text:
What is charge shift bonding?
Can someone explain to me charge shift bonding? I recognize that it is proposed as another domain of bonding - one different from ionic and covalent bonding. I am also told that $\ce{C-F}$ exhibits charge shift bonding. But why—and firstly, what exactly is charge shift bonding? All the literature I have consulted are at too high of a level for me, for example this 1992 article in the Journal of the American Chemical Society.

OK, this was a new one to me too, but I found a few resources which might be useful:

A Computational Organic Chemistry blog post
Michael Scott Long blog post
Nature Chemistry review

The basic idea is this. Most bonds are ionic (i.e., both atoms are charged, but electrostatically attracted) or covalent (i.e., the electron pair is shared).
(This is largely cribbed from the Comp. Org. Chem. blog post. Thanks to Steven Bachrach)
If we were writing a multi-configurational wavefunction to describe the bond, we might have something like:
$$\psi(\mathrm{VB}) = c_1\psi_\mathrm{covalent} + c_2\psi_{\ce{A+B-}} + c_3\psi_{\ce{A-B+}}$$
So normally, we'd look at which is the dominant configuration - maybe it's the covalent one ($c_1$) or one of the ionic ones ($c_2$ or $c_3$). Instead, sometimes the covalent form isn't even stabilizing!
From my perspective, the classic example given for this charge-shift bond is $\ce{F2}$ which of course has many $\pi^*$ anti-bonding electrons, so the classical bond order is 1, but ignores that there's a lot of antibonding character. But you can't really consider $\ce{F2}$ an ionic compound!
I think the big question is the future - can this new type of bonding show general predictability and utility. Clearly most chemists have a general intuition (if sometimes wrong) of covalent and ionic bonding. Besides the authors, I'm not sure if anyone else has an intuition of charge-shift systems.

Would a classical explanation of the instability of F2 involve lone pair-lone pair repulsions?

IIRC, that's the Lewis-style explanation. I think the question is if there is so much LP-LP repulsion (i.e., six lone pairs!) why is there a bond?

Is there lp/lp repulsion in diatomic nitrogen as well?

I am not a VB expert. I'd answer from the MO perspective, which would say such effects are minimal. $N_2$ is, after all, a very strong bond, without anti bonding $\pi$ character.

Geoff has summarized well the mathematical, VB wavefunction-based description of charge-shift bonding.  This answer aims to provide further exposition, from the perspective of the distribution of the electron density and of the electron localization in charge-shift bonded as compared to traditional covalent systems. Much of my understanding and argumentation is derived from the Shaik/Hiberty Nature Chemistry review that Geoff linked to, but I also have read through a variety of other papers on the subject (a list of much of the literature I've turned up thus far is provided at the end of the answer).
Please bear in mind that I'm no particular expert in this topic, but am just summarizing my current understanding based upon what I've read to date.  Please also bear in mind that this description is at times intentionally qualitative/descriptive, rather than technically precise.  Due to both of these aspects, I welcome any corrections/improvements.

The first thing to grasp in understanding charge-shift bonding is the difference between the electron density at a point and the electron localization at a point.  The electron density is pretty straightforward, representing the 'concentration' of electrons at each point in space.  Electron localization is more subtle, and effectively represents the extent to which the electrons contributing to the density at a given point in space also "visit" other points in the system as they go about their statistical quantum-mechanical wavefunction-y meanderings. In particular, it's critical to understand that the density and the localization are not rigorously correlated: the localization can be low even when the density is high, and vice versa.
One of the most common metrics I've seen for localization is the electron localization function (ELF), which uses the Pauli exclusion behavior of electrons to define their extent of localization on a scale from zero to one. At a high level, this exclusion behavior means that wherever in space a particular reference electron "visits" in the statistical quantum-mechanical sense, the exclusion principle requires that the likelihood of finding a different electron of the same spin in that spatial region must be correspondingly reduced. This distribution of the decreased likelihood of finding another electron due to the statistical "visiting" of a reference electron is called the Fermi hole for that reference electron.  So, if the aggregate Fermi hole associated with the electron density at a given point is compact and close to that point, the ELF will be high there (close to unity).  Conversely, if the Fermi hole is diffuse and spread far from the reference point, then the ELF will be lower.
The second thing to grasp in understanding charge-shift bonding is the phenomenon of Pauli repulsion, also called 'exchange repulsion,' which is a truly weird, quantum-mechanics-only thing.  In this case, if the wavefunction for a system "tries to push too many electrons too close together," Pauli repulsion causes them to delocalize away from the region where they're "too compressed."  In other words, the "compressed" electrons will in general end up "visiting" a much larger region of space than they would if fewer electrons were being squeezed into the area of interest.  Again, as noted above, this delocalization doesn't necessarily mean that the electron density will decrease appreciably in the Pauli-repulsed region—in many cases the density will remain relatively high even as the localization drops precipitously.

Ok, so: on to charge-shift bonding. If the localization of the electron distribution for a particular bond were a balloon and Pauli repulsion were an adorable puppy, charge-shift bonding is basically this:

In situations where a molecule "tries to cram too many electrons into a bonding region," the bonding electrons "rebel" and flee (delocalize) to other regions of the molecule.  The most straightforward way to describe this behavior in a wavefunction is via the valence bond-type mixed covalent/ionic approach presented by Geoff, and is indeed the way Shaik and Hiberty cast it in the Nature Chemistry review. To use the same $\ce{F2}$ example as Geoff, the charge-shift bonding in molecular fluorine can be represented as a resonance:
$$
\def\splrar{\enspace \longleftrightarrow \enspace}
\ce{F+\!\cdots F^- \!\splrar F-F \splrar F^-\!\cdots F^+}
$$
Just like the six-fold-symmetric electron distribution of benzene can be thought of as a resonant superposition of two structures with discrete, alternating double and single bonds, the electronic structure of molecular fluorine can be thought of as a resonant superposition of covalent and ionic electronic configurations. The two ionic resonance structures provide the 'space' for the 'fleeing electrons' to delocalize into, pushed out from the bonding region by Pauli repulsion. In this case, since $\ce{F2}$ is symmetrical about the bond, the contributions of the two ionic resonance structures are necessarily equal; this will not always be the case in general.

In order to demonstrate the nature of charge-shift bonding more explicitly, I'll show the results of a QTAIM and ELF analysis of the isoelectronic series $\ce{A2H_x}~\left(\ce{A} = \ce{\{C,N,O,F\}},~x=\{6,4,2,0\}\right)$, which is one of the prototypical examples of the spectrum between covalent and charge-shift bonding as described by Shaik and Hiberty.
I ran DFT geometry optimizations on each of the four species with ORCA v3.0.3, with input files like the following:
! UKS PBE0 def2-TZVP def2-TZVP/J RIJCOSX
! GRID5 GRIDX8 TIGHTOPT
! ANFREQ

* xyzfile 0 1 C2H6_init.xyz

The initial geometries were created in Avogadro, and I included the frequency calculation to confirm that I'd obtained minimum-energy geometries (probably overkill for such simple systems, but it never hurts to check).  I then used the ORCA tool orca_2mkl to generate MOLDEN files for import into MultiWFN v3.3.7 for QTAIM topology and ELF basin analyses, and for electron density and ELF contour plots.
The following figure shows electron density contour plots for the four species (click any of the below figures for the full size version):

For clarity, I've intentionally plotted a comparatively small number of contour lines.  I'm pretty sure the scale in each figure is the same, so visual inspection of the contours should give a reliable sense of their relative sizes. For emphasis, I've bolded the $0.2$ contour; while the various hydrogens and lone pairs obscure things somewhat on the periphery of the molecules, the electron density in the area between the two heavy atoms in each system (viz., in the bonding region) really doesn't change all that much in going from $\ce C$ to $\ce F$.  The electron density does increase somewhat in the vicinity of each heavy atom along the series (the increase in the diameter of the $1.0$ contour is the most visible indicator of this), which nicely correlates with the increase in electronegativity / nuclear charge.
This relatively modest change in the bonding electron density along the series stands in stark contrast to the trend in electron localization in the bonding region, as illustrated by ELF contour plots of the four species:

Again, for clarity I've intentionally only plotted a few contour lines, and again I've bolded one contour $(0.7)$ for emphasis. Here the differences in going down the series are stark: the localization in the $\ce{C\! -\! C}$ bonding region is large and strong, peaking well above $0.8$. The $\ce{N\! -\! N}$ bond localization is relatively strong but noticeably reduced, whereas in $\ce{H2O2}$ the $\ce{O\! -\! O}$ bond localization has fallen to the point that it doesn't even have an $0.8$ contour line.  $\ce{F\! -\! F}$ is even more dramatic, as the bond region doesn't even have an ELF contour line at $0.7$.  The ELF basins as calculated by MultiWFN show the same decreasing trend in terms of size going down the series:

In the above, the clusters of dark green dots mark the ELF basins for the bonding regions, while the light green dots mark ELF attractors for lone pairs.  The fluorine atoms in $\ce{F2}$ have "rings" of degenerate lone pair attractors since there are no hydrogen nuclei present to 'lock down' the positions of the lone pair attractors, in contrast to $\ce{H2O2}$ and $\ce{N2H4}$.
Both QTAIM analysis at the line critical points (LCPs, called bond critical points by Bader) between the two heavy atoms, as well as population/localization analysis within the ELF bonding basins, reveal illustrative trends (or the lack thereof) within this series of molecules:
$$
\begin{array}{ccccccc}
\hline
~ & ~  &  \text{Units}  &  \ce{C2H6}    &   \ce{N2H4}   &  \ce{H2O2}    &    \ce{F2}  \\
\hline
~ & \rho   &  {\mathrm{e}^- \over \mathrm{Bohr}^3} & 0.247   &  0.317   &    0.300  &  0.316  \\
\text{Bond LCP} & \nabla^2\rho  &  {\mathrm{e}^- \over \mathrm{Bohr}^5} & -0.617   &  -0.609    & -0.012    & 0.514  \\
~ & \mathrm{ELF}  &  {\small (unitless)}  & 0.956   & 0.895  &  0.769  &  0.647  \\
\hline
~  & \iiint{\rho}  & \mathrm{e}^-  & 1.816   & 1.346  &  0.608    &  0.249  \\
\text{ELF Basin}  & \text{Volume} & \mathrm{Bohr}^3 & 17.65   &  7.02   &  2.33    &   0.76   \\
~ & \text{%LI}     &   {\small (unitless)}   &     46.5\%   & 37.1\%   &  20.4\%   &   9.6\% \\
\hline
\text{Shaik %CS} & ~ &   {\small (unitless)}    & 30.2\%   &  65.7\%   &  114.3\%   &  183.9\%   \\
\hline
\end{array}
$$
The "Shaik %CS" values represent the fraction of the bond energy provided by the charge-shift character of each bond as reported in the Nature Chemistry review, in this case with the remainder provided by the "pure" covalent electronic structure for the system.  Thus, for $\ce{C2H6}$, $30.2\%$ of the bond energy derives from charge-shift behavior, while $69.8\%$ is due to the covalent contribution. As can be seen, it is possible for this value to be greater than $100\%$, in which case the purely covalent structure would be unbound. The experimentally observed stability of $\ce{H2O2}$ and $\ce{F2}$ is thus due entirely to the charge-shift phenomenon.
Consistent with the contour plots, there is minimal change in the LCP electron density through the series, whereas both (a) the ELF at the bond LCP and (b) the fraction of electrons localized within the ELF bonding basin $(\text{%LI})$ drop significantly in going from $\ce{C2H6}$ to $\ce{F2}$.  There is a reasonably significant (albeit nonlinear) trend in the Laplacian of the density ($\nabla^2\rho$), but it appears to be a quirk of the homodiatomic nature of the systems: Shaik and Hiberty have found heteroatomic systems with significant charge-shift character that still show a negative value for the Laplacian at the bond LCP. To emphasize: the sign of the Laplacian is not a reliable diagnostic of charge-shift character!
The ELF basin populations show the same drop along the series, though this is due more than anything to the precipitous decrease in the sizes of the ELF bond basins (recall that the electron density in the bonding region changes very little among the systems).

In summary, charge-shift bonding occurs when the molecular structure requires a sufficient number of electrons to reside in the bonding region of interest to result in substantial delocalization of those electrons away from that bonding region, driven by the effects of Pauli repulsion.  Per the Shaik & Hiberty Nature Chemistry review, systems where charge-shift bonding is expected to occur include: small, lone-pair rich, and/or electronegative atoms; $\pi$-bonds in doubly and triply bonded molecules; hypercoordinated compounds (oxoanions, $\ce{IF_x}$, etc.); compact cations with charge strongly localized on a single nucleus (e.g., $\ce{(CH3)3Si^+}$); and, potentially, certain organometallic and/or metal-metal bonds.

Selected Literature
$\require{begingroup}\begingroup$
This list includes all of the papers I could find by Hiberty and/or Shaik that pertain to charge-shift bonding, as well as a few I've come across written by other authors.
Papers I would regard as core reading are marked with '$\def\bp{\mathbf +}\bp$'.
Papers not authored by Shaik or Hiberty are marked with '$\def\bt{\mathbf \times}\bt$'.

"Covalent, ionic and resonating single bonds." J Mol Struct (Theochem) 229: 163 (1991) – doi:10.1016/0166-1280(91)90144-9
"The charge-shift bonding concept. Electron-pair bonds with very large ionic-covalent resonance energies." J Am Chem Soc 114: 7861 (1992) – doi:10.1021/ja00046a035 $\bp$
"The non-linear tendencies in homonuclear $\ce{X-X}$ bonds $(\ce{X = Li\text{ to }F})$ and the lone-pair bond weakening effect. An ab initio theoretical analysis." J Mol Struct (Theochem) 338: 283 (1995) – doi:10.1016/0166-1280(94)04067-3
"Comparison of $\ce{C−Cl}$ and $\ce{Si−Cl}$ Bonds. A Valence Bond Study." J Phys Chem 100: 5715 (1996) – doi:10.1021/jp960145l
"Charge-Shift Bonding in Group IVB Halides: A Valence Bond Study of $\ce{MH3−Cl (M = C, Si, Ge, Sn, Pb)}$ Molecules." J Am Chem Soc 121: 822 (1999) – doi:10.1021/ja982218f – Erratum: doi:10.1021/ja995530d
"$\pi$ Bonding in Second and Third Row Molecules: Testing the Strength of Linus's Blanket." Chem Eur J 6(13): 2425 (2000) – doi:10.1002/1521-3765(20000703)6:13<2425::AID-CHEM2425>3.0.CO;2-0
"Charge-Shift Bonding—A Class of Electron-Pair Bonds That Emerges from Valence Bond Theory and Is Supported by the Electron Localization Function Approach." Chem Eur J 11: 6358 (2005) – doi:10.1002/chem.200500265
"Barriers of hydrogen abstraction vs halogen exchange: An experimental manifestation of charge-shift bonding." J Am Chem Soc 128(9): 2836 (2006) – doi:10.1021/ja053130m
"The physical origin of large covalent-ionic resonance energies in some two-electron bonds." Faraday Discussions 135: 261 (2007) – doi:10.1039/b605161h $\bp$
"Heterolytic bond dissociation in water: Why is it so easy for $\ce{C4H9Cl}$ but not for $\ce{C3H9SiCl}$?" J Phys Chem A 112(13): 2988 (2008) – doi:10.1021/jp8004647
"Charge-shift bonding and its manifestations in chemistry." Nature Chemistry 1: 443 (2009) – doi:10.1038/nchem.327 $\bp$
"An Excursion from Normal to Inverted $\ce{C-C}$ Bonds Shows a Clear Demarcation between Covalent and Charge-Shift $\ce{C-C}$ Bonds." ChemPhysChem 10: 2658 (2009) – doi:10.1002/cphc.200900633
"The electronic structure of the $\ce{F2}$, $\ce{Cl2}$, $\ce{Br2}$ molecules: the description of charge-shift bonding within the generalized valence bond ansatz." Theor Chem Acc 122(1-2): 51 (2009) – doi:10.1007/s00214-008-0484-x $\bt$
"Topology of Electron Charge Density for Chemical Bonds from Valence Bond Theory: A Probe of Bonding Types." Chem Eur J 15(12): 2979 (2009) – doi:10.1002/chem.200802134
"Halogen Bonding: A Study based on the Electronic Charge Density." J Phys Chem A 114: 552 (2010) – doi:10.1021/jp907550k $\bt$
"The Nature of the Idealized Triple Bonds Between Principal Elements and the sigma Origins of Trans-Bent Geometries-A Valence Bond Study." J Chem Theor Comput 7(4): 955 (2011) – doi:10.1021/ct100741b
"The mechanics of charge-shift bonds: A perspective from the electronic stress tensor." Chem Phys Lett 510: 18 (2011) – doi:10.1016/j.cplett.2011.05.023 $\bt$
"An MO-Based Identification of Charge-Shift Bonds." ChemPhysChem 13: 2377 (2012) – doi:10.1002/cphc.201200147 $\bt$
"The essential role of charge-shift bonding in hypervalent prototype $\ce{XeF2}$." Nature Chem 5(5): 417 (2013) – doi:10.1038/nchem.1619
"Charge-Shift Bonding Emerges as a Distinct Electron-Pair Bonding Family from Both Valence Bond and Molecular Orbital Theories." J Chem Theor Comput 10: 2410 (2014) – doi:10.1021/ct500367s $\bp$
"A Valence Bond Model for Electron-Rich Hypervalent Species: Application to $\ce{SF_n}$ $(n=1, 2, 4)$, $\ce{PF5}$, and $\ce{ClF3}$." Chem Eur J 20(31): 9643 (2014) – doi:10.1002/chem.201402755
"Protonated Alcohols Are Examples of Complete Charge-Shift Bonds." J Org Chem 79(21): 9998 (2014) – doi:10.1021/jo501549q
"More insight in multiple bonding with valence bond theory." Comput Theor Chem 1053(SI): 180 (2015) – doi:10.1016/j.comptc.2014.09.007 $\bp$
"The Nature of Bonding in Metal-Metal Singly Bonded coinage metal dimers: $\ce{Cu2}$, $\ce{Ag2}$ and $\ce{Au2}$." Comput Theor Chem, in press doi:10.1016/j.comptc.2017.02.013

$\endgroup$

I think the two answers currently here are excellent, very detailed and a great read. I thought I would give a different perspective on charge-shift bonding, from the view of the virial theorem.
The weakness of the $\ce{F-F}$ bond in difluorine, $\ce{F2}$, is usually attributed to the lone pair bond weakening effect (LPBWE). This is classically seen as the lone pairs on $\ce{F2}$ repelling the bonding electrons, and so weakening the bond. In MO theory it is attributed to the antibonding effect of filling two antibonding π orbitals. It can also been seen in a different light, that of the balance of kinetic energy ($T$) and potential energy ($V$).
Usually when a covalent bond forms, there are 2 "phases" of bonding. Consider $\ce{H2}$. At infinite separation there is no overlap of the H 1s orbitals. Alternatively, there is no probability of the H 1s electrons tunneling from one nucleus' potential well to the other H atom's potential well. When the H atoms are brought to a finite distance apart, there becomes a finite probability of the H 1s electrons tunneling from one nucleus' potential well to the other. This means that each electron is more spread out - i.e. it occupies a larger volume. This can been seen as making the wavefunction more diffuse, less steeply varying, and so the curvature and Laplacian of the wavefunction takes smaller values. This means the kinetic energy of the electron is consequently smaller - a smaller Laplacian at all distances, averages to give an expectation kinetic energy which is also smaller. This is the primary mode of bonding interaction at long distances - a decrease in kinetic energy.
At around two times the equilibrium bond length $\ce{r0}$, the kinetic energy reaches a minimum (this is for the case of $\ce{H2}$, not generally). This puts the kinetic energy ($T$) and potential energy ($V$) out of balance:
$$\frac{\langle T \rangle}{-\langle V \rangle} < \frac{1}{2}$$
This is the virial theorem, stated for equilibrium bond lengths. The next "phase" which occurs is orbital shrinking - this lowers the potential energy term, and also increases the kinetic energy term, and so restores the balance between the two terms. Thus by the time the equilibrium bond distance is achieved the kinetic energy term is positive, and the potential energy term is twice as negative, leading to an overall binding energy:
$$E = T + V$$
Turning to the case of $\ce{F2}$, let's consider this orbital shrinkage. Due to the presence of lone pairs, Pauli repulsion - where 2 electrons of the same spin can't occupy the same space - brings an additional rise in kinetic energy. This can mean that instead of a deficit of kinetic energy, there is a surplus of kinetic energy. The most extreme example is difluorine, $\ce{F2}$, which doesn't have a covalent minimum - the lone pairs simply repel each other too much, and cancel out any decrease in kinetic energy which may occur due to usual covalent mechanism, of tunneling from one nucleus to the other.
This gives us all we need to understand the phenomena of charge-shift bonding. When bonding was first considered, a supposition of wavefunctions was assumed:
$$\psi_{\mathrm{VB}} = \psi_{\ce{A-B}} + \psi_{\ce{A+B-}} + \psi_{\ce{A^-B+}} $$
The first term is the covalent wavefunction, and the second and third are the ionic contributions. In terms of energy levels and resonance energies we get:

On the left we have the situation for a mostly covalent bond, and on the left the situation for a mostly ionic bond.
Pauling's assumption was that the resonance energy between the covalent and ionic structures was very small, and so proceeded to calculate based on the assumption the $RE_{CS}$ was zero. His second assumption was that the covalent structure was always a binding structure.
Taking again the structure of $\ce{F2}$ the covalent structure is non-bonding at all distances. However by adding in ionic contributions: $\ce{F+ F- <-> F- F+}$ with the covalent structure, a large $RE_{CS}$ is generated providing the binding energy.
Now what do these ionic structures mean? To tie into the excellent answers above, based on electron localisation functions (ELF), we can think of these ionic structures as representing the large electron fluctuation away from the binding region. These ionic structures therefore give a lowering in kinetic energy, with little effect on the potential energy - Precisely what is needed to correct the surplus of kinetic energy caused by LPBWE. In difluorine this effect accounts for the binding interaction completely. Down 1 period, $\ce{Cl2}$ this accounts for a shortening and strengthening of the bond. To tie in with the above idea of orbitals typically contracting during covalent bonding, we find that for $\ce{H2}$, $\ce{H3C-CH3}$, $\ce{H2N-NH2}$ we find significant contraction of the orbitals - IE a covalent interaction. In $\ce{H2O2}$ and $\ce{F2}$, $\ce{Cl2}$, for the purely covalent structures, the orbitals actually increase in size, due the LPBWE. However when we add in the ionic structures, we find that in all cases the orbitals have contracted to some extent. In the case of CS bonds - $\ce{H2O2}$ ,$\ce{F2}$ and $\ce{Cl2}$ - this occurs because the addition of ionic structures negates the increase in KE due to the LPBWE, and so allows for a small degree of orbital shrinking to take place. Before adding in ionic structures, orbitals actually grow, in an attempt to decrease KE by delocalisation.
This picture also ties in with the ELF diagrams shown in other answers. Difluorine looks as if the fluorine atoms have a very much "closed shell" interaction" - the disynaptic region is depleted, with minimal electron density, and a large electron fluctuation away from the site. This matches with the "closed shell" ionic structures, in which there is no "sharing" of electrons.
In summary, charge shift bonding can be seen to stem from the LPBWE whereby lone pairs excessively raise the kinetic energy. In this case the normal mechanism of covalent bond forming - orbital shrinkage - is no longer appropriate. In order to restore the balance of kinetic and potential energy, it is necessary to add in ionic structures, in a VB approach, which represent the electrons becoming more delocalised/less localised, away from the bonding site, thereby lowering the kinetic energy. Once this has happened, orbital shrinkage can occur to a small extent. In this way charge shift bonding can be characterized by small, electron rich fragments, typically electronegative - Eg. Halogens, peroxides, with minimal electron density in the binding region, and electrons which are localised only to a small extent - Pauli repulsion "pushes" them away, the ionic structures "providing space" in a sense for this to occur. Obviously the ionic structures are simply resonance forms - what they describe is the delocalisation of electron density away from the binding sites, resulting in a decrease in kinetic energy, and so restoring the balance of kinetic and potential energy dictated by the virial theorem.


text:
How to calculate the mass of a component in battery acid based on the density and mass fraction?

Battery acid has a density of $\pu{1.285 g cm^-3}$ and contains $38\%$ by weight sulfuric acid. How many grams of pure sulfuric acid are contained in a liter of battery acid?

I just took $1.285 \times 0.38$ and got how many grams of sulfuric acid are in $\pu{1 cm^3}$ of battery acid. So now I just converted it into grams per liter and got $\pu{488.3 gL^-1}$ of sulfuric acid in battery acid. I'm not sure if I did it right though.

Please have a look at this tutorial to acquaint yourself with the way math and chemical formulae can be nicely formatted on this site.

duplicate of http://math.stackexchange.com/questions/930540/math-problem-disguised-as-a-chemistry-problem/930581#930581

Please do not cross-post your questions.  It results in a duplication of efforts and wastes the time of the people that have answered it.

Yes, you did do it correctly. However, I would refer you to the homework policy of this site.
$\textrm{density} = \frac{\textrm{mass}}{\textrm{volume}}$
$\textrm{mass}=\textrm{density} \times  \textrm{volume} = 1.285 \times 1000=1285\ \textrm{g}$
Then you just multiply by 0.38 (which is the same as 38%) and you get the answer.
Just a note, next time you ought to phrase your question by asking about the specific topic of the question (e.g How do I calculate the mass given the percentage...... etcetc), rather than the question itself.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
What happens when potassium chloride, potassium dichromate and concentrated sulphuric acid react?
What happens when $\ce{KCl}$, $\ce{K2Cr2O7}$, and concentrated $\ce{H2SO4}$ reacts?
I have no idea how to approach these questions. I'd love if anybody can help me see the logic behind this.

When a mixture of  $\ce{KCl}$, $\ce{K2Cr2O7}$, conc. $\ce{H2SO4}$ and heated, then they produce Chromyl chloride ($\ce{CrO2Cl2}$) , $\ce{KHSO4}$, $\ce{H2O}$ are formed.
$$\ce{K2Cr2O7 +4KCl + 6H_2SO4 → 2CrO2Cl2 +6KHSO4 + 3H2O}$$
The above reaction is commonly known as chromyl chloride test, used for testing the presence of $\ce{Cl-}$ ion in inorganic qualitative analysis.

Have a look at the properties of the compounds involved.
Dichromate ($\ce{Cr(VI), Cr2O7^{2-}}$) is a strong oxidant and is reduced to $\ce{Cr(III)}$ if it finds a suitable partner (which then is oxidized).
Here, a potential partner in a redox reaction is chloride: $\ce{2 Cl- + 2 e^{-} -> Cl2}$.

The redox reaction doesn't takes place. This is a famous test to chloride ion, know as the chromyl chloride test. It is an non redox reaction.

I like the question above, since it gives a general strategy.
I'll point out that mixing dichromates and sulfuric acid makes "dichromix" and there are a complicated mixture of (chromic acid)[http://en.wikipedia.org/wiki/Chromic_acid] and related compounds. Depending on pH and other factors, you can have chromates, dichromates, and even sulfur-chromate mixed ions.
At one point chromic acid was used to clean glassware, since they are very strong oxidizing agents.
SAFETY NOTE The use of chromic acid (aka "chromerge") is highly unsafe and not recommended!

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Nature of spontaneous reactions
The definition of spontaneous process states that "A spontaneous process is an irreversible process and maybe reversed by some external agency"
So, can we say that all spontaneous reactions are irreversible reactions under the absence of some external force?

A spontaneous reaction, as contrasted with a non-spontaneous reaction, is a reaction quantified by a negative free energy change. A reaction is said to be spontaneous if it occurs without being driven by an outside force. So, we can say:  In any spontaenous process, the path between the reactants and products is irreversible


text:
Unequal ionization energies of methane
Why does methane have two different ionization potentials? How does this work? I understand that MO theory predicts C-H bonds of differing strength, while hybridization predicts C-H bonds of varying strength.
http://books.google.com/books?id=gY-Sxijk_tMC&pg=PA827&dq=methane+ionization&hl=en&sa=X&ei=VrMVVN3JLI-vogTv8IGABw&ved=0CCgQ6AEwAA#v=onepage&q=methane%20ionization&f=false
Is methane ionized successively and is are there two different ionization energies found? Doesn't this make intuitive sense; by ionizing methane we end up with a fundamentally different molecule which would more than likely have different ionization energies.
Also, how are these ionizations performed? The reference given above makes no mention of whether the bonds are broken homolytically or heterolytically (although ionization implies heterolytic bond cleavage). Even then, there are two ways bonds can be cleaved heterolytically.

In the photoelectron spectroscopy (PES) experiment a sample is irradiated with photons of differing energy and the kinetic energy of any ejected electrons is measured.  By knowing the energy of the incoming photon and the kinetic energy of any ejected electrons, we can calculate the energy required to ionize the electron as the difference between these two recorded values. An electron will only be ejected if the photon energy is equal to or greater than the energy required to ionize an electron in the molecule.  See the section entitled "Photoelectron Spectroscopy" in this link if you would like to read more.
In the case of methane, all of the bonds are equivalent as required by symmetry.  There is only one $\ce{C-H}$ bond strength for methane.
When we run the PES experiment with methane we see 3 signals (as pictured in your link; two of the signals are due to photoionization of electrons from the $\ce{C-H}$ bond, and the third signal, at much higher photon energy, is due to photoionization of the carbon 1s electrons).  We know that the 4 bonds in methane can be described by 8 orbitals, 4 bonding, 4 antibonding.  We can transform these 8 localized orbitals into 8 molecular orbitals; the 4 bonding MO's are pictured in the link you provided.  Notice from the picture in your link that 3 of the 4 MO's are degenerate, but there is also a different 4th MO.  All of these bonding MO's are filled and ionization can occur from either one of the 2 types - hence 2 signals from the occupied $\ce{C-H}$ MO's (in the PES experiment, these are not sequential ionizations, e.g. these are not first and second ionizations; rather they are independent ionizations that will occur only when our incoming photon energy matches the energy of the orbital containing electrons).  During the PES experiment with methane, no bonds are broken, rather an electron from an orbital is ejected and the methyl radical cation is formed.


text:
How to calculate the mass of hydrogen in a borane sample?

In a $\pu{45.9 g}$ sample of borane, a compound containing only boron and hydrogen, that's $85.63\%$ boron, what is the mass of hydrogen in this sample?

I was just going to take $\pu{45.9 g}$ and multiply by $0.1437$ to get the mass of hydrogen, but I don't think it works that way because they don't equally contribute to the mass of the compound. Then I thought, maybe I should consider the molar masses of both atoms, hydrogen being $\pu{1.008 g mol-1}$ and boron being $\pu{10.811 g mol-1}$. I am stuck at this point. How to solve the question.

The problem does not need any difficult approach, not even relative atomic masses, if you trust the one who made this exercise. Since borane (whatever type it is)* consists (by exercise's hypothesis) only of hydrogen and boron and since boron is $85.63\%$ by mass, hydrogen's mass percentage would be the remaining $100\% - 85.63\% = 14.37\%$. Therefore, we have $\pu{6.60 g}$ of hydrogen in this sample.

The compound borane, $\ce{BH3}$, is about $78.1\%$ boron per mass. However, there is a class of compounds called boranes which also only consist of boron and hydrogen. From this class pentaborane(9), $\ce{B5H9}$, with about $85.7\%$  boron by mass would fit the exercise.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why does rounding off the mol values when finding an empirical formula lead to a different result?
The question I was doing tells you that the compound consist of 49.0% carbon, 2.7% of hydrogen ,and 48.2% of chlorine. And ask you to find the empirical formula.
I did the usual procedure of dividing the percentages by each elements atomic mass. Because the result for carbon was 4.0796, I thought I should rounding to four moles of carbon. And for hydrogen and chlorine the numbers of mole for each are 3 and 1.  I want to check the formula is correct, so I make it 8 moles of carbon, but then the number of mole for hydrogen turns out to be 5 and chlorine 3.
Now there is a new empirical formula. Which one is correct?
PC: when I divided the percentages by atomic mass, I rounded the results to ten thousandth place.

Nope, not exact same

Please show us how you got 3 for Cl and 1 for H.

3 for H actually

Actually, it would be better to see all your work on the problem.

There u go, LDC3

To me, it looks like 4, 2 2/3 and 1 1/3. So after multiplying by 3 should give you the integer ratios of the 3 elements.

But it's not whole number ratio.

Please don't title your posts something like "Help me correct my mistake."  Taking a little bit of time to come up with a title that encompasses what the post is about will help people find it.

Sure they are after multiplying by 3: $4\times 3=12, 2\frac{2}{3}\times 3=8$ and $1\frac {1}{3}\times 3=4$. You should now divide by 4 to get a reduced ratio.

The concept in the linked question is the same, so it more than qualifies as a duplicate. And if you would have read the answers carefully and followed the step by step instructions, you would have arrived at LDC3's answer. 2) It would have been helpful if you at least rotated the image if you do not want to type up a few numbers again. This way no one has to stretch his/her neck out of the natural way...


text:
Density of unknown solution?

A glass vessel weighs $\pu{20.2367 g}$ when empty and $\pu{20.3102 g}$ when filled to the etched mark with water at $\pu{4 °C}$. The same vessel was then dried and filled to the same mark with a solution at $\pu{4 °C}$, the vessel was now found to weigh $\pu{20.3300 g}$. What is the density of the solution? (Assume the density of water is $\pu{1.00 g/ml}$)

Here’s what I did:
The mass of the water is just $\pu{0.0735 g}$ when you subtract the weight of (water + vessel) and (empty vessel). Then I found how many $\pu{ml}$ of water there was in the vessel by realizing that the density ($m/V$) must be equal to $1$, so $\pu{0.0735 g}$ of water must be $\pu{0.0735 ml}$ of water too. So with that, I figured out how many grams of the unknown solution were in the vessel ($\pu{0.0993 g}$), and knowing that it was filled to the same “etched mark,” we can say that the density of the unknown solution is $\pu{0.0933 g}/\pu{0.0735 ml} = \pu{1.27 g/ml}$. I went through all of that because I am not sure if my thoughts are on track with logic and I am pretty confused.

Please read this

Your answer is correct.  Density=mass/volume  In the water experiment you know density and mass, therefore you can calculate the volume of the vessel.  In the second experiment you know the mass, along with the volume from the first experiment, so you can calculate the density.

Thanks so much for taking the time to look over that. It means a lot, thank you!

This is a good example of how a HW question should be written.

What you did in your explanation is correct. First find the difference in the weight:
$$20.3102-20.2367=0.0735 \ \mathrm{g}$$
We know that from your density of water:
$\mathrm{\frac{1\ g}{1\ mL}\ \text{that }\ 0.0735 \ g = 0.0735\ mL}$
Then the difference between the unknown and the vessel is calculated as:
$$20.300-20.2367 = 0.0933\ \mathrm{g}$$
and then density is expressed as mass over volume, so:
$$\mathrm{\frac{0.0933\ g}{0.0735\ mL}=1.2693_8\ g\ mL^{-1}}$$


text:
How many Au atoms can be extracted from 0.250 L of seawater?

Gold is present in seawater to the extent of 0.15 mg/ton. Assume the density of seawater is 1.03 g/ml and determine how many gold atoms could be extracted from 0.250 L of seawater.

($\pu{1 ton}=\pu{2000lbs}, \pu{1kg}=\pu{2.205lbs}, \pu{1 mole}=\pu{6.022E23 atoms}, \text{1 mole of gold}=\pu{197g}$)
I just multiplied everything together using dimensional analysis and got $5.2\times10^{14}$ atoms of Au per liter of seawater, but they want how many atoms of Au in $\pu{0.250L}$ of seawater so I just multiplied $5.2\times10^{14}$ by $0.250$ to get $1.30\times10^{14}$ atoms of Au per $\pu{0.250L}$ of seawater. Can you simply multiply everything together like this? When can you do this and when can't you?

Note that since the rest of the units in the problem are SI, there's a good chance that "ton" is supposed to be a metric ton, which is 1000 kg instead of 2000 lb.

You also need to convert the grams of gold into moles in order to use Avogadro's number.
Atomic weight of gold = $\pu{196.967 g/mol}$
$$
\begin{align}
\frac{\pu{1.5E-04 g/ton}}{\pu{196.967 g/mol}}&= \pu{7.615E-07 mol/ton}\\
\pu{7.615E-07 mol/ton}\times \pu{6.022E+23 atoms/mol} &= \pu{4.586E+17 atoms/ton}\\
\frac{\pu{4.586E{+17} atoms/ton}}{\pu{2000 lb/ton}}&=\pu{2.293E{+14} atoms/lb}\\
\frac{\pu{2.293E{+14} atoms/lb}}{\pu{0.454 kg/lb}}=\pu{5.056E{+14} atoms/kg}&=\pu{5.056E{+11} atoms/g}\\
\pu{5.056E{+11} atoms/g} \times \pu{1.03 g/ml}&=\pu{5.208E{+11} atoms/ml}\\
\pu{5.208E{+11} atoms/ml}\times \pu{250 ml}&= \pu{1.302 E{+14} atoms}/\pu{250 ml}\\
\end{align}
$$
There would be $1.302 \times10^{14}$ atoms of gold in $\pu{0.250 L}$ of seawater.


text:
The stability of a "Di-onol" - carbonic acid - and analogies to cis-nitrous acid
The reaction of carbon dioxide with water yields carbonic acid to some extent (0.4 %) depending on pressure (and temperature).
This process, according to @ron, goes through a four-membered transition stage in which a proton transfer occurs, thereby relieving the unstable "carbonic acid" or hydrated carbon dioxide of its charge separation (The instability of hydrated carbon dioxide or "carbonic acid").

Now, the resulting species is a di-enol; we have hydroxyl groups attached to a carbon that bears a pi bond. This species is likely highly unstable due to the highly positive central carbon atom. On the other hand, neither hydroxyl group makes for a good leaving group, so I can't think of any "forward" (rather than reverse) pathways to more stable products. Also, I can't see the di-enol tautomerizing into a keto. Or can it?
Also, I was arguing that this species is analogous to cis-nitrous acid, in that both species involve a four-membered ring (or at least a psuedo four-memebred ring in the case of cis-HONO).
Why does nitrous acid exhibit (unexpectedly) different bond angles?
My points were:

Proton transfer within the charge-separated form of carbonic acid mirrors the electrostatic attraction between the hydrogen and the terminal oxygen of cis-HONO.
Because the extent of this proton transfer is 0.4%, this suggests that forming a four membered ring is unfavorable (including the case of cis-HONO). Well, of course.
This more importantly suggests that any interaction between any terminal oxygen and hydrogen in a four-membered ring is unfavorable); just look at the extent of proton transfer between an acidic oxygen bearing a positive formal charge of and an oxygen bearing a negative formal charge. In isolation, we would expect the below to be highly favorable (at least thermodynamically):

$\ce{RO^+H + RO^- ->RO + ROH}$

My professor, however, countered that the resulting product, while not charge separated, is a "di-enol". This is highly unstable; the carbon has a strong partial positive charge. So this distorts the comparison I'm trying to draw between cis-HONO and carbonic acid; the sheer instability of the resulting product from the intramolecular reaction may be hiding any thermodynamic gains to be had from the proton transfer, and thus does not rule out the possibility that the net result of hydrogen bonding in cis-HONO is unfavorable, and that H-bonding is cis-HONO is unable to close up the ONO bond angle.
The "di-enol" also reminds me of acid/acyl chlorides, which are stable enough to be isolated, but not exceedingly stable to be completely unreactive. Are acid chlorides a good analogy to carbonic acid? In both cases the central carbon bears a strong partial positive charge.
In addition, the charge-separated form is prone to disproportionation to carbon dioxide gas and water, which is highly favorable from an entropic standpoint.

Questions

Is carbonic acid a "di-enol" (if that even is a thing)?
Can true (non-charge separated) carbonic acid tautomerize into more stable forms?
Does this system provide a good parallel with regard to the case of cis-HONO?
Are acid chlorides a good analogy to carbonic acid? In both cases the central carbon bears a strong partial positive charge.
Is true carbonic acid stable?

It would be helpful if you included the links to 1) the "according to @ron" answer (cuz I don't remember advocating a 4-membered TS) and 2) the HONO question you're referring to.

http://chemistry.stackexchange.com/questions/16076/the-instability-of-hydrated-carbon-dioxide-or-carbonic-acid

http://chemistry.stackexchange.com/questions/15558/why-does-nitrous-acid-exhibit-unexpectedly-different-bond-angles

I'm still missing where I mention a 4-membered TS

That was implied when you agreed with my proton transfer mechanism.

does your university require you to get all these mechanisms for your course?

@Aditya No, but I'm curious. To all the downvoters: if you could provide some sort of constructive criticism, that would be useful.

I'm not sure if I understand what you're saying.

In this answer the issue is not carbonic acid being unstable, but carbon dioxide being especially stable as the anhydride. Solvation and electron delocalization are both invoked.

1.Is carbonic acid a "di-enol" (if that even is a thing)?

I think most folks would understand what you mean, di-onal (alkene vs. ketone), since it is a carbonyl double bond, not a carbon-carbon double bond, that is involved, might be preferred.

2.Can true (non-charge separated) carbonic acid tautomerize into more stable forms?

It can undergo keto-onol tautomerization to regenerate itself, but now with a different oxygen involved in the carbonyl and a different oxygen in one of the hydroxyls.  I can also imagine a tautomer with a 2-oxygen, 3-membered ring, but I doubt this tautomer is more stable.

3.Does this system provide a good parallel with regard to the case of cis-HONO?

My first reaction is that it is not a close analogy.

The carbonic acid situation involves proton transfer.  In this case with water around, the proton transfer is likely intermolecular, involving nearby water molecules.
The nitrous oxide problem involved a static, intramolecular hydrogen bond

4.Are acid chlorides a good analogy to carbonic acid? In both cases the central carbon bears a strong partial positive charge.

My first reaction here was - no way.  But the more I think about it, the more I see your point.  They both react via similar mechanisms, you can draw that extra resonance structure with hydroxide or chloride ionized for both of them, you have to protect them both in order to store them.  Guess I'd say there are a number of analogies between the two compounds.

5.Is true carbonic acid stable?

Yes, when stored under the appropriate conditions (just like an acid chloride). To compare it's kinetic stability to some other compound, we'd need to find out what the activation energy for decomposition is.  To compare it's thermodynamic stability, we'd need to calculate it's heat of formation and compare that to the heat of formation of water and carbon dioxide.  P.S What is untrue carbonic acid?

"My first reaction is that it is not a close analogy." What about the dihedral angle of gaseous HOOH and the ONO bond angle of cis-HONO? @ron (that was my professor's analogy; rationalize the dihedral angle of HOOH and that'll explain cis-HONO's ONO bond angle apparently).

Also what's an "onol"? Google doesn't turn up anything. I understand it's related to ketones though ..

Do you mean carbonic acid instead of HOOH, and which dihedral angle?

I mean HOOH, and this dihedral angle: http://i.stack.imgur.com/VquYQ.png. Also I added two more questions to my OP :D.

enols involve a C=C bond, this involves a C=O double bond; so not "en" (as in alkene), but rather "on" (as in ketone).

@Dissenter Let me start by saying that I value your contributions to SE Chem.  I particularly enjoy the enthusiasm and zeal you bring to learning.  However I think you would get better answers if you illustrated your questions, provided links to material you referenced and better focused your questions to remove ambiguity.  I still don't understand your HOOH question (which dihedral angle - gas or solid; and how does the HOOH dihedral angle relate to the ONO bond angle).  If you don't take the time to explain it clearly, you probably won't get as good an answer as you otherwise might.  Add to

The question is does the HOOH dihedral angle even relate to the (planar) ONO bond angle; that was a proposal I heard from my prof, and I was as taken aback as you were when I first heard it.

this that you add 2 more questions (now up to 5!) after I answered the original 3.  I see someone downvoted your question.  I thought about doing that, but thought this note might be more helpful.  I don't think it will take that much more effort on your part to write excellent questions - and sometimes you do; this one is just particularly frustrating to me.

one of the two was embedded within the post before and it was overlooked. In any case I have edited the OP to illustrate what I was talking about. I wrote this post in a hurry earlier; it popped into my mind and I had to get it out there. Always appreciate your efforts to answer my questions. I have to say that I have learned perhaps more chemistry from you and the SE members in general than I have in class.

"What is untrue carbonic acid" that's the charge-separated species initially formed when CO2 is dissolved in water.


text:
What is the chemistry behind amalgamation?
Mercury can form amalgams with metals, such as gold, however how exactly does the "reaction" if you can call it that take place?
My understanding of metal bonds is that the atoms of a metal are surrounded by a "sea" of electrons, however I don't understand how mercury can "penetrate" metals and weaken their structure or "dissolve" them.
How does it do this? Also does this apply to Gallium?

The liquid mercury atoms collide with solid metal atoms and bonds form.  The atoms in metals move when they are above 0K.

To expand on Geoff's answer below, if you are interested in knowing more I recommend MIT course 3.091 Solid State Chemistry which is about such things exactly. It is available for free on edX: https://www.edx.org/course/mitx/mitx-3-091x-introduction-solid-state-2691

An amalgam is simply an alloy of some metal with mercury.
Alloys are simply solid solutions with metals. Since the mixture of multiple elements disrupts the metal bonding, the properties are often quite different than the components. (In some cases, alloys form so-called intermetallic compounds with defined stoichiometry, but more often, the ratios are very flexible.)
I wouldn't describe the mixture as a reaction, in much the same way that I wouldn't really call dissolving sodium chloride in water as a reaction. You're forming a mixture. Water disrupts the ionic bonding in NaCl, and mercury and other metals can also accept other components or impurities.
Gallium does alloy with a lot of metals, and indium-gallium eutectic alloy is a liquid at room temperature, as is "galinstan".
As to your question about mercury dissolving the metal atoms, consider that mercury is also a metal, so it also participates in similar metal bonding.

To my knowledge aluminum treated with gallium will happily reduce water to hydrogen the way aluminum amalgam does.
But if you want to reduce a nitroalkane you have to consider hydrogen overpotential of the gallium-aluminium compared to aluminium amalgam. That is what dictates the selectivity (substrate reduction vs. hydrogen evolution).
This is the same idea as in electrolytic reductions where you have to use electrodes made of suitable materials (mercury and lead are traditional choices) to be able to actually reduce your compound instead of making hydrogen gas.


text:
How can I find the symmetry labels of atomic orbitals in a molecule?
What are the symmetry labels for the p and d orbitals of platinum in $\ce{[PtCl4]^2-}$?
I understand how to find the point group of a molecule, but am not sure how to use the character table to find the irreducible representations corresponding to the orbitals.

First, you have to know the geometry of your compound. The complex $\ce{[PtCl4]^2-}$, for example, is square planar.
The next step is to determine the point group of the compound from this geometry by identifying the symmetry elements that it posesses (a procedure on how to do this can be found here). The point group of the complex $\ce{[PtCl4]^2-}$, for example, is $D_\mathrm{4h}$.
Now you consult the character table of the point group you determined in the previous step. The character tables for many common point groups can be found here. The character table of the point group $D_\mathrm{4h}$ is shown below and can also be found here.

$$\begin{array}{c|cccccccccc|cc} \hline
D_\mathrm{4h} & E & 2C_4 & C_2 & 2C_2' & 2C_2'' & i & 2S_4 & \sigma_\mathrm{h} & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_{2g}} & 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_{1g}} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & & xy \\
\mathrm{E_g} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & (R_x,R_y) & (xz,yz) \\
\mathrm{A_{1u}} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \\
\mathrm{A_{2u}} & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\
\mathrm{B_{1u}} & 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 & & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & & \\
\mathrm{E_u} & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & (x,y) & \\ \hline
\end{array}$$

Finally you need to determine the symmetry labels of the p- and d-orbitals within the point group by identifying the irreducible representations (the labels on the far left of the character table) which match the transformational properties of the basis vectors ($\color{red}{x}$, $\color{red}{y}$, $\color{red}{z}$) and some of their quadratic combinations ($\color{blue}{xy}$, $\color{blue}{xz}$, $\color{blue}{yz}$, $\color{blue}{z^2}$, $\color{blue}{x^2 - y^2}$). These can be found in the two columns on the right columns. Again this will be shown for the example compound $\ce{[PtCl4]^2-}$ where the relevant functions are highlighted.

$$\require{bbox} \begin{array}{c|cccccccccc|cc} \hline
D_\mathrm{4h} & E & 2C_4 & C_2 & 2C_2' & 2C_2'' & i & 2S_4 & \sigma_\mathrm{h} & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,\bbox[cyan]{z^2} \\
\mathrm{A_{2g}} & 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_{1g}} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & & \bbox[cyan]{x^2-y^2} \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & & \bbox[cyan]{xy} \\
\mathrm{E_g} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & (R_x,R_y) & \bbox[cyan]{(xz,yz)} \\
\mathrm{A_{1u}} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \\
\mathrm{A_{2u}} & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & \bbox[pink]{z} & \\
\mathrm{B_{1u}} & 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 & & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & & \\
\mathrm{E_u} & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & \bbox[pink]{(x,y)} & \\ \hline
\end{array}$$
Since the three p-orbitals point directly along the $x$, $y$, and $z$ axes, they show the same transformational properties as the basis vectors, and consequently get the same symmetry labels, i.e.
$$\begin{align}
\left.\begin{array}{c}
\mathrm{p}_x \\ \mathrm{p}_y \end{array}\right\} &\rightarrow \mathrm{E_u} \\
\mathrm{p}_z &\rightarrow \mathrm{A_{2u}} \ .
\end{align}$$
The five d-orbitals show the same transformational properties as their quadratic function counterparts and thus they get the symmetry labels, i.e.
$$\begin{align}
\mathrm{d}_{xy} &\rightarrow \mathrm{B_{2g}} \\
\left.\begin{array}{c}
\mathrm{d}_{xz} \\ \mathrm{d}_{yz} \end{array}\right\} &\rightarrow \mathrm{E_g} \\
\mathrm{d}_{z^2} &\rightarrow \mathrm{A_{1g}} \\
\mathrm{d}_{x^2 - y^2} &\rightarrow \mathrm{B_{1g}} \ .
\end{align}$$
Another worked example can be found in this answer of mine.


text:
Could neutralisation of safe chemicals produce dangerous ones?
I'm basically thinking whether I could potentionally poison myself while neutralising strong acid/base by some weak base/acid when I accidentally spill it, touch it or when I'm cleaning reaction product.
Combinations of chemicals that are non-toxic in my opinion:

$\ce{HCl + NaHCO3}$
$\ce{H2SO4 + NaHCO3}$
$\ce{NaOH + CH3COOH}$

None of these are poisonous. I just thought these up, for example the first one produces table salt, $\ce{NaCl}$.
But could there be a combination of non-toxic acid and base that would produce some nasty salt when neutralised? Something worse than what I would be getting rid of?
Is there a combination where I should better just wash myself with water?

For "everyday chemistry", I cannot think of an example besides that with bleach above.
Incidentally, mixing hydrogen peroxide and sulfuric acid to make piranha solution is another unsafe mixture. Aqua regia, mixing nitric and $HCl$ would be another problem. Both reagents should be avoided. That's not what you're asking.
Consider that typical "strong" acids and bases are highly ionic compounds. That is, the definition of a strong acid or base is one that dissociates completely in water.
So typical neutralization with weak acids and weak bases will still produce soluble salts.
There are caveats.

Neutralization often is exothermic, and with common acids ($HCl$, $H_2SO_4$ and $HNO_3$ the heating effect can drive release of the gases, which are corrosive. So if you have concentrated acids, you should dilute significantly by pouring the acid into water before neutralization with weak base.
Nitric acid. The various nitrous oxides are dangerous, and care must be used to prevent them from being released. Worse, nitric acid is a strong oxidizer and shouldn't be used with organic reagents, especially organic bases.

So the usual rules apply. Dilute, then neutralize carefully.

While not an acid-base reaction, an example of mixing two relatively "safe" chemicals to form dangerous ones is bleach (sodium hypochlorite) plus ammonia - both common household chemicals. This reaction yields chloramine, dichloramine, and nitrogen trichloride, all more toxic and hazardous than bleach or ammonia.

That's a very good example!

Thank you, but I'm directly interested in acid-base neutralisation. Though a list of dangerous combinations would be very useful thing for everybody.

I'm asking about acid-base because I neutralise acid on my hands or equipment or solid product very often.

If we consider just acid-base reactions, then neutralizing a metal phosphide or arsenide would be an example - this would produce gases phosphine and arsine, which are quite toxic by inhalation. The metal phosphide or arsenide salts are toxic too, but do not pose an inhalation hazard. Same considerations apply to metal sulfides/selenides/tellurides.

I can't think of any acid base examples where you will make something more toxic than what you are trying to neutralize.
That said I think you should always be washing with copious water. There's no reliable way to know how much material is on your skin, which means you won't know how much neutralizing agent to use. You are better off trying to dilute the acid or base dramatically to avoid its adverse effects. Furthermore, this gives you an immediate procedure to follow if a spill should occur. Instead of going through the process of considering what the nature of the contamination is, finding the appropriate counter-reagent, and approximating the amount needed for neutralization, just immediately go to the sink. The time saved is critical for minimizing injury.

Indeed. I think one caveat here is $HF$. But if you're working with $HF$ you should research the safety protocols carefully.

Great point. I'm hoping that HF isn't considered a household chemical.

Mad people use it to clean car wheels: https://www.autopia.org/forums/auto-detailing-101-a/15313-death-wheel-acid.html

If you're using oxidizing acids (mainly nitric, perchloric, chloric, perbromic, bromic, periodic, iodic, also potentially selenic and arsenic acids and obviously chromic or permanganic acids and peroxyacids) and you mix them with organics, including amines, guanadines,imines, phophazenes and such, you can end with a variety of toxic fumes such as nitrosamines, cyanides, nitrogen oxides and such, an it could violently burn or just explodes. The neutralization itself wont lead to anything more toxic than the base/acid. Side reactions are what maims and kills people.


text:
Conversion of propane to methane
I had a question in my textbook to convert propane to methane. The basic idea was chlorination in presence of sunlight. However, two products are obtained in this case - namely 2-chloropropane and 1-chloropropane. So it didn't work out at all.
I came up with another idea i.e. dehydrogenation, but it has not been taught to us at all. In addition the dehydrogenation reactions I searched on google were too complex. So, I hope there's another method for this.
Can anybody help me to sort this out?

Welcome to Chemistry.SE! Are you really meaning to convert propane, i.e. $\ce{CH3-CH2-CH3}$, methane, i.e. $\ce{CH4}$ and not maybe rather propane to propene, i.e. $\ce{CH3-CH=CH2}$? If this isn't a mistake I don't understand what you are trying to achieve with the chlorination or dehydrogenation.

No the conversions may be lengthy I just could not determine what I need to do in the 1st step.

This probably defies the purpose, but still. Two-step process with steam reforming and Fischer–Tropsch synthesis
$\ce{C3H8 +O2 +H2O -> CO + H2 + CO2}$
This step is usually performed with methane over nickel catalyst, but should work with propane as well. Remaining water is condensed and carbon dioxide is separated using triethanolamine as absorbent. Then a reverse reaction is performed
$\ce{CO + 3 H2 -> CH4 + H2O}$
depending on catalyst, and exact conditions it is possible to get methanol, linear alkane/alkene mix or methane. AFAIK, $\ce{Ni}$ catalysts favor methane formation, while cobalt and iron favor linear alkanes and terminal linear alkenes.

Incidentally, this is a general case of cracking reactions.

Use:
$$\begin{array}{}
\ce{Br_2}/h\nu& \text{Bromination}\\
\ce{^{\large \circleddash} OEt/EtOH}& \text{Elimination: Dehydrohalogenation}\\
\ce{O_3/H_2O}& \text{Ozonolysis}\\
\ce{NaOH/CaO}/\Delta& \text{Decarboxylation}\\
\end{array}$$

If you are trying to do a dehydrogenation, dehydrohalogenation of both your products would get you to propene, even though it takes two steps.
Good luck getting to methane. Of course, if you were in industry you would just heat the crap out of it and watch the money roll in... Well, maybe not with methane as the product...

I wonder if it is not possible to convert propane to methane directly with a large excess of hydrogen $\ce{H2}$.
So $\ce{CH3-CH2-CH3 + 3H2 ->  3 CH4}$  where the propane/hydrogen ratio is something like 10/90 i.e. an over-excess of hydrogen under the influence of a heat source like a tungsten heated spiral ($1000{-}\pu{2500°C}$). Due to the heat, both dissociation and recombination should occur. Simple and direct.
I think that it should be possible.

I like your thinking but I don’t think it is possible.


text:
What trends affect hydrophobicity in amino acids?
I was wondering if the length of the carbon backbone affects hydrophobicity of amino acids (i.e. longer carbon backbone of side group = less polar = more hydrophobic).

A longer backbone means a larger hydrophobic surface exposed to water. That should tell you something of its hydrophobicity.

Your assumption that hydrophobicity of amino acids increases with length of the carbon chain is correct. Also, a branched chain can have a lower hydrophobicity than a linear chain with the same number of carbons, because it has a decreased exposed surface area.
When polar acidic/basic functional groups are introduced, the side chain becomes more hydrophilic. For example, phenylalanine is classified as non-polar because of the hydrophobic benzyl side chain. Attaching a hydroxy group in para-position in the phenyl ring leads to tyrosine, which is more polar. This is due to its acidic phenol group which is deprotonated at physiological pH. Note that in both cases, the number and arrangement of carbon atoms in the side chain is the same.


text:
Is Sodium metal more reactive towards an alkyl halide or an alcohol in a protic polar solvent?
Will Sodium metal donate its electrons to bromine in 1,3-dibromopropane or to the hydroxyl group in ethanol in a polar protic solvent?
This question comes from a problem:
'What is the major product formed when 1,3-dibromopropane is added to Na(sodium)/EtOH(ethanol)?
(a)cyclopropane
(b)3-bromoprop-1-ene
(c)propyne
(d)prop-1,2-diene'
The answer was (a)cyclopropane but I was not convinced as then it would have to follow Wurtz reaction- where sodium donates its electron to bromine forming a alkyl radical. But, Wurtz reaction takes place in an ether solvent and here ethanol is used as solvent.
ethanol is shown to readily react with Na to form a sodium ethoxide and liberate hydrogen. The ethoxide ion would react with 1,3-dibromopropane in either Sn1 or E2 mechanism and so it would not form cyclopropane.
If sodium donates electron to bromine however, then it would follow Wurtz reaction mechanism and form (a)cyclopropane.
So, I want to know whether an alkyl halide or alcohol reacts more readily with sodium metal and what your reasoning is to assert your statement.

What are your thoughts on the matter?  If you include what you think the answer is and your rationale, someone can help steer you in the right direction.

authors: Hëľĺ Vőļhãřđ


text:
How can I predict what dissolves what?
Regarding solvents, the only thing I remember from school is my teacher saying: "The same dissolves the same." regarding the polar/non-polar solvents.
From this I understand that water is polar solvent, because it has polar bonds and therefore it dissolves table salt, which also has a polar bond.
What I do not understand for example is, why does water perfectly mix with acetone? X
Another thing is, that non-polar organic solvents seem to have some very complicated rules. For example, acetone dissolves most paints and all ABS plastics. Ethanol or tetrachloroethylene do not.
Why does ethanol dissolve acetone-peroxide while acetone doesn't? X
Q: Are there some general rules to this, or is this big science that requires years of study?
X: these are just rhetoric questions. You can answer in comments but what I'm looking for are rules.

yeah these rules are a bit of a pain - this link might be a good start http://www.chem.sc.edu/faculty/morgan/resources/solubility/
http://www.csudh.edu/oliver/chemdata/solrules.htm
Sadly there is no real way to derive these - memorising them is the only way.

But do we know why something dissolves something else?

As you note, there are some general rules about solubility. For example, acetone is a highly polar organic molecule, so it's not surprising it has high solubility/mixibility in water. The "like dissolves like" rule is generally based on polarity and size/shape considerations, and it's a fairly good rule.
But solubility is complex because you have two big sources of driving force:

enthalpy: The molecular and intermolecular interactions between the solvent and solute are favorable. Consider methanol dissolving in water: it forms hydrogen bonds with the water, so the enthalpy is favorable.
entropy: This is often the larger driving force. You create a far more disordered result with a mixture than with two pure components.

While most chemists develop reasonably accurate intuitions of enthalpy, the entropic effects are harder to predict.
Many efforts have been made to predict solubility, both qualitatively and quantitatively. The comments above give some good links. Consider that this problem is important to chemical synthesis broadly, but also to drug delivery (i.e., can a drug be given orally and dissolve in the stomach, can a drug go through the blood-brain barrier?)
Perhaps the most successful is the Abraham general solubility model which attempts to predict solubility of organic compounds in 84 different solvents. Here’s the article.
Basically, it breaks down solubility into multiple components (grabbed from Andrew Lang and Jean-Claude Bradley):

E is the solute excess molar refractivity in units of (cubic cm per mol)/10. It represents the solute's polarizability and gives a measure of the ability of a solute to interact with a solvent through n- and $\pi$- electron pairs.
S is the solute dipolarity/polarizability. It gives a measure of the solute's ability to stabilize a charge or dipole
A is the overall (summation) hydrogen bond acidity. The hydrogen bond acidity descriptor measures the extent of hydrogen bonding by the solute in a basic solvent.
B is the overall (summation) hydrogen bond basicity. The hydrogen bond basicity descriptor measures of the extent of hydrogen bonding by the solute in an acidic solvent.
V is the McGowan characteristic volume in units of (cubic cm per mol)/100.

The nice thing about this model is that it makes sense -- you want a solute that matches lone pairs, hydrogen bond donating or accepting ability, polarizability, and volume for the solvent.
The catch is that surprisingly little data is available. The late Jean-Claude Bradley encouraged his students to make accurate solubility measurements, but also to publish them on the web. In this way, they gathered a large spreadsheet of "compound X dissolves Y amount in Z solvent."

Personally, I think it's one of those great, seemingly simple questions that point to the edge of knowledge. The more data becomes available, the easier it will be to make predictions and advance the science.


text:
Steric effects of a lone pair and piperidine
I know that in piperidine, the lone pair on the nitrogen can occupy either an axial or a equatorial position in the two chair forms. I also read that the axial position for the lone pair is preferred by a small margin in the gas phase (so intermolecular forces are minimized).

Now, this was presented as an argument against the idea that lone pairs are "sterically demanding" or "bulky" or repel other entities of negative charge more strongly than do bonding pairs, etc. The conclusion was that a lone pair must occupy less space than a hydrogen atom since it prefers the axial position in piperidine!
This argument seems correct at first, but could there be other factors affecting why the axial lone pair conformation might be more stable than the equatorial lone pair conformation? Also, are there any other examples in which the lone pair behaves "unexpectedly," at least with respect to elementary theory?
ETA: On the other hand one could make the argument that this isn't a apples to apples comparison; we are comparing a lone pair of electrons with a bonding pair of electrons plus a hydrogen atom. So it may be that this is a steric rather than an electronic (i.e. repulsive) effect.
Or could there be some sort of hyperconjugative effect going on? Carbon withdraws electron density from hydrogen; hydrogen becomes slightly partially positive, is attracted to lone pair?

There is an interesting article which looks at the idea of "bulky" lone pairs, saying it is a fix for VSEPR with no physical basis and proposing Bent´s rule as an easy alternative explanation for bond angles deviating from simple repulsion models. Source

The energy difference between the two conformations in your example is extremely small.  This makes it difficult to say that "this" is the reason why things are the way they are. Nonetheless, it is a very interesting question. Perhaps, like in the "how many angels can you draw on the head of a pin" question, the discussion around the question might be as significant as the answer.
What we need is more data.  The phosphorus analogue of piperidine, phosphacyclohexane, has also been examined and the hydrogen resides in the axial position (no evidence for the equatorial isomer was observed).

There are important differences between these nitrogen and phosphorus analogues.  To start with, phosphorus-carbon and phosphorus-hydrogen bond lengths (180-85 pm and 145 pm respectively) are typically longer than the analogous nitrogen-carbon and nitrogen-hydrogen bonds lengths (150 pm and 102 pm respectively).  More important is the fact that the phosphorus is basically unhybridized, so the lone pair of electrons around phosphorous exists in a non-directional s-orbital.  This makes conformational analysis of phosphacyclohexane easier because we don't have competing (hydrogen and lone pair) substituent effects to consider and dissect.
The observation that the hydrogen in phosphacyclohexane exists in the axial position suggests that there is an attractive 1-3 synaxial H-H interaction and/or a sterically repulsive 1-2 H-H gauche interaction.  Given the long length of bonds to phosphorous, repulsive interactions seem unlikely and the attractive 1-3 synaxial H-H explanation was favored.
Another data point comes with N-methyl piperidine where the lone pair strongly prefers the axial position.  This was attributed to a repulsive 1-3 synaxial Me-H interaction.  The fact that the methyl group adopts the equatorial position also shows that the repulsive 1-2 Me-H gauche interaction must be of little consequence and any repulsion in the corresponding H-H case would be even smaller.
Piperidine itself appears to be somewhere in the middle, in between N-methyl piperidine and phosphacyclohexane.  Certainly the repulsive 1-3 synaxial Me-H interaction observed in N-methyl piperidine has been reduced, but not eliminated.
Whether just the repulsive 1-3 synaxial H-H interaction has been reduced or whether there is now also a contribution from the attractive 1-3 synaxial H-H interaction seen in phosphacyclohexane cannot be determined.


text:
Why does the absorbance of Ca decrease in the presence of certain metals?
I'm trying to find a reason for an experimental observation, I noticed that when the absorbance of $\ce{Ca}$ is measured with FAS(flame atomic spectroscopy) that it decreases when metals such as $\ce{Al}$ and $\ce{K}$ is present(the ions). I used a $\ce{Ca}$ hallow cathode lamp.
Any idea as to why?

My guess is that the transition used for measuring calcium is very close to transitions in aluminum and potassium. You would need to talk to an AA operator.

Well whatever metal you add will interfere in the absorbance of Ca, because the metal will absorb some of the light too, that's why that of the Ca decreases and thats why the analyte is almost every time mixed only with water which does not absorb the light.( and ofcourse you can add a reagent to form the complex which helps inorganic compounds to absorb the light )

Very late to the party. In flame atomic absorption spectroscopy (Flame AAS or FAAS), apparent absorbances can differ from true absorbances because of a number of inter-related factors, such as the presence of easily ionizable elements (EIEs) or refractory elements. The latter form light scattering oxides, resulting in higher apparent absorbance. With EIEs present, the formation of ions of the desired analyte species is suppressed, so ionic absorbance is reduced.
In general, it is quite complicated to figure out exactly what causes enhancement or suppression of the absorbance of a given species in FAAS and it is even harder in inductively coupled plasma - atomic emission spectroscopy (ICP-AES). A useful paper in this regard has been published by Zaranyika et al. 1.
Note that other elements are generally not a direct spectral interference problem in terms of absorbing the very narrow hollow cathode lamp (HCL) light output. For calcium absorbance, the following figure (of mine) shows the Ca I spectral profile and the HCL emission profile.

And in FAAS and ICP-AES, the goal is to not have any multi-particle species: just atoms and ions, in their various excited or ground states.
Reference
1  M.F. Zaranyika, A.T. Chirenje, C. Mahamadi, "Interference Effects of Easily Ionizable Elements in ICP-AES and Flame AAS: Characterization in Terms of the Collisional Radiative Recombination Activation Energy", Spectroscopy Letters, 45 (2012) 1-12.

This is the correct answer, not the accepted one. Win some, lose some.


text:
How to define the atom numbers in unsaturated hydrocarbons correctly according to IUPAC?
I would like to address the compound below.

This compound on elaborate elongation would result in this structure.

But is the below given numbering correct? I am just wondering that what if the same compound is given numbering, starting from that carbon which is assigned 7th number in the first elaborate diagram (second diagram), by doing so, triple bonded substituent will also get the least number, as shown in the below given diagram.

I have a similar problem with this compound:

What if you just reverse the numbering? By reversal I mean, the carbon assigned 1st number should be given the last number, am I violating any rule? If I do that, by numbering the compound in reverse order, we get least number for the methyl chain substituent, so is it necessary that methyl group substituted in the side chain should get the highest possible number?
In short, I mean, is the below given numbering valid or not?

I think your numbering is correct.  Your choice is between including the 6-alkene or 6-alkyne in the main chain.  The rule is "tie goes to the alkene".  This link may provide useful examples starting around p. 9  http://www.csupomona.edu/~psbeauchamp/pdf/314_supp_8_nomen_2.pdf

@ron yes, i agree, but what if i start the numbering from the number seventh carbon?, is that valid too?

No, see page 5, item 7, "middle" example in the link I supplied. P.S. Good to see you again, hope things are going well.

@ron yeah, i was busy in my exams, and i am fine, but ron  i didn't found any clue in that example, it seems to be the same case as the compound i have exampled, but why is the numbering assigned as such, what if i just reverse the numbering, am i violating any rule if i reverse it?

Note how in the example I mentioned above (6-ethyl-6-methylocta-trans-2-cis-4,7-triene) they do not start numbering from the side with the terminal double bond, but rather from the side with the lowest overall double bond numbers.

@ron hurrah!, i got it, thanks ron, i have got my doubt cleared now, but what about the 2nd example in my question, which contains no double bond, i have given two structures with different numberings, which one should i consider to be correct?

the first one, it has the lower substituent numbering (1,4 vs. 2,5)

For your first structure, I think your numbering is correct. Your choice is between including the 6-alkene or 6-alkyne in the main chain. The rule is "tie goes to the alkene".  This link may provide useful examples starting around p. 9.  Also see  page 5, item 7, "middle" example in the above link for another relevant example. Note how in this example (6-ethyl-6-methylocta-trans-2-cis-4,7-triene) they do not start numbering from the side with the terminal double bond, but rather from the side with the lowest overall double bond numbers.
In your second compound, your first numbering scheme has the lower substituent numbering (1,4 vs. 2,5).

Please note that your link contains several mistakes. In particular, the numbering in the following examples is wrong: 4-methyl-1-ethylidenecyclohexane should be 1-ethylidene-4-methylcyclohexane, 6-methyl-3-methylenecyclohex-1-ene should be 3-methyl-6-methylidenecyclohex-1-ene, 1-methyl-3-(4-methylpent-3-ynyl)cyclohexa-1,3-diene should be 1-methyl-3-(4-methylpent-2-yn-1-yl)cyclohexa-1,3-diene, 4-(2,2-dimethylcycloprop-1-enyl)-6-isopropylcyclooct-1-yne should be 4-(3,3-dimethylcycloprop-1-en-1-yl)-6-(propan-2-yl)cyclooct-1-yne, 4,4-dimethylnon-8-en-1-yne should be 6,6-dimethylnon-1-en-8-yne, …

… 5-(but-2E-enyl)nona-5Z,7E-dien-1-yne should be (2E,4Z,7E)-5-(but-3-yn-1-yl)nona-2,4,7-triene, and 2-(1-methylpropyl)pent-1-en-3-yne should be 5-methyl-4-methylidenehept-2-yne.

Your first compound should be (3E)-5-ethynylhepta-1,3,6-triene. The 2 conjugated double bonds at the end determine the first carbon.
The second compound should be (4-methylpentyl)cyclobutane. The aliphatic group has less precedence than the cyclobutane group.


text:
water vapor vs air (in a balloon)
I poured some water in a balloon, inflated, and tied it up.  Over a period of couple weeks, all water in the balloon is gone.  But the balloon is still at about 2/3 of original inflated size.
I thought the the air molecules are much smaller than water vapor.  Why the water is all vaporized?

How humid is the air there?  I propose that oxygen, nitrogen and argon enter and leave the balloon at rates accounting for RMS velocity and relative pressure between the balloon and the surrounding air.  Water does as well.  In a dry environment the rate should favor water leaving the balloon.  In a very dry environment the net loss of water may be faster than the net loss of air, especially since water vapor has a higher RMS velocity than other components of air.

The air in the balloon will be saturated (with water vapour) and the air outside not. As a result there will be a vapour density (or partial pressure) gradient between the inside and outside of the balloon that will drive the mass flow of water vapour through the balloon and to the outside.

Guessing, I would say that the water diffused through the plastic. Many plastics absorb water quite well. However I'm only guessing.


text:
Sodium metal reduction of chloroform
In discussing how best to deuterate chloroform, it occurred to me that we could use sodium metal to reduce hydrogen into hydrogen gas. If this could actually happen, this would be a great thing - we wouldn't have much if any residual hydrogen in the liquid phase of the system after treatment.
Is this possible? I had a possible mechanism in mind.

The problems I foresee are:
1) Won't reduction of chlorine to chlorine gas be a competitive, side reaction? From this page (http://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)) I see that the reduction potential for chloride ion to to chlorine gas is -1.36 V (unfavorable) and reduction of hydride ion to hydrogen gas is a +2.23 V (favorable). Am I interpreting and applying the data correctly?
2) What happens to the $\ce{CCl_3}$ radicals after hydrogen is reduced? Is dimerization going to occur to a large extent?
3) How then do we introduce deuterium and effectively incorporate it with $\ce{CCl_3}$ radicals?

The charges don't balance in your first mechanistic step. There has to be an anion produced...

@jerepierre I see what you're saying, there should be a hydride anion. Thank you!

Do not for goodness sake mix chloroform and sodium metal, it can explode.

Sodium + Chloroform:
https://www.youtube.com/watch?v=du3ksrfnCcg
To prepare deuterochloroform I'd recommend the haloform reaction.

That was very, very interesting! Thank you for sharing! Perhaps the reaction can be controlled? After all the use of sodium isn't unheard of in the lab. Also the gas produced sure didn't look like chlorine gas and probably wasn't given that there were students a few feet away. Perhaps it is H2 gas.


text:
Estimating magnitude of percentage and absolute error
Can someone at least give me a hint to push me in the right direction of how to estimate the magnitude of the absolute and percentage errors when using a smaller amount in mL of a substance for a lab measurement?

Do you know what equipment you were using for measurement? Different pieces of equipment carry different relative errors.

When you take a small amount of a liquid in mL, you you usually use a pipette. On every pipette, at the top of it, the absolute error is written. As for the relative error, it is the ratio of the absolute error to the measured value. The percent error is the relative error multiplied by 100.


text:
What is the difference between polyethene and polyethane?
The monomer in polyethene seems to drop the $\ce{C=C}$ bond and end up with a $\ce{C-C}$ bond. The polymer chain then looks like $\ce{[CH2-CH2]_{n}}$. If I were to look at that polymer chain I would probably call it polyethane not polythene (or I could even call it polymethane) because it doesn't have a double bond, so I wouldn't classify it as an alkene. Am I doing something wrong with my analysis here? Or is there some sort of rule about nomenclature that change when things become polymers and drop a functional bond (in this case the $\ce{C=C}$ bond)?

The reason it is called polyethene (or polyethylene) is because it was produced from the monomer of ethylene. To produce polyethylene, a radical initiator is added to ethylene, which attacks one carbon and opens up the double bond leaving a radical at the end. This radical attacks another ethylene molecule, etc. The chain stops forming when it encounters another radical. So, from ethylene, we get polyethylene (even though it looks like polymethylene).

So we name it based on the monomer it was made from, not what the repeating unit actually looks like?

In most cases, an exception is polyvinylalcohol; it is not made from ethylaldehyde (tautomer of vinyl alcohol).

@Sam, I would think that proper IUPAC names would always derive from the monomer, as you say. Common names can be wildly variant, though--for example, while polycarbonates do contain a $-\ce{CO3}-$ type moiety, their monomers always are more complex than just plain carbonate.

don't you mean polymethylene? :P

The IUPAC name is always derived from the repeating unit, and polyethylene is the only exception where not the smallest repeating unit is taken. A (formal) CH2 diradical is called "methylene", C2H4 "ethylene" in IUPAC nomenclature.

There is not such correct term for either of the terms you've come up with. There is no polyethene. There is no polyethane, not in polymer or macromolecular chemistry and thinking.
Polyethylene (PET) a thermoset polymer (plastic) on the other hand, is ubiquitous.

Polyethene most certainly does exist, and is the polymer about which OP is talking. But correct, there is no such thing as polyethane

The common abbreviation for polyethylene is PE. PET is polyethylene terephthalate.