# Examkrackers test 1d

#### Mr. Z

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For those of you who have used examkrackers simulated exam 1d, what did you think of it? does it resemble the real thing? is the scale and scoring indicative of the real mcat?

#### SolidGold

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I just took 1D on Saturday and I think that it was pretty similar to the real one. The Verbal scale seems a bit generous but the passages themselves were right on par to the past MCAT. Sciences were fairly realistic. Overall, I thought it is more realistic than the PR and Kaplan diags that I have taken. Can't wait to see how I do on 2D.

#### Mr. Z

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Solid,

What did you think of the phyics questions 137 and 138. I flat out don't get how they are coming up with the answer to 137, even after reading their explanations.

And I really don't agree with their answer to 138, I thought of all the answers only C would be reasonable. But, not according to them. What do you think?

#### SolidGold

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For 137, you have to keep on mind that the board is on a frictionless surface (ice). When the man walks to the other side of the board he is exerting a force on the board in the opposite direction. Now, if he was on a regular surface with friction, like the ground we stand on, the board would stay still, until the force of static friction would be overcome and the board would move in the direction the force is going. Have you ever gone skiing or snow boarding or something? What happens when you put a force backwards on the ski/snowboard? It moves back right, because of the force you exert on it, and it moves because the frictional force needed to be overcome is much less on snow, the ski/snowboard would not move if the same force exerted backwards on it was on regular ground. Now back to the problem, The man is at point 0 to start with, when he walks to the right, the board begins to move backwards because of the force he exerts on the board is going backwards. By the time he gets to the right end of the board he is at point 1 and the left end is at point -1. The board is moving backward to the left as much as he walks forward on the board. The whole system comes to rest when the forces even out, and that happens when the man stops walking and therfore stops exerting a force in the opposite direction. It kind of hard to see but the whole thing happens dynamically. Read this over again if you don't get it right away. Everything has to do with Newton's First Law.

Have to get back to you on 138 though, I missed it and I will be going over it today...I'll let you know what I come up with tonight or tomorrow...I may also have a couple of questions for you too.

C

#### Chronotropic

Are the questions as hard as the ones in the lesson books?

#### Mr. Z

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Originally posted by danwsu
Are the questions as hard as the ones in the lesson books?

No not at all. I actually found them to be very comparable to aamc III (haven't taken the other aamc's yet so...?). On the EK website there are posts which address the difficulty of the lesson book questions. The EK admins say that they made them difficult to get you thinking more in depth about the topic, not to simulate the mcat.

#### Mr. Z

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Originally posted by twesting2173
i wondered if either of you (or anyone else who might be able to) woudl be willing ot send me their old EK tests. I honestly cant afford the 20 buck per exam, but hoped you might be willing ot send me your left overs when you're done. i can pay you 5 or 10 buck.....please? thanks so much, i appreciate it.

I would but I marked mine up like crazy. I have notes and calculations all over the exam. And i'm still using it to study from, sorry.

#### SolidGold

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I would too but I actually split the cost with others and I have a photocopy that is written all over. If you know others who are taking the MCAT you should try to split the cost with them.

#### SolidGold

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Mr. Z: I think I might have an explanation for 138 but I'm too tired to post it...the short version is that I think that the solvent in beaker A evaporates completely because it is volatile while the solvent in beaker B remains...it has to do with the vapor pressures of both of them being different. I think the EK explanation is crap.

#### Mr. Z

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You are right about their vapor pressures being different, that arises from the colligative properties of the solute in beaker B. But, there is no way that 100% of the solvent in beaker A could end up in beaker B. For the system to come to equilibrium the vapors from beaker A would first have to saturate the entire apparatus, once there is saturation then an equilibrium process could begin and solvent A would then start fill beaker B. But, there would always be that fraction of solvent A that was has to fill the apparatus and that fraction doesn't make it into beaker B. So based on that i thought the best answer would be C... beaker B contains 90% of all solvent.

My explanation maybe a little confusing, hopefully it makes sense. what do you think?

C

#### Chronotropic

twest try your school library. Mine had AAMC test 2 and a alot of other practice test available. Also you may want to check with your advisor. Mine had alot of material on reserve at the library.

You may also want to try the public library. Not to sure on this one but it doesn't hurt to check.

#### Mudd

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Originally posted by Mr. Z:
You are right about their vapor pressures being different, that arises from the colligative properties of the solute in beaker B. But, there is no way that 100% of the solvent in beaker A could end up in beaker B. For the system to come to equilibrium the vapors from beaker A would first have to saturate the entire apparatus, once there is saturation then an equilibrium process could begin and solvent A would then start fill beaker B. But, there would always be that fraction of solvent A that was has to fill the apparatus and that fraction doesn't make it into beaker B. So based on that i thought the best answer would be C... beaker B contains 90% of all solvent.

My explanation maybe a little confusing, hopefully it makes sense. what do you think?

I don't know what the question is, but your explanation follows sound chemical theories. If you have two beakers within an overall closed system, and one contains a pure liquid, while the other contains an impure solution with solute present, there is a driving force to dilute the impure solution. This assumes that the solute in not volatile. The solvent from the pure beaker will vaporize and migrate to the solution with impurites. However, as you mention, there will always be some vapor present above the system, due to the equilibrium between vapor and liquid (the Clausius-Clapyeron equation gives the exact amount). If they claim that all of the liquid migrates over, then that is to say that there is no vapor present, which is not physically possible. If there is no vapor, then there could not have been migration. As such, less than 100% of the pure solvent migrates from its original beaker to the beaker with solute impurities.

#### JOrsay

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The system moves toward equilibrium. Solvent molecules escape from a pure solvent faster than from a the same solvent with nonvolatile solute. The atmosphere in the enclosed container struggles to reach equilibrium with both cups. There is a net movement of solvent into the solution with nonvolatile solute.

Vapor pressure is a colligative property. The vapor pressure is independent of the amount of solvent, so the vapor pressure of the pure solvent doesn't change. The solution will always have a lower vapor pressure because it will always contain some concentration of solute. So if any of the solvent transfers to the solution, all of it must over time. Whether solvent transfers to the solution doesn't depend upon the amount of pure solvent.

The pure solvent will completely evaporate over time.

The explanation to this question posted just above this post would indicate that a puddle of water could never evaporate unless humidity were zero because there would always be some vapor pressure over the water. Although I have never taught chemistry for Berkeley Review, I would have to disagree with this supposition.

#### SolidGold

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So let me get this straight, the solvent in beaker A would end up in beaker B, but is it safe to say that the solvent in beaker A evaporates and then condenses into beaker B because of the closed system that the two beakers are in? Would it also be safe to say that if the two beakers were in an "open system" that the solvent in beaker A would evaporate much faster than beaker B, if beaker B evaporates at all? I just want to be able to see whats actually going on. I can't see how it could just "migrate" over there.

Mr. Z: When I did the problem I also choose C as the answer almost along the same line of thinking that you had. But I guess there is more going on than there first seems to be.

#### Mr. Z

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Originally posted by JOrsay
The system moves toward equilibrium. Solvent molecules escape from a pure solvent faster than from a the same solvent with nonvolatile solute. The atmosphere in the enclosed container struggles to reach equilibrium with both cups. There is a net movement of solvent into the solution with nonvolatile solute.

Vapor pressure is a colligative property. The vapor pressure is independent of the amount of solvent, so the vapor pressure of the pure solvent doesn't change. The solution will always have a lower vapor pressure because it will always contain some concentration of solute. So if any of the solvent transfers to the solution, all of it must over time. Whether solvent transfers to the solution doesn't depend upon the amount of pure solvent.

The pure solvent will completely evaporate over time.

The explanation to this question posted just above this post would indicate that a puddle of water could never evaporate unless humidity were zero because there would always be some vapor pressure over the water. Although I have never taught chemistry for Berkeley Review, I would have to disagree with this supposition.

Mr. O, you make a strong case, however...

Using your puddle analogy... Consider a puddle in a small sealed chamber with no humidity. Over time the puddle would decrease in size as the water molecules came to equlibrium with the physical space of the chamber. Once an equilibrium has been reached the puddle size will remain constant, this is a result of the air-space over the puddle becoming saturated with water vapor. However, if the chamber was extremely large, the puddle would completely evaporate.

Now consider a puddle (the same size as above) in a small sealed chamber, but this time the chamber begins at 50% humidity. The same process will occur, that is, evaporation of the puddle until the air-space over the puddle is completely saturated and equilibrium is reached. Of course, the puddle will decrease in size, but to a far less extent then it did in the first experiment. This is because far less vapor molecules are needed to saturate the environment. So a puddle in a humid environment, may evaporate and it may not, depending on the physical parameters of the environment.

Your analogy works, but, it assumes that the puddle size is very small compared to the environment. In a finite environment (such as the one in question from 1d), the extent of evaporation is very much dependant on the amount of air space available. I am just saying that in the air space over the beakers you will always find vapor from the pure solvent, and if its in the air then it is not in the beaker of solvent+solute and thus <100%.

I'm not saying this is correct, but it seems more logical.

#### JOrsay

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I am somewhat uncomfortable teaching chemistry on this forum. I prefer to leave the discussion of the matter with everyone else, which I will do after this post. I originally responded because you were being misled with respect to a question in our exam.

Your argument to me neglects to consider the solution, which is an important component of the question. If the solvent vapor were to reach equilibrium with the pure solvent, it would be at a higher vapor pressure than the nonvolatile solvent solution. Thus there would be a net movement of solvent molecules into the solution lowering the vapor pressure in the air. As the vapor pressure of the solvent in the air lowered, there would be a net movement of solvent molecules out of the pure solvent.

The solution prevents the solvent vapor from reaching equilibrium with the pure solvent.

This question is based upon a common high school science experiment. (Which, by the way, does NOT indicate that it is easy. I believe that it is a difficult MCAT question.) However, it is probably in 25% of the chemistry books in the country. This is an easily reproducible experiment with salt water. I suggest that you prove it to yourself by conducting the experiment.

I feel that I have intruded here and, if you will allow me to weasel out gracefully, I leave the discussion to the rest of you. Since I stuck my two cents in, I would be happy to discuss the matter over email with anyone who wishes.

Good Luck.

#### Mr. Z

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Mr. O, you win. You're saying the concentrated solution will behave in a similar manner to a dessicant "absorbing" the vapor from the air in an attempt for the entire system to come to equilibrium.

Its not that I didn't believe you, its just that's how I go about learning topics. I question everything, and from all possible angles until I fully understand. Your explanation makes sense.

#### Mudd

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Originally posted by JOrsay:
The system moves toward equilibrium. Solvent molecules escape from a pure solvent faster than from a the same solvent with nonvolatile solute. The atmosphere in the enclosed container struggles to reach equilibrium with both cups. There is a net movement of solvent into the solution with nonvolatile solute.

Vapor pressure is a colligative property. The vapor pressure is independent of the amount of solvent, so the vapor pressure of the pure solvent doesn't change. The solution will always have a lower vapor pressure because it will always contain some concentration of solute. So if any of the solvent transfers to the solution, all of it must over time. Whether solvent transfers to the solution doesn't depend upon the amount of pure solvent.

The pure solvent will completely evaporate over time.

The explanation to this question posted just above this post would indicate that a puddle of water could never evaporate unless humidity were zero because there would always be some vapor pressure over the water. Although I have never taught chemistry for Berkeley Review, I would have to disagree with this supposition.

Plain and simple, no matter what the source of the vapor, a beaker with pure solvent or a beaker with a non-volatile solvent, there will be some vapor present. If the system starts with no vapor pressure and is closed, then there will be a net loss of solvent to vapor. There is no physical way to have all of the solution transfer to either beaker. This follows the Clausius-Clapeyron equation. If you know of some accepted theory that predicts 100% transfer in the absence of vapor initially, I would love to hear about it.

Also, I have never seen this question and thus have not seen the question details, but if it's a closed system (which is must be to consider complete transfer) starting with just air (no vapor present prior to the addition of the two beakers), then your puddle of water theory is not comparable. First, the world is a non-homogeneous pseudo-open system. Second, in your puddle of water example, the atmosphere is starting saturated, so a 100% transfer is possible, because the puddle is not responsibe for increasing the vapor pressure of the atmosphere. The puddle of water can dry up by evaporating as long as there is another spot that absorbs moisture by condensation. However, the environment remains saturated (humid), hence 100% can only be transferred if the system has the same initial and final vapor pressure (as is the case if it was initally saturated).

If the question above is two beakers in a closed system that initially has no vapor present, then there cannot be 100% transfer. This is just fundamental chemistry. Any person with a degree in chemistry knows this. I simply assume that the author of the question probably does not specialize in chemistry. It is a good question that makes you think, but it sounds like they are only thinking of a special case, and not a more general theory.

The high school experiment will only work with 100% transfer, if the environent starts saturated with water vapor. If it is not, then there will be some loss of solvent to vapor. So my question is basically this:

"Was the solvent water and did the question address the humidity as saturated?"

Mr Z:
You are 100% correct in what you have said. Stick to your ideas here, because they are correct. You actually explain it very concisely and eloquently.

#### JOrsay

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Mr. Mudd,

I cite Brown and Lemay "Chemistry and Science" 7th edition page 469 complete with before and after picture.

or

Zumdahl "Chemistry" page 447, again complete with before and after picture.

I honestly don't know if there is any undergraduate text that doesn't have this experiment. The principles here are entirely basic. I have personally performed this experiment with my 9 year old daughter while home schooling. Any chemistry teacher would be quite familiar with this experiment.

Your posts suggest some basic misunderstanding concerning chemistry and vapor pressure. Your suggestion that a solvent CANNOT evaporate when there is solvent vapor above is incorrect. As long as the vapor pressure of the solvent is less than the pressure of the actual vapor above the solvent, there will be a net movement of solvent molecules to vapor. I suppose that you feel mentioning Clausius-Clapeyron lends credence to your argument. This would only be true if you were using the equation correctly. The Clausius-Clapeyron equation gives a relationship between temperature and vapor pressure based upon the heat of vaporization. It does not say anything about nonequilibrium conditions between vapor pressure of a solvent and the solvent vapor above the solvent. No one has mentioned either heat of vaporization or temperature, and the system is not at equilibrium. The equation is simply irrelevent to the discussion. You have made some very basic errors in chemistry.

I have noticed that this is not the first time that you have freely offered advice where you are not qualified to do so. For instance, you suggested Feynman as "a bible for common physics" when studying for the MCAT. Obviously you have never opened a book by Feynman. It is physics based upon differential calculus. Hardly a bible for common physics. One who can read Feynman and understand him is already knowledgable in both math and physics and doesn't need help to understand MCAT physics. You have suggested that the verbal passages in the "Verbal reasonig Powerbuilder" by Columbia Review are too long when they are actually shorter than the average MCAT passage. Obviously, you have never opened that book either, yet you are eager to make recommendations about it. Your past posts suggest that you believe that you are an expert at economics, MCAT, chemistry, and Examkrackers MCAT books. I am more than a little skeptical.

My concern is not for your misguided enthusiasm to assist others despite being unqualified to do so. My concern is for your adhominem attacks against Examkrackers while blatantly promoting Berkeley Review materials. In what has been all too typical of your postings, you questioned our quality without having ever opened one of our books. You based your attack on your own contrived, and as usual completely misguided, economic theory. (Admittedly, you seem to be confused about whether or not you like our books.) I don't care if you go on pretending to be an expert at every topic posted on SDN while continuing to promote Berkeley Review. I am convinced that you do more damage than good for them. However, if you insist on misinforming students about our material, I intend to correct your errors.

#### Mr. Z

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Mudd, Orsay is right. This should end the debate...

Quoting "Chemistry" by Radel and Navidi page 546 under the section for colligative properties...

"One of the more interesting vapor pressure effects is the migration of solvent molecules from a less concentrated solution to a more concentrated solution via the vapor phase. The vapor pressure of the solvent above a pure liquid or a dilute solution is greater than that above a concentrated solution. If containers of both solutions are present in a closed system, the solvent will evaporate from the dilute solution, and solvent vapor will condense into the concentrated solution. This migration will continue until both solutions achieve the same concentration and the same vapor pressure. If one container holds pure solvent rather than a dilute solution, the pure solvent will completely disapear."

the question shows a diagram of 2 beakers under a glass hood, one beaker contains a volatile solvent, the other a 10% solution of a nonvolatile solute in the same solvent. They ask which of the following is true once the system reaches equilibrium.
a) dilute contains all the solvent
b) conc. contains all the solvent
c) conc. contains 90% of all solvent
d) both contain same amount

there's nothing like a good old fashioned academic debate

Perhaps the question would be better if it asked not what ends up in the conc. beaker, but, whats left in the pure solvent beaker after equilibrium is reached.

#### Mudd

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Mr Orsay:

Thanks for the reply. I'm not sure why you need to mention your nine year old daughter, but congratulations on your ability to procreate. As for the experiment, it demonstrates priniciples of osmotic migration through vapor, which apparently you only understand at a surface level, based on your reply. While I do not own Brown and Lemay, I have a copy of Zumdahl.

You will note in the reading, no where does it say that all of the water is transferred from the beaker with pure solvent to the beaker with solute (in this case sucrose). It does say that all of the water leaves the beaker that started pure, but that is not the same as saying it all moves to the other beaker. If the system is not saturated, then some of the water from the total system must fill the container as vapor.

By your supposition, if one were to place a sucrose solution in the system, without another system present, then no vapor would escape. This is only true if it is already saturated. Your original question neglects this fact, thus there is ambiguity that makes your answer less than perfect. The vapor must come from somewhere, and that somewhere is the solvent.

Originally posted by JOrsay:
Your suggestion that a solvent CANNOT evaporate when there is solvent vapor above is incorrect.

Please show where where I said no evaporation can occur under that scenario. If the solution is saturated, then there is an equilibrium between vaporization and condensation. There will be no net loss of solvent to vapor under such an equilibrium state, but an exchange is still transpiring, given that equilibrium is dynamic. However, if the environment is not saturated, the system is not at equilibrium, so there must be a net flow of material from liquid phase to the vapor state. Your question addresses a net flow, not an equilibrium. While I appreciate your opinion, it does not follow all of the principles of equilibrium and vapor chemistry. I assume this question was written by you. What are your academic qualifications to write chemistry questions?

Originally posted by JOrsay:
I suppose that you feel mentioning Clausius-Clapeyron lends credence to your argument. This would only be true if you were using the equation correctly. The equation is simply irrelevant to the discussion.

The Clausius-Clapyeron equation points out that at any temperature within the applicable temperature range, there will exist some maximum amount of vapor. If no vapor is present, then a solution will generate vapor up to this maximum. And while you may believe that mentioning it does not apply, you are wrong. The equation very much applies, as I'm sure you addressed in your upper division or graduate thermodynamic classes. When solute is added to solution, the heat of vaporization increases, which is the root of the variation in vapor pressure. I know the idea that there are multiple routes to a correct answer requires thinking beyond entry level chemistry, but the equation is very much applicable.

Originally posted by JOrsay:
I have noticed that this is not the first time that you have freely offered advice where you are not qualified to do so. For instance, you suggested Feynman as "a bible for common physics" when studying for the MCAT. Obviously you have never opened a book by Feynman. It is physics based upon differential calculus. Hardly a bible for common physics. One who can read Feynman and understand him is already knowledgable in both math and physics and doesn't need help to understand MCAT physics.

Again Mr. Orsay, you have made an error in your reading. Here is exactly what I typed.

• The author is Dr. Feinmann [syc], although I can't recall his first name. That book and Conceptual Physics by Hewitt are considered the Bibles (or Korans) of physics from a common sense perspective.

As far as MCAT materials go?

Originally posted by JOrsay:
Your past posts suggest that you believe that you are an expert at economics, MCAT, chemistry, and Examkrackers MCAT books. I am more than a little skeptical.

You have a great deal of time on your hands, keeping track of Kaplan spies at Amazon, my posts, and running a company that never makes mistakes in chemistry. I find it odd that you have only three posts here, yet you know about messages from months before. How can that be? You seem to read a great deal without replying. Be suspiscious as you will, the basic fact is that you are wrong on this question and your lack of chemistry education shows. If you feel that personally attacking me is better than admitting your error, then so be it. Whatever makes you sleep at night.

As for me, I am no expert in economics, but I do know that a company that makes over \$100 million per year is likely not concerned over trivial book sales in one of their smaller product lines. I also understand that the value of an item is what it will bring on the open market. While ebay is not a perfect system (because of timing limitations), it is still a good model. I happen to know MCAT somewhat well, having tutored for a few years. I feel pretty strongly in chemistry, which happens after you spend every year from 17 on in chemistry, whether it be as a student, researcher, or teacher. As for my expertise on Examkrackers materials, I have never professed any such thing. I just find it odd that a medium such as this website, has such a large percentage of comments on examkrackers. I also find it hard to believe that the materials can be that much better than other items on the market. You obviously feel differently. I don't care that much to go purchase the materials, so there will be no resolve here.

Originally posted by JOrsay:
I My concern is not for your misguided enthusiasm to assist others despite being unqualified to do so. My concern is for your adhominem attacks against Examkrackers while blatantly promoting Berkeley Review materials.

Thank you for commenting on my qualifications. After your little diatribe about my comments having no merit, because I don't know the subject matter, I find it peculiar that you would comment on my qualifications without knowing anything about me personally. Care to explain Mr. Kettle?

Originally posted by JOrsay:
However, if you insist on misinforming students about our material, I intend to correct your errors.

In the meantime, congratulations on running a solid company. Good luck Mr. Orsay.

#### SolidGold

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I don't see how this needs to become a war of words here. It is a simple chemistry experiment from high school because I actually found it in my high school book. I personally am not a chemistry expert nor will I ever become one, but I'm interested in learning how to do well on the MCAT, that's why I think this exchange of ideas is a good thing. I'm actually doing this experiment just to prove to myself what will happen, despite the fact that there is like a 1% chance something like this will show up on the MCAT, but it will more likely show up in the form of some type of concept.

Mudd, I have done my own little research using various chemistry books and I have found that O's explanation is more correct. The CC equation does not apply to this problem, nor anything similar.

#### JOrsay

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Ok. Jordan told me if I posted once, I wouldn't be able to stop. This is absolutely my last post, but I have to make this post.

I finally read our question closely and your posts more closely. You guys are not disagreeing with me that the pure solvent container empties completely. Your argument is just that ALL the solvent doesn't end up in the other beaker because some of it remains in the air. You are correct. There is a tiny amount of solvent left as vapor. I understood your posts to say that the pure solvent would not totally evaporate because there was vapor pressure. Your posts were clearly written and I just misread them. It was my mistake and I apologize.

Boy, this crow sure tastes lousy.

The correct answer to the question is still B: "Beaker B will contain all the solvent." If this were on a real MCAT, AAMC would argue that MCAT asks for the best answer choice. I admit that I did not consider that anyone would interpret the question the way that you did. In fact, no one has done so before as far as I know. I believe that I wrote this question. I don't remember for sure. I believe that the question is a reasonable facsimile of MCAT as it stands. I will probably not change it. For example, if you were asked, on the MCAT, how far a baseball falls in 1 second when dropped from a tall building and given the choices:

a) 2.5 m
b) 4 m
c) 5 m
d) 10 m

The correct answer would be 5 m. On the MCAt, you wouldn't argue that it falls less than 5 m due to air resistance, so there is no correct answer, or that 4 m is a better answer. The same is true here. The amount of solvent in the vapor is likely to be negligible compared to the solvent in Beaker B.

Mudd, Feynman is not a 'common sense' physics book, the CC equation is still not useful in explaining this problem, I still believe that you are an undercover salesman for Berkeley, and your economic theory was silly. However, I did misread your post and my 'attack', in that regard, was unwarranted.

#### Mudd

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Originally posted by JOrsay:
Mudd, Feynman is not a 'common sense' physics book, the CC equation is still not useful in explaining this problem, I still believe that you are an undercover salesman for Berkeley, and your economic theory was silly. However, I did misread your post and my 'attack', in that regard, was unwarranted.

You know, a spirited conversation is one of the finer pleasures in life, so I thank you. There is absolutely no doubt that my nit-picking detail is beyond the scope of the MCAT, but basically my too long to read, even having written it essay was more of a way to vent the wrath of traffic than anything else. Nothing personal at all, and as is often the case with an argument, I walk away with a greater repsect for you. For this, I thank you.

I must agree with you on a few points, and thus admit a couple errors I have made in this exchange. Hewitt is the conceptual book I have read, along with the way things work (by McCaughley (sp?)). I got the Feynman book from an over-zealous parent as a gift while taking college physics. I looked at it long enough to discover that it wasn't my taste. I kept hearing from my physics friends (scary thought there) that it was there bible.

The CC equation is applicable in so much as it sets the idea that a vapor must be present, but in and of itself does not answer the question. If the question stated an initial vapor pressure, it could be used to determine the percentage of solvent that must be converted into vapor to saturate the system. No more and no less.

As for being an undercover salesman for Berkeley, I am neither undercover nor a salesman. I am forthright about my opinions, because I am big fan of the little guy. When a student asks me what to take, I recommend small companies. The fact that I have worked for two test prep companies and enjoyed my experience with the smaller, more personal one to a greater extent than the corporate one certainly adds to my feelings. As you may have read here, I recommend EK to anyone in your service area (which I believe to be NY and NJ from the comments here). I recommend BR to anyone in their service area for the same reason. The little guy who specializes offers a FAR better service than the conglomerate who doesn't focus. Not to mention, I have heard nothing but praise for you and Jordan from my old boss at BR. I think you share a common enemy so to speak. In a perfect world, it would be nice to see you two get all the ink and consummers, rather than Kaplan and PR. Students would certainly do better on the MCAT if that were the case.

I still stand by my economic theory. However, having tracked prices a bit longer, the discrepency between values is less. Market scarcity and timing of the auction's end time play a role, but even that cannot discount the fact that Kaplan books are available for very cheap, PR books go for a little more, your books get a price near their face value, and BR books typically go at their face value. Audio Osmosis consistantly gets a high price. While I refrain from buying much on eBay, I make it a point to let the students I know can't afford a course know about various auctions. Unfortunately, the only stuff that is dirt cheap is the Kaplan stuff.

So after all of this, I seriously hope to read more from you on this site. I hope we argue again in the future. Of course, I just wasted a great deal of the downtime I had available between steps of the reaction I'm doing, so the chance to do exposes will likely not recur. Thanks agin for the argument, and seriously, good luck in all of your endeavors.

#### Mudd

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Mr. Orsay:

Thank you for taking the time to send a PM. I greatly appreciate your candor. I plan to keep it's contents between us, but I must publicly acknowledge that you have been forthright and honorable in your message. I greatly appreciate you setting the record straight who you actually know here.

I still have some fundamental disagreements with you, but that is the beauty of a world full of differing ideas. I have no doubt the number one feature of your company is your passion. I know that is also what I admire about the company I trumpet to those who will listen. May you both kick some corporate a...!

#### blackbird03

##### Senior Member
7+ Year Member
15+ Year Member
For those of you who took EK 1D, How close was it to your real scores?

#### SolidGold

##### Florida winters are the best!
10+ Year Member
15+ Year Member
As my first diag of this time around, I scored 1 point higher than I did last year on the real thing so I think its pretty realistic. I felt the same way after it too as I did after the real MCAT last year. Do as many full lengths as you can. You need to practice, practice, and practice.

C

#### Chronotropic

I thought it was good practice but I felt the questions were a bit different from the real thing. Can't put my finger on it just a feeling I had.

Anyway on AAMC materials and tests I would score around a 10 on all sections. When I took the EK I got a 10 on verbal an 8 on PS and managed to get a 6 on bio. Hopefully bio was just a fluke. I notice this on alot of non AAMC tests. For instance on Kaplan I would jump around from 6 to 9 on all sections. So take it for practice but don't be to concerned with the score.

Good look.

C

#### Chronotropic

Good luck not look.

I need to get more sleep.

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