<p>1L O2 (1mol O2/22.4LO2) (2mol KCLO3/3mol O2) = 2/3(1/22.4)</p>
<p>35 I. Due to LeChatelier’s principle, adding H2 shifts the reaction toward products, HI
II. The reaction is endothermic, thus increasing temperature favors products
III. Decreasing pressure favors more moles, but both sides are equimolar. There’s no real effect.</p>
<p>how is pyridine polar??
[File:Pyridine-2D-full.svg</a> - Wikipedia, the free encyclopedia](<a href=“http://en.wikipedia.org/wiki/File:Pyridine-2D-full.svg]File:Pyridine-2D-full.svg”>File:Pyridine-2D-full.svg - Wikipedia) </p>
![http://en.wikipedia.org/wiki/File:Pyridine-2D-full.svg]File:Pyridine-2D-full.svg](http://en.wikipedia.org/wiki/File:Pyridine-2D-full.svg%5DFile:Pyridine-2D-full.svg){kind=link}
<p>@scholarship 1. you have to look at its structure (i.e tetrahedral). look at CH4. It’s tetrahedral. on all the ends there’s H atoms. C is more electronegative so the dipole moment(arrows) are towards the middle (C). Since the arrows cancel out - doesnt have an overall ‘pointing direction’ CH4 is not a dipole. Also, if the structure is symmetrical, it is not a dipole (asymetrical-dipole)</p>
<p>also, if there is a more electronegative element on one side/ there’s multiple electronegative elements on one side it is polar. </p>
<p>I don’t think this is a good explaination. Maybe someone else would explain better?</p>
<p>@ the O2 one, all right, I guess I just did it backwards and started with KClO3 or something.</p>
<p>@ the HI one, yes, so it should be both I and II, but the answer says it’s only I.</p>
<p>edit//
Also 42 . . . is this a typo?
It says the answer is E.
I thought it was A.</p>
<p>Are 2002 and 2008 the only MCs released? I want to try and do a whole AP (the MC section and FRQ section from one year) tonight but I’ve done both of those MCs.</p>
<p><a href=“http://i.imgur.com/YJcYP.jpg[/url]”>http://i.imgur.com/YJcYP.jpg</a> lol look what i just snapped a pic of</p>
![http://i.imgur.com/YJcYP.jpg[/url]](http://i.imgur.com/YJcYP.jpg%5B/url%5D){kind=link}
<p>^ hahaha.</p>
<p><a href=“http://griffithchem.com/AP/AP%20Tests/2008%20AP%20MC%20Test.pdf[/url]”>http://griffithchem.com/AP/AP%20Tests/2008%20AP%20MC%20Test.pdf</a></p>
<p>Again, sorry.
Could anyone explain 50? I thought it was D, but the answer is E.</p>
<p>And 54 I had no idea. I assumed that adding another substance raises vapor pressure and thus chose E, but it’s D.</p>
<p>Any why is 59 not C? It says the right answer is B.</p>
<p>This is from the 1999 mc test.
What is the final concentration of barium ions, [Ba^2+], in solution when 100 mL of .10 M BaCl(2) (aq) is mixed with 100 mL of 0.5 M H(2)SO(4) (aq)?
a) 0 M
b) .012 M
c) .025 M
d) .075 M
e) .1 M</p>
<p>can someone explain why the answer is C?</p>
<p>35 is definitely C <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;
That electronegative N has swapped out with carbon creating a net dipole moment meaning pyridine is now polar.</p>
<p>Same exact thing here abrayo, i didnt know why it was that either.</p>
<p>Osama Bin Laden is dead, no AP tests right? ;)</p>
<p>Abrayo, the answer to #35 is C, and you are correct about the reaction being endo</p>
<p>
There are 0.01 mol of BaCl2 and 0.005 mol of H2SO4.
Thus when the reaction goes to completion, there will be 0.005 of unreacted BaCl2.
The molarity of this (and therefore of Ba) is 0.005/200mL, which is 0.025 M.</p>
<p>edit//
Wow I misread for some reason . . . sorry about that, my mistake for saying it was A.</p>
<p>Can someone explain to me the reasoning behind spontaneous reactions if given dG, dS, and dH?</p>
<p>I know what they are but i don’t understand why.</p>
<p>50)
Took me a minute to see it
It’s a redox reaction, so you can either look at permanganate being reduced or bromide being oxidized. Bromide is much easier.
10Br- -> 5Br2 + 10e-
The electrons transferred = 10
Luckily they gave the equations prebalanced making the work MUCH easier.</p>
<p>Xing615: Hurrah. If only.</p>
<p>How often do you guys think a complex ion is gonna appear? I completely skipped that part because our teacher said not to worry about it ;/ </p>
<p>Thanks for answering my other question btw!</p>
<p>@ feedback411- there will probably only be 1 question on it. It may be one of the equations so quickly reviewing it will save you a couple of points. It is very simple.</p>
<p>
</p>
<p>Why do you look at the Br- and not, say, the H+ or MnO4-?</p>
<p>hey abrayo I’m looking over the same MC as you 
For #50 I was initially confused too. I still dunno why I got it right. My reasoning was that:
reactants: there’s 16 electrons from the oxygen ions(-2x4x2), 10 electrons from Br ions, 16 PROTONS from hydrogen ions. Net electrons to the products side = 10 electrons</p>
<p>dunno if this is the right way to do it, but it worked…</p>
<ol>
<li>adding a NONVOLATILE substance decrease vapor pressure bcause there’s less of the water molecules are on the surface. Thus less molecules of water can escape to the atmosphere. Lower Vp. </li>
</ol>
<p>Answer mine??
Do you know why the AP doesnt count molar mass as a determinant of London dispersion???
The rubric says
-One point is earned for recognizing that the
strength of the London forces among
molecules is proportional to the total number of
electrons in each molecule.
-.- nothing about molar mass of molecule.</p>
<p>@Godzilla
dG = dH - T*dS</p>
<p>If dH is positive and dS is negative, then there is no way for dG to become a negative number mathematically. (Note that T cannot be a negative # because it is in Kelvin)
If dH is negative and dS is positive, the opposite is true. There is no way dG can become a positive number.</p>
<p>If dH and dS are both positive, then dG can become negative as the T gets bigger. Therefore it is only spontaneous at higher T.
If dH and dS are both negative, then dG can become positive as the T gets bigger. Therefore it is only spontaneous at lower T.</p>
<p>Make sense?</p>