text:
In a battery what would happen if you had a perfectly uniform and pure cathode?
I have been watching some videos on how batteries work and they seem to talk about the cathode being the most important. I also watched another video a long time ago and while my memory is blurred it talked about impurities in the cathode being the only reason the electrical charge doesn't flow constantly. While I found this weird I also wonder how generates seem to get this unlimited supply of electrons also so it's only fair for me to assume that the battery can also.

I don't think that's accurate. Even the universe doesn't have an "unlimited supply of electrons".
Impurities in the cathode likely give rise to internal resistance inside the battery (among other things), leading to deviation from an ideal cell. I would guess that as your cathode approached purity, your battery would approach the behavior of 'ideal cells', and that discharge rates would approach those of capacitors, and efficiency would approach 100%. To get an ideal cell though, the anode and electrolyte would also have to be 'pure' and some real-world effects would have to be ignored.

Because of internal resistance, the terminal voltage of a cell that is discharging is smaller in magnitude than the open-circuit voltage and the terminal voltage of a cell that is charging exceeds the open-circuit voltage. An ideal cell has negligible internal resistance, so it would maintain a constant terminal voltage of $ \epsilon $ until exhausted, then dropping to zero. If such a cell maintained 1.5 volts and stored a charge of one coulomb then on complete discharge it would perform 1.5 joule of work. In actual cells, the internal resistance increases under discharge
Wikipedia

What do you mean by pure, exactly? Complete homogeneity and lack of any other elements? That sounds like a physics problem where you "ignore air resistence" and other real world effects for simplicity. If this is what you're after, search for things like 'Ideal Battery' and 'Ideal Electrochemical Cell'.

I think where I got confused was it will not create "unlimited" electrons (I think I meant electricity). In reality I believe that it was "a lithium ion battery without impurities will be able to hold the same amount of charge no matter how many times it is charged and drained" but it's only speculation as to what I meant.


text:
How to determine the empirical formula for an unknown iron oxide?
I have to determine the empirical formula of an oxide of iron which has $69.9\%$ iron and $30.1\%$ dioxygen by mass in it.
I have started as follows:

For our convenience we take $\pu{100 g}$ of that iron oxide:
We have $\pu{69.9 g}$ of iron in it and $\pu{30.1 g}$ dioxygen in it.
We convert into amount of substance of each element:

$\text{amount of substance of iron} = 69.9/55.8 = 1.25$
$\text{amount of substance of dioxygen} = 30.1/32 = 0.94$

Is my method and calculations right?

Please note that the proper term for "(number of) moles" is amount of substance. The former would be the same as referring to the mass as "(number of) kilograms".

First assume the mass of an oxide of iron to be $\pu{100 g}$ which implies mass of iron would be $\pu{69.9 g}$ and mass of oxygen would be $\pu{30.1 g}$. Now calculate amount of substance of both the elements by dividing the assumed mass by their atomic number.
Then calculate relative amount of substance of each element by dividing the obtained amount of substance of each element by the smallest of them. (E.g. suppose you've got $\pu{5 mol}$ of iron and $\pu{2 mol}$ of oxygen, divide 5 by 2 and 2 by 2 to get relative amounts of substance of both.)
Now just write relative amount of substances in the form of a ratio and bring the obtained ratio to least simplified form. (Remember the ratio should come in whole numbers.)
After obtaining ratio, write the empirical formula as I did in this example.
(eg. suppose I got this ratio $2:5$ from last example. So the empirical formula will be $\ce{Fe2O5}$ .)
Actually, you can skip the relative amount of substance step when you've got 2 elements in empirical formula, but you will need to perform all the steps when there are more than 2 elements as the answer comes to be different in some cases.

You've been doing everything correctly so far - what we have now is the relative amount of substance in moles of each element in this $\pu{100 g}$ compound. As you correctly calculated, we have $\pu{1.25 mol}$ of $\ce{Fe}$ and $\pu{0.94 mol}$ of $\ce{O2}$ - for the sake of simplicity we'll say it's $\pu{1.88 mol}$ of $\ce{O}$ (I multiplied by 2).
Now we just take this into a calculator! $1.88/1.25$ is approximately $1.5$, so the correct answer would be $\ce{Fe2O3}$. The reason we divide large by small is that it's easier to see whole number ratios from the values that are larger than one - $0.66$ being converted to $2/3$ may not be as obvious as $1.5$ being converted to $3/2$, for example.
Why did we divide? so we've calculated the relative ratio of the amounts of substances in this compound - this means for any compound of any mass, we'll always get the ratio of $3/2$ of $\ce{O/Fe}$, right? Which means the empirical formula is $\ce{Fe2O3}$.

source: Stack Exchange

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

sort: votes


text:
Why is NH₄⁻ an anion and NH₃ is not?
$\ce{NH4-}$ is conjugate acid and $\ce{NH4}$ is conjugate base. $\ce{NH4-}$ is conjugate acid because it has one more Hydrogen.
The thing that is difficult for me to understand is why $\ce{NH4-}$ is anion? Is negative sign refer to whole molecule or just to Hydrogen? How a molecule can be anions?
Another example of the same situation is $\ce{HCO3-}$ (conjugate base) and $\ce{H2CO3}$ (Conjugate acid). By adding one Hydrogen to the base the negative sign goes away. Why?

You seem to have quite some things mixed up. I think you mean $\ce{NH_4^+}$ instead of $\ce{NH_4^-}$ and also $\ce{NH_3}$ instead of $\ce{NH_4}$. Maybe you should read up on acids and bases first before worrying about the cation/anion question.

I'm quite confused about your post and even its title. So let's start wit title: ammonia  $\ce{NH3}$ is neutral because of electron count. Now the $\ce{NH4-}$ is tricky. You would have to add hydride $\ce{H-}$ to ammonia. As far as I'm aware it's utopia. Hydorgen atoms in ammonia are protic so they would immediately react with  $\ce{H-}$ giving $\ce{H2}$. I never heard that ammonia can react with metal hydrides as  $\ce{KH}$ or lithium triethylborohydride $\ce{Li[HBEt3]}$.
Now to the text. Ammonia is amphoteric (i.e. it can get proton or loose it; or it can act as a base or acid).
Conjugate acid of ammonia ($\ce{NH3}$) is the ammonium cation $\ce{NH4+}$. Acid, in Brønsted theory,  is a proton donor ($\ce{H+}$) and on loosing proton turns into base. In this case our acid is positively charged, so if it looses proton it turns into neutral molecule.
Conjugate base of ammonia is the amide anion $\ce{NH2-}$. If ammonia is an acid then to become base it must loose $\ce{H+}$. The final molecule is negatively charged. If you want to do it experimentally, you need to dissolve alkali metal ($\ce{Na}$) in liquid ammonia. H-radical is replaced by metal and dimerise to hydrogen $\ce{H2}$ (and salt - sodium amide is created $\ce{NaNH2}$).
And now your last question: "By adding one Hydrogen to the base the negative sign goes away". You are not adding Hydrogen (atom) but proton ($\ce{H+}$).

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


text:
Reaction mechanism of imine reduction to amine using an aluminium amalgam
I can not find any good resources on the Internet except for some shady drug syntheses. I was hoping that someone could provide me with some more educational information.
I'm interested in imine-amine reduction methods, but it was hardly discussed in my class. It would be very nice if someone could explain the exact reaction mechanism of the imine-amine reduction in the presence of an aluminium amalgam.
I also found that the very dangerous compound Mercury(II) chloride was used a lot in this method and I was wondering why for example Mercury sulfide is not used, as I thought it would be a lot safer since it is hardly soluble in water and is therefore less likely to do any physical harm.

Mercury(II) chloride is poison, but I hardly would call it very dangerous. It's not worse than a lot of other heavy elements salts. What is really dangerous is Dimethylmercury(II). See: http://www.chm.bris.ac.uk/motm/dimethylmercury/dmmh.htm As why the HgS was not used? Well, HgCl2 is water soluble, HgS is effectively insoluble in water. If you use your compound as a catalyst you need some finite concentration of it.

Yes I know that. I do not know the reaction mechanisms as I stated before, so I have no idea if solubility is required for amalgamation. What would be the best mercury salt for amalgamation in your opinion?

The mechanism is electron transfer. The imine will take one electron from metal to form a radical anion ($\ce{.CR2-NR'^{-}}$) intermediate; the $\ce{-NR'^{-}}$ will grab one proton from solvent to give you the neutral radical ($\ce{.CR2-NR'H}$); the carbon radical grab one hydrogen atom ($\ce{.H}$) from solvent to form your amine product and a solvent radical ($\ce{.OH}$ if water used as solvent); and the resulting solvent radical will take another electron from aluminum to finish the whole process. The reason to add mercury is to destroy the relatively less reactive aluminum oxide layer. Mercury also serve as the medium for electron transfer.
$\ce{HgCl2}$ is not that dangerous, since it is not volatile at all like metal mercury or alkylmercury. After reaction, simply add some sulfide or sulfur to transfer all of them to $\ce{HgS}$ which is not soluble and not reactive in non-oxidative condition. This is also the reason that you can not use $\ce{HgS}$ instead, because it is neither soluble nor reactive.


text:
Electrolysis of molten Calcium Chloride: Identifying the product
I was trying to make Calcium metal by electrolyzing molten $\ce{CaCl_2}$. I used a stainless steel spoon and a thick steel wire as electrodes. Then I attached them to a power supply. I placed some $\ce{CaCl_2}$ in the spoon and melted with a gas torch. I then placed the steel electrode in the molten pool and gradually pulled it out. A black glassy substance formed on the rod. I ran water over it and it fizzed as I would expect from Calcium metal, however, when it was exposed to air for a period of time it start to "melt" as it absorbed moisture from the air. As I was breaking it off of the electrode, it shattered and I could see that it was the glassy substance all the way through.
I am trying to figure out what I have made. Could it possibly be a mixture of unreacted $\ce{CaCl_2}$ and a small amount of Calcium?

Could you clarify, if you used freshly calcined CaCl2 or a wet one?

I would hypothesize that the "black glassy stuff" is nothing more than glassified $\ce{CaCl_2}$. Metal is an excellent heat conductor and the liquid $\ce{CaCl_2}$ likely cooled rapidly to form glass when the cold wire was inserted.
Two things could be done to avoid this:
1.) heat the wire before inserting it into the mixture of molten $\ce{CaCl_2}$.
2.) heat the $\ce{CaCl_2}$ well above its melting point.
If you are determined to make Calcium metal it might be worth electrolysing an aqueous solution of $\ce{CaCl_2}$ to evolve the $\ce{Cl2(g)}$ (do this responsibly outside/in fume hood of course as $\ce{Cl2(g)}$ is poisonous.) Calcium Hydroxide should be the result. Evaporating the water will yield solid $\ce{CaOH}$, which should be much easier to electrolize using your method since it has a melting point of 580°C vs. 772°C for $\ce{CaCl_2}$.

Better yet, just buy"slaked lime", which is $\ce{Ca(OH)2}$. You should be able to find it at home improvement stores.

Does this sound like what you made: Iron (III) Chloride. Depending on the polarity of the electrode, one constituent of the melt will combine with the material of the electrode, so it sounds like the chlorine combined with the iron in the steel wire.

It's probably not $\ce{FeCl3}$, which is yellow to green in color

Iron (III) Chloride || would brake in gas torch


text:
Halflife of hydrogen bonds
Related: Covalent Bond/Hydrogen Bond
What is the half-life of hydrogen bonds in a glass of water at room temperature, say 18 degrees Celsius? I guess values may be different for distilled water and for tab water, but both may be intersting.
My attempts to get an answer involve searching with google, google scholar, and google books. Complications were that I cannot get access to all articles using my universities database.
From this page we have

The half-life of a hydrogen bond is about 10 seconds

Is this correct?

I'm not sure whether the term "half-life" means the same as "lifetime" but I found a very interesting website about hydrogen bonding in water (with a lot of references) where it is stated that the lifetime of a hydrogen bond is in the range of 1 - 20 ps (it gives the following reference for this: F. N. Keutsch and R. J. Saykally, Water clusters: Untangling the mysteries of the liquid, one molecule at a time, Proc. Natl. Acad. Sci. U.S.A. 98 (2001) 10533-10540). That is quite a big difference to the 10 seconds from your reference. Thinking of the very dynamic nature of a hydrogen bond network and of the typical time scale for molecular processes in general I would tend to think, that the breakage and formation of hydrogen bonds on a molecular level should take much less time than 10 seconds. So if you are looking for the lifetime of a hydrogen bond on a molecular level a few picoseconds sound reasonable.
Edit: Here is another website, which also states a hydrogen bond lifetime of about a few picoseconds.

What I meant with "half-life" is indeed "lifetime". Thanks for your answer :).


text:
How to draw Lewis dot structures?
What is the most logical and easy to remember step by step method for drawing Lewis dot structures?
I'm having trouble drawing a simple one for $\ce{CO2}$.

Calculate the total number of valance shell electrons ($n_1$) in all constituent atoms or ions. For example, in $\ce{NH_4^+}$, $n_1$ will be 8 (5 of N, 4 from all H's and -1 because the molecule has +1 charge).
Now, calculate the total number of electrons that are required to fill valance shell of every constituent atom or ion ($n_2$). In other words, calculate the total number of electrons that will be required to fill valance shell of each atom or ion individually. We can use following formula to find $n_2$$n_2 = [\text{# of Hydrogen atoms} \times 2] + [\text{# of all other atoms except Hydrogen} \times 8]$
In case of $\ce{NH_4^+}$, $n_2$ will be 16 ($[4 \times 2] + [1 \times 8]$).
Now, calculate the total number of electrons that atoms or ions will need to share among each other in order to complete their octets ($n_3$).
$n_3 = n_2 - n_1$
For $\ce{NH_4^+}$, $n_3$ will be 8.
Now calculate the number of bonds that will be present in our molecule
$n_4 = \frac{n_3}{2}$
For $\ce{NH_4^+}$, $n_4$ will be 4.
Now let's find the number of unshared electrons, $n_5$
$n_5 = n_1 - n_3$For $\ce{NH_4^+}$, $n_5$ will be 0.
Now find the number of lone pairs, $n_6$, which is nothing but $n_5$ divided by 2._For $\ce{NH_4^+}$, $n_6$ will be 0.
Now find the central atom. Central atom is the one which is either least in count or more electropositive.
Always remember that Hydrogen and Fluorine can never be placed at center.
So in our case, Nitrogen gets to be the central atom.
After allocating the central atom, place the surrounding atoms such that the molecule is symmetric.
In our case, geometry of molecule will be tetragonal.
Distribute the bonds between atoms. Remember that the total number of bonds, that you distribute, should be equal to $n_4$.
In our case, we simply place a bond between Nitrogen and each Hydrogen atom.
Now use the lone pairs to complete the octet of atoms. Remember that, by now, every atom will already have 2 electrons in their valance shell because of the bond.In our case, which is $\ce{NH_4^+}$, we don't have any lone pairs. OOPS!
Now comes the Formal Charge.

Formal Charge (FC) is the charge assigned to an atom in a molecule, assuming that electrons in a chemical bond are shared equally between atoms, regardless of relative electronegativity. It is not an actual electrical charge but just a mathematical construct.

  $\text{Formal Charge of an atom} = \text{# of valence shell electrons (in Ground state)} - \text{# of unshared/non-bonded electron pairs or lone pairs} - \text{# of bonds}$

Please note that you must place formal charge on each and every atom when it is a non-zero value for that atom.
This step is little confusing so let's get back to our example. Nitrogen has 5 electrons in its valance shell in ground state. It made 4 bonds in our molecule and has none unshared electron pairs or lone pairs (please note that a single unpaired electron doesn't count here). So formal charge on Nitrogen will be 1 (5 - 0 - 4), whereas formal charge on each Hydrogen atom will be 0 as every Hydrogen atom has 1 electron in its valance shell and made one bond in our molecule.
In a sense, you actually obtain the structure in $10^{th}$ step but Formal charge, if a non-zero value, should be represented for each atom in molecule in order to complete its Lewis Dot Structure.

I will give you a simple way, but if you need to study properly, you can follow Ashutosh Gangwar's answer.
C has 4 valance electrons.
O has 6 valance electrons.
O has 6 valance electrons.
First distribute these electrons to each element and indicate the covalent bonds.
So you know carbon and oxygen need to obey octed rule, so in this case carbon makes 4 bonds with oxygens;

Rules for construction of  Lewis dot structures

Calculate the number of total valence electrons, including charges (ions) ($V$).
Calculate the number of total electrons required to complete octet/duplet ($T$) = $(2 × \ number \ of \ H \ atoms) + (8 × \ number \ of \ other \ atoms)$
Calculate the number of bonding electrons $B = (T-V)$
Calculate the number of bonds = $\dfrac{(T−V)}{2}$
Calculate the number of unshared electrons $N = (V – B)$
Calculate the number of lone pairs = $\dfrac{N}{2} = \dfrac{(V – B)}{2}  $
Distribute the bond pairs and lone pairs ensuring each atom has an octet (or for H=duplet).

If you still have trouble drawing Lewis structure for CO2, watch How to Draw the Lewis Dot Structure for CO2: Carbon dioxide by Dr. Wayne Breslyn.
Lewis Structures are important to learn because they help us understand how atoms and electrons are arranged in a molecule, such as Carbon dioxide. This can help us determine the molecular geometry, how the molecule might react with other molecules, and some of the physical properties of the molecule (like boiling point and surface tension).

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


text:
Reviving Li-ion battery in freezer?
I have read that it is possible to revive a dead Li-ion battery by putting it in the freezer for three to seven days, then letting it get back to room temperature.
Can this process work, and if so, how does it work?
Notes:

Somehow the opposite, i.e. heating (cooking) the battery, is more intuitive to me. After all, in the dead battery, there is "something stuck" (crystallized?), and heat could break that. But I don't know if it's a good idea to try that at home. ;-)
I also found that hitting a Li-ion battery, against a table, for example, has some effect. At the moment, I have a battery in an IBM ThinkPad T41 laptop that, although still working, is on it's way to death. When it runs flat and I connect the laptop to mains, then the battery is not detected properly (battery status LED continues blinking orange). So, I take it out, hit it, put it back in, and the problem is solved (LED constantly orange). Also, I have the impression that hitting gives some extra charge. It could be all just a coincidence, of course.
I am asking in part for a friend whose laptop battery has been dead, I believe, already for some weeks or months. He said that the battery is not that old, but he has let it run flat in cold weather (up in the mountains). Anyhow, whether the process works for this particular battery should not influence my decision on accepting an answer. I am primarily interested in understanding a little bit more about battery chemistry, just to broaden my general knowledge.

A note: If you want sources to back up claims, I suggest you  ask a similar question on [skeptics.se]. Be sure to check that it passes their quality standards -- you need to collect links to where unsupported claims are made.

I have no idea how this works (seems to have already been covered above), but I can confirm that this method (placed in freezer for 12 hours) has worked on two dead Li-ion batteries of mine.

Well, maybe every battery works differently, but I has a battery I had not charged for long time and was not getting charged by the charger. So I did the trick and put my battery in the freezer for 12h, then, to go back to room temperature, I had it sitting on my desk overnight and now just plugged in the charger ... it is charging!

Hitting Li-ion batteries can also cause them to leak and subsequently explode, please do not do that.

There are a number of rumors on the internet about different ways to treat your battery. Some of these involve putting batteries in the refrigerator or freezer for varying lengths of time to improve performance in some way. Snopes has concluded that there is no basis for this. However, batteries and battery aging mechanisms are very complex and so I think it's difficult to completely rule out the possibility that under some situations there could be some beneficial effect from freezing.
Nevertheless, I would recommend that you avoid using your battery in any way that it was not designed for. The reason is that internal damage to the battery (such as from extreme temperatures, moisture, or mechanical stress) could create a safety hazard. Even if cooking, freezing, or bashing does have some positive effect, it is probably not worth putting yourself at risk. The best thing to do is to recycle your old battery and buy a replacement.

Thanks, this is certainly interesting! However, the article talks about extending the performance of batteries that still work, and I presume non-rechargeable ones. What I'm talking about is reviving a dead rechargeable battery. Though unlikely, perhaps a shock like freezing can help?

Right, the snopes article doesn't specify what type of batteries they're talking about. But as far as I know, there is no direct evidence that freezing can revive a dead Li-ion battery (besides maybe a few dubious anecdotes floating around the internet). But like you say, that doesn't mean that it's not possible. I realize this answer probably doesn't completely answer your question, but that's all I know.
Also I just sent an email suggesting this idea to the MythBusters. Maybe they will test it for us!

This is a good answer from user point of view, but this is a Chemistry Q&A and I'd expect some chemistry in the answers.

Also, the article in Snopes debunks this for alkaline batteries, but OP is asking about Li-ion batteries, which are completely different!


text:
Can you color a flame and what would the effects be?
I read this question (Flame temperature for cooking food) and came up with the following questions:

Could companies color their flames so that products would heat quicker?
How would they do this?
Could you use this to make an infrared or ultraviolet flame? If
you could, how hot would they be?

Unfortunately not, I'm afraid.
I think you might have gotten cause and effect mixed up here. Blue flames are hotter than yellow flames not because of their color, but because hotter flames generate blue light and cooler ones generate yellow/red light.
To take an example of this kind of problem, I know that very red tomatoes tend to be tastier than very pale tomatoes. Does this mean that we could make tastier tomatoes by injecting food coloring into them? Well, probably not, because it's not the color that causes the flavor. Rather, tomatoes that generate more color tend to also generate more things like sugar and flavors, and so flavorful tomatoes tend to have more color.
With flames, it's even more black-and-white because while its possible to have a pale tomato that happens to taste good, the color of the flame is set by the universe: unless you add colors yourself, it is impossible to have a blue flame that is the same color as a yellow one. The color of the flame is determined by something called "blackbody radiation," and if you're curious about why flames are colored the way they are, it's definitely something to take a look at.

The flame has different color because it has different temperature. If you add something to change flames color (like source of metal ions) it will not affect temperature of the flame. So in short: they can not.

When we talk about the color of flame we should consider its source, too.  If you are burning a transition element then you will see a colored flame. Each element may give different color so the energy caused by this element will also be different. And the heat is also related to this energy.
On the page you have linked, you can see the relationships for these energies. However, if you need further informations you can study related concepts e.g. electronic spectra and quantum mechanics. Since this issue can be explained with more details.

The temperature of the flame depends (mostly) on the material being burned. Natural gas burns with a hot blue flame, while wood burns with a cooler yellow flame.


text:
Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?
I was learning about voltaic cells and came across salt bridges. If the purpose of the salt bridge is only to move electrons from an electrolyte solution to the other, then why can I not use a wire?
Also, will using $\ce{NaCl}$ instead of $\ce{KNO3}$ in making the salt bridge have any effects on voltage/current output of the cell? why?
Plus if it matters, I'm using a Zinc-Copper voltaic cell with a tissue paper soaked in $\ce{KNO3}$ as salt bridge

There's another question related to salt bridges on this site.
The purpose of a salt bridge is not to move electrons from the electrolyte, rather it's to maintain charge balance because the electrons are moving from one-half cell to the other.

The electrons flow from the anode to the cathode.  The oxidation reaction that occurs at the anode generates electrons and positively charged ions.  The electrons move through the wire (and your device, which I haven't included in the diagram), leaving the unbalanced positive charge in this vessel.  In order to maintain neutrality, the negatively charged ions in the salt bridge will migrate into the anodic half cell.  A similar (but reversed) situation is found in the cathodic cell, where $\ce{Cu^{2+}}$ ions are being consumed, and therefore electroneutrality is maintained by the migration of $\ce{K+}$ ions from the salt bridge into this half cell.
Regarding the second part of your question, it is important to use a salt with inert ions in your salt bridge. In your case, you probably won't notice a difference between $\ce{NaCl}$ and $\ce{KNO3}$ since the $\ce{Cu^{2+}}$ and $\ce{Zn^{2+}}$ salts of $\ce{Cl-}$ and $\ce{NO3-}$ are soluble. There will be a difference in the liquid junction potential, but that topic is a bit advanced for someone just starting out with voltaic/galvanic cells.

But if you consider the salt bridge as a solution that is part of the two half cells (which connect the two half cells so they are one), it doesn't really matter which side is positive or negative since the whole thing (two half cells connected by a salt bridge) Is neutral.

What'll happen if the cathode and the anode are dipped in a single chamber (like in electrolysis of water)?

@shub without a physical means to separate the two half reactions, a homogeneous electron transfer reaction can occur when the products of the two half reactions come together.  In the case of water electrolysis, the gases generated leave the cell, and therefore would be unlikely to recombine.

Without the salt bridge, the solution in the anode compartment would become positively charged and the solution in the cathode compartment would become negatively charged, because of the charge imbalance, the electrode reaction would quickly come to a halt.
It helps to maintain the flow of electrons from the oxidation half-cell to a reduction half cell, this completes the circuit.

Why it would come to a halt? Is this because the accumulation of positive charge will attract the negative electrons?

@Anton see here

@Anton read up on ELECTROLYTIC CONDUCTORS.  If you disconnect the wires, you break the circuit, and the battery current is zero.   If you break the salt bridge, same thing.   The salt bridge is a "wire," but it is an Electrolytic Conductor, same as any salt water.   Go search on:  metallic conductivity (mobile electrons,) versus electrolytic conductivity (mobile ions.) During electric currents, there are no electrons flowing in salt water, in dirt, or in human bodies.   Grade-school classes don't teach this fact, so most people incorrectly think that electrons can flow through salt water.  No.

The purpose of the salt bridge is to move ions.
If you use enough electrolyte solution on both sides, though, it doesn't matter;  in that case, the salt bridge can be neglected.

I was learning about voltaic cells and came across salt bridges. If the purpose of the salt bridge is only to move electrons from an electrolyte solution to the other, then why can I not use a wire?

If you connect the two electrodes with a wire, you will short-circuit anything else connected to the electrodes. If you connect the two electrolyte solutions with a wire, there are two things that could happen:

nothing (if there is no redox reaction happening at the wire/electrolyte interface, there will be no charge transfer. The wire can't transport ions, and the electrolyte can't transport electrons).
two more half-reactions (if there are half-reaction that the respective combinations of electrolyte and wire material can support)

In either case, this will not have the desired effect of a salt bridge, which is to balance the charge buildup that occurs when electrons travel along the wire from anode to cathode.

Also, will using $\ce{NaCl}$ instead of $\ce{KNO3}$ in making the salt bridge have any effects on voltage/current output of the cell? why?

The voltage and the current depend on what load you attach to the voltaic cell, and properties of the cell itself.
If ion transport along the salt bridge is the rate-determining step, it will affect the current (current is movement of charge per time). If you measure the voltage under load, it will also be affect by the nature of the salt bridge.


text:
Effect of vibration on wine quality
Most people that have ever visited a viticulteur (winemaker) will know that it is common wisdom that vibrations are bad for the wine quality. When I was just a kid the winemaker always told me that we should be silent because the wines needed to sleep, but later a different one explained that this must have been to stop me from running around and causing vibrations.
Even scientific research (e.g. Chung et al. 2008 Journal of Food Composition and Analysis (21)8) has shown that vibrations will indeed change the wine and, among others, accelerate the aging process.
What I am wondering is why these vibrations matter? Is it a chemical effect, in which the vibrations accelerate certain chemical reactions? Or is it more physical in nature, in which the vibrations increase mixing rates and therefore somehow change the evolution of the wine?
If it helps to get your minds working: Vibrating wine in a glass

The paper you refer to is very interesting and I think answers the question you are looking for.  Were you not able to access the full text or is there something specific from the paper that you are interested in?

I was able to access the pdf but I didn't find the answer. The type of statements made on the effect of vibration are hardly conclusive. To give an example, the following sentence attributes some chemical change to vibration, but it does not explain why the vibrations have this effect: "Consequently, the increase in refractive index in
this experiment could also be attributed to the degradation of flavonol glycodises, which was further accelerated by vibration." What I am looking for is an explanation of WHY vibration matters, not whether it matters (which is conclusively shown in the article)

One of the effects could be that vibrations increase mixing on phase boundary between wine itself and air above it.
Other thing is that vibration are generally not good for crystal growing process (instead big and nice monocrystals one gets polycrystalline solid).  Usually you can find nice crystals of tartaric acid in the bottom of bottles and barrels. I do not know chemistry behind it but certainly it will change taste of the wine if the tartaric acid stays longer in wine (slow crystallisation) instead of falling rapidly out of solution when you shake it.
Of course if there is still a sediment earlier during fermentation, certainly that will be not good if this mixes with clear wine.


text:
What elements don't observe the octet rule?
Apart from hydrogen which forms a duet, which element's don't observe the octet rule?

I plugged your question into google and this is what I got as the first link.  Do you need something more than this description?

I was hoping for a list rather than long descriptions..

see also ChemWiki:Violations of the Octet Rule

Octet rule is more advisory than a rule; It's usually obeyed by the main group elements.
Exception are paramagnetic compounds (obvious reason); first for elements in periodic system (He-configuration), 3-valent boron compounds; depending on counting system boron clusters (f.e. dodecaborane with 6-coordianted boron); transition metal complexes (these usually follow 18e rule, however there are plenty of examples with 14, 16 or even 20 electron complexes).

It would be nice if you give some examples for TM-complexes not obeying 18e rule.

Enough that you look at the metallocenes. 15e: VCp2 (Cp=C5H5-); 16e:
CrCp2; 17e: MnCp'2 (Cp'=C5H4Me-); 19e: CoCp2; 20e: NiCp2. Most of them
are commercial products.


text:
What does it mean to shift equilibrium?
What does it mean to shift a chemical equilibrium? For example,

the equilibrium shifts to the left …

I don't understand that.

Probably you are having problems with Le Chatelier's Principle.
Suppose you have an equilibrium established between four substances $\ce{A}$, $\ce{B}$, $\ce{C}$ and $\ce{D}$, such that
$$\ce{A + B <=> C + D}$$

What would happen if you changed the conditions by increasing the
concentration of $\ce{A}$?

According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the concentration of $\ce{A}$ decreases again - by reacting it with $\ce{B}$ and turning it into $\ce{C + D}$. The position of equilibrium moves to the right, making more of the products. Of course, this assumes that there is still some amount of $\ce{B}$ left in the reaction vessel.
For better understanding, refer this.

Simply it means that the equilibrium will force the conversion of certain substances to reduce the disturbance, to maintain the equilibrium itself. So, for example, $\ce{A + B <=> C + D}$ (both directions).
If you increase the concentration of the reactants — $\ce{A}$ or $\ce{B}$, then you’re increasing the conversion of reactants into products, because the rate of reaction and overall number of product formed increases. Therefore, we need to form more $\ce{C + D}$, to decrease the concentration of $\ce{A + B}$ and balance out the rate of change.
Therefore, even though the reaction occurs simultaneously as a forward and backward reaction, the movement of equilibrium to the right (to the products) means the conversion of reactants to products is further emphasised, and generally — there is more of it occurring than in a usual equilibrium, in order to balance the change. Same thing would’ve happened if you increase the concentration of products, you want the equilibrium to shift to the left (reactants) so concentrations of both reactants and products balances out, reducing the disturbance (Le Chatelier’s Principle — specifically referred to by Ashu). Simply shifting equilibrium means increased rate of conversion of substances, predicating on the change in the reaction in the first place.
Sorry, it’s quite long winded and irrelevant at times, I hope you’ve benefitted from this though. I’ve also used the idea of changing concentrations in particular from Ashu, if that’s alright.

Let’s do it simply. You start from some equilibrium. For example you have simple reaction $\ce{A <=> B}$, and in the equilibrium state there is $60~\%$ of $\ce{A}$ and $40~\%$ of $\ce{B}$. If it said that equilibrium is shifted to the right it means that you have more product $\ce{B}$ (i.e. more than $40~\%$). If you shifting to the left you gaining more substrate. Now how it's done — look on the previous answers.

“Equilibrium shift” is jargon. The equilibrium constant stays the same when you disturb an equilibrium by adding or removing species. However, the system might not be at equilibrium anymore.
When the system was at equilibrium, there was no net reaction. Now that the equilibrium is disturbed, there will be a net reaction in the direction of reactants or products (depends on the specifics of how the equilibrium was disturbed). The jargon for this net reaction is to say the equilibrium “shifted to the left” (or to the right).
It would be more formal to say equilibrium was attained again by a net reverse (or forward) reaction. This would also be easier for someone learning these concepts because it better describes what is going on.

The reaction is at equilibrium
The equilibrium is disturbed
There is a net reaction (in one or the other direction) until we are at equilibrium again.

Typically, the concentrations present at the original equilibrium state will be different from those for the second equilibrium state.
When a change in temperature causes a change in the equilibrium constant, the phrase “equilibrium shift” might be a little bit more intuitive. There will be a net reaction to adjust to the new equilibrium constant.


text:
Racemization of biphenyl compounds
How does the effect of heat and effect of substituents affect racemization of biphenyl compounds?

Preparation: Take the parent compound per se, biphenyl.  The bond keeping the two phenyl rings together leads to one carbon atom we call C1, the other C1'.  Now if you label the other carbon atoms in each  phenyl ring seperately in clockwise direction, you'll get C2, C3, C4, C5, C6 in one cycle; and C2', C3', C4', C5', C6' in the second.  (I haven't a credit to upload a picture yet.)
Now: The larger, the more sterically demanding substitutents on C2, C6, C2' and C6' are (condition), the higher the energetical barrier is to interconvert the conformers into each other.  Eventually, this may lead to atropisomerism, and the isomers are atropisomers.
Note, please, sometimes there may be already a preference for one or the other conformation.  If, for example, you substitute C2 and C2' of biphenyl by nitrogen, free molecules of this compound 2,2'-bipyridine energetically favour a conformer with the nitrogens as far as possible separated (similar to a s-trans conformation).  Offering a cation to bind, to complex to, however may lead to a preference towards the s-cis-like conformation.

Well, what if there are sone substituents likr the nitro or methoxy group present on diffrent positions? Also how does heat have an effect on racemization?

The first part of my answer was based on sterics.  In other words, the bigger the atoms / groups facing each other across space of different phenyls (i.e. not bound on adjacent/vicinal carbon atoms of one and the same phenyl ring), the more likely the two rings are even more out of a plane in common.  You are right, a second potential contribution is electrostatics. E.g., an H-donor on phenyl #1 on C2, an H-acceptor on the other ring, say on C2' may actually lower the torsion angle between the phenyls below the one only based on occupied volume by the substitutents. This later as an example.


text:
How do the stored chemicals in batteries drive electronic current from their terminal to outside devices?
I have got around to at least read on Wikipedia, and it does not clarify how this works in any sensible manner.
I wondered how the chemicals make electrons "move". How is the basic understanding of this?

A battery, at its heart, is what we call a redox cell: some compounds are stripped of their electrons, and others gain those electrons. The movement of electrons between these two compounds can be harnessed to provide electricity, but that's taking us a step too far. What we want to understand here is why the electrons move. To do this, we'll have to understand the redox cell. As much as I hate to assume things, you're going to have to understand a bit about ions and equilibrium to understand this explanation.
(Note: the explanation I'm about to use is heavily taken from the excellent Chemguide site, run by Jim Clark. I'll include a link at the end.)
Let's say I place a bar of magnesium in a water bath. I'll take a look at the magnesium bar a short while after I place it in the water. If I could visualize atoms and electrons, this is what I'd see:

Some of the atoms of magnesium leave the bar and become ions in the solution, solvated by water (and other things, but that's not crucial to this explanation). Since they become ions, they leave electrons on the metal bar. Because the ions are positively charged, they are attracted to the negative charge on the metal bar, and tend to hang around there.
If I wait a little longer, I'll see more and more ions coming off the bar. As they start to build up in the water around the metal, every once in a while, one of the ions will hit the bar and reclaim its electrons. This is what we'll see then:

If I wait long enough, eventually, the number of atoms coming off the metal will be equal to the number of ions reabsorbing onto the metal. If I take out the "bites" in the metal bar to simplify the situation, this is what I see:

Notice how there's quite a buildup of electrons on the magnesium. This is because magnesium is a very active metal, like lithium. What if we used a much less active metal, one that doesn't have a large propensity to lose atoms into solution? An example of such a metal would be copper. Here's a side-by-side of the two:

Because the magnesium is a lot more active, it tends to lose ions more, and so it has more electron buildup. If we connect the two with a wire, we give the electrons a way to flow from the high concentration to the low concentration. We can then place something along the wire in the middle (say, a lightbulb) and harvest some energy!

This, of course, is a hilariously oversimplified scheme. For one, as the electrons go to the copper, more and more magnesium ions break off to replace them. At some point, there are so many positive ions in the solution that they draw the electrons back toward them, through electrostatic attractions (which is a fancy way of saying that positive and negative charges attract each other). To solve this, we'd put a salt bridge between the solutions that lets ions migrate back and forth.
There are other problems too: for example, this setup wouldn't work in real life because there are no ions in solution: real cells use salt solutions of the metal e.g. $\ce{MgCl2}$ or $\ce{CuSO4}$. There's also the issue that this battery is liquid and wouldn't do so well if you tipped it over, the fact that the voltage (which is really the difference between the charge on the metal bars) is only about 1.1 volts, yada yada. That's not central to the explanation.
The fact that certain compounds are more likely to gain electrons and some are more likely to lose them is the central reason why things like batteries can exist. If you'd like to read more about it, I pulled this explanation from chemguide.co.uk, run by Jim Clark. I am currently ready to graduate with a B.Sc in chemistry, and his website is still one of the best resources for conceptual learning that I've come across---as a matter of fact, I learned about things on that website (eutetic mixtures and azeotropes) that I still have yet to come across in four years of university-level study. If you'd like to read more into redox reactions, the redox equilibria page is at this link right here.


text:
What is this brown liquid that I got when I boiled glycerol and water solution?
I boiled a solution of glycerol and water at about 160 degree Celsius and got a brown liquid much less in volume (boiled off the entire water I think; also not so sure about the accurate temperature).
Before mixing glycerol with water it was a colourless liquid, now it is golden brown! Why would that be? Also, I smelled burnt sugar during the last couple of minutes of boiling. So I'd like to know what that gas was. So, do I have glycerin in my hand or any other byproducts?

If your mixture reached 160 degrees C then you must have boiled off all the water. If there was some water left, it would stay at around 100 degrees C and all (most of) the heat would be used to evaporate water.
The boiling point of glycerol is about 290 degrees C and so you didn't boil that off. I disagree with @Skeptical Chymist about you making the caramel (for the sake of god, don't eat it to test it out). Not all that gets brown is caramel and there were no sugars there I presume. If i had to guess I'd say that at 160 C you've managed to completely obliterate any complicated organic structures and what's left is glycerol, charcoaled bits of any organic impurities that you happened to have in the starting material and possibly some acrolein which is a common product of dehydration of glycerol (you can recognize it by it's smell).
Overall; it's hard to tell what one gets, when they heat the hell out of something.

Apparently the process is worthy of a master's thesis.  I disagree slightly with the 100 degree comment - don't forget intermolecular forces and boiling point elevation.

@bobthechemist I never said that it's not an interesting topic, but also: compare the burner design from the thesis you linked to heating something in a pot. It is much less controlled and thus all sorts of stuff can happen. (By the way, I enjoyed reading the thesis. Thank you.) About the boiling point elevation: surely not by 60 degrees? (I don't have a vary good intuition on the topic so I admit: I don't know)

never intended to comment on your judgement of the topic, rather that I was surprised that someone wrote a thesis on it.  I agree, a 60 degree elevation due to intermolecular forces is unreasonable.  Further, since glycerol decomposes just below its boiling point, I suspect you are right in that contaminants in the glycerol are the culprit.

When heated to high temperatures, glycerol decomposes through dehydration to yield acrolein.
$$\ce{C3H3(OH)3 -> 2H2O + C3H4O}$$
Acrolein is a highly reactive unsaturated aldehyde.  It can polymerize all by itself into a yellowish material, and can probably react with any other impurities or materials in your sample, which is my guess for what caused the brown gunk.
One point of uncertainty is the temperature at which acrolein can form.  Wikipedia says decomposition starts at 280 °C, which is much hotter than the temperature you reported.
So even if there is not large-scale conversion of glycerol to acrolein, the lower temperatures you used maybe might have been enough for low levels of acrolein to form, which could react and cause browning.
A lot depends on how pure your water and your glycerol were.

In support of the line of thought «formation of acrolein», the synthesis of quinolines by the Skraup reaction (an aniline and glycerol as an easier to carry protected form of acrolein, some Bronsted acid like polymeric $\ce{H3PO4}$) basically is boiling off water, too.  (Somehow the question was shown today as «active», despite its age.)


text:
Testing for Iron-Gall Inks
I have to bottles of Lamy blue-black fountain pen ink. It turns out that some months ago the composition of the ink was changed by the manufacturer and there is no difference in the package.
Before, this ink used to contain Iron-Gall. Which makes the ink more permanent (old inks contained most of the times Iron-Gall). As Iron-Gall is an oxidizer, it has been removed from the ink because it can damage the pens if not cleaned regularly.
I don't know much about chemistry and I wonder if any of you could tell me a way to know if my ink has Iron-Gall and therefore is safe for long lifespan documents.
I was wondering if litmus paper would do the trick comparing this ink with normal black ink (which does not have Iron-Gall).

I have searched about it a lot but can only find this http://www.preservationequipment.com/Store/Products/Conservation-Materials/Other-Materials/Iron-Gall-Ink-Test-Paper .

Thanks. I tried with litmus paper comparing three inks.
1.Non Iron-Gall Ink
2.Iron-Gall Ink
3.The ink I don't know if it has Iron-Gall or not.
The result is that blue litmus paper turned red only for the second ink. So I guess the dubious ink weather has not Iron-Gall or has little quantity.

A simple and efficient method for the analysis and comparison of inks is thin layer chromatography (TLC).
In order to separate the various carboxylic and phenolic compounds that may be contained in iron gall ink, the mobile phase should be a polar solvent (e.g. ethyl acetate) or solvent mixture and made acidic (e.g. by adding formic acid).
The coloured spots can be compared under normal lighting; other spots may be detected under UV light. The phenolic compounds may be made visible by spraying with $\ce{FeCl3}$ solution.


text:
How do equilibrium shifts affect solids?
Say we have the following reversible reaction: $$\ce{NaOH{(s) }<=> Na^+{(aq) }$+ $OH^{-}{(aq)}} +10.6 \mathrm{kcal}$$
If we add $\ce{OH^-}$ and equilibrium shifts to the left, does that affect the amount of $\ce{NaOH{(s)}}$ present or does it remain constant?
I'm confused because I remember that pure solids and liquids don't affect equilibrium value, or does that only apply when we're talking about adding pure solids and liquids..not the effect on them by an equilibrium shift?

Note that your equation isn't valid, since $\ce{NaOH}$ is a strong electrolyte in water.  A better example would be something like $\ce{AgCl <=> Ag+ + Cl-}$

Strongly related question: https://chemistry.stackexchange.com/q/48600/5026

When a particular chemical process is at equilibrium, the respective rates of the forward and backward reactions are equal, meaning that no net change in the concentrations of products and reactants occurs and the composition of the reaction mixture remains stable. If $\ce{OH-}$ is added to a solution already at equilibrium, then there will be an excess of product relative to reactants and the rate of the reverse reaction will increase relative to the forward reaction until equilibrium is reestablished. This means that the ions will recombine into a crystal lattice and form a precipitate. So, to answer your first question, no, the amount of $\ce{NaOH_{(s)}}$ does not remain constant; more of it will be formed if additional ions are added to a solution already at equilibrium.
The reason why pure solids are not factored into equilibrium expressions is that they are not in fact part of the solution. Any excess precipitate, irrespective of the exact quantity, has no impact on the composition of the solution at equilibrium because it exists outside of the solution by definition. If equilibrium is disturbed due to changing conditions, then more precipitate can form, or some of the existing precipitate may dissolve back into solution. Under any given set of controlled environmental conditions, however, the product of the concentrations of the dissolved species remains constant at equilibrium, which is precisely why the equilibrium will shift if conditions or concentrations change.
For a more rigorous and mathematical explanation, you could read up on dynamic equilibria, equilibrium constants, and thermodynamic activity.


text:
What exactly is hydrogen bonding (and does it really need fluorine, oxygen, or nitrogen)?
I'm not satisfied with the rationale for the intermolecular attraction known as hydrogen bonding. In my book, it states that

Hydrogen bonding is a special type of intermolecular attraction between the hydrogen atom in a polar bond (particularly H ¬ F, H ¬ O, and H ¬ N) and non-bonding electron pair on a nearby small electronegative ion or atom usually F, O, or N (in another molecule).

It seems that chemists looked at the data and found they needed a 'fudge factor' to fit the higher boiling points of ammonia, water, and hydrogen fluoride.

Why doesn't hydrogen bonding apply to other atoms like Sulfur or Chlorine? They seem to be electronegative enough IMO. Is there an explanation that explains why this is, rather than saying that it's there? What makes Hydrogen so special?
Note: Chlorine is more electronegative than Nitrogen.

Hydrogen bonds are observed also for heavier atoms, but are generally weaker.

One thing to note: Chlorine is less electronegative than nitrogen. The $\ce{NCl3}$ molecule, when reacting with a chloride source releases $\ce{Cl2}$ showing that the chlorines in $\ce{NCl3}$ have a positive partial charge.

@Jan Wikipedia  says that the Pauling EN of Cl is more than N.

@user223679 And right underneath it says that Cl is less electronegative than N in the Allen scale. There are many electronegativity scales out there. I choose to believe the experiment, if it contradicts the theory.

The pauling scale (and even allen maybe) give only approximate values, which (my textbook says) can change depending on the element to which it is bonded

@Mithoron Can we also take an HSAB approach, and say that for functioning as an acceptor atom in an H-bonding system written as Dn–H···Ac ( where Dn represents donor and Ac represents acceptor) the proposed acceptor atom must be a hard base, as the hydrogen with which it is interacting is a hard acid? Then I guess we could say that hydrogen being a period 1 element, will be extremely hard, and can only form reasonable interactions with period 2 elements like F, O and N which are commonly prescribed as H-bond acceptors.Cl and S being period 3 elements would be too soft for us in this case.

@YusufHasan It's not so simple, as hydrogen propensity for hydrogen bonding depends also on "donor" atom - its hardness as ligand of acceptor highly depends on donor.

There are two main ways to look at hydrogen bonding.  The first is electrostatic, where the electronegativity of the atoms is used to describe the interaction.  Your argument about chlorine being more electronegative than nitrogen is a good one suggesting that the electrostatic argument is only part of the story, and there is at least one study that suggests that hydrogen bonding does occur with chlorine in polar molecules such as chloroform.
We can increase dramatically the number of atoms considered to undergo hydrogen bonding if we take a molecular orbital approach.  My reasoning here is a summary of what can be found in Inorganic Chemistry by Miessler, Fischer and Tarr.  They, in turn, rely heavily on a new definition of hydrogen bonding recommended by the IUPAC Physical and Biophysical Chemistry Division.
A hydrogen bond is formed when an $\ce{X-H}$ (where X is more electronegative than H) interacts with a donor atom, $\ce{B}$.  The attraction $\ce{X-H...B}$ can be described as consisting of the following components:

An electrostatic contribution based on the polarity of $\ce{X-H}$
A partial covalent character which arises from the donor-acceptor nature of the interaction
Dispersion forces

The first bullet is typically the only phenomenon discussed in General Chemistry classes, and this is not unreasonable since the second bullet requires the introduction of Lewis Acid/Base concepts which may not have been covered.
It is important to note that the new definition includes what I like to call "the proof is in the pudding" where the existence of hydrogen bonding requires experimental evidence, which can be found using a number of methods:

The $\ce{X-H...B}$ bond angle: a bond angle of 180 degrees indicates strong hydrogen bonding and would be accompanied by shord $\ce{H...B}$ bond distances.
A red shift in the IR frequency upon formation of $\ce{X-H...B}$
Hydrogen bonding results in deshielding of the H atom, which can be observed in NMR spectroscopy.
Thermodynamic studies should indicate that $\Delta G$ for the formation of $\ce{X-H...B}$ should be larger than the thermal energy of the system.

While electrostatics is the dominant contributor to hydrogen bonding, an analysis of frontier orbitals can also be insightful.  A thorough answer requires a fair amount of MO theory background, but for those interested, looking at the MO diagram of $\ce{FHF-}$ will be helpful.  In brief, the base ($\ce{F-}$ in this case) has a filled $p_z$ orbital that gains access to a delocalized orbital with lower energy, thus validating the formation of a strong hydrogen bond.

Is there a reason as to why hydrogen bonds are stronger than other dipole-dipole interactions?

This website (and also the definition the OP gave) mention "electron lone pairs" as an important criterion for the Hbond to be formed. Please share your comment on that. Thank you!

N, O, F have atomic numbers $Z=7, 8, 9$, respectively. Chlorine has $Z = 17$ which is much larger. Consequently, the atom is larger as well and it is more diffuse. Equivalently, the lone pairs in chlorine are at the 3-level which is too high. Because the hydrogen bond isn't a real bond but a dipole-dipole attraction and because the force between two dipoles scales like $1/r^4$, a larger atom implies a significantly weaker dipole-dipole attraction which is, in the case of chlorine, beaten by other forces. That's why hydrogen bonds are only observed with light atoms bound to hydrogen.

There could be a molecular orbital aspect to hydrogen bonding, see @bob's answer.  But this mode, too, favors more compact atoms like N instead of Cl.


text:
Cement strength by moistening
After cementing a wall, water is applied on the surface to keep it moist to maintain its strength for the next few days. Why is this necessary for maintaining the strength of the concrete?

If the surface dries prematurely, you get cracks. Then the inside of the crack dries, and it becomes deeper.

I'm not sure what you mean by "cementing" a wall. However, cement (typically Portland cement) is the paste that binds together the aggregate (gravel and sand) in concrete. The better the bond between the cement and the aggregate, the higher the resultant compressive strength of the concrete.
When cement is mixed with water in concrete it binds to the aggregate through a chemical reaction known as "hydration."
http://en.wikipedia.org/wiki/Hydration_reaction
A hydration reaction consumes water as a reactant, which is why you must keep concrete wet for several days after it is placed. If the concrete dries out then it has no water to continue the hydration reaction. The end result then is that you're left with cement that is not fully adhered to the aggregate (less compressive strength).
Careful though - you only need to keep the concrete wet after it has set up and is curing. Adding too much water to the mix (giving it a high water to cement ratio, w/c) actually drastically reduces the compressive strength of the concrete.
Curing compounds are especially helpful for properly curing concrete. (speaking from a heavy civil general contractor's point of view) We use this product all all the time:
http://www.wrmeadows.com/1100-clear-resin-water-based-concrete-curing-compound/

It's okay that you mentioned a product, but if you happen to  have an affiliation with it, please disclose that in the answer.

I have no affiliation with any products related to this post.

Awesome, thanks for clarifying and most of all for lending us your expertise!

V.shortly. Cement is a mixture of $\ce{CaO}$, $\ce{MgO}$ and $\ce{SiO2}$ (sand). The 4th component comes from air - $\ce{CO2}$. You need water for two main reasons. 1st is that you want metal oxides to react with $\ce{SiO2}$. They do it better if they are in form of hydroxide. Also some of the salts that are creted are hydrates (i.e. they have crystaline water). The 2nd reason is that hydroxides of both these metals easier react with $\ce{CO2}$ than oxides. One of the reasons is solubiltiy of $\ce{CO2}$ in water, the other, bigger renwewable reaction surface (reaction between solids are usually slower than between solid and liquid).

Thanks, but how could the reaction of metal oxides with SiO2 gives strength?


text:
Is the electronegativity value ever used in calculations, or is it just a number assigned to each element?
Does the electronegativity only qualitatively describe bond polarity, or does it have any quantitative use? Is it possible to describe electronegativity in terms of other periodic properties, like valence electrons, $Z_{eff}$, etc. (thus eliminating the need of a separately defined value)?

Actually, it does not have quantitative use. 2) No, or not easily. However, usually it is enough to know that it grows from down to top and from left to right with several additional relatiosn near top-right and top-left corners of the Periodic Table.

Aside from what permeakra said, electronegativity is not a constant property of an atom/ion. It changes in the different bonding environments and it is only defined for the use of covalent bonds. Mulliken introduced the 'absolute' EN, the mean of ionisation energy and electron affinity, thus formally getting rid of it's necessity.

True, with values of electronegativity in hand (and beside the by Pauling, there are others by Rochow, Mulliken, etc. to name some scales) you describe bond polarity. Beyond this number, infrared vibrational frequencies, chemical shifts in NMR spectroscopy and isomer shifts in Mössbauer spectroscopy are mentioned by the corresponding wikipedia article.

It totally depends at which level you are studying chemistry. If at high school, then practically you dont need the values as such. If, at a higher level then maybe, but still it doesn't have an extensive usage. As far as I have learnt, and at my level we learn Electronegativity only to understand the relative acceptance of electrons by different elements. We have the Mulliken, Pauling and Allred Rochow scale out of which the Pauling scale is famously used. You should know the varying trend of Electronegativity acrros the period and down the group. And as such, Electronegativity doesn't have exceptions as compared to the other periodic properties.
The most important use of  Electronegativity is finding out the character if compounds. Where you determine their ionic character. You also realise the Bond strength and Bond length based on the difference of Electronegativity of the two elements(delta EN). You can also understand whether the compound will have polarity/partial polarity or not. And there is a formula to it also where the Ionic character %= 3.5 (delta EN)^2 +16(deltaEN).
This is how much I know about Electronegativity. Hope your doubt has been cleared. :-)


text:
How are pure substances (elements, compounds, etc) produced?
How are large quantities of the substances we use in the lab generally produced? (As in the process from getting the stuff out of the ground to use in the lab) How 'pure' are they? Also: is it possible to create relatively pure substances with household goods?

This is really more of a Physics question - chemistry is what we do to it afterwards.

@medivh Perhaps more so engineering than physics.  The OP needs to be a bit more specific: are we talking petroleum based products? Organics? Metal salts? And regarding household goods one can be more specific as well; what household goods do you have in mind and what would be the purpose of doing so?

@medivh I strongly disagree. Chemical synthesis is certainly a major subfield of chemistry and on topic here. The problem is that this question is impossibly broad.

No lab reagent is 100% pure. But you can get really close to that.
How 'pure' are they?
Read the label or do the analysis. It varies A LOT from one reagent to another.
Is it possible to create relatively pure substances with household goods?
Depends on the substance, depends how pure you want it. For instance, you can very easily distill water at home, but pure water is hard to get.
How are large quantities of the substances we use in the lab generally produced? (As in the process from getting the stuff out of the ground to use in the lab)
You'll need to be more specific. There are entire classes you can take at the uni on the large scale production and purification methods of this and that type of reagents. I see you used the ionic-compounds tag, but that's far from specific. A more answerable question would be "How are compounds X and Y produced and purified?"

How are large quantities of the substances we use in the lab generally produced?

This is answered reasonable well on Wikipedia

Also: is it possible to create relatively pure substances with household goods?

Depends. Recrystallization can be performed at kitchen, so some substances can be purified at home. However, it is easier to buy pure substance in most cases.


text:
Why are there multiple lines in the hydrogen line spectrum?
Hydrogen has only one electron, yet it exhibits multiple lines in a spectral series, why is this?

Have a look at this link and if that doesn't answer your question, consider editing your question to make it a bit more specific.  The short answer is that the one electron can transition to/from multiple energy levels.

Yeah, another man how didn't use google before asking a question here. Have your link http://en.wikipedia.org/wiki/File:Hydrogen_transitions.svg

Related: http://codegolf.stackexchange.com/q/61371/45265

This is a 6 year late reply: in response to @permeakra, a google search brought me here :)

As the electrons fall from higher levels to lower levels, they release photons.  Different "falls" create different colors of light.  A larger transition releases higher energy (short wavelength) light, while smaller transitions release lower energies (longer wavelength).
The visible wavelengths are caused a by single electron making the different transitions shown below.  There are even more transitions that release invisible wavelengths.
 Wavelength   Transition  Color
 (nm)
---------------------------------------------
 383.5384      9 -> 2    Ultra Violet
 388.9049      8 -> 2    Ultra Violet
 397.0072      7 -> 2    Ultra Violet
 410.174       6 -> 2    Violet
 434.047       5 -> 2    Violet
 486.133       4 -> 2    Bluegreen (cyan)
 656.272       3 -> 2    Red
 656.2852      3 -> 2    Red

*Values taken from Hyperphysics: Hydrogen Energies and Spectrum
(Why are there two different 3-> 2 transitions?  See here: Hydrogen Fine Structure )

Though a hydrogen atom has only one electron, it contains a large number of shells, so when this single electron jumps from one shell to another, a photon is emitted, and the energy difference of the shells causes different wavelengths to be released... hence, mono-electronic hydrogen has many spectral lines.

Although hydrogen has only one electron, it contains many energy levels. When its electron jumps from higher energy level to a lower one, it releases a photon. Those photons cause different colours of light of different wavelengths due to the different levels. Those photons appear as lines. For this reason, though hydrogen has only one electron, more than one emission line is observed in its spectrum.

This arises due to the electron making transitions between the various energy levels in an atom . The electrons fall from a higher energy state to a lower energy state or vice versa emitting or absorbing wavelengths of different magnitude which fall directly on the spectrum of hydrogen thereby releasing colours of various wavelengths and creating a continuous spectrum .

Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the [help]. Do you want to know more about this site, please take the [tour]. || Referring to your answer: A line spectrum is of course not a continuous spectrum, as the title already says.


text:
Ca and Mg hydrated radius in clay flocculation
Ca has an atomic radius of $197\mathrm{~pm}$ and a Pauling electronegativity of $1.00$.
Mg has an atomic radius of $197\mathrm{~pm}$ and a Pauling electronegativity of $1.31$.
Since Mg is smaller and has a higher electronegativity, one would think Mg would have a smaller hydrated radius than Ca, but it doesn't.  Would someone explain to me why this isn't so?  I have read an explanation pertaining to charge density, but I'm not understanding why Ca would have a larger charge density if it is bigger with less electronegativity.
The goal of understanding this is to understand why the addition of Ca to soil will make it drain better and be looser, less sticky, etc.  And conversely, why highly respected agricultural consultants say soils high in Mg (low in Ca) are "sticky".  Any help in understanding this would be well appreciated.

Charge density is a measure of electrical charge per volume. It is independent from electronegativity, and increases with the number of charges of an ion and with decreasing radius, because a smaller ion has also a smaller volume. The number of charges for $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ is the same (+2), but $\ce{Mg^{2+}}$ has the smaller atomic radius. Its charge is therefore distributed over a smaller volume, and the charge density is higher.
The hydrated radius of an ion can somewhat differ from its atomic radius, because it is influenced by the charge density. If the ion charge is concentrated in a smaller volume, it exerts stronger attractive forces on the dipoles of water molecules, which results in the formation of a larger hydration shell. The smaller $\ce{Mg^{2+}}$ attracts more water dipoles and forms a larger hydration shell. Hence, its hydrated radius can be bigger than that of $\ce{Ca^{2+}}$, which has a larger atomic radius but a smaller hydrate shell.


text:
Soil-solution cations in competition for exchange sites
How could one determine the likelihood of one cation displacing another on a negatively charged exchange site of a clay particle?
For example, in a soil solution containing equal parts (by number of atoms) $\ce{Ca^2+}$, $\ce{Mg^2+}$, $\ce{K+}$, $\ce{Na+}$, $\ce{Al^3+}$, which cations will find the most exchange sites and which will ultimately be washed deeper into the soil? (Assuming there are not enough sites for all of them.)
Could two $\ce{Na+}$ cations displace an adsorbed $\ce{Ca^2+}$ cation simply because they "bump" into it? Or is there some mechanism where cations are dislodged? Is it easier for two $\ce{Na+}$ to replace $\ce{Ca^2+}$ or for $\ce{Ca^2+}$ to replace two $\ce{Na+}$?

Only half as many Ca2+ are needed as Na+ to neutralize the charge of the ion exchanger (here: clay).
The affinity of clay for bivalent ions is larger than that for monovalent ions:
Ca2+ > Na+
For most clays, the affinity for equally charged ions increases with the nonhydrated radius of the ions:
Cs+ > K+ > Na+ > Li+
Ba2+ > Sr2+ > Ca2+ > Mg2+

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


text:
Why do liquids and solids have constant concentrations?
I don't understand why liquids and solids have constant concentrations. Can anyone explain this to be in simple terms?

I know there are already some good answers but would you please provide a bit more context -- where will you be using this?

@EashaanGodbole The OP has not been active in a while, so maybe I'll answer: This comes up when writing equilibrium constant expressions. It is also relevant in the kinetics of reactions where the solvent doubles up as reactant.

If you think about what concentration, it usually refers to the molar concentration in $\mathrm{mol/l}$ (at least in the field of chemistry). To calculate this concentration we need to find how many moles of a certain chemical are in 1 liter of the liquid or solid.
I will start with an example and describe the general principle afterwards. Let's say we have pure water at room temperature and atmospheric pressure. This means that the density is $998\,\mathrm{g/l}$. We also know that the molar mass of water, consisting of 2 hydrogen and 1 oxygen, is $18.016\,\mathrm{g/mol}$. If you look at the units of the molar mass and the density you can already see that we can combine these into something that has units $\ce{mol/l}$. If we do that we get that the concentration of water is $\frac{998\, \mathrm{g/l}}{18.016\,\mathrm{g/mol}}=55.4\,\mathrm{mol/l}$. So the concentration of water is $55.4\,\mathrm{mol/l}$. Notice that I have only used the density and the molar mass of water, which are (at least for equal temperature and pressure) constant.
In general you can do this for any solid or liquid (or gas) and find the concentration as $c=\frac{\rho}{M}$, where $\rho$ is the density, $M$ the molar mass and $c$ the molar concentration. For a given chemical, the molar mass is constant.
Now about the ‘fact’ that the molar concentration is constant for liquids and solids: strictly speaking it is not. That is because the density of liquids and solids is temperature and pressure dependent. The reason that the molar concentration is often called constant is twofold:

the density of liquids and solids has a much weaker dependence on the temperature than gases do therefore it can be regarded as approximately constant
if you look at solutions, the variation of concentration can be
orders of magnitude, whereas for e.g. water the concentration
variation when going from $0$ to $100\,\mathrm{^\circ C}$ is only a few percent.
Therefore, again, the concentration of the liquid is approximately
constant.

I think you muddy the waters with the temperature and pressure dependence of density, especially since these concepts apply equally well to gasses.  The key is in your first paragraph: concentration is an amount divided by a volume.  For both solids and liquids, divide the amount by some number and you divide the volume by the same number (so concentration stays the same).  One can change the amount of a gas (or solute) without changing the volume it occupies.

@bobthechemist - I agree that it makes the answer a bit longer and more complex, but I didn't want to answer the question without pointing out that the premise of the question is not really correct. I am well aware that for gases the density is also dependent on (and a much stronger function of) temperature and pressure, but to assume that the density of liquids and solids is constant is plain incorrect. Strictly speaking your last statement is also true for solids and liquids as long as you go to sufficiently high pressures.

@Michiel I must say it’s one of the greatest answer I have ever seen. Your answer is really marvellous. The only thing that confuses me is why do we say that some solution is concentrated solution of $CaO$ and the other one is diluted solution of $CaO$. What does concentration means in these contexts?

In liquids and solids, particles have intermolecular interactions that keep them at a certain distance. As a result, the density of liquids and solids is not too sensitive to pressure and temperature changes.
The concentration (technically: amount of substance concentration) of a pure solid or liquid is defined as the chemical amount of the substance per volume of the sample. You can calculate it from the molar mass and the density of the substance:
$$ c = \frac{n}{V} = \frac{m / M}{m / \rho} = \frac{\rho}{M}$$
with $\rho$  the density of the sample and and $M$ the molar mass of the substance. For liquids and solids, the concentration does not depend much on pressure or temperature.
For gases, on the other hand, the concentration is proportional to the pressure and anti-proportional to the temperature (ideal gas law). For substances in solution, the concentration is variable because you can add more or less solute to the solvent.
If you consider a chemical reaction at constant temperature and pressure, pure solid or liquid reactants and products will not change their concentrations, whereas reactants or products that are gases (if the reaction is in an enclosed space) or are in solution typically will.

In simple terms, under normal conditions, one can only "fit" a definite amount of a solid or a liquid in a definite spacebut you can fit a
variable amount of a gas in a definite space. Think about the large amount of compressed gas that can be stored in deodorant spray cans, but on the other hand a glass of water, or a brick can only occupy an unchanging amount of space.
Since, in chemistry we're concerned only with molecules, we express amount of a substance by moles. Thus the concentration of a substance  [ (amount in moles)/(volume) ], pure solids and pure liquids have definite a constant concentration, for a given volume, whereas gases have a variable concentration.

My explanation is that the molarity (concentration and also the activity in case of an ideal solution) of a pure liquid (or solid) won't change considerably in a solution (in which a chemical equilibrium exists) even if the concentration of the solute in the binary solution with the said pure liquid changes. The reactants of the solvation equilibrium are the pure solvent and solute and the product would be the solvated solute. Hence their concentrations are constant and therefore their activities (a thermodynamic concept which means concentration in case of ideal solutions) can be taken as unity itself instead of its actual concentration in the binary solution. Also, the molarity of a pure solid or liquid doesn't affect the concentration of the solute/electrolyte present in the solution and remains constant giving rise to an active mass (but not the concentration and, here, the active mass is what we are actually interested in) of unity which does not affect the equilibrium of the solution.

Can you expand your answer enriched with a real example?

Vaguely,
the reason is the incompressibility of solids and liquids. They cannot have variable volume for the same amount of substance. When their amount increases, so does their volume and when amount decreases, volume decreases too in such a way that this amount to volume ratio (concentration) remains constant.
However, their concentration does change, but the change is quite small and hence negligible.


text:
Computing of strengthening of π-interaction
I need some practical advise. What would be the easiest approach to compute strengthening or weakening of π-interaction in a bond with multipole-bond character. I'm not really fond of going through coefficients of Kohn–Sham orbitals. I did it on occasion, but it's a lot of work and I just do no have time. Unfortunately the molecules I need to compare have low symmetry ($\ce{C1}$) so I can't make use of symmetry to divide $\sigma$ and $\pi$ contributions. I'm not quite sure but I remember to read that the energy decomposition analysis in ADF could be used to divide in $\sigma$ and $\pi$ components. Let me make it clear, I'm not interested in comparing bond lengths, I need quantitative measure for the strength of $\pi$ interaction: either energy component or relative contribution of $\sigma$ and $\pi$ components orbitals.
I welcome any suggestions concerning methodology and/or software that can help.

First off, definitely check out ETS-NOCV, which I believe is a fairly recent addition to ADF which seems to do exactly what you're interested in. Other NBO-esque software may also suit your needs - I seem to recall NBO5 in Gaussian can generate sigma and pi natural bonds.
Regarding doing things the energy decomposition way:
If you have not already seen it, this paper goes about doing (in ADF) what you want to do:
http://pubs.rsc.org/en/Content/ArticleLanding/2012/DT/c2dt31845h
(disclosure - I am a member of the Stranger group)
I did this for a related system a while back but have forgotten the exact details, however fragment bonding analysis is a nifty technique that allows estimation of the energy of single bonds in a molecule as well as decomposition by bond type.
Broadly, you cut your molecule into chunks on either side of the bond you're interested in and generate fragment orbitals from them. (that is, you use the TAPE21 files for these molecular fragments as a basis for the calculation of the full molecule). These fragment orbitals subsume all of the interactions within the fragments so that the total bonding energy only consists of interactions between the fragments.
From here, you can actually remove atomic orbitals from the fragment orbitals. If you prune the orbitals relevant to pi bonding, you can estimate the pi and sigma components by relative energy.
Here's where you should start: ADF manual - Analysis

Thanks. The ETS-NOCV sound interesting. I'll check the paper you cited. I was using the fragment approach but I was not aware that you can also make a decomposition by bond type. The part I do not understand is "you can actually remove atomic orbitals from the fragment orbitals". I think I need to do my homework (e.g. rtfm).

@Kris_R - Sorry, I believe what I said was incorrect - I'm pretty sure you remove molecular orbitals from the fragment orbitals. When I did this, I was looking at d to pi* donation to N2. This involved removing the relevant antibonding orbitals on the N2 fragment to preclude the interaction. I don't recall the specifics.

For clarity, NBO is NOT a quantitative analysis.  It is merely qualitative.  In fact, it would be tough to convince me that most of these energy decomposition schemes are quantitative.  The best thing you can do is use a variety of methods in tandem but don't take NBO E2 values or whatever and treat them quantitatively.


text:
Sulfuric Acid Resistance / Conductivity
I diluted sulfuric acid in deionized water in order to get a 10% $\ce{H2SO4}$ solution.
I was going to use this solution as a cathode of a porous sample to measure its electrical properties.
However, this solution has a quite high resistivity (~200 ohm.cm DC).
I checked several sources that tell me that I should be measuring a lower resistance.
Does anyone has an idea of what may be happening?

Welcome to Chemistry.SE. Could you provide more information about the "several sources" you mention? Links would great if possible. I have also edited your question to make the chemical formula of sulfuric acid look like $\ce{H2SO4}$. You can learn how to do this for yourself from the [faq].

Give all details of your set-up including electrode and the ohmmeter.  Did you check for an breaks in the wiring?


text:
Can a polymer have different resonance structures?
Several organic compounds have resonance structures, but can a polymer have resonance structures?

There's no reason why the structural subunits of polymers could not exhibit resonance, provided there are conjugated $\pi$-bonds present. A polymer such as kevlar, for example, certainly benefits from resonance stabilization within each monomeric unit.
Whether there exist any polymers in which the resonance is truly distributed over the entire molecule (if that's what you're really asking), however, I'm unsure. From the perspective of valence bond theory, the electron delocalization required for resonance necessitates a planar geometry with unhybridized $p$ orbitals that align and overlap to form $\pi$-bonds perpendicular to the plane of the $\sigma$-bond skeleton of the molecule. In molecular orbital theory, the sort of extensive electron delocalization that resonance implies requires a bonding molecular orbital, distributed over all participating atoms, constructed from a set of symmetrical, in-phase $p$ orbitals, with all planes of symmetry present in the original molecule(s) conserved in the generation of new molecular orbitals.
Given the rigidity of those criteria, I sincerely doubt that polymers of any great length could exist in which the resonance is distributed over the entire polymer chain. However, that could be just a failure of imagination or a lack of knowledge on my part, and I hope somebody else weighs in with a more conclusive answer.

In conjugated polymers, there is usually a finite conjugation length. That is, based on various properties, we know the delocalization is not infinite, even with fully planar, aligned chains like polyacetylene. Experimentally, vibrations and bond rotations make it unlikely to have very long aligned $\pi$ systems. Even in most electronic structure methods, there's a finite conjugation length, although it can be quite long.

@GeoffHutchison, thanks for your comment. At the time of my answer, my suspicion was that there were many practical reasons (variously physical, chemical, and thermodynamic) why the possible extent of $\pi$-conjugation in macromolecules is finite. On the other hand, I can't find any a priori line of reasoning as to why it must be limited (at least under some idealized set of conditions where some degrees of freedom are inaccessible and/or large energetic barriers to things like bond rotation exist).

I rather like Greg E.'s answer. I'd like to add that there is at least one polymer that has resonances distributed over the entire chain: polyacetylene. When doped with iodine, its conductivity can reach values comparable to silver.
I feel like this answer might not have been what you were looking for, since I
usually think of resonance forms as being distinct from each other, whereas for polyacetylene, the resonance forms look pretty much the same (just shifted over one bond).
https://en.wikipedia.org/wiki/Polyacetylene

The conductivity of iodine-doped polyacetylene is several orders of magnitude lower than that of silver. 4000 S/m against 60 million S/m.

I feel it's important when considering this question to remember that resonance forms are a model, not an observable phenomenon -- they serve to plug the gap between observable electronic effects and Lewis structures.
Remembering that, it should be reasonably easy to see that multiple strengths and regions of conjugation, representable by more than one resonance form, could exist at various effective ratios and levels of robustness (according to chain deformation) in polymeric structures.
Consider a simple polyacetylene, for example: ${H_2C}\left({CH}\right)_n{CH_2}$. In the ideal case, this would have perfect conjugation along the entire straight chain, representable by two resonance forms. In practice, the electron density is asymmetric where the molecular geometry is also, which can be represented as additional resonance forms with charge separation.


text:
What chemical available in 18th century would explosively react with phosphorus?
For fictional use, I'd like to "construct" a clockwork device featuring phosphorus which could reasonably be expected to:
A) glow consistently.
B) explode upon some trigger (e.g., influx of air, mixture with another chemical, a spark), with enough explosive force or incendiary strength to kill someone standing very close.
Constraints:
1) this takes place in the early 18th century, so phosphorus has been discovered, but most of the modern types of phosphorus used in weaponry have not.
2) the quantity is no larger than a couple of liters.
Progress:
I've read the relevant chapters in Emsley's 13th Element, and then moved on to the primary sources for the time – mostly records of demonstrations held for the Royal Society and Boyle's Aerial Noctiluca.
It's hard to tell exactly what sort of phosphorus any of these early chemists were working with, but it looks like I can achieve goal (A) by just letting in a bit of air now and then (based on Boyle's description / Wikipedia on oxidation).
Goal (B) is more difficult. Is just oxygen enough?
Slare added sulfuric acid ("oil of vitriol") and achieved something, but it doesn't seem to have been an explosion exactly:

being well shiken together did first heat, and then throw up such fiery Balls, which like so many stars will adhere to the sides of the Glass and continue to burn for some time

I think if you just add oxygen you'll just end up with a lump of burning phosphorus. I would suggest having a timed explosive charge at the centre of a phosphorus payload to essentially achieve something like a white phosphorus bomb.
Since white phosphorus is pyrophoric, you only need to disperse it for it to ignite. To do this, you could either use an automated striking mechanism to ignite a fuse to something like black powder, or my Improvised Munitions Handbook suggests a combination sulfuric acid/potassium chlorate or perchlorate (note: not chloride) + sugar device, with the mechanism keeping the two components separate, would be suitable.
The main problem with your design specifications as I see them is having an excuse to have the external surface of phosphorus exposed to the air: white phosphorus will only glow from the air-exposed surface, so if you put it behind glass, for example, it won't glow. A fine wire mesh as your external covering might do the job, but I'm not sure exactly what effect that has on reliability or effectiveness (and may cause the device to spontaneously and accidentally catch fire, especially if not packed properly).
Near explosion + being covered in burning white phosphorus = much sadness, probable death. See: White Phosphorus - Effect on People.
(Usual disclaimer: I am not a munitions designer.)

Re: "exposed to air" – I was thinking of packing the phosphorus inside two nested, rotating, perforated glass spheres, to allow regulated, small amounts of air in, as the pores came into alignment.

@Wajurgmitr: The reaction takes place in the gas phase above the surface of the solid, so bear in mind that you'll lose the glow anywhere there isn't a gap for the vapour/air mix to occupy.


text:
What reaction took place?
Okay, so I was fiddling about in the chemistry lab, and I added:
Sodium Nitroprusside (soln) $\ce{Na_2[Fe(CN)_5NO]}$ with Potassium Ferrocyanide (soln) $\ce{K_4[Fe(CN)_6]}$.
$\ce{Na_2[Fe(CN)_5NO]}$ was a yellowish-reddish colour - actually don't remember the exact colour, while $\ce{K_4[Fe(CN)_6]}$ was pale yellow.
Nothing really happened to the mixture - except the colours mixed giving a weird yellow colour. I assume the only thing that happened was a double decomposition reaction:
$\ce{K_4[Fe(CN)_6] + Na_2[Fe(CN)_5NO] \rightarrow K_2[Fe(CN)_5NO] + Na_4[Fe(CN)_6]}$
or something like that.
Next I added a bit of dil Hydrochloric Acid. I expected nothing to happen - nothing ever happens with the chemicals they keep in labs. :P However, the colour of the mixture turned a beautiful Emerald Green!
Now I've been trying to figure out what exactly occured to produce the green colour, but I don't know. Could anyone help me?
$P.S.$ My friend's gonna send me a picture of the green we took. I'll edit it to the question when I recieve it, maybe it will help?

It should be noted that this is a potentially very dangerous experiment.  Addition of a strong acid (note that a dilute strong acid is still a strong acid) to ferrocyanide can lead to the generation of hydrogen cyanide gas, which is extremely toxic.

Whoa. :O I did NOT know that. Thankfully, HCN wasn't formed. Phew. Thanks for the info! :D
P.S. Any idea why the green in colour was formed?

sodium ferrocyanide is green in colour. see this:-https://www.google.co.in/search?q=sodium+ferrocyanide&biw=1360&bih=667&source=lnms&tbm=isch&sa=X&sqi=2&ved=0ahUKEwiGvI6ipaLMAhXJLpQKHT3zA2IQ_AUIBigB&dpr=1


text:
sodium persulfate microwave heating cause crystallized particles
I use Sodium persulfate to etch a copper board to produce a PCB. I mix 1 dl sodium persulfate with 1 liter hot water. When the water get cold, I heat it up by microwave. After the microwave, I see such crystallized particles at the bottom of the jar. Please note that there were not any solid particles in the acid before the microwave. I wonder what are these particles? is it sodium persulfate? if it is so, why it is crystallized when it is heated up?
Solubility of water increases as the temperature of the water increases, right? Then why it causes solid particles in the car?

Btw, if I put these crystallized particles into the same solvent, they don't dissolve, however, if I put them into a new hot water cup, they dissolve.
If it is sodium persulfate and if they are crystallized because of the microwave, can't I reuse them again?
If it is copper particles, isn't it supposed to be in such color?

There are some confusing statements in your post. For example, you describe dissolving sodium persulfate in water and then you mention acid later. Did you dissolve it in water or acid (and which acid?)? That matters.

It looks to me like copper sulphate ($\ce{CuSO4·nH2O}$). If I'm right then you can't use it for etching. For this you need oxidant (persulphate). Try to solve this solid in water. You should get light blue solution. Stick into it a nail (i.e some metal iron source). If it cover itself with a brownish powder like in wiki-link you gave, then it's a copper(+2) salt.

thanks for the answer. do you have any guess why I get crystal particles after I heat up the liquid by microwave? There were no such solid particles when the solution was cold

Usually warming improves solubility of simple inorganic salts, however higher temperature generally increases reactions rate. That also includes building of crystals. If you leave your solution long enough (over night) there is a good chance that you will see the same solid. Probably the only difference would be that crystals would be bigger and nicer shaped.


text:
Lower Heating value calculation from chemcial combustion equation of Butane
QUESTION
    Butene, C4H8 is burned in an engine with a fuel-rich air-fuel ratio. Dry  analysis of the exhaust gives the following volume percents:                   $\ce{CO2} = 14.95%$, $\ce{O2} = 0%$, $\ce{CO} = 0%$, $\ce{H2} = 0%$, $\ce{C4H8} = 0.75%$, with the rest being $\ce{N2}$. Higher heating value of this fuel is 46.9 MJ/kg. Write the balanced chemical equation for one mole of this fuel at these conditions. Calculate the air-fuel ratio, equivalence ratio, lower heating value of the fuel, and energy released when one kg of this fuel is burned in the engine with a combustion efficiency of 98%.
My attempt at solution:
I write the stoichiometric equation like so:
$$ \ce{C4H8 + \frac{1}{\phi}(4+(8/4))(O2 +3.773N_2)\rightarrow n_1 C4H8 +n_2 CO2 +n_3 H2O}$$
where, $\phi$ is the equivalence ratio(a engineering term used to describe the ratio of $(F/A)_{\text{actual}}$ to $(F/A)_{\text{stoichiometric}}$.
So,now I have 3 unknowns, $n_1$, $n_2$ and $\phi$.
I unfortunately get $n_2=0$ after doing the mole balance of carbons,hydrogen and nitrogen.
I understand that the percentages given by volume are the mass fractions which should later form a separate equation. But, I still get $n_2=0$. Where have I gone wrong?(Should I still consider putting in $\ce{O2}$ and $\ce{CO}$ into the above equation even though their %by volume are zero?)

I have edited your question to make your chemical formulas equations look prettier. When using MathJax, enclosed chemistry in \ce{...} to make it look like chemistry.

I see three flaws in your reasoning.
First, butane is $\ce{C4H10}$. The formula $\ce{C4H8}$ refers to several isomers of butene. I will keep using your formula, but you should specify whether you mean butane or $\ce{C4H8}$.
Second, you know the relationship between your variables $n_1$ and $n_2$ because you have the volume percents of $\ce{C4H8}$ and $\ce{CO2}$ in the exhaust. Thus, you have fewer variables.
$$\dfrac{n_1}{n_2} = \dfrac{0.75}{14.95}$$
Third, I am not sure why $\ce{N2}$ is in the equation. It does not participate in the reaction, so it should not be there. Likewise, you do not need to add $\ce{CO}$ and $\ce{H2}$ because you know that none is formed. You can balance the equation and determine the stoichiometric coefficients of $\ce{CO2}$ and $\ce{H2O}$ from the conservation of mass:
$$\ce{C4H8 + 6O2 -> 4CO2 + 4H2O}$$
To include one of your variables, we assume that the reaction does not run to completion, and that some amount $n_1$ of $\ce{C4H8}$ remains unreacted. Since each 1 equivalent of $\ce{C4H8}$ consumes 6 equivalents of $\ce{O2}$, if $n_1$ of $\ce{C4H8}$ is not consumed, then $6n_1$ of $\ce{O2}$ is not consumed. Likewise, $4n_1$ of $\ce{CO2}$ and $\ce{H2O}$ were not produced.
$$\ce{C4H8 + (6-6n_1)O2 -> n_1 C4H8 + (4-4n_1)CO2 + (4-4n_1)H2O}$$
Now, I have rendered your problem down to one variable: $n_1$. Your other variables can be written as equations involving $n_1$ by comparing your chemical equation to mine (they are equivalent). You have enough equations in $n_1$ and $n_2$ to solve for $n_1$, $n_2$, and eventually $\phi$, which then leads to the values you really want.

authors: علي عباس لفتة