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Aamc11: 20, 23, 24.

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pfaction

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No question on #23, I'm just really ****ing pissed they made you round 50 down to 40. Are they serious? Is this the rounding that's done on the MCAT?! ****.

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HUH?! I don't get these answers.

Also a question on #1:

Are oxidation-reduction products not effected by pressure?
 

MedPR

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The 50 to 40 one you're talking about is the diopters question, correct? That one threw me off too.

For #25 read the entire sentence you underlined part of. "...proportional to the square of the speed and the area..."

It's like KE=1/2mv^2. A 50kg object moving at 1m/s has less KE than a 1kg object moving at 50m/s.

For #1, redox products are affected by pressure.

Edit: Oh, and for #20 you just need to use the stress/strain equation given in the passage.
 
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pfaction

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The 50 to 40 one you're talking about is the diopters question, correct? That one threw me off too.

For #25 read the entire sentence you underlined part of. "...proportional to the square of the speed and the area..."

It's like KE=1/2mv^2. A 50kg object moving at 1m/s has less KE than a 1kg object moving at 50m/s.

For #1, redox products are affected by pressure.

Edit: Oh, and for #20 you just need to use the stress/strain equation given in the passage.

1) Yeah, holy **** at that rounding.
2) I read that as the square of the speed || and the area. Two separate things. Like, force of centripedal acceleration is proportional to the square of velocity and mass. (Fac prop m || prop v^2)...is this just a reading error or something?
3) So then why was the answer not the one with pressure? Instead it was about doing work...
 

MedPR

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1) Yeah, holy **** at that rounding.
2) I read that as the square of the speed || and the area. Two separate things. Like, force of centripedal acceleration is proportional to the square of velocity and mass. (Fac prop m || prop v^2)...is this just a reading error or something?
3) So then why was the answer not the one with pressure? Instead it was about doing work...

1. Yea it was dumb.
2. It is proportional to the square of the air speed and the area, just like KE is proportional to the square of the velocity and the mass.
3. The answer says that redox is unaffected by pressure. Redox is affected by pressure, therefore the answer is wrong. Also, you should just know that electrochemical cells are driven by electrical work.
 

pfaction

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I just really can't ****ing read [email protected]#3.

vKLMZ.png


I guess if you increased the pressure, the reverse reaction could occur? Could it actually?


I still don't get why I was wrong on the area thing. Why, it says compact :(
 

MedPR

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I just really can't ****ing read [email protected]#3.

vKLMZ.png


I guess if you increased the pressure, the reverse reaction could occur? Could it actually?


I still don't get why I was wrong on the area thing. Why, it says compact :(

Whenever unequal molar amounts of gases are involved, pressure can play a role.

For the area thing, you were correct, but the speed answer choice is the best answer. A squared factor plays a larger role than a non-squared factor. Use the KE example and consider object A. If I increase the mass of object A by a factor of 10, I will increase its KE by a factor of 10. However, if I increase its velocity by a factor of 10, I will increase its KE by a factor of 100.
 

pfaction

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Really, wow thats awesome. I didn't know pressure gradients could affect electrolysis. Maybe just shift it so there's no transfer anymore. Thanks bro.
 

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I mean, I don't know if enough pressure could reverse the reaction completely (I doubt it can), but the important thing for this question is to know that Le chatelier applies everywhere and that electrochemical cells (as opposed to galvanic) are driven by electrical work.
 

pfaction

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I mean, I don't know if enough pressure could reverse the reaction completely (I doubt it can), but the important thing for this question is to know that Le chatelier applies everywhere and that electrochemical cells (as opposed to galvanic) are driven by electrical work.

Yeah, I was going to say that maybe it'll just stop at equilibrium or where they are at.
 
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