text:
If I have a chiral substrate, should I always assume that both enantiomers are present?
Considering the Sn1 reaction of (S)-2,3-dimethyl-3-pentanol, knowing that this molecule has a chiral carbon, should I always assume that both enantiomers are present, and therefore the products of a reaction will always have $2^n$ products for $n$ stereocenters on the substrate, assuming that each stereocenter has equal chance of reacting? Is there two products of this Sn1 reaction due to the fact that the molecule is chiral, and therefore its enantiomer is present and has equal chance of reacting?

In this SN1 reaction there would be two products even if only one enantiomer reactions.
Enantiomerically pure (S)-2,3-dimethyl-3-pentanol reacts with HBr to produce both (S)-3-bromo-2,3-dimethylpentane and (R)-3-bromo-2,3-dimethylpentane in equal proportions as a racemic mixture.
ALSO, enantiomerically pure (R)-2,3-dimethyl-3-pentanol reacts with HBr to produce both (S)-3-bromo-2,3-dimethylpentane and (R)-3-bromo-2,3-dimethylpentane in equal proportions as a racemic mixture.
When this kind of phenomenon happens, there must be an achiral intermediate that can react to produce either enantiomer with equal propensity. In this case, what is the structure (geometry and everything) of the carbocation formed when the leaving group leaves?

To add to this fine answer, reactions of this sort will not generate racemic mixtures in general, but rather only when both faces of the prochiral intermediate are equally attractive for attack. If one side of the substrate is harder to access than the other (for instance, due to steric blocking), some degree of enantiomeric excess (E.E.) will result (i.e. one enantiomeric product being produced in greater quantities than the other)

So, basically the stereocenter is rearranged to sp2 hybridization in the carbocation, and both sides of the unhybridized p orbital on the carbon are equally electrophilic.

@Leonardo, yes, that this the explanation generally given.


text:
What are physical properties and chemical properties?
As the question title says, I would like to know what the physical and chemical properties are. If possible, please include examples. Since it seems to be related, I would also like to know what physical and chemical changes are. I believe that things such as crushing a can is a physical change, but I am not sure what would constitute a chemical change (perhaps doing something that rearranges the molecules?)

Edit: I originally wanted to ask if cooking an egg constitutes as a chemical or physical change, but I thought that that would be closed as too vague. If possible, please explain which one it is too. Thanks!

This is a tricky question.
I have to concur with Paul - it's difficult to give a truly hard and fast definition of the distinction between a chemical change and a physical change as there is a fairly smooth continuum between 'physical' and 'chemical' interactions. However, a good rule of thumb is that chemical changes involve changes in electronic structure, usually in the form of chemical bonding or redox. The problem is that traditionally 'physical' properties may have big implications for chemistry - I'll give some examples.
The Van der Waals and dipole-dipole interactions are mediated respectively by 'instantaneous' and permanent polarisations of electron density with respect to positive charges. Dipole-dipole interactions such as those found in water are quite significant and give rise, for instance, to water's frankly freaky liquid range as well as water's ability to mediate a wide range of chemical processes. These interactions don't involve significant changes in electronic structure however they have chemical implications in, for example, making a choice between liquid water or liquid nitrogen as a solvent.
The atomic mass (and hence the isotope) of an element is typically considered a physical property, however it has implications for reaction kinetics. This can be illustrated (expensively) by giving living organisms nothing but heavy water to drink. They will eventually experience toxicity due to the slowing of key biochemical reactions.
The magnetic properties of an atomic nucleus are typically considered to be physical, however they can be exploited in the observation that e.g. different isotopes of uranium have slightly differing hyperfine splittings and hence different ionisation energies. This allows isotopes to be enriched by selective ionisation.
Regarding the egg:
The dramatic changes that occur when you cook an egg are the result of denaturation of the proteins in the egg. Proteins are long chains of amino acids that take on their characteristic active conformations through a process called 'folding', so called because that's what it looks like (if you've heard of Folding@Home, that simulates the folding of proteins). If you raise the temperature much higher than physiological temperature, the protein degrades through unfolding into a different conformation. This involves the Van der Waals and dipole-dipole interactions holding together the secondary/tertiary/quaternary structure of the protein being broken en mass as well as the formation and/or scission of covalent bonds.

Paul's comment and Richard's answer are correct that that chemical and physical change span a continuum. However, at the introductory level, there is a difference between physical and chemical properties.
Physical changes do not involved the conversion of one substance into another. They can be reversible (ice melting and freezing at 0 °C) or essentially irreversible (free expansion of a gas).
Chemical changes are those that involve the conversion of one substance into another. They can be reversible (acid-base equilibria) or essentially irreversible under normal conditions (combustion).
Physical Properties are those properties of a substance or mixture than seem to match with physical descriptions of objects and with the physical changes, especially changes of state.
Intensive physical properties include such characteristic data as densities, specific heat capacities, boiling points, freezing points, Henry's Law constants, etc.
Extensive physical properties vary with the sample (are not characteristic to the substance): temperature, pressure, volume, mass, size, shape, etc.
Chemical Properties are those properties which, as Richard mentions, seem to have to do with electronic transitions (aka chemical changes). These include all of the known reaction, reactivity, thermochemistry, kinetics, and equillibrium data.
In addition, (and non-intuitively) most of the properties of substances and mixture that interact with our five senses are also chemical properties: hardness, color, taste, smell, etc.
Finally, the following properties (and others like them) are also chemical properties: toxicity, flammability, conductivity, nutritional value, etc.

The problem with your division is that it takes phenomenon governed by identical laws and separates them into two categories, physical and chemical.  The ice-water equilibrium is governed by the same laws that govern the acetic acid ionization equilibrium.

@PaulJ.Gans - True, but the distinction has historically been made that way, and it is in most General Chemistry texts that way.

Physical changes are reversible, whereas chemical changes are irreversible. For example, boiling water is a physical change because you can reverse it by cooling it down, but setting fire to wood or boiling an egg cannot be undone.

I'm not sure that this is right.  Many chemical changes can be undone and a number of physical ones can not be.  In my view there is no distinction between chemical and physical changes, they both obey the same laws.  They clearly differ at the extremes, but for example, the conversion of ice to water can be looked at as a chemical change just as well as a physical one.  So can the solution of salt in water.

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text:
How can I find pKa from a titration curve?
I am fully aware that the pKa is the pH at half the equivalence point, but are there some programs or methods I could use to accurately estimate the pKa without manually drawing lines on the graph?

If you are titrating by hand, I'm afraid that there is no way out. You'll have to draw a graph.

Have you tried any curve fitting solution (Excel, Igor, Origin..)?

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text:
How does 1,3-dichloro-1,3,5-triazinane-2,4,6-trione (dichloroisocyanuric acid) generate hypochlorous acid when dissolved in water?
Can someone draw the reaction that leads to the generation of HOCl when dichloroisocyanuric acid (structure shown below) is dissolved in water?

It would be preferable if you would propose a mechanism and someone can verify it (or even better, find the portion of the reaction that's troublesome for you and ask more about its concepts).

If I could do that I wouldn't ask the question.

Well, it doesn't have to be perfect, but I'm certain you'll get a better grasp on the concepts that way.  Think about the overall reaction and what species are being oxidized and which are being reduced.

Ma'am, I have a phd in biochemistry from Rockefeller university.  I was a straight a undergrad at Cornell in biochemistry. But I've forgotten ALL the basics (as well as the rest of it). So how about just showing me the reaction.

I'll admit, I'm quite in awe of your credentials.  I gave quite a bit of thought to becoming an academic neurologist at one point.  We do try to uphold standards for users to make an attempt at a question and apply them to everyone because it's hard to draw a line at what constitutes a "homework" versus an applied problem.  I'll let it stand and see if someone has an answer for you (I don't actually have an answer for it, I'd have to sit down with it), but I think that others would be more willing to help if you had added a structure with some oxidation states or something.

Even if you could link to or include a structure of the starting compounds (dichloroisocyanuric acid), that would be helpful. A link to a source describing this compound as a precursor to HOCl would also be helpful. From what I think the structure, I would think that it would hydrolyze in water to produce chloride ion and not hypochlorite.

And after a quick Google search, I was wrong about the structure. I'll edit your question to include the structure and see if that gets us somewhere faster.

$\ce{N-Cl}$ bond is polarized toward nitrogen: $\ce{N^--Cl+}$ This means, that any chloramine can produce $\ce{HOCl}$ in water by mean of hydrolysis of $\ce{N-Cl}$ bond.
$$\ce{R2NCl + H2O -> R2NH + HOCl}$$
The mechanism is more subtle matter. Basically, it should be one of $S_N$ mechanisms, but it is hard to choose right one here ($S_N1$ or $S_N2$). The central atom in mechanism here should be $\ce{Cl}$, and $\ce{R2N-}$ is leaving group.

To build on permekra's answer, I have drawn a mechanism. I think that is would be SN2 (due to the unlikeliness of $\ce{Cl+}$).

Water attacks a chlorine atom as a nucleophile, and the chloroisocynurate anion leaves resulting in $\ce{H2OCl+}$. The chloroisocyanurate anion can act as a leaving group due to its resonance stabilization (see below). Then, the chloroisocyanurate anion depropotonates $\ce{H2OCl+}$ to form choroisocyanuric acid and hypochlorous acid.

Finally, since there is another chlorine atom, this substitution reaction occurs again to form cyanuric acid $\ce{(NCOH)3}$ and another equivalent of $\ce{HOCl}$. The net reaction for the hydrolysis is:
$$\ce{ C3HCl2N3O2 + 2H2O -> C3H3N3O3 + 2HOCl}$$
For curiosity - the sodium salt of dichloroisocyanuric acid is the active ingredient in the household cleaning product Comet.


text:
Are different enclosing mark types in chemical names needed to differentiate between substances?
What is the significance of enclosing marks in chemical names in determining the uniqueness of a substance (organic and inorganic)?
Can a simple change in the type of enclosing mark used indicate that we are dealing with a different substance or does it just mean that the names were written slightly differently (or incorrectly)? If I decided that I was always going to use parentheses and never use brackets when naming chemicals would it stop me from being able to differentiate between two substances based on their chemical name or would it just be non-standard?
This article makes me think that switching from a parentheses to a bracket has a different meaning but that a brace could be used in place of brackets just because it makes the name look nicer.
To help illustrate what I mean, the following chemical names differ only by the use of curly braces instead of brackets. Are these two different substances or are they the same substance named slightly differently?

Poly(oxy-1,2-ethanediyl),.alpha.-[4-[(1,8-dihydroxy-3,6-disulfo-2-naphthalenyl)azo]phenyl]-.ome
  Poly(oxy-1,2-ethanediyl),.alpha.-{4-{(1,8-dihydroxy-3,6-disulfo-2-naphthalenyl)azo}phenyl}-.ome

Update: The Naming and Indexing of Chemical Substances for Chemical Abstracts talks a lot about how punctuation is used to distinguish between different types of substances. It talks a lot about brackets and parentheses in these discussions but almost ignores curly braces. The publication backs up my theory that the differentiation between parentheses and brackets can be important but that curly braces are not. If I am correct then the two chemical names that I listed above are two different names for the same substance. If I can get verification on this or even opinions from chemists, that would be helpful as I am far from a chemist.

There is no real difference between the types of bracket - indeed, in online searching in the CA Registry file on STN all brackets are replaced by round () brackets. However, there are conventions with regards to square brackets in display in some names and these should be kept.

This resource does state that brackets are replaced by parentheses as you stated. Thank you for this information.


text:
Why does the hydroxide ion have a negative charge?
I've been studying the roles of hydroxide & hydronium in acids and bases, and it was mentioned that a hydroxide ion (OH-) has a negative charge. Can someone give me a layman's explanation of what causes this charge?
Since hydrogen and oxygen are sharing a covalent bond, is the charge negative because an electron is donated from oxygen, thus giving hydrogen 2 electrons?

In a sense, you are overthinking it. OH- has a negative charge, simply because it has a negative charge (it has one more electron than it has protons). There could also be an OH molecule with a negative 2 charge, but thats incredibly unlikely.
Bonds come into play after you know the number of electrons a molecule has.

A water molecule is charge neutral because there is the same number of positive
charges as there are negative charges.
In this diagram, called a Lewis structure, the dots represent electrons while the lines or dashes represent a covalent bond of two electrons.

When water ionizes one of the hydrogen atoms absconds with itself and leaves
it's electron behind, giving us the hydroxide ion. The extra electron gives hydroxide a net charge of -1.

The brackets indicate that this is an ion, charge is denoted at top right.
To go deeper down the rabbit hole on this one I recommend reading up on the Octet rule and Electronegativity.

Hydrogen exists as $\ce{H2}$, while oxygen exists as $\ce{O2}$.  They are both diatomic elements, meaning that their stable form is a pair.  There are several others...
One Hydrogen ($\ce H$) has a positive charge +1.  This has nothing to do with it having more protons than electrons, as it has one of each.  Think of it as 'willing' to donate one electron, or it has +1 electrons... see, it is on the plus side.
One Oxygen ($\ce O$) has a negative charge  of -2. Think of it as 'wanting' to borrow two electrons, or it has -2 electrons... see, it is on the plus side.
Now, put one $\ce H^+$ with one $\ce {O}$ and the $\ce H$ will donate, the $\ce O$ will borrow and the $\ce H$ will be 0, and the O will be -1.  Now the combination has a -1 charge.
Now the Hydrogen is sharing one electron with the oxygen, filling its outermost shell with 2, but the Oxygen is needing 8 in its outermost shell, and is only at 7, so the compound is at -1, or 'wanting' one more electron to make it stable.
If you add one more $\ce H^+$ the $\ce{O}$ will be satisfied, and you will get $\ce {H2O}$, a very stable compound we know as water.

Thanks for the information, that helps quite a bit! Since Hydrogen's charge has nothing to do with with having more protons than electrons, what does it have to do with? Is it a function of some atomic attribute, or is there table to reference? I'm curious about the states of other atomic ions as well. Thanks again!

The charge on hydrogen ion does have to do with it having more protons (one) than electrons (zero), but it is more common and more convenient to think of hydrogen ion having fewer electrons (zero) than hydrogen atoms (one).

Both O and H are neutral species. Do not confuse charge with oxidation number.

Warning for reader: This is a too simplistic view on the matter to be useful. It doesn't take nitpick to know that thinking of $\ce{OH-}$ as $\ce{O^2-}$ and $\ce{H+}$ is flat out wrong.

@M.A.R.ಠ_ಠ Why is this "oxidation state arithmetic" incorrect?

@H.Khan Because water isn't ionic compound! And in acid/base reactions there's no "free protons" to be found.

I may be wrong, but one way I understood the overall formal charge denoted outside the bracketed OH- molecule, was by giving OH- a backstory.
Without a backstory of how it came to form, I don't think the negative formal charge makes intuituve sense.
If we say OH- formed from the dissociation of H2O into H+ and OH-, then there is some sense.
What I mean is O and H and H came together to form water, amd did so by sharing electrons.
At this point, before they arrived together and during, even the hydrogens had 1 electron each to give to help oxygen.
The oxygen had 2 electrons to help out the hydrogens, one for each.
But when the water molecule split apart, there was a tug of war over their shared electrons.
Oxygen turned out to be stronger and yanked away the only electron one of the hydrogens had.
This hydrogen was sent loose, without an electron to help it make another bond or be useful by sharing its bond making electron with another element.
The other hydrogen stayed attached to the oxygen.
So the oxygen had an extra electron, torn from the hydrogen now detached.
This is written as O-.
But because there is an H still attached we write OH-.
(Note, the H in OH is not carrying the stolen electron but the negative charge symbol is written by it, though we could write HO- to be more clear.
(On another level however, I think this stolen electron would be whizzing around near the H in OH too, hence some negative charge of this electron woukd be exerting itself over the H in the OH from time to time.
So if ai am correct, without this backstory or creation story for OH- or HO-, it is not intuitive ...
... because if I think simply yet logically, if I take a neutral atomic O, and a neutral atomic H and make them bond, I don't think we actually have an overall formal charge.
Why ?
Because O has 6 valence electrons.
H has 1 valence electron.
When they bond, they each give eachother a valence electron.
O ends up with 7.
H ends up with 2.
In this OH or HO bond, neither atom has lost an electron.
No tug of war has been won or lost.
(Some might argue that oxygen might be pulling hydrogen's electron a bit closer to itself for a few physical reasons, and hence 'winning' - but the match has not ended, so no one can be declared a winner or loser, a taker of electrons, hence no charge, (as a prize or show of strength), can yet be denoted.
Oxygen would have seven electrons in its orbital flux aura, one away from having an octet and a complete shell.
In this way I think that in theory we coukd see O(2-).
I think this woukd refer to a single monatomic oxygen atom that has taken two extra electrons, and hence, will be somewhat stabke with a full outer shell.
It seems therefore, that when we see an atom or compound polyatimic molecule with a charge, we should be reminded to think of battle scars and spoils of war - that this atom or polyatom has a history of wars over electrons with other atoms and polyatoms, and its charge is a symbol of its current state of winnings or losings.
Some don't seem to even need to go to war.
Noble gases have full outer shells before they have even gone to battle over electrons.
They are sort of born into a position of abundance and luxury, though it may be lonely nonetheless.
It is also pretty hard for other atoms to even get into a tug of war with a noble gas, let alone win.
Some atoms like Al(3+), meaning 3 electrons lost, not 3 atoms of aluminium as in (Al3)
It is found in soil quite a lot.
I think this refers to the notion that this aluminium has lost 3 electrons.
If so, it has lost 3 valence electrons to something or many things, at some point in the past.
It is quite stable having lost its valence electrons because only had 3, which means its valence electrons are now gone.
Can it be made to react again then ?
Yes, I think so, but the chemistry eludes me.
I think Al (3+) is considered pseudo noble because it has lost its only 3 valence electrons at some prior time and has lost an oribtal, but demoted itself to a stable orbital nonetheless.
The pseudo aspect might be that this stable state is not as strong as a real noble.
One reason might be that Al (3+) now has an abundance of protons compared to electrons.
This means it will have a slight tendency to pull in electrons and create a new shell that is not full, even though it has a full outer shell now.
I think the same is true of certain pseudo noble anions, in an opposite sense.
They may have too much electron charge even if they have a full outer shell and might only weakly bind their newly acquired electron/s.
Some pseudo noble anions might really be pretty stable though.
Same with some cations.
In the soil I believe Al (3+) and Cl(-) are pretty stable pseudo noble ions.
I think the way Al (3+) starts to destabilize is when electrons start getting close in abundance around it and cause electron gains to take place upon the aluminium metal, rather than abundances cations coming and stealing even more electrons from it.
I think it is pretty hard for cations or any atoms to steal electrons from non bonding non valence orbitals like inner ones.
I think x rays might be able to temporarily knock out a inner core electron, but possibly not without causing a lot of energy to be released as well.
I could be completely wrong in my understanding of the basics though.
I hope someone will correct me if I am. I would like to know.

You're talking about the hydroxyl radical, where the oxygen indeed has 7 valence electrons (it's own 6 + one shared with hydrogen. You can see this clearly if you look up Lewis structure of hydroxyl radical and compare it with the Lewis structure of the hydroxide ion. What determines charge of the molecule, is the number of protons and all electrons. The difference between the neutral hydroxyl and the negative hydroxide anion is one electron. Look up how the hydroxide anion is made from a molecule of water (autoprotolysis of water): H2O => H+ + OH- The hydrogen there gives its electron to OH-


text:
Correct nomenclature for reaction types
What is the correct name for a reaction like this?
$$ \ce{ 4MnO2 ->[500\ ^\circ \text{C}] 2Mn2O3 + O2 ^}$$
Is it a synthesis, or a Decomposition?
I tried to find it out with http://en.wikipedia.org/wiki/Chemical_reaction but i did not get the link between the formula and the nomenclature as it seems to be both.

This is a decomposition reaction.  Synthesis requires a minimum of two reactants.  Decomposition is where a reactant breaks down into two or more products.  Here, one reactant (MnO2) goes to two products.  It seems like it may not be decomposition because MnO2 is becoming MnO3, but it is still one reactant to two products.

This reaction is definitely decomposition. Decomposition reactions involve taking one reactant and destabilizing its bonds in order to produce 2 or more products. In this case, the 500 degree Celsius condition (indicated above the reaction arrow) indicates that heat energy is being added to decompose the molecule of MnO2. This is more commonly called "thermal decomposition" (Check out this link: http://en.wikipedia.org/wiki/Thermal_decomposition).


text:
Is it possible to synthesise ethers from carboxylic acids?
Is it possible to synthesise ethers from carboxylic acids, and if so, how? By ether I mean an ether in general ($\ce{R-O-R'}$), not just specific examples.
I theorised one could perform Fischer esterification, $\ce{ROH +R'COOH -> RCOOR'}$, and then perform Wolff-Kishner reduction on the carbonyl group. The problem with this theory is that it assumes that the single-bonded oxygen between the $\ce{R}$ and $\ce{R'}$ doesn't influence the reduction process.

Wolff Kishner wouldn't work on an ester..

This is an interesting question. I can tell you that Wolff-Kishner won't work - both the hydrazine and the hydroxide required will attack the ester in a substitution fashion (with a loss of -OR'). Other reductions might work.
For example a Google search for "reduce ester to ether" brought up the following:

The reaction appears to be complete within 3 hours with moderate to good (but not stellar) yields. The abstract and key figures are available for free at www.organic-chemistry.org. The paper is published in the Journal of Organic Chemistry.
It appears that $\ce{Et3SiH}$ is the terminal reductant, but most of the work is being done by indium. The combination of $\ce{Et3SiH}$ and $\ce{InBr3}$ produces the active reductant $\ce{HInBr3}$, which provides the equivalent of $\ce{H.}$. The second $\ce{H.}$ comes from $\ce{Et3SiH}$. The proposed mechanism from the paper is as follows:

Other mild sequential one-electron reductions with a source of $\ce{H.}$ would also likely work. The two-electron reductions, like Wolff-Kishner ($\ce{NH2NH2 +NaOH}$) and hydride-transfer (e.g. $\ce{LiAlH4}$), are all heavy-handed and nucleophilic with the tendency to take apart or over-reduce the ester.

Could you try to simplify this answer? I'm a grade 12 graduate that's done an organic chemistry unit and I was hoping for something I could understand. Like, for instance, your first equation I couldn't understand for the life of me.

Oh and will these reactions be valid for the formation of 2-arachidonyl glyceryl ether (has the carbonyl group removed from 2-arachidonoylglycerol) from 2-arachidonoylglycerol. (An ester formed from the Fischer esterification of arachidonic acid with glycerol)

@BrentonHorne - The information in both of your comments needs to be in your question. Your question contained enough knowledge and proper use of vocabulary that I suspected you to have at least completed a two semester sequence in organic chemistry at the university level. The notation in my answer would be easy to understand for someone at that level.

@BrentonHorne - and secondly,  A glance at the paper suggests that it might not be. No unsaturated esters were used, and esters of secondary alcohols were over-reduced. There are no reactions in the sophomore-level Organic Chemistry book that will do what you want. Most chemists learn about these kind of reactions in graduate school.

Oh thanks, I'll take that as a compliment (that you thought I sounded knowledgeable enough to be a graduate student -- I am studying math at that level) and thanks for answering my question -- I have to learn more before I can comprehend what you're saying the solution of my problem is.

If you're willing to start with an ester, the following might serve as a side note to the excellent answer given by Ben Norris:
Matthias Beller and coworkers reported on the conversion of esters to ethers using silanes $\ce{R3SiH}$ in the presence of $\ce{Fe3(CO)12}$ as catalyst.
Reductions of esters to ethers using $\ce{LiAlH4}$ or $\ce{NaBH4}$ can be carried out in the presence of $\ce{BF3\cdot Et2O}$.
If the ester can be cyclic, i.e. a lactone, trichlorsilane in the presence of a radical initiator or under uv light will yield cyclic ethers.
If you're willing to add a carbon atom to your ester or lactone and convert it to an enol ether, the Tebbe olefination is your friend.

If you want symmetrical ethers you could do the following
Reduce carboxylic acid to alcohol using Lithium aluminium hydride and then perform bimolecular dehydration on it using concentrated sulfuric acid.
$$\begin{matrix}
\ce{R-COOH& ->[\large\ce{LiAlH4}]& R-OH\\
Carboxylic\  acid& & Alcohol\\
2\ R-OH& ->[\large\ce{conc. H2SO4}]& R-O-R\\
Alcohol& & Symmetrical\\ & & Ether}
\end{matrix}$$

For asymmetrical ethers yo could d o the following
Reduce carboxylic acid of both the constituent alkyl groups of ether using above method.
React one of the alcohols with Sodium metal to form Sodium alkoxide, and then make the alkoxide react with the other alcohol to form ether.
$$\begin{matrix}
\ce{R-COOH& ->[\large\ce{LiAlH4}]& R-OH& (reduction)\\
R'-COOH& ->[\large\ce{LiAlH4}]& R'-OH& (reduction)\\
\hline
\\
R-OH& ->[\ce{Na}]& R-O- Na+& (generation\ of\ nucleophile)\\
Alcohol& & Alkoxide&\\
\hline
\\
R'-OH\ +\ R-O- Na+& ->& R-O-R'& (attack\ of\ nucleophile)\\
Alcohol\qquad Alkoxide& & Asymmetric\ Ether&
}
\end{matrix}$$
NOTE: This is only from a theoretical point of view

I do ,not think this would work: nucleophilic substitution on a non-activated alcohol group is not very efficient (as OH- is not a good leaving group). Ether synthesis in basic conditions usually requires an alkoxide and halide alkyl (Williamson ether synthesis).

@PLD didn't think about that, but is the symmetric ether synthesis correct

@ Aditya Sriram : yes, acidic dehydrating conditions are standard for ether synthesis, and proceeds through a nucleophilic reaction on the protonated alcohol, which leaves H2O as a good leaving group.

@PLD what if I converted one reduced alcohol to alkyl halide using thionyl chloride(which gives a good yield) and then reacted the alky halide and alcohol as you said - Williamson's synthesis.

R'OH + RO- <--> R'O- + ROH by proton transfer, so your last scheme is only going to give a statistical yield of asymmetric ether: Assuming no other selectivity, 25%R'OR', 25% ROR and 50%R'OR. (If the reaction runs at all, given that OH- is a poor leaving group.) Converting R'OH to R'Cl or R'Br per your comment to PLD is a far better idea.

Take a carboxylic acid and react with an alcohol. Distill off the ester to get the formula, RCOOH + R'OH <=> RCOOR' + H2O. Decarbonylate it with a decarbonylation catalyst like Palladium or its compound. Palladium(II) Acetate is also.
Sources:
Dalle-Donne, Isabella; Aldini, Giancarlo; Carini, Marina; Colombo, Roberto; Rossi, Ranieri; Milzani, Aldo (2006). "Protein carbonylation, cellular dysfunction, and disease progression". Journal of Cellular and Molecular Medicine 10 (2): 389–406. doi:10.1111/j.1582-4934.2006.tb00407.x. PMID 16796807. Grimsrud, P. A.;
Xie, H.; Griffin, T. J.; Bernlohr, D. A. (2008).
"Oxidative Stress and Covalent Modification of Protein with Bioactive Aldehydes". Journal of Biological Chemistry 283 (32): 21837–41. doi:10.1074/jbc.R700019200. PMC 2494933. PMID 18445586.
^ Jump up to: a b c Bertleff, Werner (2000).
"Ullmann's Encyclopedia of Industrial Chemistry". doi:10.1002/14356007.a05_217. ISBN 3527306730. |chapter= ignored (help)
Jump up ^ Arpe, .J.: Industrielle organische Chemie: Bedeutende vor- und Zwischenprodukte, 2007, Wiley-VCH-Verlag, ISBN 3-527-31540-3
Jump up ^ Hartwig, J. F. Organotransition Metal Chemistry, from Bonding to Catalysis; University Science Books: New York, 2010.
Jump up ^ Zoeller, J. R.; Agreda, V. H.; Cook, S. L.; Lafferty, N. L.; Polichnowski, S. W.; Pond, D. M. (1992). "Eastman Chemical Company Acetic Anhydride Process". Catalysis Today 13: 73–91. doi:10.1016/0920-5861(92)80188-S.
Jump up ^ Beller, Matthias; Cornils, B.; Frohning, C. D.; Kohlpaintner, C. W. (1995). "Progress in hydroformylation and carbonylation". Journal of Molecular Catalysis A 104: 17–85. doi:10.1016/1381-1169(95)00130-1.
Jump up ^ W. Bertleff; M. Roeper; X. Sava (2005), "Carbonylation", Ullmann's Encyclopedia of Industrial Chemistry, Weinheim: Wiley-VCH, doi:10.1002/14356007.a05_217
Jump up ^ El Ali, B.; Alper, H. "Hydrocarboxylation and hydroesterification reactions catalyzed by transition metal complexes" In Transition Metals for Organic Synthesis, 2nd ed.; Beller, M., Bolm, C., Eds.; Wiley-VCH:Weinheim, 2004. ISBN 978-3-527-30613-8
Jump up ^ Karlheinz Miltenberger, "Hydroxycarboxylic Acids, Aliphatic" in Ullmann’s Encyclopedia of Industrial Chemistry, Wiley-VCH: Weinheim, 2003.
Jump up ^ Riemenschneider, Wilhelm; and Bolt, Hermann (2000). "Esters, Organic". Ullmann's Encyclopedia of Industrial Chemistry: 10. doi:10.1002/14356007.a09_565. Retrieved 17 December 2013.
Jump up ^ Elschenbroich, C. ”Organometallics” (2006) Wiley-VCH: Weinheim. ISBN 978-3-527-29390-2

I think you mean decarbonylation, not decarboxylation. Also, I couldn't find any reference to chromium trioxide as a catalyst for either of these processes from a quick google search. Can you provide a reference for this reaction?


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Forms of carbon in incomplete combustion
When hydrocarbons burn with little oxygen, carbon is produced. What form do these particulates of carbon come in?
Is it a simple molecule like $\ce{S8}$, or a small fragment of a larger structure, like graphite?
Why is this form more stable than others?

It heavily depends on reaction parameters (oxygen/oxidants availability, temperature, pressure, etc.)
In a free-air incomplete combustion at a (relatively) high temperature, you get mainly carbon-based molecules of various kind.
In a internal combustion engine, that usually works at high temperature and high pressure, you can also get nanoparticles of various kind, including graphite.
In a free-air, mixed, relatively cold combustion, the original small molecules, graphite and other carbon-bases compounds can further react with other molecules giving you larger molecules and/or can stick to larger particles forming nano or micro-particles of various kind. Predicting/modelling this kind of reactions can be very complicated.


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How does Palladium dissociate H₂ so easily?
If I understand correctly, $\ce{H2}$ in the presence of $\ce{Pd}$ readily dissociates as it dissolves into the metal. With the dissociation energy for the $\ce{H—H}$ bond being so large, how is this possible?
At first I thought that the $\ce{H}$ atoms were falling to a lower energy level in the $\ce{Pd}$ that was somehow only available if they were monatomic, but if that were true wouldn't some heat be generated by the absorption?

The one-line answer would be that the two Pd-H bonds that form must be lower in energy, but right now I don't have anything to back that up.

That seems a reasonable conclusion and is what I assumed until I read that Palladium hydride readily releases H₂ by the reverse process of hydrogen absorption.

There is an equilibrium process, then, that can be manipulated by changing the conditions (for example low pressure vs. high pressure). I just don't know how it is done.

Metallic (Interstitial) hydrides are quite well-knows, and proximity of Delta G (Free Energy) of formation for some hydride to zero is not that strange. What is much more interesting, is easiness, with which protons travels through solid palladium. This is a thing that hard to understand.

@BenNorris: And presumably that the transition between the Kubas-bound state $\ce{Pd(\mu -H2)}$ and the dissociated hydride must have a very low barrier and not involve much of an energy change.

$\ce{Pd}$ can dissociate $\ce{H2}$ because the resulting $\ce{Pd-H}$ bonds are more stable than the starting $\ce{H2}$.
But the reason why $\ce{Pd}$ is so good at dissociating $\ce{H2}$ is related to the energy barrier to bond formation. The dissociation of $\ce{H2}$ on a $\ce{Pd}$ surface (and on $\ce{Pt}$ and maybe several other metals) has no barrier. So you don't need to put the $\ce{H2}$ in an excited state to go over a barrier and create $\ce{Pd-H}$ bonds.
On $\ce{Cu}$ for example, the bonding is possible but there is a barrier, you need to excite $\ce{H2}$ to dissociate it.
Here is a good theoretical article (with a great title):
It shows the different barriers to dissociation of $\ce{H2}$ on $\ce{Pt}$, $\ce{Ni}$, $\ce{Cu}$ and $\ce{Au}$. It also gives an explanation for such differences (more physics than chemistry).

May you type the title of the article, please ? because the link has an error  .@Guillaume


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What are the major environmental drawbacks with using methane as a combustable fuel?
What are the biggest environmental issues with using methane, $\ce{CH4}$, (which is found as the major energy source in natural gas, shale gas, biogas, and synthetic natural gas), as a combustible fuel?

Welcome to Chemistry StackExchange @DavidWalz. Your question needs a little improvement to help us get you some good answers. First, could you tell what you know about the use of methane as a fuel? Second, this question seems to be inviting discussion, which our FAQ discourages. We would like to see an informed question that looks like it has a correct answer. Otherwise, your question may be closed for being not constructive.

@BenNorris: Btw, comment link markdown works the other way [title](URL). ANyway, I fixed it for you :)

Closed as NARQ, as per BenNorris. The question is way too broad in it's current format (the first para of Alex's answer makes this evident). It also is not constructive, due to the reasons mentioned by Ben. If you can edit it to specify the perspective and make it less of a discussion-inviting post, I'd be happy to reopen :)

Methane has many lovely properties. It's a very dense source of energy. It is plentiful, it seemed to be cheap to extract (until we worked out the real cost), and similarly seemed to be cheap to transport long distances. It can be stored for many months in vast quantities.
But.
It's a greenhouse gas - a powerful one at that. On a 100-year horizon, it has about 20 times the global warming potential of $\ce{CO2}$. That needn't be a problem, if natural-gas extraction had no leakages, and distribution systems had no leakages.  But they do have leakages, which means the supply chain worsens global warming.
Furthermore, when methane is burnt, $\ce{CO2}$ is released into the atmosphere, worsening global warming even further.
We've learnt to handle methane's other main problems: that it's colourless and odourless, so can just accumulate unnoticed waiting to explode, if you let it; and that it's also dangerous if handled carelessly in any other way.
It also generates $NO_x$ when burnt in air, and they are local as well as global pollutants. It is possible to have flue-scrubbers to remove the $NO_x$.
That doesn't mean that that methane is permanently bad.  It can be created in a carbon-neutral fashion (from biological sources, or synthesised from atmospheric  $\ce{CO2}$), and if leakages are minimised, offers a potential useful form of energy storage. (note that in these forms, it would be an energy & storage vector, not a net energy source, so its economic value would be very different - but that's economics, not chemistry)

Interesting, you brought up some really great points that were not immediately obvious. I'll have to look into ways that leakage can be minimised.

Some interesting points, but many are also true of any other hydrocarbon fuel. The issues that differentiate methane are mostly about the difficulty of transport and storage (much has to be liquefied to move it across large oceans).

From what point of view? Environmental? Industrial? Practical?
In Europe (I live in Italy) we use methane a lot, both for automotive and for house heating/cooking.
From the environmental point ov view, $\ce{CH4}$ is one of the best choices (together with LPG). When burnt (both in thermodynamics engines and in free-flames) it releases $\ce{H2O}$ and $\ce{CO2}$ only, even when burn at low temperatures and pressures (gasoline would produce a lot of pollutants in this case).
From an industrial point of view it is even better. There is a large abundance of $\ce{CH4}$ underground. We import a huge amount of it from Russia and north-Africa at a quite low price. Even better: $\ce{CH4}$ is easy to produce from other sources. For example, here in my area we use $\ce{CH4}$ (and LPG) produced (as a by-product) from petrolchemical industry and oil refineries.
From a practical point of view, it is relatively easy to manage. In my house (as in 99% of European houses) we use $\ce{CH4}$ both for heating and for cooking. We just have a high/low pressure pipeline connected to the national gas network. Nothing actually more complicated that the usual drinkable water supply pipeline.
I drove $\ce{CH4}$- and LPG-powered automobiles for years in my life. $\ce{CH4}$ forces you to have heavy, high-pressure bottles in you car (usually in the hood) but still can give a car that reach 90% of the usual performance and costs you something like 50% of a normal gasoline-powered car.
LPG is even better. My wife has a LPG car in this period. The LPG bottle is relatively  light (50% of the high-pressure bottle used for $\ce{CH4}$) and cheap (30% less than a $\ce{CH4}$ system). Her car reach 90% of the gasoline-powered equivalent car and she spend something like 60% of a corresponding gasoline-powered car. It is even competitive with a corresponding diesel car.
The version of $\ce{CH4}$ (and LPG) that you can buy for automotive and in-house uses in Europe is always added with a specific odorizer that makes very easy to spot any leakage. Our gas sensors are aimed at spotting this specific odorizer so using $\ce{CH4}$ and LPG is actually quite safe.
Of course, $\ce{CH4}$ is still a fuel gas. It still requires high-pressure bottle and high-pressure pipeline so it still is a dangerous product and still have to be handle with care. For this reason, for example, you cannot refill your car by yourself and you have to wait for the service station staff.
In the same way, every $\ce{CH4}$ pipeline used for in-house cooking or heating must be verified and certified once per year by a certified technician.
In EU and IT, there are plans to produce $\ce{CH4}$ and LPG from many new, sometime weird, sources (organic waste, biomasses, etc.) so we expect to see an even larger diffusion of $\ce{CH4}$ in the future.
Please note that many of those sources are totally renewable so there isn't any new $\ce{CO2}$ added to earth atmosphere. There is just a circulation of existing $\ce{CO2}$.


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Why do oxidation numbers work?
I understand that oxidation numbers are a method for keeping track of electrons in a reaction and how they are generally assigned (electrons in a bond are assigned to the more electronegative atom).
Why does this method work? Why does assigning these arbitrary numbers based on electronegativity work in predicting and balancing reactions?

This is a really great question. Don't end your posts with self-defeatism! The numbers are not exactly arbitrary, but they are not based on measurable phenomena. I hope that you get a good answer.

Sorry; I guess the way I put it was not the greatest, but with that last question I was wondering if people in general just don't ask this question because they don't want to know the answer (possibly if there is no good answer).

First of all they are not really arbitrary.
But the main point, regarding balancing the equations is that: You can keep track of the total number of electrons per species, which is a well defined value. Like, for one permanganate ($\ce{MnO4-}$) you have 25+4*8+1 electrons, for one $\ce{Mn(II)}$ you have 23 electrons, for one water you have 10 electrons, for one $\ce{H+}$ you have 0 electrons. So for this half-equation, you must have 5 electrons on the left side:
$$\ce{8H+ + MnO4- + 5e- -> Mn(II) + 4H2O}$$
or you can say that, here I assume all H atoms are +, all oxygens are -2, and for consistency, Mn in $\ce{MnO4-}$ is +7. I can dismiss H and O's since all are same, and all change is because of Mn, which goes from +7 to +2 and difference is 5 electrons.
You could do this like that as well: O is -4 everywhere, so H's in $\ce{H2O}$ should be +2 and they were +1 on the left side. They donated 8 electrons. Mn in $\ce{MnO4-}$ should be +15 and it is +2 on the right side, which means it accepted 13 electrons. And the difference is 5 electrons, again.
Bottomline: It is all about consistency. If all elements in each and every species sum up to the charge of this species, whatever you say their formal charges are, it will work -because this just another way of keeping track of total electron count.  So it is better to treat most of them constant like, O is -2, H is +1 almost always.

Welcome to Chemistry Stack Exchange!

The +5 electrons should had been on the left side of the equation, not the right side!

There are several ways to keep track of things.  Oxidation number is one, formal charge is another.
With oxidation numbers, the electrons in a molecule or ion are assigned to the more electronegative atom in a bond.  Thus in $\ce{H2O}$ oxygen is the most electronegative so the oxidation number of hydrogen is $+1$ (it has "lost" an electron) and oxygen is $-2$ (it has "gained" two electrons.  Using this scheme in a consistent manner one can balance equations quite nicely.
On the other hand, there are times when the formal charge is more important.  In this scheme one always splits the electrons in any bond between the two atoms.  Thus in water all the atoms have a formal charge of zero.
One can think of these two schemes as follows:  oxidation numbers pretend that all bonds are in effect totally ionic with the electrons going to the more electronegative atom.  Formal charge pretends that all bonds are covalent and the electrons in a bond are shared equally.
Of course neither one is totally correct, but either one can be used as an "electron accounting" scheme.


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Is it possible to prevent methane combining with oxygen?
The output from an anerobic kitchen waste digester is mostly $\ce{CH4}$, and $\ce{CO2}$.
If the methane is collected over water; algae/moss may use the $\ce{CO2}$ to grow. Yet oxygen is released ... which, together with methane, is a potent environment.
Will the oxygen, in this closed jar filled with water+algae+moss, recombine to form $\ce{CO2}$ and $\ce{H2O}$? Is there a means to prevent methane combining with oxygen released by the algae?

I doubt that the conditions in an Anaerobic digester (organic acids, ammonia (!), etc) allow algae to grow. If they do, the oxygen they set free will create oxic conditions and stop the AD - Methanogens are obligate anaerobe bacteria.
In nature, Methanogens and Algae inhabit different habitats - methanogens in the mud at the bottom of a swamp, algae (maybe) in the water above.
Processes for separating $\ce{CO2}$ from $\ce{CH4}$ usually work on the $\ce{CO2}$ (absorption in amines or water under high pressure), so it would be hard to pull the $\ce{CH4}$ from the headspace of your vessel - but given the biological points, this is a non-issue.

Will the oxygen, in this closed jar filled with water+algae+moss, recombine to form $\ce{CO2}$ and $\ce{H2O}$?

Very slowly, you will not see this. However, it is much worse, as explosive composition of $\ce{CH4}$ and $\ce{O2}$ will form.
In industrial applications it is possible to separate $\ce{CO2}$ from $\ce{CH4}$ and them move $\ce{CO2}$ into separate chamber with algae.


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Why does Naphthalene turn into needle-shaped crystals by sublimation?
I wonder why crystals of Naphthalene are needle-shaped after sublimation. Googling the fact, I could only confirm that it is true but failed to find the reason. Could someone please shed some light as to why this happens?

Every solid material has a crystal structure.  Sublimation tends to produce good crystals because the solid can grow slowly.  In bulk form naphthalene crystals are broken and compressed and so bulk solid naphthalene looks quite different.
The exact same situation applies to snowflakes.  They are flat hexagonal structures when seen individually.  Crunch them into a snowball and they lose that pretty structure.

Well, there are crystalline and amorphous solids...

But you don't get amorphous solids out of a one-component gas phase. Or lets say it will be very hard even get it into the gas phase. ;-)

Btw., crystals grown from the melt or solution likely form a totally different habitus that if grown from the gas phase. There's a good chance even a different phase will form.

It's the symmetry!
(Sorry, that's the default answer to all questions on crystallography.)
You get needles if your crystallite grows faster in one direction than in any perpendicular direction. Which is the fact here, because aromatic rings love to stack, but don't have much energy to gain out of the van der Waals' interactions between the H-atoms sticking out in the plane.
If a crystallite grows slower in one direction than perpendicular to it, you will get platelet-like crystals.
And if the elementary cell of the crystal has a cubic symmetry, it will likely grow equally fast in all directions, and the crystals will be cubes, like $\ce{NaCl}$. (OK, this last statement is rather weak. It's easy to predict that growth rates are very different like in naphthalene, but that they're actually identical is dure. Sodium chloride is just a nice example where this is the case. And be careful about melts. They're never isotropic near a crystal surface, so anything can happen.)


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Hypochlorous acid mechanism
When hypochlorous acid is dissolved in water, is it the chlorine atom or the oxygen that serves as the anti microbial agent?

You're talking about oxidation here, right? When you dissolve $\ce{HOCl}$ in water, it doesn't split up into the three constituent atoms. But there can be some redox reactions centered on one atom. Not sure if that's what causes the antimcrobial properties, though.

Yes I'm talking about redox reactions, not just dissociation.

The chemical disinfection (inactivation of bacteria) in fresh water can occur through a number of mechanisms, including oxidation of cell walls, inactivation of key enzymes and disruption of nucleic acids, thereby rendering them non-functional. The precise mechanism of inactivation depends upon the nature of micro-organism (bacteria, spores, viruses).
When chlorine was first used as a disinfectant in the USA in 1908 (slightly earlier in Europe), its germicidal power was commonly attributed to the liberation of 'nascent oxygen' from hypochlorous acid $(HOCl)$.
$HOCl\rightarrow HCl+\frac{1}{2}O_{2}$
Subsequent investigations (Chang, 1944, Green and Stumpf, 1946) debunked this theory and showed that chlorine reacts irreversibly with the enzymatic system of bacteria, thereby killing it.
Chang demonstrated by showing that hydrogen peroxide or potassium permanganate which release considerable amounts of nascent oxygen, nevertheless exhibit only weak germicidal activity. Chang was able to demonstrate that there is no liberation of oxygen involved in the inactivation of bacteria by chlorine.
In any case, oxygen gas $(O_{2})$ is a much weaker oxidizing agent than hypochlorous acid, as shown in the table below. Hypochlorous acid is likely to be effective as both an oxidizing agent as well as direct chlorination of the protoplasm and reaction with lipoproteins to form toxic chloro compounts that interfere with cell division (Chang, 1944).

Hypochlorous acid $(HOCl)$ dissociates in water to $H^{+}$ and hypochlorite ion $(OCl^{-})$.
$HOCl\leftrightarrow H^{+} + OCl^{-}$
Because hypochlorous acid $(HOCl)$ is uncharged, it is better able to penetrate cell walls than other chlorine species and is about 80 times more effective than hypochlorite $(OCl^{-})$ at chlorination disinfection.
The microbial disinfection due to total chlorine ($HOCl + OCl^{-}$) therefore depends upon pH and temperature. Whilst low pH and temperature favour disinfection by chlorine as $HOCl$, rate of diffusion through the cell membrane occurs faster at high temperature. The metabolic activity of the cell,rate of reaction of chlorine with enzymes and hence rate of inactivation of the cell is faster at higher temperature.

Disinfection by redox reaction using ozone gas $(O_{3})$ (a fast and powerful oxisizing agent) is sometimes used as an alternative to chlorination to reduces or eliminate the toxicity caused by residual chlorine, especially where there is concern over organic matter in the water which may produce trichloromethanes.
Sources:
Chang, S.L., Destruction of microorganisms, J. Am. Water Works Assoc., 36 (1944), pp. 1192–1206
Green, D.E. and Stumpf, P.K., The mode of action of chlorine, J. Am. Water Work Assoc., 38 (1946), pp. 1301–1305
Water Quality Control Handbook, Second Edition, by E. Roberts Alley (2007, McGraw Hill).
White's Handbook of Chlorination and Alternative Disinfectants, 5th Edition, by Black & Veatch Corp (2010, John Wiley & Sons)


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Has the molecular diatomic oxygen trianion been observed or predicted?
Has diatomic oxygen ever been observed with a 3− charge, i.e. $\ce{O_2^{3-}}$? Have any ab intio calculations been done on this ion?
Update:
Just to summarize some of the answers below and add my two cents, it seems that no studies have reported observing this ion experimentally, either in isolation or as part of a condensed phase. Ab initio calculations suggest that $\ce{O_2^{3-}}$ is unstable in isolation, meaning that the loss of an electron is exothermic. However, first-principles calculations have predicted that a molecular oxygen trianion can exist as part of a solid; see Phys. Rev. B 85, 035210 (2012).

Two things: firstly, when referring to oxidation states, the sign comes before the number, so a thing is in oxidation state +3 or -1, for example. Secondly, an atom has an oxidation state, a molecule has a charge. For charges, the sign comes after, so 2- or 2+.

One should notice that the simple existence of such a species is not obvious. Anions always have the possibility to evolve through loss of an electron, which is not the case for positively charged species.  Computation of electron affinities is not obvious, as ab initio calculations would "force" the electron to remain bound. Therefore one has to add numerous diffuse orbitals and consider that the presence of the extra electron in diffuse orbitals should be interpreted as a sign of unstability of these species.

anions form from Gain of electrons and cations form from Loss of electrons. Not the other way around.

Could you quote the section where they predict the oxygen trianion, please. I was unable to find any such reference, but I am not an expert on polarons and I might have missed it.

They don't use the word "trianion", but I think what they describe on p. 2 is equivalent: "We found that the presence of an excess electron in Li$_2$O$_2$ results in elongation and/or cleavage of one of the O–O bonds...and the excess electron is localized at the cleaved O–O site."

I do not see how this can be considered equivalent to the statement of a trianion. It should also be mentioned, that the calculations treat one polaron in a 192 atom super cell; that is one extra electron per 96 oxygen atoms and the electron is located at the site where the oxygen bond is broken or elongated. This is just a little bit more stable than having an delocalised excess electron in a perfect lithium peroxide lattice.

According to "Small Multiply Charged Anions as Building Blocks in
Chemistry" Acc. Chem. Res. 1996, 29, 497-502

it is concluded that all isolated diatomic dianions or
  other multiply charged diatomic anions are not stable
  to loss of an extra electron or fragmentation in their
  ground electronic states

So even $\ce{O2^{2-}}$, let alone $\ce{O2^{3-}}$ cannot exist in isolation.

I don't know of (and a quick search couldn't reveal) any studies on the $\ce O_2^{3-}$ ion. Note that in this ion, oxygen wouldn't have an oxidation state of –3, but –1.5.

The $\ce{O2^{3-}}$ ion is predicted hy electronic calculations in the high pressure compound $\ce{FeO2}$[1]. This oxide, believed to exist deep in Earth's mantle, is computed to have the sigma antibonding orbital between the oxygen atoms lies within the $t_{2g}$ electronic band associated with octahedral coordination of the iron atoms. This causes an electron that would normally be associated with the iron coordination sphere to be transferred to the oxygen-oxygen sigma antibond, converting the peroxide ion from $\ce{O2^{2-}}$ to $\ce{O2^{3-}}$ (and the iron from iron(II) to iron (III)).
Also noted is the impact on bond length between the oxygen atoms. With the extra electron in the dioxygen function going into an antibonding orbital, the oxygen-oxygen bond length in $\ce{FeO2}$ is found to be 1.89 Å versus approximately 1.49 Å typically found in ordinary peroxides.
Reference

Sergey S. Streltsov, Alexey O. Shorikov, Sergey L. Skornyakov, Alexander I. Poteryaev & Daniel I. Khomskii (2017). "Unexpected 3+ valence of iron in FeO2, a geologically important material lying “in between” oxides and peroxides". Sci Rep 7, 13005. https://doi.org/10.1038/s41598-017-13312-4.

Yes, it is, in $\ce{H3O2}$. This is from an a CCB bond. However, this is highly unlikely.

Could you perhaps elaborate in your answer on this?


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A thermoset or a thermoplastic?
How do you know if a polymer is a thermoset or a thermoplastic? For example:

hydroxybutanoic acid
2-amino-3- hydroxybutanoic acid

(I am not sure if these are correct in English)
I know what thermosets and thermoplastics are, and their properties, but I don't know how to see if polymers of a certain molecule is a thermoset or a thermoplastic by looking at the structural formula.

Some background, since most readers would probably need it.

A thermoplastic is a polymer that becomes pliable/moldable upon heating, and then returns to a solid (keeping its new shape) when cooled. Thermoplastics are generally high molecular weight polymers with very few or no crosslinks. Polymer chains are held together by intermolecular forces (most often van der waals) and physical entanglement, both of which can be overcome with addition of heat.
A thermoset is a polymer that is irreversibly cured, usually through covalent crosslinking (during or after polymerization). The covalent crosslinks hold the chains together in such a way that the chains cannot be separated by the addition of (small amounts of) kinetic energy. Thermosets tend to dramatically fail after the addition of lots of heat, following which they do not return to their original state. It is possible for the cumulative forces from many strong non-covalent interactions (ionic, hydrogen-bonding, and even pi stacking) to make a polymer behave like a thermoset. For example, proteins are thermosets.

Can you tell if a monomer will produce a thermoplastic or themoset based on its structure?

Yes. All polymers produced from only mono- (addition polymerizations) or bi- (condensation polymerizations) functional monomers will always produce thermoplastics. All of the hydroxybutanoic acids (e.g. $\gamma$-hydroxybutanoic acid) are difunctional monomers for condensation polymerizations. Their polymers will be thermoplastics.
Unless...
The addition of tri- or tetra- functional monomers produces crosslinks, which always produces thermosets. For example, glycerol, malic acid, or lysine could be trifunctional monomers in condensation polymerizations. 2-Amino-3-hydroxybutanoic acid (threonine) is a trifunctional monomer. It has three functional groups (alcohol, amine, carboxylic acid), that can all react in a condensation polymerization. Any synthetic polymer containing threonine is likely to contain crosslinks and be a thermoset.
Or...
Many thermoplastic polymers can be converted into thermosets by curing them afterwards. Some curing is thermal, some is irradiative (espccially using UV), and some is chemical. A prime example is polyisoprene (natural latex rubber), which is a thermoplastic (with a very low melting point). It is also an elastomer - it will stretch and return to its original shape (although melting it erases the "memory"). Rubber is converted into a thermoset by vulcanization to make it more useful while still having the elastomeric properties.


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Homework - Ideal gas law
I don't understand how to find the volume in this problem..
A flask containing 63.2 mL of neon gas was collected over water at 21C and 788 mm Hg. What is this volume of gas at STP? (WVP at 21C = 18.6 mm Hg)

Hint: what do you know about the partial pressures of the gases?  What is contributing to the 788 mmHg?

If you edit the question to reflect your initial efforts, you can flag it to be reopened.  As of now, you have just copied your homework problem in as a question.  More ideal would be if you could pinpoint the areas that are troubling you and ask a more conceptual question.

This question serves as a bad example for the homework policy.


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Why water can't mix with oil or cooking oil?
Why water can't mix with oil or cooking oil (both saturated and unsaturated)?

The basic answer is that water molecules attract each other and so clump together forcing almost all of the oil out of the clump.
The attraction exists because water is a polar molecule.  That is, it has a positive end and an negative end.  Thus the molecules tend to clump together just as a bag of small magnets tend to clump together.  Oils, on the other hand, are not polar.
This principle can be expanded.  In general polar materials tend not to dissolve in oils or, if they are liquids, dissolve in them.  The rule in chemistry is "like dissolves like".

The textbook answer for why water can't mix with oil is to use the rule 'like dissolves like'. But a rule in science is none other than summarizing a few collected facts. Answering a question with a rule is like pointing a single fact - why water can't mix with oil - to a collection of facts instead. This easily invites a follow-up question, why like dissolves like. Besides, this rule is vague and superficial; it does not address the facts that

ice is not dissolved in the same matter, i.e., water at 0 °C,
cooking oil has a polarity comparable to that of water (as measured in Debye units),
oil molecules do not clump together as strong as water, then how could they prevent water molecules, which are attracted more to oil than oil itself, from coming in? or
stuff like calcium fluoride is very polar but insoluble.

So, IMO, it's better explained in terms of energy. When water clumps with water and oil does the same, the total energy is lower than when they are mixed. The system is more stable if unmixed. But watch for a delicate point in this argument. Energy can be lowered when water molecule is moved into oil molecules. How do we account for it? The lowered energy is not enough to compensate the energy needed to pull that water molecule out of its pool.
However, the difference in energy is not the only factor of insolubility. Temperature plays a role too. When temperature is raised, the difference between dissolved and undissolved states becomes less steep, relatively speaking. When thermal energy chips in, the two substances are more miscible.

From a thermodynamic perspective, saying that a water/oil mixture is more stable when unmixed is going against the fact that this state has lower entropy. Perhaps if you focused on explaining why the total energy is lower it could improve your answer.

Entropy changes little in liquid mixing except at both ends where concentrations are very small. The entropy plot against concentration is shaped like an inverted U.

I mean to say that when attempting to explain why like dissolves like, it requires us to quantify something that is somewhat difficult to quantify. I feel like the answer you have given brings more questions while both give the same answer.

I agree with Leonardo. This simple question is hard to crack. Comparing with 'like dissolves like', the approach from energies is more understandable but so far has no quantitative result to show. Perhaps Paul can enlighten us on it.

@Leonardo While the facts may perhaps be in dispute, I think that Chin's answer does contribute additional/complementary information that may turn out to be useful to the OP.

@jonsca Yes I agree, I was just suggesting that a simple answer does justice to a simple question, rather than running the risk of confusing someone. Although, I guess if it gets them thinking about it beyond accepting that rule, it is a good thing.


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Does sublimation happen only on the surface or throughout the body?
Like evaporation happens only on the surface of a liquid, and boiling/vapourisation takes place throughout the body, how does sublimation occur? Why?

One way to think about it is whenever something leaves the solid lattice, other molecules in the air are still stopping it from moving far from the solid surface and we consider this pressure. When there is little or no pressure, nothing is stopping the molecule/atom once it leaves the solid lattice, and will keep moving away until some force stops it.

The situation is akin to boiling.  If a molecule at the surface frees itself of its attractions to its neighbors, it can move into the vapor phase easily.  Same thing with evaporation at the surface of a solid.
Can a molecule evaporate from the bulk liquid?  The answer is no.  Imagine such a thing happening.  The escaping molecule(s) form a bubble inside the liquid.  The pressure inside the bubble is the vapor pressure of the liquid at that temperature.  But that temperature is less than the boiling temperature so the vapor pressure is less than the pressure external to the liquid.  The result is that the bubble collapses.  The boiling temperature is the lowest temperature at which the bubble pressure is equal to the external pressure.
Now think of sublimation in the bulk of a solid.  Here the "bubble" has to have enough pressure to break apart the solid.  It generally won't develop that much pressure.  Indeed, one can watch "dry ice" sublime well above its freezing point and nothing happens inside the bulk of the solid.
So the general answer is no, sublimation does not take place in the bulk of a solid.

what if each molecule in the solid changes into vapour phase simultaneously. If we theoretically consider the solid to be heated uniformly, that should be happening.


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How do you Synthesise a Fatty Acid Amide?
How would you go about synthesising a fatty acid amide, preferably by means of a high-yield reaction. Please provide an explanation that I, as a grade 12 chemistry graduate, could understand.
I am familiar with:

Fatty acids
Carboxylic acid
Equilibrium reactions & constants
Amide groups
Amines
Alkanols
Esterification

If you need to use terms I may be unfamiliar with, it would be nice if you could explain them or provide a link to an explanation.

Schotten-Baumann Reaction
I found the solution to my question and it is the Schotten-Baumann reaction which takes an acyl chloride (which can be synthesised from a carboxylic acid via the methods detailed in the next section) and an amine to form an amide, a hydrogen chloride.
 (1)

[Net reaction: $ \ce{RCOCl+NH_2 R' \rightarrow RCONHR' + HCl}$]

Synthesis of Acyl Chlorides
To create acyl chlorides from carboxylic acids the following three reactions will suffice:
$ \ce{RCOOH (l) + PCl_5 (s) \rightarrow RCOCl (aq) + POCl_3 (aq) + HCl (g) }$
$ \ce{3 RCOOH (l)+ PCl_3 (s) \rightarrow 3 RCOCl (aq) + H_2 PO_3 (aq)} $
$ \ce{RCOOH (l) + SOCl_2 (s) \rightarrow RCOCl (aq) + SO_2 (g) + HCl (g)} $
(2)

Reference List (APA style)

Schotten-Baumann Reaction Schotten-Baumann Conditions (n.d.). Retrieved December 22 2012 from http://www.organic-chemistry.org/namedreactions/schotten-baumann-reaction.shtm.
Clarke, J. (2004). CONVERTING CARBOXYLIC ACIDS INTO ACYL CHLORIDES (ACID CHLORIDES). Retrieved December 22 2012 from http://www.chemguide.co.uk/organicprops/acids/pcl5.html.

You are perfectly right on that you need to convert the carboxylic acid to a more reactive species to eventually obtain the amide!
Playing with the pH will not help:

under acidic conditions, the amine is protonated and therefore unreactive
under alkaline conditions, the acid is deprotonated and the carboxylate won't react at all

The Schotten-Baumann reaction of fatty acids with $\ce{SOCl2}$ will probably work fine:

reflux 1 mol of the acid with 1.5 mol of thionyl chloride
dIstill the remaining thionyl chloride off
purify the fatty acid chloride by distillation in vacuum. Note that the bp of stearoyl chloride is about 175 °C at 2 mm Hg!

This is quite some work to obtain a fatty acid derivative that eventually reacts with an amine!
Wouldn't it be much more convenient to form the amide in a one-pot reaction, possibly at room temperature, by activating the carboxylic acid  in situ?
(In situ means: at the same spot, without isolating the reactive intermediate.)
There's a large number of coupling reagents available, such as

carbodiimides like DCC (dicyclohexylcarbodiimide)
$1H$-benzotriazoles
chloro-substituted triazines

Eric Valeur and Mark Bradley published a critical review titled Amide bond formation: beyond the myth of coupling reagents, in Chem. Soc. Rev., 2009, 38, 606-631 (DOI) on the topic.


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How does one synthesise 2-Monoglycerides from Triglycerides?
How does one synthesise 2-monoglycerides from triglycerides with additional ingredients if necessary?
My first instinct would be to add a catalyst for the reaction of $triglyceride \rightarrow 3 FA + glycerol$ (FA=fatty acid; for instance a catalyst could be sodium hydroxide) and then add additional glycerol so as to cause the reaction: $FA + glycerol \rightarrow 2$-$monoglyceride $ to occur but I just wanted to ask if this is true because to me this seems too simple to be true.

It is too simple to be true. In such way you will get mixture of 1- and 2-monoglicerides with some products of further reactions.

One way I have learnt is by taking a page out of the human body's book and using pancreatic lipase, found in pancreatic juice, to catalyse the reaction:
$triglyceride+ 2H_2 O \rightleftharpoons   2$-$monoglyceride+2 FAA$
Where FAA stands for, 'Fatty acid anion.'
If someone has a better answer I'll readily accept it over this one, this is just one method.

Just a note--we support LaTeX (with a chemistry extension) over here. See http://meta.chemistry.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here


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Classical reason for good heat conductance in diamond?
My teacher in my physics course attributed this effect to phonons, more here, but I am skeptical about this argument, it feels like he is overlooking the whole question -- what about with Silica that has similar structure to diamond with covalent bonds?
My friend attributed this to the small mass of atoms and large Young constant, considering the system to be made of springs -- $v_{sound}=\sqrt{\frac{Y}{\rho}}$. He suggested to think of heat as sound to use the formula.
Silica has less strong bonds to diamond so perhaps this is a reason why diamondis a good conductor of hear and silica is not. Air is good insulator so does it mean that carbon atoms are so closely packed and silicon atoms not? Is this the reason for the difference in conductance?

I think that there is no really good classical reason.  Heat conductivity in a solid depends on bonding in that solid and bonding is inherently a quantum mechanical phenomenon.
For instance, recall that graphite is a good conductor of heat (and electricity) in two dimensions and a fairly good insulator (heatwise and electrically) in the third.  This is due to the layered structure in graphite and that is due to the bonding in graphite.

+1 for the start but I think one could create some sort of mechanical-engineering-style approximations for the object to be counted as classical, interested how far they are off from the more real QM models.

I think the classical explanation for good heat-conductance in diamond is the spring-analogy. But it does not explain all elements such as metals where electron-gas in valence shell contribute to the good heat-conductance. In order to understand this deeper, you need to study the statistics outlined below. The phonon gas occurs between the bonds in diamond while in the electron gas with metals of blocks I/II.
Let's think about classical features such as heat capacity and heat-conductance. As said, they are different in nature depending on the location in periodic table.
I/II blocks in periodic table

Let's consider the first block and second block. The good heat-conductance in metals such as Li, Na and Mg is due to the electron-gas that forms fermi-balls according to Fermi-Dirac statistics. When you experience the heat-change, phonos are the interactive quantitative pieces of energy.

Only the electrons in the highlighted region contribute meaningfully to the specific heat capacity. The source of this picture is the page 166 here, a course booklet in 2062 Physics course in Aalto University.
In order to understand this, you must understand Fermi-Dirac statistics and Fourier Transforms by which you can get the location-frequency-spectrums (in Finnish paikkataajuus spektri): you have distribution of particles and then you use that distribution to deduce the frequency plot.

IV block elements such as Si and C

Now the heat conductance here with structures such as diamond or Silica cannot be due to electron-gas because there is no moving electrons in diamond or silica. The very good heat conductance in diamond is due to mechanical vibration and very strong bonds. Silica has much weaker bonds than carbon atoms with four covalent bonds. Young's modulus is analogical to the mechanical Hooke's constant. Einstein and Debyen explained this so that atoms act like massless springs (page 155 of the earlier lecture slide). Now you can calculate the internal energy with the help of Bose-Einstein statistics.

This plot shows the heat capacities as a function of temperature. The parameter $\theta_D$ is the Debye temperature, $R$ is the gas-constant and dots show real temperatures. The source is page 158 of the earlier lecture slides.

Now in Chemistry, people often use the Maxwell-Boltzman statistics. In Quantum mechanics, they say that Maxwell-Boltzman is like an approximation -- it only work well with very low-temperature or very high temperature -- otherwise it is a bad approximation even though useful in practice but not correct. The statistics Fermi-Dirac and Bose-Einstein are the more realistic description of nature. Depending on the spin-number of atom, you choose the statistics. If your spin is some integer (or zero with modulo 1), you have a boson. If your spin has half (or half with modulo 1), you have a fermion. Examples of fermions contain electron, neutron, proton -- and they all have spin of half.
Let's consider Rubidium 86 (atomic number 37 so 37 electrons, 37 protons and 49 neutrons) to its isotope 87 (37 electrons, 37 protons and 50 neutrons). Because 86's spin is odd and 87's spin is even, 86 is fermion while 87 is boson. Now use Boson-Einstein statistics with 87 and Fermi-Dirac with 86.
P.s. You could mathematically approach this problem this way:

1. calculate amount of particles $N$
2. derivate $N$ with respect to $E$ energy
3. find the point when the derivative $\frac{\partial N}{\partial E}$ is zero i.e. $\frac{\partial N}{\partial E}=0$.

All true, but the calculations and ideas are NOT classical.

@PaulJ.Gans that is right but I thought it to be useful to contain an in-depth explanation also -- apparently there is no simpler way to explain this. We are still missing the in-depth classical explanation. The assumption that particles as springs leads to which kind of models? Strench-compression pictures? FEM? I don't know -- perhaps needing mechanical-engineering-approximation for the classical case, investigating.

You should correct neutrnos to neutrons.


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Synthesis of nitrocellulose
I am trying to understand the synthesis of nitrocellulose.
A mixture of sulfuric acid and nitric acid is used to nitrate cellulose. The sulfuric acid acts as a catalyst, producing $\ce{NO2^{+}}$ ions from the nitric acid which bind at certain sites on the cellulose chain to nitrate it. So nitrate salts can be used instead of nitric acid because all you really need are the nitrate ions. The sulfuric acid acts by protonating nitric acid to yield nitronium and hydronium ions. If so, wouldn't any sufficiently strong acid, such as hydrochloric acid, do just as well.

As long as the Ka for that acid is higher than nitric acid's, it would work the same right?
Also, wouldn't very small amounts of nitronium ion be present without the sulfuric acid anyway?
Could one prepare nitrocellulose using only nitric acid if it were soaked for a much longer time?

The sulfuric acid acts by protonating nitric acid to yield nitronium and hydronium ions.

Electrophilic nitration reactions like the production of nitrocellulose require sulfuric acid because of the dehydrative properties of sulfuric acid. Sulfuric acid is a non-nucleophilic acid - all it can do is mediate proton transfers and dehydrate things (sulfuric acid has a large negative enthalpy change of mixing with water). Thus,$\ce{NO2+}$ is free to react with whatever you want it to.
$$1)\ \ \  \ce{H2SO4 + HNO3 <=> HSO4- + H2NO3+} $$
$$2)\ \ \ \ce{H2NO3+ <=> H2O + NO2+} $$
$$\text{overall)    }\ \ \ce{HNO3 + H2SO4 <=> H2O + NO2+ +HSO4-}$$
Hydrochloric acid and nitric acid produce aqua regia. Hydrochloric acid is a nucleophilic acid, and thus the decomposition products after proton transfer to nitric acid are different. The chloride ion is a good nucleophile, and reacts with the nitronium ion, eventually producing $\ce{NO2}(g)$ and $\ce{Cl2}(g)$. There is a big difference between $\ce{NO2+}$ and $\ce{NO2}$.
$$1) \ \ \ \ce{HCl + HNO3 <=> Cl- + H2NO3+}$$
$$2) \ \ \ \ce{H2NO3+ <=> H2O + NO2+} $$
$$3) \ \ \ \ce{2NO2+ + 2Cl- <=> 2NO2 + Cl2}$$
$$\text{overall)    }\ \ \ce{2HCl + 2HNO3 <=> 2H2O + 2NO2 + Cl2}$$

So nitrate salts can be used instead of nitric acid because all you really need are the nitrate ions.

Yes. However, starting with nitric acid shifts the equilibrium towards nitronium a little.

Also, wouldn't very small amounts of nitronium ion be present without the sulfuric acid anyway? Could one prepare nitrocellulose using only nitric acid if it were soaked for a much longer time?

Maybe, but not enough to be useful. Nitric acid is not available pure. It is only stable to a maximum concentration of 70-75% by weight (remainder water). The other reason that sulfuric acid must be used is that it is available "pure" (or at least 98% by weight) and it is so hydrophilic so as to sequester the water. Hydrochloric acid has a max concentration of 35-40% by weight remainder water. All the extra water gets in the way and shifts the equilibrium in the opposite direction (the reaction produces water).

Hey, I actually got all that! That would have made no sense had I not just completed my first proper chemistry course. God I love college. Thanks for your very complete answer. It explains everything I was wondering about and more.


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How to derive partial gas equation?

If the following gasses are to mix, what is the partial pressure of neon?

The correct answer is 0.1. I cannot figure out how my professor derived this equation. From the ideal gas law, all I can arrive to is this:
$P_x = \frac{P_tV_t}{V_x}$
This gives me an answer of 4.4, which is obviously incorrect.
I've determined the equation he used is as follows:
$P_x = \frac{P_xV_x}{V_t}$
which gives the correct answer of 0.1. How did he arrive to this answer?

you have $P_x$ twice in your second equation. Should one be $P_t$?

@BenNorris: That was done on purpose as that was the only way to get the correct answer. I listed it because it made no sense.

Mathematically, then both instances of $P_x$ would cancel and you would have $\frac{P_x}{P_x}=\frac{V_x}{V_t} \implies 1=\frac{V_x}{V_t}$, which is silly.

Right. That's why I was confused.

Prepare for an amazing feat of algebra! I sincerely hope there is a shorter way to the endpoint. (there is! See the end of my answer.)
To calculate partial pressure, $P_{\ce{He}}$, we need the total pressure $P_t$ and the fraction of the gas that is He $X_{\ce{He}}$: $$P_{\ce{He}}=X_{\ce{He}} P_t$$
You have $PV=nRT$. $T$ is not given, so assume it is constant. $n$ is unknown, but the system is closed, so $n_x$ for each gas is constant. Thus we have:
$$PV=\text{ constant}$$

Total Pressure

It is tempting to write variations of $$P_1 V_1 = P_2 V_2$$
However, for each gas, both $P$ and $V$ are changing, so we need to consider their products: $$(PV)_1 = (PV)_2$$
Thus, as trb456 suggests, $$P_t V_t = (PV)_t \implies P_t =\frac{\sum{(PV)_i}}{\sum{V_i}}$$

Fraction of He

The fraction of the mixture that is helium is determined by the ratio $\dfrac{n_{\ce{He}} }{n_t}$. Since $n=PV/RT$, and $R$ and $T$ are constant, we can write: $$X_{\ce{He}}=\frac{n_{\ce{He}}}{\sum{n_i}}=\frac{(PV)_{\ce{He}}}{\sum{(PV)_i}}$$

At last!

$$P_{\ce{He}}=X_{\text{He}} P_t = \left( \frac{(PV)_{\ce{He}}}{\sum{(PV)_i}}\right) \left( \frac{\sum{(PV)_i}}{\sum{V_i}}\right)$$
$$P_{\ce{He}_2}=\frac{(PV)_{\ce{He}_1}}{V_t}$$

Or, as it struck me as I finished:

Helium expands to fill the total volume. Now we can use $P_1 V_1 = P_2 V_2$ and completely ignore the other gasses (there is a lot empty space for the He atoms to fit in).
$$P_{\ce{He}_2}=\frac{P_{\ce{He}_1}V_{\ce{He}_1}}{V_t}$$
This equation is very similar to your second equation, except the subscripts are added to denote initial and final pressure of helium.

The total pressure of the system is:
$\frac{1.0\times0.5+1.0\times0.4+2.0\times0.2}{1.0+1.0+2.0}=1.3/4=0.325$
i.e. just the sum of the $PV$s divided by the sum of the $V$s.  Then $0.325$ is the sum of the partial pressures by the Law of Partial Pressures.  So this at least suggest that the individual partial pressures are near $0.1$.
Can you figure out how to parcel out the individual pressures from here?

Did you derive PV/Vt by using the ideal gas law? I've seen this used before and I think it's the answer to my problem.

Remember that, at constant temperature for the system, $PV$ is proportional to $n$, the moles of gas, by the Ideal Gas Law, yes. So you add up (something proportional to) moles, divide by volume, and you get pressure.  Now to allocate partial pressure, use this same idea of moles again!


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Oxidization of Copper In Electrolysis
So during school today, I am running an electrolysis experiment with water and sodium bicarbonate copper leads on both the anode and cathode.  I am noticing the usual $\ce{NaO}$ on the anode, and $\ce{H2C2}$ on the cathode.  However, on the anode, the copper is obviously oxidizing, and changing the hue of the water to a bright green-blue.  How can this be calculated into the chemical equation?
Also, is this equation correct?
$$\ce{H2O + NaC2 + energy  ->  NaO + H2C2}$$
(with the order of anode then cathode)
Yes, I am aware that this is the best possible situation, but in general, is this what is supposed to be happening?

@CodeAmdiral - you have written what looks like a valid chemical equation, but it appears unrelated to sodium bicarbonate and copper. You have sodium carbide on the left and sodium oxide or maybe peroxide on the right.

@BenNorris Yes, however, I don't get why the water is turning the colour of copper oxidization

Probably because copper is being oxidized. A good answer, which I may get around to posting, would compare the standard potentials of the various possible oxidation half reactions to show that, for example, copper is easier to oxidize than water.

Sodium compounds:

$\ce{Na2C2}$ is sodium carbide. It reacts violently with water. $\ce{Na2CO3}$ is sodium carbonate. $\ce{NaO}$ is not a valid formula for any of the oxides of sodium: $\ce{Na2O},\ \ce{Na2O2},$ and $\ce{NaO2}$ are all real compounds, and they all react with water to produce $\ce{NaOH}$, which is probably what is formed.

"Hydrocarbons"

If indeed you had $\ce{Na2C2}$, then you would form $\ce{H2C2}$ and $\ce{NaOH}$ when it was placed in water. However, you would not need electrolysis to do so. This is an acid-base reaction that is spontaneous under normal circumstances. $\ce{H2C2}$ cannot be the brown sludge on your cathode. It is a gas.
$$\ce{Na2C2 + 2H2O -> 2NaOH + H2C2 ^}$$

What actually happened?

In an electrolytic cell, the process that occurs (at least under standard conditions) is the one with the least negative standard potential $E^o_{cell}$.
The possible oxidations are either the oxidation of copper or the oxidation of water. Sodium carbonate is the supporting electrolyte.

$\ce{2Cu + 2OH- -> Cu2O + H2O +2e-}\ \ \ E^o=+0.36 \ \text{V}$
$\ce{Cu -> Cu^{2+} + 2e-} \ \ \ E^o=-0.34 \ \text{V}$
$\ce{4OH- ->O2 + 2H2O +4e-}\ \ \ E^o=-0.40 \ \text{V}$
$\ce{2H2O -> O2 + 4H+ +4e-} \ \ \ E^o=-1.23\ \text{V}$

Since the oxidation of copper to $\ce{Cu^{2+}}$ has the least negative $E^o$, it is the most likely oxidation. $\ce{Cu^{2+}}$ also explains the color, since aqueous solutions of $\ce{Cu^{2+}}$ are blue. The first equation is also likely, since it has a positive $E^o$, however it does not explain the color.
The possible reductions are the reduction of water/hydroxide and the reduction of $\ce{Cu^{2+}}$, since we know that it is being formed.

$\ce{2H2O + 2e- -> H2 + 2OH-} \ \ \ E^o=-0.83 \ \text{V}$
$\ce{Cu^{2+} + e- -> Cu+} \ \ \ E^o=+0.16 \ \text{V}$

The reduction of $\ce{Cu^{2+}}$ to $\ce{Cu^{+}}$ is most likely, but not initially. Initially, the is no $\ce{Cu^{2+}}$, so the only reduction that can happen at the cathode is the first one, which produces hydrogen gas. The redox then might be: $$\ce{Cu + 2H2O -> H2 + Cu^{2+} + 2OH-} \ \ \ E^o =-1.19 \ \text{V}$$
Once the $\ce{Cu^{2+}}$ diffuses over to the cathode, a different reduction starts to compete, with a different net redox reaction: $$\ce{2Cu + H2O -> H2 + Cu2O} \ \ \ E^o=-0.47 \ \text{V}$$
This redox reaction still produces hydrogen, but it has a much less negative $E^o_{cell}$. The other peculiarity is that $\ce{Cu2O}$ is produced at the cathode after $\ce{Cu^{2+}}$ has diffused through the cell. $\ce{Cu2O}$ is your brown sludge, since like rust, it is both brown and completely insoluble in water.


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Synthesis of hydrazine with bleach and ammonia
I am trying to make rocket fuel using hydrazine, hydrogen peroxide and home products. I want to put an action figure in orbit.
Is there any way to synthesize hydrazine with bleach and ammonia, using $\ce{H2SO4}$ as a catalyst?

On a side notes: 1) hydrazine probably can be purchased via chemical stores, if you wish. It is quite dangerous: it gives toxic fumes and can decompose violently if proper catalyst is available. 2)hydrogen peroxide in hight concentrations is extremely dangerous staff. It can ignite some substances in high concentrations and inflicts painful wounds several minutes after touch with skin. I strongly encourage you to use something less hardcore: rocket candy for initial steps and nitric acid/(kerosene or alcohol) for liquid-fueled rocket. Do not do the second without much consultations!!!!!111111

To underscore Permeakra's comment, if you don't have the proper experience, please consult someone that does.

related: https://en.wikipedia.org/wiki/Chemical_potential

I sympathize with your desire to launch an action figure. However, the chemistry you propose is really dangerous. You would be better off with black-powder based rockets.

Is there any way to synthesize hydrazine with bleach and ammonia?

Do not do this.
My answer is not "yes" or "no". My answer is Don't. Do not try this without proper safety equipment and training, and then think twice or three times before doing so, and then have an ambulance standing by. Certainly don't do it at home, or by yourself.
This particular reaction is capable of producing hydrazine. However, it also producing a whole host of additional stuff, like chloramine, hydroxylamine, nitrogen trichloride, hydrogen chloride, and chlorine gas, all of which are either gases or volatile liquids, and all of which are poisonous. Hydrazine is poisonous. Nitrogen trichloride is a shock-sensitive explosive on the order of a sneeze can set it off. The reactions are as follows.
Chloramine:
$$\ce{NH3 + HOCl -> Cl-NH2 + H2O}$$
Dichloramine:
$$\ce{Cl-NH2 + HOCl -> NHCl2 + H2O}$$
Nitrogen trichloride:
$$\ce{NHCl2 + HOCl -> NCl3 + H2O}$$
Hydrazine and HCl:
$$\ce{NH3 + NH2Cl -> H2N-NH2 + HCl}$$
Hydroxylamine and HCl
$$\ce{NH2Cl + H2O -> NH2OH + HCl}$$
Chlorine gas:
$$\ce{HCl + HOCl -> H2O + Cl2}$$
Any organic impurities in the mix can lead to other deadly stuff being produced.
This reaction is not controllable to produce a single specific product. Everything it produces is poisonous and/or explosive. I will recommend against, and I will not help you find a recipe.

Thank you so much. I will need some time for study and understand but moreover you told me I've read some information about it and I know why nasa stopped use it. Is quite dangerous and I haven't know and tools enough to manipulate those things. There will be better another alternative like black-powder based rockets.

@MirlvsMaximvs NASA has not stopped using hydrazine, and hydrazine and other propellants with similar hazards to hydrazine are still frequently used; it is still a general assumption of the space industry that hypergolic / storable propellants are toxic and/or corrosive. There has been only fairly minor experimentation with "green" nontoxic propellants. However, the space (and chemicals, and plastic foam) industries have the resources to work with this stuff safely.

Don't.  Hydrazine is toxic, explosive and a suspected carcinogen (http://www.cdc.gov/niosh/idlh/302012.html).  Nitrogen chlorides are infamous sensitive explosives (these can be side products of hydrazine synthesis); Pierre Dulong, who first prepared it, lost two fingers and an eye in two explosions associated with NCl3.  There have also been some completely unexpected explosions when bleach was mixed with an ammonia equivalent (urea):  http://www.chemaxx.com/explosion16b.htm .  Even rocket designers are starting to move to less nasty fuels than hydrazine.
And I'm guessing that you aren't working with the usual wimpy home concentration of H2O2 (i.e. 3%) but more concentrated hydrogen peroxide.  This stuff is pretty brutal, too, if you get it above 40% concentration.  I'd avoid both of these if possible.  However, if you are entertained by reading about the synthesis of really nasty propellants, you should read Ignition (http://web.gccaz.edu/~wkehowsk/ignition.pdf), it's really amusing.
I've used hydrazine in small amounts in the Wolff–Kishner reduction, but there's no way I'd try to synthesize the stuff.

Much safer than hydrazine is kerosene and liquid oxygen, especially if you build a miniature SABRE engine (SABRE stands for Synergistic Air Breathing Rocket Engine).
This is a jet engine that converts to a ramjet engine then a rocket engine as it gets higher and faster i.e. it launches its model plane as a jet engine from a runway and switches mode during flight. This would put your action figure into low earth orbit in his little plane.

It should be noted that: 1. Liquid Oxygen, while less hazardous that trying to make hydrazine, is also potentially quite dangerous due to its habit of converting anything and everything even slightly flammable into an incredibly touchy flammable or explosive mixture, 2. A SABRE engine is incredibly complicated above and beyond a simple hydrocarbon and oxygen rocket engine, which already typically is test-fired in a bunker, and 3. the design of liquid fueled rockets, quite apart from the propellants, is potentially quite expensive, tricky, and risky.

My advice would be to choose a different chemistry project, hydrazine is horrible. It is toxic and carcinogenic. In your liver it can form diazomethane which will methylate your liver.
There was a case of a man who drank a glass of what he thought was water while working in the engine room of a ship. Instead he drank a hydrazine solution, hydrazine is used in steam boilers in things like ships and power stations to manage the water chemistry. It serves to remove the dissolved oxygen.
This man ended up with serious liver damage which resulted in him going to hospital for a long time. Smaller doses of hydrazine can result in liver cancer.

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