what does this solution make....?
613688
What you have as your reactant is really just expressing a complicated equilibrium for the sodium salt of a carboxylic acid.  You have calcium ion in your products, but not your reactants, which is a no-no.  Having said that, it looks like you're replacing the sodium ion with calcium ion as the cation associated with carboxylate anion.  Separating all of the reactants from one another won't be easy, but you could try adding calcium hydroxide.  That should precipitate sodium and calcium carbonate.
290716
Unless a equation is properly balanced the amount of product and reactant will not be the same, I suppose this is like a recipe where u get the proper number of ingredients(moles) and mix to together to give the you the proper number of products.
436191
Uh, the rest of the question isn't appearing. But it's an equilibrium solubility problem. Write a reaction:\n\nMgF2 --> Mg+2 + 2F-\n\nKsp = 5.52e-9\n\nFill in the values on the ICE table (I don't know them, the numbers are missing), and set an equation up.\n\nKsp = [F-]^2 * [Mg+2]\n\nSolve with the numbers from the ICE table. Remember that those concentrations in the above equation are the equilibrated concentrations, not the initial.\n\n\nHope this helped.
500934
Combine 19 parts of solvent with 1 part solute.  Boom, 5% solution.  \n\nIf you have an already diluted substance, just figure out the ratio and try to make it 1/20.  For example, if you have a 50% solution of NaCl and water, add two parts (2*50% is 100% or 1 part) of it to 18 parts water to make a 5% NaCl and H20 solution.
959197
I have factored it please refer to.\n\nFrom this solution can be found
905059
It seems that the tripolyphosphate hydrolises to the orthophosphate\n\nThis is the only way I could see it balance BUT I am guessing and balancing ...\n\n3Na5P3O10 + 6H20 → 5Na3PO4 + 4H3PO4
974357
Yes.  Solutions are made when two a substance (solute) is dissolved in another substance (solvent).  In this case, it is a homogeneous mixture since it is consistent throughout, otherwise, it would be called a heterogeneous mixture.
624341
A solution that contains an equivalent amount of an acid and its conjugate base.
932271
yields "H2 + Zn2O3"\n\n^^!
611075
I'm assuming you are extracting from an aqueous solvent?  Preferably, you want to increase the polarity of the aqueous solution, such that the solute has less affinity for water than it would for the organic solvent.  This can be done by "salting" the water (increasing the ionic strength by adding an inert salt, a la NaCl).  You then extract the solute with your organic solvent of choice, usually using smaller aliqouts more frequently vs. larger aliqouts (deals with partition coefficient, or how much will get extracted vs. how much remains in the water).  After extraction, the organic solution is usually dried (more often than not by MgSO4 or Na2SO4) and then the solvent is distilled off to yield hopefully pure product.
712818
more soluble in what solution?
781038
I agree with birchardv here...........we need the rest of the problem, and you would learn a lot more if you did the homework yourself.  But do whatever floats your boat.  Add the rest of the problem and I'll get back to you.\n\nOkay Chemist, this problem is really long; if you want a good answer with plenty of steps, I'll need more room.  Email me and I'll send you the solution.
968613
If you want to make 1 Liter of this final solution, you would need to take 721 mL of your 52% formaldehyde solution, add 120 mL of methanol to it and bring the final volume up to 1 Liter with water.  You should be able to scale this up or down to make any particular volume you need.\n\nWhat isn't coming out right in your calculations?
805565
Ca + 2H2O yields Ca(OH)2 + H2\n\nBasically it's almost balanced already, just add a 2 in front of the H2O.
1140664
You're missing a unit. Do you mean 2.5 kg solution?
174475
mix it...let NaCl dissolve in water...then let the mmixture evaporate or perhaps bring the mixture into a boil 'til you can see the formation of crystals...   try it... i've encountered such experiment in my chemistry class... im not quite certain of the procedures though...it's two years ago...cant remember it well...^_^
737332
It's unclear what that equation says.