<p>@orangefoot I don’t think I have seen one either and it is also on the formula sheet but I will say that I think there was ONE multiple choice question on it but it was just asking about colligative properties.</p>
<p>XY2(aq) <–> X^2+ (aq) + 2 Y^-
a soluble compound XY2 dissociates in water according to the equation above. In a .050 m solution of the compound, the XY2(aq) species is 40% dissociated. In the solution the number of moles of particles of solutes per 1.0 Kg of water is closest to?</p>
<p>at johnnyhoppy, the SRP is ur best friend. i believe that the srp for Fe is in the negatives.</p>
<p>But so is Mn’s SRP (it is actually in a greater negative than Fe)</p>
<p>You have to flip one of the reactions since both half reactions are donating an electron. A cathode accepts the electron, an anode donates it. </p>
<p>This is a galvanic cell, so the reaction must be spontaneous on its own, meaning that E is greater than 0. So flip one of the half reactions, because if you flip a half reaction the E value’s sign is flipped too. Think: which reaction, if I flip this reaction, will the net E be positive or negative? You want a positive number.</p>
<p>So flip the Fe equation. It’s now
Fe 2+ -> Fe 3+ + e-. The E value is now -.77.</p>
<p>Now let’s check. If we add -.77 and 1.49, the latter of which we don’t touch in this scenario, then we have a positive value. That is good. If we had a negative value, then we flipped the wrong half reaction.</p>
<p>The cathode can be figured out from the half reactions: when you see electrons on the left side of the equation, that means that the reaction is accepting that electron in its product. In a cell like this, that is what you want. The MnO4 reaction therefore is the cathode and the Fe 3+ reaction has the anode.</p>
<p>I am 99% sure, however, that I am forgetting something about graphite and platinum… but I need someone to confirm that with me. Since the cathodes have to be solids the right side should be graphite or platinum, right? </p>
<p>But whatever. Besides the last paragraph this is basically what you have to do. Hope I helped :)</p>
<p>You look at the SRPs, whichever one is greatest/most positive is the cathode, right? So the one at the anode you have to flip, and so the sign is flipped too. So say the anode has E = 2 and the cathode is E = -0.5. What is the E of the cell? 1.5?</p>
<p>EDIT: Ah okay so when you <em>combine</em> the E’s they gotta be positive. Thats easy enough, thanks</p>
<p>@JohnnyHoppy: The half-cell on the right (the one containing Mn) is the cathode because reduction occurs there (+7 –> +2).</p>
<p>Looking at the original equation, Fe changes from +2 –> +3, which is an oxidation.</p>
<p>Ah okay, got it! Thanks!</p>
<p>still need help with my question:
XY2(aq) <–> X^2+ (aq) + 2 Y^-
a soluble compound XY2 dissociates in water according to the equation above. In a .050 molal solution of the compound, the XY2(aq) species is 40% dissociated. In the solution the number of moles of particles of solutes per 1.0 Kg of water is closest to?</p>
<p>Oh gosh I literally JUST learned cell potential yesterday because our class got behind…</p>
<p>For this problem:
According to the graph (temp vs time graph with negative slope) is the dissolution of urea in water an endothermic process or an exothermic process?</p>
<p>I said endothermic because the temperature decrease.</p>
<p>Is this a sufficient enough answer? </p>
<p>The rubric said: The process is endothermic. The decrease in temperature indicates that the process for the dissolution of urea in water requires energy.</p>
<p>ChickenOnRice, is it 0.02? Idk what the balanced equation is for but I just multiplied the percentage with 0.050. That’s the molal amount of the reactant. Multiply by kg of solvent. Ans: 0.02 mol</p>
<p>I got .11 ChickenonRice</p>
<p>orangefoot, yes that is fine.</p>
<p>For Chickenonrice’s clarification, I set up an ICE equation. Since the XY dissociated 40%, its mole amount is .02. We now know that x=.03. So X has a mole amount of .03 and y one of .06… add em all up and you get .11 moles per 1kg of water</p>
<p>For Q2d why must deltaS* be in joules?? Why not KJ? Both deltaH and deltaG was in KJ…
<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>rva2012, how did you get x=.03? Edit: nvm</p>
<p>@Chickenonrice, I got the same as rva2012</p>
<p>@orangefoot - delta S is always in J because that’s the SI unit… so just don’t forget to convert to kJ when you’re working with delta H and delta G</p>
<p>If delta S was in kJ it would always be a small number. That’s probably why we use J instead.</p>
<p>its .090 o.o</p>
<p>sheesh i even asked a few friends and they didn’t get it either. this is probably the hardest mcq i have encountered yet.</p>