Submitted by HarrisDi on Wed, 20141022 19:57
3 I2 + 6 F2 2 IF5 + I4F2
An equilibrium mixture in a 1.00L flask of the above treaction contains the following gases: 0.13M of I2, 0.26M of F2, 0.21M IF5, and 0.38M I4F2. To this mixture, some iodine was added. When equilibrium reestablished, 7.6g of F2 was found to be present. What mass of I2 was added to the system?
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Submitted by Liberty on Sun, 20141012 13:49
The molar solubility of CoCO3 in a 0.10 M Na2CO3 solution is 1.0 x 109 mol/L. What is Ksp for CoCO3? . This is my first time to encounter this question and I am stuck.
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Submitted by equilibrium2014 on Sun, 20141005 18:42
I performed a lab for the following equilibrium but there were some questions I could really use some help on explaining events in the lab through Le Chateliers Principle.
The Equilibrium Equation: CoCl4^2 + 6H20 <=> (Co(H20)6)^+2 + (4Cl)^1
1. I placed the test tube of the equilibrium solution in a hot water bath, followed by a cold water bath. The original blue solution became a darker blue in the hot water (shifted left) and it turned a lighter blue in the cold water (shifted right). As heat is exiting in this reaction I said it was an exothermic reaction. I am asked to explain my observations using Le Chatelier's Principle but I'm not too sure how I should exactly do that.
The Principle: An equilibrium system subjected to a stress will shift to partially alleviate the stress and restore equilibrium.
2. I then had to add some AgNO3 (Silver Nitrate) to the equilibrium solution. It became pink. How do I explain this too using Le Chatelier's Principle? How did this affect the concentration changes of the solution?
3. I then added some CaCl2 (solid) to the equilibrium before turning blue. How do I explain the colour change in relation to Le Chatelier's Principle? How did this affect the concentration changes of the solution?
4. I then added some water. It turned from a blue colour to a pink and then almost clear. How do I explain this color change using Le Chatelier's Principle? How did this affect the concentration changes of the solution?
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Submitted by Batman17 on Wed, 20140924 21:40
Is this question actually solvable? It was on a quiz I just took and I couldn't figure it out...
Equation is 2I_{2}+2H_{2}S <> 4HI+2S_{2}. Given: K_{C}=2.33x10^{5}, 0.250 mol HI in a 2L flask. Same with S_{2}. Want: Equilibrium concentration of I_{2}.
So I set up the ice table after putting everything in molarity.
2I_{2 }+ 2H_{2}S <> 4HI + S_{2}
0 0 .125 .125
+2x +2x 4x x
2x 2x .1254x .125x
Plugged it into ratios:
[HI]^{4}[S_{2}] / [I_{2}]^{2}[H_{2}S]^{2} = ((.1254x)^{4}(.125x)) / (16x^{4}) = 2.33x10^{5} I assumed the 4x and x in the numerator would be small so that changed it to (.125^{5})/(16x^{4}). Plugging and chugging, I end up with a value of somewhere near 0.54 = x. So 2x= 1.06. Now, that just didn't look right, but I couldn't figure how else to do this. Any help?
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Submitted by wnterhorst on Sat, 20140920 11:20
Calculate [N2O4] at equilibrium when [NO2]=2.00×10−2mol/L.?
For the reaction N2O4(g)⇌2NO2(g), the value of K at 25∘C is 7.19×10−3.
i would like to see the math behind it so i can understand it better.
thanks in advance
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Submitted by Chemslayer on Thu, 20140731 17:19
Ammonia decomposes at high temperatures. In an experiment to explore this behavior, 3.00 moles of gaseous NH3 were sealed in a rigid 1.50 L vessel. The vessel is heated at 800K and some of the NH3 decomposes in the following reaction: NH3 (g) > N2 (g) + 3H2 (g)
The system eventually reaches equilibrium and is to contain 2.18 moles of NH3. What are the values of Kc and Kp for this reaction at 700K.
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Submitted by zig5151 on Sun, 20140727 17:09
A solution was prepared such that the initial concentrations of Cu+2 and CN were 0.0120 M and 0.0400 M, respectively. these ions react according to the following equation  Cu2+ + 4CN > Cd(CN)4 ^2. Kc = 1.0 x 10^25. What will be the concentration of CN at equilibrium?
I made an ICE table and plugged everything into Kc,
so 1.0 x 10^25 = x / (0.0120  x)((.044x)^4), is that right so far?
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Submitted by Spinnerphoenix on Fri, 20140418 11:29
Mixtures of propan2ol and propanone can be spearated by distllation due to their different boiling points. Explain why these compounds have such different boiling points even though they have very similar molar masses.

2propanol

Propanone

Boiling point(°C)

82

56

Molar mass(g/mol)

60

58

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Submitted by roshani on Fri, 20140411 07:25
when we deal with the raoult's law equation for an ideal solution of 2 ideal liquids which are volatile, do we have to use the number of moles at equilibrium to calculate the mole fraction of solvent in the solution? or can we use the initial number of moles ?
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Submitted by Spinnerphoenix on Mon, 20140407 11:10
Does production of a precipitate decrease concentration?
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