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# Tag Info

54

I think your question implicates another question (which is also mentioned in some comments here), namely: Why are all energy eigenvalues of states with a different angular momentum quantum number $\ell$ but with the same principal quantum number $n$ (e.g. $3s$, $3p$, $3d$) degenerate in the hydrogen atom but non-degenerate in multi-electron atoms? Although ...

38

Well, a lot of things happen to the reactants. Some bonds stretch (and maybe eventually break), the others shrink, and your molecules morph into different molecules, which are the products. (source) As for staying at the very peak, that would be kinda unnatural, but luckily, not every peak looks like this; sometimes there is a tiny dent near the top, and ...

36

General chemistry textbooks tend to explain atomic structure exceedingly poorly using a hodgepodge of obsolete concepts. Your chemistry book provides such a typical example - the notion of penetration only makes sense in the ancient Bohr-Sommerfeld model that has been obsolete since the discovery of quantum mechanics! The idea was that orbits of electrons in ...

28

The combustion of alkanes like butane is fearsomely complicated involving dozens of transient compounds and hundreds of different reaction. If you have a few spare hours there is a dissertation that presents a nice summary of the process here (this is a 1MB PDF). A butane molecule is pretty stable and doesn't react with oxygen on contact so you need some ...

25

EZ-water is not a breakthrough. It is not new, nor is it valid. This appears to be one of the many claims about the healthful benefits of drinking "ionized water". The original "article" linked, contains a number of misleading . Water molecules make up 99% of your body. This is true, but misleading. According to various sources, the human body is between ...

19

It seems like it should be the average distance that matters No. It is the average energy that matters. Note that this stuff about "spends so much time here and so much there..." is really just a (not particularly good) way of describing a quantum-mechanical wave function's absolute square. The electron is actually never at any particular place in the ...

19

Absolutely yes. Lighting a torch in such an environment would simply be the reverse physical process (and same chemical process) of what is done in our oxygen-containing atmosphere. In the chamber or alien world of hydrogen gas, providing an ignition source to a stream of oxygen would give a flame. The chemical reaction would actually be the same as if ...

16

Most matches these days are safety matches: they're designed to need something more than ordinary levels of friction to ignite, by splitting the combustion materials between the match-head and the striking strip, i.e. the brownish paint thing down the side of the matchbox or across the front of the matchbook. Most "regular" matches now are safety matches. ...

16

Yes, $^{56}\ce{Fe}$ has the most stable nucleus, and $\ce{He}$ is the most chemically inert element. These are different and unrelated qualities, pretty much like physical fitness and intelligence in a man. As for structural stability, there is no such thing in chemistry (there is one in architecture and another in mathematics, but those are out of scope of ...

16

There is a simple (some might say simplistic) way to get an intuition about this and it involves thinking at the molecular or atomic level rather than about the bulk properties of the reaction (the thermodynamic view). For a simple reaction where molecule A has to bang into molecule B to create molecule C, the reaction will only happen if the amount of ...

15

Getting a large exotherm per mole of molecules is a bit of a bad cheat if you're allowed to make molecules arbitrarily big! Energy per mass is the only honest way to do it (or energy per mole of atoms would work too I guess). However, there really is no chance to chemically match a nuclear exotherm; nuclear energies are simply in a class of their own. ...

15

It is all about minimizing the energy of a molecule. In the case of carbon, the only molecule that adopts a perfect hexagonal geometry in its ground state is benzene (and its derivatives that possess a 6-fold rotational axis). In this case the hexagonal geometry is adopted because all of the carbons are $\ce{sp^2}$ hybridized. The ideal geometry (lowest ...

13

"Why" is a good question, and one that science has yet to fully answer. We generally have a good understanding of "how" things work at the subatomic level, based on over a century of observation followed by theory and math backed up by experimentation. However, the really basic, naively simple questions, like "where do particles get the charges, spins and ...

13

Internal energy, $U$, is the total energy contained in a thermodynamic system. However, absolute internal energy is hard to determine, and even relative internal energy and changes to internal energy are hard to determine. Here's why: Changes to internal energy usually occur through heat transfer $q$ or work done $w$. There are other ways to transfer energy ...

13

Bond formation is alway strictly exothermic in the sense of the change of enthalpy. exothermic reaction A reaction for which the overall standard enthalpy change $\Delta H^\circ$ is negative. A bond can only exist, if it needs energy to break it, i.e. the bond dissociation energy is always positive. bond-dissociation energy, $D$ The enthalpy (per mole)...

13

Of course it would break, just like you said; also, a high-energy $\beta$ particle would kill quite a lot of bystander molecules. Also, if not for other reason, the resulting molecule would no longer be DNA , since the decayed atom would no longer be $\rm P$. Also, the product would no longer be radioactive, so we wouldn't be able to detect it anyway. The ...

13

$\ce {MgO}$ ($\approx 3800$ $\pu{kJ mol^{-1}}$) has higher lattice energy than $\ce {LiF}$ ($\approx 1045$ $\pu{kJ mol^{-1}}$) mainly because of the greater charge on $\ce{Mg^2^{+}}$ ion and $\ce {O^{2-}}$ as lattice energy is directly proportional to the charges of the combing atoms. Your reasoning isn't incorrect but remember that $\ce{Li}$ and $\ce{Mg}$...

13

Authors may be sloppy about notation in this matter. I recommend considering $R_\ce{H} \approx \pu{10973 cm-1}$ and $Ry \approx \pu{2.18e-18 J}$, noting $Ry = hc \cdot R_\ce{H}$. Units of wavenumbers $(\pu{cm-1})$ and energy are often considered interchangeable in practice because they are proportional to each other by the constant value $hc$. In my notes, ...

12

The butane and oxygen in the room are in a metastable state. An example energy diagram of a metastable state is shown in the image below. Before throwing in the cigarette, the butane and oxygen molecules are in the state labeled as 1. This state is metastable because there is a lower energy state (which makes it statistically more probable) available if ...

11

I think the simplest way to explain what is going on is by way of a classical analogy. Picture an elastically bound ball that whizzes back-and-forth along the $x$ axis. Next, whack the moving ball hard with a $y$-aligned impulse. The result is an elliptical or figure-8 path, one for which the maximum radius of motion along the $x$ axis will, in a perfect ...

11

Because of the penetration, that is, because the 2s electron can spend time near the nucleus, it is less effectively shielded by the core electrons than a 2p electron. Because it is less effectively shielded, a 2s electron experiences a higher effective nuclear charge and is held closer to the nucleus than a 2p electron which gives the 2s orbital a lower ...

11

No, nickel-62 is the most stable on a binding energy per nucleon basis. Fe-58 is second and Fe-56 is third. See Fewell, M. P., "The Atomic Nuclide with the Highest Mean Binding Energy", Am. J. Phys., vol. 63, pages 653-658.

11

Dioxygen, $\ce{O2}$ is a very special molecule. A good majority of organisms on earth use dioxygen to breathe and survive. Oxygen is also the second most abundant element in the sky, as well as the most abundant elements in the earth's crust. What makes it so special? Its because of its abundance, which is obvious, and also because of it's strong oxidizing ...

10

This is not a thorough answer. It holds when comparing orbital energies of the same shell number $n$ (in your case 2) in atoms containing a low number of electrons. This question contains more information on this. The letter of the orbital ($s$, $p$, $d$, etc.) is a nameplace for the azimuthal quantum number (orbital angular momentum, $l$). The $s$ orbital ...

10

In order to answer this question, one needs to define what the absolute energy of a system is. Energy can be trapped in a system in ways not yet discovered or fully understood. Think of the energy associated with mass ($E=mc^2$), which is a result that was not known to the early founders of thermodynamics. We need to define reference points, which we can ...

10

Single point energy arises in the framework of the Born–Oppenheimer approximation and corresponds to just one point on the potential energy surface. Physically it is the total energy of the molecular system with its nuclei beeing fixed (or clamped) at some particular locations in space. In other words, it is total energy of the molecular system within the so-...

10

TL;DR Your Maxwell–Boltzmann diagram up there is not sufficient to describe the variation of rate with $E_\mathrm{a}$. Simply evaluating the shaded area alone does not reproduce the exponential part of the rate constant correctly, and therefore the shaded area should not be taken as a quantitative measure of the rate (only a qualitative one). There is a ...

10

Your question is quite broad, so I will tackle it in parts. Element abundance there is nothing as familiar to mankind (in terms of the frequency in their occurrence) as the reactions involving this compound This is because frequency depends on abundance. If you have a look to the element abundance on the earth oxygen is the first: Element Amount ...

10

As a chemist, I would agree that textbooks are not clear about the term stability and energy. It is not your fault and undergraduate organic chemistry books make the situation worse. I cannot recall the text, it was an old book, however it clearly said that stability of a compound does not mean anything. We should always ask, stability with respect to what? ...

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