text:
Are metallic/ionic bonds weaker than covalent bonds?
In mineralogy class, I was taught that metallic and ionic bonds are weaker than covalent bonds and that's why quartz and diamond have such a high hardness value. However, in organic chemistry class, I learned that covalent bonds are weaker than metallic and ionic bonds, thus organic substances have a much lower melting point than that of metals and ionic compounds.
What am I getting wrong? Are ionic and metallic bonds weaker than covalent bonds or not?

You're basically comparing an intermolecular force with an intramolecular force.

The problem with this "strength" is not well defined, and your teachers are using it in different sense.  When bond dissociation energies are compared, ionic bonds can be pretty strong, to0: https://opentextbc.ca/chemistry/chapter/7-5-strengths-of-ionic-and-covalent-bonds/

There is an additional confusion here as compounds consisting of covalent molecules don't make crystals based on covenant bonding: the crystals are formed from weaker dispersion forces (van der Waals forces) between the discrete molecules. This are definitely weaker than ionic bonds or the covalent bonds in network solids like diamond where every atom has a covalent bond to other atoms.

Quartz and diamond are stronger substances because their molecules form network covalent structures. These structures form a lattice-like structure, much the same as ionic compounds.
This molecular network is also the reason that diamond and quartz form a crystalline structures, just like you'd see in ionic substances such as NaCl. Some other structures you might want to look into are Graphite and Graphene, which are both allotropes of carbon (allotropes are, simply put, different molecular arrangements of an element).
The network structure combines to make the substance stronger than normal covalent bonded substances.
So to answer your question, substances with standard covalent bonds seem to be weaker than those with ionic bonds because the ionic bonds tend to form a lattice structure, that makes them much stronger. You can see this in the fact that the boiling points of ionic salts are much higher than that of a covalent substance like water. However, when covalent bonds form network covalent structures, atoms combine to form a singular macromolecule that is much stronger than singular covalent bonds.

Your explanation doesn't seem to explain what you are saying. You explain using the fact that ionic compounds form lattice structures in the solid state but you then write that the boiling points of ionic salts are higher. In the liquid state, ionic bonds have already been broken. So when you mention about the boiling point of ionic salts, there is no relation to strength of the ionic bond.

@TanYongBoon, I used boiling points because I felt they are most comparable to the true energy required to break the individual ionic bonds  (vs. melting points, at least). We tend to measure strength of ionic interactions by lattice energy, which is defined as the energy required per mol of ionic substance to convert the solid to constituent gaseous ions. This number, though, accounts for both sublimation energy AND individual bond enthalpy. Using boiling points, then, at least allows us to avoid comparing melting enthalpies and more directly compare bond enthalpies.

@TanYongBoon of course the lattice nature of ionic compounds necessarily complicates the comparison, though.

The last paragraph makes no sense.

What you learned in your mineralogy class was correct; bond strength decrease in the following order covalent > ionic > metallic.  The reasoning for this is as follows.  In covalent bonds such as those in methane and oxygen, the valence electrons are shared between the atoms involved in the bond and they (the electrons) spend most of their time in the region between the nuclei involved in the bond; this makes for a strong bond.  In ionic materials such as sodium chloride, the electrons are donated from one (the electropositive) atom to the other (the electronegative) atom in order for the atoms to achieve a filled shell structure.  The ionic atoms are attracted to one another through electrostatic attraction and the crystal lattices that are formed.  The bonds formed through electrostatic attraction are not as strong as those formed from covalent sharing of electrons.  Finally, in metals the outermost electrons are donated or "pooled" in the band structure that exists in metals.  The electrons are free to travel great distances (hence the conductivity of metals) and serve as a glue to hold all of the positively charged metal nuclei together.  So in the case of metals, there are no significant metal-metal bonds and these bonds are therefor the weakest.

As discussed in a different question, I respectfully disagree that ionic bonds are not as strong as covalent ones.

In the above-referenced question you note that the largest covalent bond strength is 945 kJ/mol in $\ce{N2}$.  While the ionic bond strength in $\ce{LiF}$, which should represent one of the stronger ionic bonds (see here) is only 577 kJ/mol.  So at least when comparing the stronger covalent and ionic bonds, covalent bonds are stronger.  Maybe you could provide BDE data for more average covalent and ionic bonds as well and we can see which are stronger in that realm.

I also included some comments on other covalent bonds. But I think this is a fool's errand. The great insight of Pauling was that any bond with different atoms will have at least some ionic/electrostatic component. Indeed an article I referenced in my answer suggests many bonds are strongly covalent and strongly ionic.

Some of this makes no sense. "there are no significant metal-metal bonds...so these bonds are therefore the weakest" is wrong as an observation and as an explanation.

It depends, because for covalent there are two types of bonds, network or molecular, or as I have also heard it be called, polar covalent and non polar covalent. But, network covalent consists of a vast network among the atoms and each one are connected, and they mostly made up of one element.
Take a diamond for example, it is only made up of carbon, but since the atoms are connected to each other and have not bonding between molecules, like something such as salt, which is an ionic bond, it is harder to break. However, if it were a molecular covalent bond, then the whole story is different, because they tend to be very weak bonds and easily broken like sugar or otherwise known as glucose, sucrose doesn't matter, it is still a covalent molecular bond because they have molecules while as a diamond is technically one big molecule.
But, since sugar has multiple the bond between each other molecule is weaker than the bonds between the elements themselves, then it is really weak.

Seems totally confused about why sucrose crystals are weaker than diamond.

This turns out to be a nonsense question.  Chemical bonds run the entire gamut from very strong to very weak as evidenced by the amount of energy required to break them.  Trying to say that bonds which are either ionic or covalent are stronger is a big mistake, starting with the fact that "ionic" and "covalent" are merely the hypothetical extremes of the bonding continuum and can be considered "ideal" bonds.  Real bonds lie along the continuum and have characteristics of both of the ideal bond types. Therefore, your original question has no place in the study of chemistry.

It is a returning issue due to the fact that basic courses (and the teachers of that) pretend to simply order bonds in terms of their strength. Add to this the fact that ionic bonds loose their strength in water (a working horse case for most if not all Chemists) and then you run into the fact that even here many argue that a covalent bond is in itself stronger than an ionic one.

Don't confuse the strength of bonds with the strength of the forces that hold crystalline solids together
There is a reason why the lessons you learned from chemistry are different from the lessons learned in mineralogy: they are not talking about the same things.
The problem is that mineralogy the bonds they talk about are the bonds that hold crystals together but in in chemistry what is often talked about are the bonds that hold the molecules together not the crystals made from the molecules.
This distinction is important. The vast majority of crystals in chemistry consist of discrete molecules held together by weaker intermolecular forces (sometimes called van Der Waals bonds). These are fairly weak compared to covalent bonds and result in crystals that are weak and have low melting points. So a chemist might look at compounds where the bonding in the molecules is covalent and make the generalisation that they typically form crystals much weaker than metals or ionic compounds. But that is because the bonds that make the crystals are not covalent.
A mineralogist will mostly look at compounds which are not made of discrete molecules but are made from ionic lattices or covalent networks (or both). There are no molecules of diamond, the crystal is a network solid held together by an (almost) infinite array of C-C covalent bonds, as silica is held together by an infinite array of O-Si-O bonds. Other minerals are a mix of the two with many silicates containing, for example, sheets of O-Si-O structures with a variety of ions in between. So, to a mineralogist, covalent bonds look strong compared to other types of bond. Ionic bonds are strong but not as strong as purely covalent network solids.
And the problem is further complicated by definitions of strength that are too narrow. Are metals stronger or weaker than diamond-like structures? It depends what you mean by strength. Diamond is harder than any metal but it is also more brittle. If resilience to being struck with a sharp object matters, choose a ductile metal object over a diamond any day. This happens because the crystal structure in some metals can absorb energy by reorganising crystal defects rather than by shattering bonds (pretty much the only option in silica or diamond). So, in one sense, metals are "stronger" than covalent solids like diamond.
The overall lesson is to be careful about definitions. There is no good generalisation of crystal strength based on bond types. Be careful whether you are talking about the bonds within the components of the crystal (molecules) or the bonds that hold those components together (many "covalent" compounds consist of crystals where molecules are held together by much weaker forces). Don't forget that many minerals have both ionic and covalent bonding. And be specific about what you mean by "strength" (eg hardness and resilience to impact are not the same thing).

Ionic and metallic bonds are weaker than covalent bonds. This is correct, it is why covalent crystal is much harder than ionic and metallic crystal/polycrystal.
The second statement is wrong because firstly melting point is not proportional to the strength of chemical bond. There are more factors such as flexibility of molecules. Boiling point is more proportional instead.
More important, the inter-particle forces to be compared among the organic compound vs ionic vs metallic compound is NOT among covalent bond vs ionic bond vs metallic bond. It is among intermolecular force (dipole-dipole, H-bond, Van der waals) vs ionic bond vs metallic bond. And the first one is much weaker than the second and third for sure. So the boiling point of organic compounds is much lower.

I am not sure what is the consensus that has been arrived at by chemists around the world but I would just like to offer my two cents' worth on the issue. This question has always been a question which my teachers would always tackle when they teach chemical bonding and their answer has always been the same:

It is not fair to make a comparison since these bonds are ultimately very variable in terms of strength.

I do agree with that but allow me to provide my perspective to this issue.
The strength of covalent bonds in simple molecular substances (as well as those in giant network structures) can be easily determined. Thus, the bond energies of most covalent bonds are known well and they can be easily used for such comparisons of bond energy. However, the strength of ionic bonds and metallic bonds is not so clear-cut.
By definition, the ionic bond is the electrostatic force of attraction between positively and negatively charged ions in an ionic lattice while the metallic bond is the electrostatic force of attraction between the positively charged metal ions and the surrounding electrons.
In an ionic lattice, there are so many ions interacting electrostatically with each other. How then can the strength of the ionic bond be determined? The idea of lattice energy could be used but comparisons made using lattice energy would only be of any sense when we are comparing between ionic lattice. It cannot be used to compare with covalent bonds!
Consider the bond dissociation energy of $\ce {H-H}$ and the lattice energy of sodium chloride. The BDE of $\ce {H-H}$ is $\ce {+ 436 kJ/mol}$ while the lattice energy of $\ce {NaCl}$ is $\ce {+ 786 kJ/mol}$. Both are in terms of "per mole of something". But that "something" is different in each case. In the case of hydrogen, that "something" would be the $\ce {H-H}$ bond but in the case of the ionic compound $\ce {NaCl}$, that "something" is the $\ce {NaCl}$ formula unit. And that is not the same as "per mole of ionic bonds between $\ce {Na^+}$ and $\ce {Cl^-}$". The ionic bond strength is not so easily determined because each ion is in an electrostatic environment which is influenced by all the other ions around it. The same idea can be applied to metallic bonds.
In essence, my take is that there is no basis of fair comparison between metallic, ionic and covalent bonds in terms of their bond strengths.


text:
How to calculate the number of moles present in a litre of saturated solution?
The solubility product of a salt having same molecular formula as $\ce{Ca_3(PO_4)_2}$ is $1.08\cdot10^{-28}$. Then find the number of moles of ions present in the saturated solution.
My approach: I found out the solubility of the salt i.e.: $x=10^{-6}M$
This means $x$ moles of salt are present in a litre. As 1 mole of the salt yields 5 moles of ions I thought the answer would be $5x$.
Help required.

Did the given answer not match your answer?

What is the solubility product equal to?

I don't know the answer. It is a MCQ question. The options did not match my answer.

Perhaps you have to consider hydrolysis of the phosphate anion.

@Rudstar  Your approach looks correct to me.  It would be helpful if you provided the actual question and the answer options.  For example, was $\ce{K_sp}$ given to you or did you look it up?  What salt was used in the problem?  Did they want the answer in moles/L or gm/L?

For your salt, calcium phosphate, or $\ce{Ca_3(PO_4)_2}$, this is the correct dissolution equilibrium:
$\ce{Ca_3(PO_4)_2 \leftrightharpoons 3Ca^{2+} + 2PO_{4}^{3-}}$
Upon dissolution of one unit of calcium phosphate, we get five units of ions total, true, but the solubility product is defined as follows (from the IUPAC Gold Book):
The product of the ion activities raised to appropriate powers of an ionic solute in its saturated solution expressed with due reference to the dissociation equilibria involved and the ions present.
For our purposes, we will use concentrations instead of activities. Concentrations are a rough gauge of activities in very dilute solutions (such as this one involving the semisoluble salt, calcium phosphate).
Also since you mentioned this was a multiple-choice test, I highly doubt that the following hydrolysis reaction will be relevant, even though at such low concentrations of phosphate anion, the phosphate anion should be nearly completely hydrolyzed.
$\ce{PO_{4}^{3-} + H_2O ->HOPO_3^{2-} + HO^-}$
Back to the problem. Upon dissolution of one unit of calcium phosphate, we should get 3 units of calcium ion and 2 units of phosphate ion.
Therefore, based on the IUPAC definition:
$\ce{K_{sp}~=~[Ca^{2+}]^3[PO_{4}^{3-}]^2}$
Taking $\ce{x}$ to be one unit of calcium ion:
$\ce{K_{sp}~=~[Ca^{2+}]^3[PO_{4}^{3-}]^2}=[3x]^3[2x]^2$
Solving the above yields $x = 10^{-6}$.
Therefore, with respect to your original question, the number of moles of calcium phosphate present initially can be number greater than 0 moles. For the amount of calcium phosphate present at equilibrium, then yes, $5x$ appears to be the right answer.

Doesn't that just give the same answer for x that Rudstar reported above?

I'm fairly sure it shouldn't because it appears he set $K_{sp}$ equal to $5x$. But then I haven't run through the calculations myself.

Oops, I see that I misinterpreted the question because I didn't read the title. Sorry!

source: Stack Exchange

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

sort: votes


text:
Does a patina on a knife prevent rust?
Before I begin, I think it's worth mentioning that I know little about chemistry and even less about knives.
My question is this, having seen many people talk about how patinas work and the best ways of applying them and which ones are best etc., It seemed to me like people just assumed that they work, presumably because they heard someone else do it, and them from someone else and so forth but that doesn't always make it true. If anyone has any idea If there is any reasoning behind this and if there any flaws behind the logic in it that would be greatly appreciated.
I don't exactly want to ruin or change my new knife for no reason. I assume people might think that this works similar to the thin oxide coating on aluminium, but my guess is that the corrosion acts like it would on any metal and slowly "eat away" at it.
So, simply, does it work?

In this case, the use of the word patina is a little misleading. Patina refers to the oxidation film on bronze or aged polish. I assume that you mean that the knife is coated with a substance.
First, when steel oxidizes, the iron oxide that forms does not stick well and will flake off the surface. This would not be useful for a knife since the edge will become dull quickly.
Second, if the knife is made with stainless steel, it does not need to be coated to prevent rust, only keep reasonably dry. Some stainless steel will actually rust since it was not properly formulated. Very few knifes are made of other metals since stainless steel is reasonably cheap.

If I'm not mistaken, the use of the word patina is not limited to bronze. And yes, I did mean forced oxidation (at least that's what I think it is). Sorry, I did forget to mention that I was referring to high carbon non stainless steel. While I would have thought the same too, that it flakes off, I haven't seen any of them that do that. It usually feels the same but with a weird discolouration. Don't know If any of this will affect your thinking though.

Patina in corrosion terminology is limited to pure well adhering transparent oxide layers  which is rarely formed on iron alloys- special conditions, its common for copper alloys such as bronze and zinc including galvanised steel (zinc coated).

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


text:
Clarification on vapor pressure
If I have a glass of water, and the vapor pressure is the point of equilibrium where the liquid molecules are becoming gases and the gas molecules are colliding and becoming liquid at the same rate, does that mean none of the liquid is lost? I feel as if I am misunderstanding this concept because when I leave my cup of water on the table and come back to it later on I usually notice that there's less water in the cup because it evaporated.

Yes, the vapor pressure is the point of equilibrium between the liquid and the vapor. But air moves about so the liquid never saturates the air and an equilibrium is never reached. So more and more water evaporates until it is all gone.


text:
Magical reaction tanks
I would be interested to know people's opinion on the following matter.
Background
Often, when the far-from equilibrium behaviour of a chemical reaction system is analysed mathematically (in papers or text books), the system is placed in a gradient between a fixed non-zero concentration of "resource" species, and a fixed concentration of "waste" species (which is often zero). The other species in the reaction system are designated as "intermediates", and it is their behaviour which is of prime interest.
For example, for the oscillating Lotka-Volterra reaction system:
$R + X \rightarrow 2X$
$X + Y \rightarrow 2Y$
$Y \rightarrow W$
Species $R$ and $W$ are the fixed concentration resource and waste species respectively, and species $X$ and $Y$ are the (oscillating) intermediates.
Question
It appears as if these reaction systems are being studied in a physical scenario which could not be realised in the lab, a magical reaction tank if you like.
The two reasons most often stated as to why resource and waste are held at constant concentrations, whilst the intermediates can vary, are that:
1) The reaction takes place in a large reservoir, where resource is abundant and where any waste produced by the reaction quickly diffuses away into the vastness
or
2) The reaction has resource species quickly "pumped in" by some external agent, and likewise waste species quickly "pumped away".
My problems with these explanations are as follows:
If in explanation 1, a very large reservoir is keeping resource and waste species at constant concentration, then if follows that the intermediate species can never reach any appreciable concentration, for they too exist in the same large volume. In other words, if the waste diffuses away, then why don't the intermediates?
Conversely, in explanation 2, if a pumping process is artificially maintaining resource and waste concentrations, it is hard to see how this pumping process is being "micro selective" in completely removing the waste compounds, but leaving the intermediate compounds totally unaffected.
More physically realistic models of chemical reactions can be made by also explicitly including the reactor in the model - for example by considering the reaction as taking place in semi-permeable compartments or in bulk in a CSTR. However, additionally modelling the reactor often changes the mathematical properties for the "bare" reaction reported in the literature (for example, in a CSTR, the intermediates as well as the waste are removed).

I would say it's worth investigating. My suspicion is that the answer won't change much, but the way to find out is to try it both ways and see how the assumptions affect the answer under various conditions.
Instead of a CSTR, for this particular reaction a batch reactor might be more appropriate, since that is the easiest setup for a lab. In that case, to model the reaction all we have to do is not assume that some of the rates of change are zero, and see how that changes the reaction.
Fortunately for us, this site has already done the heavy lifting and even has a solution in a mathcad file.
I haven't looked at the solution closely enough to determine if they are making the approximations you mentioned (I don't have mathcad installed), or if they are just directly integrating all four equations. They only show graphs for the concentrations of X and Y. However, it probably wouldn't be too hard to modify in either direction to compare the results under different sets of assumptions.

Thanks. Two points: First, from what I can see, a batch reactor reactor seems the same thing as a CSTR. Second, on the site you suggest they are holding species A constant for the LV reaction "A is continuously replaced from an external source as it is consumed in the reaction", and they are forgetting about waste B. Therefore, my question about how to ensure A and B have fixed concentrations in an experimental setting still stands.

To add a note about the LV reaction: the paper Properties of two‐component bimolecular and trimolecular chemical reaction systems states that the LV reaction is not "structurally stable", i.e. small changes to its equations of motion (such as those introduced by modelling the reactor) generally destroy its capacity to have conservative orbits.

The main difference between a CSTR and a batch reactor is the flow - a batch reactor has no inlet or outlet flow, a CSTR does. You are right about the solution on the site I linked assuming a fixed concentration of A - you would have to modify the mathcad file (or write your own numerical integration routine) to solve the equations without those assumptions, then compare the results. I didn't catch your question about fixed A & B in the original - I thought you were asking whether those assumptions would have an effect. My suggestion is to try it and find out.


text:
Enthalpy change of reaction
Suppose you have the following equation:
$$\ce{2NH4SCN(s) + Ba(OH)2(s) ->Ba(SCN)2(s) + 2H2O(l) + 2NH3(g)}$$
You initiate this reaction using 15.22g of the $\ce{NH4SCN}$ (and excess of the barium hydroxide) in a boiling tube resting in 50g of water. The temperature rises by 11 degrees. Therefore energy = $4.2 \times 11 \times 50 = 2.31kJ$ and moles = $\frac{15.22}{76.1} = 0.2mol$
I would then assume that it would just be $\frac{2.31}{0.2} = 11.55kJmol^{-1}$
But the actual answer is 23.1 (double my answer) which I am guessing is because there are 2 moles of $\ce{2NH4SCN}$. I do not understand why this is the case, because you are using 0.2 moles and the energy raised is 2.31 so why the multiplication by 2?

What is 15.22 g?

sorry i have updated the question

The reason lies with the definition of the Heat of Reaction. It states that it is the amount of heat when 1 mole of the product is formed. As you noted, since 2 moles of $\ce{NH4SCN}$ is needed for each mole of product, then you need to multiply by 2.
Added:
If you change the number of moles of $\ce{NH4SCN}$ to 0.1 moles (to remove the stoichiometric 2), then your calculations would give you the answer.

I'm not sure I understand then what the moles value I get when I calculate mass divided by rel. molecular mass? - surely that would be the amount of moles used in the reaction?

never mind I understand!

I know how it works, but I have a hard time to express it. What you want is to have all the stoichiometry as 1. In order to keep a balanced equation, you need to modify the number of moles in the reaction (in this case you want half the ammonium cyanide, but the barium cyanide would be the same).


text:
Why can Cu have an oxidation number of +2?
The electronic configuration for $\ce{Cu}$, adjusted for Hund's rule, is:
$[\ce{Ar}] 3d^{10} 4s^1$
So, shouldn't $\ce{Cu}$ have an oxidation number of $+1$? whereby it gives off its outermost 4s electron? Why can it also have a $+2$ charge?

Copper can form compounds with an oxidation number of +1 but you mean why does it more commonly form compounds with an oxidation number of +2.

Atomic copper has the electron configuration $\ce{[Ar] 3d^{10} 4s^1}$.  By removing one electron and producing $\ce{Cu^{+1}}$, an inert gas configuration $\ce{[Ar] 3d^{10} 4s^0}$ is produced.  While it does take a lot more energy to remove the second electron from copper (first IP=745 kJ/mol, second IP=1,958 kJ/mol), if this energy can be offset by the energy gained through bond formation and lattice energy (or hydration energy) then $\ce{Cu^{+2}}$ compounds will form.  In fact $\ce{Cu^{+2}}$ is the most common oxidation state of copper so the energetics must generally work out that the energy gained by forming more than one bond to copper and gaining additional lattice (or hydration) stabilization, more than offsets the energy cost of removing that second electron.  Removal of additional electrons can also occur with copper to form $\ce{Cu^{+3}}$ and $\ce{Cu^{+4}}$ compounds, for example, $\ce{K3CuF6}$ and $\ce{Cs2CuF6}$.

Nice answer. Is this compensation of copper hydration energy for the ionisation energy also found in other elements? Like it's group members silver and gold? Why or why not?  Thanks!

Also,is Cu(+) more stable than Cu(2+) in gaseous or non-aqueous media due to lack of hydration? Thanks


text:
Increasing mass
I'm watching a video, and the speaker states that the molality formula is better than the molarity formula because it is more constant, since molarity deals with volume and molality deals with mass. If someone adds more solvent than the volume changes, so the answer in molarity changes as well. But if more solvent is added, doesn't the mass increase as well?

You can never really say one's better than the other; they're different things. Mind providing a link to the video? In both formulas, adding solute would change the value.

I think someone may have made a slip of the tongue there: if you add more solute, the volume of the solution will change, without changing the mass of solvent present. If you just add more solvent, of course you'll change both the molarity and the molality.
Let's do a bit of a thought experiment to illustrate this: I have some toluene and I'd like to dissolve a whole bunch of tri-octylphosphine sulfur into it (don't worry if you don't know what these are, just remember that toluene is the solvent and TOPS is the solute).
Let's say I want to make a 1 molal solution. The definition of one molal is a mole of solute in a kilogram of solvent. I measure out one mole of TOPS and dissolve it in 1 kg of toluene. Done!
Now I want to make a 1 molar solution. The definition of one molar is a mole of solute in a liter of solution. I might be tempted to measure out one liter of toluene, dissolve the TOPS in it, and call it a day.
I would be wrong.
The reason is because one mole TOPS takes up a certain amount of volume. A rough estimate is that it takes up about a third of a liter. My solution is now one mole of TOPS in 1.33 liters of solution, making it 0.75 molar instead of 1 molar!
I can fix this by starting out with the TOPS, then slowly adding toluene until the total volume of everything is 1L. This will give me a 1 molar solution.
You can extend this idea a little further to see that adding solute to a solution makes its molarity a mess to calculate unless you pretend that the solution volume doesn't change. The molality, meanwhile, is easy, because the mass of solvent hasn't changed a whit.
EDIT: As Martin points out, molality has an additional benefit in that solutions naturally expand as they heat. If we heat a solution up, its volume will increase, and its molarity will change. The molality, however, will not, since the mass of solvent does not depend on temperature.

I got a bit carried away explaining this. Sorry...

It is worthwhile to state, that mass does not change with the temperature, while volume usually does.


text:
Max Ionization Extent

How is this formula derived? What does it mean; what is the use of this formula?
I know that at infinite dilution, ionization extent should be very high because of Le Chatlier's principle.
But I'm not seeing how this formula is derived and its significance in actual chemistry. I tried rewriting the formula as:
$\mathrm{\frac{K_a}{K_a+\sqrt{K_w}} \cdot 100\%}$
But this isn't elucidating where the formula came from or its significance.
I could also rewrite the formula as this for the hypothetical acid, $\ce{HA}$:
$\mathrm{\frac{K_a}{K_a+\sqrt{K_w}} = \frac{\frac{[H_3O^+][A^-]}{[HA]}}{1.0 \cdot 10^{-7}+\frac{[H_3O^+][A^-]}{[HA]}}}$
I think the trouble I'm having is defining maximum ionization extent. For example, I know that in general ionization extent is defined as (where x is the concentration of the derived species - i.e. $\ce{[A^{-}]}$ for the hypothetical acid $\ce{HA}$ and $\ce{M_{i}}$ the initial molarity of the hypothetical acid.
$\mathrm{\frac{x}{M_{i}}}=\text{ionization extent}$
I have a feeling that the $1.0*10^{-7}$ term refers to the concentration of one of the products of the auto-ionization of pure water at 25 degrees Celsius. This makes sense because at infinite dilution the concentration of solute should be negligible (=0).
Upon further thought ...

I've been looking on the web for information and came across this: http://en.wikipedia.org/wiki/Law_of_dilution

Thanks, but I don't see the relationship between that article and my question. Interesting reading though!

For every reaction with an incomplete dissociation process in aqueous solution, you have to consider water's autoprotolysis.
\begin{aligned}\ce{
(1)&&HA + H2O &<=> H3+O + A^{-}\\
(2)&&2H2O &<=> H3+O + {}^{-}OH \\\hline
(3)&&HA + 3H2O &<=> 2H3+O + A^{-} + {}^{-}OH
}\end{aligned}
The degree of dissociation is usually defined as
\begin{aligned}
&&\alpha &= \frac{c(\ce{A^{-}})}{c_0}&\text{with}&&c_0&=c(\ce{A^{-}})+c(\ce{HA})\\
\therefore&&\alpha &= \frac{c(\ce{A^{-}})}{c(\ce{A^{-}})+c(\ce{HA})}&\text{with}&&
c(\ce{A^{-}})&=K_a\frac{c(\ce{HA})}{c(\ce{H3+O})}\\
(4)\therefore&&\alpha &= \frac{K_a}{K_a + c(\ce{H3+O})}
\end{aligned}
At infinite solution, the concentration of hydronium ions will be governed by the autoprotolysis $(2)$ and can therefore be rewritten as
\begin{aligned}
&&K_w&=c(\ce{H3+O})c(\ce{{}^{-}OH})\\
(5)\therefore&&c(\ce{H3+O})&=\sqrt{K_w}
\end{aligned}
At $25~^\circ{}C$ you will find your given equation:
\begin{aligned}
\alpha[\%] &= \frac{K_a}{K_a + 1\cdot10^{-7}}\cdot100\%
\end{aligned}

This is a masterful derivation. Thanks Martin. I just showed my friend this derivation and it made more sense than a book's derivation.


text:
What is some advance reference books about alloy that gives a lot historical background and applies a lot of theoretical equation or formulas
What is some advance reference books specifically about alloy that gives a lot historical background and applies a lot of theoretical equation or formulas with explanation?
I seem to find nothing really very interesting for myself to learn general theory of alloy while searching the google books because these books is too wordy in detail but too few formula and laboratory skill to support it.

See if your library can get you something like http://www.amazon.com/Materials-Science-Engineering-William-Callister/dp/1118324579/, it's an extensive reference on a number of different materials.

You aren't going to find a lot- general metallurgy doesn't have a strong theoretical basis, its very experimentally based. Particular groups of alloy have a theoretical base but its only in expensive and university and propriety books.  A lot is fitting theories to experimental data.

The classic would be Porter and Easterling's "Phase Transformations in Metals and Alloys", apparently still available. This should give you a good basis for understanding the thermodynamics of binary (and higher) systems.
For a deeper look at how computational thermodynamics is done, like CALPHAD, the Lukas, Fries, and Sundman book "Computational Thermodynamics" might do, but that has zero historical background.
You might also try asking this in Physics using the materials-science tag.


text:
How do we know the size of atoms in body centered cubic system?
I'm told that in the normal cubic system there are $\frac18$ of an atom at each corner and these atoms' radius are such that $a = 2r$ ($a$ is the lattice constant, $r$ radius of an atom).
In bbc (body centered cubic system) they say the atoms in the diagonals are touching, and as a consequence the atoms at each corner are not big enough to satisfy the above relation $a = 2r$
Why is this? Couldn't it be like the normal cubic system but stick an atom in the center to fill up the remaining space? How do we know that they touch along the cube's diagonal and not along the edges?

The easiest answer in my opinion; try it! If each atom has the same radius, it is simply not possible to fit one of them into the center of the cubic system (known as a cubic hole). It is, however, possible if the atom is sufficiently small. CsCl is an example of such a compound.
One can show this geometrically (as is done here), but you can also convince yourself that it must be the case using equally-sized tennis balls or something similar to try it out in practise.

Consider the body centered cubic unit cell: It's either that the ions are all the same size (in that case they touch on the cube's diagonal) or that they touch along the edges (in that case the atom in the center needs to be smaller). Is my reasoning correct? Also, if we take a tetragonal unit cell, the atom in the center needs to be even smaller if we want all of them to be touching. Why is it a requirement for the atoms in a unit cell to touch? Please elaborate if you can.

It seems to me that your reasoning is correct, but I would make a distinction between the BCC and the situation where there is a smaller ion in the center of a simple cubic system. In the latter case, I would call this a cubic hole - in the first, it's just a part of the BCC. I guess this in some respects can be regarded as a convention.

When it comes to the case of atoms touching, I think one must make a distinction between a) neutral atoms in a structure and b) ions. In the latter case, touching can not be allowed, because this would lead to same-charge interactions (unfavourable energetics). In other words, the ion in the hole must be bigger than the hole itself - in must not be allowed to "rattle" around. In the former case, if the atom in the hole is bigger, the structure no longer has a closest (optimal) packing, which would not happen in practise.

Well, i still don't get what's the relationship between Bravais lattices, coordination numbers and polarization (when the size of the atoms are not identical in the unit cell - i couldn't find any reference to this one in english, all the search results are pointing to optical crystallography and polarized light, but i'm referring to the phenomenon known from organic chemistry as polarization). It's off topic, but could someone explain briefly or at least point me to a resource about the connection between Bravais lattices, unit cells and coordination numbers. It seems to me they contradict.

Can you elaborate on how you think they contradict; perhaps phrasing this as a new question? It's probably easier to follow than a comment field discussion :) That said, I guess Bravais lattices are just a way of categorizing crystal systems. The coordination number (in this context) basically refers to the number of nearest neighbors an atom/ion has in a given position (ie. an ion in a cubic hole has a CN = 8, an atom in CCP has CN=12 etc.)


text:
How can a phase be thermodynamically stable at room temperature if it only appears at high temperature?
How can a phase that appears at high temperature be thermodynamically stable at room temperature? For example, $\alpha$-alumina is a phase that is formed at temperatures above $1000^\circ C$. How is the same phase stable at room temperature?

If it is a high temperature phase then it isn't thermodynamically stable- the kinetics of transformation are very very slow for solid to solid transformations at low temperatures.

$\alpha$-alumina is thermodynamically stable at room temperature, and also at high temperature (around $1,200-1,400 \space ^{\circ}C$) - however, alpha alumina is not what is typically formed when alumina is made from the various aluminum ores (ref). Typically, one of the several other crystal structures will form, with different particle sizes and amounts of water (hydration). These and all other phases that are found along the way to $\alpha$-alumina are called transition aluminas.
This link explains what happens upon heating. In general, first the crystal is dehydrated, which causes it to form a different, dehydrated phase. Then, as the temperature continues to rise, the alumina crystal structure rearranges into more compact structures, until eventually, near the sintering point, the atoms have sufficient kinetic energy that they can "find" the most thermodynamically stable configuration - $\alpha$-alumina.
In other words, the transition alumina phases are kinetically trapped in structures that are not the most thermodynamically stable, but heating allows them to get out of that trap and into the most favorable state.

As far I have read- alpha-Alumina is stable at room temperature. They just sinter alumina powder at 1000C so the mostly physical reaction happens quickly.


text:
How to break down soap and detergents?
I am building a greywater harvesting system. It is a 3-stage system:

The 1st stage breaks down soap and detergents.
The 2nd stage filters the water.
The 3rd stage pumps water out of system.

What is the best way to break down soap and detergents in a timely manner with as few mechanical parts as possible?

When you say "break down" - do you mean chemically? What are the products that you looking for?

I highly doubt there is anything you can do in a household graywater system that would destroy detergents. Those tend to be pretty stable. You might have more luck with a skimmer if the detergents float to the top. But you'd probably have more luck asking people who do graywater systems.

|A lot of water.
|Oxygen bleach.(H2O2).
|Organic salts.
|CLR cleaner.


text:
Molality and dissociation
For $\ce{HCl(aq) -> H^{+}(aq) + Cl^{-}(aq)}$, if $\ce{HCl}$ has a concentration of 1 molar then $\ce{H+}$ and $\ce{Cl-}$ also have 1 molar. But for $\ce{HF(aq) <=> H^{+}(aq) + F^{-}(aq)}$, if $\ce{HF}$ has a concentration of 1.5 molar, then why doesn't $\ce{H+}$ and $\ce{F-}$ have 1.5 molar as well?

The short answer is because $\ce{HF}$ doesn't dissociate as much as $\ce{HCl}$.  Let's examine why this is by introducing some terms like $\ce{pH}$ and $\ce{pK_a}$.

$\ce{pH}$ does not measure the strength of a given acid, but rather the acidity of a given solution.
$\ce{pK_a}$ does measure the relative strength of an acid, look at the following equilibrium

$$\ce{H2O + HX<=>H3O^+ + X^-}$$
$$\ce{so~~~K_{eq}=\frac{[H3O^+][X^{-}]}{[H2O][HX]}}$$ but in dilute solutions the concentration of water stays constant so we can rewrite this equation as
$$\ce{K_{a}=\frac{[H3O^+][X^{-}]}{[HX]}}$$ and finally, $$\ce{pK_{a}~=~-Log_{10}K_{a}}$$ Now we can see that the more highly dissociated our acid $\ce{HX}$, the higher the concentrations of $\ce{H3O^+}$ and $\ce{X^{-}}$, and in turn the larger the value for $\ce{K_{a}}$, and the more negative the value of $\ce{pK_{a}}$.
Bottom line: highly dissociated acids have large negative $\ce{pK_{a}}$ values
Now, here is a Table from SparkNotes.  It compares the $\ce{pK_{a}s}$ (degree of ionization or strength) of the various haloacids.  You can see that as we move down the Table from $\ce{HF}$ to $\ce{HI}$ the $\ce{pK_{a}}$ becomes much more negative; our acids become much more dissociated in water.  So we see that $\ce{HF}$ is not very dissociated ($\ce{K_{a}=10^{-3.1}}$, equilibrium lies to the left). On the other hand, 1 mole of $\ce{HCl}$ will produce about 1 mole of $\ce{H^+}$ ($\ce{K_{a}=10^{+7}}$, equilibrium lies far to the right).  Note too that the $\ce{pK_{a}s}$ of these acids correlate with their bond strength which makes sense.  As the $\ce{H-X}$ bond gets weaker, it becomes easier for the acid to ionize (dissociate).

It is noteworthy, that the acidity of a HF solution strongly depends on its concentration increasing due to homoassociation $\ce{2HF <=> [FHF]- + H+}$.

HF isn't a strong acid; its dissociation isn't complete in water, unlike HCl's dissociation.
The reason HF isn't a strong acid while HCl is a strong acid has to do with charge density of the conjugate base. Charge density controls the reactivity of the conjugate base; charge is what governs chemistry and physics.
You'll find that the $\ce{F^-}$ anion is fairly reactive relative to $\ce{Cl^-}$, which, in water, is essentially inert. Both are halogens; both have the same magnitude of charge; the difference is that the fluoride anion is smaller than the chloride anion. Therefore, the real negative charge density on the $\ce{F^-}$ anion is greater than that of the chloride anion, and this effective negative charge density is what drives the reactivity of the fluoride anion. In other words, the below reactions occurs to significant extent due to the concentrated negative charge on the fluoride anion:
$\ce{F^{-} + H_2O <=> HF +HO^-}$
This above reaction is antagonistic to the dissociation of $\ce{HF}$ in water because the above reaction makes more $\ce{HF}$! Note that $\ce{K_{b}(F^{-})}$ $=6.6\cdot 10^{-4}$ while $\ce{K_{b}(Cl^{-})}$ $=8.0\cdot 10^{-21}$! So there you have it; empirical evidence that chloride ion is a very unreactive species in water solution.
You'll find that with the binary halogen acids, as you go down group 7, the acids get stronger. For example, $\ce{HBr}$ is an even stronger acid than the already strong acd, $\ce{HCl}$. This again can be rationalized using effective negative charge density; $\ce{Br^-}$ has the same magnitude of negative charge as $\ce{Cl^-}$ but in the bromide anion the negative charge is spread out over an even larger area, making bromide ion all the less reactive than chloride ion and hydrobromic acid stronger than hydrochloric acid.
I hope this answers your questions and gives you some insight into acids.


text:
How to prove that reaction is possible
I have been given task to write possible reaction equations, determine oxidizer, reducer, write corresponding electron-equations and calculate electromotive force (I do not know if that is what you call it in English) of reactions:

Iron reaction with diluted sulfuric acid
Aluminium reaction with solution of sodium hydroxide

Could you be so kind and help me with this task?

This looks like it might be homework. We don't directly answer homework questions (certainly if no effort has been attempted). Have you tried searching the web with the reagents?

Welcome to Chemistry Stack Exchange! Please add what you have attempted towards solving the problem into the body of your question.  For more information, see the site's homework policy for how to ask homework questions. You were lucky to receive an answer this time, but please adhere to this practice in the future.

The ionic equation for this reaction is:
$\ce{Fe + 2H+ ~(+\;SO4^2- ) -> Fe^2+ + H2 ~(+\;SO4^2- )}$
As two half-reactions, it is seen that the iron is oxidized:
$\ce{Fe -> Fe^2+ + 2e- }$
The reductant (iron) transfers electrons to hydrogen, and is thus itself oxidized.
And the hydrogen is reduced:
$\ce{2H+ + 2e- -> H2}$
The oxidant (hydrogen) removes electrons from iron, and is thus itself reduced.
To calculate the reduction potential you have to solve the following equation:
$\Delta E = E_\text{acceptor} - E_\text{donor} = E_{\ce{2H+/H2}}- E_{\ce{Fe/Fe^2+}}$
$\Delta E = 0\ \mathrm{V} - (-0.41\ \mathrm{V}) = 0.41\ \mathrm{V}$
The reactions runs voluntarily if $\Delta E > 0$.
You can do the same for b).


text:
Is it possible to synthesis adenine, thymine, cytosine, or guanine at home?
I'm wondering if there is any way to create adenine, thymine, cytosine, or guanine from chemicals available to anyone, using basic lab equipment.

This isn't a full answer, as you'd still have to figure out a way to break down the molecules further, but there are lots of protocols for extracting the DNA from onions, bananas, etc. http://www.sciencetheatre.org/mediawiki-1.16.5/index.php/Banana_DNA

Based on jonsca's comment, extract DNA from something. The protocol will involve detergent and isopropol alcohol. The details are available in several places and is fairly simple, but you won't get much DNA per onion, so be ready to prepare a lot of them. To do the degradation, there are several things you could try. If your detergent contains nucleases ( many contain enzymes but do a poor job of labeling which enzymes ) you could mix that in and react it at 37C for a while. Can also try high pH solutions, if you happen to have NaOH around. If you do the DNA extraction differently, you can get the RNA instead of DNA. RNA is very unstable at high pH, and RNAse enzymes are EVERYWHERE. So the RNA should degrade into nucleotides pretty readily if you just use sloppy lab technique. When I extract RNA in the lab, I use phenol/chloroform to pull the proteins and genomic DNA away from the RNA, which stays in the aqueous layer. You probably don't have access to those at home, but cooking oils may be hydrophobic enough to pull it off. Also, it helps to have a centrifuge.
Don't ask me how to purify the nucleotides from your solution and ask me even less about how to verify what you have at home. Verification would require something like a UV-Vis or HPLC or NMR. Good luck getting those at home.


text:
Lewis structure of Cl2SO
Why is it that the Lewis structure of $\ce{Cl_2SO}$ has a total of $24$ valence electrons? I thought that the number of valence electrons should be $2\times7 + 1\times6 + 1\times6 = 26$.

Could you elaborate where you found this misleading statement? Is it in a textbook or other source?

26 is correct..

@DavePhD Thanks. It must be an error in my textbook.

I'm guessing whoever wrote it did what I accidentally did the first time counting it up: forgot the lone pair on sulfur.

Sulfur is sometimes described as a hypervalent atom, or one that has more than eight valence electrons. This is possible to some extent; sulfur can indeed utilize its d-orbitals for bonding. However, recent quantum mechanical calculations suggest that the extent of this utilization is negligible. Therefore, your book is most likely representing the $\ce{Cl_2SO}$ molecule as a "charge-separated" molecule.
EDIT: Upon further thought this still makes no sense. Charge is conserved and thionyl chloride should always have 26 electrons (no matter how they are distributed). So perhaps, due to the high degree of ionic character in the $\ce{S-Cl}$ and $\ce{S-O}$ bonds, the missing two electrons are "dispersed" among these ligands. (This is speculation). Perhaps your book is genuinely in error. Perhaps they did not count a lone pair on the central sulfur atom or otherwise miscounted the number of electrons?
EDIT 2: Here's a picture of two possible Lewis structures of thionyl chloride; as we can see, the book agrees that the thionyl chloride molecule is better represented as a "charge-separated" molecule. Nonetheless, all the electrons are present.

EDIT 3: Here's a picture of thionyl chloride which makes it seem as if the molecule only has 24 valence electrons. NB: the lone pair is still there and its presence is implied through the lack of any indication of a non-zero formal charge on the sulfur atom.

As the structure of thionyl chloride is pyramidal, there cannot be a $\pi$ orbital for the double bond. The charge separated version is therefore superior in representation.

EDIT: Wait, what do you mean?

Martin I think you've misinterpreted the second picture (the second picture doesn't show the lone pair on the sulfur). The LP's presence is implied through the lack of any indication of a formal charge on the sulfur. This still makes the molecule pyramidal even though it looks planar.

To be perfectly frank: There is no double bond, because there is no $\pi$ orbital, because the molecule is not planar. The charge separated structure is the best representation. The double bonded structure is often used, because it is more convenient for organic chemists to write no charges.

Are you saying that non-planar molecules have no pi orbitals?

Yes and no. A $\pi$ orbital needs to have a nodal plane and this is only possible with (at least local) planarity. In this case the sulphur is best described as $\ce{sp^3}$ hybridised and there is now way with this orbital configuration to have a $\pi$ bond.

Let us continue this discussion in chat.

So charge separation is favored over pi bond in a non-planar system.

I think that's best decided on a case by case basis.

Lewis structure:

The properly way to determine the Lewis structure, based on this example, is:

Total valence electrons: $7\cdot2 + 6\cdot2 = 26$
Total electrons needed for octets/doublets: $8\cdot4 = 32$
Total shared/bonding electrons: $32-26=6$ (In other words, there are only three bonds.)
Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 26 - 6 = 20$ (In other words, the are only 5 pairs of lone electrons (2 pairs for $\ce{O}$, 6 pairs to $\ce{Cl}$ and the big BUT of Lewis structure analysis: the remaining lone pairs correspond to the sulfur at bonding with oxygen.)

Furhter information on how to draw Lewis structures, please follow the link


text:
Hydrolysis of ammonium cloride
As we all know ammonium chloride is a acid. And what explains this is its salt hydrolysis. It breaks down into ammonium and chloride ions. Ammonium react with water to produce hydronium or hydrogen ions.
But then why can't chlorine ions react with water to produce hydroxide ions?
Is there any way other than using an indicator to know, whether a compound is acidic or basic? We know the second reaction can't happen, because we can tell the compound is acidic by using an indicator.
What really stops the second reaction from happening?

Chloride ion does not react with the water to any significant extent. To understand this, we must first accept that all chemical reactions are equilibriums. This means that all chemical reactions are reversible to some extent. Some reactions aren't very reversible at all (at least not in the short or medium term) - such as the combustion reaction in your car's gasoline engine. But if you give it a few million years, the carbon dioxide released by your car's combustion process will be fixed (biological term) by plants and these plants might eventually turn into fossil fuels again.
So we've established that all chemical reactions are equilibriums. Now we can consider the case of the chloride ion and why this reaction does not happen, as you say, or does not happen to a significant extent, as I say:
$\ce{Cl^- + H_3^+O -> HCl + HO^-}$
The reason is that if the above reaction happened to any significant extent, then this would make HCl a weak acid. Strong acids are defined to dissociate nearly completely in water. HCl is a strong acid by any account. Think of it as a waterfall. This reaction represents going down a waterfall (this reaction is large extent; heck, it's practically 100%):
$\ce{HCl + H_2O -> Cl^- + H_3^+O}$
So if the above reaction is the going down part of the water fall, its reverse reaction must represent the going up part of the waterfall. Going up the waterfall is hard no matter how you slice it; the only difference is whether you're trying to go up a small waterfall or Niagara falls. The reaction of hydrochloric acid with water is akin to going down Niagara falls - this is a favorable reaction. This is basically a one-way street.
Therefore, chloride ion can't affect solution pH; if chloride ion could somehow react again to form more hydrochloric acid then we'd violate the definition of a strong acid - nearly complete dissociation. This is what makes hydrochloric acid a strong acid!
And this is empirically true. Chloride ion can't even be protonated effectively by the strongest acid in a solution of hydrochloric acid, which is hydronium ion, so there is no way that you can expect chloride ion to be protonated by water (which is only about a trillion times weaker than hydronium ion as an acid). Thus, your solution of ammonium chloride has a pH controlled not by the chloride ion but by the ammonium ion.

And in other words, the $\ce{H-O}$ bond is so many times stronger than the $\ce{H-Cl}$ bond. (Good answer.)

What makes the H-O bond stronger than the H-Cl bond? Size of the O vs Cl atom and resulting bond length? Thanks. I learned it from my chem prof :p.

I would argue, that a 2p orbital (O) is more compact and has lesser nodes than a 3p orbital (Cl) and makes therefore a stronger bond. One could also state, that O has a higher charge density and therefore the attraction to a proton is higher. The bond length might be an indicator - but it can be very misleading. Usually a shorter bond length results from a stronger bond, not the other way around.

Hydrogen chloride is a strong acid, which means that when HCl is mixed into water, it completely disassociates into $H^+$ and $Cl^-$. Chloride ion will only be found in solution; not as a salt (although it may precipitate). Ammonia is a weak base, so it will exists as both ammonia and ammonium ion in solution. When you have a salt like ammonium chloride, you know the chloride is unaffected, but the ammonium ion will distribute into ammonia and ammonium ion, producing $H^+$.


text:
Drinking water out of a 24 year old Ether Bottle
I have an empty aluminum Ether bottle (Shipping name diethyl Ether) that I'd like to use as a large drinking bottle as this is thick aluminum. Is it dangerous? Here is some of the info on the bottle.

Chemical Composition: (C-2 H-5)2-O f.w.74.12.
Name:  Ether Purified (For Fat Extraction)
Other: Mallinckrodt (Brand?)
   5.51 lb
0884
Diethyl Ether 60-29-7
Ethyl Alcohol 64-17-5

SAFETY FIRST - I'm sure it is "cool" to turn some such odd container into a drinking vessel, but the rule of thumb for chemical safety is never to use some container that had chemicals for food storage and visa-versa.

If it ever had something in it that's unsafe, don't use it for food purposes. Is ether safe, for the purpose at hand? Yes, but you have to be sure nobody ever used the botttle e.g. to store used ether for recycling or as a general waste bottle.

Mallinckrodt is the chemical manufacturer. The ethanol is added to the ether to prevent peroxides from forming in the ether (like an antioxidant). If you wash and rinse the container thoroughly a few times, then it should be fine.

Okay, thanks. Few? Like three? Would I die if it isn't washed properly?

Likely, all the ether (and ethanol) has evaporated so the amount in the container should not pose a problem. Washing it will decrease the amounts even more. Twice would be adequate and three times is even better.

I'd do 10. I figure if that is good enough for analytical chemistry, it is good enough to drink out of. Three is only good enough for synthetic chem.

I'm sure the bottle itself will be safe to use, but I would strongly recommend that you remove or destroy the label. Mislabeled items sometimes lead to unanticipated consequences. Admittedly, that also takes some of the fun out of it.


text:
Why doesn't a table sublimate, while ice does?
A table does not sublimate, and nor does a spoon. Ice does, however. What is the fundamental difference?

At high enough temperature and low enough pressure, a metal spoon will definitely sublimate! A wooden table is a more complicated substance, wood will tend to decompose (or burn if oxygen is present) before it sublimates.

This relates to the difference between how ice, iron spoons and wooden tables stick together.
In wood, most or all of the bonds between the individual atoms and fibre units are covalent; making them very strong. Similarly, a spoon is an alloy of iron, nickel, chrome and carbon, also a very coherent structure.
In ice, on the other hand, the units of the crystal is held together with comparatively weak hydrogen bonds, meaning that not a lot of energy is required for a surface molecule of water to escape, in a process called sublimation.
Usually, once a water molecule escapes the ice, it is, given a relatively constant temperature, very unlikely to rejoin the body of ice; meaning that especially in open air and windy conditions, ice will sublimate at sub-zero temperatures, although as the temperature of the ice drops, this effect becomes steadily less noticeable.
Edit: It is worth noting that given enough time, a table or a spoon can indeed sublimate as well, at any temperature, but the odds against it are utterly staggering.

I'm not sure whether "odds against it" is a good description here (but I'm not a native English speaker): odds against it to me sounds like a small chance that the spoon evaporates - whereas I'd describe it as an extremely slow process, which however wouldn't be subject to a chance whether it happens or not: I'd say it happens, just far to slow to play any practical role.

@cbeleites: Fair point; but given enough time the spoon will evaporate, but odds are, before that happens, it'll be destroyed by some other means; like the sun exploding.

@WillihamTotland I find this fascinating.  You note that as the temperature of the ice drops, the sublimation effect becomes less noticeable.  Presumably the effect would be greatest in a block of ice just below freezing.  What about the iron spoon? If we heated it up to just below its melting point, would we able to observe the effect?

You might want to read the prohibition on cadmium plating in space by NASA: "Cadmium is known to sublimate in a hard vacuum environment (especially at temperatures above 75°C). The sublimation products, which are conductive, can redeposit resulting in short circuits. The sublimation products may also interfere with sensitive optics."

@emory: The conditions under which steel (or iron) sublimates are fairly peculiar. Even in the extremely harsh conditions of interstellar space ablation resulting from impacts with stray hydrogen nuclei is likely to be a far bigger factor in mass loss than sublimation.

I think when you say "evaporate" (4th paragraph) you mean to say "sublime", as you did before.

@MarcvanLeeuwen: Indeed, fixed.

@WillihamTotland: That was really fast (the edit appeared just when I finished my comment)! By the way I am confused about the verb corresponding to sublimation; I was tempted to use "sublimate", but WP uses "sublime".

@MarcvanLeeuwen: As far as I know, either is fine.

One term I'm missing so far in the discussion is the vapor pressure.
You may say that the water vapor pressure above ice is far higher than the iron vapor pressure above iron  (at "normal" temperatures), or the vapor pressure of the various substances forming the table over that table.

See here for the water vapor pressure. I take from the diagram that at -20 °C, the water vapor pressure above ice is approximately 1 mbar $\approx$ 1 hPa, and at slightly below 0°C, about 7 mbar.

According to this table in Wikipedia, you need to heat iron well above the melting point to reach such iron vapor pressures. They also cite a formula for the iron vapor pressure above solid iron:

$\log (P/\mathrm{Pa}) = 12.106 - 21723 / (T/\mathrm{K}) + 0.4536~\log (T/\mathrm{K}) - 0.5846 (T/\mathrm{K})^{-3}$
with the specification that this could be used from 298 K upwards.

Plugging in 298 K, I get $p_{Fe} \approx 10^{-60}$ Pa.
This is so low that lots of other processes will lead to more substantial loss of iron from the spoon via the gas phase (e.g. according to this $\ce{FeCl3}$ has a vapor pressure of 1 mmHg $\approx$ 133 hPa at 194 °C).

<< This answer was posted before the question was completely rephrased and originally asked "why does a table not evaporate" >>
Who said a spoon doesn't evaporate?
Ice is a solid below its melting point ($0~^\circ\mathrm{C}$) but above its melting point it goes through a phase transition and melts to become a liquid (so it's liquid at room temp). Then when it's heated above room temperature to its boiling point ($100~^\circ\mathrm{C}$), it goes through another phase transition and it evaporates to become gas.
Iron (assuming an iron spoon here to keep it simple) is a solid (at room temperature) then when it's heated above room temperature to its melting point ($1538~^\circ\mathrm{C}$) it becomes a liquid.  Then when it's heated above its melting point to its boiling point ($\approx2861~^\circ\mathrm{C}$), it evaporates.

But wait, maybe I was unclear, but doesn't ice evaporate even in a fridge?.. Of course everything can evaporate given a high temperature.

Hm, maybe I use a wrong english word?.. Maybe I meen sublimation here?

@roman.brodetski: not sure about the English terms, but yes, there is a water vapor pressure above ice as well, in fridge and freezer

This answer increases the confusion created by the language issue in the question.  It would probably be best to remove it.  The question is not about evaporation of a material once it reaches that material's boiling point, but evaporation that happens during a material's normally solid state.  It should be clear that he wasn't talking about boiling water when he said that ice "evaporates".  He is talking about submlimation, where a solid goes straight to the gas phase.  A detailed explanation of what he wasn't talking about is not helpful.


text:
Does a first order reaction really get completed only at infinity?
For a $1^{\text{st}}$ order reaction $ A \rightarrow B$, the corresponding rate law is: $$\text{rate} = k[A]^{1}$$ and the integrated rate law for this is $$ \ln \left(\frac{A}{A-x}\right) = kt$$
Going by this equation, every first-order reaction must get completed only at infinity. Is this true? If not, then what is the correct explanation?

As far as I can see, the expression you have come up with is not entirely correct.
If we assume that the reaction is $\ce{A -> B}$ (i.e. saying that it is approximately irreversible, or that $k_\mathrm{reverse} \ll k_\mathrm{forward}$), we have:
$$ \mathrm{rate} = \frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k[\ce{A}] $$
Separating:
$$ \frac{\mathrm{d}[\ce{A}]}{[\ce{A}]} = -k\,\mathrm{d}t $$
Integration gives:
$$ \ln\frac{[\ce{A}]}{[\ce{A}]_0} = -kt $$
$$ [\ce{A}] = [\ce{A}]_0 \mathrm e^{-kt} $$
Here, $[\ce{A}]_0$ is the initial concentration of $\ce{A}$. It is correct that mathematically, $[\ce{A}] > 0$ for all $t<\infty$, from the last equation. In practise, however, the amount of $[\ce{A}]$ after a certain time interval $t$ will be negligibly small. How long it takes depends on $k$; if $k$ is large, the time for the reaction to go essentially to completion will be short.

To add to the other answers that only address the mathematical behavior of the first order rate equation close to $t=\infty$, let me address what actually happens physically.
Before you reach $[A]_t=0$, the first order rate equation actually breaks down. It isn't valid anymore, because one of the assumptions on which it is based, namely that the reaction rate is deterministic, breaks down.
When you get to really low quantities of (one of) your reactants, reaction events, i.e. two molecules meeting, are blind luck. Actually, for high quantities this is also the case, but then the behavior is a physical version of the law of large numbers: a probabilistic system behaves deterministic on average.
You can simulate this basically by flipping coins whether a reaction occurs or not. If you do this for a simple $A$ to $B$ reaction you get graphs like the following:

Here the chance of reaction per timestep for each molecule $A$ is 10%. You can see that the curves at low number of molecules $N$, are jumpy (and actually would look different everytime I rerun my calculation), while for $N=1000$ you already see the fairly smooth and familiar first order behavior. Just note that close to t=10 also the $N=1000$ case is behaving jumpy again as you can see from the zoom below.

Some first order reactions don't involve two molecules. The archetype being radioactive decay (OK strictly a nuclear not a chemical reaction, but the same principles apply).

No, because matter is not continuously divisible
The flaw is the mathematical reasoning is that it assumes that matter is continuously divisible. In reality it consists of finite numbers of atoms or molecules.
In a truly first order, irreversible, reaction there will eventually be just one last molecule or atom and it will react leaving nothing. The mathematical view is a statistical approximation that works when there are enough units not to worry about the lumpiness.
The classic example (though strictly a nuclear not a chemical process) is radioactive decay. The probability of any atom decaying is constant so the rate of decay depends only on the number of atoms present. But we eventually get down to just one atom. And that atom will decay. But when only small numbers are left the smooth pattern of decay will be lumpy and will no longer obey the smooth mathematics of the first order equation.
But you won't have to wait an infinite amount of time to get to zero.
If you start with a mole (about 6 x 1023 atoms) of substance you will have about one atom or molecule left after about 79 half-lives. So you can be fairly sure (never exactly sure due to the probabilistic nature of quantum things) that there will be no reaction still occurring after about 80 half-lives.

To add to Kjetil's answer, the expression given by the OP and the expression derived by Kjetil are (almost) equivalent. If we define the OP's $A$ as $[A]_0$, the original concentration of $A$ and the OP's $A-x$ as $[A]_t$, the concentration of $A$ at time $t$, then we rearrange and take the natural log of both sides...
$$[A]_t=[A]_0e^{-kt}\ \ \ \ \ (1)$$
$$\ln{\left(\dfrac{[A]_t}{[A]_0}\right)}=\ln{\left(e^{-kt}\right)}\ \ \ \ \ (2)$$
$$ln{\left(\dfrac{[A]_t}{[A]_0}\right)}=-kt\ \ \ \ \ (3)$$
$$ln{\left(\dfrac{[A]_0}{[A]_t}\right)}=kt\ \ \ \ \ (4)$$
The fundamental differences between the original equation and the one with the natural logarithms is that we have to make some assumptions. Our derivation produces an undefined term if $[A]_0=0$ (equation 2), so we must assume that $[A]_0>0$, which is logical. We also must assume $[A]_t\ne 0$ (equation 4). Only equation 1 works with $[A]_t=0$, which as was pointed out is only technically true at $t=\infty$ (or $[A]_0=0$).

Good clearification! And thanks for catching the typo :)

It is interesting to derive the first order rate expression to see why the decay continues to infinity. (It was probably first derived by von Schweidler in 1905!) The basic assumption is that the probability of reaction $p$ in a time interval $\Delta t$ is independent of the past history and depends only on the length of the time interval $\Delta t$, and when this is sufficiently short the probability is proportional to this interval as $p=k\Delta t$ where $k$ is the proportionality constant which depends on the particular atom of molecule. (This will become the decay rate constant)
The probability of no reaction in the short time interval $\Delta t$ is $1-p=1-k\Delta t$. If it survives this interval the probability of surviving the next interval is also $1-k\Delta t$ and so by the law of probabilities, the chance of surviving both intervals is $(1-k\Delta t)^2$ and so for $n$ such intervals the probability is $(1-k\Delta t)^n$. The total time is $t=n\Delta t$ and so by substituting we obtain $(1-kt/n)^n$. The chance that the molecule or atom will remain unchanged at time $t$ is just that when $\Delta t$ tends to zero, i.e. is made infinitesimally small and then we can recognize that $e^x=\lim_{n\to \infty}(1+x/n)^n$ and so we obtain the limiting value as $p=e^{-kt}$. Instead of considering just one atom let there be $N_0$ initially then the fraction remaining at unreacted at time $t$ is  $N/N_0=e^{-kt}$.
So even for a single molecule the probability of remaining unreacted can extend to infinity. In practice this is only a matter of sensitivity of your experiment. If you do time-resolved single-photon counting to measure the decay of an excited state the decays are routinely measured down to 5 orders of magnitude, transient absorption decays, remarkably, have been shown to follow a decay to at least 9 orders of magnitude in intensity. But eventually either no matter how many times the experiment is repeated or how many molecule we start with the noise inherent in the experiment will dominate and we can go no further.
Turning the equation round we have $t=\frac{1}{k}\ln(p)$ so that the chance $p$ of waiting for a long time for the molecule to react becomes increasingly small as $t$ increases. Thus if the decay has a rate constant of 1/nanosecond the chance of waiting even as short a time as a microsecond without it decaying first is tiny,  $p=e^{-10^{-6}\cdot 10^9}\approx 10^{-435}$! This is of course why we say that first order reactions comes to a halt after a given time, the chance of observing become so small and is lost in noise even if we start with a mole's worth of molecules: $10^{-435+23}$ is still utterly minute.

In your example, the reaction has been running for approx. 1443 half life periods. So the probability you calculate is $0.5^{1443} \approx 10^{-435}$. It is like flipping a coin 1443 times and see only heads. Not very likely, as can be intuitively grasped.

@Snijderfrey,  nice example but 'not very likely' is a bit of an understatement :)

Yes, certainly an understatement. If we had 1443 people flipping a coin every second and have them record their result, they would still have a ridiculously low chance to observe the described outcome within the lifespan of the universe. And the reasoning would not be much different with only 100 half life periods. The numbers involved are next to impossible to conceive.

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


text:
sp3 hybridisation in carbon
Before $\ce{sp^3}$ hybridisation, does the $\ce{C}$ -atom get excited to $\ce{1s^2~2s^1~2p^3}$ state, as it happens before $\ce{sp^2}$ hybridisation?
What is the energy of the $\ce{sp^3}$ hybridised orbital?

No it does not just like it does not in the other case. Hybridization is mumbo-jumbo that makes absolutely wrong assumptions, but somehow get correct result in some cases.

@permeakra hybridization "gets the answer right" because it is applied once the geometry is known. I find it useless for prediction.

To follow up on Swedish Architect's answer (and Martin's comment):
The energy of an orbital is a poorly defined concept, but what can be measured is the energy it takes to excite an electron from a molecular orbital (or an atomic orbital in some cases-- but that is not relevant for hybridization).  In the case of an sp3 hybridized carbon one may infer something about the 'energy' of the orbital from the bond dissociation energy of of the bond formed by such an orbital (as previously mentioned this is a fairly imprecise use of language, but the concept does have some instructional value).
This reference (http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociationenergy.pdf) provides some of these.  In particular on p4.46 of that reference in the chart, CH3CH3 (sp3) is listed at 410 kJ/mol, while CH2CH2 (sp2) is listed at 427 kJ/mol, and HCCH (sp) is listed at 523 kJ/mol.  this is the amount of energy required to abstract a hydrogen atom from that systems, which is a reflect of how 'strong' the bond is, which is a reflection of how low in energy the 'orbital' is.  In essence, the higher the bond dissociation energy, the lower the energy of the corresponding orbital is.  Since all of the molecules in question here from combinations of the hydrogen orbital with the carbon orbital, as a crude approximation, you can assign a relative energy.
Please be advised, this analysis really only has meaning for trying to teach the concept.  The actual molecular orbital includes contributions from all atoms in a molecule, and the fact that the ethane has more hydrogens than ethene or ethyne is perturbing this analysis slightly.

sp3 hybridisation is what we call the formation of hybrid atomic orbitals, by the combination of an s orbital and 3 p orbitals on the carbon atom. I think you understand this. However, like in a dative bond, even though one of p orbitals is empty before the carbon becomes hybridized, it can still combine with the other atomic orbitals, to create new hybrid atomic orbitals. So no, the atom doesn't have to get excited to 1s2 2s1 2p3 before hybridization.
Another way of thinking about sp3 hybridization, a more advanced view, is that they are merely a nice simple way of describing the bonding in atoms. If we were to use the molecular orbital approach we would have to keep going back to our computers to work out the MO's. This isn't really needed, unless we want exact energies and a detailed description of the bonding. In the case of sp3 hybridization, say in methane, the carbon s orbital interacts with ALL of the hydrogen orbitals to form a MO. Then the carbon p-orbitals interact with the hydrogen 1s orbitals. Here the case is that several AO are interacting to form one MO. Then the sp3 model can just be said to be a simplification of a more complex interaction of AOs. This hasn't directly answered your question, yet. The carbon atom doesn't have to be excited to the 1s2 2s1 2p3 state to interact like described above.
Your next question. The energy of the sp3 hybridised orbital is not one, but 4 orbitals. The energy cannot really be found because they don't really exist. After they bond, however, the energy changes based on the bond, as they form a new MO, so it may be more useful to know the energy of each MO in the individual case.

Hybridisation is not an interaction, it is only a linear combination of functions. It can therefore only be a description of a bonding situation and will always be a result of a geometric arrangement. As an hybrid orbital is only a mathematical concept, it does not have a physical meaning and therefore it does not have an energy that may be assigned to it.

@Martin, yeah I think I badly worded it in the top paragraph, so I've edited it. I also said in my original answer that hybrid orbitals don't have energies, because they don't really exist.


text:
Making a reasonable slab of alpha quartz
I want to perform a molecular dynamics simulation with alpha-quartz as my mineral surface. I want to study adsorption of water on alpha-quartz. The most dominant error that I came up with is the problem with my initial physical configuration. In other words error is: System is too far from equilibrium. I think the major problem is related to chemical structure of alpha-quartz slab.
I used the experimental data for unit cell of alpha quartz and then replicate it for making a slab of it. My question is: Is this structure stressed in some way? My cell is periodic in 3 directions and some of oxygen atoms are not in the right place! Can it be the problem?
How to replicate unit cell of alpha quartz for obtaining a reasonable structure? Where can I find a right data for unit cell of alpha quartz?

it is normal for a surface made from slicing of infinite crystal structure to be far from equilibrium. It usually undergoes significant reconstruction, with formation of new bonds and reaction with molecules from gas phase. Read relevant articles/books. 2)it is common to fix several layers of slab from side opposite to surface modeled.  3) alfa-quartz is a bad case for case study for work with crystal surfaces. It has a really, really wicked structure. I recommend to try some more studied system first, like MgO or gold surface.

Thank you permeakra, I will take your advice into consideration.

BTW, it is normal to have ~5% difference in simulated parameters from each other and real life for QC simulation.


text:
Examples of evolving networks in chemistry
I'm a master student working on networks analysis in general. A network is something that has nodes and there are links between the nodes. Nodes and links could have attributes. An evolving network is one that changes overtime (new nodes and links are added..etc). An example of that is Facebook. Nodes are users and links represent the friendship relationship. Users have attributes (gender, age ..etc). A Facebook network as you know is an example of a social network.
The issue is that so many people studied traditional evolving networks like social networks, the web, or transportation networks. Currently I'm looking for novel examples of evolving networks to study them. So I thought there might be some examples in chemistry that could represent some kind of an evolving network.
So my question: Can you give me examples in chemistry for evolving networks?

If you wouldn't mind indulging me for a little...what sort of properties of these networks are you looking to study? I can think of a few, but they'd all be fairly simple and not have a lot of complexity to them.

You've gotten a lot of negative feedback on other sites already, but I would like to (re-)remind you that cross-posting is discouraged.  While you adhered to the letter of the law and changed the question, it was still unfair to broadcast it to so many sites.  However, I do think that this subject matter is highly appropriate for our site in particular, and in the spirit of that, I would invite you to come up with an edit that asks something more specific.  This will allow you to engage our experts and perhaps get a concrete idea for an interesting project.

Synthetic Pathways
... is the only example I can think of right now.
As far back as 1971¹ people have been positing the possibility of an AI to search for novel synthesis paths, but only a couple of years ago a decent one was developed² which could interpret literature and create novel sets of steps for particular reactions -- for example, given a particular reactant and product, produce a set of steps which would produce one from the other. The particular focus of the paper for the 2012 one was to produce single-pot (multi-step) procedures to replace awkward separated multi-step syntheses, but in principle the procedure is generalisable to any path.

Sridharan, N.S., An Application of Artificial Intelligence to Organic Chemical Synthesis, Doctoral Thesis, State University of New York at Stony Brook, New York, July 1971.
C M Gothard et al, Angew. Chem., Int. Ed., 2012, 51, 7922 (DOI: 10.1002/anie.201202155) but there's a good write-up by Philip Ball in Chemistry World, August 2012 issue.

If you are looking for a biochemistry answer to this question, there are many cases in the cell structure of any living organism.
For example, the skin cells are mapped in a multi-layer system with many physical links (like gap junctions, tight junctions, etc).  Old cells are removed from the linked system and new ones added continuously.


text:
Do various soaps work better when dissolved/used in different water temperatures?
My wife and I have had an ongoing argument about various cleaning agents, and I wanted to get a Chemists' view on it.
When we are doing laundry, she will generally clean using the cold water setting and a standard amount of detergent.
If I do laundry, I will use the warm water setting with a standard amount of detergent.
Either way, the clothes come out (seemingly) clean. Using cold water saves energy from the heating process, which has been her excuse for doing such a dastardly thing. Her argument against using warm water is that it will damage/shrink the cotton. While I don't disagree on that point, I am curious if the use of warm water causes the detergent to work better or not.
In addition, there are various soaps that work best in cold water.
What is the actual mechanism at work here? Why do some soaps work better in cold (or hot) temperatures?
To make things a bit better, let's remove some more variables. We use this detergent:
Tide Pods
Unfortunately I can't find information on the normal temperatures those machines operate at.
Also who is right? (I bet it's her)

To start, soaps and detergents have two different chemical makeups, meaning that they act different under different conditions.
For general soaps, here is an explanation of how it works:
http://chemistry.about.com/od/cleanerchemistry/a/how-soap-cleans.htm
As a general chemistry note, as the temperature of a solution increases, the ease of a chemical to change (whether it is to react or change shape) increases.  This means that colder temperatures will allow a chemical to remain in the same state.  If the soap depends on a specific conformation/shape to act then colder temperatures may be preferable.  However, if a reaction needs to occur in order for the soap to act some heat may be required.
For Tide Pods, it states that it "dissolves quickly in hot & cold."
http://www.tide.com/en-US/product/tide-laundry-detergent-pacs.jspx
The chemical makeup of Tide Pods is composed primarily of salt compounds and enzymes.  Both salts and enzymes tend to operate better at higher temperatures.  So as long as you don't pass the denaturing temperature of the enzyme, higher temperatures would be preferred.

I added some clarification there, but that definitely helps me better understand some of the mechanisms at work. Thanks!

If you could offer a brief summary or a quote from the about.com article, that would make the answer self-contained.  Thanks!


text:
Which is more conductive?
There are two choices. One is Nitric acid, and the other is phosphoric acid.
The solution of each type of these acids has a concentration of 0.1 mole. I answer would be phosphoric acid, because each molecule contains more than 1 hydrogen ion. Thus it would be more conductive.
Does strong/ weak affect conductivity?

Whether the acid is a strong or weak acid does affect its conductivity.  Weaker acids are less dissociated and generate fewer ions in the equilibrium.
Nitric acid $\ce{K_{a}=24}$
Phosphoric acid $\ce{K_{a}=7.1 x 10^{-3}}$
link to Ka reference
Nitric acid is more dissociated than phosphoric acid
Electrical Conductivity κ in mS/cm at 1 Mass Percent Concentration
link to conductivity reference
Nitric acid ~ 56.1
Phosphoric acid ~ 10.1
Nitric acid is more dissociated and more conductive than phosphoric acid.

Do u memorize these things, which is dissociated more? And the constant of ionization for each acid?

But then in one mole of phosphoric acid, there are are three moles of hydrogen ions? Which is significantly more than what sulfuric acids have.

"Do u memorize these things" - No way, just google them; "one mole of phosphoric acid, there are are three moles of hydrogen ions" - Doesn't really matter if it doesn't ionize (dissociate) very much.

Why but more ions means a stable current of electricity?

Understood, but there are not more ions with phosphoric acid because it doesn't dissociate.  Above I gave the value for the first ionization constant for phosphoric acid and it was 10^-3 times smaller than the dissociation constant for nitric acid.  Look at how much SMALLER the second and third dissociation constants are for phosphoric acid;.  There is just a LOT LESS H+ present with phosphoric acid compared to nitric acid(https://www.chem.wisc.edu/deptfiles/genchem/sstutorial/Text12/Tx125/tx125.html)


text:
Types of organic chemicals?
I am looking for a list of classes of synthetic organic chemicals. I'm not very strong in chemistry. The only information I could find are from here which seem to describe SOC's are like pesticides or fuel additives, but I don't think you can call those classes.

I think this question is probably too vague to answer in a way both satisfactory to you and useful to a general audience. Can you be more specific about what you mean by "class" w/r/t chemicals? Do you mean by priority functional group (per, say, IUPAC), by industrial/consumer/medical applications, etc.? As the answer from Williham Totland suggests, organic chemistry is a huge field, and the focus of your question needs to be much narrower.

There are a lot of organic chemicals[citation needed].
Due to the incomprehensibly vast number of chemical species and variation between them, almost every classification system is domain-specific.
In terms of pure chemistry, the best you can do is usually to organize them according to what functional groups are in play; this will usually reveal something about the fundamental chemical interactions available to the species in question, but this is not necessarily useful or enlightening:
In many applications, such as as antifreeze, fuel or a solvent, methanol and ethanol are (largely) interchangeable: They both have short carbon chains attached to a single alcohol group (-OH), their boiling points, dipole moments, chemical energy and other properties are similar, and they mostly react in the same way on a chemical level.
In other applications, however, they are as  night and day: In biology, methanol is a potent poison, and ethanol is a mostly harmless intoxicant.
Organic chemistry is, in short, notoriously resilient to generally useful classification systems.


text:
Which way does this reaction between NaCl and acetic acid naturally proceed?
Here is the overall formula:
$$\ce{CH3COONa  + HCl <=> NaCl + CH3COOH}$$
(sodium acetate + hydrochloric acid $\ce{<=>}$sodium chloride + acetic acid)
Which way does the above reaction naturally proceed? I've tried Googling this, but different sources seem to suggest contradictory answers, and I can't really tell what's correct.

Mostly the reaction will proceed from left to right, unless you have ridiculous amount of acetic acid. Hydrochloric acid is much stronger, so it suppresses dissociation of acetic acid.
However, this is only part of the truth. In fact, in water solutions many compounds (partially) dissociates:
$HCl = H^+ + Cl^-$
The process is strongly endothermic in vacuum, but subsequent interaction of ions with water makes it mostly energy neutral or even highly exothermic for some compounds. This means, that in you case the solution would actually contain $H^+$, $OH^-$ (very little), $CH_3COO^-$, $CH_3COOH$, $HCl$ (almost none, may be ignored in most cases), $Na^+$. The actual reaction would be
$CH_3COO^- + H^+ = CH_3COOH$
The actual direction of the reaction would depend on initial concentraion and guarded by equilibrium constant.
The keywords for googling/textbook reading are 'electrolitic dissociation', 'ionic reaction', 'chemical equilibrium', 'equilibrium constant'. Generally, the procedure of determining which concentration of which ions the system will have requires solution of system of equations, including equilibrium constants, electroneutrality equation and material balance equations. The system often may have no analytic solutions, in which case it is usually approximated by simpler system, dropping some elements of sums in the system.

So let's say my initial conditions are reversed and I am starting with the right hand side CH3COOH and NaCl. In this scenario I should expect a small shift to the left to produce a minuscule amount of HCL and CH3COONa, but the majority of the reactants would remain as CH3COOH and NaCl?

Or more accurately, the NaCl would fully dissociate into Na+ and Cl- ions, and a small amount of CH3COOH would dissociate into H+ and CH3COO- ions, and some of the CH3COO- ions would combine with Na+ to give CH3COONa.
Thus my solution would contain Na+, Cl-, CH3COOH, H+, CH3COO-, CH3COONa, and (almost none) HCl?

@JST $CH3COONa$ is generally accepted to be fully dissociated while in solution like most (but not all, exceptions do exist) salts. Ignoring this part, you are correct.

As denoted by the equilibrium arrows, the reaction proceeds in both directions.
Initially, the net direction depends on the concentrations of the chemical species (Le Chatelier's principle). For example, if there's a lot of acetic acid, the reaction will go left. If there's lots of acetate, the reaction will go right ("lots" is a relative term, see the equilibrium constant for acetic acid: $K_a = 1.8\times10^{-5}$). The net reaction will proceed in this direction until the equilibrium concentrations are achieved. However, keep in mind that even at this stage the reaction still occurs in both directions, but there will be a net direction (one direction is favoured over the other).
Once equilibrium occurs, there is no net direction but, again, the reaction still occurs in both directions (at the same rate now).

I only put in equilibrium arrows since I don't know which way it goes or if it is indeed an equilibrium. Various online sources I read showed the arrow going either one direction or another.

Well you got lucky, because it is an equilibrium! The net reaction direction really depends on the initial conditions.

+1, but in the interest of clarity, I would specify that the equilibrium here overwhelmingly favors the formation of products. $\ce{HCl}$ is many orders of magnitude more acidic than $\ce{CH3CO2H}$, so the reverse reaction should be negligible.

Perhaps it should be in the answer, but the asker didn't specify what species they're starting with and their amounts. Or should we assume that the compounds on the left are present initially? Is it also safe to assume that the amount of HCl present would have an effect on the reaction such that it is effectively irreversible?

@canadianer, I'm not suggesting you make any assumptions about initial concentrations, I just think it would be valuable to mention that the equilibrium ratio of $\ce{CH3CO2H}$ to $\ce{HCl}$ here should be on the order of roughly $10^{12}:1$ (assuming stoichiometric amounts). The second paragraph of your answer reads to me as though the reaction from right to left is a lot more significant than it normally would be in practice. Maybe that's just me. In any case, only a friendly critique, no offense intended.

I took no offence, I was genuinely asking you :)


text:
Validity of Henry's Law for an Oxygen/Bitumen System?
Henry's Law is for a gas dissolved in a liquid and states P = kC where P is the partial pressure of the gas above the solution and C is the concentration of the gas dissolved in the liquid. The k is a "constant", that is, it is independent of P, C and time but depends on the type of gas and the type of liquid. According to Pickover, page 184 (reference below), Henry's Law assumes:
A1.  the gas/liquid system has reached equilibrium, and
A2.  the gas does not chemically react with the liquid, and
A3.  the concentration of gas in the liquid is low, and
A4.  the gas pressure is low.
I would be grateful for any help in answering the following:
Q1.  Is the list of assumptions, A1 to A4, correct and complete? I have no reason to doubt it but I would like to confirm it.
Q2.  Some authors (e.g. Dickinson E.J., Road Materials and Pavement Design, 2000, 1(3), 255-280) have applied Henry's Law to the diffusion of oxygen in air at atmospheric pressure into bitumen which is a semi-solid at room temperature. Is there any experimental evidence that Henry's Law can be extended to such a system?
Reference
Pickover, C., "Archimedes to Hawking", Oxford University Press 2008

Regarding Q1, Henry's law states:

At a constant temperature, the amount of a given gas that dissolves in
  a given type and volume of liquid is directly proportional to the
  partial pressure of that gas in equilibrium with that liquid.

A1 is answered by the definition itself: the system should be at equilibrium.
A2 is straightforward: if the gas reacted, you would not have a constant pressure as a certain (if not all) quantity of it would be removed from the system[1]. This would also keep your system from being at equilibrium, defying the previous assumption.
A3 and A4 both depend on the fact that – as the concentration or the partial pressure rise – the system deviates from an ideal model where the solute is
infinite-diluted and in such system Raoult's Law should be used
instead[2]. So you need a low concentration and partial pressure of your gas.

To answer to Q2, Henry's Law can be applied to any solvent, whether liquid, vapour or solid[3], [4]. It has been used, within certain limits, to study absorption of gas in zeolites[5] or hydrogels, but I couldn't find any precise experimental data on it. Only that in such systems it is only a limited law, adequate at low mole fractions, but useful for practical purposes when high precision is not required.
    1 Henry's Law, UC Davis University ChemWiki
    2 North Carolina State University, CH 431 / Lecture 14
    3 Francis L. Smith and Allan H. Harvey (September 2007). "Avoid Common Pitfalls When Using Henry's Law".
    4 Rosenberg, Robert M., Peticolas, Warner L., Journal of Chemical Education, Volume 81, 11, pp 1647-1652
    5 Eric G. Derouane, Zeolites as solid solvents, Journal of Molecular Catalysis A: Chemical, Volume 134, 1–3, pp 29-45

the gas/liquid system has reached equilibrium,beacause pressure is defined only when gas is in equilibrium otherwise gas will not have same pressure throughout gas.A3. the concentration of gas in the liquid is low, and A4. the gas pressure is low.because gas will deviate from ideal nature and there will be more collisions which will deviate it from following Henry's LAw.A2,the gas does not chemically react with the liquid because laws in chemistry that relates physical properties assumes that the interaction between particles is zero or negligible which is not possible in chemical reaction

Can you cite your source for this?  Also, the formatting is a bit garbled.


text:
A sample of tap water is found to be 0.025 M in Ca2+. If 105 mg of Na2SO4 is added to 100.0 mL of the tap water, will any CaSO4 precipitate?
Given:

Ksp of CaSO4 is 7.10*10^-5
Molar mass of Na2SO4 is 142.04g/mol

My Attempt:
Step 1 - figure out molar solubility of CaSO4
√7.10*10^-5 = 8.43*10^-3M
Step 2 - convert Na2SO4 from mass to moles
105mg = 0.105g
0.105/142.04 = 7.39*10^-4 mol
Step 3 - figure out moles of CaSO4 produced
Since we need 1 mol of Ca2+ & 1 mol of Na2SO4 to form 1 mol of CaSO4
And we have less Na2SO4 than Ca2+
CaSO4 produced = 7.39*10^-4 mol
Step 4 - figure out molarity of CaSO4
7.39*10^-4mol / 100ml = 7.39*10^-3 / L = 7.39*10^-3M
Step 5 - figure out if any CaSO4 precipitates
8.43*10^-3M > 7.39*10^-3M
Therefore NO CaSO4 precipitates
The answer in the book is a Yes, can anyone please explain to me where I have gone wrong?
Thanks!

$K_{sp} =[Ca^{2+}][SO_4^{2-}] = 7.10 \times 10^{-5} M^2$
$0.025M[SO_4^{2-}]  = 7.10 \times 10^{-5} M^2$
$[SO_4^{2-}] = 0.00284M$ (this is the maximum soluble concentration of $SO_4^{2-}$ in the presence of $0.025M Ca^{2+}$)
However, 0.000739 moles of $SO_4^{2-}$ are added to 100 mL of water, corresponding to a concentration of $0.00739M SO_4^{2-}$.
Precipitation occurs because 0.00739M is greater than 0.00284M.