<p>no no the form b frq was administered on the same day but at slightly different times. it says it on the collegeboard website that since ap has become much more popular, they now do form a and form b cuz of the time zones…</p>
<p>yes firered i got 2H2 + 02 –> 2H2O as well. with 1.23V. and actually if you look at the top of the page that equation is there! not sure if its a coincidence… :)</p>
<p>what did you guys get for 3f parts i and ii?</p>
<p>@ibisadoozy for 3f parts i and ii, for i I actually got 1.86 moles of electrons. and for ii, i got 299 amps. what about you?</p>
<p>firered786 / ibisadoozy, i also got 1.86 moles of electrons ^^</p>
<p>I multiplied .83 *2 and then added it to the .4 Volts…</p>
<p>Same! ok cool hopefully we can eventually work out all parts of the frq and we can all calm down a bit. any more you wanna discuss?</p>
<p>what was the equation for n2h4 + h2o?</p>
<p>who else feels like slapping them after thinking about 1) c part ii? i subtracted H+ from OH-, without even realizing it was a buffer solution. I CANT BELIEVE IT! ughhhhhhhh</p>
<p>wait when i said same i was referring to firered and apstudent sorry!</p>
<p>
</p>
<p>Uhhh for which question? 3.f)?
edit//nevermind.</p>
<p>But if you’re finding the standard cell potential, you’re adding both half reactions, right?</p>
<p>@pinkrose22, your not supposed to mess around with coefficients when working with voltages. LOL i read it in the princeton review book the day before the test (or i think actually two days) but up until them i used to multiply it by two etc etc.</p>
<p>uggggh :[ that stinks.</p>
<p>oh and for the n2h4 one, the equation was: </p>
<p>N2H4 + H20 —> N2H5+ + OH-</p>
<p>I confirmed it with my Chem teacher today :)</p>
<p>How did you guys do 3. c)?</p>
<p>@the N2H4 equation:
Noooo!
I wrote something like N2H4 + 2H20 —> 2NH3 + 2OH- or something.</p>
<p>3c… um 2(44)=88 cuz the other 2 are zeroes anyway?? I wasnt sure about that esp since it says deltaHvap for H2O(l)… shouldnt that subscript be a (g)?</p>
<p>#3 (f) (i) .93 mol H2 * (4mole e^-/2mol H2) = 1.86 ≈ 1.9mols e^- , 2 figs cause of .93 mol H2</p>
<p><a href=“f”>U</a> (ii) Amps = coulombs/sec [1.9mol e^-1 (96,500 coulombs/1mol e^-) coulombs]/[600seconds] = 305.583333 amps ≈ 310 apms , 2 sig figs again cause of 1.9mol e^-</p>
<p>^ oh man, I forgot to look back at the equation to see the moles of electrons ratio.
Great.</p>
<p>I did 88.0 as well, I believe.</p>