<p>Hi I know this may be asking a lot but can someone explain to me the quickest way to do problems 9-12 on this 1999 AP chem test? I have no clue (bad since the test is in a couple of days). I would maybe set up an equilibrium to find [H+] but no Ka is given. Maybe I’m thinking too hard. Thanks.</p>
<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>@runnerz</p>
<p>I think 9 is C because the HCl is a strong acid, so the pH will be the lowest.
I think 10 is… E? It’s weak acid + weak base.
And I’m pretty sure only A and B are buffers, because they involve an acid or base with its conjugate.
11 would be A (NH3 is a base), and 12 would be B (H3PO4 is an acid).</p>
<p>Anyone else being extremely unproductive? ■■■.</p>
<p>Hello guys. Please help me out on a AP Chem FRQ: 2010B #4(b)(i) ([Page</a> Not Found](<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board))</p>
<p>“Excess concentrated aqueous ammonia is added to a solution of nickel(II) nitrate, leading to the formation
of a complex ion.
(i) Balanced equation:
Ni2+ + 6 NH3 → [ Ni(NH3)6]2+”</p>
<p>…Where the heck did they get [Ni(NH3)6]2+ from? Are you supposed to memorize complex ions now? Please help.</p>
<p>Mordred, yes, complex ions are part of the reactions. Just like you should know acid/base, combination, redox, combustion, decomposition, etc…you should also know Complex Ions.</p>
<p>Wait, so we’re actually supposed to memorize the specific complex ion [ Ni(NH3)6]2+? Weird, i’ve never heard of it before… Could someone give me a whole list of complex ions we’re supposed to know?</p>
<p>Azyrk13: The solution was saturated before, and therefore had the highest possible [Ag+] concentration. When you gradually added more water, more silver dissolved and the solution stayed saturated the entire time, so it stayed the same concentration.</p>
<p>Mordred: You don’t have to memorize it, just know how and when complex ions form. It is usually double the charge for the number of ligands, but not in this case. My teacher said though that they gave full credit to people who did not put six ligands, and always do so for complex ions. As long as you know it is complexation, you will get the points.</p>
<p>^ so it only stayed the same because it was saturated (supersaturated?) before? Otherwise would it have lowered?</p>
<p>Many questions!</p>
<ol>
<li><p>“the atom that contains only one electron in the highest occupied energy sublevel.” The options include Br and Ga. Why is the answer Ga and not Br, if Br is only missing one electron? (I didn’t look at the periodic table for Ga since I didn’t have one with me at the time.)</p></li>
<li><p>“HCO3- + OH- <-> H2) + CO3^2- deltaH = -41.4 kJ.
When the reaction represented by the equation above is at equilibrium at 1 atm and 25C, the ratio [CO3^2-]/[HCO32] can be increased by doing which of the following:
A) Decreasing the temperature.
B) Adding acid.”</p></li>
</ol>
<p>Why is it A, and not B?</p>
<ol>
<li><p>“When solid NH4SCN (what is this called?) is mixed with solid Ba(OH)2 in a closed container, the temperature drops and a gas is produced. Which of the following indicates the correct signs for delta G, delta H, and delta S for the process?”
I thought they were all negative - because it becomes spontaneous only at a lower temperature?</p></li>
<li><p>"Which of the following is true for any substance undergoing the process represented by X(s) -> X(l) at its normal melting point?
I thought it was TdeltaS = 0, but it’s deltaH = TdeltaS.
Why? Doesn’t deltaH not change because the temperature doesn’t change?</p></li>
<li><p>"A pure, white crystalline solid dissolves in water to yield a basic solution that liberates a gas when excess acid is added to it. On the basis of this information, the solid could be:
A) KNO2, B) K2CO2, C) KOH, D) KHSO4, E) KCl
No idea why the answer is B.</p></li>
</ol>
<p>@ Abrayo - For the first question: Ga= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1 and Br= 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
Ga has only 1 e- in it’s 4p orbital, while Br has 5.</p>
<p>Oh, it’s funny because this is the multiple choice test I took for my chem final/AP practice test. Azyrk13 is correct for the first question. As for the second one, it is because the reverse reaction would be endothermic, so removing heat would make the reverse reaction slower and that means the forwards reaction is faster than the reverse one so it would become more product heavy. For three, I’m pretty sure it’s ammonium thiocyanate. I think you have misinterpreted it, the reaction is not spontaneous at a lower temperature, it CAUSES a temperature drop, meaning the delta H is negative. A gas is produced from two solids so delta S would be positive, and delta G would be negative because it just happens (and delta S should be rather largely positive). For four, I just understood this now, it is because it is at equilibrium, so delta G is 0. Move the TdeltaS over to the other side and you have deltaH=TdeltaS. Finally, for five, when dissolved it will form carbonic acid (which is a weak acid). When more acid is added, the [H+] goes up so according to le chatlier’s principle the reaction will move the other way and CO2 will be released. I’m not exactly sure if that is 100% correct, but I also memorized that when HCO3 is formed it releases a gas, so that clued me in.</p>
<p>I’m also confused about the complex ion question posted above. How do you know if the coordination number for Ni(2+) is 6 not 4??</p>
<p>In Zumdahl Nickel(II) can have the coordination numbers 4,6 but how do you know which one? My teacher said the good rule of thumb is to double the charge of metal ion, but that won’t work here. :S</p>
<p>You just have to know, but as I said, they accepted the four ligands.</p>
<p>^Oh, they did?</p>
<p>According to my AP Chem teacher, the key for complex ions is “excess” or “concentrated solution” with a (probably transition) metal and a Lewis base. The transitional metal takes on twice as many ligand molecules as its charge. So Ni2+ would, based on this explanation, take 4 ammonias. And according to the above poster, CB would have taken it.</p>
<p>So for Complex ions, even my teacher said that in Princeton Review’s book, they effed up the part of complex ions.</p>
<p>If NH3 is the reactant, then the charge is +.
If it’s OH, CN, SCN, etc, the charge is Negative, right?
Example:</p>
<p>Zn+2 + NH3 ----> Zn(NH3)4 +2
BUT
Zn +2 + SCN- —> Zn(SCN)4 -2</p>
<p>Do we need to know what X-ray Diffraction is or the Bragg equation?</p>
<p>^^You just add up the charge of the metal ion with the combined charges of the ligands you have. </p>
<p>^I don’t think so.</p>
<p>My teacher never taught half-ife, so I am confused.</p>
<p>If 87.5 percent of a pure 131-I decays in 24 days, what is the half-life of 131-I?</p>
<p>(A) 6 days
(B) 8 days
(C) 12 days
(D) 14 days
(E) 21 days</p>
<p>I tried doing a proportion…but I got it wrong. How do you do this? Thanks :D</p>
<p>Also,</p>
<p>Any guesstamates on what will be the topics of the Free Response Questions? :P</p>
<p>@mathisfun111</p>
<p>87.5% = 50% + 25% + 12.5%</p>
<p>I assume you know what half life is, the time it takes for half of it to decompose so you start out with 100%, after one half life 50%, after another half life 25%, after another 12.5%. So after 3 halflifes you have 12.5% of the compound. Hence, 87.5% has decomposed. </p>
<p>So 3 halflifes in 24 days. 24/3 = 8. The half life is 8. It takes 8 days for half of it to decompose. And if you think about it, lets say you have 100g of 131-I. After 8 days you will have 50g, after 8 more days, 25g, after 8 more days 12.5g. So 8+8+8 = 24 days you have lost 87.5g out of the original 100g. Hence 87.5 percent decayed. </p>
<p>And for FRQ, just look at the old ones on ap central. For chem, its pretty standard every year. You can count on a redox, equilib/acid, net ionic/solubility, lewis struc/thermochem questions.</p>
<p>I think it’s B.</p>
<p>Assuming 100 g of iodine…</p>
<p>100 g original - 87.5 g decayed = 12.5 g remaining</p>
<p>100(1/2)^x = 12.5
(1/2)^x = .125 or (.5)^x = .125
x = 3, so three half-lives have passed.
24 days/3 half-lives = 8 days per half-life</p>
<p>Ahh, that makes sense :D</p>
<p>Any ideas on the question topics for the FRQ section (like specific…aka Ka for the equilibrium section or something like that)?</p>