<p>Someone probably already responded but I don’t have the time to look through all the posts…</p>
<p>so can anyone tell me where I can find some released multiple choice tests? I have the 1999 one but is there any more recent ones?</p>
<p>please reply ASAP</p>
<p>THE TEST IS IN TWO WEEKS! ASDLKFJILGFSG</p>
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<p>This is the problem: Hydrogen chloride gas is bubbled through a solution of potassium cyanide.</p>
<p>You can’t split the HCl from the reactants because it’s a gas. For spectators to be cancelled, they must be aqueous on both sides of the reaction. The potassium is (aqueous KCN as reactant, aqueous KCl as product). The chlorine is part of a gas on the reactant side, so it can’t be cancelled as a spectator.</p>
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<p>There’s a magical tool on the internet called “Google”. If you’re looking for free copies of practice exams, look there. I’ve used it to find the 2003 exam before (questions and answers are 100% free; explanations not included). Otherwise, if you’re looking for a specific practice book that you can purchase in the 2 weeks or so time before the test (you’re freaking me out now), use Princeton Review for refresher and both Barron’s and PR’s practice exams.</p>
<p>To answer a question that was asked earlier:
Strong Lewis acids: Something that accepts lone pair electrons.
Strong Lewis bases: Something that donates lone pair electrons.</p>
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<p>Now to toss in something that I feel everyone should be aware of before the test (hopefully this doesn’t skew the scores XDD):</p>
<p>For a certain reaction:
Standard change in heat = +2.98kJ
Standard change in entropy = +12.3 J/K
What is the standard free energy at 25 degrees Celcius? -73C? 127C?</p>
<p>I believe it’s ΔG = ΔH - TΔS</p>
<p>So 2.98 - 298(.0123) = ΔG in kJ, for 25°C.</p>
<p>I’m SO ****ed for this test. My teacher hasn’t covered everything in depth, and I’m scoring around 50% on the MC. I’ll DIE in happiness if I score a 3. </p>
<p>****ing AP Chemistry…</p>
<p>lmao guys. started studying barrons today. sigh i never remembered that there was this much material. i completely forgot about frequency*wavelength=c.</p>
<p>btw i found out u cant use the reference table on part 1s…</p>
<p>@puggly123:
Isn’t a 50% around a 4?
Or is the curve harsher for Chem?
I thought like, 70% was a 5?</p>
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A 50% is definitely a high 3, maybe a low 4 (high/low doesn’t really matter, I’m just giving a range). You need a 68% or so on the exam to get a 5.</p>
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Yeah, I got pretty worried when I learned about that. Our teacher had provided copies of the reference table for every test, and I’ve gotten used to having it handy. At least a periodic table is provided! The reference table is only used for free response, not multiple choice.</p>
<p>Kimberlyy, you are correct! :D</p>
<p>I’m self studying for the test, but I’m in an honors Chemistry class that has done an okay job of covering a lot of material. What topics should I really hit hard? I think i have to learn: Orgo, nuclear (which isn’t hard), Redox, and Kinetics. Those are the only things I have not learned. Plus, I need to review. I really need AT VERY LEAST a 4. So, any suggestions?</p>
<p>Definitely learn Kinetics, and then Redox?</p>
<p>Kinetics, then redox, then organic, then nuclear.</p>
<p>Can someone please explain how to do this problem? I didn’t understand my teacher’s explanation…</p>
<p>What is the chromate ion concentration, [CrO42-], when 20.0 mL of 0.100 M AgNO3 are mixed with 80.0 mL of 0.0100 M Na2CrO4? (Ksp of Ag2CrO4 = 9.0x10^-12)</p>
<p>Answer is 5.6x10^-7</p>
<p>I am really confused. My teacher gave us some multiple choice practice and I can’t figure out why this one answer is correct.</p>
<p>H-C—C-H
(triple bond between the two carbon)</p>
<p>What is the hybridization of the carbon atoms in a molecule of ethyne, represented above?</p>
<p>The answer is sp.</p>
<p>Also, this may be a really simple answer but I haven’t reviewed bonding yet and it has been a few months so I might just be forgetting something.</p>
<p>It’s the areas of electron density around the atom. Triple bonds and double bonds only count as 1 area, but double bonds have 1 sigma and 1 pi bond, while triple bonds have 1 sigma and 2 pi bonds. Single bonds are just 1 sigma bond. Each carbon is bonded to one hydrogen and one carbon and thus has two bonds and has two areas, so it’s sp. (I’m probably using “area of electron density” incorrectly in this context… but I know how to answer these questions correctly so whatevs)</p>
<p>Whenever I see a triple bond I assume it’s sp hydridization. If I see a double bond, it’s always sp2 hybridization. Any exceptions to this rule?</p>
<p>Also, how do you identify a Lewis acid/base? I know a Lewis acid is an electron pair acceptor and Lewis base is an electron pair donor, but how do you identify this in a question?</p>
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<p>When you get stuff like Sulfur that can have more than 8 valence electrons, then this rule might not apply. But it’s a pretty good rule of thumb.</p>
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<p>Let’s take the first Carbon. Another Carbon and a Hydrogen are attached to it. 2 things attached+no lone pairs=2 orbitals–>one s and one p. sp hybridization.</p>
<p>If there were 2 atoms and one lone pair, sp2. If there were 3 atoms and no lone pairs, sp2. If there were 5 atoms and one lone pair (6 total orbitals), d2sp3.</p>
<p>That’s not true. Take CO2. It has double bonds around the carbon, but the carbon is still sp hybridized. If you still dont understand this, try reading through a prep book they explain things better than I.</p>
<p>Also, a Lewis acid is an electron pair acceptor and a Lewis base is an electron pair donor. I can’t explain it that well but I normally identify them based on which one looks like an acid and I look at oxidation states/charges etc… it just works for me idk it’s more intuitive.</p>
<p>What do you guys think the equilibrium will be on? My bet is Kc. </p>
<p>My teacher thinks its Kb. </p>
<p>Can’t be Ksp cause that was last year…</p>
<p>You can identify a Lewis acid/base by drawing it, I believe.
Whichever one has a lone pair (or more) would be the base, as it is the electron pair donor.
The acid would be the one that does not have its full octet, and can take on the extra pair.</p>
<p>Correct me if I’m wrong, but that’s how I thought it was.</p>