text:
What are the bond angles in cyclohexane?

In the chair conformation of cyclohexane, what are the angles of the triangles defined by:

a carbon atom, an axial hydrogen bonded to it, and the midpoint of a vicinal C-C bond ?
a carbon atom, an equatorial hydrogen bonded to it, and the midpoint of a vicinal C-C bond ?

I have unsuccessfully looked for literature values of these, and my trigonometry is not up to scratch.

The crystal structure shows a strangely distorted molecule with only $C_\mathrm{i}$ symmetry. Values have quite a wide range, $\angle(\ce{CCH})\approx 100-120^\circ$. See below for all values and xyz coordinates to read in with a molecular viewer. Source:
 R. Kahn, R. Fourme, D. André and M. Renaud, Acta Cryst. 1973, B29, 131-138.
A DF-BP86/def2-TZVPP calculation (gas phase, 0 K) gives a highl $D_\mathrm{3d}$ symmetric structure with
\begin{array}{ll}
\angle(\ce{CCH_{ax}}) & =109.1^\circ,\\
\angle(\ce{CCH_{eq}}) & =110.3^\circ,\\
\angle(\ce{CCC}) &=111.5^\circ,\\
\angle(\ce{H_{eq}CH_{ax}}) &=106.4^\circ.\\
\end{array}
 Also Bond length differ very slightly (and impossibly to determine at room temperature and pressure), with
\begin{array}{ll}
\mathbf{d}(\ce{CC}) &=1.537~\mathrm{\mathring{A}},\\
\mathbf{d}(\ce{CH_{ax}}) &=1.105~\mathrm{\mathring{A}},\\
\mathbf{d}(\ce{CH_{eq}}) &=1.103~\mathrm{\mathring{A}}.\\
\end{array}
The tiny differences can be explained with Bent's rule.
In principle the assignment Goeff already gave you is absolutely valid, especially if you take the dynamic nature of the molecules into account that they have at standard temperature and pressure.
Also see xyz coordinates for visualisation below.

Appendix
Crystal structure

\begin{array}{lr}
R(1-2)          &1.528 \\
R(1-3)          &1.519 \\
R(1-4)          &0.884 \\
R(1-5)          &1.146 \\
R(2-6)          &1.055 \\
R(2-7)          &0.929 \\
R(2-10)         &1.521 \\
R(3-8)          &1.100 \\
R(3-9)          &0.944 \\
R(3-11)         &1.521 \\
R(10-12)        &1.519 \\
R(10-13)        &1.100 \\
R(10-14)        &0.944 \\
R(11-12)        &1.528 \\
R(11-15)        &1.055 \\
R(11-16)        &0.929 \\
R(12-17)        &0.884 \\
R(12-18)        &1.146 \\
A(2-1-3)        &110.4 \\
A(2-1-4)        &106.7 \\
A(2-5)        &105.8          \\
A(1-2-6)        &106.6          \\
A(1-2-7)        &118.2          \\
A(1-2-10)       &111.3          \\
A(3-1-4)        &103.0          \\
A(3-1-5)        &114.0          \\
A(1-3-8)        &110.6          \\
A(1-3-9)        &116.9          \\
A(1-3-11)       &112.4          \\
A(4-1-5)        &116.7          \\
A(6-2-7)        &109.5          \\
A(6-2-10)       &110.6          \\
A(7-2-10)       &100.5          \\
A(2-10-12)      &112.4          \\
A(2-10-13)      &109.1          \\
A(2-10-14)      &100.4          \\
A(8-3-9)        &106.9          \\
A(8-3-11)       &109.1          \\
A(9-3-11)       &100.4          \\
A(3-11-12)      &111.3          \\
A(3-11-15)      &110.6          \\
A(3-11-16)      &100.5          \\
A(12-10-13)     &110.6          \\
A(12-10-14)     &116.9          \\
A(10-12-11)     &110.4          \\
A(10-12-17)     &103.0          \\
A(10-12-18)     &114.0          \\
A(13-10-14)     &106.9          \\
A(12-11-15)     &106.6          \\
A(12-11-16)     &118.2          \\
A(11-12-17)     &106.7          \\
A(11-12-18)     &105.8          \\
A(15-11-16)     &109.5          \\
A(17-12-18)     &116.7          \\
\end{array}
C        2.329589000      3.011988000     -0.044238000
C        3.630888000      2.632672000      0.661249000
C        1.367908000      1.836044000     -0.049671000
H        2.524936000      3.113740000     -0.900292000
H        1.933922000      3.936128000      0.506802000
H        3.388552000      2.469096000      1.674854000
H        4.348450000      3.220000000      0.605369000
H        1.030575000      1.607424000      0.972471000
H        0.592407000      1.921052000     -0.581309000
C        4.247092000      1.383956000      0.049671000
C        1.984112000      0.587328000     -0.661249000
C        3.285411000      0.208012000      0.044238000
H        4.584425000      1.612576000     -0.972471000
H        5.022593000      1.298948000      0.581309000
H        2.226448000      0.750904000     -1.674854000
H        1.266550000     -0.000000000     -0.605369000
H        3.090064000      0.106260000      0.900292000
H        3.681078000     -0.716128000     -0.506802000

Computed (DF-BP86/def2-TZVPP)

C       -1.270112000     -0.733300000      0.229506000
C        0.000000000     -1.466599000     -0.229506000
C       -1.270112000      0.733300000     -0.229506000
H       -1.328386000     -0.766944000      1.332131000
H       -2.167276000     -1.251277000     -0.145050000
H        0.000000000     -1.533888000     -1.332131000
H        0.000000000     -2.502555000      0.145050000
H       -1.328386000      0.766944000     -1.332131000
H       -2.167276000      1.251277000      0.145050000
C        1.270112000     -0.733300000      0.229506000
C        0.000000000      1.466599000      0.229506000
C        1.270112000      0.733300000     -0.229506000
H        1.328386000     -0.766944000      1.332131000
H        2.167276000     -1.251277000     -0.145050000
H        0.000000000      1.533888000      1.332131000
H        0.000000000      2.502555000     -0.145050000
H        1.328386000      0.766944000     -1.332131000
H        2.167276000      1.251277000      0.145050000

In the chair form of cyclohexane, the carbon atoms and the bonds around them are almost perfectly tetrahedral.
So the C-C-H angles will be almost exactly 109.5 degrees. (Note that while you defined the bond midpoint, the angle will be the same regardless of whether it's the midpoint of the bond or the neighboring carbon atom itself.)

So the angle between the C-H bond and the C-C bond is 109.5 degrees in both case, and I can derive the other sides/angles with sine/cosine rules ?

Personally, I'd suggest downloading http://avogadro.cc Build-->Insert->Insert Fragment, then find "cyclic alkanes" and "cyclohexane chair" and switch to the ruler/measure tool to get the angles you want.


text:
in the electrolysis of H₂SO₄ solution, when the concentration is increased why does the current increase even when OH⁻ ions reduce?
During the electrolysis of $\ce{H2SO4}$, $\ce {OH-}$ ions are more reactive than $\ce {SO4^{2-} }$ ions and so they dissociate. as $\ce {OH-}$ ions dissociate, they are the negative ions that carry current. So when the concentration of $\ce {H2SO4}$ is increased, the concentration of $\ce {OH-}$ ions actually reduces, but then the current still increases.

Near room temperature, considering solutions of water and sulphuric acid, the maximum conductivity is at about 30% sulfuric acid.  For pure water or pure sulfuric acid the conductivity is orders of magnitude lower.
In pure water, there are only $10^{-7}$M of $\ce{H+}$ and $\ce{OH-}$ to carry current.
When a relatively small amount of sulfuric acid  is added, it dissociates fully to $\ce{H+}$ and $\ce{HSO4-}$.  For example if you increase the concentration of sulfuric acid from 0M to 0.01 M, the number of ions in the solution increase by a factor of $10^5$.

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text:
How does absence of nd orbital in Oxygen affect its valency?
The original question was 'Oxygen exhibits Oxidation state -2 to +2 but other elements of grp 16 exhibit only +2,+4,+6. Explain Why'. In the solution given i can't understand meaning of this line 'It doesn't have nd-orbitals due to which its valency doesn't increase'. Please explain how?
PS:I am a chemistry noob. So please provide as much details as possible :)

I strongly advise you to acquire a copy of this article http://link.springer.com/article/10.1007%2FPL00010730 and ask your chemical teacher to comment on it.

Whether or not oxygen can reach oxidation states higher than $\mathrm{+II}$ is completely independent of the type of orbitals the second shell has. If the orbitals would matter, nitrogen would not be able to reach every single oxidation state between $\mathrm{-III}$ and $\mathrm{+V}$ but would also be restricted to <insert arbitrary number here>.
The actual reason why oxygen’s accessible oxidation states are limited is its very high electronegativity, second only to fluorine among the non-noble gas elements. There is simply no partner that can draw away electrons strongly enough to access high oxidation states. Fluorine itself, while typically being able to access high coordination numbers (higher than oxygen, compare $\ce{XeF6, NF3}$ or $\ce{SF6}$ to $\ce{XeO4, NO3-}$ or $\ce{SO3}$) is not the element of choice for accessing high oxidation states — that is oxygen. But oxygen cannot oxidise itself since it has the same electronegativity. Fluorine itself, as is evident by $\ce{NF5}$ being unknown as of today, would not be able to stabilise an oxidation state of more than $\mathrm{+II}$ on oxygen given our current knowledge.
Furthermore, the statement’s claim that group 16 elements except oxygen only exhibit positive oxidation states is bogus. $\ce{Na2S}$, $\ce{Na2Se}$, $\ce{Na2Te}$ and $\ce{Na2Po}$ are all known and in all of these compounds the chalcogen has an oxidation state of $\mathrm{-II}$.

While NF5 is still unknown, there was a report of possible NF6- anion and NF4+ is known for a long time. There's a theoretical study indicating possibility of analogous OF3+.

You need to review the concepts surrounding atomic shell structure and quantum numbers, and how they relate to the organization of the periodic table, I think.  Briefly, since oxygen is in the second row of the periodic table, we know that its valence electrons occupy the second orbital shell, corresponding to orbitals with principal quantum number $n=2$.  Recall also that for a given principal quantum number $n$, only sub-shells with orbital angular quantum number $l$ up to $n-1$ are allowed.  Therefore, the valence shell of osygen only includes s ($l=0$) and p ($l=1$) orbitals.  On the other hand, the other members of group 16 will by definition have higher principal quantum numbers, which means that their valence shells include d ($l=2$) orbitals.  It is these orbitals that can be used to rationalize the existence of higher oxidation states for those elements, for example by the $sp^3d$ and $sp^3d^2$ hybridization schemes in VSEPR (valence shell electron pair repulsion) theory.
Now, that explains why heavier elements in group 16 can have higher (i.e. more positive) oxidation numbers than +2.  Your question as stated appears to have an error however, since it says that those elements "exhibit only +2,+4,+6" ... that's not technically correct .. for example there are compounds where sulfur has an oxidation state of -2 (e.g. $H_2S$).  However I don't think it matters for the answer to your question here.

I know that Oxygen doesnt have d-orbital n other elements down the group have it but HOW exactly does that restricts oxygen OR say allows other elements to have more than +2 Oxidation State?

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text:
If the standard state symbol means that the substance is pure (and at 1 bar) how is it possible to have a standard REACTION enthalpy?
For the reaction one mole of substance $A$ in equilibrium with one mole of substance $B$ the standard reaction enthalpy is defined:
$$\triangle_rH^{\theta}(T)=H_m^{\theta}(B;T)-H_m^{\theta}(A;T)$$
However, I have two issues with this.
Firstly: the standard state of a substance (denoted by the symbol $\theta$) is the pure material at a pressure of 1 bar. How can the reaction enthalpy be standard - the reactant have to be mixed (thus are not pure) to react.
Secondly: what is $H_m^{\theta}$? Surely nothing can have a molar enthalpy because only changes in enthalpy can be measured not absolute values. Am I right in thinking that is is because the enthalpy scale has no zero value?

The standard reaction enthalpy you have describe is only for a pure substance if you are talking about a standard enthalpy of formation.  Otherwise the equation wouldn't be valid for reactions that have more than one product, right?  In the specific case of a standard enthalpy of formation, you are talking about the enthalpy change associated with the creation of one mole of a pure substance from its component elements in their standard states.  This gets at your second question ... the zero on the enthalpy scale is established by convention (and this is the reason why the definition of standard enthalpy of formation works) ... the standard enthalpy of formation of an element in its standard state is DEFINED to be zero.
Now, you can estimate the standard reaction enthalpy for any chemical reaction by subtracting the summed standard enthalpies of formation of the reactants from the summed standard enthalpies of formation of the products, taking into account the appropriate stoichiometric coefficients from the balanced chemical equation.  That process is similar to what you have written, but not precisely the same, since you omitted the stoichiometry information (and the chemical equation, for that matter).  The standard enthalpy estimate obtained will pertain to reactions carried out at the given conditions .. standard enthalpies are defined for 1 atm pressure but variable temperature, so you may need to correct tabulated standard enthalpies of formation to match the given reaction conditions.
One last point is that enthalpy is absolutely defined as a thermodynamic quantity .. it is the internal energy plus the product of pressure and volume, $H=U + pV$.  The issue is that the internal energy of a substance is hard to define absolutely .. are you going to include the binding energies of the core electrons to their atoms?  How about the strong forces holding the nuclei together?  Thus for chemical processes it almost always makes sense to deal with just changes in enthalpy, since most of the really low level stuff I mentioned doesn't change between reactants and products.  I think that may have been part of what you were getting at with your second question.


text:
Regioselectivity of alpha halogenation of ketones
Why does alpha halogenation of a ketone in acidic conditions happen at the more substituted alpha carbon atom, and in basic conditions at the less substituted alpha carbon atom?

The acidic and basic catalyzed alpha-halogenation of a carbonyl proceed through an enol and an enolate intermediate respectively.  The difference in regioselectivity arises because of the different process used to arrive at the enol/enolate in the acidic and basic cases.
In the acidic case, 1) the carbonyl oxygen is protonated and then 2) a proton is removed from the alpha carbon to form the enol.  See here for a good picture of this mechanism.  In this process, the reaction is an equilibrium so thermodynamic control will lead to formation of the most stable enol.  Therefore, a proton will be removed from the more highly substituted alpha-carbon to produce the more stable double bond (the one that is more highly substituted).
When the reaction is run under basic conditions, the process is no longer a thermodynamically controlled equilibrium, but rather proceeds under kinetic control.  The same link used above also shows the mechanism of the base-catalyzed alpha-halogenation.  The rate determining step is removal of an alpha-proton by the base.  The lowest energy proton to remove is the one on the least substituted carbon. Remember that a primary carbanion is more stable than a secondary carbanion which is more stable than a tertiary carbanion, so removing a primary proton will produce the most stable carbanion-enolate.  Hence, in the base catalyzed case the proton will be removed from the least susbstituted alpha carbon so to produce the most stable carbanion-enolate intermediate.
To summarize the acid-catalyzed halogenation proceeds under thermodynamic control to produce the most highly substituted (most stable double bond) enol.  Whereas the base-catalyzed halogenation is kinetically controlled and produces the least substituted (most stable carbanion) enolate anion.

Thank you for the nice answer. I would have one more question, but I will start a new question.

Ron's answer is not completely true.
Yes, the least substituted α-hydrogen is the more acidic one and the least hindered one (kinetic control) and the more substituted enolate is the most stable (thermodynamic control). What is not completely true is that acidic conditions favour thermodynamic control because it is reversible but that basic conditions are limited to kinetic control.
Whether or not basic conditions afford kinetic or thermodynamic control depends on the deprotonation conditions. Hence, the use of a reversible base (with associated pKa less than that of the carbonyl compound, i.e. 18-25) such as NaOH will promote the thermodynamic enolate, as will a protic solvent such as H2O. Even a very strong base such as LDA does not guarantee kinetic control, because any trace of proton or unreacted ketone will cause equilibration to the thermodynamic enolate, even at –78°C over time.
The real reason for the obtention of the least substituted halide from an unsymmetrical ketone under basic conditions is that the molecular halogen (e.g. Cl2) reacts instantly with the kinetic enolate as it forms, trapping it in a sense, and that reaction between a powerful nucleophile and a powerful electrophile is apparently faster than the enolate equilibration. Some of the reason for this is that Cl2 in OH– is actually OCl– which acts as both base and electrophile, so that a second collision is not required.
This property is the reason for the success of the haloform reaction, which occurs at the CH3 end of methyl ketones. The first halogenation facilitates the second deprotonation (by virtue of lowering the pKa) and the second halogenation facilitates the third.

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text:
How can concrete set underwater?
I'm about to teach concrete to some high school children. I'm not a chemist but as I understand it the limestone cycle is this:
$$
\begin{align}
\ce{CaCO3 &-> CaO + CO2}\tag{R1}\\
\ce{CaO + H2O &-> Ca(OH)2}\tag{R2}\\
\ce{Ca(OH)2 + CO2 &-> CaCO3 + H2O}\tag{R3}
\end{align}
$$
Two questions:

The water has to go somewhere or the concrete will stay wet. Air drying I could understand but how can concrete set underwater? And, assuming concrete dries from the outside in, wouldn't osmosis predict more water would be drawn into the concrete?

Making kilograms of calcium carbonate will take kilograms of calcium hydroxide and kilograms of carbon dioxide. How can such a large amount of carbon dioxide reach the concrete? There isn't a lot of it in water, is there? I could imagine some reaction might produce carbon dioxide in the concrete, but then wouldn't it end up frothy/honeycombed?

The chemistry you have posted is for non-hydraulic cement, which requires air drying for precisely the reasons you enumerated.  The kind of cement that sets underwater uses silicates and various oxides (primarily calcium, aluminum and iron oxides). The basic chemical process that is occurring is hydration, whereby the mixture of oxides and silicates incorporates several water molecules directly into the crystal lattice.  The fact that the setting process both uses water, and also produces a structure that is insoluble and robust to attack by water, is what makes underwater setting of hydraulic cement possible.
With regard to teaching your high school class, it's probably better to stick to the lime cycle for non-hydraulic cement as far as detailed chemical reactions are concerned. One interesting point that you can tell them about hydraulic cement is that its chemistry is still not completely understood, and even the molecular crystal structure was only very recently sorted out by MIT scientists in 2009.

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text:
Halogenation of an alpha-halogen ketone in acidic conditions
When a methyl ketone is submitted to halogenation in acidic conditions with two equivalents of a halogen, the first halogenation happens at the secondary alpha carbon atom , while the second halogenation happens at the primary alpha carbon atom. Why is it so? Why doesn't it happen on the same carbon atom (the secondary one)?

What is the specific molecule you're asking about?

Actually any ketone. Even acetone with two methyl groups, will have the halogen atoms ending up at the two different alpha carbon atoms.

The same mechanism discussed in my answer to your earlier question on acid catalyzed halogenation alpha to a carbonyl still applies here.  For reference, here is the same link supplied earlier that shows the mechanism for this acid catalyzed halogenation.
In that earlier answer I said that the enol that "produces the more stable double bond (the one that is more highly substituted)" is the enol that will form and react nucleophilically with the halogen to produce the alpha-halo carbonyl product.  In light of your new question, the statement that more highly substituted double bonds are more stable needs some qualification.
The closer in electronegativity a substituent is to carbon, the more it will stabilize a double bond.  Consequently, the more alkyl groups attached to a double bond, the more stable the double bond (barring large steric effects).  On the other hand, fluorine destabilizes a double bond.  It is so electronegative that it prefers to bond with orbitals high in p character.  When forced to bond to a double bond $\ce{sp^2}$ orbital (that is enriched in s character) destabilization results (see this earlier fluorobullvalene example for more on why fluorine "hates" to be on a double bond).
Based on its electronegativity, chlorine should be somewhere between fluorine and an alkyl group in terms of its ability to stabilize a double bond.  If we compare the heats of formation of 1-butene (-0.1 kJ/mol) and trans-2-butene (-11.2 kJ/mol) we see that placing the methyl group on the double bond stabilizes the molecule by 11.1 kJ/mol.  If we compare the heats of formation of allyl chloride (-5.6 kJ/mol) and trans-1-chloropropene (-12 kJ/mol) we see that the placing the chlorine group on the double bond stabilizes the molecule less, only 6.4 kJ/mol.
These differences mean that we will have much more of the enol without the chlorine on the enol double bond, then we would have of the enol without the methyl group on the enol double bond.  So we have more of the unsubstituted enol in the chlorine case and the enol with the chlorine on the double bond is much less nucleophilic (less reactive) due to the electron withdrawing inductive effect of the chlorine substituent.  These two factors combine to make the second alpha-chlorination take place on the unsubstituted side of the molecule.
As to steric effects, they probably don't play a significant role.  One metric commonly used to compare "steric size" is the cyclohexane A value.  The A value for chlorine is 0.43, much smaller than the A value for methyl (1.7), suggesting that a chlorine substituent is smaller than methyl substituent.

Why would then the second alpha halogenation in basic media happen at the same carbon atom?

@Marko Because the halogen substituents increase acidity.

But it increases the acidity in the first case, too.

@Marko  In the base catalyzed case it is about which proton is easiest to remove (which resulting carbanion is the most stable).  The more chlorines attached to a carbon, the weaker the remaining C-H bonds, the more stable the resulting carbanion.

Most probably due to steric reasons and the electron withdrawing effect that the halogen has on the substituted methylene group.


text:
How does photosynthesis work?
I wonder how photosynthesis works. I read the Wikipedia page, but I'm still confused.
My main gap of understanding is (I assume):
I know a photon changes the velocity of an electron. But I do not (fully) understand how a photon creates chemical reactions. Does the atom lose its electron due to this velocity increase and thereby attracting the electrons from the other atoms?
Is that all? Does this imply there is always an electron lost in photosynthesis or is that electron absorbed elsewhere (in another atom)?

A better way of thinking about it, without getting too bogged down in details, is that the plant absorbs the energy of photons and uses it to drive chemical processes - ultimately the synthesis of sugar out of carbon dioxide and water - that are energetically "uphill", and would not happen spontaneously without the energy from the photons.  The energy from the photons is used by the plant to move electrons from one place to another, but no electrons are actually generated or consumed in the process ... photosynthesis is a cycle, so you always end up back where you started, with some molecules of $\ce{CO2}$ and $\ce{H2O}$ having been consumed (along with a lot of light energy), and some extra molecules of sugar and $\ce{O2}$ having been produced.
The places where the photonic energy is used to mobilize the electrons are protein complexes called photosystem I and photosystem II, which are concentrated in the chloroplasts of green plants.  The details of how this happen are still not completely understood, but a LOT of progress has been made fairly recently using fancy equipment like laser spectrometers that can measure physicals processes that occur on timescales shorter than a nanosecond.  This has allowed researchers to map the progress of the electron as it moves from one part of the protein complex to another, following absorption of the photonic energy.
Does that help at all?

A little tiny bit. Thing is that I was more or less aware of what you wrote. Nice text anyway. Thanks.

No, photons of visible light don't ionize molecules during photosynthesis - electrons are transferred via redox reactions in electron transport chain - it seems to me that fragment of the article was maybe a little misleading.

In some steps, a red light photon has enough energy (~0.5 e.v.) to raise an electron to a higher, metastable orbit (activating it), but might not actually remove the electron from the molecule (see http://highered.mheducation.com/olcweb/cgi/pluginpop.cgi?it=swf::535::535::/sites/dl/free/0072437316/120072/bio13.swf::Photosynthetic+Electron+Transport+and+ATP+Synthesis (very clear), http://hyperphysics.phy-astr.gsu.edu/hbase/biology/psetran.html and http://hyperphysics.phy-astr.gsu.edu/hbase/biology/psetran.html).


text:
How to tell if a molecule will participate in hydrogen bonding with water?
I was recently doing some organic chemistry work and came across a set of problems.
"Which of the following compounds will participate in hydrogen bonding with water? For each compound, indicate whether it will be a hydrogen bond donor, hydrogen bond acceptor, or both."
I do not have an idea of how to approach the first part of the question. Can someone help me with the first molecule: (a) $\ce{CH3CH2OCH3}$ in order of connectivity.

I figured out the answer. If there is a lone pair of electrons belonging to nitrogen, oxygen, or fluorine, then it is possible to bond to water. Therefore, it is can accept a hydrogen. If there is a hydrogen bonded to one of the three elements listed above, then it able to donate.


text:
Can there be more solvent in a solute?
Something I know that leads to this question:

Water dissolved in ethanol.

It might sounds weird to say water as a solute. Textbooks always says that solvent is the one present in larger quantity. This leads to a contradiction with the saying that water is an universal solvent.
There can be water present in smaller quantity than ethanol in a water-ethanol solution. But which is the solute and which is the solvent?
How to identify the solute and the solvent in a solution besides locating the word "aqueous"?

The textbook statement you gave ("solvent is always the one present in larger quantity") is simply a convention that makes it easier for us to talk to each other about mixtures of substances.  We have other words for other types of mixtures.  For example, in your ethanol-water example, you are correct in stating that it doesn't make much sense to talk about solute and solvent.  So, we call these two liquids "miscible".  They mix together in all proportions.  A good demonstration that captures the essence of different types of mixtures is the phase diagram of phenol-water:

Here we have a feature that is called a "miscibility gap".  There is a region, defined by the rainbow shape in the diagram above in which phenol and water separate into two distinct phases.  If we had pure water and added just a little bit of phenol to it at room temperature, we would find that it would dissolve into the water up to about 7 or 8 wt%. So, phenol would be the solute and water the solvent.  Adding more than this amount pushes the mixture into the miscibility gap so that now the phenol and water would mostly separate into two separate phases.  Let's say we add enough phenol to get us to 40 wt%.  If we heat the two phase mixture up to 67 C then it will suddenly become miscible.  Now only one phase exists and we can add as much phenol as we want to it without having to worry about two phases forming again.  There is no solute or solvent here.  Just a mixture of two liquids that exist in a single phase.

At that temperature you can put however much you want?!

Don't you have to increase temperature as you add more to it. How can there be a solution without saturation state?

As long as the temperature is above 67 C in the phenol-water mixture you can keep adding as much phenol as you want and never have to worry about saturation. It is a miscible mixture at this temperature, just like the ethanol-water mixture. Try your best to disengage from thinking in terms of a "solution" with a certain "solute" and "solvent" in this case. Its just two liquids that mix together in all proportions to form a single liquid phase

You are saying if we heat up to 67 C then ethanol should get disappeared bcz it has low boiling point in comparison with water then why water is getting disappeared?

Ethanol and water are partially miscible liquids.

I think you may be emphasizing the term "universal solvent" a bit too hard. Water is "universal" as a solvent simply because lots of molecules are at least somewhat polar and so dissolve in polar water. But otherwise I'd say your textbook is right: the solvent is generally in greater quantity than the solute by convention.

I think you may be putting too much significance on the semantics of some mnemonics that are perhaps phrased more absolutely than they are intended to be used.  That is particularly true with the phrase that "water is the universal solvent", which is not really a scientifically significant statement at all.  Certainly gasoline, or mineral oil, will not dissolve in water, even in very small quantities, so why would it be considered "universal" in the sense you seem to be taking it?
To answer your question, the better rule of thumb is to use the statement that the solvent is always the component present in larger quantity .. but even that is not correct in all cases.  For example in a multiphase extraction, there can be multiple solvents ... components that are polar will dissolve in the aqueous phase, so for those components, water is the solvent .. while components that are non-polar will dissolve in the non-aqueous phase, which will be the solvent for those components.
A final point is that for something like a 49-51% mixture of ethanol and water, the concept of a solvent becomes largely academic ... is the ethanol dissolved in the water, or is the water dissolved in the ethanol?  The answer is: both, and neither .. they are miscible liquids, so it's a two component mixture, rather than a solution.  The physical properties of a mixture depend rather strongly on the properties of both components, while the physical properties of a solution are more likely to be dominated by the properties of the solvent.

You are saying that for non polar substances water is not solvent. Is it possible to have two liquids in the same ratio in a solution along with solute, if it is so which one would be the solvent. Plz make me understand I m not getting this....


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Obtaining chemical graph complexity index from SMILES or InChl
I wonder if anyone knows a readily available way to obtain some chemical graph complexity index (like the J index) from the SMILES or InChl representations?I'm looking for some existing implementation, e.g. in R, or OpenBabel, prior to trying to define mine if no one else exist... Thanks!

I don't know what the J index is, so it's certainly not already implemented in Open Babel. It likely wouldn't be hard to implement.

Have you looked at a package like RDKit? I'm not sure on the topological indices etc. but the package is quite extensive.

Thanks for the answers. Certainly the RDKit allows to calculate topological indeces. For anyone interested in the same thing: after some research I also found The Chemistry Development Kit, which allows for that and has an R interface (RCDK). Also found MathChem, which allows to calculate a wider variaty of indexes (https://draganstevanovic.files.wordpress.com/2013/11/mathchem-match-9.pdf). Best

@Tintanjarafe can you please add a reference for the J index? The CDK has several complexity-related descriptors, but I would not mind thinking of implementing this descriptor, but I need to know the algorithm. Was it published somewhere?

@EgonWillighagen. Sure, it's also known as the Balaban's J index. Here are links to the original publication and to the page in MathWorld: http://www.sciencedirect.com/science/article/pii/0009261482800092, http://mathworld.wolfram.com/BalabanIndex.html. Thank you!

This is a very old question, but if folks are still interested perhaps this is a useful bit of code: https://github.com/maxhodak/keras-molecules/pull/32/commits/dbbb790e74e406faa70b13e8be8104d9e938eba2


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Is there any graph that depicts the deviation of real gases from Charle's law?
I am looking for a graph that shows the basic deviation of a gas from the Charle's law. The graph should be for different values of constant pressures.This would help me evaluate how gases follow ideal behavior for low pressures and high temperatures.

Check out the van der Waals equation wikipedia page for a start ... it gives an explanation of one of the most common modifications of the ideal gas law that allows for modeling of deviations from ideality in the way you describe.  This approach is quite nice in my opinion, because it includes two coefficients (a and b) that account for specific deviations for a given gas ... a scales with the strength of intermolecular interactions, and b accounts for the actual volume of the gas molecules.
You should be able to use that information to do what you want to do regarding Charles Law.  The graph on the page shows isotherms, but you should be able to generate the isobar curves that you are after with a little bit of work.


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How does volume contraction in solvent mixing work?
I recall seeing chemistry demonstrations where a significant amount of a solute is dissolved in a solvent and the solution volume barely changes. Questions regarding this:
1) How is this possible?
2) What solvent/solute combinations best demonstrate this effect?
3) Is it possible to wind up with less solution, in other words Is it possible to have Volume(solution) < Volume(solvent) + Volume(solute) ?

Yes this happens fairly regularly. It is a deviation from ideal behavior associated with the property of partial molar volume.  For an ideal solution, the total volume always equals the summed volumes of the pure components, but ideal behavior neglects several important facts about real molecules, to wit, that they may have different shapes, and may also experience stabilizing or destabilizing interactions with each other.
It's fairly easy to see how molecular shapes and sizes could affect this property. Macroscopic objects experience similar effects.  Suppose I have a bucket "full" of golf balls and a bucket full of sand, if I work carefully, I can pour quite a lot of sand into the bucket of golf balls because the smaller size of the sand grains allows them to fit into the gaps between the golf balls.  Another example: Suppose you have a 10 liter bucket of golf balls and a 10 liter bucket of tiny cubes packed tightly together. If I then try to pour those two buckets into a 20 liter container simultaneously, it almost certainly will not be able to fit all of the balls and cubes, because they will not pack as efficiently in the 20 L container, costing some space.  These are extreme examples, but illustrate one way that different shapes can play a role.
Differences in intermolecular interactions can play an even bigger role.  If you are dissolving salt in water, the individual ions making up the salt crystal can be solvated very efficiently by the polar water molecules, which experience a strong attractive interaction with the ions.  (Recall that the O and H-atoms in water have partial negative and positive charges, respectively.)  This results in the solution having a significantly smaller volume than expected based on the summed volumes of the pure salt and water used to make the solution.
Partial molar volume is not too hard to measure experimentally. For something like aqueous sodium chloride it can be determined in a fairly straightforward way from precise measurements of the densities of a series of solutions with increasing concentrations.  However it is worth noting that the volume and mass measurements used for the calculation must be very precise, yielding measured densities with 5 or more significant digits.  This requires careful use of volumetric glassware and precision balances, and the temperature of the solutions must be kept constant throughout.
For purposes of demonstrating the qualitative effect, you could just use water and ethanol (or isopropyl alcohol), which also have a significant negative excess volume. (see the first graph on the wikipedia page for partial molar volume for details).
For comparison you could use mixtures of benzene and toluene, which behave almost ideally with respect to volume effects.

You mention dissolving salt in water. What particular salt would provide the most dramatic effects?

The most easily observable effects from commonly available ingredients would probably be mixing water and an alcohol, like ethanol or isopropanol.  You could do a demo with a couple of graduated cylinders.  50 mL each of water and ethanol should have a negative excess volume of a few mL for sure.
As far as which salt is concerned, I don't really know what would give the biggest effect there, but NaCl certainly works.  It's easier to measure the volumes liquid samples

1.Volumes do not add when you mix things together because the intermolecular forces of a mixture are different than the ones in the pure substances. The way this is handled is by using a property of mixtures called the "partial molar volume."
2.There are a couple of reasons why volumes of solution are not additive. One important one is that some chemical reaction occurs. The second is especially important if one component is water. Liquid water has an open structure, that is, the water molecules even in the liquid form molecule sized structures of lower density than the mean. The addition of ionic and/or polar solutes can "collapse" this structure resulting in an increased density. To complicate matters, the solute may also have structure that may be smaller than or greater than the average density of the solid. This will result in an expansion (or contraction) when the solute is dissolved.
There are no good theoretical models, of a general nature, to predict or estimate whether a solvent/solute combination will have a positive or negative "excess" volume of solution.
3.When $50\mathrm{mL}$ of water and $50\mathrm{mL}$ of ethanol are combined in a $100\mathrm{mL}$ graduated cylinder, the resulting mixture has a volume of $\approx 97\mathrm{mL}$.
Intermolecular interactions between water molecules in a pure sample involve hydrogen bonding, though the details of these complex interactions are not completely understood.  Some models predict transient ice-like structures within liquid water, where the local density of the sample is temporarily diminished. Ethanol and water molecules are also attracted to each other through hydrogen bonding. The ethanol likely interrupts the transient ice-like structures present in liquid water, resulting in a slight contraction of the sample.


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What kind of compound absorption the 365nm UV light on TLC
What kind of compound absorption the 365nm UV light?
compounds that has benzene ring or conjugated pi bond, make a fluorescence on 254 nm UV on TLC.
what kind of moiety(functional group) make fluorescence on 365nm UV on TLC

Though I don't think this is worth a full answer on a closed question, it might be usefull to know that some TLC plates have been functionalized to fluoresce at the wavelength used to check for compounds. Depending on the compounds that have run on the TLC plate, the fluorescence is quenched and therefore you can see a spot.

Your question is almost impossible to answer - it simply is too broad.

Excitation spectra of fluorescent organic compounds typically aren't sharp lines but consist of one (or more) bands. Even if the maximum of the excitation spectrum isn't located at $\lambda = \textrm{365 nm}$, monochromatic irradiation at that wavelength will excite a lot of different compounds (and even compound classes) which then deactivate by fluorescence.
Your light source does not emit monochromatic radiation! Hand lamps used to inspect TLCs typically have coated low-pressure mercury lamps. While the maximum of the radiation density might be centred around 360 nm, the full with at half maximum is - roughly estimated - about 30 nm or more. This means that there's even more compounds that might get excited by the emission of the hand lamp.

In summary, there's no exhaustive answer to your question, but just to name some compounds that are potential candidates:

Diphenylacetylene (1)
1,6-Diphenylhexatriene (DPH, 2)
4',6-diamidino-2-phenylindole (DAPI, 3)
6,8-difluoro-7-hydroxy-4-methylcoumarin (DIFMU, 4)


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Can state of matter from liquid to gas maintain the same molecular structure in PE or in polymers in general, like water does?
If I’ve got a reactor containing PE-Polyethylene $(\ce{{C2}$H_\text{4})n$}$ (solid) and the rest is filled with nitrogen gas. The system is closed.  Supposing we have an initial temperature of $30~\mathrm{°C}$ and gradually increase the temperature to $180~\mathrm{°C}$. We’ll observe a change in state of the PE, passing from solid to liquid, at this point the molecular structure is the same.
If temperature is increased beyond $250~\mathrm{°C}$ pyrolysis will occur on PE and syngas will be produced, at this point the gas molecules will have different structures compared to $(\ce{{C2}$H_\text{4})n$}$, thus "we’ll not observe" a change of phase from liquid to gas maintaining  the same molecular structure. Gas molecules will nave n types of molecular structures.

Is it possible by varying PVT to have a change of phase, from liquid to gas, of PE and maintain the same molecular structure. Just like with water->vapor?
If this is not possible, why? what are the topics to study?

There are many types of polyethylene, melting point you added suggests HDPE, but I think your asking generally?

It is generally assumed that polymers are unable to be in gaseous form. In fact, they even don't have a truly liquid and solid states, being in that respect similar to glasses: there is no strict border between liquid and solid states. Moreover, some kind of polymeric materials doesn't have liquid state either, being cross-linked. || However, deeply theoretically speaking, all compounds have non-zero, even if negligible vapor pressure, so at extremely low pressure polymers should be gaseos... except nobody was able to detect it.

@Mithoron yes, I'm talking in general about plastic polymers, the example was to make my question clear.


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Gibbs free energy can be expressed as a function of P,T and n but are enthalpy and internal energy also (partially) functions of n?
$$G=G(P,T,n)$$
$$\mathrm dG=V\,\mathrm dP-S\,\mathrm dT+\mu\,\mathrm  dn=\left(\frac{\partial G}{\partial P}\right)_{T,n}\,\mathrm dP+\left(\frac{\partial G}{\partial T}\right)_{P,n}\,\mathrm dT+\left(\frac{\partial G}{\partial n}\right)_{T,P}\,\mathrm dn$$
This allows open systems to be considered (where $\mathrm dn$ does not equal zero). However, can enthalpy, internal energy and Helmholtz free energy also be treated in this way to allow for open systems?

I'm not really sure what you are asking.  Do you mean, is it as simple as adding a $(\frac {\partial X}{\partial n})dn$ term, where $X$ is the state function of interest?  In that case no .. for example, Helmholz free energy is defined as the useful work that can be obtained from a closed thermodynamic system at constant T.  The definition of Gibbs free energy includes the open nature of the system, and the state function has the corresponding "natural" thermodynamic variables, P,T and n.

Yes that's exactly what I mean. Why can't you do this for state functions like helmholtz free energy and enthalpy? Or perhaps it's more useful to think about the converse: why does gibbs free energy allow for an open system to be considered whilst enthalpy and so on do not?

Because that's specifically how it is defined.  Also, it's not that enthalpy (and other function) don't allow for open systems to be considered .. the enthalpy is part of the Gibbs free energy, after all.  It's more that open systems are not part of the natural definitions of those other state functions.  You can define the exact differential of helmholz free energy for an open systems .. you just add $\sum_i\mu_idn_i$, so the definition is still in terms of chemical potential, which relates to changes in the Gibbs free energy with composition.

What exactly do you mean by the natural definitions? I don't see anything particularly different between $dG=VdP-SdT$ and $dH=TdS+VdP$ that would cause gibbs free energy to naturally take into account open systems. Also, on the note of just adding $\sum_i\mu_idn_i$ I mean add on the appropriate equivalent to $\mu$ molar enthalpy for example or molar helmholtz free energy.

What I mean is that the derivation of the Gibbs free energy was specifically done to include the effects of composition changes .. this is not true of the other functions.  See here for a short explanation of how the natural thermodynamic variables are obtained.  Regarding the other point, I understand .. that's why my original comment said "no, you can't just do that" ... you can still account for open systems, you just can't do it that way.

Thank you. Although it does show that one can do it for internal energy. Also, searching enthalpy in wikipedia I was able to find the expression that involved the $\sum_i\mu_idn_i$ term. With all due respect, are you absolutely sure you are right?

I am not sure why you seem to find my comments inconsistent.  I have said that you CAN account for the effects of open systems on thermodynamic state functions by adding $\sum_i\mu_idn_i$ to the exact differential.  I have also said (although not directly) that $\mu_i=(\frac {\partial G}{\partial n_i}){T,P}$ is always the definition of the chemical potential.  I said you CANNOT add the analogous quantity $\sum_i(\frac {\partial A}{\partial n_i}){T,V}dn_i$ to the exact differential of Helmholz free energy (for example) to allow for open systems.  Yes, I am sure that is all correct.

Oh, I see. However, why does adding a partial derivative of G work for the total derivatives of A or H? Thanks for your patience, I am sorry for all the questions.

I'm not completely certain (I actually hadn't thought about it that way), but I think it's because the natural variables of those other state functions include extensive properties, like volume, or entropy, that depend on the total number of particles in the system.  Thus when you add the $\sum_i\mu_idn_i$ term to the exact differential of one of the other state functions, it is not independent of the other terms.  For Gibbs free energy, the other natural variables are the intensive properties p and T, so the $n_i$ are independent variables.

I think I have made too much of the number of particles being a natural variable of ONLY the Gibbs free energy.  Actually, it is a natural variable of any thermodynamic potential (internal energy, enthalpy, Helmholtz free energy, or Gibbs free energy).  The more important consideration for your question is the intensive vs. extensive nature of the other state function variables.  I am working on a response to your other question on this topic that I hope makes this clear.

Let us continue this discussion in chat.

EVERYONE PLEASE DISREGARD MY OTHER COMMENTS ON THIS THREAD. I made a dumb mistake right at the start, and then compounded it on down the line.  I think we have it all sorted now.

Yes, you can define chemical potentials as partial molar enthalpies, or partial molar internal energies, or partial molar Helmholz free energies.  (Sorry for my initial confusion on this point ... see also my answer on your other question on this topic, here).
The definitions as partial molar enthalpies and partial molar internal energies are not particularly useful, since they involve working at constant entropy, which is practically impossible, but they are still mathematically valid.
The definition of chemical potential as partial molar Helmholz free energy at constant temperature and volume may actually have some practical applications, (e,g. perhaps for reactions that evolve gases in closed containers? although I have never come across any.

The definition $\mu_i = \left(\frac{\partial{A}}{\partial{n_i}}\right){T,V,n{j \neq i}}$ is very useful for gases in general - they are frequently described by equations of state (EoS) in (T,V,N)-coordinates (e.g. ideal gas law, SRK etc.), which makes Helmholtz free energy the most suitable thermodynamic potential. A practical example is the calculation of phase equilibria, but there are many others.


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How to calculate a reaction's DeltaH using a MM forcefield calculator software?
I want to calculate $\Delta H$ for hydrogenation of benzene and producing cyclohexane using MM+ Force Field Energy. I am using Hyperchem for energy calculation and optimization. I calculated the energy for cyclohexane and then I calculated the energy for benzene. But the difference of two energies is totally false. The calculated energy is $8~\mathrm{kcal/mol}$ however the real value is $-49~\mathrm{kcal/mol}$.
Is it a true way to calculate $\Delta H$? If not, how can I do it ?

What you are forgetting is that the reaction will also involve hydrogen.
$\ce{C6H6 + 3 H2 -> C6H12}$
In principle, the $\Delta H$ for $\ce{H2}$ should be 0, but in practice, that may not be true with a force field. The energies from molecular mechanics are seldom accurate enough to compute reaction energies. They are parameterized on a set of molecules, but the errors are usually large, both systematic and random errors.

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


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Salt analysis using spectroscopy
A few days ago, I learned the lengthy process of salt analysis during my practicals. The process is based on deductive reasoning but is too lengthy.
What I am thinking is, can't it be done using the fact that all molecules have unique emission spectra? I searched for it and found that it's the same thing as the flame test which was covered in my class.
But a flame test is only used for metal ions. So, my question is: Is there a way in which I can determine the metal ion as well as the radical part(anion part) of the salt using spectroscopic methods? What equipment would I need for it?

Not sure what you mean by "the radical part of the salt" ... I am assuming you mean the anions?  Beyond that, it's a good enough idea, but not one that you could implement very easily.  To cover all kinds of salts, you'd need a wide range of spectrometers, since different ions might only give detectable signals in specific spectral regions.

Yes. I mean the anions. I don't know why don't they show up during flame test.

Well, a full answer to that is pretty complicated, but it's easier if you start by thinking about why the metal ions do show up.  They show up because the high-energy environment of the flame excites electronic transitions of the metal ions that relax by emitting light that shows up in the visible region of the spectrum.  So, by inference then, the anions either don't have electronic transitions in this region of the spectrum, or in cases where they do have such transitions, the emission isn't intense enough to be detected by your eye.

Well, if I pass the light through a prism and observe the emission spectrum. Would that work?

It might, but you'd need to know which specific spectral patterns you are looking for to identify each ion .. atomic emission spectroscopy essentially works this way, but you still might need a more sensitive detector than your eye.  There is another issue as well .. suppose you see the spectral pattern for sulfur .. which anion would that indicate?  How would you tell between sulfide, sulfate, sulfite, bisulfate and bisulfite, just to name a few fairly common possibilities?

Isn't the emission spectrum of a sulphide ion is different from the emission of a sulphate or sulphite ion? Doesn't the spectrum change when atoms bond?

You are assuming that the atoms will stay bonded in the flame, which will probably happen for some strong bonds, while other weaker bonds will break.  The chemistry of flames is quite complicated.  That is why this technique of atomic emission spectroscopy/flame analysis is usually restricted to elemental analysis.  You put the sample in an extremely high energy environment that is guaranteed to break almost every bond, so that only atomic ions are present.  See inductively coupled plasma AES for example.

Let us continue this discussion in chat.

Two of the methods are:
Atomic Absorption Spectroscopy - similar to the flame test, but is usually quantitative. The sample is heated (in a flame or furnace) and the absorbance at a defined wavelength is measured against standardized solutions. It will work for any atom if you use a tunable laser.
ICP/MS (Inductively Coupled Plasma/Mass Spectrometry) - a solution of the compound is ionized and passed into a mass spectrometer. Some atoms (especially halogens) cannot be properly ionized and passed into the mass spectrometer. Oxygen and nitrogen (and a few others) would be difficult to quantify since it may come from impurities in the sample.

X-ray fluorescence is another possibility. Best for Na, and elements above Na in atomic number. (Z>= 11)


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Catenane optical activity
Would a [2]catenate whose each ring consist of a different number of methylene groups only, be optically active if one hydrogen atom of a random methylene group is replaced with e.g. chlorine? I think yes, but I need a second opinion.
The compounds attached seem to be enantiomers. I don't see how, by rotation, they could be super-imposable.

If a molecule, or any conformation of a molecule, contains a plane of symmetry, then the molecule is not chiral (achiral).
The all hydrocarbon [2]catenane will be achiral. The molecule contains a plane of symmetry; picture a plane that contains one of the rings and bisects the other ring. In the following image, you can see that the plane of the screen contains the ring on the right and bisects the ring on the left.

You might say that if we replaced the circles with cyclohexane rings, because of their chair-like shape, the plane of symmetry is lost.  That is true, but you can flatten one of the cyclohexane rings and the plane of symmetry is restored.  Remember, if any thermally accessible conformation of a molecule is achiral, then the molecule is achiral.
Sometimes it helps to build a model or use a Newman projection to help with the stereochemical analysis.  Here is such a projection for your chloro-[2]catenane.  Note that it still possesses a plane of symmetry (a plane perpendicular to the screen).

If we go further and add another chlorine atom, but now to the back ring, then the molecule becomes chiral — just like 1,3-dichloroallene.

Regarding:

The compounds attached seem to be enantiomers. I don't see how by rotation they could be super-imposable.

Look at the bottom figure in my second drawing.  I've taken your interlocking 5- and 6-membered catenane and rotated the 6-membered ring a bit further along (a conformational isomer; remember, as I mentioned above, if a molecule, or any conformation of a molecule, contains a plane of symmetry, then the molecule is not chiral [achiral]).  The plane of the screen now contains the 5-membered ring and it bisects the cyclohexane ring.  The plane of the screen also contains the methyl group.  The plane of the screen is a symmetry element of this molecule; therefore this molecule is achiral.

But what if the rings consist of different number of C atoms?

@Marko  Doesn't matter.  Consider the simple case of a [2]catenane composed of a flat cyclopropane and flat cyclobutane.  As discussed above, a plane exists for the all hydrocarbon case and the monochloro case  - remember, you might have to rotate the ring containing the chlorine so that it (the chlorine) is contained in the plane.

@ron I think you should consider rotation of the substituted rings - if it can turn one isomer into other compound is optically inactive

@Mithoron  Yes, that's absolutely correct. My statement, "any conformation of a molecule" was meant to include the rotational case you're presenting.  I'll mention here that a plane of symmetry is a sufficient, but not a necessary condition to make a molecule achiral.  More complex catenanes without a plane of symmetry but belonging to the $\ce{S_{n}}$ point group would also be achiral.

It has a plane of symmetry, but because the two rings are connected, the two "enantiomers" can't inter convert. I don't see how it would be possible to do that.

@Marko  Sorry Marko, I'm confused by your last comment.  Which two enantiomers are you talking about?

I edited my question. Check it.

@Marko  I've edited my answer to address your comment.


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Comparing adamantane carbon types for accepting OH using MM ForceField
I want to compare two type of carbons accepting OH in adamantane which I've shown below.

I know the blue position is preferred. But I wanna prove it with energy optimization and MM Forcefield. I thought it is related to transition state and corresponding carbocations. But MM+ ignores all charges.
How can I test it using ForceField softwares ?

One thing to try would be just to build the two isomers and compare their total energies .. that would at least show which position was preferred thermodynamically (i.e. more stable).  It sounds like you are more interested in the kinetic consideration though (i.e. which reacts faster) ... in that case, looking at the relative energies of the transition states will give you some basis for rationalizing the preference.  I am not sure how good MM is at dealing with transition states, however ... you would need to rely rather heavily on your chemical intuition to find the correct structures, and even then, the bond angles and bond lengths may be distorted to such an extent that your forcefield is not well parameterized to deal with them.


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Enantiomerism and Diasteriomerism
I am really confused between them. I know that an enantiomer is an isomer having optical activity, non super-imposability of image on original configuration, same molecular formula(Correct me if I am wrong). The thing that is causing me trouble is what diffrentiates this from diasteriomerism. It would be appreciated if you provide an example to clarify like a fischer projection

Your definition of enantiomer is correct.
This simple further classification might be helpful.  Let's take 2 molecules with the same molecular formula.  These molecules are either constitutional (or structural) isomers or stereoisomers.  If they have different structures (or connectivity patterns)  and are separated by an energy barrier, then they are constitutional isomers.  If they have the same structure or connectivity pattern, then they are stereoisomers.
Among stereoisomers there are 3 classes that we can place a pair of molecules into.  Take any two stereoisomers and compare them, they are either

the same (2 identical molecules)
mirror images of one another (enantiomers)

If they are not the same or enantiomers, then they are

different, and we call them diastereomers.

One additional thing to add to @ron's answer is that diastereomers have different chemical properties .. they can be distinguished using conventional spectroscopic techniques, or even by measuring thermodynamic parameters (at least in principle .. sometimes the differences are very small and hard to resolve).  They can also have different chemical reactivities.  Apart from their different optical activities, enantiomers have identical chemical properties ... that is to say, you have to do an experiment that exploits their chiral natures to tell them apart.
Another point is that diasteromers always have at least two chiral centers ... a molecule with a single chiral center will have two enantiomers.  A molecule having two chiral centers will have two enantiomers and two disasteromers.  There is a nomenclature convention that can be used to index the various choices: each stereocenter is labeled "R" or "S" depending on whether the components are arranged clockwise (R) or counter-clockwise (S) around it, according to a conventional order of preference.  In the case of a molecule with two stereo centers, the (R,R) and (S,S) stereoisomers are mirror images, and therefore are enantiomers, and the (R,S) and (S,R) stereoisomers are not mirror images, and therefore are usually diastereomers. As pointed out by @ron in the comments, if the molecule is highly symmetrical, the (S,R) and (R,S) forms may actually be indistinguishable, which would be a "meso" compound.

Good point. When I said diastereomers were different, that means in every way imaginable, physical properties, spectrally, reactivity, etc.  BTW, your sentence, "A molecule having two chiral centers will have two enantiomers and two disasteromers" (and some of what follows after that) is not quite correct.  It is possible that the (R,S) and (S,R) forms are identical - this gives rise to a "meso" compound.

Good point about mess compounds .. I had neglected that special case .. I have edited my answer accordingly.

@dtmoore, +1, Your answer helped me in understanding ron's answer well


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Hooke's Law for absorption spectra
Does anyone happen to know who was the first to use Hooke's law to model vibrational absorption peak frequencies/wavenumbers?
I know Hooke's law is first outlined here:
R. Hooke, Potentia restitutiva, or spring, in Lectiones cutlerianae, or a Collection of lectures: physical, mechanical, geographical, and astronomical, made before the Royal Society on several occasions ... to which are added divers miscellaneous discourses, ed. by J. Martyn (John Martyn, London, 1679), p. 4
But I'd like to know about whose idea it was to use it for spectra.

Okay, I think I found an answer. It's a set of four papers which were published consecutively in Physical Review 36 (1930). Interestingly, each was submitted on June 23rd, 1930, which makes me wonder about the back story.
The first comes from General Motors Corporation Research Laboratories in Detroit (Kettering, Shutts, and Andrews). The other three papers are out of Johns Hopkins (Donald Andrews, from Chemistry, and Yates, from Mathematics). I suppose Andrews could have graduated from Hopkins and found that GMCRL was working on the same thing; or perhaps he had a dual appointment.
As a side note, there appears to be a patent suggesting that L. W. Shutts invented shock absorbers, so it makes immediate sense that he would propose the spring law as an explanation.
So Donald Andrews (or at least Johns Hopkins) is the obvious link between GMCRL and Robert C. Yates. Did Andrews know Yates? Was someone about to be scooped? The answer to that is probably beyond the scope of this question.


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Luminescent materials under IR radiation?
is there a database somewhere of materials that are luminescent  when exposed to Infra Red radiation?

Various optics companies still make the old classic near-IR detector (viewing) cards.  They were 'charged' with visible light, then would fluoresce under IR exposure (although you had to keep moving them to keep the spot bright).  I have no idea what the material was, however.

@jOh  Are you asking about luminescence in the visible region (e.g. 2 photon IR absorption followed by visible emission)?

Visible would be great. Alternatively, if it were viewable with some form of photo-detector, that could work.

There are two types of materials I'm aware of that produce light in the way I think you want: two-photon fluorescence and photon upconversion. Two-photon fluorescence is basically the same as normal fluorescence but two photons are absorbed simultaneously, emitting a single photon when returning to the ground state. Photon upconversion similarly involves the absorption of multiple photons, but sequentially (by various mechanisms), followed by emission of a single photon.
Two-photon fluorescence is mostly seen used for two-photon excitation microscopy and useful dyes tend to be big conjugated organic systems. Materials used for photon upconversion are generally lanthanide-doped nanoparticles and they've found many uses in applications traditionally served by quantum dots and organic fluorophores. They're both particularly useful in medical imaging as IR can penetrate tissue deeply, but upconverting nanoparticles have a big advantage in that two-photon fluorescence requires very high excitation intensity (pretty much just lasers) as both photons must be absorbed simultaneously, but upconversion processes are sequential and can thus occur at lower intensity. Though, for some applications, the intensity requirements for two-photon absorption are useful as one can play tricks with focusing such that only a small region receives enough excitation light to emit, making imaging small areas possible.
As for a database of compounds, Thermo-Fisher has a catalogue of two-photon dyes, but I don't know that there are a lot of commercial manufacturers of upconverting nanoparticles, though Sigma-Aldrich formerly carried up converting nanocrystal under the Sunstone brand.

Your last link gives 404 now.

@Ruslan They seem to have stopped carrying them, but I fixed the link for the dyes (Life Technologies was acquired by Thermo-Fisher)

I am not aware of any materials that have this property.  There are certainly molecules that emit fluorescence in the IR, when irradiated with shorter wavelength light, but I can't think of any that would emit in the IR after being irradiated in the IR in a manner that can be called "luminescence".  Luminescence describes the specific phenomenon where a photon of a particular wavelength had excited a molecule to a higher-energy electronic state, the molecule then relaxes to a lower energy state, with emission of a second photon, which is always of longer wavelength than the original photon.  Because molecular electronic transitions are typically in the UV-visible region of the spectrum, the phenomenon of luminescence is generally restricted to those spectral regions.
One particular issue with IR luminescence is that it would be pretty difficult to tell apart from normal thermal IR emission ... luminescence refers specifically to light that is emitted outside of the thermal spectrum of an object.  That means that an object would need to be pretty cold initially, in order to distinguish luminescence in the IR region.  And again, since the emitted wavelength is necessarily longer than that of the absorbed wavelength, a molecule that was luminescent under IR irradiation would emit deeper in the IR than where it absorbed, making all of this harder.
Having said that, there are certainly molecules with electronic transitions that absorb in the IR, and if you got them cold enough and used an appropriate spectrometer, you might be able to observe IR luminescence from them.  It's something that would be restricted to pretty highly specified research labs though.

yes, these things exist. Im aware of some, but they cost crazy$$$ because they use platinum.

In addition to Michael D. M. Dryden's excellent answer:
Freshly prepared graphene oxide seemingly shows two-photon absorption in the IR and deactivation by fluorescence, as described in an 2014 article in
Nature Science Reports. (DOI)
Fig. 2D in that article shows a linear correlation of the emission intensity at $\lambda_{\textrm{em}} = \textrm{680 nm}$ with the square of the intensity of the excitation at $\lambda_{\textrm{exc}} = \textrm{1140 nm}$.
Fig. 2C shows that the emission wavelength can be tuned by variation of the excitation wavelength!


text:
Why should the redox reaction happen between the copper ions and iron atoms?
I know that when $\ce{Cu^{2+}}$ ions react to $\ce{Fe}$ atoms, it becomes $\ce{Cu}$ + $\ce{Fe^{2+}}$. My question is, why?
Copper atoms normally have 29 electrons. This means that in the K-shell there are 2 electrons, in $L = 8$ electrons, in $M = 18$ electrons and in $N = 1$ electron. (Yes I know that there is a system with s and other letters, but I haven't been taught in that way at my school). Therefore $\ce{Cu^{2+}}$ should have 27 electrons, which means that the shells are assigned like this: $K = 2, L = 8, M = 17$.
So when the $\ce{Cu^{2+}}$ ions react with $\ce{Fe}$, why does it want to become $\ce{Cu}$? Because $\ce{Cu}$ doesn't have a complete electron shell at the end. Why doesn't it want to become $\ce{Cu+}$, because then it would have a complete shell at the end ($K=2, L=8, M=18$).
I know that when you take away electrons from copper, it does it in a special way. I'm 15 years old and not english so I'm really not an expert or anything, I'm just wondering why $\ce{Cu^{2+}}$ would want to become $\ce{Cu}$ when it meets iron ($\ce{Fe}$).

This is a good question, but you may have gotten off on the wrong track by thinking about the electronic shell structure (although what you have written seems correct, if a bit old-fashioned).  The relevant phenomenon here is the relative abilities of the metal ions to oxidize (or remove electrons from) each other.  For your specific example, the copper ion is a better oxidizing agent than the iron atom.  Therefore it is able to take two electrons away from the iron atom, oxidizing it and leaving the iron 2+ ion behind.
As for why this happens, at your current stage of education the best answer is probably "because iron is easier to oxidize than copper".  You can look up the electrochemical activity series to find out how various metals relate to each other in terms of how easy they are to oxidize.  Copper is one of the harder ones to oxidize, but if you were to mix copper metal with silver ions, you would end up with silver metal and copper ions, because silver is even harder to oxidize than copper.
Does that help?


text:
Experiments in chemistry involving use of stereo microscope?
I recently got a decent stereo microscope. Besides using the scope to teach my children, I have been thinking of picking up something new to learn myself. I have been interested about chemistry for a long time but never studied it after upper secondary school. I was thinking would there be some experiments that would allow me to learn more microscopy and chemistry at the same time? I have ~10  years of experience in machine vision, signal processing, neuroscience, programming etc., so research-grade problems are welcome too.

Welcome to chemistry.SE! Please take some of your time to visit the help center or take a tour of the site.

For starters, it might be fun to grow different kinds of crystals in shallow dishes and watch them grow in real time under a microscope. Either a low melting solid, melted and resolidified, or growing from a saturated solution. Also, looking at how a metal surface is etched by acid using a microscope would be fascinating.

Could you elaborate on the home-experiment part (see tags in the question)? Anyway, nice answer.

For crystals, sodium acetate is a good one to grow quickly by cooling a saturated solution. You can make it from vinegar and baking soda. Alum (aluminum potassium sulfate) and potassium sodium tartrate produce spectacular crystals with striking geometric shapes. Sodium chloride produces cubic crystals that often have indented faces.

Common metals that can be etched might include copper (that requires nitric acid - might be hard to get), iron, aluminum, and magnesium. You can get hydrochloric "muriatic" acid at Home Depot and sulfuric acid at an auto parts store ( battery acid). You might get better results etching the metal first, then rinsing it off and looking at it under the microscope.


text:
How does the chemical potential allow for open systems to be considered?
Gibbs free energy can be defined as: $$dG=VdP-SdT+\sum_i\mu_idn_i$$ where $\mu=(\frac{\partial G}{\partial n})_{P,T}$. Last term, $\sum_i\mu_idn_i$ allows for open systems to be considered (where $dn$ does not equal zero). However, this can also be used for state functions like enthalpy too:
$$dH=TdS+VdP+\sum_i\mu_idn_i$$
But my question is why can this be used for enthalpy, considering that $\mu=(\frac{\partial G}{\partial n})_{P,T}$ (i.e it's a partial derivative of $G$ not $H$). Perhaps it's clearer for me to write the total derivative out like this: $$dH=(\frac{\partial H}{\partial S})_{P,n}dS+(\frac{\partial H}{\partial P})_{S,n}dP+(\frac{\partial G}{\partial n})_{P,T}dn$$ where I have considered a single-component open system (the component may enter or leave the system)

I am sorry .. I am afraid my errors in our comment discussion on your other post led you to ask this question (which reflects an incorrect assumption that I fear you got from me).  It is in fact perfectly ok to define chemical potential as a partial derivative of other thermodynamic potentials with respect to composition.  Those definitions don't have much practical use (since they require constant entropy), so I had forgotten about them, but they are still completely valid.

Thanks, Why do they require constant entropy?

Actually, there is a completely valid definition of chemical potential in terms of enthalpy (I realize your confusion on this point may be my fault, since I initially told you the opposite on another post.)
The correct form for chemical potential as a partial molar enthalpy is (as I think you already suspected):
$$
\mu_i=(\frac {\partial H}{\partial n_i})_{p,S,n_{j\neq i}}
$$
So the total exact differential is expressible as you had originally expected:
$$
dH=TdS+VdP+\sum_i\mu_idn_i=(\frac {\partial H}{\partial S})_{p,n_i}dS+(\frac {\partial H}{\partial p})_{S,n_i}dS+\sum_i(\frac {\partial H}{\partial n_i})_{p,S,n_{j\neq i}}dn_i
$$
  The definition of chemical potential as a partial molar Gibbs free energy is only valid at constant temperature and pressure, so it wouldn't make any sense to write it that way in the expression for the exact differential of enthalpy.
Sorry for any confusion my initial errors may have caused you.  Hope this clears it up.

Thank you very much. I think I understand - the chemical potential $\mu$ is a quantity that defines how a state variable changes with composition of the system but the state variable that it describes is dependent on the variables that you hold constant. Do you agree?

Please see above - comment edited.

Well, I'd phrase it a little differently, but I think you have it essentially right.  Chemical potential can be defined as the partial molar form of any thermodynamic potential, and the variables held constant determine the relevant thermodynamic potential. So if you hold entropy and pressure constant, chemical potential is partial molar enthalpy, and so on.  The Gibbs free energy form (constant T,p) is the most commonly used since those are the most relevant conditions for a wide range of chemical experiments.

Great, that is what I meant. It makes perfect sense now. Thanks very much.


text:
Will the water at the bottom of the tube boil at the same temperature as the water at the top of the tube?
Consider a long, vertical tube of water, as shown below:

Where $P_\mathrm{atm}$ is the atmospheric pressure and $P_\mathrm{bottom}$ is the pressure at the bottom of the tube.
The tube is long enough such that $P_\mathrm{atm}$ is much less than $P_\mathrm{bottom}$.
It is known that water at the surface (top of the tube) will boil at a temperature such that the water's vapor pressure equals $P_\mathrm{atm}$. Let's call this temperature $T$. For example, $T=100~^\circ\mathrm{C}$ when $P_\mathrm{atm}$ is about $1\mathrm~{atm}.
Question: At what temperature will the water at the bottom boil? Will it be higher than $T$ or equal to $T$? Why?
References, if available, would also be greatly appreciated.

Your setup, if heated from below, is actually a pretty good working model of a geyser. Water at the bottom will reach 100∘C and above, and still remain liquid. But as soon as the bubbles start to form, and some of the liquid spills over the top, the pressure decreases, and suddenly the whole bulk of water would start boiling explosively, throwing everything out.

Here's a cool example that should help illustrate the correct answer to your question.  As for why it works that way, you might think about it in terms of bubble formation .. in order for a liquid to boil, it needs to form bubbles.  Is it harder or easier to do that when the ambient pressure is high?
As for the calculation, you can use the Clausius-Claperyon equation to calculate the correction as described here.  You will need some values for $P_0$, $T_0$ and $\Delta H_{vap}$; you can use 1 atm, 373 K and 40.68 kJ/mol.

Thanks for the cool example - 464 °C! Wow. It is now clear from the 2-D phase diagram that the boiling temperature is higher.

Just to followup, I read that "A liquid boils at a temperature at which its vapor pressure is equal to the pressure of the gas above it." This only applies to the water on top surface right? Because the water on the bottom boils at a much higher temperature at which the vapor pressure is much higher than $P_{atm}$, which is the "pressure of the gas above it".

I think a better statement would be "A liquid boils at a temperature at which its vapor pressure is equal to the external pressure." FYI the original statement is here (https://www.chem.purdue.edu/gchelp/liquids/boil.html). (For some reason that site automatically downloaded a useless file to my computer).

Look up "black smokers", undersea volcanoes spouting water at 200 C or higher that does not boil because of the immense pressure at depth. At high enough temperature and pressure, though, it can become a supercritical fluid, q.v.


text:
Why does the freezing point of a sample gas occur at a certain temperature?
From "North Carolina Measures of Student Learning:
NC’s Common Exams Chemistry" (Source, Number 20):

The graph below shows a cooling curve for a sample of gas that is uniformly
  cooled from $155~^\circ\mathrm{C}$.

  Why does the freezing point of the substance occur at $–20^\circ\mathrm{C}$?

because the latent heat energy is absorbed by the substance as it is
  converted from a liquid to a solid
because the latent heat energy is released into the air as the substance is
  converted from a liquid to a solid
because the average kinetic energy is increasing for the substance as it is
  converted from a solid to a liquid
because the average kinetic energy is decreasing for the substance as it is
  converted from a solid to a liquid

Its really just a simple questioning and I think the answer is 3, because the average kinetic energy is increases when converted from solid to liquid. Any reason why I'm wrong and the answer says it is 2?

Could you move the question from the file into the above? Also, please provide a more descriptive title.

@AwesomeFlame123 As you can see, it was neither a problem to move the question into the posting nor was it difficult to give it a more descriptive title. Please invest some more time in asking your questions, people replying here do that on their own free time and they deserve some appreciation for that they are helping you. Not even trying to type the question into the post, I would consider rather rude.

The first thing is that it has to be going from liquid to solid because the substance is being cooled. Like tap water put in the freezer: liquid to solid.
On that note, the average kinetic energy goes down (directly assumed because that's what temperature is, and it's going down.)
Which leaves us with two answer choices: latent heat absorbed or latent heat released, where latent heat is the heat of a (thermodynamic system) at a constant temperature (like a phase change!)
If you're gaining energy, particles are going to have more velocity, like those virtual diagrams of gases with fast moving balls. Since we're going to a solid, losing energy makes sense because particles can't move as fast and are trapped in a fixed location (like an ice cube, which has a definite cuboid shape).
So liquid to solid, latent heat because the temperature is constant, and losing energy to become a solid - B is the answer.

Thanks, didn't know what Latent Heat Energy was. Also I should've thought that temperature doesn't necessarily have to change at a phase change. Sorry if I wasted your time with such a simple question.


text:
Why doesn't HCl form when you dissolve NaCl in water?
People say it dissolves, but shouldn't the acid form?
$$
\ce{NaCl + H2O ->[?] HCl + ???}
$$

If you dissolve NaCl in water you will get some HCl molecules but there's definitely not going to be a significant concentration of HCl formed.
The reaction that you propose -
$\ce{Cl- + H2O -> HCl + HO-}$
is highly thermodynamically unfavorable.
We can ascertain this fact through consultation of any pKa/pKb table. In the equation above, the product acid (HCl) is a much (as in almost a trillion trillion times) stronger acid than water.
Given that HCl is several trillion times stronger than water as a acid, then naturally, HCl will want to protonate hydroxide ion, a byproduct of HCl formation from chloride ion. This is ignoring the fact that hydroxide ion is also a strong base in water, so it has a high proton affinity in water.
So even if the products were formed - again, very unfavorable from a thermodynamic standpoint because the reactant base and reactant acid are both so weak - then the products would certainly react with each other and form the reactants again, resulting in no net change in solution contents and pH.

As a result, the reaction that you propose is more like this (except that the bottom/reverse arrow should be a lot bigger).
$\ce{Cl- + H2O  <<=> HCl + HO-}$

Ka of water is 1.8*10^-16 ... Ka of HCl is >55 that's for sure but I'm not sure exactly what. Likely less than 10000 if I remember correctly. A trillion is what, 10^12? So perhaps not a trillion trillion.

pKa of water is ~16. pKa of HCl is ~-7 or so. pKeq then would be ~23. That's just about a trillion trillion.

@Dissenter Thanks for your insightful answer, so what really happens is that HCL will donate the H to HO and then water will form again and that just leaves us with Cl. And the reason Cl and Na won't bond again is because they are dissolved (broke apart) by the $\ce{H2O}$

@AwesomeFlame123 what will really happen isn't that ... the process of Cl- reacting with water won't even happen much if at all because that process is like trying to climb Mt. Everest ... it's energetically VERY unfavorable.

Wouldn't the sodium also help to keep the solution relatively pH neutral?

@CJDennis the sodium ion is hydrated by water to some extent but doesn't really affect the pH.

It does form HCl, the only question is how much? Dissociation constant of hydrochloric acid is (reference):
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{Cl-}]}{[\ce{HCl}]} = \pu{1.3e6 M} \tag{1}$$
$$K_\mathrm{w} = [\ce{H+}][\ce{OH-}] = \pu{1e-14 M^2} \tag{2}$$
In neutral water, $[\ce{H+}] = \pu{1e-7 M}$.
So in $\pu{1 M}$ $\ce{NaCl}$, assuming neutral pH and substituting $[\ce{H+}] = \pu{1e-7 M}$ in equation (1), we find $\pu{1e-13 M}$ $\ce{HCl}$ is present.
Not very much, but still billions of molecules in a 1 liter sample.
If you want to calculate rigorously, instead of assuming neutral pH solution, to the above two equations you would add two more:

Electrical neutrality requirement:
$$\ce{[Na+] + [H+] = [Cl-] + [OH-]}$$
Number of chlorine atoms is conserved:
$$\ce{[HCl] + [Cl-] = [Na+]}$$

And you would have 4 equations and 4 unknowns ($\ce{[HCl]}$, $\ce{[Cl-]}$, $\ce{[H+]}$, and $\ce{[OH-]}$), with $\ce{[Na+]}$ being known from the amount of $\ce{NaCl}$ added to the water.

The above is a simplification of the real situation.
As explained in Essentials of Chemistry by Hessler and Smith at page 327, which even though it is over 100 years old I still think is quite accurate:

in an aqueous solution of sodium chloride [are] eight distinct things: the ions hydrogen, hydroxyl, sodium, and chlorine, and the undissociated substances: water, sodium chloride, sodium hydroxide and hydrochloric acid.

Yes you are right, I like your view into this problem, the math makes sense.


text:
Why is benzene called "aromatic"? Why not "circular" instead?
I wish to know why aromatic hydrocarbons like benzene are called aromatic when they can be just called circular to avoid the confusion?

Molecules must be more than circular to be considered aromatic.

The ones which smell good are called aromatic I guess.

Aromatic compounds are more stable due to the resonance in a single plane. I guess they have a very important meaning and can't just be called off circular. Only particular cyclo compounds are aromatic not all. And on top of that, we have other things like anti-aromaticity. I guess the word aromatic has a different and unique meaning for itself.

Aromaticity is a electronic property of a molecule, not a description of its shape. An aromatic molecule contains a circular system of conjugated π bonds and p orbitals that contains 4n+2 electrons, where n is an integer.
Benzene is aromatic: it has 6 electrons in a circular conjugated system (4n+2=6, n=1).
Cyclooctatetraene is not aromatic: it appears to have a circular conjugated system, but it contains 8 electrons (4n+2=8 would require n=3/2). This molecule actually adopts a non-planar geometry to avoid antiaromaticity.
Cyclohexane is not aromatic, even though it is circular: it does not have a conjugated electron system.

As @Kyle said aromaticity is an electronic property, not a description of shape and so circular would not be an appropriate description, particularly seeing as these molecules are not at all circular (benzene is hexagonal, for example).
The name 'aromatic' is a historical quirk. Supposedly it was used because many of the chemically aromatic substances know at the time happened to have notable aromas although wikipedia seems to dispute this. Several other reputable sites like this and this do support this view though so I'm not sure what the right answer is (or even if anybody knows).


text:
Why do two atomic orbitals form two molecular orbitals?
According to molecular orbital theory, two atomic orbitals form two molecular orbitals analogous to waves combining constructively or destructively but how can a wave combine destructively and constructively at the same time to produce two molecular orbitals?
Why don't two atomic orbitals combine to form a molecular orbital that can be BMO or ABMO?

Ultimately I think it has to do with the fact that the Schrödinger equation is a linear differential equation, though I lack the mathematical knowledge to expand on it.

This is a good question that a lot of people struggle with. The answer has to do with the conservation laws, especially angular momentum and energy. I would be happy to see a thorough answer that is understandable without the use of linear differential equations.

It's also worth noting that the word usually used to describe the mathematics of combining atomic orbitals into molecular orbitals is "mixing". Each resultant "orbital" is made of parts of the atomic orbitals that went in.

Well, one important thing to keep in mind is that atomic and molecular orbitals are not "real" things, they are mathematical constructions that are useful in helping understand and predict how electrons behave in atoms and molecules.  They are like a set of rules that tell us all of the ways in which it is possible for the electrons with a particular set of quantum numbers to behave.  For example, a H-atom in the ground state has only one single electron in a $1s$ orbital (quantum numbers: $n=1$, $l=0$, $m_l=0$), but we also know that if the H-atom absorbs some energy, the electron can end up in a higher electronic state, for example the $3\mathrm p_z$ orbital (quantum numbers: $n=3$, $l=1$, $m_l=0$).
Now, when you are dealing with molecular orbital theory, you are talking about the possible ways in which electrons in atomic orbitals can combine in a covalent interaction.  So it's not that the two atomic orbitals combine "constructively and destructively at the same time", it's just that those are the two ways in which its possible for the two orbitals to combine: in the region where they overlap, they can either add their amplitudes, or subtract them.  The "constructive" case where they add produces an molecular orbital that is lower in energy (that is to say, an MO that describes a state where the electrons will have lower energies) than the "destructive" case, so by the aufbau principle, electrons will occupy the lower energy "bonding" orbital first.  In a case like $\ce{H2}$, where there are only two electrons, only the bonding orbital is occupied, but because the Pauli exclusion principle tells us we can only have two electrons in a given orbital, if you add another electron (as in $\ce{He2+}$), it will occupy the higher-lying "anti bonding" orbital.


text:
Is the fact that table salt is composed of Na & Cl a chemical or physical property?
So I came across this question asking whether something is physical or chemical.

Table salt is composed of $\ce{Na}$ & $\ce{Cl}$. Is it a physical property or a
  chemical property?

I really don't understand why the answer says its a chemical property. I asked the others and they did not give me an understandable answer.
Edit: Now as I think about it, I really don't think there is an accurate answer, even if the book said it was chemical. Bad question from the book.

Why do you think it should be physical?

@MARamezani I think it should be physical because you are looking at what the compound is physically composed of. There is nothing there that tells me that it is chemical. Physical because you are looking at the physical atoms it self. And since there is no evidence of any chemical change or property, I go with physical. Whats your view into this?

There is not enough information provided to give an informed response to the question. All I see is a compound being described. It's similar to saying limestone is made from seashells. It doesn't indicate any properties. If I said how the seashells are transformed, that would be a chemical property. It is not a description of a physical property (as the temperature, or a solid). It is not a description of a chemical property (as in how reactive it is, or a chemical reaction).

@LDC3, Yes I think that is true. I have also arrived at that conclusion, but I do not know why such a question would be asked.

I do have painful experiences of these crazy-driving questions. GGGoooosssshhhhh!

The problem rises when we are mistaken about what a property is.

The different types of matter can be distinguished through two components: composition and properties. UCDavis

Do a search about chemical property, and you get:

a property or characteristic of a substance that is observed during a reaction in which the chemical composition or identity of the substance is changed. Dictionary.com
A chemical property is any of a materials properties that becomes evident during a chemical reaction; that is, any quality that can be established only by changing a substance's chemical identity. Simply speaking, chemical properties cannot be determined just by viewing or touching the substance; the substance's internal structure must be affected for its chemical properties to be investigated. However a catalytic property would also be a chemical property. Wikipedia
Any characteristic that can be determined only by changing a substance's molecular structure.
   Boundless

What's a physical property?

A physical property is any property that is measurable whose value describes a state of a physical system. The changes in the physical properties of a system can be used to describe its transformations or evolutions between its momentary states. Physical properties are often referred to as observables. They are not modal properties. - Physical property in wikipedia.

Basically, a physical property is what that doesn't deal with the composition of matter; but other factors, such as it's physical state. "Physical" isn't at its literal meaning.
Table salt is never pure $\ce{NaCl}$; because simply there's nothing pure in the nature. It's mainly composed of Sodium Chloride crystals. But if, and only if, we consider it to be only made from $\ce{NaCl}$, we are talking about its chemical composition and therefore, it must be a chemical property.
I understand that sometimes there are wrong questions for us to answer, so if composition is to be a property, it must be a chemical one.

But everywhere online I've seen it saying, a chemical property is visible in only a reaction. Why is this?

According to UCDavis, talk about matter has two "components". First is composition, the other is its properties. Composition is supposedly very different from properties. And yes, chemical properties are only visible in chemical reactions, and that's because they change.

First, table salt is composed of $\ce {NaCl}$ wich is different from $\ce {Na}$ & $\ce{Cl}$.
Second, actually table salt isn't exclusively composed of $\ce {NaCl}$ depending of the soure but contains also a fair amount of $\ce {KCl}$
Third, composition isn't what you can call a "property". I mean, it's a definition, compound containing only $\ce {NaCl}$ or other $\ce {KCl}$ is table salt that's all.

Yes but that's not an option for a question. Why does it take it as a chemical property since you are saying it as 'composition'?

Composition isn't a "property".

@AwesomeFlame123 Yes, your problem stems from that the composition isn't a property - so it's neither physical nor chemical...