<p>5.b) I said no, because they’re in a tetrahedral structure.
Is that right?</p>
<p>I also said the vapor pressures were equal.</p>
<p>I think that’s right.
The statement in the box was true right?
Because combustions are always highly exothermic</p>
<p>I said false because the heat came from the breaking of the intermolecular bonds, not just the “strong bonds”.
(as you can tell, I didn’t know what to write. Seeing as they didn’t specify whether or not they’re intermolecular or intramolecular forces . . .)
Damn, I definitely got that one wrong.</p>
<p>@Abrayo - technically it’s trigonal pyramidal due to the unbonded electron pair.</p>
<p>The answer for FRQ 5 b was that they do not lie on the same plane because there is a lone pair on each of the N atoms, and thus the shape of the overall molecule is two connected trigonal pyramids and thus not all six atoms lie on the same plane. At least, I think that is the answer.</p>
<p>. . . NOOO.
I don’t remember if I wrote molecular structure or electron geometry!
Molecular structure would be tetrahedral and electron geometry would be trigonal planar?
Or the other way around?</p>
<p>@apstudent12345, The statement was false because breaking bonds requires the input of energy. Breaking bonds is usually endothermic.</p>
<p>Thanks firered786.
I forgot about that one equation with the bonds broken/formed.</p>
<p>@Ivysaur - The explanation in 5g is false. Breaking bonds takes in energy, while forming them releases it. Since a triple bond is formed in the product N2, a lot of energy is released and therefore the reaction was exothermic.</p>
<p>Edit - Bah firered</p>
<p>Sorry, but I asked my Chemistry teacher and he said my answer was right. Oh well, at least I got something right! </p>
<p>And the first free response question about the acid/base, I can’t believe how easy it was. I cant believe i messed up on something so simple. ughh</p>
<p>If it’s any consolation to anyone, I heard this test was the hardest one that the CollegeBoard has put out in 10 years.
Did anyone know what was up with those laboratory procedure questions? Or the one involving physics? I used Faraday’s constant but I’m not really sure if I did it correctly.</p>
<p>That stupid electron one, I had no idea.
I think I multiplied Faraday’s constant by the 0.93 moles for part a), and then for part b) I divided it by 600.</p>
<p>How do you actually do it?</p>
<p>For FRQ 2
(a-iii) I said that volumetric flasks are used when trying to dissolve a solid to prepare a solution? I think it’s probably wrong, since it mentioned sig figs and all…</p>
<p>What did you guys put?</p>
<p>^ I wrote a long paragraph because I wasn’t exactly sure . . .
basically my main point was that we didn’t need to be that accurate, because the values given didn’t even have any decimal places.</p>
<p>wait what was 2.a)i.? Did you round it to 19 mL?</p>
<p>What did people get for 3 d. and 3 e.?</p>
<p>@abrayo,
I guess that makes sense, but accuracy with sig figs always seems like a benefit. I rounded to 19ml for 2.a)i as well</p>
<p>@firered786,
for 3 d- i just redid it in my head, and I think i got:
2H20 + 4OH- + O2 + 2H2 –> 4OH- + 4H2O
and 1.23 V i think.</p>
<p>I just realized what a stupid mistake i made in that eqn…
i didn’t simplify.</p>
<p>Was I the only one that noticed that the form b frq was linked to? That test isn’t supposed to be given until the 19th of May…</p>
<p>I confused electrochemical with electrolytic/whatever the non-galvanic one is.
So I ended up with a negative cell potential and a wrong equation.</p>
<p>^Yeah I saw that too. . .
WOW CB is stupid if that’s the real thing.</p>
<p>for the equation i got: 2H2 + 02 –> 2H20…
is that right?</p>