Sciencemadness Discussion Board - AP Exam in Three Days??!?

AP Exam in Three Days??!?

Rich_Insane - 8-5-2009 at 20:28

Hey everybody. I am typing, really, really scared now. So i was studying for my AP Chemistry (I have Bio too) Exam, and I took a look at the sample test. On one page, I could only get 20% of them. That's pretty bad.

Here's what I did not know:

Molten NaCl is electrolyzed at a constant current of 1 A. How long will it take to generate 1 mol of Na metal. How can you do this without a given rate.

Memorizing gas laws. They are easy. I can do it. But I need a good way to memorize them.

The worst of all

I completely forgot almost everything about atoms. Almost. I forgot: Pi, sigma bonds (what are they), sp3 hybridization, and calculating shape (I found a really good book, but right now, it's just guessing the structure), and quantum numbers, spin values, the quantum stuff.

Enthalpy change

How would you know which solution had the highest boiling point? It gives a list of various salts. (NaOH, HF, Na2SO4, KC2H3O2, NH4NO3)

Acidity and basic constants ka etc. (what are they?)

Finding the rate of an equation (the rate law)

There's a lot more, but I don't just want answers or anything. I just need a way to keep the memory. I studied, but unfortunately my memories been cleaned multiple times... and now the exam is 3 days away. I'm aiming for a 5. I know your jaw is dropping at how stupid I am

, but I need your help. Any website will help. These books are trash. WOn't ever use them again.

bfesser - 8-5-2009 at 21:16

What books? If you don't want them, I'm sure someone here wouldn't mind picking them up.

Rich_Insane - 9-5-2009 at 07:29

Study books especially for the AP exam. I got the best there are from the library. I'm going to hold on to them just in case I can't find any other good sources, and I will continue practicing on them.

Siddy - 9-5-2009 at 23:40

What does AP mean?

chemrox - 10-5-2009 at 01:15

advanced placement

Lambda-Eyde - 10-5-2009 at 05:44

http://www.chemguide.co.uk

This site has learned me just about everything I know about bonding.

Rich_Insane - 10-5-2009 at 10:40

I did go to that site. It was pretty good. I got so much info on bonding!. But now the test is.... tomorrow. I still don't have a clue how to do the electrolysis rate, and I need to find a good site for memorizing gas laws and all that entropy. I also need to know the rate equations, but I am looking at a site right now for that.

[Edited on 10-5-2009 by Rich_Insane]

Ok, that site for rate laws was useless. I couldn't understand a thing there.

[Edited on 10-5-2009 by Rich_Insane]

Saerynide - 10-5-2009 at 16:39

The NaCl problem is solved by the fact that 1 A is a rate of charge being moved at 1 C/s.

Rich_Insane - 10-5-2009 at 20:31

Ok, so one Coulumb per second. How do we relate that to the amount of sodium being produced? How much time till 1 Mol is formed?

Ramiel - 10-5-2009 at 22:51

Lambda-Eye, the trick there was to pass it off as sarcasam, but the moment's passed.

Play nice people :cool:

Saerynide - 10-5-2009 at 23:43

The charge of an electron is 1.602176487 × 10-19 C. How many electrons are needed to make a mole of Na from Na+? At a rate of 1 C/s, how many seconds would it take?

Edit:

Btw, there is only one gas law you need to know: PV = nRT. All the others can be derived from PV=nRT because n and R are constant.

[Edited on 5/11/2009 by Saerynide]

Rich_Insane - 11-5-2009 at 17:14

I feel so stupid

I totally knew that.

Ozone - 11-5-2009 at 18:46

If you don't know this stuff by now, you won't know it by tomorrow. Get some rest so that the info trapped in your brain can come out when prompted. Try to relax.

Good luck,

O3

Rich_Insane - 11-5-2009 at 20:40

Ha ha never mind guys. I took the AP Bio today instead. I will take the AP Chem next year, in 8th grade, since I am lacking proper preparation.

Saerynide - 11-5-2009 at 22:51

In 8th grade??? I like your ambition, but I think you should wait a bit. Take the AP Chem class first, or at least an intro highschool chem class.

Memorizing this stuff from the Baron's Study Guide will not help - you need to understand it like the back of your hand and be able to do it with your eyes closed to get an easy 5. You know you are prepared when you remember that the mw of Cl is 35.453 and that Fe is 55.85 without looking at the AP periodic table

.

AP Chem is supposedly the hardest AP exam... Most people don't take it until they are juniors or seniors.

But dont let that discourage you. Getting a 5 is easy (trust me

). But it doesn't come from memorizing. If you understand the concepts, it's a breeze - you can do the whole exam in half the time and use the remaining time to snooze while everyone else stares at you in contempt

Edit: Try practicising with the SAT II Chem - it's easier.

Ah... AP Bio... I wrote until my hand fell off. I contemplated stopping after securing the 5, but decided I should finish the test just out of academic sportsmanship.

[Edited on 5/12/2009 by Saerynide]

Rich_Insane - 12-5-2009 at 15:13

AP Bio was the most fail exam ever. Everyone was writing pages and pages but the answer was so simple!

I want to take classes at my local university in high school, because the classes here are crap. I have a professor who's really nice to me who might let me take O Chem Year 1 right now, in 7th grade, but I really, really don't want to considering how bad my memory at chemistry is. I get everything but the atoms and the gasses.

The thing is that my memory fails on me often.... and there are hundreds of different reactions in O Chem.... and if my memory fails on an exam, I'm screwed.

I took a class for the stuff in the AP Exam from an online course. Most boring course ever. It was essentially a slideshow explaining all of this. But I learned a lot of it. The problem is keeping that info inside my head.

[Edited on 12-5-2009 by Rich_Insane]

It's test week again!

Rich_Insane - 27-4-2010 at 08:55

Hey guys,

So last year I canceled taking the AP Chem exam because I felt like I didn't know much. So I ended up taking AP Chemistry, and I feel a lot better. But I'm sort of worried; I have serious trouble with math. It's not even funny, I just can't do math. I do fine in Algebra II/Precal, but for some reason when math is really prevalent in a certain chapter of chemistry, I do terribly in that chapter. For example, I do great on acids and bases because it's simple mathematics and formulas; there's not much to remember (I have terrible....TERRIBLE memory), but on reaction kinetics, I just don't get it. Could someone explain how to diffrentiate between different rate laws and their graphs? I'm also wondering just how much thermodynamic I need. I can do Hess's law and BDE stuff easily, but I have issues with more complex matters. Anyways, I just need to know what to do. I'm using the Princeton Review book to study, and I really need to get a 5 on this test.

Magpie - 27-4-2010 at 12:06

In what year in school are you? How old are you?

zed - 27-4-2010 at 17:43

True, there are a lot of reactions. But, generally speaking, organic chemistry is simple.

Most reaction are driven by the addition or removal of a small molecule. Often H2, O2, or H2O.

As for Math....Most folks don't understand it, at least not initially. According to my one-time buddy Carl Abrams, folks generally start to grasp the actual implications of the mathematical formulas they have memorized.....Sometime in graduate school!

Carl had devised fairly simple methods of muddling through chemical mathematics. Some of those methods might have been included in his book, which was very well received when it was first published.
A Course in Experimental Chemistry (Hardcover)
by Manus Monroe (Author), Karl Abrams (Author)

There might be a copy lurking in your school library. Also, Amazon has used copies for 8 bucks or so. Haven't actually seen a copy of the book, but Carl had a genius for simplification. The convoluted and complex, became straightforward and easily absorbed under his guidance.

[Edited on 28-4-2010 by zed]

Magpie - 27-4-2010 at 20:25

Quote: Originally posted by zed

As for Math....Most folks don't understand it, at least not initially. According to my one-time buddy Carl Abrams, folks generally start to grasp the actual implications of the mathematical formulas they have memorized.....Sometime in graduate school!
[Edited on 28-4-2010 by zed]

This is exactly the point of my questions to Rich about his age.

I know it is fashionable for high schools to offer AP courses nowdays and to allow students to take college courses in their senior year in high school, but I have often wondered about the sanity of this. My concern is that if the student does not have the physical maturity and experience he really shouldn't be taking these advanced courses. When I ask this question of parents who have such students I usually get replies that indicate he/she is doing OK in the class, so, I guess young people are just smarter today than we were. I've never agreed with that answer.

This parallels the rampant grade inflation of todays high schools where if you don't have at least a 3.895 grade point you are not doing all that well. In my day only 1 or 2 peolpe had GPAs above 3.5. Again, the idea that students are smarter or work harder seems unreasonable.

Sending high school juniors out to calculus summer camp at Johns Hopkins just sounds dumb: just a way for the parents to brag and the college to make money. The students probably detest it.

Rich_Insane - 27-4-2010 at 20:40

Quote:

This is exactly the point of my questions to Rich about his age.

Huh?

Quote:

I know it is fashionable for high schools to offer AP courses nowdays and to allow students to take college courses in their senior year in high school, but I have often wondered about the sanity of this. My concern is that if the student does not have the physical maturity and experience he really shouldn't be taking these advanced courses. When I ask this question of parents who have such students I usually get replies that indicate he/she is doing OK in the class, so, I guess young people are just smarter today than we were. I've never agreed with that answer.

I'm not even in high school yet. I'm 14 now. They dumbed down all chemistry up till college level. Not just a little, but dumbed down to the point where you spend 3 weeks on chemical formulas. Since chemistry and biology are my favorite subjects, I'd rather not waste time with elementary stuff and be completely unprepared for harder material later. That's why I want to take AP. I am also pursuing some research which will soon require me to apply a lot of biology and a lot of chemistry (it already does). I don't think anyone's smarter than a previous generation; they just dumbed down classes to a terrible point.

I just really hate how linear school is, I'm not a big fan of memorizing tons and tons of information; I like to see the concept and learn formulas on my own.

Quote:

True, there are a lot of reactions. But, generally speaking, organic chemistry is simple.

AP level O Chem is very easy for me. That's not an issue. It's mainly reaction kinetics, thermodynamics, and scary looking redox reaction (which are more than doable, just daunting for me).

I'm not sure if it's the math I don't get. It's more of the amount of math there is, making it so complex to understand. Especially reaction kinetics.

Magpie - 27-4-2010 at 21:02

Quote: Originally posted by Rich_Insane

I'm not even in high school yet. I'm 14 now.
AP level O Chem is very easy for me.

It's mainly reaction kinetics, thermodynamics, and scary looking redox reaction (which are more than doable, just daunting for me).

I'm not sure if it's the math I don't get. It's more of the amount of math there is, making it so complex to understand. Especially reaction kinetics.

Uh, I'd forgotten how young you are. You are obviously very talented/driven.

My point is that learning and understanding math, and doing lengthy calculations, takes much practice and maturity. At least it did for me.

densest - 27-4-2010 at 21:11

Hmmm... most of reaction kinetics comes down to probability and 1st year calculus. If you can do algebra, the only hard part of calculus is the idea of a limit. Once you wrap your mind around limits or the cookbook rules that apply to limits (I didn't really understand limits until I took a graduate level pure mathematics course - the most terrifying year and, in retrospect, the most valuable course I took in college) a huge amount becomes either remembering a key equation or deducing it on the spot.

Once you have calculus rates become, well, not necessarily obvious, but "simple" matters of calculation. You get a feeling for which forms of equations "look right". A good first year calculus course also teaches you how to handle series of computations like 1/2 + 1/3 + 1/4 + 1/5 with a line or two of equations. The algebra tells you how to solve sets of equations with multiple unknowns by rules which handles problems where multiple reactions are taking place. Calculus (maybe a bit more advanced) tells you how to take those rules and apply them to rates and rates of change of rates and .... all by rules.

Redox is algebra. Juggle the numbers until left charges = right charges = 0. The electrochemical series tells you which reactions go which way giving or taking how many electrons.

I don't know if this helps or not. The most important thing about math is that it all hangs together. If you aren't too stressed, if you know 2/3 of the problem you can often deduce the rest by rules and logic. Sort of like sudoku

[Edited on 28-4-2010 by densest]

JohnWW - 27-4-2010 at 22:05

"1 + 1/2 + 1/3 + 1/4 + 1/5", mentioned above, the "harmonic series" is a DIVERGENT series, with no finite sum as the number of terms approaches infinity; although it diverges very slowly, with the sum of a million terms being less than 20. It is, however, related to the logarithmic function, with the sum to N terms approaching, with increasing terms, ln(N) - y where y = Euler's constant = 0.57721 56649 01532 86060 65120 90082 40243 10421 59335 93992 ... approx.
This approximate sum to N terms is, of course, the finite-difference equivalent of the integral of the function 1/x from 0 to N = ln(N) + constant of integration.
See: http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_consta...

However, noting also the relationship to the Riemann zeta function, if the terms in this series are replaced by any power of the same terms greater than 1, it becomes absolutely convergent as the zeta function, with the sum decreasing with increasing such power.

If the terms are all multiplied by (-1)^(N+1), so that the terms alternate in sign, the series becomes "conditionally convergent", in this case with a sum of ln(2) = 0.6931...., which is a consequence of Taylor's Theorem from which the series for natural logarithms ln(x) is derived. Alternating signs also result in convergence if the terms are raised to any positive power, although the convergence is "conditional" for powers less than 1.

[Edited on 28-4-10 by JohnWW]

Rich_Insane - 28-4-2010 at 20:07

Well for me it's not really the essence of the math that bothers me; It's the complexity of the process that is required. I did look into reaction kinetics. Turns out that I was probably overreacting about how little I know on it. I just can't find a solid way to find exponents in the rate law. How would one do as such? Isn't it something like (change on concentration)^(order; the x) = (change in rate)? I have terrible, terrible attention span, so that's why things like long mathematical process trouble me. I tend to wander off in the middle of a long process.

I know I could just look this up in my textbook, but what exactly does H+ addition in a reaction do (as well as OH-)? I know the fact that you form water somehow, is it to balance out water or something?

DJF90 - 28-4-2010 at 20:51

If you have [H+] or [OH-] in your rate equation it means that there is some form of proton transfer in the rate determining step. Take for instance the reaction of acetone with iodine under acidic conditions. By varying the concentration of acid, acetone and iodine (one variable at a time please!), we can show after a little bit of interpretation of the data that the rate equation is:

k = [Me2C=O][H+]

Notice how there's no [I2] in the rate equation... this is because iodine does not participate in the rate determining step, i.e. no matter how much iodine you add to the reaction, it will not change the rate at which the reaction proceeds. If you consider the mechanism for the reaction, the reason becomes obvious:

The first step of the reaction is enolisation of the ketone, in which acid participates. This is the slowest step of the whole reaction and as such is the rate determining step. The second step is a very rapid nucleophilic attack by the enol onto iodine.

I don't know how useful this will be to you, but I really think you're getting too far ahead of yourself. I'm all up for people being keen chemists (I was one such person myself) but be careful that you learn everything correctly as making a mistake whilst teaching yourself will make like more difficult at a later date, as opposed to making life easy as is probably half the motivation for teaching yourself in the first place.

Rich_Insane - 30-4-2010 at 09:16

Well I understand what the rate law is and how some things participate in a reaction. I just need confirmation on how to find the exponent (aka [A]^2[B] = 3rd order reaction). I have an idea, but I'm not sure I'm right. I also still am confused at how to determine whether something is a second order or third order, or whatever just my looking at the graph. As far as I know, 1st order has something to do with ln, and 2nd order is exponential.

I'm pretty shaky on redox reactions. Especially ones that don't involve metals in elemental form becoming cations. So to verify, please tell me if I'm correct witht he following reaction:

NH3 + H2O --> (NH4+) + OH-

So the H2O has a charge of 2- and the NH3 has a charge of 0. What happens is that the NH3 loses an electron and is protonated by water to form NH4+ (which would involve one electron being transfered to H2O to drop off the proton), and that electron goes to OH. Basically NH3 is getting oxidized and water is being reduced... right?

In most reactions, the H+ is present to balance out the oxygens right? (for example, with Cr2O7, those 7 oxygens are balanced out by 14 H+ to form 7H2O). What is the purpose of a hydroxide balanced redox reaction then?

@ DF90: I'm actually taking the class. However since none of the high schools nearby willt ake me as they are grossly overcrowded, I am forced to take it online. I'm more worried about memorizing all these formulas and remembering them on the test day. Do you seriously have to memorize them all?

Lambda-Eyde - 1-5-2010 at 02:08

That's not a redox reaction, that's an acid-base reaction. The N in ammonia has a -3 charge on both sides of the arrow. H is +1 and O is -2.

Rich_Insane - 1-5-2010 at 21:48

The book describes it as one, I'm aware it isn't one. I think they want to know the transfer in charge.

Alright, so my class is finished (took the final exam). I got a B overall, but my exam was.....OK. It seems as though I need to memorize some important equations. Is there anyone who can tell me what formulas I should definitely memorize for the AP Exam (which is next week) ?
[Edited on 2-5-2010 by Rich_Insane]

[Edited on 2-5-2010 by Rich_Insane]

Rich_Insane - 10-5-2010 at 09:16

Quick question here, it's a day before the exam, but I have to just open my mind up to some information that I forgot. I know that one can find the spontaneity of a reaction with Gibb's free energy (dG = dH -TdS), but when you have an equilibrium reaction like the Haber Bosch process, how would one find which side of the reaction is favored? Doesn't it have something to do with the equilibrium constant? I'm pretty sure I know this, but I can't recall how one would find that out. Does anyone have some general tips for the exam? How difficult is it?

JohnWW - 11-5-2010 at 00:14

The pressure and temperature greatly affect the yield, as per equilibrium constants, of NH3 in the Haber process for producing NH3 from reaction of N2 with H2. Higher pressures favor NH3, as the result of Avogadro's law.

Skyjumper - 11-5-2010 at 11:30

Test was today for americans. The chemistry department was empty today.

Rich_Insane - 12-5-2010 at 09:19

Well, I took the test. I'm not sure whether it was easy or hard, but all the equilibrium questions were easy. NH3 is favored because nature favors higher pressures right? So the side that creates the most mols of gas will be favored. Al higher volumes and lower pressures, the reactants will be favored.... right?

The free response was fairly easy. I sort of forgot all my calorimeter stuff so one of them was difficult, but all the stoichiometry and reactions were simple. Sucks that they didn't have a question about synthesis. I'm adamant I got at least 3, but I'm really hoping for a 5.

Magpie - 12-5-2010 at 16:42

Quote: Originally posted by Rich_Insane

NH3 is favored because nature favors higher pressures right?

Where did you get this idea? What is your basis for saying this?

entropy51 - 12-5-2010 at 16:51

Magpie, since Nature abhors a vacuum she must favor high pressures, no?

Quote:

I have a professor who's really nice to me who might let me take O Chem Year 1 right now, in 7th grade, but I really, really don't want to considering how bad my memory at chemistry is.

But Richie, a year ago you said that you were already taking organic chem in college, didn't you????

Ah, here's some of it:

Quote: Originally posted by Rich_Insane

I'm supposed to take the SAT every year after this. I took it in 7th grade, and from now on I have to take it every year. It pisses me off.

I might just study until college and do experimentation there. I mean at least I'll have the knowledge right? I dropped out of O Chem in the 2nd term (I'd be taking it in High school anyways). It was actually pretty good, because now I have ways to make my life easier when it comes by.

You said you were past seventh grade a year ago, but above you said you're in seventh grade. You're full of it, aren't you? Are you insane, or just a pathological liar??

[Edited on 13-5-2010 by entropy51]

Rich_Insane - 12-5-2010 at 17:59

Quote:

You said you were past seventh grade a year ago, but above you said you're in seventh grade. You're full of it, aren't you? Are you insane, or just a pathological liar??

Summer between 7th and 8th. The SAT was beginning of the year in 7th grade. My bad if that was confusing. It's not a big deal, a lot of people I knew took it too. Also I didn't take O Chem, I was auditing the class to see what it was like. I took that in the summer. The O Chem was a summer session which I mentioned before. Therefore I was in between 7th and 8th, so I wasn't sure what to say. That's the background. That's not the issue here either.

I learned that the Haber Bosch process was volume dependent, or rather pressure dependent, and if I remembered correctly, the higher pressure if favored. Lower volume = higher pressure and more moles of gas = higher pressure.

[Edited on 13-5-2010 by Rich_Insane]

Magpie - 12-5-2010 at 18:36

Quote: Originally posted by Rich_Insane

I learned that the Haber Bosch process was volume dependent, or rather pressure dependent, and if I remembered correctly, the higher pressure if favored. Lower volume = higher pressure and more moles of gas = higher pressure.

Isn't this the Haber process?:

N2 + 3H2 <---> 2NH3

If so, it is turning 4 moles of gas into 2 moles of gas. According to Le Chatelier's principle higher pressure would favor the formation of less volume, ie, the NH3.

Nature likes entropy (if it "likes" anything at all), which would favor higher volumes and lower pressures. But it is man's machinery and energy input that is providing the pressure for the Haber process, not Nature.

Rich_Insane - 14-5-2010 at 09:01

Thanks Magpie. I probably got that backwards. I always thought higher pressure was higher entropy. I'd assume that man-made equipment does something to react N2 and H2 to form NH3, because if you just released some H2 gas into the air, I highly doubt that ammonia would form all of a sudden.

DJF90 - 14-5-2010 at 09:20

Rich_Insane: Yes, you have to learn by heart alot of equations. If you're clever enough you can just learn the fundamentals and work the rest out from first principles etc. Thermodynamics, quantum mechanics, statistical thermodynamics, kinetics, spectroscopy, electrochemistry etc all have fairly large numbers of equations to learn. Its a pretty tough life being a chemist.

Rich_Insane - 14-5-2010 at 09:25

The only thing I have real trouble on at this point is thermodynamics, because I get confused and mix up properties of entropy with enthalpy and vice versa. I also need to figure out how to find which way an equilibrium tips. If I am right, k is derived from the law of mass action (products concentration/reactants concentration), which is derived from experimental data. If I am not mistaken, when k is larger than 1, the products are favored. When it is less than one, the reactants are favored. At one, both exist in the same amounts. Is that all? What equations are absolutely essential to memorize (I've already got Gibb's free energy, the Nernst equation, and that one equation used for buffers, I can also do most equilibrium constants)?

Magpie - 14-5-2010 at 10:07

Quote: Originally posted by Magpie

Nature likes entropy (if it "likes" anything at all), which would favor higher volumes and lower pressures.

Actually I don't think this statement is accurate. For reversible (frictionless/equilibrium) processes entropy does not change with either compression or expansion.