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43

I'm a physicist, so apologies if the answer below is in a foreign language; but this was too interesting of a problem to pass up. I'm going to focus on a particular question: If we have oxygen and nothing else in a box, how strong does the magnetic field need to be to concentrate the gas in a region? The TL;DR is that thermal effects are going to make ...

38

Physics says yes. And at the US Patent Office, it certainly looks like the answer is yes, according to (at least) this US patent application. The abstract reads: A process for separating O$_2$ from air, that includes the steps effecting an increase in pressure of an air stream, magnetically concentrating O$_2$ in one portion of the pressurized air stream, ...

18

No, to separate oxygen from air you need extremely high gradients of field strength. in this paper http://link.springer.com/article/10.1007%2Fs11630-007-0079-1#page-1 they used about 0.4T per mm, but to cause breathing problems it is enough to get the O2 concentration below 17%, so lets say 0.1T/mm is needed. To sustain such gradients over a whole room (lets ...

14

TL;DR : There's no reason for it to show any periodicity. It doesn't depend in a simple way on the outer electron configuration of atoms. The specifics of crystal structure are important in ferrmoganetic (also in antiferromagnetic, and ferrimagnetic) materials. This by no means is a complete description of the phenomenon. If you're interested, consult ...

11

I believe GATE is a university entrance exam in India, so they will not expect you to solve an extremely complicated equation to predict ferromagnetism or to memorize an infinite list of substances. Sadly, there is some element of rote memorization still lingering like a pest in the educational testing system. Good news for you is that relatively few ...

10

Some matchstick heads contain iron(III) oxide as a colorant. The yellowish color of the burning match indicates that it has low oxygen, i.e. a reducing flame. It reduces the iron oxide to iron which is attracted by the magnet. The reduction reactions that occur are probably quite complex. Below are two simplified possible reaction equations. $\ce{(CH2)_n}$ ...

8

The ur-piezoelectric crystal is quartz. Being composed of $\ce{SiO2}$, it more or less epitomizes chemical/biological inertness. It has the additional advantage of being comparatively cheap and common. There are a number of other natural minerals which are piezoelectric (e.g., various perovskites). Most of those minerals are, as far as I know, relatively ...

8

Can PBE (and LDA) actually be a better choice sometimes? Of course, they can. This is in fact one of the major problems with DFT: there is no systematic way of improving a functional, so we never know a priori which will work better for some particular problem. The only thing we know is that some functionals perform systematically better that the others for ...

8

Simplistically speaking orbital angular momentum is present when some conditions are satisfied: A set of orbitals are degenerate; These orbitals can be "interconverted" by rotation about a certain axis; The set of orbitals is not empty, half-filled, or fully filled. Physically speaking, the "rotation" of an electron from one orbital to the next generates ...

8

The reasons may possibly be: T1shortening Relatively less adverse reactions Increased contrast enhancement The most commonly used clinically approved contrast agents for MR imaging are gadolinium-based compounds that produce T1 shortening. Tissue relaxation results from interactions between the unpaired electron of gadolinium and tissue hydrogen ...

8

Don't use a mixture, just use pure Hg. Chill to 4.2 K, conveniently reached with liquid He, and all magnetic fields are excluded. Levitate on a magnet, sans yogi. Mercury in a static magnetic field, with direct current, forms a simple homopolar electric motor, creating a vortex. A safer way (sans neurotoxic Hg) to demonstrate a homopolar motor requires just ...

7

For the same reason $\ce{O2}$ is paramagnetic: degeneracy. The filled orbitals are $\sigma(\mathrm{1s})^2$, $\sigma^*(\mathrm{1s})^2$, $\sigma(\mathrm{2s})^2$, $\sigma^*(\mathrm{2s})^2$. The half-filled orbitals are $\pi(\mathrm{2p}_x)^1$ and $\pi(\mathrm{2p}_y)^1$. But you're right to think that this is unexpected: why are the $\pi(\mathrm{2p})$ ...

7

Maybe OP has already contacted the authors of this paper by now, but this was interesting. I'm no expert but I can Google things and this was too long for a comment. Antisymmetric exchange: At first I thought it was simply an exchange interaction where the wave function's sign is changed during exchange, now I don't think it's so simple. Antisymmetric ...

7

Magnetite has a spinel structure with two types of $\ce{Fe}$ sites: octahedral and tetrahedral, respectively in green and orange below. (in this structure, oxygen atoms are the small red balls). This structure is called a spinel, with $\ce{Fe^{2+}}$ ion in tetrahedral coordination and $\ce{Fe^{3+}}$> ions in octahedral sites. The coupling between atoms ...

7

If all the electrons are spin paired then the molecule will be in a singlet state$^*$ and will be diamagnetic, if they are not so paired then it will be will be paramagnetic. The rules are the same whether or not the molecule is 'organic'. Most 'normal' organic compounds, e.g. methane, benzene, etc. have paired electrons in each of the orbitals from the ...

7

Neon and argon are correctly eliminated, no questions asked. Henceforth, we do indeed need to pay attention to question semantics and the phrase ‘regardless of its electronic configuration’. The key here is that following Hund’s rule will only give you one possible electronic configuration. For example for carbon, Hund’s rule will give $\mathrm{1s^2\, 2s^2\,... 6$\ce{O_2}is paramagnetic. source: http://www.chem1.com/acad/webtext/chembond/CB-images/liquidoxygeninmagnet.jpg The stable Xenon isotope Xe-129 has been used in MRI studies: Hyperpolarized Xenon MR Imaging of the Brain A radically new magnetic resonance imaging (MRI) technique, using hyperpolarized 3He and 129Xe, is being developed to produce high-... 6 First of all, in your question i feel a confusion about the nature of magnetism and paramagnetism. These magnetic interactions have nothing to do with attraction/repulsion of electric charges, and magnets do not attract or repel Coulomb charges (when we discuss the magnetic part of the interaction). In short static, magnetic and electric properties are two ... 6 Firstly I recommend you to understand the concept of "Magnetic Susceptibility". My source for this is Miessler,Tarr Inorganic Chemistry Book 2nd edition. Consider Ni2+, its electronic structure is 4s03d8. This is a d8 ion. Then it has 8 electrons in d orbitals. d orbitals have 5 degenerate level of orbitals. Also you should know gyromagnetic ratio which ... 6 This is from Shriver and Atkins' Inorganic Chemistry (p.586, 2009 ed) : The magnetic moment of many d-metal ions can be calculated by using the spin-only approximation because the strong ligand field quenches the orbital contribution. But, for the lanthanoids, where the spin orbital coupling is strong, the orbital angular momentum contributes to the ... 6 Historically, the term ferromagnetism was used for any material that could exhibit spontaneous magnetization: a net magnetic moment in the absence of an external magnetic field (Wikipedia). However, in 1948, Louis Néel showed that there are two levels of magnetic alignment that result in this behavior (Ref.1): One is ferromagnetism in the strict sense, ... 5 You seem to be confusing the elements with the pure simple substances of the elements. In atomic terms, both the oxygen atom and the sulfur atom are paramagnetic, since they contain two unpaired electrons in the ground state. However, the substances oxygen (\ce{O2}$) and sulfur (generally intended as$\ce{S8}$) are not the same as the atoms. In the simple ... 5 Table salt (NaCl) is diamagnetic. For paramagnetism, one needs free unpaired spins. The NaCl lattice is built up from closed shell cations and anion, there is no such unpaired spin pure NaCl. Also: experimentally it is difficult to measure paramagnetism, and you generally cannot just try it with a magnet and observe it without some tricks. 5 I see your point of confusion! The above is the structure of tertiary butyl cation. It is clear that if we consider the electronic configuration of the central carbon it is$\mathrm{1s^2\,2s^2\,2p^1}$. Well yes but no. Remember that you don’t have a lone atom in space but a carbon bound to three other atoms. These three σ bonds must originate from ... 5 For somebody who is not well-versed in quantum mechanics, it might be natural to guess that this is the case. But if I were you, I wouldn't assume so. Both electrons share the same p-orbital, and are therefore still paired. They both occupy the entire volume of the p-orbital. What they do not do, is to divide the p orbital up into two half-orbitals, with ... 4 In short this is the way I imagine them, as slight directional changes in spins propagating through the system. The Heisenberg Hamiltonian for the exchange energy associated with magnetic coupling is a pairwise sum over an exchange integral$J_{ij}$for two sites$i$and$j$, with spin moments$\hat S_i$and$\hat S_j$. H_{ex}=-\sum _{i&... 4 Magnetism appears in many very different forms that have very different origins. All elements show some diamagnetism, even though often overshadowed by other, stronger, forms of magnetism, because it is a property of the bound electrons. The diamagnetic contribution caused by the inner electrons that are not part of a chemical bond will therefore remain ... 4 This is about the spin Hamiltonian of the system, and more like naming an empirical observation then giving an actual physical explanation. So in the most simple (regular) spin systems a Hamiltonian with H=JS1S2 where J is a scalar handles pretty well the behaviour os spin. Note, in this case there is no orientation preference for the coupled spin compared ... 4 Yes it has to do with the number of electrons. Consider a simple model of the electron on a ring, then the electron has to follow a circular path and its wavelength has to 'fit' exactly round the ring. Thus the wavelength has to satisfy the condition$L=q\lambda$where$q$is a quantum number$q=0, \pm 1, \pm 2 ...\$ characterising each eigenstate. The ...

4

I always suggest students to try Google Scholar (scholar.google.com) when a simple Google search fails. I just searched three keywords : alkali metals ammonia solutions and the third result is highly relevant. When your book talks about "in concentrated solution", it means more alkali metal in liquid ammonia. This paper, which you should search in Google ...

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