text:
Why does the Oxidizing agent have to be higher than Reducing agent for a redox reaction to spontaneously occur?
On the Standard Reduction Potentials of Half-Cells table, why does the Oxidizing agent have to be higher than Reducing agent for a redox reaction to spontaneously occur?

I thought that the only requirement was: if $E^{\circ}$ is positive = the redox is spontaneous. I thought that it did not matter which one is more positive.

You're right. It has to be positive and I get that. However, that's the mathematical explanation. But how can you explain it scientifically?

Need some clarification here based on your comment. Are you interested in why a positive cell potential means a negative deltaG or are you interested in why a negative deltaG means spontaneous reaction?

The sign convention may be considered to be  'back to front' since $\Delta G=-nF\Delta E$ with  $\Delta E$ as the difference in standard redox potentials for the reaction. $\Delta G$ is negative for a spontaneous reaction. Oxidising agents have more positive redox $E^0$ in tables of redox potentials than reducing agents do. It is the difference in redox that determines if a reaction is spontaneous.  'Scientifically'  a reducing agent is an electron donor and so an oxidising agent is an electron acceptor.


text:
Is ZnO still diamagnetic after you dope it with aluminium?
There are no unpaired electrons in $\ce{ZnO}$, making $\ce{ZnO}$ diamagnetic.
Aluminium has an unpaired electron, which would mean it is paramagnetic.
Will $\ce{ZnO}$ become paramagnetic after it is doped with aluminium?

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text:
What can be products when aluminum sulfate is dissolved in water
This is a follow up from an answer to one of my questions (here is the link)
I tried dissolving aluminum sulfate in pure distilled water. I used the same concentration: 12 mL of aluminum sulfate in 100 g of water, yet I saw a precipitate. Can someone please let me know what is happening. My best guess is that the aluminum sulfate is reacting with water and forming aluminum hydroxide.

Most likely is that some aluminium sulphate just wouldn’t dissolve.

Explain the origin of the Alum . Post the content of the Sulfate and post in details here. An observation is that you apply 12 mL of Aluminum Sulfate. Therefore, you are using a solvated solution or just measured solid in a volumetric flask?

How is it supposed to be different from your original question?

@Mithoron  It is clarified the used of  pure distilled water to discard the possibility of precipitates others that saturation limits.

@Mithoron I am using PURE WATER. That's how this question is different.

I am assuming that you are measuring 12 mL in a flask.  This is very unusual because Sulfate is in a solid form therefore you are best to weight the sample in a balance on order to determine the solubility component to hold in the water volume.
The density of Alum assumed to be the unhydrate is 2.572 grams per mL  therefore you have 32.064 g of the anhydrous form. The solubility i water if 31 to 36 g per 100 mL.  You are adding in 100 Ml of water in laboratory analytical conditions. Therefore, you are closing too close to the solubility lists at room temperature.
Another effect to consider is the dramatic increase in specific gravity. The water volume in the solution will shrink to compensate the final  specific gravity of 1.33. This will saturate the solution and will precipitate to accommodate specific gravity and normal solution solubility at the temperature.
This chart will help you to evaluate how the ratios of sulfate weight and water volume impacts the specific density the Aluminum Sulfate solution.
http://pubs.acs.org/doi/abs/10.1021/ie50430a030

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text:
Peptide at pH=1
In the following peptide:

I want to know the net charge of the peptide at pH=1.
According to me the as the solution is acidic so every nitrogen will get protonated.
Hence the number of charge = number of Nitrogen atom
Am I doing it in correct way?

You are on the right track, but not every nitrogen atom will be protonated - only those that are basic. None of the nitrogen atoms in amide groups are basic. Neither is the nitrogen atom in the indole side chain on tryptophan. Only one nitrogen is basic for each of the imidazole side chain of histidine and the guanidine side chain of arginine.
Thus, the positive charge in acid is more correctly determmine by the number of basic side chains plus the N-terminus amine.

Why they are not basic

why these are not basic http://i.imgur.com/H2knPtC.png

@help Amides exist in resonance ($\ce{RC(=O)N(H)R' <-> RC(-O^-)=N+(H)R'}$) which delocalizes the lone pair, making it not basic.


text:
Is the amount of cubic inches in a cubic foot exact up to infinite significant figures?
I know we are supposed to use SI units and I know they are defined very well.
But does that mean that Imperial units of length/volume aren't exact? Don't they  have infinite significant figures in the equations?
I couldn't find it in the SI brochure.
I'm asking because I need to convert from cubic feet to liters and trying to decide how many significant figures to use. My first thought is that the conversion factors are exact and I should use all the numbers my calculator comes up with. But the conversion factor wasn't in the back of my book so maybe cubic inches in a cubic foot aren't exact?

Cubic inches in a cubic foot is an exact conversion.  There may be some uncertainty in the conversion between SI and imperial -- I'm not 100% sure -- but if there is it's probably in the 8th or 9th decimal place.

@CurtF.  I guess the question boils down to: Is the amount of Liters in a cubic foot exact? Or if not, how many significant figures should we use?

A short time before the French revolution, the royal rent collectors showed up at farmsteads with the largest 'bushel' measures ever seen in that country, to measure out the royal share of the crops.  The problem with non-SI measurements is the multiple disagreeing sources of authority.

Dealing with imperial measures in science is problematic. The main reason for that is that you have (at least) two different definitions (fixed values) for a lot of them, one based on the laws of the United States and one based on the laws of the United Kingdom. While the differences are quite insignificant for everyday use (as in baking a cake), in the science world these small deviations can cause further inaccuracies. On Wikipedia you can find more, but for the argument, let's look at the values for the yard:
United States    0.914401829 m
United Kingdom   0.9143993   m
International    0.9144      m

In principle the definition of cubic inch to cubic foot is exact, while the conversion to the SI system may carry an error.
Unit        Abbreviation    Definition  SI equivalent (rounded 4sf)
Cubic inch  cu in or in³                16.38 cm³
Cubic foot  cu ft or ft³    1728 cu in  0.02831 m³
Cubic yard  cu yd or yd³    27 cu ft    0.7646 m³

Most of the time this shouldn't affect your measurements more than the use of significant figures, which is used as stating an error implicitly. That means it is in itself inaccurate. In any scientific capacity you would state the error explicitly.
When a value is stated as $1.00$ it has three significant figures it actually means that the measurement falls within the interval $0.99$ to $1.01$, or $1.00\pm0.01$.
Let's have a quick example. You measure the volume of water in a beaker to be $0.35~\pu{in^3}$. The beaker doesn't specify how accurate the measurement is, so we simply assume $\pm0.01~\pu{in^3}$. You would state that value as $0.35\pm0.01~\pu{in^3}$, or $0.34-0.36~\pu{in^3}$, or simply $0.35~\pu{in^3}$ with implied error. On the other hand, when we know significant figures are used, we would interpret a value of $0.35~\pu{in^3}$ as $0.35\pm0.01~\pu{in^3}$.
When you convert this to the SI system, your value doesn't become more or less accurate. You cannot use significant figures to express the error in one system in the same way you would do it in the other, because the implicit error is different.
The exact conversion (of the international yard) is
$$\frac{0.9144~\pu{m}}{1~\pu{yd}}=\frac{0.9144~\pu{m}}{3~\pu{ft}}=\frac{0.9144~\pu{m}}{36~\pu{in}}\implies1~\pu{in^3}=\left(\frac{0.9144}{36}\right)^3~\pu{m^3}.$$
The actual trust of the value is $\pm0.01~\pu{in^3}=\pm0.16387064~\pu{cm^3}$. When you convert your value $0.35~\pu{in^3}=5.74~\pu{cm^3}$ (keeping the same number of significant figures since the conversion is exact), you would imply an error of only $0.01~\pu{cm^3}$, which is far less than the actual trust. Instead your measured value falls somewhere between $5.58-5.90~\pu{cm^3}$. So the correct way of implicitly stating the error would be $6~\pu{cm^3}$.
TL;DR
The conversions are exact, the system of significant figures is not. In doubt, state error margins explicitly.

@Nicolau That is pretty much what I said in the last sentence. I also use the exact (international) definition (of the yard) in my answer. I only try to make the point that the implied error is completely different when using imperial units compared to SI units.

Note that the value you list for the US yard-to-meter conversion is archaic. To make conversions precise, (and as noted in the Wiki article), the inch was redefined in 1960 to be 2.54 centimeters, exactly.

@JonCuster on Wikipedia it says this value is exact. However, this exactness does not extrapolate to the implied error.

Yes and no. Using "significant figures" is a sloppy method of doing error propagation. Typically two significant figures in centimeters gives two significant figures in inches. // We could debate if 0.35 cm is +/- 0.01 cm or +/- 0.005 cm too.

@Max I agree and I don't like the significant figure method either. I think implying an error is ignoring the error. However, it is used extensively and one should be aware of its inaccuracies.

@Martin-マーチン Thank you for your answer. Can you join this chat? http://chat.stackexchange.com/rooms/51938/dimensional-analysis

Yes, there are exactly $12 \times 12 \times 12$ = 1728 cubic inches in a cubic foot.
Also, an inch is exactly 2.54 centimeters.
A foot is exactly 0.3048 meters.
A yard is exactly 0.9144 meters.
See NIST Special Publication 1038 for more information.

Within a certain system of measures, conversion factors are typically exact.
In imperial units, this means that a foot is always twelve inches, a yard is always three feet and a mile is always 1760 yards. With the exact conversion, we can use multiplication to see that:
$$1~\mathrm{yd} = 36'' \pm 0''\\
1~\mathrm{m} = 5280' \pm 0'\\
1~\mathrm{m} = 63360'' \pm 0''$$
Likewise, the conversion from cubic imperial units — which are defined as $\text{imperial unit}\times\text{imperial unit}\times\text{imperial unit}$ — to other cubic imperial units is equally exact.
$$1~\mathrm{yd^3} = 1~\mathrm{yd}\times 1~\mathrm{yd}\times 1~\mathrm{yd} = 3' \times 3' \times 3' = 27~\mathrm{ft^3}$$
This is akin to the exactness of the conversion of metric units, except that the metric factors are always multiples of 10. But there are also semi-metric units, such as the German ‘metric pound’, by definition $500~\mathrm{g}$. Conversion from grams to German metric pounds is as exact as conversion from grams to kilograms.

This becomes different if you leave your system of measures and compare different systems. Except in a few very special cases, where two systems chose physically identical starting and/or ending points but with different step sizes,[1] the conversion factors will be nonexact.
Any nonexact conversion factor will introduce an error of its own into the equation and how to deal with these is best shown in Martin’s answer.
Note that even if the transformation is exact at a starting and ending point, measured values will still have an inherent uncertainty in them, as Martin pointed out. The uncertainty is transformed across the conversion and does not change magnitude.

[1]: See for example temperature. A measured temperature of $15~\mathrm{^\circ C}$ can be transformed into an exact value in kelvin because the kelvin and celsius scales are identical except for an addition factor. Thankfully, also the error is identical.
$$(15.00 \pm 0.005)~\mathrm{^\circ C} = (288.16 \pm 0.005)~\mathrm{K}$$
Similarly, the Réamur temperature scale is defined such that the boiling point of water is $80~\mathrm{^\circ r}$ while the zero-point is identical to the celsius scale. Therefore, the step is different and $1~\mathrm{^\circ C} = 0.8~\mathrm{^\circ r}$. Upon converting, we get a similarly exact value but a different error.
$$(15.00 \pm 0.005)~\mathrm{^\circ C} = (12.00 \pm 0.004)~\mathrm{^\circ r}$$

Thanks for your answer and example, could you stop be the chat too? http://chat.stackexchange.com/rooms/51938/dimensional-analysis

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text:
Corrosion of metal
Why would a metal corrode more in a lower concentration of nitric acid compared to a higher concentration sulfuric acid? Isn't sulfuric acid suppose to be a stronger acid compared to nitric acid?

Sulfuric acid is not as strong of an oxidizer.  Nitric acid, particularly when hot, is very strongly oxidizing.
Note that both acids are "strong acids" in that they completely dissociate (at least one proton in the case of sulfuric acid) in water.  However, this does not completely determine a compound's ability to oxidize metals.  Oxygen and chlorine for example will oxidize many metals very well, and they certainly are not strong acids!

Thanks @airhuff !! I read on a similar post that nitric acid is a better oxidizing agent because unlike other metal dissolution reactions, the H+ of HNO3 isn't reduced-- the NO3− is. And it was correlated with the reduction potential. How does the reduction potential determine which ion is being reduced?

It depends on the electronegativity of the metal relative to $\ce{NO3^-}$. For example, Mg, Mn and Zn all reduce  $\ce{H+}$ to  $\ce{H2}$.  The more electronegative metals Ag however reduces $\ce{NO3^-}$ to NO.  $\ce{NO3^-}$ is in it's most oxidized state reasonably possible, and more prone than some other acids to being reduced, thus making it a relatively strong oxidezer.


text:
Formation of symmetrical ether or dehydration of primary alcohol?
My textbook gives the formation of diethyl ether from dehydration of ethyl alcohol:

$\ce{\underset{\text{ethyl alcohol}}{2C2H5-OH} ->[\text{conc.} H2SO4][413~K] \underset{\text{diethyl ether}}{C2H5-O-C2H5} + H2O}$

In the reaction where it is said that concentrated $\ce{H2SO4}$ at temperature $413~\mathrm K$ is used as a catalyst, will the primary alcohol not dehydrate to form alkene?
Concentrated sulfuric acid is a strong dehydrating agent, and is used as $95~\%$ concentrated when dehydrating primary alcohol. So will the alcohol dehydrate or will the diethyl ether be formed?

When forming an alkene this way you have a carbocation intermediate.  This isn't very stable on a terminal carbon, so that pathway is not very favorable.

Still, if we push harder, it will dehydrate.

Indeed, $413~\mathrm K$ ($140~^\circ \mathrm C$) is used for this reaction while $453~\mathrm K$ ($180~^\circ \mathrm C$) is used for making ethene from ethanol. We might have to resort to the concept of thermodynamic and kinetic control.

Related: http://chemistry.stackexchange.com/questions/41836/reactions-of-alcohol-with-sulfuric-acid


text:
Why is ammonia more basic than acetonitrile

Why is $\ce{NH_3}$ more basic than $\ce{CH3CN}$?

I tried to reason it out with the hybridization of nitrogen in both the molecules. In $\ce{NH_3}$ nitrogen is $\mathrm{sp^3}$ hybridized while in $\ce{CH3CN}$ nitrogen is $\mathrm{sp}$ hybridized. Hence it follows that nitrogen in $\ce{CH3CN}$ is more electronegative as compared to that in $\ce{NH_3}$, which makes $\ce{NH_3}$ more basic. Is this the main reason for $\ce{NH_3}$ to be more basic or is there any other reason?

Electronegativity is a property of atoms, not molecules.

p orbitals are higher in energy than s orbitals.
sp3 orbitals with 75% p-character are therefore higher in energy than sp orbitals with 50% p-character.
This means that the lone pair in ammonia is higher in energy than that in acetonitrile.
Higher energy lone pair corresponds to greater basicity.

No need to bring electronegativity into the picture.
For a similar question see Relative acidities of alkanes, alkenes, and alkynes


text:
How is a counterpoise corrected geometry optimization done?
I understand the problem of basis set superposition error (BSSE) and I know how the counterpoise correction for single point energies is calculated.
Today I found out that many software packages allow for counterpoise correction during optimization calculations, but how does this actually work, especially for methods where analyical gradients are used for optimization?
During geometry optimization we calculate the first derivative of the energy to get energy gradients which we follow down to our minima. I understand that I could use counterpoise correction if I calculate those gradients numerically, which is quite easy to understand but very expensive to do, but it seems that counterpoise can also be used in combination with analytical gradients. How is the counterpoise implemented to get counterpoise corrected gradients?

Background
For a system consisting of two molecules (monomers or fragments are also used) X and Y, the binding energy is
$$
\Delta E_{\text{bind}} = E^{\ce{XY}}(\ce{XY}) - [E^{\ce{X}}(\ce{X}) + E^{\ce{Y}}(\ce{Y})]
\label{eq:sherrill-1} \tag{Sherrill 1}
$$
where the letters in the parentheses refer to the atoms present in the calculation and the letters in the superscript refer to the (atomic orbital, AO) basis present in the calculation. The first term is the energy calculated for the combined X + Y complex (the dimer) with basis functions, and the next two terms are energy calculations for each isolated monomer with only their respective basis functions. The remainder of this discussion will make more sense if the complex geometry is used for each monomer, rather than the isolated fragment geometry.
The counterpoise-corrected (CP-corrected) binding energy [1] to correct for basis set superposition error (BSSE) [2] is defined as
$$
\Delta E_{\text{bind}}^{\text{CP}} = E^{\ce{XY}}(\ce{XY}) - [E^{\ce{XY}}(\ce{X}) + E^{\ce{XY}}(\ce{Y})]
\label{eq:sherrill-3} \tag{Sherrill 3}
$$
where the monomer calculations are now performed in the dimer/complex basis. Let's explicitly state how this works for the $E_{\ce{XY}}(\ce{X})$ term. The first molecule X contributes nuclei with charges, basis functions (AOs) centered on those nuclei, and electrons that will count to the final occupied molecular orbital (MO) index into the MO coefficient array. There is no reason why additional AOs that are not centered on atoms can't be added to a calculation. Depending on their spatial location, if they're close enough to have significant overlap, they may combine with atom-centered MOs, increasing the variational flexibility of the calculation and lowering the overall energy. Put another way, place the AOs that would correspond to molecule Y at their correct positions, but don't put the nuclei there, and don't consider the number of electrons they would contribute to the total number of occupied orbitals. This means that for the full electronic Hamiltonian
$$
\hat{H}_{\text{elec}} = \hat{T}_{e} + \hat{V}_{eN} + \hat{V}_{ee}
$$
calculating the electron-nuclear attraction $\hat{V}_{eN}$ term is now different. Considered explicitly in matrix form in the AO basis,
$$
\begin{align*}
V_{\mu\nu} &= \int \mathop{d\mathbf{r}_{i}} \chi_{\mu}(\mathbf{r}_{i}) \left( \sum_{A}^{N_{\text{atoms}}} \frac{Z_{A}}{|\mathbf{r}_{i} - \mathbf{R}_{A}|} \right) \chi_{\nu}(\mathbf{r}_{i}) \\
&=\sum_{A}^{N_{\text{atoms}}} Z_{A} \left< \chi_{\mu} \middle| \frac{1}{r_{A}} \middle| \chi_{\nu} \right>
\end{align*}
$$
there are now fewer terms in the summation, since the nuclear charges from molecule Y are zero (the atoms just aren't there), but the number of $\mu\nu$ are the same as for the XY complex. This and the $\hat{T}_{e}, \hat{V}_{ee}$ terms aren't really mathematically or functionally different then; this is more to show where the additional basis functions enter, or to show where nuclei appear in the equations [3].
These atoms that don't have nuclei or electrons, only basis functions, are called ghost atoms. Sometimes you also see the term ghost functions, ghost basis, or ghost {something} calculation. Adding the basis of monomer Y to make the full "dimer basis" means taking the monomer X and including basis functions at the nuclear positions for Y.
Geometry optimization
Now to calculate the molecular gradient, that is, the derivative of the energy with respect to the $3N$ nuclear coordinates. This is the central quantity in any geometry optimization. For the sake of simplicity, consider a steepest descent-type update of the nuclear coordinates
$$
R_{A,x}^{(n+1)} = R_{A,x}^{(n)} - \alpha \frac{\partial E_{\text{total}}^{(n)}}{\partial R_{A,x}}
\label{eq:steepest-descent} \tag{Steepest Descent}
$$
where $n$ is the optimization iteration number, $\alpha$ is some small step size with units [length2][energy], and the last term is the derivative of the total (not just electronic) energy with respect to a change in atom $A$'s $x$-coordinate. Even Newton-Raphson-type updates with approximate Hessians (2nd derivative of the energy with respect to nuclear coordinates, rather than the 1st) need the gradient, so we must formulate it.
Formulation of the energy
We're in a bit of trouble, because we want to replace $E_{\text{total}}$ in the gradient with $E_{\text{total}}^{\text{CP}}$, but all we have is $\Delta E_{\text{bind}}^{\text{CP}}$. The concept of CP correction can still be applied to a total energy, but the BSSE must be removed from each monomer. The BSSE correction itself for each monomer is
$$
\begin{split}
E_{\text{BSSE}}(\ce{X}) &= E^{\ce{XY}}(\ce{X}) - E^{\ce{X}}(\ce{X}), \\
E_{\text{BSSE}}(\ce{Y}) &= E^{\ce{XY}}(\ce{Y}) - E^{\ce{Y}}(\ce{Y}),
\end{split}
\label{eq:2}
$$
which, when subtracted from $\eqref{eq:sherrill-1}$, gives $\eqref{eq:sherrill-3}$. More correctly, considering that the geometry for each step is at the final cluster geometry and not the isolated geometry, the above is [4]
$$
\begin{split}
E_{\text{BSSE}}(\ce{X}) &= E_{\ce{XY}}^{\ce{XY}}(\ce{X}) - E_{\ce{XY}}^{\ce{X}}(\ce{X}), \\
E_{\text{BSSE}}(\ce{Y}) &= E_{\ce{XY}}^{\ce{XY}}(\ce{Y}) - E_{\ce{XY}}^{\ce{Y}}(\ce{Y}).
\end{split}
\label{eq:sherrill-10} \tag{Sherrill 10}
$$
The CP-corrected total energy is the full dimer energy with BSSE removed from each monomer is then
$$
\begin{split}
E_{\text{tot}, \ce{\widetilde{XY}}}^{\text{CP}} &= E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY}) - E_{\text{BSSE}}(\ce{X}) - E_{\text{BSSE}}(\ce{Y}), \\
&= E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY}) - \left[ E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{X}) - E_{\ce{\widetilde{XY}}}^{\ce{X}}(\ce{X}) \right] - \left[ E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{Y}) - E_{\ce{\widetilde{XY}}}^{\ce{Y}}(\ce{Y}) \right].
\end{split}
\label{eq:sherrill-15} \tag{Sherrill 15}
$$
Note that I have modified which geometry is used for each monomer in $\eqref{eq:sherrill-15}$. All monomers are calculated at the supermolecule geometry. This is convenient for two reasons: 1. We are only interested in removing the BSSE, not the effect of monomer deformation, and 2. a isolated monomer geometry without deformation doesn't make sense in the context of a geometry optimization. I also added the tilde to signify that the supermolecular/dimer geometry used may not be the final or minimum-energy geometry, as would be the case during a geometry optimization. We simply extract all structures consistently from a given geometry iteration. Perhaps $\ce{XY}(n)$ would be better notation.
Formulation of the gradient
As Pedro correctly states, the differentiation operator is a linear operator. Because there are no products in $\eqref{eq:sherrill-15}$, the total gradient needed for $\eqref{eq:steepest-descent}$ will be a sum of gradients [5]:
$$
\frac{\partial E_{\text{tot}, \ce{\widetilde{XY}}}^{\text{CP}}}{\partial R_{A,x}} = \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY})}{\partial R_{A,x}} - \left[ \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{X})}{\partial R_{A,x}} - \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{X}}(\ce{X})}{\partial R_{A,x}} \right] - \left[ \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{Y})}{\partial R_{A,x}} - \frac{\partial E_{\ce{\widetilde{XY}}}^{\ce{Y}}(\ce{Y})}{\partial R_{A,x}} \right],
$$
so each step of a CP-corrected geometry optimization will require 5 gradient calculations rather than 1. Note that the nuclear gradient should be included for each term as well, which is a trivial calculation.
Extension to other molecular properties
Although not commonly done, counterpoise correction can be applied to any molecular property, not just energies or gradients. Simply replace $E$ or $\partial E/\partial R$ with the property of interest. For example, the CP-corrected polarizability $\alpha$ of two fragments is
$$
\alpha_{\text{tot}, \ce{\widetilde{XY}}}^{\text{CP}} = \alpha_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{XY}) - \left[ \alpha_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{X}) - \alpha_{\ce{\widetilde{XY}}}^{\ce{X}}(\ce{X}) \right] - \left[ \alpha_{\ce{\widetilde{XY}}}^{\ce{XY}}(\ce{Y}) - \alpha_{\ce{\widetilde{XY}}}^{\ce{Y}}(\ce{Y}) \right]
$$
where I believe it now makes even less sense to have each individual fragment calculation not be at the cluster geometry. In papers that calculate CP-corrected properties, no mention is usually made of which geometry the individual calculations are performed at for this reason.

References

Boys, S. Francis; Bernardi, F. The calculation of small molecular interactions by the differences of separate total energies. Some procedures with reduced errors. Mol. Phys. 1970, 19, 553-566.
Sherrill, C. David. Counterpoise Correction and Basis Set Superposition Error. 2010, 1-6.
One implementation note: Most common quantum chemistry packages should allow for the usage of ghost atoms in energy and gradient calculations. However, as Sherrill states, they do not properly allow for composing the full gradient expression to perform CP-corrected geometry optimizations. Gaussian can, and Psi4 may. For programs that can calculate gradients with ghost atoms, Cuby can be used to drive the calculation of CP-corrected geometries and frequencies.
There is a typo in the Sherrill paper; the subscripts for all 4 energy terms should be $AB$, which here are $\ce{XY}$.
Simon, S.; Bertran, J.; Sodupe, M. Effect of Counterpoise Correction on the Geometries and Vibrational Frequencies of Hydrogen Bonded Systems. J. Chem. Phys. A 2001, 105,, 4359-4364.

The CP-correction can be seen as a sum of a correction term to the total energy of the complex. That correction term is essentialy the diference between the energies of the fragments with their basis set and with the whole basis set at the complex's geometry. The total CP-corrected energy is thus a sum of different complex and monomer's energies.
Then, as the derivative is a linear operator, any derivative of the total CP-corrected energy is a sum of derivatives of the individual energies.. as simple as that... You can obtain CP-corrected 1st derivatives wrt nuclear coordinates and get optimized structures, but also CP-corrected armonic (and anarmonic) frequencies from 2nd (and 3rd) derivatives, dipole moments, etc...

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text:
Is this Newman projection energy profile for 2,3-dihydroxypropanal correct?
Is this Newman projection energy profile for 2,3-dihydroxypropanal correct?
Also, I was to identify R/S configuration for the initial Newman projection. I am pretty sure it is R, but I need help here as well.

This is too tough to answer as is. There may be hydrogen bonding, so the energy profile could be highly dependent on solvent. (R)- is correct though.

I think fifth structure (in which OH is gauch to both CHO and OH) is most stable and has minimum energy . It is due to H bonding of OH with both O of CHO and H of OH.

There seems to be only one problem. The 240° conformer* will be lower in energy as compared to the 120° conformer as there will be hydrogen bonding between the two hydroxyl groups
.
Rest all looks good.

*a conformer is also called a rotamer or rotational isomer. Anyways R is correct

Perfect, my professor had not mentioned anything apart from steric conditions, so I had not even considered it! Would you mind telling me what steps I should go through to assign the relative energies when plotting the diagram? I really struggle with more complicated examples and I am afraid I will not have time on the final exam. All comments appreciated!

Actually I think your professor is quite right. Steric and torsional strains are the fundamental factors that affect energy. Things such as H bonds arise only as exceptions. The standard approach would be to do what you have done above in your drawing and then look for any additional factors that could change your graph


text:
When does hydrogen peroxide act as a reducing and when as an oxidising agent?
My chemistry books details out both, the oxidizing and reducing properties of hydrogen peroxide, $\ce{H2O2}$.

Its reducing properties:

It reduces $\ce{Cl}$ to $\ce{HCl}$
It reduces $\ce{Ag2O}$ to $\ce{Ag}$
It reduces $\ce{K3[Fe(CN)6]}$ (potassium ferricyanide) to $\ce{K4[Fe(CN)6]}$ (potassium ferrocyanide)
Etc.

Half equations for reduction:

In acidic medium: $\ce{H2O2 -> 2H+ + O2 + 2e-}$
In basic medium: $\ce{H2O2 + 2OH- -> 2H2O + O2 + 2e-}$

Its oxidizing properties:

It oxidizes sulphites to sulphates
It oxidizes arsenites to arsenates
It oxidizes nitrites to nitrates
It oxidizes potassium ferricyanide to potassium ferrocyanide
Etc.

Half equations for oxidation:

In acidic medium: $\ce{H2O2 + 2H+ + 2e- -> 2H2O}$
In basic medium: $\ce{H2O2 + 2e- -> 2OH-}$

I was unable to find any pattern in all of the reactions I studied above (except for potassium ferrocyanide, for which the medium triggers the electron addition/removal) to recognize when hydrogen peroxide behaves as an oxidising agent and when it behaves as a reducing agent.

Are you familiar with reduction potentials?

@gsurfer04 So you are trying to say it depends on the relative reduction potential ?Is this the only reason(I find so many exceptions in chemisty that I suspect there might be some exception here too)

The standard electrode potential ($\pu{1M}$ solution at STP) of $\ce{H2O2}$ to  $\ce{H2O}$ is $\pu{+1.78V}$. The $\ce{O2}/\ce{H2O2}$ potential is $\pu{+0.70V}$. As always, changing the reaction conditions will change the thermodynamics. So if you get water, $\ce{H2O2}$ is reduced and if you get oxygen it has been oxidised. I've just noticed that you've put the ferri/ferrocyanide reaction as both reduction and oxidation.

@gsurfer04 I fixed your comment (almost two years later :D). However, you generally have two options here: delete and repost, custom flag for moderator.

@gsurfer04 What about the reactions in which H2O2 is converting to both water and oxygen? As in the reaction of hydrogen peroxide with ozone or silver oxide.

@Apurvium In the case of AgO (and K3Fe(CN)6, NaOCl, K2Cr2O7 etc.), H2O won't show up in the half reactions.

Let us say that $\ce{H2O2}$ reacts with some ion $\ce{Y+}$. If $+1$ is yttrium's highest oxidation state then it will have no other option but to to get reduced to one of its lower oxidation states, and hydrogen peroxide acts as reducing agent. If this is not the case it will get oxidized to some higher oxidation state, and hydrogen peroxide will act as oxidizing agent. This is because $\ce{H2O2}$ is both an oxidizing and reducing agent.

In a Fenton system, $\ce{H2O2}$ can act as an oxidizing and reducing agent.
\begin{align}
\ce{Fe(II) + H2O2 &-> Fe(III) + OH^. + OH-}\\
\ce{Fe(III) + H2O2 &-> Fe(II) + HO2^. + H+}
\end{align}
See reactions (14) and (15) in the following publication:
 Muruganandham, M.; Suri, R. P. S.; Jafari, S.; Sillanpää, M.; Lee, G.; Wu, J. J.; Swaminathan, M. Recent Developments in Homogeneous Advanced Oxidation Processes for Water and Wastewater Treatment. Int. J. Photoenergy 2014, 2014, 1–21 DOI: 10.1155/2014/821674.

Hydrogen peroxide is a redox substance. Its behaviour as oxidising agent and reducing agent depends upon nature of substance that reacts with hydrogen peroxide. If a substance is oxidisable then H2O2 will acts as oxidising agent.

Then what decided whether subject is oxidisable or not,and if we go along the lines of modern definition of Oxidation,pretty much every substance can be oxidized?


text:
Effervescence with NaHCO3
How To find which organic compound gives effervescence when reacts with $\ce{NaHCO3}$? Specially in case of phenols, which one reacts and which does not? According to my professor, all acids stronger than $\ce{H2CO3}$ evolve $\ce{CO2}$ when treated with $\ce{NaHCO3}$. But I do not understand how to find which acid is stronger than $\ce{H2CO3}$, further how to find same for $\ce{Na}$ and $\ce{NaOH}$  especially when the pKa values are not given in question.

You can assume that the phenols are less acidic than $\ce{H2CO3}$.

What about substituted phenols like nitro Phenols etc.?

Good point. 2,4-dinitrophenol is a stronger acid than $\ce{H2CO3}$.

Are there many such compounds that I will have to remember or just few exceptions?

What exactly does your question look like?

I mean to say that if your are given a list of different organic compounds in test and told to find out how many of the compounds will show effervescence with NaHCO3 then what will be the approach to solve the question?

Under an examination context, I do not know what your examiner expects you to know.

Phenol having one or more nitro groups on o and p position are stronger than H2CO3 and gives effervascence with sodium bicarbonate.

Only nitro groups matter or other groups which show electron withdraw movement matter too?

Yes all electron withdrawing groups on ortho and para positions of phenol are stronger acids than phenol.

Can you please tell the answer of above match up so that I can cross check it with my answer

a)pqs b)pqrs c)pqrs d)pqrs


text:
Is optical activity of a given compound a statistical picture?
When optical activity is taught it is done in terms of single molecules. But when it is actually measured it is done on a large collection of such molecules. So, how does the asymmetry manifest itself as the rotation of plane polarised light. Is it a collective effect of all the molecules put together in a sample or does each molecule play a role?

Please have a look at the wikipedia article on specific rotation for the effect of concentration and path length on the outcome of the measurement.

This earlier question and answer, "How does polarimetry account for the orientation of the molecules?" may be helpful.


text:
Why does a complete discharge destroy a Li-Ion-Battery?
I read that completely discharging a Lithium Ion Battery is a very bad idea because it will lose plenty of capacity. But why?
I know that the reaction (fot the LiFePO4 battery) is:
$$ \ce{LiFePO4 + C6 <=>[1 e-] FePO4 + LiC6 } $$
So at a complete discharge, we only have our electrodes made of $\ce{LiFePO4}$ and $\ce{C6}$. Why shouldn't this react to the products of the reactions above?
When overcharging the battery I could find a solution, why it destroys the latter. Maybe all the graphite is ionized and so, the electrode is in solution. This would be a very big problem, but the graphite will only ionize if it can react with a cation (like lithium), so a small excess of graphite should prevent this.
What about other Li-Ion-Batteries?
This is what I read on the internet:

The discharge voltage is around 2.5 V depending on the cell type; It
  must not fall below, otherwise, the cell will be destroyed by
  irreversible chemical reactions. However, many electronic devices
  already switch off at significantly higher voltages, eg 3.0 V.

What do they mean with the irreversible chemical reactions? Which ones?

Oxygen is evolved as the cobalt gets more oxidised since at 0.5 Li you get overlap of the cobalt and oxide bands. Oxide ions are oxidised and oxygen is evolved

True, but oxygen evolution happens at high potentials, not low ones.

In a Lithium ion cell, the anode material can dissolve in the electrolyte, and then on recharge, precipitate in the midst of the electrolyte and insulating membrane, short-circuiting the cell. Further, the cathode material can release oxygen, which migrates away and does not get reincorporated on charging.
Another problem with most secondary (storage) cells, Pb-acid as well as Li, is physical destruction of the structure on over-charging and under-charging. During the charge-discharge cycle, there is some change in volume of the electrodes, which causes them to spall, shedding pieces of the active material. This leads to loss of capacity and sometimes a short-circuit.
Even a fully charged cell deteriorates gradually... but increased discharge increases the growth of precipitates. Li-ion cells self-discharge about 2%/mo, (half-life of ~2 years), so a charged cell is quite stable, with no perceptible growth of whiskers or precipitate for years.

May I ask why such destruction happens only at total (100%) discharge and not gradually during use of the Pb-acid battery?

oxygen loss occurs primarily at high states of charge, at least for layered materials.

I know from bitter experience that over discharging a NiCd battery pack ruins it, it will greatly lower the storage capacity. The same thing happens with lithium ion batteries.
What happens during over discharge is that it is possible to reverse charge at least one cell in the battery. This will cause a lot of damage to the cell which is reverse charged. The reverse charging can be regarded in some ways as an extreme version of discharging the cell.
What will happen is that new and unwanted reactions can occur in the cell.

Well it kind of depends on the materials in the battery. One consideration would be the strain or rearrangement of the crystal structure of the cathode material. There's not really room to insert more than 1 Li+ per formula unit into LFP, but if the potential is below 2.5V (or whatever the pristine cell potential is) then there will still be a driving force until 0V to further reduce the iron. Whatever reaction happened in response to that driving force will most likely result in a different crystal structure. Raising the potential doesn't reverse this process, and the new material is probably not going to make a very good electrode. Alternately, the crystal structure could expand so much to accommodate more Li+ that the strain causes the particles to fracture, disconnecting them from the conductive matrix that they're embedded it.


text:
Whose solubility increases with increase in temperature?

Solubility of which among the following substances in water increases slightly with rise in temperature?
(a)$\ce{KBr}$
(b)$\ce{KCl}$
(c)$\ce{KNO3}$
(d)$\ce{NaNO3}$

Answer is (b).
I don't know why.
According to me if temperature is increased then Bonds between any of these four must be broken due to supplied energy & solubility must decrease. And what about the other three. Do they too also show the same trend.

All four substances have higher solubility at higher temperatures in the range from 0 to 100 degrees Celsius.
$\ce{KBr}$ and $\ce{KCl}$ both roughly double their solubility from 0 to 100 degrees Celsius.
To me the question does not make sense.

In case of KCl there is stronger ionic lattice and at lower temperature entire lattice will not break during dissolution at lower temperature and at higher temperature this lattice  absorb energy and get dissociated. So there is increase in solubility with increase in temperature. In others lattice is weaker due to larger size of anion.


text:
Calculating the pH of a mixture of Na2HPO4 and Na3PO4?
I know there are more questions about this on the forum, but I was just wondering: when we mix both solutions, would we have to consider the equilibria corresponding to $\mathrm{p}K_\mathrm{a2}$ and $\mathrm{p}K_\mathrm{a3}$, or just the latter?

Actually you need $\ce{{pKa}_3}$.

Sorry, I wrote the wrong values! I just updated the question.

I'll give a round about answer based on significant figures.
The whole truth is that any time that you add any phosphate ion into an aqueous solution, then you will have all four phosphate species ($\ce{H3PO4}$, $\ce{H2PO4^-}$, $\ce{HPO4^{2-}}$, and $\ce{PO4^{3-}}$) in solution. The relative amounts depend on the 3 equilibrium equations, and the total concentration of all of the phosphate species.
For phosphoric acid the three pKa's are different enough so that only two phosphate species will have a "significant" concentration at whatever pH the solution is at. Here is where the answer gets fuzzy. What is "significant"? To account for 99% of the species (2 significant figures) is typically good enough and at most two species would need to be considered. However if you want to account for 99.99999999999% of the species (12 significant figures), then you're going to have to consider all four phosphate species. So the gist is how many significant figures do you need to consider in the calculations?
What is typically done is to simply the four equilibrium equations to the two "significant" ones (maybe only 1 species at high or low pH's), and then calculate the concentrations of the last two species using the found concentrations of the first two.
In reality there is another consideration. If you look at the pKa values for phosphoric acid, Wikipedia lists pKa1 = 2.148, pKa2 = 7.198, and pKa3 = 12.319. There are only three significant figures in each of these equilibrium constants. (Only the mantissa counts, not the characteristic.) So you can only have three significant figures for any given phosphate species.
So, to three significant figures, for any sort of mixture of $\ce{Na2HPO4}$ and $\ce{Na3PO4}$ salts you'll need to consider both $pKa_2$ and $pKa_3$ and you'll end up with a quadratic equation to solve. The simplifying assumption is that $$\ce{[H3PO4] + [H2PO4^-] << [HPO4^{2-}] + [PO4^{3-}]}$$
An "exact" iterative solution, considering all four species, can easily be solved via a computer program, but it is really messy to do such a calculation by hand.
As a final check you should calculate the concentrations for $\ce{H3PO4}$ and $\ce{H2PO4^-}$ and verify that the assumption holds. (For phosphoric acid if the assumption doesn't hold to three significant figures, then normalizing the calculated values would yield a "good enough" result to 3 significant figures.)

This mixture can be considered as phosphate buffer as pKa3 of phosphoric acid is very high ( a weak acid). In  Henderson equation for acidic buffer we will use pKa3 because Na2HPO4 acting as weak acid and Na3PO4 as its salt with strong base.


text:
Which chemical reaction to use to remove carbon/peat from sieved earth/gravel?
I produce gravel for my aquarium by sieving and rinsing earth from my garden. After boiling and disinfecting it (with a solution against parasites) the rest of the organic material are composted or become invisible duff inside the gravel during the initial cycling of the aquarium. The only thing remaining visible are peat and other carbon particles of the size of the gravel which I currently need to remove mechanically.
Is there a way to remove them /shrink or reduce them to small pieces with a (chain of) chemical reaction(s) which

can be performed with a maximum of chemicals and equipment which can be bought at a supermarket or building supplies store
doesn't harm the fish after introducing them into the aquarium some weeks after the treatment - both through toxines and an instable PH value (more than +/- 0.5)
leaves the color and the size of the gravel untouched because my motivation to use custom gravel are mostly aesthetic and reducing the size would risk to block my undergravel filter (has a filter chamber with a separator for gravel of a certain size)

I thought about:

burning the particles, but that seems wasteful in terms of energy and requires some equipment (burner, fire proof permeable net, etc.)

I'm looking for a solution for some kg of 1mm to 3mm gravel in a 100 l aquarium filled with drinking water with a PH value of 7.2 to 8.
In case it's necessary to have more exact info what the particles in the gravel consist of, I'll ask a separate question how I can figure this out.

It's easier to use flotation or froth flotation to separate out the less-dense peat (silt?) and carbon (I assume from the activated-carbon filter) than to use a chemical reaction..
Simple flotation would depend on making a dense solution (e.g. saturated Epsom salt, $\ce{MgSO4}$) and shaking the gravel mix in it... if the liquid's density is greater than the average density of silt with its entrained air, the gravel will sink and silt float, to be skimmed off. You can reuse the liquid for more separation.
Froth flotation relies on a surfactant wetting the silt and keeping it suspended while the gravel settles out. Try a bit of hand dish-washing
detergent.

I'll give it a try, thank you. I'm not sure if the minimal amount of dish-washing detergent I can use without risking to harm the fish (even after intensive rinsing) will have an effect. MgSO4 seems very affordable (< 1€/kg).

Soap should also work in froth flotation -- more quickly biodegradable than most detergents.

Feedback after try: I got fertilizer granulate "Compo Bittersalz", meaning bitter salt which is a synonym for MgSO4·7H2O according to the german Wikipedia and saturated Epsom salt, I think. It contained 16% MgO and 13% S which I assume became MgSO4  because it made much more bio mass float than tab water as expected. I added about 1 kg of the fertilizer to 1 l water until it no longer dissolved. Warming seemed to much of an effort because I couldn't maintain the temperatur during the separation anyway. I successfully removed about 95 % of bio mass which remained after intensive rinsing.

Thank you! Can you confirm that the MgO and S became MgSO4 after putting it into the water? If not, would the effect be stronger if I'd get real MgSO4?

That solved http://pets.stackexchange.com/questions/13549/how-to-remove-carbon-particles-peat-from-my-self-made-gravel/16154#16154.

You bought MgSO4... the chemical analysis is based on breakdown into other chemicals, e.g. MgO and S, to provide a percentage basis.

It turned out that a gold washing/placer mining technique helps a lot, no matter whether MgSO4 is used or not.


text:
What is the relationship between the values of n and ml?
Example question: ml=+5 what is the lowest value of n?
I can't seem to remember the easy trick to finding the relationship if someone could explain that would be great. -Thank you in advance.

Just work it out the proper way, in my opinion you shouldn't have to resort to shortcuts for such a question.
The magnetic quantum number $m_l$ takes values between $-l, -l+1, \cdots, l-1, l$. As such, as long as $l \geq 5$ then $m_l = 5$ will be valid.
The orbital angular momentum quantum number $l$ takes values between $0, 1, 2, \cdots, n-1$. Consequently if $n \geq 6$, then $l \geq 5$ and $m_l = 5$ is permissible. The smallest possible value of $n$ is therefore $6$.


text:
Convert .car file PBC to basis vectors?
I have a .car file that contains dimensions (periodic boundary conditions) for the unit-cell of a hexagonal crystal material I'm working with:
PBC   27.1979   27.1979   15.4999   90.0000   90.0000  120.0000 (P1)

That's the a, b, c lengths and $\alpha$, $\beta$, $\gamma$ angles.
I need to convert this information into basis vectors in cartesian space, e.g.
basis1 27.1979 ?       0.0
 basis2 0.0     27.1979 0.0
 basis3 0.0     0.0     15.4999
This may be better suited as a math question but it's computational-chemistry specific.

I found the way to convert the hexagonal parameters:
I have a,b,c and $\alpha$, $\beta$, $\gamma$. The solution in basis vectors is:
basis1 = (a, 0, 0)
basis2 = (xy, yx, 0)
basis3 = (xz, yz, c)

where
\begin{align}
xy &= b\cdot\cos(\gamma)\\
yx &= b\cdot\sin(\gamma)\\
xz &= c\cdot\cos(\beta)\\
yz &= \frac{b\cdot c\cdot \cos(\alpha) - xy \cdot xz}{\sqrt{b^2 - xy^2}}\\
\end{align}
Thus my solution was:
basis1 = (27.1979,  0,       0     )
basis2 = (-13.5990, 23.5541, 0     )
basis3 = (0,        0,       15.499)

Since $\alpha=\beta=90^\circ$, you may simplify this a great deal. Your third vector is just $(0,0,c)$.


text:
The magic of ink tubes - Hydrophilic polymers?
First Chemistry SE post. I'm trying to make a capillary ink tube similar to the ones they used to have in chart recorders (here's an example of a capillary tube patent).
That is, an ink reservoir and a writing nib of some sort (probably felt tipped) connected with a long thin tube. What is the nature of that tube?
Physioscribe Chart Recorder - We found one at a local tech salvage place, but minus the critical ink tube portion
I tried some plastic tube experiments to see if the tube would draw water-based ink up out of a reservoir, but it did not. Afterward, I research the material, and found the tube was made of teflon. Teflon is hydrophobic and so apparently discouraged capillary action.
So here are a series of questions:

Is there a thin tube material that will draw water-based ink up out of a reservoir?
What would this thin tube be made of? Some hydrophilic polymer? Something such as polyvinyl alcohol, poly vinyl pyrrolidone, or cellulose?
Are there common suppliers that make tube of this material? And where?
Finally, is there something I am totally missing that would work better?

Thanks for the help.

Welcome!  What was the inner diameter of the tubing used in your experiments and what was the tubing made of?  Do you know the inner diameter of the Teflon tubing used in the chart recorder?

I don't know what the tubes in the chart recorder were made of since they were missing, but it is unlikely they were teflon since it is hydrophobic and resists capillary action.  The ones I experimented with were 7mm to probably 0.5 mm. But they were teflon and didn't work.

In the sited patent, they used "... Saran which is the trade-mark given to a group of tough, flexible, thermo-plastics, containing polymeric vinylidene. "  I would browse through the online catalog of Sigma-Aldrich or of a capillary tubing specialty company like Paradigm Polymers


text:
Can you theoretically hold more of a volume of water the colder it is?
I have basic knowledge of chemistry and how it all works (will be taking the class next year) but I understand density and mass. Since the hotter a liquid, in this case, the molecules are more everywhere and not contained so to say. In a way, a $\ce{200 cm^3}$ cup will hold less water if say brought up to a temp of 90°C (194°F) and will hold more water if brought to a temp of 2°C(35.6°F) due to the molecules being more contained and not bouncing everywhere. Does this make sense or did I underestimate my understanding of this whole thing?
*I am talking about mass not volume.

Water reaches its maximum density at about $\pu{4 ^\circ C}$.  As its cooled from higher temperatures the density decreases as you would expect as the molecules move more slowly and they can thus "settle" in closer together.
But, as it approaches the freezing point, the molecules are moving slowly enough that they start combining together into a more crystalline, ice-like conformation.  It's still completely liquid, but is starting to organize into the ice structure, and ice is about $\pu{7\%}$ less dense than liquid water.
So in the case of water, you can "fit" the most water into a given volume at about  $\pu{4 ^\circ C}$.

The solid is less dense than the liquid for (at least) Si, Ge, Sb, and Bi (and perhaps more I've forgotten). Bismuth is technologically important as an additive in welding and brazing alloys for just this reason. Not a bad percentage for the elements. Any good synthetic chemist can do similar things with more complex molecules.

Very good point @JonCuster.  My statement should have been more clear and I will edit it.  I meant to say something like "it is unusual for liquids found in nature".  Since it was an aside not really relevant to the question I think I will just remove the statement.  I appreciate the clarification and interesting examples.

No problem! There is a very persistent meme that water is unique in having the liquid denser than the solid. I try to fight the good fight...

Ah. I have got it now. Thank you very much @airhuff !

@Jon_Custer also Gallium, which is fun. It melts in the palm of your hand (low toxicity, but stains your skin black), and will then crack a glass bottle if you pour it into one.

For reference, you can store more water in a closed container by storing it in solid form with a different crystalline lattice: https://en.wikipedia.org/wiki/Ice#Phases

And if you have a properly equipped lab, as in the ability to form ice under 10,000 atmospheres or so, there are even non-crystalline types of ice that are more dense than liquid water (density around 1.2 or so).

The volume is the amount of space in the container, and no matter what temperature, a liter of water at that temperature occupies one liter.
What I think you mean to say is, "Does a liter of cold water have more mass than a liter of hot water?" The answer is yes, it does.
BTW, an open container (e.g. a beaker or drinking glass), can hold a larger volume of cold water than of hot water because the cold water has higher surface tension and therefore a higher meniscus... it's over the top.

Ah yes. This does make sense. I didn't think about the surface tension of the water. That is what I mean too. I am not the best at wording questions. Thank you for your explanation!

Yes this does make some sense, though probably less than you expect. Look up water density at different temperatures. Indeed, water at 90°C is about 4% less dense than water at 2°C. Whether or not this counts as significant is up to you.
Also, welcome to Chem.SE.

Depends on your unit of measure; volume of water, or count of water molecules. As you've noted, as water heats up, the amount of space per molecule increases. Thus, 1) the same amount of molecules will either occupy more space, or 2) the same volume will hold fewer molecules of water.
When chemists talk about water, they talk about the chemical compound H2O, and can be a bit unclear on whether they mean gaseous, liquid, or solid water. (Solid water being ice). H2O is a bit strange as far as molecules go; almost all of them take up LESS space as they get colder. Water, as noted elsewhere, takes up more space when it freezes. (Which is why ice floats). The reason which is somewhat complicated and has to do with the strength of electron bonds and crystalline structures.

Yes, the water would expand due to it's crystalized structure when frozen. Above, it was stated that 4°C was the ideal temp. for holding the most water in a given volume.


text:
How does dissolving a hydrate affect concentration?
Say I have 0.72 grams of calcium chloride dihydrate. If I dissolve it in water, the 2 water molecules attached onto it dissociates just like the $\ce{Ca^{2+}}$ and $\ce{Cl-}$, right?
And if so, wouldn't it lower the concentration of the calcium chloride, as compared to making an anhydrous calcium chloride solution?

Instead of our answering this question, why don't you compute the difference assuming the two waters of hydration just sum into the total volume. Try this for a few different target concentrations: 0.001 M, 0.1 M, 10 M.

There are two methods to dissolve the calcium chloride salt.
For the first method you'd dissolve the salt in a "small" amount of water and dilute to some particular volume. So if the solution was 0.100 molar, the amount of water would be the same regardless if you started with $\ce{CaCl2}$ or $\ce{CaCl2\cdot2H2O}$.
The other method would be to make say a 1.00 % solution by weight of the calcium chloride salt. Here it does matter. Do you want 1.00 % solution of $\ce{CaCl2}$ or a 1.00 % solution of $\ce{CaCl2\cdot2H2O}$?

I think CaCl2, as I want to react it with another compound, and the hydrate will not participate in the reaction.


text:
Electrons in different shells
I know this is a correct image, but I am a bit confused because I have seen different versions. If the 3rd orbital can take 18 electrons, why does it suddenly have 8 valence electrons on the 4th orbital, and then another electron on the 5th. I have seen other examples where the 4th orbital goes up to 32 electrons. How do I know when to use which ? I believe it might have to do with the complexity of the atom. Could someone please explain it in simpler terms
Couldn't attach the picture so I have the link here: https://commons.wikimedia.org/wiki/File:37_rubidium_(Rb)_enhanced_Bohr_model.png

The diagonal principle can be used to memorize electronic configuration.

The name for the diagonal filling process is typically given as the Aufbau principle. https://en.wikipedia.org/wiki/Electron_configuration#Atoms:_Aufbau_principle_and_Madelung_rule

I think images have to be jpg not png.

see http://chemistry.stackexchange.com/questions/18466/difference-between-shells-subshells-and-orbitals

The electron configuration of an Rubidium atom can be written like this:

You are correct, the 4th Shell may hold a maximum of 32 electrons. However, electrons are arranged in a way so they have the least energy. As you can see above, the 4th shell is not completely filled because orbital 5s is a lower energy orbital, thus preferred over 4d. This principle applies to all atoms.
A easy method to remember is via the Aufbau principle:

https://en.m.wikipedia.org/wiki/Electron_configuration#Atoms:_Aufbau_principle_and_Madelung_rule


text:
What home products are reactive with glass?
Glass seems to be one of the most stable solid substances - which chemicals, particularly home products that react with glass?

Drain cleaner (concentrated solutions of NaOH) but very slowly.

Hydrofluoric acid (HF) is an effective substance to dissolve Glass.
Its also very effective in killing you in a very slow and very painful way (by basically dissolving your bones).

Actually, HF kills you by depositing CaF2 in your microcapillaries. Blood flow in extremities diminishes to zero over a few hours, leading to cell death and gangrene. It is possible to reverse the process if it has not progressed too far, but it is a hospital-type procedure. Definitely painful.

Well, HF can hardly be considered as a home product, frequently used/found in households.

Some rust and stain removers (like Whink) contain hydrofluoric acid* at low concentrations (1-2.3%) Here is the MSDS for Whink: https://www.whink.com/cmssites/ws0811www.whink.com/uploads/Documents/Rust%20Stain%20Remover%20SDS.pdf
Now, at those concentrations it would probably take a while to dissolve silica, but it would be arguably quicker than a $\ce{NaOH}$ cleaning product.
*I do not recommend playing around with $\ce{HF}$ and its products unless you are experienced with them.


text:
How to Prepare a Buffer Solution?
Problem
I am given the task of preparing three buffer solutions at pH $10$, $9.5$, and $9.0$ . I have available concentrated ammonia and $\pu{3M}$ hydrochloric acid. The buffer capacity desired is $\pu{0.1 M}$.
Attempt at Solving

By adding $\ce{HCl}$ to a solution containing ammonia, it will completely consume the strong acid. Resulting solution will contain ammonia and it's conjugate acid ammonium ion, $\ce{NH4+}$.
$$\ce{NH3 + H+ -> NH4+}$$
Where $$[{\ce{NH4+}]_\mathrm {final}}= [{\ce{HCl}]_\mathrm{initial}}$$
and
$$[{\ce{NH3}]_\mathrm{final}} = [{\ce{NH3}]_\mathrm {initial}} - [{\ce{HCl}]_\mathrm {initial}}$$
I think at this point I should make some assumptions:
$\pu{500 mL}$ buffer to be prepared, $\mathrm pK_\mathrm {a,\ce{NH4+}}$=$9.25$
$[\ce{NH3}] = \pu{2 M}$, ammonia in ethanol solution
From here I think, from the Henderson-Hasselbalch equation.,
$$\mathrm{pH} = \mathrm pK_\mathrm a + \log \left(\frac{[\ce{A^-}]}{[\ce{HA}]}\right)$$
$$\mathrm{pH} = \mathrm pK_\mathrm a + \log \left (\frac{[\ce{NH3]}}{[\ce{NH4+}]}\right)$$
$$\implies \mathrm{pH} = \mathrm pK_\mathrm a + \log \left(\frac{[\ce{NH3}]}{[\ce{NH3}] - [\ce{HCl}]}\right)$$
Making the pH = 10 solution
$$10 = 9.25 + \log\left(\frac{[\ce{NH3}]}{[\ce{NH4+}]}\right)$$

It is pretty visible at this point though that I don't exactly know how to proceed. I have reached one variable but through processes that would not really hold up under scrutiny. The $[\ce{NH3}]$ that I have as my last variable is either $\ce{[NH3]}_\mathrm {initial}$ or $\ce{[NH3]}_\mathrm {final}$ but I have no idea how to determine which it is. I also feel I may have applied the wrong method from the start.
If there is anything I can do to add clarity or stop this post from being thrown out please let me know so I can remedy the situation asap.

You have some of the chemistry wrong.
Looking at your assumptions, concentrated ammonia is typically in aqueous solution, as is concentrated hydrochloric acid. You could have ethanol solutions saturated with $\ce{HCl}$ and $\ce{NH3}$ but that would be unusual.
Concentrated ammonia in aqueous solution is about 18 molar and is usually notated as concentrated ammonium hydroxide.
Since all three buffer solutions are in alkaline solution all the $\ce{HCl}$ will be reacted according to the following reaction:
$\ce{NH3 + HCl -> NH4^+ + Cl^-}$
The $\ce{Cl^-}$ anion is a spectator anion and will not effect the pH.
$\ce{\text{pKa}_{ammonia}= 9.25}$ is sound.
Using the given pKa, at pH = 9.25 then $\ce{[NH3] = [NH4^+]}$. So:

at pH 9.0 there will be a bit more $\ce{[NH4^+]}$ than $\ce{[NH3]}$.
at pH 9.5 there will be a bit more $\ce{[NH3]}$ than $\ce{[NH4^+]}$.
at pH 10.0 there will be a even more $\ce{[NH3]}$ than there was $\ce{[NH4^+]}$ at ph 9.5.

The ratio of $\ce{[NH3]}$ to $\ce{[NH4^+]}$ can be calculated based on the pH using the Henderson-Halbach equation which for the ammonia/ammonium equilibrium is:
pH = 9.25 + log ([NH3]/[NH4+])

The problem as stated however seems incomplete.
If you really must make the buffer solution using only concentrated ammonium hydroxide and 3 molar HCl, then there is a unique solution for each buffer.
However if you can add additional water, then there is no unique solution and you'd need to know the buffer capacity for each of the buffers.

It seems really odd to have solutions as buffer solutions as strong as 18 molar.
I'll explain further that the equilibrium equation
$\text{K}_a = \dfrac{\ce{[NH3][H^+]}}{\ce{[NH4^+]}}$
should really be written as the activities of the species not the concentrations. Below about 0.1 molar the activity and concentration are pretty equal. But in as concentrated solutions as concentrated as 18 molar the assumption is dicey and some correction would need to be made since the activity of the various species would be less than their actual concentrations.

Working towards a solution for 0.10 molar buffers
Edit 1/19/2017, noon
So now the problem is that 0.1 molar buffers are needed for pH values of pH 10, 9.5, and 9.0. (I'll assume that ph 10 is 10.0 so 1 significant figure in concentration.)
Even though we are going to work with pKa of $\ce{[NH4^+]}$, let's nt forget that we are starting with "pure" ammonium hydroxide which ionizes as:
$\ce{NH3 +H2O <--> NH4^+ + OH^-}$
In "pure" ammonium hydroxide not much of the ammonia will ionize to ammonium, so we can assume
$\ce{[OH^-] = [NH4^+]}$
and the pH will be 11+. So we can add various amounts of HCl and create the needed buffers.
However there is a consideration. A 0.1 molar buffer means that given 1 liter of solution then:

If 0.1 moles of a strong acid is added, the the pH drops no more than 1 pH unit.
If 0.1 moles of a strong base is added the the pH increases no more than 1 pH unit.

Now a buffer would be made "best" at the pKa or pKb value of a chemical so that the buffer capacity was the same for either a strong acid or a strong base.
The buffers at 9.0 and 9.5 are reasonably close to the pKa of ammonium which is 9.25. So these buffers will guard against nearly the same amount of acid as base.
However the buffer at pH 10.0 is a considerable distance from the pKa of ammonium. So it will take much less of a strong base like NaOH to change the pH to 11.0 than a strong acid like HCl to change the pH to 9.0.
So assume that:

The 0.1 molar buffer capacity means "at least 0.1 molar".
You're making 1 liter of each of the solutions
You only need volumes to about 5% for the reagents.

The last assumption is a "gotcha" of sorts. I haven't worked through the solution, but I don't think there is a way to solve all the constraints directly in one pass. I think the "exact" solution would have to be calculated iteratively. No problem for a computer, but it is painful by hand. So overall this is trying to avoid iterating using the quadratic equation.

You're right about the strangeness of the conc. solution. I could not find "concentrated ammonia pH' anywhere and I do not have access to the materials so I just looked on sigma aldrich for something that looked right. This was an incorrect assumption.
I didn't think of the inability to use HCl above the pKa but now that you've said it, it immediately makes sense.
I can certainly use water so I will investigate that trail of options as soon as I'm done writing this.

The buffer solutions don't have to be 18 molar at all, and I would agree that seems very over-the-top. I was just given those ingredients from which to prepare a buffer for my reaction.
I think I might have to look into your buffer capacity thing more before I continue.
From what I'm gathering it seems like it's a simple HH problem but I can't seem to grasp the mechanics of it.
You've stated my questions seems incomplete. Could you please elaborate on that?

Incomplete in the sense that the concentration needs to be specified somehow. Either as overall ammonia species, or as the buffer capacity.

The buffer capacity  desired is 0.1M and this was just gained as additional information. Should I update my original post with this information or update it in general?

Edit 1/19/2017, noon Can you finish working the problem with the additional information? // Sorry I mislead you about higher pH buffers earlier. I had a massive brain fart.

I believe I have worked it out. Would you like to review my new edit and see if it is up to snuff?

Here is what you would do in the lab (2 recipes, depending on your needs):

You start by diluting the concentrated ammonia to about $\pu{100 m}\mathrm{\small M}$. Then, while measuring the pH, you add sufficient $\ce{HCl}$ to get the pH to $10.0$, and put some of the buffer into a bottle. Then you add some more $\ce{HCl}$ to the remaining buffer until the pH is $9.5$, and put some of it into a second bottle. Finally, you add some more $\ce{HCl}$ to the remaining buffer until the pH is $9.0$, and put that into a third bottle. Now you have the required 3 buffers. This recipe does not give you buffers of defined concentration (and not even of equal concentrations), but your question initially didn't ask for a specific concentration.

If you need a specific concentration, take $\pu{100 mL}$ of $\pu{200 m}\mathrm{\small M}$ ammonia, set the pH with concentrated $\ce{HCl}$ and fill up to $\pu{200 mL}$ with water. That will give you a buffer with a concentration of $\pu{100 m}\mathrm{\small M}$ (total concentration of $\ce{NH3}$/$\ce{NH4+}$) at the desired pH.

The reason you do it this way is because buffer solutions are not ideal (by design, you want high concentrations of buffer substances, whereas calculations are accurate for infinitely dilute solutions). Especially if your buffer contains other substances such as high concentrations of $\ce{NaCl}$, it is better to set the pH experimentally rather than do a calculation. Also, it is much faster.


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Filtration methods at removing ions present in water
I'm doing a lab where I have to first find out what ions are present in the water (easy part). The next part is to filter out those ions from the water. Here are some methods that are stated on the paper: Coffee filter, Your own design, Britta filter. I've done my research and just can't seem to find what ions coffee filters and Brita filters filter out.
My question is: What are some own designs that i can create to filter out some ions ($\ce{Ca^{2+}}$, $\ce{Pb^{2+}}$, $\ce{Ba^{2+}}$, $\ce{Zn^{2+}}$, $\ce{SO4^{2-}}$, $\ce{PO4^{3-}}$, $\ce{Cl-}$, and $\ce{OH-}$).
Or do those filters actually filter any of those ions?

What do you need your final product to be? You can always distill to get pure water or use a dialysis bag to get the ions out.

There are different kinds of Britta filters. See https://www.brita.com/why-brita/what-we-filter/

You can't filter out $\ce{OH-}$. If you filter out one of them, more shall come to take its place.

Cations and anions of water-soluble salts cannot be removed from water using only mechanical filters (which rely on particle size) such as coffee filters.
In order to remove such ions, ion exchange resins can be used. These resins consist of an insoluble matrix, e.g. polystyrene that is crosslinked with divinylbenzene to vary the porosity of the resin. To convert the resin into an ion exchanger, the polymer structure is functionalized with many functional groups. Typical functional groups are

sulfonate ($\ce{R-SO3-}$) for strong acid cation exchange resins,
quarternary ammonium (e.g. $\ce{R-N(CH3)3+}$) for strong base anion exchange resins
tertiary amine (e.g. $\ce{R-N(CH3)2}$) for weak base anion exchange resins

The counterion of new cation exchange resins used for water treatment is usually $\ce{H+}$ or $\ce{Na+}$ (other forms can be used for special applications, e.g. $\ce{Li+}$ in pressurized water reactors), i.e. $\ce{R-SO3- H+}$ or $\ce{R-SO3- Na+}$. These counterions can be replaced by another cation from the water. Thus, a cation exchange resin can remove various cations from the water and replace them with an equivalent amount of $\ce{H+}$ and $\ce{Na+}$, respectively, e.g.
$$\ce{2 (R-SO3- H+) + Ca^2+ <=> (2 R-SO3- Ca^2+) + 2 H+}$$
The counterion of new anion exchange resins is usually $\ce{OH-}$ or $\ce{Cl-}$. Thus, an anion exchange resin can remove various anions from the water and replace them with an equivalent amount of $\ce{OH-}$ and $\ce{Cl-}$, respectively.
In order to remove both kinds of ions, the water has to be passed through a cation exchanger and through an anion exchanger. Alternatively, a mixed-bed ion exchanger (which contains cation exchange resins and anion exchange resins) can be used. They are even more effective but cannot be easily regenerated.
Typical Brita water filter cartridges contain ion exchange resin beads to remove undesired ions and activated carbon to absorb other undesired compounds.

This is pretty good againt Pb

You may consider getting a hold of an ion-exchange resin of some kind. These actually replace ions in the filtered sample, e.g. $\ce{Ca^{+2}}$, with another (more innocuous) ion, e.g. $\ce{Na+}$.
I used to do a lot of research on an IMLIG (immobilized ligand) material (basically organic polymer strands stuck on silica gel) called Octolig that removed anions (phosphate, nitrate, fluoride, arsenate, etc.) and heavy metals from water.
Also look up solubility rules for your selected ions and try to find a counter-ion that will form an insoluble salt. You could treat your sample with another salt and precipitate out the unwanted ions physically. e.g. to remove $\ce{Cl-}$, you could add a lead or silver salt.
FYI a coffee filter will be essentially useless for removing aqueous ions. It's much more suitable for capturing solid particles. The pores are much larger than the ions and basically made of paper that doesn't have any substantial ionic interaction.


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Why are NBr3 and NI3 unstable, but NF3 is stable?
Why are NBr3 and NI3 unstable, but NF3 is stable? Is there any reasoning behind the stability? I found that one nitrogen can accommodate fluorine easily, but not the others. Also, I don't understand what is meant by "accommodate."

Imagine taking a ping-pong ball and three basketballs, then trying to get all three basketballs to touch the ping-pong ball at the same time. Do you think there would be enough space around the ping-pong ball to fit the basketballs?  In other words, can a ping-pong ball accommodate three basketballs at once? Probably not. A similar thing happens when large atoms try to surround small atoms in molecules, with some complications due to the fact that atoms are not hard spheres.

‘Why is A unstable even though B is stable?’ is not a question that gets us far …


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Freezing of vinegar
Concentrated acetic acid freezes around seventeen degree Celsius. Water freezes at zero degree Celsius. But Vinegar has a very low freezing point , that is around minus two degree celcius.
Why is it so? How does mixing of substances change their  melting/freezing points?

It is pretty common for a mixture to have freezing point much lower than both pure components. Read about eutectic systems.

Think of ice as a highly ordered crystalline solid in which the water molecules are able to bind strongly with each other.  How strongly the molecules bind to each other determines how much energy, in the form of heat, will be required to break them apart.
Whether its ice or an altogether different type of solid, what happens when you disrupt these inter-molecular forces (i.e. hydrogen bonding, van der Walls forces, etc.) by introducing a second chemical substance?  It requires less energy, less heat, to break apart the solid.  Thus, it’s melting point is lowered.  This is commonly known as "freezing point depression".

A small point but depression of freezing point and elevation of boiling point only strictly apply to non-volatile solutes.

@porphyrin: Elevation of boiling point only applies to nonvolatile (or less-volatile) solutes, but depression of freezing point applies no matter what the volatility of the solute.

@Sean, yes you are correct and I was wrong about the depression of the freezing point being restricted to involatile solutes. The depression is usually derived assuming an involatile solute via lowering of free energy but the final equations depend only on the solute's mole fraction not on its nature.


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Workfunction and Bond Energy
Workfunction is the minimum energy required to remove an electron from the surface of the material. If the striking photons does not have the required energy then it won't be able to eject the electron even though many photons can be striking the surface i.e. many low energy photons cannot be combined to get the required energy. In contrast to this many particles can combine to break a bond. Silicon for example has bond energy of 1.1eV, with an average energy of 0.026eV 42 (1.1/0.062) particles would be required to break the bond and set the electron free. Why is it that a similar thing not happen with photons?. If many photons strike the the surface than all their energy should add up and remove the electron
EDIT
I am watching lectures on Youtube on Chemistry and Semiconductor devices. In this lecture at 4:45 the instructor talks about workfunction. And in this lecture at 42:00 the instructor talks about the number of particles required to break a bond

By "many particles combine to break a bond" are you referring to chemical ionization? Also, what are the sources for the silicon bond energies and the "42 particles"?

Photons behave as 'particles' in this type of experiment. What is the experimental evidence that shows that ' many particles', presumably not photons, break the bond?

@zer0c00l If you have two accounts that you would like to join together, please sign into either account, visit the contact form and select ‘I need to merge user profiles’.

@quantum-americcino, I have added the sources in my original post. I tried searching more on this and it seems like many photons with low energy cannot combine to add up their energy, because is no mechanism for the absorption of light (ASU) but many electrons can combine.


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Polysaccharides from non-cyclic sugars?
I have only come across polysaccharides from monosaccharides that has undergone intramolecular cyclization reaction.  I was wondering if it is possible for polysaccharides to form from linear sugars? I think that the answer lies in how the other glycosidic bond is formed, however my knowledge of chemistry is limited so I don't know if bond formation is possible with a linear sugar.
I know a carboxylic acid group can react with a hydroxyl group, and aldehyde can be oxidised to carboxylic acid. So my guess is that it is possible to create a long ester chain from aldoses but not ketoses (as ketone group cannot react and is fully oxidised). Although then I think it would no longer classify as a polysaccharide but a poly(ester)? I'm sure these exist — I have heard of polyester of course! — although I have not come across any others specifically.

In theory and especially synthetically there is no real reason why one should restrict oneself to polysaccharides where all monomers are in cyclic forms. A linear monosaccharide could, for example, act as a bifurcating branching site with two different sugars attached to the aldehyde carbon to build up the required O,O-acetal. Another variant may be to have a methyl group block the second acetal oxygen. In any case, you want to avoid the aldehyde becoming a hemiacetal because these are too reactive and would degrade quickly.
$$\begin{align}\ce{\underset{aldehyde}{R-CH=O}} && \ce{\underset{hemiacetal}{R-CH(OH)(OR)}} && \ce{\underset{acetal}{R-CH(OR)(OR)}}\end{align}$$
You may want to look up acetal and acetalisation in the general organic chemistry textbook of your choice to learn more about why hemiacetals are much less stable than acetals.
In practice, however, linearised monomers in polysaccharides are not interesting. To the best of my knowledge, they do not occur in nature due to the labile hemiacetal intermediate that would be required — but to a much greater extent due to the fact that nature optimised the connection of cyclic monosaccharides into polysaccharides and is fully content with what it has. There is no need to evolve a synthase that is able to shield the highly reactive intermediates from the surroundings and thus it didn’t happen. While the formation of oligosaccharides can also be a spontaneous process given a catalyst, linear monomers in oligosaccharides will not form that way.
For most natural product or bioorganic chemists, it is by far more interesting to synthesise species that are known in nature or logical extensions thereof. While a linearised monosaccharide might qualify as an extension, it is not that logic an extension and thus of much lesser value.


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Why are hydrocarbon flames yellow or blue?
A candle gives a yellow flame while kerosene gives a blue flame on burning.
Now, saturated hydrocarbons give a blue flame but unsaturated ones give a yellow flame. I want to know why the colour of the flame is yellow and blue respectively?

why do you not answer his *actual question! about colour * - yellow is due to hot carbon-as-amorphous-graphite ( Cx  ) while the blue-green colour is due to hot  C-H ( carbogen) blue-.... and hot  C2 (carbon dimer)-green
______ref >http://lup.lub.lu.se/luur/download?func=downloadFile&recordOId=4695221&fileOId=4695231<
2.2.3 CH Emission
The excited carbogen (CH∗) emits radiation in a narrow band at around 431 nm, in theindigo region. CH∗
is formed by the following reaction mechanism [14]:
C2 + OH → CO + CH∗
(R3)
O + C2H → CH∗ + CO (R4)
O2 + C2H → CH∗ + CO2 (R5)
CH∗
is usually used as an indicator of temperature in ame studies. It is a radical thathighly depends on the temperature thus it has a strong presence at slightly rich flames(φ ' 1.1) [15], where the flame reaches the highest temperature.
2.2.4 C2 Emission
The excited diatomic carbon (C*2) is a radical that is not present at lean flame conditions.
It is detected at equivalence ratios Φ ≥ 1.0. The reaction mechanism that describes the formation of C∗2
is given bellow [14]:
CH2 + C → C∗2 + H2 (R6)
C∗2 :emits in a narrow band at around 513 nm, in the cyan/green region. Another-characteristic of this radical is its strong second side band at around 465 nm (blue region).

It depends on the amount of oxygen available and on the ratio of hydrogen to carbon in the fuel.
Hydrogen burns with an almost invisible flame no matter how much oxygen is present, but solid carbon burns with a yellowish (black-body) flame. If the fuel combusts in a multi-step process where all reactants are gaseous, you get a bluish flame. For example, the combustion of methane produces methanol, then formaldehyde, formic acid, carbon monoxide etc.
The combustion of acetylene at lower temperature, with little oxygen, however, starts with cracking of the C-C triple bond, releasing carbon particles that have a luminous flame.
So the greater the hydrogen:carbon ratio, the easier it is to get a blue flame.

The colour of the flame does not necessarily have to do with the type of hydrocarbon (saturated or unsaturated). In most introductory lab courses and even at many schools, students learn how to use a Bunsen burner (or related burners) in one of the first practical exercises. These burners burn natural gas which is mostly methane and have inlets at the bottom to allow the addition of air. If the flame is lit when the inlets are closed, the flame is yellow but if the inlets are open and air is allowed to pass in, the flame becomes bluish and transparent. (Safety instructions always remind you to light the flame first and open the inlets afterwards; when putting out the flame, close the inlets first and then extinguish.)
Methane, the principal component of natural gas, is saturated; it only has one carbon atom thus it cannot be unsaturated. Yet we still get both flame colours. As you may have noticed from my description, the difference is the amount of air that is allowed to mix with the gas. And since we are dealing with combustion, the key difference boils down to oxygen content.
The bluish flame tells us that complete combustion is taking place. The actual mechanism is not important at this point, but ‘complete combustion’ means that the only products are carbon dioxide and water vapour. Thus, we can easily write a reaction equation as follows:
$$\ce{CH4 + 2 O2 -> 2 H2O + CO2}\tag{1}$$
The difference between the two flames (mixing in air or not) tells us that additional air is the key to complete combustion. Probably, it is necessary for methane and oxygen to be well-mixed. Complete combustion also means that the reaction is most exothermic, i.e. this type of flame reaches the highest temperature because all possible half-reactions are strongly exothermic, and all atoms undergo these complete half-reactions.
What happens if we reduce the air content? Suddenly, the reaction becomes a lot less defined. In fact, practically anything can happen. Yet we have a strong indication that one of the products in a non-neglegible amount is elementary carbon, because it is that that gives the flame its yellow colour. We will still be burning some methane completely to carbon dioxide and we could have some incomplete burning to carbon monoxide or other compounds. Equation $(2)$ is one of the may possible equations and a particularly simplified one.
$$\ce{2CH4 + 3O2 -> 4 H2O + CO2 + C}\tag{2}$$
This is the key difference between blue and yellow flames; whether it is a candle or kerosene or anything. In some cases (e.g. candle) we actually want incomplete combustion to occur because without it there would be hardly any light. In other cases (e.g. heating) we want complete combustion to occur because it liberates more energy and less harmful products.
Note that methane is not the only compound that can be burnt in two different ways by modifying the oxygen content. Acetylene is another. If it is simply burnt normally like in a Bunsen burner, the flame will be yellow and a lot of soot (i.e. elementary carbon) will be produced. But if oxygen gas is added, the flame will become white-hot (not bluish because of the heat) and can be used for welding.

Hmmm... But why does complete combustion give a blue flame?


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What is the mechanism for the oxidation of a lactol to a lactone?
Recently, I came to know that lactols can be converted to lactones but I wasn't able to find the suitable reagent and the reaction mechanism. Can someone elaborate the same?

Can tollen's reagent also be used for this?

Are you oxidizing for analysis or for synthesis? Tollen's reagent is unsuitable for medium or large scale applications due to risk of explosion.

I came across this in a question where one of the steps involved using tollens reagent on a lactol, so I was just curious if it could be used or not, is there like a restriction on oxidation by tollens reagent like in case of fehling solution (which does not oxidise aromatic aldehydes).

Methyl(trifluoromethyl)dioxirane (TFDO) in 1,1,1‐trifluoro‐2‐propanone (TFP)may be able to do your sought oxidation (see image [1] below): In there, cyclic ethers (shown example is tetrahydropyran) yield lactones by subsequent oxidation of a stable cyclic hemiacetal intermediate such as lactol in your case.

(source: literatumonline.com)
It is considered as selective and reactive oxidizing agent, capable of epoxidation of unreactive alkenes and arenes, oxyfunctionalization of alkanes, and oxidation of alcohols and ethers in following reference: Encyclopedia of Reagents for Organic Synthesis (1; March 2016); 11 pages. (https://doi.org/10.1002/047084289X.rm267.pub3)
You may also read: Oxidations by methyl(trifluoromethyl)dioxirane. 5. Conversion of alcohols into carbonyl compounds: J. Am. Chem. Soc., 1991, 113(6), 2205–2208 (DOI: 10.1021/ja00006a044).

Lactols are basically aldehydes in ring formation.
What happens is the lactols unwind in solution into chain formation and it is here that they are oxidized to carboxylic acid by something like Tollen's reagent. Then when they reform, they do so in lactone form.

The lactol does not have to open to the hydroxy aldehyde to oxidize. The aldehyde hydrates before oxidation. The lactol is in effect a hydrated aldehyde. PCC oxidations of primary alcohols often give esters as by-products. PCC and Collins oxidation of 1,4-butanediol afford butyrolactone and not succindialdeyde.


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Why isn't the ionic radius of Mg2+ bigger than F-?
They will both have the same number of valence electrons, but $\ce{F-}$ is more electronegative. Why is it not the case that $\ce{Mg^2+}$ has a bigger radius than $\ce{F-}$?

Strongly related if not dupe: http://chemistry.stackexchange.com/q/64335

These two species are isoelectronic (have the same number of electrons). In fact, they're electronically identical with the except that the magnesium ion has 3 more protons. Since all other electronic factors are equal, the net effect is that the three extra protons pull all of the electrons in closer to the nucleus. I think you understand this part.
Electronegativity is not correctly used in this context. One way to think about it is that you should only be using that concept on a neutral atom. The fact the fluoride anion is negative and magnesium cation is positive means you've already applied electronegativity to the system in order to ionize them.

So because it has more particles in pulling in the e- than Mg+ will have a smaller radius! Thanks I understand this now.

so is this also why is the 1st ionisation energy of Na+ bigger than Ne (more protons than Ne so bigger attraction and capability to gain an electron than Ne)

Correct with regard to $\ce{Na+}$ vs. $\ce{Ne}$.

The number of electrons decreases by 2 in $\ce{Mg^2+}$ compared to neutral Mg atom, but the number of protons inside the nucleus remains the same. The effective nuclear charge hence increases causing the electrons to be pulled towards the nucleus, thus, smaller ionic radius.
But, in case of $\ce{F^-}$ one electron is extra compared to neutral F atom but the number of protons inside the nucleus remains the same. The effective nuclear charge hence decreases causing the electrons to move away from the nucleus, thus, larger ionic radius.
One very important point to be noted is that in case of $\ce{Mg^2+}$, the the outermost orbital n is 2 instead of 3 in case of neutral Mg atom. This absence of the outermost orbital significantly reduces the ionic radius of $\ce{Mg^2+}$.


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Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why?

Is it (a) para-bromonitrobenzene and sodium methoxide or (b) sodium para-nitrophenoxide and bromomethane?
Or are they both appropriate? (And is one more appropriate than the other?)
One book (NCERT Chemistry Class 12 [India]) and at least one source Free NCERT solution of this question mention the latter as the appropriate. [However, I doubt the credibility of this source, as most likely, it is made to show how the NCERT book is correct (for CBSE Boards' exam) which, though is credible, is still a human-made book, which may have mistakes/errors.] NCERT only gives the answer as (b) without explanation. Another book (abc of Chemistry [India]) mentions "chemically, both are equally probable" as -nitro group activates the bromine site on benzene for substitution. And IMO, this book is more credible.
Any expert opinion?

In principle, both reactions will work.
(a) is a nucleophilic aromatic substitution, in which is $\ce{Br-}$ is displaced by $\ce{CH3O-}$
(b) is a methylation of a phenoxide
I'd probably go for (a) for two reasons:

The nitro group in para position decreases the charge density on the oxygen atom of the phenol. As a result, this anion will be less reactive than an unsubstituted phenoxide.
Methyl bromide with a boiling point of just 4 °C is difficult to handle and a poor choice for a methylation reagent. I'd rather try dimethyl sulfate if I had to.


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Why is the first ionisation energy of a sodium cation bigger than that of a neon atom?
I understand that the 1st ionization energy gets bigger for elements along a period from left to right and along a group from down to up.
But why is the 1st ionization energy of $\ce{Na+}$ bigger than $\ce{Ne}$?
I would have guessed the opposite.

$\ce{Na+}$ has more protons than $\ce{Ne}$, so there would be a greater pull on the electrons towards the nucleus, meaning that you would have to supply more energy to remove the electron. Remember, $\ce{Na+}$ has a noble-gas configuration and is already very stable!

$\ce{Na+}$ is the same as $\ce{Ne}$, only with more protons. Guess what does that mean for electrons.

A sodium cation $\ce{Na+}$ and a neutral neon atom $\ce{Ne}$ are isoelectronic species; meaning they have the same number of electrons and also the same electronic configurations.
Yet, the two have different nuclei: most prominently, sodium always has an additional proton when compared to neon. Therefore, only one of the isoelectronic species (neon) is neutral, the other is positively charged.
It may now become obvious why the ionisation enthalpy of $\ce{Na+}$ is much larger than that of neon. Different ways to express this all boil down to the fact that it is harder to remove a negatively charged electron from a positively charged species than it is to remove one from a neutral species.

$\ce{Na+}$ and $\ce{Ne}$ are isoelectronic species and size of cationic species is smaller among isoelectronic species. Ionization energy increases with decrease in size.


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Clemmensen Reduction
If 4-hydroxy-3-methoxybenzaldehyde undergoes Clemmensen reduction, will the hydroxyl group (OH) be converted to a chloro group (Cl)?
I know that it happens in cyclohexanes, and I am not sure if this is true for aromatic compounds as well.

You're right about the side reaction (hydroxy-halogen exchange) in the case of cyclohexanols. It is nucleophilic substitution on a (secondary) alkanol under acidic conditions.
In the case of vanillin, your starting material, the situation is different. Under the conditions of the Clemmensen reaction, that is in the presence of (amalgamated) zinc and hydrochloric acid, a nucleophilic aromatic substitution can be ruled out.

Thanks . Also does clemmensen reduction affect methoxy group too like hydroxy ( in cycloalkanes ) ?


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Bromination of cyclohexene in presence of UV light or heat

The reaction between hexene and bromine in presence of light gives 3-bromocyclohexene. Why is 1,2-dibromocyclohexane not formed instead?

Please, do not use short forms such as "rxn" for "reaction".

Where is the source that the major product is 1-bromohex-2-ene instead of 1,2-dibromohexane?

NCERT (Indian government Book for Chemistry)

You can find the answer here.

The question is find the major monohalo product. But i wbt to know if . It's a general question then. Which is major product and why??

I am writing an answer now.

Both products are formed although 3-bromocyclohexene is the major product.
Formation of major product: 3-bromocyclohexene
Under UV light, $\ce{Br2}$ undergoes homolytic splitting to generate $\ce{Br*}$ radicals:
$$\ce{Br2 ->[hv] 2Br*}$$
The formation of 3-bromocyclohexene is an example of substitution of alkanes, which require the free-radical mechanism:

In the first step of the upper mechanism, which is also the rate-determining step, a stable allyl radical is generated, which is stabilized by resonance:

As a result, the activation energy of the first step is significantly lowered.
Formation of minor product: 1,2-dibromocyclohexane
Individual bromine radicals are not electrophilic enough to attack the double bond in the cyclohexene, so the formation of 1,2-dibromocyclohexane requires the ions mechanism, typical for addition reactions (the lower mechanism in the following diagram).

The first step in this mechanism is the rate-determining step. In this step, bromine is ionized, which requires a moderate amount of activation energy, albeit still much higher than the rate-determining step of the upper mechanism.
Conclusion
Therefore, the upper mechanism occurs at a much faster rate than the lower mechanism, which makes the major product 3-bromocyclohexene and the minor product 1,2-dibromocyclohexane.

PS: Many people think that addition reaction is very fast. It is only true in water, where the bromine ion is stabilized by solvation in water.

Disclaimer: The mechanism I used for the addition reaction probably contradicts with your book. However, it does not matter. The point is that an ion is formed which makes the activation energy high.

"Many people think that addition reaction is very fast. It is only true in water, where the bromine ion is stabilized by solvation in water." It might be relatively slow but even in aprotic solvents with polarity in the mid-range you can get quite a lot of by-product. That's exactly why you want to use NBS for allylic bromation, since the the amount of bromine at every time is very limited and this side reaction can be surpressed.


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How to color bonds by the strain?
I need to create a figure of a molecule in which the bonds are colored in a RGB scale according to the values read from a data file. The data file contains, for example, the increase or decrease in the bond length compared to some reference structure. In this way, the intensity in a color range red-white-blue would be given by the difference between the relaxed and unrelaxed bond lengths, similarly to the figure attached.
I have tried to make this figure using jmol, but I didn't have success so far. Could anyone suggest a program that is able to do such figure? Or could someone point out any suggestion of how to do this color maps on chemical bonds?

I will be grateful for any help you can provide.
All the best

Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using LATEX syntax. If you receive useful answers, consider accepting one.

This is quite a special request and you probably have better luck contacting the authors of the paper (could you include the reference please) how they obtained this image. None of the programs I work with (molden, chemcraft) seem to be able to do so.

I guess it should be possible to search for certain distances between two atoms and color those bonds differently with PyMol, but it would involve some scripting.

Thank you for the comments and suggestions. The problem has been solved using Jmol program. After some discussion in the Jmol mailing list, prof. Robert Hanson suggested a very simple script that can assign colors to the bonds according to the respective length.
PS.: The figure attached to my comment was taken from the book "Nanodevices and Nanomaterials for Ecological Security, Ed. Springer", chapter 5, author Arkady V. Krasheninnikov.

@Max Pinheiro you should make your comment an answer to your question so it shows up as answered.

StrainViz is the program that might be useful [1].
Reference

Colwell, C. E.; Price, T. W.; Stauch, T.; Jasti, R. Strain Visualization for Strained Macrocycles. Chem. Sci. 2020, 11 (15), 3923–3930. DOI: 10.1039/D0SC00629G.