account for both to determine [H+], i believe you use the above equation…</p>
<p>the answer would then be 2.5 x 10^-5 or (D)</p>
<p>Just a heads-up, every question you help others with makes it more likely that they will out-perform you, this is a cuthroat competiton, not trying to come off as rude…</p>
<p>I’m pretty sure it’s 2.4 * 10^-2, because the second dissociation is basically negligible. Unless D is 2.5 * 10^-2, which I doubt (they’re too similiar).</p>
<p>i’d like to think i can be fine even helping others</p>
<p>Hey, guys!</p>
<p>I live in GA and I still did not receive the scores…does it mean that I am disqualified?</p>
<p>@ChemistryTutor
You are right that kids who make camp prepare for 1 -2 years. At least that’s what I have seen in my school.</p>
<p>Do you need organic chemistry to get to top 50? Apart from Zumdahl isn’t it better to work with Atkins too? Other than Insight there are two more bonks by Atkins on Physical Chemistry and Inorganic Chemistry. Do you think these are needed or it is too much detail for the national exam?</p>
<p>I can’t comment on book selection because I am not that familiar with what exactly in each of them. Many like Atkins. I have looked through Zumdahl and I would say it contains enough knowledge to solve GChem problems National and International level. The problem is that one needs to solve something like >80% questions on part I and II to get to top 50. At that level each extra question is a super-hard General Chemistry question. At the same time there are several easy Organic Chem questions just sitting there to be picked in part I. and in Part II there is bound to be an OChem problem (15%). Basically, getting to top 50 is pretty hard without OChem.</p>
<p>Well the answer is D (which is 2.5 x10 -5, sorry about that typo)</p>
<p>I still did not receive my score…anybody from GA?</p>
<p>I got my score back today–a 49/60! I’m pleasantly surprised, considering I hardly studied. Also, I think my teacher got an email that said I’d be taking the nat’l exam, so I guess I made semis! Woohoo!</p>
<p>Would you not be considered as a national qualifier if you are invited to take the national exam but you choose not to take it instead?</p>
<p>I have a quick question. Obviously, I’m not going to make camp because the test is in less than two weeks, and I haven’t studied much at all. However, I’m wondering if it’s possible (or at all likely) for me to make Top 150 Honors for tha nat’l exam. How difficult is it to achieve Top 150? Does two weeks of hardcore studying afford me a chance to do so?</p>
<p>I ask because I have a conflict on the day of the test, and if the test isn’t likely to go well for me, I’ll probably skip it.</p>
<p>43-44 on national MC used to be the cut-off for 150. if you are scoring somewhere close to that on the old tests you should give it a shot.</p>
<p>From what I have seen, 49/60 is too low to score >40 on national. But definitely try the old tests and review whatever you have time to.</p>
<p>I have Raymond Chang’s 8th edition. is this good enoough? or does it lack anything?</p>
<p>How is final placement determined? I know that a good MC score is the only criterion for top 150, but is there some sort of equation that gives the final score that decides who advances?</p>
<p>Top 20 score about 90% correctly on part I and on part II</p>
<p>I haven’t the faintest idea how to solve this problem, number 23 on part 1 from 2011. I don’t even know how to begin.</p>
<p>When MgO reacts with H2O at 25 °C and 1 atm, the volume change is –4.6 mL•mol–1.
MgO(s) + H2O(l) → Mg(OH)2(s)
What is the value of ΔH - ΔE for this reaction?
(A)
–4.7 × 10–1 J•mol–1
(B)
–4.7 × 102 J•mol–1
(C)
4.7 × 102 J•mol–1
(D)
4.7 × 10–1 J•mol–1</p>
<p>Any help would be appreciated.</p>