<p>Oh… Awkward… I could’ve sworn I pasted the study thread link. Anyway, it’s here:
<a href=“http://talk.collegeconfidential.com/sciences/1492199-official-ap-chemistry-exam-study-thread-2012-2013-a-10.html[/url]”>http://talk.collegeconfidential.com/sciences/1492199-official-ap-chemistry-exam-study-thread-2012-2013-a-10.html</a>
Yeah, you should know the bond angles and structures. It’s best to memorize them because there’s usually a FRQ on that stuff almost every year. And graphite is definitely the answer to that question.</p>
<p>Could someone explain the answer to the following question? This is from the 2013 Chem Course Description. (<a href=“http://media.collegeboard.com/digitalServices/pdf/ap/2013advances/AAP-ChemistryCED_Effective_Fall_2013.pdf[/url]”>http://media.collegeboard.com/digitalServices/pdf/ap/2013advances/AAP-ChemistryCED_Effective_Fall_2013.pdf</a>)</p>
<ol>
<li>A student carries out the same titration but uses an indicator instead of a pH
meter. If the indicator changes color slightly past the equivalence point, what will
the student obtain for the calculated concentration of the acid?
(A) Slightly less than 0.0800 M
(B) Slightly more than 0.0800 M
(C) Slightly less than 0.125 M
(D) Slightly more than 0.125 M</li>
</ol>
<p>hey everybody, I’m totally new here and just starting, but I’m freaking about my AP Chem exam tomorrow, mostly about Redox reactions. I missed almost the entire unit and I’m still not really clear on them- especially using the reduction potentials chart, how to pick the best reducing/oxidizing agents, and how to predict the products of a reaction. If anyone can help and explain that, you’ll be my new bestest friend.</p>
<p>Does CollegeBoard penalize for round-off error?</p>
<p>Say their answer is 40.1, and I get 40.2. Do I get full credit?</p>
<p>My teacher says they -1 for wrong sigfigs. :(</p>
<p>I don’t think so, usually, it has to be like, if the answer is 40.1 you have to be off by a certain amount, and it’ll generally be okay if you’re between like 39.6 and 40.6 or something around there.</p>
<p>I’ve heard that too. But comparably, that’s not too much. But that’s different from a rounding error.</p>
<p>My teacher says like -.1 or.2 for sigfigs, but not a whole point unless it’s a lab question.</p>
<p>Is question 3 always thermo?</p>
<p>@Par2014</p>
<p>It should be B) Slightly more than 0.0800 M.
Looking at the titration graph, the equivalence point occurs at 40 mL, meaning 0.04 L * 0.100 M = 0.004 mol of titrant has been added. The molarity of the acid is therefore 0.004 mol / 0.0500 L = 0.0800 M. However, since the student used a pH indicator and the indicator changed AFTER the equivalence point, the student has added a bit more titrant than was necessary to neutralize the solution.</p>
<p>Also, if anyone wants to briefly explain vapor pressure/boiling point elevation/freezing point depression, I’d be very thankful!</p>
<p>They don’t take off points if their answer is 40.1 and yours in 40.2 and your calculations work out to 40.2. </p>
<p>Question 3 is not always thermo. Only question four is a certainty.</p>
<p>Freezing/Boilin Point elevation uses an equation. </p>
<p>ChangeT=i x kb/f x m</p>
<p>i=# of ions
kf/b=given number (should know that water is 1.86 for freezing and .51 for boiling)</p>
<h2>m=molality</h2>
<p>So on a pretty reliable website I saw this:
4. When [H+] in solution is > [OH-] the solution is basic (and vice-versa).</p>
<p>Awkward.</p>
<p>How do you find the order for a single reactant when the other reactant doesnt have two experiments in which the concentration is the same?</p>
<p>vapor pressure = (moles of solvent)/(moles total) * initial vapor pressure</p>
<p>@elmo You have to do it algebrically. </p>
<p>Rate 3/Rate 1= [A]^x**^y/[A]^x**^y</p>
<p>Use the concentrations for the respective rates and solve.</p>
<p>@ohhaiitslily</p>
<p>Redox reactions are reactions that involve the transfer of electrons (so they can be used to make batteries)</p>
<p>If you aren’t clear on oxidation numbers, they’re basically a way to keep track of electrons in a compound. Oxidation numbers take into account an atom’s electronegativity by assuming that an atom with higher electronegativity will “steal” the other atom’s valence electrons, so in CO2, each O will have -1, because O has
a higher electronegativity, and the C will be -2 because it had its two electrons stolen by O. If you notice, oxidation numbers of each atom will add up to the charge on the molecule. (i.e. OH-: oxygen is -2, hydrogen is +1; -2 + 1 = -1) I draw lewis structures to help determine oxidation number. </p>
<p>Another quality similar to oxidation numbers is formal charge, which is like oxidation numbers, but instead of assuming that atoms steal electrons based on electronegativity, formal charges assume that atoms perfectly share electrons. The equation for formal charge of an atom is:
FC = valence<em>electrons - single</em>bonds - lone<em>pair</em>electrons
Double bonds count as two single bonds, triple bonds count as three.
For example, in CO2, the formal charge on C is:
FC = 4 - 4 - 0 = 0
Carbon has 4 valence electrons, it has two double bonds (2*2=4), and it has 0 lone pair electrons.
The formal charge on each O is:
FC = 6 - 2 - (2+2) = 0
Oxygen has 6 valence electrons, and in CO2, it is doubly bonded to C, and has two lone pairs (totalling 4 electrons in lone pairs). The sum of the formal charges in a molecule should also add up to the charge.</p>
<p>In real life, the actual charge on an atom is somewhere between formal charge and oxidation number, because atoms don’t always completely steal electrons or share them.</p>
<p>Oxidizing a species is basically taking electrons away.
Reduction is gaining electrons.</p>
<p>An oxidizing agent is a reactant that causes the oxidation of another molecule, while a reducing agent is a reactant that causes the reduction of another. Usually, the oxidizing agent is the molecule that gets reduced, and the reducing agent is the molecule that gets oxidized.</p>
<p>A popular mnemonic for this is:
LEO says GER (lose electrons oxidation, gain electrons reduction)</p>
<p>In electrochemistry, you can remember:
A LEO says CGER (anode lose electrons oxidation, cathode gain electrons reduction)</p>
<p>Balancing redox reactions is then quite simple; the transfer of electrons must be balanced.</p>
<p>For example, in the reaction:
Na + 2Cl → 2NaCl</p>
<p>The oxidation state of Na is 0.
The oxidation state of Cl is also 0.
In NaCl, Na has an oxidation state of +1, while Cl has an oxidation state of -1.
Since the reaction involves changes in oxidation numbers, it’s a redox reaction.</p>
<p>Na has been reduced, while Cl has been oxidized. Since the transfer of electrons balances out (+1 and -1), the equation is balanced.
The reducing agent is therefore Cl, and the oxidizing agent is Na.
(that was just a quick summary, if you need help on anything in particular, let me know)</p>
<p>@tbradsworth: thanks for your explanation.</p>
<p>More on a general note, on the question set I had posted (<a href=“http://media.collegeboard.com/digitalServices/pdf/ap/2013advances/AAP-ChemistryCED_Effective_Fall_2013.pdf[/url]”>http://media.collegeboard.com/digitalServices/pdf/ap/2013advances/AAP-ChemistryCED_Effective_Fall_2013.pdf</a>), could someone go through questions 8 through 10 with explanation? Thanks in advance.</p>
<p>This really sucks! I’m so not ready for tomorrow. I’ve been so stressed and panicked the entire weekend.</p>
<p>If some nuclear craps comes up on the FRQ, I will stop what I’m doing and sob uncontrollably. I was gone the week we did nuclear in first year chem, and my teacher decided to not go over it this year.</p>
<p>LOL same here. I really don’t want any nuclear or electro chem on the FR. I’m hoping there is an easy equilibrium question, a gas law question, and then also a thermodynamic question.</p>
<p>@Par2014
For #8. The average kinetic energies are the same in each container because temperature is the same as average kinetic energy and temperature is the same in each container.</p>