Official DAT Destroyer Q&A Thread

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

densaugeo

Full Member
7+ Year Member
Joined
Dec 6, 2014
Messages
45
Reaction score
110
Hi guys, since I'm currently going through the DAT Destroyer and I'm sure many of you guys are as well I felt this thread would be helpful. I usually have various questions while going through Destroyer regarding why a certain answer choice is correct or other times I am in need of a more detailed explanation.

Instead of making multiple threads each time we have questions, I thought we could just post them here and anyone can chime in with an explanation or further clarification.

Please don't post entire questions as that would be a violation of copyright. You can ask specific questions regarding a particular problem in the Destroyers.

Hopefully others find this a useful thread. Thanks!

Members don't see this ad.
 
  • Like
Reactions: 1 users
So on Gen Chem Destroyer, pg 166, #105, i understand that delta H of formation is delta H products - delta H reactants. I understand that we reverse the second reaction and not the first because we need CuO in the product. However, when we reverse the second rxn, why do we not reverse the delta H, making it -10 instead of 10? In that scenario it would be -140 - (-10) which equals -130. But the answer for this problem is -150. Please help me understand why we dont reverse the second delta H! Thanks!
For this one, we are using Hess's Law and we want a formation reaction. The key to a formation reaction is that we only want 1 mol of product and can only use elements in their standard states as reactants. Based on the given information, we only need to reverse equation 2. If you write both out with the reversal you would get
Cu2O + 1/2 O2 -> 2 CuO
CuO + Cu -> Cu2O
Based on that reversal, we can see that the Cu2O would cancel from both equations and we would get 1 mol of CuO (which is what we want!) Additionally, we are only left with elements in their standard states as reactants (O2 and Cu metal). Perfect. Adding those together gives -150. Let me know if you have more questions on this!

Hess's Law is not products - reactants. Hess's Law is simply the sum of delta H's that give you the desired reaction. You use products - reactants when you are given enthalpy values only. If you are given several equations and asked to find some rearrangement of those, use Hess's Law.
 
Last edited:
Dr. Romano,
Can you explain #149 destroyer QR. I'm totally lost

Thanks!

Here's another way to do the problem. it's called a binomial expansion.
See attached picture
 

Attachments

  • sdn.jpg
    sdn.jpg
    204.4 KB · Views: 63
  • Like
Reactions: 1 user
For this one, we are using Hess's Law and we want a formation reaction. The key to a formation reaction is that we only want 1 mol of product and can only use elements in their standard states as reactants. Based on the given information, we only need to reverse equation 2. If you write both out with the reversal you would get
Cu2O + 1/2 O2 -> 2 CuO
CuO + Cu -> Cu2O
Based on that reversal, we can see that the Cu2O would cancel from both equations and we would get 1 mol of CuO (which is what we want!) Additionally, we are only left with elements in their standard states as reactants (O2 and Cu metal). Perfect. Adding those together gives -150. Let me know if you have more questions on this!

Hess's Law is not products - reactants. Hess's Law is simply the sum of delta H's that give you the desired reaction. You use products - reactants when you are given enthalpy values only. If you are given several equations and asked to find some rearrangement of those, use Hess's Law.
So based off of what your saying, we did reverse the Delta H of the second reaction? But because it's hess' law we are adding not subtracting..
Why do we not use Delta H products - Delta H reactants. Is it because we weren't given Delta H's of formation for individual compounds, but for separate rxns?
 
Members don't see this ad :)
So based off of what your saying, we did reverse the Delta H of the second reaction? But because it's hess' law we are adding not subtracting..
Why do we not use Delta H products - Delta H reactants. Is it because we weren't given Delta H's of formation for individual compounds, but for separate rxns?
Yes!
 
  • Like
Reactions: 1 users
Dr Romano, when doing the DAT Destroyer, do you recommend/suggest that we time ourselves? So far I haven't been timing myself, but i thought i would give it a try - 40 Gen Chem Q's in 40 minutes, but i only got through about 2/3 of the questions!
 
Dr Romano, when doing the DAT Destroyer, do you recommend/suggest that we time ourselves? So far I haven't been timing myself, but i thought i would give it a try - 40 Gen Chem Q's in 40 minutes, but i only got through about 2/3 of the questions!
He is going to respond to you by saying: No, the DAT Destroyer book is not meant to be a timed test. It is meant to drill the concepts into your brain. Then he will probably say: I don't understand why some people would ever suggest that these problems should be taken as timed tests.
 
He is going to respond to you by saying: No, the DAT Destroyer book is not meant to be a timed test. It is meant to drill the concepts into your brain. Then he will probably say: I don't understand why some people would ever suggest that these problems should be taken as timed tests.
haha ok, well thanks for letting me know. I just felt like due to the complexity of some questions (for good reasons of drilling) it is very hard to do in timed conditions, some questions take me 3-4 minutes alone!
 
Hello Dr Romano,

For question #244 DAT Destroyer Gen Chem (2017). Why is a buret used to measure that volume as opposed to the buret? A similar question was asked in one of the DAT Bootcamps and the answer was to used a pipet.
This was the question "Which of the following pieces of laboratory equipment would be best for a chemist to use to measure and transfer 8.7 mL of a solution from an Erlenmeyer into a test tube?" its quite similar to Q244, but the correct answer here was a pipet rather than a buret.
 
Hello Dr Romano,

For question #244 DAT Destroyer Gen Chem (2017). Why is a buret used to measure that volume as opposed to the buret? A similar question was asked in one of the DAT Bootcamps and the answer was to used a pipet.
This was the question "Which of the following pieces of laboratory equipment would be best for a chemist to use to measure and transfer 8.7 mL of a solution from an Erlenmeyer into a test tube?" its quite similar to Q244, but the correct answer here was a pipet rather than a buret.

@gkhoda

I actually just did the BC question you are talking about today. And the reason that was a pipet is because a pipet is more portable as oppose to a buret. But to me, one big reason why the answer for the destroyer problem was that it specified NaOH, which is a strong base. I think the two problems have fundamental differences.

Hope that helps!
 
@gkhoda

I actually just did the BC question you are talking about today. And the reason that was a pipet is because a pipet is more portable as oppose to a buret. But to me, one big reason why the answer for the destroyer problem was that it specified NaOH, which is a strong base. I think the two problems have fundamental differences.

Hope that helps!
Yeah, i agree that makes a lot of sense! its just that the solutions in destroyer didnt mention the reason being because of base (though i definitely see your point). It just says that 'since the student wanted to measure exactly 12.3ml, an odd volume, a buret would be best. Pipets allow transfer of accurately known volumes from one container to another. The precision you get from a buret is much greater than a pipet" So maybe in the BC question pipet was correct being the volume was being transferred into a erlenmeyer flask?
 
Yeah, i agree that makes a lot of sense! its just that the solutions in destroyer didnt mention the reason being because of base (though i definitely see your point). It just says that 'since the student wanted to measure exactly 12.3ml, an odd volume, a buret would be best. Pipets allow transfer of accurately known volumes from one container to another. The precision you get from a buret is much greater than a pipet" So maybe in the BC question pipet was correct being the volume was being transferred into a erlenmeyer flask?

Well the thing is, there are some pipets that can be very accurate. You have the plastic ones, but you also have the glass ones, which can also be very accurate (just depending on which you use). I mean, I definitely want to hear what Dr. Romano has to say about it, but thats how I initially approached that problem for destroyer. And yes, totally, that is why pipet was the one for BC, the fact that it was a quick transfer (I don't remember what it was exactly but something about a simple quick transfer of liquid), which is why pipet was correct for the one on BC.
 
Well the thing is, there are some pipets that can be very accurate. You have the plastic ones, but you also have the glass ones, which can also be very accurate (just depending on which you use). I mean, I definitely want to hear what Dr. Romano has to say about it, but thats how I initially approached that problem for destroyer. And yes, totally, that is why pipet was the one for BC, the fact that it was a quick transfer (I don't remember what it was exactly but something about a simple quick transfer of liquid), which is why pipet was correct for the one on BC.
Yeah the gilson pipettes can be used to measure out micrometers!!
 
  • Like
Reactions: 1 user
Members don't see this ad :)
GCHEM #124 and #187 in 2016 destroyer have what seems to be a contradiction. I need some help understanding.

The solution to 124 states "Since C4H10 has more atoms and therefore a greater molecular complexity than the smaller C3H8, the delta S would be higher for C4H10."

This implies that the "more ordered" (less disorder) molecule has a higher value of S. However, the solution to #187 says "Dissolving increases the amount of disorder, + delta S."

So which one is it? Is a positive S value indicating more DISorder, like the solution to 187 states, or does a positive S value indicate more complexity aka more order, as in problem 124?


On second thought, the statement in 124 is just plain wrong. Something with a larger value of S should be more disordered and less complex, since S is a measure of disorder after all. What am I missing here?
 
Last edited:
@mrdeez Allow me take a shot at this.

Delta S signifies entropy. When DeltaS + , signifies more entropy. And DeltaS - , signifies less entropy (it really depends on the value, but you can deduce if there is more or less). The universe always wants to go in the direction of more entropy, or +S.

As what the solution says, C4H10 has more atoms and therefore a greater molecular complexity. This means that the DeltaS is MORE + for C4H10 than C3H8, which means that there is MORE disorder, NOT more order. For 186, if you dissolve... that is go from a solid to a liquid, there is MORE disorder, which is why delta S is MORE +. (I say more + because IDK what the actual values of the DeltaS are).

I think you may be just confusing the definition of delta S. Let me know if that helps.
 
GCHEM #124 and #187 in 2016 destroyer have what seems to be a contradiction. I need some help understanding.

The solution to 124 states "Since C4H10 has more atoms and therefore a greater molecular complexity than the smaller C3H8, the delta S would be higher for C4H10."

This implies that the "more ordered" (less disorder) molecule has a higher value of S. However, the solution to #187 says "Dissolving increases the amount of disorder, + delta S."

So which one is it? Is a positive S value indicating more DISorder, like the solution to 187 states, or does a positive S value indicate more complexity aka more order, as in problem 124?


On second thought, the statement in 124 is just plain wrong. Something with a larger value of S should be more disordered and less complex, since S is a measure of disorder after all. What am I missing here?
For #124 and # 187.....you seem very confused. If you compare 2 solids...say Carbon and Molybdenum, you will see that the Molybdenum has the far greater disorder. It has a greater molecular complexity...which means more protons, neutrons, electrons, and atomic orbitals. The same reasoning would be for gases when we compare them. C4H10 would indeed have a greater complexity than C3H8. Thus it would have a greater entropy value. Check this in any General Chem text. In advanced organic chemistry.....it is not such an easy task analyzing all these orbitals !!!! Dissolving is totally different.....Here we have a PROCESS.....by which order is decreasing. We are getting more disordered !!! For example if KCl is dissolved in water....1 particle becomes 2 particles. I hope you see this now. If not, you need to return to a text book like Chang or Zumdahl for further clarity.

Hope this helps.

Dr. Romano
 
Hello Dr. Romano,

For OCHEM DAT Destroyer, #37, you said that option B and C are electron withdrawing groups, but BC and online resources consider them as electron donating groups. So, i'm a bit confused as to which it is.

Also, is there a way to help me memorize a quick way to identify if a substituent is either donating or a withdrawing group? I tried memorize them, but mix them up sometimes.

Thank you for your help!
 
@chicagodentist123 I will try to take a shot.

Yes I agree that in any other situation, the NH2 on a benzene WOULD BE electron donating. But the differences of electron donating and withdrawing, in the concept of benzene, refer to benzene reactions (nuc or ele). However, in this question, @orgoman22 is referring to BASE strength, and not any specific reaction involving benzene. NH2 is the molecule he is referring to that will act as a base, and so he is testing your knowledge about how the different substituents will affect NH2 base strength. Benzene is electron withdrawing because it can stabilize via resonance, and thus in BCE, there is a benzene that is attached which will pull the electron density FROM NH2, thus ultimately decreasing the base strength of the NH2.

Please let me know if that helps.
 
  • Like
Reactions: 1 user
@chicagodentist123

I completely missed your second question, but it is fairly easy to remember IF you understand the concept of withdrawing and donating. I would HIGHLY recommend looking and studying at the structures of each.

In donating, there is usually a lone pair that can directly contribute to the resonance (N or O, etc). Others are alkyl groups that don't retain electrons that easily (like CH3). Now you may ask why C=O (which is withdrawing) when connected to benzene is withdrawing... and this is because of the HIGHLY electron withdrawign effect of O (electroneg is high).

A benzene-C=O to a benzene is DIFFERENT from a benzene-O-C because the electrons ON THE O OR N can DIRECTLY contribute to the resonance of benzene.

Another thing that you should notice is NH3 or NR3 VS something like NH2..... NH2 is don. and NH3 AND NR3 is with. WHY??? Well this is because the if you looked at the lewis/structure of the NH3 or NR3.... there is a + on the N and no + on the NH2.....The + will want to be stabilized by electron which is RIGHT NEXT to benzene and will basically take some of the density from benzene.

I hope this kind of helps...
 
  • Like
Reactions: 1 user
Hello Dr. Romano,

For OCHEM DAT Destroyer, #37, you said that option B and C are electron withdrawing groups, but BC and online resources consider them as electron donating groups. So, i'm a bit confused as to which it is.

Also, is there a way to help me memorize a quick way to identify if a substituent is either donating or a withdrawing group? I tried memorize them, but mix them up sometimes.

Thank you for your help!
No....No.....No.....You are missing my point. When I use the word electron withdrawing.......I am referring to the BENZENE ring. It is withdrawing electron density AWAY FROM THE NITROGEN........thus, the amine is less able to share it's electrons with a proton. Aromatic amines are not very basic, because, the aromatic portion removes electron density off the Nitrogen atom. This renders the amine less basic by magnitudes !! The bottom line is this....Amines are BASES ,, especially if they have sp3 Nitrogen atoms. If you see an sp2 as in choices, B, C, D or E......be cautious. An sp2 N most certainly will have a group withdrawing electron density. Do not confuse up the terminology. Once again.....when I use the term " electron withdrawing ".....I am referring to the influence the benzene ring has on the electron density on the Nitrogen atom.

Also.....I think you need to memorize the ortho - para directors.......and the meta directors.. Groups such as N02, CN, COOH, SO3H, and Carbonyl groups are the most popular meta directors.......OH, NH2, R groups, OR groups, and halogens are the most popular ortho para directors. Halogens direct o/p but slightly deactivate the ring.

Hope this helps.

Dr. Jim Romano
 
  • Like
Reactions: 1 user
@chicagodentist123 I will try to take a shot.

Yes I agree that in any other situation, the NH2 on a benzene WOULD BE electron donating. But the differences of electron donating and withdrawing, in the concept of benzene, refer to benzene reactions (nuc or ele). However, in this question, @orgoman22 is referring to BASE strength, and not any specific reaction involving benzene. NH2 is the molecule he is referring to that will act as a base, and so he is testing your knowledge about how the different substituents will affect NH2 base strength. Benzene is electron withdrawing because it can stabilize via resonance, and thus in BCE, there is a benzene that is attached which will pull the electron density FROM NH2, thus ultimately decreasing the base strength of the NH2.

Please let me know if that helps.
No....No.....No.....You are missing my point. When I use the word electron withdrawing.......I am referring to the BENZENE ring. It is withdrawing electron density AWAY FROM THE NITROGEN........thus, the amine is less able to share it's electrons with a proton. Aromatic amines are not very basic, because, the aromatic portion removes electron density off the Nitrogen atom. This renders the amine less basic by magnitudes !! The bottom line is this....Amines are BASES ,, especially if they have sp3 Nitrogen atoms. If you see an sp2 as in choices, B, C, D or E......be cautious. An sp2 N most certainly will have a group withdrawing electron density. Do not confuse up the terminology. Once again.....when I use the term " electron withdrawing ".....I am referring to the influence the benzene ring has on the electron density on the Nitrogen atom.

Also.....I think you need to memorize the ortho - para directors.......and the meta directors.. Groups such as N02, CN, COOH, SO3H, and Carbonyl groups are the most popular meta directors.......OH, NH2, R groups, OR groups, and halogens are the most popular ortho para directors. Halogens direct o/p but slightly deactivate the ring.

Hope this helps.

Dr. Jim Romano
AHH, that clears it up. Yeah, i do have most of them down. Just thought there was a 'trick' that could help me remember them, and @dl250 helped out with it.

Thank you very much Dr. Romano and @dl250! I really appreciate it your help!
 
  • Like
Reactions: 1 user
DAT Math destroyer 2017 Test 9 Question #37..

So on the explanation, I am not understanding which rule we are using to convert ((sinx-cosx)/(cosx)) = tanx - 1

Can anyone please explain how to get tanx - 1 from there?
 
DAT Math destroyer 2017 Test 9 Question #37..

So on the explanation, I am not understanding which rule we are using to convert ((sinx-cosx)/(cosx)) = tanx - 1

Can anyone please explain how to get tanx - 1 from there?
Split the fraction;
Sin(x)/cos(x) - sin(x)/sin(x) = tan(x) - 1

Hope this helps!
 
  • Like
Reactions: 1 users
Hello Dr Romano,

Q289 DAT Destroyer, Organic Chemistry, (2017). Although the question isnt asking this, a Nitro group is electron withdrawing/benzene deactivating meta director, so why have the chlorines been placed in the para position when comparing the different resonance structures?
 
  • Like
Reactions: 1 user
Hello Dr Romano,

Q289 DAT Destroyer, Organic Chemistry, (2017). Although the question isnt asking this, a Nitro group is electron withdrawing/benzene deactivating meta director, so why have the chlorines been placed in the para position when comparing the different resonance structures?
Contrary to what is presented in undergraduate books, a meta isomer is NOT produced in exclusivity.... but rather ..often in yields of 85%. The other 15% is split between the minor ortho and para products, In this problem,,,,,you needed to understand why the para pathway would yield a VERY LOW para isomer.....possibly like 2 %......

Hope this helps.

Dr. Romano
 
  • Like
Reactions: 1 users
Hello Dr Romano,

Q289 DAT Destroyer, Organic Chemistry, (2017). Although the question isnt asking this, a Nitro group is electron withdrawing/benzene deactivating meta director, so why have the chlorines been placed in the para position when comparing the different resonance structures?
Came here to ask the same question lol!
 
  • Like
Reactions: 1 users
For #124 and # 187.....you seem very confused. If you compare 2 solids...say Carbon and Molybdenum, you will see that the Molybdenum has the far greater disorder. It has a greater molecular complexity...which means more protons, neutrons, electrons, and atomic orbitals. The same reasoning would be for gases when we compare them. C4H10 would indeed have a greater complexity than C3H8. Thus it would have a greater entropy value. Check this in any General Chem text. In advanced organic chemistry.....it is not such an easy task analyzing all these orbitals !!!! Dissolving is totally different.....Here we have a PROCESS.....by which order is decreasing. We are getting more disordered !!! For example if KCl is dissolved in water....1 particle becomes 2 particles. I hope you see this now. If not, you need to return to a text book like Chang or Zumdahl for further clarity.

Hope this helps.

Dr. Romano

C4H10 is obviously more complex than C3H8, that is not the issue. For something to be more complex, the arrangement is less disordered. This would mean the value of delta S is MORE NEGATIVE. In your book, you used the word HIGHER. Since in your book, you said "HIGHER ENTROPY", this indicates a more POSITIVE value of S.

If by HIGHER you mean MORE NEGATIVE, then you are absolutely right. I just don't know what variant of the english language someone would have to use to come to that conclusion. Thanks for your help. s/
 
Dear Dr. Romano,

For question 78 in the 2017 Orgo Destroyer, I was wondering what happens when you put the s.m. in acid instead of in base?
 
Hello, I am using the 2017 version.
For Gen Chem #60, the solution says that “Water will form a concave meniscus because the forces between water and glass are stronger than between water molecules.” I thought that INTRA molecular forces were always stronger than INTER molecular forces, so why are the forces between glass and water greater than between water molecules?
Thanks!
 
C4H10 is obviously more complex than C3H8, that is not the issue. For something to be more complex, the arrangement is less disordered. This would mean the value of delta S is MORE NEGATIVE. In your book, you used the word HIGHER. Since in your book, you said "HIGHER ENTROPY", this indicates a more POSITIVE value of S.

If by HIGHER you mean MORE NEGATIVE, then you are absolutely right. I just don't know what variant of the english language someone would have to use to come to that conclusion. Thanks for your help. s/
In chemistry parlance, higher entropy means more disorder. From the literature values,,,,,carbon graphite has Entropy value of about 6 J/moleK...while Mo is 182.....not even close or debatable...With increasing molecular complexity we see MORE disorder, just the OPPOSITE of what you stated.

Hope this helps.

Dr. Romano
 
Hello, I am using the 2017 version.
For Gen Chem #60, the solution says that “Water will form a concave meniscus because the forces between water and glass are stronger than between water molecules.” I thought that INTRA molecular forces were always stronger than INTER molecular forces, so why are the forces between glass and water greater than between water molecules?
Thanks!
First of all, attraction between water molecules is an INTERmolecular attraction, not intramolecular. Intramolecular forces are within the same molecules, not between molecules.
This question is about forces of adhesion (between 2 different molecules) being greater than forces of cohesion ( between the same 2 molecules). This is what contributes to the concave meniscus.
The forces of ADHESION between the Silica in glass DOMINATE over the COHESION forces between the water molecules. If Mercury or a Hydrocarbon was used, we see a convex meniscus. The forces of COHESION dominate over the forces of ADHESION.

Hope this helps.
 
Dr. Romano,
#316 GC (2017)... Can you explain the difference between a transition state and an activated complex? I've tried googling it but nothing I have found has helped with my confusion.
Thanks!
 
Dr. Romano,
#316 GC (2017)... Can you explain the difference between a transition state and an activated complex? I've tried googling it but nothing I have found has helped with my confusion.
Thanks!
Many teachers interchange transition state and activated complex, but should not. Let me simplify this for you.A transition state represents an energy maximum.....it is the point of highest energy. The ACTIVATED COMPLEX is the specie present WITHIN this high energy transition state. In other words, it is the arrangement of atoms within the transition state itself.

Hope this helps.

Dr. Jim Romano
 
  • Like
Reactions: 1 users
GC # 244: A student wanted to transfer exactly 12.3 mL of a 0.20M NaOH solution. Which should she use? Buret
OC #226: An organic chemist wanted to deliver 15.6 mL of a liquid in an experiment. Which should she employ? Pipet
wtf???? what's the difference?? lol help
 
OC #243: The answer is C. I thought E would work as well because there is only one E2 product that can be formed. Is the reason we use tert-butoxide on this primary halide instead of EtOK because EtOK would produce an Sn2 as the major product? If this is the case, both reagents would work, however C would be considered the best one?
 
OC #243: The answer is C. I thought E would work as well because there is only one E2 product that can be formed. Is the reason we use tert-butoxide on this primary halide instead of EtOK because EtOK would produce an Sn2 as the major product? If this is the case, both reagents would work, however C would be considered the best one?
Correct. Since EtO is both a strong nuc and a strong base, and the substrate is primary, the major product would be Sn2. In a case like this, always think about the major product. I would think the amount of elimination product that you would get from using option E would be VERY minimal.
 
  • Like
Reactions: 1 user
OC #243: The answer is C. I thought E would work as well because there is only one E2 product that can be formed. Is the reason we use tert-butoxide on this primary halide instead of EtOK because EtOK would produce an Sn2 as the major product? If this is the case, both reagents would work, however C would be considered the best one?
This is a MAJOR concept. A primary halide will do SN2 if EtOK or EtONa, or any strong base/strong nucleophile is employed. It is not so easy to get any elimination product on a primary halide UNLESS a large base such as t-butoxide is employed. For this particular reaction in #243, the yield would be over 93 % in favor of the E2 product. !!!!! Any textbook.....will show you more examples.

Hope this helps.

Dr. Romano
 
  • Like
Reactions: 1 users
In chemistry parlance, higher entropy means more disorder. From the literature values,,,,,carbon graphite has Entropy value of about 6 J/moleK...while Mo is 182.....not even close or debatable...With increasing molecular complexity we see MORE disorder, just the OPPOSITE of what you stated.

Hope this helps.

Dr. Romano

I just found it in an old thread. The larger the molecule gets, the more microstates it can exist in. In this case, more rotation around single bonds, increasing the amount of disorder. This problem is just outright nasty, I learned a lot. Thank you, sorry for getting frustrated.
 
  • Like
Reactions: 1 user
I just found it in an old thread. The larger the molecule gets, the more microstates it can exist in. In this case, more rotation around single bonds, increasing the amount of disorder. This problem is just outright nasty, I learned a lot. Thank you, sorry for getting frustrated.
Keep problems coming always happy to help.

Dr. Romano
 
DAT Math Destroyer Test #11, Question 12.

So in this problem it gives us a circle inscribed equilateral triangle with side length 12 (hypotenuse). So when the line bisects the 60 degree angle at the top of the equilateral triangle, it gives us two angles of 30 and the bases opposite the angle of 6. So to find the radius, I solved for the missing side (height) and got 6/root 3 (diameter) and divided this by 2 to get root 3/3. I am not understanding the explanation of this problem as it says the answer is 2/root 3.. Please explain
 
DAT Math Destroyer Test #11, Question 12.

So in this problem it gives us a circle inscribed equilateral triangle with side length 12 (hypotenuse). So when the line bisects the 60 degree angle at the top of the equilateral triangle, it gives us two angles of 30 and the bases opposite the angle of 6. So to find the radius, I solved for the missing side (height) and got 6/root 3 (diameter) and divided this by 2 to get root 3/3. I am not understanding the explanation of this problem as it says the answer is 2/root 3.. Please explain
It's an equilateral triangle, which means all the sides are equal. There's no hypotenuse. We end up with a 30° 60° 90° triangle. The opposite side to 30° is r and the adjacent side to 30° is 6.
The adjacent side to 30° is equal to the opposite side times sqrt(3)
See the attached picture

sdn.jpg
 
  • Like
Reactions: 1 user
Dr. Romano,
For Bio #533, since there is a generation skip, why wouldn't the mode of inheritance be autosomal recessive? Also, why are we assuming that the mother is a carrier of the disease?

Thank you!!
 
Dr. Romano,
For Bio #533, since there is a generation skip, why wouldn't the mode of inheritance be autosomal recessive? Also, why are we assuming that the mother is a carrier of the disease?

Thank you!!
We assume that the mother is a carrier because she is NOT affected; however, 2 of her 4 sons are. This leads me to think sex-linked recessive. Recessive because she is not affected, but SOME sons are. So, she gave the recessive allele to 2 of her sons. The other 2 got the dominant allele. I hope that helps.
 
  • Like
Reactions: 1 user
Hello Dr Romano,

For Q.330 DAT Destroyer 2017, Gen Chem... it asks for the pH of very dilute solution of acetic acid...I was under the impression that when an aqueous acid is very dilute, that the pH of the solution is ~7 (of water) because its the concentration is so low, and thus only the dissociation of water takes place. However, the correct answer for this question is that the pH would be ~5. Obviously I would assume this initially, but i thought the wording of 'very dilute' was implying something else.
 
Hello Dr Romano,

For Q.330 DAT Destroyer 2017, Gen Chem... it asks for the pH of very dilute solution of acetic acid...I was under the impression that when an aqueous acid is very dilute, that the pH of the solution is ~7 (of water) because its the concentration is so low, and thus only the dissociation of water takes place. However, the correct answer for this question is that the pH would be ~5. Obviously I would assume this initially, but i thought the wording of 'very dilute' was implying something else.
The key to this question was "Vinegar taste". If it has Vinegar taste, it means that you have more protons than comes from water. Therefore weak acid would have a pH in the range of 3-6. Hope it helps.
 
  • Like
Reactions: 1 user
Hello,

For these questions from 2015 DAT Destroyer General Chemistry:
27. What would be the approximate pH of a 1*10^-12M HCL solution. Why do you need to take into account the amount of H+ ions from water? Is this done for every solution that is diluted?

30. Which of the following ions has the largest radius. S-, F-, Ne, K+, Ca++. Is this not talking about atomic radius trend on the periodic table?
 
Hello,

For these questions from 2015 DAT Destroyer General Chemistry:
27. What would be the approximate pH of a 1*10^-12M HCL solution. Why do you need to take into account the amount of H+ ions from water? Is this done for every solution that is diluted?

30. Which of the following ions has the largest radius. S-, F-, Ne, K+, Ca++. Is this not talking about atomic radius trend on the periodic table?
For 27. Consider the Kw @25 degrees Celsius =[H+][OH-] both of which are 1*10-7. Notice that this value (for the H+ concentration) is FAR GREATER than the given [h+] of the HCl solution. Thus, this is the value that you would use to find the pH. It is BECAUSE the concentration of the solution is too dilute.
For 30 - Not quite. Here, we are looking at an isoelectronic species (with the exception of F-). All this means, is that the species with the greater negative charge has the larger radius.
 
  • Like
Reactions: 1 user
Math destroyer 17 #5 wouldn't the condition be minimum $212, maximum $428 because Carlos got at least both $10 and $22

So either 19(10)+22 or 19(22)+10
 
Hello Dr Romano,

for Q2 on the Math Destroyer (2017) test 5... If Ted is mowing the lawn in 2 hours and thats equaly to 2/5 of the lawn, then eliot mows 3/5 of the lawn in 4hr 48mins, why arent we adding the two times together (i.e 2 hours + 2hr 48 mins) im really struggling to understand this even after reading the solutions!
 
Top