text:
How to make 0.3% ammonia water from 28-30% ammonia hydroxide
Can someone tell me how to make 0.3% ammonia water from 28-30% Ammonia hydroxide?
I am trying to make a gallon or 3.78 Liters.

Hint : Ammonia water is actually diluted Ammonium Hydroxide ruled by the following equilibrium : $$\ce{NH3 + H2O <=> NH4+ + HO-}$$

Is it a question for homework/study or you need it for an experiment? I am asking because ammonia is volatile: if you need this solution with accurate concentration/pH, you also should be able to measure somehow what you get at hand.

Ammonium hydroxide does not exist. $\ce{NH4OH}$ is a formula that was invented in the 19th century to explain why $\ce{NH3}$ is so easily and exothermically dissolved into water, as if there was a chemical reaction with water. But it has been proven that as soon as an ion like $\ce{NH4^+}$ meets the ion $\ce{OH^-}$, the following reaction occurs : $$\ce{NH4^+ + OH- <=> NH3 + H2O}$$ But this equilibrium is strongly displaced to the right hand side. So that a solution of $\ce{NH3}$ contains only a few percent of ammonium ions.
Anyway the concentration of the so-called "concentrated $\ce{NH4OH}$" does not contain $29$% $\ce{NH4OH}$, but it contains $29$% of $\ce{NH3}$.
Now to produce a solution containing $\pu{0.3}$%  $\ce{NH3}$ with this concentrated solution (density : $\pu{0.91 g/mL}$), it is necessary to dilute it $100$ times. In other words, take $\pu{37.8 g = 41.5 mL}$ into one gallon water ($\pu{3.78 L}$)

What about density ?


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Comparing the strength of metallic bonds
Why is the boiling point of iron higher than magnesium?

There are two main reasons:

Metallic bonding is more prevalent for transition metal elements because there are more electrons available for delocalization (from s and d orbitals), while Magnesium has only s electrons
Fe has a higher atomic mass than Mg --> van der Waals forces are stronger.

Is there any impact on the atomic radius too?

It is an unrelated property, transition metals tend to have smaller atomic radii than group 1 or 2 elements because of d-metal contraction due to poor shielding of d-electrons

vdW forces aren't important here. Also related http://chemistry.stackexchange.com/questions/23960/how-many-delocalised-electrons-in-gold

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text:
What type of organic compounds are used in OLED technology?
I know that there are organic compounds that exhibit electroluminescence, and are those types of materials difficult to synthesize?

The short answer is a variety of pi-conjugated materials.
A huge variety of compounds, both small molecules, oligomers, polymers, and organometallic species have been used. Key requirements for the emissive material is a high efficiency emission process, either fluorescence or phosphorescence.
Some of the first compounds were simply conjugated aromatic molecules and dyes. Martin Pope at NYU observed luminescence from anthracene (e.g. Wikipedia) in the 60s.
A wide variety of conjugated dyes have been tried with varying success.
In general, there are 2-3 different types of compounds used in an OLED:

Hole-transport materials: Easily oxidized and conduct positive charges (holes).
Electron-transport materials: Easily reduced and conduct negative charges (electrons).
Emissive materials: Holes and electrons recombine and emit light.

In many cases, these are made in a sandwich device: ITO anode with hole transport layer on top, emissive layer, then electron transport layer, then cathode. (I'm leaving out various interface agents, "blocking layers," adhesion layers, etc.)
You're asking about the emissive component. Basically, you want materials that will efficiently emit (fluoresce or phosphoresce) at different wavelengths / colors.
The biggest real breakthrough, IMHO was from $\ce{Alq3}$ which is an aluminum organometallic compound with remarkably bright emission. But emits green, so for displays, you need other things.
I've been out of the field for a while, but here are some examples. Chemical companies like lumtec sell various components.

Polyfluorenes: different copolymers have tailored emission colors. Usually the "bridge" carbon is substituted with two alkyl substituents for solubility and processability (and to avoid aggregation).

In OLEDs, polyfluorenes are desirable because they are the only family of conjugated polymers that can emit colors spanning the entire visible range with high efficiency and low operating voltage. Furthermore, polyfluorenes are relatively soluble in most solvents, making them ideal for general applications[...]

Various vinyl carbazoles: Similar reasons to the polyfluorenes.
Phosphorescent metal complexes (e.g., iridium): If you work out statistics, you get a lot of electron-hole combinations that are triplets, so phosphorescent emission can be very efficient, particularly with transition metals with a high spin-orbit coupling.

I think most leading emissive materials still largely fall into those categories. Small molecule purely organic targets exist, but aren't as efficient.
A bigger question is usually how you intend to deposit the compounds. Vapor deposition (i.e., evaporation or sputtering) can be done with almost anything that sublimates, but if you want to do drop-casting, spin-coating, or inkjet printing, you'll want soluble compounds.

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Comparing the nature of ionic compounds
I want to know how to compare the strength or stability of ionic compounds.
Example : $\ce{Mg(HCO3)2}$ and $\ce{NaHCO3}$
I know that polarity and polarizability affect the ionic and covalent nature.

"Strength". Hmm, what firstly comes to mind is you want a comparison of lattice energies (or more precisely, lattice enthalpies)
Lattice energy is a [crude] way of estimating how strong the electrostatic attractions are between ions in an ionic compound. Let's take the formation of $\ce{NaCl}$ into account, as I see it's the most common "ionic" compound that teachers start teaching the lesson with.
According to Born-Haber cycle this is the reaction that results in the formation of pure table salt:
$$\ce{Na+(g) + Cl- (g) \rightarrow NaCl(s)}$$
Forming the salt crystals is exothermic and this is called the lattice enthalpy of formation.

Note that some textbooks mean to define lattice energy as the energy that is obtained by the thermodynamic system in the process of turning the solid compound to its gaseous ions.

To make a long story short, the more the lattice enthalpy is, the stronger "bonds" you get.
There are two main factors that affect the lattice formation enthalpy:

Charge of the ions
Radius of the ions

Charge of the ions:

Source
Look at the diagram above. $\ce{NaCl}$ is known to release approximately 822 kilojoules upon formation, while $\ce{MgO}$ releases 3890 KJ. And what does that mean? It means that Magnesium Oxide is way more "stable" than Sodium Chloride. So, complete the rest:

The more the charge of the ions is in an ionic compound, the _____ is its lattice enthalpy and the ______ its stability.

Radius of the ions:

Source
As you see, and could've imagined if you were on the right track, bigger ions (literally) mean weaker attraction. I wanted the answer to be fairly complete, so I did an addition to the main point that answers your question. Thus, there's no need of expanding this.
In your case:
You might as well have reached a conclusion. Mg ion has a 2+ charge, while Na has only a single electron missing. The anion is pretty much the same, $\ce{HCO3-}$ (hydrogen carbonate), and its priorly covalent nature shouldn't mislead you in doubting your answer. There is a case of comparison of the charge of the ions, and I would say Magnesium ion wins.
In case of additional study, you can visit:
http://www.chemguide.co.uk/physical/energetics/lattice.html
http://www.knockhardy.org.uk/sci_htm_files/08le.pdf
http://www.rsc.org/learn-chemistry/resources/gridlocks/downloads/answers/Born%20Haber%20NaCl%20ans.pdf

Addendum:
The charge and the radius of ions are by no means the only factors that affect the lattice enthalpy. It's just that the other factors are considered "trivial" by current textbooks.
The lattice enthalpy and lattice energy aren't the same. They do have close meanings, though.

The strength of an ionic bond is determined by two factors:
The size of the ions (the smaller the ions the stronger the bond) NaF stronger than NaCl because F- is smaller than Cl-
The charge of the ions (the higher the charge the stronger the bond) Na2O stronger than NaCl, because O is a double negatively charged ion


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Would C bond with H or D?
Is a C-D bond better than C-H bond?  I mean if carbon has a chance to either for a bond with deuterium or hydrogen what would it choose and why?  What if tritium competes too?

C-H would be the weakest bond, then C-D, then C-T would be the strongest.
This is due to the zero point energy of the bond and the fact the zero point energy is proportional to:
$$\sqrt{\frac{m_C + m_H}{m_C m_H}}$$
Zero point energy is the minimum energy the quantum harmonic oscillator can have.
Because the greatest mass isotope has the smallest zero point energy, the energy of the bond is greatest.

Does this apply for other pairs, say C-O and C-O(18)?

@Koyomi-chan yes, it's just a small effect

Kinetically, hydrogen will react at the fastest rate, followed by deuterium and then tritium (given a reaction such as a carbon atom or radical plus molecular hydrogen giving a C-H bond plus an •H atom). The most quickly formed product may not necessarily be the most stable.


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How to calculate the parts-per-million concentration from a percent concentration?
I am asked to find the concentration of magnesium ions in ppm for a $0.025\ \%$ solution of $\ce{MgCl2}$. I can use this formula:
$$\text{concentration in ppm} = \frac{\text{mass solute}_\mathrm{(g)}}{\text{mass solution}_\mathrm{(g)}}\times10^6$$
Assuming $100\ \mathrm g$ of solution, $2.5 \mathrm g$ solute/$\ce{MgCl2}$, and $97.5\ \mathrm g$ solvent, I get $\pu{95.205u}$ for $\ce{MgCl2}$, and then $0.026259\ \mathrm{mol}$ of $\ce{MgCl2}$, and thus $0.026259\ \mathrm{mol}$ of $\ce{Mg}$. That times $24.305\ \mathrm{g\ mol}^{-1}$ for $ \ce{Mg}$ gives me $0.63823\ \mathrm g$ of $\ce{Mg}$. $\frac{0.63823\ \mathrm g}{100\ \mathrm g}$ solution gives me $0.00638$, so that multiplied by $10^6$ would be $6382.28$. But the only options for the questions don't go over $500$. Like $64\ \mathrm{ppm}$ or $500\ \mathrm{ppm}$. Nothing even close to as high as I'm getting.

100 ppm = 1/100 th %

You're using $2.5\ \mathrm g$ of solute out of $100\ \mathrm g$ of solution, which is $2.5\ \%$, but the initial question is asking about a $0.025\ \%$ solution, which is why your answer is two orders of magnitude too large. You should be using $0.025\ \mathrm g$ of solute and $99.975\ \mathrm g$ of solvent.
As a general helpful hint for answering multiple choice questions, when you notice that one of the answers has the same coefficient but is off by a couple orders of magnitude (in this case the $64\ \mathrm{ppm}$ answer) go and double check all your initial concentrations and volumes, you probably forgot to account for a "$\%$" or a "$\mathrm k$" somewhere.

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Why is water considered ionic in fuel cells but otherwise covalent?
If water, $\ce{H2O}$ is a covalent compound formed by sharing of electrons, why is it said (in case of fuel cells) that formation of water from hydrogen and oxygen is a redox reaction (transfer of electrons, not sharing). This would make it an ionic compound.
Any insights?
Note that, the concept of water being formed as a result of redox reaction is highlighted in fuel cells (which is the basic principle of fuel cells - generating electricity by exploiting this transfer of electrons).
I know how fuel cells work, and it is indeed a redox reaction; but my doubt is - can water be formed by both covalent/ ionic means?  (In textbooks, or in exams when asked to identify bonding present in water); when both covalent and ionic type can be achieved in its formation? Is both "covalent" and "ionic" correct while describing the formation of $\ce{H2O}$ molecule?

Why can't you have covalent compound formed via redox reaction ?

Perhapse it is referring to the polar nature? (vs organic solvents which are nonpolar)

Water is a covalent compound. But as we know no compound is 100% ionic or 100% covalent. Since the electronegative difference (END) between hydrogen and oxygen is very high it is a strongly polarised covalent compound therefore the electron cloud is shifted towards the oxygen atom and therefore water has a very high % ionic character.


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LiAlH4 / H2O reaction energy
I'm dealing with a free-energy scam on another forum. It is well-known that $\ce{LiAlH4}$ reacts explosively with water. My chemistry days are long behind me, so I thought I'd come here to ask:
$1~\text{mol}$ of $\ce{LiAlH4}$ reacted with $4~\text{mol}$ of water gives how much energy?
I'm not about to do this, mind you. I was just curious about the specific numbers.

So you have the reaction :
$$\ce{LiAlH4 + 4H2O <=> LiOH + Al(OH)3 + 4H2}$$
You want to get the "energy" given by this reaction, which is nothing but the reaction enthalpy.
$\Delta_rH^0(T) = \sum_{i}\nu_i.\Delta H_{i,f}^0(T)$
Now you get this, just look for the theorical values of each $\Delta H_f^0$ you need for the 5 compounds you have (they are tabulated) and sum them up pondered with the stoechiometric coefficients.


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Why is oxygenated water not H2O2?
Oxygenated water has become popular for a variety of uses recently - for health purposes internally as well as enhanced wound healing.
Anyhow, my dad asked me recently if I knew how it is that oxygenated water does not become hydrogen peroxide. I don't know the answer to it; but it got me thinking... How does this work? Does the oxygen not bond to the water molecule? If so, how does the bottle of water not just out gas when ever it is opened until the amount of oxygen contained in the bottle is equal to that of the room air.

Oh, if oxygenated water did change to hydrogen peroxide, I shouldn't have been alive now! Some reactions, just don't like to happen.

Sorry but anyone foolish enough to buy oxygenated water is an idiot wasting money.

@MaxW I would have to agree! I just was more curious about the chemistry of the whole thing :)

The oxygen is dissolved in the water, just as salt can be dissolved. It does not (to any appreciable degree) combine with water molecules to form hydrogen peroxide.
The reason that oxygenated water is not fizzy like soda water (CO2) is the solubility of oxygen in water is about 2% that of CO2. So the amount of bubbles formed when the cap is removed is only about 1/50 as much as for soft drinks. Or champagne.

When CO2 is dissolved in water, part of it reacts with H2O to produce H2CO3. Similarly part of the dissolved oxygen reacts with H2O to produce H2O2, doesn't it? If not why? If yes, what's the equilibrium?

@PetrPudlák Not similar at all. The formation of peroxide is very unfavourable and requires a lot of energy. So it doesn't happen easily or spontaneously, totally unlike carbonic acid.

I think the problem is that the reaction:
$\ce{2H2O2-> 2H2O + O2}$
is spontaneous, meaning that it proceeds (slowly) in standard conditions with no other inputs required. So far from oxygen combining with water, you have hydrogen peroxide spontaneously decomposing into water and oxygen.

First you need to know what is oxygenated water. It's actually form by adding additional oxygen to the water under pressure. It means it's a mixture, not a compound.
We know that a mixture is two or more substances mix together physically. However, hydrogen peroxide bonds chemically which is a compound.
So, we can conclude that oxygenated cannot be classified as hydrogen peroxide.

With respect to the question, 'how it is that oxygenated water does not become hydrogen peroxide', one answer is that in natural sunlight waters it actual does create H2O2, albeit even then, in only small amounts.
More precisely,  the presence of H2O2 and other reactive oxygen species (ROS) have been cited in the literature as occurring in illuminated natural waters (see, for example, ‘Reactive Oxygen Species in Natural Waters ‘, by Neil V. Blough and Richard G. Zepp). To quote from an online two-page preview :

The presence of ROS in aquatic ecosystems was first reported in 1966 by Van Baalen and Marler, who detected the presence of hydrogen peroxide (H202) in tens to hundreds of nanomolar concentrations in surface seawaters... Although their data were inadequate to specify the source of the H202, they postulated that photochemical reactions, biological processes, or atmospheric deposition might contribute to its presence in seawaters. Some years later, Swallow (1969) suggested that the hydrated electron ($\ce{e-(aq)}$) could be produced in seawater by the action of...photoionization of endogenous phenolic compounds. He further proposed that rapid reaction of $\ce{e-(aq)}$ with dioxygen (02) would yield superoxide ($\ce{.O2-}$), ultimately leading to the formation of H202 via  disproportionation….

My updated understanding of these reactions is presented below which starts with a solvated electron, $\ce{e-(aq)}$ and $\ce{H+}$ per the reactions:
$\ce{O2 + e-(aq) ⇌ .O2- (aq)}$  (superoxide radical anion)
$\ce{H+ + .O2- (aq)  ⇌ .HO2 (aq)}$  (hydroperoxyl radical with a pKa = 4.88)
$\ce{.HO2 (aq) + .HO2 (aq)  -> H2O2 + O2 (aq)}$ Source: Equation (1)
In natural waters in sunlight, dissolved organic matter (DOM) can be a source of solvated electrons (see 'Photoproduction of hydrated electrons from natural organic solutes in aquatic environments') and electron holes:
$\ce{DOM  + hv -> e- + h+}$
Also, the presence of transition metal ions and dioxygen via so-called metal auto-oxidation can be a potential source of superoxide:
$\ce{Fe(2+)/Cu(+)  + O2 ⇌ Fe(3+)/Cu(2+) +  .O2-}$ (Source example: Eq (Vl))
The cited work by Blough and Zepp further notes that since the 1980’s:

...environmental scientists have acquired evidence for the near-ubiquitous occurrence in surface waters of not only H202 and 02" but also singlet dioxygen..., the hydroxyl radical (OH), and organic peroxyl radicals (R02), as well as other transient intermediates that are either immediate precursors or products of the ROS. It is now recognized that the production of this diverse array of species is driven primarily by abiotic photochemical reactions involving naturally occurring organic (and sometimes inorganic) chromophores (Zepp, 1991; Zika, 1987; Zafiriou et al., 1984). Because of its high concentration in surface waters (-250 nanoM), O2 dominates the photophysics and photochemistry of these materials. The resulting reaction sequences are readily interpreted within the well established concepts of direct and sensitized photo-oxidations, potentially coupled to thermal autoxidation processes.

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Why does potassium form peroxides but sodium does not?
As we go down the periodic table, Group 1 alkali metals hold their single outer valence electron more loosely, and so reactivity increases. Below potassium (Na, Li), we can store the metal in oil with no difficulty. Above potassium (Rb, Cs), we need to exclude all oxygen to prevent reaction in air. Potassium (K) is the unique element where we can technically store it in oil, but it is highly likely to react with any available oxygen to form explosive peroxides. I'd like an explanation of why this is; i.e. why is the reactivity favorable for the formation of peroxides. Thanks!

If you look at the standard Gibbs free energies for the alkali metal peroxides, actually, lithium should form it most easily. The standard Gibbs free energies of formation are -564.8, -430.1, -418.4, -351.5, and -359.8 kJ/mol for $\rm Li_2O_2, Na_2O_2, K_2O_2, Rb_2O_2,$ and $\rm Cs_2O_2$, respectively.
If we're looking at reaction of an alkali metal with oxygen to form explosive compounds, perhaps we should look at superoxides. The superoxides are powerful oxidizing agents, and they can explode when mixed with water, acids, organics, or powdered graphite.
At room temperature, oxygen readily reacts with K, Rb, and Cs to form superoxides ($\rm MO_2$), while Li and Na don't. There is a direct correlation between superoxide stability and electropositivity of the metal concerned (source: Cotton & Wilkinson's Advanced Inorganic Chemistry). $\rm NaO_2$ can be obtained by high pressure and temperature reaction of $\rm Na_2O_2$ with $\rm O_2$, and $\rm LiO_2$ won't form at room temperature.

Nice answer -- here's a vid of exactly how high-temperature you need to get to make $\ce{Na2O2}$: https://www.youtube.com/watch?v=ZqEWUw6sgpA

You may want to compare the quoted free energies with those of the normal oxides. For example, with lithium you have $-564.8$ kJ/mol for the peroxide versus $-562.1$ from WP for the normal oxide. These are so close that they are likely within experimental error: the peroxide may or may not be possible at room temperature. Only sodium and heavier in Group 1 have the free energies far enough apart to definitively favor the peroxide.


text:
How can I draw 2,3-diethylheptane?
This question was given by my teacher in class:

Draw the structure of 2,3-diethylheptane.

So I drew the following:

But now if I name the compound according to the longest chain, it comes out to be 3-methyl-4-ethyloctane! How can I properly draw these structures?

What I usually do when I start numbering is look for branching and start numbering at the end with the closest branch. Also, keep in mind that with branching you'll want to consider numbering from one of the branches (if it'll yield a long chain.)
I'm using the structure you made as an example:
Here I numbered from the side with the most branching. But then I numbered it in two different ways. One way was starting with the straight chain which gave me a 7 carbon chain. Then I renumbered it (the image on the right) starting from one of the branches. This gave me an 8 carbon chain. (You could also number starting from the other branch, but that gives you 7 as well.)
Choose the longest chain. Since stuff that looks like a side chain can actually be part of the main chain, seriously keep this in mind during exams. It's totally okay to number the same alkane in different ways and play around with it to find which is the longest chain.
4-ethyl-3-methyloctane

Now, if you're having trouble drawing them:
I would draw just a zig-zag pattern (like I drew above) containing the number of carbons you need. You need a heptane, so start out with a 7 carbon chain.
This will just be your bare, skeleton structure. Then number the carbons, then add the ethyls to carbons 2 and 3.
HOWEVER, the question is flawed. You actually can't have an ethyl on carbon number 2 here, because it just makes the chain longer.

Oh so we must have ethyl on at least at 3? and methyl at 2.

Yeah, you couldn't have a 1 methyl hexane, for example. It would just be a heptane. I think you've pretty much got it.

You actually can have an ethyl on carbon number 2, for example in 2-ethylbutanedioic acid/2-ethylsuccinic acid.

First, according to the name that you were provided, 2,3-diethylheptane, you did draw the correct structure. The problem actually lies with the name: it does not follow the standard IUPAC nomenclature. While it is possible to draw the structure, as you have, it's not the correct name for it.
Any name derived from an organic alkane structure should start with the longest chain from start to end. From your structure the "2-ethyl" branch is actually part of the longest chain, which then includes 6 other carbons leading to the furthest end. Now you have to deal with a methyl group and an ethyl group on that octane chain: in this case, both lie on the same side, so there is no competition for which has priority in determining the numbering. So from the end that they're both closest, the methyl is on the 3-carbon; the ethyl is on the 4-carbon.
It is possible to get correct structures from incorrect names; the system is degenerate in that sense. However there is (ideally) only 1 correct IUPAC name for any organic structure.


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Doubts about IUPAC nomenclature of 4-ethyl-1-fluoro-2-nitrobenzene
Why is 4-ethyl-1-fluoro-2-nitrobenzene (structure is in the attached picture) called so ?
Shouldn't it be called as 1-ethyl-4-fluoro-3-nitrobenzene since the precedence of ethyl is greater than halogens?

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

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g. Also note that all prefixes (i.e. ethyl, fluoro, and nitro) are considered together; i.e., there isn’t any order of functional group priorities at this point.
Furthermore,

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
other locant sets, each cited in order of increasing value, has the lowest term at the first
point of difference; (…)

Therefore, the example is named as 4-ethyl-1-fluoro-2-nitrobenzene rather than 1-ethyl-4-fluoro-3-nitrobenzene since the locant set ‘1,2,4’ is lower than ‘1,3,4’.

I see that the locant set of 1,2,4 is lower than 1,3,4 but the precedence of ethyl is greater than fluorine .. so shouldn't ethyl get the lowest number ?

@user126190 The prefixes (‘ethyl’, ‘fluoro’, and ‘nitro’) are ‘all considered together in a series of increasing numerical order’.

But this website link and many other websites too say alkanes are before halogens ..

@user126190 Your link is about the seniority of functional classes in functional class nomenclature, which is not related to your question. In your example, the functional class is already given; it is neither alkane nor alkyl halide – it is benzene.

@Loong is correct. As far as numbering of locants is concerned, alkyls, alkoxys, halos, and nitros all has the lowest priority because they are substituent only functional groups.


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Why Ca 2+ and not Ca +2
I have been learning chemistry for a year or two but I still can't understand this. According to maths we should place the plus or minus sign in front of a number like -2. However, In chemistry we always put the sign behind a number when writing ions like $\ce{Ba^2+}$ ,$\ce{Cu^2+}$, $\ce{O^2-}$ etc. So, I want to know why do we write it that way?

Chemistry has an organization called the International Union of Pure and Applied Chemistry (IUPAC), which sets standards for nomenclature (among other things). Chemists usually follow its recommendations.
IUPAC says

Ionic charge is indicated by means of a right upper index, as in $\ce{A^n+}$ or $\ce{A^n-}$ (not $\ce{A^$+n$}$ or $\ce{A^$-n$}$).

and

The ionic charge number is denoted by a right superscript, by the sign alone when the charge number is equal to plus one or minus one.
...
$\ce{Al^3+}$ is commonly used in chemistry and recommended by [74 = see quote above]. The forms $\ce{Al^$+3$}$ and $\ce{S^$-2$}$, although widely used, are obsolete [74], as well as the old notations $\ce{Al^{+++}}$, $\ce{S^=}$, and $\ce{S^{--}}$.

No reason is given, but you'll find a historical account of sign conventions for charge and a defense of IUPAC's choice in this article. He says,

[IUPAC's convention] avoids confusion with the conventional symbolism for inherently positive and negative numbers and maintains consistency in how we count physical entities. Thus, in counting apples, we say two apples, three apples, etc., not apples two, apples three – that is, the number always precedes the name of the entity being counted. Likewise, when counting charges, we should say two positive charges or three negative charges, not positive charges two or negative charges three. The IUPAC ruling was intended to make the charge number symbolism consistent with this verbal convention.

Personally, I think the charge should be written like any other signed number. Formulas typed from the keyboard without superscripting can be ambiguous; is X2+ to be interpreted as $\rm X^{2+}$ or $\rm X_2^+$? (You should type X(2+) or X2(+), but few people actually do that.) But it's pretty clear what X+2 would mean. Putting the sign first also makes it trivial to parse the charge from a formula typed this way without the need for parentheses.

A previous version was quoting IUPAC talking about charges in written long names. I modified the answer with better quotes.

I understand Jensen's defense of IUPAC, but don't agree.  Of course, this is all arbitrary, but here's my view:  If the Ca ion were Ca w/ 2 pos. chgs., it would make sense to write "2+" (like "2 apples").  However, that's not the case— both the ion and the neutral atom have 20 pos. chgs.  Rather, the Ca ion is in a net +2 state.  And to indicate this state, we first indicate what's most important (the sign)—and then, within that big category, the subcategory: the magnitude.  I.e., the proper convention to apply here is what we use to indicate the value of numbers: e.g., "–3", not "3–".

$\ce{Ba^{2+}}$ means that you have a Barium ion with two positive charges.
so the 2 counts the positive charges and the + is not the symbol to add something, the + means only a positive charge.

I'm adding this answer as maybe reasoning of why IUPAC proposed putting the charge symbol after the number. Consider it a supplement to the @fred-senese's answer of 'cause IUPAC says so, and the linked article.
We put the charge symbol (+ or -) after the number because charges are entities, not numbers. Placing the number before the symbol helps to keep the idea that charges (more specifically multiples of the charge on an electron) are things we are counting clear.
It is the same convention as when there is multiple of an entity in algebra. If you had two $a$'s, that is $a+a$, it is written as $2a$, not $a2$. What is confusing (and unfortunate) is that the symbols chosen, for units of charge are the same as those for positive and negative (signed) numbers; + and -.
Even though we use the positive and negative symbols for charge, they are not positive and negative numbers. In the context of charges, 2+ means two positive things, 3- means three negative things. (In the case of ions, the charge represents the difference from a net zero, neutral, charge, but that discussion is getting away from the point of the question of "why does the charge symbol come last in the subscript?".)
To get philosophical, we, historically, could have chosen any other two names for charges: red and green, apple and orange, foo and bar, Romeo and Juliet, the article linked in Fred Senese's answer proposed primes and dots as the symbols! In the end we chose positive and negative, and use the associated symbols from positive and negative numbers, + and -. We can use the arithmetic of signed numbers to do computations involving charge, balancing ions, redox, conservation of charge and so on; but, this is because the properties of charge mirror that of signed numbers, not because charges are positive and negative numbers.

I have 2 cars, 5 bikes and 3 cats. I don't say I have cars 2, bikes 5, and cats 3.
Barium ions have 2 positive (+) charges, oxides 2 negative (-) charges. Don't say Barium has + charges 2.

History based customs beat logic in many areas, from chemistry to English.

As an old school adherent, I am not at all a fan of this over-formalism. It sacrifices intuitiveness for minor, irrelevant details and helps to turn people away from chemistry.
And it is not true that you could just as well use Romeo and Juliet, for example. The fact is if you combine a -1 charge with a +1 charge you get a zero charge. I don't know what you would get if you combined a Romeo with a Juliet.
And if a cation with a +3 charge gains two electrons it now has a +1 charge. So the simple arithmetic of whole numbers still applies.
Sorry, I think the way chemists had been doing it for 200 years worked just fine.
I sometimes think that the IUPAC, or whatever organizations made this change, have way too much time on their hands.

This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review

If one stops seeing +/- as unary operators and starts seeing them as symbols for signed elementary charges, it is easy. It could be conventionally denoted as "p" and "m" instead and we would have $\ce{Ca^{2p}}$ and $\ce{SO4^{2m}}$. // By other words, instead of "+2"/"-2", it is 2 times +/- or p/m.


text:
Single line equilibria notation
I'm reading a paper with a reaction scheme where an enzyme binds loosely to inhibitor reversibly and then the Enzyme-Inhibitor complex undergoes another reversible conformational change to a tightly bound complex. My question is about the notation used (which I cannot type because I can't find the appropriate symbols). The second reaction uses the regular equilibria symbol which is fine - no problem. However, the first reaction uses a modified form of the equilibria symbol which is one line rather than two with 'half arrows' at either end. What does this symbol mean?

You mean this: ↔

No - similar but with 'half arrow heads' at either end

You should provide a link to that paper, otherwise it would be difficult to say anything conclusively. I think that must have been a printing/typesetting mistake.

...equilibria symbol which is one line rather than two with 'half arrows' at either end

I'm pretty confident you mean "⥋" arrow within the reaction scheme for enzyme $\ce{E}$ and inhibitor $\ce{I}$ which I tried to recreate from your description:
$$\ce{E + I ⥋ \underset{\rm loose\\bond}{EI} <=> \underset{\rm strong\\bond}{EI'}}$$
I've seen this arrow in literature before, and in general it's been also utilized to denote equilibrium. This notation is not officially supported by IUPAC standards. However, there is a couple of references tangentially mentioning it.
From the article "The Chemical Equation Part I: Simple Reactions" [1, p. 189] (emphasis mine):

Sometimes reactants interact quickly, but the reaction never goes to completion because the reverse reaction is competing with the forward reaction, and the system ends up in a state of chemical equilibrium. A pair of arrows pointing in both directions indicates an equilibrium system, but the second arrow need not be there. Even when two arrows are used, the equation still cannot identify the point of equilibrium.

There is also a recent outstandingly well-written review with the title that speaks for itself: "Chemistry: A Panoply of Arrows" [2, p. 592], which pretty much says the same (emphasis mine):

Van't Hoff's double arrow was modified in 1902 by Hugh Marshall,... who removed the inner barb of each arrow. This created the half-headed double arrows, ⇌, which have since been further simplified to ⥋. In justifying his proposal, he noted that single arrows were being used in organic chemistry
  “to indicate merely the stages and methods by which a substance can be produced from some other substance as a starting point. In such cases, as no attention is paid by the bye products (sic), there is nothing of the nature of an equation
  involved.”

My humble opinion is that authors' intent was to visually separate those two processes, e.g. due to exceptional difference in rates, or maybe they wanted to denote conformational change aside from the chemical equilibrium.
Notes

"⥋" arrow belongs to the Supplemental Arrows-B Unicode block under the name "Left barb down right barb up harpoon" and is invoked via U+294B.
This and many other symbols you are having troubles identifying can be recognized by services like Shapecatcher or Detexify.
I also asked a question on Mathematics.SX hoping that someone could provide more definitive answer regarding the proper usage of this arrow (no answer yet).

References

Kolb, D. Journal of Chemical Education 1978, 55 (3), 184. DOI: 10.1021/ed055p184.
Alvarez, S. Angewandte Chemie International Edition 2012, 51 (3), 590–600. DOI: 10.1002/anie.201101767.


text:
Is it possible to chemically treat grass to make it digestible?
Is it possible to chemically treat grass after "harvesting" it to make it digestible by humans?  I know we can "eat" grass, but we can't break it down in our stomachs. If there is some chemical treatment to partially break grass down into something we can digest, I see it possibly being used to end world hunger!  I know we already treat bone to make it into gelatin, and I am wondering if something similar can be done for grass.

Comestible grass is already available. It's called salad...

@Babounet However, we don't grow salad in our lawns typically.  However, this does bring up an interesting point in that maybe we can grow something in our lawns that is edible and suitable for a lawn (if there is such a thing).  If it doesn't exist yet, perhaps it could be made?

Lots of plants growing on meadows are edible. Btw this question is rather for Biology SE or for Seasoned Advice SE - things that you eat are generally cooked not thrown into chemical reactor although it's not so big difference as on might think.

"Edible" and "digestible" are two different things. The latter implies the ingested matter can be mostly metabolised into nutrients/energy storage. Presumably a large portion of a plant's mass could be converted to digestible sugars for humans with the use of enzymes such as cellulase, or acid/base catalyzed hydrolysis of polysaccharides.

I take it you're not looking for the answer, "feed it to a cow"? ;-)

How about that grass is low in nutrition and thus making it human edible doesn't mean a lot. Cows need to eat a lot of grass to get their nutrition.

The indigestible component of grasses is cellulose, which is polymerized glucose.
Yes, it is possible to hydrolyze the cellulose to glucose, which is of course digestible.
The hydrolysis can be catalyzed enzymatically with a cellulase, or using acid hydrolysis.
The process of hydrolyzing cellulose is often discussed in the context of so called cellulosic ethanol, as the glucose is then fermented to ethanol subsequent to hydrolysis.


text:
Reliable data describing acidity of 2,4,6-trinitrobenzenesulfonic acid
I wonder what is the $\mathrm{p}K_\mathrm{a}$ or the value of Hammett's $H_0$ function of 2,4,6-trinitrobenzenesulfonic acid? PubChem gives $\mathrm{p}K_\mathrm{a} = 0.13,$ but doesn't specify the solvent. Even if it's real value I doubt it's the value for water. Why? For starter even aliphatic sulfonic acids are strong, and with electron withdrawing groups, even stronger.
Guthrie reports gives $\mathrm{p}K_\mathrm{a} = -4.0$ for p-nitrosulfonic acid [1]. Even if resonance of o-nitro groups is blocked, it doesn't negate inductive effects. Reasoning with solvation effects also seems dubious — nitro groups aren't hydrophobic at all. Is there some reliable data for this, rather obscure, acid?
Reference

Guthrie, J. P. Hydrolysis of Esters of Oxy Acids: $\mathrm{p}K_\mathrm{a}$ Values for Strong Acids; Brønsted Relationship for Attack of Water at Methyl; Free Energies of Hydrolysis of Esters of Oxy Acids; and a Linear Relationship between Free Energy of Hydrolysis and $\mathrm{p}K_\mathrm{a}$ Holding over a Range of 20 $\mathrm{p}K$ Units. E. Can. J. Chem. 1978, 56, 2342–2354. DOI: 10.1139/v78-385@cjc-csc.issue01. (PDF)

The reference for the PubChem entry for picryl sulfonic acid is:

Serjeant, E.P., Dempsey B.; Ionisation Constants of Organic Acids in Aqueous Solution. International Union of Pure and Applied Chemistry (IUPAC). IUPAC Chemical Data Series No. 23, 1979. New York, New York: Pergamon Press, Inc., p. 40

So, yes that is the value of the pKa for picryl sulfonic acid (2,4,6-trinitrobenzenesulfonic acid) in water.
The pKa of benzenesulfonic acid is 2.554, although the reference does not specify a solvent. The pKa of methanesulfonic acid is -1.86, with the same reference as picrylsulfonic acid.
The difference between methanesulfonic acid and the arylsulfonic acids might be the ability of water to solvate the anion. The aryl groups are less polar and more hydrophobic. They are also large. This same reason (decreased solvation of the anion) is invoked as the explanation for the difference in acidity between primary, secondary, and tertiary alcohols. The magnitude of the effect in alcohols is a few pKa units, which appears to be of a similar scale for the same effect in sulfonic acids.

While I appreciate your effort, there's virtually no way this constant would be so high. Strong acids are notoriously difficult to pinpoint with pKa and values obtained with different methods vary very much, but not that much. Check out for example this analysis. It clearly indicates that aryls only make sulfonic acids stronger.


text:
Can alloys or ceramics have higher density than elemental solids?
Are there any examples of alloys or ceramics that have higher density than their densest elemental constituent?  If so, do they have unusual or difficult synthesis processes?  If not, is it provable that this is not possible, or demonstrably unlikely.
Since homogeneous metals are essentially crystals in their solid phase it seems plausible that, in theory, there are elements that could fit within the crystal structure without altering it to increase their density.
This would be particularly interesting for metals that are used in applications primarily for their density: e.g., lead, tungsten, and uranium.

Yes, for both alloys and ceramics it is possible have a density that is higher than their densest elemental constituent. This is because every lattice structure of the denser elemental constituent contains interstitial void spaces, or holes, typically tetrahedral holes (surrounded by 4 atoms) or octahedral holes (surrounded by 6 atoms). These holes exist despite the crystal having an optimal packing structure, thus a maximum density for the pure solid.
The densest elemental solids are typically made up of larger atoms toward the bottom of the periodic table, that have larger atomic radii due to electron-electron repulsion, but which have much higher atomic mass due to their nuclei. This produces larger interstitial voids that can fit smaller atoms without significantly disturbing the lattice constants of the larger, denser component. An increase in mass without concomitant increase in volume will produce a denser composite.
Specific examples are difficult to find, as densities are not always reported. One common material which is denser than the corresponding "densest elemental constituent" is steel. Steel (maximum density: 8.05 g·cm⁻³) is an alloy of iron (density 7.874 g·cm⁻³) and carbon (diamond density: 3.5 g·cm⁻³), although occasionally some chromium (density: 7.19 g·cm⁻³) and nickel (density: 8.91 g·cm⁻³) may also be included.
Another possible, although less clear-cut example is the alloy heavymet, which is used in ATLAS. It is composed of nickel (9.9 atom %), copper (4.0 atom %), and tungsten (86.1 atom %). Heavymet has a density of 19.3 g·cm⁻³, which despite the less dense diluents, may be slightly larger than tungsten's density which is 19.25 g·cm⁻³.

Excellent answer so far.  Can we put a theoretical upper limit on the size and/or classes of atoms that can fill the interstitial voids of heavier metals?  The alloy examples you cite are helpful but not dramatic.  I'm wondering if there are or could be examples where the alloy density is significantly greater than the elemental density.

In a somewhat different vein are compounds where the density of hydrogen is greater than that of pure liquid hydrogen. Water is one such compound, and several metal hydrides that are being considered for hydrogen storage also qualify. All these compounds sport 0.11 to 0.15 gram hydrogen per cubic centimeter,versus only about 0.071 for pure liquid hydrogen.
In such cases we are dealing with hydrogen atoms that are relatively large for their mass on their own, but lacking inner core electrons they can be compressed with favorable energy balance to gain stronger covalent bonds with other relatively small atoms (water) or ionic bonds small ions (magnesium hydride and other hydrifldes listed in the answer referenced above).


text:
How exactly are molecules detected in space?
Recently the sugar glycoaldehyde was detected in a star system 400 light years from Earth. How exactly are molecules detected in space? I am aware that spectroscopy is used to detect them, but I don't know the details.

This is a very interesting question. My basic understanding that the molecule has a spectral line within the electromagnetic spectrum that identifies it, like a thumbprint. When a telescope is directed at such star you will receive the spectral line but at a different frequency due to the phenomena of redshift.

You may ask this on the astronomy stack exchange and get an answer!

yeah, try astronomy or physics stackexchange

I'm not an astrochemist, but the key technology apparently is infrared spectroscopy. One instrument currently in use is the HIFI (Heterodyne Instrument for the Far Infrared) aboard the Herschel probe. Please have a look at the site of ESA and the JPL for further information.

The research article corresponding to the observation to which the question is referring is Detection of the simplest sugar, glycolaldehyde, in a solar-type protostar with ALMA. Note that the molecule is glycolaldehyde rather than glycoaldehyde.
ALMA is the Atacama Large Millimeter/submillimeter Array, a radio telescope array in the Atacama desert.  The extremely low humidity of the Atacama desert minimizes water vapor interference.
This observation of glycolaldehyde in space is not the first.  It was earlier observed by Greenbank radio telescope, in West Virginia, USA, for example.  The Greenbank observation is reported in INTERSTELLAR GLYCOLALDEHYDE: THE FIRST SUGAR.  Glycolaldehyde was detected through radio signals in the 71-103 GHz range that are photons emitted when the molecule transitions from an excited rotation state to a lower energy rotational state.
In the ALMA observation, the molecule was detected based upon radio signals near 220 GHz and 690 GHz.  These correspond to photons emitted by the molecule when transitioning from an excited rotational state to a lower energy rotational state, but are a different set of transitions from those observed at Greenbank, including some for excited vibrational states.
I'll try to add more information later, there are a wide variety of spectroscopic techniques: absorbance vs. emission; rotation, vibration and electronics transitions, and many different frequency ranges.

How does rotation and vibration lead to emission. How do you map the frequency of light received uniquely to a molecule. How do you know this frequency of light comes from a molecule and not some energy from a star or other source. I guess I'm concerned with how the actual spectroscopy works.

The rotational states are quantized.  Quantum mechanics dictates that a molecule can not rotate arbitrarily, but only in discreet energy amounts.  See http://en.wikipedia.org/wiki/Rotational_spectroscopy  On Earth in laboratories libraries of spectra of molecules are compiled such as: http://spec.jpl.nasa.gov/ftp/pub/catalog/catdir.html  Then a comparison is made between the signal from space and the library of spectra


text:
Why is DNA negatively charged and what makes it so?

What part in the strand contributes to the overall non neutral charge?
DNA is not isolated in the body, so what keeps it stable while being charged?
Why is it important for DNA to be charged?

related http://biology.stackexchange.com/questions/3793/does-dna-react-in-all-of-the-ways-most-other-acids-do, http://biology.stackexchange.com/questions/20109/dna-is-charged-negative-where-is-all-the-positive-charge-in-my-body/20111#20111

As cibrail said, DNA is a polymer of nucleotides. They join themselves through phosphodiester bonds (a specific kind of covalent bond) that can grow to as much as millions of nucleotides.
What part in the strand contributes to the overall non neutral charge?
The reason why DNA is negatively charged is the phosphate group that makes up every nucleotide (pentose + nitrogenous base + phosphate). When forming part of the phosphodiester bond, they retain 1 of 2 negative charges (the other being lost to form the other ester bond to a new pentose, that's why the bond is called "phospho-di-ester").

Lehninger 5th edition. Part of figure 25-5
DNA is not isolated in the body, so what keeps it stable while being charged?
The main molecules in charge of that are proteins called histones. They have alkali features and positive charge, and so, they neutralize these negative charges. On the other hand, that's not the main point of histones. Histones play a really important key role in the regulation and compaction of DNA within the nucleus of the cell, and are important targets of diverse molecules that want to alter DNA structure and function:

Lehninger 5th edition.Figure 24-26
Why is it important for DNA to be charged?
It has great importance because cell machinery is adapted to that property. If DNA wasn't negatively charged, the structural fundamentals of interactions between DNA and transcription factors would be completely different.
Thanks to the fact that it is negatively charged, it becomes spontaneously pretty straight when denaturalised, because negative charges repel themselves. It forms very wide curves.
And of course, histones wouldn't be able to join DNA and exert their key regulating functions.

Great answer! Welcome to the site, hope to see you around!

DNA is basically a polymer of nucleotides. These are held together by covalent bonds formed between the phosphate groups, each of which forms and ester with a hydroxyl group of the pentose of the NEXT nucleotide. This uses two of the three acidic 'OH' groups of the acid, leaving the last free to ionize. This ionization leaves a negative charge on each phosphate group.
These charges allow the DNA strands to band to protein molecules which stabilize the structure, giving a nucleoprotein complex.
This complex helps stabilize the DNA—which as important in several ways
– The DNA molecules are VERY long and easily broken
– particularly when being moved etc at cell division

DNA contain negative charge because the presence of phosphate group. Generally, hydrogen is not shown in DNA structure.
DNA
There are 3 hydrogens in phosphoric acid, each of them leaves as:
1 H joins with 3' -OH of previous nucleotide to form $\ce{H2O}$ during DNA replication.
1 OH joins with H of next nucleotide to form $\ce{H2O}$ during DNA replication.
1 H is released as $\ce{H+}$ (remember it was phosphoric acid), hence giving rise to the acidic character of DNA (remember it is deoxyriboNucleic acid).


text:
What safety measures should I take when operating a backyard aluminum furnace
I have made a DIY furnaces out of cement. Before I do any smelting, I would just like to take any suggestions in regards to any safety measures I should take.
I plan to only melt aluminum cans and small parts. I then plan to pour them onto a flat plane to make aluminum plates. Later I plan to make casts.
My furnace is very similar to the one pictured below.

I plan to connect a hair drier to the hole (with an extension metal rod) in order to increase air flow. It will operate by burning charcoal brickets.

Set this up in an area free of leaves and debris.
I recommend wearing safety gear: heat resistant goggles and fire-resistance clothing and gloves that will protect you from heat, especially when pouring.
Get a couple practice runs in pouring before using it so you get a feel for the best angle to hold in and learn what NOT to do. Even better, you could assemble something that would do the pouring for you. (I'm thinking of something along the lines of this, but it doesn't have to be as elaborate.)
You may also want to scrape the paint off the cans before putting them in. Otherwise, you'll  get a lot of crud from the paint in your molten aluminum.


text:
What is a triglyceride?
I'm confused on what a triglyceride is, from what my text book it says its a type of gylercide, then from a website it said

Glycerides can be subdivided into two categories. The first group, the neutral glycerides are nonionic and nonpolar. The second group, the phosphoglycerides contain a polar region, the phosphoryl group.

Which makes triglyceride not a directly under a glyceride but under one of these groups. Then it says

Esterification may occur at one, two or all three positions producing monoglycerides (monoacylglycerols), diglycerides (diacylglycerols), or triglycerides (triacylglycerols). The most prevalent and most important are the triglycerides.

So does that mean that triglyceride can be produced by either of the two groups phosphoglycerides and neutral glycerides. And if that's correct then how are monoglycerides and diglycerides produced.
Source

There is some minor argument as far as nomenclature goes for these lipids. Most sources I can find, including IUPAC have glycerides as only including esters of glycerol with fatty acids. Monoglycerides have a single fatty acid ester with two free OH groups, diglycerides have two fatty acid esters, and triglycerides have three. Phosphoglycerides are defined separately as esterified derivatives of phosphatidic acids, so you can think of them like glycerides where for one of the substituents, instead of a fatty acid, some derivative of phosphoric acid is present. However, the way the two terms are defined, they don't encompass any common molecules, i.e. there are no molecules that are both glycerides and phosphoglicerides and neither group is a subset of the other.
So: mono-, di-, and triglycerides are subsets of glycerides while phosphoglycerides are a different group altogether, though structurally similar.

Oh I see that makes sense so phophoglycerides are not a type of glycerin but is completely different from each other and triglycerides are a type of glyercides. That makes so much sense now thank you so much.


text:
Calculating the ionization energy for hydrogen

What energy would be needed to remove the electron from the $n = 4$ level of the hydrogen atom?

$\pu{−3.49 * 10^{−17} J}$
$\pu{−1.36 * 10^{−19} J}$
$\pu{+2.18 * 10^{−18} J}$
$\pu{+1.36 * 10^{−19} J}$

I assume that the way to do this is to start from the Rydberg formula,
$$
E = \mathcal{R}Z^2 \left( \frac{1}{n_i^2}-\frac{1}{n_f^2} \right),
$$
set the initial level of the electron as $n_i = 4$, and the final level corresponding to removing the electron (ionization) as $n_f = \infty \implies \frac{1}{\infty^2} = 0$, leading to
$$
E = \pu{13.6*\frac{1}{4^2} eV = 0.85 eV = 1.36*10^{-19} J}.
$$

I assume you refer to Bohr's atomic model.
The energy of an electron in Bohr’s orbit of Hydrogen atom is given by the expression:
$$
\begin{align}
E_{n} &= \frac{2\pi^{2}me^{4}Z^{2}}{n^{2}h^{2}\left(4\pi\epsilon_{0}\right)^{2}} \\
&= \color{\navy}{-13.6\frac{Z^2}{n^2}\pu{ eV}}
\end{align}
$$
Since $Z = 1$ for hydrogen, the above equation can be further simplified. Now just plug in the required values.


text:
Carbon and dilute sulfuric acid
I have to distinguish between copper oxide and carbon. I thought of dilute sulfuric acid because it forms copper sulfate with copper oxide. But I don't know how carbon reacts with dilute sulfuric acid. I know how carbon reacts with concentrated sulfuric acid but dilute?

see this:-http://chemistry.stackexchange.com/questions/42880/what-is-the-product-of-the-reaction-of-carbon-with-sulfuric-acid/50763#50763

Sulfuric acid is  unreactive with carbon- indeed only hot concentrated sulfuric acid reacts with carbon to produce carbon dioxide.

Then how can i distinguish between copper oxide and carbon?


text:
How to produce black smoke using chemical reaction
Tell me the requirements and procedure for producing black smoke using chemical  reactions.

is it a theoretical question or practical one? If second, what is the situation we are talking about?

When ignited, mixtures of potassium chlorate, charcoal and anthracene (or naphthalene) produce black smoke.
However, note that milling and grinding dry mixtures of potassium chlorate and charcoal is a well-known method to cause an explosion and severely harm or kill yourself!


text:
How can I Create an explosive with everyday used chemicals?
I would like to know How can i create a simple (Maybe Complex) Chemical that can be used as an explosive.
I would like to use it to blast A regular cardboard box.

Ammonium nitrate maybe?

Trying to synthesize explosives without the proper training and equipment is a really, really bad idea. You probably won't blow up a cardboard box, but your own hand or worse. I don't think questions about explosives from household chemicals belong here.

@MadScientist Actually I'm glad that these questions are asked (and answered) here, rather than on some obscure amateur pyromaniac sites.

If you don't mind a personal advice: Don't! First aid on explosion victims isn't fun for anyone involved.

Use potassium nitrate.

Note that, in addition to being extremely dangerous, in most jurisdictions it is a serious crime to prepare or use explosives without a permit. In some jurisdictions, it is even a crime to give instruction on preparation of explosives to an unauthorised person.

It depends!
Can you regrow lost organs?
If you're genetically close to a newt and can easily regrow organs lost in an explosion, you should give triacetone triperoxide (TATP) a try. Cool an equimolar mixture of acetone and 50% hydrogen peroxide to 0 °C, add some drops of concentrated sulfuric acid and let it come to room temperature without stirring. After 24h, collect the precipitate and dry it carefully.
TATP is extremely shock sensitive - one careless scratching and you're done - and has a Trauzl value similar to that of TNT!
If you feel by now that you better shouldn't play with it: You are right!
What about a something less suicidal?
On contact with water, calcium carbide forms calcium hydroxide and releases ethyne:
$$\ce{CaC2 + 2 H2O -> Ca(OH)2 + C2H2 ^ }$$
This might be enough to burst a sealed container. Alternatively, you might consider to ignite a mixture of ethyne and air. In any case, take care!

Acetone peroxide is probably the most shock sensitive and dangerous-to-prepare of all the explosives that can be made according to the the constraints outlined so eloquently above. I find your suggestion of it, along with the notion of drying it, to be in poor judgment. If your looking to blow up a cardboard box and make a loud noise, why not consider containing dry ice inside a 2-liter, PETE bottle until it ruptures, using water as a catalyst. A louder, deeper boom and you still get to keep your hand if the 'explosion' goes of in it. Explosion can be defined as a rapid expansion of gasses.


text:
How do I synthesize 3-hydroxydihydrofuran-2-one from formaldehyde?
I want to synthesize 3-hydroxydihydrofuran-2-one from formaldehyde but I do not know of a method. I tried a Mannich reaction and lactonization after dehydration.

Starting from pyruvic acid which is readily available and not very expensive: react pyruvic acid (only one enolizable side) with a Mannich base, then react the resulting amide with $\ce{MeI}$ and base to obtain the exo-methylene product. Brominate the exo-methylene. Now cyclization will prefer the five-membered ring over the strained four-membered ring. Reduce the ketone with a reletively mild reducing agent like $\ce{NaBH4}$ and then eliminate $\ce{HBr}$ in a E1cB type reaction, yielding 3-hydroxydihydrofuran-2-one.
This synthesis involves a Mannich reaction as you asked, but it can be done much simpler without it I think.
Note: I'm not an expert, just an enthusiastic amateur. Before trying anything in the lab  make sure your synthesis is checked by a supervisor.

Thank you so much!
Your explanation is very clear.
Fortunately, I don't make anything in laboratory, just I write this synthesis in my homework, but thanks for the prompt.

Jori, can you show me the other method without the Mannich reaction?
Thanks a lot for the help


text:
Can an emulsion be emulsified?
Can clusters of an emulsion be emulsified into another emulsion?
If so is there any common product or natural substance that would have such a structure?
Finally is there a name for this? I looked around but the question gets drowned in all emulsion topics.

Yes, they're called multiple emulsions (and more specifically: double, triple emulsions, etc.). Usually they're made from alternating polarity phases—water inside oil inside water, for example. There are bulk ways of making them though there isn't great control over how many droplets end up in each encompassing droplet without using microfluidic techniques. David Weitz's lab at Harvard has some interesting work in building microfluidic devices to make multiple emulsions.
I don't know if there are any common natural occurrences but the pharmaceutical and cosmetics industries seem to be the the main parties interested in this stuff as they can be used for encapsulating drugs for time release and the like.


text:
What are some popular science books in chemistry?
I have been reading the books The Elements by Theodore Gray and Stuff Matters by Mark Miodownik, both being very popular chemistry science books. Are there some other easy-to-read, popular but scientifically accurate chemistry books available?
The answer to the question should not be "primarily opinion based", since "popular" can be measured.

Napoleon's Buttons by Penny Le Couteur and Jay Burreson.  Salt by Mark Kurlansky

Molecules that Changed the World by  K. C. Nicolaou

Oxygen by Nick Lane.

1) Curiosity: How Science Became Interested in Everything
by Philip Ball
 (https://www.goodreads.com/book/show/13650233-curiosity)
2) Molecules: The Elements and the Architecture of Everything
by Theodore Gray, Nick Mann
(https://www.goodreads.com/book/show/20729580-molecules)

Are you familiar with these books? What do you like about them? What topics do they cover? Why should the asker look into them?


text:
How pure NaCl is typical table salt? How is it purified?
How pure $\ce{NaCl}$ is typical table salt? Is it pure enough for research-level experiments in chemistry? How is $\ce{NaCl}$ purified if need to be?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

Table salt is often iodized (iodine is an essential micronutrient, and this is one way to introduce it into the diet in the right amount). According to Wikipedia,

Edible salt can be iodised by spraying it with a potassium iodate or potassium iodide solution. 60 ml of potassium iodate, costing about US $1.15 (in 2006), is required to iodise a ton of salt. Dextrose is added as a stabilizer to prevent potassium iodide from oxidizing and evaporating. Anti-caking agents such as calcium silicate are commonly added to table salt to prevent clumping.

Elsewhere, Wikipedia mentions other anti-caking agents, like sodium ferrocyanide, sodium aluminosilicate, and magnesium carbonate. And they point out that salt may also be "fortified" with iron or other micronutrients.
The certificate of analysis for "ultra pure NaCl" from American Bio shows TRACE amounts of aluminum, arsenic, bromides, heavy metals, iron, magnesium, phosphates, and sulfates. Those impurities may well be expected in table salt, too.
From all that, you'd think that table salt is fairly impure. Let's look at the numbers to put the amounts of these impurities in perspective. The box of Morton iodized salt in my kitchen gives me the amount of sodium per serving. If all of the sodium is in the sodium chloride, I estimate the purity as
$$\frac{0.590~\mathrm{g}~\ce{Na+}}{1.5~\mathrm{g}~\text{table salt}}\frac{1~\mathrm{mol}~  \ce{Na+}}{22.989769280~\mathrm{g}~\ce{Na+}}
\frac{1~\mathrm{mol}~\ce{NaCl}}{1~\mathrm{mol}~\ce{Na+}}\frac{58.44277~\mathrm{g}~\ce{NaCl}}{1~\mathrm{mol}~\ce{NaCl}}\cdot 100\% = \fbox{99.99% NaCl}$$
where I'm assuming the label information is actually a bit more precise than reported (exactly 1.5 g of table salt contains approximately 590 mg Na). This is just a BOE (Back of the Envelope) calculation; I should round my answer to three or possibly two significant figures.
In fact there are sometimes other sodium salts present (anti-caking estimates such as Tivolex contain sodium) so this purity is a slight overestimate. One salt company (WA Salt supply) lists "typical" purities for the table salt it provides at 99.72%; this salt contains uses Tivolex. Morton Salt claims that their iodized table salt contains 0.04% dextrose and "less than 0.5% calcium silicate", so the mg Na per serving on the label is a little bit higher than it actually is; I should get something above 99.46%, considering that additional information.
Still, that's pretty pure. The assay for "ultra pure" NaCl linked above gives 99% to 100.5% NaCl. Sigma-Aldrich has reagent-grade NaCl (>98%), ACS grade NaCl (>99.5%) and TraceSelect NaCl for doing heavy metal analysis that is more than 99.999% pure.
Is 99% purity "pure enough for high level experiments in chemistry?" There is no one answer. It depends on what those experiments are. How many significant digits do you need in your results? Are the results going to be affected by the presence of tiny amounts of iodide, iodate, glucose, calcium, or silicate, or the other impurities listed above? Maybe, maybe not. If a specific impurity interferes, find a way to lock up, remove or otherwise neutralize its effects.

A sense of social responsibility dictates removal of the (!) next to arsenic unless you have verified that it is in a harmful form and present in a harmful concentration (or the addition of a ! to all other compounds listed). Similar to the way that brussel sprouts could land you in the hospital is an irresponsible tag line for an article.

@Jason C, you're right. What was I thinking? Exclamation point REMOVED.

@FredSenese no way table salt is 99.99% NaCl see http://www.wasalt.com.au/Table.html and http://www.spexcertiprep.com/knowledge-base/files/AppNote_GourmetSalts.pdf

@DavePhD The first reference gives the purity as 99.72% for their product, and they use an anti-caking agent that contributes sodium to the mix. The second reference deals mainly with "Gourmet Salts" which contain many minerals. The table salt they analyze is quite clearly not at all like the Morton Iodized salt I'm discussing above; they don't give %NaCl at all, did I miss that? The top contaminant in table salt they list is potassium at 0.0006 - 0.029 mg/g.

I have to ask as it is a bit of a bugbear of mine but how can anything be 100.5% pure - does it contain negative amounts of some impurities?  It reminds me of a certain avert "Not less than 100% beef, (also contains..)"

@FredSenese here is an analysis of a Morton's food grade salt http://www.mortonsalt.com/content/pdfs/PDS%202200%20Wht%20Pretz%20Crse%201013%20Rev%203.pdf and Morton's FAQs says "Morton® Iodized Table Salt contains 0.04 percent dextrose or 40 milligrams per 100 grams of salt" and "Calcium silicate ... is added at less than one half-percent." http://www.mortonsalt.com/faqs/food-salt-faqs

also, the potassium range in the report is 0.6 to 29 mg/g, not 0.0006 - 0.029 mg/g.

@DavePhD, you're right about the potassium---sorry about that.

@DavePhD, The first link is for pretzel salt, which is a bit different from table salt; it has larger grains and a different texture, so the anti-caking requirements are different. Thanks for the second link, though, clearly the label information can only be trusted to three sigfigs at most. I'll add that to my answer!

Salt purity is measured on "dry salt" - which is to say, water does not count towards the purity. In the UK, 99% pure NaCl is used for ice-treating areas that might be subject to staining by less pure gritting methods.
99.9% purity is used for food:
http://www.british-salt.co.uk/Products/PureDriedVacuum.aspx
This is the same purity as pharmaceutical-grade salt, except that pharm. grade also requires a moisture content of 0.01% or less.
For export markets, British Salt can then add the required impurities for that market's requirements.
The British Salt company provides about 2/3 of domestic cooking salt in the UK.

Some table salt sold for consumption contain additives which address a variety of health concerns, especially in the developing world. The identities and amounts of additives vary widely from country to country. Wikipedia

Table salt is hardly even close to pure.

In summary, a variety of additives is used in common table salt. In addition to potassium iodide, the additives can be categorized as either iodide stabilizers or anticaking agents. Iodide stabilizers act to prevent the oxidation of iodide to iodine, while anticaking agents keep salt free-flowing. Abstract of this article

Thus, table salt can result in an approximation in the experiment you're conducting. You can look for kosher salt though.

I did not know there's a thing such as Kosher salt. Thanks for the tip!

See Analysis of Gourmet Salts for the Presence of Heavy Metals which reports analysis of table salt (noniodized), 12 gourmet salts, and reagent grade NaCl using inductively coupled plasma mass spectrometry.
Considering metals, other than sodium what is present, by far the greatest other metal ion present in table salt was potassium at 1000 ppm (1000 micrograms per gram of sample).  Next was calcium at 218 ppm.  Considering that pure NaCl is 393000 ppm Na, the presence of K and Ca alone brings the purity of table salt down to 99.7%
There is no indication that the above study tested for carbon, oxygen, silicon or hydrogen, so levels of water, sugar such as dextrose, silicates or carbonates should be considered unknown.
The Pennsylvania State University agricultural extension service advises against using table salt for canning because up to 0.04% dextrose and up to 0.5% calcium silicate are added.
CODEX Alimentarius is an international set of food standards endorsed by the World Health Organization and Food and Agriculture Organization of the Unitied Nations.
The CODEX STANDARD FOR FOOD GRADE SALT is linked and requires

The content of NaCl shall not be less than 97% on a dry matter basis, exclusive of additives.
The remainder comprises natural secondary products, which are present in varying amounts depending on the origin and the method of production of the salt, and which are composed mainly of calcium, potassium, magnesium and sodium sulphates, carbonates, bromides, and of calcium, potassium, magnesium chlorides as well. Natural contaminants may also be present in amounts varying with the origin and the method of production of the salt. Copper shall not exceed 2 mg/kg
The name "salt" shall have in its close proximity a declaration of either "Food Grade" or "Cooking Salt" or "Table Salt"

So "Table Salt" as internationally recognized can have impurities of up to 3%, plus additives (such as for anti-caking or iodine supplementing), plus moisture.


text:
Can you find the mass of solvent with mass of solute, volume of solution, and solution density?
If you have, for example, $\pu{10 g}$ of a solute dissolved into $\pu{100 ml}$ of a solution, and the solution density is $\pu{1.5 g ml^-1}$, can you just multiply $100\ \mathrm{ml} \times 1.5\ \mathrm{\frac{g}{ml}}$, then subtract the $\pu{10 g}$ of solute from the result to get the mass of the solvent in grams?
Does that work or am I trying to do do this wrong?

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

That works!  In most chemistry (but not nuclear reactions!),  mass is always conserved, so
$$m_\text{solution} = m_\text{solvent} + m_\text{solutes}$$
Mass is also equal to density times volume, so
$$m_\text{solution} = \rho_\text{solution} V_\text{solution} = m_\text{solvent} + m_\text{solutes}$$
Solving for $m_\text{solvent}$:
$$m_\text{solvent} = \rho_\text{solution} V_\text{solution} - m_\text{solutes}$$
Which is exactly the calculation you did.


text:
Are there economical ways of synthesizing nanotubes?
I've heard Nanotubes will replace Li-ion batteries in the future and it is much more relevant. I just wanted to know could it really replace Li-ion batteries? Is synthesizing nanotubes economic?

I'd say a really important factor is if people want it or not. It was predicted that people will wear plastics in 21st century, while what really happened was, most ended up with comfy cotton clothing.

Sure, But people may need many things also such as quartz.

You might wanna take a look at this.

You would need equipment in order to handle something that is less than a nm. You could easily isolate graphene by taking scotch tape and sticking it onto graphite (tip of pencil) but of course you can't see it much less handle it with traditional tools

Depends what you call economic - nanotubes are expensive, but they are already used commercially.

If carbon nanotubes with additional 'bumps' (nanobuds) are counted, then the answer would be yes. See for example a Finnish company called Canatu. They make touch sensors from the fullerene functionalised carbon nanotubes. If you are interested in the manufactory process, you might like to check this Patent of them, for example. Unfortunately I do not know about the possible uses of CNTs in batteries.


text:
What is a barrier-less reaction in Quantum Chemistry?
After finding a reaction "Transitional State" by GAMESS, and finding Hessian Eigenvalues, my project instructor said:

seems your imaginary frequency (negative eigenvalue) is not that big (it is around -13.01), which means your reaction is barrier-less.

As I am a beginner in Quantum Chemistry, I could not figure out what is the relation between a small imaginary frequency and being barrier-less.
As I remember from the Arrhenius equation, being barrier-less has something to do with higher temperature leading to a slower reaction, but I could not understand its relation to imaginary frequencies.

See for example this related question

Small magnitudes in imaginary vibrational modes could be due to numerical noise if using DFT approaches.  You might want to tighten up your convergence or try another method to verify.  That being said, I don't agree that this would be barrier-less at all.  Of course you could try to take into account zero-point vibrational energy correction to the TS energy (though how does one ZPVE-correct a TS is beyond me) which would raise the energy even more.  Just compare the electronic energies of the TS and corresponding minima to see how big the barrier is.

@GeoffHutchison This Marcus theory was awesome !  Thank you

@LordStryker it was a great suggestion to try another method. Is there any cutoff for saying this is a barrier-less reaction ? for example can we say if the imaginary eigenvalue is larger than a specific number ( or a range of numbers ), it is assumed barrier-less ?

@Aug I have little experience with barrierless pathways. I only know more orthodox mechanisms where you characterize a minimum energy pathway for a reaction.  I've read into this a bit and it looks interesting, however.  You may want to read this (http://www.math.uni-leipzig.de/~quapp/JKQCPaper2.pdf) as it may be useful.  Also, I wouldn't correlate an imaginary mode of vibration with a barrierless reaction.  It sounds very wrong indeed to do that.

Usually, in the reactions with a barrier, the saddle point with one imaginary frequency is taken as TS. However, many unimolecular dissociation reactions are barrierless: the reaction path on potential surface, connecting (minimum of) reactant with (minima of) products, is such that the energy along the path does not have a local maximum. Kinetic treatments of such reactions are more difficult: it is still possible to assign a TS, but it no longer a saddle point. In order to find it, one can use variational TST, but it is somewhat expensive.
A general recommendation after location of a saddle point: run reaction path calculation to see that this saddle point indeed connects the products and reactants that you expect.
About this small eigenvalue: this eigenvalue can easily pop up instead of zero eigenvalues of rotations and translations. Even if your calculation explicitly projects them out, numerical noise can be this order of magnitude (as pointed out above).


text:
Thionyl Chloride and enols, carboxylic acids, alcohols
\begin{align}
\ce{Propanol + SOCl2 &-> Expected~ alkyl~ halide~ product}\\
\ce{Propanoic~ acid + SOCl2 &-> Expected ~acid ~chloride ~product}\\
\ce{Acetone ~enol + SOCl2 &-> doesn't~ work ~in~ the ~expected ~manner}\\
\end{align}
My prof says the $\ce{SOCl2}$ only works on $\ce{sp^3}$ substrates.
But ... both propanoic acid and acetone enol involve an $\ce{sp^2}$ carbon bound to an alcohol group ... why does $\ce{SOCl2}$ work in one case but not in the other?
I think it instead has to do with the nucleophilicity of the electrons involved with oxygen ... the mechanism starts with electrons associated with oxygen attacking thionyl chloride. A regular alcohol has an oxygen with lone pairs that aren't fairly "free," so to speak. An carboxylic acid has resonance donation enhancing the nucleophilicity of one oxygen's lone pairs. In acetone, however, there is no resonance donation ....... (inductive donation, sure).
So does the $\ce{sp^2}$ vs. $\ce{sp^3}$ explanation hold water?

http://www.sciencemadness.org/talk/files.php?pid=106928&aid=3936

Obviously it's not true that thionyl chloride only works with sp3 substrates. As you say, a carboxylic acid, an sp2 hybridized substrate, can be chlorinated with thionyl chloride.
Ultimately, the difference is that the carbon of an enol is not readily attacked by a nucleophile, while the carbons in the other substrates can be. Each of the reactions go through the same first step(s): addition of the hydroxyl to sulfur with chloride leaving sulfur (likely through addition/elimination, not shown below). However the next steps are different depending on the functional group of the substrate. For an alcohol, chloride attacks the activated alcohol (looks very much like a sulfonate) in an SN2 reaction to give the alkyl chloride. For a carboxylic acid, chloride attacks the carbonyl carbon, and after addition/elimination (not shown in detail), the acid chloride is produced. Starting from an enol, reaction with thionyl chloride gives an enol chloro sulfate. Unlike the sp2-carboxylic acid case, the pi bond doesn't have an effective electron sink (such as oxygen), which means that the reaction would have to be a direct SN2 displacement. As I'm sure you learned, SN2 at sp2 carbons is very slow.

A SciFinder search shows that other reagents, notably phosphorus pentachloride can accomplish the transformation.

thank you for the thorough answer as always. "As I'm sure you learned, SN2 at sp2 carbons is very slow." Well, what I've been taught and what I know are two disparate entities ;).


text:
Why does (atmospheric) pressure have an effect on solubility?
The Wikipedia article titled Solubility table states that

The table below provides information on the variation of solubility of different substances (mostly inorganic compounds) in water with temperature, at 1 atmosphere pressure.

Why does pressure have an effect of on solubility?

It comes from a combination of Henry's law and Le Châtelier's principle. Henry's law is
$$p=k_hc$$
where $p$ is the partial pressure of the gas, $k_h$ is the Henry's law constant and $c$ is the concentration of the gas (because liquids and solids are not affected as much as gases, if at all). This means that with a greater pressure comes a greater concentration, and with a lower pressure comes a lesser concentration, and vice versa
Le Châtelier's principle states that if the pressure, concentration, temperature or volume of a system changes, something else will change to compensate and swing the whole thing back into equilibrium. If there is a greater pressure, there will therefore be a grater concentration. A greater concentration means there is a greater solubility.
The UC Davis Chem Wiki (I didn't see that DrMoishe Pippik had linked to it until after I posted this!) is rather helpful - much more helpful than Wikipedia, in this case.

Pressure has a strong effect on the solubility of gases, of course, "forcing" them into solution; see http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Solubilty/Solubility_and_Factors_Affecting_Solubility.
But even solution of solids in liquids are affected somewhat by pressure: for example, proteins may be denatured and made less soluble by high pressure, as in ocean deeps. See http://www.desy.de/news/@@news-view?id=4561&lang=eng for information on that.


text:
Are Na and Cl ions from NaCl in water separate charged particles?
Suppose I pour some sodium chlorine into water. So, what happens is that
$$
\ce{NaCl (s) -> Na+ (aq) + Cl- (aq)}
$$
meaning that the ionic bond between Na and Cl breaks up (correct this far?)
Now, does this mean that the water actually contains separate charged Na and Cl particles? So... since chlorine boils at −34.04 °C according to Wikipedia, why is there then no chlorine gas evaporating? Because it is chlorine ions there, not chlorine atoms?
If I feed electrons some how into the solution, will chlorine gas start forming?
Also, could I use this so that I pour NaCl into water and get Na and Cl ions, and then (since they are separate) add something more to create Na[something] or Cl[something]? Something that would not form if I just mixed NaCl + [something else]. I hope I am not completely lost here..

I think you got it mostly right, but as there are anions of chlorine not cations one should take electrons away (via electrolysis for example) to obtain chlorine gas

The effects of diffusion and electrostatic attraction of ions with water overcome ionic bonding between sodium and chlorine atoms -- so yes, the charged sodium and chlorine ions separate. This is a computer simulation which includes a model for van der Waals and electrostatic forces in real time (a tiny $\ce{NaCl}$ cube dissolving in 4 ps): https://www.youtube.com/watch?v=IPQzY34eCt4

Most of what you wrote is correct. It is difficult to feed many electrons into the salt water because the Coulomb forces are extremely strong, see 1 and 2, but you can feed electrons in if you drain them out elsewhere so there is no net gain or loss of charge. that is called an electric current.
You can use that to electrolyze salt water, and get chlorine as you surmise; NASA has a nice lab. BTW, you would expect to get Cl gas at the positive electrode, and do get it, but you do not get Na metal at the negative, because hydrogen bubbles out before Na. You can electrolyze molten salt, though, to get a bit of Na (or at least you see it form and then catch fire with a yellow flame).

Suppose I pour some sodium chlorine into water. So, what happens is that
NaCl(s) → Na+(aq) + Cl-(aq)
meaning that the ionic bond between Na and Cl breaks up (correct this far?)

That would be completely correct in an infinitely dilute solution of NaCl.
However, in real solutions of NaCl there are ion pairs in addition to the separately solvated ions.
There is an equilibrium between the aqueous ion pairs and the aqueous seperately solvated ions:
$$ \ce{NaCl(aq) <=> Na+(aq) + Cl-(aq)}$$
For a detailed consideration see Physical Electrochemistry of Strong Electrolytes Based on Partial Dissociation and Hydration: Quantitative Interpretation of the Thermodynamic Properties of NaCl(aq) from "Zero to Saturation" Journal of Electrochemical Society, Vol. 143, No. 6, June 1996, pp. 1789 - 1793. (no paywall link).
In summary, in a relatively concentrated solution of NaCl, there will be about 20-30% ion pairs with the remainder being seperately solvated ions.

If I feed electrons some how into the solution, will chlorine gas start forming?

You need to remove an electron from each of two Cl- to get chlorine gas, but yes you can form chlorine gas through electrolysis of NaCl solution.

Also, could I use this so that I pour NaCl into water and get Na and Cl ions, and then (since they are separate) add something more to create Na[something] or Cl[something]? Something that would not form if I just mixed NaCl + .

You could added silver nitrate solution to an NaCl solution and get AgCl precipitate.
For a video of the reaction see this University of Illinois site: http://www.chem.uiuc.edu/clcwebsite/AgCI.html


text:
Calibration Curve Error Propagation
I am using ED-XRF for geochemical analysis for a geology dissertation. The ED-XRF provides a ppm reading for elements and a 2sigma error, for example 472 +- 5.11ppm Sr.
An example calibration curve made for Sr comprising 37 standards has a linear regression of y=1.26x + 32.1 (R^2 0.9991), which I can easily apply to correct my unknown analyses. I need to plot my corrected readings with their error bars, so how is the 2sigma error associated with this unknown calibrated?
Does this error value also simply get calibrated with the curve equation, or just the slope value, or is this error propagation far more complex? I have searched for a solution and keep finding an equation for uncertainty propagation with calibration curves, but this only calculates the uncertainty caused by the curve, and has no relation to the uncertainty of each analysis? So is the total error the original 2sigma error provided by the machine + calibration curve uncertainty?
Thanks

I will try to offer a detailed way to propagate errors here, but before beginning it's worth asking what your audience is likely to expect "error bars" to mean.  In your field are error bars usually determined simply by replicate analyses of the same sample?  Or replicate samples of the same experiment or field site?  If so, it's probably best just to stick with what your audience is likely to expect.  That said...

The theoretical best way to fit your calibration curve would be with weighted linear least-squares algorithms.  In that algorithm, in addition to the {x, y} data, which in your case is $\{\ce{Sr_{known}, Sr_{meas}}\}$, you also supply weights corresponding to the uncertainty in the y values.  The weight for each y data point would be $\frac{1}{\sigma^2}$, which you could calculate from the $2\sigma$ uncertainty given by your instrumentation for that data point.  In practice, if you did this, I doubt that the parameters of your calibration would change very much at all from what you did previously.  But formally the weighted variant of least-squares is required if the uncertainty in y depends on y.  See wikipedia for more.
After doing the (weighted) least squares calibration regression, you can find the uncertainty in the estimation using the calibration equation.  $\sigma_{est} = \sqrt{\frac{\sum{\left(y_{pred}-y_{meas} \right)^2}}{N}}$.  In that equation $y_{pred}$ is the calibration-predicted Sr concentration and $y_{meas}$ is the measured Sr concentration, and for you $N$, the number of data points, is 37.  This uncertainty is one part of the uncertainty you'll eventually report, and reflects the uncertainty in your calibration curve.  Based on the R-squared you reported, my guess is that this uncertainty will be low.
After all that (!), you need to estimate the other major source of uncertainty, the uncertainty in the measurement of your unknowns ($\sigma_{meas}$).  There are several OK methods for doing this, the best of which would be replicate analysis of several samples.  It might be OK to use the $1\sigma$ value calculated from the $2\sigma$ uncertainty reported by your instrument too.
Finally, combine the uncertainties with the standard equation for combining multiple sources of uncertainty $\sigma_{total} = \sqrt{(\sigma_{meas})^2 + (\sigma_{est})^2}$.


text:
Rate of Reaction Graph - Gas Produced
I did a reaction and I measured the volume of gas produced against time. I need to find the rate constant of the reaction.
I am trying to plot a graph of ln(vol of gas produced) against time so that the gradient will be equal to -k so i can calculate the rate constant k. I am using the equation ln[A]t = -kt + ln[A]o
The problem is I can't plot ln0. I timed the reaction every 5 mins and gas wasn't produced until 15mins had passed. This means that the volume of gas produced for time values 0, 5 and 10 mins is 0cm^3, but ln0 doesn't exist.
How do I use my data to create a graph to calculate the rate constant?
My results were:
Time 0___5___10___15___20___25___30___35___40___45___50___55___60 ____(in mins)
Vol _0___0___0_____5___10___18____22___28___36___44___51____60____70___(in ml produced)

What was the reaction? And what does the rest of the data look like? You can do a linear fit to get the y-intercept, or if the reaction truly didn't start until 15 minutes (weren't the reactants mixed?) then you might make 15 minutes time zero.

Decomposition of what? $\rm H_2O_2$, maybe?

The decomposition of sodium hypochlorite

One last question, was it just sodium hypochlorite and water, or did you add anything else? I mean, you were producing oxygen gas, right?

Yep I think so. The main problem I'm having is how to put the vol of gas produced on the graph in ln form

It is unclear what "A" refers to in your question.  If you assume rate constants must be positive, then your equation only allows for decreasing amounts of A in time.  But the gas formation you observed was increasing with time.  So A must be something other than gas formed.  But what is it?

I think the first bit of gas produced probably dissolved in the water, so you didn't see it for a little while.
You can skip your three initial "points" with zero volume, and you're not going to have $\ln(0)$, which is what's hanging you up. Note that you did this in your report, and why!
Do a linear fit with the remaining points. Your y-intercept will be $\ln \mathrm{[A]}_0$ and your slope will be $-k$. It's not a terribly good fit.

Fred Senese's post is an excellent starting point.  I wanted to add that transforming data in this way is conceptually very useful but distorts parameter estimates because of unequal weighting of data point uncertainty.
The uncertainty of your "zero" data points is probably not negligible.  How exactly could you read your gas measurement apparatus?  To within 0.1 mL, or 1 mL, or ...?  Say it was 0.1 mL, then your "zero" data point in mL is $0 \pm 0.1 \text{ mL}$.  And your 70 mL data point is $70 \pm 0.1 \text{ mL}$.  When you transform the data by taking the logarithm, however, the uncertainty in your data is unevenly affected.  The uncertainty of $\ln \left( 0 \pm 0.1 \text{ mL} \right)$ is approximately $\ln{0.1} - \ln{0}$ which as you are discovering is infinite.  The uncertainty in log-space of your 70 mL data point is $\ln{70.1} - \ln{70}$, which is 0.0014.  So even though the uncertainty on each data point was the "same" before logarithms, say 0.1 mL, after transformation the uncertainty is as unequal as can be (infinity vs. 0.0014)!
Direct non-linear fitting can fit parameters to data without needing transformations.  Here is some R code to do the fitting on your data.  Because of the issue related to my comment, I had to include an offset parameter.  (An exponential equation tends to a positive value as t approaches zero, not to zero...unless there's an offset.)  This method is probably overkill for your particular problem or course, but I wanted to include it in case it's useful to other chem.se readers.
#read the data from the question into R and put it in a data frame
g <- as.numeric(str_split('_0___0___0_____5___10___18____22___28___36___44___51____60____70___', '_{1,5}')[[1]]); g <- g[!is.na(g)]
t <- as.numeric(str_split('0___5___10___15___20___25___30___35___40___45___50___55___60 ____', '___')[[1]]); t <- t[!is.na(t)]

dat <- data.frame(g = g, t = t)

# scale the data to avoid numeric problems taking the exponential of large numbers
dat$g.scaled <- dat$g/max(dat$g)
dat$t.scaled <- dat$t/max(dat$t)

#compare to fitting result by neglecting first three zero points and taking logarithm
dat$ln.g[4:13] <- log(dat$g)
lm(formula = ln.g ~ t, data = dat)

# show data
dat
        g  t   g.scaled   t.scaled
1   0  0 0.00000000 0.00000000
2   0  5 0.00000000 0.08333333
3   0 10 0.00000000 0.16666667
4   5 15 0.07142857 0.25000000
5  10 20 0.14285714 0.33333333
6  18 25 0.25714286 0.41666667
7  22 30 0.31428571 0.50000000
8  28 35 0.40000000 0.58333333
9  36 40 0.51428571 0.66666667
10 44 45 0.62857143 0.75000000
11 51 50 0.72857143 0.83333333
12 60 55 0.85714286 0.91666667
13 70 60 1.00000000 1.00000000

#do direct nonlinear fitting
fit <- nls(g.scaled ~ A0*exp(k*t.scaled)+offset,
           data = dat, start = list(A0 = 1, k = 1, offset = 1))

#do logspace fitting to linear equation by transformation and neglecting first three data points
dat$ln.g[4:13] <- log(dat$g[4:13])
logfit <- lm(ln.g ~ t, data = dat)

#compare predicted values from fit to real data
dat$pred.scaled <- predict(fit)
dat$pred <- dat$pred.scaled * max(g)
dat$logpred[4:13] <- exp(predict(logfit))
require(ggplot2)
ggplot(data = dat, aes(x = t)) + geom_point(aes(y = g)) + geom_line(aes(y=pred), color = 'blue') + geom_line(aes(y=logpred), color = 'red')

The blue curve is the direct nonlinear fit and the red curve is the linear fit of log-transformed data (dropping the first three data points).
The obtained rate constant is quite different in the two cases!  For the direct nonlinear fit, the rate constant is $1.4 \text{ hr}^{-1}$, and for the log-transformed data with linear fit, the rate constant is $0.053 \text{ min}^{-1}$, or $3.2 \text{ hr}^{-1}$!