Official DAT Destroyer Q&A Thread

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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!

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Can anyone please help me understand the difference between Gene Flow and Founder effect? I feel like they're exactly the same. Thank you!
 
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Hi Dr. Romano, I came across a question in Odyssey that said that internal alkynes don't react with Ag(NH3)2+ and the lab tests page in Destroyer says that terminal alkynes will react with this: Ag(NH3)OH. Is this also called Tollen's test? As in, would it be correct to say that Tollen's test tests positive not only for aldehydes and alpha-hydroxy ketones, but also for terminal alkynes?
 
Hi Dr. Romano, I came across a question in Odyssey that said that internal alkynes don't react with Ag(NH3)2+ and the lab tests page in Destroyer says that terminal alkynes will react with this: Ag(NH3)OH. Is this also called Tollen's test? As in, would it be correct to say that Tollen's test tests positive not only for aldehydes and alpha-hydroxy ketones, but also for terminal alkynes?
You are correct that ammoniacal silver nitrate is Tollen's reagent. I am glad to see that you are recognizing the reagents that we use in Organic Chemistry. The Tollen's reagent is mostly used for its reaction with aldehydes to produce a silver mirror, but it also reacts with several other types of compounds, such as alpha-hydroxy ketones and terminal alkynes. I also want you to focus on reactions of alkynes such as hydration , ozonolysis, and reduction reactions using reagents such as Lindlar's catalyst and Na in liquid Ammonia.

Hope this helps.

Dr. Romano
 
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Hey guys!

I am currently doing the Destroyer 2016 and I am having trouble with General Chemistry problem 247.

247.) If 30ml of .02 M Hbr solution is mized with 40 ml of .03 M KOH, what is the pH of the mixture?

I get that you need to calculate the moles for each one by using the molarity formula. I read the solution and what I don't get is how did the calculate 6x10^-4 = moles base.

How exactly did you get this number? It says 1.2x10^-3 - 6x10^-4 = 6x10^-4 moles base. How did you know to use that? How did you we know we are determining a base and not looking for an acid?

Please let me know as soon as you can.
 
Hey guys!

I am currently doing the Destroyer 2016 and I am having trouble with General Chemistry problem 247.

247.) If 30ml of .02 M Hbr solution is mized with 40 ml of .03 M KOH, what is the pH of the mixture?

I get that you need to calculate the moles for each one by using the molarity formula. I read the solution and what I don't get is how did the calculate 6x10^-4 = moles base.

How exactly did you get this number? It says 1.2x10^-3 - 6x10^-4 = 6x10^-4 moles base. How did you know to use that? How did you we know we are determining a base and not looking for an acid?

Please let me know as soon as you can.

Once you get the moles of Acid and moles of base, you need to see which js in excess. In order to titrate 6*10^-4 moles of HBr, you need 6*10^-4 moles of base. We have more base here, so, we see how much base os left over, which is 6*10^-4 moles once you subtract it from 1.2*10^-3.
That is what will determine our pH.

Hope this helps.

Dr. Romano
 
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Question 8B, pg. 314 in the general chemistry destroyer 2016; we are asked to find the molality. Dr. Romano reminds us that density is a good way to find the kg of the solvent; however this only works when we can assume 1 liter solution, and he does not state this explicitly for one of the values we start with.

Dr Romano says there is 1280 g solution, based off the density provided (doesn't mention that this is per liter, but it is). We are also told that the solution contains 200g of Mn(NO3)2, which he subtracts from the total to find the solvent mass. We are never told exactly how many liters of solution we have started with, so it's impossible to know if the 200g value provided is the mass in one liter of solution or in many liters of solution (he seems to assume 1L). I believe one can't assume this, however, and thus we cannot simply subtract it from the 1280g/liter value.

Is this question wrong? I think dr Rimano forgot to mention that it was a 1 liter solution.
 
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Question 8B, pg. 314 in the general chemistry destroyer 2016; we are asked to find the molality. Dr. Romano reminds us that density is a good way to find the kg of the solvent; however this only works when we can assume 1 liter solution, and he does not state this explicitly for one of the values we start with.

Dr Romano says there is 1280 g solution, based off the density provided (doesn't mention that this is per liter, but it is). We are also told that the solution contains 200g of Mn(NO3)2, which he subtracts from the total to find the solvent mass. We are never told exactly how many liters of solution we have started with, so it's impossible to know if the 200g value provided is the mass in one liter of solution or in many liters of solution (he seems to assume 1L). I believe one can't assume this, however, and thus we cannot simply subtract it from the 1280g/liter value.

Is this question wrong? I think dr Rimano forgot to mention that it was a 1 liter solution.


You have missed the key point. If you have the density of a solution, the mass of that solution is EASILY obtained by considering 1 liter of solution......this automatically gives you the grams of solution. This is a commonly asked question on many many exams, and many students get it wrong. Once the grams are known you are set to go for the kill. If you need any further clarification any General Chemistry teacher can help you.

Hope this helps.

Dr. Romano
 
You have missed the key point. If you have the density of a solution, the mass of that solution is EASILY obtained by considering 1 liter of solution......this automatically gives you the grams of solution. This is a commonly asked question on many many exams, and many students get it wrong. Once the grams are known you are set to go for the kill. If you need any further clarification any General Chemistry teacher can help you.

Hope this helps.

Dr. Romano

Thank you for the very prompt reply. Certainly what you say is true, and it makes sense both in this question (pg311 #8B) and earlier questions. I do not dispute the conversion of the density to mass (1.28g/cm3 --> 1280g); however, the question also states that a solution (with density of 1.28g/cm3) contains 200g of Mn(NO3)2. My problem is with this 200g value that was provided; We do not know how much solution we began with, and so we can't determine the density (and ultimately the mass) of the solute (is it 200g/L? 200g/20L?). As such, it seems presumptuous to subtract 200g from the mass we determined via density (1280g/L).
 
Thank you for the very prompt reply. Certainly what you say is true, and it makes sense both in this question (pg311 #8B) and earlier questions. I do not dispute the conversion of the density to mass (1.28g/cm3 --> 1280g); however, the question also states that a solution (with density of 1.28g/cm3) contains 200g of Mn(NO3)2. My problem is with this 200g value that was provided; We do not know how much solution we began with, and so we can't determine the density (and ultimately the mass) of the solute (is it 200g/L? 200g/20L?). As such, it seems presumptuous to subtract 200g from the mass we determined via density (1280g/L).
In the parlance of chemistry, it is always assumed a 1 liter solution unless otherwise noted. I have found a problem for you from an analytical chemistry problem set. Note how the final portion of the calculation is carried out. I hope this helps.........

Problem Example 9 A solution prepared by dissolving 66.0 g of urea (NH2)2CO in 950 g of water had a density of 1.018 g mL–1. Express the concentration of urea in a) weight-percent; b) mole fraction; c) molarity; d) molality

Solution: a) The weight-percent of solute is (100%) –1 (66.0 g) / (950 g) = 6.9% The molar mass of urea is 60, so the number of moles is (66 g) /(60 g mol–1) = 1.1 mol. The number of moles of H2O is (950 g) / (18 g mol–1) = 52.8 mol. b) Mole fraction of urea: (1.1 mol) / (1.1 + 52.8 mol) = 0.020 c) molarity of urea: the volume of 1 L of solution is (66 + 950)g / (1018 g L–1) = 998 mL. The number of moles of urea (from a) is 1.1 mol. Its molarity is then (1.1 mol) / (0.998 L) = 1.1 mol L–1. d) The molality of urea is (1.1 mol) / (.066 + .950) kg = 1.08 mol kg–1.
 
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Dat destroyer 2016, o Chem number 71.

I thought nitrogen is sp3 hybridized because it's attached to 3 atoms plus the lone pair. Shouldn't that be an sp3 hybridized? Why is the answer sp2 instead of sp3?


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Dat destroyer 2016, o Chem number 71.

I thought nitrogen is sp3 hybridized because it's attached to 3 atoms plus the lone pair. Shouldn't that be an sp3 hybridized? Why is the answer sp2 instead of sp3?


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Indeed NOT. You have missed a very critical point. This nitrogen is involved in a resonance system. Draw a few resonance forms and you will see that a double bond can be drawn to the N atom. Since a double bond is associated with N.....it is sp2 !

I hope this helps.

Dr. Romano

Resonance-imidazole.png
 
For Bio question #398 it says that lysosomes are found in both plant and animals cells. All the resources I've studied so far state that only animal cells have lysosomes and that the central vacuole in plants fulfills some of the lysosomes duties in a plant cell. Could you clarify? Thanks!
 
Indeed NOT. You have missed a very critical point. This nitrogen is involved in a resonance system. Draw a few resonance forms and you will see that a double bond can be drawn to the N atom. Since a double bond is associated with N.....it is sp2 !

I hope this helps.

Dr. Romano

View attachment 208683

Thank you so much for this, this helped me understand so much better!


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For Bio question #398 it says that lysosomes are found in both plant and animals cells. All the resources I've studied so far state that only animal cells have lysosomes and that the central vacuole in plants fulfills some of the lysosomes duties in a plant cell. Could you clarify? Thanks!
You are correct. Plants do have vacuoles.
However, new microscopic techniques have shown that plant vacuoles have enzymes (acidic hydrolases) that are also found in animal lysosomes.

Hope it helps.
 
Hey Dr. Romano!

I am having trouble with how to complete this DAT Destroyer Gen Chem problem(2016 version) #66

I have attached a pic of the problem so you can see what I am talking about.

How should I go about attacking this problem. I know I have to find the intermediate first, but I just get lost after that. Please help!
As stated in a very very detailed solution, you need to write the slow step rate law FIRST. If you see that you have an intermediate, which you do.....a simple substitution is performed. The solution shows all the details.

Hope this helps.
 
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As stated in a very very detailed solution, you need to write the slow step rate law FIRST. If you see that you have an intermediate, which you do.....a simple substitution is performed. The solution shows all the details.

Hope this helps.

Ok, thanks for your help.


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Doctor Romano for gen Chem 2015 edition question 214 can you please explain why do we consider the volume as the mass in KG?


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Im considering buying DATDestroyer for chemistry but im wondering if there are any discount codes. And can I only buy the books from their website?
 
Queen thank you!!! I end up asking MR khan academy because I am assuming the 2016 destroyer is so much different! Lol


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Doctor Romano for gen Chem 2015 edition question 214 can you please explain why do we consider the volume as the mass in KG?


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Because if you look at the DEFINITION of molality......it is strictly defined as moles of solute divided by kg of solvent.

Hope this helps.

Dr. Romano
 
Hi Dr. Romano,

On the Math Destroyer 2016, test 11, problem 12, it asks to find the radius of a circle inscribed within an equilateral triangle whose side is of length 12. I can't seem to get the answer of 2(sqrt3). I attached a picture to help show what I did. In the answers, it says the lengths of the sides would be 6, 6/(sqrt3), and 12(sqrt3). However, if the 60 degree angle is opposite a length of 6, then the 30 degree angle should be opposite a length of 6/(sqrt3), and the hypotenuse should be twice that at 12/(sqrt3). The hypotenuse should also be the one that is the radius of the circle. I am confused.

Thank you!!
IMG_0829.JPG
 
Hi Dr. Romano,

On the Math Destroyer 2016, test 11, problem 12, it asks to find the radius of a circle inscribed within an equilateral triangle whose side is of length 12. I can't seem to get the answer of 2(sqrt3). I attached a picture to help show what I did. In the answers, it says the lengths of the sides would be 6, 6/(sqrt3), and 12(sqrt3). However, if the 60 degree angle is opposite a length of 6, then the 30 degree angle should be opposite a length of 6/(sqrt3), and the hypotenuse should be twice that at 12/(sqrt3). The hypotenuse should also be the one that is the radius of the circle. I am confused.

Thank you!!
View attachment 209725
Your picture is incorrect.
It's a circle inscribed inside a triangle not the other way around.
Hope this helps.
Attached is the correct picture.

sdn.jpg
 
Thank you so much!

I have another question, in the explanation for Bio question #385 it says that "Protists are eukaryotic organisms such as hydra, paramecium, euglena..."

I don't understand, why are Hydra listed with the protists? I thought they were in the Animal kingdom under Cnidaria.

Thank you!
 
Hi I have a question on #34 Orgo (2015 ed)

Which set of reagents would synthesize m-bromobenzoic acid from benzene?

Answer: CH3Cl, AlCl3; K2CrO7, H3O+; Br2, FeBr3

Why cant the Br be added first and then the methyl and COOH? Or why cant the Methyl be added, Bromone, then COOH??

I was thinking that the COOH should be added last since it is an electron withdrawing group more so then Br is. Can someone please explain this order. Also if we used in step 1 C2H5Cl, AlCl3 we would get the same product right?
 
Because BR is slightly activating on benzene ring so if you add benzene it will activate ring and you will not be able to place the group on meta position


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Hi,
Math destroyer_Practice 16_#28
Circumference of the cylinder given is 2(pi)r; I thought the circumference of cylinder should be 2[2(pi)r+h] and 2(pi)r is circumference of circle (one of the base of the cylinder only). Please correct me if my concept is incorrect. Thanks in advance.
 
Hi,
Math destroyer_Practice 16_#28
Circumference of the cylinder given is 2(pi)r; I thought the circumference of cylinder should be 2[2(pi)r+h] and 2(pi)r is circumference of circle (one of the base of the cylinder only). Please correct me if my concept is incorrect. Thanks in advance.
No.
The circumference of a cylinder is 2πr.
The circumference of a cylinder is the same as the circumference of a circle.
Your equation works out to be: 4πr +2h.
I'm just curious where did you find that equation.
 
No.
The circumference of a cylinder is 2πr.
The circumference of a cylinder is the same as the circumference of a circle.
Your equation works out to be: 4πr +2h.
I'm just curious where did you find that equation.

Thank you for your quick reply. I see. I just thought that a circumference is something similar to a perimeter. So I added the perimeter of the base/circle (top and bottom) and two sides of the height.

Is this true that 3D shape model such as cylinder, triangular or rectangular prism don't have/ can't find the perimeter but we can find the perimeter of its base?
 
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Orgo destroyer question 225

It says "all resonance contributors must have the same number of paired and unpaired electrons"

Does this statement mean total electrons (paired + unpaired)?


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Orgo destroyer question 225

It says "all resonance contributors must have the same number of paired and unpaired electrons"

Does this statement mean total electrons (paired + unpaired)?


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The total number of electrons in the molecule do not change and neither do the number of paired and unpaired electrons.
 
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The total number of electrons in the molecule do not change and neither do the number of paired and unpaired electrons.

Thank you! I'm being nitpicky here but in this example here, don't you have 4 unpaired electrons in figure A and 6 unpaired electrons in figure B? Sorry if this is kind of bad question but I wanted to make sure.

ImageUploadedBySDN1477507046.297975.jpg



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Destroyer 2016 OC Question #4:

It asks for the conjugate base of the molecule. The oxygen gets deprotonated instead of nitrogen. I was wondering why can't the nitrogen be deprotonated instead? What rules determine this?
 
Destroyer 2016 OC Question #4:

It asks for the conjugate base of the molecule. The oxygen gets deprotonated instead of nitrogen. I was wondering why can't the nitrogen be deprotonated instead? What rules determine this?
Thanks for a great question. When looking at acidity, we must EXAMINE THE ANION. I must say that to my students 50 times during my acid-base lecture ! An O can carry the negative charge better than a N, hence is greater stabilized !!! Thus we prefer to deprotonate the OH over the NH2 group. Here is the anion stabilization rule of thumb......1. Across the table from left to right......electronegativity dominates ! Hence HF is more acidic than H2O which is more acidic than NH3....because when a hydrogen is removed, it results in a MORE STABILIZED ANION !!!! Rule #2....Down a Group anion stabilization is dominated by size !!! Therefore....H2Te is more acidic than H2Se which is more stable than H2S which is more stable than H2O. I hope this helps.

Dr. Romano
 
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Thanks for a great question. When looking at acidity, we must EXAMINE THE ANION. I must say that to my students 50 times during my acid-base lecture ! An O can carry the negative charge better than a N, hence is greater stabilized !!! Thus we prefer to deprotonate the OH over the NH2 group. Here is the anion stabilization rule of thumb......1. Across the table from left to right......electronegativity dominates ! Hence HF is more acidic than H2O which is more acidic than NH3....because when a hydrogen is removed, it results in a MORE STABILIZED ANION !!!! Rule #2....Down a Group anion stabilization is dominated by size !!! Therefore....H2Te is more acidic than H2Se which is more stable than H2S which is more stable than H2O. I hope this helps.

Dr. Romano

Thank you for taking the time to answer Dr. Romano. I really appreciate it!
 
Question 39 in dat destroyer gen chem. The question says that the statement "C-C bond is stronger than an Si-O bond" is false. But isn't diamond made out of C-C bond? Dr. Romano, do you mind explaining this one? @orgoman22
 
Another question from gen chem destroyer.
#75 : A student heated Na2SO4 10H2O to determine the % of water. Which statement is incorrect?
A) heating must be controlled to avoid decomposition of the anhydrous dry salt.
B) A sample that is not sufficient;y cooled will weigh less
C) If the % of Na2SO4 was calculated to be lower than expected, a possible reason would be loss of gas from the anhydrous salt.
D) If a salt binds the waters so tightly, it is possible for decomposition to occur
E) A dessiccator is employed to prevent loss of mass due to decomposition during heating

Why is it that binding water tightly cause decomposition?
Also, by heating the Na2SO4, it will become gas, so if it wasn't cooled, would it be floating in air? Is that why the % Na2SO4 would be lower than expected?
 
Lets say you had a hydrated salt such as a nitrate salt. These salts bind water tightly. If we were trying to heat up the salt to dehydrate it, we might actually decompose the salt before all the water is actually removed. In the case of nitrate salts, the nitrate portion can decompose into nitrate oxides.

For your second question, if you weigh a sample that is not at ambient (room) temperature i.e. a sample much hotter or colder than the surrounding air, this sample will cause air currents to form. These air currents can actually affect the reading of the scale and.
 
Question 39 in dat destroyer gen chem. The question says that the statement "C-C bond is stronger than an Si-O bond" is false. But isn't diamond made out of C-C bond? Dr. Romano, do you mind explaining this one? @orgoman22
You are way OVERTHINKING this question. As a great general rule, C-C bonds are much much weaker than a Si-O bond. There are exceptions to any rule. For example,,,carbon has 4 bonds as a general rule......BUT, I bet you never knew it could have 5 !!!!!! The methanonium ion is CH5+ !!!!!!!! On questions like the DAT exam, never think about an isolated case. Yes....A diamond would need over 4000 degrees C to melt,,,,,,,but most ....like 99.9 % of all organic compounds would be only a few HUNDRED degrees !!!!! I hope this helps

Dr. Romano
 
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Hello,

I had a question on Organic Chem #204

3 step reaction of 2-butanol
1.) SoCl2
2.) (CH3)3ONa in (CH3)3OH
3.) HBr, ROOR

The answer is 1-bromobutane.
However, despite (CH3)3O being a bulky base, wouldn't it still deprotonate @ the secondary carbon and form 2,3-butene as opposed to the answers 1,2-butene

And after the 3rd reaction with HBr and peroxide, form 2-bromobutane?

Thank you!
 
You are way OVERTHINKING this question. As a great general rule, C-C bonds are much much weaker than a Si-O bond. There are exceptions to any rule. For example,,,carbon has 4 bonds as a general rule......BUT, I bet you never knew it could have 5 !!!!!! The methanonium ion is CH5+ !!!!!!!! On questions like the DAT exam, never think about an isolated case. Yes....A diamond would need over 4000 degrees C to melt,,,,,,,but most ....like 99.9 % of all organic compounds would be only a few HUNDRED degrees !!!!! I hope this helps

Dr. Romano

Ok thank you! So why is c-c bond weaker than si-o bond in general?


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Hello,

I had a question on Organic Chem #204

3 step reaction of 2-butanol
1.) SoCl2
2.) (CH3)3ONa in (CH3)3OH
3.) HBr, ROOR

The answer is 1-bromobutane.
However, despite (CH3)3O being a bulky base, wouldn't it still deprotonate @ the secondary carbon and form 2,3-butene as opposed to the answers 1,2-butene

And after the 3rd reaction with HBr and peroxide, form 2-bromobutane?

Thank you!

Like you said, we are using a bulky base and it will be more favorable to abstract a proton from C1 compared to C2. We mainly form the Hoffman product, a double bond between C1 and C2. And therefore, our final product will be 1-bromobutane

C-kotbu.png
 
Ok thank you! So why is c-c bond weaker than si-o bond in general?


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Great question. The enhanced strength of a Si-O bond is largely due to the fact that Silicon is more electronegative than carbon. As a consequence, there is a substantial coulombic component to the bond energy. Typical values of a C-C bond is 335kJ/mol while a Si-O bond is about 530 kJ/mol !!!!! Thus ...in other words,,,Silicon is quite positive relative to the negative carbon,,,,this bond polarization translates into BOND STRENGTH !!!! Hope this helps.
 
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