What is the melting points of 1,2-diphenylethyne ?
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*
1079104
depression of melting point
363124
detlaH=H(products) - H(Reactants)\n\n200KJ/mol
383259
go to\nwww.chemistry.com
231898
There are a few isomers possible.\n\nEssentially there are two C's double-bonded, with either 2H's off one C and 2Br's off the other C.\nOr 1H and 1Br off each C
484467
Under 1 atm of pressure,\n-The melting point / freezing point of Oxygen gas is: 54.8 Kelvin.\n-The boiling point of Oxygen gas is: 90.2 Kelvin.\n\nOxygen gas would be a liquid (under 1 atm) anywhere between 54.8 and 90.2 Kelvin (-218.3 degrees C and -182.9 degrees C).
975732
Substance: Water\nLatent Heat Melting J/g:  335\nMelting Temp°C: 0\nLatent Heat Boiling J/g:  2272\nBoiling Temp °C: 100 \nHeat of vaporization (kJ/mol): 40.65
1093944
i) add NaNH2(1 eq.)  then CH3I(1 eq.) (I=iodine)then NaNH2(1 eq.)  then C2H5I then add hydrogen in presence of palladium and BaSO4.\nii)add NaNH2(1 eq.) then C3H7I(1 eq.)  then add NaNH2(1 eq.)  then again C3H7I(1 eq.) then add ZnBr2 or MgBr2.
357611
Sharp melting temperatures are a characteristics of those substances which have a definite fixed structure, with always the same number of atoms and bonds throughout. This description would includes almost all substances we know because all have a fixed molecular mass; but this doesn't apply to polymers, because the polymers chains produced in the industry are of varying lengths - they would have the same characteristic repeating unit throughout the chain; but their chain lenghts can vary from a few hundred single units (joined together) to tens of thousands of them. A polymer with a short chain length will have less surface area - therefore less contact - and consequently less number of weak vanderwaal forces among them. A less quantity of these forces in a polymer would require less energy to be broken and hence have a less melting temperature (and vice versa). So polypropene would "soften over a range of temperatures" until all the long and short- chained polymer structures melt out. \n\nAs for the empirical formula, yes I guess you're right - though it's better to write it as (C3H6)n, where 'n' denotes many repeating units. I thought it would be CH2 if we simplify C3H6 further, but then CH2 would be the empirical formula for propene. (as in any homologous series)
867788
Liquid.  Melting point is 600.65K
983521
I find in a handbook a different Bp for H2Se, it is, -41.25 degrees C. The boiling point for H2Te is given as the same as the previous answer. This difference may be because the Handbook is of a different year or some difference in STP.
768102
p,p-dichloro diphenyl trichloroethane
358448
specific heat? that would be represented by Cp.\nHowever if you are using delta H then it would either be: delta H (fus) which is the enthalpy of fusion, used in problems concerning melting point (indicating a state change from solid to liquid)\nOr it could be delta H (vap) which is the enthalpy of vaporization, used in problems concerning a boiling point (indicating a state change from liquid to gas)\n\nHeat calculations are done using the formulas:\nq=m x (delta T) Cp <-- Used when there is a temperature change, but no state change\ng= m x (delta H fus) <-- Used for state changes from solid to liquid\nq= m x (delta h vap) <-- Used for state changes from liquid to gas
144775
The melting point of titanium is approx. 1660 Celcius
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its cis 1,4 polyisoprene(polymer)\n\nH2C=C(CH2)-CH=CH2
441962
At first glance you might make this assumption - but pyrene is an exception to Huckel's rule.  Huckel's rule of 4N + 2 is valid for mono cyclic ring structures (not in the case of the 4-fused benzene structure of pyrene).
1068373
If you know you're dealing with an organic compound, you take the formula weight and divide by 12.  This gives the number of carbon atoms.  You look at the remainder and determine if it could be the number of hydrogen atoms.  Alkanes have 2n + 2 hydrogen atoms (where n = number of carbons).  Alkenes have 2n and alkynes have 2n - 1.