would 10.00 mL of saturated solution increase the molar solubility and Ksp instead of 20.00 mL? Expain.?
486970
The volume of a solution will have no effect on its Ksp: the only difference is that 20mL of a solution of X will have twice as much X as 10 mL.\n\nOn the other hand, if you DILUTE a saturated solution from 10 mL to 20mL, then a lot of things can change.
304287
This is called Base line correction.\nYour solvent may be having some absorption, or the impurities in it may be having some absorption..so to measure your compunds absorption exactly, you need to take the solvent (n whcih you are going to dissolve your compound later) and do this base line correction.\nIn a way you are telling instruments "NOT TO WORRY ABOUT THESE ABSORPTIONS"\n\nThats it\nYou got it?
1039581
Which method of deaeration are you using?  It makes a big difference.  Definitely less than 0.1 mg/L.
911906
....Sorry, but your question is unclear.  \n\nRephrase?\n\nMethyl Alchohol is soluble in water, so there is usually some slight energy expended as the solute dissolves in the solvent.  But other than that, I can't see what effect could occur related to pH or buffer capacity.
768648
Actually it is not a solubility problem, but a buffer.\n\nLet's use the Henderson-Hasselbalch equation for simplicity\n\npH=pKa+log[CH3COONa]/[CH3COOH]\n\nIf V is the volume of the solution in L then\n[CH3COOH]= mole/V= 0.010/V\n[CH3COONa]= mole/V =x/V\n\nso substitute in the equation and you get\n\n5.37=4.74+ log((x/V)/(0.01/V))=>\n0.63=log(x/0.01) =>\nx/0.01= 10^0.63 =>\nx =0.01*10^0.63= 0.0427 mole =42.7 mmole or since MW=82\nm= 0.0427*82= 3.50 g
528373
It has to do with the "activiation energy" involved in the process, increasing the stirring rate in a certain manner, increases the rate of collisions which increases the rate of reactions.  Solubility has to do with the free energy which are state functions with respect to temperature; the reactant and product states. Rates of reactions pertain more towards transitions states.
900560
Most solutes (sugar in this case) have limited solubility (ability to dissolve) in most solvents (water in this case). This solubility varies with temperature - the hotter the solution, the higher the solubility. As mentioned in an earlier answer, this means more energy must be added to dissolve more solute. On the other hand, if energy is taken out of a solution, the solute will usually  come out of the solution, depending on the concentration.\n\nThis physical law that governs this is the 2nd law of thermodyamics.
709994
If the solution is unsaturated, then more sugar will dissolve in it and raise the ratio of sugar to water.
1167833
sounds like an AP question.  So you are trying to calculate Ksp for silver acetate. \n\nAgC2H3O2 <--> Ag+  + C2H3O2-\n\nSolubility will be darn low so from there set up your ice chart and go to town.
982289
low solubility; im sure
454203
The Solubility Product, Ksp, is defined as (for the hypothetical compound MmAn, which dissolves in water according to the equation:\n\nMmAn (s) ---->  mM(+)(aq) + nA(-)(aq)\n\nis:  Ksp = [M(+)]^m * [A(-)]^n\n\nSo, for the dissolution of Barium fluoride:\n\nBaF2 ---> Ba(+2)  + 2 F(-)\n\nthus Ksp = [Ba]*[F]^2\n\nIf [F] = 1.5E-02, then [Ba] = 1/2*[F], or 7.5E-03\n\nKsp  = [7.5E-03]*[1.5E-02]^2 = 1.7E-06\n\nThe textbook value of Ksp for barium fluoride is 1.84E-07, so if your value for the concentration of fluoride in solution is from an experiment, it is a bit high.
371354
The Ksp for a salt is a constant and should not change in response to solution volume.  However, we must be careful not to forget the definition of Ksp.  The Ksp of an alkaline earth hydroxide that dissociates according to the following reaction is:\n\nX(OH)2 <----> X(+2)  + 2OH(-1)\n\nKsp = [X][OH]^2\n\nIn other words we simply multiply the concentration of the metal ion by the concentration of the hydroxide ion squared, but ONLY IF THE DISSOCIATED IONS ARE IN EQUILIBRIUM WITH SOME UNDISSOCIATED SOLID!!  So there must be some undissolved solid that remains.  If we add so much solvent that the solubility limit of our salt is not reached, and it dissolves entirely, then we can NOT make any measurement of the Ksp.\n\nIn short the answer is:  This will have no affect on the calculated Ksp as long as the additional volume of solvent added did not cause all of the salt to dissolve.
882396
ummmm, maybe increase by a factor of 5 or so. Its an exothermic reaction to begin with and you're not limited by the amount of water( i think, if not not all your drano will dissolve) so it should be about five times the energy released compared to just two grams.
470567
To do this problem we need to do some convertions.  Convertion factors are simply ratios that allow us to change the units of a certial quantity.  For example: the molar concentration (molarity) is a convertion factor that is a ratio of the moles of dissolved substance to the volume of solution (moles/L). Using the concentration  we can convbert moles to liters or liters to moles depending on what we have to start with.  Another convertion factor that we will need is the molar mass of KOH.  This is the ratio of the mass of KOH to the moles of KOH (this is usually measeured in grams per mole (g/mol)).  We can find this number by looking up the molar masses of K, O, and H on a periodic table and adding them together. K =39.0983g/mol, O = 15.9994g/mol, and H = 1.00794g/mol. \nAdding these together we have a molar mass of 56.10564g/mol for KOH.\nSchematicaly this is the road map we will follow to find our answer:\n\n-------(moles/L)-------(g/mol)\nvolume <------> mole <-------> mass\n\nWhere volume, moles, and mass are actuall ammounts and the convertion factors between them are on the arrows.\n\nNow for your specific problem we will start with the given volume 750mL.  First we will convert it from mL to L then convert it to moles of KOH then finally to grams of KOH.\n\n750ml(L/1000mL)(15.0mol KOH/L)(56.1056g KOH/mol KOH)= 631.188g KOH\n\nTo three significant figures our answer is 631g KOH.  This is a big number but a 15.0 M solution is very concentrated so it is not surprizing.
793017
Take 1.875 L of the .2 molar stock solution of NaBr and add .625 L of water to it and you get 2.5 L of a .15 molar solution of NaBr this just use the c1v1=c2v2 equation.\n\nNote:\nThis is the only way it could work. You do not want to add more NaBr because your concentration is already higher then desired.   If you just added 0.5L of water to the 2.0 L of .2 molar stock solution your molarity would still be to high at .16 meaning you have to dilute a smaller portion of your 2L .2 molar stock solution.
921065
you know you have got 152g total. So 1.8g/ml means you have got 152/1.8 for volume.Now just use MW KOH. Molarity will be moles KOH per litre, molality will be moles KOH/kg of solvent and Normality in this case will be the same as molarity but written  with symbol N.\n\nAs an aside the density of this solution would not be anywhere near 1.8g/ml, are you sure you have this right?
684681
C because the solution is SATURATED with solute, meaning no more solute can be dissolved by the solvent in the solution.