CHEMICAL ENGINEERING QUESTION: Binary Mixture- how are bottom and distillate volatilities calculated?
543031
I hate to use this evil word, but you have to use Thermodynamics.  IIRC, the alpha value is k_heavykey / k_lightkey.  Thus, you need the k values for top and bottom.  The k values will depend on temperature and composition.  Recall from thermo that for an ideal solution k_i = y_i / x_i = P_i_sat / P, where y = vapor phase mole fraction and x = liquid phase mole fraction.  So, if you know the total pressure and the vapor pressure at the temperature, you should be able to calculate the alpha values.\n\nIf you use a more complicated solution model, you'll include the fugacity (yes, evil word) as well.
827418
First, you have to calculate the mass of each component by multiplying their volumes by their densities.  You should now have weights for the ethanol and for the water.\nNext, calculate the number of moles of each solvent by dividing those weights by their atomic weights.\n\nNow you can begin answering the questions.  Molarity is defined as the number of moles of a substance per liter of solution.  The total volume of the solution, due to the assumption that the volumes are additive, is 100 ml + 200 ml, giving you 300 ml or 0.3 liters.  So, to get the molarity with respect to ethanol, divide the number of moles of ethanol by 0.3 L.\n\nMolality is the number of moles per kilogram of solvent.  Do exactly the same as above, except this time, you have to calculate the total Kg of solvent instead of the total volume of solvent.  You have to add the mass of ethanol and the mass of water you have calculated from the volumes and the densities.  That's your total mass but don't forget to divide by 1000 to get Kg.  Then you calculate the molality of ethanol by dividing the number of moles of ethanol by the weight of your total solution in Kg.\n\nThe mass percent of ethanol would be the mass of ethanol divided by the total mass of the solution multiplied by 100%.  This is just like any other percentage calculation.\n\nThe mole fraction of ethanol can be calculated by dividing the number of moles of ethanol by the sum of the moles of ethanol and water.\n\nGood luck.
476736
=======\n\ngem is right.\n\nYou will get a mixture of two isomers of bromopropane.
509763
Try using the calculator at the webs site listed below. \nEthanol at 57.5 deg C has a heat of vaporization = 40.23 KJ/mole.\n\nThis web site has calculators for a number of other physical properties.\n\nBest wishes !
816920
Call p1 = p(Cl2), p2 = p(O2)\n\np1 = 0.15 atm, p = p1 + p2 = 0.365 atm\n\n-> p2 = 0.365 - 0.15 = 0.215 atm\n\nV = 0.538 l\n\nT = 273 + 32 = 305 K\n\nR = 0.082 L*atm / K\n\nFor a mixture, all gases share the same V and T:\np2.V = n2.RT -> n2 = p2.V/(R.T) = 0.004625 mol\n\nTherefore, the mass of O2 is:\n0.004625 * 32g = 0.148 g
857152
I'll agree that you need help.  Density is mass/vol.  You've given the density, you need another value to calculate the remaining value.
1113315
The equations are wrong as written.\n12.0Mx V1 = 0.600Mx0.300L   is correct , solve for V1\n\n15%xV1 = 0.10%x0.0200  is correct.,  solve for V1 \n\nNotice that when you multiply concentration times volume the result is an amount.  The units of the amount depend on the units you use for concentration and volume.  In the first case the amount is moles.\n\nSo the equation says  moles1 = moles2  \nThe moles on the left side must equal the moles on the right.\nOtherwise the equation would violate the principle of conservation of matter.\n\nIn the second example, % is strictly speaking not a unit but a label.  The label, in this case, tells you how the number was arrived at.  That is to say, to get % you take a dimensionless fraction (dimensionless because units in the numerator and denominator cancel one another) and multiply it by 100.  Therefore all percents and parts per labels denote dimensionless numbers.  In the second example then, the units on the left and right side of the equation are both Liters.  The amounts on the left and right sides, therefore,  are specified as liters. \n\nYour problem seems to have more to do with algebra than chemistry.  Also, when working numerical problems like these always analyse the units involved.  All the science is in the units.  The numbers are meaningless without units or labels.  Therefore study all equations from this viewpoint and you will soon begin to have a much deeper understanding.
735251
I'm not sure they mix and if they don't I don't know which one floats on the other.\nGet a jar and pour in one then the other. Hope there is a colour difference. If they do not separate then your idea should work.\nIf they do separate it depends on which fuel will be at the level of the fuel pick up in the tank.
845233
Indeed; when different liquids are mixed, they can be separated by this process. Each liquid has its own boiling point. In fractional distillation, the temperature is brought to the temperature of the substance of the lowest boiling point, then distilled off.  The temperature is raised and the process is repeated.  Oil refineries use this process; called cracking, wherein a tall reaction chamber is used and the low boiling point substances are removed from the top and the lowest at the bottom.  Hope this helps.
422875
U have not mentioned the gas gravity to determone deviation factor to have an exact answer...generally..taking z=1\npsia ( absolute value) will also be used \nPsc*Vsc/Tsc=PV/T\n\nso: Vsc/V=PTsc/PscT=53.1/T \n\nThis is the simplest correct answer.... make is easier but worse that the process is isothermal so\n\nVsc/V=102 --> 1cu ft at TVD= 102 SCF\n\nThe most exact solution can be provided if u want..with taking temperatire and deviation,composition in account
215160
http://www-ec.njit.edu/~skawinsk/chem243/chap11/chap11.htm\nthis may help–near the bottom
852232
Pv = nRT\nn = m/M (M Molecular weight)\nPV = (m/M)RT\nP = (m/V)RT/M\nPM/RT = m/v = Density.\nPlug the numbers, watch the units and get the density.\n\nOh a numerical answer, yes, ...42.
569322
Use the ideal gas law \n\nPV/T = constant for the final and initial states.
240972
If you have a system which has, say, A, B, C and D mols of substances 1, 2, 3, and 4, respectively, then:\nmol fraction of substance 1 is A/(A+B+C+D) and so on for the other 3 substances, e.g. for substance 2 it's B/(A+B+C+D)
564798
Ba(NO3)2 + Na2CO3-->BaCO3 +2NaNO3\n\nto do a double displacement follow this formula\n\nAB + CD --> AD + CB
5197
More information is needed.  What is the concentration is it a percentage or a molarity.\n\nDo you need to find the concentration or just the amount of alchohal?
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