text:
Converting a PDB file to XYZ file
I am trying to convert a pdb file of gold nanoparticle to the xyz file format to get the atomic coordinates. But when I try to use openbabel to do this, I get the following error.
*** Open Babel Error  in ReadMolecule
ERROR: not a valid PDB file

My command is obabel shelxpro.pdb -O shelxpro.xyz
(shelxpro.pdb is at http://www.sciencemag.org/content/318/5849/430/suppl/DC1
inside the .zip folder 1148624_data.zip)
Also when I try to visualize this structure with Jmol or VMD, I get only part of the expected structure.
Is there a way to fix either one of these issues?

Point of clarification: A CIF file is not a PDB file - they're completely different formats. (The file you point to is a PDB file, though.)

The conversion issue you're seeing is because the PDB you're using is not technically well-formatted. If you take a look at the output, you see a bunch of the following warnings.
==============================
*** Open Babel Warning  in parseAtomRecord
  WARNING: Problems reading a PDB file
  Problems reading a HETATM or ATOM record.
  According to the PDB specification,
  columns 77-78 should contain the element symbol of an atom.
  but OpenBabel found '  ' (atom 276)
==============================

This thread from the OpenBabel mailing list goes into more details. Basically, the official PDB specification requires that the atom names be present off to the right-hand side of each ATOM/HETATM line. Older PDBs (and lazy PDB-outputting software) don't do that.
For the most part that works, as the atom name entry normally contains the element designation. Parsing this is tricky, though, as it doesn't always work.
That's what's causing Open Babel to choke. Your PDB contains gold atoms, and Open Babel doesn't know how to handle the "AU" atoms. If you manually edit the ATOM lines to contain the letters "Au" in columns 77-78, then Open Babel is able to make the conversion, but still warns about the C/N/O atoms which don't have their elements correctly designated. - But for C/N/O it's okay, as Open Babel is able to figure out the element from the atom name.
If you don't want to manually edit the lines, you just need to pass the PDB file through a program that can make sense of the AU atoms, and will output the element name in the appropriate colum. VMD doesn't seem to help (it reads things fine, but doesn't add the element annotation), but PyMol is able to read in the file and when re-saving the molecule will put the correct values in columns 77-78.
Regarding the "missing atoms", the PDB file contains a crystallographic symmetry record in the first four lines. (CRYST1, SCALE1, SCALE2 and SCALE3 records). I'm guessing the atoms you're missing are related by crystallographic symmetry to the ones that are present. To generate those atoms, use a program which can use the PDB symmetry records to generate the symmetric partners. (PyMol is one such program.)


text:
How to predict one equilibrium constant given two others?
$$\begin{alignat}{2}
\ce {CO2 (g) + 3H2 (g) \;& <=> CH3OH (g) + H2O (g)}\qquad&&{k_1=\;?} \\
\ce {CO (g) + H2O (g) \;& <=> CO2(g) + H2(g)}\qquad&&{k_2= 1.0\times10^5} \\
\ce {CO (g) + 2H2 (g) \;& <=> CH3OH(g)}\qquad&&{k_3= 1.4\times10^7}
\end{alignat}$$
What is the value of $k_1$? I have tried reversing equation two and then multiplying $k_2$ and $k_3$ together, but I am not sure how to work this problem out.

Remember that when you reverse the reaction the equilibrium constant changes. For the general gas-phase reaction $\ce{A(g) + B(g) <=> C(g) + D(g)}$ the equilibrium constant expression is $$K_\text{f} = \frac{p_\ce{C} p_\ce{D} }{p_\ce{A} p_\ce{B}}$$ (Strictly, it's activities, not partial pressures, but the principle is the same.)
The reversed reaction $\ce{C(g) + D(g) <=> A(g) + B(g)}$ has the equilibrium constant expression $$K_\text{r} = \frac{p_\ce{A} p_\ce{B} }{p_\ce{C} p_\ce{D}} = \frac{1}{K_\text{f}}$$
You are correct in multiplying the equilibrium constants to get the equilibrium constant for the combined reaction.

do I multiply k2 and k3 together to get kf?

I figured it out. The answer is 140.

@DaveLaRumbe That's right.

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text:
Is graphene a polymer?
A polymer is sometimes defined as a macromolecule with MW larger than 1000 g/mol. So can graphene be considered a polymer?

If you have access, this review article might be a useful read  - "Two-Dimensional Polymers: Just a Dream of Synthetic Chemists?". The definition of what constitutes a 2D polymer is still up for debate, but you can defend that graphene is indeed an example of one.

According to your logic, every defect-free piece of metal (above the nanoscale) would also be a polymer… It consists of many (!) repeating units and its macroscopic order can be described very well. If this doesn't count for you, take silicon as it's used for computer chips in wafers. This would be a polymer either. Do you get my point?

I have heard of graphite being called a polymer and this describes it as such https://shankharajsarkar.wordpress.com/2017/01/23/in-short-carbon-different-allotropes/   "graphite is a polymer of graphene."

Wow. Good question. Graphene is not considered a polymer in the traditional sense. While it is used a doping agent for other polymers it does not have the designation of being one itself, the allotropes of carbon, at least in my experience and that of my adviser, are typically in a class of their own. Unique structures such as buckminsterfullerene and diamond differ from graphene only by the angles between the carbon atoms, 3-d shape, (and number of carbons present). All of which are definitely not considered polymers. Also remember that a macromolecule is composed of polymers.
http://www.macrogroup.org.uk/schools/polymer_chemistry.php
https://answers.yahoo.com/question/index?qid=20100906193125AAzb9T0
So the short answer would be no, graphene is not considered a polymer.

What is so special about the allotropes of carbon? Also, where would you put elementary silicon? Is it a polymer, or does it also form a class of its own?

The common usage of the word "polymer" does not quite match the definition of it as "a macromolecule composed of repeated subunits". Looks like that the definition contains a specification note, tacitly assumed by everybody but never stated explicitly, which says that the subunit in question (i.e., monomer) must exist as a molecule in its own right. That would include all plastics, but exclude covalent crystals like graphene, diamond, silicon, red phosphorus, boron nitride, silicon carbide, quartz, etc.
On the other hand, buckminsterfullerene is a molecule with double bonds, and if we break some of them and connect the molecules together, we obtain a thing which (while still being a carbon allotrope) is rightfully called polymerized fullerene.

I think it boils down to semantics at the end of the day. While it does seem a bit unnatural, I don't really see a problem with saying that graphene is a 2D polymer of repeating "C" units.

The version of definitions to molecules, macromolecules and polymers that I accept is the following.
Macromolecules are defined as molecules with molecular weight larger than $10^4$ D. The limiting value $10^4$ is not meant to be strictly sharp. This definition emphasizes more on the following points

Macromolecules should be molecules. There are other types of pure substances that do not consist of molecules, and their "molecular weight" cannot be defined. Graphene fails at this point when it wants to be a macromolecule. This will be explained under the definition of molecules.
The molecular weight of macromolecules should exceed some very large value so that their physical properties are qualitatively different, which require additional theories and calls for a new field of study. For example, solutions of macromolecules often exhibit non-Newtonian flowing behavior, which requires an independent field called rheology to the already existing field of fluid mechanics. This idea clear up many other arguments. For example, I saw someone asking whether benzene is an oligomer of ethylene. But distinguishing benzene as such does not cultivate any new study.

Molecules are electronically neutral entities formed by one or more atoms solely by covalent bonds. A molecule should be the smallest unit of a pure substance that can exist (at least theoretically) independently and can retain all the physical and chemical properties of that substance.
Although allotropes of elemental substances of carbon (e.g., diamond, graphite, fullerene, carbon nanotubes, graphene...) are all composed of covalent bonds, only a few types are molecular substances.
For example, graphene failed to be a molecule because its smallest unit that preserves the physical/chemical property of graphene (which may be three
sp3 carbons cut from any graphene sheet) do not exist in the form of electronically neutral entity. Graphene does not "melt" into a liquid of such a thing.
In contrast, a molecule of H2O is the smallest existing unit that maintains the physical chemical property of the substance water.
This "smallest property-maintaining unit" constraint is not seen everywhere, but I think this is quite important and useful.
One can think of still many ways to increase the molecular weight of a molecule to the extent that it becomes a macromolecule, without making it not a molecule anymore. But the most feasible way is forming a covalently bonding chain of atoms long enough for its molecular weight to exceed $10^4$ (which defines the concept of long-chain molecules). DNA molecules (single-strand DNA) are often examples. But they are not polymers.
Polymers are long-chain molecules formed by repeating units.
And therefore single-strand DNA molecules are not polymers. DNA molecules are formed not by repeating, but according to a code of, the four nucleobases.
But let's see from an example why long-chain molecules are molecules in the sense that they are "the smallest unit of a pure substance that can exist independently and can retain all the physical and chemical properties" of a polymeric substance:
What smallest unit determines the melting point of a polymer? Does it need to be a whole chain of molecules? Or only the nature of its repeating units? The latter is correct at the first thought and excludes polymers from being molecules. But when we say that low-density polyethylene has a lower melting point (106°C to 112°C) than its high-density counterpart (120 to 130 °C), no difference in repeating unit is taking effect here.
However, the convention that polymeric "pure substance" does not strictly require all molecules to be the same molecular weight still creates some conceptual embarrassment in keeping polymers belonging to molecules.

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text:
How does the Lennard-Jones potential relate to energy?
Considering the Lennard-Jones potential,
$$V(r)=\frac{A}{r^{12}} − \frac{B}{r^{6}}$$
How does this relate to force or energy? I suspect that there is some physics equation I'm forgetting. I know that $F=−\nabla(V)$, but that's a partial derivative with respect to time, and I don't see how time comes into effect here. Also, the derivative of energy (work) with respect to displacement is force, but I'm not sure how to connect that to potential.
An example application that stimulated this question: given the interaction potential above, how would you find the interaction energy between two systems?

Tricky naming schemes come back to bite us once again. In my intro physics class, the "potential" was the amount of energy per unit something (e.g. energy per coulomb, or energy per kilogram). In this case though, "potential" simply refers to potential energy.

This helps, thanks so much! So for two systems, I guess it follows that one would just have to integrate to "add up" the interaction potential energy from all of system1 to all of system2. Is there I can rate this comment as best answer?

No, but I can convert it to an answer if you'd like. Quick question though (I didn't really think too much about this on my first reading): what do you mean by the interaction energy between two systems? Could you give an example system? When I think of a "system," I think of two atoms separated by $r$, in which case the LJ potential completely describes the energy of the entire system. I don't fully understand what you mean by "two systems."If we had two of those systems, this potential becomes meaningless (because we have 4 atoms, and 10 different distances ($r$) to consider)

We have some nomenclature issues here.
In physics, the electric potential and gravitational potential refer to energy per unit something. In the gravitational potential, it's usually energy per unit mass, e.g. kJ/kg. In the electric potential, it's usually J/C (more commonly known as a Volt).
Here, though, we're not talking about a potential in that sense. Rather, I suspect (though I may be wrong) that potential was once a shorthand for potential energy surface, which is a way to think about the potential energy of a system with respect to various parameters. In the case of an L-J potential, there is only one parameter, the distance between the two particles of the system.
As mentioned in my comment though, I don't understand what you mean by two systems. The Lennard-Jones potential describes the energy of one system. If you put two of these systems together, the potential becomes ill-defined.

Added later: I found today that the Wiki page on thermodynamic potential says the same thing, namely that the "thermodynamic potential" is actually an energy instead of a potential. It quotes an ISO/IEC document that I don't have access to. I'm still not sure of the origin of this nomenclature.

About two systems, I was thinking about two slabs of some specified number densities that faced attraction to each other. Referring to your example of four atoms, why would the potential become meaningless? Couldn't you just add up the relationships, i.e. (AB+AC+AD+BC+BD+CD), for each of the six distances? What are you referring to when you say 10 different r distances?

Oh, I see what you mean now. Yes, I suppose that's possible, but you'd be assuming that there's no other effects that come into play when you introduce multiple particles, and I don't know if that holds or not.

Also, for the 10 distances--I miscounted and included self-distances by mistake (AA,BB,etc.) . You're correct, it should be six.

Potential is the same as energy. As to the force, you got the formula right, it's $F=−\nabla(V)$. This is not a partial derivative with respect to time, but one with respect to coordinates, see the definition.

My mistake on the del operator, thanks for pointing that out. In general though, I remember learning that potential isn't the same thing as potential energy. Unless this case is, as @chipbuster noted, where potential refers to potential energy. Is that correct?

Well, if you'd remember what reasons were called to justify this point of view (that potential isn't the same as potential energy), I would be able to either argue or agree with those. As it stands now, I can do neither.


text:
Why does rain have that smell?
What chemical reactions in the soil cause that smell when it rains? In particular, what reactions in the soil causes the petrichor smell?

Interesting question. I actually had to look up the definition of petrichor.  After getting a little nostalgic about rainfall, I decided to go ahead and forage around for an answer for you.  I found a nice summary article here for those that are interested, and then a really nice bit of history here with a link to a great video about how rainfall helps to produce aerosols, which can contain the petrichor compounds.
Your question regards the reactions in the soil that give the petrichor smell, so let me get to that:
The main constituents of petrichor are volatile plant oils and geosmin. Oxidation of the fatty acids produced by plants can lead to some of the volatile compounds that constitute petrichor. Geosmin is a metabolite of actinobacteria and is produced through a biological pathway.

https://en.wikipedia.org/wiki/Geosmin - a brief intro to Geosmin
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3013058/ - A free paper highlighting the biosynthesis of geosmin

To all, be sure to see the high speed video of aerosols released from rain drops associated with the 2nd link. Pretty cool.


text:
N-alkylation of amides with alkyl halides?
Can amides be N-alkylated with alkyl halides? Say I was trying to make N-propyl acetamide. Would iodopropane and acetamide react to form it? Would they react to form N,N-dipropyl acetamide? If so, what catalysts would I need, and if not, why not and how would it be possible?

Anecdotic evidence is no evidence, but trying to methylate two free hydroxides on a $\ce{C5}$-amide I got full methylation, i.e. on the nitrogen, too. Conditions used were $\ce{Ag2O}$, $\ce{MeI}$ and stirring over the weekend at room temperature.

Sure, there are a number of ways to do this with copper, iridium, or palladium catalysts.  This link should hopefully be helpful.
http://www.organic-chemistry.org/synthesis/C1N/amides2.shtm

The amide nitrogen is not a good nucleophile to react. But it can become nucleophilic in the form of $\mathrm N^-$ as a conjugate base of a weaker acid. This means that you need a strong enough base to deprotonate the amide. Some standard conditions are NaH in THF or DMF, or n-BuLi, LDA and the like in THF/ether etc.

I've read of several related sn2 reactions. Generally the procced via an alkyl bromide, I2 catalyst and CsCO3 to drive the reaction to the right.
They all seems to be sensitive to the solvent employed. Many solvents seem to work, but the difference in yield can be very large.
From CH2Cl2 to CH3OCH(CH2) to formamide. I presume that the solvent system is the key and thus within the sphere of chemical engineering.

A better route could be to form the alkylated amine first and use an activated carbonyl group such as an anhydride or an acyl halide.
Although stericly hindered, the amine should react with good yield to form the amide.
There is a good chance a primary amine could simply be reacted with carboxylic acid under heat to form the amide, especially if reaction water can be drawn off, and even a secondary amine would be worth a try.


text:
Why can NCl3 be hydrolyzed but CCl4 cannot?
While explaining why $\ce{CCl4}$ cannot be hydrolyzed, we say carbon atom doesn't have any d-orbitals and therefore a water molecule (lone pair electrons of $\ce{O}$ atom) cannot form coordinate bond with carbon. So, $\ce{CCl4}$ cannot be hydrolyzed.
In $\ce{NCl3}$, nitrogen also does not have any d-orbital.
Why it can be hydrolyzed?
Is hydrolysis mechanism for $\ce{CCl4}$ and $\ce{NCl3}$ different? If so how they are different?
Again, in some sources I have found, in the case of $\ce{NCl3}$ water forms coordinate bonds with the $\ce{Cl}$ atom and hence the molecule gets hydrolyzed. Why does the same thing thing not happen in $\ce{CCl4}$ which also has a $\ce{Cl}$ atom?

Related question: http://chemistry.stackexchange.com/questions/37073/why-ccl4-does-not-dissolve-in-water-while-sicl4-does/

$\ce{CCl4}$ can be hydrolysed. Also, welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site and visit the [help] if questions about using it arise. Also note that you can improve your posts with MathJax; see this meta page for more information. I'll do it to this post of your's right away.

There are a few (mis-) conceptions in the question that require addressing other than giving the simple (yet correct) answer that permeakra gave.
$\ce{NCl3}$
The hydrolysis of this compound is different from the hydrolysis of the other two. This is because the $\ce{N-Cl}$ bonds are polarised towards nitrogen, not towards chlorine as they would be for $\ce{SiCl4}$ and $\ce{CCl4}$. Since the oxygen in water usually attacks nucleophilicly, the attack happens on the chlorine; it can proceed nicely with an $\mathrm{S_N2}$ mechanism (the rear side of all three chlorines are pointing into the solution). Note that another water molecule or the same will likely protonate the nitrogen either before or concertedly, so the leaving group won’t be $\ce{NCl2-}$ but rather $\ce{NHCl2}$. The other reaction product is $\ce{HOCl}$.
Other concerted mechanisms including only one water molecule might be possible, but sound less likely.
$\ce{SiCl4}$
Silicon atoms are rather large compared to both carbon and nitrogen: They have an additional electron shell. (Layman’s simplification, but it works.) As such, even in crowded cases like tri-iso-propylsilyl chloride $\ce{((CH3)2CH)3-SiCl}$ with three bulky iso-propyl groups, the silicon atom can be rear-side attacked by water (or catalysts such as 4-(N,N-dimethylamino)pyridine) liberating a chloride atom in a traditional $\mathrm{S_N2}$ mechanism.
No d-orbital contribution is required to explain it. In fact, silicon does not have ‘low-lying’ or ‘accessable’ d-orbitals; that is an old, superceded explanation.
The hydrolysis products of $\ce{SiCl4}$ are $\ce{Si(OH)4}$ (which will likely condensate to some polysilicate or $\ce{SiO2}$) and $\ce{HCl}$.
$\ce{CCl4}$
Technically, this could also proceed via an $\mathrm{S_N2}$ mechanism. However, carbon is much smaller and the chlorine atoms much larger in relation, so the rear-side attack is disfavoured sterically.
Instead, this molecule has access to the $\mathrm{S_N1}$ pathway, which can be described as follows:
$$\ce{CCl4 -> CCl3+ + Cl-} \\
\ce{CCl3+ + H2O -> Cl3C-OH2+} \\
\ce{Cl3C-OH2+ + Cl- -> CCl3-OH + HCl}$$
And so on, displacing the remaining chlorine atoms, ultimately liberating $\ce{H2CO3}$. (After the second chlorine displacement, phosgene will form which can be attacked directly by water with the $\ce{C=O}$ bond acting as an electrophile.)
Note that here again we gain $\ce{HCl}$, an indirect proof of chlorine being the more electronegative element.
Conclusion
The three molecules we considered are so different in their steric and electronic properties, that we need to consider three different mechanisms that lead us to two different sets of products (depending on which side gains the hydrogens and which formally gains hydroxides). Nowhere do we need d-orbitals.

$\ce{CCl4}$ can be hydrolysed under pressure at elevated temperature. Seems the reaction proceeds via $S_N1$ mechanism. See the source.


text:
Why don't polar and non-polar compounds dissolve each other?
Dipoles can also be induced in polar and non polar compounds, then why don't they dissolve?

Think of it like two groups of people who speak different languages, it's possible for them to try to mix around and socialise but it's just easier for them to stick to their own groups. Now chemicals are pretty rude and have no idea about social conventions, so they prefer to keep within their own group.

By the way dipoles aren't induced in polar compounds. They're already there. That's why you call them a permanent dipole. Their magnitudes will fluctuate but that's a different matter entirely

Shouldn't non polar dissolve in polar the way non polar dissolve in non polar?

No they should not. The disruption of the solvent is very different.

Very simply, you explain the reason for this solubility rule by taking in consideration the energy requirements for the breaking of intermolecular forces between the molecules in the solute and the solvent.
Note: this is only a simplified explanation as it also depends on other factors such as change in entropy
Here is some background information on intermolecular forces. In non-polar substances, there are dispersion forces between each molecule. These dispersion forces are relatively weak and hence only require little energy to break them. In polar substances, there are dipole dipole and hydrogen bonding (depending on the substance) between each molecule. These forces are much stronger than dispersion forces and require more energy to break.
Now lets consider the following cases:
Non-polar Solute and Solvent
For the solute to dissolve, the dispersion forces between the molecules in the solute and solvent need to break. This only requires very little energy. However when the solute dissolves into the solvent, they are able to be make dispersion forces with each other. The making of these forces releases very little energy. So simply put, very little energy is required to break the forces and very little energy released when making the forces. Hence overall everything balances out and the process occurs.
Non-polar Solute and Polar Solvent (and vice versa)
For the solute to dissolve into the solvent, both dispersion forces and dipole dipole forces are broken which require large amount of energy. However the molecules in the solute and solvent are only able to make dispersion forces with each other (as they aren't both polar). This only releases very little energy. Therefore overall, more energy is required than released and hence the process won't happen.
Polar Solute and Solvent
For the solute to dissolve into the solvent, dipole dipole forces are broken which require large amount of energy. However when they do dissolve, the molecules in the solute and solvent are able to form dipole dipole forces which releases large amount of energy. Therefore overall, everything balances out and the process occurs.

I was thinking the same but in my textbook there are some orders given like NaI is less soluble than LiI in Non polar solvent  because as per fajan's rule LiI is more covalent but its very vague statement because  I think dipole moment of NaI is   greater than LiI because Na^+ IS BIGGER than Li^+ and also polarizing power of Na+ Is also less so positive and negative charge density is well separated so its dipole -induced dipole interaction with solvent will be high and also its lattice energy is less than LiI so it should dissolve more than LiI but it doesn't how?

TL;DR
It's because the enthalpy changes of a solution generally don't favor dissolution.
A longer version:
To explain this, usually the enthalpy change explanation is given. For the sake of understandability, let's see what happens when two compounds dissolve. Take ethanol dissolving in water as an example. Here's the gist of what happens:

The intermolecular forces (i.e. hydrogen bonds in this case) in water break apart. $\rm \color{green}{(endothermic)}$
The intermolecular forces in ethanol break apart. $\rm \color{green}{(endothermic)}$
A new force and attraction is formed between the ethanol and water molecules. $\rm \color{red}{(exothermic)}$
$\hspace{12ex}$
$\hspace{22ex}$ Ethanol and water forming hydrogen bonds; Source

This happens for any two species that will dissolve in each other. For ionic solutes, the 'bond cleavage' is actually the lattice breaking apart. So you'd expect an endothermic process with a gaining of energy equal to the enthalpy of formation of lattice.
You have to be careful about two things:

Solubility isn't binary. We usually have to indicate it with short and comprehensible (i.e. by a large audience) words, and that's why we use it. Is gypsum as insoluble in water as calcium carbonate or as soluble as propanoic acid? (Propanoic acid is miscible in water in RTP and STP)
That "nonpolar doesn't dissolve in polar" isn't accurate. Nonpolar solutes are generally insoluble in polar solvents. We can easily think of exceptions. Bromine water is an example for a start, but certainly not the most remarkable example.

So the question that comes to mind is,

If the process for dissolution is going to be the same for polar or nonpolar molecules and crudely the same for molecules with hydrogen bond and for ionic compounds, why are some solutes insoluble in some solvents?

As we saw, there are two endothermic processes and one exothermic process involved. Very simply put, a rudimentary answer to your question is that "because induced dipoles are known as one of the weakest intermolecular interactions and thus the solvent-solute interactions wouldn't release enough energy while being formed, so $\Delta H > 0$. This would mean that thermodynamically it's more favorable for the solvent-solvent interaction not to be broken and hence, no dissolution".
In a conclusion, I would say "yes, the polarity of the solute/solvent plays an important rule in determining solubility or insolubility. But that's not the half of it." A small review of the matter can be found below.
$\color{gray}{\textit{I don't even know why I'm doing this.}\\\ \textit{A general, prescriptive rule can be explained mainly by another simplistic view of the matter.}\\\ \textit{Don't read on if you're not interested.}}$
However, just taking enthalpy into account isn't scientifically accurate. At least, that's not what's happening in real life. The meaning of enthalpy is associated with constant temperature and pressure. That's not what is happening in real-life chemistry.
For a start, it's best if we consider the hydrophobic effect (related to entropy), size of the solute species, rate of dissolution, the common ion effect, ionic strength.
The hydrophobic effect:
Just like a system favors the least potential energy, it favors disorder. The hydrophobic effect can better explain why some nonpolar molecules can't dissolve in water:

The hydrophobic effect is the observed tendency of nonpolar substances to aggregate in aqueous solution and exclude water molecules. This occurs because interactions between the hydrophobic molecules allow water molecules to bond more freely, increasing the entropy of the system. The word hydrophobic literally means "water-fearing," and it describes the segregation and apparent repulsion between water and nonpolar substances. - The hydrophobic effect, Wikipedia

Simply put, the reason for this isn't well-understood. A simplified explanation is that the structure of water allows it to have three degrees of freedom and it can form four hydrogen bonds. If it does so, it can't orientate as easily as it could and thus the entropy would decrease. So, to favor entropy, this needs to happen minimally.
If you want to study entropy of mixing, the linked Wikipedia article is very nice.
The size of species:
$\ce{AgCl}$ is less soluble in water than $\ce{AgNO3}$. This could best be described with the fact that silver and chlorine ions are almost the same size, and hence they may be packed tighter together. i.e. they're harder to break apart and 'dissolve'.
Note that "like dissolves like" or any similar rules can't explain this. Actually, the impressive delocalization of the electron in the nitrate ion can explain the massive range of soluble nitrates.
Rate of dissolution:
Would you still call a species soluble if it dissolves in the solvent in a wide enough time span? I'd imagine not, since the radioactive species with long enough half-life periods are commonly attributed the 'stable' medal.
Rate of dissolution is not a thermodynamic property, but a kinetical one.

Dissolution is not always an instantaneous process. It is fast when salt and sugar dissolve in water but much slower for a tablet of aspirin or a large crystal of hydrated copper(II) sulfate. These observations are the consequence of two factors: the rate of solubilization (in kg/s) is related to the solubility product (dependent on temperature) and the surface area of the material. The speed at which a solid dissolves may depend on its crystallinity or lack thereof in the case of amorphous solids and the surface area (crystallite size) and the presence of polymorphism. - Rate of dissolution, Wikipedia

Additives (Dispersants):
There's no obligation that we only consider the existence of solute and the solvent. What would you do if you needed to dissolve a fatty acid (which is hydrophobic) in water?
Getting help from micelles is a way. This is crudely the same way fats are being transported in the blood, and the same mechanism soaps use to cleanse oil from your skin.

Solubilization is distinct from dissolution because the resulting fluid is a colloidal dispersion involving an association colloid. This suspension is distinct from a true solution, and the amount of the solubilizate in the micellar system can be different (often higher) than the regular solubility of the solubilizate in the solvent. - Micellar solubilization, Wikipedia

$\hspace{3ex}$
$\hspace{7ex}$Micellar solubilization of fatty substance in water with the use of a dispersant -
Andreas Dries; Source
Additives (The common ion effect):
The common ion effect uses the Le Chatelier's principle to explain the lesser solubility of a certain precipitate because of the existence of a similar ion in the solution.
For instance, a dilute solution of magnesium sulfate is less soluble if some copper(II) sulfate is dissolved.
Ionic strength:
To expand on the concepts related to the common ion effect, ionic strength is defined:

The ionic strength of a solution is a measure of the concentration of ions in that solution. Ionic compounds, when dissolved in water, dissociate into ions. The total electrolyte concentration in solution will affect important properties such as the dissociation or the solubility of different salts. One of the main characteristics of a solution with dissolved ions is the ionic strength.
The ionic strength, $I$, of a solution is a function of the concentration of all ions present in that solution.$$I = \frac{1}{2} \sum\limits^n_{i=1} c_iz_i^2$$
where $c_i$ is the molar concentration of ion $i$ (M, mol/L), $z_i$ is the charge number of that ion, and the sum is taken over all ions in the solution. - Ionic strength, Wikipedia (Emphasis mine)

I thought I should expands on activity coefficients and how they lead to deviations from ideal behaviors of ideal solutions, esp. from Raoult's law, but I think that's enough for now. :P Still I didn't expand on pressure's effect, since it would require explanations regarding partial pressures.


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How high is the variance of the shore D hardness for a synthetical resin?
I realize that the variance of hardness depends on the homogenity of the polymer throughtout a specific sample, the polymer itself and on the measurement, but maybe there are typical values.
To be a bit more specific: Im interested in the order of magnitudes of the variance of hardness an adhesive or resin, mixed exactly as prescribed in the data sheet should have.


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Do we neglect the temperature changes during a reaction while calculating delta(H)?
Suppose for a reaction: $$\ce{$a$A + $b$B -> $c$C + $d$D }$$
we know the $\Delta U$ as $\mathrm dU$ and all the components of the reaction are gas.
Then to calculate the $\Delta H$ at a temperature $T$ we use the expression $\mathrm dH=\mathrm dU+\left(c+d-a-b\right)RT$.
Then while using this expression did we assume the temperature to remain constant as it wouldn't really be constant as the actual expression for $\mathrm dH$ is: $\mathrm dH=\mathrm dU+\left(pV\right)$ and so we can write $\mathrm dH=\mathrm dU+\mathrm d(nRT)$ and for $\mathrm dH=\mathrm dU+\left(c+d-a-b\right)RT$, $T$ has to be constant?

Yes. By saying that $\Delta (pV) = \Delta (nRT) = RT\Delta n$ you have made the assumption that $T$ is constant. When you take anything out of a $\Delta$ sign, you are saying that the initial and final values are the same. Otherwise you would have to write $\Delta (nRT) = R \Delta (nT) = R(n_2T_2 - n_1T_1)$ - this only simplifies to $RT(n_2 - n_1)$ if you assume $T_1 = T_2 = T$.

By the way, I want to be picky about the formulae that you are using. Look at $$\mathrm{d}H = \mathrm{d}U + (c + d - a - b)RT$$
The problem with this is that $\mathrm{d}H$ and $\mathrm{d}U$ are infinitesimals, but $(c + d - a - b)RT$ is a finite quantity. Do not ever add or subtract these two in the same equation. It is like trying to add a mass to a length - what is $30 \mathrm{~kg} + 25 \mathrm{~cm}$? It doesn't make sense.
Same thing applies for $\mathrm{d}H = \mathrm{d}U + (pV)$.

In addition, the equation $dH = dU + d(pV)$ already assumes that temperature is constant before anything involving $\Delta$ terms.  The fundamental equation is $dU = T dS - p dV$ or if you prefer $dH = T dS + V dP$ .  The differential entropy $dS$ of an ideal gas depends on both temperature and volume.  To neglect that term requires assuming $T$ is constant.

When the heat of reaction is referred to, it means the enthalpy of the pure products (in stoichiometric proportions) minus the enthalpy or the pure reactants (in corresponding stoichiometric proportions), all at the same temperature (usually 25C) and pressure (usually 1 atm.).


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Resin curing and adhesion materials
I have a question that applies to material science and chemistry, I have no idea how to answer it (my chem is weak) so I thought I would ask it...
I'm trying to make an SLA/DLP type 3D printer that uses DLP light to cure resins in UV spectrum...
I'm good to go mechanically and electronically... But I'm at the point now where I need to decide what build plate material to use... I need to resin to adhere well to the build plate for the printer to work properly... I've been doing research, but havent found much... Most people use aluminum or other metal plates, but I know there is a better material out there, I just haven't found it yet...
I found the MSDS stats on a certain resin, and thought if I figure out what the resin is made of, I can figure out what materials it will want to stick to given it's chemical properties... So here it is
Isobornyl acrylate (15-30%)
Acrylic monomer (15-30%)
Urethane acrylate (10-30%)
Acrylic monomer (5-10%)
Epoxy acrylate (5-10%)
Acrylate oligomer (2-5%)
Photoinitiator (<1%)
http://www.sculpteo.com/media/imagecontent/DOC-06119-Objet-VeroClear-RGD810-US.pdf
Do you have any ideas of what this would naturally want to stick to?


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Origin of diastereoselectivity in Simmons–Smith cyclopropanation of allylic alcohol
In Organic Chemistry by Clayden et al., the following reaction is shown:

The cyclopropane ring is formed on the same side as the alcohol group. Is this because of the interaction between Zn and the OH group, which directs the reagent to approach from the same side as the OH group?

Yes. Right after Clayden mentions this stereoselectivity, he proceeds to state that this is due to "coordination between the zinc atom and the hydroxyl group in the transition state" (pp. 1067-8, 1st ed):

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


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Nomenclature of tetrahydropyran ring
What's the logic behind the following nomenclature: 3,4,5,6-tetrahydro-2H-pyran?
Why not simply 2,3,4,5,6-pentahydropyran? Isn't the hydrogenation of the heterocycle obvious by including hydro- prefix? Specifying 2H-pyran seems to be redundant, unless it carries information about which position was hydrogenated first (which isn't the case).

There are two isomers of pyran, 2H-pyran and 4H-pyran.

In principle, it is possible to describe the corresponding saturated compounds using the prefix ‘hydro’, which yields the systematic names ‘3,4,5,6-tetrahydro-2H-pyran’ and ‘2,3,5,6-tetrahydro-4H-pyran’. However, these two names actually describe the same compound. In this case, the correct choice is ‘3,4,5,6-tetrahydro-2H-pyran’ because ‘2H’ is lower than ‘4H’ (note that indicated hydrogen atoms have priority over ‘hydro’ prefixes for low locants).
All locants are omitted in compounds or substituent groups in which all substitutable positions are completely substituted or modified, for example, by hydro, in the same way. Thus, in this case, the systematic name is simplified to ‘tetrahydro-2H-pyran’. Furthermore, omission of indicated hydrogen is permitted in general nomenclature if no ambiguity would result. Thus, in this case, a correct name is just ‘tetrahydropyran’.
Anyway, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), preferred IUPAC names of saturated heteromonocyclic compounds are formed by the extended Hantzsch–Widman system. Therefore, the preferred IUPAC name of the given compound is ‘oxane’.


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Does the temperature during this reaction remain constant?
Considering all gases to be ideal, the enthapy change for the reaction
$$\ce{I2 (g) + H2 (g) -> 2 HI (g)}$$
is $\pu{−106.78 kJ}$ at $\pu{350 K}$.
To find the change in internal energy the formula we use is
$$\Delta H = \Delta U + (\Delta n)RT \, .$$
To derive this equation we must have assumed the temperature during the reaction to be constant in the expression
$$\Delta H = \Delta U + \Delta(nRT).$$
I know that $\Delta U$ depends on temperature and the amount of substance, but if we assumed the temperature to be constant then $\Delta U$ should be $0$,
as the amount of substances of reactants and products are also the same, but it has a non-zero value. Is this possible?

Explain to me, why should $\Delta U$ be 0 for a constant-temperature reaction?

If the change in temperature during a reaction is 0 then ∆U=nCv∆T and ∆T is 0 so ∆U should be 0.

Under what conditions does that equation apply? This is the danger with thermodynamics. If you don't remember what equations apply when you will end up using them in entirely inappropriate situations.

Doesnt the  change in internal energy is equal to nCv∆T always?

Nope. Go and reread your notes or whatever material you have.

No idea why you'd use adiabatic - things are getting cooled normally, aren't they?

Can you please tell me when this is applicable an when it is not as there is no conditions given in my study material, if it does not mind you?

If things are cooled normally then the there should some temperature change during the reaction and so ∆(nRT) should not be equal to (∆n)RT.

In short, this is applicable when there is no reaction.

I have changed the reaction, please consider my question again.

Refer to my answer here. http://chemistry.stackexchange.com/a/39719/16683

It is often said that $U = 5nRT/2$ for a diatomic gas (c.f. $U = 3nRT/2$ for a monatomic gas). According to this formula, the internal energy should not change in the reaction you considered, provided that the temperature is held constant. The reason is that in this reaction, one always has the same number of diatomic molecules in the system, regardless of how far the reaction has proceeded.
However, $5nRT/2$ is not the entire internal energy of the gas! This value reflects only the (translational and rotational) kinetic energy associated with the overall motion of each molecule. The total internal energy  should also include the energy associated with the relative motion of atoms and electrons within each molecule. The energy of the relative motion manifests itself in the bonding energy. In this particular case, the total bonding energy of two HI molecules is different from that of an H2 molecule and an I2 molecule, so the internal energy should change in the reaction.

The internal energy of an ideal gas mixture is determined not only by the temperature, but also by the chemical composition of the mixture, and, in the case of a chemical reaction, the composition changes.  For a real gas, the internal energy also becomes pressure-dependent at higher pressures.

What if the gas is ideal and temperature and the no of moles of reactant and product remains the same as in case of the reaction in my new question?


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What is the best Electronegativity cutoff for my calculator?
I have written a Chemical Bond Polarity Calculator application, which many students use for their homework.  I sometimes get questions from these students as to why the cutoff I use is 2.0, and not some other number between 1.7 and 2.1.  I understand that there is no hard and fast rule as to what the cutoff should be.  I just want to know what a real chemist (I'm a computer programmer, not a chemist) would want this particular application to use for the cutoff.  Conversely, would a real chemist want to be able to input the cutoff themselves?
This post indicates that 2.1 is a good cutoff, and this post indicates the number should be between 1.7 and 1.9.  How would the real chemists out there recommend I modify this application?

I don't know to what extent this answers your question, but I'll just drop my 2 cents about "getting questions from these students". Just be honest. Tell them it's arbitrary. Tell them that someone just threw darts and decided that 2.0 is a good cutoff and that in reality there's a whole spectrum between ionic and covalent and if they want to use 1.7 as their cutoff, so be it. Just that the teacher will be using 2.0 so it's probably easier for them to stick to 2.0 so that they're on the same page as the rest of the class.


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Stability of furanose vs. pyranose form of glucose?
I am trying to imagine the cyclization of the Fischer projection of glucose as it forms an $\unicode[Times]{x3B1}$-D-glucofuranose and an $\unicode[Times]{x3B1}$-D-glucopyranose (right now I am trying to understand the Haworth projection, and as my 3D visualization gets better, I will extend to the more thermodynamically stable chair representations of these monosaccharides). Now, if I understand correctly, in the pyranose form, the alcohol on $\mathrm{C5}$ attacks the carbonyl carbon on $\mathrm{C1}$. In the furanose form, the alcohol on $\mathrm{C4}$ attacks the carbonyl carbon on $\mathrm{C1}$. These results in 6-membered and 5-membered rings, respectively.
I understand in nature, these are the most stable rings. However, what thermodynamic incentive does the furanose form have? That is, why would $\mathrm{C4}$'s alcohol attack and form a 5-membered ring at the expense of the $\ce{CH2OH}$ group and the $\ce{OH}$ group being near each other as a side chain on the 5-membered ring? Doesn't this introduce considerable strain? My idea is that the 6-memebered ring is more thermodynamically stable and it doesn't have the problem of a hydroxymethyl and a hydroxyl group on a single carbon of the 5-membered ring.

If you look up the statistical distribution of glucose isomers in aquaeous solution, you will find $36\,\%$ $\unicode[Times]{x3B1}$-pyranose and $64\,\%$ $\unicode[Times]{x3B2}$-pyranose. All the remaining forms lumped together only account for less than $1\,\%$ of glucose in solution; that includes both furanoses, the open-chain form and the (possible, but unlikely) seven- and four-membered rings.
Neither five-membered rings nor six-membered rings contain any relevant amount of angle strain. You can easily verify this buy getting yourself a molecular model kit and attempting to bild these rings from four-bond sp³ carbons: both can be built smoothly, like a charm. So angular strain is not a reason why the pyranose forms are preferred.
Considering reaction mechanisms, the formation of five-membered rings is usually kinetically preferred due to easier reachable transition states. So this also cannot be the reason for preferring six-membered rings.
The actual reason is not so much angular strain, but rather dihedral angle strain: the same strain that prevents ecliptic conformations in alkanes. The pyranose form has the perfect $60^\circ$ dihedral angle between two non-ring atoms which minimises this type of strain. Furanoses must choose between envelope or twist conformations to at least relieve some of this strain.
All this is, however, happening at an equilibrium, which is semi-rapid in adjusting itself. So there will always be a certain percentage of furanose form in solution being formed and breaking apart. If any kind of reaction selectively only happens with furanose anomers, then this will cause the formation of new furanose which again will be captured away. As I wrote before, the attack which creates a furanose is preferred over that which creates a pyranose. (But the breaking apart of pyranose is slower than that of furanose.)

The notion of a statistical distribution's area being <1% sold me quite well. I hadn't thought about that. Also, thank you for eloquently explaining the difference between preference of attack and preference of intermediates.

Then what is the reason that Furanose is more stable in case of Fructose ?

@AdityaShrivastav According to the Wikipedia article, pyranose is the predominant configuration ($70~%$) of fructose in solution. Please state your source.

Thank you for the fact ,I never came across it. I always seen only furanose structure of the fructose in many books. Then why Sucrose consist of furanose form of Fructose ?
https://en.m.wikipedia.org/wiki/Sucrose#/media/File%3ASaccharose2.svg

@AdityaShrivastav In sucrose, you are no longer dealing with a hemiketal function in the fructose moiety that can react back and forth rapidly. Instead, you have a kinetically rather inert full ketal that is not subject to equilibration like free fructose is.


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How does absorption spectroscopy work?
So a beam of white-light photons passes through a gas and various photons, having exactly the same energy as the gas's molecular energy transition state deltas, get absorbed. We then get a characteristic absorption pattern. But wait a minute, don't the energised electrons fall back to their ground states within a nanosecond and create photons of precisely the same energy as those that caused the disturbance? How come we still see the absorption band?

Basically, our beam was aimed at one direction. The re-emitted photons will go to all directions, so most will perish without us seeing them.

Plus, who says the will drop directly back down - there are other (radiative and not) channels for the electrons to decay.

Thank you Ivan. However if the emitted photon goes in ALL directions how can you explain Fluorescence? And thank you Jon. You appear to be suggesting that in some cases the atom's relaxation from an energised state can be reconciled by something other than photon energy. In my self-taught reading I haven't come across this and would appreciate some guidance. Of course I understand that there are both vibrational and rotational quantised energy states but these are considerably lower in energy than electronic energy levels.

What about fluorescence? It's like I said: the re-emitted photons go to all directions, and some of them just happen to go in the right direction to get seen by us.

In an absorption experiment a collimated beam of light is passed through the sample. The intensity of this is measured before and after the sample. The absorption is measured as optical density which is $D=[c]\epsilon _{\lambda} l$, where $[c]$ is the concentration, $\epsilon_\lambda$ the extinction coefficient at wavelength $\lambda$ (a measure of the strength of absorption), and l the path length through the sample. At each wavelength the measured light intensities are $I=I_0\exp(-D)$ where $I_0$ is the initial intensity. This is called the Beer–Lambert law.
Now as you mention the sample is placed into its excited state where it remains for some period of time. This depends on the molecule being studies but may be from picoseconds (relatively rare) to several nanoseconds to microseconds (also rare). In an absorption spectrophotometer the light intensity used is low, so the excited state population is always very small. As any fluorescence produced is emitted over $4\pi $ radians (in solution) the amount captured by the detector will be small, just on geometric grounds as the detector is placed some distance away from the sample and the light passes through slits before entering the detector.(The slits & grating give spectral resolution). Also note that the fluorescence is mainly at a longer wavelength than absorption but the zero zero line in absorption and emission can overlap. The geometric reason just given suffices to make absorption accurate even in this case.
In laser experiments the situation can be different, particularly when pulsed lasers are used. It is quite easy to observe 'gain' in  the sample, this means that transient signals that are more intense than the measuring (probe) beam are observed. This is due to stimulated emission caused by the probe beam. It is also quite easy to see amplified stimulated emission (ASE) from a cell containing a fluorescent dye if the laser beam is focussed.
EDIT:
An explanation as to why stimulated emission is not normally observed in an absorption experiment and why fluorescence is observed under similar conditions is given below.
In the simplest case of two levels $1$ and $2$ , where $1$ is the lowest level the rate of absorption from $1$ is $N_1B_{12}\rho(\nu)$ where $N_1$ is the population of level $1$, $B_{12}$ the Einstein coefficient for (stimulated) absorption from level $1$ to $2$ and $\rho (\nu)$ the energy density per unit frequency interval ($\pu{Jm^{-3}/s^{-1}}$). The rate of transitions from $2$ to $1$ is $N_2[B_{21}\rho(\nu) +A_{21}]$ where $A_{21}$ is the rate of spontaneous emission from level $2$ and $B_{21}$ the coefficient of induced (stimulated) emission. Note that $A$ does not depend on the number of photons present. In is convenient to assume that the molecule is bathed in thermal radiation that follows  Planck's blackbody law at temperature $T$,
$$\rho(\nu)= \frac{8\pi h\nu^3}{c^3}[e^{h\nu/(k_BT}-1]^{-1}$$
At equilibrium the rate of change $dN_1/dt=dN_2/dt=0$ then
$$ \frac{N_1}{N_2}= \frac{B_{12}\rho(\nu)}{B_{21}\rho(\nu) +A_{21}}$$
The ratio of populations is also given by the Boltzmann distribution, also at temperature $T$
$$ \frac{N_2}{N_1} = \frac{g_2}{g_1}e^{-h\nu/(k_BT)} $$
where $g_1,\,g_2$ are the degeneracies of the two levels.
Combining these last two equations and solving for $\rho$ gives
$$\rho(\nu) = \frac{A_{21}/\,B_{21} }{ \frac{g_1B_{12}}{g_2B_{21}}e^{h\nu/(k_BT)}-1}$$
For this relationship to be true $g_1B_{12}=g_2B_{21}$ and $A_{21} = n(\nu)B_{21}h\nu$ where
$$n(\nu)= 8\pi\nu^3/c^3$$
and is the number of radiation modes per unit volume per unit frequency.
This relationship means that the ratio of induced to the spontaneous emission rate (leading to fluorescence) is equal to the number of photons in this mode.  The figure shows that in the visible part of the spectrum,  $\approx 400$ to $700$ nm at at thermal emission at room temperature the number of photons mode is somewhere between  $10^{-10}$ and $10^{-100}$, the exact number does not matter much as it is so small. Thus this means that the spontaneous emission rate (i.e. fluorescence) far exceeds that of stimulated emission. Under normal circumstances where the radiation is weak, as in a spectrophometer, and $N_2 \ll N_1$, the linear absorption spectrum is obtained.  If many photons can be concentrated into a few modes,  then for these modes, stimulated emission may exceed spontaneous emission; the result is a laser amplifier.

The figure shows the average number of photons/ mode with temperature, assuming Planck black body distribution vs. the photon frequency. The vertical lines are approximately the range of visible light. The horizontal dashed line is at $300$ K.

First consider only a two state system with states A and B:
Before radiation they are in equilibrium and we set the groundstate energy of A to zero. This means ratio of occupation between the states is just:
$$
p_B=\exp(-E_B/(k_\mathrm{B}T))
$$
You excite from state A to state B with the right frequency/energy and disturb this equilibrium. If we assume a steady beam, we can apply Einstein radiation theory.
You will see that there is spontaneous and stimulated emission.
Stimulated emission is coherent and anisotropic and leads to your paradoxon.
When looking at the equations in the wikipedia link you see that it appears the more the equilibrium is distorted. If you have an occupation ratio of one only stimulated emission is happening. (In lasers you are using exactly this property)
Spontaneous emission is isotropic. So if you put your detector in a sphere around your absorbing molecule you get again your paradox. Usually you put the detectors parallel to the the beam in a small area. The ratio between the detector area and the virtual sphere with the same distance/radius gives the fraction of spontaneous emission that reaches the detector.
This applies to all quantum states and is true for rovibrational, UV-Vis... spectroscopy.
Now since you asked for fluorescence in your comments. This is more involved since you need to know about selection rules and at minimum a three state system. Especially it is not quite connected with your "paradoxon". A good start would be Jablonski diagrams.

The absorption spectroscopy you are referring to is a steady-state technique, which means that a continuous light source is used for the excitation. Usually, this is a lamp in a UV-VIS molecular spectroscopy experiment.
To answer your question, in a steady-state technique the medium (atoms or molecules) that you want to measure is constantly excited by the light source during the experiment, i.e. the electrons are always in the excited states and the overall process results in an absorption, i.e. attenuation of the light excitation.
The experiment where you should be concerned about excited electrons falling back to the ground state is a time-resolved one. Here the light source that perturbs the medium is pulsed (usually a laser with pulse duration ranging from ns to fs) and the medium is excited very shortly (depending on the pulse duration) , than another light pulse is used to follow the evolution of the excited states, that in the easiest case consists in excited electrons going back to the ground state.


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Is this reaction with H+ and OH− a possible reaction?
I am wondering about the reaction $\ce{H+ + O^2- -> OH-}$. Is it possible?
I think this is not a possible reaction, because here a strong acid plus another particle reacts to a strong base. If it is possible, what would this mean for $\ce{O^2-}$ in the acid-base system? Would it be a very, very strong base?

Don't use mathjax in title.

Well, let's look from this POV. You take a strong acid $\ce{HCl}$, subtract a proton, and get $\ce{Cl-}$, which is not an acid at all. You take a weak acid $\ce{CH3COOH}$, subtract a proton, and get $\ce{CH3COO-}$, which is in fact a weak base. You take a very weak acid $\ce{H2O}$, subtract a proton, and get $\ce{OH-}$, which is a strong base. See the trend?

So, yes it's very strong - one of strongest common bases.

You can try this in the lab: Take some $\ce{CaO}$, add $\ce{HCl}$ and see what happens. Actually, no don’t. Don’t try this unless you know exactly what you are doing.
A single proton does not exist, especially not in aquaeous solution. Using $\ce{H3O+}$ as a proton-donating species is a much better way to formulate reactions. A professor once said, the smallest actually proven proton-equivalent be $\ce{H9O4+}$, so four water molecules that shift the additional proton between them, but since it was first-year, he didn’t go into details. Also in organic solutions or others, protons will always be bound to something, however weakly.
Now that I got that out of the way, let’s take a look at the $\ce{H3O+}$ particle I just postulated. It has three hydrogens connected to an oxygen and in a very, very, very broad way of arguing, every hydrogen can be considered a potentially acidic proton. Therefore, we can think of hydronium as a triprotic acid with the following conjugate bases:
$$\ce{H3O+ ->C[- H+] H2O ->C[- H+] HO- ->C[- H+] O^2-}$$
Aha! Oxide ions can be considered as fully deprotonated water. Of course, the entire reaction series is reversible, so add ‘protons’ to oxide ions (whatever the source) and find that they get protonated to hydroxide and finally to water. You knew of the second step (hydroxide to water). And here’s the big surprise: The first one is identical, only the equilibrium is much, much further on the hydroxide’s side.
Adding excess acid to oxides would just create water. However, the acid-base neutralisation reaction is exothermic. Very exothermic. Which is precisely why I wrote do not try this unless you know exactly what you are doing at the top. If you have ever neutralised semi-concentrated hydrochloric acid and sodium hydroxide, you know how much heat can be created. Think more than twice that.
In fact, water has what is termed a nivelling effect: Water itself is too strong an acid to allow oxide bases to exist in its vicinity; so $\ce{O^2-}$ cannot be dissolved as is, it will immediately get protonated to hydroxide. Find some $\ce{CaO}$, dissolve it in water, and you would immediately get $\ce{Ca(OH)2}$. Oxides are only able to exist in aquaeous solutions if they are either complexed entirely by cations ($\ce{Al}$ can form oxido-hydroxido multi-core complexes at $\ce{pH 7}$ which include internal oxide ions) or if they are in a bound state, so not technically oxides, such as in the vanadyl cation $\ce{(VO)^2+}$.
Finally, you stated something in your question saying that a strong acid would react with something to a strong base which you think is impossible. That is not the case; if the product of the reaction of something with a strong acid is a strong base, then ‘something’ just needs to be an even stronger base. Which is precisely the case of oxide ions. Just to recall for a second that strong acids and even stronger bases react very, very exothermicly. I feel I have not stressed it enough: Do not try this unless you know exactly what you are doing!


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What's causing my plastic fork to break?
This has been a real mystery for me at work. For the past three days I've eaten salads for lunch. I used a plastic bowl with a lid and when I was finished with the salad, I didn't rinse out the bowl and left my plastic fork in the bowl with the lid sealed over night. Well, the next day, when I open the lid, the prongs of the fork are always broken off... like they melted or something. This has happened three days in a row.
Is it something to do with the sealed container, or the dressing residue?  Each day I used a different salad dressing  but the result was the same. The fork was just laying in the bowl so there's no reason it should be broken in pieces the next day...

We'll need more info about those salads in order to properly answer this query, and for sure need to find out about that dressing in detail. Please amend your post with more information and we'll hopefully get some resolution on the fork issue before any further negative utensil-related incidents occur.

If it's at work, using a communal fridge, suspect a coworker with odd sense of "humor".

DrMoishe Pippik has a point. Try it at home on a weekend. // The only other idea is that some oil could be dissolving into the surface layers of the plastic creating a lot of stress.  That would seem really weird though.

Those cheap plastic forks carry a lot of internal tension from the production process, check google for pictures on "plastic fork polarised light".

It may be due to acids present in the salad.

It's highly probable the fork is made of some kind of biodegradable or compostable material designed to mimic plastic. Most plastics, especially ones used to make utensils, are stable under the conditions you described. I would carefully inspect the fork for any numbers molded on the plastic shown on the chart.

Biodegradable "plastics" are intentionally designed to become unstable and breakdown overtime. It's possible the salad dressing is made with vinegar or citrus juice which is acidic enough to breakdown or at least react with the material. Water can have the same effect, but may take longer, since you don't want the fork disintegrate while you're eating.


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Are there any substances that react very slowly with water to create heat?
I know there are things like the alkali metals that react very violently with water. I was wondering if there is anything that produces a slower more controlled reaction?
Also, is there anything that reacts with water to form heat that is safe for humans to touch?
I'm trying to think of ways to melt snow besides salt.

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site and visit the [help] if you have any questions about it. I know that there are species that will react slow enough to liberate heat in a controlled and safe manner, but I think that none of those will come close to the cost-efficiency of salt (let alone top it).

Hand- and foot-warmers are made containing iron filings, with a bit of paper pulp and salt, and sometimes a sealed water packet. These rely on the (surprisingly rapid) rusting of iron in contact with air, catalyzed by the water and salt (the pulp is just to hold the water and keep it in contact with iron). This reaction is a fairly safe way to melt a bit of snow, with some caveats:

It's more expensive than using freezing point depression of salts such as $\ce{CaCl2}$, $\ce{Mg(C2H3O2)2}$, or $\ce{NaCl}$ (table salt). Even adulterated wine containing ethylene glycol has been used for runway deicing.
The rust will eventually leave brown streaks on any surface below the deicer, such as the walls of a building if used to melt snow on the roof.


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Using gas laws to find density

An unknown gas at $\mathrm{63.1\ ^\circ C}$ and $\mathrm{1.05\ atm}$ has a molar mass of $\mathrm{16.04\ g/mol}$.
  What is the density of the gas?

I know I need to use the $pV=nRT$ equation but it gives me g/mol and not moles. I'm not really sure what to do here.
$$1.05\ V = n\ 0.08206 \cdot (63.1+273.15)$$
Ok that's as far as I got. I have no idea how would I even go about converting g/mol back to moles so I can solve for V, and after that I would also need to figure out a way to find grams so I can divide g/v to get the density.

Put 1 L for V so you can see how many moles there’d be in a liter. Convert between mass and moles by the equation mass = moles * molar mass. Divide your mass by the volume (1 L) and you’ll get the density in grams per liter.

@lightweaver Thank you very much, so if a variable in any equation isn't specified is it automatically assumed to be 1?

In this case you can set one of the variables to 1 as density is known as an intensive  variable, meaning that its value is independent of how much material there is. So it could be any number, but the value will remain the same. We just set it to 1 as it simplifies the calculations.

In my answer I have done the same thing, but I made the number of moles 1 and then see how many litres there are in 1 mole. Either way, you will get the same answer

@Nanoputian I tried the method suggested by lightweaver and it got me .61 which was the correct answer for online homework, maybe you did some calculations wrong? Either way since no one else answer and you tried to help I will award you the best answer, thanks.

Thanks for notifying me. Sorry I was rushing when I was typing up my answer and plug in some of the values wrong. I will edit my answer. Happy to know that I could help.

The molar density is n/V, so the mass density is Mn/V, where M is the molar mass, n is the number of moles, and V is the volume.  So Mn/V= (pM)/RT.

Okay, to find the density of the gas, to need to know its mass and its volume. So to do this, lets say we have one mole of this substance. So its mass will be $16.04~\mathrm g$. Now all we need to do is find its volume.
This can be done by rearranging the ideal gas equation: $$V = \frac{nRT}{p}$$ Now all we have to do is plug in the values. However you have be careful that you use the write units. This is a common mistake, especially when using the ideal gas law however it is good to see that you have used the correct units: $$\begin{align}V &= \mathrm{\frac{1~mol\times0.08206~L~ atm~ mol^{-1}~K^{-1}\times336.25~K}{1.05~atm} }\\
&=\mathrm{26.28~L}\end{align}$$
As you can see, all the units cancel nicely so that you get the final answer in litres which is the correct unit for volume.
Now, density can be calculated: $$\rho =\mathrm{\frac{16.04~g}{26.28~L} = 0.61~g~L^{-1}}$$


text:
Why does electronegativity increase as effective nuclear charge increases?
I know that electronegativity is the ability to attract shared electrons and that effective nuclear charge is the pull of the nucleus on outer electrons based on my notes.
But I'm not really sure how to explain the connection. Please help.

I would like to point out that electronegativity does not necessarily increase with increasing effective nuclear charge. Take for example, the elements magnesium and calcium. Ptable clearly shows that magnesium has a higher electronegativity than calcium (1.31 vs 1.0), despite its lower effective nuclear charge (Zeff) acting on its valence electrons (3.308e vs 4.398e), as provided by Wikipedia. (Just as a side note, the e in the Zeff values represent the elementary charge of a proton, or 1.60 x 10-19 Coulombs to 3 s.f.)
As Jan mentioned, electrostatic forces are dependent both on the charge of the two bodies in question, as well as their distance of separation. I think it's useful to include the formula for Coulomb's force to better illustrate this point:
$$F=k_e\frac{q_1q_2}{r^2}$$
Because electronegativity, like others have mentioned, is a measure of an atom's tendency to attract external electrons, it is dependent mainly on the net electrostatic force of attraction exerted by the atom. As such, the Coulomb's Force comes in handy to explain the magnesium and calcium disparity; say we take q1 as being the Zeff value (which takes into account inter-electronic repulsion between the electron shells), q2 as being an external electron with charge -e, and r as being the atomic radius (from Ptable, calculated), we can substitute the relevant values for calcium and magnesium, such that:
$$F_\mathrm{Ca}=k_\mathrm{e}\frac{-4.398|e^2|}{(\pu{194E-12})^2}=k_\mathrm{e}\frac{-4.398|e^2|10^{24}}{37636}$$
$$F_\mathrm{Mg}=k_\mathrm{e}\frac{-3.308|e^2|}{(\pu{145E-12})^2}=k_\mathrm{e}\frac{-3.308|e^2|10^{24}}{21025}$$
From the output magnitudes, the higher electronegativity of magnesium than calcium is justified. Numerically speaking, the distance of separation (i.e. atomic radius) as a (squared) denominator term influences the final electrostatic force of attraction much more than effective nuclear charge does. This actually also applies to the electric potential energy term derived from the Coulomb Force. Using atomic radius as a more significant indicator of an atom's electronegativity would account for the general decrease in electronegativity down a group, as well as its increase across a period (left to right), even though effective nuclear charge acting on an atom's valence electrons technically increases in both cases (since the addition of protons with increasing atomic number will always outweigh any shielding done by the addition of a similar number of electrons to the surrounding shells).
Note: The example here uses Zeff values assuming the corresponding valence subshells are not filled, so the actual electric force acting on a mobile electron may be further influenced by interelectronic repulsion with the valence subshell electrons, as well as how many valence subshell electrons there are in the first place. The point is just to raise a method of visualising the electronegativity trend.
Initially, I was also confused as to why a higher Zeff could, in fact, yield larger atomic radii down periodic groups; a simple explanation would be that the increase in the number of electron energy levels (in the form of principal shells) would inherently cause a corresponding rise in atomic radii. This doesn't apply across a period because the number of principal shells doesn't change then; the number of subshells can, change, which is why the trend in atomic radii across a period is also not perfectly "neat".
P.S. Apologies for the 7 year late reply, thought I'd elaborate further on the points brought up by Jan and better illustrate them with some calculations.

The effective nuclear charge can be thought of the charge of the nucleus minus the charge of the core electrons. For an element such as fluorine, the nuclear charge is $+9$ and the core electrons have a charge of $-2$ so the effective nuclear charge is $+7$. Similarly for carbon it would be $+6 - 2 = + 4$.
Now let’s assume a $\ce{C-F}$ bond. There are two electrons in this bond, each charged $-1$. On the left, from the carbon atom, they ‘see’ a charge of $+4$; on their right, from the fluorine atom, they ‘see’ s charge of $+7$.
Electrostatic interactions are greater at shorter distance and at greater charge separation, so the interaction (i.e. force) the bonding electrons experience from fluorine’s side is almost double that from carbon’s side. (The carbon and fluorine atoms can be considered to be roughly of the same size.) It should now be clear why the electrons are drawn to the fluorine.
Electronegativity is just a name humans gave the observed effect of ‘this atom can pull electrons towards it better than others’. It takes a look at the result of the above process and rationalises it. The effective nuclear charge approach is the underlying physical force.

Note: This explanation is extremely simplified, does not take any quantum effects or the like into consideration and should thus only be used at introductory levels.

Electronegativity can be thought of how much an atom "wants" electrons.
An atom that wants electrons is one that has a high positive charge, or effective nuclear charge. Effective nuclear charge is the pull the nucleus has on outer electrons (taking into account the repulsion of electrons in the atom.)
The higher the effective nuclear charge, the more the nucleus wants electrons, which is why atoms high in one is high in the other and vice versa.

I think Lewars textbook gives a good outline as to what electronegativity is. The thing with electronegativity is that that it is vague. Anyhow, here is what Lewars says:

From this viewpoint electronegativity of a species is the drop in energy when an infinitessimal amount (infinitessimal so that it reamins the same species) of electronic charge enters it. It is a measure of how hospitable an atom or ion, or a group or an atom in a molecule, is to the ingress of electronic charge, which fits in with our intuitive concept of electronegativity.

You could put it this way: the nucleus attracts the electron more if it has a high effective nuclear charge. Remember that nuclear charge is positive, and that electrons are negative https://en.wikipedia.org/wiki/Electric_potential_energy. Thus an electronegative atom/ion/etc "gains more energy" when it pulls an electron towards it. And the reason it "gains more energy" is because it has high effective nuclear charge.

What's "Lewars textbook"?

"An Introduction to Computational Chemistry" Errol G. Lewars


text:
What metals can be melted inertly in alumina crucible?
Specifically, a crucible made from 90+% Aluminum Oxide.  What metals can be melted in it without any weird reaction taking place?

Could you elaborate as to what you think would be a "weird reaction"?

I would suppose that the manufacturer would provide details as to what the melting point of the crucible is....and it probably isn't a good idea to find for yourself. Anyhow for sure you can melt gallium.
https://en.wikipedia.org/wiki/Gallium
But it probably cannot melt tungsten.
https://en.wikipedia.org/wiki/Tungsten

I meant common engineering metals, like metals bicycles or cars would be made out of... :D


text:
Calculating volume of gas produced

For the following reaction, what volume of $\ce{SO3}$ can be produced from $4.2\ \mathrm{L}$ of $\ce{O2}$ (measured at the same temperature and pressure), assuming an excess of $\ce{SO2}$?
  $$\ce{2SO2(g) + O2(g) -> 2SO3 (g)}$$

I demonstrate what I tried here but I have no idea what to try here, anyone have any suggestions on what to do?

TL;DR There is twice  the amount of $\ce{SO3}$ produced compared to oxygen, thus the volume will be twice as much as oxygen. Therefore $8.4$ litres of $\ce{SO3}$ will be produced.
Since $\ce{SO2}$ is in excess, the amount of $\ce{SO3}$ produced will be twice as the amount of moles of oxygen. So if there is originally x moles of oxygen gas, 2x moles of $\ce{SO2}$ is produced. It is important to note that temperature and pressure remain constant throughout the reaction.
Note one of gas laws state, the same number of moles of any gaseous substance will have the same volume at constant temperature and pressure. So x moles of $\ce{SO3}$ will hold the same volume as x moles of oxygen gas, so 4.2 litres.
Now there is also another gas law which states that the volume of a gas is directly proportional to the number of moles of the substance. So 2x moles of $\ce{SO3}$ will have twice the volume of x moles of $\ce{SO3}$. Therefore the volume of 2x moles of $\ce{SO3}$ will be 8.4 moles.
Therefore the answer is $8.4$ litres


text:
In the condensation reaction of an aldehyde with a hydroxyl amine, why is an oxime formed and not an amide?
Condensation reactions are those that involve removal of a water molecule. When you analyze the reaction between an aldehyde and hydroxyl amine, I don't understand why it's not possible for the acidic hydrogen of the aldehyde to be abstracted by the $\ce{OH-}$ radical of the amine, thus leaving behind an amide ($\ce{R-CO-NH2}$) rather than an oxime being formed?

The aldehyde hydrogen (the H in -CHO) is not acidic, so it can't leave as H+. Furthermore the hydride ion is an exceedingly poor leaving group, so it can't leave as H-. Point is, there's no way of getting that hydrogen away from the carbon.

I don't understand why it's not possible for the acidic hydrogen of the aldehyde to be abstracted by the $\ce{OH-}$ radical of the amine

There is a lot wrong in this sentence and that’s the key to the answer. For one, there is no hydroxide radical (neither in hydroxyl amines nor anywhere else); if anything it is a hydroxyl radical $\ce{HO^{.}}$ — but it is not present in hydroxyl amine. If you check out the mechanism of the condensation, you will realise that no basic hydroxide species occurs:

If any basic species ever exists in here, it is the oxygen anion in the first step. Note especially that the hydroxyl group on nitrogen never loses its proton and thus never becomes basic in this mechanism.
I can hear you asking ‘so why doesn’t the oxyanion in step two abstract the neighbouring hydrogen?’ And this is really the key:
That hydrogen is not acidic!
Never, never, never ever deprotonate this hydrogen on an aldehyde. The only way to make it moderately acidic (but you would still need very strong bases to do so) is something along the lines of the Corey-Seebach umpolung. It generates a cyclic S,S-acetal out of your aldehyde which is stable. Its $\mathrm{p}K_\mathrm{a}$ is low enough that it is actually accessable. That of an aldehyde is not.
Also, to finally generate an amide, you would need some kind of migration of the hydroxyl oxygen to the aldehyde carbon after hypothetic deprotonation. I’m not saying it’s impossible, but rather unlikely (but that is difficult to judge due to the unlikelyness of the previous step). Note that an amide’s carbon has an oxidation state of $\mathrm{+ III}$ while the oxime’s carbon has $\mathrm{+I}$. You need a redox mechanism to get there.

It is not the $\ce{-OH}$  which would attack the $\ce{-C=O}$ bond i.e., the carbonyl bond of aldehyde, but its the lone pair of electrons on $N$ which will attack it.
To make it simple, first of all you need to know whats happenings here, the carbonyl bond $\ce{-C=O}$ is not electrons rich because of the EN $O$ bonded to it, so the best neucleophile will be the one to attack the $\ce{-C=O}$ bond.
General important method to remember
Any carbonyl group, aldehyde or ketone when reacts with (excess of) $\ce{NH2-Z}$ will give $\ce{H2O}$ and $\ce{C=NZ}$, where Z can be any thing $H$ or $-OH$, etc.In an acidic Medium.
Basically what you need to remember is $\ce{H2}$ of $\ce{C=NZ}$ reacts with $=O$ of $\ce{-C=O}$ to form water. this is an easy way to remember but this is not really happenings, not directly, there are few steps occuring in between.
Here,
$\ce{R-HC=O}$ + $\ce{NH2-OH ->R-HC=N-OH     + H2O}$
$\ce{R-HC=N-OH}$ will oxidise to $\ce{R-HC=N=O}$

The last statement (R−HC=N−OH will tautomerise to R−HC=N=O) is incorrect. As written, that's an oxidation, not a tautomerization.


text:
What is the difference between steric strain and torsional strain?
I know that in ethane, the extra energy present in the eclipsed conformer is caused by torsional strain.
In butane, the gauche conformation experiences steric strain. But the eclipsed conformation at 0 degrees has substantial amounts of both steric strain and torsional strain.
What is the difference?

TL;DR Torsional strain can be thought as the repulsion due to electrostatice forces between electrons in adjacent MOs. Meanwhile steric strain (also known as van der Waals strain) can be thought as the repulsion when two bulky groups which are not directly bonded to each other become too close to each other and hence there isn't enough space for them.
Here is the more detailed version.
Torsional Strain
Lets consider an ethane molecule. The C-C sigma bond is free to rotate and in principle there are an infinite number of possible conformations. However only 2 are significant, these are staggered and eclipse conformations. Different conformers are usually drawn as Newman projections as they can easily be compared with each other. Below are the Newman projections for the eclipsed and staggered conformer:

The staggered conformer is the most stable conformer while the eclipsed conformer is the least stable conformer. The staggered conformer is approximately $\mathrm{12~kJ~mol^{-1}}$ more stable than the eclipsed conformer. This energy difference between this maxima and minima is known as the torsional barrier.
So what is torsional strain? The reason for the eclipse conformer being higher energy than any other conformers is due to the destabilising electrostatic repulsion between the electron pairs of the C-H sigma bonds on the two carbons. Furthermore, there also exists a stabilising feature which is greater in the staggered conformer. In the staggered conformer there is a constructive orbital interaction involving the bonding and anti-bonding MOs of the adjacent H atoms. This results in hyper conjugation which stabilises the compound.

The two effects that I have mentioned above is what torsional strain refer to. So you can think of torsional strain as the strain that is the result of electrostatic forces.
Steric Strain
Now lets consider butane. Rotation of the $\mathrm{C_2-C_3}$ sigma bond also lead to infinite possible conformers. However there are 4 main conformers shown below:

Here there are 2 types of eclipsed conformers which are the least stable of the conformers. The most unstable one is known as the eclipsed syn form. The syn form is approximately $\mathrm{20~kJ~mol^{-1}}$ higher in energy than the staggered conformer. The reason for this can partly be attributed to torsional strain as there is repulsion between the electrons in the sigma bonding orbitals. However also it can largely be attributed to the repulsion of the two relatively bulky methyl groups as they become too close to each other and there isn't enough space for them.
This repulsion is known as steric strain. Therefore steric strain can be define as the repulsion which occurs when non-bonded groups which are not directly bonded approach each other too closely. This repulsion only exists for bulky substituents, such as methyl or ethyl groups. So in ethane there is no steric hindrance as the hydrogen atoms are not that bulky.

Nice answer and nice dashed line in the bottom right of the Newman projection image ;)

It's probably helpful to point out a few more things too. Torsional strain is defined as not existing when the conformations are staggered, so there is technically no torsional strain at 60, 120, and 180. Torsional strain also only ever exists between atoms separated by three bonds, such as the hydrogens in CH3-CH3, more precisely those three bonds are H-CH2-CH2-H. Steric strain exists only in atoms separated by four or more bonds, such as in propane (H-CH2-CH2-CH2-H). It's probably more helpful to consider bonds of more than four, however, such as butane.

Excluding the effect of MO constructive interference, do both torsional strain and steric strain arise from electrostatic repulsion? When you say that steric strain comes from the repulsion of bulky groups because there isn’t enough space for them, it’s also a repulsion between the electrons of those groups, correct? If you ignored the MO interference part, would it be suitable to consider steric strain as, in a way, additional torsional strain, as it were?

Thanks for the fantastic answer. To answer lightweaver's question, yes, there will be electron repulsion as well because the two bulky groups' filled sigma bond orbitals will approach each other and feel a "filled-filled" 4 e- interaction which is similar to the MO illustration above.

For simplicity's sake, torsional strain is defined as as the strain experienced by the bonds when conformations are not staggered. So, at any angle other than 60, 120, or 180, there is torsional strain. Additionally, torsional strain can only ever exist in atoms separated by only three bonds.
Therefore, the hydrogens in ethane only ever experiences torsional strain, and it is zero (technically minimized) when it's staggered.

Steric strain exists only in molecules who have four or more bonds, since steric strain is defined as the repulsion felt between atoms at four or more bonds separated from each other forced closer than their van der Waals radius would typically allow.
In the butane below, there is torsional strain between the two central carbons and hydrogens from each central methylene group. However, atoms separated by four or more bonds experience steric strain. Steric hindrance will never be zero, but it can be minimized when the atoms are separated by as much space as possible.

How energy is increased if nothing adds energy to the molecule ? It violates the law of conservation of energy.


text:
How to find the change in internal energy (q+w) for an expanding monoatomc ideal gas?

If a monoatomic ideal gas simultaneously expands against a constant external pressure and drops in temperature, how do you find the internal energy change?
Known values are: $T_1$, $T_2$, $p_1$, $p_2$, $p_\text{external}$ and $T_\text{external}$.

Finding the work done by the expansion is easy, but I don't understand how to change in heat.
It is not a reversible process, so it can't be found from the work, and it's not isothermic, so it isn't zero. It can be found if I have the heat capacity but that is not given.
Edit:
The pressure changes from $p_1$ to $p_2 = p_\text{external}$

The internal energy $U$ is a state function that can be expressed as a function of temperature $T$, pressure $p$, and amount of substance $n$:
$$U=U(T,p,n)$$
The change in internal energy $\Delta U$ is independent of the path by which the new state was reached. Hence, you do not necessarily have to calculate the exchanged heat $Q$ and the work $W$ done during a particular process.
The ideal gas gives simpler equations. According to Joule’s second law, the internal energy $U$ of an ideal gas depends only on the amount of gas and the temperature, and is independent of the volume or the pressure:
$$U=U(T,n)$$
For a monoatomic ideal gas (i.e. a model system of point-mass molecules) we find that the internal energy is proportional to the temperature $T$ and number of molecules $N$ or amount of substance $n$:
$$U=\tfrac32NkT=\tfrac32nRT$$
where $k$ is the Boltzmann constant and $R$ is the molar gas constant.
Since you know the temperatures $T_1$ and $T_2$, and since the amount of substance is constant $(n_1=n_2)$, you should be able to calculate the change in internal energy $\Delta U=U_2-U_1$.
Note, if you don’t know the number of molecules $N$ or amount of substance $n$, you may express the result as molar internal energy $U_\mathrm m$ (i.e. in $\mathrm{J/mol}$):
$$U_\mathrm m=\frac Un$$

This is an adiabatic irreversible expansion so $q=0$.  The  change in internal energy is $\Delta U=\int_{T_1}^{T_2}C_V dT$ where $C_V$ is the heat capacity at constant volume for the monoatomic gas (=3R/2). The work done is $w=-\int_{V_1}^{V_2} pdV$ at the expense of the internal energy. As the temperature drops as no heat added or lost and then $\Delta U =w$. As heat capacity in this case is independent of temperature  and the gas is ideal $\Delta U=C_V\Delta T =-p\Delta V $


text:
Difference between state of matter and phase
What is the difference between state of matter and phase?
 This site  said that:

Phases are different from states of matter. The states of matter (e.g., liquid, solid, gas) are phases, but matter can exist in different phases yet the same state of matter.

I don't really understand this. Could anyone start from scratch and explain this? Or send me a link to go to?

Well the answer is a bit more complicated than either of the two answers below. For example we'd say that wood is a solid, but it can never have a solid phase. If you have a oil and vinegar salad dressing which is mixed into an emulsion we'd say that the salad dressing is a liquid, but it has two liquid phases.

A phase is a region of space where all physical and chemical properties are uniform. This means that for example the density, the chemical composition and the temperature is equal everywhere in the region. This will also mean that the state of matter (solid, liquid, gas, etc.) is equal everywhere.
A state of matter is a form that matter can take. Examples are solid, liquid, gas, plasma. There are some others that appear under extreme conditions.
For example, nitrogen and water at 50K are both solid, so they have the same state of matter. However the chemical composition is not the same and hence they don't have the same phase.

So oil and water would be different phases because they have different chemical compositions? (Even though both are liquid).

@JoeAcademia Yes: It's because they have different compositions, but also that even when you put them in contact with each other, they don't merge together into one indistinguishable liquid with a single composition.

"A phase is a region of space where all physical and chemical properties are uniform." this isn't absolutely true. Think of a metal alloy with two components. As the metal solidifies the chemical composition can change.

@MaxW: Metal alloys are great example for more phases in one state of matter. Steel is probably the most studied example. Of course, the composition of phases changes during  annealing, quenching, and tempering, but the resulting (meta-)stable piece of iron is a multi-phase solid.

@wythagoras In your definition you state that physical properties must be uniform. Do you mean homogeneous? Because a material can be homogeneous and not isotropic and vice versa.

Vegetable oil such as olive oil is a liquid.  Water is also a liquid.  Thus, vegetable oil and water have the same "state".
But, as anyone who has tried to mix them together knows, they aren't the same "phase".
You could also add liquid mercury to the system, and probably also a fluorinated liquid such as perfluorodecalin.  But the different liquid phases would not mix, despite all being the same state (liquid).

This isn't really true. In air, I seal a 1 liter bottle with 250 ml of water and 250 ml of benzene. and shake it thoroughly. I let it sit. I now have 3 phases. (1) A gas phase with air/benzene/water (2) A liquid phase with mostly benzene and a little water and (3) a liquid phase with mostly water and a little benzene. So it is wrong to say that benzene and water don't mix - the implication being at all.

I stand by my answer.  I think you are drawing extreme implications from it and then attacking those implications.  What sentence in my answer do you think is not true?

The definition of both state and phase depend highly on the context of the discussion.  Knowing the context of the discussion is the responsibility of anyone involved in the discussion.
State generally refers to a condition which the matter is exposed.  When exposed to certain conditions matter may take on a particular form, which the form is sometimes referred to as a state depending on the context of the discussion.
Phase when discussing chemistry generally refers to portions of matter which are physically distinctive.  Phase may also refer to a state within a cycle depending on the context of the discussion.
Phase and state can be completely synonymous, again, depending on the context of the discussion.
Example:

A chemist working on a batch reactor may refer to the state as the condition of the reactor and refer to the phase of the matter in the reactor.
Another chemist working on a similar batch reactor may refer to the phase of a given sequence in the reactor's operation and refer to the state of the matter in the reactor.
Yet another chemist with a similar batch reactor may interchangeably refer to the 'state' or 'phase' of matter in the reactor.

Alone, both 'state' and 'phase' can be ambiguous; however, this ambiguity can be avoided through clear communication of the context of the discussion.

Intuitive answer
Different states are best demonstrated on water. You have three different states:

Solid = ice
Liquid = water
Gas = water vapour

Phase is more general term. These three states are also three phases, but you can have different phases within one state too!
For example iron has different properties under different conditions. Under atmospherical pressure it has these phases dependent on temperature:

$\alpha$-iron: from -273 K to 912 K
$\gamma$-iron: from 912 K to 1394 K
$\delta$-iron: from 1394 K to 1539 K

The portrait of different phases dependent on external conditions (f.e. pressure and temperature) is called phase diagram. This one shows phases of iron:

Image is from https://en.wikipedia.org/wiki/Allotropes_of_iron.
Additional information
I think states and phases has good definition on Wikipedia.

Below is a definition from the IUPAC Gold Book for a liquid crystal.
mesomorphic phase
Also contains definition of: nematic state
Often formed by concentrated systems of surfactants. These are states of matter in which anisometric molecules (or particles) are regularly arranged in one (nematic state) or two (smectic state) directions, but randomly arranged in the remaining direction(s).
It seems that state and phase have specific definitions as well as the common understanding of solid, liquid and gas with interchangeable use of states or phase. The context is all.


text:
Questions about van't Hoff factor in colligative property
I learned from many websites that $i$ of strong acids and bases and some weak bases could be determined by how many ions are dissociated. But, I realized that I need to calculate the $n$ in $i=1+(n-1)a$ ($a$ is the degree of dissociation), not the $i$ in my problems.
Is there any rule to calculate the $n$ for strong acids and bases and weak acids and bases? Do I also need to check solubility table to solve colligative property problems that using van't Hoff factor?

Strong acids and bases dissociate completely. Why'd you put them in the same problem? And what's this with solubility? Of course precipitation changes the colligative properties. Please rethink and rephrase your question.

I didn't ask about that. I asked for the $n$ value. Please look my questions carefully. I saw that $/ce{H2SO4}$ has $i=3$, but is that the $n$ also 3?

The formula you cite is only useful if alpha is not one, i.e. a weakly dissociation compound. And n is the number of aggregates your original compound dissociates into. And no, HSO4- is a rather weak acid, so i is not 3 for sulfuric acid.

In general, 'n' is the number of species the compound dissociates into, and such formulas are applicable only when the acid itself is monobasic, i.e.only one dissociaton is involved.For sulphuric acid,this is not applicable as there are two dissociations with two dissociation constants.The best way to find the van't hoff factor is to find the ratio of the final number of particles to the initial number of particles.

You must know what van't hota factor is right. Its the ratio of observed colligative property and calculated or theoritical colligative property. And there are few more formulas like this.
n is the no. Of mole after dissociation or association what ever according to the problem.
Generally this formula is used to find degree of dissociation or association i.e, alpha.

From what i have mentioned, i=3, but n=2

Did you mean any dissociation of any electrolyte, no matter weak or strong electrolyte? Will you put $n=3$ or $n=1$ for $\ce{H3PO4}$?

$\ce{H3PO4 <---> H+ + H2PO4-}$  is the equation for dissociationo

Here both n and i are equal to 2

Why not 3H+ + (PO4)3-

If you want to know about the colligative properties of compounds with more than one acidic proton, please say so in your question.

What i wrote is after first ionisation of it, and what you wrote is after Three ionisation of it


text:
Ion pair definintion in chemistry
I read in the journal of organic chemistry:

A less striking, but nonetheless significant, attenuation of the reactivity of phenyl selenide anion can be achieved by varying the counterion and/or the relative degree of solvation of the ion pair. For example, when valerolactone (3) is allowed to reflux with sodium phenyl selenide in THF/HMPA for $3\,\mathrm{h}$, one obtains an $85\,\%$ yield of the $\unicode[Times]{x3C9}$-phenylselenenyl carboxylic acid 4.

My question is what is the meaning of the bolded expression (solvation of the ion pair) above?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] and visit the [help] for questions about the site. I have improved the formatting of your post, more details can also be found in the [help]. As for the question you are asking: An ion pair is merely a pair of ions (i.e. two). They can be in solution and are there solvated by the solvent.

see also ion pair in the IUPAC Compendium of Chemical Terminology (Gold Book)

As already said, ion pair is a pair of two ions like NaCl. When the pair separates completely into its constituent ions in the solvent (like Na+ & Cl- in water), it is called dissociation.
After dissociation, if the solvent associates itself and surrounds the ions like this-

(Sodium ion surrounded by water molecules), it is called Solvation.
Solvation stabilises the compound and makes it either more reactive or less reactive.

Could the ion form a complex with the water molecules?