<p>No, you just got the first part wrong, and then after that they base your answers to what you did preceding them.</p>
<p>yeah oops i forgot to put the X100 (it shud be there)
good thing i put it on the test though</p>
<p>For the one that asked about a color change, was it from purple to clear or purple to some other color?</p>
<p>Clear to purple</p>
<p>"wait i thought it would produced carbonic acid, which dissolves limestone. It produces CO2 and H2O? why?"</p>
<p>Limestone IS (mainly) CaCO3. You're right that acid dissolves limestone... but HNO3 is an acid, too! Would it make sense that HNO3 would react with limestone to create an acid that would then react with remaining limestone? Not really =)</p>
<p>Well, when acids and bases react, they form H20 and a salt, right? The CaCO3 is the base. Think about when CaO reacts with an acid- the O reacts with 2H+ to form H2O. CaCO3 reacts the same way... but only one O reacts, "leaving" CO2, which also forms. Then, Ca(NO3)2 (the salt) is form... and H2O of course. Carbonates and bicarbonates always form CO2 when they react with an acid.</p>
<p>i said the same thing (Carbonic acid)
Our teacher still hasnt given us our FR yet</p>
<p>Have we decided definitively whether or not the pH was 3.32?</p>
<p>(I hope so =P)</p>
<p>i got 3.32</p>
<p>clear to purple?! no.. because the reactant, MnO4 2+ is purple and is reduced to Mn2+ which isn't purple... right?</p>
<p>When you titrate a solution past the end point, you use up all the Fe2+ (colorless) and have excess MnO4 is in solution, which is purple. No more Mn2+ is being made because of no more Fe2+, the reaction no longer occurs in a sense.</p>
<p>i'm just wondering... how many pts. was each FRQ worth and also what %?</p>
<p>for question 5, the reactions, were you supposed to write the net ionic equations? cuz i know in previous tests you had to, but this one they changed it so i wrote balanced reactions but not net ionic cuz i thought i didn't have to (i read the directions twice just to make sure), but i am still doubtful, so can someone confirm? please and thank you.</p>
<p>It says "Omit formulas for any ions or molecules that are unchanged by the reaction." So, yes, it has to be net ionic.</p>
<p>How many FRQ's do I need with a 50 MC Raw Score to get a 5?</p>
<p>Definitly net ionic, it doesn't say the word net, but tells you to cancel out spectator ions.</p>
<p>anyone wanna post a compiled one?</p>
<p>yeah can someone post the answers that are definitely right?</p>
<p>ON Number 4, what happens if we included spectator ions on one of them (#2 I think)..do we get a 0, or still get credit?</p>
<p>Also, for the electrolytic cell....if we said delta g is negative, so e0 is positive, then based the explanations off of that and the rest of the problems (getting the wrong signs, but right answers), do we lose ONE point or ALL the points? </p>
<p>Also, number 3f: is it 21.06 Liters (1.16)<em>V=1</em>0.08*298</p>
<p>And 3e: 1.5 amps<em>40 min</em>60sec/min=3600 Columbs=3600 grams?</p>
<p>Compiled answers (not official, but generally agreed upon by several AP teachers.)
1a K=[H3O+][F-]/[HF]
1b .017
1c .004 moles
1d 0.15M (depending on sig fig treatment, may be 0.2)
1e 3.32 (depending on sig fig treatment, may be 3.3)</p>
<p>2a -181 kJ
2b 950. K
2c -33.8 kJ
2d 6
2e 141 kJ/mol</p>
<p>3a counterclockwise (toward the copper electrode)
3b 2Cu2+ + 2H2O --> 2Cu + O2 + 4H+
3c Since E is negative, G is positive
3d +344 kJ
3e 1.19 g
3f .197 L</p>
<p>Continued in next message</p>
<p>Compiled answers (not official, but generally agreed upon by several AP teachers.)
4a 2OH- + Pb2+ --> Pb(OH)2 1/2 mole of Pb(OH)2 is made
4b CaCO3 + 2H+ --> Ca2+ + H2O + CO2 Acid rain converts insoluble CaCO3 to soluble Ca2+ ions, causing the statues to dissolve.
4c Ag+ + Fe2+ --> Ag + Fe3+ Silver metal</p>
<p>5a +7
5b Fe2+
5c ignoring the Fe ions, clear to purple. There is debate among some teachers whether the Fe ions should be considered. Fe2+ would cause the solution to start out greenish, but Fe3+ would cause the solution to be a mixture of colors after the endpoint.
5di MxVx5
5dii 5di x 55.85
5diii 5dii/g x 100
5e Since V would be too large in 5di, answer would be too large</p>
<p>6a Too hard to draw: I with 3 F's attached, and 2 lone pairs on the I
6b T-shaped
6c O=S-O <--> O-S=O also a lone pair on the S in both structures.
Bonds are equal because with resonance structures, the "real" structure is an average of the resonance structures
6d sp2
6e 2 horizontal lines, first one high, second one low. 2 hills drawn between the lines, lower hill is for catalyzed reaction, higher hill is uncatalyzed.
6f ratio increases (Le Chatelier's)
6g ratio doesn't change. Catalyst affects rate, not eq. positions.</p>
<p>Official answer key won't be posted at college board until later this summer.</p>