DAT in 5 Days 2009 scores (TIPS welcomed)

[MolarBear541]

Bio- 25
Gen Chem- 24
O chem- 21
PAT- 22
What is the general consensus to the real thing? I keep hearing mixed opinions on it. Also does anyone have a link to download the test? I got mine from a friend and the RC section was half washed off. I didn't bother to do QR since the QR section has changed, but i did a bootcamp test instead.
Finally does anyone have any tips on RC? I have been focusing on mostly everything but RC. I took around 7 RC tests and scored 20s and above but i have not taken a RC test since last week.
Any last minute tips are much appreciated!!!

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[Pearl E. White]

Bio- 25
Gen Chem- 24
O chem- 21
PAT- 22
What is the general consensus to the real thing? I keep hearing mixed opinions on it. Also does anyone have a link to download the test? I got mine from a friend and the RC section was half washed off. I didn't bother to do QR since the QR section has changed, but i did a bootcamp test instead.
Finally does anyone have any tips on RC? I have been focusing on mostly everything but RC. I took around 7 RC tests and scored 20s and above but i have not taken a RC test since last week.
Any last minute tips are much appreciated!!!

Nice scores! Here is a link to the exam. http://www.predent.org.vt.edu/admissions/dat-sample-test-2009.pdf
Good luck!

 

[MolarBear541]

Nice scores! Here is a link to the exam. http://www.predent.org.vt.edu/admissions/dat-sample-test-2009.pdf
Good luck!

Thanks!!!

 

[MolarBear541]

Also could anyone help me understand this question? The answer is A

 

[Pearl E. White]

Also could anyone help me understand this question? The answer is A

The reaction they are asking about has been reversed and halved in comparison to the equilibrium reaction. Therefore K must be inverted (due to the reversal) and raised to the power of 1/2 (since the equilibrium reaction has been reduced by half). Since K was 2 initially, the sum of the changes would be (1/2)^(1/2). I hope that makes sense.

In summary and in broad terms, if you reverse a reaction you invert K. If you multiply or divide a reaction by some factor, you raise K to that factor.

 

[MolarBear541]

The reaction they are asking about has been reversed and halved in comparison to the equilibrium reaction. Therefore K must be inverted (due to the reversal) and raised to the power of 1/2 (since the equilibrium reaction has been reduced by half). Since K was 2 initially, the sum of the changes would be (1/2)^(1/2). I hope that makes sense.

In summary and in broad terms, if you reverse a reaction you invert K. If you multiply or divide a reaction by some factor, you raise K to that factor.

Ohhhh okay that makes sense. Thanks again!