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Is there a simple field test for heavy water?
Thought experiment
Given two glasses of water, how would one detect which glass contains heavy water, and which contains potable water without using complicated laboratory equipment?
Something like the way a piece of litmus paper (yes, it is laboratory equipment and easily available to the layman at that!) may be used to test whether a solution is an acid, or alkali.

heavy water has higher boiling and melting points and is slightly toxic for eukaryotes. It is also heavier (1.1 g/ml) then normal water.

This is just a hunch, but I imagine if you filled a small waterbomb from each glass, and placed it in the opposite glass, you would see one sinking (in the normal water) and one floating (in the heavy water)

Not sure how big your glass is, but would a 7 dollar hydrometer be "complicated laboratory equipment"?  A small volumetric flask (10 ml) and a balance would be less than $50.

Why would you oppose heavy water and potable water, to begin with?

There's a very simple test for $\ce{D_{2}O}$ that springs to mind - ice cubes made with heavy water sink in light water. I assume you're talking about differentiating a glass of essentially pure heavy water from, say, tap water, in which case this test should work rather well and requires equipment no more sophisticated than a freezer, ice cube tray and some glasses.

Cool (+: But could there be something still simpler - akin to the boiled egg floats in salt-water? As an extreme example, a person out trekking may not have a freezer ^ power-source handy.

@Everyone - You could make a custom float that floats on heavy water but sinks in light water. Quantitative floats similar to this are used, for instance, to estimate alcohol concentrations in distillates from hobby stills.

There is a much simpler test. Just weigh each one, or even lift up each one and feel which is heavier.

This would not work if the regular water has enough dissolved impurities...

If you don't have access to a freezer, you could allow sun light to pass through
your glass vessels and then through a prism. The heavy water will absorb less orange red light and the difference might be observable.

One of these might help (not complicated).
P.S. I did not want to post this as an answer because I'm not sure if it's       realistic but I can not see were to add a comment.

You can't add comments unless you have earned enough reputation. Besides that, I think this qualifies as an answer, although you will need a considerable layer thickness to see the differences.

@Antimon Thanks for the clarification Sb. I was uncertain if the naked eye would be able to tell the difference between H2O and D2O.  Do you think this might work better if one substituted the sun with an artificial light source that emitted at a wavelength that was absorbed, Like a 660nm LED?

The problem lies in the Beer-Lambert law and the low extinction coefficient of water, not in the light source. As you can see in the spectra you posted, there is no significant absorption or characteristic bands in the visible region.

Pity one could not make use of that bump at around 950 nm, like with a long darkened tube with a InGaAs photodiode at one end and a GaAs IR LED at the other.

Why couldn't you?

In a sufficiently long cell (sufficient meaning several meters), you can see the overtone of the O-H stretch.  Light water looks visible blue.  Lookie here: http://www.dartmouth.edu/~etrnsfer/water.htm

@AbelFriedman Link is dead now...

Where did you find the absorbtion data?  I'd like to try this with a spectrometer, which can numerically measure the 950nm response...

Mass. Heavy water is noticeably heavier then light water. Simply weigh an equal amount of each water and see which weights more.
Light water weights 1 g/mL (0.99700 g/mL at 25 °C)
Heavy water weights 1.107 g/mL at 25 °C
So if you have 100 mL you will see an 11 g difference, assuming pure samples.
(The Science and Technology museum in Ottawa actually has samples you can lift to feel this).

Old question, but in the interim Cody's lab has released this video (among others) on tasting heavy water. His conclusion was that $\ce{^2H_2}\ce{^{16}O}$ tasted sweeter than $\ce{^1H_2}\ce{^{16}O}$ but $\ce{^1H_2}\ce{^{18}O}$ didn't. He wasn't able to tell the difference after eating. Of course, chemists absolutely freak out at the suggestion that one attempt to identify chemicals by tasting them, normally an extremely dangerous practice.

(Completely untested since, sadly, I have no access to heavy water.) Slowly pour the contents of one glass into the other. The index of refraction of heavy water is 1.328, that of light water is 1.333. Not much, but maybe just enough to see the inhomogeneities. If the decanted glass's contents sink, it had the heavy water. Otherwise, light water.
But if you have even a cheap kitchen scale, that would probably be the better test, since the densities differ by about 10%, as opposed to the index of refraction, which differs by far less.
Or, just add a drop of food colouring to one of the glasses. Or a drop of milk. Just to help out with the contrast.
Do the test with just a small sample from each glass if you want to avoid contaminating the expensive heavy water.

What would you be looking for?

The simplest way of checking the nature of your water is related to the difference in density. If you dissolve $80$ g NaCl in $1$ liter ordinary water $H_2O$, you obtain a solution with a density $1.056$ g/mL. You may add a little bit of an ink to make the demonstration more visual. Now you pick up maybe 1 mL of this NaCl solution with a pipette, and let fall one drop in the water sample. One drop of this solution will go to the bottom of pure water $H_2O$. But it will float on $D_2O$ as it has a density $1.107$ g/mL.

No, it won't float -- it will get mixed in the other glass either way.

If you have two glasses—knowing that one contains deuterated water, the other contains potable water; one might determine the contents by freezing both samples in a freezer at $\pu{0 ^\circ C}$. Since the melting point of the potable water is $\pu{0 ^\circ C}$ and the melting point of the deuterated water, $\ce{D2O}$, is $\pu{3.8 ^\circ C}$, the $\ce{D2O}$ will take longer to melt assuming they are side-by-side.
I would add, having done this,  practically speaking: this means you can hold the glass of frozen $\ce{D2O}$ in one hand and $\ce{H2O}$ in the other and while the regular water melts, the frozen $\ce{D2O}$ takes noticeably longer to melt.  And on a side note: imagine some of the toxicity in Eukaryotes is due to the subtle differences in thermal properties between these two water species.
Be aware: $\ce{D2O}$ must be in an enclosed container, otherwise it will exchange heavy hydrogen in the water for atmospheric light hydrogen in atmospheric moisture. If left uncovered, the experiment may work a couple times at first; however, it will become less effective (as the deuterium exchanges and evaporates).

Why was this downvoted? If you're doing so can you atleast give a reason in comments, so that other people can understand why this answer was not helpful

probably because it's yet-another abstract principles-answer, largely repeating what's already been answered, and still not addressing the question of a field test for two glasses of water, one of which contains more D2O than the other.  i.e. without a 'pure specimen body of D2O in vessel #2'

Probably because you can just put the two into a refrigerator, at a temperature where one would melt and the other one not -- this way you don't have to go through the trouble of covering it, observing it while it melts, etc.


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Which has one monochlorination possibility?
Which would have one monochlorination possibility?
2,2-dimethylpropane:

or 2-methylpropane:

It seems I'm missing something because I thought both would only have one monochlorination possibility. (unless I am allowed to substitute the additional H on the 2-methylpropane with a Cl?)

Replacing Hydrogen is the only way how chlorination (or halogenation) works.
So in 2,2-dimethylpropane the central carbon (carbon-number 2) is quaternary, meaning all four of its valencies are satisfied by carbon atoms, which implies that there are no hydrogen atoms which can be substituted by chlorine (or any halogen), whereas all the remaining 4 carbon atoms are identical, so substitution on any one of those will result in the same compound (1-chloro-2,2-dimethylpropane).
Whereas in 2-methylpropane(commonly known as isobutane) the carbon-number 2 is only tertiary meaning only 3 of its valencies are satisfied by carbon-atoms and it has one hydrogen atom on it. So on substitution of this hydrogen atom by chlorine the product is 2-chloro-2-methylpropane and on substitution of any one of the terminal (last) carbon atoms would all result in 1-chloro-2-methylpropane. That is two different products.

“unless I am allowed to substitute the additional H on the 2-methylpropane with a Cl”
Why wouldn't you be? I think that's the key to the answer.

“unless I am allowed to substitute the additional H on the 2-methylpropane with a Cl”
I assume your confusion lies on the fact that Cl substitutes preferably the H that bonds with a primary carbon, whereas Br would substitute preferably the H that bonds with a tertiary carbon. This is of course correct and this would be the main product. However, a small amount of the second product will indeed be produced and since the question of the exercise is so clear you would pick 2,2-dimethylpropane as your answer.

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What do graphene "armchair(5,5)" and "zigzag(5,0)" refer to?
I am reading some articles about the electronic properties of graphene and while reading it I came across graphene armchair and zigzag structures, which have different bond structures.
The two different orientations (armchair and zigzag) are due to the arrangement of carbon atoms in the periphery (wiki), but I am unable to understand what armchair(5,5) and zigzag(5,0) mean.

For those who are still looking for an answer i found this website found this website  which provides a visual represantation of armchair and zigzag struture

The third picture, subpic (d) here shows you the geometric armchair(red) and zigzag(blue) structure:

It's derived from  structure description of Carbon nanotubes similar to surface reconstruction/overlayer structures:

The integers n and m denote the number of unit vectors along two directions in the honeycomb crystal lattice of graphene. If m = 0, the nanotubes are called zigzag nanotubes, and if n = m, the nanotubes are called armchair nanotubes. Otherwise, they are called chiral. The diameter of an ideal nanotube can be calculated from its (n,m) indices

I hope you have some background, otherwise here is a nice visualisation, or look the lecture linked above. It's important to understand what the unit cell of a 3D crystal or 2D surface is.

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Find mass of CO₂ and heat released per minute in a combustion reaction
I am not sure where to start with this problem. I know that $\small q = mc(T_f-T_i)$, but that seems like it would not help here. I think I need to balance the equation, but I am not sure how the rate factors into the problem. Anyway, if anyone could help me find the relevant equations/method for solving this problem, it would be greatly appreciated:
A natural gas mixture is burned in a furnace at a power-generating station at a rate of 13.0 mol per minute.

If the fuel consists of 9.3 mole $\small\ce{CH4}$, 3.1 mole $\small\ce{C2H6}$, 0.40 mol $\small\ce{C3H8}$, and 0.20 mole $\small\ce{C4H10}$, what mass of $\small\ce{CO2} (g)$ is produced per minute?
How much heat is released per minute?

I think it's safe to treat each of those reactions separately (so CH4 will reaction with O2 to form CO2 and H2O without being affected by C2H6, which will itself react with O2, etc.).  In terms of heat released, you probably have a table of enthalpies (deltaH) of formation for each of those reactions in a table or something.

You know the amounts of the four gases burned in a minute.  Just write combustion reactions for each of these and work out the heat produced by each reaction as written. That's your answer to the second part.  Adding up the carbon dioxide produced in each of the four combustion reactions is the answer to the first part.


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Fitting titration curve
To simulate the action of gastric $\ce{HCl}$, we have titrated 100 ml of a meal with 0.1 N $\ce{HCl}$. To plot the data, use (R-code)
vol = read.table("http://menne-biomed.de/uni/volph.txt",sep="\t",header=TRUE)
plot(vol$vol,vol$pH,pch=16,type="l",main="Titration of 100 ml meal with 0.1 N HCl")

I am trying to determine, for example, how much 1N $\ce{HCl}$ I need to add to get a pH of 1.5; if I understand correctly, I have to fit the Hasselbalch equation.
Being a statistics guy, the numerics is no problem, I just do not understand how to get the terms in that equation together with my data.

I don't quite understand what your problem is. Could you specify your question a bit further?

In the Hasselbalch equation, I know the pH, I want to know the pK, but how do I associate the [HA] and [A-] (or their ratio) with my partial titration curve? Or, more naively: how do I get pK from my titration curve (which remains in the acid all the time, the substance is not stable in alcaline).

Could you also give us information about the concentration of your inital sample? Did you add any acid before the titration in order to make the solution acidic?

No, that's a standard meal, chocolate drink with some Ca that acts as a weak buffer, so initial pH is 5.4 as indicated in the data. The final question is inverted: we have measured the pH of the meal in the stomach (in addition to actual volume), and want to know the amount of protons "pumped" (see drug proton-pump-inhibitor).

The best thing would be if you went back to the mass action law, considered all reactions that take place ($\ce{Ca}$ buffer!) and derived a formula for your specific problem. The Henderson-Hasselbalch equation actually does exactly that for a one-component acid/base system with a single protonation stage. Remember that the concentrations that occur in the mass-action law are those at chemical equilibrium, not the initial ones; i.e. for any component $C$, the eq. concentration is $[C] = [C]_0 - \Delta_C$. The problem at hand is then to find how the changes in concentration depend on the added volume of titer.
If you just took the Henderson-Hasselbalch equation (equating the change in concentrations with the added concentration of $\ce{HCl}$), you do not include the $\ce{Ca}$ buffer, which will probably play an role in this. (You get a $pK_a$ of around 0.7, which I guess is a bit on the strong side ...)
One more thing I just noticed: You apparently added 100 mL of titer to 100 mL of sample. That's kind of bad, because you'll also have to consider how the change of overall volume affects your concentrations (effectively the pH) during titration.

Thanks, yes, I am aware that the concentration has to be corrected, but that's the easy part, because it is standard numerics. If I understand you (and this is in agreement with what I read), it's hopeless to get anything from this titration curve, if I do not know what is going on inside. The Ca buffering is one I know of, but there may be others.

There are basically two components in your system: The Ca buffer and your sample. (The HCl dissociation can be considered complete, so you won't have to bother with that.) So you can get a $pK_a$ from your titration, if you also include the Ca buffer. However, if there is indeed more than one component partaking in the reactions, you only get "effective" values and concentrations out, valid only for the pH region you have investigated. Do you have any information about the composition of that meal?

Effective values are fine; I believe we do know the concentrations, but the required range is not covered. I think one should redo the titration using 1N HCl so that the whole required range will be covered.

Sounds like a good idea, experiment always beats extrapolation. Just remember to pay attention to the Ca concentrations, and you should be fine. (Unless I am forgetting something, which may always happen ...)


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Compounds that vary water's superheated and supercritical limits
Are there any compounds that reduce water's sensitivity to extreme heat change and/or volatility to high heat (over 800 degrees Fahrenheit), whether organic or inorganic?

Please read the tag wikis before putting tags like "computational-chemistry" on your post, this is a completely unrelated question.


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Why do single, double and triple bonds repel each other equal amounts?
I'm here to share with you something that totally confuses me, as I can't see the logic behind it, and my teacher doesn't know either.
Let's take a set of bonds that's trigonal pyramidal, with a lone pair of electrons and three bonds. The angle between the bonds is 107 degrees.
Now, the part where I get confused is if, say, we have one double and two single bonds (I know this makes the atom have 10 electrons, but just ignore that. It's not the point).
Now, in my brain, it would be logical that the double bonds, containing four electrons, would repel more than a single bond, containing just two. However, according to what I've been taught, it doesn't. What is that?

Does this answer your question? Do any of the geometry names(VSEPR) change if you use a double or triple bond instead of single bonds?

The way you have learnt to predict molecular geometries is called the VSEPR theory. As all theories, it makes assumptions and has a certain number of limitations. The issue you are asking about is one such limitation.
When predicting the geometry of molecules with VSEPR, one should treat multiple bonds as a single entity. However, the geometry obtained is just a first approximation, or “idealized” geometry. In fact, multiple bonds induce distortions from this ideal geometry, pretty much like lone pairs do (I assume you know that already. If not, well, check the above-linked Wikipedia page for “lone pair”.) Citing from this online course:

Multiple bonds contain a higher electronic-charge density than do single bonds, so multiple bonds also represent larger electron domains ("fatter balloons"). Consider the structure of phosgene, Cl2CO, which is shown below.

From the Lewis structure of phosgene, we might expect a trigonal planar geometry with 120°-bond angles. However, the double bond seems to act much like a nonbonding pair of electrons, reducing the ClCCl bond angle from 120° to 111°. In general, electron domains for multiple bonds exert a greater repulsive force on adjacent electron domains than do single bonds.

So: yes, lone pairs and multiple bonds give rise to deviations from the ideal VSEPR geometries. It's nice that you figured it out on your own!

The VSEPR theory I have been taught doesn't even state that different bond types and lone pairs all introduce the same "amount" of repulsion, but that they are indeed different, with the "spatial need" following the trend LP>triple>double>single bonds. I'm a bit confused why any form of theory should assume otherwise ...

@Antimon well, in order to get the regular patterns as a starting base, you need to say that variations in “repulsion” (or “spatial need”) are a second-order quantity. Otherwise, you cannot get the regular 3D constructions which emerge as first-order approximation of the molecular geometry.

Hmm. I don't see the problem there, since the underlying principles of bonds repelling each other and molecular geometries minimizing those repulsion still remain the same, regardless of whether you assign the same repulsion to them or not. But anyway, in the end, there is no difference between either getting it "right" from the start, or doing it in a two-step process where you apply corrections to a first approximation.

Thank you, F'x, you just cleared up a lot of stuff in my mind! :)
However, I guess I'll just have to stick to 120 degrees because it's what the syllabus teaches (and for simplicity), but it's nice to know what actually happens.

While we're at it, @F'x , you seem knowledgeable on the subject, so I'm just wondering, say you have a tetrahedral shaped molecule with only single bonds. Can the angles vary slightly from 109.5 degrees if the other atoms aren't all the same, or isn't that possible?

@DLA yes, variation of angles is definitely possible if different atom types are present


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Why do some materials turn black and others white after combustion?
Most solids objects turn black after they are burnt.
But some turn white.
Why is that ?

The objects which turn black are usually carbon based/organic compounds. When burned/combusted they follow one of the following unbalanced reactions:
\begin{align}
\ce C_x(\ce H_y\ce O_z)_\text{(solid/liquid)} + {\ce{O2}}_\text{(gas)} &\ce{->} \ce C_\text{(solid)} + \ce{H_2O}_\text{(liquid)} + {\ce{CO_2}}_\text{(gas)}\\
\ce C_x(\ce H_y\ce O_z)_\text{(solid/liquid)} + {\ce{O2}}_\text{(gas)} &\ce{->}  \ce{H2O}_\text{(liquid)} + {\ce{CO2}}_\text{(gas)}\\
\ce C_x(\ce H_y\ce O_z)_\text{(solid/liquid)} + {\ce{O2}}_\text{(gas)} &\ce{->}  \ce {H2O}_\text{(liquid)} + {\ce{CO2}}_\text{(gas)} + \ce{CO}_\text{(gas)}
\end{align}
You'll notice that some of the reactions will produce solid carbon, which is the black thing you see.
On the other hand, inorganic compounds begin to decompose into their ions and/or release their  "trapped" water molecules. As you'll notice, most ionic compounds are white solid salts.

Thanks for the info. How do you know (theoretically) how to do the states of the matter in the 'computation' ?

@mick Usually you need to check for intermolecular bonds, molecule size, etc.

And if you pay close attention, you'll notice that some materials turn black when partially burned and then white (well, gray) when fully used up (e.g. wood to charcoal to ash).


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What chemical compounds were used in CRT phosphors?
In the old days computer monitors and televisions were built from cathode ray tubes (CRTs) where the image was generated by an electron beam hitting phosphors which emitted light approximating the three primary colours.
What were the phosphors made from and what were the key reasons why the particular substances were chosen?

"In the old days"? Like 5-10 years ago? I still own the CRT television I bought in 2006.

@BenNorris It certainly feels like the old days even to me and I remember when digital watches used LEDs and you had to press a button to see the time. And portable computers had built in CRTs but took two people to carry them.

You should ask yourself: "For which application?". In fact there are different types of "phosphors" (most of them have no "Phosphorus" inside) for different applications and different compound were used during the evolution of CRT monitor. Some kind of phosphors may have advantages in terms of:

Efficiency of the luminescence induced.
Cheaper price.
Longer life.
Better color.
Shorter decay time.

Note however that in many application longer decay times are prefer because allow to the user to see transient signal one of this application is radar in this case is used the GB phosphors (it usually used the WTDS symbol to refer to a specific compound. In Phosphor Handbook
 by Shigeo Shionoya,William M. Yen,Hajime Yamamoto p. 607 you can find the relative compound in this case is  $(Zn,Cd)S:Cu + (Ca,Mg)SiO_3:Ti)$) that has a time decay between 100 ms and 1 s.
If you are  interested in RGB monitor and you want a list of the phosphors manly used.
I've copied the the following table from Practical Applications of Phosphors by William M. Yen,Shigeo Shionoya,Hajime Yamamoto p. 232 showing how phosphors compound for every RGB channel changed during time.
\begin{array}{c c c  c}\hline
&Blue  & Green  & Red \\ \hline
1               & ZnS:Ag^{2+}   &  Zn_2SiO_4:Mn^{2+}   & Zn_3(PO_4)_2:Mn^{2+}\\
2 & ZnS:Ag^{2+} &  (Zn,Cd)S:Ag    &  (Zn,Cd)S:Ag  \\
3& ZnS:Ag^{2+} & (Zn,Cd)S:Ag & YVO_4:Eu^{3+}\\
4 & ZnS:Ag^{2+} &   (Zn,Cd)S:Cu,Al & Y_2O_2S:Eu^{3+}or\space Y_2O_3:Eu^{3+} \\
Current & ZnS:Ag^{2+}       & ZnS:Au,Cu,Al  \space or \space ZnS:CU,Al    &Y_2O_2S:Eu^{3+}    \\ \hline
 \hline
\end{array}

I've read in a book that Europium is used in the red phosphors...
See more at this wikipedia page :
http://en.wikipedia.org/wiki/Phosphor
on this part of the webpage you will find info about used substances :
http://en.wikipedia.org/wiki/Phosphor#Phosphor_degradation
http://en.wikipedia.org/wiki/Phosphor#Materials

I was hoping for a little more detain inside the answer.


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Thought Experiment: Applying a high frequency high voltage to methane
I am in the process of constructing a kitchen waste digester (vegetable/fruits) from which, hopefully, over time adequate methane may be collected to burn. The digester won't be complete for several months yet - a few minutes each week or so.
Anyway, if a high frequency high-voltage field were to be applied to the methane (taking due care to avoid sparks) would the methane decompose into carbon, and hydrogen? A google search seems to indicate that methane may decompose, or form a plasma; I understand there may not be a simple/straight answer for this - but a general direction would be nice to have.

Sounds like a neat project.  I think it's definitely on-topic for us here.

To decompose methane molecule you have to apply energy about several electron-volts directly to molecule. Since microwave radiation wavelength measured in millimeters, to really decompose methane you'll need ridiculous voltage, much more them mere kilovolts needed to ignite electrical discharge.
And btw, electrical discharge is the easiest way to supply the energy needed for molecule to break.

What if instead of increasing the potential we were to pump up the frequency so it is in the RF range? I seem to recall that Energy is directly proportional to the frequency - albeit that last may be out of context ...

@Everyone this is point of photochemistry. They usually use ultraviolet range of the electromagnetic spectrum.


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Is alkene reaction with hydrogen to form an alkane an addition reaction?
I have seen in some textbooks that such a reaction is listed under reduction of alkenes.
I understand that the double bond is getting reduced to single bond and one hydrogen atom is added to each of the carbons.
Should I classify this reaction as an addition reaction or a reduction reaction?

It is both. $%$

Yes, the hydrogenation reaction of alkenes is typically classified as an addition reaction. See here and there, for example.

OK the double bond gets reduced through addition mechanism. Its the underlying mechanism that dictates the classification. I am new here (<15 rep) can't up vote you. But thanks. I understood the idea.

Yes it is an addition reaction... No doubt! And as there are more hydrogen now after addition so according to classical definition alkene got reduced!

This doesn't really add much to what was already in the question and what was already in the other answer.  Can you elaborate a bit?

Addition reaction means simply addition of any species without substituting whatever was there in the molecule before addition...the addition reaction here is across a double bond... the nature of the bond is changed after adding H2 but there was no substitution of any species...


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Determining enthalpy and internal energy of formation for potassium superoxide

Given $2.14\ \mathrm g$ of $\ce{K(s)}$, determine $\Delta H^\circ_\mathrm{f,m}$ and $\Delta U^\circ_\mathrm{f,m}$ for $\ce{K2O}$.
We know:

The calorimeter's constant: $1849\ \mathrm{J\cdot K^{-1}}$
The mass of water inside it: $1450\ \mathrm g$
The change in temperature: $2.62\ \mathrm K$
The end product is $\ce{K2O}$

The process should be: determining mol of $\ce{K(s)}$, which is $0.054\ \mathrm{mol}$. Then we obtain the amount of energy absorbed by the calorimeter/water. Here are the issues. I don't know which one changed its temperature by $2.62\ \mathrm K$ or if the calorimeter's constant already considers the water. Either way, the amount of energy released by each mol of potassium is extravagant; according to the result it is $322\ \mathrm{kJ}$ or $96.8\ \mathrm{kJ}$.
Continuing with this reasoning, the $\Delta U^\circ_\mathrm{f,m}$ should be $755\ \mathrm{kJ}$ or $193\ \mathrm{kJ}$, right? What do we need to obtain $\Delta H^\circ_\mathrm{f,m}$?

Now, when we say that the temperature rose by $\pu{2.62K}$, this is the temperature of the calorimeter and the water together. We are assuming that they have been given time to reach thermal equilibrium (and generally water and copper calorimeters take relatively less time to reach equilibrium, it's pretty safe to neglect the time). Calorimeters can be filled with varied amounts of water, so the calorimeter's constant is for just the calorimeter, not the water in it.
Note that if we were to assume that the specific heat of potassium/oxide were not negligible, we would have to know the specific heats of the two, as well as the temperature at the start of the reaction and the temperature at which they want $\Delta H^o$ to be calculated at. "Standard state" doesn't fix temperature, though when you say $\Delta H^o$, you generally mean $\Delta H^o_{\pu{298K}}$. Using these, you can form a reaction pathand calculate the answer (I can elaborate on this if you want, it's pretty interesting)
With these assumptions, one gets the heat released to the order of 10 kJ, and the energy is to the order of 100 kJ/mol. Not too extravagant, remember that a joule is a rather tiny unit of energy. And formation enthalpies/energies are generally to the order of 100 or 1000 kJ.
Finding the enthalpy change now is pretty easy. Since $\Delta H=\Delta U+ \Delta(PV)=\Delta U+\Delta(nRT)$. Using this and looking at the overall reaction, it shouldn't be too hard :)


text:
Is there a household method to decolorize potentially contaminated water?
I am living in my dorms at the moment and apparently it is an accepted thing that the water in the building is brown and yellow. According to my RA, this is a characteristic of the town I live in.
The only "clean" water we get is a water filter at the zeroth floor of our building where transparent water comes out.
Other than that the water that comes out of the tap, the kitchen, the shower, and the toilet bowl is light yellowish-brown water. I haven't done enough research as the cause of this problem.
Boiling water doesn't seem to change the colour either (is this even possible?). I for one cannot live like this, so I am asking the community here if they know a household method to remove the discoloration from the water.

If you write to/email your local water company, they might be able/willing to tell you exactly what the local significant impurities are, which would help you devise a strategy or reassure you.

If you want to try be a scientist, do some research on which compounds are present in the ground near you. Take copper for example. If you let a sample of water with cupric compounds it will leave a cyan smudge. Tell us what you find out.

Not enough info. I personally recommend to ask you water company - they should keep track on what impurities are present in the water. After that, post their comments and we shall see, what can be done. I personally remember one lake I camped nearby in youth with yellow water which resulted from dominant bog water feeding of the lake. This yellowish color was very persistent.

Alternatively, if you have access to a chemistry lab, figuring out what's in your tapwater would make a nice exercise in qualitative inorganic analysis; certainly more fun than the random blue-green goop we had to analyze in first-year chem. Personally, my money would be on suspended iron(III) oxide (i.e. rust) from old iron pipes as the main colorant.

A simple test you can do is to boil a large amount of the water and see if a sediment is left in the vessel. This will tell you certain things about you contaminant.

Is it particulate? We expect so because a filter is removing the contaminant.
Is it's boiling point below that of water?

If you get a residue what is it's state? I'm expecting a solid, I dont anticipate a coloured liquid that is miscible in water with a boiling point higher than water.

Does it burn? How does it burn? Colour of flame? Further residue?

I'd also try adding a colourless surfactant to the water, does it remain coloured?

I believe, there are two possible options, that are not immediately harmful for a drinker

fine rust particles in water. I suspect it is not your case, but if it is, fine mechanical filter should do the job. To check, add rust-removing chemicals, such as oxalic or citric acid, to warm colored water. If the coloration will vanish, you probably can remove the rust with fine mechanical filter (beware, I have absolutely no idea how fine the filter may be in your case). The reason is bad condition of water pipe system and should be dealt with accordingly.
organic compounds, most likely tannin contamination, is very possible if your water source is swamp-fed. The contamination may be very durable, and, honestly, even being natural for local water, still may be harmful. To check, find bleaching agent based on any hypochlorite, and see if it will destroy/change the coloration. If it will, most likely the source is organic and may be slightly harmful if used for drinking for a long time.

There should be also some government subdivision, concerned with water quality regulation. I believe, you will find more information if consult with them.

My guess would be that the yellow-brown color in your tapwater is mainly caused by small suspended particles of rust (i.e. hydrated iron(III) oxides) from old iron pipes.
Even in places with well-maintained plumbing and good drinking water, such rust contamination can often appear temporarily after maintenance work or some other interruption in the water supply dislodges built-up rust deposits from the pipes.  If you're getting rusty water all the time, it suggests either that the water distribution network in your building / town is in pretty poor shape, and/or that the raw water supply used to produce your tapwater is already quite rich in dissolved iron, which will precipitate out into rust when the water is aerated and chlorinated at the water processing plant.
Alas, I can't think of any simple "household chemistry" method to get rid of the rust suspension in a way that would be practical for drinking water.

In theory, if you simply filled a large plastic container with the rusty water and let it sit long enough, the rust should settle to the bottom.  However, if the particles are very fine (and they likely are), that may take a long time.  You could speed the process up with a centrifuge, but that takes it way out of the "easy and practical" range.
If you just wanted to get rid of the visible coloration, you could add a chelating agent like citric acid, polyphosphate or EDTA to dissolve the rust.  However, this will leave you with a solution of e.g. ferric citrate, which, while indeed drinkable in principle, may not quite have the neutral flavor you'd expect of clean water.  (This could still be worth a try, if only to check if the color is indeed caused by rust; citric acid can be found in many stores, both as a food additive and as "biodegradable rust / scale remover".  Do make sure it's pure, if you intend to drink the result, and don't overuse it.)
The most practical solution may simply be to get your own water filter.  You'll want to get one that's suitable for removing rust particles, or whatever it is that's actually coloring your water.  If this is indeed a common problem in your town, a local home improvement store can likely recommend a suitable filter system for you.


text:
What is the difference between saturated and unsaturated fatty acids?
Chemically speaking, what are the biggest differences between saturated fatty acids and unsaturated fatty acids? What are the effects of those differences on the geometry of the molecule? And what is the effect of that on our body?

Saturated fatty acids are fatty acids with no double bonds and unsaturated fatty acids are fatty acid with one or more double bonds. This is the biggest difference between a saturated and unsaturated fatty acid.

There is two types of structural configuration for unsaturated fatty acids due to the presence of double bond; cis (kink/bend in fatty acid struture) and trans(No kink/bend in fatty acid struture).
.

cis fatty acid (wikipedia)
A cis configuration means that adjacent hydrogen atoms are on the same side of the double bond. The rigidity of the double bond
  freezes its conformation and, in the case of the cis isomer, causes
  the chain to bend and restricts the conformational freedom of the
  fatty acid. The more double bonds the chain has in the cis
  configuration, the less flexibility it has. When a chain has many cis
  bonds, it becomes quite curved in its most accessible conformations.
  For example, oleic acid, with one double bond, has a "kink" in it,
  whereas linoleic acid, with two double bonds, has a more pronounced
  bend. Alpha-linolenic acid, with three double bonds, favors a hooked
  shape. The effect of this is that, in restricted environments, such as
  when fatty acids are part of a phospholipid in a lipid bilayer, or
  triglycerides in lipid droplets, cis bonds limit the ability of fatty
  acids to be closely packed, and therefore could affect the melting
  temperature of the membrane or of the fat.

Iodine value test is used to determine degree of unsaturation in a unsaturated fatty acids; that is to measure number of double bonds present in a unsaturated fatty acids.
This link will explain the health benifits of saturated and unsaturated fatty acids


text:
What is the product formed in the reaction of dimedone and benzaldehyde?
I'm having a hard time deducing the structure of the product formed when dimedone and benzaldehyde react in the presence of catalytic piperidine. I think that a Knoevenagel reaction will get me to this product:

but the answer says that the product has a molar mass of 368 g/mol. The product above is only 246 g/mol. What else can have occurred?

You did an aldol-type reaction. The Knoevenagel condensation goes further. What does condensation mean? There is no mention of stoichiometry. Finally, think Michael reaction.

This  image  should  help  you:

The conditions in this reaction are clearly not the same as those stipulated in the question. What makes you think that the product will be the same? It probably is, but this line of reasoning is highly unclear.


text:
What is the opposite of a surfactant?
Surfactants decrease surface tension between two fluids.  What do you call a substance that increases the surface tension between two fluids?  How is this effect acheived?

Briefly, the cause for surface tension is the difference in the energetic state between a molecule at the surface compared to a molecule inside the liquid phase: the molecule inside has lower energetic state as it has more inermolecular interactions (or, to get to the surface, a molecule needs to break intermolecular bonds, which needs energy). This is more pronounced with higher polarity or when ions (charges) are around as the interactions are then stronger.
You can increase the polarity of a (polar) solvent to get higher surface tension, by dissolving salts.

The traditional specific term for a substance that, when added to a fluid, increases its surface tension is tensoionic. It is derived from the fact that it was first observed in solutions of ionic salts on water.
However, the recent trend is to consider all solutes which have an effect on surface tension as surfactants, i.e. surface agents, whether their contribution to surface tension is positive or negative. This usage can however be ambiguous.
The effects driving this phenomenon are complex and thermodynamic in nature: at low concentrations it seems to be driven mostly by the Gibbs isotherm behaviour of mixtures, but at higher concentration, as solvation takes over an important fraction of the solvent, an entropic effect dominates, and all strong electrolytes tends to the same, entropic-driven asymptotic behaviour.
For a discussion on the thermodynamics involved see Levin, Y. Interfacial tension of electrolyte solutions. J. Chem. Phys. 2000, 113, 9722-9726


text:
Is there a standard symbol for molar density?
Is there a standard symbol for molar density, i.e. $n/V$, where $n$ is the total number of moles of every species in a system?
I realise that there is standard term "molar volume", which is the reciprocal of what I'm asking for, but my equations are going to look pretty silly with $1/V_m$ written everywhere. As a simple example, I want to say that if my system consists of a mixture of $\ce{A}$ and $\ce{B}$ with the reaction $\ce{A <-> B}$, the total concentration of $\ce{A}$ and $\ce{B}$ will be constant and we can write
$$
\ce{[B]} = c - \ce{[A]},
$$
where $c$ is the thing I'm looking for the symbol for.
I didn't find "molar density" in the IUPAC green book, but I don't know my way around it very well, and if it has a different name it might be hiding somewhere I didn't think to look.

To my knowledge, n/V is known as molar concentration and uses symbol $C$ and it seems okay for me to use $C_t$ i.e. total concentration to mark sum of concentrations of all particles in system.

To my mind this is the best answer.  If the OP agrees that "concentration" or at least "molar concentration" is the correct term, I'd recommend accepting this one.  If not, I wonder if they would mind explaining a bit more why "concentration" is not the word they are looking for.

I don't know of a standard symbol for molar density, but rho ($\rho$) is a standard symbol for density in general, so if there's no better symbol, I'd go for something like $\rho_m$. As long as you explicitly define it somewhere, it should be okay.

I agree with Aesin, but I have another comment to make.  The term "molar density" has a confused meaning.  That is because the density of one mole of a substance is identical to the density of two moles of the substance, or any number of moles of the substance.
The term "molar" is most often used to make an extensive quantity into an intensive one.
An extensive quantity is one that doubles if you take two identical copies of a system and put them together.  Volume is extensive, temperature is not, it is intensive.  Most importantly, density is already intensive, so that "molar density" is essentially devoid of any meaning other than that the question states, the inverse of the molar volume.

That's a fair point. I intended "molar density" to be read as "particle density expressed in (pseudo)units of moles", but I can see that it could just as easily be read as "density per mole", which would make no sense as you say.

In enzyme kinetics, there is a similar situation, where the symbol $E_t$ is used to indicate the "total enzyme concentration".
In other words, there can be free enzyme with concentration [E], substrate bound enzyme with concentration [ES], and inhibitor bound enzyme [EI], but for the total concentration of all forms of the enzyme the symbol $E_t$ with no brackets is often used.
See for example page 354 of Essentials of Chemical Reaction Engineering by H. Scott Fogler where he has a note about using that symbol.


text:
Is an explosion merely an exothermic reaction?
An exothermic reaction is apparently one that releases energy to it's environment. According to Wikipedia, an explosion is a rapid increase in volume & energy.
Does this mean that an explosion is basically an exothermic reaction?

The distinguishing feature of an explosion is how fast it happens. There are plenty of slow exothermic reactions, like digestion of food.

The key parts to the definition of an explosion are "rapid" and "increase in volume".
Exothermic reactions do not necessarily meet these requirements. Exothermic reactions need only release energy to the surroundings by converting chemical potential energy into kinetic energy.
For example, the combustion of methane is very exothermic, with a standard enthalpy change of -890 kJ/mol.
$\ce{CH4(g) + 2O2(g)->CO2(g) +2H2O(g)} \space  \space \space \space \Delta_c H^o=-890 \text{ kJ/mol}$
However, at room temperature this reaction is very slow. Mixtures of oxygen and methane do not spontaneously ignite without a spark source. Also, there is no change in pressure with this reaction, since there are 3 moles of gas on both sides of the equation.
On the other hand, consider nitroglycerin. Yes, it's combustion is significantly exothermic. More importantly, the combustion happens rapidly, needs only the heat of friction or the energy of impact to ignite, and produces significantly more gas than it consumes:
$\ce{4C3H5O9}(l) \ce{+1O2(g)->12CO2(g) +10H2O(g) +6N2(g)} \space  \space \space \space \Delta_c H^o=-5660 \text{ kJ/mol}$
All thermodynamic data were mined from the NIST Chemistry Webbook.

May we then say an explosion is a special case of an exothermic reaction?

In some rare cases, substances may be explosive without evolving gas. An example is silver acetylide (here's a demo of its detonation). Furthermore, some substances are still explosive though their decomposition is only slightly exothermic, or conceivably even slightly endothermic, so long as the resulting reaction results in a large enough entropic gain. These are termed entropic explosions.


text:
What are the key chemical characteristics that determine the octane rating of fuel?
The octane rating of petrol (gasoline) is a measure of the how much compression a fuel air mixture can have before detonating (which in petrol engines is a bad thing as you want the mix to combust only when the spark plug arcs).
What are the key chemical characteristics that determine this rating? What is the molecular-level explanation? And how do the various non-hydrocarbon additives work to modify the result?

Combustion
 is (usually) chain radical reaction with heavy branching. More
branching means faster combustion and less branching is slower
combustion. And detonation is wave of super-fast combustion.
So, if you can suppress branching or deaden some branches, you can
increase durability of air-fuel mixture. Actually, this means, that some
 radical trap is needed. The oldest trap was tetraethyllead, but it is
currently dropped (lead oxides as fine dust is not a health friendly
substance). Non-saturated hydrocarbons of different types also can trap
active radicals, transforming them into more stable allyl radicals, so
fuel companies tries to keep as much of them in fuel as possible. Some
other compounds, like ferrocene, also can be used to increase octane
rating.

What do you mean by branching and suppressing branching?

@BenNorris: When people talk about radical-based reactions, often three types of step are referred to: initiation, a reaction which makes a radical; propagation, a reaction which preserves the number of radical species, and termination, a reaction which consumes a radical. But, if you have a reaction which produces more radicals than it consumes, you get 'branching', which accelerates the rate of the overall reaction since now more propagation and termination steps can happen.

Ah, okay. Since the question was about octane rating, I, as an organic chemist, was trying to relate this answer to branched molecules, for example isooctane (defined octane rating of 100) is a more branched structure than heptane (which is defined as octane rating of 0).

@BenNorris Branched hydrocarbons usually have higher octane rating as they can product tertiary radical: most stable of alkyl radicals.


text:
How can scientists be certain that there are no undiscovered elements missing from the periodic table from hydrogen to lawrencium?
Scientists are certain that there are no undiscovered elements missing from the periodic table from hydrogen to lawrencium. How? and why?

You may find it of interest that there are atoms which don't belong in the periodic table. They're called Exotic atoms, and one rather crazy type is an -onium. There may be some unsynthesized or even undiscovered(this corresponds to an undiscovered particle) ones here...

@Manishearth: Interesting stuff!  I think from a chemist's standpoint, exotic atoms would be viewed like isotopes and ions; i.e. proton count would still define "element".

trb456: Exotic elements often don't have protons (For example, there is one that is 2 electrons and 2 positrons orbiting one another). Or an 'isotope 0' of hydrogen that is an electron and a muon.

@Canageek: Again interesting!  But can we really consider these "elements" then, or just unusual structures of subatomic particles?  For example, a chemist would want to consider chemical bonds.  Can we get bonds with these structures?  Presumably if they could share electrons (or positrons), then perhaps you would have to come up with a way to make them fit the current system (i.e. your H0 isotope).

@trb456 You can get H-X with the mueon one. However, the scientist presenting it considered it to be an isotope of hydrogen, just the '0 mass' element. Its reactivity for the microsecond it exists is apparently quite similar. I wish I could remember the name of the scientist doing that work; he is at TRIUMF in BC.

@Canageek: Thanks for the reply. It sounds like the duration of these structures is like very heavy synthetic elements.  If any one of these exotic atoms could form stable compounds, I think chemists would be very interested in thinking about them as "real" elements.

@trb456 Yeah, I was very disappointed when I found that out; it does live long enough to get a reaction time measurement though.

In order to seperate the element from the the atomic table #92 the man made ele
ment and the atomic table #92 the both ellements have to have the same distinctive varation order to be seperated from oneanother. P.s. In futher conclusion both elements have have the same number of atomic nucleos.

@Canageek I know David Walker (U British Columbia) is working in this area (muonium).

@EricBrown That could well be him then.

You are talking about the first 103 elements.  An element is by definition identified with the number of protons in its nucleus, the so-called atomic number.  Any atom with 6 protons is carbon, it cannot be anything else.  So there cannot be any missing elements out to 103, as all of these elements have been confirm discovered by accepted scientific methods.
For example, here is a link to a CHEM Study film showing how transuranium elements (i.e. atomic number > 92 and man-made) can be separated from one another and verified as distinct.
http://archive.org/details/transuranium_elements


text:
Maximum temperature attainable by thermite reaction

In the thermite reaction
  $$\ce{2Al(s) + Fe2O3(s) -> 2Fe(l) + Al2O3(s)}$$
  what is the maximum temperature attainable by the products? Assume reactants at $25\ \mathrm{^\circ C}$ and $1\ \mathrm{atm}$ and that all heat capacities remain constant.

I started by using Hess's law to find the enthalpy change of the reaction. I couldn't find a heat of formation for liquid iron so I assume I have to find that out. I just wrote out the equation
$$\ce{Fe(s) + heat + enthalpy of fusion -> Fe(s)}$$
and figured the heat of formation for liquid iron at $25\ \mathrm{^\circ C}$ is around $-302.71\ \mathrm{kJ/mol}$.
To find the max temperature I used
$$\Delta H_\mathrm r=\Delta T\cdot C_\mathrm p(\text{products})$$
and came out with an equation like
$$-14445\ \mathrm{kJ/mol}=0.147\ \mathrm{kJ/(mol\ K)}\cdot T_\text{max}$$
This gives me a negative temperature, so I assume it is wrong. Even if I were to take the absolute value, the answer seems awfully high. I was expecting something around a few thousand degrees not near ten thousand.

Thermodynamics will never give a complete picture here: the actual heat will be dominated by the kinetics of the reaction and the kinetics of heat transfer to the environment. Thermodynamic calculations will give an absolute upper limit at best.

According to Wikipedia the maximum temperature is approximately 2500 °C, limited by the boiling point of aluminum, which is 2519 °C.  The article provides some discussion of how the temperature is limited, and how to calculate it using Hesse's law.

This gives me a negative temperature, so I assume it is wrong.

Remember that the energy given off as an exothermic reaction progresses is absorbed by the surroundings. The chemical potential energy stored in aluminum and rust is converted into kinetic energy. You are interested in $\Delta H_{surr}$, which should be $-\Delta H_{sys}$. The "system" in this case is the electrons of the materials engaged in various bonding patterns, and the "surroundings" are the bulk materials.

I was expecting something around a few thousand degrees not near ten thousand.

I would agree. Both aluminum oxide and iron melt and then boil below 10000 oC. Are you sure you are determining the $\Delta _f H^o$ of $\ce{Fe}(l)$ correctly?  Melting is an endothermic process (absorbs heat from the surroundings).

well for one I used a ridiculous number for the melting point of iron. That combined with fixing what is negative and what is not in each circumstance I am getting around 5000C which is more reasonable but still seems high. I could have just made another simple error

I am coming out with an equation like -14445 kJ/mol = 0.147 kJ/(mol K) * maxT

It is wrong. $C_p$ is temperature dependent, so you have to use a more sophisticated equation with integrals. Moreover, parts for heating liquid and solid phases will be different.

This question is assuming Cp is not temperature dependent in the given temperature ranges

Well... It is wrong. And again, one have to count at least the fact, that the reaction gives partially melted products, that also reduces overall effect.

There are many different types of thermite compositions although the cheapest form is aluminium based, while you could have magnesium, titanium, zinc, silicon and boron as the fuels source. The oxidzers can also vary from bismuth (III) oxide, boron (III) oxide, silicon (IV) oxide, Chromium (III) oxide, manganese (IV) oxide, iron (III) oxide,iron(I,III) oxide, copper (II) oxide or lead (II, IV) oxide.
Most thermite compositions burn at around 4,000 degrees, however once started the reaction can't be stopped until all the fuel and oxidizers have been spent.


text:
Which theta should be taken from x-ray diffraction pattern for calculating Scherrer's equation?
We use Scherrer's equation to calculate crystalline size of a particle:
$$d = \frac{Bλ}{β\cos{θ}}$$
But which theta should we take from the XRD pattern for this calculation?
For example, the following plot shows the diffraction pattern of nano (above) and bulk (below) silicon and it has two diffraction peaks at $2θ \approx 69^\circ$ and $2θ \approx 76^\circ$. Which theta should we take for calculating crystalline size of nano silicon and why?

for the polycrystalline materials the instrumental error can be eliminated by subtracting the peak width of polycrystalline materials by single crystals peak broadening in same machine.

You should take the theta for the peak for which you measured the peak width. For an ideal case, the peak widths should be Bragg angle (i.e. resolution)-dependent. Or you could refine beta for the entire data set (Michael Green's answer) keeping in mind that this might be overkill given the approximations and caveats (answer by F'x).

Scherrer’s formula is a simple equation for the simple case of a single peak broadened only due to crystalline particle size. Application of the formula to other, more complex cases, gives only an estimate. Quoting Wikipedia:

It is important to realize that the Scherrer formula provides a lower bound on the particle size. The reason for this is that a variety of factors can contribute to the width of a diffraction peak; besides crystallite size, the most important of these are usually inhomogeneous strain and instrumental effects. If all of these other contributions to the peak width were zero, then the peak width would be determined solely by the crystallite size and the Scherrer formula would apply. If the other contributions to the width are non-zero, then the crystallite size can be larger than that predicted by the Scherrer formula, with the "extra" peak width coming from the other factors.

Scherrer formula gives an estimate, an order of magnitude. Here, calculate the particle sizes for the two peaks, and they should be close. The differences are covered by the large uncertainty of your resulting value.

You can use a simple mathematical manipulation to adjust the scherrer equation to take data from an XRD plot directly...
$$d = \frac{B \lambda}{\beta \cdot cos( \frac{2\theta}{2}))}$$
...where $2\theta$ is the bragg angle.
For the lower XRD pattern set, you'll need to deconvolute one of the peaks. If you know how to use python, the easiest way this can be done is that you can use a least-squares regression analysis to fit a set of pseudo-vogit functions to the XRD data and then use one of the functions to determine the FWHM and $2\theta$ value, where the pseudo-vogit is...
$$PV(x,A,\mu,\sigma,\alpha) = \frac{(1-\alpha)A}{\frac{\sigma}{2\cdot ln(2)}\cdot\sqrt{2\pi}}\cdot e^{\frac{-1}{2}\cdot \frac{(x-\mu)^2}{(\frac{\sigma}{2 \cdot ln(2)})^2}}+\frac{\alpha A}{\pi}\cdot[\frac{\sigma}{(x-\mu)^2 + \sigma^2}]$$
...and the minimizing condition is...
$$PV_{sum} = \sum_{i=1}^{n_{peaks}} PV_i(x_i,A_i,\mu_i,\sigma_i,\alpha_i) $$
...finding the condition which...
$$(data_i - PV_{sum}(data_i))^2 \rightarrow 0 $$
which is something a computer can do pretty easily. This'll output the set of parameters which fit $PV_{sum}$ to the entire data set - just use one of the PV conditions to calculate FWHM and $2\theta$ to determine the particle size.

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


text:
Is the half-full rule and full rule followed in the 6th and 7th periods?
Is the half-full rule and full rule followed in the 6th and 7th periods? (Note: Half-full rules is Hund's rule)
Example: What is the correct electron configuration?
$$\ce{W = [Xe] 6s^2 4f^{14} 5d^4}$$ doesn't use half full rule, or
$$\ce{W = [Xe] 6s^1 4f^{14} 5d^5}$$ uses half full rule.

Just a note: here at my place, it's more common to call it "fully-filled" instead of just "full".

Short answer: no! The first known exception to occur in the periodic table is in period 5:  niobium’s ground state electronic configuration is

$\ce{Nb: [Kr] 5s^1 4d^4}$

which is not warranted by the “usual” rules for determining electronic configuration. So, the “half-full rule” is not sufficient in periods 5 and higher.

Let’s look at Wikipedia’s list of exceptions to Madelung’s rule. In period 4, the only exceptions are $\ce{Cr}$ and $\ce{Cu}$, which are accounted for by the “half-full rule”. (Don’t bother too much about the dispute for Ni.)
In period 5, the exceptions to Madelung’s rule are classified in three groups:

$\ce{Mo}$ and $\ce{Ag}$: they are the analogues of Cu and Cr
$\ce{Pd}$ is $\ce{5s^0 4d^10}$: a different type of application of the explanation that “fully filled shells are particularly stable”, because in this case two electrons from the s shell were pulled into the d shell. Not what you usually learn.
$\ce{Nb}$ is $\ce{5s^1 4d^4}$,  and $\ce{Ru}$ is $\ce{5s^1 4d^7}$: these cannot be explained by simple rules, and one has to perform complex quantum chemistry calculations to understand these electronic configurations.

Period 6 contains further inconsistencies, because of the introduction of $\ce{f}$ orbitals into the mix.

F'x, I'm sure this was just a slight oversight, but another important exception in period 5 is rhodium ($\ce{5s^1 4d^8}$), as given on that Wikipedia page. Also, a very important supplementary read: How can one explain niobium’s weird electronic configuration?


text:
Which conformer of beta D-glucose is more stable?
I drew the two chair conformers of  $ \beta $-D-glucose:

Which one is more stable?
I think it's the one with all of the OH groups in the equatorial position because of less steric hindrance, but I also know that the anomeric effect makes the axial one favored as well.  So I'm unsure which is more stable.

I also know that the anomeric effect makes the axial one favored as well.

The anomeric effect does stabilize an axial group, but only at the anomeric position. Other axial groups are not similarly stabilized. Thus $\alpha$-D-glucose is more stable than you think it should be based on steric considerations alone.
Knowing this, which is the more stable?
PS - very pretty chairs.

isn't the beta d-glucose the one with the equitorial bonds?

Both of your structures are beta-D-glucose, just different conformations. alpha-D-glucose is the anomer, and would have the OH at the anomeric position different. Otherwise, the two are the same.

right.  so i'm going to say the one on the left is more stable because the anomeric effic is still higher in energy than having all groups equitorial

And I would agree.


text:
Treatment of copper (0) with sodium hydroxide
I recently made self-assembled monolayers (SAMs) of alkanethiols on a copper substrate. After scrubbing off the $\ce{Cu2O}$ top layer and rubbing the surface with ether and ethanol, I let my copper plates stand in a 5% (w/w) $\ce{NaOH}$ solution for 5 minutes, which I then throughly rinsed off with deionized water.
I'm wondering if I could skip this step. I'm not sure what it is needed for. My knowledge of chemistry tells me that $\ce{Cu(0)}$ should not react with sodium hydroxide. I'm also positive that any grease or other contaminants were washed off in the previous steps.
What do you think?

$\ce{Cu(0)}$ will not be attacked by $\ce{NaOH}$, and will not decompose in this aqueous environment at room temperature. So, this is not to create some reaction with the copper. Two uses I can think of:

$\ce{NaOH}$ increases the solubility of $\ce{Cu2O}$, so maybe this is a way of getting rid of the last remaining copper oxide.
To bind the thiols with copper, you'd better be in alkaline conditions rather than acidic conditions, right? If so, then maybe this is the way of making sure you don't have any acidic groups or molecules remaining adsorbed on the copper surface, to improve bonding at the next step.

Your first point might be the reason why... The protocol isn't clear on why we have to do this NaOH step. What I forgot to mention is that I thoroughly rinsed with distilled water after, so pH should be around 7.


text:
How can one explain niobium’s weird electronic configuration?
As cited in an answer to this question, the ground state electronic configuration of niobium is:

$\ce{Nb: [Kr] 5s^1 4d^4}$

Why is that so? What factors stabilize this configuration, compared to the obvious $\ce{5s^2 4d^3}$ (Aufbau principle), or the otherwise possible $\ce{5s^0 4d^5}$ (half-filled shell)?

5s subshell here has lower energy, then 4d subshell. However, there is non-zero energy of electron pairing on same orbital. So, we have one electron on 5s subshell and another is pushed onto 4d subshell. ||| do not overestimate these so-known 'principles', they work in most cases, true, but can fail dramatically on border cases, like this one where energy gap between two orbitals is not so big. ||| I do not post this as answer as I cannot prove my point by math or reference, it is a guess.

@permeakra yeah, but if you follow this simple logic, there's no reason not to push two electrons into the d shell :)  Hence the reason I ask how this can explained, which will most possibly involve quantum chemistry calculations.

@F'x It is repulsion of two electrons on same orbital that pushes one of them from 5S subshell. After one is pushed away, the other have no reason to move.

@F'x This question is very close to my heart.  My favorite element and current chemical obsession, palladium, does this same thing to a further extreme.  I've never really been able to understand the explanation I've found as they involve high level quantum stuff. I would humbly suggest that the answer to this lay more with the physicists then chemists.

I was about to ask "Why are there exceptions to the Aufbau principle/Madelung rule?" but found this and I assume it would be considered a duplicate. Yet, I think this question deserves being re-titled since it's about many elements and not just Niobium.

There is an explanation to this that can be generalized, which dips a little into quantum chemistry, which is known as the idea of pairing energy. I'm sure you can look up the specifics, but basically in comparing the possible configurations of $\ce{Nb}$, we see the choice of either pairing electrons at a lower energy, or of separating them at higher energy, as seen below:
d:   ↿ ↿ ↿ _ _        ↿ ↿ ↿ ↿ _        ↿ ↿ ↿ ↿ ↿     ^
                 OR               OR               |
s:   ⥮               ↿                _            Energy gap (E)

The top row is for the d-orbitals, which are higher in energy, and the bottom row is for the s-orbital, which is lower in energy. There is a quantifiable energy gap between the two as denoted on the side (unique for every element). As you may know, electrons like to get in the configuration that is lowest in energy. At first glance, that might suggest putting as many electrons in the s-orbital (lower energy) as possible, and then filling the rest in the d-orbital. This is known as the Aufbau principle and is widely taught in chemistry classes. It's not wrong, and works most of the time, but the story doesn't end there. There is a cost to pairing the electrons in the lower orbital, two costs actually, which I will define now:
Repulsion energy: Pretty simple, the idea that e- repel, and having two of them in the same orbital will cost some energy. Normally counted as 1 C for every pair of electrons.
Exchange energy: This is a little tricky, and probably the main reason this isn't taught until later in your chemistry education. Basically (due to quantum chemistry which I won't bore you with), there is a beneficial energy associated with having pairs of like energy, like spin electrons. Basically, for every pair of electrons at the same energy level (or same orbital shell in this case) and same spin (so, if you had 2 e- in the same orbital, no dice, since they have to be opposite spin), you accrue 1 K exchange energy, which is a stabilizing energy. (This is very simplified, but really "stabilizing energy" is nothing more than negative energy. I hope your thermodynamics is in good shape!) The thing with exchange (or K) energy is that you get one for every pair, so in the case:
↿ ↿ ↿

from say a p-subshell, you would get 3 K, for each pair, while from this example:
⥮ ↿ ↿ ↿ ↿

from a $\ce{d^6}$, you would get 10 K (for each unique pair, and none for the opposite spin e-)
This K is quantifiable as well (and like the repulsion energy is unique for each atom).
Thus, the combination of these two energies when compared to the band gap determines the state of the electron configuration. Using the example we started with:
d:   ↿ ↿ ↿ _ _         ↿ ↿ ↿ ↿ _         ↿ ↿ ↿ ↿ ↿    ^
s:   ⥮           OR   ↿           OR   _            |
PE:  3K + 1C          6K + 0C          10K + 0C     Energy gap (E)

You can see from the example that shoving 1 e- up from the s to the d-subshell results in a loss of 1C (losing positive or "destabilizing" repulsive energy) and gaining 3K (gaining negative or "stabilizing" exchange energy). Therefore, if the sum of these two is greater than the energy gap (i.e. 3K - 1C > E) then the electron will indeed be found in the d shell in $\ce{Nb}$'s ground state. Which is indeed the case for $\ce{Nb}$.
Next, lets look at perhaps exciting the second s e- up to the d-subshell. We gain 4 additional K but don't lose any C, and we must again overcome the energy gap for this electron to be found in the d-subshell.
It turns out that for $\ce{Nb}$: 4K + 0C < E  (remember that C is considered a negative value, which we're not losing any of), so $\ce{Nb}$ is ultimately found in the $\ce{5s^1 4d^4}$ configuration.

Just in case it wasn't obvious for any reader: "from say a p-subshell, you would get 3 K" that's because there are $^3C_2=3$ electron pairs. Similarly, "from a $\ce{d6}$, you would get 10 K" that's because there are $^5C_2=10$ electron pairs. The author calculated this by simple combinatorics ;)

Well it turns out Atoms are a way more scientific than all best scientist put together, lol so much complexity just to be in low energy state. Great answer at @Gavin kramar , could have added the actual figures for (E) i.e the energy gap

The question of anomalous electronic configurations, meaning $\mathrm{s^1}$ or $\mathrm{s^0}$ in one case (Pd) is very badly explained in textbooks.
For example, the anomalous configuration of Cr ($\mathrm{3d^5~4s^1}$) is typically explained as being due to "half-filled subshell stability".  This is wrong for several reasons. First of all there is nothing especially stable about half-filled subshells. Secondly it does not explain the fact that many second transition series elements show anomalous configurations even though they do not possess half-filled subshells. The possession of a half-filled subshell (henceforth abbreviated as hfss) is neither necessary nor sufficient for there to be an anomalous configuration. There are atoms that have hfss but do not have anomalous configurations and there are atoms that have anomalous configurations but do not have hfss.
Conclusion is, the use of the hfss explanation is ad hoc and should be avoided. It so happens that Cr and just one other atom have both hfss AND show an anomalous configuration. But it cannot be generalized.

Unfortunately, the Aufbau rule cannot predict all electron configuration as it doesn't take into account electron-electron interactions. In the end the Aufbau is only a rule of thumb. Electronic levels have to be found using quantum calculations taking into account electron-electron interactions (not to mention spin orbit coupling). Therefore there is no simple or rational explanation for this.
References:

http://www.physicsforums.com/showthread.php?t=239512
Other exceptions to Madelung's rule paragraph

Just as in a seesaw similarly in Zr equal number of 2 electrons are present each for 6s and 5d orbitals which balances each others forces; however, in Nb as the new entrant electron is bound to enter into the vacant 5d orbital it helps in pulling 1 of the 2 electrons of 6s orbital into the 5d orbital due to sheer electronic number superiority and the following electrons similarly are successively pulled into the vacant 5d orbitals of the subsequent elements of this second series of transition elements period.
This only happens as the two 6s and 5d orbitals have almost negligible energy gap and it is not possible to prevent electron entering the 5d orbital from the 6s orbital.


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Adenine precursor synthesis
I am currently exploring the synthesis of xRNA, which contains benzo-expanded purine and pyrimidines. I have found this paper which deals with exactly this subject, and provides the following synthetic pathway for the xA ribonucleotide.

Unfortunately, the first 5 synthetic steps are omitted from the paper, and are likely to be found in one of the following three papers, none of which are publicly available.
1. 2. 3.
Nonetheless, I am interested in any synthetic pathway which will lead me from the first to the second molecule shown above. I understand the presented material, but as an undergraduate cell biologist, I am a little rusty on my synthetic organic chem.
EDIT: The first thing that came to mind was the condensation of a biscarbonyl electrophile and bisamine nucleophile like in pyrimidine synthesis.

Well, you are right, the synthesis of your second compound is described in DOI:10.1021/ja038384r:


text:
Number of hydrogen molecules necessary in acetonitrile/butanediamine polymerisation
The beginning of the picture, from the word 'Acrylonitril' to the word 'structuurformule:', means the following: Acrylotitrile reacts with 1,4-butanediamine. When you use a lot of acrylonitrile, under certain circumstances you almost exclusively get the following: Then you can see the first of the 2 structural formulas. The text between the 2 structural formulas, says This 'bond' can react with hydrogen. You'll get addition. When this addition finished, you get a bond with the following structural formula: And then you see the picture. The final blob of text: We can make polymer molecules which keep branching on and on because of reactions with hydrogen and acrylonitrile.
The final question is: We start with 1 molecule of 1,4-butanediamine. Explain how many molecules of hydrogen are needed to make a polymer molecule with 16 $\ce{NH2}$ groups, assume that all reactions are complete.
The answer should be 56, but I have no clue why, can someone explain? I have half of 56, I just did 14x2, because you get 2 $\ce{H2}$ molecules per added acrylonitrile, and you have to get 14 acrylonitrile molecules to get to 16 $\ce{NH2}$ groups, so that would equal 2x14=28.

no, there isn’t… I also think your chance of getting an answer rapidly depends on question clarity and formatting, and this question isn't very attractive. You've dumped your question as a big image with non-English text, provided piecewise translation above it making it hard to read through, and it's not much chemistry as counting atoms…

Even though I provided an answer, I agree with @F'x. If I was not already familiar with this dendrimer system, this question would have been very challenging.

This question appears to be about the preparation of polyacrylonitrile dendrimers. Each addition of acrylonitrile completely reacts away the NH2 groups previously present:

assume that all reactions are complete.

Your answer of 28 assumes that the reactions could be controllable to allow partial consumption of NH2 groups, which requires less addition of acrylonitrile and fewer equivalents of H2.
You need 16 NH2 groups, which is 8 on each side. This is the third generation dendrimer:

Generation 0 is butanediamine (2 NH2 groups). Let $G_n$ represent the interior region of the $n^{th}$ generation.
$$\ce{NH2CH2CH2CH2NH2} \equiv \ce{NH2}G_0\ce{NH2}$$
Generation 1 involves reaction of 4 acrylonitriles followed by 8 molecules of hydrogen to make 4 NH2 groups.
$$\ce{NH2}G_0\ce{NH2 +4CH2=CHCN->(NCCH2CH2)2N}G_0\ce{N(CH2CH2CN)2} $$
$$\ce{(NCCH2CH2)2N}G_0\ce{N(CH2CH2CN)2 +8H2-> (NH2CH2CH2CH2)2N}G_0\ce{N(CH2CH2CH2NH2)2} \equiv \ce{(NH2)2}G_1\ce{(NH2)2} $$
Generation 2 involves reaction of 8 acrylonitriles followed by 16 molecules of hydrogen to make 8 NH2 groups.
$$ \ce{(NH2)2}G_1\ce{(NH2)2 +8CH2=CHCN-> [(NCCH2CH2)2N]2}G_1\ce{[N(CH2CH2CN)2]2} $$
$$\ce{[(NCCH2CH2)2N]2}G_1\ce{[N(CH2CH2CN)2]2 +16H2 -> [(NH2CH2CH2CH2)2N]2}G_1\ce{[N(CH2CH2CH2NH2)2]2} \equiv \ce{(NH2)4}G_2\ce{(NH2)4}$$
Generation 3 involves reaction of 16 acrylonitriles followed by 32 molecules of hydrogen to make 8 NH2 groups.
$$ \ce{(NH2)4}G_2\ce{(NH2)4 +16CH2=CHCN-> [(NCCH2CH2)2N]4}G_2\ce{[N(CH2CH2CN)2]4} $$
$$\ce{[(NCCH2CH2)2N]4}G_2\ce{[N(CH2CH2CN)2]4 +32H2 -> [(NH2CH2CH2CH2)2N]4}G_2\ce{[N(CH2CH2CH2NH2)2]4} \equiv \ce{(NH2)8}G_3\ce{(NH2)8}$$

Total number of H2 molecules = 8 + 16 + 32 = 56 starting at $G_0$.


text:
Tests on sodium peroxoborate
When I prepare sodium perborate by mixing sodium hydroxide with metaborate ($\ce{NaBO}$) and hydrogen peroxide and cooling it I applied a test to it by acidifying it with $\ce{HCl}$ and adding $\ce{KMnO4}$. A precipitate appeared and it turned black. What is the reason?
I tested another sample of sodium perborate by acidifying it with $\ce{NH4Cl}$ and after that KI and this turned white. Again, why?

In your first test, the $\ce{KMnO4}$ probably reacted with the $\ce{H2O2}$ excess remaining after your preparation. The ionic equation for the reaction is
$$\ce{2MnO4- + 3H2O2 -> 2MnO2 + 2H2O +3O2 + 2OH-}$$
The $\ce{MnO2}$ formed is an insoluble black precipitate. Although there is an $\ce{OH-}$, the reaction apparently occurs in neutral and faintly acidic medium as well.
The result of your second test doesn't seem to be very clear. Sodium perborate apparently reacts very quickly with any water so if you added any water during the second test, it must have reacted to form oxygen gas bubbles (which appear white).

The second may be pure boric acid, that is weakly soluble

@permeakra Thanks! I was not quite sure what that one was. Do you want to make that into another answer or shall I edit it into mine?

Edit: The reaction also occurs in neutral and faintly acidic medium not only basic medium.


text:
What is deuterium-depleted water actually used for?
I noted, flipping through the paperback Alfa Aesar catalogue today, that they sell deuterium-depleted water. Under the usage category, they list 'NMR'. I'm trying to work out what specific use this water has. Unfortunately, on the basis of very limited clinical evidence of improved cancer outcomes I've been bombarded mostly with companies recommending deuterium depleted water for drinking, which trips my snake oil sensor and is something I doubt Alfa has in mind (at US$1930 per liter, and a US RDI of 3.7 L/day for an adult male, that would seem to be a comically expensive drinking habit.)
I suspect biophysics experiments are a small part of the market for this water, but that there's some other NMR-related use I can't think of, possibly calibration related. What is it?

“I've been bombarded mostly with companies recommending deuterium depleted water for drinking” — Really? I so wish everyone would see how getting their medical advice from medical experts actually makes sense…

@F'x - you know what people are like.

When you perform 1H-NMR spectroscopy in solution, there are many cases where you want to use deuterated solvent, so that signals coming from solvent hydrogen nuclei don’t interfere with the signal from your target molecule. Hence the frequent use of deuterated water, deuterated acetone, deuterated methanol, and deuterated chloroform. Aprotic solvents are also in common use, including carbon tetrachloride.
The situation is reversed when you want to perform 2D-NMR: you then need deuterium-free solvents, so their signal doesn’t mess with the molecule you're interested in. Hence, you will be interested in buying deuterium-depleted water.
(If you're wondering about real use cases for solution deuterium NMR, you can browse the literature for articles, there are plenty. One recent exemple is this one.)

There's an additional reason for deuterated solvents in NMR, you usually lock on a deuterium signal from the solvent to compensate for drift of the magnetic field. Even when you measure in plain H2O, you usually add 5-10% D2O for the lock.

@MadScientist the “you lock on a deuterium signal […] for drift of the magnetic field” sounds like a line from Doctor Who :)

@F'x It is, however, accurate.

@Canageek oh yeah, I didn't want to cast doubt about that… I just thought it would be a great line out of context, very sciency-sounding


text:
What happens during cleaning silverware?
I know two distinct recipes for cleaning silver:

Put them into a pot with some club soda and a piece of aluminium foil and pour over a hot water.
Gently clean silverware with a fine chalk powder (obtained from a pharmacy).

I'd be happy to know:

What happens chemically during these procedure - why does it work, and what are the byproducts of these reactions?
How toxic (if at all) are the byproducts?
How much silver is lost in such a procedure?
Are these procedures "idempotent"? I mean, if I apply one of these procedures to an already cleaned item, will it consume some more silver, or will it remain intact? In other words, do the reactions consume just tarnish or do they consume pure silver too?

I like to add a few comments to the silver cleaning process. I don't dispute the overall reaction, but aluminium sulfide is unstable and reacts with water to produce Al-hydroxide and H2S which is only slightly soluble in water. It gets driven off gradually. Next, aluminium sulfide (which exists as a solid) is white, like all inorganic aluminium salts I know. I don't know what causes the aluminium to turn black. Aluminium metal doesn't react easily with acids but is dissolved more easily in basic solutions. I expect that aluminium foil contains a skin of the metal oxide which hampers the electr

The tarnish on silver objects is often silver sulfide with some oxide. Most processes involve removing that chemically. Since tarnishing only happens on the surface, there is very little loss of silver in this process.

Put them into a pot with some club soda and a piece of aluminium foil and pour over a hot water.

I am familiar with this process. I am not familiar with the other. Perhaps someone else can provide an answer for it.

What happens chemically during these procedure - why does it work, and what are the byproducts of these reactions?

This process cleans silver by a redox process. Aluminum is a more active metal than silver. This means that aluminum metal, Al(s), will reduce silver salts (salts of Ag+). The most common type of tarnish on silver is silver oxide, Ag2O.
The reaction is:
$$\ce{3Ag2O(s) + 2Al(s) -> 3Ag(s) + Al2O3(s)}$$
The club soda - containing sodium bicarbonate - plays two roles. First it acts as a supporting electrolyte so that electrons can be conducted through the solution. Second, the carbonic acid in it dissolves the passivation layer of aluminum oxide that protects the aluminum foil. This produces aluminum carbonate, which is unstable in water and decomposes into amphoteric aluminum hydroxide, which is soluble through the formation of a number of complex ions.
$$\ce{Al2O3(s) + 3H2CO3(aq) -> Al2(CO3)2(s) + 3H2O(l)-> 2Al(OH)3 (aq) + 3CO2(g)}$$
The net reaction is then:
$$\ce{3Ag2O(s) + 2Al(s) +3H2CO3(aq) -> 3Ag(s) + 2Al(OH)3(aq) + 3CO2(g)}$$
The byproducts are carbon dioxide and aluminum hydroxide.

How toxic (if all) are the byproducts?

Not very. Aluminum salts have very low health impact on you or the environment. Additionally, you produce only very small amounts of aluminum salts this way.

How much silver is lost in such a procedure?

None (Almost). The process coverts solid silver oxide into solid silver metal, both of which are negligibly soluble in club soda (or any other aqueous solution). The $K_{sp}$ of AgOH (which is what the surface of Ag2 looks like when exposed to water) is $1.52\times10^{-8}$. All of the action happens at the surface of both metals. The only common silver salts of appreciable solubility in aqueous solutions are silver nitrate and silver sulfate.

Is this procedure "idempotent"?

Yes. This procedure converts silver oxide into metallic silver. It has no effect on metallic silver: metallic silver does not react with most acids; it cannot be reduced farther by aluminum metal; and it is not soluble in water or most aqueous solutions.

Thanks for a thorough answer. I observed that the aluminium foil gets black during the process, so I thought they are some silver compounds. But if there are no silver compounds as the result and everything reduces atomic silver, what is it then? (Just a minor thing, formatting got wrong at * How much silver is lost in such a.)

The black stuff is aluminum salts: aluminum carbonate, aluminum hydroxide, or some other salts depending on what else is in the club soda. Maybe aluminum phosphate.

Or aluminum sulfide if the tarnish on the silver is silver sulfide.

@PaulJ.Gans - Of course, if the tarnish is silver sulfide, the chemistry changes a little, but the most important parts of the answer - no silver is consumed and the procedure does nothing to untarnished silver - remain the same.


text:
Is it okay for a sulfite ion to have 10 electrons around the sulfur?

Possible Duplicate:
Can an atom have more than 8 valence electrons? If not, why is 8 the limit?

I was watching this youtube clip ( http://www.youtube.com/watch?v=XY1XLgHUAIQ ) in which a lewis diagram was drawn for a sulfite ion.
Sulfur ends up with four bonds as well as a lone pair of electrons. This gives sulfur 10 electrons. Isn't this too many?

Hi DLA, it is a nice question. It has been asked already, in a somewhat more generic manner, so I am closing yours as a duplicate. The short answer is: yes, it is possible! This behavior of elements from the third period and lower is called hypervalence.


text:
What undergoes reduction in a zinc-copper voltaic cell?
In a $\pu{Zn-Cu}$ voltaic cell, the electrolyte is sodium chloride solution. The question was:

In this cell, which of the following undergoes reduction?

My answer was copper, but why is the exact answer "a hydrogen ion"?

Copper cannot undergo reduction because, in the list of components of the cell you listed, it is only present in its most reduced state, Cu(0). Thus, it cannot be copper that is reduced. See here for additional hints about your system.
You may have been confused by the fact that, in the historical (and widely taught) Daniell cell, copper is indeed reduced: it is present as Cu(+II) in a solution of copper sulfate.

@F'x Hey, thanks for your answer! So, I wish to confirm, is copper ion a spectator ion in this redox reaction?


text:
Is this the lowest chair conformer of Beta D-galactose
I drew the conformer (below) and I'm wondering if this is the most stable chair conformation.
Three of the bonds are equatorial so it looks like the most stable to me.

See this related question http://chemistry.stackexchange.com/questions/2462/which-conformer-of-beta-d-glucose-is-more-stable

that's glucose.  this is galactose

I know that; I am not suggesting the questions are duplicates. However, the issue of the anomeric effect raised in that question is relevant to this question. Someone interested in one question will be interested in the other.

The chair conformation that you have drawn (4C1) is likely to be the most stable one, as it minimizes the number of heavy axial groups.
To determine the chair conformation of a hexose, it is generally easiest to draw it and compare it with β-D-glucose, where all heavy groups are equatorial and the conformation is 4C1. If the number of heavy axial groups becomes smaller when the conformation is changed to 1C4 (all equatorial groups in 4C1 become axial and vice versa), then it is likely that the conformation is 1C4.

The conformation shown loses the anomeric effect, worth roughly 17 kcal/mol, and about 0.5 kcamol  for the axial OH.  This is opposed by the penalty for two axial OH and an axial CH2OH.  The penalty for these interaction sis harder to estimate, but it probably amounts to 5 kcal/mol.  Thus, on the basis of sterics alone, I expect you have drawn the least stable isomer.
Hydrogen bonding is probably not a factor because in both conformations, the same number of hydrogen bonds is available.


text:
What is the name of the "primary" carbon in a benzene derivative?
If we have a benzene with a single substituent, such as chlorobenzene, what do we call the carbon atom to which the group is attached? I'm familiar with the ortho- meta- para- nomenclature, but the last one escapes me.

It's ipso-.

Wikipedia page
IUPAC Gold Book definition of the ipso-attack


text:
Naming of quantum chemistry basis sets
For some DFT calculations I performed with localized (Gaussian) basis sets, I wonder how the basis sets I used are properly named. The names I could gather are:

H, C and O: 6-311G(d,p)
Al: 8-511G*
Ga: 86-4111d41G

(They all come from this website.)
However, I am not sure that the above names are consistent. Specifically:

Shouldn't I write 6-311G** instead of 6-311G(d,p), for consistency with 8-511G*?
86-4111d41G looks weird to me, but I don't know how it could otherwise be named. My knowledge of basis sets naming conventions didn't really cover anything beyond the third period of the periodic table!

Unfortunately the link went dead :(

Regarding 6-311G**:
The * notation (one for polarization functions on the heavy atoms and ** for polarization functions on the heavy and hydrogen atoms) is an older notation belonging to the split-valence basis sets of John Pople. Many people still use this notation today but if you want to be more precise and more descriptive, the (d,p) notation is what you would want to use.  In fact, we only ever use the explicit (d,p) type notation and never use the *'s in my lab.
X-YZg
Split-valence basis sets have the aforementioned structure for us to quickly and easily determine how that basis set was constructed. The X simply is the number of primitive Gaussians that describe the core electron shells of an atom. Most of the time you will be freezing core in your computations so you split the core and valence regions of your atoms and assign various basis functions to each one.  The Y and Z components mean you have two sets of primitive Gaussian functions which describe your valence region. Because you have two sets here, you can denote this basis set as a double-zeta. In your case, a 6-311G** is in fact a triple-zeta quality basis set, which is better than a double-zeta.
For the sake of interest, a 6-31+G* includes a + symbol. This simply means that diffuse functions would be added to the heavy atoms.  A ++ symbol would extend these diffuse functions to the hydrogen atoms. Of course we would never want to use the '*' notation so I should have written this as 6-31+G(d,p).  Of course polarization functions can come in sets as well such as 6-31+G(2df,2pd), and other combinations exist.
Lastly, when using split-valence basis sets such as 6-311G*, you must be very careful to check the manual and see what function space the program is going to use by default.  I will use Gaussian 09 as an example.  By default, Gaussian 09 uses spherical harmonics as its default... EXCEPT FOR A FEW DIFFERENT TYPES OF BASIS SETS including the family of 6-31G* type basis sets which uses the CARTESIAN SPACE BY DEFAULT.  This means that rather than the 5D, 7F stuff that you may be used to using on a regular basis, invoking the 6-31G* type basis set will automatically use the 6D, 10F function set which inevitably changes the size of your basis set which means that you may have data that is not readily comparable to other things.  I always opt out for the 5D, 7F type functions.

Okay, thanks, so I'll rather use “8-511G(d)” instead of “8-511G*”. What about 86-4111d41G, do you have any insight on that one?

Metals is something we comp chemists try to avoid like the plague.  Again this basis set takes on that split-valence form so I would assume that you have 86 primitive Gaussians available to describe your core (which makes sense  because metals have a larger core than 2nd row atoms). It also looks like a quadrulple-zeta quality basis set but I am unsure as to what the d41 stands for thought I would imagine it relates to some optimized coefficients on some valence orbitals. I'll check around and post if I find out more.

The Ga basis set (here) has 8 shells, respectively: s-type, 8 Gaussians; sp-type, 6 Gaussians; sp-type, 4 Gaussians; 3 more s shells with one Gaussian; one d shell with 4 Gaussians; 1 more d shell with 1 Gaussian. As such, maybe it should be noted 8-64111G(d,d)?

The notation isn't really complete enough to suggest a fully informative label, I fear. What I've seen people do with lesser-known heavy element basis sets is, after the label provided, just say how many functions you started with of what types, and how many you contracted them to, so, for example, this is 8s13(sp)5d contracted to 1s5(sp)2d.


text:
Is combustion considered a redox reaction?
When carbon combusts with oxygen, is this considered a redox reaction since the oxygen atoms gain electrons and the carbon atoms lose them?

In general, yes.  If the reaction involves oxygen going from the oxidation state of zero in $\ce{O_2}$ to an oxidation state of -2, then there is oxidation and the reaction is a redox reaction.

Could you elaborate a bit? When you say "in general", does that mean that there are some special cases? If so, you may want to add them to your answer, it'll make it more interesting...

@ManishEarth There's nothing 'in general' about it.  a combustion reaction requires an oxidizer; therefore it is always a redox reaction.  (Late comment, I know.)


text:
What is the difference between distilled water and rainwater?
I am studying different means of collecting water for consumption. From resources available on Internet, I understand that distilled water is not advised for regular consumption because of the lack of minerals and salts, while rainwater can be consumed if not overloaded with unwanted chemicals.
What confuses me is that I know water to be pouring down from clouds, which is water vapor turning into droplets through condensation. To me, this looks to be exactly the same as the process of distillation, i.e. evaporation and condensation of water.
So is my assumption about rainwater formation wrong? What is the difference between rainwater and distilled water? What causes this difference to happen?

While there are some internet memes that advise against drinking distilled water, there is no real problem with doing so. The minerals in most water are only ever present in very small amounts with barely measurable effects on health (very hard water is known to marginally lower heart attack rates) so their lack is not going to be significant either.

@matt_black: perhaps you would be interested in this discussion then, http://skeptics.stackexchange.com/questions/5484/does-drinking-distilled-water-remove-needed-minerals-from-your-body

Not everything on skeptics is reliable. Read Rex Kerr's answer and the comment about the provenance of the study that supposedly proved it. Or read my answer to this question and see some proper statistics on the small differences to people caused by different mineral contents.

Agreed. Nutrition is a tough science, because it's just tough to experiment, since manifold ingredients end up to make big soup in the body. Not even mentioning metabolism are not all the same.
BTW Do you know a way of remineralising water other than adding a package of minerals?

""way of remineralising water other than adding a package of minerals?"" ROFL

Rain water passes through the atmosphere and picks up whatever is present there, including particulates such as dust, trace amounts of atmospheric gases, nitrates and nitrites, and so forth.
Rain water is relatively low in mineral content.  In practice, people drink rainwater from streams and wells, where it has had ample opportunity to pick up minerals from soil and rock.
https://www.google.com/search?q=rain+water+composition

The YouTube channel minute earth even explains why it is impossible for rain to form without dust or minerals in the middle of each drop, but sadly I can not post the link at the moment.

Do you know of better sources for this than "lmgtfy"?  Well-regarded published papers, for instance.  It would also help for this answer to include some quantitative information - both rainwater and distilled water will contain some amount of impurities in practice, so what we really care about is how much.

@Nate: There are several well-regarded studies in that Google Search.  I'm happy to enumerate all of them in my answer, but the Google Search is maintained and updated automatically.

Indeed.  But part of the purpose of SE is to get answers from experts.  Google can tell us what studies there are, but it can't give us an expert opinion as to whether they are relevant, reliable, appropriately conducted, etc - that's where you come in.  There's also a preference for answers which are not "link-only" but give a summary of the conclusions, thus helping to avoid link rot.

W@Nate: Well, I'm not actually an expert, but I do know enough about the rainwater ecosystem to answer the OP's question at his level of expertise.  If the OP wants detailed composition analyses (he didn't ask for them), then someone else is probably better qualified to render those.  Properly distilled water really shouldn't contain anything in significant amounts except water molecules, though I suppose you could contemplate copper atoms picked up from the condensation tubing.

Rain water contains very low amounts of salts and other nitrates but it takes in any gas present in air. That is why acid rain occurs when the air is polluted. The air contains oxygen so people recommend the rain to be consumable as it has lots of oxygen contents in them. But distilled water does not contain any minerals in them, so it is not recommended for consumption because of lack of minerals.