Top Score Solubility Problem

[krazchikin]

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65. What is the expression for the solubility product constant, Ksp, for BaCl2?

a. [Ba2+] [Cl2-]
b. [Ba2+]^2 [Cl-]^2
c. [Ba2+] [Cl-]^2
d. [Ba+]^2 [Cl-]^2
e. [Ba+] [Cl-]^2

I put C, but Top score says the answer is A, which doesnt make sense to me...BaCl2 should dissociate to form one Ba with a charge of plus 2 and two Cl with a charge of -1....especially since halogens usually form -1 anions....not sure if there's something im missing or if this is a top score error...hopefully it's the latter...lol

Top scores solution explains that the solubility product is a description, not of solubility, but simply of the ionic concentrations in the saturated solutions.

 

[Sugafoot79]

are you sure? mines says the answer is C. u might wanna reinstall or call itsupport

 

[krazchikin]

could be b/c mine is from december....but that's a relief....do you think you could write out what the solutions says? thanks!!

 

[smurf528]

yeah mine said C too. I would look it over to make sure.

 

[krazchikin]

What about question 73 on ochem of the first test? It asks for the 1HNMR spectrum that shows only two singlets.

a. H3C-H2C-O-CH2-CH3
b. H3CO-benzene ring-OCH3
c. cyclopentane-CH3
d. H3C-CO-CH3 (acetone)
e. epoxide ring-CH3

Mine says B is the answer, but wouldnt the Hydrogens on the benzene ring show up on the NMR? They're not singlets, but they would at least show up on the NMR as triplets...and the question asks for ONLY two singlets?

 

[krazchikin]

Sorry, but what about question 76 where you have benzaldehyde reacting with acetone in presence of NaOH and then H3O+ ? I put B, but Top score says the answer is A, which is benzene-CH=CHC=O(ketone)-CH3, which doesnt have contain the carbonyl Oxygen. But the solution does say that the answer is an aldol, so does the acidification with H3O+ turn the O- into an OH instead of =O? FYI, I might be adding a some more questions if I'm getting more discrepancies...hopefully you guys will still be around...lol thanks!

 

[doc toothache]

What about question 73 on ochem of the first test? It asks for the 1HNMR spectrum that shows only two singlets.

a. H3C-H2C-O-CH2-CH3
b. H3CO-benzene ring-OCH3
c. cyclopentane-CH3
d. H3C-CO-CH3 (acetone)
e. epoxide ring-CH3

Mine says B is the answer, but wouldnt the Hydrogens on the benzene ring show up on the NMR? They're not singlets, but they would at least show up on the NMR as triplets...and the question asks for ONLY two singlets?

For 1,4 dimethoxybenzene the hydrogens on C2, C3, C5 and C6 are equivalent and will show as a singlet.

http://www.nmrdb.org/predictor?smiles=C1(=CC=C(C=C1)OC)(OC)

 

[krazchikin]

For 1,4 dimethoxybenzene the hydrogens on C2, C3, C5 and C6 are equivalent and will show as a singlet.

http://www.nmrdb.org/predictor?smiles=C1(%3DCC%3DC(C%3DC1)OC)(OC)

AHHHHHH...okay.....so acetone will only show ONE singlet since they're also equivalent? makes more sense...i didnt even remember what equivalent hydrogens were...lol....thanks dr. TA! any takers on the other problem?

 

[Sugafoot79]

yeah, how about question 56? Care to explain where that 2.854 came from in the solution? I'm lost. Everything else makes sense except that number. Where did it come from?

 

[krazchikin]

yeah, how about question 56? Care to explain where that 2.854 came from in the solution? I'm lost. Everything else makes sense except that number. Where did it come from?

lol yeah i almost forgot about that one....what a weird problem

 

[Sugafoot79]

Bump

 

[krazchikin]

i was thinking about bumping, but i wasnt sure if ppl did that on sdn...lol