text:
Dipole moment of cis-2-butene
I need someone to back me up on this before I go confront my teacher:
I was doing some analysis of the dipole moment of cis-2-butene.
Let's say that the alkyl groups are both on top.
Would the dipole moment be a vector that has its tail on the top and its head on the bottom?
How would one calculate that?
I made some calculations assuming all C-H bonds have the same dipole moment vector and came to the conclusion that the three H's attached to the alkyl groups actually end up overcoming the dipole moments of the C-H bonds of the two unsubstituted H's on the double bond.
Am I right?

Sounds like a good 'research project'.  I will have some results that I will post for you once I get them.

Mini Research Project Time
Updates

Added CCSD(T) $n_i$ and dipole moments and tweaked discussion (the delay was caused by a system-wide storage upgrade on the machines which took nearly a week to complete).

Preamble
This response is in no way meant to be contrary to what Geoff has already posted.  I happen to enjoy these types of questions and I like to tackle them as little research projects for fun.  Therefore, my answer will be brutally detailed and in-depth.  That said, this should be a good introduction to how I would approach this problem if I were going to do this at the 'production-level' of research but this definitely goes far beyond the purpose of any reasonable response for an SE answer.
Its All About the Dipoles Baby
Okay so I wouldn't use THAT heading in a paper but maybe a talk depending on who my audience was...
We can easily determine the dipole moment of cis-2-butene via electronic structure theory.  I have modeled two conformers of cis-2-butene as shown below.  I will refer to the geometry on the left as StructA and the geometry on the right as StructB.

A couple of things to note since I don't include any captions to tables.  Energies are always reported in $\mathrm{kcal\ mol}^{-1}$ and dipole moments ($\mu$) are given in Debye.
Computational Methods
Full geometry optimizations and corresponding harmonic vibrational frequency computations were performed with second-order Moller-Plesset perturbation (MP2) theory and a variety of density functional theory (DFT) methods using the Gaussian 09 software package for the conformers of cis-2-butene.  Both conformers were characterized in $C_{2v}$ symmetry.  The DFT methods implemented include B3LYP, B3LYP-GD3(BJ), M06-2X, MN12SX, N12SX, and APFD.  The B3LYP-GD3(BJ) method employs Grimme's 3rd generation dispersion correction as well as the Becke-Johnson damping function. All DFT computations employed a pruned numerical integration grid having 90 radial shells and 590 angular points per shell.  The heavy-aug-cc-pVTZ basis set was employed for these computations where the heavy (non-hydrogen) atoms were augmented with diffuse functions (e.g. cc-pVTZ for H and aug-cc-pVTZ for carbon).  This basis set is abbreviated as haTZ.  The CCSD(T) (i.e. coupled-cluster method that includes all single and double substitutions as well as a perturbative treatment of the connected triple excitations) was similarly employed using the CFOUR software package.  The magnitudes of the components of the residual Cartesian gradients of the optimized geometries were less than $6.8\times 10^{-6} E_h\ a_0^{-1}$.
Single point energies were computed with the explicitly correlated MP2-F12 [specifically MP2-F12 3C(FIX)] and CCSD(T)-F12 [specifically CCSD(T)-F12b with unscaled triples contributions] methods in conjunction with the haTZ.  These computations were performed with the Molpro 2010.1 software package using the default density fitting (DF) and resolution of the identity (RI) basis sets.
Natural bond orbital (NBO) analyses were performed for the MP2 optimized structures using the haTZ basis set and the SCF density.
All computations employed the frozen core approximation (i.e. 1s$^2$ electrons frozen in carbon).
Computational Methods in English
Two different conformations of cis-2-butene were characterized with a variety of cheap (but usually okay) approximations (i.e. DFT methods) as well as reliable (but generally more expensive) wave function methods (i.e. MP2 and CCSD(T)).  The wave function methods are necessary to validate the DFT results.  CCSD(T) is the gold-standard and gives very good results for single-reference closed-shell well-behaved systems so we will use this as our 'best estimate'.  We use a variety of methods in order to look for agreement in the results.  If we see good agreement across the board, we can be confident in our results.  If we see massive discrepancies then we will have to be careful when we analyze the results.  Geometries have been converged to a tight threshold (i.e. we have good molecules) and our DFT computations use a relatively dense integration grid (which leads to more accurate results).
Notice that I employ a heavy-aug-cc-pVTZ (haTZ) basis set.  Why leave the diffuse functions off hydrogen?  The purpose of diffuse functions is to describe electron density far away from the nucleus of an atom.  Therefore we slap these functions onto carbon which are relatively large atoms compared to hydrogen.  Hydrogen, on the other hand, has only one electron and therefore has a small electron density when isolated.  In cis-2-butene, hydrogen is bonded to carbon via a rather small bond distance.  The electron density around the hydrogen is even more reduced than an isolated hydrogen atom in the gas phase.  Therefore, it would be impractical to include diffuse functions on hydrogen.  Doing so may even lead to erroneous results since we will be trying to describe electron density far away from the hydrogen nucleus when in reality there is virtually none to be found.
Finally, we perform single point energies using explicitly correlated methods.  Because the CCSD(T) opts and freqs will likely not be done in time for this posting, we can gauge how the resulting geometry from each optimization procedure will vary from another geometry given by a different method.  If all of the energies are similar (within a few tenths of a $\mathrm{kcal\ mol}^{-1}$), then we can be confident that our geometries are not only very similar but that small deviations in the geometry will have little effect on the corresponding energies at least in this region of the potential energy surface (PES). Large deviations will usually mean that the method which produced the 'outlier' is not a good approximation for the system (I do not expect cis-2-butene to be problematic at all).  Explicitly correlated methods accelerate convergence to the CBS limit.  These methods have been shown to give results that a large basis set and a canonical method would provide but with a much smaller basis set.  For example,  the result that I get with a regular CCSD(T)/aug-cc-pV5Z basis set could be obtained using CCSD(T)-F12/aug-cc-pVTZ.  This makes the computations less intensive and much more feasible.
Results
The number of imaginary frequencies ($n_i$), relative MP2-F12 and CCSD(T)-F12 energetic  ($\Delta E^{\mathrm{MP2-F12}}$ and $\Delta E^{\mathrm{CC-F12}}$, respectively, in $\mathrm{kcal\ mol^{-1}}$) and dipole moment ($\mu_z$ in Debye) are given in the following table for both conformers of cis-2-butene for a variety of a methods.
The relative energies were determined by taking the difference of the respective geometry and the reference CCSD(T) geometry [e.g. E(CCSD(T))-E(MP2)].

StructB is a second-order saddle point ($n_i = 2$) on every single PES considered and therefore is not a minimum energy structure.  StructA is, however, a minimum ($n_i = 0$) on every PES considered.  The characterization of the nature of the stationary point is consistent between CCSD(T), our best estimate, MP2, and DFT methods.
Single point energies reveal negligible differences in the optimized geometries.  The energies associated with the MP2 optimized structures is the reference point for all other relative energies.  Deviations grow no larger than 0.27 $\mathrm{kcal\ mol}^{-1}$ for the MP2-F12 and CCSD(T)-F12 relative energies.  In addition, there is good agreement between the MP2-F12 and CCSD(T)-F12 relative energies, suggesting that higher-order correlation effects are small.
The dipoles for StructA and StructB are very similar with very small magnitudes, on the order of a couple tenths of a Debye.  To put these quantities into perspective, the dipole moment of water is 1.85 D.  We all know that water has a pretty large dipole moment so by comparison, cis-2-butene has a very WEAK dipole moment.  You can compare to the dipole moments of other molecules by referring to this NIST reference.  Clearly, the rotation of the methyl groups have a very small effect on the dipole moments of each conformer.  The MP2 and DFT dipole moments deviate from the best estimate by no more than 0.03 D, but remains in qualitative agreement for both StructA and StructB.
The figure below shows the directionality of the dipole.  The head of the (unscaled) arrow points toward the negative pole while the tail of the (unscaled) arrow is oriented to the positive pole.  The numbers on the atoms represent 'natural charges' from an Natural Bond Orbital (NBO) analysis of the MP2 optimized Struct A.  Clearly the carbon atoms have a small negative charge to them as these atoms are sucking (great scientific term here) electron density away from the neighboring hydrogen atoms.  This is because the nuclear charge (i.e. the number of protons) on carbon is much greater than that of hydrogen (6 vs. 1).

It was suggested that I add some data highlighting the energy difference between the two different conformers.  The following table presents the energy difference (where $\Delta E_{\mathrm{A-B}}$ is equivalent to E(A)-E(B)) of each optimized geometry using the MP2-F12 and CCSD(T)-F12 single point energies (in $\mathrm{kcal\ mol}^{-1}$).  We can see that StructA is about 1.5 $\mathrm{kcal\ mol}^{-1}$ lower in energy than StructB.  This makes sense because StructB is a higher order saddle point and StructA is a minimum on the PES. (I'm actually quite surprised that the energy difference is this large for a couple of methyl rotations...)

Conclusions
The dipole moment for two conformers of cis-2-butene has been examined using seven different computational approaches and a triple-$\zeta$ quality basis set.  The performance of these methods have been tested by evaluating the energies of each geometry with the 'gold standard' CCSD(T) method (the explicitly correlated variant).  Good agreement is seen across the board with respect to the Hessian indices, energies, and dipole moments. Only StructA was a minimum on each PES.  Both conformers of cis-2-butene have a very weak dipole moment on the order of 0.2 -- 0.3 D.  The positive pole is in the vicinity of the methyl groups whereas the negative pole is centered around the sp2 hybridized carbons.
FAQ
So you may be asking yourself (or rather, should be asking yourself) questions such as these listed below.  I will tackle them one at a time.
1.) Why did we look at two conformers of cis-2-butene and why does it matter?
StructA is a minimum which means that if you were to characterize this guy in the gas phase, you'd find StructA rather than StructB.  This is important because if we were to report these results to other scientists, they will want to know what they can expect to find without wondering.  Therefore, StructA is going to be our conformer of interest rather than StructB.  The latter still will provide insightful results but that is about it for the purposes of this examination.
2.)  Why did we use a variety of methods to characterize these molecular systems?
Computational methods are simply approximations.  They are not guaranteed to give the 'correct' answer.  So we try to address this by using a variety of methods (i.e. approximations) and we analyze the results accordingly.  If the results are in agreement, then we can feel confident that the results are correct since we tested them against a list of methods.  You can never rely on just one method unless it is rigorous, well-tested and has been shown repeatedly to perform well in the literature.  When we say 'perform well', this usually means that the computational results are in reasonable agreement with experimental results.  This is important because experimental results is the LAW (for all intents and purposes).  If the computations disagree with experiment, 99.9% of the time this means that your computational approach sucked and that your approximation was either flawed or misapplied.  By using a host of methods, we can put a little more faith into the results we observe because the chances of massive disagreement between the results of the methods is very unlikely in a normal situation.
3.) What is the point of doing a bunch of energy points?
Again, because we used a slew of methods to characterize the geometries of cis-2-butene, we end up with non-identical geometries each time we use a new approximation.  For instance, the methyl C-H bond lengths from B3LYP are going to be a little bit different from those obtained with MP2.  So then that begs the question, "How do these small differences effect the property of the system that we are interested in?"  Generally, minute differences will have little effect on the resulting energies of each molecule under consideration.  Energy is a very important property that chemists love to look at.  So, if we take each geometry (and each one is unique from the other) and we evaluate the energy of the molecule at the same level of theory (in this case, MP2-F12 and CCSD(T)-F12), then we can quickly see how 'resolved' each geometry was.  There should be very good agreement between the relative energies of each geometry (probably to within a few tenths of a $\mathrm{kcal\ mol}^{-1}$).
4.) Okay, so why MP2-F12 AND CCSD(T)-F12?
We use MP2-F12 AND CCSD(T)-F12 to test for 'higher-order correlation effects'.  MP2 methods are much cheaper than CCSD(T) methods but MP2 is not as rigorous and can be error prone in a host of molecular systems.  Therefore, we test the performance of MP2 by busting out our 'gold-standard' which is the CCSD(T) method.  If MP2 agrees well with CCSD(T), then we can feel confident in our MP2 results and never have to revisit the difficult, time-consuming CCSD(T) computations ever again.  ALso, CCSD(T) will also tell us how DFT performed as well.  DFT methods must always be calibrated against something more rigorous since DFT is known for 'getting the right answer for the wrong reasons' and it isn't always right.
The 'F12' bit just means that these are 'explicitly-correlated' methods.  Rather than give an introduction to what this means, you should understand why we use it instead.  You may have noticed that whenever we do a computational job, we specify a method AND a basis set (e.g. heavy-aug-cc-pVTZ).  These basis sets can be measured by how many atomic orbitals (or functions) are given in the set.  The more that are given, the better the 'basis-set approximation'.  Think of this in terms of Riemann sums where you try to approximate the area under a curve using a set of rectangles.  Each rectangle is a basis function and the number of rectangles you use forms a basis set.  The more rectangles you use, the better your approximation of the area under that curve will be.  Basis sets in computational chemistry behave the same way.  When you approach an infinite set of rectangles, you approach the exact answer.  When you approach an infinite number of basis functions, you approach what is called the CBS (complete basis set) limit.  At the CBS limit, you have an exact answer.  We cannot implement an infinite basis set in chemistry (for obvious reasons), and very large basis sets are cost prohibitive.  Therefore, people have devised these F12 approximations that are constructed in such a way to give results that are comparable to those that you'd get with a large basis set, but you can get them by using a relatively small basis set instead!  This is a powerful approach to convergent quantum chemistry that saves you a lot of time while maintaining a set of very good results.
5.) Why didn't you provide more pretty pictures?
That's just the nature of the beast.  Computational chemistry is usually short on graphics but very dense on spreadsheets.  Plus... I'm no artist.  I actually spent a good couple hours trying to get some electrostatic potentials posted but the new version of G09 hates the molecular viewer programs I currently use so I ditched that idea.

Out of curiosity, could you check the sign on the dipole moment for B3LYP? For me, the magnitude was the same, but the direction was different than MP2 or CCSD. This was matched by the Mulliken charges from B3LYP (and PM6) which showed the opposite directions.

This is very good research and nice results. I miss BP86, though (it's my favourite child :P), just for a comparison it is always nice to have a pure GGA functional. Could you also include $\Delta E(\ce{A;B})$, I would assume that free rotation is feasible at room temperature.

Wow! Really great answer, both you and @GeoffHutchison , that goes way past the contents of my course. Thank you for introducing me to computer models.
I'll be honest with you, the way I tried to solve this problem was using bond angles, trigonometry, and algebra. I did get the same answer as you two, but it was due to luck. I was hoping there could be a way of just looking at the molecule's structure, assuming there is no dipole in C-C bonds,assuming all C-H bonds had equal dipoles, and calculate the direction of the overall dipole. I will wait for more votes to decide on the best answer.

@GeoffHutchison All the dipole moments were consistently in the -Z direction.  But this really is just dependent on how the program orients your molecule in space.  I know that CFOUR will orient opposite to what G09 usually does so I end up having to take that into account when comparing certain data.

This was actually an interesting problem.
Well, I ran a quick calculation using Avogadro and GAMESS, although other packages would work. This is a CCSD/aug-cc-pVTZ calculation, pretty much the gold standard in quantum chemistry. (CCSD(T) and a larger basis set might be better, but are unlikely to differ much here and I can't run them on my laptop.)
The overall dipole moment is 0.305 Debye for cis-2-butene, so it's weakly polar. Here's the picture.

As you probably know, the standard in chemistry is that the dipole moment is drawn from + pointing to - (e.g. +-->) so the dipole arrow is as drawn in the figure, with the "tail" of the arrow from the methyl groups to the double bond.
So yes, it's very similar to your thought process. Also, there are some comments below with links very similar to your discussion.
Now the interesting part. The differences in partial charges are very small. My initial calculations used the PM6 and PM7 semiempirical methods. I also tried B3LYP, a density functional method.
These get the size of the dipole moment right, but the direction opposite. So who knows, maybe your teacher did the same thing I did at first.

Geoff, sp3 releases electron density towards the lower energy sp2.  Isn't the direction of the dipole moment the opposite (plus up by the methyls and negative by the double bond) of what you've drawn ?

Nope. PM6 and B3LYP/6-31G* both indicate the methyl carbons are negative relative to the C=C. If I take the 3xH atoms added into the methyl carbons, I still get -0.023500$e^-$ for the CH3 relative to +0.0235 for the sp2 carbons.

And incidentally, I promise it's not a drawing. I ran two quick calculations on my laptop.

for example, see p. 127 at  http://www.as.utexas.edu/astronomy/education/spring07/scalo/secure/LoudonCh4Alkenes.pdf

or here  http://nptel.ac.in/courses/104103071/module4/lec7/4.html

@GeoffHutchison CCSD is the gold standard in chemistry????  (I think you meant CCSD(T))

OK, fair enough. I thought the "pretty much" would satisfy folks like you. Certainly I don't expect CCSD(T) to vary significantly from CCSD here, nor do I expect a large basis set effect. Clearly these are more reliable than B3LYP or PM6.

@GeoffHutchison (T) changes everything for a variety of systems.  I just couldn't not say anything when I saw CCSD being called a gold standard.  At that level I'd just resort to MP2.  Thats all.  I didn't mean to sound overreacting.

@LordStryker I'll try to be more precise with "gold standard". I doubted there would be much difference in dipole moment and without an F12 code on my laptop, I stuck to something that wouldn't take too long.

@GeoffHutchison How would you rationalize the direction of the dipole moment?

@Dissenter I think the other rationalizations here are good, although the effects are very subtle, since a C-H bond in any form doesn't have much of a bond dipole.

@GeoffHutchison so did everyone agree that the direction of the dipole moment is from the sp3 carbon to the sp2 carbon or vice versa? Sorry I got a bit lost.

@Dissenter Yes. At lower levels of theory, I get the reverse, but I agree that with quality calculations, the dipole moment all match all other answers here.

@GeoffHutchison Cool, because the intuitive answer makes sense - the DM should be running from sp3 to sp2 because sp2 carbon is more EN than sp3 carbon.

According to the informations from LordStryker, you are showing the transition state (StructB) - shouldn't it be mentioned somewhere?


text:
Which chemical is responsible for making marker ink permanent?
As asked in the question and also, a bonus question: which property of a chemical would make it 'permanent' in terms of markers?

The "permanence" is just a reference to it being waterproof.

In general, the ink comprises a main carrier solvent, a glyceride, a pyrrolidone, a resin and a colorant.

This is mostly the result of solubility. There are other markers sold on the market with labels such as "buff proof" (slang for removal proof), that are really just pens that use a different carrier solvent. With those, typically organic solvents would be the correct solvent for removal.

This composition is taken from a patent, isn't it?

The quote is from the Wikipedia.

I added the second part about the other class of markers from personal experience.

Basically, ink is a liquid or paste made up of pigments or dyes that is used to color a surface to create printed words or pictures. Modern inks are often made of complex combinations of various chemicals, including solvents, pigments, dyes, resins and other materials.

(From here)

The chemicals found in permanent markers--xylene, toluene and urethane resin--are what give these markers their characteristic abilities to handle unusual surfaces and leave long-lasting messages. While substantial or prolonged exposure to these chemicals is harmful and toxic, the amount present in a permanent marker is minimal for the casual user.

(From here)
This is what I got from google search. The above mentioned chemicals are most oftenly used, which make ink permanent. However If you have notice you would realise that the ink actually is not permanent, it fade away after certain period of time!

This answer is just plain wrong. Permanent markers are permanent because they use ink that is not water-soluble (that's where the xylene and toluene come in) and pigments or dyes that don't fade. Adding xylene, toluene and urethane to a non-permanent ink wouldn't make it permanent.

@DavidRicherby I don't get were it is written in the answer that permanent ink is prepared from non-permanent ink. In answer Hey has tried to explain how permanent ink differ from normal ink.

@RutvikSutaria The answer implies that permanence comes from the inclusion of substances such as xylene, toluene and urethane but just including those substances doesn't make an ink permanent. Apologies for the rather opaque phrasing of my original comment.

@DavidRicherby I think you should give answer to this question.

@DavidRicherby, I have worked with "Permanent" markers in my job for several years. They worked pretty well until the toluene and xylene were removed by the manufacturers. They used to withstand removal from vinyl products when I used plain isopropyl alcohol. Now, they can be easily removed with it, without any effort. Allot of them will state "Xylene Free". We are down to one product, from one manufacturer, that stains in the manner we require. All others we have tested, wipe off with a rag and alcohol.

@DaleRomine OK but, still, it isn't the solvent that makes the ink permanent: the solvent is merely a carrier for the pigment. Presumably the point is that using toluene or xylene allows you to use pigments that are more resistant to attack by other solvents but it's those pigments that make the ink permanent.

In my experience, ink of a permanent marker can be removed by drawing over it with a non-permanent marker and wiping it off (success depends on surface). The same goes for non-permanent writing that can't be removed because it is too old. Expo (apparently a company that sells markers and related stuff) confirms this in their FAQ

the big question:
how do I get permanent marker off a whiteboard?
No worries if those pesky permanent markers make their way into your whitespace. Just draw over it with an Expo Dry Erase Marker and poof! It's gone.

From this I would conclude that non-permanent markers contain some solvent that permanent markers lack. But obviously, permanent markers need some solvent, too. Otherwise, you wouldn't be able to write in the first place. So it must be some solvent that evaporates quickly while the solvent in whiteboard markers doesn't.
If you search for ingredients of whiteboard markers, the organic solvents Xylene and Toluene keep popping up, so that's obviously not the answer.
Whiteboard cleaners contain all kinds of organic solvents that evaporate much faster:

the main ingredient is isopropyl alcohol, which is sometimes called "rubbing alcohol." Other ingredients may include 2-butoxy ethanol and (di)chloroacetic acid.
  ~~~
  ISOPROPANOL (ISOPROPYL ALCOHOL), 2-PROPANOL, DIMETHYL CARBINOL
  2-BUTOXYETHANOL (ETHYLENEGLYCOL MONOBUTYL ETHER), BUTYL CELLOSOLVE, BUTYL GLYCOL, GLYCOL ETHER EB

Given the wide range of possibilities, the actual chemicals used probably depend on the marker's brand.  Isopropyl alcohol seems to be a good bet, though.

It also evaporates quickly, leaves nearly zero oil traces, compared to ethanol, and is relatively non-toxic, compared to alternative solvents.

To answer your more generic "bonus" question:

which property of a chemical would make it 'permanent' in terms of markers?

The ink needs to waterproof, i.e. not water soluble
The solvent needs to vaporize quickly

This appears to be speculation rather than an actual answer. There is no race to answer questions here. If you don't know the answer, at least wait a couple of days to see if somebody posts something authoritative before posting possibly-helpful speculation.

Aren't Isopropanol and 2-propanol the same thing?

yes, it's also the same as dimethyl carbinol. The 4 chemicals in the line below are also synonyms for each other.

Regarding the bonus question: Permanence will depend on resistance of the resin to water and solvents, which may be obtained using a resin that self-crosslinks at room temperature.  Alkyd paints contain this type of resin, but there are many other possibilities. You also need to protect the bond between resin and substrate, which gets tricky, because of so many possibilities for the substrate's chemical structure -- relatively easy to stick to wood or acrylic, but hard for polyolefins, especially if fluorinated.  An ideal resin would be of low molecular weight and a good solvating power for the substrate, so it would penetrate somewhat into the surface, and then chemically attach as well as self-crosslinking.

source: Stack Exchange

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

sort: votes


text:
Calculating collision diameter with viscosity, but not having the mass given
I've got a homework question i'm stumped on.

The viscosity of the ethylene gas was measured at $99.8\times10^{-7}\ \mathrm{N\ s\ m^{-2}}$ at $288\ \mathrm K$. Calculate the collision diameter in $\overset{\circ}{\mathrm{A}}$ for this gas molecule.

I started with
$$\eta=\frac{m\overline v}{3\sqrt2\pi d^2} $$
then i said
$$\overline v=\sqrt{\frac{8RT}{\pi M}}$$
Molar mass $M=28.05\ \mathrm{g\ mol^{-1}}$
So my equation looks like
$$\eta=\frac{m}{3\sqrt2\pi d^2}\sqrt{\frac{8RT}{\pi M}} $$
but $m$ is still a variable. What should I do from here? Is there another substitution I can do?

So apparently my confusion about solving this problem, came from a lack of understanding what the $m$ variable was supposed to represent. I had thought that it was the total mass of the system, but it is actually the mass of one molecule.

I just asked my professor and he speedily said something about taking the molar mass and multiplying by $10^{26} \mathrm{kg}$.

Answering my own question here (just received the answer to the problem.)
the value for $(\mathrm{m})$ was supposed to be in $\mathrm{kg}$, and could be found by taking the Molar mass and converting:
$\mathrm{M} \cdot 1.661 x10^{-27}\mathrm{kg\cdot amu^{-1}}$
The solution came out looking like
$\overset{-}{v} = \sqrt{\frac{8x8.314 \mathrm{J \cdot K^{-1} \cdot mol^{-1}} x 288\mathrm{K}}{3.14x28.05x10^{-3}kg \cdot mol^{-1}}}$
$d=\sqrt{\frac{m \overset{-}{v}}{3 \sqrt{2} \pi \eta}}$
$d=\sqrt{\frac{(28.05 \mathrm{amu} \space x \space 1.661x10^{-27} \mathrm{kg} \cdot \mathrm{amu}^{-1}) \space x \space 466.23 \mathrm{m} \cdot \mathrm{s}^{-1}}{3 \sqrt{2} \pi \space x \space 99.8x10^{-7} \mathrm{N} \cdot \mathrm{s} \cdot \mathrm{m}^{-2}}}$
$=4.04x10^{-10} \mathrm{m} = 4.04 \overset {\circ}{A}$

Your numerical calculation is correct, but your units are wrong. The conversion factor from molar mass $M$ (e.g. in $\mathrm{g/mol}$) to mass $m$ (e.g. in $\mathrm g$) is simply given by the Avogadro constant $N_\mathrm A=6.022,140,857(74)\times10^{23}\ \mathrm{mol^{-1}}$. Note that $1/N_\mathrm A=1.661\times10^{-24}\ \mathrm{mol}$, which is equivalent to the numerical value that you have used.  You also converted $\mathrm g$ to $\mathrm{kg}$; that’s why your value is $1.661\times10^{-27}\ \mathrm{kg\ mol/g}$.

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


text:
Determining the amount of chlorine in a swimming pool
I need to determine the amount of chlorine in a swimming pool. How do I get started? What kind of experiment would I need to carry out?
What equipment would I need for this experiment?

If you need a no-frills solution, get some quick-test paper, such as this.
A slightly more complex solution would be a test kit like this, which makes use of the formation of Bandrowski's base from 1,4-phenylenediamine in the presence of oxidants, such as chlorine.


text:
How would one express pemoline in terms of phenethylamine?
Pemoline is a psychostimulant that is no longer used in most countries. Similarly to the psychostimulants amfetamine and methylphenidate I can clearly see a phenethylamine backbone to this compound. I know it is also an oxazoline, but I am writing an article on my Wiki about substituted phenethylamines and their pharmacologic and structural properties so I would like to know what its chemical name would be in terms of its phenethylamine backbone. If you want an example of what I mean, see the current revision of my Wiki article on said drug [1] and look at the chemical names I have written. For example for methylphenidate the chemical name is N,α-butyl-β-methoxycarbonylphenethylamine.
My chemistry knowledge is only first-year university chemistry level, hence why I need your help as I don't know all the adjective forms of functional groups.
Below is the 2D structure of pemoline if you would like it in order to deduce its name in terms of phenethylamine.

What you are really looking for is the concept of pharmacophore, that's structural and electronic features in a molecule that allow it to bind to a certain receptor.
If you squint at pemoline just right you can see an aromatic group, and two or three carbnos away you see a hydrogen bond acceptor, just as in phenethylamine, so you could argue that the pharmacophores are the same.  At least both compounds act in some way on the dopaminergic system.
Another important concept to remember is polypharmacology.  That means that one compound can act on different targets. A medical chemist would call these "off-target effects", and a physician would say "side effects".  It's also that any biological system has many components - there are not only dopamine receptors, there are also dopamine transporters that regulate agonist levels in the synaptic cleft and allow for agonist recycling, and there is dopamine metabolism.  Any compound that acts on any system might affect these components differently, consequently target identification is an important step in medicinal chemistry.
Your question is perhaps ill-posed (well, you are a first year student), but it probes really deep levels in med. chem.

I would like to know what its chemical name would be in terms of its phenethylamine backbone.

Artificial numbering and naming schemes that only exist in ones own cloud cuckoo land are pointless.
IUPAC nomenclature exists for a reason.
Besides the IUPAC nomenclature, other naming schemes - based on established (trivial) names exists and are widely used - for good reason and with acceptance by the scientific community. Typical examples are found in the field of natural products chemistry (steroids, alkaloids, etc.)
It is however crucial to understand that in these cases, prefixes and/or suffixes denote a variation of the underlying core structure, such as

additional substituents
lack of methyl groups
an elongated side chain, etc.

In the current case, the central structure is a completely different heterocyclus, so markedly different from the original phenethylamine that it would not make any sense to apply a phenethylamine-based name here.
Neither would one try to apply such a naming scheme to the morphine skeleton - despite the fact that one might spot a six-membered ring with $\ce{CH2-CH2-NCH3}$ side chain here.
This does not mean that a member of the phenethylamine family is not among the ancestors of morphine in the biosynthesis! In fact, reticuline, a morphine precursor, most likely is a grandchild of dopamine.
But again, at a certain point structures have become so markedly different that a valid nomenclature (= valid for a particular compound class) no longer is applicable.
Trying to describe similarities between structures is not wrong, but trying to apply an inappropriate naming scheme is not helpful here.
Similarities between molecular entities should be expressed otherwise.
For sequences (DNA, RNA, etc.), Hamming and Levenshtein distances can be used. For the title molecule and the whole dopamine/amphetamine class, I'd consider the comparison of their fingerprints (Tanimoto coefficients).

I am doing this for the sake of my article as it is about substituted phenethylamines. While this might be your opinion if you want me to accept your answer you may wish to actually answer my question. I am not arguing against IUPAC nomenclature, otherwise I'd start off by badmouthing IUPAC naming, I am doing this for a specific purpose.

Renaming a substituted oxazolinone as "some kind of phenethylamine" is amateurish. As a reviewer, I'd reject any manuscript that use this nonsense. I have no problem if you cannot accept my answer. I am not willing to make up arbitrary names names following an unscientific and dilettantish rule set.

I agree that it will be difficult to devise a systematic name for pemoline as a substituted phenethylamine. The structure is just too different. However, I disagree with the notion that non-IUPAC naming schemes are necessarily less useful than IUPAC. The descriptor 'phenethylamine' is non-IUPAC, but the audience for the Wiki page likely understands (and expects!) that nomenclature. To the greatest extent possible, numbering schemes should follow established conventions, which likely exist for this important class of compounds.

@jerepierre I'm fully aware of non-IUPAC naming schemes based on established names  and the use of (homo, nor, abeo and other modifiers. 'phenethylamine' or 'amphetamine' were widely used in the literature for a large number of compounds to denote various arene substitution patterns and even branching on the side chain. In this case however, such an approach bends the nameing scheme far beyond reason and is pointless. Neither would one name a tetrahydroisoquinoline alkaloid as [...]phenethylamine

@KlausWarzecha I'm sure that you are aware of that, but the original questioner, as a first-year university student, might not be. The statement "IUPAC nomenclature exists for a reason" implies to me that the questioner should ignore other naming conventions.

@jerepierre Fixed. Other naming conventions should not be ignored - but they should not be abused either.


text:
Quick way to find magnetic nature of compounds or ions
According to the Molecular Orbital Theory, the method to check whether a given compound is paramagnetic or diamagnetic is to draw the molecular orbitals and then to see whether we have all the electrons paired or is there some unpaired electron. However, I want to know, is there a quick way to find the magnetic nature without drawing the orbital diagrams?

Well, for some compounds it's easy, because there are an odd number of electrons - thus there's no way to pair them all.
Otherwise, there is no quick-fire way without doing the MO diagram.
Consider $O_2$ which has an even number of electrons, but is a ground-state triplet with two unpaired electrons. The only way to know is to look at the MO diagram.
If you practice, you'll realize certain patterns and develop more of an intuition.

Just a small addition: If you have a completely unsymmetric molecule with an even number of electrons, it is usually a singlet. And then again, a qualitative MO scheme will not really help with compounds with a small HOMO-LUMO gap.

To clarify the last bit from @Martin, compounds with small HOMO LUMO gap are tricky because unpaired electrons can go into the LUMO or other low-lying orbitals.


text:
What are chiral environments?
I keep seeing the term cropping up, but cannot seem to find a definition for it. My understanding is that it is an environment in which, when the isomers interact with the environment, the two enantiomers can be distinguished. The only example I can think of is with the human body, where one type of enantiomer can form an enzyme-substrate complex and the other cannot. I can't think of any other examples though. Can you create a chiral environment in a lab?

My understanding is that it is an environment in which, when the
  isomers interact with the environment, the two enantiomers can be
  distinguished.

That's exactly right.  Some examples of chiral environments that could be created in a lab would include

Running a reaction in a chiral solvent
Running a reaction in a chiral cavity
Running a reaction with a chiral reagent
Running a photochemical reaction with a polarized light source
Carrying out a reaction on a surface to which chiral molecules have been attached (certain clays have this property)

All of these items are like a glove, and each enantiomer will fit differently.  This different "fit" leads to transition states of different energy as each enantiomer moves along the reaction coordinate towards the product(s).

Thank you for your answer! I just wanted to clarify: in a chiral environment, is only one of the enantiomers produced, or are both produced but only one of them interacts, or can it be a combination of these two i.e. you have a chiral environment just as long as there is some distinction between the enantiomers, whether it be that only one is created or that both are created but only one interacts with the environment?

If you run a reaction that can produce 2 enantiomers in a chiral environment, then one enantiomer will be preferentially produced.  Because in the chiral environment the two possible transition states will be diastereomeric (e.g. of different energy).  If the chiral environment is "strong" enough then one enantiomer may be produced to the exclusion of the other enantiomer.  I'm not sure I understand the second part of your comment, but both enantiomers and the transition states leading to them are diasteromeric in a chiral environment,

so the transition states will have different energies leading to the preferred formation of one enantiomer over the other.


text:
Simple explanation of dative bonding
I need a simple explanation of dative bonding. I am a student that has learnt simple covalent bonding and ionic bonding, but I need help to learn dative bonding. This if for an urgent assignment, so please help. Thanks.

what you need in ADDITION to what you can read in wikipedia article http://en.wikipedia.org/wiki/Dipolar_bond ?

Alright, I will keep this extremely brief. You know the following things I presume:
Ionic bonding - formed when electrons are transferred from one atom to another, forming a cation and anion.
Covalent bonding - formed when two atoms share an electron pair
Dative covalent bonding - formed when two atoms share an electron pair, but this electron pair comes entirely from one of the atoms
For example, in a Hydrogen to Hydrogen covalent bond:
H-H
They share a pair of electrons, one electron coming from each.
However,in the carbon monoxide molecule, there are two normal covalent bonds and one dative bond. In the dative bond, both electrons are initially from the oxygen, and are then shared between the two.

Hope this helps


text:
Why does zinc react with sodium hydroxide?
If zinc is less reactive than sodium, then why does it still react with sodium hydroxide?

Indeed, $\ce{Zn}$ is lower than $\ce{Na}$ in activity series of metals, so the following reaction won't take place
$$\require{cancel}\ce{Zn + 2NaOH \cancel{→} Zn(OH)2 + 2Na}$$
Remember, however, that $\ce{Zn}$ is amphoteric, so it can reacts with a strong base such as $\ce{NaOH}$ as an acid forming sodium zincate
$$\ce{Zn + 2 H2O + 2 NaOH -> Na2Zn(OH)4 + H2}$$
P.S. I forget a lot of good ol' chemistry, and I have just realized that usually we do not call metals themselves amphoteric, but rather use this term for their oxides/hydroxides. For metals we just say that some of them (such as $\ce{Zn}$) dissolve both in acids and in bases. The point is that $\ce{Zn}$ does not react as an acid in the reaction above since the reaction is not an acid-base one, but rather oxidation-reduction. So, the answer then is that $\ce{Zn}$ reacts with aqueous solution of sodium hydroxide because it can be oxidized in these condition in accordance with the reaction mentioned.

does this explain why? so zinc can react because it's amphoteric?

@Nick, chemistry (especially at school level) very rarely answers "why" questions. It is almost all about remembering answers to "what" questions, i.e. what happens when A and B are mixed together. What do you want me to do beyond that? Do you want me to calculate Gibbs energy for this reaction and show that it is negative so the reaction is favoured and then (somehow) estimate the reaction rate to show that it will indeed proceed in a reasonable amount of time?

@Wildcat. The closest to real answers to why-questions you get is from computational chemistry using first-principles methods. Unfortunately we are not yet able to fully simulate chemistry in the bottle with this, and its doubtful that we ever will be.

In effect zinc is not displacing sodium. It's displacing hydrogen. We expect a metal to do this from a hydroxide base if

the metal is more reactive than hydrogen and
the metal's own hydroxide can act as acid  towards this base (amphoteric hydroxide).

Zinc passes both tests and so, party time!

Zn is a amphoteric element. Therefore Zn can react with both acids and alkalis eliminating hydrogen. Learn more about amphoteric elements and compounds
https://www.chemistryscl.com/chemistry_theories/amphoteric_compounds/main.php

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


text:
How do we know that hydrogen gas consists of two hydrogen atoms?
Of course I know that both 1s orbitals will combine and that the electrons will occupy the bonding MO and that the two atoms will reach a lower energy state – but how do we really know that is the case?

Well, we can measure the molar mas of hydrogen gas and it will be close to 2 g/mol which means that it indeed consist of $\ce{H2}$ molecules rather than $\ce{H}$ atoms.

@Wildcat You mistakenly put your answer into the comment box and not the answer box further below. :P

@LordStryker, yes, but OP mentioned orbitals and labeled equation with "quantum-chemistry" tag, so maybe his question actually is how do we now that $\ce{H}$ atomic orbitals do combine to form $\ce{H2}$ molecular orbitals? The answer would be different then. :D

@Wildcat It would seem as though a better title to the question is merited then.

Assuming ideal gas behavior we can determine the molar mass $M$ of the hydrogen gas from its density $\rho$ (which can be measured) by the following formula
$$
    M = {{RT\rho}\over{p}}\ \, .
$$
At STP $\rho \approx 0.09 \, \mathrm{g} \, \mathrm{L}^{−1}$ so the following back-of-the-envelope calculation
$$
    M = {{8.31 \, \mathrm{L} \, \mathrm{kPa} \, \mathrm{K}^{−1} \, \mathrm{mol}^{−1} \cdot 273.15 \, \mathrm{K} \cdot 0.09 \, \mathrm{g} \, \mathrm{L}^{-1}}\over{101.325 \, \mathrm{kPa}}}\ = 2.02 \, \mathrm{g} \, \mathrm{mol}^{−1} \, .
$$
shows that $M$ is indeed close to 2 g/mol which means that hydrogen gas consist of $\ce{H2}$ molecules rather than $\ce{H}$ atoms

I think we should try to answer this question outside the context of our current consensus reality of chemistry. There was a time when water was written HO and oxygen was thought to have atomic weight of 8. Experimentally, how did early chemists first discover that H is actually H2?

Hmmm... But why should we answer the OP's question outside the context of our current consensus reality? The question you asked is an interesting one without a doubt, but its a different one. OP knows the electronic structure of $\ce{H}$ atom, so he surely knows its atomic mass.

I think you're asking a historical question too. This largely comes from a similar question of how we get the modern periodic table and atomic masses. The Chemical Heritage Foundation has a nice outline.
In my opinion, the big breakthrough here was due to John Dalton (the Wikipedia page is also good).
Dalton really pushed the atomic theory to molecular formulas. At the time, they knew from various analyses the various compositions - there is 8 times the mass of oxygen in water than hydrogen. So Dalton reasoned if everything is made of atoms, they must combine to form different compounds (molecules) in whole number ratios. For example, if an atom can't be divided, we can't have $H_{1.5}O$ -- that violates the idea of indivisible atoms.
So he guessed and reasoned through the different compounds and compositions known and published a table of atomic masses. As mentioned above, he was wrong about some things, assuming $\ce{HO}$ instead of $\ce{H2O}$, etc. But you can make some fairly good reasoning of the law of multiple proportions to determine stoichiometry.
Avogadro was the next step. He reasoned that equal volumes of gases should have the same number of molecules. So if you weigh equal volumes, you can get the (molar) mass. Unfortunately, his work was largely ignored. (Interesting note, Loschmidt is actually the person who worked out $N_A$, not Avogadro.)
Other people, notably Cannizzaro helped to clarify and expand Avogadro's work at the Karlsruhe Congress.
Avogadro believed that most gases were binary, i.e., $\ce{H2, O2, N2}$ not $\ce{H, O, N}$ as assumed by Dalton.
So if you carefully measure the reaction of hydrogen gas and oxygen gas to get water, you realize it's $\ce{H2O}$ based on the proportions, and then the atomic masses fall into place.
We get further proof of $\ce{H2}$ from other compounds. Sodium bicarbonate clearly has a small amount of hydrogen and you can work out $\ce{NaHCO3}$.
Put simply, this all required many people and a lot of careful measurements and deduction. But the law of multiple proportions and Avogadro's theory will eventually give you $\ce{H2}$.

See also http://chemistry.stackexchange.com/questions/4137/basic-understanding-of-relative-atomic-mass

Nice answer, I like the historical perspective.  One thing I'd add, in addition to burning hydrogen and oxygen to produce water, the reaction of hydrogen and chlorine to produce HCl was also needed (two  equations, two unknowns if you will) to ascertain that the hydrogen molecule was really a diatomic.  For example, see this link


text:
Why can't this mixture be used as a buffer solution?

Explain, using a chemical equation to aid your answer, why a mixture of $50~\mathrm{cm^3}$ of $0.10~\mathrm{mol\:dm^{-3}}$ ammonia solution and $50~\mathrm{cm^3}$ of $0.10~\mathrm{mol\:dm^{-3}}$ hydrochloric acid cannot be used as a buffer solution.

So writing the equation down: $$\ce{NH4OH + HCl -> NH4Cl + H2O}$$
there are $0.005~\mathrm{mol}$ of each reactant. From my notes, I think there has to be double the amount of the weak base, but I don't really understand why. Even after googling explanations, I can't find one that answers my question. Why do you need double the amount (if this is correct) of the weak base? An answer to the question above would also be appreciated.

You have an example of an alkaline buffer solutions here. They are commonly made from a weak base ($\ce{NH3}$ aqueous solution) and one of its salts ($\ce{NH4Cl}$) and work in a little simplified picture as follows.
Ammonia is a weak base, so the position of the equilibrium for the following reaction is well to the left
$$
    \ce{NH3_{(aq)} + H2O_{(aq)} <<=> NH4+_{(aq)} + OH^{-}_{(aq)}} \, ,
$$
and once you add some $\ce{NH4Cl}$ it pushes the equilibrium of the reaction above even more to the left (see Le Chatelier's Principle). As the result you have both lots of ammonia (due to the equilibrium of the above reaction being well to the left) and of ammonium ions (coming from $\ce{NH4Cl}$). And you will still have some hydroxide ions due to the reaction above, so that solution will be alkaline. The actual $pH$ depends on in which molar proportions the ingredients were mixed. Usually, $\ce{NH3_{(aq)}}$ and $\ce{NH4Cl}$ mixed in equal molar proportions, the solution would have a pH of 9.25.
When some acid is added to such buffer it will mostly be removed by the following process
$$
    \ce{NH3_{(aq)} + H+_{(aq)} <=>> NH4+_{(aq)}} \, ,
$$
and when some base is added it will mostly be consumed by the following reaction with ammonium ions
$$
    \ce{NH4+_{(aq)} + OH^{-}_{(aq)} <=>> NH3_{(aq)} + H2O_{(aq)}} \, ,
$$
where in both cases the equilibrium is well to the right.

So, as you see, both the weak base and its salt should present in the solution. Thus, there should be some excess of weak base when you mix it with a strong acid, otherwise there will be no weak base at the end of neutralization reaction. The idea that one should have double amount of a weak base comes from the fact that, as I mentioned, usually a weak base and its salt are mixed in equal molar proportions.
$%edit$


text:
Why doesn't copper react with hydrochloric acid while the other metals do?

5) Which one of the following metals does not react with hydrochloric acid?
A. zinc
  B. magnesium
  C. iron
  D. copper
  E. aluminium

The right answer must be copper, but why doesn't copper react with hydrochloric acid while the other metals do?

You might want to look up some terms, such as

noble and less noble metals
reduction potential
galvanic series

Here, reaction means that

hydrogen gas is formed
the metal is dissolved

In order to form hydrogen, protons need to be reduced to hydrogen atoms which then combine to $\ce{H2}$.
$$\ce{2 H+ + 2 e- -> H2}$$
The metal serves as an electron donor and is oxidized, e.g.
$$\ce{Zn -> Zn^{2+} + 2 e- }$$
The more noble a metal is, the more reluctant it is to lose electrons. This is the case for copper, which is therefore not oxidized under these conditions.

In principle, non-oxidizing acids cannot directly oxidize copper since the redox potentials $E$ for $\mathrm{pH} = 0$ show that $\ce{H+}$ cannot oxidize $\ce{Cu}$ to $\ce{Cu^2+}$ or to $\ce{Cu+}$:
$$\begin{alignat}{2}
\ce{2H+ + 2e- \;&<=> H2}\quad &&E^\circ = +0.000\ \mathrm{V}\\
\ce{Cu^2+ + 2e- \;&<=> Cu}\quad &&E^\circ = +0.340\ \mathrm{V}\\
\ce{Cu+ + e- \;&<=> Cu}\quad &&E^\circ = +0.521\ \mathrm{V}
\end{alignat}$$
Note that copper(II) is usually more stable than copper(I) in aqueous solutions.
However, the situation is slightly complicated because of the low solubility of $\ce{CuCl}$ in dilute hydrochloric acid (in concentrated hydrochloric acid, copper forms chlorido complexes such as $\ce{[CuCl2]-}$ and $\ce{[CuCl4]^2-}$):
$$K_\mathrm{sp} = \left[\ce{Cu+}\right] \cdot \left[\ce{Cl-}\right] = 1.72 \times 10^{-7}$$
The corresponding effective redox potential may be estimated as follows:
$$\begin{aligned}
E&=E_{\ce{Cu+}}^\circ+\frac{RT}{F}\cdot\ln\left[\ce{Cu+}\right]\\
&=E_{\ce{Cu+}}^\circ+\frac{RT}{F}\cdot\ln\frac{K_\mathrm{sp}}{\left[\ce{Cl-}\right]} \\
E_{\ce{CuCl}}^\circ&=E_{\ce{Cu+}}^\circ+\frac{RT}{F}\cdot\ln K_\mathrm{sp}\\
&= E_{\ce{Cu+}}^\circ+0.0592\ \mathrm{V}\cdot\log K_\mathrm{sp}\\
&=0.521\ \mathrm{V}+0.0592\ \mathrm{V}\cdot\log\left(1.72 \times 10^{-7}\right)\\
&=0.121\ \mathrm{V}
\end{aligned} $$
Therefore, oxidation of $\ce{Cu}$ to $\ce{Cu+}$ is favoured in dilute hydrochloric acid. Nevertheless, $\ce{H+}$ is still not strong enough to oxidize $\ce{Cu}$.
However, $\ce{Cu}$ can be oxidized by $\ce{O2}$:
$$\ce{O2 + 4H+ + 4e- <=> 2H2O}\quad E^\circ = +1.229\ \mathrm{V}$$
Therefore, copper is slowly oxidized in dilute hydrochloric acid in contact with air.

Although you should take a look at what Klaus said, copper does in fact react with hydrochloric acid, it just takes a week until all the copper is converted into copper chloride (green) and another week or so until it forms crystals and you can dissolve them in water to form copper chloride again (but depending on the amount of chloride it has, it'll be blue or green).
$$\ce{2HCl (aq) + Cu(s) -> CuCl2(aq) + H2(g)}$$
Copper chloride reacts with aluminium to form copper (metal) and aluminium chloride.
$$\ce{3CuCl2(aq) + 2Al(s) -> 3Cu(s) + 2AlCl3(aq)}$$

Is your acid exposed to air?

Only if there is an oxidizing agent present, or electrolysis is used, can copper metal react with hydrochloric acid. So this answer tacitly assumes oxygen from the air is the oxidizing agent.

Sorry for Jaime's statement. Metallic copper DOES NOT react with $\ce{HCl}$ and even if it does with the help of Oxygen, it does not produce $\ce{H2}$ : No $\ce{H2}$ !


text:
Which of the following elements forms a water-soluble, alkaline oxide?
An element burns in oxygen to form a solid oxide which dissolves in water. The pH of the resulting solution is greater than seven. Which one of the following could the element be?
A. sulphur
B. phosphorus
C. magnesium
D. sodium
E. carbon
The right answer is sodium. But I don't really understand why it can't be magnesium as well.

solid oxide which dissolves in water

Magnesium oxide $\ce{MgO}$ is (practically) insoluble in water.
You might be interested to know that burning of magnesium was the classical flash light in the early days of photography:  Wikipedia.

If you burn it under water (in the absence of oxygen) then it is of course reactive enough to form magnesium hydroxide (and hydrogen).

The answer is sodium. In fact, burning sodium in oxygen atmosphere will produce sodium oxide. When water is added to sodium oxide $\ce{NaOH}$ is produced. The $pH$ of the resulting solution is greater than seven. It can not be $\ce{Mg}$, because $\ce{MgO}$ is insoluble in water.


text:
If the electron of a hydrogen atom transitions from energy level n=5 to n=3, which level corresponds to n1 and which level corresponds to n2
which level corresponds to n1 and which level corresponds to n2 in the Rydberg equation?

Welcome to Chemistry Stack Exchange! Please add what you have attempted towards solving the problem into the body of your question.  For more information, see the site's homework policy for how to ask homework questions. Thanks!

$\bar\nu=\bar R(1/3^2-1/5^2){}{}{}{}{}{}{}{}$


text:
Freezing point depression to find molecular weight
0.845g of the unknown dissolved in 4.250 grams of camphor had a freezing point of 1.36C. The freezing point depression constant for camphor is 40C*kg/Mol and the melting point of pure camphor is 129C.
Calculate the freezing temperature depression, T? Tf=Kf*m
Calulate the molality? Which i know is mol/kg solvent
Calculate the number of moles present of the unknown compound?
Calculate the molar mass of the unknown compound?
Thank you in advance and please lists steps if possible. I really want to understand and to get this!! :)

$T_f$ is really $\Delta T_f$, and you have K. See if you can get a bit further and then someone can help steer you in the right direction.  Remember you're solving for g/mol

Okay here is what I got   Part 1: 1.36-40 is 38.64kg/mol= 0.966    Then for molality .875moles/0.00425kg =205.9 hg/mol Then the number of moles present of unknown compound i got: 0.845g/ 0.966= .875 Then for the fianal one i got for molar mass was 0.845g/.875mol= 1g/mol. please let me know if i am correct

Mass of solute $W_b=0.845\times10^{-3}\;kgm$
  Mass of solvent $W_a=4.250\times10^{-3}\;kgm$
  Depression in freezing point $\Delta T_f=1.36\;K$
  Depression constant for solvent $K_f=40\;K.kg.mol^{-1}$
  Freezing point of Camphor $T_f=129\;^\circ C$

Part I: $\Delta T_f$
Part II: Molality($m$): $$\Delta T_f=K_fm\implies m=\frac{\Delta T_f}{K_f}$$
Part III: Moles($n_b$): $$m=\frac{n_b}{W_a}\implies n_b=W_am=W_a\frac{\Delta T_f}{K_f}$$
Part IV: Molecular Mass of solvent($M_b$): $$n_b=\frac{W_b}{M_b}\implies M_b=\frac{W_b}{n_b}=\frac{W_bK_f}{W_a\Delta T_f}$$

Okay here is what I got Part 1: 1.36-40 is 38.64kg/mol= 0.966 Then for molality .875moles/0.00425kg =205.9 hg/mol Then the number of moles present of unknown compound i got: 0.845g/ 0.966= .875 Then for the fianal one i got for molar mass was 0.845g/.875mol= 1g/mol. please let me know if i am correct –


text:
Periodic trend in difference of energy between the s and p orbitals
Why does the difference of energy between the 2s and 2p orbitals of the second period elements increase with increasing atomic number? Does this difference increases by moving down a group, e.g. is the difference between 3s and 3p bigger than 2p and 2p and why?

Atomic Orbital Ionization Energies, eV
$$
\begin{array}{lccccc}
\hline
\text{Atom} & \mathrm{1s} & \mathrm{2s} & \mathrm{2p} & \mathrm{3s} & \mathrm{3p} \\
\hline
\ce{H} & 13.6 & & & & \\
\ce{He} & 24.5 & & & & \\
\ce{Li} & & 5.45 & & & \\
\ce{Be} & & 9.30 & & & \\
\ce{B} & & 14.0 & 8.30 & & \\
\ce{C} & & 19.5 & 10.7 & & \\
\ce{N} & & 25.5 & 13.1 & & \\
\ce{O} & & 32.3 & 15.9 & & \\
\ce{F} & & 46.4 & 18.7 & & \\
\ce{Ne} & & 48.5 & 21.5 & & \\
\ce{Na} & & & & 5.21 & \\
\ce{Mg} & & & & 7.68 & \\
\ce{Al} & & & & 11.3 & 5.95 \\
\ce{Si} & & & & 15.0 & 7.81 \\
\ce{P} & & & & 18.7 & 10.2 \\
\ce{S} & & & & 20.7 & 11.7 \\
\ce{Cl} & & & & 25.3 & 13.8 \\
\ce{Ar} & & & & 29.2 & 15.9 \\
\hline
\end{array}
$$

Data source: www.colby.edu

Oh, this. Well, in short... The difference in ionization energies is not equal to the difference in orbital energies. Carefully read ron's answer to understand what is going on here.

Imagine that you have, for simplicity, $\ce{B}$ atom. It has one $2p$ electron and one $2s$ electron. You first ionize it by removing $2p$ electron and then continue by removing $2s$ electron. Note, however, that when you perform the second ionization, you already have $\ce{B^+}$ ion, rather then $\ce{B}$ atom, so it has different orbitals.

That is why such ionization energies often called successive ionization energies.

Let's start with neon, the attractive force of 10 protons, screened to varying degrees by the 9 other electrons, must be overcome in order to remove the outermost p electron, this requires 21.5 eV.  There are only 9 protons holding the outermost electron in fluorine in place, again screening plays a role, but with only 9 protons it is easier to remove fluorine's outermost electron (18.7 eV).  This trend of decreasing IP continues for the remaining elements with 2p electrons as we move to the left across this row of elements.
The corresponding 2s IP's are much higher.  For example, the neon 2s IP=48.5 eV.  To measure this IP, we have already removed neon's 6 outermost 2p electrons.  The 2s electron is still held in place by 10 protons, but now only 3 other electrons remain to screen these protons.  Consequently it is much more difficult to remove one of neon's inner 2s electrons.  As we move to the left across this row of elements looking at the 2s IP's we see a trend similar to what we saw with the 2p IP's, but 2 factors are now involved. It get's easier to remove the 2s electron as the number of protons decreases (reduced nuclear attraction) and as the number of already removed outer 2p electrons becomes smaller (more screening).
The same trends are observed in the n=3 series, but the IP's are lower to start with and differences between successive ionizations are compressed because the number of protons, and therefore the number of screening electrons, has increased by 8.
Everytime we add a proton to the nucleus, we also add an electron to keep the charge balanced.  In the case of neon, that 10th electron does not feel the full attraction of the 10 protons - the other electrons screen this last electron from the nucleus.  The bigger and bigger we make our nucleus, the more electrons there are between the last electron we added and the nucleus, therefore that last electron feels less and less of a full proton's attraction, as the number of electrons between it and the nucleus increases.
Here is a good example of compression.  The first IP of neon is 21.5 ev, the IP to pull out a fifth electron (after the first 4 electrons have already been removed) is 126.2 eV - about 6 times more energy is required.  Now consider radon, its first IP is only 10.7 eV.  That electron in radon has so many more electrons between it and the nucleus (76 more to be exact) that screen that last electron from the nucleus, that it is much easier to remove that outermost electron in radon.  The fifth IP for radon is 55 eV - only about 5 times more energy (less than 6) is required here, that is compression.
Here is a link to a nice, thorough discussion on ionization energies and their trends that you may find of interest.
Finally here is a pictorial representation of what we have been discussing.

Note how the spacing between the lines decreases as we move to higher atomic number; also note that the slope of the lines tends to increase as the atomic numbers increases.  Both of these observations illustrate what we have been discussing.  As the atomic number increases there are more and more electrons between the outermost electron and the nucleus making it easier to remove that last electron and making the both the absolute and relative differences between successive ionizations (e.g. 1st IP and 2nd IP, 1st IP and 5th IP, etc.) smaller.

The absolute difference is bigger for the 2s-2p but the relative difference is bigger for 3s-3p...?

There are several relative differences we can compare, to me they appear roughly the same or significantly compressed: 1) 2p(Ne)/2p(B)=21.5/8.3=2.59, 3p(Ar)/3p(Al)=15.9/5.95=2.67 about the same; 2) 2s(Ne)/2s(Li)=48.5/5.45=8.9, 3s(Ar)/3s(Na)=29.2/5.21=5.6 significantly compressed; 3) 2s(Ne)/2p(Ne)=48.5/21.5=2.26, 3s(Ar)/3p(Ar)=29.2/15.9=1.84 significantly compressed

How does the increase of protons and electrons by 8 cause the compression?

I'll edit my answer and try to clarify this.

Ok. You have said: "The first IP of neon is 21.5 ev, the IP to pull out a fifth electron (after the first 4 electrons have already been removed) is 126.2 eV". That electron is still in the p orbital, but in the picture I uploaded, the ionization energy of 2s electron is 48.5eV. I don't understand how. And in the picture I uploaded nitrogen has a smaller IP than oxygen, how?

Your 2s IP data for elements that have 2p electrons does not match what my tables show.  Do you have a link to your data?  I don't think the 2s data (for elements that have 2p electrons) is IP data.  Maybe it is photoelectron  data, dunno.  The O/N anomaly is explained by Hund's rule.  There is a paired pair of electrons with O, not with N, the extra electron-electron repulsion due to pairing makes it easier to remove that first electron from O then we would have expected.

I know that, only it wasn't in agreement with the data I provided. Here is the link http://butane.chem.uiuc.edu/pshapley/GenChem2/Tables.html All in all the energy difference between a ns and np orbital increases in the same period but decreases in the group as n increases.

Marko, see here, it says your numbers (the 2s numbers that don't match the expected IP) are, "one-electron ionization energies of the valence orbitals calculated by finding the
average energies of both the ground-state and
ionized-state configurations. (Harry Gray,
“Electrons and Chemical Bonding,” Benjamin, 1964, Appendix)".  I don't have the book, so not sure how Gray determined them.

In naive terms I think this could be explained as follows:

$2s$ orbital is lower in energy than $2p$ orbitals because it is closer to the nucleus, and thus, experiences a bigger attractive force to the positive nucleus. The so-called effective nuclear charge is bigger for an electron on $2s$ orbital than on one of $2p$ orbitals since in the later case there will be an additional shielding from electrons on $2s$ orbital.
As the nuclear charge increases, both $2s$ orbital and $2p$ orbitals become lower in energy due to the increased attractive force to the nucleus, but the $2s$ orbital is affected to a larger degree for the same reason: it is closer to the nucleus and experiences a bigger gain in attractive force to it.

Thus, the energy gap between $2s$ orbital and $2p$ orbitals becomes larger as the atomic number increases.

That is all what they can tell you in school. But, of course, the situation is more complicated, and i don't think it can be addressed without doing electronic structure calculations. Yes, effective nuclear charge experienced by an electron is often expressed as $Z_{\mathrm{eff}}= Z - s$, where $Z$ is the real nuclear charge and $s$ is the so-called screening constant; $s$ is different for electrons at orbitals corresponding to different values of the principal quantum number $n$ and there are some empirical Slater's rules which provide numerical values of $s$ for different values of $n$. But these picture is "not quite right".
First, as I already pointed out and as one indeed can see from the Wikipedia page linked above, $s_{2s} = s_{2p}$ (same value of $n=2$). But surely if one calculate these screening constant by solving the corresponding Schrödinger equation he/she will indeed found that $s_{2s} \neq s_{2p}$. But even without solving the Schrödinger equation the naive argument given above in (1) favour this inequality. So, $s$ is not just a function of $n$, but also of $l$.
Secondly, screening constants in Slater's rules does not change their values as atomic number increases which is also wrong. Loosely speaking, the more nuclear charge electron can screen, the more will it screen. Below is the related graph from the article with screening constant calculated at the Hartree-Fock level linked above.

So, $s$ is also a function of $Z$, and besides for different $n$ and $l$ it depends on $Z$ in different ways. The "real" picture (in quotes since orbitals aren't real anyway) as we can see is much more complicated, and, honestly speaking, I do not even believe that there is a trend you are looking an explanation for.

Okay, what's the case when comparing the energy difference between 2s-2p versus 3s-3p?

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


text:
Conductance and Conductivity of acid base titration?
I have a question with $\ce{NaOH}$ initially present with given concentration and conductivity, then an equal volume of given concentration of HCl is added and the final conductivity is given, again a same volume of HCl of same concentration is added, the conductivity of which is again given. Now I need to find molar conductivity of $\ce{HCl ,NaOH, NaCl, H2O}$.
Molar conductivity($\lambda_m$), conductivity($\kappa$), Molarity($M$)
Using the formula $\lambda_m=\kappa/M$,firstly I found the molar conductivity of NaOH, afterwards when HCl was added, there will be only NaCl remaining, the rest being used up, so I again found molarity of NaCl and put in the formula and found molar conductivity, but in the last part I am confused, there is only HCl and NaCl remaining and I think one of these must be true:
$$\kappa_{solution}=\kappa_{\ce{NaCl}}+\kappa_{\ce{HCl}}\\\lambda_{solution}=\lambda_{\ce{NaCl}}+\lambda_{\ce{HCl}}$$
Probably the second one due to Kolrauch's Law? Also I am not aware of what is going on here because I'm just using a pretty formula. Please explain in detail the physical situation and the processes involved.

Edit:
The question is:

The conductivity of $0.1$ M $\ce{NaOH}$ is $0.0221\;\Omega^{-1}cm^{-1}$. When an equal volume of $0.1$ M $\ce{HCl}$ solution is added, the conductivity decreases to $0.0056\;\Omega^{-1}cm^{-1}$. A further addition of $\ce{HCl}$ solution, the volume of which is equal to that of first portion added, the conductivity increases to $0.017\;\Omega^{-1}cm^{-1}$. Calculate $\Lambda_m(\ce{NaOH}),\Lambda_m(\ce{NaCl}),\Lambda_m(\ce{HCl}),\Lambda_m(\ce{H^+,OH^-})$

Your formula for molar conductivity is wrong, it is λ=κ/M. And you should better write the whole problem with all the data given.

@Marko edited...

As the concentrations of $\ce{HCl}$ and $\ce{NaCl}$ are equal, that concentration can be regarded as the "concentration" of the solution. If you divide both sides of your first equation (which includes $κ$) with that concentration you would get your second equation (which includes $λ$). So both equations are valid. The problem would arise if the concentrations of $\ce{HCl}$ and $\ce{NaCl}$ are different, you couldn't define the concentration of the solution, and you couldn't calculate it's molar conductivity, therefore only the first equation would be usable. Note that the Kolrauch's Law can also be expressed through $κ$.


text:
GAMESS crash course?
I'm still learning about MO theory – and I thought that I would do some calculations with GAMESS to become more familiar with the concept. Even though I look forward to delving into the realm of quantum math in the not-so-distant future – it's very difficult to get a grip on all of the terminology. I'm using the GAMESS Avogadro extension to write some input files, but I don't know which options to use. Which options should I use if I want to study the orbitals of, say, shikimic acid? RHF, MP2, AM1 / MINI 6-31G(d)? Computer specs: 3.6GHz i5 with 16GB of RAM

I strongly advice you to not enter this dangerous path. First things first. Learn theory, then apply it. Otherwise 99% of your calculations will be GIGO (garbage in, garbage out). Programs can crunch almost any garbage you input, but the output numbers won't have the meaning you expect.

gamess-US source code comes with good enough example jobs (afaik, in test directory), though making it work is a bit of challenge. For learning purposes, i.e. molecules up to 16 atoms with 1-2 d-element atoms it is enough to use mp2/6-31G(d). Thouhg in case your job starts to crunch your hard drive, you should consider reducing amount of atoms.

@Wildcat I do agree that makes no sense of doing e.g. MP2 without knowing much about theory, but wouldn't discourage OP from experimenting. There is always a subset of calculations that makes sense to try, even with limited familiarity with the theory. E.g. doing Huckel or extended-Huckel calculations can make sense and has strong pedagogic value. I think we do our job best here, if we help OP to find those problems/calculations. Also, testing methods agains each other and seeing how they break down even "simple" cases like bond breaking in diatomic molecules, teaches a lot.

@Greg, that was just my own opinion based on personal experience. I mean, I was there many years ago, and I would not recommend anyone to take the same road.

First off, welcome to Chemistry.SE! For formatting help visit the help center and for more information about this page, take the tour.
While I do think you should learn some theory, it's definitely possible to learn some practical computational chemistry through experimentation. And yes, there are a lot of abbreviations to follow.
One thing you should track while you explore is how long different calculations take. In general, more accurate quantum chemical methods are also more computationally costly. This means that they take longer, but also as systems get bigger, the cost rises faster. You might hear some people talk about "linear scaling" (i.e., double the size of the system, twice as much time) or $N^2$, $N^3$, etc.
A good resource, IMHO, is Molecular Modeling Basics by Jensen. It's available through CRC Press but Jan also blogs extensively. He uses GAMESS and Avogadro but also Jmol and other programs.
While the Avogadro project is working on some more detailed tutorials, I'd suggest starting some calculations with HF/6-31G* or B3LYP/6-31G* and see how things go for different molecules. I strongly suggest running some calculations and then consider how the results compare to experiment. Benchmarking is important. Computational models all have systematic and other errors, and for real work it's critical to run a set of molecules and carefully choose methods that work well.
Here are a few key terms to consider:

Method: There are several types of methods.

Semiempirical methods: These include AM1, PM3, RM1, (and PM6, PM7, and others that I don't think are in GAMESS-US). Such methods are fast, since they approximate some integrals and use empirical parameters. They also have an implied basis set which is usually very small. (Avogadro doesn't support viewing semiempirical calculations with GAMESS yet.)
Wavefunction methods: These include HF and relatives. These are similar to typical quantum chemistry methods taught in school.

MP2, MP3, MP4: These are perturbation methods that improve on Hartree Fock and introduce some level of electron correlation.
CCSD, CCSD(T), etc.: These are "coupled cluster" methods that are highly accurate, but often slow, and do not handle large molecules.

Density Functional: This includes another large set of methods, including BLYP, B3LYP, and newer methods like M06-X. Rather than approach things through the Schrödinger equation,

Basis: Most methods (except AM1, RM1, PM3, etc.) also require a "basis set," a set of functions to describe the atomic orbitals. Wikipedia article There are a few families, including:

Pople Type: 3-21G(d), 6-31G(d), 6-311+G(d), etc. These are older, more traditional basis sets. Generally 6-31G(d) or 6-31G* is considered minimally acceptable today. The "*" or (p,d,f) included at the end indicate that higher polarization is included on some or all atoms.
Correlation Consistent: cc-pVDZ, aug-cc-pVTZ, etc. These are newer basis sets, designed to function better with methods that treat electron correlation. The "aug-" includes various diffuse functions, which are useful particularly for anions and electron density that is particularly, well, diffuse. (In Pople basis sets, these are indicated with "+".

Thank you for your reply! Now I know where to start. One question though: to what extent is the hard drive used in QM calculations? I understand that temporary files are stored during the calculations, but do I have to worry about fragmentation or even exceeding my disk space for relatively small molecules like shikimic acid?

@Ravaru, it depends on your HDD size. But if you have something around 1TB (more or less standard this days), you don't have to worry about exceeding the disk space. Fragmentation could be a potential problem, although, I don't think that it will considerably affect performance.

Sorry, 1TB free? I think I have around 600GB free. Will that suffice? The data that takes up space is only temporary, right?

@Ravaru For small molecules, a few GB free will be just fine. Certain types of calculations (MP2, coupled-cluster) will chew through disk space quickly. For 10-20 heavy atoms, you'll have no problems.

@GeoffHutchison, you mean light atoms, not heavy one, aren't you?

@Ravaru, yes, 600GB will be enough for your purposes. And yes, its just for temporary data.

@wildcat in the context of quantum calculations, "heavy atoms" means "not hydrogen," like C and O.

@GeoffHutchison, ah, you mean this tradition I hate and never use. :D Besides, I doubt OP is aware of this computational chemistry notion of "heavy elements" which differs essentially from one often used in general chemistry.

@GeoffHutchison, I prefer to use "light elements" for first-, second-, and third-period elements, i.e. for which non-relativistic approximation is usually a good one.


text:
Using MestReNova: An Introduction
Can anyone help me with a quick question regarding MestReNova 9.0. I have been fiddling with this thing all day and before you ask yes, I have consulted the users guide to no avail.
I want to do three things:
1)Rescale the chemical shift axis to focus only on the relevant sections of my spectra.
2)Set a threshold so that only peaks above a certain intensity will automatically label.
3)how on earth do you get the parameter box to fit onto the page? Ive managed to get it up but it is about three times larger than my spectra and rescaling using the little green buttons leaves it the same size but just cuts loads out.
Any help is much appreciated!

Testing this on version 9 will have to wait until tomorrow, but on version 8 (which I have in front of me, but which has limited differences to version 9 by memory)
1)Rescale the chemical shift axis to focus only on the relevant sections of my spectra.
Do a manual zoom.
->View >Zoom >Manual Zoom (or just type M, or click the manual zoom button)
Type in the values you want
2)Set a threshold so that only peaks above a certain intensity will automatically label.
Do Pick Picking via a Manual Threshold
->Analysis >Peak Picking >Manual Threshold (or just type K)
Left click and hold to define a box within which you will have your peaks picked
3)how on earth do you get the parameter box to fit onto the page? Ive managed to get it up but it is about three times larger than my spectra and rescaling using the little green buttons leaves it the same size but just cuts loads out.
You need to resize the text to a smaller font size

First select the parameters box.
Left click the parameters (which should have the green selected boxes shown) and choose properties. Make sure under the geometry tab, the Auto box is selected for size.
Then from the text toolbar (by default is usually at the bottom of the screen, but if not shown go to >View>Toolbars>Text) change the font size to something small enough to fit on the screen.

Try that for your version - if no luck I'll check on one of my other processing stations tomorrow which has version 9 installed. Let me know how you go.

Brilliant answer thank you very much! It is slightly different but mostly what you say sounds exactly the same as the previous version. thanks again


text:
How many moles of aspartame are present in 4.00mg of aspartame? (3 sig figs)
This is a part of my online chemistry homework (mastering chemistry)
I know how to do it but when I enter $\pu{1.36E-5}$ into the program, it says that it's incorrect and that the correct answer does not depend on $\times10$. I'm not sure what it wants from me. The answer should be given in moles with $3$ sig figs.

What if you put in 0.0000136? The way you ask your question gives very limited scope to understand what is happening. What program are you using? What is the exact answer given from the program? Other information would be helpful.

I will add that as an answer and hopefully you will accept it, and close off the question

Try putting in 0.0000136. Remember that zeroes between the decimal point and the last numbers do not contribute to significant figures. Good luck.


text:
Why is disilyne bent?
Why is disilyne $\ce{Si2H2}$ bent when the steric number of silicon is 2?

Short answer is: because VSEPR theory is oversimplified model of reality.

"The equilibrium geometry of disilyne is not linear, but is twisted. The potential surfaces of acetylene and disilyne have a critical internuclear distance between the central atoms, where the stable geometry changes from linear to twisted forms. The R-dependence of the valence-shell electron energy causes the difference in the structure of the molecules." dx.doi.org/10.1016/0009-2614(82)87044-9

Could you elaborate that in an answer? Why is that particular form more abundant than linear?

So electronic structure calculations cited above predict twisted equilibrium geometry for $\ce{H2Si2}$ with the only issue that the calculations are rather crude (HF/STO-3G). :D

Is there any intuitive explanation?

The twisted form of $\ce{H2Si2}$ is more energetically stable, and thus, it is more likely to find $\ce{H2Si2}$ in this form. The question why the twisted form is more stable does not have an answer you are looking for, except for that this is just the way nature works. :D It can be explained in terms of electronic structure theory as in the paper linked in one of my previous comments.

Nope. I doubt that there exist any "intuitive" explanation for this, where by intuitive you mean "without using electronic structure theory".

You can actually explain this in terms of a second-order Jahn-Teller distortion and associated correlation diagrams. Basically the HOMO becomes more lone-pair like.

Could you elaborate that?

The following disilyne has been prepared and found to be stable to ~100 C.  An X-ray crystal structure found that the two silicon atoms in the triple bond, Si(1) and Si(1'), along with the two attached silicon atoms, Si(2) and Si(2') are coplanar and the Si(1)-Si(1')-Si(2') angle is 137.44 degrees ("trans-bent").

The full text of the article describing the preparation, characterization and reactivity of the compound can be found here.  See figure 4 in the paper for a MO explanation as to why the trans-bent geometry occurs. Basically, the authors suggest that, "the bending is thought to be the result of the mixing of an in-plane $\ce{\pi}$-orbital with a $\ce{\sigma^{\ast}}$ orbital whose energies are close enough to cause the interaction."

Very interesting, but it's hard to imagine the overlap.

I was impressed by the fact (above link, Section 4, paragraph 2) that the bond length contraction going from $\ce{Si=Si}$ to $\ce{Si#Si}$ is half that for $\ce{C=C}$ to $\ce{C#C}$.  That seems significant to me.


text:
Are higher branches of mathematics used in sub-fields of chemistry?
I'm wondering is there any branches of Chemistry that make use of manifolds, Riemann geometry, tensors etc. So far I have not come across anything but hoped that the wider community would know of something?
Forgive me if this is too vague a question but I'm just looking for some topics I can research myself.

How many chemist you think know differential geometry? So, unless you try to explain what the hell is manifold and all other stuff you mentioned, your question is very unlikely to be answered.

The only thing I could come up by reading the Wikipedia article on manifold, is that potential energy surfaces do have some topological properties, thus, there should be some use of differential geometry for analysis of potential energy surfaces. Then I just googled and found something, say, http://dx.doi.org/10.1007/978-94-009-2313-3_3 amongst many other papers.

The author of the paper linked above also wrote a book with an interesting title "Shape in Chemistry: An Introduction to Molecular Shape and Topology" http://eu.wiley.com/WileyCDA/WileyTitle/productCd-0471187410.html

Aha, there is the so-called chemical topology. There is even a short article in Wiki about it (http://en.wikipedia.org/wiki/Topology_%28chemistry%29) and some books with exactly this title (http://www.crcpress.com/product/isbn/9789056991746).

And I hope, the information above is relevant, since I have no idea what these papers and books are about. :D

Thank you very much Wildcat. After all the internet searching I have done (and trust me it is extensive) I haven't come across either of the books you kindly linked! Thanks once again :D

Check out the book "When Topology Meets Chemistry" by Erica Flapan. In the light of new research on topologically interesting molecules (including molecular machines), the field is very interesting and relevant.

We use tensors in my research lab to analyze solid state NMR.
In my research laboratory we use manifolds all the time for stimulating annealing calculations.

There are several research groups in physical chemistry which utilize what you are describing, but it will never be an undergraduate topic. One particular group that comes to mind that uses this sort of stuff a lot is the Hirata group at the University of Illinois. They actually made a piece of software called the Tensor Contraction Engine. http://www.scs.illinois.edu/~sohirata/software.html

Also tensors in EPR and solution state paramagnetic NMR


text:
pH of a buffer solution calculation

Calculate the pH of a mixture of $75\:\mathrm{cm^3}$ of methanoic acid solution of concentration $0.10\:\mathrm{mol\:dm^{-3}}$ and $75\:\mathrm{cm^3}$ of sodium hydroxide solution of concentration $0.050\:\mathrm{mol\:dm^{-3}}$. The $\mathrm{p}K_\mathrm{a}$ of methanoic acid is $3.75$.

I got $0.00375\:\mathrm{mol}$ of $\ce{HCOONa}$, the salt, made. The concentration of that is $\dfrac{0.00375}{150/1000}=0.025\:\mathrm{mol\:dm^{-3}}$(?)
Using Henderson–Hasselbalch: $\mathrm{pH}=3.75+\log\dfrac{0.025}{0.10}=3.15$
The answer says $3.75$, is that a typo?

I don't know if it is a typo, but you do have an error in the Henderson Hasselbach equation. You forgot that the total concentration of methanoic acid changes when you add the sodium hydroxide.

Thanks, had a look at it and tried it again, got 3.45 as pH instead. Do you mean that the concentration changed because the methanoic acid was in excess, so the concentration would be 0.00375/(75/1000)? Because that's what I did.

Yes, the methanoic acid is in excess. Since you have equal volumes of methanoic acid and sodium hydroxide, I took a shortcut to determine that you start with twice as much methanoic acid than sodium hydroxide. You still must have something wrong with the equation, since the calculation of $pK_a$ is when there is the same amount of acid as the salt.

Upon reading about the equation, I realized that it is not a typo (and you should get the value 3.75). So there is still an error in the equation. Determine what each variable represents in the HH equation. http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3747999/

@Jim The only problem remaining now is that your concentration of methanoic acid is 0.00376/(150/1000).  You forgot that its volume increases when the base is added.  Lots of places to make small mistakes in these questions.

One will need the Henderson Hasselbach equation which is given as
$$\text{pH}=pK_a+\log\frac{\ce{A-}}{\ce{HA}}$$
where HA and A- are the concentrations of the conjugate acid and conjugate base.
First we find our moles of each.
$\mathrm{{mol_ {acid}}=75~cm^3\times(0.10~mol~dm^{-3})\times(10cm)^{-3}\ dm^{3} = 0.0075\ mol\ \times \dfrac{1000\ mmol}{1\ mol} = \ 750~mmol}$
same thing for base: $\mathrm {mol_{base} = 375} $ mmol
Set up an ICA (not ICE) to handle your limiting reagent: (Write down the chemical equation if this is hard to see. One should react acid with the best base available)

$\ce{HA}$: $\pu{750 mmol -375 mmol} = \pu{375 mmol}$
$\ce{OH-}$: $\pu{375 mmol - 375 mmol} = \pu{0}$
$\ce{H2O}$: --------------------
$\ce{A-}$: $\pu{0 + 375 mmol} = \pu{375 mmol}$

In the Henderson Hasselbach equation moles can be used instead of molarity because both conjugates are in the same volume so they cancel:
$$\mathrm{pH}=3.75+\log\dfrac{375\ \mathrm{mmol\ \ce{A-}}}{375\ \mathrm{mmol\ } \ce{HA}}$$
$\mathrm{pH}=3.75$

What's an ICA table? I can't find that on the internet.

ICA table is just a way to organize limiting reagent stoichiometry. It is typically done in mol or mmol. It stands for initial, change, after.

I think you have miscalculated the no. of moles for acid and base. In both cases, the volume times strength will be divided by 1000 and not multiplied.

$\mathrm{{mol_ {acid}}=75~cm^3\times(0.10~mol~dm^{-3})\times(10~cm)^{-3}~\pu{dm^3} =\pu{ 0.0075mol}}$

Thanks, good catch! I did not clarify mmol vs mol; the answer is updated.


text:
Resonance of 4-isopropylbenzenethiol
Can someone draw what the missing structure is? I can't find it.

If someone could find the resonance structures of phenol in a book or online, could you upload it? I want to compare my answers for that too.

It seems you didn't draw the thiol group ($\ce{-SH}$) in any of your canonical structures, which is why some are missing. Remember that the sulphur atom has non-bonding pairs next to a $\pi$-conjugated system. What can this mean?

The missing resonance structure is

If you replace "SH" with "OH" in the 4 structures you've drawn (your 5th structure is the same as the one pictured on the exam) and the one I've drawn, you'll have all of the analogous resonance structures for phenol.
EDIT : response to OP's comment
In my view, it doesn't really matter how you order the structures, but your suggestion (between the 3rd and 4th structures from the left) seems logical.  I've redrawn the figure to show the electron movement.

Perfect! Ron, I still have a question, where should I include this resonance structure? My teacher  wants to see the electron pushing. Say I were to put your structure between the THIRD and FOURTH ORGANIC STRUCTURE FROM LEFT, would I have to resend the electrons originally on the sulfur atom to the sulfur atom and then transport the double bond onto the + charge formed?

yolo123, I've edited my post to address your comment, let me know if you still have questions.


text:
What to do about a un-dissolved substance?
To exterminate a coackroach infestation, I bought solid boric acid (in powder form). Rather than awaiting the cockroaches' encounter with the boric acid, I thought about spraying boric acid onto affected areas or cockroaches. I only took high school chemistry, but remembered solubility so checked its solubility in water. I then compared it with the solubility of other compounds, to see if this were an effective idea:

Boric acid: At 100 Celsius, 27.53 g/100 g of water.
Sodium chloride: At 100 Celsius, 38.99 g/100 g of water.

This looks fine, so I poured boiling water into a bottle with the boric acid powder. I shook it and saw the boric acid dissolve. Yet two weeks later, the boric acid seems to have reversed dissolution. Now, it's separated from the water and sits at the bottom of the water.
Can substances do this? What should I do now? I'd need a pot to boil the mixture over the stove. But then, in the interest of health and safety, I couldn't use this pot anymore, so don't want to sacrifice one.
Update: With thanks to and in view of the instructive answers, would it help to refill the bottle, to benefit from the low solubility of boric acid at room temperature? There's only around 10% of water left, with the boric acid sitting at the bottom.

Never pour water into acid, either solid or liquid. Heat is released when acids dissolve in water. This could lead to local boiling and an acid splashing into your face.

You could put the bottle (of boric acid and water) into a pot of water to heat it up. But why bother. You can use the spray as is and when the liquid is mostly gone, you can add more water, mix it every five minutes for about an hour and you can use that to spray the cockroaches.

I think what you forgot is that the solubility of boric acid (or sodium chloride.. or most solids) is lower at room temperature.
So when you dissolved the boric acid in boiling water, there's no problem at first. As it cooled, you will have a supersaturated solution until eventually the boric acid will recrystallize or crash out.
I'd create a new mixture using room temperature water and add only a small amount of boric acid (as mentioned by @K_P, about 5g / 100 mL of water). Let it dissolve slowly while you mix.
Then if you spray, you shouldn't leave large amounts of boric acid around.
Updated Answer
Yes, if there's only ~10% water, you will want to add more (room temperature) water. This will dissolve more boric acid. Stir/shake to dissolve as much as possible, wait a bit and then you can spray again.
As mentioned, you don't want to add water to a hot acid. In this case, adding more room temperature water to a room temperature acid will probably produce heat, but add slowly and carefully.

Usually, when you want to create saturated solution the approach is following:

You heat the water (does not need to boil though)
You dissolve as much solid substance as possible
Let the water cool down
Pour the water through filter to filter out any recrystallised solid compound

The amount you can dissolve is the amount that can be ever dissolved. If the temperature decreases, some of the dissolved compound will recrystallize.

Looking at your link, the solubility of boric acid at room temperature is around 5g / 100g of water. You can either heat it to boiling point to redissolve or you can filter it and this way you will get a 5g / 100g solution of boric acid that you can still use and also recover solid boric acid. You may need to experiment with different DIY filtering devices depending on the particle size of your precipitated boric acid but I guess you made some nice crystals that shouldn't be very hard to filter.


text:
Can assumptions made during calculation of ka of buffer solutions be justified?
We make many assumptions in sciences which are not always true but as they make little difference we ignore them for simplicity. But today I have decided that I can no longer be lured into using expression of $K_{a}$ and $K_{w}$ by assumptions which I deem to make a significant difference.

We say that $[H_{2}O]$ is constant. Why I think it is not justified?
$$\ce{H2O + HCL <=> H3O+ + Cl-}$$
$[H_{2}O]$ is almost $ 55.6 \:\mathrm{mol\:dm^-3}{}$.
At around $20^{o}C$ $20\:\mathrm{mol}$ of $HCl_{aq}$ can be dissolved. Let's say we dissolve $20 \:\mathrm{mol}$ of $HCl$ and try to find it's WE ASSUME that % disassociation of water is $100 \%$. So we must take $[HCl]\approx0$ that means that $20\:\mathrm{mol}$ of $H_{2}O$ has reacted and $[H_{2}O]$ is no longer a constant but has rather changed to $35.6\mathrm{mol}$.
So this expression can no longer be used. $$K_{a}=\frac{[H_{3}O^{+}].[Cl^{-}]} {[HCl]}=K_{c}.[H_{2}O]$$
Note: you might point out that $K_{a}$ is not used for strong acids but I just used $HCl$ as an example but make my point easier to explain.
Assumption made about concentrations when calculating $Ka$ of buffer solutions.
$$CH_{3}COOH_{aq} \rightleftharpoons CH_{3}COO^{-}_{aq} + H^{+}_{aq}$$
We have added $CH_{3}COONa$  to make a buffer solution.
$$CH_{3}COONa_{aq} \rightarrow CH_{3}COO^{-}_{aq} + Na^{+}_{aq}$$
So,
$$[CH_{3}COO^{-}]_{total}=[CH_{3}COO^{-}]_{CH_{3}COONa} + [CH_{3}COO^{-}]_{CH_{3}COOH}$$
$$[CH_{3}COO^{-}]_{CH_{3}COOH} = [H^{+}_{aq}]$$
BUT WE ASSUME that $[CH_{3}COO^{-}]_{total}=[CH_{3}COO^{-}]_{CH_{3}COONa}$ because we say $[CH_{3}COO^{-}]_{CH_{3}COOH}$ is negligible yet we include [$H_{aq}^{+}]$ in the expression which has same value as $[CH_{3}COO^{-}]_{CH_{3}COOH}$. Same value is considered negligible at one place and not negligible at another in the same expression. How is that justified?

ThThis might help:  http://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation#Derivation  Think about how large the value of the acetate from sodium acetate is compared to the value of acetate from acetic acid.

You neglected that Ka and Kw are only meaningful when the concentration is low (usually below 1 M). For HCl, you have a solution near 12 M. In the second case, as explained below, it is ignored because it make less than 0.1% difference in the calculation (sometimes it would make the calculation very difficult to solve if it was not ignored).

Answering your first question, $[\ce {H2O}]$ is constant because the square brackets indicate a concentration. Concentration is the number of moles of the solute on the volume of the solvent.
Except in this case, water is the solvent. So the concentration will always be constant, since the volume is proportional to the number of moles.
Besides, while dealing with the law of mass action, for liquids we don't take concentration, we take activity. Which is approximately one for water.
As for your second question, we basically have the following product:
$$\ce{([CH3COO- ]_{CH3COOH} + [CH3COO- ]_{CH3COONa})[H+ ]}$$
which can be rewritten as $(a + b)b$, where $b << a$. Let's say $a$ was $1$ and $b$ was $0.0001$.
The actual value of the expression is $1.0001\times 10^{-4}$. Neglecting only the first $b$ will give $1\times 10^{-4}$, which is not too much different.
Neglecting only the second $b$ will give $1.0001$, and neglecting both will give $1$. Both of these answers are very wrong.
When a small number is added to a large one, the small number can be ignored. But not when it is multiplied.

The first part failed to explain the reason behind concentration of water being constant. The reason I had figured out and written in the answer below and your answer is being marked for it explains second part beautifully. For this reason first part had to be removed.

@Suchal I clarified it a bit.

For the first part I was forgetting this simple this:
$$[H_{2}O]=\frac{n(H_{2}O)}{Volume_{H_{2}O}} mol.dm^{-3}$$
So as the $n(H_{2}O)$ will decrease the $Volume_{H_{2}O}$ will decrease too making $[H_{2}O]$ a constant.
The second part have been answered by ManishEarth.