# do you know any of these answers?

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#### lola

##### Bovine Member
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OH MY GOD, i think i actually know something for once

2)If 1 mol of H2 and 1 mol of O2 are mixed and allowed to react according to the question: 2H2 + O2 ----> 2H2O, the maximum number of moles of H2O that could be produced is:

the answer was 1. I did this kind of logically, but is there a better way?

I did it logically too. I don't know a better way other than to say that since you only have 1 mole of H2, you can't make more than 1 mol of H20 b/c it's a 1:1 ratio

3) For the equilibrium (NH4)3PO4 <==>3NH34+ +PO4^3-, the solubility product expression (Ksp) is:

the answer is [NH4+]^3[PO4^3-]. Just checking why do you not put the reactant on the bottom here in your answer? Are you supposed to only have ions in the Ksp? This might be a stupid question Im asking!

You only put aqueous & gases in Ksp expressions. For all K expressions, you DO NOT put solids or liquids

5) The equilibrium for the following rxn:

Cl2 (g) + PCl3 <====> PCl5 (g) delta H= -88 kJ/mol
can be shifted to the right by:

A) decreasing the volume
B) decreasing the pressure
C) increasing the temp
E) emoving chlorine from the container

you use le chatelier's principle here. if you decrease the pressure, it goes to the side with the most number of moles. since pressure and volume are inversely related, decreasing the volume is the same as increasing the pressure. when you increase the pressure it goes to the side with the least number of moles -- to the right.

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hey thanks lola!

#### lola

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ha ha... i just saw some of them were already answered on your other post. oh well!

just to add one more thing... for Ksp, you will never have the reactant in the expression b/c it's always a solid dissolving. for most other K expressions (e.g. Ka), the reactants will go in b/c they are usually not solids or liquids -- but could be so watch out!

#### vixen

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bump, I put some more questions on that link. Sorry, this one practice exam I have doesnt have explanations!

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bump again

#### freakazoid

##### Guy Friend Extraordinaire
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Vixen:

Could you post your questions numerically increasing so that 1) and 2) don't get mixed up from 1) and 2) from before?

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When an oxygen nucleus containing nine neutrons emits an alpha particle, the other product of the balanced rxn:

O (17 on top, 8 on bottom) -------> He (4 on top, 2 on bottom) + ____ is:

The answer is C (13 on top, 6 on bottom)....but Al (13 on top, 6 on bottom) was also a choice.

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I know this has been answered before, but I also thought it would be useful to note that the bottom number is the atomic number, and Al has an atomic number of 13 (diff from the 6 of C). When in doubt, check the periodic table.

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1) How many moles of AgIO3 (Ksp=3.1 x 10^-8) will dissolve in one liter of a 10^-5M soln of NaIO3?

the answer is ([3.1x10^-8]^1/2) - (10^-5). How'd they get that?

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Since AgIO3 --> Ag(+1) + IO3(-1), we get a solubility expression of Ksp=[Ag(+1)][IO3(-1)]. Since we know that every molecule of Ag(+1) dissociated came from AgIO3, then [AgIO3]=[Ag(+1)]. We also know that every molecule of IO3(-1) dissociated (not in the presence of NaIO3) there was an Ag(+1), so that [Ag(+1)]=[IO3(-1)], and if we set that equal to x, Ksp=x^2, so that x=(Ksp)^(1/2). We subtract the amount of [IO3(-1)] because AgIO3 can't dissociate anymore once one of its ions have reached max concentration.

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2) what is the standard Enthalpy change for the following rxn:

2SO2 + O2 ---> 2SO3

substance delta Hf (kJ/mol)
SO2 -297
O2 0
SO3 -396

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Portlander had the right idea, it's simply deltaHf = deltaHf(products) - deltaHf(reactants)

Just make sure you multiply in the appropriate number of moles and it's done.

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1) Which of the following cmpds is not capable of hydrogen bonding?

A) C4H11N
B) C4H9NO
C) C4H7F3
D) C4H8O2
E) C4H10O

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Remember hydrogen bonding occurs between
1) an H attached to FON and
2) the lone pair of another FON atom
of a molecule, so a molecule has to have these two requirements to hydrogen bond with itself.

So if you tried to draw the molecules out and fulfill the octect rule for all the atoms (besides H), you'll find that you'll get these two requirements, except C. An easier way of noting this is that F has a valence number of (-1) and so can typically attach to one atom at a time, unlike N and O. Since F can only attach to C, it fulfills the second requirement but does not have another bond with H, so does not fulfill the first requirement. Actually, now that I think about it, 1) is the only requirement--regardless, FON will have a lone pair whether or not an hydrogen is attached to it (think of water). 2) is sort of accessory--it is nothing w/o the first requirement.

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2) Which cmpd would not be expected to react w/sodium metal?
A) CH3OCH3
B) CH3CH2OH
C) CH3COOH
D) C6H5OH
E) CH3I

The answer is A. Is this becaue its an ether, which really wont react with the Na, and then rest are more likely to react w/Na?

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I'm pretty stumped on this one. The only possible reason I could come up with is that out of all of those (alcohol, carboxylic acid, phenol, haloalkane), I would guess the ether to be least reactive. Maybe Na, being a metal, wants to be oxidized and lose an electron. I could see hydrogens off the phenol, alcohol, and carboxylic acid grabbing the electron and turning into H2(gas), as well as the iodine off the haloalkane and bubbling out I2(gas). The ether has nothing, really, at least nothing that would dissociate normally in an aqueous solution . . . (but then again I dunno if CH3I dissociates)

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3) Na+ is NOT an electrophile because:

A) It does not readily form a covalent bond with a nucleophile
B) It is an ion
C) It has a noble gas configuration
D) It has an empty orbital
E) It is electron deficient

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Tough. Only thing I can think of is pick the most relevant one.

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4) Choose the letter corresponding to the INCORRECT completion. In the nitration of benzene by conc H2SO4 and conc HNO3:

A) the rxn is irreversible
B) the rate determining step is the removal of H+ from the intermediate
C) the attacking electrophile is NO2+
D) The second nitro group enters the benzene ring meta to the first nitro group
E) If the conc H2SO4 is replaced by some other strong acid such as HF, the rxn will still proceed at a similar rate.

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Weird, weird, question. I don't think this will be tested, but . . . as far as I know, this rxn is called electrophilic aromatic substitution . . . it's asking for the INCORRECT completion, which means B) isn't right, meaning H+ leaving isn't the rate-determining step. A) and C) are just qualities of the rxn, D) because NO2 is deactivating and meta-directing, and E) I don't agree with because HF is NOT a strong acid.

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Also, it has a question with stability and it says in the answer that 2,2-dimethyl butene is more stable than cyclohexene, does anyone know why?

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Same thing. The substances are so dissimilar that I don't think the MCAT will test you on that. (Which is also a good way of saying I don't know)

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Which of the following alkenes will react with Bromine to yield a meso cmpd?

A) trans-2-butene
B) 1-butene
C) cis-2-pentene
D) 2,3-dimethyl-2-butene
E) cis-2-butene

answer is A. How do you get a meso cmpd from that? I dont see how there could be symmetry if they are trans?

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B) and C) don't have any symmetry across the double bond, and D) seems too sterically hindered to react. That leaves A) and E), and you just have to go through the rxn to tell that A) a plane of symmetry. Draw it out: the double bond grabs one of the bromine, and temporarily binds both carbons of the double bond, leaving that side too sterically hindered to be attacked by the other bromine, which attacks from the other side at one carbon, causing the other bromine to *leave* from that carbon (like nucleophilic substitution, SN2).

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I'm not sure I helped much on some of the problems, but hope this helps.

#### freakazoid

##### Guy Friend Extraordinaire
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On the last question, if D) did react anyhow, it wouldn't yield any chiral centers, and meso cmpds need at least two, I think . . .

#### vixen

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freakazoid, THANK you sooooo much! I really appreciate it!!

#### STACM

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I tried solving #34 GCHEM in destroyer which is comparable to GCHEM Kaplan Online test #59 (the one with AgIO3 in freakazoids post) stated in this thread, but for some reason it doesnt work out, could someone do #59 please?

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