<p>The bronsted base question… Which was which?</p>
<p>@Waitwutmyname yeah thats what i did. I just wrote like NO3- or whatever the ion was and drew an arrow in the direction of where the oxidation was occuring</p>
<p>@acttest the net ionic was Sn + Cu2+ -> Sn2+ + Cu
@Emokittycat I put H30+ as the acid and H20 as the base. I didn’t like this question because both the water and the other acid were lewis bases and acids…</p>
<p>After calculating my score, best case scenario, I got a 42/46 and if the test takers are mean and don’t accept some of the explanations I wrote, then I got a 38/46. I’m scared about the MC though… Don’t know how well I did on that…</p>
<p>Here is #1-6, still working on #7. I’ve never done a problem like 7A and B before:</p>
<p><a href=“AP Chemistry 2014 FRQ - Google Docs”>AP Chemistry 2014 FRQ - Google Docs;
<p>See I used the weird acids and based they gave… Cuz technically they donated a proton and accepted too…</p>
<p>For 7A you could tell it was first order because the half lives were the same for trials 1 and 2.</p>
<p>trial 1 had 300 torr of substance and only took 100 seconds to react 150 so the rate was 150/100. </p>
<p>trial 2 had 600 torr substance had also took 100 seconds but it had to react 300 therefore the rate was 300/100 which is the double of the rate with 300 torr substance. </p>
<p>Like the PVC question I only wrote that the half-lives were the same between trial 1 and 2 and didn’t get into detail like this. I hope they accept my answer…</p>
<p>@Emokittycat water and the hydronium ion did the same, accepted a proton and gave a proton.</p>
<p>Sorry hypnotoad you messed up number 3, it is Cu(NO3)2 in solution meaning its Cu+2 ions not cu+ so the reduction potential is .34 not .52</p>
<p>Hopefully they’ll accept what I wrote… That FRQ section was brutal. I’m kind of angry. Because the secured material stuff our classroom received wasn’t anywhere near as bad as this</p>
<p>I’m going to write answers here for you @hypnotoad107 because some people are trolls… 7B there’s a equation half life = .0693/k. Plug in half-life and you get k</p>
<p>@hypnotoad107 For 5D, the amount of F the halogen can bond with depends on the orbitals available to it. If the halogen has a f orbital like Iodine then it can have 7 F atoms bond, if it has a d orbital like Br then 5 F atoms, if it has a p orbital like Cl then 3 F and if a s orbital then 1 F.</p>
<p>@hypnotoad107 for 2C, I think the pH will be 7 because the weak acid is the definition of a buffer. The strong base will react with the weak acid and keep the pH at 7.</p>
<p>Ok if I messed up my pKa and got one that differed a lot from the original so based on such I said that you’d need a different indicator? Because I proved I had the knowledge on the indicator and when it’s used but I made a miscalculation</p>
<p>3 E i. should be Sn(s) + Cu2+(aq) → Sn2+(aq) + Cu(s)
V = 0.48</p>
<p>ii. delta G = -(2)(96,500)(.48)
= -9.24 x 10^4 j/mol</p>
<p>4C. It wouldn’t be equal. After being increased to 1.5, it would shift back to equilibrium, and be at a value between 1.04 and 1.5. So it would be greater than 1.04.</p>
<p>I agree with Daiymo Except! i used the reference table value of Faradays constant and used 96,485J will that be considered right or wrong?</p>
<p>I got the wrong voltage for 3e, and I used that to solve for the ii part… Do you think they’ll still give me credit since I worked it right, just used different numbers?</p>
<p>If you do the right measurements, with the wrong numbers, i assume you’ll get credit off in the first part for getting the wrong numbers, then they’ll give you credit for the rest of it.</p>
<p>And you won’t get points off for using the legit value of Faraday’s Constant lol</p>
<p>@dbnitzz, I said that the reaction was first order because a graph of ln[cis-2-butene] over time would yield a straight line. Do you think I would get the point, since there was enough information in the problem to calculate moles of the gas, molarity of the gas and time of the reaction (half-life)?</p>
<p>It would be equal for 4c since Kp = P(CO2). At the same temperature, this is constant so it would be equal to 1.04 atm. </p>
<p>Thanks for the comments and observations. I made corrections where I felt necessary. </p>