text:
How to balance more complex redox reactions?
I'm in an advanced level chemistry course, and so far we've been introduced to balancing redox reactions. The basic ones like
$$\ce{MnO4- + Cl- -> Mn^2+ + Cl2}\tag1$$
are pretty easy for me, but I'm having trouble with equations where it's not apparent what I exactly need to put in my half reactions. Some examples of these kind of equations are:
\begin{align}
\ce{Pb + PbO2 + SO4^2- &-> PbSO4}\tag2\\
\ce{Ag(s) + CN- (aq) + O2(g) &-> [Ag(CN)2]- (aq)}\tag3\\
\ce{Cu + HNO3(aq) &-> Cu(NO3)2(aq) + NO2(g) + H2O(l)}\tag4
\end{align}
I'm not looking for solutions to any of these problems. I'm just looking for ways to correctly construct my half reactions before balancing, adding water and hydrogen ions, and balancing the charges.
Like in the first reaction on the reactants side, the first $\ce{Pb}$ has a charge of $0$ while the next one has a charge of $+4$. The $\ce{Pb}$ on the products side however has a charge of $+2$. It seems like one is being oxidized and the other reduced. And I'm also not sure what to do with the sulfate.

I wrote an article about this not too long ago... Check out the manuscript at the following link: http://www.sciepub.com/portal/downloads?doi=10.12691/wjce-3-3-4&filename=wjce-3-3-4.pdf

Your first reaction is an example of a comproportionation reaction. These are a little tricky to recognize and balance--you're absolutely correct in saying that one is oxidized and the other reduced.

Once you know the method, balancing any redox equation that you will get in AP chem will be quite easy. All you have to do is follow this simple method:

Balance all the elements in the equation except for $\ce{O}$ and $\ce{H}$. For polyatomic ions that are spectator ions, just think as them as one big element.
Check if $\ce{O}$ are balanced. If they aren't, add water molecules to the side, witch doesn't have enough $\ce{O}$.
Check if $\ce{H}$ are balanced. If they aren't, add $\ce{H+}$ ions to the side, witch doesn't have enough $\ce{H}$.
Now that all the elements are balanced, make sure that the charges are balanced. If not balanced, add electrons to the side which is more positive.
Check your equation now to make sure that everything is balanced correctly.

I will do the first equation as an example for you. You should be able to balance rest of the equations by following this procedure.

Balanced elements except for O and H: Here the sulfate ion is a spectator ion, so you can imagine it as one big element: $$\ce{Pb + PbO2 + 2SO4^2- -> 2PbSO4 }$$
Balance oxygen: $$\ce{Pb + PbO2 + 2SO4^2- -> 2PbSO4 + 2H2O }$$
Balance hydrogen: $$\ce{Pb + PbO2 + 2SO4^2- + 4H+ -> 2PbSO4 + 2H2O }$$
Balance charge: charge is already balanced. Don't need to add electrons.
Check equation to make sure everything is balanced correctly: all the elements and charges are balanced correctly.

So your balanced redox equation for the first example is: $$\ce{Pb + PbO2 + 2SO4^2- + 4H+ -> 2PbSO4 + 2H2O }$$
Note: If the question asks to provide the redox equation in basic conditions there are two more steps.

Add the same number of $\ce{OH-}$ ions as $\ce{H+}$ to the products and reactants. The $\ce{OH-}$ ions and $\ce{H+}$ will react to form water.
Cancel out the water molecules from both sides (a bit like algebra).

That should give the redox equation in basic conditions.

Great answer.  Balancing reactions is just a way to express that mass, charge, and moles of .atoms are conserved

What do we do with the leftover electrons?

@Aditya From the equations in the question, you should not get leftover elections since these represent redox reactions. However, if you were asked to write the half-equation for an oxidation or reduction equation, then it is perfectly fine to have leftover elections, in fact you must have leftover elections, otherwise you are doing something wrong, E.g. half-equation for oxidation of iron is: $\ce{Fe} \rightarrow \ce{Fe}^{2+} + 2e^-$.

Yes, in this particular case balancing was not required since charges were equal. But what if the charges weren’t equal? You said we have to add electrons to balance charges, but how do we eliminate them?

@Aditya In that case, you will need to use the above process to write the 2 half-equations (which should have electrons on the opposite side). Once you have done that, you will need to multiply each half-equation by an appropriate number so that when you had the 2 half-equations, the electrons cancel each other out.

A quick example to illustrate: Consider the reduction of lead ions by aluminium. The oxidation and reduction half-equations are respectively, $\ce{Pb}^{2+} + 2e^- \rightarrow \ce{Pb}$ and $\ce{Al} \rightarrow \ce{Al}^{3+} + 3e^-$. As you can see, the electrons are on the opposite sides as expected. However we cannot add the 2 half-equations right away since the electrons will not cancel out. Instead, we need to multiply the first half-equation by $3$ and the second half-equation by $2$ so that the number of electrons for both half-equations are $6$. Now you can add them together.

For $Fe+NH_3\rightarrow Fe(NO_3)_2+NH_4NO_3+H_2O$, Fe is getting oxides and N is getting reduced. So the half reaction for Fe will be $Fe\rightarrow Fe(NO_3)_2$, but then there is no $NO_3^{-1}$ term on the LHS. How do we balance that then? The number of electrons cannot be determined without it

@Aditya A useful thing to do before trying to balance an equation, and in fact when you see any equation, is to ignore all the spectator ions. This is because they do not participate in the reaction, so by ignoring them, you can better see what is actually happening. In this case, nitrate is the spectator ion, so by ignoring them, you should easily see that what is happening is that $\ce{Fe}$ is getting oxidised to $\ce{Fe}^{2+}$. So the half-equation you have written is incorrect, there should not be any nitrate ions in it.

Then how do we compensate for the missing nitrate ion?

@Aditya sorry not sure what you mean by "missing nitrate ion". If you ignore all the spectator ions, there will not be any nitrate ions at all, so there is no need to try to balance it. If you still have any questions, I suggest it will be better if you ask this as a separate question. That way you will be able to get a comprehensive answer rather than a brief explanation in a comment.

We eliminate the nitrate ion, but we still need it in the final reaction


text:
Replacement for cyclohexane
I am attempting to isolate pure iodine from kelp using the method described here
but I do not have any way of possibly obtaining cyclohexane.  I was wondering if any easier to obtain solvents would work as well to isolate the iodine from the rest of the mixture.  I was hoping to use a solvent that is easy to obtain in a grocery or hardware store.

@ K_P Methylene chloride and chloroform are polar and both immiscible with water but actually have a very minimal solubilty .

Cyclohexane is used for this extraction because it is nonpolar, as is iodine. A more polar solvent like acetone or ethanol will also extract a whole bunch of other stuff like essential oils, proteins, and other polar compounds.
You want Mineral Spirits which you should be able to purchase at any quality hardware store or Walmart. Mineral spirits, mineral turpenine, or whatever you want to call it, is a mixture of saturated hydrocarbons distilled from petroleum. It is a general purpose paint thinner for oil-based paints. Since it is mainly C7-C12 alkanes, it will have a higher boiling point than cyclohexane $(140\ ^\circ\mathrm{C}$ vs $81\ ^\circ\mathrm{C})$., but it will have the same general polarity as cyclohexane.
If you can find products labeled as petroluem ether, light naphtha, or ligroin, all are alkane mixtures with lower boiling points than mineral spirits.
Another common low polarity paint thinner is toluene, which has a lower boiling point at $110\ ^\circ\mathrm{C}$.

n-Pentane might work. It is very discriminating for nonpolar substances and its low BP, 36C, means it will be easily distilled from the iodine and recaptured. You would have to buy it from a chemical supply firm or online. Cyclohexane, BP 80C, is used because it will evaporate relatively easily and cleanly; slower evaporation rate could give larger iodine crystals. Mineral paint thinner is higher boiling and might be difficult to remove. Shop around, different brands could have different boiling ranges. Saturated hydrocarbons also have lower toxicity than aromatics like toluene or benzene, but inhalation and skin contact should be minimized and avoided. These solvents are all flammable.

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text:
How does the "kreosan" Coca-Cola rocket work?
A YouTube channel called "kreosan" (made by Russian-speaking people who I believe live in Ukraine) posted a video showing rockets made from bottles of Coca-Cola.
Here's what I've figured out from the video.

First step: pour out some Coca-Cola to make room in the bottle to add gas.
Second step: add butane (looks like a canister for refueling butane lighters)
Third step: this is unclear.  He adds a gas from a large container that appears to say "GAS-1".  The YouTube description says "Coca Cola + propane = turbo ROCKET" so I guess it's propane.

Unknown: Does the rocket require both butane and propane?  I think the second
   step was just to show that you don't get very much gas out of a
   little butane canister, and all you really need is the propane from
   the last step.

Final step: invert the bottle, and some reaction pushes Coca-Cola out the mouth of the bottle with enough force that the whole bottle flies quickly as a rocket.

What is going on here?  How does inverting the bottle trigger the reaction?  Is this like the "Mentos and Diet Coke" thing where the trick was to get the carbon dioxide dissolved in the drink to come out quickly?
My understanding is that the surface of Mentos candy provides a bunch of nucleation sites for bubbles, causing a lot of carbon dioxide to come out of solution quickly.
So, does the "kreosan" rocket fly on just carbon dioxide or is there some other reaction going on here?

You should ask your question in PSE. Obviously its also chemistry but I don't think chemistry will get you the entire answer. :)

The physics is rather obvious: gas expands at high speed, bottle flyes. What's a lot more interesting is the chemical reaction causing such a high increase in pressure...

Here's a paper written about propane forming small ice crystals when sprayed into a cloud of water vapor: James R. Hicks and Gabor Vali, J. Appl. Meteor., 1973, 12, 1025–1034.
I'm no expert, but I'm guessing the propane forms small ice crystals as it passes through the Coca-Cola, and when flipped, the Coca-Cola warms up the liquid propane, which causes it to expand as it travels upwards. This provides an immense surface of nucleating sites, similar to a Mentos reaction, but with millions of microscopic Mentos (this means larger surface area). The pressure waves caused by the violent nucleation further drives the reaction, as the liquid propane expands further and freezes more ice crystals and so on. This all happens quite fast, and with one easy way out, everything shoots out the mouth of the bottle.
Of course this was written by a Materials Engineering undergrad, so don't take this explanation too seriously.

Seems plausible. Also in the video there is some white semi-solid layer visible above the Coca-cola surface.
... and probably only brave Russians would launch rockets from hand ;)

This behavior has little to do with using a carbonated beverage. Try search terms butane water rocket to see plenty of other examples of this.
The phenomenon that is happening is the large volume difference between liquidus and gaseous butane (with maybe some $\ce{CO2}$ to boot) forces the dense liquid water out of the vessel at a high pressure, and Newton's First Law takes care of the motion of the plastic bottle relative to its liquid contents.
The Steps:
Adding butane
When the butane is introduced into the container, it is an expanding vapor, which loses some energy from the expansion. Butane condenses into a liquid at about $0°\!\mathrm{C}$, and so when you add butane from the compressed can some of it condenses into the liquid from the heat loss due to expansion, aided by contact with the presumably cold (but still $>0°\!\mathrm{C}$) surface of the cola. Butane's density is about $0.6\,\mathrm{g}\,\mathrm{cm}^{-3}$ and so it floats on top of the sugar solution.
Releasing the container
When the vessel containing the two liquids (cola on bottom, butane nearer the mouth) is inverted, the (relatively) warm water (or cola, or otherwise) comes in intimate contact with the liquid butane. The high surface area of the interaction causes the temperature of the butane to be immediately raised above $0°\!\mathrm{C}$, and is given plenty of energy to enter the gas phase. In the case of cola, I suppose it is possible that during all of this, droplets of butane are providing nucleation sites for $\ce{CO2}$ as well, but as the whole effect can be shown to happen with water, it is rather beside the point.
The change in volume
As mentioned, the density of liquid $\ce{C4H10}$ is about $0.6\,\mathrm{g}\,\mathrm{cm}^{-3}$, so with a molecular weight of about $58\,\mathrm{g}\,\mathrm{mol}^{-1}$ and an ideal gas volume of $22.4\times10^3\,\mathrm{cm}^3\,\mathrm{mol}^{-1}$. Each $\mathrm{mL}$ of butane contains about $0.01\,\mathrm{mol}$ of $\ce{C4H10}$, and the liquid butane will expand to $230$ times the volume of the liquid that is added initially (using ideal gas approximation). If the liquid is about $1\,\mathrm{cm}$ deep and the diameter of the bottle is about $7\,\mathrm{cm}$, that is about $40\,\mathrm{cm}^{3}$ liquid, or about $10\,\mathrm{L}$ of gas.
If we take a look at the answer to another question about soda, we see that there is about $0.14\,\mathrm{mol}\text{ of carbon dioxide per }\mathrm{liter}\text{ of soda}$, and so if all of the $\ce{CO2}$ is released simultaneously (like the memetic Mentos example), it adds about $3\,\mathrm{L}$ of gas to our expansion.
The upshot:
The gas is aching to get out of the cola bottle, however now the liquid in the inverted bottle is standing between the gas and its equilibrium (atmospheric) pressure. The liquid, being a fluid, is forced out of the mouth of the bottle. Presumably the user had a loose grip on the vessel, which is then propelled in the opposite direction from the liquid plug with the equal and opposite momentum. For reference, since conservation of momentum states $\mathrm{\mathbf{p}} = 0 = m_1\cdot\mathrm{\mathbf{v_1}} + m_2\cdot\mathrm{\mathbf{v_2}}$ a $10\,\mathrm{g}$ soda bottle would travel fifty times faster than $500\,\mathrm{g}$ of liquid it is being forced to release, as in a partially empty bottle of cola.
Referring to the Wikipedia article on Water rockets, pressures from bike pumps ($75\,\mathrm{psi}$) and air compressors ($\mathop{>}200\,\mathrm{psi}$, said to be dangerous) are used to launch this type of rocket. Since we are starting from a non-enclosed vessel, we wouldn't really have static pressure but the $13\,\mathrm{L}$ of gas that we are generating (at final pressure $1\,\mathrm{atm}\approx 14.7\,\mathrm{psi}$) would have some equivalent. If confined to say, a $250\,\mathrm{cm}^3$ headspace, this would correspond to something like $750\,\mathrm{psi}$ to $1000\,\mathrm{psi}$!

I have no professional knowledge about chemistry, so what's following are just my personal thoughts on the matter.
My first reaction to the video was "that's nonsense" but I have since seen other people repeating this, so I'm not so sure about that anymore.
Propane, Butane and Ethane have boiling points around or below $\pu{0 ^{\circ}C}$, so if there was any liquid (or solid) lighter fluid in the bottle before the reaction takes place it should be boiling. I haven't seen anything of that sort. Furthermore these compounds are nearly insoluble in water especially in the arrangement they used with a very small contact surface between the Cola and the gas. That's why I doubt that the explanation for the reaction is any kind of chemical reaction between the lighter fluid and the Cola.
If this is actually real, my best guess would be, that there is some kind of reaction taking place between the bottle wall and the gas (thermal shock or chemical reaction?) which then gives a similar reaction to the Mentos - Cola reaction just with a much bigger contact surface. I doubt however that you would get one liter of $\ce{CO2}$ from one liter of Cola (under normal conditions) and I'm not sure whether that would be enough to produce such a violent reaction, if you can't even replace all the liquid with gas.
It would be interesting (and dangerous, I suppose) to see this repeated with different kinds of containers (glass, aluminium can, etc.).

Propane’s boiling point is around $-40~\mathrm{^\circ C}$, ethane’s around $-80~\mathrm{^\circ C}$.

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text:
Is liquid oxygen non flammable?
Every one knows that liquid oxygen is flammable, but why this tank labeled like this:

so is it really liquid oxygen is non flammable?

Liquid oxygen is non-flammable, that's why. See this for instance, or this.

Because "flammable" is not a synonym to "increases the risk of devastating fire", though "everyone" (by your definition) thinks it is, and this question just keeps popping up every now and then.
Come to think of it, would you call the flame itself flammable?

To quote one of my professors: "Liquid oxygen is, by itself, not all that dangerous. Of course, if you accidentally left a little bit of oil on one of the surfaces you pour it onto, it'll do it's best impression of the Challenger and take your head off."

It is flammable in hydrogene atmosphere :-)

Oxygen is not flammable, as it does not burn, it supports burning. For burning to happen, a strong oxidizer (for example, oxygen) and a strong reducer (for example, carbon) must be present.
No, the problem with oxygen is, that while not being flammable, it is much more efficient in supporting burning. In air oxygen is mixed with nitrogen, meaning that for the same amount of reducer a quadruple amount of oxidizer is required and a lot more heat is lost with inert gases (nitrogen). For example, iron wire can burn in oxygen, but not in air. It is also a lot easier to lit things up in oxygen. A classic experiment is to move a smoldering splinter into a vial with pure oxygen. It immediately starts to burn with a bright flame
Now, liquid oxygen is an entirely new level of crazy. In addition to constantly releasing gaseous oxygen... If an organic material (for example, sawdust) is mixed with liquid oxygen, it becomes an explosive.

Technically, Oxygen is almost never flammable. It oxidizes other materials quite easily and many materials will burn in the presence of oxygen.  But oxygen will never actually burn itself except in the presence of fluorine.

As many comments point out, liquid oxygen is not (in)flammable.  But you might see the label "Oxidizer".  That means it strongly supports or could set off combustion of things that are flammable.


text:
Why molten lithium burns through the glass but sodium not?
It's well-known in laboratory practice that one shouldn't melt lithium in a glass equipment and there isn't such rule about other alkali metals.
You may damage the glass with sodium but lithium will make a hole in your flask.
So, why does this rule only hold for lithium?

...  I definitely recall that one of methods of preaparation of buthyllithium includes melting of lithium in boiling dodecan. I even participated in making of one bunch using this method. Worked like a charm with no problems.


text:
Does the shared electron in Covalent bonds revolve around nucleus?
We know that electrons are charges that revolve around the nucleus. Then, when in covalent bonds the electron is shared; does the electron obey the rule?

We don't know that electrons "revolve" around the nucleus. That is a pre-quantum mechanics view of atomic structure.

@matt_black Not exactly, it may be not compatible with Copenhagen interpretation, but it's not so bad for Bohmian mechanics.

I am guessing that you have an understanding of the atomic orbitals of an atom such as 1s and 3p atomic orbitals. Atomic orbitals are the probability distribution of where an electron is going to be $90$% of the time. Technically, electrons don't revolve around the nucleus but are rather quantum particles that are superimposed in several positions at the same time. However the description of electrons revolving randomly around the nucleus is suffice at this level.
When a covalent bond is formed, the electrons do still continue to revolve around the nucleus in orbitals, however a new kind of orbital is formed which are called molecular orbitals (rather than atomic orbitals). Here is a simple explanation of what molecular orbitals are.
Molecular orbitals are formed by constructively and destructively overlapping the atomic orbitals together. The simplest example of this is when $\ce{H2}$ is formed. Here, each H atom has a 1s atomic orbital. When they form, these atomic orbitals form 2 molecular orbitals. Where one molecular orbital is due to constructively overlapping the 2 atomic orbitals and the other molecular orbital is due to destructively overlapping the 2 atomic orbitals. This shown in the diagram below:

As you see that these 2 molecular orbitals that are formed are quite different to each other. The MO that is formed due to constructive overlap has greater electron density between the 2 positive nucleus. Meanwhile the MO that is formed due to destructive overlapping has decreases electron density between the 2 positive nucleus and in fact have a node between them. Therefore the first MO is lower in energy than the second MO.
Like AOs, the same rules apply to MOs. Each MO is able to hold 2 electrons and the electrons fill the lowest energy orbital first. So in $\ce{H2}$ the 2 electrons will fill the first MO and will revolve around the nucleus in that orbital (well technically only for $90$% of the time).
This theory can be applied to any compound, however it gets more complicated as the number of AOs increases and the number of atoms increases. However essentially, when atoms form covalent bonds, the electrons continue to revolve around the nucleus in MOs rather than AOs.

Electrons are essentially quantum particles, and are controlled by the laws of quantum mechanics. There is absolutely no way to reconcile these laws with our physical intuition of macroscopic world. Electrons do not revolve; rather, they occupy certain atomic orbitals. To put it simply, they just stay there, delocalized over some space around the nucleus, everywhere at once. In a covalently bound molecule, electrons would occupy molecular orbitals and thus get delocalized all over the molecule.
The description of electrons as tiny charged solid particles that revolve around the nucleus, much like planets revolve around the sun, is the approximation known as Bohr atom. Significant for historical reasons as it is, one should not mistake it for coherent description of electron's behavior. Also, I doubt if you can extend it consistently to model covalent bonds.

The best answer is that the electron does not "revolve" around or "orbit" the nucleus the same way that the planets orbit the sun.
Pictures like the following (from Wikipedia user Halfdan) are compelling, but refer to a model of the atom that was beginning to be supplanted by quantum mechanics a century ago.

The modern quantum-mechanical model of the atom incorporates things like the Heisenberg Uncertainty Principle and wave-particle duality. We now consider electrons in atoms to behave much more like waves. We can determine their energy very precisely, but as a consequence of the uncertainty principle, we cannot determine their position or momentum very precisely.
The atomic orbital pictures we are used to seeing for $s$, $p$, $d$, etc. orbitals are constructs describing regions of space where the electron is. Typically, we plot the 90% surface. In one interpretation that means 90% of the time the electron can be found in that region of space. In another interpretation, 90% of the electron-wave is confined to that space.
When atoms come together to form bonds, the electron-waves overlap to produce new orbitals shared by the atoms. Nanoputian's answer provides a graphical representation of how this works.

There was no much advance in quantum mechanics “a century ago”. Planck’s patches to save physics from the UV catastrophe? Einstein’s theory of photoelectric effect? Even first Schrödinger’s successes were as late as in 1920. What do you refer to?

@IncnisMrsi 1920 is almost a century ago - I think this is only indicated as a rough timescale.

The Bohr model of the atom, which included quantized energy levels (but also still included electrons as particles) replaced the model in my picture in 1913, and it was quickly apparent that it only worked for hydrogen. So, a century ago, classical models of the atom were already on their way out.\


text:
How do meth lab test kits detect the presence of a meth lab?
Many houses in the USA are illegally used as meth labs to produce methamphetamines, which can cause many negative health effects in their inhabitants, even after the house is sold to the next unwitting owner.
The Denver Post estimates that 7% of single family and 11% of multi-family units that are foreclosed have traces of meth residue.
With these risks, many commercial test kits for meth lab residue have appeared. However, many of them are intentionally vague about their chemical mechanisms.
What chemical mechanisms do meth lab test kits use to detect the presence of a meth lab?

It seems rather sketchy to me. Cannot really imagine that there is a "significant" health hazard associated with being in a room in which meth has been smoked on some occasions. Especially if it has been cleaned up properly. But I might be wrong. Interesting question +1

Simple self-made or commercially available tests use the Marquis reagent, which is a mixture of sulfuric acid and formaldehyde. For example, Marquis reagent can be prepared by adding 100 ml of concentrated sulfuric acid to 5 ml of a saturated aqueous solution of formaldehyde (40 % formaldehyde by volume or 37 % by mass).[reference]

The Marquis reagent reacts with many alkaloid drugs to form a variety of colours. Methamphetamine creates an orange dye (deep reddish orange to dark reddish brown).

After reading the paper, it seems that Simon's Reagent (A12) is more specific towards methamphetamines.


text:
Redox Reaction Equation with elusive oxygen
I've been balancing redox equations for the last couple of weeks and have become quite accustomed to the procedure however I've come across an equation that eludes me every which way when it comes to balancing oxygen. I come to the next to last step to balancing my redox equation but am unable to write my net ionic equation.

Liquid hydrazine reacts with an aqueous solution of sodium bromate. Nitrogen gas and bromide ions are formed.

I am asked to balance this reaction in an acidic solution.
I first write out my equation:
$$\ce{N2H4(l) + NaBrO3(aq) -> N2(g) + Br- (aq)}$$
I break my equation up into two half equations, one being my oxidant equation and the other my reductant. I balance them both:
$$\ce{N2H4(l) -> N2(g) + 4H+ (aq) + 4e-}$$
The other half of the equation is where I get stuck with balancing the oxygen. I end up balancing the hydrogen on both sides to be 10 but I come up short of one oxygen on my product side.
$$\ce{NaBrO3 (aq) -> Br- (aq)}$$
I use the algebraic method to assign an oxidation number to $\ce{Br}$ on reactant side of half equation.
$\ce{Br + Na + 3O = 0}$
$\ce{Br + 1 + 3(-2) = 0}$
$\ce{Br - 5 = 0}$
$\ce{Br = 5^{+}}$
$\ce{Br}$ on the product side has an oxidation number of -1 equal to its net charge
In order to balance the oxidation charge of both sides, I add $\ce{6e-}$ to the reactant side. This then brings the oxidation charge to -1 on both sides of the half reaction.
$$\ce{NaBrO3(aq) + 6e- -> Br- (aq)}$$
I then compensate net charge of the acidic reaction by adding 5 hydrogen ions to the reactant side to bring the net charge on both sides to 1-.
$$\ce{NaBrO3(aq) + 6e- + 5H+ (aq) -> Br- (aq)}$$
I then must add water to the product side in order to balance Hydrogen and oxygen atoms in the half reaction. I have 5 hydrogen for reactants so I must add 2.5 water molecules on the product side to match hydrogen.
$$\ce{NaBrO3(aq) + 6e- + 5H+(aq)->Br- (aq) + $2.5$H2O(l)}$$
Staying customary to the law of combining volumes and Avogadro's hypothesis I multiply the half equation by 2 to get a whole number for water
$$\ce{2NaBrO3(aq) + 12e- + 10H+(aq) -> 2Br- (aq) + 5H2O(l)}$$
The oxygen atoms here elude me. I can't figure out how to balance them. I figured I took a wrong turn on the way but my troubleshooting comes to no avail thus far. Can anyone fill me in on this little secret?

You have Na on the left, but none of it on the right. You'd better fix this one way or another (preferably by not having it on either side, but that does not really matter much, just be consistent).

Now if you need to use $\ce{OH-}$ ions. Use the same scheme:
$$\ce{BrO3- -> Br-}$$
There is no oxygen in the right part so we need to add $\ce{H2O}$ and/or $\ce{OH-}$ ions somewhere. Look closely at $\ce{H2O}$ and $\ce{OH-}$. Water have 2 hydrogen per oxygen atom and $\ce{OH-}$ have 1 hydrogen per oxygen atom. Problem here is that we can't just add element to one side. When we add $\ce{H2O}$ or $\ce{OH-}$ we change both oxygen and hydrogen amounts. We will need to solve system of equation for that. First equation will be hydrogen balance and second one will be oxygen balance. What we need to add to the left part and what we need to add to the right part? As you can see $\ce{OH-}$ contains the same amount of oxygen and less hydrogen compared to $\ce{H2O}$. The only way to balance those two equations is by adding $\ce{OH-}$ to the oxygen poor (right) part. It means that we will add water to the left part. Now lets make equations, $a$ will be the amount of water, $b$ is the amount of $\ce{OH-}$.
Hydrogen: $2a=b$
Hydrogen can be contained in $\ce{H2O}$ or $\ce{OH-}$ only, but one $\ce{H2O}$ molecule contains two times more then $\ce{OH-}$. It means we need to use 2      multiplier.
Oxygen: $3+a=b$
Because we already have 3 in left part.
Solve this system. We get $a=3$ and $b=6$.
$$\ce{BrO3- + 3H2O -> Br- + 6OH-}$$ Wow, this works! Now equalize charges.
$$\ce{BrO3- + 3H2O +6e- -> Br- + 6OH-}$$
P.S. Adding water to the right side will make our system unsolvable in natural numbers. Take a look.
Hydrogen: $b=2a$
Oxygen: $3+b=a$

Thank you @Cosmodruid. System of equations is a great tool to use.

First of all you need to write equations with real ions and molecules that do exist in solution. Your oxidation agent here is $\ce{BrO3-}$. In neutral media you can use $\ce{H+}$, $\ce{OH-}$ and $\ce{H2O}$. Your half-reaction can be made from this scheme:
$$\ce{BrO3- -> Br-}$$
There is no oxygen in the right part so we need to add $\ce{H2O}$ there, we need 3 molecules.
$$\ce{BrO3- -> Br- + 3H2O}$$
Now there are no H atoms in the left part and 6 H atoms in the right part. Fix this by adding 6 protons, $\ce{H+}$, in the left part:
$$\ce{BrO3- + 6H+ -> Br- + 3H2O}$$
Charge of left part is +5 and charge of right part is -1. Now equalize charges in left and right parts by adding 6 electrons to the left part.
$$\ce{BrO3- + 6H+ + 6e- -> Br- + 3H2O}$$
Half-reaction is ready.

I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here. Additionally, you can visit this chatroom for more assistance.

This scheme is very useful for solving this acid redox problem. It works and the process is considerably faster than the textbook definition I used above. How would you apply this scheme to a redox reaction in a basic medium dealing with Hydroxide instead of Hydrogen ions? I've tried to apply this scheme to several other redox problems in basic solution but I have been unsuccessful. Thank you for the help you have given me.


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Name for HxOy compounds?
Is there a name for the class of stand alone molecules that contains oxygen, water, hydrogen peroxide, and ozone? ($\ce{O2}$, $\ce{H2O}$, $\ce{H2O2}$, $\ce{O3}$) I would not want to include hydroxyls $\ce{OH}$ because they aren't neutral molecules that I can buy.
I think oxide and oxygenate is too generic.
Thanks for your help!

Oxygen and ozone are oxygen allotropes. Water and peroxide are both hydrogen oxides (dihydrogen di/mono oxides). I don't think there's a name for the four of them.

For C and H, there are hydrocarbons. These do not include the elements and their allotropes, though.

Not only I've never heard of a name for such a class, but the very choice of compounds seems quite artificial to me. What feature do they have in common?

Consisting of H and O? That would exclude oxygen and ozone.
Consisting of H and/or O? That would include $\mathrm H_2$ as
well.
Containing "active oxygen", whatever that might mean? That
would exclude water, but include a bunch of inorganic and organic
peroxides.
Oxide is indeed too generic; also, it does not include oxygen and ozone. Oxygenate is even farther off.

Why would you want to have such a class, really?

The idea was indeed to have a class of compounds that contain an oxygen that can be or is already active. Adsorbing water, ozone, oxygen, or hydrogen peroxide on a TiO2(110) surface for example will all interact with oxygen vacancies present on the surface. (I guess this is more of a surface science kind of question.) I wanted a way to refer to them all without explicitly listing them out each time. I guess there isn't a good way to classify them, thanks for the response!

Now that makes things a lot clearer. But is the class really this narrow? Not being quite familiar with surface science, I still have a strong feeling that any alcohol (and maybe almost any other oxygen-containing molecule) would do just as well. Would it not?

Right, I can call organic compounds oxygenates to encompass alcohols, aldehydes, and ketones but I wasn't sure if there was a term I could use as easily for those without carbon essentially. Thanks for the input!

Probably the best generic description is given by Wikipedia:

Hydrogen polyoxides (also known as oxidanes, oxohydrogens, or oxyhydrogens) are chemical compounds that consist only of hydrogen and oxygen atoms, are bonded exclusively by single bonds (i.e., they are saturated), and are acyclic (have molecular structures containing no cycles or loops). They can, therefore, be classed as hydrogen chalcogenides.

Examples are known experimentally up to $\ce{HO3}$ and $\ce{H2O5}$, and neutral $\ce{H3O}$ has been proposed on theoretical grounds. The polyoxygen species enter into the mechanisms of reactions involving $\ce{O2}$, ozone and hydrogen peroxide, and $\ce{H3O}$ at high pressures may exist in ice-giant planetary interiors.
Molecules that contain a divalent oxygen atom are sometimes named as derivatives of "oxidane", the parent being $\ce{H2O}$, even though that compound per se is of course named "water".


text:
Does an electrically conductive gas exist at room conditions?
As I know, the gases are insulators, because their particles are mainly electrically neutral, and thus there is no movable charges in them.
But maybe it shouldn't be always so, for example if a gas has a significant proportion of charged particles, I think the electrical conductivity isn't impossible. If at least one of the electrons of its particles bounds very weakly, maybe even at room conditions is it possible to have enough low resistance to be considered as a conductor.
Am I right? Does a such gas already exist? Or there is some mechanism which prohibits this idea?
Extensions:

Maybe a such gas could be considered as plasma, but I think from this aspect we could see the plasma as gas as well.
I am thinking at around 1 atm pressure and 300 K temperature, in the gas, not around the box around it. Electrostatic discharge is also problematic, because in most gases it can happen only far from the room conditions and isn't an equilibrium state.
If there is none, I am looking for the gas which is the possible nearest to the these 3 conditions (300 K, 1 atm, conductive).

Related: http://physics.stackexchange.com/questions/38631/can-gases-conduct-electricity.

http://chemistry.stackexchange.com/questions/26326/is-it-possible-to-combust-a-plasma

@Mithoron This question is about room conditions (300K, 1atm).

The gas inside discharge neon lamp is not really hot. Also, think of St. Elmo's fires - these definitely do appear at ambient pressure and are "cold", i.e., normally they would not burn anything. But there is a catch: here we are looking at strongly non-equilibrium situations, caused by strong electric fields.
As for the gases which would behave like that in normal conditions, I haven't heard of such. Some compounds turn into plasma easier than others, but still that requires many hundreds °C. Among the easiest are complex halogenides like $\ce{KBF}_4$ or $\ce{KAlCl}_4$. At 1000K they may be not vaporized yet, but have some vapor pressure. Also, it's not like the vapors consist entirely of ions - no, far from that, the ions are still a minority, but that's enough to create some significant conductivity.

Mercury boils already at 357$^{\circ}\mathrm{C}$! Maybe some metal mixture do even lower! Is it conductive? (Btw, it is not about mercury vapor lamps, I am thinking here on the ohmic resistance of 1atm mercury vapor in equillibric state, without any discharge effect.)

No; why would it? The mercury vapor is atomic.

Because mercury is a conductor, although not a really good one. I think, if the fluid mercury has enough moveable charges, maybe as a vapor should it have as well.

No, it should not. Liquid mercury has some free electrons due to the fact that it is a metal, with metallic bond and all that comes with it. Gaseous mercury does not have metallic bond, or indeed any bond whatsoever; surely it should not have any free charges.

A plasma  is very hot if we consider the distribution of the energies of the different atoms which are present. A plamsa might have very little heat in it compared with a normal gas of the same volume as in many plasmas the density of gas atoms is very low.

@NuclearChemist Plasma inside a gas discharge has strongly non-equilibrium distribution of particle energies, hence the words "hot" or "cold" are not quite applicable to it. If you measure the effective temperature by the average energy, it may pretty well be cold. Think again of  St. Elmo's fires. Their density is definitely not much lower than that of the ambient air.

Depends on over what volume / area you measure / estimate the temperture by average energy. If you consider the whole of the room it will be low. If you consider just at the tip of the spike at which St Elmo's fire is occuring then it will be much higher there

@NuclearChemist Higher than in the whole room, that's for sure, but still not hot enough to cause a fire.

Well to start a fire that is an interesting thought, I know some people use a tesla coil to light bunsen burners in the lab. Regarding the St-Elmo's fire the discharge is able to cause some chemical reactions to occur. I know that methyl iodide is decomposed at a corrona discharge in air. Fire is just a series of chemical reactions so in theroy we could have a corrona combustion. Keep in mind that there are plenty of flameless combustions. Consider hydrogen on a Pt gauze and acetone on hot copper metal, these are two examples of flameless combustions.

It depends on what you are prepared to consider a gas and what you are prepared to consider room conditions.
The gas inside all discharge lamps (fluorescent lamps and neon lamps in shop signs, for example) conducts electricity and the lamps work under normal room conditions. However, the gases are often at low pressure and are, strictly speaking, plasmas not room temperature neutral gases. Some lamps, especially high intensity ones like the xenon discharge lamps in floodlights) exist with high pressure gas when working so low pressure isn't the key.
And air itself will conduct electricity given the right conditions. Lightning couldn't exist if it didn't. Nor could you get the (much safer) discharges from Tesla Coils or Van de Graaff generators. But, again, the effect relies on the creation of a plasma from the breakdown of the neutral molecules in the gas.
So the answer is "yes" but only if you allow plasmas created from the medium to count as a normal gas.

Ok, but although a lamps works obviously on room conditions, the filling gas therein doesn't: it has a high temperature and small pressure for that. The case is similar in lightning: it is only a short-term process and has also very high temperature. The filling gas mix in the halogen lamps doesn't have significantly lower excitation energy as any other, they are only used because the halogen-cycle improves the lifetime of the lamp.

@peterh I think that's halogen filament lamps where the halogen is used to regenerate the tungsten filament. Halogen discharge lamps use doping by metal halide salts to alter the colour of the gas discharge in the gas plasma (some transition metal halides have specific emission lines that can make white light when combined).

Air at STP does conduct a tiny bit due to ionization by cosmic rays; this might even provide a path for lightning leaders. "Alex V. Gurevich of the Lebedev Physical Institute [et al] suggest that... cosmic ray... might provide a conductive path that initiates lightning."
In addition, ionization-type smoke detectors use a little radioactive material (e.g. $\ce{^241Am}$) to increase conductivity, which can be detected with a high-impedance amplifier.

Although I was looking for equilibrium-state gas, thank you very much this answer, too!

+1 for radioactive ionisation.  I expect that in a sufficiently high radiation flux air (or perhaps even easier, some other gasses) could become significantly ionised.  Searching for papers on breakdown voltages and corona discharge inside nuclear reactors might give further clues.

We need to distingish between conduction mechanisms at very high electric field strengths such as that in St-Elmmo's fire and the conduction which occurs at low electric field strengths.
When the electric field is very high, the breakdown of the dielectric occurs in some sort of casscade. What will happen is a fast moving electron or other particle will cause ionization of molecules. The electric field will accelerate these charged particles and this will allow the process to occur again. In this way the density of charged particles will increase, this conversion of unionizised molecules / atoms into electrons and cations will increase the temperture of the gas / plasma.
Some of the ionized gas atoms will emit photons, these photons can sometimes cause ionization of other gas molecules / atoms. Or they will take energy away from the spark.
As more and more current flows through the spark the number of electrons and ions will become higher. For a DC spark to be stable in a given system then there will be a given current which must flow. In an AC system when the frequency is low (like 50 Hz) the spark may snuff out as the current drops to zero during the cycle. The cations and the electrons during this time may be able to recombine thus lowering the charge carrier density so far that the spark is snuffed out.
In a RF plasma the charge carriers have too little time to recombine during the zero current momments in the cycle. Here the plasma / gas remains conductive during the whole cycle.
Now with a low electric field the casscasde of reactions caused by the rapid acceleration of the electrons does not occur. Here the conduction of the gas is dictated by the radiation level (either gamma, beta, alpha, X rays or hard UV) which creates electrons and cations.
It is interesting to note that during the Fukushima accident that the radioactive release at one point modified the earth's electric field. You can read an academic paper about this here.


text:
How can it be proved that (S)-bis-[2-(t-butylsulfinyl)phenyl] diselenide is a chiral non-racemic product?
I read in paper that:

The aim of this part of the work was to synthesise new aromatic diselenides with chiral phosphorous or sulfoxide centres in coordinative distance to the selenium as precursors for new chiral selenenylating reagents.

How can it be proved that (S)-bis-[2-(t-butylsulfinyl)phenyl]diselenide is a chiral non-racemic product?

What exactly are you asking? Are you asking how, experimentally they determined that the product is chiral, or why, theoretically, it should be so?

Chould you please add the paper, that your are citing? And welcome to chemistry.se!

That's some hardcore you've got there ;) , it seems not googlable so you really should provide link or more details

I want to know how I can assign that  this product has (S) or (R) chirality

This'll be easier if we can see the structure.

How can I insert the structure of this product  in this site , please ???

@Nado: You can upload a picture containing the structure.

@Nado AFAIK, you can't upload an image because your reputation doesn't allow it. If it is so, you could put a link to it and we'll edit and add the image for you.

Compounds with the general structure $\ce{R-SO-R'}$, i.e. containing a sulfinyl $\ce{-SO-{}}$ functional group attached to two carbon atoms, are called generically ‘sulfoxides’. Sulfoxides may be considered as a tetrahedral system composed of a central atom, ligands, and a lone pair of electrons (or phantom atom).

When the two organic residues $\ce{R}$ and $\ce{R'}$ are dissimilar, the sulfur atom is a stereogenic center.
The compound bis[2-(tert-butylsulfinyl)phenyl]diselenide contains two chiral sulfinyl groups and thus two stereogenic centers.

Therefore, three stereoisomers exist:

bis[2-((R)-tert-butylsulfinyl)phenyl]diselenide,

bis[2-((S)-tert-butylsulfinyl)phenyl]diselenide,

and the meso compound [2-((R)-tert-butylsulfinyl)phenyl][2-((S)-tert-butylsulfinyl)phenyl]diselenide

Now it's very clear . Thank you so much Prof with my best regards  to  you .


text:
Relative levels of the two most prominant glutamate species at pH 4.7

Calculate the relative levels (i.e. ratios of concentrations) of the two most prominant glutamate species at pH 4.7.

I know that the Henderson–Hasselbalch equation is to be used which is
$$\mathrm{pH=p}K_\mathrm a + \log_{10}\mathrm{\frac{[A^-]}{[HA]}}$$
For one glutamate species, this would change to
$$\mathrm{4.7=2.19 + \log_{10}\frac{[A^-]}{[HA]}}$$
and for the other species it would be
$$\mathrm{4.7=9.67 + \log_{10}\frac{[A^-]}{[HA]}}$$
I know there are some other variables I need to figure out to solve the equation, but I'm not sure how to go about finding them with the information given.

Welcome to Chemistry.SE. Take the [tour] to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that we aren't doing your homework for you.

It might help to start think about which glutamate species could exist.

I know that the Henderson–Hasselbalch equation is to be used which is pH= pKa + log [A-]/[HA]. What I'm first stumped on is how I would go about finding the pKa and then calculating the conjugate base and acid.

Please edit the question with your thinking.

What do you mean by other species?

The two species are related. They are conjugated acid and base to each other. Assuming one species $\ce{A^2-}$ with concentration $x \,\pu{mol/L}$, the second species $\ce{HA-}$ with concentration $y\,\pu{mol/L}$. There is a third species $\ce{H2A}$ with concentration $z \pu{mol/L}$.
$$\frac{10^{-4.7}\times x}{y}=10^{-9.67} \tag{1}$$
$$\frac{10^{-4.7}\times y}{z}=10^{-2.19} \ \ \tag{2}$$
Based on above two equations, the two most prominant species are $\ce{HA-}$ and $\ce{H2A}$ and ratio should be $y/z$. I utilized definition of $K_\mathrm a$ instead of $\mathrm pK_\mathrm a$ to get the above two equations. $10^{-4.7}$ is simply $[\ce{H+}]$ at $\mathrm{pH} = 4.7$
Therefore the final answer is:
$$\frac{y}{z} = 10^{2.51}$$


text:
Water molecule bent shape and electron configuration
The hydrogen atoms in a water molecule are negatively charged, so they should repulse each other completely due to the electromagnetic force, forcing the hydrogen atoms to be on opposite sides of the oxygen atom. However, this is not the case, as the H-O-H bond angle is $104.5 ^\circ$.
Why does the lone electron pairs repulse the hydrogen atoms to an angle of $104.5 ^\circ$ in $\ce{H2O}$ instead of $180 ^\circ$?

Y'know, it gets a bit disappointing to see someone with more than 10 posts who still doesn't format their posts with Mathjax properly.

@PhyCS Not sure what you're talking about... linear water in excited state?

Possible duplicate of Why is H₂O V shaped?

According to VSEPR theory (there are other models, but they give the same qualitative results), since there are four electron pairs around oxygen they will be in a tetrahedral configuration. Assume the hydrogen atoms are at A and B as you've described them in your question; C and D (the lone pairs) are in a plane perpendicular to the one containing A and B. Below is a ball-and-stick model of water, showing the lone-pair electrons as yellow "atoms".
 (Public domain image from Wikimedia Commons)
It sounds as if you're thinking in two dimensions with the hydrogen atoms and lone pairs in the same plane. If water existed in only two dimensions it could conceivably be linear, but it really is in three dimensions. (The molecule itself is flat because the lone pairs affect the shape but are not counted as part of it.)
All of the above applies to water under normal conditions. I'm not aware of any excited or exotic states of water in which the VSEPR prediction doesn't apply, but I can't guarantee that there aren't any.

Thank you so much! I have had a lot of good answers in the past, but they usually involved difficult (to me) concepts and sometimes calculus and a bunch of stuff that I did not even begin to understand. I don't know if this question was that easy, but you definitely gave me understanding to this problem and introduced to me new concepts that should serve me well. Thanks!


text:
food grade diatomaceous earth
Is all unaltered diatomaceous earth (DE) food grade?
I am aware that all have not spent the money to have it certified and labeled as food grade, but I'm thinking that is just a way for the government to get more money and also have more control. All countries do not have an FDA, and therefore have no need to certify it, and have used it for thousands of years.
I just want to know the answer without somebody being afraid to say it is still food grade, labeled or not. I know the DE exposed to extreme heat is not safe and is for pool filters.

Is all water food grade? would you drink water out of a mountain stream? what about an unspecified large river? Or would you play it safe and dring water that's been certified as safe? It's the same thing.

Is all unaltered diatomaceous earth (DE) food grade?

No, and that's because it's not regulated and in general trying to categorize "all" of something (regardless of source, and especially when unregulated) as "food grade" isn't possible. Here in the U.S., the Food and Drug Administration (FDA) has weighed in on the issue and determined that the components of diatomaceous earth are "generally regarded as safe (GRAS)" when used as a filter aid.
From the FDA's report entitled Select Committee on GRAS Substances (SCOGS) Opinion: Diatomaceous earth (filter aid):

Silicon dioxide and various silicates occur abundantly in the earth's crust, are present in practically all natural waters,animals, and plants, and are part of the normal human diet. The question of whether or not silicon is an essential human nutrient remains unresolved. Silicon compounds consumed as added food ingredients contribute only a minor proportion of the total dietary silicon intake. The estimated possible human intake of sodium aluminosilicate, the predominant silicate added to foods in this country, is approximately 0.3mg per kg body weight per day. Silicon compounds that are GRAS for use as direct food ingredient, except potassium and sodium silicates, are insoluble or very slightly soluble in water and appear to be biologically inert. The water-soluble silicates are also of low acute toxicity. The acute oral LD50 in rats of sodium aluminosilicate is >1g per kg. No significant tissue accumulation, pathology, or toxicity has been reported from the ingestion of those insoluble or very slightly soluble GRAS silicon compounds for which data are available. Of the five substances that were reported as added to foods in the NRC survey, biologic effects and safety data are available for all except sodium calcium aluminosilicate, and there is no reason to suspect that the toxicity of the latter would differ from those for which there are data. The results of two studies (1967-1970) in which various silicon compounds were fed to laboratory animals for 1 mo at a level of 0.8g per kg body weight (as silicon dioxide) and for 3 mo at levels of 6 to 30 mg per kg body weight suggests there may be a species-related susceptibility to renal damage from ingestion of sodium silicate, magnesium trisilicate, and finely ground quartz.
No substantiating reports of these effects have appeared. Magnesium trisilicate was recognized as safe for prolonged use in human ingestion in large amount as a component of antacid preparations by the Advisory Review Panel on Over-the-Counter Drugs, and the available evidence on the acute toxicity of sodium silicate indicates that it is low. Consumption data are lacking for aluminium calcium silicate and tricalcium silicate, two compounds that are listed as GRAS for use as anticaking agents. However, their use in keeping with good manufacturing practice and in currently regulated amounts would be of the same order of magnitude as the other GRAS silicates. In addition, the Select Committee has limited information on the amounts of talc that are currently used in foods. However a major food use of talc is in the coating of rice. Assuming package label statements are ignored and coated rice is not washed prior to cooking and no other losses occur, maximun per capita intake of talc from this source appears to be 0.5g per day. With respect to paper and cotton food- packaging products, the possibility is remote that biologically significant amounts of talc, diatomaceous earth, or sodium silicate migrate to food from packaging, materials containing these substances. Perlite, a naturally occuring polysilicate substance, has an oral LD50 in the rat of >10g per kg body weight. Estimates of the maximun quantities of minerals that might be extracted from perlite and diatomaceous earth used as filteraids in food processing indicates no hazard to public health. There are no food grade specifications for aluminum calcium silicate, sodium calcium aluminosilicate, and talc. Such specifications for the substances used in foods are desirable.
Specification for food grade talc should limit the content of asbestos fibers even though the potential hazard of ingested asbestos fibers even though the potential hazard of ingested asbestos is not clearly established. The Food and Drug Administration is sponsoring the development of an assay method for asbestos fibers. An upper limit for cadmium should be added to the specifications for food-grade perlite, and consideration should be given to the need for limitation of cadmium content of other silicates. In the light of all of the foregoing, the Select Committee concludes that: It is essential to establish food-grade specifications for aluminum calcium silicate, sodium calcium aluminosilicate, and talc, with provision for an upper limit of asbestos fibers in talc. There is no evidence in the available information on aluminum calcium silicate, calcium silicate, magnesium silicate, potassium silicate, sodium silicate, sodium aluminosilicate, sodium calcium aluminosilicate, tricalcium silicate, silica aerogel, and talc that demonstrates or suggests reasonable grounds to suspect a hazard to the public when they are used at levels that are now current or that might reasonably be expected in the future. There is no evidence in the available information on diatomaceous earth, silicon dioxides, sodium silicate, and talc that demonstrates or suggests reasonable grounds to suspect a hazard to the public when they are used as ingredients of paper and paperboard products used in food packaging in accordance with current practice. There is no evidence in the available information on sodium silicate and talc that demonstrates or suggests reasonable grounds to suspect a hazard to the public when they are used as ingredients of cotton and cotton fabrics used in dry food packaging in accordance with current practice. There is no evidence in the available information on diatomaceous earth and perlite that demonstrates or suggests reasonable ground to suspect a hazard to the public when they are used as filteraids in food processing at levels that are now current or that might reasonably be expected in the future.
*Complete reports containing details of the safety studies that formed the basis of the opinions and conclusions and are available from the National Technical Information Service (NTIS), 5285 Port Royal Road, Springfield, VA 22161 (703) 605-6000.

That said, "food grade" appears to apply to a mean particle size of less than or equal to 12 $\mu$m (Wikipedia), and only applies to diatomaceous earth labeled that way.

I very much doubt that all diatomaceous earth is food grade.  For some reasons why, consider the Wikipedia page for diatomaceous earth: "Kieselgur" was discovered when folks in the 1800s were trying to dig a well.  Since knowledge of hygiene was some decades away, it's not implausible that those people may have been trying to dig a well that happened to be near an outhouse.  I'd doubt that diatomaceous earth recovered from a site near an outhouse would be food grade.
Also, Wikipedia lists "salt-water-derived" diatomaceous earth and specifically mentions that it is:

not suitable for human consumption
not effective as an insecticide
often calcined, which converts the safer amorphous silica present in most natural diatomaceous earths into a far more dangerous crystalline silica.  This can lead to silicosis if long-term exposure to airborne microparticles of crystalline silica occurs.

Thus, any diatomaceous earth that contains lots of salts or which has been calcined is unlikely to be food grade.


text:
Chemicals that Decrease fertility
Is there any known substance that can decrease fertility of a male?
Less specifically what about one time chemical castrations (ignoring a perfect sterilization)?
I heard about them before once but wasn't sure really.

Not planning anything, I hope? ;)

@santiago I'm not looking for the chemical exactly, but just sources/confirmation that it exists.

RISUG claims to be perfect male contraception that can be reversed at any time. As for a one-time chemical castration, I'd be surprised if something like that exists, because you're basically asking to completely destroy the functioning of a single organ without affecting any others. That's a pretty tall order.

@chipbuster I think that's a good start - my reading of the question made me think about RISUG as well. Since there are few working chemical male contraceptives it's clear it's not a super-easy problem.

It appears to be used for sex offenders.

Look at the wikipedia page on chemical castration. It mentions some drugs. These however, do not have an irreversible effect.

@WYSIWYG I have had confirmation from a doctor that although a chemical isn't public known, such a chemical does exist.  Again I can't site this, but I don't think we should should eliminate because "it doesn't appear on Wikipedia."

Numerous chemicals are linked with decreasing male fertility.
In fact, studies suggest that unhealthy lifestyles and environmental factors (e.g. plasticisers in the water system) may be causing this decrease in fertility.
See: http://www.telegraph.co.uk/men/active/mens-health/10695991/Why-are-mens-sperm-rates-falling.html etc
Obviously conclusive studies are limited at this stage.
But if you want to be more specific, I recall that chloroform is known to cause miscarriages and sterility in mice. Googling that to find a citation led me to this article:
http://www.ncbi.nlm.nih.gov/pubmed/10969727
Which found evidence of an otherwise non-toxic antifertility agent for male rats.
To be honest, I would be surprised to find studies which have tested a anti-fertility drug on men. Such a thing should not make it past an ethics board (given the ability to prevent impregnation by far less drastic measures, e.g. condoms).

If however you were talking about the sort of chemical castration they did to someone like Alan Turing, they were not designed to sterilise him (heck, his 'crime' was homosexuality and children were not a likely outcome of that), but to alter his hormones and personality. Anyway, modern drugs might be something like Cyproterone acetate or Depot medroxyprogesterone acetate - though that is also used as a birth control drug for women.

+1 Thank you sir.  I guess Chemical Castration is more of what I was looking for.  Thank you for posting both.

Do you know how effective a chemical castration is for sterilization?

Testing an anti-fertility drug would not be impossible to get past an ethics board. For example, find volunteers (e.g., men looking for a vasectomy).

@GeoffHutchison although vasectomies are reversiblish.

Be it surgical or pharmacological, sterilization is a way to prevent conception, whereas castration is a suppression of the testes (emasculation). Infertility can be achieved with an intervention on the testes or on other reproductive organs (prostate, seminal vesicle, vas deferens).
Testes not only produce sperm but also androgen hormones (testosterone), which have a plethora of effects on the body, the behaviour, etc. Sperm is made from germ cells and testosterone is produced in Leydig cells.
Antiandrogenic interventions (hormonal or metabolic) will block testosterone (possibly irreversibly, see finasteride).
This will affect sperm production, but will also result in androgen deficiency.
Cytotoxic chemicals used in chemotherapy (e.g. cisplatin) affect sperm production, which involves many germ cell divisions.
But this is also likely to damage the Leydig cells.
Papaya extract action on sperm motility is reversible.

There are several ongoing studies for male contraceptives (have a look on Wikipedia). One of the big problems though is to reverse the effects.
VasalGel is thought to hit the market by 2018. It is a gel that physically blocks the sperm once injected in the vas deferens but can be washed out to revert its effects.

"but can be washed out to revert its effects." How?


text:
How to evaluate the heat of formation with computational chemistry?
I'm sorry if this is a stupid question, but since I'm a beginner with computational chemistry, please, be patient.
Let's consider the reaction of formation of water:
$$\ce{1/2 O2(g) + H2(g) -> H2O(g)}$$
This reaction shoud give a $\Delta_\mathrm fH^\circ = -241.83~\mathrm{kJ/mol}$. The reaction is intentionally simple to let me run fast calculations and tests with single point energies.
First, I've performed a geometry optimization on $\ce{H2O}$, $\ce{O2}$ and $\ce{H2}$ (HF/6-31G*). Then, I've used M06/Def2-QZVPP for the single points:

$E_\mathrm{SP}(\ce{H2O}) = -76.431519932791~E_\mathrm h$
$E_\mathrm{SP}(\ce{O2}) = -150.262365177069~E_\mathrm h$
$E_\mathrm{SP}(\ce{H2}) = -1.171819656537~E_\mathrm h$

In this case, doing $$E_\mathrm{SP}(\ce{H2O}) - \left(\frac12 E_\mathrm{SP}(\ce{O2}) + E_\mathrm{SP}(\ce{H2})\right) = \Delta E$$ I'm getting $\Delta E = -0.1285176877195084~E_\mathrm h$ that is equal to $-337.42312874~\mathrm{kJ/mol}$, almost $96.6~\mathrm{kJ/mol}$ higher than the reference of $-241.83~\mathrm{kJ/mol}$.
Is this the correct approach for a case like this? Or is this discrepancy normal?

States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different. And -337 is lower than -241. I'll answer soon-ish, your approach is only incomplete, but not wrong and discrepancies are normal.

Your approach is quite correct, but as Jan already pointed out, it is incomplete. What you calculated is the difference in the electronic energy of the reaction $$\ce{H2 (g) + 1/2 O2 (g,\,{}^1\Delta_{g}) -> H2O (g)}.$$
Let's turn this into some kind of a tutorial and a little exercise, that you can try to reproduce at home. For a more detailed description on how it works in Gaussian, you can read the paper by Joseph W. Ochterski. I will use Gaussian09 rev. D for the calculations involved.
The enthalpy of formation of water from the standard state of the elements involved, is basically the reaction enthalpy of the combustion of hydrogen and oxygen. You are in luck, because most of the quantum chemical programs shall provide you with the necessary data for reactions straight away. What you do need to do is calculate the thermal corrections to the reaction. Essentially this is done by calculating the Hessian, i.e. a frequency calculation. The second derivatives only have a interpretable meaning if they are performed at a stationary point. Hence they must be performed with the same method, that you used to optimise the geometry.
Jan already pointed out, that HF/6-31G(d) is probably not a well enough fit for geometries and frequencies. I'd always consider some level of theory with at least some coverage of correlation energy. I often prefer BP86 or M06L and the def2-SVP basis set as a first anchor point. The smaller the system, the more accurate the methods you can use (or afford).
Jan also pointed out, that especially for oxygen, the spin state is important. (I did a quick check and am almost 100% certain you calculated it in the singlet state; see above equation.)
For this exercise I have matched your suggestion with using M06/def2-QVZPP as it is still feasible. Keep in mind, that most of the times density functional approximations have to be calibrated. It's all about what you can or cannot afford. You should perform calculations to the best of your abilities.
Okay, let's dive into some mathematics. First, we consider the reaction equation
$$\ce{H2 (g) + 1/2 O2 (g,\,{}^3\Sigma_{g}^{-}) -> H2O (g)}.$$
And we would like to know the enthalpy of formation for water at standard state (which we will use as $298.15~\mathrm{K}$ and $1~\mathrm{atm}$)
$$\begin{align}
\Delta_{f}H^\circ(\ce{H2O}) &= H^\circ(\ce{H2O})
                               - (H^\circ(\ce{H2})
                               +\frac12 H^\circ(\ce{O2}))\\
\Delta_{f}H^\circ(\ce{H2O}) &=
   [E_\mathrm{el}(\ce{H2O}) + H_\mathrm{corr}(\ce{H2O})]\\
&\qquad
 - \left([E_\mathrm{el}(\ce{H2}) + H_\mathrm{corr}(\ce{H2})]
 +\frac12 [E_\mathrm{el}(\ce{O2}) + H_\mathrm{corr}(\ce{O2})]\right)\\
\end{align}$$
The data in the expanded form we can directly take from the output of the frequency calculation:
Oxygen  mo6qzvpp.freq.log/.com
 SCF Done:  E(UM06) =  -150.322374247     A.U. after    1 cycles
 Zero-point correction=                           0.003897 (Hartree/Particle)
 Thermal correction to Enthalpy=                  0.007203
 Thermal correction to Gibbs Free Energy=        -0.016049

Hydrogen mo6qzvpp.freq.log/.com
 SCF Done:  E(RM06) =  -1.17221845130     A.U. after    1 cycles
 Zero-point correction=                           0.009902 (Hartree/Particle)
 Thermal correction to Enthalpy=                  0.013206
 Thermal correction to Gibbs Free Energy=        -0.001590

Water mo6qzvpp.freq.log/.com
 SCF Done:  E(RM06) =  -76.4321360023     A.U. after    1 cycles
 Zero-point correction=                           0.021635 (Hartree/Particle)
 Thermal correction to Enthalpy=                  0.025415
 Thermal correction to Gibbs Free Energy=         0.004010

Therefore we will obtain
$$\Delta_{f}H^\circ(\ce{H2O}) = -236.6~\mathrm{kJ/mol}$$
which I think is reasonably close to the experimental value.

Appendix
The following input files for the computation of above values.
%chk=mo6qzvpp.chk
#p M06/def2QZVPP
opt
int(ultrafinegrid)

oxygen

0 3
O
O 1 1.2
 (blank)

%chk=mo6qzvpp.freq.chk
%oldchk=mo6qzvpp.chk
#p M06/def2QZVPP
freq
geom=check guess=read
int(ultrafinegrid)

oxygen freq

0 3
 (blank)

%chk=mo6qzvpp.chk
#p M06/def2QZVPP
opt
int(ultrafinegrid)

hydrogen

0 1
H
H 1 1.2
 (blank)

%chk=mo6qzvpp.freq.chk
%oldchk=mo6qzvpp.chk
#p M06/def2QZVPP
freq
geom=check guess=read
int(ultrafinegrid)

hydrogen freq

0 1
 (blank)

%chk=mo6qzvpp.chk
#p M06/def2QZVPP
opt
int(ultrafinegrid)

water

0 1
O
H 1 1.2
H 1 1.2 2 109.0
 (blank)

%chk=mo6qzvpp.freq.chk
%oldchk=mo6qzvpp.chk
#p M06/def2QZVPP
freq
geom=check guess=read
int(ultrafinegrid)

water freq

0 1
 (blank)

Have a lot of fun.

Thank you for the edits and the amazing explanation: I can not ask for a better answer.

@Henrique Thank you for the compliments - I am very happy to help (And I have not forgotten about your other question). Since you are probably still somewhat new to the site: Please consider accepting an answer.

you need to perform a Hessian calculation and add the enthalpy correction to each energy
what spin state are you using for $\ce{O2}$?
HF/6-31G(d) is probably not good enough for geometries and frequencies.
Depending on what accuracy you're after, M06/Def2-QZVPP may not be good enough

Thank you for the tips, Jan. As Martin guessed, my O2 was a singlet.

Don't forget to reference Gaussian and all the authors ;p !!!!

The problem is that such a simple calculation has several associated errors (geometries, energies, vibrational frequencies and, therefore, thermodynamic functions) which render the result very inaccurate.
In the case of water, using the same M06 method you used, one obtains -212.9 kJ/mol with the cc-pVDZ basis set and -246.0 kJ/mol with the much larger aug-cc-pV6Z basis set. Using a much more accurate method, CCSD(T), the enthalpy of formation with the same previous two basis sets are -200.19 and -241.89 kJ/mol. The latter result is nearly identical to the experimental one.
Incidentally, notice that quoting enthalpies of formation with more than one decimal digit is only justified if you are using highly accurate methods. You will very rarely have accuracies of better than 1 kJ/mol and certainly not using DFT methods.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. While it may seem strange at first we do not use salutations in our Questions and Answers. Please use the upvote button to say thanks, or leave a comment how the post can be improved.

Thanks, Oscar, I wasn't even considering CCSD(T). I'm still learning, so, any hint is very valuable to me.


text:
I have 1 litre of water at 45 °C. How long would it take to evaporate when ambient temperature is 25 °C?
As a fun experiment, I'm trying to evaporate water in my kitchen. I've heated the mixture to 45 °C, and I'm wondering how long it might take to completely evaporate.
Assume that the water is 2 cm deep in a flat pan (if that is relevant).

My guess is that we are speaking about days. Just pour the solution into something shallow and leave for several days in place with strong ventilation, I used this scheme for evaporation of solutions of some salts.

Might also depend on ambient humidity. If the relative humidity in the air is 100%, then the air is already saturated with water vapor and the water will never evaporate.

Why, if the relative humidity in the air is 100%, but the air is at 25 °C, and your water is heated to 45 °C, then it will evaporate. Though of course that would not be as fast as it could be in a less humid air, and you'll get moisture all over the place, too.

The evaporation rate $E_r$ can be approximated by:
$$E_r = C \times A \times (x_s - x) = 0.013$$ kg/h
$C$... evaporation coefficient $ = 25 + 19 \times v$
$A$... water surface $= 0.03 $ m$^2$
$x_s$... saturation humidity $= 0.027$ kg/kg (at 30 °C)
$x$... absolute humidity $= 0.011 $ kg/kg (at 30 °C and 40 % rel. humidity)
$v$... air speed above water surface $= 0.08$ m/s
If you mesure the air temperature above the water surface - which I approximated with 30 °C - you can get a way better approximation for your case. Also having the rel. humidity you can then find the values for $x_s$ and $x$ in a Mollier diagram (h,x-diagram).
PS: You will waste a lot of energy when you want to obtain the water temperature at 45 °C. Heating the air instead should be more effective.
Reference: Link (in German)

I'm curious if you'd care to respond to a similar question I posted in SE Earth Science, here: http://earthscience.stackexchange.com/questions/5071/evaporation-loss-formulas. I never did get a response and don't want to cross-post.

@http://earthscience.stackexchange.com/questions/5071/evaporation-loss-formulas: Since your 1. Formula also takes the surface temperature into account, when calculating the evaporation coefficient, i´d say its more accurate. But if you were trying to calculate evaporation rates at room conditions, like in this post, then both of them should be inprecise. The only way to really be sure is to find out more about the specific test conditions, under which both formulas had been constructed.

Formula 2 does use the pond surface temperature indirectly from the saturation pressure calculated in B7.

Thank you for the answer, @AstronAUT. So according to your formula (and given those approximations), it evaporates at a rate of 13grams per hour and will take a little over 3 days.
Also, just to clarify... are you saying that the water will evaporate faster in an oven than on a stove (over direct flame)?

At the same temperatures, turned on convection and if you allow the oven to constantly receive fresh air, yes of course.

I copied some data when I was trying to predict the drying time of a paint:
"The evaporation of water does not depend on the depth of a lake.
It depends on the temperature and the wind.
At 10°C, the level of the water decreases by 0.9mm/day if the wind is zero, by 1.2mm/day with a 2m/s wind, and by 1.7mm/day with a 9 m/s wind.
At 20° C, those values are respectively 3.2 mm/day, 4.5 mm/day and 6.1mm/day, when the wind blows at 0 m/s, 2 m/s and 9 m/s.
At 30° C, the same values are respectively 6.8 mm/day, 9.6 mm/day, 12.9mm/day, when the wind blows at 0, 2m/s and 9m/s."
2 meters/second = 4.5 miles/hour; 9 m/s = 20.25 mph.
These values are taken from a book by M. Dorsey, Properties of the Water Substance, Reinhold Co., New York 1940.
And there is a detailed online calculator at https://www.engineeringtoolbox.com/evaporation-water-surface-d_690.html  which starts out: "Evaporation of water from a water surface - like a swimming pool or an open tank - depends on water temperature, air temperature, air humidity and air velocity above the water surface."
It seems that air velocity across the surface is as controlling as temperature. There are always some water molecules just above the surface of the water - air blowing across the surface removes them. No air speed, no evaporation. Well, that's not a perfect approximation, but if you had a very tall container, the last centimeter of water would have to evaporate and then diffuse up to the top and then mix with air to get away, and that could take a very long time.


text:
What kind of structure could give you a HNMR spectrum consisting of a singlet integrating to 5 protons?
A clean spectrum was given to me with the integrations filled in. I just need to assign peaks.
I just don't see how it's possible to have a singlet integrate to 5 protons.
I've considered symmetry, but still, 2 or 3 way symmetry wouldn't give me anything.
The chemical formula is $\ce{C11H14O2}$, with 5 degrees of unsaturation.
The spectrum is as follows:

5H, singlet, 7.5 ppm
2H, singlet, 5.1 ppm
2H, triplet, 2.4 ppm
2H, quartet, (or sextet?) 1.8 ppm
3H, triplet, 1 ppm

I'm just looking for a hint on what a 5H singlet could be.

Unlikely to be a singlet in the aromatic region- expand the spectrum and you will see overlapping signals

@long If the coupling between the peaks is stronger than the CS difference, they coalesce and cannot be resolved, unless you go to significantly higher fields.

The 5H singlet is likely due to five protons on a monosubstituted benzene ring.
The peak comes in at 7.5 ppm which is the aromatic region, immediately telling you that they are hydrogens on an aromatic ring. Benzene rings are the most common of these.
But why are they all equivalent? They actually aren't, but their shifts are so close that they appear to be equivalent. Sometimes they will appear as different peaks (usually one peak for the 2 hydrogens close to the substituent and one peak for the 3 hydrogens away from it) if there are strong enough electron withdrawing groups on the substituent that are close to the ring. It is likely that the singlet you're looking at isn't clean cut and looks a bit like a multiplet (see this question).


text:
Why doesn't my stainless steel Thermos get really really hot?
Why doesn't my stainless steel Thermos get really really hot? Seems like a fairly straightforward question.
A few notes:
I was looking and found (obviously different values because steel is an alloy) that $$0.35~\mathrm{\frac{J}{g\,K}}\leq C_{p,\,\mathrm{steel}}\leq 0.55~\mathrm{\frac{J}{g\,K}}.$$
Obviously, that is just a rough range, but that is a very small heat capacity as is usual of metals.
Why then, does my 200 degree Fahrenheit coffee not heat up the stainless steel mug super quickly!
And even more baffling to me: Why in the world would it be a really good insulator?
It's entirely possible that I am devoid of any understanding of how insulation works, but I thought things which are bad conductors of electricity made good insulators. While stainless steel isn't praised for its electrical conductivity, I thought a material had to be about as bad as a gas at conducting electricity to be crowned with the title of "good insulator." I would just imagine that stainless steel, being metallic in many ways, would be a better conductor than that.
So what's going on? Why isn't my Thermos reaching 200 degree Fahrenheit?

I have a solid steel pint glass and I can tell you that it pretty much instantly heats up to the temperature of the liquid you put inside. You probably have some sort of double-wall vacuum flask, which insulates by using a vacuum between the steel.

Huh... I just looked at the thermos I have and it is that exact thing. Well that was anticlimactic! I wanted Chemistry and all I got was technology!

Maybe you could make an answer out of that just for completeness.

That technology is a very nice demonstration of chemical properties

Very true. It's quite interesting. I was just hoping for some kind of crazy unusual behavior of steel.

I have a solid steel pint glass and I can tell you that it pretty much instantly heats up to the temperature of the liquid you put inside. You probably have some sort of double-wall vacuum flask, which insulates by using a vacuum between the steel.

As you can see, there are two walls to the flask: an inner wall and an outer wall. The inner wall is probably very close to the temperature of whatever is inside the flask. However, since the two walls are separated by a vacuum (which is a fairly good insulator), the outer wall, i.e. the one that you touch, remains pleasantly near ambient temperature*.
*Unpleasantly near ambient temperature, if you've been outside in a hot summer for a while.


text:
Azo coupling with arenediazonium salts - why isn't nitrogen gas lost?
Arenediazonium compounds contain a very good leaving group - nitrogen gas - and thus undergo nucleophilic substitution reactions rapidly. With water, for instance, they give phenols.
In the formation of azo dyes from arenediazonium salts, however, the arenediazonium cation acts as an electrophile to carry out a substitution reaction on activated benzene rings like those of phenols and anilines. Why doesn't displacement of $\ce{N2}$ happen here as well? Normally the carbon attached to $\ce{-N2+}$ acts as the electrophilic centre, so phenols and anilines should react at that $\ce{C}$ rather than at the $\ce{-N2+}$ group. Moreover, phenols and anilines have more nucleophilic centres at their $\ce{O}$ and $\ce{N}$ than at the carbon atoms of their benzene ring - so why aren't the final products be diphenyl ether with phenol, and N-phenylaniline with aniline?

A simple explanation would be that the displacement of a diazonium group by oxygen to form a phenol or phenyl ether linkage typically requires heating, indicating that it is a slower process than the very fast room temperature coupling reaction at aromatic carbon to form an azo dye. Why? One explanation is that the diazonium cation and the phenol or amine being coupled are a "soft" acid and base respectively, whereas water is a "hard" base. But a substituted phenol or aniline that could not form an azo dye might well form a phenyl ether with a diazonium salt, as you propose.

What are soft and hard acids and bases?

See the response to this question: http://chemistry.stackexchange.com/questions/10273/hardness-and-softness-are-terms-used-to-describe-which-aspect-of-acidic-behaviou/11055?s=6|0.1429#11055

Another possibility is that the aryl nucleophiles that form azo dyes are bulky and attack on the aromatic ring is stetrically hindered.  The exposed nitrogen in the diazonium function offers an alternative that is reasonably electrophilic ($sp$ hybridization + nitrogen instead of carbon --> more electronegative) and has more room.


text:
Addition of inert gas to equilibrium mixture under constant pressure
Adding inert gas under constant volume will not disturb equilibrium. But I want to know what would happen if inert gases are added at constant pressure.
If total pressure remains constant and inert gases are added, then mole fraction of any gas participating in equilibrium decreases and hence partial pressure decreases.
If $\Delta n$ is non zero, then $K$ changes. But how can $K$ change if temperature remains constant?

Try to reason backwards: you know K does not change when the temperature remains constant, what should happen to the equilibrium position and the partial pressure to make it so? Does the system still remain at equilibrium?

I was using the android app. It did not suggest any results bases on the title. I will delete the thread.

How do I delete it?

Under your post there should be a delete button. I'm accessing the question via a computer but there should be something similar on Android.


text:
What closure relations can we use in molecular modelling?
Following my question here about the BBGKY hierarchy I was wondering does anyone have any experience in using the method outlined. In particular how would I take an equation off the page and for some reduced phase space density or distribution function and start obtaining meaningful results and simulations? I understand that it is necessary to implement closure relations along the way. If anyone has experience in doing this sort of thing, could you outline (qualitatively) what steps I need to take etc.


text:
Acidic strength of oxides of nitrogen
Why do we compare acidic strength of oxides of nitrogen even though they don't have hydrogen?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. || It would be helpful if you tell us what your background is. Have you studied Lewis or Bronsted theories of acids and bases?

I am a student, Yes I have studied about the acids you mentioned but don't understand them

Compounds containing hydrogen are not only acidic but we have other concepts also. Like in your case it can be explained through Lewis acid-base concept.
Lewis concept: A base is defined as a substance which can furnish a pair of electrons to form a coordinate bond whereas an acid is a substance which can accept a pair of electron.
For example, $\ce{NH3}$ is a Lewis base since on nitrogen we have one electron pair which easily be donated to form a coordinate bond. As for acid, Take any two oxides of nitrogen for example $\ce{NO2}$ and $\ce{N2O5}$ . First check the oxidation state of nitrogen in two cases, here it is +4 and +5 . Since a substance which can accept a pair of electrons is an acid, so the nitrogen with more positive oxidation state will relatively accept pair of electrons with more easily, therefore $\ce{N2O5}$ is a better acid then $\ce{NO2}$.

It would be helpful if you elaborate a bit about what Bronsted and Lewis definitions are, possibly with some references.

This answer is wrong you'd need Lux-Flood theory for oxides, not Lewis

@Mithoron  lux-flood theory is applicable only if both the reacting species are oxides right?

Don't see why you need Lux-Flood as when an oxide ion is accepted it can also be seen as accepting an electron pair from the oxide ion. Isn't Lux-Flood just a subset of Lewis?


text:
Is it wrong to call carbon dioxide "methanedial"?
Since dialdehydes have a carbonyl group on either end of an alkane, is it reasonable to call carbon dioxide "methanedial", or is it that ethanedial (glyoxal) is the first dialdehyde?

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-66.6.1.1.1 Mono- and dialdehydes derived from alkanes are named substitutively using the suffix ‘al’ added to the name of the parent hydride with elision of the final letter ‘e’ of the parent hydride before ‘a’.

Examples:
pentanal ($\ce{CH3-CH2-CH2-CH2-CHO}$)
pentanedial ($\ce{OHC-CH2-CH2-CH2-CHO}$)
In this context, the term ‘aldehyde’ is unambiguously defined:

P-66.6.0 Introduction.
The class name ‘aldehyde’ traditionally refers to compounds containing the $\ce{-CH=O}$ group attached to a carbon atom. However, nomenclature for aldehydes has been extended to describe a $\ce{-CHO}$ group attached to a heteroatom.

Since carbon dioxide ($\ce{O=C=O}$) does not contain a $\ce{-CH=O}$ group, it is not considered an aldehyde and thus Section P-66.6.1.1.1 does not apply to carbon dioxide.
Therefore, the shortest dialdehyde is ethanedial ($\ce{OHC-CHO}$). (Note that the preferred IUPAC name is ‘oxaldehyde’, which is derived from the retained name of the corresponding carboxylic acid.)

What about a diketone nomenclature? Does the blue book specify two adjacent $\ce{R}$ groups, and thus methanedione would also be invalid?

@Brian To call a carbonyl group a ketone, the oxygen has to be double bonded to a carbon atom bonded to two other carbons. So the ketone series begins with propanone and the diketone series, with butanedione.

@JayJay Gotcha, that's what I'd assumed. Thanks!


text:
Are all carbohydrates a subset of alcohols?
According to the definition,

an alcohol is any organic compound in which the hydroxyl functional
  group (-OH) is bound to a saturated carbon atom.

Monosaccharides such as glucose contain several hydroxyl groups. Does that mean all carbohydrates are a subset of alcohols?

Yes, it does, why the doubt?

Whenever you see -OH is bound to a saturated carbon atom in organic chemistry it is Alcohol group.
So answer is YES. All carbohydrates are  subset of alcohols.
Though in linear forms they also have aldehyde or ketone group.

Roughly speaking, nearly any sufficiently large bioorganic molecule is a member of all classes at once.


text:
Why the H+ ion remains with the molecule even though there is no bond between it and hyperconjugative structure?
In hyperconjugation, it is known that form an $\alpha$ hydrogen, one $\sigma$ electron goes to the corresponding carbon carbon bond, and there is apparently or practically, no bond between the hydrogen and carbon. I read in a textbook,

although one $\ce{C-H}$ bond of the alkyl is shown to be broken in each hyperconjugative structure, $\ce{H+}$ is never free from the rest of the molecule.

But then, still there is no bond. Is there any specific reason why the $\ce{H+}$ ion remains with the molecule though there is no bond attaching the ion to the rest of molecule?

Maybe too much chat recently ;) I don't  like viewing of hyperconjugation with mesomeric structures - it confused you - and hyperconjugation as a concept altogether - it's only tries to describe things as they are, not causes anything.

@Aneek Have you studied molecular orbital theory. As Mithoron is suggesting this would give you a better understanding of how hyperconjugation occurs.

More to the point - if you have resonant structures where in one of them bond is broken it simply means it's weaker than it would be if this structure wasn't present.

I believe what you saw in your textbook was a not-so-good but clear way of explaining the idea of hyperconjugation. By simplifying the idea of hyperconjugation to resonance is not chemically correct but it effectively certainly conveys the idea of the side-to-side overlap of orbitals.

Refer to this post for a clearer understanding of hyperconjugation. Interestingly, the most upvoted answer also used your textbook explanation. May I suggest that you look at the second most upvoted answer as well. https://chemistry.stackexchange.com/questions/8412/what-is-hyperconjugation

Resonance structures do not exist on their own! Resonance is not a rapid equilibrium between different structures. It is a description of a single delocalised structure using contributing forms which only involve 2-centre-2-electron bonds. It is therefore meaningless to think about what would happen to an atom in a resonance contributor if it were left around.
In your example the two structures are resonance contributors to the single delocalised structure and so the hydrogen ion cannot move away from the molecule because it never exists on its own as a hydrogen ion. It is always bonded to the rest of the molecule.
Hyperconjugation can be explained through resonance but it is much more clearly explained through molecular orbital theory. In this picture you have donation of electron density from the $\ce{C-H~\sigma}$ orbitals into the empty $p$ orbital of the positively charged carbon. This is possible because the two orbitals can be aligned in a plane due to rotation around the C-C bonds.

(1) The hyperconugation extension of the Lewis-structure model simply suggests to take more than the traditional mesomeric partial formulae into account. (2) even if those hyperconjugated forms would be the only ones, that would mean that there would be still be the (purely) ionic interaction. (For Na$^+$Cl$^-$ you also would not assume that the sodium ion would leave the "molecule").


text:
Why is the bond order of CO+ 3.5?
According to my books the bond order of $\ce{CO+}$ is  $3.5$. But shouldn't it be $2.5$? On googling this, I found the following answer that is on Stack Exchange but its only talks about the bond length.
I am unable to understand why it is $3.5$ as I am in class 11.

Very similar question here you might wanna check out: Why is the bond length of CO+ less than that of CO?

For a long time it was taught in school and universities that the HOMO of carbon monoxide is anti-bonding. Without more context it was also often taught that the bond order in CO is three, since there are eight electrons in bonding orbitals and two in anti-bonding orbitals. $$\text{Bond order} = \frac12(\text{bonding} - \text{anti-bonding})$$
By assuming that the HOMO is anti-bonding (it is not!) and taking away one electron, the  bond order has to increase to 3.5. This is wrong.
When we have a look at the MO diagram, a calculated version can be found here, we know that the HOMO, i.e. 3σ, is a bonding orbital, while the anti-bonding orbital is the 2σ. Upon ionisation, we would indeed remove one bonding electron and therefore the bond order has to decrease to 2.5 as you suggested.
However, it is not that easy. Strictly speaking the below MO scheme is, as well as MO theory itself, an approximation, and only one possible configuration. While we do not have to use resonance structures with MO theory, we have to consider other configurations (analogous to excited states). So naturally the bond order of CO is not strictly 3. And removing an electron does not mean we are removing it from only one orbital, rather than decreasing the electron density. Therefore we cannot accurately predict the bond order with these simple considerations.
Experimental observations and theoretical calculations suggest that the bond indeed becomes stronger when removing an electron. See the linked question and Philipp's answer within for more detail. (Don't look at the other answers, they are as wrong as they could be.)
In short: The bond order of $\ce{CO}$ is not exactly 3 and removing an electron will not increase the bond order to 3.5. In both cases, the observed bond order is probably closer to 2.5, while experiments suggest that the bond is stronger in $\ce{CO+}$.

An orbital with bonding character has no node perpendicular to the bond axis; an orbital with anti-bonding character has at least one node perpendicular to the bond axis (electron density is zero). Strictly speaking there are no non-bonding orbitals.

Shouldn’t the last sentence be ‘… in carbon monoxide’? Iirc, there are a few nonbonding orbitals (due to symmetry) e.g. in $\ce{HCl}$.

@Jan That's why I said strictly, one of the two categories will fit any orbital. What we usually classify as non-bonding orbitals are linear combinations that "don't change in energy". That's simply not possible due to an external field. In HCl the non-bonding orbitals don't have a node perpendicular to the bond axis (you caught that I forgot that), so they are classifiable as bonding.

@Martin-マーチン as shown, doesn't the $\mathrm{3\sigma}$ orbital have 2 nodes perpendicular to bond axis? (or maybe they are not nodes; just wanted to clarify)

@Aniruddha you are right, and my wording is maybe a bit off. The perpendicular nodes pass through the nuclei, therefore there is no node on the bonding axis between the respective nuclei that span it. If you look at the infinity of that axis, you are absolutely correct. I should probably retired that. Thanks for catching it.


text:
What does a square mean in a chemical formula?
We find in the book Minerals and their Localities by Jan H Bernard and Jaroslav Hyrsl   several chemical formulas containing a square. What does this square means? An example: Franciscanite $\ce{Mn3 (V,\square)(O,OH)3 (SiO4)}$
Thanks for your answer for  I have been  looking for answer for a long time.

That square refers to a vacancy (or framework vacancy) in the crystal structure. Here's one reference from Advances in Electronic Ceramics. Ceramic Engineering and Science Proceedings. Volume 28, Issue 8. Ed. C. Randall ${\it et\;al.}$
In short, a framework vacancy is a feature of some inorganic materials in which a "guest atom" (or more than one) can fit into the structure or denotes a structure that is missing interstitial or packing atoms (as outlined here).
Here's the info on your mineral where you can see that the square is instead denoted by square brackets [].

Note that, according to IUPAC recommendations, a square box ($\square$) is indeed used for vacancies in certain contexts, but the use of italicized $V$ is preferred, the element vanadium being written with the upright letter $\ce{V}$.


text:
Why must the movement of electrons through the wire be balanced by a movement of ions in the solution?
I have read through Barron's SAT chemistry prep book, in the topic of electrochemistry, the book says

The movement of electrons through the wire must be balanced by a movement of ions in the solution.

It explains that

Anions move toward the anode to replace the negatively charged electrons that are moving away.

Can someone help explain these two statements? I don't quite get this concept.

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If the movement was not balanced, a charge buildup would happen at the electrodes, instantly stopping the flow also in the other medium. That's exactly what happens if you cut the wire between the electrodes.

Karl Ratzch is basically correct, but I will try to provide a little more detail.
Electric charges, when separated, exhibit a very large tendency to try to get back together again.  This is represented in basic Physics by Coulombs Law and the large value of the constant in that relationship.
I presume we are talking about a battery here. If the flow of electrons out of the wire is different from the rate of ions arriving at the plate, the net charge on the plate will change.  Depending on which flow rate is greater, the voltage of the plate will instantly go up or down.  This will then change the rate of both charge flows such that the two almost instantly get back to the same value.
More detail: Assume that the flow of electrons out of the plate through the wire is larger than the flow of anions into the plate from solution.  This reduces the net negative charge on the plate.  This reduces the "force" that is pushing electrons down the wire.  The actual name of that "force" is voltage.  The voltage will change such that it doesn't push electrons out the wire as fast and will repel anions in solution less aggressively allowing more of them to reach the plate (per second).  The opposite is also true.  As a result, the flow in and out of the plate will always remain almost exactly the same, with small differences existing for very brief periods of time when the voltage changes.