<p>Wait, are you sure, for that one, the formula was CF4... Hm... I don't remember a CF4 - Tetrahedral one... Uh Oh?</p>
<p>:)</p>
<p>CF4 is see-saw</p>
<p>No... CF4 is Tetrahedral (4 + 7 (4) = 32). But I do remember putting See-saw for one of the answers. This leads me to believe that CF4 wasnt the thing it was asking for... Maybe?</p>
<p>:)</p>
<p>It was a F-C-F bond and the angle is 109.5 degrees because it is tetrahedral (not see-saw varience - SF4 is seesaw due to the extra electron pair in S compared to C).</p>
<p>Also, since I'm an uber newb I'm gonna do my answers to the equilibrium FR right now, so I'll post them and people can compare.</p>
<p>Equilibrium (FR1) responses:</p>
<p>a) Ka = [H+][C2H5O2-]/[HC3H5O2]</p>
<p>b) 1.34<em>10^-5 = [x][x]/.265M
x^2 = 3.66</em>10^-6
x = [H+] = 1.88<em>10^-3M
pH = -log([H+]) = -log(1.88</em>10^-3M) = 2.73</p>
<p>c) i) (.496g NaC3H5O2) * (1 mol/96g) = 5.17<em>10^-3 mol NaC3H5O2 = 5.17</em>10^-3 mol C3H5O2-</p>
<p>Concentration: 5.17<em>10^-3 mol C3H5O2-/.05L = .103M C3H5O2-
.103M C3H5O2- + 1.88</em>10^-3 M C3H5O2- (from part a; = [x]) = .105M C2H5O2-</p>
<p>ii) Ka = [H+][C2H5O2-]/[HC3H5O2]
1.34<em>10^-5 = [H+][.105M]/[.265M]
1.34</em>10^-5 = <a href=".396">H+</a>
[H+] = 3.38*10^-5M</p>
<p><strong><em>This answer makes sense because a common anion was added to the solution in part b. When a common anion is added, the [H+] decreases, which we have from part b to part c. To check even further, the pH of the solution in part c is -log(3.38</em>10^-5M) = 4.47, which makes sense considering less H+ will be in solution.</strong>*</p>
<p>d) i) Kb = [OH-][HCO2H]/[HCO2-]
Kb = [4.18<em>10^-6][4.18</em>10^-6]/[.309]
Kb = [4.18<em>10^-6]^2/[.309]
Kb = 5.65</em>10^-11
ii) Kw = Kb<em>Ka; Ka = Kw/Kb
Ka = 1</em>10^-14/5.65<em>10^-11
Ka = 1.77</em>10^-4
<strong><em>This is an exact match to the Ka value of methanoic acid as posted previously (from a link) so this pretty much asserts that I did this FR correctly (or at least the last two questions)</em></strong></p>
<p>e) Since the Ka value of methanoic acid is larger than propanoic acid, methanoic acid is stronger. A larger Ka value means more dissociation of the acid; thus, a larger Ka value means more H+ ions in solution, making the pH lower, and again, thus the acid is stronger.</p>
<p>Anyways, pretty sure I got full credit on this. Comments, questions, corrections always welcome.</p>
<p>do they care a lot about sig figs? i couldve sworn the EQ problem had three sig figs, if so, the pH should be 2.720 because pH has a special sig fig rule</p>
<p>hmmm... good point. I thought you had to do the special rule but I wasn't 100% sure so I did it to normal 3 sig figs. I'm not sure what the ruling is on that from the graders, but I hope they let it slide. Otherwise, my answers are correct, though, right??</p>
<p>did anyone do number 2?</p>
<p>SWEET my answers match exactly.. i already have my green insert back, and when i finished i wrote my answers in the green insert..
100% on question one.. things are looking good :)</p>
<p>Question 3</p>
<p>a. Both are first-order.</p>
<p>b. Rate = k[I-][ClO-]
k = 6.1 x 10^2 L mol^-1 s^-1</p>
<p>c. ln[H2O2]
mol L^-1 min^-1
Uncatalyzed graph begins at (0.0 min, 1.00 mol L^-1) and has a less negative slope.</p>
<p>Question 4</p>
<p>a. Zn (s) + Ni2+ (aq) --> Zn2+ (aq) + Ni (s)
c. C2H2 (g) + O2 (g) --> CO2 (g) + H2O (l)
d. CaCO3 (s) + H+ (aq) --> Ca2+ (aq) + H2O (l) + CO2 (g)
e. Li (s) + N2 (g) --> Li3N (s)
h. Pb2+ (aq) + I- (aq) --> PbI2 (s)</p>
<p>Question 5
a. Nitrogen: Splint goes out completely (extinguishes)
Oxygen: Splint reignites or glows brighter.
Hydrogen: Splint explodes/"pops"</p>
<p>b. CaO + H2O --> Ca(OH)2, a strong base. Hence, pH > 7.
Silicon oxide is a neutral compound. pH = 7.
CO2 + H2O --> H2CO3 --> H+ <-> HCO3-
HCO3- <-> H+ + CO3^2-
Since H2CO3 can ionize twice, its pH < 7.</p>
<p>c. Black ppt is Ag2S (s)
Mixing solution 1 with soln 3 yields the AgCl (s), a white ppt.
Solution 1: AgNO3 (Silver nitrate)
Solution 2: Na2S (Sodium sulfide)
Solution 3: KCl (Potassium chloride)</p>
<p>Question 6</p>
<p>a. CF4: C is central atom. 4 single bonds.
PF5: P is central atom; 5 single bonds.
SF4: S is central atom; 4 single bonds. S atom has 2 additional electrons.</p>
<p>b. Approx. 109.5º
PF5 has sp3d hybrid.
SF4 has a seesaw molecular geometry.</p>
<p>c. 4 sigma + 1 pi (from the double bond)
Structure 1 represents the molecule better because each atom has a formal charge of 0, hence making it more stable.</p>
<p>Question 8</p>
<p>a. Change in G is negative because the dissolving of AgNO3 into its component ions in water is a spontaneous process. Any compound with the NO3 ion is always soluble in water.</p>
<p>b. Change in S is positive because a solid object's dissolution increases its disorder.</p>
<p>c. The sign of H is positive because the reaction is endothermic: heat must be absorbed.
Yes, it is consistent. However, TS needs to be more positive than H in order to yield a negative value for G.</p>
<p>d. Ox: Na (s) --> Na + e-
Red: O2 + 4H+ + 4e- --> 2H2O
Oxidation rxn takes place at the anode
G is always positive for electrolysis rxns because work must be applied.</p>
<p>Hope this helps, guys!</p>
<p>Crypto86, I think you need to use the Henderson-Hasselbalch equation (for buffers) in c-ii. Adding sodium propanoate yields the NaC3H5O2-HC3H5O2 buffer (a weak acid and its salt). I may be wrong, though.</p>
<p>everything else in question 1 looks right to me.</p>
<p>Question 4, continued
b. Al(OH)3 + KOH --> KAlO2 + 2H2O
f. BF3 (g) + NH3 (g) --> H3NBF3 (s)
g. SO3 + 2 NaOH --> Na2SO4 + H2O</p>
<p>These were indeed hard!</p>
<p>
[quote]
c) i) (.496g NaC3H5O2) * (1 mol/96g) = 5.17<em>10^-3 mol NaC3H5O2 = 5.17</em>10^-3 mol C3H5O2-</p>
<p>Concentration: 5.17<em>10^-3 mol C3H5O2-/.05L = .103M C3H5O2-
.103M C3H5O2- + 1.88</em>10^-3 M C3H5O2- (from part a; = [x]) = .105M C2H5O2
[/quote]
Dude, why did you add the hydrogen concentration to your answer at the end of this part Crypto??? I am a 100% sure you're noit supposed to do that. My AP Chem teacher did this part of the problem on the board and he didn't add the hydrogen concentration(from part a) to the end of this part of the problem. Why would you do that??? Whatever, it doesn't really change the answer much.</p>
<p>question 4...
CaCO3+HC2H3O2 ---> Ca(C2H3O2)2 + CO2 + H2O</p>
<p>Cant write H+ for a weak acid.. could prolly also be written Ca2+ + C2H3O2... but dont know for sure</p>
<p>if anyone wants to discuss the exam w/ me..
IM silvachik08 on AIM...</p>
<p>we can discuss it fast!!</p>
<p>man on fire - remember, the original solution has [x] concentration of the C3H5O2- ion in it already. Remember - it's [x]^2 that equals 3.66<em>10^-6; the [H+] times [C3H5O2] = 3.55</em>10^-3, so [H+] = [C3H5O2-] = 1.88*10^-3M</p>
<p>Therefore, you need to consider the already existing C3H5O2- ion concentration already in solution; you can't just disregard it from part a. That's why I added 1.88*10^-3 (meant for C3H5O2- concentration, not H+) to the ion added from the NaC3H5O2, because some ion is already in solution before you add the solid. I'm pretty sure I'm right on that, but I'll double check again.</p>
<p>Also, student, for the solid Al(OH)3 and KOH solution reaction, I had:</p>
<p>Al(OH)3 + OH- --> [Al(OH)4]-</p>
<p>KOH is a very strong base (Kb > 1) therefore it is always ionized. So the first ionic equation would be:</p>
<p>Al(OH)3 + K+ + OH- --> [Al(OH)4]- + K+; and the K+ would cancel out on each side, leaving: Al(OH)3 + OH- --> [Al(OH)4]-</p>
<p>Just a little comment about the reaction questions. I personally thought they were the easiest of any reaction Q I've seen, but then again, my teacher did prepare us with weekly quizzes for a little. I mean, a electrochem rxn, a combustion rxn, and a few others were really simple.</p>
<p>I chose:</p>
<p>Zn + Ni+2 => Zn+2 + Ni (battery)
C2H2 + O2 => CO2 + H2O (combustion)
Li + N2 => Li3N (appeared on a previous test)
BF3 + NH3 => BF3NH3 (appeared on a previous test)
Pb+2 + I- => PbI2 (common lead iodide demo rxn)</p>
<p>And, with the previous answers given for equilibrium, I agree completely.</p>
<p>
</p>
<p>I agree except for the units in (c). The units would just be minutes^-1, since the rxn is first order, rate = k[H2O2].</p>