How do I calculate osmotic pressure?
823368
Of what?\n\nπ= i*C*R*T\n\nT= absolute temperature (in Kelvin)\nR= the gas constant in whatever units you need to express osmotic pressure (e.g. if you want π in atm then R=0.082 L*atm/(mole*K))\nC= the concentration of your solute in mole/L\n\nand...\n\ni is the van't Hoff coefficient.\nFor non-electrolytes i=1\nFor strong electrolytes i= the number of ions that are produced by the dissociation according to the molecular formula e.g for NaCl you have 2 ions (1 Na+ and 1 Cl-) so i=2, for CaCl2 3 ions (1 Ca+2 and 2 Cl-) so i=3.\nFor weak electrolytes, if n is the number of ions coming from the 100% dissociation according to the molecular formula and a the degree of dissociation then i=(1-a)+na. E.g. if we assume for CH3COOH a=80% i=(1-0.8)+2*0.8= 0.2+1.6=1.8
558840
When you dissolve a salt in water, the freezing point of the resulting solution is lower than the freezing point of the pure solvent (water).\nThe amount of the freezing point depressure depends on two factors, the concentration of the solute (salt) and the freezing point depression constant of the solvent.\n\nThe formula to calculate the freezing point depression in therms of the change in the original freezing point (delta T) is,\n(delta T) = k_f * m\nwhere k_f is the freezing point depression constant and m is the concentration of the solute in terms of molarity.\n\nIt is mathmatically possible for the concentration of the solute to be so greate at to create a change in the freezing point of water of more than 32 degrees F.\n\nThe freezing point depression constant of water is -1.86 degrees C per molality.  So mathmatically, if one had a solute concentration of about 9.57 molality or greater, the freezing point of water would be below 0 degree F.
433864
Use the General Gas Equation:\n\nPV = nRT\n\nwhere\nP = pressure \nV = volume \nn = number of moles\nR = Universal Gas Constant (value depends on units of P,V, n, T eg R = 62.3638 when V is in litres, P is in mmHg, T in Kelvins and n in moles.)\n\nUse the above equation to solve for n. If you want the answer in grams, simply multiply the number of moles by the molar mass of H2.\n\nIf you require a more precise answer, then you should use the Van der Waals equation (a more sophisticated version of the General Gas Equation that takes into account interactions between particles and the volume of those particles).
48128
d = density\nM = molar mass or mass per mol\nV = volume\nn = number of mols\nR = gas constant\nT = temperature\nP = pressure\n\nd=nM/V \n\nV=nRT/P\n\nd=nMP/nRT\n\nd= MP/RT \n= (28x10^-3 kg/mol x 300x10^3 N/m^2)/(8.314 J/(mol K)x450K) \n\n= 2.25 kg/m^3
525014
Let's make this very simple.\n\nTo measure the volume of a cylinder, you find the diameter of the interior portion of the cylinder.  Take that number and divide by two...that's your radius (r).  Now measure the height of the interior of your cylinder...that's your height (h).\n\nNow, take the radius (r) and the height (h) and plug them into the formula below:\n\nV = π*(r)*(r)*(h)\n\n(π equals roughly 3.1516)\n\nNow if your answer is in cubic inches, multiply V by .0036 for British gallons or .00433 for US gallons (yes, they are different).  If V is in cubic centimeters, multiply by .00022 (British) or .00026 (US).\n\nI don't know how to make it any clearer...good luck.
641078
I think this is how you would solve this problem. Use the equation PV=nRT, re-arrange to solve for V. \n\n(STP = standard temperature and pressure = 1 atm of pressure and 273 Kelvins)\n\nV = nRT/P = (1 mol)(0.0821 L*atm/k*mol)(273)/1 atm = 22.4 L
897586
X2O\n\nExamples:\nH2O => Water
898533
Assuming it acts as an ideal gas you can use the equation PV=nRT.\n\nV is in L so 61 mL = 0.061 L\nn = 1.08 * 10-3 mol\nR = 0.0821\nT is in K so 45 C = 318 K\n\nso P (0.061 L) = (1.08*10-3 mols)(0.8021) (318 K)\n\nP = 4.52 atm \nNext convert atm to torr   1 atm = 760 torr\nso multiply 4.52 by 760\n\nP = 3432 torr
22880
There's a formula for volumes of gasses:\n\nPV = nRT.\n\nIf I remember right, P = pressure, V = volume, n = number of moles, T = temperature, and R = molar density, or something molar that is specific to the gas.
268708
Wow, this question is harder than I thought lol, I thought it would just be a plug and chug but its not u actually have to think!  I havent quite figured it out but I will give you a start:)  Maybe I'll come back to it.  \n\nSo the formula is: \nOsmotic Pressure in atm. = Molarity ( R ) (Kelvin Temp.)\n\nOsmotic Pressure = 3.05 atm\n\nR = 0.0821 liter-atm/mole kelvin \n\nConvert the temperature from Celsius to Kelvin. \nK = 10 + 273 = 283 K \n\nWe dont have the Molarity so first find that, then since we know Molarity = (g/ molecular weight)/liter and in the problem we have the relationship 4g/liter and 2g/liter.  Figure it out!:)\n\nGood luck!
877652
These are GAS LAW problems, so you need to use PV=nRT.  However, for these problems, nRT will be constant so you only need to use P(pressure) and V(volume).  Therefore, for both of these questions, use the equation:\nP1V1 + P2V2 = PtVt   where P1 is the pressure in the first vessel, P2 is the pressure in the second vessel, and Pt is the total pressure of the connected system.\n\nProblem #1\nP1V1 + P2V2 = PtVt  \n(7.0atm)(15.0L) + (P2)(8.4L) = (10.0atm)(23.3L)  \nP2 = 15.4 atm\n\nProblem#2\n(7.2atm)(15.0L) + (4.1atm)(8.3L) = (Pt)(23.3L)  \nPt = 6.1 atm
34599
Osmotic pressure or turgor (also called turgor pressure) is the pressure produced by a solution in a space that is enclosed by a differentially permeable membrane.\n\nWhen a biological cell is in a hypotonic environment (the cell interior contains a lower concentration of water than its exterior), water flows across the cell membrane into the cell, causing it to expand. The membrane (or, in plant cells, the cell wall) restricts the expansion, which causes an increase in pressure. The resulting pressure is called turgor. This pressure is what prevents more water from flowing into the cell, thus creating a pressure equilibrium between water flowing down the concentration gradient and the taut membrane pushing back. In this example, the equilibrium prevents the cell from ever becoming isotonic to its environment. Cells not adapted to hypotonic environments, with the flow of water into them but no strong membrane or cell wall, will burst. The osmotic pressure π of a dilute solution can be calculated using the formula\n\n pi = MRT\n \nwhere\n\nM is the molarity \nR is the gas constant \nT is the absolute temperature \nNote the similarity of the above formula to the ideal gas law, and also that osmotic pressure is not dependant on particle charge.
860365
use the equation  pH= -lgKa+ lg( salt/acid)\n(this equation only applies for buffer system)
1172866
this can be a problem to you if you don't know anything about simultaneous equation or if u don't know how to form simultaneous equation. another thing is u should at  least have an  idea about chemistry. look\n  \nm1v1+m2v2=MV.............1\nm1+m2=total volume after dilution=150Lt..............2\n now\n              take m1 is the initial conc of saline soln=32%=0.32\n                   ''  v1                      volume                 =?\n                 m2 is the conc of water =0\n                 v2 is the volume of water =?\n               M is the final conc of the soln = 20%=0.20\n                V is the final volume of the soln obtained after mixing=150Lt\n\nnow\n             0.32v1+0*v2=0.2*150......................1\n\n             v1+v2=150.................2\nfrom eqn 1\n                      0.32v1=30\n           so v1= 93.75Lt's\n   so the remaining amount is of water\n                v2=150-93.75=56.25lts\nso you need 56.25litres of water and 93.75litres of 32%saline to prepare the solution you need.\n\nif you have these solution just try to mix at that proportion then u will tell me.
265173
You are correct using the formula:\n\nC1*V1 = C2 * V2\n\nwhere C is the concentration and V is the volume.\n\n\nNow, you know that you want a concentration of 0.39M solution of NaCl in a 100 ml sample.\n\nSo, the equation should read:\n\n0.82 * V1 = 0.39 * .1 L\n\n0.82*V1 = 0.039\n\nV1 = .047L\n\nso, take 47 ml of the .82M NaCl and dilute it up to 100 mL.  That will give you the 0.39 M NaCl.
894614
PV=nRT\nP= (nRT)/V  (1st eq)\n\nn,R,T u know.\n\nV >>>\nDensity = mass/ volume\n2.3 = .00446 *10^3/Volume\n\nrearrange and ull get volume.............and then put that into the first equation.............and voila
444216
Write the Kc and the Kp\n\nChange PV=nRT to P/RT=(n/V) to P/RT=(concentration) note n/V is concentration.\n\n\nSubstitute P/RT for each concentration in the Kc.  You will see the Kp in the equation with extra R's and T's.  Finaih solving.