text:
How does the litmus pH indicator work?
I'm wondering about how does the litmus solution work as a pH indicator.
And another question:
When you put drops of litmus solution into an colorless acid (e.g. $\ce{HCl}$), it turns red. But then if you add colorless base (e.g. $\ce{NaOH}$) to the mixture, it turns blue. How come?
I thought that the litmus solution doesn't change color after it's changed color once already after a chemical change with an acid or a base. In this case, I thought that the mixture wouldn't turn blue after adding the $\ce{NaOH}$ into the mixture. But it seems that it still works well after used once.
How come?

This explanation is true for most organic indicators.  Litmus is a complex mixture of organic molecules.  The molecule responsible for the color change (the chromophore), is 7-hydroxyphenoxazone (image taken from Wikipedia, public domain):

See that $\ce{OH}$ group at the lower left?  That's where the so-called "acidic proton" is.  This molecule is actually a weak acid, meaning that in the presence of a strong base, the $\ce{OH}$ group can lose a proton, which reacts with an hydroxide ion ($\ce{OH^{-}}$) to form a water.  The 7-hydroxyphenoxazone is now a negatively charged ion: it looks the same except it is missing the $\ce{H}$ on the $\ce{OH}$ group, and it leaves behind an electron.
To think about the equation that describes what's going on, pretend that the entire 7-hydroxyphenoxazone molecule is represented by the formula $HLit$, where $\ce{H}$ is the hydrogen on the $\ce{OH}$ group and $Lit$ is everything else (this is common procedure: only the hydrogen matters in the acid-base equation).  Then the acid-base equation is:
$$\ce{OH^{-} + HLit <=>H2O + Lit^{-}}$$
As more strong base is added, more hydroxide ion is present, driving the equilibrium to the right and producing more $Lit^-$ ion.  And now, the color change: $HLit$ is red, but $Lit^-$ is bue.  So the more acidic your solution, the more the equilibrium is to the left and thus there is more $HLit$ than $Lit^-$.
Litmus is pretty good for checking rough acidity or basicity, because its midpoint pH is close to 7 (neutral).  Litmus is as red as it gets at a pH of 4.5, and is as blue as it gets at a pH of 8.3, so the midpoint is about 6.4, a bit more acidic than pure water.

Superb explanation. Can you help me with the source of your information? I want to know more.


text:
Is diamond an allotrope or polymorph of carbon?
Is diamond an allotrope or polymorph of carbon? When is something considered an allotrope vs a polymorph?

Allotropy is a special case of polymorphism. Polymorphism is the phenomenon of a substance exhibiting different crystal structures. Allotropism is the same phenomenon limited to the subset of all substances that contains only the chemical elements. According to the Wikipedia article on polymorphism:

Polymorphism ... is the ability of a solid material to exist in more than one form or crystal structure. Polymorphism can potentially be found in any crystalline material ...

The allotropy article defines allotropy:

Allotropy or allotropism (coined from Greek "other" + "form") is the property of some chemical elements to exist in two or more different forms ... The term allotropy is used for elements only, not for compounds. The more general term, used for any crystalline material, is polymorphism.

Thus, diamond and graphite represent two different allotropes of carbon, which also makes them different polymorphs of carbon.
Water is well-known for having multiple solid phases. These phases are polymorphs, but, because water is not an element, they are not allotropes.

Allotropy is not a subset of polymorphism. We talk about polymorphism in case of crystalline solids, however, allotropy exist in liquids and gases, too, eg ozone vs normal diatomic oxygen.

Diamond is an Allotrope of Carbon.
As cleared by @Ben Norris,
Allotropy is property of any element to exist in two or more different forms.
Whereas the term Polymorphism meant the ability of a solid material to exist in more than one form or crystal structure.
In diamond, each carbon (an element)  is bonded to four Carbon atoms forming a rigid 3-dimensional structure.

The difference between allotropy and polymorphism is simple, the former talks about elements while the latter talks about anything material, other than an element, a molecule, a compound or even mixture, existing in different crystalline structures due to the conditions the material is placed of temperature and pressure .

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text:
What trends exist in the Periodic Table?
I wish to know the trends in the periodic table and why that trend exists, for the following parameters:

Atomic size
Ionization energy
Metallic character
Positive ion size
Negative ion size
Electronegativity

My progress (if my reasoning is wrong for any of these please correct me):

Atomic size - I memorized from my class lecture that the greatest atomic size is towards the bottom-left of the periodic table. I think the reason for this is because as you go down the groups (IA, IIA, etc.) you add more electrons. Adding more electrons means the atom will logically grow bigger. I am not sure why the answer is not bottom-right, though.
Ionization energy - This is the energy required to remove an electron from an atom. If there are more electrons then the electron will be further from the nucleus and by Coulomb's law, since the distance increases, the force attraction that keeps the electron in the nucleus's orbit is weak. Thus, the ionization energy is the highest toward the top-left.
I have no clue what to think for metallic character.
Positive ion size I know the answer is that it increases towards the bottom left, but I don't know why.
Negative ion size I know the answer is that it increases towards the bottom left, but I don't know why.
Electronegativity I am pretty sure is highest at F which is 4.0 and is lowest towards the bottom left. So the trend increases towards the top-right.

You are correct on atomic size being due to number of electrons
and their shells and the reason why atomic size decreases from
left to right is due to the number of the protons being greater
on the right than on the left of the periodic table with same number of shells due to shell theory. (Reference).
The concept you are forgetting is
Valence Electrons.
Atoms are stabilized most often when there outer most shell is
filled with 8 valence electrons (except for hydrogen and helium with
0 or 2). The elements on the left side of the periodic table lose
electrons easily since they have a very large atomic radius allowing
for lower ionization energy. on the right side of the periodic
table(nitrogen, oxygen , and halogens) have a smaller atomic radius
creating a stronger pull making it harder to lose an electron thus
they will steal an electron to become stable.
The best way to summarize this is to quote Chemistry.about.com

"There are trends in
  metallic character as you move across and down the periodic table.
  Metallic character decreases as you move across a period in the
  periodic table from left to right. This occurs as atoms more readily
  accept electrons to fill a valence shell than lose them to remove
  the unfilled shell. Metallic character increases as you move down an
  element group in the periodic table. This is because electrons
  become easier to lose as the atomic radius increases, where there is
  less attraction between the nucleus and the valence electrons
  because of the increased distance between them."
(Reference)
Positive ion size decreases due to the lack of electrons which allows for the protons to pull stronger on fewer electrons.
The inverse applies for negative ion size. Due to the fact that there are more electrons, the size will increase since they will occupy more space.
Electronegativity is the pull a elements has on electrons. This determines how it bonds with other elements i.e.(whether it shares or looses electrons). If a element has over a 1.7 Electronegativity difference then these the element with the smaller electronegativity will give away several electrons 1-4. Elements on the left side of the periodic table have the smallest Electronegativity due to their atomic size and inability to gain enough electrons to be stable ions. Elements from left to right in a period show increasing elctronegativity, and electrons form up to down in a group will have a lower electronegativity. Also fluorine has a 3.98 elctronegativity. (Reference).

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text:
Lennard Jones liquid vapor phase transition
I am trying to generate a phase diagram of pressure versus temperature in a NPT ensemble using molecular dynamics method. I am using the Lennard Jones potential with 108 particles and Berendsen thermostat and barostat to fix the parameter.
In order to observe phase transition I plot the radial distribution function (RDF) and also density versus temperature graph for various fixed temperature. However it is hard to differentiate liquid and gaseous phase from RDF.
Another peculiar thing is that there should be a discontinuity in density as the transition temperature is reached, which I observe only for solid liquid transition but not for the liquid vapor transition. The change in density should be huge for liquid vapor transition. Any idea on what possibly goes wrong?
The parameters:
Periodic boundary
cut off radius = 2.5
Truncated and shifted Lennard Jones potential
number of particle = 108
number of steps = 2000
timestep = 0.004
Berendsen barostat of scale coeefficient of 0.1 is used to scale the size of simulation box

The density versus temperature graph for pressure of 1.0MPa:

The solid liquid transition can be seen but not the liquid vapor transition.

I agree with Richard that with 108 particles, you're not going to see anything on the RDF. Also, you give us a pressure in units of MPa, rather than a reduced pressure; thus, we cannot check your simulations. Finally, the reduced critical pressure for LJ fluids is around 0.14; make sure you're below that, or you might simply have a supercritical fluid, as Richard (again) suggested.

I think your transition, if present, may be drowned out by noise, as your RDF is only produced by 108 particles. As the Lennard-Jones interaction is quite simple and you are using cutoffs (good idea!), is it possible to increase the number of particles in your simulation somewhat, or else take the mean of some large number of trials?
This said, it does not take much imagination to see a continuous PV plot from about 0.7 onwards, which might mean you have something analogous to a supercritical fluid. Maybe try a number of trials at different pressures?


text:
What is the oxidation state of Calcium in teeth?
What is the oxidation state of Calcium found in human teeth?

Welcome to Chemistry SE, Kinnard Hockenkull. You can make your question more answerable by telling us what progress you yourself have made in finding the answer to the question.

Ca3(PO4)2 is the compound found in teeth ; the calcium has an oxidation state of 2+

As a group II metal, you will be hard pressed to find calcium in the 1+ oxidation state (a quick google search yields an isolated report of calcium(I) chloride). Calcium's willingness to oxidise is evinced by its vigorous reaction with water. As you have noted, the calcium in tooth enamel is hydroxyapatite, often with varying degrees of fluorination. Which brings me to your next question, which hopefully you or a mod can slipstream into the main body -
I would definitely warn against eating chalk without good reason as it will evolve $\ce{CO2}$ in your tummy and raise your stomach pH, and is associated with a variety of negative health outcomes:

Milk-Alkali Syndrome
Hypercalcaemia
Antacid side effects

As always, all things in moderation. On a personal note eating chalk sounds like exactly the thing certain sectors of the 'alternative health' crowd would be into (namely, those who assert that the body is 'too acidic' (there's another crowd who claim the opposite and will recommend, say, drinking $\ce{HCl}$ (through a glass straw, natch)), ignoring for a moment the fact that the human body is pretty good at maintaining its pH and will up and have a cardiac arrest or something if the pH drifts more than a bit)

What about gargling?

@KinnardHockenhull - Take it up with an accredited medical professional (which I am not :D)

sometimes you have as a result of frequent heartburn and ulcers too little HCl in your stomach and sometimes you have no HCl at all in your stomach. In that case you need to take HCl pills since they are much safer than 1M HCl(aq)

The answer to the question asked is the +2 oxidation state.


text:
Ambiguity in synthesis (crystallization)
A paper says the following,

Dichloroglyoxime was prepared by chlorination of glyoxime in 10 % HCl at 0 deg C and crystallized from toluene.

When they say "crystallized from toluene" do they unambiguously mean recrystallized from boiling toluene at standard pressure? Thermally these compounds are unstable and instead of a white nice product I've ended up with a purple one much to my dismay and I'm not sure what actually causes the purple.
I've had problems with this synthesis because the $^1$H NMR always shows more than one oxime environment.

I'm sorry to ask, but would you be so kind to post full synthesis? It is relevant to one of my interests

No.  This could also mean applying a vacuum to the toluene-dichloroglyoxime solution, and letting the solvent evaporate-away at room temperature.


text:
Balancing and combining the half equations in redox
I can understand redox equations of the following form, I break these down into half equations and combine them.
$$\ce{MnO4^- + C2O4^{2-} -> Mn^{2+} + CO2}$$
Where I have issue is with the following (unbalanced) equation:
$$\ce{H2 + NO -> NH_3 + H2O}$$
I am asked to show balanced half equations and the final combined equation.
I can see the following changes in oxidation states
reduction: $$\ce{ H -> H^{1+} + e^{1-} }$$ (for both the $\ce{H}$ in water and ammonia)
oxidation: $$\ce{ N^{2+} + 5e- -> N^{3-} }$$

My issue is then with balancing these and combining them, I am not sure if the half equation involving $\ce{H}$ should have $\ce{H2O}$ as the product or the $\ce{H3}$ from $\ce{NH3}$.

How should I go about breaking this into half equations?
Is it ok for both the half-equations to have molecules of the same compound in the product?
Should I ever have water as the product when trying to show a half equation (ie $\ce{H -> H_2O}$)?

Assuming this reaction is taking place in aqueous phase, you can follow the conventional 7 steps for balancing redox reactions although this one does require some extra thoughts.
I will be writing the equation as it should be after each step.

$$\ce{H2 + NO -> NH3 + H2O}$$
Step 1: Ionise the required compounds and remove spectator ions
None of the compounds on either side are ionic other than $\small\ce{H2O}$ also we have no spectator ions as well.*
$$\small\ce{H2 + NO -> NH3 + H+ + OH-}$$
*-Spectator ions are ions which don't participate in the reaction or retain their oxidation state.
Step 2: Split into Oxidation and reduction halves
This is where the thought is required, if you decide of taking the oxidation half as $\ce{H2 -> NH3}$  (you can't simply take $\ce{H3^3-}$ as $\ce{NH3}$ is not ionic rather covalent) you would never be able to balance the nitrogen as the only source of nitrogen is $\ce{NO}$ if you observe the left hand side. Also, since you know that Nitrogen is being reduced (+2 -> -3) the other element, Hydrogen must get oxidised (0 -> +1).
So it is decided that hydrogen cannot convert into ammonia, leaving us with only one option $\ce{H2 -> H2O}$.
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 -> H+ + OH-} & \ce{NO -> NH3}
\end{array}
$$
Step 3: Balance only those atoms undergoing redox
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 -> H+ + OH-} & \ce{NO -> NH3} \\
\text{(because Hydrogen is } & \text{(because only Nitrogen } \\
\text{undergoing redox but} & \text{is undergoing redox,} \\
\text{Oxygen is not.)} & \text{it's already balanced)} \\
\end{array}
$$
Step 4: Balance Oxygen by adding water($\ \small\ce{H2O}\ $)
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 + H2O -> H+ + OH-} & \ce{NO -> NH3 + H2O}
\end{array}
$$
Step 5a: Balance Hydrogen by adding $\ \small\ce{H+}$
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 + H2O -> H+ + OH- + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\
\ce{H2 + H2O -> [H+ + OH- ] + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\
\ce{H2 + H2O -> \qquad H2O \qquad + 2H+} & \ce{NO + 5H+ -> NH3 + H2O} \\
\hline
\ce{H2 + \qquad-> \qquad \qquad \qquad \quad 2H+} & \ce{NO + 5H+ -> NH3 + H2O}\\
\hline
\end{array}
$$
Step 5b: (Only in case of basic medium) Adding $\ \small\ce{OH-}$ to $\ \small\ce{H+}$
This step must follow Step 5a and means nothing on it's own.
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
\ce{H2 \qquad -> 2H+} & \ce{NO + 5H+ \qquad \qquad -> NH3 + H2O \qquad \qquad} \\
\ce{H2 + 2OH- -> 2H+ + 2OH-} & \ce{NO + 5H+ + 5OH- -> NH3 + H2O + 5OH-} \\
\ce{H2 + 2OH- -> [2H+ + 2OH- ]} & \ce{NO + [5H+ + 5OH- ]-> NH3 + H2O + 5OH-} \\
\ce{H2 + 2OH- -> \quad \quad 2H2O} & \ce{NO + \qquad 5H2O \qquad -> NH3 + H2O + 5OH-} \\
\hline
\ce{H2 + 2OH- -> \quad \quad 2H2O} & \ce{NO + \qquad 4H2O \qquad -> NH3 + 5OH- \quad \qquad  }\\
\hline
\end{array}
$$
The medium being acidic or basic merely determines whether $\ \small\ce{H+}$ or $\ \small\ce{OH-}$ will be seen in the final redox equation.

Till the above steps mass has been balanced, now charge balancing.

Step 6: Balance charge in each side of each half by adding electrons
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half}  \\
\hline
(acidic)\quad \ce{H2  -> 2H+ + 2e-} & \ce{NO + 5H+ + 5e- -> NH3 + H2O} \\
\hline
(basic) \quad \ce{H2 + 2OH- -> 2H2O + 2e-} & \ce{NO + 4H2O + 5e- -> NH3 + 5OH-} \\
\hline
\end{array}
$$
→ Verify correctness by ensuring electrons are on right hand side in oxidation half and on left hand side in reduction half.
→ Also verify if net change in oxidation state is equal to the number of electrons appearing in each half.
eg. Hydrogen is going from 0 to +1 oxidation state and there are 2 Hydrogen atoms in the left side of the oxidation half undergoing oxidation(don't count $\small\ce{H}$ from $\small\ce{OH-}$), meaning a net change of '2', which is equal to the number of electrons.
Similarly Nitrogen is going from +2 to -3 and there is just one Nitrogen atom in the left side of the reduction half meaning a net change of '5' which is equal to the number of electrons appearing in the equation. (This is what goes on to be later called as the 'n-factor' or the 'valency-factor')
Step 7: Make 'net electrons produced' equal to 'net electrons consumed'
Multiply oxidation half by 5 and reduction half by 2 (this is basically like taking the LCM and multiplying both halves to make the number of electrons equal to the LCM, here 10.)
$$
\begin{array}{l|r}
\text{Oxidation Half} & \text{Reduction half} \\
\hline
(acidic) \ce{5H2 -> 10H+ + 10e-} & \ce{2NO + 10H+ + 10e- -> 2NH3 + 2H2O}\\
\hline
(basic) \ce{5H2 + 10OH- -> 10H2O + 10e-} & \ce{2NO + 8H2O + 10e- -> 2NH3 + 10OH-} \\
\hline
\end{array}
$$

Finally add the two halves and cancel out the common cmopounds and you get the balanced equation
Acidic Medium
$$\begin{matrix}
\ce{&5H2 + &2NO + &10H+ + &10e- &-> &10H+ + &10e- + &2NH3 + &2H2O} \\
\ce{&5H2 + &2NO & & &-> & & &2NH3 + &2H2O}\\
\end{matrix}$$
Basic Medium
$$\begin{matrix}
\ce{&5H2 + &10OH- + &2NO + &8H2O + &10e- &-> &10H2O + &10e- + &2NH3- + &10OH-} \\
\ce{&5H2 + & &2NO & & &-> &2H2O + & &2NH3}\\
\end{matrix}$$

In this case the oxidation half could have been directly/trivially stated as $\small\ce{H2 -> 2H+}$
You can also carry on $\small\ce{H+}$ ion till the end net reaction and then perform Step 5b which is less tedious.

I sympathise with your problem using electron half equations when one of the reactants produces two different products. For this reason and the fact that (as the previous answer points out) there are no ionic compounds invoved, I prefer to use oxidation number half equations.
Oxidation half equation: $\ce{H2  ->  2H+}$
Hydrogen oxidation numer goes up 2
Reduction half equation: $\ce{NO -> NH3}$
Nitrogen's oxidation number goes down by 5 (+2 to -3)
Oxygen is unchanged (-2)
Therefor to balance the oxidation numbers $5~\ce{H2}$s require $2~\ce{NO}$s and the equation becomes: $$\ce{5H2 + 2NO  -> 2NH3 + 2H2O}$$


text:
Clarification on Photodissociation
Can someone please explain what photodissociation is and if it is more efficient than electrolysis while splitting water into its components?
Thanks in advance.

Photodissociation is a unimolecular reaction triggered by absorption of light. You start with one molecule, a bond (or more) breaks and you end up with two (or more) products. In water you can get $\ce{OH}$ and $\ce{H}$ (etc), usually in excited electronic states. You can not get the products of the electrolysis.
Water electrolysis is a more complex reaction. The final balance is:
$\ce{2H2O -> 2H2 + O2}$,
so you break four bonds and make three bonds.
Both reactions are not spontaneous (in the electronic ground state of water). The net balance is not energetically favorable (you start with more energy in bonds than what you end up). However, the way you add this energy is also different. In photodissociation the energy is given by light; in electrolysis the energy is given by forcing an electron current.


text:
What makes Coke acidic?
What makes Coke and other soft drinks acidic?  These drinks are carbonated, so a freshly-opened can should have a lot of dissolved carbonic acid, but the ingredients also lists phosphoric acid.  Would flat Coke be significantly less acidic than carbonated Coke?

This is an interesting question.  Off-handedly, I'd say it was mainly the phosphoric acid (which is added intentionally for "bite"), but if someone sits down and calculates given the pressure of the carbonation and the concentration of phosphoric acid, it would be a great answer.

Interestingly, the 'other coke' is a base.

I think coke is a Acid, because if you put a coin in a glass of coke it cleans and strips the coin. If it were a alkalis or neutral it would not do that, were as a acid would.

If you look at the list of ingredients, you will find that Coca-Cola and many other brands contain phosphoric acid (food grade - don't worry). This is probably the dominant cause of acidity. Citric acid is another common component. Of course, the carbon dioxide is also acidic, but weakly, so there should be very little difference between carbonated and flat.

Please see @Chemistry answer. My oral mucosa would agree, but of course that's not very scientific.

I am of the opinion that this is the correct answer. My reasoning is that the pH of most colas is on the order of 2-4, and the pKa of carbonic acid is about 6. You'd need a hell of a lot of carbonic acid to get to that pH, while the pKa of phosphoric acid is 2.1, so it'd be quite easy to get to a pH of, say, 3.

If I put baking soda in cola, will it become less acidic?

Flat coke is less acidic than carbonated coke. I did a lab report on this topic, and found that Carbonated coke had a concentration of around 0.131M whereas flat coke had a concentration of 0.064M. I guess we can assume that the carbonic acid that's dissolved in Coke makes up a large proportion of the acidity.

pK of carbonic acid is around 3.5, it is not a factor at pH=2.5

Mostly Carbonic acid, probably. Because it is the only species in coke, that can escape in the form of gas.
Just shake a coke, or heat it and see how much of the acidity is gone. It is the part that is due to $\ce{CO2}$.

Could you back this up? This seems to be a guess at the moment. I don't see the connection between "it being a gas" and "it is the largest contributor to acidicity". Phosphoric acid doesn't easily form a gas, that's all.

Yes, this is mostly a guess. The connection that you don't see is that: When you shake or heat coke, its acidity goes away almost totally and this change is due to CO2 being undissolved.


text:
What defines matter's phase at room temperature?
So, a simple question: what makes neon a gas; what makes water liquid; what makes plutonium-238 a solid at room temperature? Of course, the phase is determined by the excitation of the molecules, but what determines that excitation?

It's all in the intermolecular forces (and the air pressure). Intermolecular forces are the forces of attraction and repulsion between molecules to neighboring particles. To understand how they determine physical properties like enthalpies of vaporization, fusion, etc..., you need to think about what states of matter are and how they affect the molecules. When you heat up water to a temperature (average kinetic energy) of 90 °C, the energy within the molecules isn't strong enough to overcome the intermolecular bonds (and the atmospheric pressure above the sample) and become a gas. However, it does have enough energy to overcome the bonds to a certain extent and thus can remain in a liquid state instead of in a solid state. This principle applies to any substance. In order for there to be a change in state of matter, the energy within the molecules has to overcome the intermolecular bonds and atmospheric pressure.
As mentioned several times above, another factor is the atmospheric pressure. A common experiment is to put water in a vacuum and watch it boil, something many first-time-experimenters wouldn't expect. The reason water is able to vaporize at room temperature is because it doesn't have the atmosphere to prevent it from vaporizing and therefore, in order to reach an equilibrium, the water is able to vaporize with much less energy than normally required at STP (standard temperature/pressure).
As a direct response to your question about neon, the reason that the noble gases are usually gases unless at extreme temperatures has to do with the weak intermolecular forces they have (van der Waals forces, or more specifically, london dispersion forces). Because the noble gases don't usually bond intramolecularly with anything else, including themselves, and have full energy levels, they have extremely weak intermolecular forces allowing them to almost always be in the gas phase.
I've listed some helpful links below if you want to take your research to the next step.

http://en.wikipedia.org/wiki/Intermolecular_force
http://wps.prenhall.com/esm_hillpetrucci_genchem_4/16/4219/1080299.cw/index.html
http://www.youtube.com/watch?v=A30lguyG0WM
http://chemed.chem.purdue.edu/genchem/topicreview/bp/intermol/intermol.html
http://en.wikipedia.org/wiki/Van_der_Waals_force


text:
Can magnesium chloride absorb water and ammonia at the same time?
Anhydrous magnesium chloride will absorb ammonia to the tune of ($\ce{Mg(NH3)6Cl2}$).
Incidentally, magnesium chloride is usually a hexahydrate ($\ce{MgCl2(H2O)6}$) at STP.
My question is, could magnesium chloride be induced to absorb both ammonia and water at the same time? Since both of the absorbed molecules in question are close in shape, size, structure and since six seems to be the occupancy limit for either one (at STP), would they be able to coexsist?
For example, could I expose some magnesium chloride dihydrate to ammonia and get
($\ce{Mg(NH3)4Cl2(H2O)2}$)?

anhydrous magnesium chloride will adsorb both, however composition of the phase formed is not easily predictable from general principle


text:
Completely chemically inert metal alloy?
Background:
I was a chemistry major and have recently started watching this series of videos by NurdRage.
I learned that Platinum dissolves in Aqua Regia but Ruthenium is only attacked by high temperature halogens and some alkalis.
Question:
Would a Ruthenium and Platinum alloy be resistant to both acid and alkali?
But more importantly, is there any alloy or pure metal that is completely chemically inert to every strong oxidizing reagent? Is there any research about the most chemically inert metal alloys?

There is no completely inert material. However, in case you work with relatively small temperatures (<200 celsium) teflon is considered (almost) inert.

Not an alloy, but this came up in a question about noble gas fluorides - fluorine forms a passivated layer on nickel which is tremendously corrosion resistant. As such, fluorine-passivated nickel vessels are used for fluorine photolysis experiments.

My best guess would be a combination Pt, Ru and Ni.

Why should some alloy be more resistant to oxidation than the "best" of its constituents? And do not mix up passivation (like aluminium) and chemical inertness!

Alloys have special properties that can exceed those of their constituents.

Due to electronegativities no alloy would be thermodynamically stable in the presence of fluorine (F$_2$). Even in an alloy the constituent metal atoms are either "sharing" electrons or receiving/donating electrons. Fluorine is the element with the highest known electronegativity and would "pry" the electrons from the electron-accepting metal atoms.

The problem is that in different media different metals / alloys are so corrosion resistant than we regard them as being "inert".
It is interesting that aluminium in strong nitric acid is passive as an oxide layer which clings very tight to the aluminium forms. Equally titanium, zirconium and halfmium in nitric acid are very passive. Even while they are from an thermodynamic point of view very reactive metals.
In alkali aluminium dissolves well as it forms soluble aluminium complexes.
Titanium in anhydrous methanol suffers dire corrosion if you scratch it. What happens is that you form soluble titanium methoxide which then removes metal from where the scratch is.
In nitric acid palladium and ruthenium will be very slow to corrode, but if you add chloride then the formation of the chloride compelxes such as PdCl4 2- greatly assists the nitric acid in dissolving the palladium and also the ruthenium. I think that if you want a super corrosion resisting metal or alloy you need to state what medium it will be in.
One good all rounder is tantalum, this is poorly attacked by all the common acids. While it will tolerate aquea regia it is attacked by HF and other flouride containing solutions.

According to the Wikipedia article on the chemical reactivity of fluorine, fluorine (a very strong oxidizing agent) reacts with every element except helium and neon.  So it looks like no metal has the property you want.  The reason helium and neon are so unreactive is that their outer electron shells are completely filled, with no hidden subshell available to accept an electron.  No metal (or any other element, for that matter) has this property.

Good answer, but what about alloys?

I'm less certain here, but I don't think alloying would change the chemical properties very much.  It definitely changes mechanical properties (like hardness) because of the different crystalline structure of the alloy. For example, stainless steel is more corrosion-resistant because the addition of chromium allows a passivating layer of chromium oxide to form (like with aluminum on its own). So in stainless, iron borrows a chemical property from chromium, but the elements are still as they were chemically.

Okay. But anyone else can back this up or refute it?

Not the best source but Wikipedia says that Iridium is one of the most corrosive resistant metal elements (I believe I have seen it referred to as "the most" elsewhere). Only some molten salts and halogens are corrosive to it. However, this does not mean that an alloy could not be more resistant, although comments above suggest alloying doesn't change the chemical properties of the constituents, which makes sense to me. One critical point for what I have mentioned though, is: are corrosion resistance and inertness the same thing?

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text:
What is a pure substance?
I found as I was preparing to teach "Introduction to thermodynamics" that there are varying definitions or examples of pure substances. For instance the text book I have to use "Thermodynamics an engineering approach, Cengel and Boles" defines a pure substance as one that has the same chemical composition throughout. Air in the gaseous form is cited as an example. Air in its liquefied form is NOT since it seperates into it's various components.
The Dummies book on Chemistry suggests that sugar can be considered a pure substance as it has the same chemical composition throughout. And water can be considered a pure substance for the same reason.
So would a homogeneous mixture of sugar and water be considered a pure substance? After all, a homogeneous mixture has "identical properties" throughout it's phases.
Yet another definition of a pure substance suggests that any material made of the same kind of atoms is a pure substance. So individual elements in the periodic table and stuff made "purely" of them are pure substances. Other stuff is just homogeneous or heterogeneous mixtures.
So are pure substances (irrespective of whether we view them as elements, compounds or mixtures) just very difficult to break down into their individual components?
Or would the definition of a pure substance change depending on who you speak to (chemists, phycisists, thermodynamicists)?

Air is not a pure substance; it is a mixture.

@Nick but the constituents of air cannot be extracted without a lot of effort. So in case a lot of energy is required to extract constituents, can it be described as a pure substance?

no, a pure substance is not a mixture. If you can separate substances from a mixture using a centrifuge, chromatography, or any other physical method then they are not pure.

I am unable to choose a right answer to this rather subjective question. Any advice?

@BenNorris 's answer seems correct to me. Engineers might use a different definition if it suits their purposes, but Chemistry defines a pure substance precisely.

@Nick This was the very reason I was circumspect about migrating this question from physics to chem.SE. It was a troublesome (to categorize) question in physics.SE! :P As an engineer, I don't always agree with chemists!

"Thermodynamics an engineering approach, Cengel and Boles" defines a pure substance as one that has the same chemical composition throughout.

This is the correct definition of a pure substance. However, air, regardless of phase, is not a pure substance.
All matter is categorized as either a "pure" substance or a mixture. The word "pure" in front of "substance" is unnecessary, since the definition of substance implies purity. A substance is a sample that is chemically uniform in composition. In other words, all particles of that sample are the same chemically (whether they are atoms or molecules or ionic lattices or what have you). Substances cannot be separated into simpler components by any physical process (i.e. without breaking chemical bonds). Mixtures are samples that contain two or more substances. All mixtures can be physically separated, although the means to do so may be neither easy nor practical.
Air is a mixture of nitrogen ($\ce{N2}$), oxygen ($\ce{O2}$), carbon dioxide ($\ce{CO2}$), argon ($\ce{Ar}$), water vapor, and traces of other gases. Each of these gases separately is a substance. Air can be easily (but not in the comfort of your own home) separated into its constituent parts. Cooling air will cause the constituent components to condense out of it at progressively lower temperature: first water vapor, and then $\ce{CO2}$ (195 K), followed by $\ce{Kr}$ (120 K), $\ce{O2}$ and $\ce{CH4}$ (90 K), $\ce{Ar}$ (87 K), $\ce{N2}$, $\ce{Ne}$ (27 K), $\ce{H2}$ (20 K), and $\ce{He}$ (4 K).
As each substance condenses out of air it can be physically separated from the remaining gas, reducing the complexity of the mixture. Each substance, if returned to the gaseous state separate from the original mixture is physically and chemically distinct from the original mixture (or whatever is left of it). Such physical separations do not need to be easy to make air a mixture, they need to be possible.
A pure substance does not exhibit this behavior. When pure water is cooled, it undergoes a phase change to ice, which is physically separable from liquid water. However, the solid and liquid samples are still water. You have not reduced the complexity of the sample, just changed its form. If you allow the separated solid water to melt, it will be physically and chemically indistinguishable from the original water sample. Pure water cannot be separated into other chemically distinct substances without performing a chemical reaction, which means breaking and forming bonds between atoms.
The majority of matter that we interact with in our everyday lives is comprised of mixtures.
We encounter a few pure substances in the kitchen. Table sugar is a single compound (sucrose). So is table salt (sodium chloride), baking soda (sodium bicarbonate), and cream of tartar (potassium bitartrate). Water is a pure substance, so long as it has been purified. Most water that we encounter has other substances dissolved in it.
Gold (as long as it is 24K) is a pure elemental substance. So is diamond (carbon), graphite (also carbon), and the neon in neon lights. All minerals are pure substances, including asbestos. Glass is a pure substance. Most precious stones are pure substances.

Van Wylen in his textbook "Fundamentals of classical thermodynamics" defines air as a pure substance as long as it is in it's gaseous form. When liquified, it seperates into it's components.

Air is a homogeneous mixture of several gaseous substances. If air is a pure substance, so too is diet cola (also a homogeneous mixture).

Yes, the punch line of Wan Vylen's definition is that air CANNOT BE SEPARATED into it's constituents easily in the gaseous form. I Think just opening a can of diet cola/regular cola starts releasing CO2 bubbles...

Yes, but the caffeine, phosphoric acid, caramel coloring, artificial sweetener, etc. are much harder to separate, but they are separable. The human body does it all of the time. Van Wylen's inclusion of air as a pure substance is not correct. I have updated my answer to be better explain why.

I guess the definition is quite subjective.

The definition is not subjective. A pure substance has all particles being of the same chemical identity. This definition is on the first or second page of all introductory chemistry texts. It's it the Wikipedia article for substance. Air fails this criterion regardless of phase.

let us continue this discussion in chat

Only the definition of "all particles being of the same chemical identity" is no good, because that would render all ionic compounds mixtures. Even purified water is strictly a mixture of $\ce{H_2O}$ but also $\ce{H_3O^+}$ and $\ce{OH^-}$ (even you describe those, quantum mechanically, in terms of excitation particles). You don't get around the definition of being not easily physically separable, which is indeed subjective – but so are many definitions in chemistry.

That is why my definition does not use the word "easily". The ions in NaCl cannot be physically separated, easily or otherwise.

Is water not a pure substance if it contains oxygen-18 out hydrogen-2 isotopes? Is acetone not a pure substance if some of it equilibrates into its enol form?

Simply, a pure substance is a substance which is composed by molecules or directly by atoms of only one kind. Diamond is a pure substance because it is directly made up of carbon atoms. Salt water is not because it contains sodium chloride and water.
But there is almost no 100% pure substance in this world (Exceptions exist, as there can be only one or two molecule and you can call that a substance), thus we normally recognize relatively pure substances as pure substances.

The definition of a "pure substance" seems to be quite subjective. I have found a couple of these varying instances now. One is like I have noted in Cengel and Boles. Van Wylen in his text book says air, for instance, can be treated as a pure substance as it cannot be separated into it's constituents in the gaseous form. In liquified form, air is NOT a pure substance.

It is quite possible for a material consisting entirely of one element to be "impure" - for example, a mixture of carbon nanotubes, graphite and buckyballs could hardly be called "pure" despite being made of carbon only. For some applications, even a mixture of different types or lengths of carbon nanotube isn't "pure" enough.
On the other hand, you can buy "pure" orange juice in the shops that is neither entirely one element nor entirely one molecule. In fact, "pure (from concentrate)" orange juice isn't even entirely made up of stuff that was squeezed out of an orange. So "purity" is in effect a matter of how you define it.
For thermodynamics purposes, "purity" is a kind of order - if this order is lost by mixing two pure substances together, it takes work to un-mix them. Conversely, with a carefully designed machine it is possible to harness the entropic "force" in favour of mixing to generate useful work.
ATPase, the enzyme which performs the final stage of converting food energy into ATP, the internal energy "currency" used by chemical processes in cells, is the most widespread example (although in this case it was evolved rather than designed). As a crude summary, it uses the energy released by mixing acid and water to drive a rotor, which as it turns forces ADP and phosphate together making ATP. (The acid is generated in earlier stages of the process.)

In the lab, we refer to a pure substance as any substance with no impurities.
That is to say: pure water is 100 % $\ce{H2O}$ with nothing else in it; pure acetone is 100 % acetone.
We do however frequently use the term with respect to known mixtures. One can say we have pure clay despite the clay containing several KNOWN components because there are (practically) no impurities that have not been identified or would upset our experiments.
I would be careful describing a mixture as pure, however, unless you know everyone is on the same page. Pure tap water sounds absurd but pure air, oil, product, or rubber doesn’t.

Basically, a pure substance in this context means a substance that is always available. As in the expression of rate law, if a solvent is always available,say water. Then rate won't depend on the concentration of water , as it is always available. Hence its activity can be considered as 1.
Certain textbooks refer to pure substances in physical chemistry in this sense.


text:
What makes Kleenex Cool Touch tissues cool to the touch, and is it safe?
I love Kleenex Cool Touch tissues.  I'm just wondering what makes them actually feel cool.  For some reason it makes me think of refrigerant but I'm sure that's not in them.
Is it the combination of common ingredients that somehow gives it a moist feel, making it feel cooler than normal tissues?
Are the tissues safe to put to your face, if you are potentially inhaling the ingredient(s)?
Ingredients:  Stearyl heptanoate, stearyl alcohol, polyethylene, mineral oil, coconut oil, aloe

It's definitely safe, yes. All of the ingredients you've listed above are non-toxic. Aloe verra is often used to cool burning areas (though it's treatment is questionable).

Stearyl alcohol make it cool to touch. Because alcohol evaporates easily. It is safe to use.

@lambda23: It makes sense that evaporation would make the cool-to-touch feeling, but by that logic, by the next day all of the alcohol would be gone. Are you sure that's right?

@lambda23: Maybe body heat warms up the stearyl alcohol which is what causes it to evaporate. I don't have access to a Cool Touch tissue right now, but maybe the OP does. OP: Try touching the whole tissue for so long that it's not cold anymore. Then come back and do it again. And again. And again. See if it ever stops feeling cold.

@Nile: Yes, body warms the stearyl alcohol and it will evaporate.

@lambda23 - Consult the liquid range of stearyl alcohol and you will find it is a solid at room temperature and has a boiling point of 210 C. Not the best candidate for evaporative cooling...

@RichardTerrett At the wikipedia site, the boiling point of the stearyl alcohol is 210 C at 15 mmHg. The pressure isn't n the standard state (760 mmHg).

@lambda23 - Well spotted - I missed that entirely. However it lends further support to my point that a liquid that boils at 210 C at a pressure not much higher than Mars' atmosphere is fairly nonvolatile. It's virtually wax.

@lambda23 - a bit of googling around suggests that the ingredient might be 'proprietary' and unlisted (I'm not sure how the laws work). Nothing solid though. If indeed there are extra ingredients it could be something that creates a mere sensation of coolness, similar to menthol.

@Nile: tried repeatedly touching and rubbing the tissue several times over the past few days.  It doesn't seem to be losing any coolness.  Even my wife confirmed it.  I'll try a bit more but it looks like whatever it is, it's not evaporating.  Seems more likely to be something that is quick to absorb heat.

@trpt4him It probably absorbs moisture in the air and evaporates when you begin to transfer energy

I had one of the cool touch Kleenex for a week and no matter how much I touched it after awhile I went to touch it again and it was still cold.

I thought they contained menthol or something like it. This doesn't work by affecting heat transfer but by altering the heat sensitive cells in the skin (it's the opposite effect to chilli).

Don't guess, don't read advertising. Look it up in the primary literature. US patents 7879191; 7879190; 7879189; 7879188; 7883604; 7837831; 7807023; 7820010; 5415320.

@trpt4him Volatile liquids don't all evaporate if you leave them out. You can keep a bottle of hand sanitizer out and it still has that chilling effect. Repeatedly touching would eventually get rid of the liquid, though.

Uncle Al gave a very good hint, and Leonardo came close, but then gets lost in things like metal heat conductivity, which he does not understand. The answer is simply: heat of melting! Its the same as some choolates with a high content of hardened cocos fat.

My first guess as to the nature of this behaviour is similar to water. Hypothermia is a realistic risk if you are drenched in water. This is because water can absorb large amounts of energy (it has a large specific heat capacity) to only rise in temperature slightly. Water itself evaporates into steam spontaneously, if some water is spilled then it will evaporate readily, regardless of the temperature. Likewise, the solids in the tissue are probably coming closer to melting temperature, which I believe is higher than your body temperature, but quickly absorbing a lot of energy from your body in the process.
Something similar to properties of water are probably characteristic to the tissue. Leave the tissue in the open in a cool room, along with a glass of water. After awhile, touching the glass of water and the tissue should invoke the same sense of coolness. This is because your body is at a higher temperature than the items in the room.
There is likely another property of matter besides specific heat capacity that is a measurement of how quickly heat is transferred, this is likely to play a role as well. I do not know what this property is called and it was never covered in my classes so far, but take metal for example. Metal has a very small specific heat capacity, it can not hold much energy. However, touching a metal object invokes a sense of coolness. I believe this is because heat is transferred very quickly through metal, although it warms very quickly due to its limited heat capacity. If you put a metal cup and a porcelain cup on the stove, the metal will burn you much quicker if you grab it than grabbing the porcelain/ceramic cup.
If you were very interested and inquistive you could test the specific heat of the tissue through experimentation. Also, try holding the napkin for a few seconds to try and find out how quickly the tissue warms to the touch. Chances are, it warms relatively fast and can absorb a lot of energy, giving you the impression that it is 'cooler' than most objects, to which marketers will be overwrought with the joy of hearing.

Saying "how quickly heat is transferred" is incorrect. Heat is the transfer of energy.

@Nile http://en.wikipedia.org/wiki/Heat_transfer "Heat ...and is synonymous with heat flow and heat transfer.

@sencer Reading the heat wikipedia page, "Heat ... is synonymous with heat flow and heat transfer." Sorry, I stand corrected.

@sencer: I see you've updated your post. Just as an FYI, it doesn't say that on the link you've sent. It says that on http://en.wikipedia.org/wiki/Heat

@Nile yes, sorry. It doesn't let me fix it.

@Nile Semantics is nothing to worry about, as long as the person asking the question does not get confused. Conventionally we might describe it as $\Delta H$

Yes, it does warm quickly, but your first, immediate, impression is that it is considerably cooler than your hands.

@trpt4him Thats because it probably is a lot colder than your hands. If it is ever really hot outside, like hotter than 100 degrees Fahrenheit, your Kleenex will likely fail to impress you if my answer is at all correct. It probably only feels cold because entropy saps the heat from you but quickly matches its surroundings.

Here is the answer, but I don't understand it.
http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&p=2&u=%2Fnetahtml%2FPTO%2Fsearch-bool.html&r=86&f=G&l=50&co1=AND&d=PTXT&s1=%22kimberly+clark%22&OS=%22kimberly+clark%22&RS=%22kimberly+clark%22

While this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference.


text:
What is the definition of of 'compound', 'mixture', 'element' and 'molecule'?
I am looking for the precise definitions, as I am very confused as to what they are exactly because although I mostly understand what they mean, I have encountered some conflicting definitions that confused me.
As it stands, this is what I understand them to mean:

Compound: two or more different atoms bonded together.
Mixture: two or more different atoms together but not joined.
Molecule: two particles (same or different) bonded together.
Element: only 1 type of atom; this definition is applied to things both bonded and not to itself.

I don't know whether these terms apply microscopically, macroscopically, or both. For example, I think 1 $\ce{H2O}$ molecule is a compound but is a bathtub of them called a mixture (as it contains more than 1 atom type), a compound, or both? That is, to be a compound, do all the atoms in the compound have to be bonded physically together? Generally, to what extent can these four terms overlap?
I would very much like a systematic way of thinking about these terms.

If one atom, then not molecule and not mixture and not compound. If two atoms and same element, then molecule and not compound and not mixture. If two atoms and not same element, then compound and molecule and not mixture. If two different molecules and evenly mixed, then homogeneous mixture. If two different substances and differentiable on the molecular level, then heterogeneous mixture. The air is a homogeneous mixture. Your oatmeal is a heterogeneous mixture.

Have a look at the IUPAC Gold book. It defines what Chemists consider these things to be. goldbook.iupac.org

I can't really provide a systematic approach, but I can attempt to clarify (as a student myself).

Elements are classes of atoms. Atoms of the same element are similar (if not identical) in their physical and chemical properties (but be aware of Isotopes which are physical variations among atoms of the same element).
A definite (I suppose, systematic) way to distinguish elements is that their corresponding atoms have different numbers of electrons orbiting the nucleus.
The term is also used to describe a collective of the same atom (element). A bar of gold (Gold being an 'element'; a chemical class) is said to be an element itself.

Molecules are a group of atoms covalently bonded to each other (which can be considered a 'direct connection', if you will). The molecule can consist of atoms of the same element, or atoms of different elements.
If you joined to molecules via covalent bonding, you'd have created a new molecule (it's nothing special).

Compounds are made of two or more elements bound together. This means a lot of phases made of molecules (like H2O) are considered compounds (though H2 is not). While in mixtures proportions of constituents are highly variable, in compounds they are not. If a compound is molecular, all it's molecules are virtually identical (this doesn't account for different isotopes, or molecular conformations).

Mixtures are broad group created when you mix different substances, as long as they don't react to make a single pure compound in the process. For example juice is a mixture. It contains some Vitamin C, some water, some sugar, etc.

Here is a horribly drawn diagram to help clarify the distinctions.

Reflecting on this, here is what in your understanding needs specific attention.
Compound: A molecule or a group of identical molecules.
Mixture: two or more different atoms or molecules together but not as a single compound
Molecule: two or more atoms (same or different) bonded together.

What about ionic substances? Is NaCl a compound? I always thought it was but if we take your definition there is no such thing as an ionic compound.

@AntiEarth we can still have monoatomic molecules, e.g. neon atoms. I think the definition of a molecule is that its the smallest possible particle capable of independent existence or something.

@SherlockHolmes It is my understanding that single atoms are not molecules; the wikipedia definition reads "A molecule /ˈmɒlɪkjuːl/ is an electrically neutral group of two or more atoms held together by chemical bonds"

I think there is an error in your definition of atoms: "...way to distinguish elements is that their corresponding atoms have different numbers of electrons orbiting the nucleus" (emphasis mine). By your definition, $\ce{O^{-2}}$, $\ce{F-}$ and $\ce{Ne^0}$ (all isoelectronic) would be equal atoms, which is not true. Atoms are defined by their atomic number (number of protons), not the number of electrons.

Another thing: "Molecules are a group of atoms covalently bonded to each other." Although this seems to be the accepted definition, this is being challenged by the existence of mechanical bonds (and people seem to consider those things as molecules).

@FelipeS.S.Schneider yea I recoil at most of what my highschool self has written here

In a sample of water, you'll find the compound $\ce{H2O}$. In a sample of hydrogen gas, you'll find the element hydrogen, though it'll take the form of the molecule $\ce{H2}$. In a mixture, you may find different molecules, compounds, and elements.
At this point, you may find it helpful to look at the following classification of matter.

The left side of the tree denotes chemical bonds/changes. The right is physical. (i.e., you can get a heterogeneous mixture to convert to a homogeneous mixture by changing physical things about the mixture like adding in another element. You can't do the same to elements and compounds.)

I was also confused when I learned this all stuff first time. When you get any element, molecule, compound or mixture don't think about complex stuff but think on the basic level.
I have given the most basic definition (properties) of an element, molecule, compound or mixture.
Element:

A substance which cannot be broken down chemically is called elements.
All substances listed in Periodic Table are elements.

Molecule:

Two or more atoms or elements combined together are called a molecule.

Compound:

Two or more elements combine together to form a compound.
The smallest unit of a compound is a molecule.
Compounds possess different characteristics compared to constituent elements.
Example: $\ce{H2O}$.

$\ce{H}$ is highly combustible
$\ce{O}$ helps in combustion
$\ce{H2O}$ is used in fire extinguishers (different characteristics than $\ce{H}$ and $\ce{O}$  )

Mixture:

When two or more matter mix without changing their original property this is called a mixture.
Mixtures can be separated by a simple process, like filtration.

Compound: a pure substance formed from the bonding of two or more different elements together. It can be decomposed into its constituent elements by chemical means.
Element: a pure substance consisting of one type of atom only.
Mixture: two or more different substances are mixed but they are not chemically bonded together and may be separated by a physical process such as filtration, distillation and so on.
Molecule: I'm actually not sure about this - I always read that it's a group of atoms (same or different) bonded together and is the smallest particle of an element/compound capable of independent existence; however, we treat monoatomic gases as a molecule, e.g. Neon.

I'm only a student but I just did some research on it (for myself) but I thought I could maybe help you so this is what I understand.
Compound: two or more different atoms chemically bonded together.
Molecule: two or more different or same atoms chemically bonded together.
Element: one single atom.
A mixture contains two or more substances (elements, lattices, molecules, compounds), that are not chemically bonded together.
So basically a mixture is like you said a bath tub with elements, compounds, lattices, and/or molecules in it but they are not bonded together chemically. Otherwise, it would be a molecule.
I just think of a mixture as a box with things chucked in it like toys can be molecules and books can be elements, you can separate them easily.
This is what I understood so it might not be 100 % right.

I disagree with your definition of element — I think you mean only atoms of a single type. I would call a single atom an atom.

Everything in the universe is made up of something, and when we try to refine or filter and figure out what those are at the very basic level we come to a certain number of things.
These certain number of things when combined in different ways and quantities make up anything and everything matter. Humans, plants, air, water everything. These things are well classified, studied and placed in the modern periodic table and have a well-defined number of them and are called the elements' of the modern periodic table. And the tiniest possible particle of an element is called an 'atom' of that element. why tinniest?.
Here is a better explanation for all your queries.
Hope it helps.


text:
Difference between thermodynamic and kinetic stability
What is the difference between thermodynamic and kinetic stability? I'd like a basic explanation, but not too simple. For example, methane does not burn until lit -- why?

What are your initial impressions about it?  It will be helpful for those answering to be able to gauge your present understanding.

To understand the difference between kinetic and thermodynamic stability, you first have to understand potential energy surfaces, and how they are related to the state of a system.
A potential energy surface is a representation of the potential energy of a system as a function of one or more of the other dimensions of a system. Most commonly, the other dimensions are spatial. Potential energy surfaces for chemical systems are usually very complex and hard to draw and visualize. Fortunately, we can make life easier by starting with simple 2-d models, and then extend that understanding to the generalized N-d case.
So, we will start with the easiest type of potential energy to understand: gravitational potential energy. This is easy for us because we live on Earth and are affected by it every day. We have developed an intuitive sense that things tend to move from higher places to lower places, if given the opportunity. For example, if I show you this picture:

You can guess that the rock is eventually going to roll downhill, and eventually come to rest at the bottom of the valley.
However, you also intuitively know that it is not going to move unless something moves it. In other words, it needs some kinetic energy to get going.
I could make it even harder for the rock to get moving by changing the surface a little bit:

Now it is really obvious that the rock isn't going anywhere until it gains enough kinetic energy to overcome the little hill between the valley it is in, and the deeper valley to the right.
We call the first valley a local minimum in the potential energy surface. In mathematical terms, this means that the first derivative of potential energy with respect to position is zero:
$$\frac{\mathrm dE}{\mathrm dx} = 0$$
and the second derivative is positive:
$$\frac{\mathrm d^2E}{\mathrm dx^2} \gt 0$$
In other words, the slope is zero and the shape is concave up (or convex).
The deeper valley to the right is the global minimum (at least as far as we can tell). It has the same mathematical properties, but the magnitude of the energy is lower – the valley is deeper.
If you put all of this together, (and can tolerate a little anthropomorphization) you could say that the rock wants to get to the global minimum, but whether or not it can get there is determined by the amount of kinetic energy it has.
It needs at least enough kinetic energy to overcome all of the local maxima along the path between its current local minimum and the global minimum.
If it doesn't have enough kinetic energy to move out of its current position, we say that it is kinetically stable or kinetically trapped. If it has reached the global minimum, we say it is thermodynamically stable.
To apply this concept to chemical systems, we have to change the potential energy that we use to describe the system. Gravitational potential energy is too weak to play much of a role at the molecular level. For large systems of reacting molecules, we instead look at one of several thermodynamic potential energies. The one we choose depends on which state variables are constant. For macroscopic chemical reactions, there is usually a constant number of particles, constant temperature, and either constant pressure or volume (NPT or NVT), and so we use the Gibbs Free Energy ($G$ for NPT systems) or the Helmholtz Free Energy ($A$ for NVT systems).
Each of these is a thermodynamic potential under the appropriate conditions, which means that it does the same thing that gravitational potential energy does: it allows us to predict where the system will go, if it gets the opportunity to do so.
For kinetic energy, we don't have to change much - the main difference between the kinetic energy of a rock on a hill and the kinetic energy of a large collection of molecules is how we measure it. For single particles, we can measure it using the velocity, but for large groups of molecules, we have to measure it using temperature. In other words, increasing the temperature increases the kinetic energy of all molecules in a system.
If we can describe the thermodynamic potential energy of a system in different states, we can figure out whether a transition between two states is thermodynamically favorable – we can calculate whether the potential energy would increase, decrease, or stay the same.
If we look at all accessible states and decide that the one we are in has the lowest thermodynamic potential energy, then we are in a thermodynamically stable state.
In your example using methane gas, we can look at Gibbs free energy for the reactants and products and decide that the products are more thermodynamically stable than the reactants, and therefore methane gas in the presence of oxygen at 1 atm and 298 K is thermodynamically unstable.
However, you would have to wait a very long time for methane to react without some outside help. The reason is that the transition states along the lowest-energy reaction path have a much higher thermodynamic potential energy than the average kinetic energy of the reactants. The reactants are kinetically trapped - or stable just because they are stuck in a local minimum. The minimum amount of energy that you would need to provide in the form of heat (a lit match) to overcome that barrier is called the activation energy.
We can apply this to lots of other systems as well. One of the most famous and still extensively researched examples is glasses.
Glasses are interesting because they are examples of kinetic stability in physical phases. Usually, phase changes are governed by thermodynamic stability. In glassy solids, the molecules would have a lower potential energy if they were arranged in a crystalline structure, but because they don't have the energy needed to get out of the local minimum, they are "stuck" with a liquid-like disordered structure, even though the phase is a solid.

This is an exceptional answer!

The best answer that I have come accross!

Legendary answer. Better explanation of these topics than I've seen in any textbook

Kinetics deals with all things that impact the rate at which a reaction occurs. One implication of that is that reaction kinetics are partially a function of the stability of the intermediate species (and transition states) that the pathway of a chemical reaction necessarily has to pass through en route to the formation of the final product. If those intermediates are highly unstable (i.e., highly energetic) relative to the initial reactants, then the initial barrier of activation energy that must be surpassed to set the reaction in motion will be relatively high, and the reaction will tend to proceed slowly (if at all).
Thermodynamic stability, on the other hand, is strictly a function of the change in free energy (ΔG), which is a state function, meaning that its value is determined exclusively by the difference between the initial state and the final state (or the free energy of the initial products and the final products). In other words, thermodynamic stability is totally independent of the pathway between reactants and products.
While it's true that thermodynamic stability is the ultimate arbiter of the spontaneity of a reaction, if the kinetics render the reaction too slow then, in practice, the reaction may not actually occur at all (under some given set of conditions). The classic textbook example is the conversion of diamond to graphite, which is thermodynamically favorable because the free energy of graphite is lower, but doesn't occur under ordinary conditions because the kinetics of the reaction (in the form of the immense activation energy required) are extremely unfavorable.
Imagine you have some set of reactants, A + B, and there are two different pathways through which A + B can react, both of which lead to final products that are lower in free energy than the reactants A + B. Pathway #1 has much higher activation energy than Pathway #2, but the final products at the end of Pathway #1 are lower in free energy (i.e., more thermodynamically stable) than those at the end of Pathway #2. This means that Pathway #1 is thermodynamically favored, but Pathway #2 is kinetically favored (that is, it poses a lower barrier in terms of activation energy and the reaction will proceed more rapidly via this pathway). The consequence of this is that the kinetic pathway, Pathway #2, will be the favored one, unless environmental conditions (temperature, pressure, concentrations of catalysts, etc.) are tuned in a such a way that the higher activation energy of Pathway #1 can be surpassed.
In summary, thermodynamic stability (in terms of free energy difference between reactants and products) will determine whether a given reaction could be theoretically spontaneous, but kinetic factors will decide whether the reaction occurs in practice under the given environmental conditions, as well as which pathway for a given reaction takes primacy.

The first paragraph has points that are a good complement to Thomij's post and weren't explicitly stated there. +1

@Greg E. "One implication of that is that reaction kinetics are partially a function ...the final product." So activation energy partially depends on stability. Can we say that kinetics depends on thermodynamics whereas thermodynamics does not?

There is often a "reaction barrier" that needs to be overcome.  Thus one must supply energy to get the reaction started, but once started it becomes self-sustaining.
One can think of it this way:  one has to break bonds in a reaction before one can make bonds.  Energy is needed to break the first bonds.  Then making new bonds can supply enough energy to keep the reaction going.
Note that I'm talking about thermodynamically allowed reactions.  If a reaction is not thermodynamically allowed, then even if some bonds are broken by adding energy, the reaction will not spontaneously continue.

So you mean activation energy for example is kinetic stability. However I still do not understand Thermodynamic stability, does it have anything to do with entropy as I know which reactions are spontaneous by working out their Delta S Total.

@Cyrus: Yes, activation energy has to do with kinetic stability.  For example, diamonds are thermodynamically unstable relative to graphite, but don't hold your breath waiting for the spontaneous conversion.  The second half is more complex.  The thermodynamic quantity predicting the direction of a reaction depends on the conditions of the process being considered.  For example at constant $T$ and $p$, the Gibbs Free energy $\Delta G$, must be negative.  At constant $E$ and $V$, the entropy change $\Delta S$ must be positive.

Kinetic - how unwillingly something reacts.
Thermodynamic - the energies involved, amount of gradient.
For example, aluminium metal is kinetically much more stable than iron metal in damp air (it doesn't rust), but thermodynamically much less stable (if it did oxidise, it would give out more energy)
You are thermodynamically unstable in air, but don't worry, you're not going to spontaneously combust, which is another way of saying that you are kinetically stable.

Kinetic stability refer to a pre equilibrium conditions. For example the equilibrium constant K may be very large meaning the formation of the product is favorable, but the condition is such that the reaction does not go (or goes extremely slowly)
Thermodynamic stability refer to a system in which products formation is not favourable. For example we have larger reaction quotient Q.
A reaction can be thermodynamically unstable but kinetically stable because the condition are such that the rate of the reaction is zero.  A mixture of coal and oxygen are kinetically stable because the condition for combustion are not obtained until the coal is heated to its ignition temperature. The system is thermodynamically unstable because at equilibrium the coal oxygen reaction favors the formation of $\ce{CO2}$.


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What is charge density?
For example, Magnesium has a higher charge density compared to Sodium, therefore its metallic lattice is stronger and it has a higher melting point. Does that mean that Magnesium has more protons and a smaller atomic radius or is it about its charge (2+), that makes its lattice stronger?
What does charge density show?

There are really two separate things here - "does having a higher ionic charge make magnesium's metallic bonding stronger", and "what is charge density". You might want to clarify exactly what your real question is.

If it was about the number of protons in the nucleus, then Mercury would hardly be liquid at room temperature, would it?
Actually, it is about the density of free electrons, i.e. the density of the "electron gas" that makes up the metallic bonding. Obviously, this density has to be the negative of the ions' charge, so you might equivalently say it's about magnesium's (2+) charge, though that doesn't make much sense physically unless you also consider the conduction electrons, as the ions themselves of course repulse each other.

So at my level it means the charge, like Magnesium's charge density is 2+ and sodium's in 1+.

Charge density
Charge density is the density of charge around the ion. Silberberg (Chemistry: The Molecular Nature of Matter and Change. 4th ed. 2006) define it as:

"The ratio of the charge of an ion to its volume."

The factors playing into the strength of an ion's charge density are therefore the charge of the ion (e.g. 2+ for Mg, 1+ for Na) and the effective volume which that charge acts over - which is where the ionic radius comes in. (Note this is different to atomic radius.)
Charge density = charge/volume

As the magnesium ion and sodium ion both have the same electronic configuration i.e. that of neon (because Mg has lost 2 electrons and Na has lost 1 and they now both have 10 electrons in total) the charge of the electrons around the ions is the same. But - magnesium has one more proton in its nucleus. This means magnesium has a smaller radius (think more condensed due to the greater pull of the nucleus - see effective nuclear charge). $Na^+$ has an ionic radius of 102 pm and $Mg^{2+}$ has an ionic radius of 72 pm. (Also from Silberberg)
Since the volume is less and the charge is greater for magnesium, the charge density will be greater for magnesium than for sodium.
Periodic trend
The periodic trend is that charge density will decrease going down the periodic table as the charge stays the same but the size increases. Going across (as for Mg and Na) the charge density increases going to the right as the charge increases and the size decreases.
...
What do you mean by "What does charge density show?"

In coordination  compounds, charge density can be defined as the ratio of the charge to the radius of the metal ion. Charge  density  is directly  proportional  to the stability  of Coordination  Compounds.  E.g The ionic radius of metals $\ce{Cu^2+}$ and $\ce{Cd^2+}$ are $\pu{69 pm}$ and $\pu{97 pm}$ respectively.  Stability = $\frac{2}{69}$ and $\frac{2}{97}$ . The stability of $\ce{Cu^2+}$ > $\ce{Cd^2+}$.

@Another.Chemist Thanks for your edits! For ions with multiple charges, you need to explicitly superscript the charge, or else mhchem will get confused: so, $\ce{O^2-}$ produces $\ce{O^2-}$ (oxide ion), but $\ce{O2-}$ (no caret ^) produces $\ce{O2-}$ (superoxide ion).


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Why shouldn't Uranium come in contact with water?
In the documentary about Chernobyl, it was mentioned that Uranium should not  come in contact with water, otherwise an explosion occurs. What is the reason for that? What kind of reaction makes it explode?

They didn't use Uranium- it was uranium oxide(I can't be sure but its usually UO2) . Inherently its the zirconium alloy that contains UO2 that is the hazard forming hydrogen gas reacting with hot water.

Interesting question. The documentary was almost certainly referring to the extremely hot lava generated by the nuclear meltdown at Chernobyl coming into contact with water. This lava is made of a substance colloquially known as 'corium' and is a combination of molten fuel rods, moderator, reactor walls and whatever else is melted by the incredible temperature of the reactor failure. This is as far as I know why the failure mode is called a 'meltdown', because the corium melts and pools in the bottom of the reactor vessel. In a molten pool, the fissile material may achieve uncontrolled criticality and just get hotter and hotter.
If the temperature is high enough the corium may melt through the bottom of the reactor. This is really bad, because at least hypothetically the lava may melt down to the water table and cause a steam explosion. This is quite a bit like what happens if you try to extinguish an oil fire by throwing water on it (something you should never do) - the water violently boils and expands and sprays burning oil everywhere. However in this case the burning oil is radioactive lava.
This is not a chemical reaction, per se, but rather a case of extremly vigorous heating of water and associated violent expansion.
Additional
If you remember the Fukushima disaster, at least one of the buildings was destroyed in a hydrogen explosion. As I understand it, in the event of a meltdown, the fissile material can get hot enough to effect the thermolysis of water. (The wikipedia article of the event seems to confirm this) Inside a sealed building, the water will develop a headspace filled with hydrogen and oxygen gas. This mixture is very explosive.

I am greatly appriciated sir. Thanks a lot. I was wondering it for days but couldn't find any good answer for it. Thanks again.

Unfortunately, this answer is mostly wrong. The problem is with red hot or even molten metal coming into contact with water, producing metal oxide and hydrogen. Not thermolysis of water. The "steam explosion" kitchen analogy is also nonsense.

Another possible explanation: water can act as a neutron reflector, at least under some conditions.  Therefore, it might be possible for a subcritical mass of uranium to become supercritical if immersed in water, as some of the emitted neutrons could be reflected back into the uranium.  This could lead to a criticality accident, with large amount of ionizing radiation emitted and possibly some type of explosion.
EDIT: I found a reference to this in Richard Feynman's essay Los Alamos From Below:

So [Emilio Segrè] finally went down [to the Oak Ridge uranium enrichment plant] to see what they were doing, and as he was walking through he saw them wheeling a tank carboy of water, green water—which is uranium nitrate solution.
He says, “Uh, you're going to handle it like that when it's purified too? Is that what you're going to do?"
They said, “Sure—why not?"
"Won't it explode?" he says.
Huh! Explode?
And so the Army said, “You see! We shouldn't have let any information get to them! Now they are all upset.”
Well, it turned out that the Army had realized how much stuff we needed to make a bomb—20 kilograms or whatever it was - and they realized that this much material, purified, would never be in the plant, so there was no danger. But they did not know that the neutrons were enormously more effective when they are slowed down in water. And so in water it takes less than a tenth - no, a hundredth - as much material to make a reaction that makes radioactivity. It kills people around and so on. So, it was very dangerous, and they had not paid any attention to the safety at all.

This is more about water being a neutron moderator than a reflector, but the idea is the same - uranium in water could lead to a criticality accident.

+1 You raise a good point here. In addition to Feynmann's anecdote, I recall reading that there was a criticality accident at Mayak involving a solution of a plutonium compound. Ah here it is I was under the impression that the accident occurred due to changes in the solution geometry on stirring, but wikipedia suggests that it was due to a bad container geometry.

It appears that uranium metal reacts with water Uranium Metal Reaction Behavior in
Water, Sludge, and Grout Matrices - when molten, steam will be generated by the heat of the molten metal and by the exotherm. Not a healthy situation in an enclosed environment.

The equivalent of a phreatomagnatic eruption.


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What is the appropriate method to visualize electrostatic potentials?
Using a molecular viewer like PyMOL, I understand that its often incorrect to simply look at positive and negative residues and use those residues to calculate the charge density and instead, you should be using something like APBS to calculate the electrostatic potential.
What is the correct way of going about this?

One way to do it is to perform a quantum chemistry calculation of the electrostatic field generated by the molecule. Another is to determine (or assume) partial atomic charges for each of the atoms of the system, and use a Poisson solver (as I believe APBS is) to calculate the electrostatic field generated by this point charge distribution.


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Calculating heat capacity of a calorimeter
I am struggling on this question :
The heat of neutralization of $\ce{NaOH}$ and $\ce{HCl}$ is $-57.1\:\mathrm{kJ}$, calculate the numbers of moles of $\ce{HCl}$ and if $x$ moles of $\ce{HCl}$ were used how many $\:\mathrm{kJ}$ would be given out if one mole results in $57.1 \:\mathrm{kJ}$
I already got the moles as $0.05$ by doing $n=\frac{v}{1000}  c$ where I had $50\:\mathrm{mL}$ of $2\:\mathrm{M}$ $\ce{HCl}$

I suspect this is a homework question. Given that, please see the guidelines at http://meta.chemistry.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange/142#142. I will tell you that your calculation of the number of moles is incorrect given the molarity of the solution you've specified. Once you have the correct number of moles, think about how you can construct a conversion factor relating the units of kJ and moles of your reactant.

First, is this a question from your book, and do you have the correct answer to check your work.
Do you know how to use dimensional analysis?
Start with 50 mL. You have 2 Molarity of some subtance, how do you write that as a quotient and what are the units of the denominator and numerator? You will also need to know how many mL are in a liter and you should get the correct moles of that substance. It is very important that you write this down on paper correctly and make sure your units will cancel leaving only moles. It is hard to be wrong when your units cancel correctly.
Do you mean the heat of neutralization is $-51.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}$
The more information you give us, the more help you will get.

First lets work out moles of $\ce{HCl}$. The concentration is $2\ \mathrm M$, or in other words, 2 moles per liter. To figure out how many moles you have then, you need to figure out how many liters you have. $50\ \mathrm{mL} = 5\times10^{-2}\ \mathrm L$.
Multiplying concentration and volume, you are saying $\frac{\mathrm{mol}}{\mathrm L}\cdot\mathrm L=\mathrm{mol}$. So you have $2\ \mathrm M\times5\times10^{-2}\ \mathrm L=0.1\ \mathrm{mol}$
You then return to some information you've been given. The heat of neutralization is: $-51.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}$. So to get the total energy in kJ, you can see by the unit that you have to multiply by the number of moles. That cancels the units like we did with the volume: $$-51.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}\times0.1\ \mathrm{mol}=-5.17\ \mathrm{kJ}$$


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Hund's rule & different H₂ molecules
Does Hund's rule allow both of the following scenarios?

Filling each orbital with a single electron, so that a sub-shell, at first, only electrons with a negative spin
Filling each orbital with a single electron, such that a sub-shell, at first, only electrons with a positive spin?

I assume two H atoms whose electrons have different spins would form $\ce{H-H}$ and atoms where the spins are the same would form $\ce{H:H}$. Is that correct?

Perhaps you can look at ptable.com and click the orbitals tab at the top and observe the various ways electrons will fill the different orbitals. Maybe Chromimum has a similar configuration as to what you are asking about?

@Leonardo: If you took some screenshots and expanded it into an answer (with explanation), that would be great! :)

For H$_2$ there are only two electrons and thus you can only have singlet (electrons with opposite spin projection) and triplet states (electrons with equal spin projection).
The spin wave function of the singlet state is a linear combination (a superposition state) of the first electron with spin projection $\alpha$ and the second electron with spin projection $\beta$ and the opposite situation. Both spin states in the quantum superposition have opposite phase:
$$|\uparrow\downarrow\rangle -|\downarrow\uparrow \rangle$$
If the electrons are in the same molecular orbital then (by the Pauli principle) only the singlet state exists. If they are filling $\pi$ molecular orbitals then they could be in different degenerate orbitals (with angular momentum projection on the molecular axis $M_z = \pm \hbar$) and thus both singlet and triplet terms would exist.
There are three degenerate states belonging to the triplet term, that is, there are three wave functions that have equal energy. There spin components are:
$$|\uparrow \uparrow \rangle$$
$$|\downarrow \downarrow \rangle$$
$$|\uparrow \downarrow\rangle + | \downarrow\uparrow \rangle$$
So you can see that it doesn't matter if you fill one molecular orbital with one electron in spin $\alpha$ and a different molecular orbital with the electron with $\alpha$ spin or you do so with both $\beta$ and $\beta$ spins. However you can not put both electrons with the same spin projection in the same molecular orbital. That is against the Pauli principle, that is, against the anti-symmetry of the electronic wave function.


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Is it more reasonable to use stratification to separate similar compounds rather than using temperature?
In a simple reaction between butane and chlorine, it is more likely for the secondary carbons of the butane to lose a hydrogen and gain a chlorine. The carbocation is a more stable intermediate in that situation compared to a primary carbon losing the hydrogen. So if the desired synthesis is 1-chlorobutane as opposed to 2-chlorobutane then it seems reasonable that a chemist might use stratification. The reason that I believe it is less desirable to use temperature as a method of separation is because that, at least in this example, both compounds will move between the gas and liquid phase randomly providing the possibility of still having mixed samples. Is this true, and will stratification work more reliably if purification is the desired result?
Furthermore, I never really learned why mono-chlorinated products form. Will there not also be compounds that have been chlorinated twice? Or will, over time due to entropy, all compounds be mono-chlorinated assuming that the reagents are in a ratio of 1 chlorine atom per molecule of butane?

What do you mean by stratification?

@RichardTerrett Dividing into distinct layers. Density stratification.

Density stratification will only work if the substances are immiscible (mutually insoluble). For example, water and hexane could be separated this way because they are immiscible. Water will settle to the bottom (density = 1.0 g/mL) and hexane will float on top (0.65 g/mL).
Methanol ($\ce{CH3OH}$) and ethanol ($\ce{CH3CH2OH}$) cannot be separated this way because they are miscible with each other. They form a homogenous solution when mixed. One of the properties of homogeneous solutions is that they do not separate into their components without a transfer of heat to or from the solution. You do not see bottles of soda separating into their components on the shelf because they are homogenous mixtures. You do see bottles of salad dressing separate because they are not homogeneous mixtures.
While the various monochlorinated, dichlorinated, trichlorinated and so forth derivatives of butane have different densities, they are likely all miscible in each other, which means the homogeneous mixture will not separate by density.
If you are looking for an alternative to the fractional distillation, you could try fractional freezing, in which you cool down the mixture, and the substances crystallize out according to the order of their freezing points. To achieve high purity, you would need to repeat the procedure several times for each substance. A good fractional distillation apparatus already has the repeated distillations built in, so it would probably be the fastest technique.

This is a good answer because it describes why stratification is not always a reliable choice.

The answer to your second question:
The halogenation of saturated hydrocarbons occurs by free radical substitution in the presence of UV light or heat and not by formation of carbocations. There's a very clear representation of the reaction mechanism on this Wikipedia page.
Since free radical intermediates are involved, this is a chain reaction and so what you said is correct, a large number of substituted products can be formed depending on the reaction conditions.
For example: If you consider the simple reaction between Methane and Chlorine in the presence of sunlight, Chloromethane, Dichloromethane, Chloroform, Carbon tetrachloride and a small amount of Ethane (by combination of Methyl radicals) are formed depending on the amount of chlorine available.
All these products will be formed even if the reactants are in a 1:1 ratio but they will be present in low concentration since there is less chlorine available.
In the case of higher alkanes like Butane and Pentane, the secondary carbons are substituted faster since the secondary free radical intermediates are more stable (reasons are the same as that for secondary carbocations).

I can't seem to find anything about density stratification being used for that sort of synthesis. I think 1-chlorobutane is most usually synthesised from 1-butanol by reaction with $HCl$.


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What is a 'Trace Crossing'?
I'm reading about electrodeposition of conductive polymers in a chemistry textbook, and the author mentions 'trace-crossing' in the 'potentiodynamic response'. From the context I can figure out that 'potentiodynamic response' means cyclic voltammogram.  But what is the 'trace crossing'?

'Trace crossing' is in the literal sense, where 'trace' is another word for 'line'. 'Trace crossing' basically means that the line in the voltammogram crosses itself.

It would be great if you edited your answer to include a short explanation of when the voltammogram crosses itself and why it is important :)


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What is a "lagged bath"?
First question, and this is one I've searched for in vain.
In the context of setting up cooling baths, I've come across the term "lagged" bath.  I cannot extract what this means from any available context.  I assume it has something to do with time passing (i.e. a lag), but I'm not certain.  If I had to guess, I'd assume it means allowing the cooling agent to come to equilibrium temperature, but again, not certain.
Any direct technical experience that can explain this term would be greatly appreciated.

I'm not 100% certain on this, but my best guess on this would be referring to 'lagged' as in 'lagging', as in insulation material. So, a lagged bath is an insulated bath.

Duh! I'll bet that's right.  I'll wait a bit for potential additional commentary, but that jives with the context I read.  So I'll likely accept this--thanks!

Yes, this term is also used to describe insulating material for pipes:  https://www.environmentcentre.com/wp-content/uploads/2016/08/5_Pipe_lagging_new_branding.pdf


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Determine max of 2 parameter functions
I have 12 experimental tables with OD(substrate concentration, time), where OD is optical density and one table with $\mu$ (substrate), where $\mu$ is growth speed. OD and $\mu$ are functions.
The experiment is "growth of proteins on a methanol substrate".
With passage of time, the concentration of substrate and oxygen is decreasing (substrate to zero, and $\ce{O2}$ to 10% from 100% and the process is terminated).
With substrate decrease the number of particles in bioreactor is growthing. I must determine the parameters $(s, \ce{O2})$ when the growth speed is $\rm max$ to $\mathrm{max }- x\%$. $x$ may be $10\%-15\%$ of $\rm max$.
I need a model to determine the functions $\mu(s)$, $\mu(\ce{O2})$ and $\mu(s, \ce{O2})$ or the max value of miu(s,o2), where $\mu$ = growth speed, $s$ = substrate concentration and $\ce{O2}$ = oxygen concentration.
I must simulate the function $\mu(s,\ce{O2})$, to observe the maximization surface(sized by $s$ and $\ce{O2}$) through variation of parameters $s$ and $\ce{O2}$. The surface formed by $s$ and $\ce{O2}$ when $\mu$ values are between $\mathrm{max}(\mu) - 10\%$ and $\mathrm{max}(\mu)$.
The volume of bioreactor is $15\:\mathrm{L}$ and the density inside is $\approx 1$.

Have you tried methods to fit your data into an equation (linear/exponential/logistic etc.) - there are few applications which can be useful - Maxima - http://en.wikipedia.org/wiki/Maxima_%28software%29 , R - http://en.wikipedia.org/wiki/R_%28programming_language%29 or even simple solver of Excel or LibreOffice can do it's job

Have you looked into Michaelis Menten kinetics? With this you should be able to come up with some differential equations for your growth rates


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What is the proper nomenclature for CH₂O₂?
Can you tell me what's the nomenclature of $\ce{C H_2 O_2}$?
What is its IUPAC name?

This can be a useful site for such queries if you know the structure of the compound or atleast one name, for example to answer your query click this

The compound you provided can be written more clearly as $\ce{HCOOH}$
IUPAC name : Formic acid
Systematic name :Methanoic Acid
Other names:
Aminic acid, Formylic acid, Hydrogen carboxylic acid, Hydroxymethanone, Hydroxy(oxo)methane, Metacarbonoic acid, Oxocarbinic acid, Oxomethanol

Well it could also be dioxirane. Maybe Bharat Khanna could precise his question and the context in which he came up to it.

It could also be dihydroxymethylidene.


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Why are nitrogenous hydrocarbons sometimes colorful?
Alanin and dyes contain nitrogen, is being colorful a special feature in nitrogenous substances?

Introduction from Color Introduction

"What causes a compound to appear as being colored? The answer to this question lies in the way that electromagnetic energy interacts with matter. The portion of the electromagnetic spectrum that is of interest to this discussion is the visible portion, or what we call light. This visible light includes all the colors that we perceive: violet, blue, green, yellow, orange, red, and the various combinations of these. Although light can interact with matter in several different ways, this discussion will limit the modes of interaction to absorption and reflection."
In most cases color comes from the excitation of electrons due to an absorption of energy performed by the chemical. But - there is still ... :)
"(...) The vast majority of simple inorganic (e.g. sodium chloride) and organic compounds (e.g. ethanol) are colorless. Transition metal compounds are often colored because of transitions of electrons between d-orbitals of different energy.
Organic compounds tend to be colored when there is extensive conjugation, causing the energy gap between the HOMO and LUMO to decrease, bringing the absorption band from the UV to the visible region. Similarly, color is due to the energy absorbed by the compound, when an electron transitions from the HOMO to the LUMO.(...)" Wiki - color of chemicals

You can also assume that if a compound has double or triple conjugated (2 or more) bonds it can be colorful.
To answer your question: in alanine there is almost no chance to make double bond in amine group (e.g. $\ce{=NH}$), but nitrogen group ($\ce{O-N=O}$) has a capability to conjunct with other double bonds (can be also delocalized electrons or pi electrons - e.g., nitrobenzene)

I expect the OP meant aniline, not alanine. Anyway, pure aniline is still colourless. It is used in the manufacture of dyes such as Aniline Purple, where several molecules become linked to form a large system of conjugated double bonds as you describe.


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Is it possible, chemically, to change a dextrorotatory compound to its levorotatory counterpart?
Is it possible, chemically, to change a dextrorotatory compound to its levorotatory counterpart? E.g., is it possible to change dextrodopa to levodopa?

Related, but not a duplicate: http://chemistry.stackexchange.com/questions/105/what-are-known-examples-of-drugs-that-racemize-stereoconvert-in-vivo-and-how-ar

Yes,this process is called stereoconversion. Both SN1 and SN2 reactions can invert the chirality of a carbon center. In the case of the SN1 mechanism, departure of a leaving group generates a prochiral intermediate which is then open to nucleophilic attack on either face. If both faces are equally acessible to the nucleophile this will result in racemisation of the molecule. In the case of the SN2 reaction the nucleophile will enter a trigonal bipyramidal intermediate with the molecule. Assuming the nucleophile has the same stereochemical precedence as the leaving group, the stereochemistry will be inverted.

Doesn't "dextrorotatory compound to levorotatory counterpart" mean we have to preserve the groups on the chiral carbon, rather than substituting one of the groups with another.

@AdityaSriram - Yes, but if the leaving group and nucleophile are the same species, you're in the clear. This can happen in the case of keto-enol tautomerism at a stereocenter.

Adding an example to your answer will really help a lot.

@RichardTerrett I think Aditya Sriram is right you need to add an example not just of how you would change the chirality from dex to levo but how you would preserve the groups on the chiral carbon.

I disagree with this. Under both SN1 or SN2 conditions, treatment of an alkyl bromide with bromide ion (for example) can at best only lead to racemisation of the compound and not complete inversion, which the question is asking for.

The following example(s) might help to illustrate the stereoconversion described by Richard Terrett:

In the case of alcohols, the "classical" method is the Mitsunobu reaction, using triphenylphosphine ($\ce{Ph3P})$ and diethyl azodicarboxylate (DEAD, $\ce{EtO2C-N=N-CO2Et})$ as reagents, and water as a nucleophile to quench the intermediate with inversion of configuration.
In the course of the reaction, $\ce{Ph3P}$ adds to DEAD in the presence of a proton donor to form an intermediate.

The latter adds to the alcohol, forming a stable $\ce{P-O}$ bond.

The addition of a nucleophile proceeds according to $S_N2$ with inversion of configuration, the formation of triphenylphosphine oxide renders the reaction irreversible.