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The study of chemical systems using the laws and concepts of physics. This usually requires the techniques of thermodynamics, statistical mechanics and quantum mechanics.

5
votes
1answer
Explaining the reasons behind electron repulsion, Radiochemistry asserts that, Interelectronic repulsion in related not just to electron pairing, but also to [the] angular momentum of the electron …
asked Jun 14 '16 by Linear Christmas
13
votes
2answers
Summary of IUPAC definition $\def\d{\mathrm{d}}$In IUPAC recommendations from 1994[1, 1166–1167], the authors discuss the process $$\ce{A <-->[$k_1$][$k_{-1}$] X\\ X + C ->[$k_2$] D}$$ A kinetic st …
asked Nov 26 '17 by Linear Christmas
3
votes
3answers
Stuck proving the relation $$\mu_i=\mu_i^\circ + RT\ln a_i,\tag{1}$$ I looked up the definiton of activity. It is $$a_i=e^{\frac{\mu_i-\mu_i^\circ}{RT}},\tag{2}$$ but the the intuition/motivation f …
asked Oct 25 '16 by Linear Christmas
5
votes
1answer
I was reading a chemisty textbook when I came across the following statement. $\ce{Fe(OH)2}$ is not stable, even more so in basic conditions. [---] $\ce{Fe^2+}$ ions are also less likely to oxidis …
asked Jun 1 '16 by Linear Christmas
3
votes
1answer
If the ideal gas law $PV=nRT$ is not enough, as a second approximation the van der Waals equation $$\left(P+a\frac{n^2}{V}\right)\left(V-nb\right)=nRT$$ is applied. The constants $a$ and $b$ are van …
asked Jun 10 '16 by Linear Christmas
6
votes
2answers
In simple terms, the collision of two atoms $\ce{A}$ and $\ce{B}$ will result in ions $\ce{A^+}$ and $\ce{B^-}$ if $$I_a(\ce{A})+E_a(\ce{A})<I_a(\ce{B})+E_a(\ce{B})$$ where $I_a$ and $E_a$ are the ion …
asked Jun 29 '16 by Linear Christmas
4
votes
0answers
It seems to me that f-orbitals for lanthanide metals are treated as 'core-like' when a certain number of electrons have been removed. Or, as Radiochemistry puts it, The 4f binding energy is so g …
asked Jun 14 '16 by Linear Christmas
3
votes
0answers
For most lanthanide metals, the stable oxidation state is III [*]. The general electronic structure is $$\ce{[Xe] 4f^{0}^{-14} 5s^2 5p^6 5d^{0}^{-1} 6s^2}\ \ [**].$$ Elements that have the d-electron …
asked Jun 12 '16 by Linear Christmas
8
votes
1answer
For most lanthanide metals$^{[1]}$, the stable oxidation state is III. The general electronic structure$^{[2]}$ is $$\ce{[Xe] 4f^{0}^{-14} 5s^2 5p^6 5d^{0}^{-1} 6s^2}.$$ Elements that have the d-elec …
asked Jun 14 '16 by Linear Christmas