<p>Ugh I knew the answer to the calorimetry one, its so simple. So frustrating, I think this is what is going to keep me from getting a 5. I got MgO as limiting reactant and crossed out trial 1. Then part c i messed up by not counting HCl in the mass.
with q=mcdeltaT formula, you get 100.5g<em>4.18</em>-4=-1680j
With the -1680, you convert to kj by dividing by 1000j, and diving by moles of MgO (.5g/40.1g)=-135kj/mol. Which is fairly close to the number you get when you do products - reactant formula. So frustrating…</p>
<p>If you bonded two oxygens with one hydrogen I don’t think that acceptable…</p>
<p>I agree. I screwed up with part c calorimetry too -_- maybe they’ll give us a point for the work. Maybe if we’re lucky.</p>
<p>Dragonmite, i did that at first, I connected the oxygen in water with H, but i scratched it out and changed to H from water to oxygen in methanal. My reasoning was that H-C was non polar however I think you would get the points because the difference between the electronegativities of hydrogen and oxygen is still big which allows attraction, maybe my second reasoning is wrong since electronegativity only matters between the atoms?</p>
<p>Did anyone else have a different test?</p>
<p>@niquii77 my bonding was like this O-H2-C=O. i added the O(from water) on the left side</p>
<p>Does anyone remember getting -151kJ on part (e) of #3</p>
<p>Yup collegedreams29, I thought the heat transfer was between the MgO and HCl (stupid), but didnt realize that it was HCl solution and that the heat transfer is between the reaction and water(hence the 4.18SH). What did you get for part f. I said no since the experimental delta h recieved was higher (even though my calculation was a mistake) </p>
<p>PS. I got -151kj in e</p>
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<p>Is that not right? O.o
Isn’t MgO the limiting reactant?</p>
<p>Do you want me to solve all of the FRQ’s out on a seperate piece of paper and upload the solutions? It would benefit me as well so it’s a win-win situation.</p>
<p>@kilerskullz</p>
<p>That would be awesome :D</p>
<p>randomandweird, yes i agree.
guys for the g= -nFe, we could have used 96.5 right?</p>
<p>Alright, give me about 30 minutes or so.</p>
<p>For 5a, what two reasons did you guys put?</p>
<p>Was (e)(i) of #2 1.52V??</p>
<p>Thanks killerskullz!!!</p>
<p>For q on 3.c., did anyone get the units to be kilojoules or joules? Or was it kJ/mol or J/mol.</p>
<p>It was kJ/mol.</p>
<p>@randomandweird, yes I also got 1.52V.</p>
<p>@collegedreams29, I put that the frequency and force of collisions increased (not sure if that’s right).</p>
<p>HIGH FIVE, Flying! </p>
<p>I put that as well!</p>
<p>Or well…I put that the collisions with the wall would increase and due to the KE increase the force at which they collided</p>
<p>@college I mentioned the temperature difference and compare that with Trial 2 and 4 and explained what would be expected of Trial 1 and 3.</p>