50 cubic centimeter of propane and 50 cubic centimeter of butane are mixed.?
80657
if STP = 22.41L\n\n mass = 88.36%\n\nmoles = 90.91%\n \nI am not sure of this ok
521266
This sounds like homework, so I'll give you the ideas but not the data.\n\nThe enthalpy of the mixture at standard conditions is the sum of standard heats of formation of the components of the mix. It will be given in kJ/mol (Ar is 0 kJ/mol being an element, the other two you can look up in tables or on internet). \n\nThen you need to adjust for the sensible energy difference compared to standard conditions : you will also need to look up the specific heat of each component (usually expressed in kJ / mol ºK) and multiply the specific heat of each component by the amount and by the difference in temperature. Make sure you have everything in the same units.\n\nRemember if you are working in moles its easier because to a first approximation, one mole of any gas occupies the same volume (if you undrestand how to allow for deviations from the ideal gas laws, you wouldn't be asking this question) and I assume that the percentages are volume percent, not by weight\n\nEnjoy
311555
that's bad...  Methane is CH4, propane is C3H8, and octane is C8H18... the rule to find the number of H in an alkane (single bonded Carbons) is 2N+2 where N is the number of carbon atoms... Methane is 1, ethane is 2, propane 3, butane 4, pentane 5, hexane 6, septane 7, octane 8, nonane 9, decane 10, etc....
1108696
Propane is a by-product from two sources: natural gas processing and crude oil refining. Most of the LPG used in the United States is produced domestically. When natural gas is produced, it contains methane and other light hydrocarbons that are separated in a gas processing plant. Because propane boils at -44°F and ethane boils at -127°F, it is separated from methane by combining increasing pressure and decreasing temperature.\n\nThe natural gas liquid components recovered during processing include ethane, propane, and butane, as well as heavier hydrocarbons.\n\nPropane and butane, along with other gases, are also produced during crude refining as by-products of the processes that rearrange or break down molecular structure to obtain more desirable petroleum compounds.
634933
Density and phase 0.626 g/cm3, liquid
895333
1,1,1,2-Tetrafluoroethane (R134a)\nA 50/50 blend of R-32 and R-125 is marketed under the trade name Puron®.
726522
The correct answer would be hydrocarbons because propane, C3H8 is an alkane, Benzene, C6H6, is an aromatic hydrocarbon or a cyclic hydrocarbon, Acetylene (I'm assuming that is what you meant be Actylene), C2H2 is an alkyne, and 2-butene,C4H8, is an alkene, but all of the compounds are hydrocarbons as they only contain carbon and hydrogen.
226482
Here goes: your 80/20 gas mixture is flowrate x 1.8\n\nyour 70/30 gas mixture is flowrate x 1.6\n\n***Dana Oakes is a great book to use***
166213
50 kg x 700 m x 3 x 3600 s x 1/3 x 10-4 m3/sec x 2/3 x 103 J/sec = 645.67 Joule
816907
It varies with the grade produced.  This example is is 94.5 nitromethane with nitroethane and nitropropane.  I doubt there is any measurable amount of propane left in the commercial product.
702782
CH2OHCl: 66,45g/mol\nDo you mean 4dm^3 or a cube with 4dm length  edges?\nIf you mean 4dm^3 then\n4dm^3= 4*10^3 cm^3 = 4*10^3 mL= 4L\n2M= ?/4L\n?= 8 moles of CH2OHCl \nif you meant 4dm edges then\n4dm= 4*10^1cm\n(4*10^1cm)^3= 6*10^4cm^3 (actually it is 6,4 but because of the lack of figures you must report 6 instead of 6,4)\n6*10^4cm^3= 6*10^4 mL = 6*10^1 L\nTherefore\n2M= ?/6*10^1L\n?= 1,2*10^2 moles of CH2OHCl \n\nPart II\nCH4 + 2O2 → CO2 + 2H2O\n7dm of O (again: 7dm edges or 7dm^3???)\nfor 7dm sides\n(7*10^1cm)^3= 3*10^2 L \nO density = 1,43 g/L\nTherefore: 5*10^2g or 0,5 kg of Oxygen\n4g CH4*(1mol CH4/16g CH4)*(1mol CO2/1mol CH4)= 1*10^1g of CO2\nIf you compare the amount of Oxygen and methane and the molar ratios there is no need to do the same for O. Obviously CH4 is the limiting reagent\nCO2 density: 1,98*10^-3 g/mL\n1,98kg/m^3\nIf 5dm (again the same problem) were produced, then\nfor 5dm edges: 2*10^2g of CO2 (obviously not this one)\nfor 5dm^3: 5*10^3 cm^3 of CO2: 1*10^1g of CO2 this is it!)\nNow the yield\nIf you would have added more figures we would have had\n(9,9g/11g)*100= 90% yield\nBut because of the lack of figures we end up with\n(1*10^1g /1*10^1g)*100= 1*10^2 % (you got the maximum yield)
562353
pure butane is neutral for sure. And butane, no matter isobutane or just butane, is alkane. Alkane does not dissolve in water since no ionic bond exists in the compound, so measuring its pH does not make much sense.\n\nCarbon and hydrogen have very similar electronegativity, so it's very difficult to remove hydrogen from the carbon chain. It's pka value is around 50.
875372
Well, since you assume the heat capacity and density of the mix are the same as water, remember that water conveniently has a Cp of 1 cal/gm-C and density of 1 gm/ml.\n\n  m = (33 ml + 42 ml) * 1 gm/ml = 75 gm\n\n  dH = m Cp dT = 75 * 1 * (31.8 - 25) = 510 cal\n\nThe HCl molarity = 1.20, meaning a liter of water has 1.20 moles of HCl.  So the number of moles reacting in the mix is:\n\n   M = 1.20 * (33/1000) = 0.0396 moles HCl\n\nFinally, the dH/mole HCl = 510 / 0.0396 = 12879 cal/mole\n\nNow, you try the other 9.\n\nedit:  Believe the 12879 cal/mole should be negative, since heat is released.
395302
mixture is combining substances, solution is dissolving a substance in another
559355
If the 20 ml of propane is measured at STP, then:\n\n1.0 mole of gas at STP has a volume of 22.4 liters = 22400 ml\n\n20 ml/22400ml = 8.93 x 10^-4 moles of propane.\n\nPropane is C3H8, so the equation is:\n\nC3H8  + 5 02  ======>  3 CO2  +  4 H2O\n\nso, you would need 5 * 8.93x10^-4 moles = 4.46 x 10^-3 moles of oxygen gas, which, at STP, would occupy:\n\n22400 ml/mole * 4.46x10^-3 moles = 100 ml\n\nAn easier way to calculate the same answer would be to realize that, from the equation:\n\nC3H8  + 5 O2  ======>  3 CO2  +  4 H2O\n\nIf both gases are measured at STP, then you need five times the volume of oxygen to react one volume of propane, thus:\n\n5 x 20 ml = 100 ml of oxygen.
183956
Half the diameter, or circumference divided by 2.pi
764912
If I understand correctly what you need then this is what you do:\n(by the way we will take He as 83.7 because it is over 100% other wise.)\nmultiply the molecular or atomic weight of each species by the fraction (percent divided by 100).\n\nMW(CO2) = 44\nMW(N2) = 28\nAW(He) = 4\n\n44*0.11 + 28*0.053 + 4*0.837 = 9.67 g/mol\n\nThis is the equivalent "molar mass" of the mixture.