1pt  9floz. & then 2pt 10 fl oz from a container of film developer that holds 3qt How much of the dev. remains
411953
in the UK 20 fl. oz. to the pt. 2pts. to the qt.\n\nso adding up gives 3pts 19fl. oz. there are 3qts - 6pts in the container\n\nso the answer is 2pts 1 fl. oz.
634933
Density and phase 0.626 g/cm3, liquid
780066
V = nRT/P\nV = (25g/44g/mol)(0.08206Latm/molK)(298K)/1atm\nV = 138.9 L
750075
Ask a pharmacist...it would depend on what was in the formula
143234
I think you at least need to know the density of pure acetic acid to solve your problem. We can assume density of water at 1.0 g/mL.\nIf we assume that the density of your 40% w/v solution is 1.0 g/mL then:\n\n    V = volume of 40% solution required would be: \n\n        V x .40  = 0.0225 x 250 \n\n  V = 14.1 mL \n\nYou place the 14.1 mL in a flask and add water to complete 250 mL and voila... You are done. \n\nLet me know if this is wrong.
261798
The aqueous solubility of AgNO3 at 20C is 67.8 (solubility is 100*m2/(m1+m2) where m2 = mass of solute and m1 = mass of solvent).\n\nRearraging the equation gives us m2 = (m1*67.8/100)/(1-67.8/100).  or\nm2 = (m1*.678)/(1-.678) = 210.56g of AgNO3 @ 20C\n\nAt 50C the aqueous solubility of AgNO3 is 79.2 so\nm2 = (m1*.792)/(1-.792) = 380.77g of AgNO3 @ 50C\n\nTo find the amount needed to make the second solution saturated we must subtract answer 1 from answer 2.\n\nWe find that 170.21g of AgNO3 must be added to make the solution saturated at 50C.
245982
Sorry, can't help. You use different units than I do.\n but concentration(mol per dm3) = no. of moles(mol) divided by volume(dm3)
741771
0.5M = 500mM\nv x c=v'xc'\nv x 500 = 50 x 1\n0.1mL of the 0.5M solution diluted to 50mL will be 1mM
279761
i'm going to take a guess - 2 oz.
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.
872765
It won't matter how much you started with or how much higher the solution was heated.  When the solution cools to 20C it will hold 0.22g/100mL of solvent (water).\n\nSo if you start with 1.75g and 0.22g remain in solution, the amount crystallized at 20C will be the difference, 1.53g.
69568
None, unless you pour something into it, then it is a measure of 1.5 oz. \n\nHope it is something yummy.
820519
First write the balanced chemical equation:\n\nPb(NO3)2 + 2NaBr --->  PbBr2  +  2NaNO3 \n\nThen, find the total volume of the solution: 200mL + 200mL = 400mL \n\nFrom the density, the mass is:   density = mass/volume   mass = density * volume = 1.0 g/mL * 400 mL = 400 g\n\nQ = mC(Tf - Ti) ; where Q = heat released, m = mass of solution,, C = specific heat capacity, and (Tf - Ti) = change in temperature\n\nSo,  Q = 400g * 4.2 J/gK * (2.44 + 273)  (Since C is J/gKelvin, the temperature rise also has to be in Kelvin)\n\nQ = 462.5 KJoules\n\nHope this helps...
955266
1.3-.9=.4kg for half a bottle of oil no bottle so 1 bottle of oil with no bottle should wiegh .8kg so the bottle should wiegh.5kg
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.
676978
you can only dilute it w/v as it is a solid. Just take 10gr and dissolve it in about 50ml distilled water, then make up to 100ml when it has cooled.
410035
good news...10.63 is correct. Now how to get there.  \nMorp + H20 ---> Morp+ + OH-\ninitial concentration of Morp=.115M  everything else is 0.\nEquilibrium concentration of Morp=.115M-X, H20 ignore, Morp+=X, OH-=X.   Set up your equilibrium equation.  Kb=[OH-][Morp+]/[Morp]  pKb=-log Kb  so Kb=1.62e-6.  1.62e-6=[X][X]/[.115-X]  We can ignore the X subtracted from .115 since it is very small when compared.  so solving for x we finally get X=4.32e-4 which is the OH- concentration.  pOH=-log OH-  pOH = 3.36.  pH = 14-pOH .....pH=10.63.