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matt.shanley22

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Hi people,

I am going through the destroyer books and I was wondering if anybody else who is going through the 2016 edition or is very familiar with it could help me with a few problems? Once I get a reply I'll post the problem or we can chat directly if you want. Thanks! Test is in 16 days by the way and I am starting to feel nervous/unprepared.
 

matt.shanley22

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Here's one of them:
In Gen Chem
120.) CaCl2 has a weight of 111g/mol. How many grams of CaCl2 x 2H2O would be required to make a 1M solution?
 

jz123

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Here's one of them:
In Gen Chem
120.) CaCl2 has a weight of 111g/mol. How many grams of CaCl2 x 2H2O would be required to make a 1M solution?
to make 1M of a solution, you need 1 mole of solute in 1 L of solvent. So let say now you have 1L of water, and you need to figure out how many grams of solute you need. Basically, in this question, you just need to calculate how many grams are there in 1 mole of Cacl2*H2O. So it is just the mass of 1 mole of CaCl2 + 2 moles of water = 111+ 2(18) = 147.
 

matt.shanley22

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GC Destroyer
136.) How much heat is needed to convert 20g of ice at -40 degree Celcius to steam at 300 degree Celcius?

Then they give you a table of "substances" and "Specific Heat" respectively..
Ice = 2.0 J/g degree C
Water = 4.2 J/g degree C
Steam= 1.8 J/g degree C
deltaH_fusion= 6.0 KJ/mole
deltaH_vaporization= 41 KJ/mole
 

matt.shanley22

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to make 1M of a solution, you need 1 mole of solute in 1 L of solvent. So let say now you have 1L of water, and you need to figure out how many grams of solute you need. Basically, in this question, you just need to calculate how many grams are there in 1 mole of Cacl2*H2O. So it is just the mass of 1 mole of CaCl2 + 2 moles of water = 111+ 2(18) = 147.

Okay that clears it up thank you.
 

rolltide15

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You would need to figure out how much the 2H20 weighs and add it to the
GC Destroyer
136.) How much heat is needed to convert 20g of ice at -40 degree Celcius to steam at 300 degree Celcius?

Then they give you a table of "substances" and "Specific Heat" respectively..
Ice = 2.0 J/g degree C
Water = 4.2 J/g degree C
Steam= 1.8 J/g degree C
deltaH_fusion= 6.0 KJ/mole
deltaH_vaporization= 41 KJ/mole


you would need to take into account the phase changes so the first calculation would be q=mcat to go from 0 to -40. Then you would have the fusion phase change and then you would heat again from 0 to 100 using q=mcat then from 100 to 300 you would use delta H of vaporization and add up all the Delta H values to get the heat needed.
 

matt.shanley22

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Heres another one,

ORGO destroyer #166.)

Which compound nitrates most rapidly?

Then it gives you: nitrobenzene, ethylbenzene, benzene with 0ch3 on top, benzene with cooh on top, and benzene with so3h on top
 

super frank

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Heres another one,

ORGO destroyer #166.)

Which compound nitrates most rapidly?

Then it gives you: nitrobenzene, ethylbenzene, benzene with 0ch3 on top, benzene with cooh on top, and benzene with so3h on top
Whichever is the strongest donating group.
 

matt.shanley22

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GC Destroyed #172:

If 20 mL of 0.012M solution of Ca(OH)2 is added to 48 mL of HBr, what is the concentration of the HBr?

Since M=mol/L why cant you find the moles of Ca(OH)2 and then convert those to moles of HBr to find its molarity? Wouldn't it be a 1:1 mol ratio?
 

Lnguyen7

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GC Destroyed #172:

If 20 mL of 0.012M solution of Ca(OH)2 is added to 48 mL of HBr, what is the concentration of the HBr?

Since M=mol/L why cant you find the moles of Ca(OH)2 and then convert those to moles of HBr to find its molarity? Wouldn't it be a 1:1 mol ratio?

Of course you can. However, it's not 1:1, but 1:2.
Ca(OH)2 + 2HBr -> CaBr2 + 2H2O
 

orgoman22

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    What's an easy way to determine which is the strongest donating group?
    The strength of an electron donating group is determined by experiment. For the DAT exam.....just know that the most reactive is the NH2 group. This would be followed by OH group, OR group such as methoxy, and then R groups. The most important thing is to recall which groups are activating and direct o/p...and which are deactivating and directing meta. be able to do problems which involve using the directing influences of these groups. For example, if toluene is oxidized with KMNO4, and acidified,,,we get benzoic acid. Now, if we brominate benzoic acid we get 3-bromobenzoic acid because the COOH group is a meta director. Meta directors ALL deactivate the ring, whereas an activating group USUALLY activate the ring. Exceptions include halogens as well as the Nitroso group, NO.

    Hope this helps.

    Dr. Romano
     

    Alpha Centauri

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    The strength of an electron donating group is determined by experiment. For the DAT exam.....just know that the most reactive is the NH2 group. This would be followed by OH group, OR group such as methoxy, and then R groups. The most important thing is to recall which groups are activating and direct o/p...and which are deactivating and directing meta. be able to do problems which involve using the directing influences of these groups. For example, if toluene is oxidized with KMNO4, and acidified,,,we get benzoic acid. Now, if we brominate benzoic acid we get 3-bromobenzoic acid because the COOH group is a meta director. Meta directors ALL deactivate the ring, whereas an activating group USUALLY activate the ring. Exceptions include halogens as well as the Nitroso group, NO.

    Hope this helps.

    Dr. Romano
    With this logic, would aniline be more basic than phenol?
     

    orgoman22

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    With this logic, would aniline be more basic than phenol?
    You are comparing two different things. Aniline is a AMINE.....and as such is a base. However.....since the amine has a benzene ring that would slightly withdraw electrons, and also the electrons being part of the aromatic ring system, this would render aniline a poor base. We must compare pKa values of the compounds in their protonated states.... The Pka of Phenol is 10...the Pka of Anilinium ion is 4.6.......thus Phenol is the weaker acid,,,,, since it has a smaller pKa...hence would be the STRONGER base. You are not going to be asked something this detailed on the DAT...however.....you should understand that amines are usually good bases,,,,especially if we increase the amount of methyl or ethyl groups on the N atom. Groups like Benzene significantly lower the ability to function as base since the electrons are less available to capture a proton.

    Hope this helps.

    Dr. Romano

    This picture clearly shows that these electrons on the N are DELOCALIZED and shared with the ring system......not very available to capture a proton !!!

    sdn.gif
     

    matt.shanley22

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    General Question concerning DAT studying:

    For those who have in recent years (after 2010) followed the 2010 8-week study schedule that is on SDN, upon completing the number of destroyer problems they suggest, has anybody done the remainder of the problems that are in the books? Or have you just gone back over the ones you already did like the study schedule suggests?
     

    matt.shanley22

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    Math Destroyer: Test #7 problem #6

    A queen size mattress costs $180 more than a twin-sized mattress. If the twin-sized mattress is 3/4 the cost of the queen-sized mattress, what is the cost of the Queen sized mattress?
     

    matt.shanley22

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    Math Destroyer Test #7 problem #19:

    What is the largest distance between any two points inside a rectangular box whose dimensions are 3, 4, and 5?
     

    matt.shanley22

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    Math Destroyer Test #7 problem #28:

    What is the smallest positive number that is divisible by both 14 and 49?

    I do not understand their explanation.
     

    Lnguyen7

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    #6
    A queen size mattress costs $180 more than a twin-sized mattress.
    Q = T + 180

    If the twin-sized mattress is 3/4 the cost of the queen-sized mattress
    T = (3/4)Q

    what is the cost of the Queen sized mattress?
    Q = $720

    #19
    14by14.jpg
    Pink can be calculated by yellow
    Green can be calculated by pink and red.

    #28
    I don't know what they explain so...
    But is it 98?
     

    matt.shanley22

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    #6
    A queen size mattress costs $180 more than a twin-sized mattress.
    Q = T + 180

    If the twin-sized mattress is 3/4 the cost of the queen-sized mattress
    T = (3/4)Q

    what is the cost of the Queen sized mattress?
    Q = $720

    #19
    View attachment 206466
    Pink can be calculated by yellow
    Green can be calculated by pink and red.

    #28
    I don't know what they explain so...
    But is it 98?

    For #28 yes it is 98 but I do not understand why. I know 14= 7x2 and 49=7x7 by prime factorization, but why would that equal 98?

    Also, for #6, I understand that that is the equation you set up but I do not see how you end up with $720. For some reason I just cant get the math right in that problem.
     

    Lnguyen7

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    For #28 yes it is 98 but I do not understand why. I know 14= 7x2 and 49=7x7 by prime factorization, but why would that equal 98?

    Also, for #6, I understand that that is the equation you set up but I do not see how you end up with $720. For some reason I just cant get the math right in that problem.

    Q = T + 180 (1)
    T = (3/4)Q (2)

    Substitue (2) to (1)
    (1): Q = (3/4)Q + 180
    Q - (3/4)Q = 180
    (1/4)Q = 180
    Q = 180*4 = 720

    It's just that 49*2 = 98 which is divisible by 49 and 14. Sometimes you can also use the answers to find the correct answer too.
     

    matt.shanley22

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    Q = T + 180 (1)
    T = (3/4)Q (2)

    Substitue (2) to (1)
    (1): Q = (3/4)Q + 180
    Q - (3/4)Q = 180
    (1/4)Q = 180
    Q = 180*4 = 720

    It's just that 49*2 = 98 which is divisible by 49 and 14. Sometimes you can also use the answers to find the correct answer too.
    Okay both of those responses made it much clearer to me thank you very much.
     

    matt.shanley22

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    MATH Destroyer Test #8 Question #5:

    If it takes "a" pounds of grain to feed "b" pigs, how many pounds of grain will it take to feed "c" pigs?

    Can somebody give me a logical answer to this?
     

    matt.shanley22

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    Math Destroyer Test #8 Problem #7:

    A bicyclist travels 56 miles to a Trading Post at a constant speed. On his return trip he travels at half the speed. If the round trip takes 6 hours, what is his speed on the way to the Trading Post?
     

    matt.shanley22

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    You've posted a lot of questions to this thread. Have you attempted to understand the solutions in the back of the book first? If so, you should include what you don't understand from the solution in your question. I don't want to say the exact same thing.
    Yes, I also started this thread and yes I have tried to understand it in the back of the book but obviously that didn't work with me for specific problems which is why I'm asking them here with a week until my exam. I could post what I didn't understand about their answer but it's easier for me to ask the question and see if somebody can answer it in a less confusing way so I understand that way instead of the destroyers complicated method.
     

    Lnguyen7

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    MATH Destroyer Test #8 Question #5:

    If it takes "a" pounds of grain to feed "b" pigs, how many pounds of grain will it take to feed "c" pigs?

    Can somebody give me a logical answer to this?

    One of the powerful ways to solve a problem is to change the problem and find the logic behind it.

    #5
    If it takes "1" pounds of grain to feed "2" pigs, how many pounds of grain will it take to feed "3" pigs?
    The answer is 1.5. Easy enough? But how did I come up with 1.5 is the question. Well I can say 1 pound feed 2 pigs, so 0.5 lb feed 1 pig and since there are 3 pigs, 0.5x3 = 1.5 lbs. Or everything in one step: 3 pigs*1 lb/2 pigs = 1.5 lbs.

    So If it takes "a" pounds of grain to feed "b" pigs, how many pounds of grain will it take to feed "c" pigs?
    The answer is c*a/b lbs!

    #7 A bicyclist travels 56 miles to a Trading Post at a constant speed. On his return trip he travels at half the speed. If the round trip takes 6 hours, what is his speed on the way to the Trading Post?

    Lets assign some variables
    Time/velocity for first trip: t1, v1
    Time/velocity for second trip: t2, v2


    first trip = constant speed and second trip = half speed
    These tell you the time he takes for his second trip is twice as much as the first trip or t2 = 2t1

    The round trip takes 6 hours or t1 + t2 = 6

    Solving for t1 = 2 hours
    v1 = d1/t1 = 56/2 = 28 miles/hr
     
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