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05-14-2005, 10:58 AM
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#76 | | Junior Member
Join Date: Aug 2004
Threads: 22
Posts: 223
| can somebody please post solutions to all of them? | 
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05-14-2005, 11:16 AM
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#77 | | Member
Join Date: Mar 2005 Location: Chapel Hill, NC
Threads: 96
Posts: 777
| most of them are already posted..
and people here are too lazy to redo equilibrium | |
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05-14-2005, 11:35 AM
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#78 | | Junior Member
Join Date: Jan 2005 Location: Reno, NV
Threads: 21
Posts: 279
| If I missed the part about the carboxyl group being responsible for the pH of 2.6, how many points will be taken off? | |
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05-14-2005, 02:03 PM
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#79 | | Member
Join Date: Aug 2004 Location: NJ, USA
Threads: 27
Posts: 501
| Probably only 1. That's a question where either you know it or you don't - not like a math one where you can tinker around a bit. I would definitely say only 1 point for that. | |
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05-14-2005, 03:16 PM
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#80 | | New Member
Join Date: Aug 2004
Threads: 6
Posts: 22
| OK, question. Im probably the only idiot who made this mistake. For the bonding question (#6 i think) it asked for the bond angle of "F-C-F". I know that tetrahedrals have 109.5 but at the last second i changed it to 180 because it stated specifically F-C-F instead of showing the actual tetrahedral structure or instead of asking for it "around the C atom". I figured it was a trick question, lol. Yeh, i know theres electron repulsions so it wouldnt be exactly 180...etc so i put approximately 180. Is there anyway i could get credit for this or was this completely insanely stoooooooopid? | |
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05-14-2005, 03:32 PM
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#81 | | Senior Member
Join Date: Mar 2005 Location: Minnesota!
Threads: 18
Posts: 1,077
| Wait, are you sure, for that one, the formula was CF4... Hm... I don't remember a CF4 - Tetrahedral one... Uh Oh?  | |
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05-14-2005, 04:54 PM
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#82 | | Member
Join Date: Mar 2005 Location: Chapel Hill, NC
Threads: 96
Posts: 777
| CF4 is see-saw | |
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05-14-2005, 04:57 PM
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#83 | | Senior Member
Join Date: Mar 2005 Location: Minnesota!
Threads: 18
Posts: 1,077
| 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? | |
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05-14-2005, 04:58 PM
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#84 | | Member
Join Date: Aug 2004 Location: NJ, USA
Threads: 27
Posts: 501
| 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).
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. | |
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05-14-2005, 05:27 PM
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#85 | | Member
Join Date: Aug 2004 Location: NJ, USA
Threads: 27
Posts: 501
| Equilibrium (FR1) responses:
a) Ka = [H+][C2H5O2-]/[HC3H5O2]
b) 1.34*10^-5 = [x][x]/.265M
x^2 = 3.66*10^-6
x = [H+] = 1.88*10^-3M
pH = -log([H+]) = -log(1.88*10^-3M) = 2.73
c) i) (.496g NaC3H5O2) * (1 mol/96g) = 5.17*10^-3 mol NaC3H5O2 = 5.17*10^-3 mol C3H5O2-
Concentration: 5.17*10^-3 mol C3H5O2-/.05L = .103M C3H5O2-
.103M C3H5O2- + 1.88*10^-3 M C3H5O2- (from part a; = [x]) = .105M C2H5O2-
ii) Ka = [H+][C2H5O2-]/[HC3H5O2]
1.34*10^-5 = [H+][.105M]/[.265M]
1.34*10^-5 = [H+](.396)
[H+] = 3.38*10^-5M
***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*10^-5M) = 4.47, which makes sense considering less H+ will be in solution.***
d) i) Kb = [OH-][HCO2H]/[HCO2-]
Kb = [4.18*10^-6][4.18*10^-6]/[.309]
Kb = [4.18*10^-6]^2/[.309]
Kb = 5.65*10^-11
ii) Kw = Kb*Ka; Ka = Kw/Kb
Ka = 1*10^-14/5.65*10^-11
Ka = 1.77*10^-4
***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)***
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.
Anyways, pretty sure I got full credit on this. Comments, questions, corrections always welcome. | |
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05-14-2005, 05:44 PM
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#86 | | Member
Join Date: Dec 2004 Location: Williams College
Threads: 105
Posts: 698
| 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 | |
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05-14-2005, 05:48 PM
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#87 | | Member
Join Date: Aug 2004 Location: NJ, USA
Threads: 27
Posts: 501
| 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?? | |
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05-14-2005, 06:59 PM
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#88 | | Junior Member
Join Date: Aug 2004
Threads: 22
Posts: 223
| did anyone do number 2? | |
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05-14-2005, 07:47 PM
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#89 | | Senior Member
Join Date: Oct 2004 Location: Warrenton, VA
Threads: 39
Posts: 1,560
| 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 | |
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05-14-2005, 08:56 PM
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#90 | | Member
Join Date: Nov 2004
Threads: 54
Posts: 635
| Question 3
a. Both are first-order.
b. Rate = k[I-][ClO-]
k = 6.1 x 10^2 L mol^-1 s^-1
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.
Question 4
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) | |
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