<p>
</p>
<p>Was this a typo?</p>
<p>If it’s an INTRAmolecular force, look at 6.c) here: <a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>It says in part ii. that the broken bonds (i.e. hydrogen bonds) are intermolecular forces . . . and not intramolecular forces.</p>
<p>@godzilla…Kp is the same thing as Kc or Keq. It’s just you use pressure instead of concentration 
Kp = [ (P BR2)(P NO)^2 ]/ [(P NOBr)^2]</p>
<p>(2 x4)/(16) = 1/2 Is that a the right answer? O.o</p>
<p>G0DZILLA, is the answer D?
If it is, I’ll explain, otherwise I won’t bother explaining a wrong answer. xD</p>
<p>Kp is just the Keq for partial pressures (gases). Since the question supplies data in partial pressures you don’t need the Kp =Kc equation. Just set up like a normal equilibrium problem. Also, is there a value for partial pressure of NO (g)? It doesn’t seem to me like you can do that problem without it…</p>
<h2>also, the intermolecular forces aren’t actual bonds - the molecules aren’t bonded to each other, they are just forces of attraction (at least that’s what my teacher said) C:</h2>
<p>When comparing london dispersion forces, (eg. h2o and h2S) can you say that since H2S has a greater molar mass it is heavier, therefore easier to induce a temporary dipole? Therefore it has stronger London forces? </p>
<p>The rubric said H2S has more electrons in its electron cloud so it’s more polarizable.</p>
<p>Yes it’s D.</p>
<p>omchi23:
first the first answer u gave me, what is conc? can u elaborate on that one.</p>
<p>Hybridization: this part always messes me up.
Compare CH4 with NH3.
Do they both have sp3 hybridizations? I thought CH4 had sp3 and NH3 had sp2?</p>
<p>Yes, what does “polarizable” even mean?
I also would have compared the masses and said that since it’s bigger, the forces would be stronger. </p>
<hr>
<p>Regarding the partial pressures question,
I assumed that the partial pressure of NO was the same as NOBr because they have the same coefficient in the balanced equation . . . alternatively it would twice as much as Br2 when comparing their coefficients as well.
Then I did the Kp thing, and NO and NOBr cancel, leaving you with 2.
Is this a correct assumption to make?</p>
<p>yeah i think ur right abrayo. ch4 has a different hybridization then NH3.</p>
<p>um to compare them why not say that they both form covalent bonds?</p>
<p>@Godzilla - The partial pressure of NO (g) is needed to actually solve the problem.</p>
<p>Kp = (P(Br2) P(NO)^2)/(P(NOBr)^2)</p>
<p>@chickenonrice…
In an aq soluition with a ph of 11.5 at 25 Celsius the molar concentration of OH- is approximately?
Hmm. okay. so, pH + pOH = 14.
You have the pH. Therefore, you can find the pOH : 14- 11.5 = 2.5
Since pOH = -log [OH], you take the antilog of 2.5, which also the same thing as 10^-2.5. and you get 3.16 x 10^-3 :D</p>
<p>I apologize, NO is 4 atm. Answer is still D.</p>
<p>So do I just do</p>
<p>([NO]^2 x [Br2])/([NOBr]^2)</p>
<p>And plug in the pressures?</p>
<p>for the guy with the partial pressure question. i basically did 4 = x + 2 and 2 is ur answer. i forgot which law but it says that pressure( i think raoult’s law …) that total pressure is equal to partial pressure of a plus partial pressure of b.</p>
<p>The term “Hydrogen bonding” is INTERmolecular while the actual bond of hydrogen to oxygen in water in INTRAmolecular.</p>
<p>In that free response is is talking about boiling which is a change of state, so INTERmolecular forces are involved, which is talking about the term “hydrogen bond” rather than the actual bond between hydrogen and oxygen.</p>
<p>the easiest way that I think about hybridization is that for </p>
<p>single bonds (4 bonds) -> sp3
double bonds -> sp2
triple bonds -> sp hybridization</p>
<p>But I remember it’s not always this case, only most of the time.</p>
<p>@Godzilla - Yep, you got it.</p>
<p>AP question:
Which of the following measures of concentration changes with temp?
a) mass percentage
b) mole fraction
c) molarity
d) molarity
e) parts per million by mass </p>
<p>The answer is C but why?</p>
<p>I feel like I am prepared, but then I make stupid mistakes on FR.</p>
<p>M=mol/V and V depends on T</p>