text:
Correlation between Coulomb's law and VSEPR theory
I've always been told in chemistry lessons that lone pairs repel more due to a greater charge density than bonding pairs. And that makes sense when steric effects are involved when bond orbitals overlap.
But this conflicts with what I have learnt in physics i.e. Coulomb's law, which states that the only thing affecting the force between two charged bodies is the magnitude of the charge and the distance between them. Therefore it seems weird to me that a lone pair should "repel" more because of greater charge density, despite having the same charge magnitude as a bonding pair.
What am I neglecting or if I'm right, does that mean for large atoms (such as iodine, where electron pairs are further apart) the lone pairs and bonding pairs repel equally as per Coulomb's law?

Let me elaborate on my question. Coloumbs law states that two charged particles of charge +q and distance d repel with force (k * q^2 / d^2 ) where k = 1 / (4pi8.85*10^-11). The 'volume' of the two charged bodies and thus their charge densities are irrelevant provided the distance between their centres are constant .

I guess you meant VSEPR. It is a very useful tool to explain certain geometries. For predictions I would not rely on that theory as one has to use other crude approximations like hybridisation.

In bonded pairs of electrons, the repulsion of the negative charges is somewhat reduced by the positive charge of the bonded atom's nuclei. Since lone pairs don't have to deal with this positive charge, naturally their repulsion is stronger.

Makes perfect sense and fits with Coloumbs law! Thank you for clearing up this matter of confusion and I can't believe I didn't think of that.

The charge magnitude is the same in both cases; however, the sizes of the orbitals are different. The lone pair electrons are in the field of one nucleus, while the two electrons of a bonding pair are in the field of two nuclei. As a result, the orbital with the bonding pair tends to be smaller, and it requires less space than the more diffuse orbital of a lone pair. Therefore, the distance between bonding pair orbitals is larger. This reduces the repulsive force between them, because according to Coulomb's law, the electrostatic force $F$ between two charges $q_1$ and $q_2$ is inversely proportional to the distance $r$ between them.
$$|F|=k_{e}\frac{|q_1q_2|}{r^2}$$

Interesting, @Jannis, you state that the bonding pair has a higher density and the OP state the lone pair has the higher density. I think a reference is needed here.

This is what in found in my textbook (Riedel, Inorganic chemistry) and what seems logical to me.

Everything you have stated is fundamental to VSPR theory but you haven't explained why lone pairs repel more since as i explained in the OP the repulsion is subject to coulombs Law which suggests charge 'density' to be irrelevant in repulsion.

The lone pairs repel more because their electrons are contained in larger orbitals. Larger orbitals tend to be closer to each other, especially when they belong to the same atom. Since the distance between charges is considered in Coulomb's Law, this would explain the higher repulsion.

Lone pair-Lone pair repulsion $>$ Lone pair-Bonding pair repulsion $>$ Bonding pair-Bonding pair repulsion.

Lone pairs are attached to same atom, they are free. As expected, they should repulse to greater extent. Bond pairs are attached to two different atoms, the heavy atoms attached to them might make them not repulse to greater extent. Now, lone pair-bond pair repulsion should be intermediate, as they have one lone pair and one bond pair.

For better understanding, look at the images of this page, and remember I have just guessed that this might be reason, as SE aims to have exact answer, I would be happy to know any errors, and will for sure try to correct it.

Your answer states the fundamentals of VSPR theory but doesn't explain why lone pairs repel more. Being 'free' doesn't suggest to me any reason to repel more. Please refer to Coloumb's Law which would suggest that charge 'density' should be irrelevant.


text:
C + O2 is equal to C + O, how is that possible
C + O = CO2 . This is because carbon has valency of 4 while oxygen has valency of 2. When they react the valencies are criss crossed which means we will get C2O4 but this is simplified to CO2.
But my textbook say that C + O2 = CO2.
So how is it possible that C + O2 and C + O both get the same product of CO2?
Also you can say that the reaction of C + O = CO2 is not balanced but the balanced chemical reaction would be C + 2O = CO2 not C + O2 = CO2 which is the reaction that my textbook says.

Your question is a jumble of text, and it really is not quite clear what you're asking right now. I suggest you make some edits to it.

I have explained a bit further. hope it is more clear now.

@AbhishekMhatre Only diaatomic oxygen and triatomic, ozone, exist in nature.

As stated Jun-Goo is not true that $C+O =CO_{2}$ you should rearrange the question...

@GM Heh, thanks. :) Will do when I am done with classes in a couple of hours. There's been a lot of good questions popping up lately.

@Jun-Goo Kwak, This is not true. Atomic Oxygen O exists in nature (atmosphere) but it is strongly reactive. Dioxygen O2 is more stable but it is still reactive. N2 at standard condition is quite stable and unreactive. Anyway to answer to Abhishek Mhatre, you cannot use equal sign if you do not weight compounds to ensure atomic balance. This is what means an = in a chemical reaction.

@jlandercy If it is so reactive, wouldn't it just react with itself to form dioxygen?

From NASA, "atomic oxygen is an elemental form of oxygen that does not exist in Earth's atmosphere." Link: http://www.nasa.gov/vision/earth/everydaylife/AtomicOxRestoration.html

@Jun-Goo Kwak, because a compound it strongly reactive does not mean it does not exist. It implies that it is difficult to isolate it or store it. According to JPL Data Evaluation (NASA,http://jpldataeval.jpl.nasa.gov/), there are several reactions including Atomic Oxygen (radical).

@jlandercy Well yes, it can be synthesized in the laboratory. But it can't be found and isolated in nature is what I was trying to say.

@Jun-Goo Kwak, yes natural is a difficult word in that situation. Anywya the answer bellow is quite well written even if it does not exaclty answer the first question.

@jlandercy Ah yes! Thank you very much. I added a small section to the end which I hope will resolve the issue. :)

Atomic oxygen definitely exists in nature- its a problem in space exploration. At low pressures many reactive chemical species exist.

Is it state level textbook or NCERT? I am asking because if it is state level then it mistake and if it is NCERT then there should be some reason :)

When thinking about chemical reactions it is very important to know which chemicals may react with each other. Jun-Goo Kwak already pointed out the nature of oxygen.
A quick reminder: The ground state of elementary oxygen is the triplet biradical $\ce{o2}$, which is a gas. This is what we have on the surface of the earth. Carbon however comes in many  different forms in nature. The most popular and often is graphite. Other forms include diamond, fullerenes and graphene.
At one point in your life you haw almost certainly come in contact with graphite: coal. As the principle repeating unit is carbon itself, its formula will be written as $\ce{C}$.
As for the binary combination of oxygen and carbon, there are also many different modifications. The most important of them are carbonmonooxide ($\ce{CO}$) and carbondioxide ($\ce{CO2}$). Like Uncle Al stated there are also suboxides known, which are usually byproducts of incomplete combustion (if not targeted explicitly).
Having said all that, if you burn coal, the following main reaction will happen (1):
$$\ce{C + O2 -> CO2}$$
However, given the right conditions (excess carbon) also carbonmonoxide  may be formed (sum of reaction, 2):
$$\ce{2C + O2 -> 2CO}$$
The reaction itself will detour via the Boudouard reaction, which is very important in blast furnace processes. First forming carbondioxide via 2 and then convertin excess carbon to carbonmonoxide via 3:
$$\ce{C + CO2 <=> 2CO}$$

Let's first take a look at the allotropes of oxygen, and look into dioxygen more in depth.

Atomic oxygen ($\ce{O1}$, a free radical)
Singlet oxygen ($\ce{O2}$), either of two metastable states of molecular oxygen
Tetraoxygen ($\ce{O4}$), another metastable form

From NASA, http://www.nasa.gov/topics/technology/features/atomic_oxygen.html, regarding atomic oxygen:

Atomic oxygen doesn't exist naturally for very long on the surface of Earth, as it is very reactive. But in space, where there is plenty of ultraviolet radiation, $\ce{O2}$ molecules are more easily broken apart to create atomic oxygen. The atmosphere in low Earth orbit is comprised of about 96% atomic oxygen. In the early days of NASA's space shuttle missions, the presence of atomic oxygen caused problems.

Dioxygen, or triplet oxygen, is the most commonly known allotrope of oxygen. It has the molecular formula $\ce{O2}$. Oxygen has 8 electrons with 2 in the 1s, 2 in the 2s, 4 in the 3p orbitals. Alternatively, there are 6 valence electrons. If there are another oxygen molecules, oxygen will pair up, form a double bond with bond order of two. In short, the potential energy of dioxygen is far less than that of atomic oxygen.
One interesting aspect of oxygen is that it exhibits paramagnetism unlike $\ce{N2}$ and it can exist in two different electronic states called singlet oxygen. The picture of the molecular orbital (MO) diagram of oxygen makes this more clear:

The MO diagrams above are for the singlet oxygen $a^1\Delta g$ excited state, the singlet oxygen $b^1\Sigma \text{g+}$ excited state, and the triplet ground state $X^3\Sigma \text{g-}$ respectively.
What you may notice is a spin flip in the $b^1\Sigma \text{g+}$ excited state.
This definition taken from Purdue University summarizes Hund's rule of maximum simplicity nicely: every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin.
The two first diagrams are in violation of the 1.) spin-selection rule: spin-flips are foridden, and 2.) Laporte selection rule: trasnsitions between orbitals of the same parity are forbidden, where parity means symmetry with respect to inversion. There is a german notation, gerade - which refers to symmetric with respect to inversion and ungerade - antisymmetric with respect to inversion.
There are many ways to produce ozone. https://en.wikipedia.org/wiki/Ozone#Production
Ozone is a triatomic molecule with 3 oxygens. It is much less stable than dioxygen and often breaks down into dioxygen.
What you may have been confused with is Dalton's incorrect "rule of greatest simplicity." Dalton was trying to resolve the issue of the correct ratio and number of atoms in respect to the chemical formula.
He assumed that:
$$ \ce{H + O -> H2O}$$
However, we know that:
$$ \ce{H2 + O2 -> 2H2O}$$
It wasn't until Avogadro and Gay-Lussac who stated the Law of Multiple Proportions and postulated the existence of diatomic molecules, that we can now resolve Dalton's incorrect hypothesis.

When two elements form a series of compounds, the masses of one element that
  combine with a fixed mass of the other element are in the ratio of small integers
  to each other.

H + O = H2O, O has valency of 2 and H has the valency of 1. If they react with each other, by the criss cross valency rule we will get H2O but since it is not balanced we will get 2H + O = H2O

@AbhishekMhatre There seems to be a fundamental misunderstanding you have with chemical reactions. Please try reading all of what I wrote above. You are making the exact same mistake Dalton did with his rule of greatest simplicity. Of course, I didn't balance the example above, but Dalton would balance it the way you did.

This text is quite enlightening, but unfortunately it does not answer the question.

@Martin What would you add? Abhishek still seems to not understand the nature of homonuclear diatomic molecules, and I tried doing an overview of allotropes of oxygen by finally summarizing with Dalton's 6th incorrect assumption in his atomic hypothesis.

I wrote something myself - I still am not quite sure I accomplished an answer that completely covered the question. You can read it and if you like migrate it to your post, I would then delete my attempt.

@Martin It's fine. I like how you touched on different reactions with carbon, oxygen, carbon monoxide, and carbon dioxide. It shows the relationship between the reactants and products while at the same time showing the necessity in preserving the homonouclear diatomic oxygen. I think we have all contributed towards the question. I still feel as if the OP does not have an understanding why some molecules form diatomic molecules. However, he does not seem to be replying back.

You might be right about that, there is some understanding missing here. We will see if he states something here. I guess with the answers we pretty much covered the question for further references.

@Martin Thanks for all your feedback. It really helps catch mistakes that myself and others overlook, and helps improve the answer overall.

A great answer +1. But I believe he has just started taking chemistry as a subject. So I don't think he understood half of your text and maybe I presume he is in that stage where he has just started to know about valencies and reactions and stuff.

The common products of carbonaceous materials' combustion are carbon monoxide and carbon dioxide. The products determine the equation.  The equation does not dictate the products.
Carbon suboxide $\ce{C2O3}$ is known.   Benzene hexacarboxylic acid trianhydride is a carbon oxide.  Combustion generally outputs simple molecules in deep thermodynamic holes.

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text:
the perfect campfire
As far as chemistry goes, there are laws or observations that can be useful to determine the perfect shape and disposition of the wood in a campfire ?
For example what chemistry says about the 2 most popular "shapes"

teepee
log cabin

or even other variations such as the swedish stove ( 1 log with the top splitted in multiple segments )
The properties that I'm looking for:
 - easy to start
 - long lasting
The properties that I would like to have but I can discard:
 - significant heat generation

From many years as a trained firefighter, I can tell you that there is certainly science involved. There are a number of very heavy calculations for calculating things like solid combustible burn time and heat release rates for combustible materials, which might possibly be useful to predict the perfect shape, fuel size and arrangement for a campfire, but are probably well beyond the scope of producing the perfect sausage sizzle. However, a number of key factors that influence fire behaviour, and which must be considered in building the best campfire include:

Size of fuels. For a given fuel type, the larger the fuel piece, the greater the resident burn time, but the greater the energy input required to obtain ignition. So small fuels will ignite and burn first, and provide energy through preheating to ignite larger fuels.
Availability of fine fuels (typically those under 5mm diameter). For ignition to occur, you need to have fine fuels, especially in the absence of other freely burning ignition agents (fire starters, a cup full of petroleum etc). As heat rises, the fine fuels need to be placed at/near the base to preheat larger available fuels. Plenty of fine fuels is critical - you can't have too much.
Available air capacity. Compact fuels that do not have sufficient air flow are poor propagators of fire. Oxygen is one of the critical elements of the fire triangle (or quadrilateral if you choose), and if not being positively injected (bellows etc), there must be sufficient air flow between fuels to allow drawing in of fresh oxygen supplies (As fire burns, the hot gases rise, and fresh oxygen is drawn in at the base, and must be able to freely disperse.
Exposure to radiant and convection heat energy. Heat energy passes via radiant, convection and conductance pathways. Solid fuels need to be preheated to release the combustible gases, and this requires sufficent exposure to radiant and convection energy pathways, which is usually above the burning fuel source.
Fuel moisture. Critical factor. Many fuels with greater than about 15% fuel moisture will have difficulty igniting; around 25% most fuels have difficulty in sustaining burn. Fine fuels exchange moisture with the atmosphere far more readily, and so weather conditions must form a real factor in your calculations (temperature, relative humidity, wind). On hotter, drier days, larger diameter fine fuels can be used for ignition. Drier fuels also burn hotter and provide greater energy output. Moist fuels release lower energy flames, taking longer to ignite other fuel.
Resident burn time. Different materials (including different types of wood) have different resident burn times, and so your choice of fuels will be significant. Iron bark, for instance, has a greater resident burn time, and greater energy output than radiata pine. This derives from the density and composition of the fuels themselves. Fine fuels, of course, have a very short resident burn time, so it is important to have enough to ensure sufficient preheating of larger fuels. How much will depend on fuel moisture!
Surface area of fuels. Smooth, polished wood will take longer to ignite than rough sawn timber
In addition, you may have to stipulate some other limitations:
Availability of renewable fuels. Is you campfire fuels source static, or are you able to add more fuels as the campfire burns. This clearly influences how long it lasts, and heat generation
Stability of design. As the finer fuels burn, there is a point of critical collapse, where the top fuels collapse on the sub-layers. This will influence a number of factors, including availability of air and exposure to preheating.
My design for my favourite campfire is somewhat similar to your log cabin design. It is a crib design, which means it is layered, quite often with the bottom layer of fine fuels (tinder) sitting on a spaced grid of larger sticks to allow air flow underneath. The mix of large fuel and smaller fuels (kindling) is staggered to give stability and access to preheating. The overall shape gets smaller as the layers go up, so the final shape is somewhat square-pyramidal. Separation of fuels needs to be close, while still ensuring maximum air flow.

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Is this reaction to produce ammonium sulfide safe?
For my chemistry experiment, I was assigned to produce ammonium sulfide and test the chemical properties of the compound. From my online research, I found out that I can mix household ammonia and match heads to produce Ammonia Sulfide.
I think this will satisfy the purpose of my assignment, but I have a few concerns.
I'm not sure what proportions I should mix these materials in order to produce an ideal solution. I am also unsure about the exact results of mixing ammonia and match heads, since match heads consists of more than just sulfer, but also potassium chlorate, antimony trisulfide, and some oxidizing agents. Will the end product potentially be dangerous? I plan to mix the two chemicals in a enclosed test tube, and seal it for a few days.

The name of the compound you are planning to make is ammonium sulfide. Ammonia ($\ce{NH3}$) is a gas under standard conditions. Household ammonia is a solution of this gas in water. If you increase the temperature, the gas will boil off. Keeping the mixture in a sealed test tube might not be the best idea then. If you are uncertain about possible health effects, it is always a good idea to have a look at the material safety datasheet (MSDS)!

You didn't mention whether you just want a solution or your product as a solid. Have you considered an alternative route: Generate hydrogen sulfide in a Kipp's apparatus and run it into household ammonia? In any case: Fume hoods are a great invention!

I just want to make the solution and observe the characteristics of the solution.

General rule: If you have to ask "Is it safe?", it's probably not.

$$ \ce{H2S(aq) + 2 NH3(aq) -> (NH4)2S(aq)} $$
Upon mixing ammonia with potassium chlorate:
$$ \ce{KClO3(aq) + NH3(aq) -> NH4ClO3(aq)} $$
$$\ce{Sb2S3 (aq) + NH3(aq) -> Sb(OH)3(s)}$$
Sb(III) reacts with aqueous ammonia to precipitate white Sb(OH)3. MSDS sheets say that antimony trisulfide is:

Toxic to aquatic life with long lasting effects.

Furthermore:
$$ \ce{Sb2S3(aq) + (NH4)2S(aq) -> (NH4)3SbS3 (aq)} $$

You used potassium chloride instead of potassium chlorate.

@LDC3 Ah, thank you. I'll go ahead and fix that.

I go to a charter's school, and the curriculum is very hands on and open. I actually volunteered to do this experiment because I actually can get my hands on these materials.
What do you mean taht it's toxic to aquatic life? It will only be exposed to me when I test it. I'm just hoping it won't harm my health in some serious ways. I will make sure to work in a ventilated area in case anything happens. Do you think if that's enough precaution?

Honestly, I don't think if anything serious will happen because of the low concentration of the chemicals that I'm messing with. And the potassium chlorate is only a small part of the match heads, and there won't be any artificial heating up in this experiment because I will simply mix ammonia and match heads together.

I am downvoting this because most of the presented equations are wrong. Only starting with your ammonia has an extra $\ce{H}$. Check this and edit it.

@Martin I used the ammonium ion instead of ammonia. Thanks for the clarification.

For your ammonium you were missing a charge ;) I would strongly suggest that you again revisit your equations and balance them. (I'm sorry, I just want to help.) I think it is okay to sometimes leave out byproducts, but since this is clearly a question asked by a non-expert, I would not leave room for mistakes.

I think that you are looking at the mote but not at the beam the problem is not the clorine or the $\ce{Sb2S3}$ but hydrogen sulfide...

In fact there are many types of matches so you should be aware of what you are using. In home-made experiments the problem is that many times you don't know the composition or the concentration of your reactants.
Take a look a The Preparatory Manual of Black Powder and Pyrotechnics
by Jared Ledgard for understand the composition of your matches.
We can assume that you have 20-50% of potassium chlorate and if present 1-15% of sulfur. I don't know the weight of a match's head (if someone have a laboratory balance you could determine it) however I think surely it weights no more then 4 mg. So in 50 pics matches box you have max 0,2 g of heads matches. So 0,1 g of potassium chlorate and 0.06 g of sulfur.
Hydrogen sulfide
Hydrogen sulfide is toxic and so can be dangerous but assuming that all the sulfur became hydrogen sulfide you have $0.06g_{Sulfur}~/ 32 \frac{g}{mol}=0.001875~mol$ of $H_{2}S$ hence $$0.001875~mol \times 34.08\frac{g}{mol}= 0.0639 g$$ of Hydrogen sulfide. So if you work in a room of $4~m \times 4~m \times 2~m $ so $ 32 m^{3}$ you have 1.9 $\frac{mg}{m^{3}}$ of hydrogen sulfide according to EPA is above the LOAEL (HEC): 1.9 $\frac{mg}{m^{3}}$ dose. However I hope that you will to the experiment in a well ventilated ambient. In this condition I think that is safe, however I don't think is a good experiment to perform.
Chlorine
Assuming that the chlorine in potassium chlorate is release as chlorine with a 100% yield (I'm quite sure that this can't happen) you have:
$$0.1g_{KClO_{3}} / 122.5495 \frac{g}{mol}=0.00082 mol_{KClO_{3}}$$
Hence, 0.00041 mol of $Cl_{2}$ and so if 70.9 is the molecular weight 0.029069 g of chlorine.
So in a $32 m^{3}$ "room" you have 0.90625 $\frac{mg}{m^{3}}$ of chlorine according to EPA is below the LOAEL 1.2$\frac{mg}{m^{3}}$ dose.
P.S. Up to my knowledge $\ce{Sb2S3}$ is no more used.

It is also to my understanding that Antimony is not used any more. People understood the hazardous nature of Arsenic and Antimony is many times worse...


text:
Do lithium and beryllium want to lose electrons to be like helium?
I can't seem to remember. I know that they all want to fill their valence shells, but, aren't these different?

Yes, lithium wants to lose electrons to be like helium because full valence shells are more stable states and all noble gases have full valence shells. So alkali metals lose one electron to achieve nearest noble gas configuration.
But beryllium is exceptional because of its small atomic and ionic radii. Due to this, its ionization potential is very high. So it mostly forms covalent bonds.

Yes, they do. All alkali metals, in group 1 of the periodic table, display the tendency to lose electrons as well as group 2 alkaline earth metals. Instead of losing one electron like alkali metals, alkaline earth metals such as beryllium lose two!
The loss of an electron to some other element, say a halogen, is very favorable as it gives the potential for alkali metals or alkaline earth metals to form a full valence shell or stable octet.
Therefore it is very common to see ions such as $\ce{H+}$ as well as $\ce{Li^{+}}$ as well as $\ce{Be^{2+}}$. Thus, salts such as $\ce{NaCl}$ as well as other compounds such as hydroxides, $\ce{Li(OH)}$ are the product of these metals once they lose their electrons and react with another element.
First ionization energies:

In general this is true. However, Beryllium is very special. Like no other (Main group) Metal it will form covalent bonds with non-metalls. This is due to the fact that the electron conformation is $\ce{1s^2 2s^2}$ and hence has already a full shell. The most stable oxidation state is +2 though.

@Martin although the electron configuration for beryllium is stable and the atmo is quite small, the first and second ionization energies 9.32 eV, 18.21 eV respectively.


text:
Would this recipe on stink bomb work?
I found this recipe on how to make a stink bomb:
The recipe basically says to mix ammonia and match heads to create ammonium sulfide. However, I am skeptical, because

It seems that match heads have mostly phosphorus, and only a negligible amount of sulfur.
I can't find any reference for synthesizing ammonium sulfide from ammonia and sulfur; rather, I found that it is synthesized from hydrogen sulfide and ammonia.

Would this recipe still work?
Would it also make malodorants other than ammonium sulfide?
If so, would these other malodorants be safe? Or would it be better to use a different process to create a stink bomb, like using actual sulfur, or something  entirely different?

possible duplicate of Is this reaction safe?

@Michiel possible duplicate of a comment :P

@Martin - Interesting, apparently if you flag it as a duplicate it also puts the comment there. I didn't know that

@Michiel In my opinion the questions are quite different, this is a better question covering multiple aspects of the procedure. Are related but not duplicates.

Not all the matches have sulfur but only some types in The Preparatory Manual of Black Powder and Pyrotechnics
by Jared Ledgard see page 191, you can find the best matches for your needs, classic matches are okay.
You are right, no reaction is reported between sulfur and ammonia. However if you put in a closed box a compound containing sulfur and a little bit of organic matter Sulfur-reducing bacteria (SRB) reduce sulfur in sulfide anion.
$$\ce{S + 2e- -> S^{2-}}$$
$S^{2-}$ is a base and reacts with water to produce hydrogen sulfide:
$$\ce{S^{2-} +2 H2O -> H2S + 2OH-}$$
$H_{2}S$ is a conjugate acid of $S^{2-}$ so now can react with ammonia with an acid-base reaction:
$$\ce{H2S + NH3 <=> NH4SH}$$
or:
$$\ce{H2S + 2NH3 <=> (NH4)2S}$$
Of course like many ammonium salts it can easily decompose into the two odorous gasses.
Others odorous substances are  skatoles, mercaptans, indoles however I don't think that these are present in great quantities in this case. See this site for further informations.

(+1) Since you have a water based solution, it is probably more likely to form ammonium first and react with sulfide later: $$\ce{NH3 + H2O <=> H4N+ + {}^-OH}\\ce{2H4N+ + S^{2-} -> (H4N)2S}$$ Those strike anywhere matches usually have $\ce{P4S3}$ so the sulfur is already partially reduced (decomposes with formation of $\ce{H2S}$).

@Martin thanks a lot Martin, you are right, I think you should write a little answer...

I think it is sufficient to have it in the comments as an extension to your answer.


text:
Find the pH of a sodium chloride solution using the extended Debye-Huckel equation

Question
  Use activities to calculate the pH of each of the following solutions, being sure to use $\mathrm{\alpha}$ values and the extended Debye-Huckel equation.
The first one the concentration is $0.020 \mathrm{M}$ $\ce{NaCl}$
  with the answer being : $\mathrm{pH}$ = $6.99$
The second one the concentration is $0.10 \mathrm{M}$ $\ce{NaCl}$
  with the answer being : $\mathrm{pH}$ = $6.98$

The thing is i have no idea how to find the activity coefficient of $[\ce{H}^+]$
I know ionic strength $u$ is calculated as
$\mathrm{\frac{1}{2}\left(\sum C_iZ_i^2\right)}$
but how does $[\ce{H+}]$ work into the charge balance?
I also know that the activity coefficient is calculated as
$$\log\gamma=\frac{-0.51(z^2)(\sqrt{u})}{1+(\alpha(\frac{\sqrt{u}}{305})}$$
The best I have is:
$$\ce{Na+ + Cl- + H+ + OH-}$$
and that raises the question of how do I know what $[\ce{H+}]$ is and $[\ce{OH-}]$ for calculating the ionic strength? Am I approaching this wrong? Where do I start?

$\ce{NaOH + HCl -> NaCl + H2O}$. It is possible to find the $\ce{[H+]}$ from the pH by $10^{-pH}$ and $\ce{[OH-]}$ by $10^{-pOH}$

I'm sorry maybe i worded the question wrong. We're solving for the pH and i already know the answer. I just don't know how to arrive at it.

Is it $u$ ($u$) or is it $\mu$ ($\mu$)?

Let's first look at ionic strength:
Ionic strength is a measure of the total ion concentration in solution, however, species with higher charge can also have elctrostatic interactions with other species.
$$\mu = \frac{1}{2} \sum_i C_iZ_i^2 $$
What is the strength of 0.02 M NaCl solution?
$$\mu = \frac{1}{2}([\ce{Na+}]z^2 + [\ce{Cl-}]z^2) = \frac{1}{2}([0.02](1)^2 + [0.02](-1)^2 ) = 0.02 M$$
With the ionic strength, we can now calculate the activity coefficients ($\gamma_i$) of ions in solution. The activity is merely the product of the concentration of each species and the activity coefficient.
$$\textrm{activity} = a_i = \gamma_i[i]$$
Let's look at a typical equilibrium expression:
$$\ce{A + B <=> C + D} $$
$$K_{eq} = \frac{[C]\gamma_c[D]\gamma_d}{[A]\gamma_a[B]\gamma_b}$$
The Extended Debye-Huckel equation is used to find the effect of ionic strength on the activity coefficent. It is given as above:
$$\log_{\gamma \ce{H+}}=\frac{-.51(z_i^2)(\sqrt{\mu})}{1+(\alpha(\frac{\sqrt{\mu}}{305})}$$
where $\alpha$ is the size of the hydrated ion in pm.
The Extended Debye-Huckel equation is valid from $\mu = 0 \ \text{to} \ 0.1 \ \text{M}.$
To find pH, you must first find $\mu$ of the hydrogen ion. You can then use the Extended Debye-Huckel equation to find the $log_{\gamma \ce{H+}}$. Take the antilog.
$$a_{\ce{H+}} = \gamma_{\ce{H+}}[\ce{H+}] $$
Take the negative log and you will get pH.
$$ pH = -\log( a_{\ce{H+}} ) = -\log( \gamma_{\ce{H+}}[\ce{H+}] ) $$
In context with the problem:
$$\log_{\gamma_{\ce{H+}}} = \frac{-.51(1^2)(\sqrt{0.02})}{1+(900 (\frac{\sqrt{0.02}}{305})}$$
$$\log_{\gamma_{\ce{H+}}} = -0.0508886 $$
$$\gamma_{\ce{H+}} = 1.1243166 $$
$$\textrm{pH} = -\log a_{\ce{H+}} = -\log (\gamma_{\ce{H+}} \cdot 1.0 \cdot 10^{-7}) = 6.95 $$
Your error is when you take the antilog of the $\log_{\gamma \ce{H+}}$. You raised it to the $10^{\log_{\gamma \ce{H+}}}$ which is $0.89429197$. It is actually supposed to be $10^{-(-0.0508886)}$ which comes out to be $1.124316$.
For the 0.10 M of $\ce{NaCl}$ I calculated the pH to be $6.92$.

Okay so here is where I'm having trouble, I am solving the debye-huckel equation with $u$= $0.02$ and $\alpha$ = $900$ and am getting 0.889 as my activity coefficient. I took this and multiplied it by $[H^+]$ which I'm assuming comes from $kw$=$[H^+][Oh^-]$ which is $1*10^{-7}$ multiplying that by the coefficient and then taking the $-log$ i am getting $7.05$.

@JohnSnow I see. Of course, in a neutral solution $\ce{[H+][OH-]} = K_w$, let me use those values and see where you are going wrong.

@JohnSnow You just made a small sign error in taking the antilog.

Okay thank you for that, but now the pH is $6.95$ which is closer to the right number but we're still not arriving at $6.99$ which is the answer from the answer key.

@JohnSnow I reviewed my calculations. I carried out the calculation and did not drop any significant figures. It is possible the answer key may have truncation errors. Did you try doing the calculation over? What value are you arriving at?

I'm now arriving at the same value as you. It's possible that the answer key has the wrong answer.

Using the formula as:
$$\log \gamma_\ce{H+} = \frac{- 0.51(z_i^2)\sqrt{\mu}}{1 + \sqrt{\mu}}$$
instead of your formula , this formula gives more accurate answer:
$$\frac{-0.51 \times \sqrt{0.02}}{1+\sqrt{0.02}}=-0.063188664981$$
then
$$10^{-(-0.063188664981)}
=1.063188664981$$
then
$$\mathrm{pH}=-\log(1.063188664981 \times 10^{-7})\\
\mathrm{pH}=6.973389662203$$

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What does this provide that the other answer doesn’t?


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Is it possible to compress an atom to infinite density?
I know, by nature, normally atoms are stable. But I wonder, if it is possible to compress an atom, say the Hydrogen atom, to infinite density by applying external force? After all, electrons and protons would attract each other by electrostatic forces and nuclear forces are also there at least up-to a certain distance...
Has it been done?
The reason I thought of it, was because I was trying to find the ratio of the amount vacuum to the amount of matter in the whole universe...

I would suggest you ask this question in the physics.sx

@AwalGarg Black hole. :) Prior to this, the nuclear reaction $\ce{Proton + Electron -> Neutron + Neutrino}$ occurs. As you keep applying pressure, electron degeneracy takes place. If you keep applying more pressure, you observe the phenomenon neutron drip where neutrons are expelled freely. This phenomena can be observed in neutron stars. If you keep continuing, you go through neutron degeneracy and finally create a black hole.

I'm voting to close this question as off-topic because it's not really about chemistry.

@Jun-Goo Kwak don't we get electron degeneracy first, before proton-electron coalescence?  Thought that was what white dwarf stars are made of.

We do not know. Physicists THINK that there ought to be a fundamental limit in scale for space-time that occurs near $10^{-33}$ centimeters and $10^{-43}$ seconds; often called the Planck Scale. There is also a unit of mass associated with this scale which is about $10^{-5}$ GRAMS or $10^{19}$ Billion Electron Volts (BeV or GeV). It is a simple matter to construct a 'Planck density' by computing $10^{-5}$ grams / $\text{(} 10^{-33} \text{centimeters)}^{3} = 10^{94}$ grams/cc. Many physicists think that this represents the highest density for any natural compression mechanism, because at higher densities, space-time itself transmutes into some type of quantum mechanical system for which 'density' is not a meaningful concept. So, if fundamental particles are truly without any structure larger than the Planck scale, then they can be packed together into a 'system' with a density of $10^{94}$ grams; the highest achievable density for energy/mass in this physical universe. Even within black holes, the famous 'Singularity' may not have infinite density, but merely a finite density near the Planck value. This is how 'quantum gravity' is thought to regulate, or eliminate real physical singularities; something which ordinary general relativity is not able to do.
I read it in a book Elegant Universe, by Brian Greene. This book has a whole seperate chapter for this.

No, because under enough compression atoms stop being atoms.
After a certain point, a proton will - under enough pressure - react with an electron and become a neutron and an electron-neutrino. This happens in the formation of a neutron star; the atoms are compressed so much by gravity that all the protons absorb all the electrons leaving an extremely dense ball consisting of only neutrons.*
*The electron-neutrinos can fly straight through most matter, so they just escape into space.
There is, however a bigger problem with your question.
In asking about the 'vacuum/matter ratio' you make the implicit assumption that matter has a set volume, when in reality no such value can be set. "Matter" isn't even that well defined a term in the first place, but taking it to mean 'stuff that has rest mass' doesn't get you very far. Even for relatively "normal" atomic/molecular matter, gasses don't have a set volume.
However you may think that when we get into the super-dense matter of neutron stars (called neutron degenerate matter) there is a set density; all that's left is neutrons and they can't be compressed further, right? Wrong. Two neutrons can occupy the same position if they have different momenta - a good-ish way to think of this is that in addition to normal "position space" particles also occupy a kind of "momentum space" - and if pressure increases they can just use more momentum space and less position space.
So does this ever stop? Is there something that happens when you compress neutron degenerate matter more and more? Well yes, but it still doesn't help. Eventually, after a lot of compression, the neutron degenerate matter will collapse into a black hole. What density does that happen at? The answer - as you probably expect at this point - is that there is no set density for a black hole to form.
Instead of depending on the mass per volume (A.K.A. density) of an object, black hole formation depends on the mass per radius of an object. For something with the mass of the earth to become a black hole it needs to be compressed down to about 1.8 centimeters in diameter - which is extremely dense. Something the mass of a whole galaxy on the other hand will become a black hole if compressed into a space about half a lightyear wide, giving a density significantly less than that of air!
And all of this is to say nothing of dark matter, which outmasses all "normal" matter in the universe by almost an order of magnitude. All in all, there's no meaningful way to define the volume of most things in the universe - so "trying to find the ratio of the amount vacuum to the amount of matter in the whole universe" is a bit like trying to find the amount of blue in the time 3:15. The problem itself doesn't make sense.

But we know Nuetron stars can be compressed down to extreme Densities,and extreme spherical  radius diameters, caused by the extreme compression,via Gravitational collapse,we know that any Mass therefore can be compressed to these extremes, because these very compressed masses already exist in the partial vacum of space,don't they, so some force has compressed there mass,into an extreme density. I suggest (E=MC2) has a new meaning, and that is, ("That Energy is equal to Mass Compressed Squared,or Spherically compressed"). You must Spherically compress, as you do when Bubble Collapse accurs,during implosion, with it's extreme tempatures, and extreme pressures created,one would expect that extreme density ocurred momentarially,before explosion or rebound ocurred then the transfer of energy,takes place via Photon emission, which ocurrs during implosion.So one would expect a chandresaker limit exists for a mass compressed or a Proton's Nuclei, to be Compressed,the BEC of the meson behavior is a Super Critical Fluid, at these extreme Pressures and Tempatures which is BOTH a Gas and a Liquid behavior, so it is Compressable,one only needs to read the work of the scientist/inventor who has discovered Hydrinos,which are below the normal energy level's,and are entirely new what i call Negative energy levels, only they are an absorbent only, or energy sink energy level's, all 9 of them,which at the last Jump from  the lowest energy level, one would have to assume the Chandresaker limit would take over. So Implosion and the Centripedal compression of the Vortice,or mini Black Hole compressession, compresses mass,and compresse's the Nuclie. Just as it does in a super massive black hole,only this on the micrpscopic level,the mini black hole on the Atoms Shell, allows for the Phonon resonance,to cavitate the Nucleons into Resonace,and if one Cavitates with Coherent optical metals like Tantalum, to resupply back onto itself the resonace in a loppback circuit,a closed loop,totally reflective/Optical properties of reflection, one creates a Laser Waveguide tube, with an end Mirror which momentarily PIN'S charge, zero time, zero mass time/Zero spacetime.Zero energy amplification, no loss of time, no loss of energy,Pure Potential only,this is why Resonance amplify's,is because of NO Time Loss Zero Time Loss,the evanescent wave, is transductive only, in one direction, down into,and the Phonon resonance does exacltly this,then couples to the evanescant field,resonates with the evanescent field, it Pins, and collimates, or lenses optically, to be re- supplied back onto itself,to create a coherent beam, of extreme energy which is coherently tuned, to spacetime/masstime,and also Compressed Nuclei,which is now a new energy level in the case of a Hydrino, it's tuned,and Amplified, or coherent  resonant energy,Created by (MASS COMPRESSION) via cavitation, and Implosion,the Vortice centripedal compression. So (E=MC2) Really always Meant this, and that was what Einstein Always meant, by this equation. That (Energy is equal to Mass Compressed Squared,or Spherically compressed Mass) just like the Atomic shape charge used to compress a Nucler Bomb first,before the other reactions take place, the Direct Mass to energy conversion,or (the Missing Mass) is Simply the Compressed Mass, the new Radius,and the new Density, the Loss was because it was compressed,then converted to Photon energy, and sent away and Photon emitted the excess,  as Highly energized,extreme energy of those extreme emitted Photon energies, as electron Volts Ev, extreme energy Photon emission, created by the MassTime Spacetime conversion compression/Pinning of, which then gets converted to, the the new compressed energy Level.

This is nearly impossible to read at this point.  There's a lot of extra info here and no paragraphs in sight.


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Safety concerns - Heating PMMA (acrylic glass) and fluorinated polymers
I have some plastic optical fiber, which consists of a thin core of PMMA (polymethyl methacrylate), covered by a fluorinated polymer.
I would like to heat a bundle of these fibers to ~130 degrees, in order to form a simple image guide.
However, I won't start until I know this is safe. Can you help me find out what types of gases I may encounter by heating these materials? Anything known dangerous? What other safety precautions should I take?

One problem is bonding of the bundle.  The contact surface is the fluororesin cladding that allows more nearly perpendicular rays to be totally internally reflected.  If you nick the cladding you get a light leak.  Will the cladding fuse when heated?
If you ligthly bond a bunch of cylinders you get air gaps where they touch.  Those will be a triangular grid off dark spots, 9.3% of the area of the image.
http://en.wikipedia.org/wiki/Circle_packing

(you've probably done this by now, but...)
Apply some common sense to the quantity you are working with. Are you making this on an industrial scale, or one small item to use in your lab? If the latter, don't worry about it. Unless something massively toxic is expected to come out (VX, for example) just don't go out of your way to breathe the products of the process. Even arsenic won't hurt you in small quantities.

PMMA melts at 160 °C and ignites in air at 460 °C, forming CO and formaldehyde as toxic gases when burning. Since you are working with a much lower temperature, PMMA should be safe. What type of fluorinated polymer do you want to use?

Thank you, the product I have only states it is made of PMMA and "fluororesin". I have sent them an email to get more detailed information.


text:
What exactly happens to mass during deuterium fusion?
I know that the mass of Helium-4 is slightly less than the mass of 2x Deuterium it may be made of and that it is the reason fusion gives a lot of energy.
But where exactly is this missing mass taken from? Aren't those 2 protons, 2 neutrons and 2 electrons forming Helium-4 atom the same as they were in previous Deuterium atoms?

Hi I think this is more a physic question!

Fusion of H-2 does NOT give He-4.  It gives ~50:50 H-3 (1.01 MeV) + p (3.02 MeV) or He-3 (0.82 meV) + n (2.05 MeV).  Either way, the missing "mass" is the mass equivalent of binding energy, E/c^2.  Binding energy is negative.  A bound system masses less than its separated constituent parts.  You must input energy (that being mass-equivalent) to unbind it.
Fusion converts weakly bound nuclei into more strongly bound nuclei.  The net difference in binding energies comes out as the kinetic energies of the product nuclei and particles.  Count neutrons and protons going in versus neutrons and protons coming out.  If one convetrs to the other, there must be elecron neutrinos or electron antineutrinos emitted to balance the books for energy, momentum and lepton generation number.
http://en.wikipedia.org/wiki/Nuclear_fusion
   Middle


text:
Is there any way to use syringe pumpe in pumping flow
I want to use syringe pump to pump water at 0.82 ml/min for 30 min ? Is that possible ( I read some paper use that but don't know how ?)

It depends which syringe pump do you have?

This should be no problem when you use a syringe pump model with adjustable flow rate and sufficient volume (at least 30 ml).

If it's 30 ml how I cam use it for 30 min ?? Could you explain for me please

When you pump with a flow rate of 0.82 ml/min for 30 minutes, then the volume of liquid consumed from the syringe is 0.82 * 30 = 24.6 ml. So you need a syringe of at least this volume if you want to pump continuously for 30 min, without the need to refill the syringe in the meantime (which you want to avoid if possible, I think). And 30 ml should be a common size for syringes.


text:
Test for glyphosate in water, boolean style
Testing for contamination of the ever so popular glyphosate based herbicide "Roundup" in water is a service that many labs provide, one of them is the Microbe Inotec Laboratory . They use an enzyme to check for levels of of contamination in ppb, which, I'm sure, is rather a complicated process.
My question is, is there any way to test for it more easily, without finding the actual amount of glyphosate but only checking whether it's present or not?

The proposed method above most likely uses ELISA (enzyme-linked immunoabsorbent assay). In most modern labs, ELISA is quite a standard procedure, however, each assay is easily worth a couple hundred dollars.
Another method of detection is using HPLC (high-performance liquid chromatography), however those machines can run upwards of a couple hundred thousand dollars and each test is expensive as well.
The EPA has a protocol which they used to detect glyophosphate contamination in water:
https://www.bucksci.com/catalogs/EPA%20Method%20-%20547%20-%20GLYPHOSATE%20IN%20DRINKING%20water%20by%20isocratic%20HPLC%20with%20flourecense%20detector.pdf
What may be of interest to you is an article from the journal of the Royal Chemical Society in which they published an article where they used gold nanoparticles to detect presence of glyphosphate using the naked eye or UV light.
http://pubs.rsc.org/en/content/articlelanding/2013/AY/c2ay26391b#!divAbstract

Cyteamine functionalized gold particles sounds very interesting, do you know anything more about it? or, rather, could you tell me more about it? Can't access the study in its entirety.


text:
Why are we not supposed to heat Erlenmyer flasks?
Chemistry says that we are are we supposed to heat boiling flasks, and not Erlenmeyer flasks.  However, I can't find any information about this on Google.  Is there a particular reason for this?
Is it just because we want to separate the purposes for our glassware, or is there an actual valid reason?

Can you cite the "Chemistry" which says it? Erlenmeyer flasks are routinely used for heating.

This really depends on what you mean by heating.
One of the main uses for Erlenmeyer flasks is for recrystallization, which of course involves heating the container and contents. Typically this is only moderate heating, such as over a steam bath or with a hair dryer. Some manufacturers specify that their Erlenmeyer flasks are ideal for heating applications.
Traditionally, boiling flasks have more uniform and thicker walls, which allow for much greater heating limits, and a more uniform heating distribution. They are designed to handle more intense localised heating, and manage thermal and physical shock distribution. Erlenmeyer flasks usually have thinner walls and greater wall thickness variation, and are less able to handle sudden thermal changes, such as heating with a high energy source like a Bunsen, or heat gun. Using them in this way may well lead to cracking, fracturing or worse, and is unsafe.
So, although Erlenmeyer flasks can be heated, they are not specifically designed for high temperature, constant heating applications. For this, you should use a boiling flask.

In Essential Environmental Science: Methods & Techniques
by Simon Watts,Lyndsay Halliwell I've found this:

This because his conical shape does not allow an uniform expansion and so the glass could break. Furthermore if you are heating substances that will leave a residue will be very hard to clean it.

The conical shape of an erlenmeyer flask creates a sharp angle between the bottom and sides of the flask.  Stress, be it from heating, cooling, mechanical shock, etc., is not as uniformly distributed across such an angular structure as it is in a spherically shaped flask.  Hence, the erlenmeyer flask is more prone to cracking under stress than a spherical flask.

Since most Erlenmeyer flasks are made with Pyrex glass, there should be no problem heating them up. I have been told not to heat volumetric flasks since the volume could change.


text:
Hygroscopic behaviour of DMSO - how bad is it?
For a fluid flow experiment I am using DMSO (dimethylsulfoxide), because of its low volatility, reasonably high surface tension, low viscosity and relative safety.
In the MSDS of DMSO I find that the liquid is hygroscopic, meaning it adsorbs moisture from the surrounding environment. Unfortunately, I cannot find anywhere how much water will be absorbed (mass%) or how quickly this will happen, is it saturated in minutes, hours, days?
Therefore, my question is twofold: (i) how much water will DMSO at maximum absorb and what is the rate at which this absorption occurs? (ii) with the absorbed water, how badly do the physical properties (viscosity, surface tension, density) of the (now) mixture change?

Being curious about the hygroscopicity I have tried the experiment using pure, dried DMSO in our lab where the temperature was $21^\circ \text{C}$ and the relative humidity about $60\%$.
Have a look at the two images below. This is a DMSO droplet originally with $V=50\; \mu\text{L}$. You can clearly see that the volume of the droplet has increased substantially in 20 minutes from the different height.

Using a crude approximation, assuming that both shapes are caps of a sphere, I calculated that the change in height from 100 px, to 112 px is equivalent to a volume increase of $15\%$, which is much too big for an accurate flow experiment.

I only used DMSO in other contexts and I have no idea how fast the water uptake is or how it changes the physical properties, but

DMSO is miscible with water in any ratio
it does not form an azeotrop with water

The typical protocol for drying is:

Keep over night over anhydrous $\ce{CaSO4}$ or powdered $\ce{BaO}$
Decant off
Distill over $\ce{CaH2}$ (about 10 g/liter) at reduced pressure (bp is around 75 °C at 16 hPa)
Store over 4 A molecular sieve in a dark and tightly closed bottle

And before you do that you should be very certain, that you need it "that" dry. Did it once, hated it, turned theoretician. :D

@Martin :D I mostly did it for Swern oxidations or in fluorescence spectroscopy. It is a bit tedious and I don't know if it is necessary in Michiel's case, but it might be helpful do have somewhat defined conditions initially.

For these reactions it is absolutely necessary. For everyday reactions stirring it over night over $\ce{CaSO4, BaO, NaH, CaH2}$ is sufficient. btw, you are missing an $\ce{H}$.

It doesn't necessarily have to be very dry. My main concern is that the properties (in particular mass) of the liquid shouldn't change significantly over the course of about 1 hour of exposure to air at about 50% RH.

The liquid in my case, is a droplet of 50 $\mu L$ forming a roughly hemispherical cap.

@Martin and Klaus, just curious - if you add NaH to DMSO, wouldn't the DMSO also be deprotonated?

@orthocresol I doubt it. The $\mathrm{pK_b}$ of DMSO is around 35, NaH isn't a strong enough base. LDA should work fine.


text:
Are acid chlorides really acidic?
If I test a pure solution of an acid chloride with a pH-meter, will I get a reading of pH 7?
When I add water to it, I know $\ce{HCl}$ gas will form, but this gas will dissolve in water to give an acidic solution.
My main confusion is that we call ethanoyl chloride an "acid" chloride, but is its pure solution acidic?

As long as it is water free it should not be acidic as there are no free protons.

Just think of it as the chloride derivative of a carboxylic acid ;)

@Klaus Warzecha: Hmm... so that's where the "acid" part came from :P

@Eliza Exactly :)

@G M: when acid chloride reacts with water, carboxylic acid and hydrogen chloride gas is formed... well that's what I learned

@Eliza I've misunderstood... yes is right!

I think Klaus is right, is better to call it acyl chloride, in fact I think is a nucleophile so acts more like a base...

An acid chloride is a covalent molecule.  It has no pH.  If you put a pH electrode into a reactive acid chloride you will dehydrate and acylate its hydrated glass surface ($\ce{#Si-OH}$) and damage its future response when properly used.  pH electrodes are stored sitting in buffer or saline solution.
An acid chloride is an oxo-acid whose acidic $\ce{-OH}$ has been replaced with $\ce{-Cl}$.  Hydrolysis restores the oxo-acid, displacement makes derivatives. Hydrochloric acid is not an acid chloride.

pH-values only truely make sense in a solution whose solvent has acidic (as in ‘abstractable’) hydrogens. Apart from the obvious water this includes alcohols, enoliseable ketones and more. This means that a pure acid chloride cannot have a pH value, because there is no $\ce{[H+]}$ around.
However, all those solvents that can liberate a proton in a Brønsted manner have then created an anion. This anion, whatever its composition may be, is likely to be nucleophilic enough to attack the acid chloride and liberate $\ce{HCl}$ thereby reducing the pH value of the solution.
So a pure acid chloride or one dissolved in non-protic solvents has no pH as the solvent has none. Those that are dissolved in protic solvents are likely already liberating $\ce{HCl}$ due to solvolysis.

Is there such a thing as a completely pure solution ? Would there not always be some water contamination and hence would you not always have some protons floating around ?
However, as a thought experiment I imagine that in a solution of 'pure' ethanoyl chloride you would have a pH of 7 as there are no free protons moving around in solution.

pH = 7 says [H+] = 10^(-7) molar.  Your statement is incorrect.  There is no water in a bottle of acyl chloride.  If there were, it would react to give HCl and the acyl anhydride.


text:
Different coloured chromium salts
Evaporating cold aqueous chromium(III) chloride produces violet crystals. However when hot aqueous chromium(III) chloride crystallizes, green crystals result.
Could someone please explain the chemistry behind this... I mean both the crystals contain $Cr^{3+}$, so the difference in colour cannot be due to two different oxidation states. Also, they contain the same ligand $Cl^-$. Can someone please tell me where I am going wrong?

Well, actually when we evaporate aqueous chromium(III) chloride, the hydrate of the crystal will be gone, so we get anhydrous chromium(III) chloride ($\ce {CrCl3}$). This type of anhydrous produce violet color.
But if we crystallizes aqueous of the chromium(III) chloride, the hydrate won't go, so we get hydrous chromium(III) chloride (usually hexahydrate, like $\ce {CrCl3.6H2O}$). This crystal produce dark green color.

lambda23: could it also be due to the $Cl-$ ion acting as both a ligand and an anion?


text:
Does removing solution affect a supersaturated solution
Given a mixture of: a saturated solution of a solute in a solvent together with excess of the solute compound as crystalline material.
If one were to change the bulk composition of the mixture by removing some of the solution could this cause dissolution or precipitation of the solid in the remaining mixture, or the concentration of the remaining solution?
Presumably one should expect some crystallisation in the portion of the solution that is removed from the mixture?

If one were to change the bulk composition of the mixture by removing some of the solution [...]

This does not change the composition of the solution above the solid. Since the concentration of the solute remains constant, I do not expect a change.

Presumably one should expect some crystallisation in the portion of the solution that is removed from the mixture?

It still is a saturated solution. Unless you evaporate the solvent (= increase the concentration of the solute) or change the temperature, nothing will change.

When a solution is saturated, the remaining solid is in equilibrium with the solvent:
$$\ce{AB<=>[solv.]A+_{(solv)} + B-_{(solv)}}$$
That means that there is as much of your solid solved. If you take out some of the solution then the concentration remains the same, in your original flask as well as in the portion you withdraw.
If you remove only solvent, i.e. by heating your solution or diffusion over time. In this withdrawn portion crystals will form. The concentration will not change, only the amount of substance $n$ in your solution will change as the volume shrinks ($c=\frac{n}{V}$).


text:
Why does having equivalent resonance structures give more stability?
Although the phenolate ion has more resonance structures (4) compared to acetate ion (2), acetate is more stable because it has two equivalent resonance structures of same energy.
Why does having equivalent resonance structures give more stability? Shouldn't a compound with more resonance structures be more stable?

Hückel theory might already give a nice picture without dwelling on QM.

Preamble
One important thing to know is that what we call "resonance structure" is a byproduct of our chemical notation which can't describe the structure of some compounds effectively using only one chemical structure. Personally I find the old term mesomeric structure more appropriate (meso- Greek mésos in the middle;  -merism from Gk. merismos "dividing, partition") because resonance is already used for a phenomena, a real phenomena, in physics that has nothing to do with resonance in chemistry. Furthermore the compound is not continuously changing it's structure between the possible resonance structures, like the term resonance seems to suggest, but is a "normal steady" compound that simply has an electron distribution that cannot be described with a single chemical structure.
Why does having equivalent resonating structures give more stability?
The resonance energy is due to electron delocalization, so more the electrons are delocalized in the molecule higher is the resonance energy ( this means a decrease in the overall molecule energy). This is clear in the acetate ion - we can draw a resonance hybrid  like this:

Here the electrons are equally distributed (delocalized) between the two oxygen atoms.
However for some compounds the real structure is not always a structure right in the middle between the different resonance structures; the first assumption is true only if the two structures are two equivalent structures. So it can occur that the electrons are not well delocalized between all the atoms hence the resonance energy is lower and so the overall stability.
In the case of phenolate ion I think the first structure in the figure [2] gives a greater contribution to the final structure of the phenolate ion so the electrons are not well delocalized, but if I'm not wrong are mainly in the oxygen atom. Unfortunately  I can't find the energy of the different resonance structures.

Finally, however, I think that you should study case by case if the number of the resonance structures or their contributions are determinant to the overall resonance energy taking into account valency and electronegativity of the atoms involved.

By delocalizing the charge into the phenyl ring, you would put electrons in an antibonding (aromatic) orbital. This is of course possible, but since the electronegativity of Oxygen is greater than in carbon, not very likely. If however you put an external field around the molecule ([point] charges or non-protic solvent) you could see more negative charges at the indicated atoms. Resonance energy is a mere mathematical concept which is not measureable and strongly dependent on the definition. {I'd give additional +1 for your preamble.}

@Martin thanks a lot for the comment always very useful, I think that resonance energy can be consider as the difference between the true energy of the molecule and the energy calculated without taking in account the delocalization.

@GM So isn't neccesary that more resonance structures will to greater stability right? is also about the stability (energy) of each resonance structure?

@adosar correct

@GM But isn't this contradictory to the fact that the greater the delocalization of electrons (number of resonance structures) the greater the stability?

@adosar no, they both influence the stability but it is not simple to understand the contribution of each other maybe you can post another question regarding this aspect.

@GM So it is quality over quantity right?

Although the phenolate ion has more resonance structures (4) compared to acetate ion (2), acetate is more stable because it has two equivalent resonance structures of same energy.

I would argue the phenolate ion has five mesomeric structures, two with the charge assigned to oxygen (analogous to the two resonance structures of phenol), and three with the charge assigned to the ring carbon in ortho and para positions (see G M's answer).

Why does having equivalent resonance structures give more stability? Shouldn't a compound with more resonance structures be more stable?

I don't think the statement is true in this general form. It is always possible to make mesomeric structures a little bit less equivalent (for instance, through isotope-labeling of one of the oxygen atoms in acetate). This does not change the acidity of acetic acid much.
To give another example, we can compare the acidity of propene, acetaldehyde and formic acid. When you remove a proton from each of these, you get three anions that have multiple mesomeric structures.

The ones for propene and formic acid are symmetric, the one for acetaldehyde is not. Formic acid is most acidic, followed by acetaldehyde with propene least acidic. I would observe that for formate, the negative charge is assigned to oxygen for both mesomeric structures whereas for the deprotonated propene, it is assigned to a carbon atom. The enolate anion has the charge on oxygen for the more significant mesomeric structure (shown) and on carbon for the less significant one (not shown).
The argument for phenol vs acetic acid is similar. So it matters what the significant mesomeric structures are and which atom type carries the negative charge. When mesomeric structures are symmetric, it ensures that they have the same significance. If you compare apples to apples (i.e. anions where the negative charge is assigned to carbon for all mesomeric structures), it helps to distribute the charge (having the carbons with assigned charge at large distances), and having many possible locations for the charge.
For phenolate, the large number of mesomeric structures is less relevant because the ones with the charges assigned to oxygen are the most significant (the only ones that don't force us to write a structure lacking alternate double bonds in the benzene ring).

Yes, exactly! Having multiple equivalent resonance structures isn't itself a reason for stability; it's because it leads to greater delocalisation or distribution of charge, and that's merely one factor in the grand scheme of things.

No quantum, no bounty... @orthocresol

True — but an answer nonetheless, and a useful one at that...

@MilanPaul, that's kind of you to say but it is misplaced: you are well within your rights to award the bounty to any answer you wish (I'm not sure if the system lets it give it to yourself), but not to my tiny comment, which hardly stands on its own and is a mere restatement of Karsten's (and others') points. Just as well the system definitely does not allow that. :-)

@orthocresol To be fair, I did base my answer on https://chemistry.stackexchange.com/a/6160 and https://chemistry.stackexchange.com/a/41810, which is the cool thing about a community such as the one contributing to this site.

I would like to give a very rough answer as to why equivalent resonance structures have a stabilizing effect.
To begin, we assume that resonance can be represented as a discrete n-state system. Feynman has used this simple approximation to explain many of the essential features of chemical bonding and resonance in the Feynman Lectures on Physics Vol III, Chapters 8 - 10.
To simplify we begin with a molecule which can have 2 different canonical structure based on the usual rules. Let the energies of the two states be $E_1$ and $E_2$ .
The Hamiltonian of this two state system will be:
$$H = \pmatrix{E_1& 0\\0&E_2}$$
The canonical structure are themselves eigen-states here. However in the real world we find more symmetry to the molecule/different bond lengths/angles than that is predicted by  our canonical structures. This is because the two canonical structures are not the actual eigen states.
We assume that the new eigen states however can be represented as a superposition of the 'old' states.
Therefore in the most general case the Hamiltonian expressed in the 'old' base states become:
$$H = \pmatrix{E_1 & H_{12}\\H_{12}^*&E_2}$$
We find the energy eigenvalues from this matrix.
The energy eigenvalue with the lower energy in this case is:
$$
E^{'}_1 =
  \frac{E_1 + E_2}{2}
  - \frac{\sqrt{(E_1 - E_2)^2 + 4H_{12}H_{12}^*}}{2}
$$
Here without loss of generality let us assume that $E_2 \ge E_1$
So taking the difference $D = E_1 - E^{'}_1$ which is the energy difference between the old base state and the new eigenstate with lower energy:
$$
D =
E_1 - E^{'}_1 =
  \frac{E_1-E_2}{2}
  + \frac{\sqrt{(E_1 - E_2)^2 + 4H_{12}H_{12}^*}}{2}
$$
This $D$ is a measure of the stabilization due to resonance.
Setting $E_1 - E_2 = x \le 0$, we get
$$D = \frac{x}{2} + \frac{\sqrt{x^2 + 4H_{12}H_{12}^*}}{2}.$$
The graph of this function is:

So we see that for $x \le 0$ , $D$ is maximum when $x = 0$, i.e. $E_1 =  E_2$. One way this can happen is when our base states are symmetrical or in other words when the molecule has equivalent resonance structures.
However one should be careful of the comparison being made here. We are not comparing the molecule to some other molecule. All that is said here is that the stabilization of the molecule due to resonance is maximum when the canonical structures are equivalent (for a 2-state system).

It should be $D = E_1' - E_1$. It's always final minus initial state. It will give you a negative value, which makes sense, since a stabilization will formally release energy.

"The answer given here is not based on the Molecular orbital theory which should actually be used." There is no resonance in MOT, so you cannot answer this question in that way. What you are describing here reminds me a bit about quantitative Valence Bond Theory, but I have a feeling that the reasoning here is circular. If $E_1 \approx E_2$, then neither can. structure leading to $E_1$, nor $E_1$ is a good approximation to the el. structure, hence you need resonance. The truth still remains any 'resonance energy' is a purely hypothetical value which only seems to carry any physical meaning.

@Martin-マーチン Resonance energy need not be "real" (ie an observable) but rather explains why some structures are observed instead of simple eg Lewis/basic VB theory structures that do not invoke resonance. Isn't your statement akin to saying that  perturbation theory gives "imaginary" results, or that linear combinations of orbitals are not "real" solutions?

I'm not really sure why we're spending all of this effort on the situations where resonance structures are more or less stable. Resonance structures are just a model for how things work and not a great one at that. Insofar as it doesn't represent reality, any attempt to explain a fake thing is also fake. There's no reason to explain that equivalent resonance structures are "good" because that's the setup for the fake model that most agrees with reality.

I'm amazed how ppl are struggling with appreciating this approach, when it's not all to different from LCAO. Hell, I could even say MOs are fake, since they are just an artifact of the Slater determinant approximation. In that sense the only thing that really matters are the multielectron wave functions, which are the solution to the electronic Schrodinger equation.

@Kexanone You actually should say that MO are fake, they are just a mathematical solution, which can be interpreted physically. There is also Valence bond theory, which is equally wrong (or right). I'm not really seeing the link to LCAO though. The reasoning in the answer still is curricular and based on a false premise.

@BuckThorn I don't understand what you are saying unfortunately. Perturbation theory assumes that the reference system is already a quite good solution to the problem. (That is of course a stretch coming from HF. And we can see this failure regularly with MP2.) So this relationship really doesn't apply here. One Lewis-like structure is not a well enough description for a system that needs the resonance treatment. It's actually the opposite you want to prove in the question and what this answer aims at.

@Martin-マーチン If I understand correctly what the answer presents is akin to Huckel theory: "We assume that the new eigen states however can be represented as a superposition of the 'old' states." MO also mixes AOs from all atoms, the net result is similar. Your main point in the comment seems to be about the physicality of the resonance energy. I mentioned perturbation theory (and could be extension mention the application of the variation principle) only to bring attention to the fact that new theories sometimes provide methods of sequential approximation.

The absence of MOs in VB theory is improved upon in Huckel's theory by mixing the "resonating" forms. I could spin this further but I think I would need to refer to original articles to see what Huckel and Pauling etc meant.

@BuckThorn This is developing into a discussion, and not that I am not interested in having this, it should probably be had in chat. Just a final clarification here: My main issue is not the unphysical nature of the resonance energy. I object to the general reasoning: It proves that the resonance energy is greatest when the initial assumption breaks down. This is essentially a similar reason why MP2 gives completely bogus values for systems with a very small HOMO-LUMO gap. Sequential approximation only works when your initial state is reasonably close to the actual solution. Here it is not.

@Martin-マーチン I cede to your expertise :) Thanks for the input, I have much to learn.

But just for clarity, should anyone stumble on this conversation (and it still be up), what exactly is the "original assumption" ? @Martin-マーチン

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


text:
Do materials exist that absorb Visible light and emit Near InfraRed light?
I am thinking about the possibility of making a hat or other clothing item that might show up funny on cameras, as they are more sensitive to infrared light than human eyes.
Therefore, I was wondering if there are any (non hazardous) materials that emit InfraRed light when they are exposed to visible light.
Do these exist, and would they be suited for this kind of application?

Yes, there are some minerals that shows VIS-NIR luminescence here a couple:

Synthetic ruby $\ce{Al2O3:Cr}$
Kyanite ($\ce{Al2SiO5}$)

(Reference here)
A more sophisticated material is $\ce{LiSrPO4:Eu2+, Pr3+}$ you can read the full article in 11 February 2013 / Vol. 21, No. 3 / OPTICS EXPRESS 3161.
There are also some pigments like egyptian blue or cadmium yellow (CI pigment yellow 37, that is cadmium sulfide) that were also used in the past. Egyptian blue (calcium copper silicate ($\ce{CaCuSi4O10}$))  is safe, other pigments with cadmium are more toxic.
One of the most spread pigment shows VIS-NIR luminescence: Chlorophyll!
Have a look to this site for a nice overview.

All the luminesce processes have not a high quantum yield, so to record them, most of the times, you have to avoid reflection. Tungsten, halogen lamps and sun emit a lot of NIR radiation so is possible that reflected light overhang the luminescence. When I performed NIR luminescence I used VIS LEDs and a long pass filter at 850 nm.  If you don't have result, first thing you have to try is to remove the ICF (IR-Cut off-Filter) filter from your camera so NIR light can reach the sensor more easily.

NIR-emitting long-persistent luminescent nanoparticles have been prepared from $\ce{Cr(III)}$- and $\ce{Pr(III)}$-doped gallogermanates. According to the authors, these materials exhibit very long afterglows up to 360h (= 15 days).
Source: DOI1, DOI2

interesting! Is this a case of NIR phosphorescence?

@GM Yes! Isn't ist amazing? For the answer to this question, I I thought in commercially available organic fluorescent dyes (Cy7 Cy7.5, and the long-wave Alexa dyes) first, but their Stokes shifts are in the 20-30 nm range only. This inorganic stuff is weird :)

I've worked in NIR luminescence imaging a little bit, if I find where I've the file I'll add an image of $CdS$ luminescence but I've never seen NIR phosphorescence! Thanks for the articles!

CCD cameras typically have an IR-absorbing filter.  Short wavelength scattering (varies as inverse fourth power of wavelength) renders white cloth opaque, cellulose versus air refractive index.  If the scattering is quenched by wetting or avoided by IR-viewing, transparency results.  The champion visible-absorber, IR-fluorescent mateiral is chlorophyll.  Military camouflage extends into the IR.
IR-fluorescent dyes are common as biochemistry tags,
http://www.bruker.com/products/mr/nmr/avance-iii-hd/overview.html
http://www.sigmaaldrich.com/life-science/cell-biology/antibodies/cf-dyes.html

+1 I like those dyes, however the scattering in this case doesn't matter because the light is absorbed in the visible, NIR transparent materials are still okay.


text:
Chemical Formulas of Acids and Bases
How can we identify bases and acids by their chemical formulas?
Take for example this reaction. I know that HNO3 is an acid and KOH is a base, but if I didn't know that how would I be able to identify that using the chemical formulas
$\ce{HNO3(aq) + KOH(aq) \rightarrow KNO3(aq) + H2O(l)}$

I'm going to assume you are in high school, thus talking about Brønsted acids/bases.
A Brønsted acid is a compound that can donate a proton $\ce{H^+}$ whereas the base can accept a proton. For some chemicals it can be easily seen that they can donate a proton, like in the example of $\ce{HNO3}$ you give. There is obviously a proton in there so it makes sense that it is an acid. For chemicals with $\ce{O^-}$ in their structure (like the $\ce{KOH}$ in your example which dissociates into $\ce{K^+}$ and $\ce{OH^-}$) it is also clear that they can accept a proton and thus are obviously a base.
Now the difficult part is when a molecule can do both: these are called amphiprotic substances, the most famous being water $\ce{H2O}$, but there are many more of these, think about $\ce{H2PO4^-}$, it both can accept and donate a proton. In these situations molecules can indeed be considered both a base and an acid and the way they behave depends on the other molecules they come into contact with. If the other chemical is a stronger acid, as measured by the acid dissociation constant then the current molecule will act as base and vice versa.

Look at The Periodic Table.  Metal (Group I and II) hydroxides are bases, non-metal "hydroxides" are acids (especially with high positive charges on the non-metal.)  Then we have amphoteric hydroxides like zinc and aluminum that afford zincates and aluminates in very alkaline media.  $\ce{Mn(OH)3}$ is a base but $\ce{HOMnO3}$ is a strong acid.  Look at core oxidation state, charge polarization, and oxophilicity of that ion.
Arrhenius, Brønsted-Lowry, Lewis; hard, soft; solvo-cation, solvo-anion.  How do you classify $\ce{Al(OMe3)}$ that is a Brønsted base for methoxide and a Lewis acid for aluminum(III)?
$\ce{NH4}{^+}$ $\ce{NH3}$ $\ce{NH2}{^-}$ $\ce{NH}{^{2-}}$  $\ce{N}{^{3-}}$.  Ammonium is definitely the acid.  Nitride is definitely the base.  What are the species in-between?


text:
Double Displacement Reactions that produce gas
What type of double displacement reactions can form a gas. How can i determine if a double displacement reaction will form a gas?

I can't give you a general answer but just looking at this wiki page you'll find at least two examples: http://en.wikipedia.org/wiki/Salt_metathesis_reaction

"How can i determine if a double displacement reaction will form a gas?" I think the only thing to to is watch if one of the product is a gas...

Reacting acids with carbonates will produce carbon dioxide gas:
$$\ce{CaCO3(s) + 2 HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)}$$
With double replacement reactions you generally either get:
1.) A neutralization reaction
$$\ce{HCl(aq) + NaOH(aq) -> NaCl (aq) + H2O(l)} $$
2.) Gas displacement
$$\ce{CaCO3(s) + 2 HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l)}$$
3.) Precipitation
$$\ce{NaCl(aq) + AgNO3 (aq) -> NaNO3(aq) + AgCl(s)}$$
Thus it is important to know your solubility rules.

Do redox reactions count? What about a some gas other than carbon dioxide then?
Transition metal halides are reduced by $\ce{CN-}$ and form cyano complexes, while a part of the cyanide is oxidized and released as dicyan:
$\ce{2 CuCl2 + 10 KCN -> 2 K3[Cu(CN)4] + (CN)2 ^  + 4 KCl}$
$\ce{AuCl3 + 4 KCN ->  K[Au(CN)2] + (CN)2 ^ + 3 KCl}$


text:
Adjusting the ph of salicylic acid using sodium hydroxide

Question
  How many milliliters of $0.202\ \mathrm M$ NaOH should be added to $25.0\ \mathrm{mL}$ of $0.0233\ \mathrm M$ salicylic acid to adjust the $\mathrm{pH}$ to $3.50$?
  The answer from the text book is $2.2\ \mathrm{mL}$ NaOH
  My attempt
  I believe the correct equation to use for this is the Henderson–Hasselbalch equation

$$\mathrm{pH}=\mathrm pK_\mathrm a + \log\left(\frac{[\ce B]}{[\ce{BH}]}\right)$$
$\mathrm pK_\mathrm a = 2.972$
So I set it up as:
$$3.50=2.972+\log\left(\frac{x}{0.0233}\right)$$
Solving for $x$ I get $0.0785$
I don't know where to go from here
since $x$ is the concentration I figure that if I multiply by $0.025$ (the volume) and then divide by the molarity ($0.202$) I would get the answer but that's not right.

The Henderson-Hasselbalch equation is the correct equation, but ask yourself what you are putting in there:
$$pH=pK_a + \log\left(\frac{c(A^-)}{c(AH)}\right)$$
Since $A^-$ and $HA$ occupy the same volume, you can rewrite:
$$pH=pK_a + \log\left(\frac{n(A^-)}{n(AH)}\right)$$
When adding Sodium Hydroxide consider the following equation
$$\ce{AH + NaOH <=> A- + H2O + Na+}$$
Given that you will find
$$ n(A^-) = n(\ce{NaOH})\\
   n(HA) = n_0(HA)-n(\ce{NaOH})$$
Then you can substitute and solve for $n(\ce{NaOH})$:
$$pH=pK_a + \log\left(\frac{n(\ce{NaOH})}{n_0(HA)-n(\ce{NaOH})}\right)$$
You then divide by the concentration, yielding your volume.

Intermediate results:
Initial amount of $HA$:

     $n_0(HA) = 0.5825\ \text{mmol}$

Amount of $\ce{NaOH}$:

 $n(\ce{NaOH}) \approx 0.4993\ \text{mmol}$

Final Volume:

 $V(\ce{NaOH}) = 3.22\ \text{mL}\ \text{(3sf)}$


text:
Why isn't ΔG equal to zero for this equilibrium reaction?
In the reaction $\ce{E(g) <=> P(g)}$ at 25 °C, equilibrium is reached when the pressure of the product is 0.100 that of the reactant. What is $\Delta G$ in joules?
Using $\Delta G = RT\ln K$, my answer is $-8.31 \times 298 \times \ln (1.00) \ \text{J}$.
However, the answer is $-8.31 \times 298 \times \ln (0.100) \ \text{J}$. Wouldn't this mean that $\Delta G > 0$, therefore favoring the reactants? (As an aside, why would the reaction even be shifted to the left since it has a higher pressure?)

You need to look again. $\Delta G$ is negative.

Wouldn't the negative value from ln(0.100) cancel with the negative in front?

Sorry, I didn't notice you changed from 1.0 to 0.1. But as it is stated, pressure of the products = 0.1 times the pressure of the reactants, which means there are more reactants.

Okay, that part makes sense. But then how does that correlate with the system being in equilibrium?

The value of $\Delta G$ does not indicate anything about equilibrium. But an example might be CH4 + 2H2O <==> CO2 + 4H2 under extreme pressure.

I don't understand; I thought $ΔG = 0$ indicated equilibrium, such as described here: http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch21/gibbs.php#relation

It's been too long and I tired. You're absolutely right.

The equation is different, the web page has $\Delta G° = \Delta G + RT$ ln $K$. $\Delta G$ is 0 at equilibrium, so $\Delta G°$ is $-RT$ ln $K$ or −8.31∗298∗ln(0.100) J.

You are mixing up two different quantities. The equation you are using describes the standard Gibbs free energy of reaction $\Delta G^{0}$, i.e. $\Delta G$ for standard conditions without any requirement of the reaction being in equilibrium.
And this quantity can very well be different from zero.
For non-standard conditions the Gibbs free energy of reaction is
\begin{equation}
 \Delta G = \Delta G^{0} + RT \log K
\end{equation}
and as you stated correctly this quantity must (per definition) be equal to zero for equilibrium conditions.
For a derivation see this answer of mine.


text:
What is the minimum distance of Rubidium and Iod ions in a centered cube?
I have a problem to visualize the following assignment:
What is the minimum distance between $\ce{Rb^+}$ and $\ce{I^-}$ ions if radius of  $\ce{Rb^+}$ is ${1.49\cdot10^{-10}}~\mathrm{m}$ and radius of $\ce{I^-}$ is ${2.17\cdot10^{-10}}~\mathrm{m}$ if they are arranged in centered cube?

I guess you have a problem visualizing the "centered cube". The problem is that the expression is ambiguous: You can either have a face-centered cube or a body-centered cube.

The distance between $\ce{Rb^{+}}$ and $\ce{I^{-}}$ in the fcc lattice would be $(d(\ce{Rb^{+}})/2) + (d(\ce{I^{-}})/2) = 3.76 \times 10^{-10} \mathrm{m}$

Thank you, can you please clarify the solution? How can we explaint that ions do not touch each other? Why is there a gap between them?

There is not really a gap between the ions since they are closely packed in the structure. The calculated distance is the distance between the "centers" of the ions when you imagine them as spheres, that means, the distance between the nuclei.

Thank you. I guess it was my interpretation of the question.

Unfortunately this is wrong. The distance between the centres of the ions is the sum of the radii, not only half of the radii.

@Martin You are right. In the formula above, the ion diameter $d$ should be used instead of $r$. I have edited this accordingly.

@Jannis You are presenting me with the wonderful opportunity to use the delete button :)

@Martin It was my pleasure. :)


text:
Indonesian Chemical Olympiad: Calculating mass from predicted reaction
This problem is taken from the Indonesian Chemical Olympiad.

A sample containing barium peroxide was added to a glass full of sulfuric acid. It produced an insoluble product and its filtrate. The filtrate was reacted with potassium iodide in acidic conditions. The product was titrated with $11.8 \,\mathrm{mL}$ of $0.1 \,\mathrm{M}$ sodium thiosulfate.

Write the reaction above.
What is the mass of the barium peroxide in the sample?

Actually I know if the first reaction will make H2O2, but when this peroxide reacted with potassium iodide, I'm confused what is the product.

@lambda23 start considering the following: first obtained $\ce{H2O2}$, then yellow $\ce{I2}$, then titration yielding $\ce{S4O6 2-}$ and $\ce{I-}$, both colorless.

In sulfuric acid, bariumperoxide generates hydrogen peroxide and barium sulfate (= precipitate).
$\ce{BaO2 + H2SO4 -> H2O2 + BaSO4 v}$
Under acidic conditions, iodide is oxidized to iodine.
$\ce{H2O2 + 2I- + 2H3O+ -> I2 + 4H2O}$
With an excess of iodide, triodide is formed.
$\ce{I- + I2 -> I3-}$
The amount of iodine (in triodide) is determined by titration with thiosulfate:
$\ce{I3- + 2 S2O3^2- -> S4O6^2- + 3 I-}$
I leave the math for you :)


text:
Why is filter paper fluorescent?
Typical filter paper used in chemistry lab (e.g. Whatman #1) exhibits autofluorescence in near UV with excitation maximum at 365 nm and emission maximum at 440 nm. It was reported in literature1 that fluorophores from filter paper can be extracted by PBS buffer in wide range of pH, and thus they are at least partially soluble in water. However, cellulose, the major component of filter paper, is not fluorescent in this wavelengths range and is insoluble.
What is the source of filter paper fluorescence? I assume it comes from paper pretreatment and bleaching, but what is chemical identity of fluorescent compounds in it?

Danielson, T. L. et al.; Impact of filter paper on fluorescence measurements of buffered saline filtrates. Talanta 2003, 59, 601–604.

In the production of white paper, the bleached pulp is often treated with optical brightening additives (OBA), that absorb in the uv and emit in the blue range. If I remember correctly, a typical example is 4,4'-diaminostilbene-2,2'-disulfonic acid.

Here's an article from The Open Textile journal (OA) in which the compound mentioned above is used as a starting material in the synthesis of another optical brightener for textiles.

I asked this question when looking for background information for one of my PhD projects and I gained a lot of insight since, which I wanted to share. If you are interested, the project outcome was published recently$^1$. In short - autofluorescence of filter paper originates both from OBAs as well as from cellulose fibrils itself.
There is several sources of fluorescence in papers and other lignocellulosic materials including both optical brightening additives (OBA) as @Klaus pointed out as well as cellulose itself. In case of biomass, lignin is another strongly fluorescent component, that is however removed in manufacturing process of paper.
There exist few earlier literature accounts proving that pure cellulose itself exhibits weak near UV fluorescence, including this paper by Olmstead and Gray$^2$. Here Emission spectra of various cellulose samples at $\lambda_{ex}=320$ nm was recorded (Fig. 1). Bacterial cellulose is know to be fully lignin-free, so there is no doubt that the spectra comes from cellulose fibrils themselves.

The absorbance spectrum of multi-component biomass materials is more complicated. I measured the Raman spectra of filter paper and two unbleached biomass samples and confirmed the cellulose was present in all of them (Fig. 2C). Lignin was absent in filter paper, but present in biomass samples. Next, I performed excitation/emission autofluorescence spectral scans of same samples in UV-VIS range (Fig. 2D). I found dominant peak at 360/420 $\lambda_{ex} / \lambda_{em}$ present in all specimen that can only be attributed to cellulose. Filter paper had a a few more sharp fluorescence peaks at lower wavelengths, absent in unbleached biomass, that I attributed to OBAs. Biomass sampled exhibited a broad peak shoulder at higher wavelengths that originates from lignin and was absent in delignified filter paper.

Fig. 1 Pretreated biomass substrate characterization. Side and top view of cylindrical discs (6 mm in diameter) produced from (A) napier and (B) miscanthus perennial grass. (C) Raman spectra of pretreated biomass substrates using 568 nm excitation. Bands at 380, 435, 1095, 1123, 1377, and 2985 cm$^{−1}$ were attributed to cellulose, with lignin contribution visible at 1600 cm$^{−1}$. Spectra were background-corrected and vertically offset for clarity. Excitation/emission autofluorescence spectral scans of (D) filter paper, (E) miscanthus, and (F) napier grass exhibited a prominent cellulose peak at ≈365/430 nm $\lambda_{ex} / \lambda_{em}$. A lignin shoulder at longer wavelengths was present in miscanthus and napier grass samples.
One question I was unable to answer is what is the molecular origin of cellulose fluorescence ? The origin of lignin fluorescence is well understood $^{3,4}$, with distinct fluorophores formed by aromatic moieties in lignocellulose structure. Cellulose, on the other hand, does not have an aromatic character. I came across a hypothesis that fluorophores are formed when glucose rings in cellulose chains interact with each other via van der Waals and stacking interactions forming cellulose nanocrystals. That would suggest that purely amorphous cellulose should not be fluorescent. I was, however, unable to find studies on that topic.

Malinowska K, Rind T, Verdorfer T, Gaub HE, Nash MA (2015) Quantifying Synergy, Thermostability, and Targeting of Cellulolytic Enzymes and Cellulosomes with Polymerization-Based Amplification. Anal Chem 87:7133–7140.
Olmstead JA, Gray DG (1993) Fluorescence emission from mechanical pulp sheets. J Photochem Photobiol A 73:59–65.
Albinsson B, Li S, Lundquist K, Stomberg R (1999) The origin of lignin fluorescence. Journal of Molecular Structure.
Radotić K et al. (2006) Component analysis of the fluorescence spectra of a lignin model compound. J Photochem Photobiol B 83:1–10.

You might want to check out the following links:
https://link.springer.com/article/10.1007/s11426-013-4923-8
https://onlinelibrary.wiley.com/doi/abs/10.1002/adma.201401356
https://www.sciencedirect.com/science/article/pii/S1001841716301498?via%3Dihub
https://onlinelibrary.wiley.com/doi/abs/10.1002/pola.28420
The first is a paper that reports on the fluorescence/phosphorescence of carbohydrates, whilst the others are all reviews that at least discuss compounds that are luminescent upon aggregation and possible reasons why the phenomena occurs.
I realise that I'm a few years too late here, but hope it as of interest to anyone that comes across this thread (as I did) when trying to understand the phenomena!

Could you please add the relevant information from the papers into the answer itself?


text:
Meaning of "mono-N-demethylation"
I know what demethylation is.
I have no idea what mono- means, the N is nitrogen of course, I know that from the line structure image in front of me if I could not guess it - and after quite a bit of google searching I gave up and came here.
Chemistry is not my field but out of curiosity and for no practical purpose whatsoever I started learning a few things outside my field, and I came across this tiny little naming issue. I thought this is so simple I could easily find something, but apparently it is even MORE simple than I thought - meaning everybody uses the term but nobody bothers to explain it. :-(

If the starting material were colloquially named "stuff," the product would be called N-desmethyl stuff.

@UncleAl Thanks, but that wasn't the question. I already knew how structures are named. The question was the naming modifier of a process, the "mono" ("demethylation" was clear). Now I learned it means that ONE methyl group was removed (and not more). So, not a chemical structure naming question but a chemical process naming one.

Mono-N-demethylation means the removal of a single (mono) methyl group from a nitrogen(N)-containing functional group. For example, a reaction which transforms the alkylated amine group $\ce{-NHCH3}$ into $\ce{-NH2}$.

You might know the demethylation from removing a methyl group from an oxygen atom. A methoxy group $\ce{-OCH3}$ is converted to a hydroxy group $\ce{-OH}$.
This is correctly called a $\ce{O}$-demethylation.
The $\ce{N}$-demethylation is the same - for a nitrogen atom.
The mono prefix denotes that only one (of several) methyl groups is removed from a nitrogen atom.

In organic chemistry, when you speak of mono-N-demethylation, normaly you are refering to an amine bearing 2 or more methyl groups. Let's say you have N,N-dimethylaniline and you want to get N-methylaniline, then you are talking about removing ONE methyl group thus mono-N-demethylation. For organic chemists out there... Yes I know you cant do that one easily and specifically but one keeps dreaming...