What are the procedure to determine the melting and freezing point of napthalene?
103100
To determine the melting point of it, you put a small sample in a closed-end tube and use a melting point apparatus. (No I'm not being sarcastic, that is what it's called. It's a device with a thermometer and a heating element that you use to controllably heat the sample until it melts)\n\nYou don't need to determine the freezing point ... Napthalene is a solid at room temperature.
670047
Melting point 933.47 K\n(660.32 °C, 1220.58 °F)\n\nBoiling point 2792 K\n(2519 °C, 4566 °F)
325629
It might help if you spelled "naphthalene" correctly. \nhttp://www.madsci.org/posts/archives/may98/894550073.Gb.r.html \nThis website has the structural pictures you can use to do your Lewis dot structure.  \nNaphthalene is an aromatic compound.  It's two aromatic rings sharing a side.
131859
melting point is 3550 (Degree celcious) and boiling point is 4827 (degree celcious)
883310
At temperatures above 220 K the rxn would be spontaneous. To determine spontaneity, first determine the delta G value by using the formula deltaG=deltaH-Tdelta S. Now find out what temperature value would make the delta G negative which would indicate a spontaneous rxn, and in this case it is spontaneous above 220 K.
131196
The melting point is 1068 kelvin and 795 celsius. The boiling point is 3633 kelvin and 3360 celsius.
24553
Oxygen is is difficult to liquefy, but has a boiling point of -183 degC, and the liquid is pale blue in colour and is appreciably magnetic. At still lower temperatures, light-blue solid oxygen can be obtained, which has a melting point of -218.4 degC.
1164458
Napalm commonly refers to any number of flammable liquids used in warfare, and is often used to mean 'jellied gasoline'. Napalm is actually the thickener in such liquids, which when mixed with gasoline makes a sticky incendiary gel. It was developed by the U.S. in World War II by a team of chemists led by Louis F. Feiser, and the name comes from the use of the original chemicals of naphthalene and palmitate, which were added to gasoline to cause it to gel.\n\nNapthalene melts at 80 degrees, so it would freeze below that temperature.\nPalmitate melts at 62 degrees, so it would be frozen below that temperature.\nOnce mixed with gasoline or any number of flammable liquids, the freezing point would depend on the amount of flammable liquid added and the freezing point of that flammable liquid.
718253
Melting point 1811 K\n(1538 °C, 2800 °F)
857462
log10(P) = A − (B / (T + C))\n    P = vapor pressure (bar)\n    T = temperature (K)\n\nAntoine Equation Parameters\nTo get the References, go to the website below, search by compound name using Phase Change checkmarked, hyperlink to complete journal reference\n\nEthane: (two sets for two temperature ranges)\nTemperature (K) 91.33 - 144.13\nA  4.50706\nB  791.300\nC  -6.422\nReference: Carruth and Kobayashi, 1973\n\nTemperature (K) 135.74 - 199.91\nA  3.93835\nB  659.739\nC  -16.719\nReference: Loomis and Walters, 1926\n\n\nEthylene:\n\nTemperature (K) 149.37 - 188.57\nA  3.87261\nB  584.146\nC  -18.307\nReference: Michels and Wassenaar, 1950
105897
Boiling point of oxygen 90.20 K (-182.95 °C, -297.31 °F)\n\nadditionally Melting point is54.36 K (-218.79 °C, -361.82 °F)
776170
1) Raise the temperature from -20°C to 0°C:\nMass of ice * specific heat of ice * temperature rise\n10.0g * 2.1 J/g K * 20° = 420 J\n2) Melt the ice:\nMass of ice * Heat of Fusion\n10g * 333J/g = 3330 J\n3) Raise the temperature from 0°C to 100°C:\nMass of water * specific heat of water * temperature rise\n10.0g * 4.2 J/g K * 100° = 4200 J\n4) Evaporate the water:\nMass of water * Heat of Vaporization\n10.0g * 2260J/g = 22600 J\n5) Raise the temperature from 100°C to 120°C:\nMass of steam * specific heat of steam * temperature rise\n10.0g * 2.0 J/g K * 20° = 400 J\n\n6) Addition: 420 + 3330 + 4200 + 22600 + 400 = 30950 J
926085
1,2-diphenylethyne: 60 C\nhttp://ptcl.chem.ox.ac.uk/MSDS/DI/diphenylacetylene.html\n\nThe others: at least 120 C, which is the melting point of dibromodiphenyl*methane*
481495
1. Perform vacuum distillation.  Place the solution in a vacuum flask; apply vacuum (removing air from vessel); apply heat and "boil" the water off.  Collect the condensate and measure its volume.  Weigh the remaining sugar that is left behind in the vacuum flask.\n  -or -\n2. There may be a way to measure the refractive index of the sugar solution.\n-or-\n3. Heat a sample of the solution to determine the boiling point elevation; if you have a table that compares molarity vs. boiling point elevation, you may be able to determine the molarity of the solution by knowing at what temp. the solution boils.\n\n4. You may be able to extrapolate by weighing exactly one cubic centimeter (cc) (volume) of the solution at 25 degrees C.  There may be a table comparing weight of a 1 cc solution at that temp (as the density of 1 cc of water is different than a 1cc sugar solution.
103857
THE CORRECT BOILING POINT OF AU OR GOLD IS 2807.0 DEGREE CELCIUS ( 3080.15K,5084.6 DEGREE F)
968993
Color, shape, hardness, melting point, boiling point, density
761477
Actually you have an electrolyte, so the drop in the freezing point  is\n\nΔT=i*Kf*m\n\nwhere i is the van't Hoff coefficient and (for simplicity) for strong electrolytes it is equal to the number of ions that are produced by its dissociation, according to the molecular formula.\nNaCl dissociates completely to two ions (Na+ and Cl-) thus i=2\n\nfor water Kf=1.86\n\nm is the molaLity and not molaRity. \nYour solution is 2 M = 2mole solute/L of solution \nmolality is mole solute/kg of solvent. You need additional info in order to convert molarity to molality\n\nIf you had 2 molaL and not molaR solution then m=2, or if we assume due to lack of more info that in this case molaRity=molaLity...\n\nΔT=i*Kf*m =2*1.86*2 = 7.44\n For pure water Tfwater=0 C, so for the solution \nTf= Twater-ΔT= 0-7.44= -7.44 C