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ADA OAT practice test questions

HoHOhoh

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Can anyone explain these questions from ADA OAT practice test? My test is in two days!! so if anyone can help explaining these questions that would be great!
http://www.ada.org/~/media/oat/files/oat_sample_test.ashx

Gen Chem Questions
44. If the mass percent of oxygen in a nitrogenoxygen compound is known, given the mass percent of oxygen, all of the following are needed to determine the molecular formula of a nitrogen-oxygen compound EXCEPT one. Which one is this EXCEPTION?
A. Atomic mass of nitrogen B. Atomic mass of oxygen C. Avogadro's number D. Empirical formula E. Molar mass of the compound
The answer is D. (This might be an error and the answer is C..

61. A 0.60 M solution is made by dissolving solid compound X in water. After ten seconds, the concentration of X is 0.40 M. All of the following could account for these results EXCEPT one. Which one is this EXCEPTION? A. Precipitation B. Neutralization C. Evaporation D. Decomposition E. Disproportionation
The answer is A. (The answer might be C...?)

69. Carbon tetrachloride, CCl4 , is observed to have a higher boiling point than chloroform, CHCl3 . All of the following statements are true base on this observation EXCEPT one. Which is this EXCEPTION?
A. Dispersion forces are experienced by CCl4 and CHCl3 .
B. At a given temperature, the vapor pressure of CCl4 is less than that of CHCl3 .
C. The molar heat of vaporization of CCl4 is greater than that of CHCl3 .
D. Dipole-dipole interactions are responsible for the larger boiling point of CCl4 .
The answer is D. (Why not C?)

OChem

76. Nitration of toluene (Ph–CH3 ) with HNO3 /H2 SO4 occurs
A. faster than nitration of benzene and produces mostly ortho and para products.
B. slower than nitration of benzene and produces mostly meta product.
C. faster than nitration of benzene and produces mostly meta product.
D. slower than nitration of benzene and produces mostly ortho and para products.
E. at the same rate as nitration of benzene and produces mostly meta product.
The answer is D (I don't understand why it is faster. How do you know if it is faster or slower?)

84. Which of the following compounds best fits the 1H NMR spectral data listed below?
(Sorry I cannot post the image.)
The answer is B. (This is very confusing to me. please help!)

88. Which of the following statements is true regarding a pair of compounds that are diastereomers of each other?
A. They are configurational isomers.
B. They have the same physical properties.
C. They are mirror images.
D. They are always optically active.
E. They have equal but opposite alpha values.
The answer is A. (why??)

93. What is the hybridization of a nitrogen atom if it forms two sigma two pi bonds?
A. sp B. sp2 C. sp3 D. sp3d 2
The answer is C. (or A?)

Physics
#26 The water in a swimming pool is 3.0 m deep. During the day, the atmospheric pressure increases by 2.0 x 103 N/m2 . During this same period, the pressure at the bottom of the pool, in N/m2 , will
A. not change. B. increase by 1.0 x 10^7 . C. increase by 6.0 x 10^3 . D. increase by 6.0 x 10^4 . E. increase by 2.0 x 10^3 . The answer is E.

#39. What is the focal length in cm of a lens that produces an image 30 cm behind it when the object is placed 6 cm in front of it? A. 7.5 B. 36 C. 5.0 D. 24 E. 18.0
The answer is C. (I guess an image behind lens would be positive to get this answer. Can anyone explain when you use positive and negative? )

Thanks!!
 
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sweetsweetcan

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I just finished reviewing physics so I can answer 39 for you. Just use the simple lens formula, you'll get 5 cm as focal length, it's plug and chug and not a trick question. Since it says the image forms behind the lens, it means it's a converging lens, so focal length is positive. Image forms in front of lens if it's a diverging lens, which makes the focal length negative.
 
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HoHOhoh

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Did you get help on the questions yet?
No, not yet. Still need help on those other questions!! Please help if you know any of them.
Thanks for the help, Sweetsweetcan!!
So image behind the lens means, the image formed where your eyeball is, correct?
 

knioette

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Alrighty, I'll do my best!

44: Not sure, I agree with you though.

61: No clue.

69: The question is asking you to pinpoint the WRONG reason that CCl4 has a higher boiling point than CHCl3. Having a higher molar heat of vaporization is the same thing as having a higher boiling point, but the question is asking for the WRONG answer, not the correct one. A, B and C can explain why the statement in the stem is correct, but D is wrong because CCl4 is a tetrahedral where all the 'edges' are the same atoms (Cl), therefore the dipole-dipole balances out and you actually end up with no net dipole moment (pretty cool, huh?). CHCl3 on the other hand, the net dipole is 'down' (if you put the H on top and all the Cl's down). See this video for an explanation why the CCl4 tetrahedral has no net dipole: https://www.khanacademy.org/science...ew/electronegativity-polarity/v/dipole-moment

76: Ortho/para directing molecules are deactivating, as in, they discourage a reaction from happening (therefore make it slower), whereas meta directing molecules are activating, as in, encourage a reaction to happen (as in make it faster). The exception is halogens, they are ortho/para directors that are activators. Just memorize it, and memorize as many molecules as you can of the ortho/para versus meta directors. If you have time to see why (it's kinda cool! :D), watch this video:

84: If you don't understand this question, you have got to watch O2012 5.6 NMR Spectroscopy by Chad: http://www.coursesaver.com/videos/category/o5-lab-techniques-and-spectroscopy-133/ you will need to pay for it, but it's worth it. Might as well watch 5.5 IR Spec while you're at it.

88: Alright, so we know that diastereomers have some, not all, of the chiral centres switched. They certainly don't always have the same physical properties. Because not all the chiral centres are switched, therefore they can't be mirror images (for example, if I switch around my pinky finger on the right hand to face the opposite way, but the other fingers on that hand are facing the same way, that hand is no longer a mirror image of my left hand). They're not always optically active, meso compounds are diastereomers that are optically inactive. They can't have equal and opposite alpha values (the optical rotation), because they need to be non-superimposable mirror images (again, frankestein pinky).
Now look at this chart here: http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-1.html#constitutional that will tell you what's up with the whole constitutional/conformational/configurational crap.

93: I thought it would be A too, but no idea why they chose C.

Physics:
26: Assume you have a U-shaped tube with water in it. The pressure you apply on one side will be equal to the pressure the water exerts on the air on the other side. That's the case with this question, doesn't matter if the other end is thicker or tighter, or going down a straight 'tube' (aka swimming pool) versus a curved one. Pressure will have to be the same (where else would it go or come from?!). Make sure you pay attention to pressure versus force, P1=F1/Area1=P2=F2/Area2. Force changes if the area changes, whereas pressure is simply the force per metres (or whatever) squared.

Let me know if you want me to clarify anything! (and if you figure out the questions I don't know, for the love of God tell me about them or I'll go crazy. Not urgent though, I did my OAT like 2 weeks ago anyways).

Good luck!
 
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HoHOhoh

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Alrighty, I'll do my best!

44: Not sure, I agree with you though.

61: No clue.

69: The question is asking you to pinpoint the WRONG reason that CCl4 has a higher boiling point than CHCl3. Having a higher molar heat of vaporization is the same thing as having a higher boiling point, but the question is asking for the WRONG answer, not the correct one. A, B and C can explain why the statement in the stem is correct, but D is wrong because CCl4 is a tetrahedral where all the 'edges' are the same atoms (Cl), therefore the dipole-dipole balances out and you actually end up with no net dipole moment (pretty cool, huh?). CHCl3 on the other hand, the net dipole is 'down' (if you put the H on top and all the Cl's down). See this video for an explanation why the CCl4 tetrahedral has no net dipole: https://www.khanacademy.org/science...ew/electronegativity-polarity/v/dipole-moment

76: Ortho/para directing molecules are deactivating, as in, they discourage a reaction from happening (therefore make it slower), whereas meta directing molecules are activating, as in, encourage a reaction to happen (as in make it faster). The exception is halogens, they are ortho/para directors that are activators. Just memorize it, and memorize as many molecules as you can of the ortho/para versus meta directors. If you have time to see why (it's kinda cool! :D), watch this video:

84: If you don't understand this question, you have got to watch O2012 5.6 NMR Spectroscopy by Chad: http://www.coursesaver.com/videos/category/o5-lab-techniques-and-spectroscopy-133/ you will need to pay for it, but it's worth it. Might as well watch 5.5 IR Spec while you're at it.

88: Alright, so we know that diastereomers have some, not all, of the chiral centres switched. They certainly don't always have the same physical properties. Because not all the chiral centres are switched, therefore they can't be mirror images (for example, if I switch around my pinky finger on the right hand to face the opposite way, but the other fingers on that hand are facing the same way, that hand is no longer a mirror image of my left hand). They're not always optically active, meso compounds are diastereomers that are optically inactive. They can't have equal and opposite alpha values (the optical rotation), because they need to be non-superimposable mirror images (again, frankestein pinky).
Now look at this chart here: http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-1.html#constitutional that will tell you what's up with the whole constitutional/conformational/configurational crap.

93: I thought it would be A too, but no idea why they chose C.

Physics:
26: Assume you have a U-shaped tube with water in it. The pressure you apply on one side will be equal to the pressure the water exerts on the air on the other side. That's the case with this question, doesn't matter if the other end is thicker or tighter, or going down a straight 'tube' (aka swimming pool) versus a curved one. Pressure will have to be the same (where else would it go or come from?!). Make sure you pay attention to pressure versus force, P1=F1/Area1=P2=F2/Area2. Force changes if the area changes, whereas pressure is simply the force per metres (or whatever) squared.

Let me know if you want me to clarify anything! (and if you figure out the questions I don't know, for the love of God tell me about them or I'll go crazy. Not urgent though, I did my OAT like 2 weeks ago anyways).

Good luck!


Thanks a lot for your help! Really appreciate it!!!
 
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fittycent

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I realize this is an old post, but I came across this when I googled something like "61 opted sdn" (I couldn't figure out what the hell was going on with #61).

And... I'm pretty confused about one of these if I'm wrong. I'm pretty sure that #76 is A.

First, nucleophilicity is a measure of rate/kinetics. So, stronger nucleophiles react faster.

In EAS reactions, benzene is the nucleophile. but, benzene isn't an amazing nucleophile on its own; aromaticity makes 'em very stable and therefore relatively unreactive. So, in EAS reactions, the idea is that donating electron density to benzene ring will make benzene a stronger nucleophile (i.e., will "activate" the benzene ring as a nucleophile). And, withdrawing electron density from benzene ring will make benzene a weaker nucleophile (i.e., will "deactivate" the benzene ring as a nucleophile)

-Nucleophiles are electron donors, so they're better nucleophiles when they have more electrons
-C groups for benzene reactions are "activating". To be activating, you make the benzene, the nucleophile, more nucleophilic by donating electrons (increase electron density of benzene)
-Activating groups are ortho/para directors

-"Deactivating" groups (WITHDRAW electron density - DECREASE nucleophile strength) are meta directors
*Except, in the case of Halides (-X) [I don't remember the reason, sorry)
*Halides are DEACTIVATING, but they direct ortho/para
--> As far as the OAT is concerned, it's good to just know this as an exception to the activators->ortho/para directing and deactivators->meta directors

If I'm wrong, someone please correct me. I wouldn't be surprised if I'm wrong; organic goes over my head more than any other subject I've ever studied.
 
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