TopScore 1 Scores + Random Weird Questions

[2PacClone23]

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Top Score 1:

Bio/Chem/OChem/PAT/RC/QR/TS/AA

21/20/19/21/22/20/20/20



10 More days until DAT. Are these good scores? Some of those biology questions (their topics I mean), I haven't seen or read anywhere. A few stupid mistakes for GC and OC too.

Anyway, the following questions really had my balls in a salad tosser:

1)

Where in the world are they getting the 2.854 number? I get where 1.896 comes from but the other one is just WTF.

2)

I thought signal # = neighbors (H's on C's) + 1; in that case, wouldn't "D" also work? Both sets of H's have no neighbors (the C's don't have H's) + 1 = 1 singlet for each.

3)

Is this supposed to be treated like a Fischer projection where horizontal means coming TOWARDS you? Just looking at these bonds wouldn't make sense because Choice A is "S" but if you look at it in terms of a Fischer projection, then it's an R

4)

Maybe I'm being too technical here but since they didn't mention HBr @ 50 Degree Celsius, I just assumed it would be 1,2 addition. I thought 1,4 addition only occurs at high temperatures (which wasn't specified in the question)

5)

Why is the angle HALF? I thought it'd be 120.

6)

After watching the Chad QR videos on probability and permutations, I thought I had this down. The explanation is way too complicated for me to understand. Chad made it sound so much easier. What would be the "Chad way" of explaining this problem?

 

[Demps]

You almost have mirror image of my topscore#1. You should be fine. Just review the questions that you had gotten wrong. You will be in good shape.

1. total M in the beginning = 14.25 mol / 3 L = 4.75M

It says it decomposes 40% meaning 60% of it remains as reactant.

4.75 x 0.6 = 2.85 (reactant)

4.75 x 0.4 = 1.9 (product)

Now I hope it makes sense that choice A is correct.

2. Choice D does not work. It will only give a singlet. There is a symmetry and those hydrogens will not split. They are all identical. Identical hydrogens do not split. Only B works.

3. Though it is hard to explain with just words, I will try. The best way to assign S and R for these kinds of problems is that just follow along the regular way of assigning S and R except reverse it if the last priority group is either on left or right. I don't think this should confuse you in general.

4. You are correct. Both thermodynamically and kinetically controlled products will be seen because the question did not specify.

5. Refer to Math Destroyer. It has the same kind of problems. But I think that's just the general formula: intercepting arc divided by two.

6. WOW I hoped I never see something like this question on my test! Sorry I can't help here. indigenoustw could probably help you here.

Good luck!

 

[2PacClone23]

You almost have mirror image of my topscore#1. You should be fine. Just review the questions that you had gotten wrong. You will be in good shape.

1. total M in the beginning = 14.25 mol / 3 L = 4.75M

It says it decomposes 40% meaning 60% of it remains as reactant.

4.75 x 0.6 = 2.85 (reactant)

4.75 x 0.4 = 1.9 (product)

Now I hope it makes sense that choice A is correct.

2. Choice D does not work. It will only give a singlet. There is a symmetry and those hydrogens will not split. They are all identical. Identical hydrogens do not split. Only B works.

3. Though it is hard to explain with just words, I will try. The best way to assign S and R for these kinds of problems is that just follow along the regular way of assigning S and R except reverse it if the last priority group is either on left or right. I don't think this should confuse you in general.

4. You are correct. Both thermodynamically and kinetically controlled products will be seen because the question did not specify.

5. Refer to Math Destroyer. It has the same kind of problems. But I think that's just the general formula: intercepting arc divided by two.

6. WOW I hoped I never see something like this question on my test! Sorry I can't help here. indigenoustw could probably help you here.

Good luck!

Thanks so much dawg. I would love to get your real DAT scores 🙂

 

[jwnichols21]

Anyone got an answer to his #6? I'm looking for the "Chad Way" to solve this also!

 

[LuckyTangerines]

Anyone got an answer to his #6? I'm looking for the "Chad Way" to solve this also!

Number of combinations to get 6 heads out of 9 tosses (and order does not matter):

9!/(6!*3!) = 84 combinations

The 9! is the 9 tosses. The 6! Is the number of tosses as heads and the 3! Is the number of tosses as tails. If you write out the whole factorial above, you cancel out some terms and get 84.

Probability of 6 heads: (1/2)^6
Probability of the the remaining tosses as tails: (1/2)^3

Multiply everything for the answer: 84*(1/2)^6*(1/2)^3

 

[jwnichols21]

Number of combinations to get 6 heads out of 9 tosses (and order does not matter):

9!/(6!*3!) = 84 combinations

The 9! is the 9 tosses. The 6! Is the number of tosses as heads and the 3! Is the number of tosses as tails. If you write out the whole factorial above, you cancel out some terms and get 84.

Probability of 6 heads: (1/2)^6
Probability of the the remaining tosses as tails: (1/2)^3

Multiply everything for the answer: 84*(1/2)^6*(1/2)^3

Thanks..but this is the equation way. Did you happen to see the way Chad solves his permutation/combination problems?

 

[jevan]

It's been a while since I've watched chads vid's but I think this is how he'd explain it (or maybe this is how he'd explain it if he was hit with the ******ed bus)...
Make 9 spaces, 6 marked heads and 3 marked tails. to get the likelihood of this exact combination (meaning-this order, where the first 6 were heads and the last 3 were tails) we multiply the chances of each individual event occurring on its own, by each other. so each individual flip has a 1/2 chance of being heads or tails so we have a (1/2)^9 chance of this specific order occurring. so this specific order actually has the same chance of occuring as getting exactly 9 heads.

But there are many ways to get exactly 6 heads and 3 tails...for instance if the first 2 were heads, the following 3 tails, and and the next 4 heads.
so to figure this out we take the smaller number (just because its its easier to work with) which in this case is the 3 tails, and see how many different positions these 3 tails could occupy...

so the first tail could occupy 9 spots, which leaves only 8 spots for the next tails, and now only 7 possible spots for the 3rd tails...or 7*8*9 ways to scramble these 3 tails among the 9 spots.
But...it doesnt matter if its the so called "first tails" or "second tails" that occupies the first tails flip in the sequence, meaning order doesnt matter-or there is no order within the tails,( i'm sorry i know this last explanation kinda sucks).
heres an example- lets say the heads/tails were boys/girls ...we dont care whether the first 3 'flips' are ordered stacy, jane then mary or jane, mary then stacy- it still only represents one sequence/way to get 3 girls and 6 boys. so the 7*8*9 statement has some overlapping identical sequences that we need to correct for. in this case there are 3 diff girls, so there will be 3*2*1 ways to order each specific turnout, or sequence.


So...we take (7*8*9)/(3*2*1) which is the number of possible sequences of 6 heads and 3 tails, and multiply by the probability of 6 heads and 3 tails (1/2)^9..and viola

 

[FROGGBUSTER]

This reminds me of a problem on Topscore 1:

A florist purchased 3 yellow roses, 2 pink roses, and 5 red roses. How many 3-rose arrangements are possible?

The answer according to them is just a simple 10C3 = 10 * 9 * 8 / 3!

But I checked this with my statistician friend, and the answer is 3*3*3 - 1 = 26. A problem this difficult won't show up on the actual DAT, but if you were suspecting something was fishy about that problem, it is.