If the entire unknown does not vaporize in the 125 mL Erlenmeyer flask, will the reported molecular weight be?
961353
You realize that simply copying the questions from your homework *without* also providing the rest of the experiment conditions -- what was being done, how and so on -- is making answering any of your questions totally impossible?
878037
Suspect this is what you mean...\n\nSuppose you have 0.500 g of a substance you believe is pure...\nyou analyse it and determine that it contains 0.300 g of the substance you are looking for..\nso your sample is (0.300/0.500) x 100% pure  = 60%
423892
I'm sure that's what you're supposed to determine in the lab you're working on. Good luck.\n\nI've been able to find a value for the heat capacity of n-hexadecanol (cetyl alcohol) in my copy of _Lange's Handbook of Chemistry, 13th Edition_.  When you've come up with your value from the experiment, post it here and I'll check to see if you're close.
872856
It will eventually melt at 2572C, and it will boil at 2850C.\n\nHeat does not cause it to react, typically.
720671
Excellent question.  Unfortunately, I don't have an answer and don't know where to find an answer.\n\nInteresting to note that on one website, the specification for some sets of balls is that each individual ball within a set weigh within one gram of the average of all the balls of that set.  That seems like a pretty high acceptable deviation, and enough to create at least a minute difference in its density, thus creating a higher probability of being at the bottom of the mixing container.
869862
I dont understand what r u doing with "p*v/t"; but if assume ideal gas behavior, PV = nRT, so P = nRT/V [n is number of moles, R is ideal gas constant, T is temperature, V is volume of the flask, P is the pressure]
137529
No, I dont think so.\n\nA liquids vapor pressure is the pressure being exerted on the liquid by its molecules in the vapor/gas phase.  If there are no molcules in the gas phase, there would be no vapor pressure, but I cannot think of a situation in which a substance would exert a negative pressure on itself.
872765
It won't matter how much you started with or how much higher the solution was heated.  When the solution cools to 20C it will hold 0.22g/100mL of solvent (water).\n\nSo if you start with 1.75g and 0.22g remain in solution, the amount crystallized at 20C will be the difference, 1.53g.
670103
Assuming that NO2 is an ideal gas is these conditions (which is a reasonable assumption at such a low P), the number of moles into that flask is: n = PV/(RT) where P = 432 mmHg = 0.568 atm, V = 0.5 L, T = 60 C = 333 K, R = 0.082 L atm/(mol K).\nThen: n = 0.01 mol and, taking into account the Avogadro Number (in number of particles/mol) the number of molecules is NM = 0.01*6.022x10^23 = 6.022x10^21.
767855
you didn't leave moles of H2 nor CO ....so this can't be answer...
961444
We will need to find if there is a limiting material.\n\nCO + 2H2 ------> CH3OH\n\n(1.61 g H2)*(1 mole H2/2.016 g) = 0.7986 mole H2\n\n(0.7986 mole H2)*(0.5 mole CH3OH/mole H2) = 0.3993 mole CH3OH at most\n\n(15.95 g CO)*(1 mole CO/28g) = 0.5696 mole CO\n\n(0.5696 mole CO)*(1 mole CH3OH/1mole CO) = 0.5696 mole CH3OH at most\n\nH2 is the limiting reagent, therefore:\n0.3993 mole CH3OH * (32.032 g/mole CH3OH) = 12.79 g\n\nPlease check for errors, I went by memory on what the atomic and molecular weights were, you may want to check this.
696357
51.9961 is the _atomic_ weight.  Molecular weight would be the overall weight of a mole of some molecule of which one or more chromium atoms is a part.
708158
well aside from that you said the vapor pressure is 18.6 and put 18.9 in your calculations, it looks great! Besides, even with that difference, it doesn't change the answer at all, its still .877 atm with both values.
949133
OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
1156050
I would visit the site cited below.  The discussion includes:\n\n"Irving Langmuir developed a way to measure vapor pressure \nby measuring the evaporation rate, which we will now flip around backwards. His reasoning is that the rate at which molecues are lost due to evaporation to a gas with no partial pressure of the evaporating substance is the same as the rate at which molecules of the substance would hit the surface if it were in equilibrium with the vapor (because in equilibrium the evaporation rate and re-condensation rate cancel each other out). His expression is: \n\n(mass loss rate)/(unit area) = (vapor pressure - ambient partial pressure)*sqrt( (molecular weight)/(2*pi*R*T) ) "
870952
To answer this Question you need to use the Conservation of Energy Law.   This takes its simplest form in the idea that:\n\nHeat Lost by the Steam    =   Heat Gained by the cold water\n\nYou will need to say:\nLet L be the unknown Quantity and form an Equation:\n\nHeat Lost by the Steam:\nThe steam (one assumes at 100Celsius) condenses to water at 100Celsius.  In doing this it gives up its Latent Heat of Vaporization (mL)   The water it forms is then cooled to the final temperature giving up this Internal Energy to the System.  The Internal Energy(Heat) gained is given by m.s.dT where dT is the change in temperature.\nNote that we have two  Energies to Add Together here.\n\n\nHeat Gained by the cold water:\nThe Internal Energy(Heat) gained is given by m.s.dt where dt is the change in temperature.\nBoth the cold water and the Calorimeter are heated from their starting temperature(15C) to the final temperature(70.6C)\nNote that we have two  Energies to Add Together here.\n\nNow write down the equation:\nHeat Lost by the Steam (steam + hot water)     =   Heat Gained by the cold water (cold water + calorimeter)\n              \nm1.L + m1.s.dT  =  m2.s2.dt + m3.s3.dt\nAnd solve for the Unknown L
827958
The first part is correct.  If the numbers in the second part are correct, it is true only for an ideal gas.  Other materials have various densities in their liquid and solid states.