how do I prepare a 1.25 CsCl density solution in 10mM Tris and 1mM MgCl2?
83196
I've usually made up a concentrated stock of various buffers and dilute it to the concentration I need.  From a 1M Tris stock it's a 1/100 dilution to 10mM and from a 1M MgCLs it's a 1/1000 dilution.  I would decide how many milliters I wanted to make, measure out the cesium powder then pipette out the amount of each Tris (1/100 of the total solution vulume) and Mg (1/1000 of the volume) then bring to the final volume with distilled water.  This works fine if the concentrated solutions don't interact and cause precipitates or something else like being nearly a saturated solution.
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps
931815
Your KCl example is a good one.  A 0.5 molar would have 1/2 mole/liter or 37.75g KCl in a liter. A 5% w/v KCl would be 50 g/liter; so the 5% is more concentrated.\n\nThe 2.5% w/v sodium bisulfate is 25g/liter and the .025M is 3 g/liter.\n\nIt all depends on the molecular mass.
342141
ok well, the equation is balanced so you need equal moles of Ag and Cl to make the products. They give you the concentration and volume of Ag, so you can find the amount of moles by multiplying 0.104mol/L * 0.0272L = 0.0028288 mols of Ag. so 0.0028288 mols of Cl were there originally. and you have the volume of water, so you divide moles by volume and you get the concentration. 0.0028288 mols/0.025L = 0.113152 M. \n\nanother way to do it is the easy way, use c1v1 = c2v2. you have c1 = 0.104M, v1 = 27.2mL, v2 = 25mL, so you can find c2. you dont have to worry about units in this case so its easier. \n\nhope this helped. let me know if u need more help. laterz
990586
You cant.\nThe key word here being "completely".\n\nNo matter how insoluble a given precipitate is, it is always soluble to some, very minor, degree.\n\nHowever, by choosing to precipitate a compound with a very low, practically zero, solubility, the ion is (for all intensive purposes) removed from solution.\nWhen dealing with such low-low solubilities, the Ksp value for the compound becomes important...the lower the Ksp, the lower the solubility.
607804
The easiest way to prepare an insoluble salt is to produce it from two aqueous solutions each containing an ion present in the salt!\n\nSo, you'll just mix two solutions, and the insoluble salt appears! Of course, if you intend to prepare a certain amount/ mass of the salt, you may need to carry out some calculations to determine the necessary concentrations/ volumes of the 2 initial solutions. \n\nYou'll also get another product in the reaction, but it should be soluble and will not contaminate the desired product (the insoluble salt).  \n\nYou may refer to this website to gain a better understanding...\nhttp://www.wpbschoolhouse.btinternet.co.uk/page01/AqueousChem/AqueousChem.htm#salts [Refer to point (4)]
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
361205
1,000,000 millimoles= 1 mole \ntherefore u'd need 50/1,000,000 * 147kg of calcium chloride \nie. 0.00735kg and since 1000g=1kg u multiply this by 1000 and that gives u 7.35 grams.\nhope this helps... cheers
1038474
http://www.chem.purdue.edu/gchelp/howtosolveit/Solutions/concentrations.html
746627
Cl2 + 2KI --> 2KCl + I2\n\nsingle replacement
843345
I will do the Na+ for you. All the others are similarly done.\n\nin 3 M of NaCl you have 3 mol/liter.  In 3 liters you have 9 mol.\nIf you mix both volumes you will have 3 + 2 = 5 liters\n\nso your Na+ concentration will be 9/5 = 1.8M
95602
2:3:1/2 cement sand water\n2:1:2:1/2  cement gravel sand water
836016
Here is a good guide\n\nhttp://dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry/Limiting-Reagent.html
969868
What parameters do you already have?  I guess you have an a/c weight? The CL equation is a standard with CL= L/(1/2 density*V^2*S)  where S is the Area.
833855
1 cubic centimeter = 1 milliliter.  Take the weight of the item in question in grams and divide it by the density of the material to get equivalent cubic centimeters and then translate to milliliters.\n\nFor water, density is 1g/cc which is the equivalent of 1 g/ml
831841
Let g = grams of salt added.\n\nThere are already 3 g of salt in there.  So there will be 3+g grams of salt when you are done, and there will be 25+g grams of total solution.\n\nWrite a proportion:  (3+g)/(25+g) = 20/100 and solve
446884
to calculate the density you don't need to worry about the temprature and pressure if you already know the volume, as you do.\n\ndensity = mass per unit volume   \n\nit's standard units are g/cm^3\nyou volume is in mL, but that is fine since 1mL=1cm\n\ndensity = 1.6/15000\n=1.07x10^-4 g/cm^3