Equilibrium Concentration question?
Use ICE table -----------------I2----------- = ---------2I Initial----0.00854 M Change------(-x)--------------------(+... Equilibrium-(0.00854-x)-----------2x K = [I]^2/[I2]= (2x)^2/(0.00854-x) = 4x^2/(0.00854-x) = 3.76 x 10^-3 You cannot use approximation here since initial concentration/K is not greater than K. SO you will use the quadratic equation: 3.76 x 10^-3 = 4x^2/(0.00854-x) (3.21104 x 10^-5) - (3.76 x 10^-3)x = 4x^2 4x^2 + (3.76 x 10^-3)x - (3.21104 x 10^-5) = 0 --> Solve for x using the Equation Mode of your calculator or by solving manually using quadratic equation x = 2.4 x 10^-3 M At equilibrium [I2] = 0.00854 - x = 0.00854 - (2.4 x 10^-3) = 6.14 x 10^-3 M [I] = 2x = 2(2.4 x 10^-3) = 4.2 x 10^-3 M
Find the concentration of iodine remaining in the aqueous layer after extraction?
The distribution coefficient for iodine between an organic solvent and water is 85. Find the concentration of iodine remaining in the aqueous layer after extraction of 50.0 ml of 1.00 x 10-3 M iodine with the following quantities of the organic solvent: a. Single extraction with 50.0 ml of the organic solvent. b. Two successive extraction with 25.0-ml portions of the organic solvent. c. Five successive extration with 10.0-ml portions of the organic solvent. What is the percentage of iodine extracted?
The amount of I3–(aq) in a solution can be determined by titration?
The amount of I3–(aq) in a solution can be determined by titration with a solution containing a known concentration of S2O32–(aq) (thiosulfate ion). The determination is based on the net ionic equation: 2S2O3{-} (aq) + I3{-} (aq) → S4O6{2-} (aq) + 3I{-} (aq) Given that it requires 43.0 mL of 0.330 M Na2S2O3(aq) to titrate a 30.0-mL sample of I3–(aq), calculate the molarity of I3–(aq) in the solution. I'm so lost I don't even know where to start with this.
When an aqueous iodine solution is shaken with an equal volume of tetrachloromethane?
Yes, you are correct in that both iodine and CCl4 are non-polar substances, and so the iodine will readily dissolve in the CCl4 ("like dissolves like.") However, becuase CCl4 is non-polar, it is not miscible (doesn't mix with) water, which is polar. So it sort of looks like oil and water...two layers form, one on top of the other. In this particular case, the CCl4 layer with the iodine dissolved in it will be the bottom layer because CCl4 is considerably denser than water (1.59 g/mL vs. 1.00 g/mL for water). Does that help?
What is the colour of copper sulphate solution after dipping magnesium ribbon in it?
Another homework question?Aqueous copper sulfate solution is blue from the colour of the hydrated Cu++ cations and since the sulfate anion is colourless. The colour of the copper sulfate does not change during the reaction with magnesium metal.Explanation:IF, there is sufficient Mg ribbon and IF the reaction is allowed to complete, then there would be a displacement reaction in which aqueous Mg++ cations, which are colourless, replace the aqueous Cu++ cations, which are blue, in solution whilst simultaneously, copper cations are reduced to copper metal and the magnesium metal is oxidised to magnesium cations.Between the two extremes, while the reaction proceeds, the intensity of the blue colour from the aqueous Cu++ solution will fade (the intensity is proportional to the concentration of aqueous Cu++ cations remaining in solution), but the colour (that is the specific wavelengths of light reflected from the aqueous Cu++ cations - or to be more precise, the wavelengths of light not absorbed by the aqueous Cu++ cations) will remain exactly the same. The colour is constant, but the intensity decreases making the colour fade, NOT change. If the colour was not constant, then the science behind spectroscopy would be pointless. Those with chemistry degrees should know better, but this is a common mistake, like using mass and weight as though they were the same thing.This is an example of imprecise language used in chemistry. However, I acknowledge that when learning the subject some approximations need to be made in order for pupils to have some sort of foundation. Teachers should always emphasise the correct science using the correct words (they don’t - even in universities) otherwise the pupils find the ambiguities make the subject more confusing than it really is and they miss the opportunity to learn how to express themselves precisely.Clarity of language is essential to avoid confusion and ambiguity. Science is an excellent way of teaching this precision of words - providing it is done properly. The unambiguous use of language is one of the reasons that science is a core subject in the school curriculum - not simply to teach scientific facts / concepts.
What is the colour of iodine and iodide ion?
Iodine is a dark purpley-black solid at room temperature. It is purple in a hydrocarbon solvent. It is yellowish-brown in water The dependence of the colour of iodine solutions upon the nature of the solvent is generally believed to be due to the presence of loosely bound iodine-solvent complexes, or 'solvates', in the brown solutions (for example, in ethanol) and of 'unsolvated' diatomic iodine molecules in the violet solutions. a solvation serve to account for the marked difference in chemical properties between the violet and brown solutions AndIodide ions are colourless, so for example a solution of sodium iodide is colourless. Iodide solutions will only be coloured if the positive counterion were to be coloured.
Why does the color of iodine solution turn blue when starch is added to it?
I hope this will help :What properties of starch (given its chemical structure) allow it to be used as an indicator? Davender Khera, Yale UniversityWhen starch is mixed with iodine in water, an intensely colored starch/iodine complex is formed. Many of the details of the reaction are still unknown. But it seems that the iodine (in the form of I5- ions) gets stuck in the coils of beta amylose molecules (beta amylose is a soluble starch). The starch forces the iodine atoms into a linear arrangement in the central groove of the amylose coil. There is some transfer of charge between the starch and the iodine. That changes the way electrons are confined, and so, changes spacing of the energy levels. The iodine/starch complex has energy level spacings that are just so for absorbing visible light- giving the complex its intense blue color.The complex is very useful for indicating redox titrations that involve iodine because the color change is very sharp. It can also be used as a general redox indicator: when there is excess oxidizing agent, the complex is blue; when there is excess reducing agent, the I5- breaks up into iodine and iodide and the color disappears.BTW...there is something known as GOOGLE. Next time you have such kind of doubts, please feel free to use it
Chlorine water + Potassium iodide + cyclohexane?
i previously asked a question similar to this, but my results were quite different to the answers given. after carrying out this reaction I was left with a multi-coloured solution. it was yellow at the bottom and pink/ light purple at the top. am i correct in thinking that the chlorine has displaced the iodine to form potassium chloride and iodine? if so would this mean that the yellow colour is the potassium Chloride and the pink is the iodine? what confuses me is the addition of cyclohexane. does it partake in any of the reaction or does it simply act as a way of splitting the solution in two, in order to see the colours produced?