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My version, versus, another version, interchangeable? Oct. 24th, 2007 @ 10:35 pm

Hello All,
I'm in a college Chem 2 class and I've just recently taken a quiz.  The problem is, is that- I ulitmately got the correct answer, however, the approach of solving this problem is incorrect.  I was marked off points in the structure of solving this problem, however, my prof. didn't have time to go over the quiz with me and I want to be sure that my version, is theoretically and quantitavely, correct, and somehow my version can be interchangeable with the prof. version.  Any help, explanation, would be great.  Thank you for your time:

The Question:

A solution is prepared to be .20M HC2H3O2 and .4M KC2H3O2.  What is the PH of this solution? (Ka= 1.7 x 10^ -5 for acetic acid)

My version of solving this problem:

KC2H3O2 >>>>>  K  +   C2H3O2                                                                C2H3O2 + H20>>>>>HC2H3O2 +  OH
      0.4                    0.4          0.4                                                                          0.4            ----                  0.2               0

INTIAL:                           0.4          0.2              0
CHANGE:                      -x             +x              +x
EQUIL:                        0.4-x        0.2+x            x   

Kb=     Kw/Ka  =    1.00 x 10^ -14/ 1.7x 10^-5  =   5.88 x 10^ -10
Kb=    [HC2H3O2] [OH] /  [C2H3O2]

5.88 x 10^-10 = (0.2 + x) (x) /  (0.4 - x)

2.35 x 10 ^-10 = 0.2x

x= 1.18 x 10^ -9

-log(1.18x 10- -9) =   8.93  =  POH           PH= 14- 8.93 = 5.07


HC2H3O2 + H2O >>>>>>  H3O + C2H3O2
     0.2             ----                      0            0.4

INTIAL:               0.2         0          0.4
CHANGE:           -x          +x          +x 
EQUIL:            0.2-x         x            0.4+x

Ka=    [H3O] [C2H3O2] / [HC2H3O2]

1.7 x 10^ -5 =   (x) (0.4 + x) / (0.2 -x)

8.5 x 10^-6 = x = [H]

-log(8.5 x 10^ -6) = 5.07 = PH 

Conversions Dec. 24th, 2006 @ 06:35 pm
I'm taking chem for the 1st time. So this question might be stupid but. I need steps written out for me on how to solve these problems and what not. My teacher just went over it quickly. I need to have the steps written out for me so I can do problems myself. Here are some of the problems, I have a quiz on it tommrow any help would be wonderful. Thank you.

1. 270mm = m
2. 1.2km = cm
3. 3420mg = kg
4. 720ml = l
5. 2.4 l = dl
6. 8qts = ml
7. 10.0 lbs= g

Jan. 19th, 2007 @ 11:33 am
oh i hope someone out there knows something about structural isomers. i am having trouble finding the isomers of heptene. there seems to be so many, but i don't know if i am doing it right. i think i've found like 8 so far, but my brain it telling me no. i also have to find out if any of them can exzibit cis/trans isomers. help please!
Currently feeling: frustratedfrustrated

quanti-frig-ization Sep. 20th, 2005 @ 05:44 pm
The existence of discrete (quantized) energy levels in an atom may be inferred from:

a. visible spectrum
b. experiments on the photoelectric effect
c. atomic line spectra
d. diffraction of electrons by crystals
e. x-ray diffraction by crystals


thanks much

Solutions Feb. 9th, 2005 @ 08:35 pm
This one is stumping me because we haven't covered it yet but somehow, it got into the packet.

You dissolve 2.0 g of solid MX in 250. g of water. You find the freezing point to be 0.028 degrees C. Calculate the Ksp of the solid.
A) 5.6 x 10^-5
B) 7.1 x 10^-7
C) 2.2 x 10^-4
D) 6.3 x 10^-5
E) none of these

My gut inclinations are towards B or E, but I'm really not sure. Any help would be greatly appreciated.

hello! Dec. 30th, 2004 @ 01:03 am
When given a problem like Cu + H2O(g)-> how do I solve it? Is there a process to find out what type of reaction it is? I also had the problem of balancing correctly, but the reaction I wrote couldn't happen in real life (e.g.: I thought KClO3-> became 2KClO3->K2+2ClO3, so I got it wrong because i found out that never happens in nature). Which rules would help me avoid that from happening again? even some good links would be greatly appreciated, because I'm so confused! Please help me! :(

Here's some examples from a packet that could be used if needed. the answers are below, but i couldn't understand how to be able to find them on a test:

Write correctly balanced equations for these expressions:
1) KClO3 ->
Ans: 2KClO3 -> 2KCl+3O2

2) KBr + MnO2 + H2SO4 ->
Ans: 2KBr + MnO2 + 2H2SO4 -> 2K^(+1) + Br2 + Mn^(2+) + 2H2O + 2SO4^(2-)

3) AgCl(s) + NH3(aq)
Ans: AgCl(s) + 2NH3(aq) -> Ag(NH3)2^(+1) + Cl^(-1)(aq)

4) Cu + H2O(g)
Ans: Cu + H2O(g) -> Cu^(2+)(aq)
Currently feeling: confusedconfused!!

I hope someone still reads this.. Oct. 13th, 2004 @ 10:23 pm
OK, basic chem question.

Aspirin is prepared by the reaction of salicylic acid(C7H6O3) with acetic anhydride(C4H6O3) according to the following equation:

C7H6O3 + C4H6O3 ----> C9H8O4 + CH3CO2H

How many grams of acetic anhydride are needed to react with 4.50 grams of salicylic acid? How many grams of aspirin are formed? How many grams of acetic acid are formed as a by-product?

I found the weight of acetic anhydride(3.32 grams) but I cannot for the life of me figure out how to find the weights of the aspirin and the acetic acid. Any help?

Jul. 2nd, 2004 @ 12:00 am
I cannot figure these two problems out for the life of me, so I'm hoping someone here might be able to help.

Compare the amount of cooling experienced by an individual who drinks 400.0 mL of water (0°C) with the amount of cooling experienced by an individual who sweats out 400.0 mL of water. Assume all sweat evaporates. The density of water is 1.00 g/mL, the specific heat of water is 4.18 J/g°C and the heat of vaporization of water is 2.41 kJ/g at 37°C.

9. A 1.80-gram mixture of potassium chlorate, KClO3, and potassium chloride, KCl, was heated until all of the KClO3 had decomposed. The liberated oxygen, after drying, occupied 405 milliliters at 25.0°C when the barometric pressure was 745 torr.

a. How many moles of O2 were produced?
b. What percent of the mixture was KClO3? KCl? (I can do the calculations for these, I just need help with writing out the reaction)

success! Jun. 8th, 2004 @ 04:51 pm
hi all, i would like to say that this round of postings has been quite a success. i would like to thank you all for making use of each other’s knowledge; in particular i would like to thank those that have helped others: masella, tarheelzz, also dannnnnnnnnnn. it is only a shame that this didn’t start earlier.
keep in mind that while the year is winding down, any further questions you have can still be posted here. in addition, the few of you taking AP Chem next year are welcome to use the board (next year) as well.
Currently feeling: happyhappy

#11 Jun. 6th, 2004 @ 10:34 pm
my good man, scroll down the page and look for a #11 how-to-do explanation by paul. it's there. you can't miss it.
Currently feeling: complacentcomplacent
Currently playing: Very Best of Jethro Tull
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