Thanks!! Yeah for 23 I thought that was the kind of logic you had to apply and it wouldn’t be exactly 2.4, sort of weird though.
Im not sure at all but I think you can easily narrow it down to A or B due to what an ionic crystal is, not completely sure about the rest but I was able to get it right from an educated guess
@JuicyMango that exact diagram is in the Barron’s book.
A) As the positive and negative ions slide past each other, the + line up with + and the - line up with -. Like charges repel, which causes the solid to fall apart (note that the question asks about cleavage and brittleness)
B) makes absolutely no sense
C) this is the electron sea model, which applies to only metals. Notice the positive nuclei surrounded by a sea of mobile electrons?
D) I’m pretty sure this is a chain of covalent compounds
17
you have to look at their structures.
The higher bond order the more stable the bond and shorter the bond length (shorter internuclear distance).
More stable = less energy.
I unstand O2 vs. N2 but not sure for H2 for #17 on 2014. I guess that H2 is where it is because it is a few energy levels smaller than O2 or N2.
21
yeah that’s true
23: (not sure about this explanation)
What I think is that you use P1V1 = P2V2 to find the INITIAL total pressure. The system is now at that new pressure without having established equilibrium yet.
Since pressure increases, equilibrium shifts to the right (least number of moles)
When the system moves towards equilibrium, 1 mol of CO and 2 mol of H2 are consumed for 1 mol of CH3OH, reducing the total number of moles (change in moles is -2). Based on this, it can not be C or D.
But, since the K value of the reaction was so high, the change that results from decreasing the partial pressures of reactants is so minimal that it does not bring it below 1.2 atm again. So, the answer is B.
30
A C-H bond is non polar, so it can’t have h-bonding since the dipole moment is not strong enough.
39
The Heat of fusion is .94kJ/mol (energy needed to melt or energy released to freeze)
64g / 16g = 4 mol methane
.94kJ/mol * 4 mol = 3.8kJ
40
In the gas phase, the intermolecular attractions are diminished greatly since the molecules are further apart (this allows the gas to move freely). Solids and liquids are closer together and still have strong IMFs in comparison to eacother
What practice test are you guys talking about?
That all makes a ton of sense, thank you so much!!
2014 MC
I fear that the multiple choice on the actual exam will be a lot harder than the one in the practice book. That happened with AP Human last year. So does anyone know where I can see actual multiple choice questions from the test? I know they have FRQs online but I want to see what the MC is like.
does 100.0 have 4 significant figures?
HELP I HAVE A QUESTION ABOUT HYBRIDIZATION
So to find it, you count the number of sigma bonds and lone pairs right? So 2 sigma bonds and 1 lone pair would be sp2? 2 sigma bonds and 2 lone pairs would be sp3, and 3 sigma bonds at 1 lone pair would also be sp3?
^
1
@chem1998 you got a 45/50 on the multiple choice? That’s incredible! You’re set to do this exam with your eyes closed
@BassGuitar yes, that’s how you do it. Another way to think about it is lone pairs + number of atoms the central atom is bonded to. So in CH4 C is bonded to 4 other atoms and has 0 lone pairs, so it’s sp3. And notice that in sp3 for example s is to one and p is to 3 so they add up to 4 and that’s equal, again, to lone pairs+ number of atoms. Sorry if that’s obvious, it was never explained to me when I first learned hybridization so I just thought it might help in case
@APScholar18 yup, 4 sig figs!
@stemscholar Same. Hybridization was explained to me in the worst, most complex way possible. Is there any way to know the hybridization just by looking at the compound, or do you have to do a lewis structure? Or least think of the lewis structure in your head?
MODERATOR’S NOTE:
@dsi411 Sample MCQ’s start on p. 118
http://apcentral.collegeboard.com/apc/public/exam/exam_information/221837.html
Do not ask for any exams not found on the CB website as it constitutes a violation of ToS.
Anyone else think the 2013 MC is like twice as hard as the 2014 MC?
@BassGuitar yeah, I think the lewis structure is the only way
@bassguitar how i do it is i count “s p 2 3 d 2” and whatever it lands on is what it is, and i count the bonds and lone pairs from the structure
@alecdtatum - I actually thought the 2014 MC was 2x harder than the 2013 MC…