CHCl3 boils at 61.7C and a heat of vaporization of 31.4kJ/mol.  What's it's vapor pressure at 36.2C.  Explain.
969517
Use the clausius-clapeyron equation for this one.\n\nln (P/Po) = (delta H/R ) (1/To - 1/T)\n\nPo is one atm at To = 61.7 C based on the info in the question.  T is 36.2C.  This formula requires absolute temps, so convert to K.\n\nSo:\n\nln (P/1 atm) = 31400/8.314 (1/334.8K - 1/309.3K)\n\nIf I didn't fat finger the calculator P = 0.39 atm.\n\nHope it helps.
915075
I don't have the constant for the change of steam to water.  But 373.15 is the boiling point of water.  So simply multiply 42.3 g/h20 by the constant for the conversion of steam to liquid and you will get it.
968707
NH3  -33.34 °C\nCH3F -78°C\nCH4 −161.6 °C\nHe -268.93 °C
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)
984988
(CH3)2 CH - CH = CH - CH - (CH3)2   you will dehydrate to remove water upon heating.
836776
PV=nRT  n=PV/RT   V=nRT/P\n\nR= 0.08206 L atm mol^-1 k^-1\n\n922 torr/760 torr= 1.213 atm\n\nn= (1.213 atm x 35.3 L)/(0.08206 Latm/molK x 315K)\n\nn=1.6565 mol Propane\n\nSTP = 1 atm, 273K\n\nV = (1.6565 mol x 0.08206 Latm/molK x 273K)/(1 atm)\nV = 37.1095 L\nV = 37.1 L (sig figs)
797227
You have Pressure (being asked) You have the volume (5.43L). You have the number of moles (1.82)You have temperature (69.5C or 342.5K). I assume you know R.\nSo yes it is PV=nRT
738802
V1/T1 = V2/T2, where T is measured in Kelvins!!!\n\nV2 = V1*T2/T1\n\nV1 = 69.0 mL\nT1 = 49.0°C = 322.2K\nT2 = 89.0 °C = 362.2K\n\nNow you can do the arithmetic! (Hint, the balloon will be larger!)
209192
Boiling Point: \n204C (399F) \nMelting Point: \n100C (212F) \n\nCHECK IT OUT BUT I THINK ITS CORRECT
585380
Assuming the temperature is 298K (25oC) The equation is P1V1/T1 = P2V2/T2\n\nWith temperature in Kelvin and Pressure in dm-3\n\nSo P2=1*0.50*333/(298*0.66)\n=0.847Pa
130715
Adding table salt to water causes the water to boil at a higher temperature.\nThe three important concepts are the vapor pressure, the boiling point, and the freezing point.  For water at standard pressure (1 atmosphere), the boiling point occurs at 100 degrees Celsius, and the freezing point occurs at 0 degrees Celsius.   A Liquid will boil when the vapor pressure is equal to the atmospheric pressure. If we start with just barely boiling water and add some salt, the water will stop boiling.  The salt has lowered the vapor pressure of the solution.  To increase the pressure we need to add heat.  To add heat, we turn up the temperature.  It turns out that in the dilute solution case, the change in boiling temperature is proportional to the concentration of the add solute (salt).
1098900
The specific heat at constant volume for solid KCl as a function of temperature between 298 and 900 K is given by  (see 1st source):\n\nCp° (J/mol/K) = A + B*t + C*t^2 + D*t^3 + E/t^2\n\nwhere \n\nA = 35.41597 \nB = 70.03472 \nC = -91.38233 \nD = 52.52426 \nE = 0.153460 \nF = -449.1357 \nG = 108.9982 \nH = -436.6841\nand\nt = Temperature in K/1000\n\ndelta-H of formation of KCl at 298.15 = -436.68 kJ/mol (1st source)\n\ndelta-H of formation of KCl at 298.15 = -436.47 kJ/mol (2nd source\n\ndelta-H of formation of K+ (aq) at 298.15 = -252.17 kJ/mol (2nd source)\n\ndelta-H of formation of Cl- (aq) at 298.15 = -167.08 kJ/mol (2nd source)\n\ndelta-H of solution (using data from second source for KCl(s)):\n\n-252.17 +(-167.08) -(--436.47) = +17.22 kJ/mol\n\ni.e., this is the enthalpy change for the reaction:\n\nKCl (s) -> K+ (aq) + Cl- (aq)\n\nCan't help you with the value for Cv, but if you happen to know the molar volume, molar expansivity, and isothermal compressibility of solid KCl, at the temperature of interest, you can calculate Cv from the thermodynamic relationship:\n\nCp - Cv = T*V*a^2/K\n\nwhere T is the absolute temperature, V is the molar volume, a is the isobaric expansivity, and K is the isothermal compressibility.
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.
775809
I believe the units of your supposed answer are wrong.  They should be kJ/mol and not kJ.\n\nHeat = m * Cp * DeltaT = 60 * 4.186 * (13.2 - 27.2) = - 3516 J\n\nThis heat is absorbed by 28.7 g of KI.  Per mole (MW(KI) = 166 g/mol), the heat will then be:\n\n28.7 g ===> 3516 J\n166 g ===> x\n\nx = 3516 * 166 / 28.7 = 20,340 J/mol = 20.3 kJ/mol
825292
i guess I'm wrong   but I thought that water would  freeze at 0  degrees celsius.  also it would boil at 100 degrees C. Even in a vacuum.  my guess!  btw at absolute 0  kelvin,  does that only exist in a vacuum?  If we could reach  - 450 degrees F.  would our cpu's operate at an infinite speed.....got me on this one.  I'm not sure on how atmospheric pressure relates to these conditions.  But I know someone knows!
288242
The reaction should be written as follows,\n2 KClO3(s) --> 2 KCl(s) + 3 O2(g) H = -89.4 kJ\n\nYou either did not include the arrow or it is not displaying properly.\n\nPotassium Chlorate decomposes with heat to form Potassium Chloride and Oxygen gas, no where in this reaction is elemental Potassium involved.\n\nThe standard enthalpy of formation of any pure element in its natural state is zero. Oxygen is a diatomic molecule and thus, Oxygen gas (O2) has a enthalpy of formation of 0 kJ in this reaction, no matter how much of it you have, it is still zero.\n\nSince the Potassium Chloride is the only product left that has any non-zero enthalpy of formation, we know that the over all change in enthalpy is equal to the enthalpy of formation of KCl * the number of moles produced.\nIf the over all change in enthalpy is -89.4 kJ and 2 moles of KCl are produced, then the enthapy of formation of the KCl must be 44.7 kJ/mole less than the enthalpy of formation of the Potassium Chlorate.
552193
Can be solved by simply plugging values into the Ideal Gas equation without using two states.  Pressure must be converted to atmospheres, however, and temperature to Kelvin.  P = 1.243 atm, T = 306 K\n\nn = PV/RT = 1.243 * 2.74 / (0.08206 * 306) = 0.136 mol\n\nmass = moles * MM = 0.136 * 28.01 = 3.81 g