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Tungsten's melting point of 3422 °C is the highest of all metals and second only to carbon (3550 °C) among the elements. This is why tungsten is used in rocket nozzles and reactor linings. There are refractory ceramics and alloys that have higher melting points, notably $\ce{Ta4HfC5}$ with a melting point of 4215 °C, hafnium carbide at 3900 °C and tantalum ...
50
Sorry, can't comment here, but I wanted to more directly answer your question.
Blacksmiths avoid melting their forges because the "heat" that can melt or oxidize iron and steel is actually contained in a ball in the center of the coal. In fact, maintaining coal "structure" is an important skill in blacksmithing.
To clarify better, imagine ...
29
Because $\ce{PCl5}$ does something which is not immediately obvious from its molecular formula: it autoionizes and becomes an ionic solid $\ce{PCl4+PCl6-}$. As such, it has much stronger interactions than $\ce{PCl3}$ with its mere dipole-dipole attractions, hence the higher melting point.
If not for that fact, you deduction should have worked just fine.
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Good question. Let's assume the container is infinitely strong, non-deformable, and constant in volume. Let's also assume that cooling the water is an equilibrium process -- that way, we won't have any supercooling.
At equilibrium, the first tiny bit of ice that freezes will take up more volume than the water it froze from. This will raise the pressure ...
23
Honey is indeed a complex mixture containing more than hundred compounds.
As for Wikipedia and depending on the point of view it is
a supersatured liquid solution
a viscous supercooled liquid (in the sense that it can get so viscous as to appear solid, without affecting its status of being a supersatured solution, and undergoes glass transition).
https://...
22
Diamond (carbon) does not melt at 1 atm. It sublimes to vapor.
Using carbon's theoretical phase diagram below (from Wikimedia), "liquid diamond" could be achieved at about 10 GPa (99 thousand atmospheres) and 5000 K (4700 °C).
Edit:
In fact, heating up diamonds at 1 atm turns them into graphite first. Only further heating would cause sublimation, as ...
22
We use an levitating furnace to heat samples of refractory ceramics up to approx $3000~^\circ\mathrm{C}$. Its for research purposes, so the samples are small (2 mm) beads. These are balanced on a jet of argon and heated with $\ce{CO2}$ lasers.
Here is a paper which talks about the technique:
D. Langstaff, M. Gunn, G. N. Greaves, A. Marsing, and F. Kargl, ...
21
Generally, the melting point of para-isomer is quite higher than that of ortho- or meta-isomers. This is due to the fact that it has a symmetrical structure and therefore, its molecules can easily pack closely in the crystal lattice. As a result, intermolecular forces of attraction are stronger and therefore, greater energy is required to break its lattice ...
20
Wolfram company doesn't conduct any experimental determinations of physical constants for chemical compounds and uses literature data sources.
The webpage for ChemicalData Source Information lists numerous sources of chemical information used by the company's products, including Wolfram Alpha.
Wolfram Alpha Knowledge Database is linked with 87th ed. (2006) ...
19
The melting and boiling points of noble gases are very low in comparison to those of other substances of comparable atomic and molecular masses. This indicates that only weak van der Waals forces or weak London dispersion forces are present between the atoms of the noble gases in the liquid or the solid state.
The van der Waals force increases with the ...
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High symmetry molecules fit into crystal lattices especially well (higher m.p.), but are volatile for having fewer van der Waals interactions (lower b.p.).
$$\begin{array}{lrr}
\hline
\text{Compound} & \text{m.p.}/\pu{°C} & \text{b.p.}/\pu{°C} \\
\hline
\text{pentane} & −130 & 36.1 \\
\text{isopentane} & −160 & 27.2 \\
\text{...
18
Melting and dissolving are all the same when you look at mixtures close to saturation.
You can say water lowers the melting point of the sugar, or that the solubility of sugar increases with temperature. Different description, same fact.
What makes this seem different from e.g. a salt water solution is that the molten (i.e. non-crystalline) sugar is fully ...
17
Gallium melts at 30 °C but doesn't boil until 2200 °C. If 30 °C is a bit too warm to count as "room temperature" or "normally" for you, I found an old paper that recommends tetralkyl silanes such as tetradodecyl silane as lubricants that are liquid over very wide temperatures.
Addendum:
Dowtherm A is a eutectic mixture of biphenyl and diphenyl ...
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Yes . You are right that structural symmetry comes into play .
Boiling point depends upon intermolecular interactions which over here is more in cis due to its net dipole moment . The dipole moment enables electronic interactions which hold molecules together . This shows some general factors of boiling point . Also the below link to the google book ...
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source
One way to raise the melting point of water is to increase pressure beyond about 635 MPa. By raising pressure you could get the melting point to be even greater than the normal boiling point.
A second way is to lower the pressure, but this can only increase the melting point by 0.01 K.
A third way would be to added enough of a high melting point ...
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I don't really like parts of this explanation (was hoping to find a better one here actually), but it's the best I know. I'll build on suggestions by @michielim and @vrtcl1dvoshun.
(Note: I think most of this argument can be transposed to the orbital overlap picture of bonding, but it might be slightly trickier to describe.)
For a backdrop to this answer, ...
13
I don't believe there is an equation that you can use for melting points of a general substance as a function of pressure (since the melting phase transition has a lot to do with the geometry of the molecule and the structure of the solid), but there is one for the boiling point of any pure substance when you are not near the critical point.
The liquid-...
13
Obsidian
This is a volcanic glass, formed when rhyolitic lavas cool too quickly for crystals to form. From a chemistry point of view, it's a mixture of silicon dioxide, aluminium oxide, with sodium ,potassium, calcium and iron oxides in various quantities. These lavas will also contain significant volatiles (water and $\ce{CO2}$) held in solution by ...
12
The strength of the ionic bond depends on Coulomb's law for the force acting between two charged particles where larger force translates to a stronger ionic bond. The equation is
$$F = \frac{-k\cdot q_1 \cdot q_2}{r^2}$$
$k$ is a constant; $q_1$ and $q_2$ are the charges on the ions and $r$ is the distance between the ions.
So the larger the charge the ...
12
One could melt them floating on a pool of high-boiling point denser metal, or in space where they can readily be contained. Or one could create a thick actively cooled shell and melt them inside it, melting part of the shell as well. Finally, it's probably not very practical, but one could use an air jet to keep then suspended away from other matter and then ...
12
The polar aprotic solvents (for example dimethylformamide, mp −61 °C, bp 153 °C, or hexamethylphosphoramide, mp 7 °C, bp 230–232 °C) would be a place to start. Silicone oil is often used in heating baths – one product in the Aldrich catalog is advertised as having a working range of −40 °C to +230 °C.
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You've figured out one half of the Hall-Heroult process. Just melting alumina on its own won't liberate the oxygen. The reason is difficult to explain without getting into chemical thermodynamics - one way to think about it is that the lowest energy state is one where oxygen is bound to alumina - where the electrons from alumina have been "taken" by the ...
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There are two alternatives to the other answers here, though whether they can be used on a large scale is open to question.
The first is to use an actively cooled vessel to hold the metal and a method of getting energy into the metal not based on the heat of the crucible. Many metal-vapour reactions (used for small scale chemistry research) do this and ...
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Now that's a great question indeed! Evidently, at 0K all elements except helium are solids, at 10000K they are all gases, so someplace in between the number of liquids must reach a maximum; what and where might that be?
Well, there is no formula or theorem that says liquid hydrogen must boil at 20K, nor is there such a thing for any other element, so this ...
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You seem to approach the separation of alloys much like the separation of water-soluble salts, probably based on the fact that both involve the liquid/solid phase diagrams, which may even look alike. This is wrong; these systems are fundamentally different.
Say, you have a water solution of some well-soluble salt which you need to recrystallize. You cool it ...
10
To make an argument based on electron configuration, one could see that the orbital occupation does affect the melting point:
Element Z Electron Config. Melting Point / K
-----------------------------------------------------
Vanadium 23 [Ar] 4s2 3d3 2183
Chromium 24 [Ar] 4s1 3d5 2180
Manganese 25 [Ar] 4s2 3d5 1519
...
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Melting point depends not only on the structure of the molecule itself, but also on the structure of the solid (crystalline) phase and molten/liquid state of the substance. That is a much bigger computational problem than calculating energetics for a small molecule or ensemble. It is not easy to predict how a molecule (especially a complex one) will ...
10
Sure. Even discounting non-equilibrium conditions, any substance which expands upon freezing will by symmetry melt under pressure, even if the pressure is caused by supporting its own weight. There's a simple example: water! A sufficiently tall self-supporting block of ice standing on a flat surface will create enough pressure at its bottom that it could ...
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Anhydrous caffeine occurs in two different polymorphic crystal forms, with a transition between the two forms occurring at $\pu{136 °C}.$ The high temperature form has a melting point of $\pu{236 °C}.$
The vapor pressure of the high temperature solid form is described by the equation:
$$\log P[\pu{Pa}] = -\frac{5477}{T[\pu{K}]} + 14.395,\tag{1}$$
and for the ...
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