The Official 4/26/13 MCAT Thread

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In lieu of the other threads popping up, I thought it'd be good for those of us taking the test to identify ourselves 👍

I'm not planning on following a rigid study schedule until January, but I think I'm going to watch a single video off of wikipremed daily to lightly review.

Good luck to everyone... we will crush this exam..
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ktanner777 said:
i took it to mean that it was less than .5 ohms. so it has to be 4.0A or more. 4 was the largest number in the choices so i went with it. but the question is worded terribly.
Click to expand...


My math skills. I was thinking as the denominator decreased, the value would end up more like the numerator. No idea where I pulled that one from.
 
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Ibn Alnafis MD said:
Regarding Q. 38 of the same passage:

It was easy to eliminate options B and D simply becaues the charges on both sides don't cancel out. I chose the right answer "C" but used different rationale than the one they provided in the solution. I thought, since Cu is one of those elements that don't oxidize easily (money elements), it must be reduced and Al will be oxidized instead. Was I luck? or does my rationale also make sense?
Click to expand...

For this question, all you need is the table. Don't rely on outside information that much. Looking at the table, when you have Cu2+ and Al(s) together, you get a new solid (precipitate) -> meaning that Cu2+ is getting reduced to Cu(s). However, if you have Al3+ and Cu(s) together, there is not change in the composition of the metal strip -> meaning that the metal strip remains copper. Therefore, Cu2+ is the one being reduced and Al is the one being oxidized. Hope that helps!
 
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This question refers to the chemical equation:

4C3H5N3O9(l)---> 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)

The pressure inside a sealed 3-L tank containing 25g of which product Reaction 1 would be the highest at 300K? (Assume no condensation.)

A) CO2
B) H2O
C) N2
D) O2

I choose A because I see from the chemical reaction that it has the greatest number of moles. Perhaps I did not understand the question or passage correctly?

I sorta understand the answer that AAMC gave but I don;t see how I was supposed to interpret the question properly to get that answer. Does the tank only contain H2O and how was I supposed to know that it only contained one product?

Also, I know PV=nRT and I agree that if all other variables are held constant that if n increases so does P... It does not make any sense that increasing P would increase n... or does it?
 
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tinylilron said:
This question refers to the chemical equation:

4C3H5N3O9(l)---> 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)

The pressure inside a sealed 3-L tank containing 25g of which product Reaction 1 would be the highest at 300K? (Assume no condensation.)

A) CO2
B) H2O
C) N2
D) O2

I choose A because I see from the chemical reaction that it has the greatest number of moles. Perhaps I did not understand the question or passage correctly?

I sorta understand the answer that AAMC gave but I don;t see how I was supposed to interpret the question properly to get that answer. Does the tank only contain H2O and how was I supposed to know that it only contained one product?

Also, I know PV=nRT and I agree that if all other variables are held constant that if n increases so does P... It does not make any sense that increasing P would increase n... or does it?
Click to expand...

I believe the answer is H2O because you are looking for the greatest number of moles since everything else is constant. Therefore, you are looking for the greatest ratio of (stoichiometric coefficient / MW) and that is H2O:

H2O: 10/18
CO2: 12/44
N2: 6/28
O2: 1/32
 
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osprey099 said:
I believe the answer is H2O because you are looking for the greatest number of moles since everything else is constant. Therefore, you are looking for the greatest ratio of (stoichiometric coefficient / MW) and that is H2O:

H2O: 10/18
CO2: 12/44
N2: 6/28
O2: 1/32
Click to expand...

Correct. Sealed 3L tank with a set amount of 25g of each substance. 25g of H2O will lead to the most moles of H2O, therefore, the highest pressure.
 
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osprey099 said:
I believe the answer is H2O because you are looking for the greatest number of moles since everything else is constant. Therefore, you are looking for the greatest ratio of (stoichiometric coefficient / MW) and that is H2O:

H2O: 10/18
CO2: 12/44
N2: 6/28
O2: 1/32
Click to expand...

Okay perhaps I misread the question... I am supposed to assume that there is only 1 product in the tank... therefore because H2O has the lowest MW then it would have the greatest number of moles would produce the highest pressure..

Why do we need these ratios? Why can't we just get our information from the MW?
 
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tinylilron said:
This question refers to the chemical equation:

4C3H5N3O9(l)---> 12CO2(g) + 10H2O(g) + 6N2(g) + O2(g)

The pressure inside a sealed 3-L tank containing 25g of which product Reaction 1 would be the highest at 300K? (Assume no condensation.)

A) CO2
B) H2O
C) N2
D) O2

I choose A because I see from the chemical reaction that it has the greatest number of moles. Perhaps I did not understand the question or passage correctly?

I sorta understand the answer that AAMC gave but I don;t see how I was supposed to interpret the question properly to get that answer. Does the tank only contain H2O and how was I supposed to know that it only contained one product?

Also, I know PV=nRT and I agree that if all other variables are held constant that if n increases so does P... It does not make any sense that increasing P would increase n... or does it?
Click to expand...

I missed this question, too, for the same exact reason.
 
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Just finished re-doing the BS section of AAMC 11. I was done with 18mins to spare. Got 14, but most importantly, I got every question right on the Ebola passage🙂. Although when I checked my performance yesterday I knew how many questions I missed on that passage, I didn't pay attention to which questions I missed. Making another attempt at this passage gave me a totally new prespective. The questions I missed yesterday were because I didn't read the graph right. This time, I spent more than a whole minute examining the graph and I discovered things I wasn't aware of.

On the other hand, I missed two questions on the secretory lysosome passage 🙁. I still did better than yesterday on this passage, but still not satisfied with my performance given that this is my second attempt. In my opinion, the lysosome passage is harder than the Ebola one.
 
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I have a question regarding Q.#50 on the PS of AAMC11.

I ruled out A and C, immediately, but was debating between B and D. Ultimately, I picked B because I was more confident about it. But, why D is wrong?

Tracing my way through the graph, I find that the lines intercept at the Fe(OH)2 region.
 
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Ibn Alnafis MD said:
Just finished re-doing the BS section of AAMC 11. I was done with 18mins to spare. Got 14, but most importantly, I got every question right on the Ebola passage🙂. Although when I checked my performance yesterday I knew how many questions I missed on that passage, I didn't pay attention to which questions I missed. Making another attempt at this passage gave me a totally new prespective. The questions I missed yesterday were because I didn't read the graph right. This time, I spent more than a whole minute examining the graph and I discovered things I wasn't aware of.

On the other hand, I missed two questions on the secretory lysosome passage 🙁. I still did better than yesterday on this passage, but still not satisfied with my performance given that this is my second attempt. In my opinion, the lysosome passage is harder than the Ebola one.
Click to expand...

dont waste your time re-doing old problems. either do new problems, watch youtube videos on certain topics ( carbonyl reactions, sn1,sn2,e1,e2) or just go over notecards...
 
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Ibn Alnafis MD said:
Just finished re-doing the BS section of AAMC 11. I was done with 18mins to spare. Got 14, but most importantly, I got every question right on the Ebola passage🙂. Although when I checked my performance yesterday I knew how many questions I missed on that passage, I didn't pay attention to which questions I missed. Making another attempt at this passage gave me a totally new prespective. The questions I missed yesterday were because I didn't read the graph right. This time, I spent more than a whole minute examining the graph and I discovered things I wasn't aware of.

On the other hand, I missed two questions on the secretory lysosome passage 🙁. I still did better than yesterday on this passage, but still not satisfied with my performance given that this is my second attempt. In my opinion, the lysosome passage is harder than the Ebola one.
Click to expand...

18 minutes to spare really? I understand that it was a re-do but still... I re-read all the passages if/when I re-attempt a test... at most I will have 5 minutes to spare... maybe 7 minutes as a high
 
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Ibn Alnafis MD said:
Just finished re-doing the BS section of AAMC 11. I was done with 18mins to spare. Got 14, but most importantly, I got every question right on the Ebola passage🙂. Although when I checked my performance yesterday I knew how many questions I missed on that passage, I didn't pay attention to which questions I missed. Making another attempt at this passage gave me a totally new prespective. The questions I missed yesterday were because I didn't read the graph right. This time, I spent more than a whole minute examining the graph and I discovered things I wasn't aware of.

On the other hand, I missed two questions on the secretory lysosome passage 🙁. I still did better than yesterday on this passage, but still not satisfied with my performance given that this is my second attempt. In my opinion, the lysosome passage is harder than the Ebola one.
Click to expand...

graphs are important... i hope if there are any on friday that they will be easy to understand
 
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PS #23 tripped me up. Since when is Di = F? I was trying to solve it as 1/f = 1/Do + 1/Di but had no value for Do. The explanation they give says that F = distance to retina but i assumed that when consider the lense of the eye, the distance to the retina is simply the image distance of the converging lense. can someone explain?
 
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ktanner777 said:
PS #23 tripped me up. Since when is Di = F? I was trying to solve it as 1/f = 1/Do + 1/Di but had no value for Do. The explanation they give says that F = distance to retina but i assumed that when consider the lense of the eye, the distance to the retina is simply the image distance of the converging lense. can someone explain?
Click to expand...

You over thought the problem. Everything they gave you was there. P is equal to 1/f. You just had to make sure to convert the units to meters and do the inverse.

This had nothing to do with image or object distance. Power is inversely related to the focal length and power of a diopter involves nothing else. It was a discrete in a passage.
 
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Albein said:
You over thought the problem. Everything they gave you was there. P is equal to 1/f. You just had to make sure to convert the units to meters and do the inverse.

This had nothing to do with image or object distance. Power is inversely related to the focal length and power of a diopter involves nothing else. It was a discrete in a passage.
Click to expand...

i made this mistake as well.

they gave us the distance between the retina and the lens, but i didnt know that equaled focal length...
 
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Assume that a circuit similar to that in Figure 2 is set up in which X=Al, X(n+)=Al(3+), Y=Cu and Y(m+)=Cu(2+). Which of the following reactions will occur?

I chose C because I knew from reaction 2 that reducing Cu(2+)--->Cu(s) is spontaneous...

Isn't there some rule about Anode being on the left and Cathode on the right? Do you think if there is that it is safe to use on MCAT? In addition, I just wanted to make sure that I understood AAMC's answer correctly...we should have looked at table 1... in table 1 we see that when Al--->Al(3+), Cu(2+)---->Cu and this is a spontaneous reaction... and the reverse reaction does NOT occur with Cu--->Cu(2+), and Al(3+)---->Al(s) since it shows that there is no change in table 1.

Thus the answer has to be Al--->Al(3+), Cu(2+)---->Cu as it is spontaneous... thus not balanced:

Al(s) + Cu(2+)---->Al(3+) + Cu(s) then I can balance it to get answer C.

Is how I solved it correct?
 
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Albein said:
You over thought the problem. Everything they gave you was there. P is equal to 1/f. You just had to make sure to convert the units to meters and do the inverse.

This had nothing to do with image or object distance. Power is inversely related to the focal length and power of a diopter involves nothing else. It was a discrete in a passage.
Click to expand...

But since when is distance from lense to retinal = focal length. Distance to retina is distance to image which is = to the inverse of focal length - inverse of distance to object.
 
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Ibn Alnafis MD said:
I have a question regarding Q.#50 on the PS of AAMC11.

I ruled out A and C, immediately, but was debating between B and D. Ultimately, I picked B because I was more confident about it. But, why D is wrong?

Tracing my way through the graph, I find that the lines intercept at the Fe(OH)2 region.
Click to expand...
They do not. They intersect at the Fe(OH)3 region. Use your pencil to make the straight line on the screen and you will see.
 
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tinylilron said:
18 minutes to spare really? I understand that it was a re-do but still... I re-read all the passages if/when I re-attempt a test... at most I will have 5 minutes to spare... maybe 7 minutes as a high
Click to expand...

I have never had a problem finishing early on the BS section (knocking on wood). I hope this doesn't change on Friday. Yesterday, I finished the BS section with ~5 mins to spare. Today, since I had a recent exposure to these passages I was able to go through them faster.

Not until yesterday was I really crunched on timing in regards to PS. I remember I had about 9 mins left when I got to the last passage, so I skipped it, did the discretes at the end, and returned to it. Thanks God, it was an easy passage, so I skimmed in a minute and did the short questions first and the long one I left to the end.

Verbal on the other hand has always been a problem for me. I always arrive to the final passages with less than 8 mins. In fact, when I was taking AAMC 11 yesteday, I arrived to the 6th passage with only 12.5 mins left. I skipped it, did the 7th one (a natural science one), came back to it with only 3.5 mins left. I bubbled everything B at first. Then I read only the intro and the conclusion, and tried to answer the questions to the best of my ability. Luckily, I got half of them right!
 
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GandalfTheWhite said:
i made this mistake as well.

they gave us the distance between the retina and the lens, but i didnt know that equaled focal length...
Click to expand...
You can know it's the focal length because that is where the image is formed. Light enters the eye and hit the retina. That's how we see. It was a weird hybrid between a Bio and Physics discreet. Needed to recall your eye anatomy and physiology.
 
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Back 5 said:
They do not. They intersect at the Fe(OH)3 region. Use your pencil to make the straight line on the screen and you will see.
Click to expand...

I see now. I didn't pay attention that the graph had Fe(OH)3 while the answer choice had Fe(OH)2 in it. Gotta pay more attention to such subtle differences.
 
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Back 5 said:
You can know it's the focal length because that is where the image is formed. Light enters the eye and hit the retina. That's how we see. It was a weird hybrid between a Bio and Physics discreet. Needed to recall your eye anatomy and physiology.
Click to expand...

i knew that the retina is where the image formed, but they didnt say where the object was so i was stumped
 
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That's only for galvanic cells. In electrolytic cells, they often have the anode on the right. And yes you have the right understanding.
 
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GandalfTheWhite said:
What reaction requires strong bases vs strong BULKY bases? (SN1, SN2, E1, E2)
Click to expand...
The easy way I remember SN1 vs SN2:

3->2->1. Tertiary and secondary = SN1. 1->2. Methyl and primary = SN2.

Tertiary and secondary substitution reactions aren't going to happen very well with large bulky bases because of steric hinderance. So, bulky bases, like tert-butoxide are going to be used for SN2 and E2 reactions.
 
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tinylilron said:
Assume that a circuit similar to that in Figure 2 is set up in which X=Al, X(n+)=Al(3+), Y=Cu and Y(m+)=Cu(2+). Which of the following reactions will occur?

I chose C because I knew from reaction 2 that reducing Cu(2+)--->Cu(s) is spontaneous...

Isn't there some rule about Anode being on the left and Cathode on the right? Do you think if there is that it is safe to use on MCAT? In addition, I just wanted to make sure that I understood AAMC's answer correctly...we should have looked at table 1... in table 1 we see that when Al--->Al(3+), Cu(2+)---->Cu and this is a spontaneous reaction... and the reverse reaction does NOT occur with Cu--->Cu(2+), and Al(3+)---->Al(s) since it shows that there is no change in table 1.

Thus the answer has to be Al--->Al(3+), Cu(2+)---->Cu as it is spontaneous... thus not balanced:

Al(s) + Cu(2+)---->Al(3+) + Cu(s) then I can balance it to get answer C.

Is how I solved it correct?
Click to expand...

I'm not sure if such a rule exists, but making this assumption can be dangerous. Just remember oxidation occurs at the anode and redoxtion at the cathod (an ox, red cat). This will eliminate B and D. In regards to choosing between A and C, I also neglected the table and reasoned that Cu (a money metal) doesn't oxidize easily, so it must be reduced and Al should be oxidized. However, the correct way to solve this problem is by examining the table provided.
 
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Ibn Alnafis MD said:
I think strong bases favor E2 while strong, bulky bases favor E1. SN1 and SN2 favor strong nucleophiles, but the latter favors smaller nucleophiles.
Click to expand...

nucleophiles dont matter for SN1 and E1 since they dont affect the rate.

E2 favors strong bulky bases whereas SN2 just favors strong bases (bulky bases such at tert-but can hinder rate)


correct me anyone if you think I'm wrong.
 
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Albein said:
Can anyone help me understand 44 in PS on AAMC 11? I can't seem to grasp it
Click to expand...

I don't have a "scientific" way of solving this problem, but the way I though of it is comparing it to a pendulum. The pendulum as its minimum KE when it's making a 90 degree angle with the its position at rest. Therefore maximum KE occurs at zero angle and minimum KE occurs at 90 degree angle (pi).
 
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Albein said:
I have them memorized. I just don't see why it's 90 as opposed to 45
Click to expand...

It's a pendulum.

Think back when you were a kid (or an adult, I don't know how your nights are) and you spun in a circle with a ball on a string. The faster you go, the more horizontal the string/ball became. The string would ultimately get to 90* and that is where the tension was highest, the ball was feeling the highest effects of gravity, or G forces. The angle will not exceed 90*.

If the ball/string is only at 45*, it's not at it's maximum.
 
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Back 5 said:
It's a pendulum.

Think back when you were a kid (or an adult, I don't know how your nights are) and you spun in a circle with a ball on a string. The faster you go, the more horizontal the string/ball became. The string would ultimately get to 90* and that is where the tension was highest, the ball was feeling the highest effects of gravity, or G forces. The angle will not exceed 90*.

If the ball/string is only at 45*, it's not at it's maximum.
Click to expand...

I get that too lol. I don't see why we approach it as a pendulum. I'm feeling really dumb because it seems obvious.
 
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Albein said:
I get that too lol. I don't see why we approach it as a pendulum. I'm feeling really dumb because it seems obvious.
Click to expand...
Because a pendulum is attached at 1 point, It won't exceed 90* relative to it's anchor.

Anyway...the good news is we won't see this question again 🙂
 
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I'm reviewing AAMC 9, and question 106 is really bothering me. I know why I missed the question... But did anyone else think that the question was asking to compare isomers?
 
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What are you opinions on Passage III of the PS of AAMC? Three random unrelated experiments... then a discrete in the questions that had NOTHING to do with the three random unrelated experiments? I was like WTF? This was not similar to the other practice MCATs
 
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tinylilron said:
What are you opinions on Passage III of the PS of AAMC? Three random unrelated experiments... then a discrete in the questions that had NOTHING to do with the three random unrelated experiments? I was like WTF? This was not similar to the other practice MCATs
Click to expand...

Hmm... i felt like the types of questions were different from what I was used to. But they weren't over the top difficult. I think as long as you understood the figure and the main idea of the passage, the questions weren't terrible.

I did miss the optical power question... 50 is not close enough to 40 for me. Haha
 
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GandalfTheWhite said:
i knew that the retina is where the image formed, but they didnt say where the object was so i was stumped
Click to expand...

1/f=1/o + 1/i

Compared to the distance from retina to lens, we can assume that distances from lens to object are much much bigger. Hence we assume i=f
 
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