text:
Why does Ga improve conductivity of Ge?
This is for my AP Chemistry course and the question is:

Which of the following elements is important in the semiconductor industry to improve the conductivity of Ge?

K
Ga
Al
Mg

It seems like any of the above could be a correct answer. The answer given was Ga. What about gallium would make it the most appropriate answer?

It is a dopant. But then, Al is also. So, bad question by the AP folks...

I've heard that $\ce {Na}$ poisons any semiconductor by, loosely speaking, giving the outermost electron away too easily to far away sites, countering p-doping. This would even more be true to $\ce K$. I would relay like to see some answers that are more substantial than my hearsay .

Most likely the key factor favoring Ga is just a similar size, electronegativity valence shell structure, which the other choices do not have.  To get the band structure you want you just want to modify the electron count without geometric distortions or giving up good covalent overlap in the diamond-like structure.

This questions has two parts if you look closely:

Which of the following elements is important in the semiconductor industry to improve the conductivity of Ge?

All of the above will improve the conductivity of $\ce{Ge}$.  This is because germanium is a semiconductor which conducts electricity via the movement of holes and electrons.  Since germanium is tetravalent and gallium is trivalent, doping germanium with gallium creates a p-type semiconductor which is exponentially rich in holes, thus increasing the conductivity. This mechanism holds true for aluminum, potassium and magnesium meaning all of them can increase the conductivity of germanium

Which of the following elements is important in the semiconductor industry to improve the conductivity of Ge?

This part is what makes gallium special.  Gallium and germanium are sequential elements in atomic number.  They have very similar atomic radii.  Gallium is also very much less reactive than the other elements mentioned. These characteristics allow gallium to dope the germanium to a higher concentration without forming a new  undesired solid phase.  Sodium and magnesium form germanides with germanium and will not dope as shown by the phase diagram.  Aluminum will dope, but its smaller atomic radius can alter and strain the crystal structure of germanium much more than gallium which can affect band gap.

Thank you for the good answer to the question, but I see that you know a whole lot more about it than was asked in this question. Where can I (and others) learn more about that topic?


text:
Name of an unsaturated cyclic compound

I deduced the name as cyclodeca-1,6-dien-3-yne. Is it correct?

Related: Why is it -en-yne and not yn-ene? What is the explanation for this IUPAC specification?

The sketch has been replaced by a diagram. However, the accepted answer still refers to the sketch.

You have the right name unless you need to specify the geometric isomer.
If I may stretch you a little, consider whether to specify the relative positions of the atoms at the double-bonds, e.g. (1Z,6Z)-cyclodeca-1,6-dien-3-yne. It's not possible to tell which geometric isomer it is from your drawing, because you would need a clear depiction of the bond angles around the double-bonds. If this drawing is all you have to go on, you can't specify the isomer and your answer is spot on.
The below image is a model of your molecule showing the (1Z,6E)-cyclodeca-1,6-dien-3-yne form. You can see the difference between E ("opposite" or in this case "trans"), on the left, and Z ("together" or in this case "cis"), on the right, by looking at the shape of the chain of the four carbons nearest each double-bond. Of course, there is more to geometric isomerism than this but I hope this is a useful introduction if you've not encountered it before.

Your answer is correct. In case of cyclic unsaturated hydrocarbon organic compound with no side group attached, then numbering of multiple bonds should be done in such a way that there will be minimum number to multiple bonds.

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text:
Calculating Final Temperature and Volume of Adiabatic Expansion

A $\mathrm{5.00\ L}$ sample of $\ce{CO2}$ at $800 \ \mathrm{kPa}$ underwent a one-step (irreversible) adiabatic expansion against a constant external pressure of $100\ \mathrm{kPa}$. The initial temperature of the gas was $300\ \mathrm{K}$.
An alternative path between the initial and final states consists of a reversible isothermal expansion from $5.00\ \mathrm{L}$ to the final volume $V$, followed by (reversible) constant volume cooling to the final temperature $T$.
a) Give equations (in terms of $V$ and $T$, the final volume and temperature of the gas) for $\Delta U$, $Q$ and $W$
  for all three processes.
b) Briefly state why $\Delta U$ is the same for both paths.
c) Hence or otherwise calculate the final volume and temperature of the gas.

My Attempt
Part A
I can do this part. I am pretty sure I got the correct answers except for $\Delta U$ for the cooling process. Could you please check if my answers are correct.
Adiabatic expansion: $\Delta U = 33.33(T-300)$, $Q = 0$ and $W = 33.33(T-300)$
Isothermal expansion: $\Delta U = 0$, $Q = 4000\ \ln\frac{V}{5}$ and $W = -4000\ \ln\frac{V}{5}$
Cooling: $\Delta U = 33.33(T-300)$, $Q = 33.33(T-300)$ and $W = 0$
Part B
That is just because internal energy is a state function, independent of the path taken.
Part C
What confuses me here is the word 'hence', which implies that I need to use the fact that internal energy are equal for the two process. However I have no idea how to use this to find the final temperature and volume. I suspect that I might gotten the expression for $\Delta U$ wrong for the cooling process.

That 33.33 is low.  The constant volume molar heat capacity of carbon dioxide at these temperatures is 3.47R.

The final volume V and final temperature T can be determined exclusively from the information for the irreversible path if we assume that the gas continues to expand until it equilibrates at a final uniform pressure of $P=P_{ext}=100$kPa and a final uniform temperature of T.  Let $T_0$ be the initial temperature (300 K), $P_0$ be the initial pressure (800 kPa), and $V_0$ be the initial volume (5 L).  Then the number of moles is $$n=\frac{P_0V_0}{RT_0}$$
The work done on the surroundingsis given by:$$W=P_{ext}(V-V_0)$$
The change in internal energy is $$\Delta U=nC_v(T-T_0)$$
So, from the first law,$$nC_v(T-T_0)=-P_{ext}(V-V_0)$$
The final volume is given by: $$V=\frac{nRT}{P_{ext}}$$
These equations are sufficient to solve uniquely for the final temperature T and the final volume V exclusively in terms of the input data values.

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text:
What would you recommend to isolate these compounds from cooking oil?
Recycled cooking oil, aka "gutter oil", is a major problem in countries such as China. I am doing some research to see if there's a reliable and quick way to determine if something is gutter oil.
I came across a Chinese paper that uses Solid Phase Micro Extraction with Gas Chromatography/Mass Spectroscopy. In this paper, they listed out a list of chemicals that are commonly found in their samples of gutter oil.
Originally, I was planning to use Surface Enhanced Raman Spectroscopy for detecting these chemicals inside a matrix but it seems like the fluorescence of the oil is masking the raman signals of these chemicals.
All of these chemicals are dissolved in the sample oil in trace quantities. The are (listed in order of priority):
Methyl-3-butenenitrile
Hexane
5-Cyano-1-Pentene
1- decyne
Cyclohexane
Allyl cyanide
Acetic acid
benzene
Methyl cyclopentane
3-butenyl isothiocyanate
Allyl isothiocyanate
N-heptane
Toluene
1-hexanol
Xylene
Heptanone

Are there any simple extraction/filtration techniques that could isolate some of the above for the purpose of Surface Enhanced Raman Spectroscopy? I am looking for a sample preparation method that is less involved as solid phase extraction.

I have to admit I had never heard of 'gutter oil' - sounds nasty! However the description of your problem sounds very much like it is a textbook candidate for NMR fingerprinting and multivariate analysis. Almost no sample preparation is needed. A quick search shows that it is a method already being developed.

I would suggest high performance liquid chromatography for such a medley of chemicals followed by $\ce{^{13}C}$-NMR.


text:
Which complexes are stable: ammine complexes or carbonyl complexes?
According to Wikipedia,

A spectrochemical series is a list of ligands ordered on ligand
strength and a list of metal ions based on oxidation number, group and
its identity.
$$\ce{I- < Br- < SCN- < Cl- < S^2- < F- < OH- < C2O4^2- < H2O < NCS- < edta^4- < NH3 < en < CN- < CO}$$
Ligands arranged on the left end of this spectrochemical series are generally regarded as weaker ligands and cannot cause forcible pairing of electrons within 3d level and thus form outer orbital octahedral complexes. On the other hand ligands lying at the right end are stronger ligands and form inner orbital octahedral complexes after forcible pairing of electrons within 3d level.

According to above passage $\ce{CO}$ is a stronger ligand than $\ce{NH3}$. So, carbonyl complexes are more stable than ammine complexes.
Now, there are many other factors which determine the stability of complex:

Nature of central metal atom
Nature of ligand
Crystal field effects

Nature of ligand has further classification:

Basic character of ligands
Charge on ligands
Chelate effects

Classification source: see here and here
Basic character of ligands states that the more basic a ligand is, the greater the ease of ligation with the central metal atom is and the more stable the complex is. Now, $\ce{NH3}$ is more or less a basic compound and $\ce{CO}$ is a neutral molecule.
So, according to above assumption, $\ce{NH3}$ is a stronger ligand than $\ce{CO}$. So, ammine complexes are more stable than complex complexes.
So, which one is the stronger ligand: $\ce{CO}$ or $\ce{NH3}$? or which complexes are stable: ammine complexes or carbonyl complexes?

which complexes are stable : ammine complexes or carbonyl complexes? || The answer: it depends. One of parameters it depends on is the charge of central atom. Carbon in CO is electron-deficient, so carbonyls are significantly stabilized by back-donation, which is not a thing for strongly positively charged ions. As result, good luck finding a mononuclear complex with +3 charge and CO as a ligand and good luck finding a complex with -2 charge and NH3 as a ligand.

You're conflating way too many things together to the point that your question hardly makes any sense. The positions of NH3 and CO on the spectrochemical series has little, if any, correlation with the stability of their complexes as measured by formation constants $K_\mathrm{f}$. When wikipedia talks about "stronger" and "weaker" ligands it is referring to stronger- and weaker-field ligands, which is terminology for describing the position of a ligand on the spectrochemical series, not the stability of any complex that it forms.

That quotation from Wikipedia is bad and should be reworded (I can’t think of good ways to edit it, though). Ligands in a spectrochemical series are not ordered by bonding strength or anything, they are ordered solely by the field split they induce.
Both $\ce{CO}$ and $\ce{NH3}$ are great at forming complexes — as is iodide. The main difference is that they won’t form similarly strong complexes with the same cores. Just to give a few examples:

Carbon monoxide is neutral and a π-acid: It has an empty orbital with π symmetry which can take part in metal-to-ligand back-bonding. As such, it stabilises low oxidation states of metals. The neutral carbonyl complexes such as $\ce{[Fe(CO)5]}$ are well known, but also complexes such as $\ce{[Fe(CO)4]^2-}$ (yes, that’s iron($\mathrm{-II}$)!) are stable.
Ammonia is also neutral and a very weak π-base. However, it is a much stronger Brønsted base than carbonyl is. Therefore, it is much better at stabilising medium to high oxidation states such as in $\ce{[Cu(NH3)4(H2O)2]^2+}$.
The halides, finally, are anionic and good π bases and thus able to stabilise high oxidation states. They are classed as weak-field ligands because they generally generate a very low field split. An example of a halide complex would be the bright yellow $\ce{[FeCl6]^3-}$.

You cannot take ligand $\ce{X}$ and call it strong or weak and thus expect it to make stable or unstable complexes. You cannot even take the same ligand and expect it to make inert or labile complexes all the time. One ligand-metal combination may be stable, another may be very unstable — $\ce{[Fe(CO)6]^2+}$ can only be synthesised under a $\ce{CO}$ atmosphere in $\ce{SbF5}$ from $\ce{FeF2}$ where the antimony pentafluoride abstracts any free fluoride ions from the solution leaving a ‘naked’ iron ion which will grab the only ligands it can reach. Iron pentacarbonyl is stable. The same goes for ammonia: Some will readily accept it as a ligand, others won’t. There is no black and no white, only shades of grey.
(Although since coordination compounds are generally colourful let me rephrase that: There is no black and no white, only a wheel of colours.)

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text:
Is there a way to use free software to convert SMILES strings to structures?
So far I've been unsuccessful in finding an open source tool that will convert a large number (>100k) of SMILES strings to a chemical structure.
Is this really only solved by commercial applications?
I would not mind converting through a number of intermediate formats.
Ideally I'd do it with R, but that is the least of my worries.
It's also neither for personal nor academic use.

I think my question would be what tools you found that you didn't think could convert SMILES to structures and how you were searching. Using "structure generation SMILES" on Google seems to work fine for me - obabel was the 6th result.

Open Babel worked for me. @GeoffHutchison Avogadro did not install on Windows 10 for me.

According to the website, Open Babel should do the trick: Documentation - SMILES, Sourceforge.
For example, the following code will give you a neat SVG file of the molecule benzene:
obabel -:"c1ccccc1" -O benzen.svg

If you experience problems using it, you are welcome to ask more specifically.
Alternatively, you can use a web-query from the national cancer institute. It is easily accessible by the following code
http://cactus.nci.nih.gov/chemical/structure/"structure identifier"/"representation"

For example: benzene, "structure identifier"=c1ccccc1, "representation"=image.
Another open source solution, where you can directly export the structure into a molecular editor is Avogadro. (It uses Open Babel though.)
Depending on the actual problem, however, there might already be more advanced routines.

Thank you, that should do the trick for me! Suddenly I feel silly for asking. I'm at a loss how I missed that.

In addition to the other good answers, I'd recommend rdkit, an open-source, freely available software for chemoinformatics.  Most people use rdkit via its Python interface.
Here are some rdkit basics:

The code base is available in GitHub, here.
The license is quite permissive; you don't need to worry about what type of work (commercial, personal, or academic) you are doing.
The Python API makes using rdkit easy, but all the core functions are written C++, making it fast and efficient.  The Python API provides access to these functions in Python, making it flexible and easy to learn.  If you happen to be fluent in C++, a C++ API is available.
It does a whole lot more than convert SMILES to structures; see some examples here.

Here is one way to convert a SMILES to a structure in rdkit.
from rdkit import Chem
from rdkit.Chem import Draw

import matplotlib.pyplot as plt
%matplotlib inline

penicillin_g_smiles = 'CC1([C@@H](N2[C@H](S1)[C@@H](C2=O)NC(=O)Cc3ccccc3)C(=O)O)C'

penicillin_g = Chem.MolFromSmiles(penicillin_g_smiles)

Draw.MolToMPL(penicillin_g, size=(200, 200))

Here's a picture of the code and the resulting image.

I got <matplotlib.figure.Figure object at 0x7fb4fa67a490> from this code ...?

For those who want to convert a few SMILES strings to images, you can also use the CDK 1.5-based Depict utility from John May (www.simolecule.com/cdkdepict/, GitHub). It provides various options and outputs Scalable Vector Graphics (which can be easily converted into other formats).
For example, caffeine with title: https://www.simolecule.com/cdkdepict/depict/bow/svg?smi=CN1C%3DNC2%3DC1C(%3DO)N(C(%3DO)N2C)C%20caffeine&abbr=on&hdisp=bridgehead&showtitle=true&zoom=1.6&annotate=none

Thus, with the basic web API you can create a script to convert all SMILES strings too, e.g. using the RCurl package. This StackOverflow post explains how you convert the SVG to other formats.
However, since you probably prefer a pure R-based solution, please do have a look at the rcdk package.

The CDK Smiles Depict tool converts from SMILES to 2D Structure (as mentioned above).
The URl in a previous answer looks obsolete, but the tool can be found here https://www.simolecule.com/cdkdepict/depict.html

I'm surprised that you've had difficulty finding a toolkit - is it that the licence must be MIT or as permissive? I guess that you will be using this in software you are making, rather than a one-off data conversion?
For example, OpenBabel (C++), Chemistry Development Kit (Java), etc - in addition, the CDK can interface with R - would seem to suit your needs?

at the moment it is just for a one-off data conversion. at some point later it might be in the software, but then it would be okay to spend some money.
also I'm suprise that you're surprised - may I ask what search terms you used?

@BootstrapBill - Gilleain helps to develop the CDK.

@BootstrapBill Hah! Geoff is right - probably it is easier for me to find them since I already know about them! :) That's the difficulty of answering questions, of course. I was really just checking that you hadn't already found and rejected the ones on the wikipedia list.

In Python you could try pysmiles.
Starting from the SMILES description you should be able to create a NetworkX graph object with code along the lines of
from pysmiles import read_smiles
import networkx as nx

smiles = 'C1CC[13CH2]CC1C1CCCCC1'
mol = read_smiles(smiles)


text:
What is the difference between Fukui FMO approach and Woodward-Hoffmann rules?
If I understand correctly Frontier MO-theory is based on the perturbation theory of reactivity and the Woodward-Hoffmann rules consider conservation of orbital symmetry.
Does FMO use symmetry considerations?
Do the Woodward-Hoffman rules consider differences in HOMO/LUMO energies?

There are no orbital phases in the W-H approach.

The FMO approach, based on perturbation theory, focuses on analyzing the relative energies and populations of the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO). It predicts that reactions are most favorable when the attacking species interacts with the region of highest electron density, which corresponds to the HOMO of the reacting molecule. Symmetry considerations are not a central aspect of FMO theory.
The Woodward-Hoffmann rules are rooted in the conversation of orbital symmetry. They provide guidelines for understanding the stereospecificity and regioselectivity of pericyclic reactions. For example electrocyclizations and [2+2] cycloadditions. These rules consider the symmetries of the orbitals involved in the reaction, specifically the frontier orbitals (HOMO & LUMO), and predict whether a given reaction is allowed or forbidden based on the conservation of these symmetries.
Regarding your second question, the Woodward-Hoffman rules do not directly consider differences in HOMO/LUMO energies. Instead, they focus of the symmetry of the frontier orbitals and their overlap in the reacting system to determine the feasibility of a particular pericyclic reaction. Please note that while both FMO theory and the Woodward-Hoffmann rules provide insights into chemical reactivity, they approach the topic from different perspectives and use different applications.


text:
How to calculate the composition of a mixture of iron oxides containing inert impurities with titration data?

An $\pu{8 g}$ sample of $\ce{Fe3O4}$ and $\ce{Fe2O3}$ containing some inert impurity was treated with excess of aqueous $\ce{KI}$ in acidic medium, which converted all iron to $\ce{Fe^2+}$. The resulting solution was then diluted to $\pu{50 mL}$.
$\pu{10 mL}$ of it was taken and the liberated iodine required $\pu{7.2 mL}$ of $\pu{1 M}$ sodium thiosulphate to reduce all iodine.
Another $\pu{25 mL}$ was taken and the iodine was removed. The remaining solution required  $\pu{4.2 mL}$ of $\pu{1 M}$ $\ce{KMnO4}$ to oxidize all $\ce{Fe^2+}$.
Calculate the percentage of composition of the mixture.

The initial solution already contains $\ce{Fe^2+}$ from $\ce{FeO}$, which does not react until the second titration.
Working backwards, milliequivalents (m.eq) of $\ce{Fe^2+}$ is $5\times 4.2\times 1 = 21$ in $\pu{25 mL}$ solution. So in $\pu{50 mL}$, the amount of $\ce{Fe^2+}$ is $42$ m.eq.
Similarly, the m.eq of iodine from titration with thiosulphate is $36$.
From the first reaction (with $\ce{KI}$),
$$\text{m.eq of iodine liberated = m.eq of $\ce{Fe^2+}$ formed.}$$
The excess $\ce{Fe^2+}$ is from $\ce{FeO}$ which is present in $\ce{Fe3O4}$.
The amount of $\ce{FeO}$ is $6$ m.eq.
Hence its weight is $\frac{6}{1000}\times\frac{72}{2}=\pu{0.216 g}$ since $n$-factor is $\mathrm{2}$.
Since $\pu{1 mol}$ of $\ce{Fe3O4}$ contains $\pu{72 g}$ of $\ce{FeO}$, the mass of $\ce{Fe3O4}$ is $\pu{0.6945 g}$. But this answer is wrong. Is there anything wrong in my procedure?


text:
Al2Cl6 - Different length of bonds
How can we justify the longer length of the dative bond by bridge Cl with Al?
Is it because to prevent steric hindrance or repulsions the Cl atom is farther away from other Al as compared to normal Cl?

We know that the number of electrons in both the bonds are same. One of the bond involves a covalent bond $\ce{Al-Cl}$ where as the other is a 3-centred bond $\ce{Al-Cl\bond{->}Al}$. Thus while providing a dative bond, the effective atomic charge of bridged chlorine decreases in comparison with that at the terminal position. Due to this difference in atomic charge, the terminal ones attract more and closer that the bridges.


text:
Relative etch rates of nitric acid on metals and oxides
I am trying to either find or calculate the etch rate of nitric acid ($\ce{HNO3}$) for various different metals and oxides. The ones I am interested in are titanium ($\ce{Ti}$), zirconium ($\ce{Zr}$), and lead ($\ce{Pb}$); and their oxides $\ce{TiO2}$, $\ce{ZrO2}$, and $\ce{PbO/PbO2}$.
I have checked lots of sources, including the CRC Handbook of Metal Etchants, and have not been able to find anything that describes comparative etch rates for those six materials. Where can I find this information or, alternatively, how can I estimate these quantitites?

https://www.google.co.in/url?sa=t&source=web&rct=j&url=http://engineering.dartmouth.edu/microeng/processing/etching/Miscellaneous%2520Etchants.pdf&ved=0ahUKEwivwpSygafKAhWDcI4KHc_dA58QFggZMAA&usg=AFQjCNF8sWBMUoyiGyij78SanpYYzFnuxg&sig2=3zP6ANYmFvCK8rK2jj2qIw

https://www.google.co.in/url?sa=t&source=web&rct=j&url=http://www.chemcut.net/wp-content/uploads/2015/02/Etching-Titanium-with-HF-and-Nitric-Acid-Solutions-Part1.pdf&ved=0ahUKEwjvhaClgafKAhURjo4KHU5FA6MQFggZMAA&usg=AFQjCNHgsxipDmDaCMz8LvtrjPkjYJNa1Q&sig2=GiK_mnUAtpSpZygJegy2Hg


text:
Hydrophobic sand in water

Can you please explain this experiment? Why is this happening?

Have you done any research? This is supposed to be a serious chemistry site not a guess-what-reaction-this-is online multiplayer game.

true but i searched on internet , and didn't get anything good. . This is an academic site and most of the members are very dedicated in chemistry things. Therefore I can have a precise answer. If I had any clue over the picture, i would add that definitely. I hope you would not discourage people to learn something very interesting. I am not here  to do a comedy show. :)

There's no reaction, it's only mixing, as you should see in the end - if there was reaction they wouldn't have starting powder in the end, also you clearly have two phases all the time

Dupe of https://chemistry.stackexchange.com/questions/43319/hydrophobic-sand-in-water/43322#43322

Directly from Wikipedia under Magic Sand:

Hydrophobic Sand/Magic Sand is a toy made from sand coated with a hydrophobic compound. The presence of this hydrophobic compound causes the grains of sand to adhere to one another and form cylinders (to minimize surface area) when exposed to water.[1] When the sand is removed from water, it is completely dry and free flowing. Magic sand is also known as Aqua Sand.
These properties are achieved with ordinary beach sand, which contains tiny particles of pure silica, and exposing it to vapors of trimethylsilanol $\ce{(CH3)3SiOH}$, an organosilicon compound. Upon exposure, the trimethylsilane compound bonds to the silica particles while forming water. The exteriors of the sand grains are thus coated with hydrophobic groups.
Magic sand was originally developed to trap ocean oil spills near the shore. This would be done by sprinkling magic sand on floating petroleum, which would then mix with the oil and make it heavy enough to sink. Due to the expense of production, it is not being used for this purpose. It has also been tested by utility companies in Arctic regions as a foundation for junction boxes, as it never freezes. It can be also used as an aerating medium for potted plants.
Magic sand is made in blue, green, or red in colors but appears silvery in water because of a layer of air that forms around the sand, making it unable to get wet.

Also, see this informative video with explanations and demonstrations.


text:
Why do we consider all gasses to be in one phase?
Why do we take the phase of all gases 1. If all gases exist in 1 phase then why do we say that some gases are heavier than air, some are light also we collects some gases in lab by air displacement method. Is this all possible if gases are completely miscible.

Just because two fluids are completely miscible does not guarantee that they are uniformly mixed at all times.
For example, if you have a glass half full with sugar water solution, you can gently pour pure water to form a layer on top.  This is the basis for sucrose density gradient centrifugal separation and layered drinks.
So the same can occur for gases, a non-equilibrium state can exist for considerable time where the gases aren't mixed even though they are completely miscible.


text:
Reduced Graphene Oxide (RGO) Conductivity vs Everything Else
Reduced Graphene Oxide (RGO), is the reduced version of Graphene Oxide (GO). Both RGO & GO are easily manufactured when compared to the complex process used to manufacture the 'atom-thick' Graphene. And because RGO & GO are arranged in the same fashion as graphene, they keep similar properties; but nothing in comparison. Nevertheless, it is worth taking a look into the characterization of these properties because we can manufacture them quite easily.
But here's my problem, is it a waste of time to create thin sheets of RGO if there is a material that is more conductive?
So I arrive at my question, is a small sheet RGO's conductivity greater than that of, say, a small sheet of Copper? I have tried researching this every where but have not came up with a solid answer.


text:
Why is contact angle higher when I let the glass to be dried in silan solution?
I've read in literature that the silanization reaction is completed in a second. However, I realized an interesting thing; after one hour that my cleaned glass beads are deep coated in a silan bottle two things might happen:
1) If I take out the beads from the silan bottle and then wash them with water and then acetone, contact angle is ~102 deg and the beads are still transparent.
2) But if I take out the beads and then put them on a flat surface (in one layer) and pour some silan solution on them and let them to be dried in silan, then the contact angle is ~118 deg and glass beads are not transparent anymore (they're milky white).
Has anybody had a similar experience or know the reason of this?

What is in the silan solution?  What is its solvent?  In method 2, perhaps the solvent evaporates, leading to a coating of the glass beads not only with the chemically reactive silanization reagent, but also whatever stabilizers, emulsifiers, and who knows what else might be in there.  That could explain the color and the contact angle, but without more info on the composition of your silan solution, this is just speculation.

You can try to measure thickness of the coating. It should by different and I suppose that the dipped layer is thicker. The molecules have more time to create smoother layer than the molecules prepared by the second approach.

@JaroslavKotowski But in both methods, first I deep coat the glass beads for 60 min which gives them enough time to be coated properly. After this stage, I either let them to be dried while they are smeary in a thin silan cover (which gives a higher CA) OR wash them instantly (gives lower CA)

@CurtF. Silan solution was Octadecyltrichlorosilane >90% and I used it without any other additives or solvents. Also, the point that you mentioned for creating a coating of whatever is in silan solution on the beads surface after drying out, is very reasonable but you think they could not be cleaned up even by acetone?


text:
Why would the effect of fluoride ions Protecting egg shells from acid erosion vary with different acids?
I did an experiment reacting egg shells and acid. The egg shells were soaked in varying concentrations of fluoride and it is seen the rate of reaction between egg shells and acid decreases with higher concentrations of fluoride ions.
I performed this with hydrochloric acid, citric acid and ethanoic acid.  However the percentage decrease in the rate of reaction from the egg shells soaked in distilled water to the egg shells soaked in the highest concentration of fluoride ions Is calculated, it is largest for HCl, then citric and ethanoic with a 10-20% difference each time
This showed that fluoride had a greater effect for HCl than citric.
Why would this happen?

Can you post your procedure and data for us?


text:
Does sodium cyanide react with water, and can it cause an explosion?
According to some conspiracy theory, the explosion that happened last year in Tianjin (China) wasn't caused by sodium cyanide. Can sodium cyanide react with water, and can it cause an explosion if you have 700 tons of it?

It normally simply dissolves.

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Just checking... according to the conspiracy theory, the explosion was, or wasn't caused by NaCN?

Usually, I dislike conspiracy theories.1 However, this one (if it even is one) is true. Looking at the list presented on Wikipedia of the chemicals that were stored in that warehouse, why would you even suspect sodium cyanide as being a cause of the explosion? It is a stable ionic compound that can be put on the shelf, wet with water, dissolved in water, and won’t react with air with any measurable half-life. (Except that you possibly shouldn’t put it on a shelf because it is slightly volatile and highly poisonous.)
What was also in the warehouse and is an explosive was ammonium nitrate. In the 1921 explosion in Oppau (now part of Ludwigshafen), BASF workers tried to use dynamite to loosen a plaster of ammonium nitrate (it is hygroscopic and baked together). Mixtures containing at least $40~\%$ ammonium sulphate were considered stable and the mixture in question was a $50~\%$ mixture of the two according to the production data. After approximately 20,000 uses of dynamite where nothing happened, the entire 4,500 tonne lot exploded killing some 500 or 600 people and leaving thousands injured. The theory $40~\% = \text{safe}$ was later disproven in careful examinations. Apparantly, the explosion could be heard in Munich, some $300~\mathrm{km}$ away.
Another substance stored there and dangerous is calcium carbide $\ce{CaC2}$. When calcium carbide gets wet, it produces acetylene gas (and calcium hydroxide). The acetylene is flammable and can also explode in certain acetylene–oxygen mixtures.2 Luckily, no acetylene or carbide explosions are known enough to make it onto Wikipedia.
So far more likely than sodium cyanide having any primary effect would be one the following theory:

Fire (something must have ignited because the compounds I read there aren’t self-flammable).
Attempts to extinguish fire with water by the fire brigade.
Water reacts with calcium carbide liberating acetylene.
Acetylene burns.
The critical heat is reached, ammonium nitrate explodes.
Toxic sodium cyanide is liberated into the environment causing further severe health risks for man and animal alike.

1: Okay, that was a lie. I love listening to them and making fun of them.
2: A teacher at school talked about his time at uni and mentioned a professor that let a balloon of acetylene explode — he made a fuse long enough to ignite it from outside of the lecture hall (nobody was allowed inside), had the doors closed and my teacher said ‘it was the loudest bang I ever heard.’


text:
Why do chlorinated silanes have lower boiling points than their methane analogs?
The boiling points of the chlorinated silanes and methanes are given below:
$$\begin{array}{ccc}
\hline
\text{Species} & \text{Boiling point (X = Si) / }\mathrm{^\circ C} & \text{Boiling point (X = C) / }\mathrm{^\circ C} \\
\hline
\ce{XH3Cl} & -30.4 & -24.2 \\
\ce{XH2Cl2} & 8 & 39.6 \\
\ce{XHCl3} & 31.8 & 69.2 \\
\ce{XCl4} & 57.6 & 76.7 \\
\hline
\end{array}$$
The chlorosilanes are expected to have larger dipole moments and stronger dispersion forces.  Why do the silanes instead have lower boiling points than the methanes?

Out of curiosity, I checked the bromides (most data from this source plus Wikipedia), and the same pattern holds. I also added silane and methane for comparison, which challenges the trend:
$$\begin{array}{ccc}
\hline
\text{Species} & \text{Boiling point (X = Si) / }\mathrm{^\circ C} & \text{Boiling point (X = C) / }\mathrm{^\circ C} \
\hline
\ce{XH4} & -112 & -161.5 \
\hline
\ce{XH3Br} & 1.9 & 4 \
\ce{XH2Br2} & 66 & 96-98 \
\ce{XHBr3} & 109 & 147-151 \
\ce{XBr4} & 153 & 189.7 \
\hline
\end{array}$$

And here are the fluorides, which seem to be more complicated (data less certain):
$$\begin{array}{ccc}
\hline
\text{Species} & \text{Boiling point (X = Si) / }\mathrm{^\circ C} & \text{Boiling point (X = C) / }\mathrm{^\circ C} \
\hline
\ce{XH3F} & ? & -78.4 \
\ce{XH2F2} & -77.8 & -52 \
\ce{XHF3} & -97.5? & -82.2 \
\ce{XF4} & -86 & -127.8 \
\hline
\end{array}$$

I don't see how one expects weaker dispersion forces for bulky halogens around carbon. Where does this expectation originate?


text:
The color black
If a perfectly black object were present and it had a rough texture with bumps and crinkles, would it reflect any light? And can black reflect any light?

Welcome to Chemistry.SE.  I think such a question is more appropriate on Physics.SE.

I'm voting to close this question as off-topic because ^

The perfect black body is called a "black body radiator." It is a theoretical concept which doesn't exist in reality.

I am closing this as it is cross-posted to physics.se.

This question contains a tautology. If we define something as "perfectly black" by definition we mean something that has an absence of color and hence can neither reflect nor emit.
On the other hand, structure on the surface of an object can have a complex effect on the resulting field. For example, much of the coloring in morpho butterfly wings is due to interference. Other relatively bulk effects exist that can trap light, such as the reverse laser. And obviously a black hole.


text:
Can gaseous hydrogen and gaseous oxygen be compressed without reaction?
I'm looking into building a clean Bunsen-style burner (or rocket engine - I haven't decided yet). I will be using electrolysis to extract hydrogen and oxygen from water, but the apparatus I am thinking about building will render gaseous hydrogen and gaseous oxygen, but in a mixture. Preferably I would store them compressed, and not have spontaneous combustion.
My question is this:
Will gaseous hydrogen and gaseous oxygen react with each other if compressed in the same chamber?If so, how much pressure is required before it blows up?

As far as I know they need a spark to blow. However, such a spark can probably easily be created by electrostatics. I probably wouldn’t risk it if I could avoid it.

But the pressure alone wouldn't do it?

When you mix hydrogen and oxygen you have a bomb. Don't do it.

@MaxW ,so questions involving dangerous experiments are allowed on SE? Shouldn't you stop promoting such questions? Like a flag for "dangerous experiment"? (Just an opinion.)

@AdityaDev - Not really a problem. After his "experiment," we'll give Daniel a new nickname - "One eyed lefty."

Nice one. But still. Prevention is better than cure. I don't want someone posting questions like "how to make explosives". We already have terrorist problems :D

I'm not looking to store a whole lot of it.  Just enough to buffer for the burner.

You won't make a real rocket from this. Real H2-O2 rockets use liquid gasses and are typically huge.

Well, I happen to have seen first-hand and tested a working prototype of a rocket engine that burns gaseous H2 and O2.  It was the size of a football (American football that is).  It produced about 10lb of thrust and was quite efficient.

@MaxW: At least he isn't trying to mix liquid hydrogen with liquid oxygen.

A demo in college had the prof blowing bubbles with H2 gas and igniting them with a long firestick. Then he mixed some O2 in with the H2, and the pops became louder - and louder - until a bubble about 8 inches in diameter made an unbelievable POP in an auditorium with 300 people. We loved that guy!

The study Explosion Characteristics of Hydrogen-Air and Hydrogen-Oxygen Mixtures at Elevated Pressures includes data for pressures up to 200 bar.
Data were collected for Hydrogen-Oxygen mixtures at both 20 and 80 degrees C at pressures ranging from 1 to 200 bar.  A high voltage spark was still needed to cause explosion.
Storing a compressed mixture of hydrogen and oxygen still seems extremely unsafe to me.  I definitely wouldn't do it.

I built something similar to use as oxy /hydrogen welding and cutting. Basically I generated the gasses over water by electrolysis using graphite electrodes and a solar panel. The oxygen and hydrogen were collected separately in tanks made from plastic buckets, in, and separated by, a water bath. The gasses were led out via taps in the bottom of each bucket and only came in contact with each other within a welding handpiece. Pressure was regulated by putting bricks on top of the open bottomed containment tanks and varying the water level above them.
I only ever used pressures up to about 20 psi without any problem.
I wouldn't think that this would be acceptable on a worksite though!!!!
Have fun!


text:
Will freezing water still expand width-wise in a pipe (possibly breaking it) if it's free to expand lengthwise?
Will freezing water still expand width-wise in a pipe (possibly breaking it) if it's free to expand lengthwise?

It surely will, if the pipe is long enough and both ends get blocked by the ice.

It might do.
Imagine you have a U-shaped pipe. The two ends are above ground. The pipe runs into the ground for a short length in the middle, then back out again. It's free at both ends but filled with water.
You have a problem there because in cold temperatures, the two above-ground parts are going to freeze first. Then, if it's really cold, the underground part will freeze also. When that happens, the already-frozen above-ground ice won't move and your underground pipe may well go pop. What was initially "free at both ends" became blocked.

So the pressure needed to move the blocking ice is greater than the pressure needed to bend/break the pipe?

Strictly, it is going to depend the physical characteristics of the pipe material (e.g. elasticity, strength), its thickness and shape (e.g. opening vs closing funnel, further bends in the pipe), the amount of ice and any chemical bonding at the ice/pipe surface. I'm afraid I don't have figures to help you calculate it for any particular scenario.

Yes, if you do it right and this is indeed a technique used to maintain pipes see: 1,2.
That being said, there are obvious problems when you have two frozen chunks as ice in a pipe is able to withstand significant PSI without moving.

What if the water is frozen by nature (i.e. the pipe is outside)?


text:
Corrosive/ Acid cabinet gets wet from time to time
I have a metallic acid cabinet in a chemistry-based lab which I use store hydrochloric acid, sulfuric acid and ammonia. I also store nitric acid together inside but they are physically separated from the rest by an additional PP plastic case inserted into the acid cabinet.
The issue is that from time to time, I realized that the doors and secondary trays in my acid cabinet will get wet, especially the area near the sulfuric acid, and white powdery crystals sometimes form on the walls of my HCl bottles. I have ensured that I tightly sealed each of the bottles. Can anyone help me out? Is this dangerous? I'm trying to seek a solution before anything serious that might happen.
Also I have a few bottles of unopened, still sealed sulfuric acid which have turned black but the acid is still clear and usable when poured out, is it still alright to use for experiments? Will the results be affected?

Put a tray near the sulfuric acid, and sterilize/clean (with DI-water) the HCl bottle so you can wait to collect some of that too. Analyze them both with with IR, and maybe NMR. All of those should have distinct absorption properties.

The white powdery crystals are probably $\ce{NH4Cl}$, that is formed thanks to leakage form amonia. You can try to measure the pH of the wet layer, should be acidic because of the leakgae from other acids bottles that condensate on inner walls. Most experiment results should be ok, but if you are working in an analytical lab, the results can be shifted.


text:
What is the product after treating ethene successively with chlorine and water, sulfuric acid, and potassium hydroxide?
Question:

Ethene reacts with $\ce{Cl2}$ in $\ce{H2O}$ to form A. A on reaction with $\ce{H2SO4}$ followed by heating gives B. B on reaction with alcoholic $\ce{KOH}$ gives C. Find C.

My answer is:
A: $\ce{CH2(OH)CH2Cl}$
B: $\ce{CH2=CHCl}$
C: ethyne.
Explanation:
Ethene on reaction with chlorine water undergoes addition reaction to give A. A on reaction with $\ce{H2SO4}$ followed by heating undergoes dehydration to give B. B on reaction with alcoholic $\ce{KOH}$, undergoes elimination to give ethyne.
But the answer given is $\ce{CH2=CH-O-CH=CH2}$.
Where I am going wrong and what are the products?

I think diethyl ether is thermodynamically favourable because ethylene is not that stable. (less substituted alkene). Try this: https://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cad=rja&uact=8&sqi=2&ved=0ahUKEwjRqbe04JTKAhXQGo4KHVHsAIAQFgggMAA&url=https%3A%2F%2Fweb.anl.gov%2FPCS%2Facsfuel%2Fpreprint%2520archive%2FFiles%2F49_2_Philadelphia_10-04_1181.pdf&usg=AFQjCNFfQBSE_MRg0g1UG-Y9YR7MjSjI8Q&sig2=egb0mopsHHPvA0GNTXkKqg&bvm=bv.110151844,d.c2E

I don't know about this question, feels way too prone to "my school syllabus teaches that this reaction occurs with conc. H2SO4/ethanolic KOH, so only this reaction happens".

@Aditya Dev thank you very much for replying but can you please show the mechanism  of how it happens?(just tell me what happens step by step).

@orthocresol thank you very much for replying,do you mean to say only ethyne is formed?

It's the B product that's confusing. To get the required product, eliminate water from two molecules of A. Then alcoholic KOH will give the product that's given in your book. But, the more important part is "why intermolecular reaction takes place".

@ Aditya Dev ,thank you very much for replying,actually this is MCQ and ethyne is not present in option so to derive the answer we can carry out intermolecular reaction.

@Aditya Dev can you please elaborate on the mechanism?

http://www.masterorganicchemistry.com/2014/11/14/ether-synthesis-via-alcohols-and-acid/

yes,now i understand it,by adjusting the temparature we can enable ether formation instead of dehydration,but temperature is not given ,anyhow ethyne is not there in the option so we can assume that ether is formed in second step which undergoes elimination to produce CH2=CH−O−CH=CH2CH2=CH−O−CH=CH2.

The first reaction is a hydrochlorination reaction:

The second reaction is an acid catalyzed ether synthesis. The the solution is full of other 2-chloroethanol, which is capable of acting as a nucleophile in an $\mathrm{S_N2}$ reaction to eject $\ce{H2O}$. This is reaction will have a much lower energy barrier than the dehydration you proposed, which requires the extraction of an aliphatic hydrogen. The ether synthesis mechanism can be seen here:

The final step is a series of two $\mathrm{E2}$ reactions (shown condensed here):
$\hspace{4cm}$


text:
Trend in the boiling point of the hydrogen halides
The boiling points of the hydrogen halides are as follows:
$$\begin{array}{cc}
\hline
\text{Species} & \text{Boiling point / }\mathrm{^\circ C} \\
\hline
\ce{HCl} & -85.1 \\
\ce{HBr} & -67.1 \\
\ce{HI} & -35.1 \\
\hline
\end{array}$$
(source: Greenwood & Earnshaw, Chemistry of the Elements 2nd ed., p 813)
My textbook writes that this is due to an increase in the number of electrons, which increases the van der Waals forces between molecules.
But the $\ce{HX}$ molecules are polar. Shouldn't the boiling point be determined instead by permanent dipole-dipole attractions instead? Since the dipole moment decreases going from $\ce{HCl}$ to $\ce{HI}$, the boiling points should actually decrease. Why isn't this the case?

There are formulae you can use to calculate the strength of interactions between two molecules, which are derived from first principles. You can find an overview here on Chemistry LibreTexts. The two relevant ones are:
$$U_{\mathrm{dipole}} = -\frac{2}{3kT} \frac{p^4}{(4\pi\varepsilon_0)^2r^6}$$
$$U_{\mathrm{dispersion}} = -\frac{3}{4} \frac{\alpha^2 I}{(4\pi\varepsilon_0)^2r^6}$$

$p$ is the molecular dipole moment
$r$ is the average separation between neighbouring molecules
$\alpha$ is the polarisability of the molecule, which in a nutshell, measures how easy it is to create an induced dipole in the molecule
$I$ is the first ionisation energy of the molecule

If you plug all the numbers in for the hydrogen halides $\ce{HX}$, you will find that the magnitude of the dispersion forces is much larger than the magnitude of the dipole-dipole attractions, i.e. $\left|U_{\mathrm{dispersion}}\right| \gg \left|U_{\mathrm{dipole}}\right|$.
Unfortunately, I don't have all the necessary data on hand, but I can quote some results that were given in first-year lecture notes at Oxford (which should be legitimate). For $\ce{HCl}$, dispersion forces contribute $86\%$ to the intermolecular attractions, and for $\ce{HI}$, they contribute $99\%$.
Considering this fact, it is not surprising that variations in the magnitude of dispersion forces affect the boiling point much more than variations in the magnitude of the dipole-dipole attractions. If you had the all the data required and evaluated the interactions using the two formulae above, you would reach the same conclusion quantitatively.
In fact, this trend is hardly unique to the hydrogen halides. The hydrides of the Group 15 and Group 16 elements behave exactly the same way:
$$\begin{array}{|cc|cc|}
\hline
\text{Species} & \text{Boiling point / }\mathrm{^\circ C} & \text{Species} & \text{Boiling point / }\mathrm{^\circ C} \\
\hline
\ce{PH3} & -87.5 & \ce{H2S} & -60.3 \\
\ce{AsH3} & -62.4 & \ce{H2Se} & -41.3 \\
\ce{SbH3} & -18.4 & \ce{H2Te} & -4 \\
\hline
\end{array}$$
(source: Greenwood & Earnshaw, Chemistry of the Elements 2nd ed., pp 557, 767)
Of course, the first-row hydrides are left out of the discussion because of hydrogen bonding, which makes their boiling points anomalously high.


text:
Why is the Haber process carried out at such high temperatures?
On a large scale, ammonia is prepared via the Haber process:
$$\ce{N2(g) + 3H2(g)->2NH3(g)} \qquad \Delta _\mathrm{f}H^\circ = -46.1~\mathrm{kJ \cdot mol^{-1}}$$
The optimum conditions for the productions of ammonia are a pressure of $200~\mathrm{atm}$ and a temperature of about $700~\mathrm{K}$.
The process obviously is exothermic and $700~\mathrm{K}$ is, by no means, a low temperature. Shouldn't the temperature be much lower for optimum production of ammonia?

Related: Exothermic reactions at high temperature and the chemical equilibrium

As others have pointed out, it is purely kinetics, but you may still wonder, why.
For a reaction to actually occur (in both directions) and thus for an equilibrium to be reached, you need to overcome the activation energy. In the case of the Haber-Bosch process, this involves breaking the highly stable $\ce{N#N}$ triple bond. Even with the catalysts used, the energy required to break apart $\ce{N2}$ is still enormous. Therefore, a lower temperature may give a better yield of ammonia theoretically (i.e. based on equilibrium and Le Châtelier considerations) but the reaction speed would be a lot slower.
Even if you consider a batch-wise process of generating ammonia (which, as orthocresol points out, isn’t the case), it is more efficient to run two batches in half the time for two sets of $15~\%$ than to run a single batch for twice the time to get an overall yield of maybe (note: This yield is a ballpark estimate) $25~\%$ — with a batch size where $1~\%$ yield is equivalent to $100~\mathrm{kg}~\ce{NH3}$, two runs at higher temperature give $3~\mathrm{t}\ \ce{NH3}$ and one run at lower temperature $2.5~\mathrm{t}\ \ce{NH3}$ in the same time frame.

I think that last part may not necessarily accurately reflect the numbers. If you assume the percent yield is of the form $B e^{C/T}$, the time to run a batch is $A e^{E/T}$, and the desired final percent yield is some given number $p$, then the whole process takes time $\frac{p A e^{E/T}}{B e^{C/T}}$. The dependence here on $T$ depends on how $E$ and $C$ compare to one another. (I know my formulae are oversimplified.)

@Ian The last part most certainly does not accurately reflect numbers, since it is a ballpark estimate. If you have good numbers, do please share!

Sorry, I don't mean exact numbers, I mean the way the calculation would go if you had the exact numbers. There is a nontrivial competition between the yield per run and the time to make a run. Specifically, if $\Delta H$ is sufficiently negative compared to $E_a$, then (at least based on my heuristics above), the best thing to do would be to make a single very cold run. (Of course, once we start talking about "very cold", the very assumptions that make these heuristics make sense start to break down.)

@Ian Okay … All I know about the Haber-Bosch process is that it has been practised for about a century now and that the conditions used are probably already optimised to a tenth of a degree centigrade … So as I said, I was really just throwing numbers out there to show that going for highest batch yield is not necessarily a good thing ^^

So let me summarise this. If we take the temperature higher, then according to Le Chatelier's principle, equilibrium will shift backwards, though the reaction rate increases. And on making the temperature lower, according to the Le Chatelier's principle, equilibrium will shift forward, but the rate will be too slow, right?

@ReeshabhRanjan Yes.

@Jan Why to break the triple requires high activation energy? I mean couldn't the bonds be broken one after the other i.e. first the triple then the double and last the single bond?

The answer mainly has to do with kinetic considerations, as aml points out. I want to point out another thing.
In a typical industrial setting, you don't just mix the $\ce{N2}$ with the $\ce{H2}$ at a certain $T$ and $p$, collect the ammonia, and throw the unused reactants away. That would be horribly inefficient.
Greenwood & Earnshaw, Chemistry of the Elements (2nd ed) writes:

The gas leaving the catalyst beds contains about $15\%~\ce{NH3}$; this is condensed by refrigeration and the remaining gas mixed with more incoming synthesis gas and recycled.

The "synthesis gas" that they speak of is a mixture of $\ce{H2}$ and $\ce{N2}$ that is freshly produced via a series of reactions.
So, at first glance, the "yield" is absolutely rubbish. But reusing the reactants allows you to get high overall yields (albeit with more steps).

A schematic visualization of the process from Chemguide: http://www.chemguide.co.uk/physical/equilibria/haberflow.gif

While it is true that the Haber Process would be much more efficient at a lower temperature it's carried out at a higher one because it happens much faster at the higher temperature and while the industry wants an efficient process they also need it to happen fast enough to be commercially viable.
So you are correct but it's an example of practical realities coming before theoretical ideals.  The use of the word optimum is misleading in this context


text:
Understanding how to calculate the pH of a buffer with ice tables
Let's say you are asked to calculate the pH of a solution with:

$\mathrm{0.1~M}~\ce{HCl}$
$\mathrm{0.2~M}~\ce{CH3COOH}$.
$K_\mathrm a = 2\times 10^{-5}$

This is how I would do it:
$$\ce{CH3COOH + H2O <=> H3O + CH3COO-}$$
The corresponding ICE table:
\begin{array}{|c|cccc|}\hline ~ & \ce{CH_3COOH}& \ce{H_2O}& \ce{H_3O^+}& \ce{CH_3COO^-}\\ \hline  \mathrm I & 0.2 & & & 0.1\\\Delta  & -x & & x & x\\ \mathrm E& 0.2-x & & x & 0.1 +x\\\hline\end{array}
calculate the pH:
$$\begin{align}
 K_\mathrm a &= \frac{\ce{[H3O+][CH3COO^{-}]}}{[\ce{CH3COOH}]}\\
  &= \frac{x \cdot (0.1+ x)}{(0.2-x)}
\end{align}$$
as $x$ is very small, we leave it out:
$$\begin{align}
 \implies 2\times 10^{-5} &= \frac{0.1x}{0.2}\\
 \implies 2\times 10^{-5} &= \frac{0.1x}{0.2}\\
\implies~~~~~~~~~~~~ x &= 4\times 10^{-5}\\ \implies
 -\log(x) &= 4.40~= \mathrm{pH}\\
\end{align}$$

First of all. Is this correct?
Where do you include $\ce{HCl}$? I didn't write it anywhere. And I see most of the people not writing the second molecule. How/where is it included?
What is the molecule in the first column of the ice table? What does it represent? Do you always write the acid in that column?

I made that exercise up. The original exercise has $0.1$ molar $\ce{HCl}$ and $0.1$ molar $\ce{CH3COOH}$. I changed it in order to clearly see which value is which one at which place in stead of all having the same ones. This is the original exercise:
\begin{array}{|c|ccc|}\hline & \ce{CH_3COOH} & \ce{H_3O^+} & \ce{CH_3COO^-}\\ \hline \mathrm I & 0.1 & & 0.1\\ \Delta& -x & x & x\\ \mathrm E &0.1-x & x & 0.1 +x\\ \hline\end{array}
\begin{align}K_\mathrm a &= \frac{x\cdot (0.1 +x)}{(0.1-x)}\\ &\approx \frac{x\cdot 0.1}{0.1}\\ \implies x&= 2\times 10^{-5}\\\therefore -\log(x) &= 4.70\;.  \end{align}

You seem to have some command of LaTeX formatting please review the edits to understand that you don't need '\ce{}' to make the the equations do subscripts and multiplication.

@A.K. Please make sure when you edit, that you don't accidentally convert multiplication sings to decimal delimiters.

If ...

[...] you are asked to calculate the pH of a solution with:

0.1 M HCl
0.2 M AcOH ($K_a = 2 \cdot 10^{−5}$)

it would be a tremendous waste of time to assume it being a buffer solution and think about how you can get everything to work.
Hydrochloric acid is at least 1000000 times stronger than acetic acid which tells you that the resulting pH will be the pH of your 0.1 M HCl which then again is a very easy question (namely what is the pH of a 0.1 M HCl solution?).

Even if the question would have been to calculate the pH of a solution of

0.1 M HCl
0.2 M NaOAc

the question would have been anything else than hard to solve, too.
Adding 0.1 mol of HCl to 0.2 mol of NaOAc will give you 0.1 mol acetic acid and 0.1 mol acetate which only needs to be entered into the Henderson-Hasselbalch equation.
$$pH = pK_a + \log_{10} \frac{[A^-]}{[HA]}$$

The answer given is totally wrong. With 0.1 molar HCl, then the pH would be 1.0 from the HCl and virtually none of the acetic acid would be dissociated.
REVISED PROBLEM

I made that exercice up. The original exercice has $0.1$ molar $\ce{HCl}$ and $0.1$ molar $\ce{CH3COOH}$.

This makes no sense.
The work shown in the second image is for 0.1 molar $\ce{CH3COOH}$ and 0.1 molar $\ce{CH3COO^{-}}$. That work gets the right answer, but the wrong way. At equilibrium: $$[\ce{CH3COOH}] = 0.1 - x$$
but $x$ is so small that $$[\ce{CH3COOH}] \approx 0.1$$ and $$[\ce{CH3COO^{-}}] \approx 0.1$$ which simplifies the mathematics considerably.

ok I made that exercice up. The original exercice has 0.1 molar HCl and 0.1 molar $\ce{CH3COOH}$. I changed it in order to clearly see which value is which one at which place in stead of all having the same ones. This is the original exercice: http://imgur.com/d3URXIk .
Could you please tell what is wrong? (I've seen some people even use two ice tables to solve it)

Or could you please show/explain how to solve it correctly?


text:
Suggestions for reaction conditions to prepare methyl 2,4,6-trimethylbenzenesulfonate
I'm a beginner in organic chemistry and I wish to produce methyl 2,4,6-trimethylbenzenesulfonate.
My reactants are 2,4,6-trimethylbenzenesulfonyl chloride (solid crystals) and methanol (solvent as well as reactant). The product of this reaction should be methyl 2,4,6-trimethylbenzenesulfonate and HCl.
Now I've tried these many conditions: distilling the methanol, nitrogen inert atmosphere, slight heating about 40 °C, room temperature, closed system.
After trying these conditions and their combinations I'm not getting the product in the desired amount.
Kindly help me out and suggest a few tweaks and optimizations if possible. As far as my knowledge goes the problem seems to be with methoxy group.
This is the entire synthesis process, please check it out. I am stuck on step two of this synthesis.

Can you add structures or link to structures? Maybe I'm missing something, but the names of these compounds don't make sense. Trimethyltoluene sulfonate would be a tetramethylbenzene sulfonate, of which there are a couple isomers.

I guess, you want to prepare methyl 2,4,6-trimethylbenzene-1-sulfonate from 2,4,6-trimethylbenzene-1-sulfonyl chloride. Could you please check and edit your question?

@jerepierre and others, guys sorry for the delay. I have added the structures and the synthesis process please have a look.

@AVyas I've changed the name of the compound to methyl 2,4,6-trimethylbenzenesulfonate, which is the product of reaction 1. I am still confused though because your edit says that you are having problems with step 2, which is the transfer of the methyl to a pyridine. Could you clarify which step you are having problems with and what exactly the problem is? "Not getting the product in the desired amount" could be anything.

I'm closing this as unclear because I cannot tell which step there is a problem in and what the problem actually is, cf @jerepierre's comment above.

Are you following a literature procedure? It looks like you should be able to find a reliable method that you can reproduce. Some general hints for optimisation:

Use a base to mop up the $\ce{HCl}$ that is produced. Either an
organic base like pyridine (preferably) or triethylamine or you can use an
inorganic base like $\ce{NaOH}$ or $\ce{Na2CO3}$ under
Shotten-Bauman conditions  i.e. as a biphasic aqueous-organic system
Increase the reactivity of the nucleophile. Methoxide is a much better nucleophile. You can use commercially available $\ce{MeONa}$ or you can create it by something like $\mathrm{^tBuOK/MeOH}$

EDIT: I am assuming OP tries to make methyl 2,4,6-trimethyl toluenesulfonate since he states that MeOH is a reagent.


text:
Dissolving sodium bicarbonate?
I would like to dissolve or equally blend sodium bicarbonate in liquid coconut oil. Is there a way this can safely be done at home?

Why do you think it is not safe?


text:
Why is Aluminum's 2nd ionization energy higher than Silicon's?
According to the periodic table trend for I.E. Silicon should have a higher 2nd ionization energy than Aluminum so I'm confused.

Let us look at the electronic configurations of aluminum and silicon and their respective cations.
$\ce{Al}: [\ce{Ne}]\mathrm{3s^2\, 3p^1}$
$\ce{Si}: [\ce{Ne}]\mathrm{3s^2\, 3p^2}$
$\ce{Al^2+}: [\ce{Ne}]\mathrm{3s^1}$ It involves removing from $\mathrm{3s}$ sub shell
$\ce{Si^2+}: [\ce{Ne}]\mathrm{3s^2}$ It involves removing from $\mathrm{3p}$ sub shell
A $\mathrm{3p^1}$ electron has lower binding energy than that of $\mathrm{3s^2}$ electron.
It is due to more screening effect (as the predecessor sub-shells are higher in $\ce{Si}$) and higher stability attained by having a completely filled $\mathrm{3s}$ sub-shell


text:
Creating Boric Acid from Borax and Hydrochloric Acid
I'm working on an experiment in school and I need to create 200 mL of 2 mol/L boric acid. According to a formula I found on wikipedia, boric acid can be created with a reaction with hydrochloric acid.

$$\ce{Na2B4O7·10H2O + 2 HCl -> 4 H3BO3 + 2 NaCl + 5 H2O}$$

So to calculate the amount of moles of boric acid needed, I multiplied the volume with the concentration to get 0.4 mol. After using the molar ratios in the equation, the molar mass of borax decahydrate and the concentration of HCl, I calculated that I would need 38.14 g of borax and 33.56 mL of 5.96 mol/L HCl in order for a complete reaction.
After mixing these together, I ended up with about 40 mL of a really thick solution, which is not the intended 200 mL. What did I do wrong and how do I actually create boric acid of this volume and concentration? Am I missing steps in between?

Boric acid is not soluble to make a $2~M$ solution at room temperature.

At what concentration would it be soluble at? I can change the concentrations as long as I'm getting the right volume for each concentration . And is it supposed to look like a thick paste or is it supposed to dissolve?

Refer to the wikipedia page:  https://en.wikipedia.org/wiki/Boric_acid

I never noticed that table before, thanks. So if i were to dissolve the right amount at room temperature, would ny process be fine?

Assuming that boric acid is completely soluble in water, you did everything correctly so far, you simply missed the last step: dilution.
First you calculated the amount of substance of boric acid you would need to produce a two molar solution. With that you basically assume, that you dissolve solid boric acid in water.
You then calculated the amount of borax you need for complete reaction with hydrochloric acid. Since this is already hydrogen chloride dissolved in water, you bring some of the water you assumed before. But this is as your calculation says only about 40 mL.
Still assuming that it is completely soluble, you would have produced a solution that is about ten molar. You still need to dilute it to the final volume, to have the desired concentration.
Now let's go one step further. Boric acid is not very soluble in water under standard conditions, see Wikipedia. As A.K. pointed out, solubility is 47.2 g/L at 20 °C. That translates to 0.76 mol/L. As you can see solubility increases with temperature, in boiling water the maximum concentration is about 4.5 mol/L.
So what you are trying to attempt is not possible, at least not in pure water at room temperature. Try heating it up a bit, somewhere around 70 °C should be enough.
Alternatively PubChem has kindly provided us with a little more statistics about solubility.
In methanol you can obtain the desired two molar solution at room temperature, as solubility is 173.9 g/L at 25 °C.

Just for clarification, do i add 200 mL of water to the solution or 160 mL to the existing 40mL?

@BobMarshall you dilute to 200  mL as this is your final volume, hence you only have to add about 160 mL.


text:
How to know if acid is polyprotic?
I know that some acids for sure and for certain are polyprotic. For example sulfuric acid $\ce{H2SO4}$ is diprotic giving you disulfate $\ce{HSO4-}$ and sulfate $\ce{SO4^2-}$
However not all acids with multiple hydrogens are polyprotic, even if 2 or more hydrogens are attached to very electronegative atoms. So if I am given an acid and it's chemical structure how would I know if it is polyprotic or not? Acid base reaction with something like sodium hydroxide $\ce{NaOH}$  and then distillation to separate the product/products from the water that is produced? Or is there a better way to prove whether or not an acid is polyprotic?

Could you give an example for "even if 2 or more hydrogens are attached to very electronegative atoms" which is not polyprotic?

I can't really find an example but I would think most of these would be of the form $\ce{HaNbXc}$ where X is a halogen and easily gives up a hydrogen and nitrogen(a very electronegative element) does not easily give up a hydrogen simply because it is more electronegative. This would require that the halogen be hypervalent of which the smallest is chlorine.

I don't understand that. Why should chlorine be hypervalent - what does hypervalent even mean to you? Chlorine is a little more electronegative than nitrogen. Does such a compound as you propose even exist? Your question must have come from a place where you had a doubt about the polyprotic nature of a compound, I was just trying to find out about your confusion.

Chlorine is hypervalent because it is able to accept more electrons than just 8. Why else would Cl2O7 or ClO4 be able to form in the first place if chlorine wasn't hypervalent without chlorine becoming super positively charged which chlorine just hates so much?  And I do believe that a compound like what I proposed exists even if it hasn't been discovered yet.

You may believe whatever you like, but that's just not how chemistry works.

Chlorine cannot have more than 8 electrons. That is just impossible. There are various reasons at play, why compounds like ClO or ClO3 exist, hypervalency was used until there was a better understanding. In these compounds, the chlorine is indeed very positive.

Third period elements can have up to 18 electrons in the outer shell. Since chlorine is a third period element it makes sense that it can have more than an octet. Plus most chemistry courses still mention Hypervalency and how chlorine can have more electrons than just the 8 electrons in chloride. In fact chlorine is the most common example of Hypervalency

This is going straight into off-topic territory, but if you think that chlorine can accept 18 electrons, then you are wrong. Hypervalency is an outdated, revoked theory that still lingers in many books and get taught to innocent people. I am sorry that you are a sufferer of this. Related: http://chemistry.stackexchange.com/q/29142/

That being said, there are indeed quite a few compounds in which only part of their hydrogens are acidic. But organic chemistry would be a huge offtopic here.

And the current model of Hypervalency has MO theory as part of why atoms like chlorine can accept so many electrons and why VSEPR theory still holds for hypervalent atoms

So how do you know if an acid is polyprotic??

As discussed in your question and ensuing comments, there are ways to predict via theory whether a particular compound would be a polyprotic acid, but your last sentence asks how to prove that an acid is polyprotic. This pretty much limits us, as you proposed, to laboratory experiment, namely acid-base titration.
Although your suggestion "Acid base reaction with something like sodium hydroxide NaOH and then distillation to separate the product/products from the water that is produced" is on track, the experiment needn't be nearly so complex.
The experimental procedure is as simple as preparing a solution of the unkown acid, then titrating with a strong base solution while monitoring the pH.
The following is a titration curve for the diprotic oxalic acid, using sodium hydroxide as the strong base titrant. (Image taken from this boundless.com post):

Note that there are two equivalence points, at 25 and 50 ml of added titrant. These equivalence points are the features to look for when titrating the unknown acid. A single equivalence point indicates a monoprotic acid, two indicate a diprotic acid, etc.
There are a couple drawbacks to this experimental procedure. First, the equivalence points are not always as well defined as that shown in the figure. For instance, two acidic moieties may have similar Ka values and thus may almost appear as one.  Making a derivative plot of the titration curve can help to distinguish the equivalence points. In other words, you plot the change in pH vs volume of titrant added rather than just the pH of the titrant added.
The second drawback of this procedure is that it doesn't tell you which protons were titrated by the base. This would require a far more complex experiment, for instance by synthesizing the acid using deuterium in place of hydrogen at specific sites, then analyzing the compound after each equivalence point to see whether the deuterium was pulled off.
I hope I understood your question correctly and that my explanation of the titration procedure to determine polyprotic acids was clear. Don't hesitate to ask for clarification in the comments.

authors: Назарій Бойчук


text:
Why can anthracene, but not phenanthrene, take part in Diels–Alder reactions?
Both anthracene and phenanthrene seem to be structurally similar, with three aromatic rings. The middle rings in particular seem to be similarly activated towards cycloadditions.

I wouldn't be entirely surprised if phenanthrene could undergo D-A reaction, but as the dienophile instead of the diene.

Have a look at the reaction mechanism of the Diels-Alder Reaction, e.g. at Wikipedia.
We have talked about this reaction before on this site, and concluded that a sufficient explanation is only possible with the help of molecular orbitals: How accurate is this polar mechanism for the Diels-Alder reaction?
Borrowing from Wikipedia,[1] here is the most important part of the mechanism, the HOMO of the diene and the dienophile:

Now the most important part of this picture is the symmetry of the HOMO/LUMO of the diene, since this is the subject of your question. We need either a HOMO or a LUMO of these symmetries and then could match a suitable dienophile.
In anthracene we find the appropriate symmetry of the HOMO:

The reason for the regioselectivity has also been discussed before on this site: Diels Alder with Anthracene and Maleic anhydride.
We could also tune our reaction to be of inverse demand, as the LUMO has also appropriate symmetry:

In phenanthracene we do not have the appropriate orbital symmetry in the HOMO:

The LUMO is also not suitable for an inverse Diels-Alder reaction.

Reference and Notes

"FMO of Diels-Alder reaction" by Chrito23 - Own work. Licensed under CC BY-SA 3.0 via Commons.
Calculations on the DF-BP86/def2-SVP level of theory employing Gaussian09 rev. D. Contour values for the orbitals are at 0.05 e/rBohr. Occupied orbitals are depicted in blue and orange and unoccupied orbitals are depicted in yellow and red. Phases are chosen at random.

HOMOs and LUMOs don't explain everything.   Polyaromatic hydrocarbons also have their reactivities explained by Clar's aromatic sextet rule.   There is an excellent reference here from the Journal of the American Chemical Society that explains both the rule and reactivities of polyaromatics (http://pubs.acs.org/doi/abs/10.1021/ja00529a046).  In short, not all rings in an polyaromatic have equal aromatic character.   Clar's rule explains how to determine this for a given polyhex, which is a good idea since there are 22 possible structures for five ring polyaromatics and doing the HOMO/LUMO calculations can be tedious.  BTW, there are 82 possible six ring polyaromatics, and the numbers keep going up fast beyond that.


text:
What is the oxidation state of C in this lewis structure?
What is the oxidation state of $\ce{C}$ in this lewis structure?

I know how to find it: based on the number of electrons leaving/coming to the most/least electronegative element. But it s the $\ce{OH}$ that gets me confused...

What's confusing about OH? Clearly, O is the most electronegative element here. That being said, though, the very idea of oxidation states is rarely applied to organics.

@IvanNeretin, is OH more or less electronegative than C? This is what I don't know

O is more electronegative than C. Electronegativity is a property of atoms, not groups.

Reopening this because the duplicate was deleted.

This answer uses electronegativities for the calculation of oxidation states as proposed in the Expanded Definition of the Oxidation State by Hans-Peter Loock in 2011.
Foreword: As mentioned in the OPs comments, electronegativity and oxidation states are atom properties, not group properties.
Comparing Electronegativities
This table shows an excerpt from Pauling electronegativities ($\chi_{\mathrm{Pauling}}$):
$$\begin{array}{cc}
\hline
\text{Element} & \chi_{\mathrm{Pauling}}\\
\hline
\ce{H} & 2.20\\
\ce{C} & 2.55\\
\ce{O} & 3.44\\
\hline
\end{array}$$
Hence, the following applies:
$$\chi_{\mathrm{Pauling}}(\ce{H}) < \chi_{\mathrm{Pauling}}(\ce{C}) < \chi_{\mathrm{Pauling}}(\ce{O})$$
Looking at heteroatomic Bonds ($\ce{A-B}$, $\ce{A=B}$, and $\ce{A#B}$)
Since the oxygen atom has a higher electronegativity than the carbon atom, it attracts the bonds atoms more, resulting in an polarized bond. Presuming one would apply pulling force to the two atoms in a $\ce{C-O}$ or $\ce{C=O}$ bond until the atoms separate, it is assumed that the bond cleavage is heterolytic, thus allocating the bonding electrons to the oxygen atom.
The same applies to $\ce{C-H}$, $\ce{O-H}$, or any other heteroatomic bond ($\ce{A-B}$, $\ce{A=B}$, and $\ce{A#B}$).

Putting the jigsaw puzzle together
Now, having a look on the complete structure of the molecule in question, and applying the aforementioned rule to each bond:

Last thing to do is calculating the atoms hypothetical charge after separation, which is to be equatable with the oxidation state:
$$ \text{Oxidation state} = N_\mathrm{i}(\ce{e-}) - N_\mathrm{f}(\ce{e-}) $$
With $ N_\mathrm{i}(\ce{e-}) $ representating the number of electrons in a free atom, and $ N_\mathrm{f}(\ce{e-}) $ the one after separation (Never forget the lone pairs).
This will lead to the following oxidation states for each atom:

$$\begin{array}{cccc}
\hline
\text{Atom} & N_\mathrm{i}(\ce{e-}) & N_\mathrm{f}(\ce{e-}) &\text{Oxidation state}\\
\hline
\ce{H-1} & 1 & 0 & +1\\
\ce{H-5} & 1 & 0 & +1\\
\hline
\ce{C-2} & 4 & 2 & +2\\
\hline
\ce{O-3} & 6 & 8 & -2\\
\ce{O-4} & 6 & 8 & -2\\
\hline
\end{array}$$
The structure with assigned oxidation states:

The oxidation state of the carbon is $\mathrm{+II}$.

When determining the oxidation state of carbon in the given compound (formic acid), the electronegativities of the atoms directly bonded to the carbon are taken into account, because, as already mentioned in the comments, electronegativity is a property of individual atoms.
Oxygen is more electronegative than carbon (3.44 for $\ce{O}$ versus 2.55 for $\ce{C}$ on the Pauling scale), and the 2 electrons of the $\ce{C-O}$ single bond are formally assigned to oxygen, giving it an oxidation state of -II (the same happens with the $\ce{O-H}$ bond, leaving the hydroxy $\ce{H}$ with an oxidation state of +I). Likewise, all valence electrons of the carbonyl $\ce{C=O}$ bond are assigned to the oxygen, which therefore ends up with -II as well.
Carbon is, however, more electronegative than hydrogen (2.55 versus 2.20 for hydrogen on the scale mentioned above), so it "gets" at least the 2 valence electrons of the $\ce{C-H}$ bond.
After all, carbon is left with two valence electrons out of its original four, which results in an oxidation state of +II. The result can be verified by summing up the oxidation states of all atoms, which in this case gives 0, the correct value for a neutral molecule with no net charge. The image below schematically illustrates the formal assignment of valence electrons to the individual atoms and the resulting oxidation states.


text:
How to calculate a molar mass based on density?
I have been asked to calculate the molar mass and mole fraction of a 35 w/w% HClO4 solution with a density of 1.3g per ml.
This is the equation I used to try to solve it:
$\frac{1}{M} = \frac{0.35}{100.46} + \frac{0.65}{18.02}$
Is this correct?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This appears to be a homework question, please share your thoughts and attempts towards the solution. This question has some information that can help you.

What we are talking here are 3 measures of concentration:

mass fraction is about $m_{solute}/m_{solution}$
mole fraction is about $n_{solute}/n_{solution}$
molar mass is about the mass of one mole solution $m_{solution}/n_{solution}$

What we know that the mass fraction is $0.35w/w$ and the density is $1.3g/mL$. You can count the molar mass of water and HClO4 I assume, let's call them $M_{solvent} (M_{H_2O})$ and $M_{solute} (M_{HClO_4})$.
I will use the following equations by the calculations:
$$m_{solvent} = m_{solution} - m_{solute}$$
$$n_{solute} = \frac{m_{solute}}{M_{solute}}$$
$$n_{solvent} = \frac{m_{solvent}}{M_{solvent}}$$
$$n_{solution} = n_{solute} + n_{solvent} = \frac{m_{solute}}{M_{solute}} + \frac{m_{solvent}}{M_{solvent}}$$
From the mass fraction you can get the mole fraction this way:
$$\frac{n_{solute}}{n_{solution}} = \frac{\frac{m_{solute}}{M_{solute}}}{\frac{m_{solute}}{M_{solute}} + \frac{m_{solvent}}{M_{solvent}}} = \frac{1}{1+\frac{\frac{m_{solvent}}{M_{solvent}}}{\frac{m_{solute}}{M_{solute}}}} = \frac{1}{1+\frac{m_{solvent} \cdot M_{solute}}{M_{solvent} \cdot m_{solute}}} = \frac{1}{1+\frac{(m_{solution} - m_{solute}) \cdot M_{solute}}{M_{solvent} \cdot m_{solute}}} = \frac{1}{1+\frac{(1 - \frac{m_{solute}}{m_{solution}}) \cdot M_{solute}}{M_{solvent} \cdot \frac{m_{solute}}{m_{solution}}}}$$
You can count the molar mass the following way:
$$\frac{m_{solution}}{n_{solution}} = \frac{m_{solution}}{\frac{m_{solute}}{M_{solute}} + \frac{m_{solvent}}{M_{solvent}}} = \frac{m_{solution}}{\frac{m_{solute}}{M_{solute}} + \frac{m_{solution} - m_{solute}}{M_{solvent}}} = \frac{1}{\frac{\frac{m_{solute}}{m_{solution}}}{M_{solute}} + \frac{1 - \frac{m_{solute}}{m_{solution}}}{M_{solvent}}}$$
You don't need the density for these calculations.
$$M_{HClO_4} = 100.46g/mol$$
$$M_{H_2O} = 18.02g/mol$$
$$m_{HClO_4}/m_{solution} = 0.35$$
$$\frac{n_{HClO_4}}{n_{solution}} = \frac{1}{1+\frac{(1 - \frac{m_{HClO_4}}{m_{solution}}) \cdot M_{HClO_4}}{M_{H_2O} \cdot \frac{m_{HClO_4}}{m_{solution}}}} = \frac{1}{1+\frac{(1 - 0.35) \cdot 100.46g/mol}{18.02g/mol \cdot 0.35}} = 0.088$$
$$M_{solution} = \frac{m_{solution}}{n_{solution}} = \frac{1}{\frac{\frac{m_{HClO_4}}{m_{solution}}}{M_{HClO_4}} + \frac{1 - \frac{m_{HClO_4}}{m_{solution}}}{M_{H_2O}}} = \frac{1}{\frac{0.35}{100.46g/mol} + \frac{1 - 0.35}{18.02g/mol}} = 25.28g/mol$$

First of all thank you for your answer. Could you please confirm the correctness of this equation I used to solve my problem:
$\frac{1}{M} = \frac{0.35}{100.46} + \frac{0.65}{18.02}$
?

@JaqcuesMartin It's correct.


text:
Water splitting ways
Do you know the frequency you can split hydrogen from the water bonds? I have doing at now but i need to absorb more than 3 liters gas hydrogen per minute. My device working only with one ampere by some 40..50...100..dc volts. It is not fuel cell, not dry fuel cell , no membranes no antenna... but, it's working as well many years at now. So my question is to make it better to produce 100 liters gas hydrogen per minute with the same results of power input.

Eh, um, er, em, um, uh, I dunno what you're talking about.

The electrolysis of water in standard conditions requires a
  theoretical minimum of 237 kJ of electrical energy input to dissociate
  each mole of water, which is the standard Gibbs free energy of
  formation of water. It also requires energy to overcome the change in
  entropy of the reaction. Therefore, the process cannot proceed below
  286 kJ per mol if no external heat/energy is added.

https://en.wikipedia.org/wiki/Electrolysis_of_water

So there is a theoretical minimum energy required to split the water. To count the energy your device can use for the conversion, you need the ampere and volt numbers. You can count the energy with $E = U \cdot I \cdot t$, where we measure the E (energy) in Joule the U (voltage) in Volts the I (electric current) in Amperes, t (time) in seconds. With your parameters $E = 100V \cdot 1A \cdot 60s = 6000J = 6kJ$ energy can be used to convert hydrogen on the maximum settings.
To produce $100L=0,1m^3$ volume of hydrogen in atmospheric pressure you can count the minimum energy required the following way. According to the ideal gas law $p \cdot V = n \cdot R \cdot T$. So in $25°C=298.15K$ temperature at $1atm=10^5Pa$ pressure (standard conditions) the amount of the hydrogen will be $n = \frac{p \cdot V}{R \cdot T} = \frac{10^5Pa \cdot 0.1m^3}{8.314\frac{J}{mol \cdot K} \cdot 298.15K}=4.105 mol$. To produce that much gas you need at least $E = 4.034 mol \cdot 286 kJ/mol = 1153kJ$ energy, which is not possible with your device. The maximum it can do in this conditions is $\frac{6kJ}{1153kJ} \cdot {100L/min} = 0.52L/min$. On higher temperatures and smaller pressures you may produce somewhat more than this in volume (but not in mass).


text:
How can I make something like fire which is safe?
I want to know a magic trick that helps me put  my hand on fire without burning. My teacher told me you can use acetone but I should fire it fast in my hand because of volatility. Also I want to have fire that doesn't get put out fast; can I fire acetone for 5 minutes?!
If I fire acetone in my hand it may burn my hand. Are there any safe materials for my experiment?
Is there anyway to convert this heat into cold or Protectival material from burning?
I am looking for cold fire(fake fire)!
I read an article about cool flame;can I use it for solving my problem?

Of course there's always the heat when you say "fire". I wonder though if there are chemicals that produce something visually similar to fire, which would ideally be what you're looking for.

Of course, you can use lycopodium powder.


text:
Formation of Coordination Compounds
While reading applications of coordination compounds, I read that Hardness of water can be checked using Na2EDTA in hard water containing Ca+2 and Mg+2 which form complexes with EDTA.
Also, Lead (Pb), a  p block element is removed by using EDTA in case of lead poisoning.
Can elements other than those of d block form complex compounds? Like Ca(II) or Mg(II).
How are such complexes stable?
Can we desrcibe crystal splitting in them?

I know there are elements not from the d-block that can form coordination complexes. I don't remember exactly which ones (maybe the bigger ones, block p like Al (III), maybe Tl (I), etc., not sure if someones of group II). My initial guess for describing crystal splitting: depends, generally dissolved in water, must consider hydration enthalpy change and lattice energy breaking, etc.

But which orbitals will split? np orbitals?

Not sure, that will depend on the theory that describes the orbital overlap: more calssical CFT (https://en.wikipedia.org/wiki/Crystal_field_theory) or more quantum- orbital molecular LFT (https://en.wikipedia.org/wiki/Ligand_field_theory). Elements from 4th period and after [maybe Tl (I, III)?] may have p-d overlaping, thus allowing coordination.

Most of the transition metal form edta complex. See: https://pubs.acs.org/doi/abs/10.1021/jp904296m


text:
What happens when deodorants or perfumes expire
I know that deodorants and perfumes have alcohol as their main ingredient so I was thinking what happens when their expiry date passes?
I tried to search on google but got mixed results. Some say that they do not ever expire but some say that an expired deodorant can burn the skin.
What i think is  that as they contain mainly alcohol so maybe a major part of the alcohol gets oxidised to ketones or acids which may in turn harm the skin. Does anyone of you know what happens when deodorants expire?

I don't think ethanol gets oxidized in time, it won't react with air. Microbes can't use it either on higher concentrations (12%>) they will die. You won't do keton from it either, just aldehyde. The decay of the other components can cause the problems or maybe microbes if there is not enough ethanol in it.

why won't it get oxidized? I mean in India the expiry date is 2 years from the day we unpack the bottle(I don't know if its same in other countries too) so if ethanol gets oxidized to aldehyde which are highly prone to oxidation, it will easily get oxidized to ethanoic acid. And one of the perfumes(not body deodorant) that i have has 80% alcohol as per the label on it.

I have a different experience. I drank many years old ethanol several times... Only microbes can ferment it to aldehyde or lactic acid with enzymes by low concentrations. By standard conditions without these enzymes the transformation is very-very slow. You won't recognize it after decades or hundreds of years.

The meaning of expiry dates
The date of expiry is widely interpreted by consumers as the date after which a product is known to detiorate.  However, the meaning of expiry dates varies widely from product to product, and from country to country.  Often times, the date of expiry is not a "first known bad" date, but a "last known good" date.  In that latter case, the dates indicates that manufacturers know their product will last up until the expiration date, but after that, they don't know what happen.  Their product could last 1000s of years, completely stable and undeteriorated.  Or it could degrade instantaneously the day after expiry.
I don't know for sure what perfume manufacturers mean by their expiry dates.  You could try asking the manufacturer for clarification.  Ask for their data from their product stability studies.
Likely routes of perfume degradation
All of the above notwithstanding, let's suppose that perfumes do degrade over a time span of several years.  Subject to that assumption, what could be happening happening?  First, I agree with the commenters that ethanol oxidation to acetaldehyde or acetic acid is unlikely to be a major contributor to perfume degradation.
However, the same is not true of other perfume components.
Oxidation

Aldehydes such as octanal, benzaldehyde, citral, are an important ingredients in many perfurmes.  Citral degradation has been studied and is accelerated by acid and oxygen, with the main ultimate product being p-cymene.  Benzaldehyde is oxidized in air to benzoic acid, and octanal is oxidized to octanoic acid.
Terpenes are also important pefume components.   Limonene is a component of many perfume blends and is a representative terpene.  It is also oxidatively unstable, with products such as limonene oxide, carveols, and carvones being formed.  Some of these oxidation products can trigger allergic reactions, which while not precisely "burning the skin", could lead to such a perception.

Other possible degradations
Perfume, especially if opened frequently, could also change by absorbing water from the atmosphere, leading to dilution and an alteration of its olfactory properties.  Frequent opening or exposure to the atmosphere could also result in differential evaporation of the more volatile components, which would also change its olfactory properties.
The bottom line
Expired perfume or deodorant might be just as good as regular perfume.  Or it might be degraded.  The degradation products might possibly include compounds that have been reported to form rashes.  It is unlikely, however, that the chemical mechanism for this is the formation of large amounts of acid.

Also, many components in a perfume can react to form acetals with alcohols that may change its scent after the expiry date. Known examples are the hydroxycitronellal diethyl acetal and the acetals of lyral and helional. Another unpleasant reaction is the aldol reaction, for instance aldehyde C12 lauric and phenylacetaldehyde. And in some cases, the products of reaction can change also the color of the perfume.


text:
Estimate inflammability of ABS?
I am studying some electrical components and particularly Raspberry Pi 2 official cases. Their manufacturer says that they consists of ABS (Acrylonitrile butadiene styrene) when I ask if their components are inflammable.
This answer proposes me that their components are inflammable.
Wikipedia about ABS

Generally ABS would have useful characteristics within a temperature range from −20 to 80 °C

I think I can say little about the inflammability without extra pieces of information about their ABS.

How can you estimate how inflammable some ABS is?

see if this helps: http://fire.nist.gov/bfrlpubs/fire86/art017.html


text:
Is there a chemical reaction between tungsten carbide and acrylic?
My company machines acrylic (Poly(methyl methacrylate) $\ce{(C5O2H8)_n}$), and notices that tungsten carbide tools break down quickly compared to high-speed steel, even though carbide is harder and lasts longer than steel when cutting most any other material.
Obviously, there is mechanical breakdown because of friction, but is it possible that there is also a chemical reaction between the acrylic and the carbide that is causing this faster breakdown?

Tungsten carbide drills are very brittle. If the plastic melts during drilling, the drag on the bit from the melted blob is enough to shatter the bit. AFAIK, that is not an issue with tungsten carbide blades, though, which have a tough (ductile) metal backing.


text:
Could a safe water ocean exist on a world with predominantly chlorine atmosphere?
I am researching for a Si-Fi book.  Would a water ocean on a world with a predominantly chlorine ($\ce{Cl2}$) atmosphere be safe to touch?  Would the chlorine over time saturate the water creating a "laundry bleach' solution, or would it create a ocean of hydrocloric acid ($\ce{HCl}$) and hypoclorous acid ($\ce{HOCl}$), or would the chlorine just not react with the water that much?

Chlorine gas is dissolved in drinking water at low concentrations for disinfection purposes.  Chlorine gas has also be dissolved in somewhat higher concentrations in swimming pools.
The amount of chlorine dissolved will strongly depend upon pressure.  If the atmospheric pressure on the planet is low the concentration of chlorine gas in the water will be low.
Yes, most of the dissolved chlorine will reversibly react to form HCl and HOCl.
For quantitative information about the equilibrium constant and solubility as a function of pressure and temperature see Solubility of Chlorine in Water Industrial and Engineering Chemistry vol. 33, pages 741–744.

DavePhD.  Thank you, That is exactly what I needed to know.  I think for the situation in the book the water will be considered safe to touch and use as a disinfectant but probably too Chlorinated to drink without processing.


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What makes these particles hygroscopic?
Hygroscopy is the ability of a substance to attract and hold water molecules from the surrounding environment.
For cotton there is capillary action.
Hydrates like KOH attract water due to lower energy.
What are other possible reasons?
Particles like - diesel fuel, glycerol, salt, sugar and H2SO4 are also hygroscopic?
I have tried to find the reason, but I cant?

Any reference about diesel fuel? (The others are polar.)

@inf3rno That's apparently a popular idea in some circles, e.g. this forum thread, though someone sets the record straight further down by pointing out that diesel isn't extremely hygroscopic. Seems that diesel might be just hygroscopic enough that it becomes a problem, unlike gasoline (which you can leave sitting in a tank for months and have no problems).

@chipbuster I would not accept a forum thread as fact. I'll search, maybe I find some evidence.

@chipbuster http://www.sciencedirect.com/science/article/pii/S0043164813005619 http://pubs.acs.org/doi/abs/10.1021/je300279c Probably not the diesel fuel itself (which consists of apolar components https://en.wikipedia.org/wiki/Diesel_fuel#Chemical_composition), but contaminants are making it hygroscopic. I think there is a significant connection between polarity and hygroscopy.

@inf3rno apologies, didn't mean to imply that the forum thread was fact, but as a suggestion for where the idea might have come from. I'd be quite surprised if diesel was hygroscopic to anywhere near the extent that, say, KOH is/

@chipbuster Me too. I was surprised it was hygroscopic at all. :-)

As to your question a to whether a hydrophobic compound can also be
  hygroscopic, the simple answer is "yes to a point." Organic compounds
  can elicit both behaviors simultaneously when they have varying
  functional groups with a wide enough separation. BioDiesel is an ester
  with a very short side and a long side, the short side can exhibit
  polar qualities due to the presence of oxygen while the long chain
  exhibits nonpolar qualities. The polar side is what attracts the
  water, while the long chain repels the water. Cells of living
  organisms utilize a similar method to control the amount of materials
  crossing the membrane. I hope this explanation helps. Das Nerd 20:08,
  28 October 2006 (UTC)

https://en.wikipedia.org/wiki/Talk%3AHygroscopy

The composition of petrol diesel is very different, that's why it is not as hygroscopic as biodiesel.
As far as I know hygroscopy depends mostly on polarity. The water molecule is a polar, so it likes to mix with other polar compounds, like KOH, glycerol, salt, sugar, H2SO4, etc. Apolar compounds repel water. Air consists mostly of apolar components, so when the water in the air meets with polar things like sugar, it starts to move to that phase, because it has lower energy in there.
E.g. silicagel which is a common desiccant has this structure:

where you have polar $-OH$ chemical groups on the surface, that's why it is very hygroscopic.
While silicone oil contains molecules with this structure:

where you have apolar $-CH_3$ chemical groups on the surface, which makes it not hygroscopic.

authors: زهراء علاء الدين