<p>OK S is now stressing out just reading a tiny little bit on here. Does anyone know if you start with the wrong equation and use that as a basis of subsequent answers if all answers are wrong, even if the work was correct on the subsequent answers… referring to free response.</p>
<p>Probably you get partial credit. The exam says, “partial credit may be given.”</p>
<p>Wow sorry everyone for posting bad info–hydroxylamine’s definitely a weaker base from numerous other reliable sources. From Chemical Principles pg 437, the pKbs in increasing order are methylamine (3.44) < ammonia (4.75) < hydrazine (5.77) < hydroxylamine (7.97).</p>
<p>Thanks for clearing that up. that makes more sense because H’s attached to O’s that are bonded to electronegative elements are more acidic</p>
<p>Wait so the answer was HOCl?</p>
<p>Or am I asking about a completely different question?</p>
<p>Completely different question. I don’t think there was even a question with answer choice HOCl. We’re talking about the amine alkalinity question, and the answer was methylamine :)</p>
<p>could somebody post MgO problem?
thanks</p>
<p>In an experiment to determine the empirical formula of magnesium oxide, a student weighs and empty crucible then adds a strip of magnesium and reweighs the crucible. The crucible and magnesium are heated with a burner flame, which ignites the magnesium and forms a gray-white solid. After cooling, the crucible and solid are reweighed and the data are analyzed to give an empirical formula of Mg5O4. Which could account for the observed Mg5O4 result rather than the expected MgO?</p>
<p>(A) Some of the magnesium reacts with atmospheric nitrogen to produce magnesium nitride.</p>
<p>(B) A mixture of magnesium oxide and magnesium peroxide forms during combustion.</p>
<p>as somebody pointed out before it can’t be peroxide since Mg:O is more than 1:1.
Mg3N2 then.</p>
<p>This is an experiment so the mole ratio or mass % of MgO versus MgO2 wouldn’t really matter since in this experiment, the mass of oxygen would be determined by [Mass of white grey solid (MgO) - mass of magnesium ribbon].
(Because you would have to assume the grey solid is just magnesium oxide and no side products are formed). Then you would just convert grams of O and Mg to moles.
In order to get an empirical formula of Mg5O4, that means some oxygen must have been lost so the mass difference between the solid and Mg is smaller. So that means you’ve got a smaller mass of white solid then expected for a empirical formula of MgO.</p>
<p>Not sure whether MgO2 or Mg3N2 as a side product would cause you to get less solid. But Mg3N2 is a light yellow-green color, and the problem said white grey?</p>
<p>However, I’ve never heard of peroxides forming from combustion, but my teacher said there’s a possibility that they can.</p>
<p>@Sophia7X–this is how I thought about the question. Assume that you conveniently have 24.31 grams of Mg ribbons, then you burn it to get the white/grey solid. </p>
<p>-To have gotten Mg5O4, you must have calculated moles Mg:O = 5:4, so working backwards we have (4/5) of a mole of oxygen. We must have measured 24.31 + (4/5)*16=37.11 g product.</p>
<p>Scenarios:
(1) If you got pure MgO as the product, you’d have 24.31 + 16 = 40.31 g.
(2) Say you got pure Mg3N2 as a product–you’d have (2/3) mole of nitrogen. The scale would measure 24.31 + (2/3)<em>14=33.64 g.
(3) Say you got pure MgO2 as the product, you’d have 2 moles of oxygen. The scale would measure 24.31 + 2</em>16=56.31 g. </p>
<p>Comparing what we actually got (37.11 g) with scenario 1 (40.31 g), we see that we’ve gotten less mass than we’d expect. Of scenarios 2 and 3, the one the gives an amount of product less than the 40.31 g of scenario 1 is (2), that of magnesium nitride. </p>
<p>I think the focus of this question was on stoichiometry, so they wanted us to compare the masses of the potential products. Besides, I think Mg3N2 might be a legitimate side product as there is a bunch of N2 in the air. Hope that clears things up a bit.</p>
<p>Oh I see. Thanks. The color sort of threw me off</p>
<p>So nitride was the correct answer?</p>
<p>@thenerdyjew Yes</p>
<p>For your information: sodium forms a mixture of sodium peroxide and oxide when burned in oxygen. Potassium forms a mixture containing potassium superoxide when burned in oxygen. Sodium peroxide is extremely reactive and dangerous. Reactive = fun!!! :D</p>
<p>Okay so I know we’ve kind of beat this topic to death by now, but what do you guys expect camp scores to be this year?</p>
<p>@Sophia7X-- no problem. I didn’t even consider the colors, as I didn’t know any…</p>
<p>^^hmm. The USAMO takers this year started a google doc so that everyone could record their likely scores and predict MOP cutoffs.</p>
<p>I suppose a similar thing can be done with this USNCO. Except there is not a geeky forum where all the people taking USNCO visit, so that would be hard to coordinate.</p>
<p>Personally I think I have a chance, hopefully the stupid mistakes I made on the FR will be okay because of this year’s difficulty</p>
<p>I think that this year’s cutoff will be 55-56 for MC, as per the usual. The Free Response might be around 85-90, although I have little to back that up with. The Lab portion will probably require average/above-average for both problems. </p>
<p>There were some trickier MC, including some high level topics in p-chem, but not so much in orgo. The free response was nothing special for the first half, but some reactions in part 5 were new, especially Al(OH)3 + H+ → Al(H2O)6. Problem 7 was mostly straightforward, but it seems a number of people didn’t have enough time (spent too much on part 1). Personally, I made the unfortunate error of mistaking uni- vs. bi- for the organic mechanisms. The lab portion was nothing too unique, although they provided less guidance for solving the problems than I would’ve expected. </p>
<p>Overall, it seems to me that they really wanted to separate the top-notch kids from the rest. IChO is, after all, in Washington D.C. and the U.S. needs to represent itself appropriately.</p>