# Tag Info

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This may be obvious to Fermi, but not to the rest of us. One has to keep in mind the context of the derivation, Fermi is talking here about ideal gases. In other words, gases which are very dilute, such that the interparticle distance is much much larger than the range of the interparticle interaction. This means that particles don't see each other ...

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I'm not sure I would focus on the the liquid to gas phase change. Sublimation, seems better behaved. I've heard that some alloys do not have well defined melting points. I don't know if something like that occurs in sublimation. But imagine a crystal lattice consisting of two fairly different substances. In this case it seems the material that sublimes at a ...

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Your question is about the factors of $2$ in the expression for $dB \over dt$. The first term, $2k_+A$, has a factor of $2$ because for every reaction of an $A$ molecule two $B$ molecules are formed. The second term, $-2k_- B^2$ has a factor of $2$ because in the reaction of $B$ with $B$ to form $A$ two $B$ molecules are lost. Hope this helps.

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Ok so this is a over a year later, but no answer yet so here goes. One approach would be to use a Boltzmann type/Arrhenius type approach to collisions to see if they lead to reaction. So you could figure out a probability of reaction, $p$, which would be equal to $1$ if the reaction is 'Exothermic' (where energy is released) or equal to $p=e^{-{\Delta E ... 2 Computational chemistry methods have advanced significantly in the last 5-10 years, including much more accurate DFT methods, quantum mechanical dynamical methods (like Car Parrinello MD, and better classical molecular dynamics techniques. That said, dealing with the dynamics of molecular reactions is an active area of research. Perhaps the most promising ... 2 It depends on the level of theory you want to apply to a simulation. For example, the current state-of-the-art ab-initio calculations for a single low ernergy ($\lt 10~eV\$) electron approaching and interacting with a molecule can cope with perhaps 20 to 40 electrons in the target molecule. Note that ab-initio calculations contain, in principle, no ...

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magnesium burns hot enough to separate water molecules and because hydrogen is flammable and oxygen is what fire needs to get bigger your hose becomes a nice flame thrower :)

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i will give another answer (or maybe two related answers). Smell and taste as raw signals relate to specific molecules and their bio-chemical properties which cannot be simulated electronicaly. Not to mention the brain processes involved (do we actually see the same color or just agree on same name of a color?). On the other hand, audio or visual signals, ...

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I think this solves everything, it is slightly adapted from the wikipedia. "The closely related concept of matter conservation was found to hold good in chemistry to such high approximation that it failed only for the high energies treated by the later refinements of relativity theory, but otherwise remains useful and sufficiently accurate for most chemical ...

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Would the mass of burnt firewood be equal to the mass of firewood before burning? You won't get a good answer by simply looking at the "burnt firewood". The combustion is using oxygen from the air, and it is creating carbon dioxide and many volatilized materials that will disperse in the air. But we can imagine combustion happening in a box that is ...

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If we add the energy and mass we will find that it adds to the total mass of the original wood. The total energy of the system is conserved, this will always be true. If you only look at the weird and burnt wood you will find a discrepancy. This is because some of the mass was lost in smoke. If you measured the mass of the smoke and the wood you will be in ...

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The mass of the original material equals the mass of the combustion product. You have to take the carbondioxide and the oxygen into account, not just wood and ash. The energy is stored chemicaly before released. Not every time energy is stored it is in form of additional mass. For the subject of mass loss have a look at weak interaction, Einstein is off ...

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The law of the conservation of mass was superceded by the more general law of conservation of energy when it was realized that mass and energy were equivalent. Anyway, you are correct. The mass of the combustion products will always be less than the mass of the original materials. The difference being equivalent to the energy produced.

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The issue is not "equilibrium" - the issue is boiling. During boiling, there is explicitly NO equilibrium: the water wants to get out of the liquid phase, and into the vapor phase. The temperature of the liquid is sufficiently high that liquid can evaporate below the surface (strictly speaking the temperature must be slightly above boiling for that, as the ...

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