how can you determin the rate of diffusion f 2 gases?
155279
By measuring the change in pressure.
562274
Diffusion of a Dye in Hot and Cold Water\n\nPurpose:  The purpose of this experiment is to demonstrate the principle of diffusion.  Diffusion is the net movement of particles (atoms, ions, or molecules) from an area of greater concentration to areas of lesser concentration.\n\nMaterials and Methods:  For this experiment, the student will require a projector, two petri dishes, two 150 ml beakers, hot plate, potassium permanganate (purple dye), and a spatula.  Obtain about 50 ml of cold water in one beaker and about 50 ml or boiling water in the other.  Place the petri dishes on the projector and add cold water to one dish and boiling water to the other.  Allow the water in the dishes to become still.  Using the spatula, add a crystal of potassium permanganate to the center of each dish\n\nExperiment:  Observe the experiment for several minutes and note what happens to the purple dye in each petri dish.\n\nConclusion:  Do you observe diffusion in each?\n\nHow does the rate of diffusion differ? \n\nWhy is there a greater rate of diffusion in one
921703
Just let the units tell you what to do.  H=mass* T*sp heat.  with T in K
355529
V(gas)=0,2 m3,T1=333 K,T2=533 K,V2=?\nidel gas law: pV=nRT,p=const. so nRT1/V1=nRT2/V2\nT1/V1=T2/V2, V2=T2V1/T1\nV2=533 K x 0,2 m3 /333 K =0.32 m3
955797
Use the Arrhenius equation to predict the rate:\n\nhttp://en.wikipedia.org/wiki/Arrhenius_equation\n\nK = Ae -E/RT\n\nIt all boils down to your reactants, the activation of energy, and the product that is being synthesized.
48084
F = ma    \nNewton's second law.
936208
Geez, that's a beast of an assignment if you haven't even heard of these laws before.\n\nNone of these laws are recent, so any good encyclopedia should have lots of information about them.  Even Wikipedia, but always take wikipedia with a grain of salt, since it's nonprofessional and many professors don't accept citations from it.\n\nThe gas laws should be pretty well covered in an elementary chemistry textbook.  I really can't recall if such a book would cover Fickian diffusion, but Dalton's law and Henry's law should be in there as well.  If you have a chapter in your course textbook about blood chemistry, that might be a place to look, as Dalton's law deals with the partial pressure of gases, and Henry's law with the solubility of gases in liquids.  Fick's law would certainly be covered in any references about the lungs.
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.
888307
In the absense of additional information or context the basis for this question rests in the size of the molecule. The smaller molecule will have a greater rate of diffusion than the larger molecule.  In general size of molecules is directly proportional to their molecular weight. When all molecules are the same molecular weight you will need to consider which ones are smaller based on the attractive forces between them. \n\na)  CO and N2 - N2 has a triple bond (the Ns are held closely to each other therefore small) and is very stable and I would bet that it will diffuse faster than CO since its actual size is smaller and because it is stable it is less likely to react with any medium through which it is diffusing.\nb) No2 and SO2 - this is likely to be based on size with the larger SO2 diffusing more slowly.\n\nHope this helps.
788892
OF2 (~104), NF3(~107), CF4 (~109.5),  BF3 (120), BeF2(~180)
1138290
Look up stuff like Arrhenius equations. I don't have my phys chem text book to hand.\n\nFor part two, I think it's D, as I'm fairly certain it's not any of the others.
624128
his applies only to transfer pumps since we deliberate add a gas load to the chamber and measure the quantity of gas issuing from the mechanical pump's exhaust. The method works as long as the gas chosen is not condensable, not captured in the foreline trap, does not dissolve in oil and is present in sufficient quantity to be measured.\n\nIf the mechanical pump has gas ballast, make sure it is off. Note the chamber's steady state pressure under the gas load. Put a bubble flow meter on the pump's exhaust, and measure the flow rate in cc per sec at atmospheric pressure. Using the gas in = gas out (at constant chamber pressure) principle, the gas entering must equal the measured throughput. If more precise numbers are needed, measure the exhaust gas temperature and the barometric pressure and re-calculate the quantity of gas at STP with Boyle's and Charles' law combined:\n\n(P1 x V1) / T1 = (P2 x V2) / T2\n\nExample The measured flow of gas at the pump's exhaust is 2.3 atm cc per minute from a chamber at a measured pressure of 1.4 x 10-5 torr. The exhaust gas temperature is 32°C and the barometric pressure is 750 mm Hg. What is the (corrected) effective pumping speed from the chamber?\n\n2.3 atm cc per min = (2.3/60) atm cc per sec\n= (2.3/60) x (760/1000) torr.liter per sec\n= (2.3/60)x (760/1000) torr.liter per sec at 750 torr and 32 °C\n= (2.3/60) x (760/1000) x (750/760) x (273/305) torr.liter per sec at STP\n= 0.0257 torr.liter per sec\nEffective pumping speed\n= (0.0257/1.4x10-5) liter per sec\n= 1838 liter per sec
736070
If you know the volume and density of the liquid you can calculate the number of moles n.\n\nOnce you have n and if you assume that the gas will behave like and ideal gas and you fix the temperature and pressure, the volume will be given by:\n\nV = nRT/P.
415862
R is the gas constant (= 8.31 J/mol*K).  The value should be with the list of constants in your book.\n\nOk Nic, you seem to be a bit confused on this issue so lets go back and start from scratch.\n\nFirst you obtained you values for T and K (experimentally or otherwise).  MAKE SURE YOU CONVERT YOUR TEMP VALUES TO KELVIN.  It's important for the units of the constant (R) are in Kelvin.  So then you calculate 1/T and lnK and plot the values for each set to generate your graph.\n\nYou calculate slope by taking two of your points (x1,y1) and (x2,y2).  From the data you added, let's say we'll use your first (0.002976, -3.079) and last points (0.003165, -4.225).\n\nslope= [-4.225- (-3.079)]/[0.003165-0.002976]= -6063.49 kelvin\n\nSince slope = -delta H/R\nThen -delta H = slope * R\nSo -delta H = -6063.49kelvin * 8.31 J/molkelvin\nAnd -delta H= -50387.62 J/mol\nRemoving the negatives from both sides\nDelta H= 50387.62 J/mol\n\nFor the second part of the question, use one (x,y) experimental point to find the y-intercept.  They don't want you to determine it from your plot, they want you to calculate the exact value.\n\nThe equation of a line is y=mx+b where "m" is the slope and "b" is the intercept.  So take another experimental point from readings (ie 0.003012, -3.751) and calculate the intercept by using the slope value calculated above.\n\nIf y=mx+b, then b= y-mx\nPlugging in the values\nb= -3.751 - (-6063.49kelvin * 0.003012/kelvin)\nb= -3.751 - (-18.26)\nb= 14.51\n\nTo calculate delta S for that same point (for that's what is requested), first you have to calculate delta G for that point.  Use the  following equation;\n\ndelta G = -RTlnK\n\nYou know the value for R (8.31 J/molkelvin), and you have the temp value and lnK value for that point for it was expermientally determined\n\n1/T= 0.003012/kelvin so T=332.01 kelvin\nlnK= -3.751\n\nSo delta G= -8.31 J/molkelvin * 332.01 kelvin * -3.751\nand delta G = 10349.02 J/mol\n\nYou can then calculate delta S for\nintercept = delta S/ delta G (given in your original data)\n\nTherefore delta S = intercept * delta G\nSo delta S = 14.51 * 10349.02 J/mol = 150164.28 J/mol\n\nHope this clarifies things.
1165015
The diffential flow rate betwwen air and fluid depends on the following factors. An exact equation is very difficult.\n1. type of fluid used because of the viscosity of the fluid.\n2. altitude and barometric pressure of the test location.\n3. test pressure\n4. temperature of the test air.\n5. many more\n\nAs you can see the formula for this would be very long. A rough formula you may try would take into account.\n1. volume of air at what rate.  in2/sec\n2. pressure for the air to obtain that volume\n3. viscosity of the fluid\n4. use 1 and 2 as baseline with air density\n5. compare 4 with fluid viscosity at desired pressure\n\nI know I couldn't give you a formula but I hope this may help you.\nYou may try math2.org for formulas or search for math formulas.
899533
Diffusion is inversely proportional to molecular weight.\nHigh MW = low diffusion rate\nLow MW = high diffusion rate\n\nSo calculate their MW and see which one is heavier.
755048
The ideal gas equation is pV = nRT.\n\nSo if you double n on one side, you must double something on the other. In your case it is p.