what does equilibrant do?
813186
a counterbalancing force or system of forces.
551829
EDTA has very little selectivity so I doubt you will find much more.Masking/demasking agents such as cyanide complexes,phosphate anions.. are nothing else than complexation examples.Another method is called kinetic masking; for example determination of Cr and Fe in steel sample solution (under cool condition,the EDTA will react with Fe first and after you heat the solution the EDTA will react with Cr).The choice of indicator might increase selectivity (different indicators have different reaction rate)but I'm not sure about that one...
423379
let me see if i can at least help you with part of your problem, solubility in chemistry is simple, basically the equation will always be:\n\n[Solubility equalibrium constant] = concentration of product divided by concentration of reagents (concentration of each reagents multiplying by each other).\n\nthe easiest way to understand this is to simplify things for you so as you know, a chemical equation always go like this:\n\nreagents <==> products\n\nJust know this from the above, the dynamic here is that when you have an excess of either the reagents or the product, the balance of the equation tend to shift in the opposite direction (ie excess concentration of product will cause more reagents to form and vice versa), in a chemical equation such as the one from above, the reaction can actually go either way, products or reagents can form.  What really matters is the equalibrium constant that I talked about in the first equation. This constant represent the concentration of the products and the reagents when they are in equailibrium or when they are balance, meaning that the rate at which products or reagent form stays the same and there are no change in the concentration of either the products or the reagents so the whole thing is static at the point.   Think about this, more concentration of products wouldn't form unless you have extra reagents to spare that can shift the balance.  When something is saturated it means that it can longer dissolve in a solvent (let's use water to simplify the explanation)so it just accumulates at the bottom, water molecules can only separate so many ions from each other (the reagents) and preventing them from forming a solid product since water itself has limited molecules (if you mix enough table salt in a glass of water, it'll get to the point where it'll accumulate at the bottom and won't dissolve).  If remember correctly, in a chemical equation, when there's different amount of reagents needed to form the product, there's a slight modification to the equation I gave you, for example:\n\n2 CaCO3 + 4 ZnO2 <----> 3 ZnCO3 + 3 CaO2\n\nthen  [equilibrium constant] = [Cao2]^3 x [ZnCO3]^3 divided by [CaCO3]^2 x [ZnO2]^2\n\nnote that you should balance the chemical equations first, I am just giving an example.\n\nhere's a summary of how to compare numbers, if you plug given concentrations of reagents and product into the equation and the number you get is larger than the real equilibrium constant, then the concentration of products is higher than it should be and the equation will shift towards forming more reagents (in the case of the solid product, it'll dissolve) and if there's more reagents, more solid product is gonna form.  Hope that helps. Remember the Chemistry textbook is still your best reference.
453196
Actually let me explain a bit further.  You have added a know amount of one of the substances.  and you titrate the other against it.  In the presence of your indicator.\n\nNow it doesnt matter how much water you add you still have the same amount of the thing you measured out and so it will take the same amount of titrant to titrate it.\n\nhope that makes sense.
361668
A) When temperature is increased the equilibrium shifts right because the left is is warm an unstable so by shifting right it will balance among both sides.\n\nB) Barium nitrate is 100% soluble in water and will therefore act like the products side of the equilibrium (the reactant is solid). The equilibrium shifts left to balance the amount of barium.\n\nC) When Ba(ClO3)2(s) is added the equilibrium shifts right for a similar reason as the first.
155966
yeah, it means the reaction reaches equilibrium faster
521974
Briefly,\na. The equilibrium constant is constant for a particular reaction, hence when more reagent is added, more products are formed. (Equilibrium constant is (product of conc of products/product of concentration of reactants)\n\nb. Each time reacting species collide, they might not have the right orientation or sufficient energy and hence reaction does not always occur.\n\nc. Increasing the temperature by 10 C nearly doubles the rate of a reaction. This is because the fraction of reacting molecules possessing energy higher than the activation energy increases exponentially with rise in temperature. Also the speed of the molecules increases.
150817
What Is It??\nHello, everyone, and welcome to the wonderful world of chromatography! What is chromatography, you ask?? Well, quite simply, it is a broad range of physical methods used to separate and or to analyze complex mixtures. The components to be separated are distributed between two phases: a stationary phase bed and a mobile phase which percolates through the stationary bed. \n  \n  \nHow Does It Work? Like Magic!\nA mixture of various components enters a chromatography process, and the different components are flushed through the system at different rates. These differential rates of migration as the mixture moves over adsorptive materials provide separation. Repeated sorption/desorption acts that take place during the movement of the sample over the stationary bed determine the rates. The smaller the affinity a molecule has for the stationary phase, the shorter the time spent in a column.
929685
When you have weak electrolytes, (equillibria in general) you should make the following table.\n\n1. write the reaction\n\nHNO2 <=> H+ + NO2-\n\n2.Put the initial amounts (expressed as concentration) of each molecule. In this case we have only the concentration of HNO2, so\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\n\n3.Assume that an unknown amount "x" of the reactants react.\nThen write also how much of each product you expect to be produced according to the stoichiometry. In this case 1 mole of HNO2 gives one mole of H+  and NO2-, so x mole/L HNO2 will  give x mole/L of each and our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\n\n4. Ka is an expression for the concentrations at equilibrium. So let's see what are those. For the reactants you will have initial-what reacted. We have only HNO2 with 0.5-x. For the products it will be the initial+what was produced. We didn't have any intial so it is only what was produced (x for both of the products). Thus the table in its final form is \n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. .. .. x .. .. .x\n\nKa= [H+][NO2-]/[HNO2] = x^2/(0.5-x)\nNow you either solve the quadratic equation and find x, or we assume that 0.5 >>x and practically 0.5-x=0.5.\nThen Ka=x^2/0.5=4*10^-14 =>\nx=squareroot(2*10^-14)= 1.4*10^-7. It is much smaller than 0.5 so our assumption is valid (otherwise we would have to solve the quadratic and get an exact solution).\n\nFor a simple level of chemistry we would now say\npH=-log[H+]= -logx =-log(1.4*10^-7) = 6.85\n\nHowever for more advanced and accurate expectations we see that the H+ coming from HNO2 is found to be in the same order of magnitude as that coming from the self dissociation of water (10^-7 for distilled water). Therefore in this case we cannot ignore the self dissociation of water (as we usually do) for calculating the pH. \nThere are 2 approaches for this case, here is one of them.\nRemember that we didn't put an initial amount for H+ since in most cases we ignore the self dissociation of water. Now we will consider that we had pure water, so the initial concentration of H+ is 10^-7 and we want to see how much the HNO2 will dissociate under these conditions so our table becomes\n\n.. .. .. .. .. .. HNO2 <=> H+ + NO2-\nInitial .. .. .. .. 0.5 .. .. .. 10^-7\nDissociate .. ..x\nProduce .. .. .. .. .. .. .. ..x .. .. .x\nAt Equil. .. .. 0.5-x .. 10^-7+x. .x\n\nand Ka=(10^-7+x)x/(0.5-x)\nNow we have to solve the quadratic so we can't do any simplification\n\n0.5Ka-Kax=10^-7x +x^2 =>\nx^2+ (10^-7+Ka)x-0.5Ka=0\n\nThe only acceptable solution is the positive one\nx=0.5(- (10^-7+Ka)+ squareroot (D))\nwhere D=(10^-7+Ka)^2-4*(-0.5Ka)=\n=(10^-7+4*10^-14)^2-4*(-0.5*4*10^-14)\n=9*10^-14\n\nand x=0.5*(-((10^-7))+4*10^-14) +3*10^-7) =10^-7\n\nso now pH=-log(10^-7+x) =-log(2*10^-7) = 6.70
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5.  K = [NH3][HCl] = (0.066 M)^2 = 4.36 x 10^-3\n\n9.  I'm assuming that equation should be H2(g) + CO2(g) <---> H2O(g) + CO (g)\na.  Add H2O will drive equlibrium to the left\nb.  Removing CO will drive equilibrium to the right\nc.  Adding H2 will drive equilibrium to the right\nd.  Adding something to absorb CO2 will drive equilibrium to the left\n\n6. \na.  Increasing the volume of the reaction vessel will reduce the pressure, favoring the production of more gaseous products, so the equilibrium will go to the right\nb.  Equal number of moles of gas on both sides, so increasing the volume will have no effect.\n\n1.  \na.  Decreasing the volume increases the pressure, favoring the side with fewer moles of gas.  Equilibrium would go to the left.\nb.  Equilibrium would go to the left, as there are fewer moles of gas on that side.
181021
It is a method used to separate a solid organic substance from other solid impurities.\n-The impure solid is dissolved in a suitable hot solvent (to dissolve only the organic solid, not the impurites)\n-The impurities that did not dissolve in the solvent are removed by filtration.\n-The filtrate is left to cool, whereby any soluble impurities remain in the solution and the pure product crystallizes out.\n-The crystals formed are filtered using a vaccum filtration apparatus to help them to dry out quickly.\n-The crystals are washed with the cold solvent to remove impurities on the surface of the crystals.\n-Final drying of the crystals is carried out in a desicator at low pressure.
4764
You can't balance it because you don't know what it yields.\n\n\nP.S. This needs to be in the chemistry category. You really must not know much about chemistry and physics.
63716
It's called the bore evacuator. It softens the backward recoil energy of the gun when it fires by providing an enlarged space for the proppelant gases to expand into.
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if you increase the pressure inside the vessel, which has an opening, vapour will immediately leave thru the opening keeping the pressure inside the same as outside.\n\nIf it was sealed and pressure increased inside, then u opened it, vapour would leave for the same reason to create a pressure equilibrium.\n\nThere is not formulae that i know of but pressure equilibrium is a known phenomenon
1022205
ok.. first off the equilibrium expression is the product of concentration of products over the product  concentration of reactants, with any coefficients taken to the power of the of the coefficient For\n\nA+2B <=> C \n\nthe expression would be [C] /(over) [A] x [B](to the second)\n\n(that's C divided by A times  B-squared)\n\n____________________________________________________\n\nNow, in this particular problems the solutions are given to you. (Btw, in this solution i'm assuming your carbon concentration is carbon dioxide because there is no carbon the equation).\n\nYou should have [H2O] X [CO] /  [H2] X [CO2]\n\nSub in your contrations and you should get: .7864\n\nWARNING: I HAVEN'T DONE THIS IN A LITTLE WHILE... I'M A CHEM MAJOR, BUT I'M NOT GOING TO 100% GUARANTEE THAT
459648
Themes > Science > Chemistry > Miscellenous > Help file Index > Precipitation Equilibrium > Ksp and Ion Concentration \n\nThe solubility product constant can be used to compute the equilibrium concentration of ions in solution in the same way that you use other equilibrium constant expressions to compute equilibrium concentrations. \n\nUse the Initial, Change, Equilibrium method to compute the concentrations: write the equilibrium expression, the initial amounts of each thing in the expression, how much they change by, then the equilibrium amounts. From that, you can compute the concentrations of all species. \n\nExample: BaF2 has a Ksp of 1.8*10-7. What is the concentration of barium in a saturated solution of BaF2? \n\nSolution:. First, write the balanced equation and the solubility constant expression: \n\nBaF2(s) < = > Ba+2(aq) + 2F-(aq) \nKsp = [Ba+2][F-]2\nWe start with no barium or fluoride ions in solution. If we set the amount of barium ions formed to x, then the number of flouride ions formed is just 2x \n\n [Ba+2] (M) [F-] (M) \nInitial 0 0 \nChange +x +2x \nEquilibrium x 2x \n\n\nNow plug in the equilibrium values shown above and solve for x \nKsp = [Ba+2][F-]2 \nKsp = (x)(2x)2 \nx = 3.6*10-3\nSince we defined the concentration of Ba+2 to be x, we've solved the problem \n[Ba+2] = 3.6*10-3 \n\nInformation provided by: http://learn.chem.vt.edu\n \n**try to visit the site below..** You can calculate the Ksp based on the above example..I'm sure the answer is in your books and notes..have fun learning..
506705
Provides inertia in order to stabilize a trajectory.  If an object is still, it tends to stay still.  If moving, it tends to keep moving.  Helps to compensate for the effects of gravity.