text:
Determining the Equivalence Ratio
Determine the equivalence ratio of the following mixture: 4g of Butane, $\ce{C4H10}$, and 75 grams of air.
Answer is 0.82
Ok so I try converting everything into mols
$4~\mathrm{g}~(\ce{C4H10}) \times \frac{1~\mathrm{mol}}{58~\mathrm{g}} = \frac{4}{58}~\mathrm{mol}$
$75~\mathrm{g~(Air)} \times \frac{1~\mathrm{mol}}{28.966~\mathrm{g}} = \frac{75}{28.966}~\mathrm{mol}$
Air-fuel ratio = $\frac{\frac{75}{28.966}}{\frac{4}{58}} = 37.54$
This is just a guess that the air-fuel ratio is some sort of equivalence ratio.
I saw that equivalence ratio =  actual air-to-fuel ratio divided by the theoretical. I don't know how to get either actual or theoretical? Is the data in the problem sufficient enough to get both?
I also see a formula in the reviewer: Fuel + Air = Products of combustion and the appropriate reaction, but it only works for finding the exact amount of mols like: $\ce{C4H10 + aO2 + 3.76aN2 = bCO2 + CH2O + 3.76aN2}$
This is for a mechanical engineering board exam review; unfortunately, university only tapped briefly on the chemistry side of fuels.
Any hint on what I should be doing? Point me to the right direction?

I have improved the formatting of your post using $\LaTeX$. For more information on how to do this yourself please see here and here.

$$\begin{align}
\text{AFR (Air to Fuel ratio)} &= \frac{n_{\text{air}}}{n_{\text{fuel}}} \\
\text{Equivalence ratio} &= \frac{\text{AFR}}{\text{AFR}_{\text{stoich}}} \\
\end{align}$$
According to the problem statement, if you did everything right,
$$\begin{align}
\text{AFR} &= 37.54 \\
\text{AFR}_{\text{stoich}} &= \frac{\nu_{\text{oxygen}}}{\nu_{\text{fuel}}}\times 4.76 \\
\ce{C4H10 + $6.5$\,O2 &-> 4 CO2 + 5 H2O} \\
\text{AFR}_{\text{stoich}} &= 4.76 \times 6.5 = 30.94 \\
\text{Equivalence ratio} &= \frac {37.54}{30.94}=1.21
\end{align}$$
Now there are two types of equivalence ratio. One is air to fuel and another is fuel to air. I suspect in the answer they meant fuel to air equivalence ratio:
$$\text{FAR equivalence ratio} = \frac{1}{\text{AFR equivalence ratio}} = \frac{1}{1.21}=0.82$$

where'd you get the 4.76?

In air, for every mole of Oxygen you have 3.76 mole of Nitrogen or 4.76 mole of total (Oxygen+Nitrogen) (neglecting other trace compounds)

$\frac{79}{21}=3.76$

I followed this procedure. Note that "oxygen" here refers to ${\rm O}_{2}$.
They define the equivalence ratio as: $${\rm F/A\over (F/A)_s}$$ where F/A is your fuel to oxygen ratio (moles of fuel divided by moles of oxygen), and (F/A)$_{\rm s}$ is the "stoichiometric ratio" defined as the number of moles of oxygen needed to combust 1 mole of fuel.
For the numerator, you have 4/58 moles of fuel (F) and [0.21 $\times$ (75/29)] moles of oxygen (A) (0.21 because 21% of air is oxygen). Your numerator (F/A) is equal to 0.126.
For the denominator (F/A)$_{\rm s}$, we know that for each mole of fuel ${\rm C_mH_n}$ you need ${\rm (m + n/4)}$ moles of oxygen (see linked text for that). For butane (your fuel) that means you need 6.5 moles of oxygen to combust 1 mole of fuel. Your denominator is 0.153 (that's just 1/6.5).
Dividing 0.126/0.153 = 0.82


text:
Finding large databases of sphere packings
Part of my research is investigating sphere packings and their properties.I really would like a large database of sphere packings that I could investigate using a program I have written that tells me certain things about the packings. After doing some research I found some online databases but a lot of them haven't been taken care of so they're riddled with dead links.  I was wondering if anyone knows of any books or websites that have a lot of sphere packings that I could investigate.  Or if there is a way to interpret crystal structures as a sphere packing, seeing how there is a lot of crystal structure databases.
If anyone is confused as to what kind of data I'm looking I just need coordinates of the centers of spheres and their corresponding radii.  These need to form a packing where the spheres are touching like in this paper where the authors have found new bi-disperse packings.  Thanks!
http://pubs.acs.org/doi/abs/10.1021/jp206115p

there are several open crystallographic databases with various liberty of access, but selecting and extracting sphere packings would be up to programmer.

That is exactly what I would like to do, permeakra! I guess I'm unsure as to how I would take one of these databases like,
http://www.ccdc.cam.ac.uk/pages/Home.aspx
or
http://webmineral.com/
and begin extracting sphere packings?  Sorry for the late response I should have checked this sooner, thank you for your reply!

It is not clear what you want to do. Molecules are not spheres, so only crystallography database I would call "spherepacking" would be crystals of metals and simple salts. Of course, it is arguable how far they are spherical, too.

I think it's very clear from the post what I want to do, I want a database of sphere packings.  Nowhere in my post do I say that molecules are spheres or insinuate that.  Perhaps you meant to put this on my answer?  Either way, it's straight forward to interpret crystallography data as a sphere packing.  Take the Cartesian positions of the atoms and then take those positions for the center of spheres.

I have found a partial answer to finding a large database of sphere packings, thanks to the comment from permeakra!
So far I have been obtaining .cif files from the large databases of crystal structures online such as, http://www.crystallography.net/.  Once I have the .cif files I use openbabel to convert those files to .xyz files.  From here I have a table that describes a crystal's structure by the x,y, and z coordinates of each particular element for example,
\begin{array}{c|c c c}
\ce{Mg}      &   1.01054     &   1.01054   &     1.01054\\\hline
\ce{Al}      &   1.01054     & 1.01054     &   1.01054\\\hline
\ce{Al}      &   4.04217     &   4.04217   &     4.04217\\\hline
\ce{Mg}      &   4.04217     &   4.04217    &    4.04217\\\hline
\ce{O}       &   2.11390     &   2.11390    & 2.11390
\end{array}
But my problem from here is that babel has failed to return the ionic radius which I suppose can be retrieved online via wikipedia or something.  I posted another question that is more specific to .cif files which expands on this problem which can be found here, Extracting "Crystal Radii" from a CIF file.\
Limitations to this approach include that I can only find 3D sphere packings and it's a rather long and clunky procedure for obtaining lots of sphere packings.

It's not clear to me what kind of radius you want - certainly Open Babel has van der Waals and covalent radii as I indicated in the other question.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Will oxidation/reduction of copper still occur in an electrochemical cell if it is not in an aqueous environment?
Notation of the reduction and oxidation of copper typically includes the aqueous state. Illustrated below:
$$\ce{Cu(s) -> Cu^2+(aq) + 2e-}$$
This event usually occurs at 0.34 V. What if the copper was in an organic environment instead? I ask because I am doing cyclic voltammetry of a capacitor with copper as the current collector and need to measure the non-faradaic capacitance. Our current setup is using $\ce{KOH}$ as the electrolyte and we are basically just seeing a non-reversible/quasi-reversible CV like below:

It is certainly possible for metals, including copper, to corrode in non-aqueous environments.  The details depend on the precise nature of the solvent, electrolytes, and other potentially complexing species present.  For example, in polar protic solvents (like water) that don't have any complexing ligands for copper, the equilibrium potential for copper oxidation will likely be the same as in water.  But oxidation potentials will likely shift in solvents that are non-polar, in ways that depend strongly on the composition of the electrolyte.
However, if you are using potassium hydroxide as the electrolyte, your conditions might be more aqueous than you think.  You can think of "hydroxide" as a type of water that likes to exist at high pHs.  And hydroxide itself generates high pHs.
Elecrochemical water oxidation occurs via this reaction:
$$\ce{2 H2O -> O2 + 4 H+ + 4e-}$$
The pH-dependent equilibrium potential is $1.23~\rm{V} - 0.059\frac{\rm V}{pH}\times pH $, which at a pH of 14 comes out to 0.4 volts.  So you might also possibly be irreversibly oxidizing hydroxide to oxygen during your electrolysis.

I would agree that oxygen is probably being produced under these conditions, especially since the forward sweep extends so far. One should also note that potassium hydroxide is not particularly soluble in many organic solvents (apart from short-chain alcohols) and that things like lithium perchlorate or tetra-alkyl ammonium salts are usually used as electrolytes for non-aqueous electrochemistry.

Out of curiosity, how can you infer that oxygen is being produced from the forward sweep? I'm having trouble understanding how to interpret our CVs and could use any advice.

Hi curt,  I wanted to check out the pH dependency you were talking about. But I looked at the equation you gave and it doesn't make sense the way it's written. The pH cancels out and doesn't have an effect. Is this a typo?

Sorry, I'm meaning a bit confusing with my nomenclature.  The "V/pH" bit is a unit, volts per pH unit.  The "pH" by itself means the actual pH.  So at a pH of 0 (standard state), the potential is 1.23 V, but for each change of 1 pH unit, the voltage goes down by 59 mV.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Color of precipitate formed on reaction of acetylene with Tollen's Reagent
I am asked to find the color of the precipitate formed by the passage of acetylene gas into an aqueous solution of ammoniacal silver nitrate.
I thought ammoniacal silver nitrate was Tollen's reagent, but acetylene is neither a ketone nor an aldehyde so I don't know how to figure out the color of the precipitate.

You are correct that ammoniacal silver nitrate is Tollen's reagent. Tollen's reagent is most notable for its reaction with aldehydes to produce a silver mirror, but it also reacts with several other types of compounds, notably alpha-hydroxy ketones and terminal alkynes.
With terminal alkynes, the alkyne reacts as a acid to form a silver acetylide which precipitates. In the case of acetylene, both hydrogens can react: $$\ce{HC#CH + 2AgNO3 + 2NH4OH -> AgC#CAg + 2NH4NO3 + 2H2O}$$
The colour of the precipitate is variously described as yellow, gray-white, or white. The actual colour you observe is probably strongly dependent on the purity of the sample and the size of the precipitated particles.

Please, note!!!! Silver acetilenide formed  is a sensitive, even if relatively weak explosive !!! Do NOT produce any significant quantities of it and do not dry it, as drying increases its sensitivity dramatically!

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


text:
Functional Group compatibility in organic chemistry
As a continuation to the question:
Predicting reaction among compounds
I found the list of functional group on wikipaedia:
https://en.wikibooks.org/wiki/Organic_Chemistry/Overview_of_Functional_Groups
Now in the previously mentioned question, I got a clue that if the functional groups are compatible then reaction can occur, can you please tell me which are compatible with which?

I realize that you're genuinely curious; but as for the downvotes, I can understand them. This question is far too broad. It can't be answered with 100 books.

so there is no such compatibility rule?

There is no such compatibility. It's different from what you think. You might search for how chemists go about explaining reaction mechanisms in organic chemistry; terms like nucleophile, electrophile, leaving group, addition etc.

Being a computer scientist you would better stop "getting clues" on how things work in chemistry because your clues will likely be wrong. Just ask what actually you want to do and we will tell you is it even possible in the first place or not.

Four of your six questions have been closed, some of them with a negative score. The system will notice that and if you continue to ask these kinds of questions, and get closed/downvoted, you will likely run into a block/ban  again.

@Wildcat I did, but again that question was closed too

As I mentioned in my answer to this question, you are asking for something about which my two-semester organic chemistry course sequence just scratches the surface.

After covering the basics of organic structure and reactivity, most introductory organic chemistry texts have 15-20 chapters organized by functional group containing the reactions of and syntheses of said functional groups

In some cases, we can be reasonably certain that two functional groups will always react (primary amines and acyl halides), but in many, many other cases additional reagents, conditions, catalysts, etc. are needed. Different reagents and conditions can cause different outcomes from the same combination of reactants.
Here is a short sample of what I mean from among the earliest reactions we tend to study in the first semester organic chemistry course.
Alcohol + alkyl halide
In most cases, this combination of compounds will not amount to much, but there are some special cases:

Tertiary halides in primary alcohols undergo solvolysis

$$\ce{(CH3)3CBr + CH3CH2OH -> (CH3)3COCH2CH3 + HBr}$$

Addition of strong base promotes elimination in any alkyl halide, especially with heat.

$$\ce{CH3CH2CH2Cl + KOCH2CH3 ->[\ce{CH3CH2OH}][\Delta] CH3CH=CH2 + KCl + HOCH2CH3}$$

The major regioisomeric product can be controlled by which base is present.
$$\ce{CH3CH(Cl)CH2CH3 +NaOCH2CH3 ->[\ce{CH3CH2OH}][\Delta] CH3CH=CHCH3 + NaCl + HOCH2CH3}$$
$$\ce{CH3CH(Cl)CH2CH3 +KOC(CH3)3->[\ce{(CH3)3COH}][\Delta] CH2=CHCH2CH3 + KCl + (CH3)3COH}$$

If we take our primary alkyl halide and strong base, but the reaction is in DMSO (dimethylsulfoxide) instead of ethanol, and the reaction is kept cold, subsitution prevails.
$$\ce{CH3CH2CH2Br + NaOCH2CH3 ->[\ce{DMSO}] CH3CH2CH2OCH2CH3 + NaBr}$$
The reaction space in organic chemistry is very large and, while functional group analysis is always a good place to start learning, the effects of reagents and conditions, sterics and conformation, and other factors are also important. Rather than ask a question here that has an answer that amounts to whole textbooks worth of information, maybe you should sign up for a free MOOC on organic chemistry. That will give you a sense of what types of things are and are not possible (as per M.A.Ramezani's comment), and then you can ask more focused questions here.


text:
Why does color of halogen deepens and the boiling point decreases down the group?
Why does the color of the halogens deepens as we go down the group? also why there is a change in physical state of halogens down the group?
\begin{array}{c|c}
\mathbf{Gas} & \mathbf{Color} \\\hline
\text{fluorine} & \text{very pale yellow (almost invisible to naked eye)}\\
\text{chlorine} & \text{yellow-green} \\
\text{bromine} & \text{red-brown}\\
\text{iodine} & \text{violet-black}\end{array}

Just a thingy: In case of $\ce{Cl}$ , the color is also almost invisible... You sometimes have to place a white paper in the background to see the slight color.

BTW a quora link : http://www.quora.com/Why-are-halogens-coloured

Fun Fact : Possible color of $5^{\text{th}}$ halogen : $\ce{At}$ is Black! (black solid)

@NeilRoy Maybe At would be black, if there was an isotope that wasn't too radioactive to observe that.

As for the BP:
The attraction of covalent molecules is due 'dispersion' interaction. It is a quite complex effect, commonly described as interaction of 'instantaneous dipoles', i.e. interaction of dipoles formed by nucleus and electrons around it. Its efficiency generally increases with number of electrons (due to sheer numbers) and size of electron shell. Since outer electron shells vastly grow in size in fluorine-iodine row, the efficiency of dispersion interactions grow. In fact, close halogen-halogen contacts for heavier halogens (bromine and iodine) are very common in crystal structures of organic compounds.
As for color. As a guess, I'd say that the active transition is lone pair orbital $\rightarrow$ $\sigma$-antibonding orbital (see any introduction to MO theory for the terminology). The energy of this transition should lower from fluorine to iodine because of great decrease in bonding energy and increase of size of atoms. According to color wheel, the absorption in violet spectrum (with highest energy of photons in visible spectrum) should give yellowish apparent color (case of fluorine and chlorine), red apparent color to adsorption in lower-energy and violet color for adsorption at even lower-energy red spectrum.

The boiling point trend can also be accounted for by stronger halogen bonding going from difluorine to diiodine. Also, I believe the colour is due to a pi* to sigma* transition.

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


text:
What is the chemistry behind smell of human perspiration(sweat)?
What is the chemistry behind  the bad smell of human perspiration? also what is its composition?

-1 however a common question, this doesn't show any research effort. A simple Google search can lead you to very comprehensive articles.

Why does sweat smell?!

Sweat released from sweat glands odourless. Bad smell of sweat is due to certain bacteria on your skin mix with sweat(That is why sweat smell bad after sometime) and also due to certain hormones released during puberty.
I have tried my best to keep this part chemistry point of view(Actually this belongs to biology).

Chemical composition of sweat

The main reason people perspire is so evaporation of water can cool the body. Thus water is main component.
The chemical composition of perspiration varies between individuals and also depends on what they have been eating and drinking, why they are sweating, how long they have been perspiring and several other factors.
On average, the mineral composition is:

sodium (0.9 gram/liter)
potassium (0.2 g/l)
calcium (0.015 g/l)
magnesium (0.0013 g/l)

Trace metals that the body excretes in sweat include:

zinc (0.4 milligrams/liter)
copper (0.3–0.8 mg/l)
iron (1 mg/l)
chromium (0.1 mg/l)
nickel (0.05 mg/l)
lead (0.05 mg/l)

Source: ncbi and chemistry.about

In addition to simpler lower carboxylic acids contributing rather to socks smell (see Q: Why do socks stink?), some notable representatives contributing to the human axilliary (armpit) odor are

carboxylic acids

(I) 3-hydroxy-3-methylhexanoic acid,
HMHA;
(II) (2E)-3-methylhex-2-enoic acid,
3M2H.

I enantiomers (S)-I and (R)-I occur in approx. ratio 2:1, where (S)-I detection threshold by human smell sense was measured 0.08 ng/L air, cumin like smell; (R)-I detection threshold 0.2 ng/L air, weak animalic odor.
thiols

(III) 3-sulfanylhexan-1-ol,
SH;
(IV) 3-methyl-3-sulfanylhexan-1-ol,
transpirol, MSH;
(V) 1-methoxyhexane-3-thiol.

These can be smelled in much lower concentrations, and stereochemistry matters – lowest measured (S)-V 4×10−5 ng/L, compared to (R)-V 1.09×10−3 ng/L.
steroids

(VI) 5α-androst-16-en-3α-ol,
androstenol;
(VII) 5α-androst-16-en-3-on,
androstenone.

These are real human products unlike bacterial decomposition products above. Their role is not completely understood. VII is oxidation product of VI and has stronger odor. It is claimed some people cannot smell it because of mutation in one of olfactory receptors gene. It is sometimes also claimed, that there are sexual pheromones detectable by (in humans controversial) vomeronasal organ. They appear not only in humans, but also other animals like pigs.

References:

Wikipedia – Body odor
Wikipedia – Biochemistry of body odor
Buettner, A. Springer Handbook of Odor; Springer, 2017.


text:
Effect of enatiomers in pharmaceuticals
Often (or always?) one enantiomer of a certain drug will be effective and the other ineffective or harmful. The famous example is thalidomide, where one enantiomer caused mutilation of the unborn child and the other relieved morning sickness.
Are there any other examples where the other enantiomer of a drug that is used today or in history causes some adverse side effect, but has obviously been separated from the product to leave the effective drug, in addition what is(are) that side effect(s)?
Does absolute configuration at chiral/stereogenic centres affect the effectiveness of a drug molecule or perhaps cause unwanted side effects?

This wikipedia page has a list of some drugs with adverse side effects from different enantiomers. I can't answer the second part of the question definitively but I suspect that it depends on the exact structure of the drug and its mechanism of action.

If the second part of the question is " if a chiral centre exists in a drug, does it need a level of enantiopurity to function effectively",  my answer would be no. Look at ibuprofen.  It has a chiral centre but ibuprofen sold is racemic. It racemises in the body anyway, so neither enantiomer would seem to confer an advantage.

I'd say that most steroisomers would need to be in a specific form to be physiological active.  // The (S)-(+)-ibuprofen (dexibuprofen) was found to be the active form.  There is an isomerase which converts the (r) isomer to the (S). So there is no racemization of the moecule.  https://en.wikipedia.org/wiki/Ibuprofen#Physical_and_chemical_properties

For reference, you cannot give someone one enantiomer of thalidomide. It racemizes in vivo to provide both enantiomers.

Answers and length on site are irrelevant to whether the question should be closed. It is too broad, and therefore closed.

In pharmaceutical industries, $56\%$ of the drugs currently in use are chiral molecules and $88\%$ of the last ones are marketed as racemates (or racemic mixtures), consisting of an equimolar mixture of two enantiomers.

More recently, drugs originally marketed as racemic mixtures are reintroduced using the active isomer. Examples include racemic citalopram (Brand Name: Celexa among others) and its S-enantiomer, escitalopram (Brand Name: Cipralex and Lexapro among others); racemic omeprazole and its S-enantiomer esomeprazole; and racemic modafinil and its R-enantiomer armodafinil.
Sometimes, the difference in pharmacological activity between stereoisomers is dramatic. The dextrorotatory isomers in the morphine series are cough suppressants with less risk of substance abuse, whereas the levorotatory isomers contain the analgesic activity and significant risk of substance abuse. Although the direction of optical rotation is opposite to that of the morphine series, dextropropoxyphene contains the analgesic activity, and the levo-isomer contains antitussive activity.

Some were originally approved as racemic mixtures, and later a specific isomer was marketed with claims of having fewer adverse reactions in patients.
An example of the latter is the local anesthetic levobupivacaine, which is the S-isomer of bupivacaine. Both the R- and S-isomers have good local anesthetic activity, but the R-isomer may cause depression of the myocardium leading to decreased cardiac output, heart block hypotension, bradycardia, and ventricular arrhythmias. In contrast, the S-isomer shows less cardiotoxic responses but still good local anesthetic activity. Escitalopram is the S-isomer of the antidepressant citalopram. There is some evidence that the R-isomer, which contains little of the desired selective serotonin reuptake inhibition, contributes more to the adverse reactions than does the S-isomer.

The reintroduced S-enantionmer of citalopram with less pronounced side effects
The second part is a bit tricky, but there are different possibilities.
i. Absolute configuration
The two enantiomers of a chiral drug are best identified on the basis of their absolute configuration or their optical rotation. The absolute configuration at a chiral center is designated as R- or S- to unambiguously describe the 3-dimensional structure of the molecule. A chiral drug may have more than one chiral center, and in such cases it is necessary to assign an absolute configuration to each chiral center within the molecule.
Although the enantiomers of chiral drugs have the same chemical connectivity of atoms, they exhibit marked differences in their pharmacology, toxicology, pharmacokinetics, metabolism etc. Therefore, when chiral drugs are synthesized, as much effort goes towards the rigorous separation of the two enantiomers. This ensures that only the biologically active enantiomer is present in the final drug preparation.

The difference between two enantiomers of a drug is illustrated below using a hypothetical interaction between a chiral drug and its chiral binding site. In this case, one enantiomer is biologically active while the other enantiomer is not. The portions of the drug labeled A, B, and C must interact with the corresponding regions of the binding site labeled a, b, and c for the drug to have its pharmacologic effect. The active enantiomer of the drug has a 3-dimensional structure that can be aligned with the binding site to allow A to interact with a, B to interact with b, and C to interact with c. In contrast, the inactive enantiomer cannot bind in the same way no matter how it is rotated in space. Although the inactive enantiomer possesses all of the same groups A, B, C, and D as the active enantiomer, they cannot all be simultaneously aligned with the corresponding regions of the binding site:

Here is another example of epinephrine:

A postulated fit to epinephrine’s receptor can explain why (-)-epinephrine exhibits 12 to 15 times more vasoconstrictor activity than (+)-epinephrine.
This is the classical three-point attachment model. For epinephrine, the benzene ring, benzylic hydroxyl, and protonated amine must have the stereochemistry seen with the (-) isomer to match up with the hydrophobic or aromatic region, anionic site, and a hydrogen-bonding center on the receptor. The (+) isomer (the mirror image) will not align properly on the receptor.
ii. Another possibility
It is difficult to conclude that one isomer is superior to the other. For instance, S-verapamil is a more active pharmacological stereoisomer than R-verapamil, but the former is more rapidly metabolized by the first-pass effect. First-pass refers to orally administered drugs that are extensively metabolized as they pass through the liver.

This illustrates that sometimes absolute configuration has little impact on pharmacological activity and sometimes makes little difference whether a racemic mixture or one isomer is administered.
The popular nonsteroidal anti-inflammatory drug (NSAID) ibuprofen is sold as the racemic mixture. The S-enantiomer contains the anti-inflammatory activity by inhibiting cyclooxygenase. The R-isomer does have centrally acting analgesic activity, but it is converted to the S-form in vivo.

In some cases, the portion of a molecule containing the chiral center(s) may be in a region that does not play a role in the molecule's ability to interact with its target. In these instances, the individual enantiomers may display very similar or even equivalent pharmacology at their target site. Even in these cases, the enantiomers may differ in their metabolic profiles as well as their affinities for other receptors, transporters, or enzymes.

The body with its numerous homochiral compounds being chiral selective, will interact with each racemic drug differently and metabolize each enantiomer by a separate pathway to generate different pharmacological activity. Thus, one isomer may produce the desired therapeutic activities, while the other may be inactive or, in worst cases, produce undesired or toxic effects.
Although slightly off topic, it is important to note that in pharmacology area, activity of racemic drugs can be divided into three main groups:

Group 1. Racemic drugs with one major bioactive enantiomer

Group 2. Racemic drugs with equally bioactive enantiomers

Group 3. Racemic drugs with chiral inversion

Overall, it is critical to distinguish the single enantiomer from the racemic form because they may differ in their dosages, efficacies, side effect profiles, or even indicated use.
The decision to use a single enantiomer versus a mixture of enantiomers of a particular drug should be made on the basis of the data from clinical trials and clinical experience.
The two enantiomers of a chiral drug may differ significantly in their bioavailability, rate of metabolism, metabolites, excretion, potency and selectivity for receptors, transporters and/or enzymes, and toxicity. The use of single-enantiomer drugs can potentially lead to simpler and more selective pharmacologic profiles, improved therapeutic indices, simpler pharmacokinetics due to different rates of metabolism of the different enantiomers, and decreased drug interactions.

References

Chiral drugs

Pharmaceutical Analysis: David C. Lee, Michael L. Webb, Eds,, Pharmaceutical Analysis; 1st Edn., CRC Press LLC (Behalf of Blackwell Publishing Ltd.): Boca Raton, FL, 2003.

Chemistry for the Biosciences: Jonathan Crowe and Tony Bradshaw, In Chemistry for the Biosciences: The Essential Concepts; 3rd Edn., Oxford University Press: Oxford, United Kingdom, 2014.

Chiral Drugs: Lien Ai Nguyen1, Hua He, Chuong Pham-Huy, "Chiral Drugs: An Overview," International Journal of Biomedical Science 2006, 2(2), 85-100 (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3614593/); Also availble at: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3614593/).

Stereochemistry in Drug Action: Jonathan McConathy, Michael J. Owens, "Stereochemistry in Drug Action," Primary Care Companion to The Journal of Clinical Psychiatry 2003, 5(2), 70–73 (doi:10.4088/pcc.v05n0202); Also available at: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC353039/.

Organic Medicinal Chemistry and Pharmaceutical Chemistry: John M. Beale, Jr., John H. Block (Eds.), Wilson and Gisvold's textbook of Organic Medicinal Chemistry and Pharmaceutical Chemistry; 12th Edn., Lippincott Williams & Wilkins (a Wolters Kluwer business): Baltimore, MD, 2011.

When designing a total synthesis of some pharmaceutical, it is much easier to do a racemic synthesis compared to a total synthesis for one enantionmer, although this is not always possible.  In most cases the molecule must have the correct absolute configuration to work to the highest potential.  An example of this is oseltamivir (Tamiflu), it has 3 chiral centers.  There are 8 possible isomers of this molecule, yet only one is clinically significant.  For most compounds, two enantiomers are of clinical significance and one of these is just the other with all the chiral centers inverted.  There are various drugs that are sold as various enantiomers that include, Ibuprofen, Amphetamine, Omeprazole, Amlodipine, etc., there are various that are single enantiomers too, Levamlodipine, Escitalopram, Arformoterol, etc.
There is a drug called Ethambutol.  This drug, one enantiomer is used to treat tuberculosis.  Another enantiomer causes blindness.  There is also a drug called Naproxen, one enantiomer is used to treat arthritis, the other induces liver poisoning.


text:
Using equilibrium constant in gibbs equation
In my chemistry textbook, it is written that :

Gibbs energy for a reaction in which all reactants and products  are in standard state, $\Delta_\mathrm{r}G^\circ$ is related to the equilibrium constant of the reaction as follows:
  $$\Delta_\mathrm{r}G^\circ=-RT\ln K$$

A solved example was given there for the above equation.

Calculate $\Delta_\mathrm{r}G^\circ$ for conversion of of oxygen to ozone, $$\ce{3/2O2(g)->O3(g)} $$ at $298\ \mathrm{K}$ if $K_p$ for this conversion is $2.47\times 10^{-29}$.

The answer to above problem is simple. Just substitute the values in the gibbs equation and get the value of $\Delta_\mathrm{r}G^\circ$ but the thing that I can not understand is that in the equation they substituted the value of $K_p$ directly in the equation without converting into $K_c$ for equilibrium constant.
So my question is:
 In the gibbs equation what does equilibrium constant stand for? ($K_c$ or $K_p$).

Found it, http://chemistry.stackexchange.com/q/5811/16683 It's a very good answer but it might be a bit too detailed, not sure.

Well the answer is really good even though I have to take time to understand it. I am sure i will sort out my problem. Thank you very much for the link.

No problem, tell me if you want any explanation. Thermodynamics is one of the (unfortunately many) topics where you learn that whatever you were taught before was a lie...


text:
Can d-orbitals participate in bonding?
My textbook says:

It is appropriate to remark that the trick of rationalizing apparent violations of the octet rule by invoking the participation of d-orbitals in the bonding scheme is a matter of some controversy. Some scientists feel that the d-orbitals lie so high in energy that they should not be treated as valence orbitals, but as excited orbitals which cannot confer any appreciable stability to chemical bonds. Others feel that while d-orbitals are of high energy in free atoms, their energy decreases as other atoms approach to make bonds.
This author has not been convinced of the absolute validity of either side of this argument, and interprets the controversy merely as more evidence that the description of chemical bonds in terms of a few atomic orbitals is a highly approximate procedure. The most convincing reason for invoking d-orbitals in chemical bonding is that it is a simple idea which works, if it is used carefully and not over-interpreted.

Since this book is quite an old edition, I wanted to know whether there are any newer developments in this direction. Which of the two views is generally acceptable?

Surely they can. Even f-orbitals do so, see e.g. this question and references therein.

Please visit this page, this page and this one on how to make your future posts better.

Text in this book is about hypervalence of p-block and is all right. In d-block it's the other way round - d orbitals are used

this may answer it http://chemistry.stackexchange.com/a/5242/9961

Are you specifically interested in d-orbital participation in hypervalent/expanded octet compounds or are you looking for example of d-orbital bonding in general. If it is the former then there have been a number of excellently answered questions on this topic before. Here and here are good places to start.

Mmmm. Let me put it this way: the results of QC calculations for main group compounds are more precise if we include d-orbitals in used basis set. However, the resulting orbitals have only minor d-characters. Still, use of d-orbitals in the basis set is critical for chemically precise results, since said results are usually a difference of two large values, so even a small relative improvement of precision may have significant impact on the results.

@permeakra The d-orbitals you are talking about and their contributions are usually polarisation functions, their inclusion is important even in single-determinant methods. That doesn't mean that they represent the vacant d-orbitals of the main group elements. Also their contribution is usually below 1% for these.

Yes it works, according to result they have with organometallic chemistry. You may not know $e_g$ and $t_{2g}$ orbitals. Have a look at this diagram of orbitals :
Its diagram for an octaedric complex, where M is the metal. Positions of $e_g$ and $t_{2g}$ depends of what L is.


text:
Heat capacity of mixtures
After reading this recent question I was interested in how to calculate the specific heat capacity of a mixture based on the specific heat capacities of its components.
According to this website the specific heat capacity of an ideal mixture is given by

$$C_{p(\mathrm{mixture})} = \sum\limits_iC_{p(i)}x_i$$
where $x_i$ is the mole fraction of each component.

Where does this come from and what are the assumptions involved in treating it as an ideal mixture?
What deviations from this are observed in non-ideal mixtures? Can you easily predict the deviations based on the mixture and/or predict the actual specific heat capacity?

Suppose you have $n_1$ mole of A and $n_2$ mole of B. If they are at temp. $T_1$ and you want to raise the temp. to $T_2$ then the amount of heat you need to add is
$$\Delta H~= n_1 \times c_{pA} \times \Delta T + n_2 \times c_{pB} \times \Delta T$$
$$\Delta H~= (n_1 \times c_{pA} + n_2 \times c_{pB}) \times \Delta T$$
$$\Delta H~= (n_1 +n_2)\times \frac{(n_1 \times c_{pA} + n_2 \times c_{pB})}{(n_1+n_2)} \times \Delta T$$
$$\Delta H~= n_{total} \times(y_1 \times c_{pA} + y_2 \times c_{pB}) \times \Delta T$$
$$\Delta H~= n_{total} \times c_{p,avg} \times \Delta T$$
So basically you are not losing any information in case of ideal situation where molecular interaction of two different species has no effect on each other's heat capacity. But from statistical mechanics we know,
$$C_v=\frac { <E^2>-<E>^2}{k_B \times T^2}$$
Now if we solve the schrodinger equation for individual species, we must include an extra potential energy term to account the interaction from other species, which in turn will result in fluctuation of $C_v$ from it's previous value.
Slightly off topic: When I see the question first things that came to my mind is kopp's rule. It's used to find heat capacity of molecule using atom's heat capacity or to find alloy's heat capacity using the constituents heat capacity.

I have data that suggests a mixture of components (C1,C2,C3 and N2) has a higher heat capacity than their pure counterparts. Is this possible with regards to the schrodinger equation and molecular interactions?

It's possible from statistical mechanics point of view. I found this paper which discussed about this issue (http://nopr.niscair.res.in/bitstream/123456789/25014/1/IJPAP%2041%284%29%20280-289.pdf).


text:
What is the final temperature of each substance upon absorbing 2.25 kJ of heat?

Suppose that $\pu{28 g}$ of silver is initially at $\pu{29.0^\circ C}$. What is its temperature upon absorbing $\pu{2.25 kJ}$ of heat?

My work: $2250 = 0.240 \times 28 \times t - 29$
Solve for $t$ and I get $\pu{340 ^\circ C}$. But the correct answer is $\pu{370 ^\circ C}$ and I have no clue how to get there.

You are almost correct, however, to calculate these values you need to put a bracket around he $t-29$ term (which is the $\Delta T$ term from $Q = mc \Delta T$). Hence, from your working out step:
$$2250 = 0.240 \times 28 \times (t - 29)$$
You'll need to expand the brackets.

$\pu{2250 J} = \pu{28.0 g} \times 0.240 \times (T_\text{final} - \pu{29^\circ C})$ [$q = mC_\mathrm{s}\Delta T$]

Convert kJ to J as $\pu{2.25 kJ} \times 1000 = \pu{2250 J}$
The 0.240 is the specific heat capacity for silver (will most times be given to you)
The m is given
$\Delta T = T_\text{final} - T_\text{initial}$; We have T initial so plug in
Do the math. :)


text:
How can glycine act as a buffer at pH 6.00 and why?

How can glycine act as a buffer at pH 6.00 and why? Please use
chemical equations to show how it can act as a buffer at this pH
specifically?

My work:
At pH 6.00 glycine acts as a buffer, because at pH 6.00 the condition is acidic and glycine is an amino acid, so its zwitterion forms an equilibrium, which means there is a positive charge on the N of the amino group and a negative charge on the carboxylate. So there are no sudden pH changes and we know buffer solutions resist sudden pH changes, so at pH 6.00 glycine acts as a buffer solution.
Equation:

What I need explained:
HOW does glycine act as a buffer at pH 6.00 with chemical equations, meaning to show an example of how it specifically buffers another species at pH 6.00 with the proper chemical equations.

I rolled back your edit, as it just becomes a "homework-dump" (off topic) otherwise.

Your equation is not correct for pH 6; it should be for pH 1. For pH 6, the equation should be:

If an acid is added, then the equation becomes:

and if a base is added, then the equation is:

Glycine behaves as worst buffering agent along isoelectric point(pH = 6.1). Because In Zwitterion form, if there is a sudden addition of H+ ions then it behaves as a strong acidic compound because of NH3+ and COOH group. However, during an increase in pH, it behaves as Strong Basic Group due to COO- and NH2 radicals. Therefore it has its least buffering action around pH = 6.1 or pI point


text:
What is actually the difference between valence bond theory and molecular orbital theory?
Recently I have read about both of the concepts in my book (Physical Chemistry by Atkins, Paula). It was literally a reading; though I could understand the language and superposition of orbitals, inversion symmetry, and so on, I never could really comprehend how these two theories differ (or agree) with each other. I have googled it but couldn't gain a lucid explanation between the differences in their approaches. So, now my question is:
How do these two theories differ from each other? What are the differences between their approaches?
Also, in VB theory, resonance plays a pivoting role; why isn't such concept needed in MO theory? Why it is told that "MO theory provides a global, delocalized perspective on chemical bonding"? This portion of the question was split to a separate question: Why is the resonance concept not required in molecular orbital theory?
Edit: After viewing some of the answers, I think IMHO, I had to clear what I was really asking. By saying "difference between MOT & VBT", I am not intending to mean that just write the different predictions the two theories can separately succeed in making. What I really mean is the difference between their approaches & their way of explanation. Of course, the case is not like "Hey, look, one theory is outdated; just follow the other one..."; they are both important and hence do want to know the difference between their pathways of analyzing the cases.

It is a good question and a lot of books only scratch the surface of the theories anyway. Understanding both on a conceptual basis is necessary to distinguish the difference and similarities. There are plenty of resources of comparison available online, maybe UC Davis helps, I have not read it, but the following might also help: http://www.chem.wisc.edu/courses/562/fall08/PDFfiles/Hoffmann.pdf

UC Davis archive link. :-)

TL;DR
VB theory treats atomic orbitals (including hybridized orbitals) as providing a good mathematical/physical description of the true form of the molecular wavefunction. MO theory uses atomic orbitals (with Gaussian radial functions) as a tool of computational convenience in an effort to define a molecular wavefunction that in its final form often bears little resemblance to the original atomic orbitals, or to the VB theory description.

Long Version
The key difference between the VB and MO theories lies in their respective assumptions of the underlying mathematical/physical form of the electronic wavefunction.
VB theory holds tightly to the mathematical structure of atomic orbitals (AOs), assuming that the electronic structure of a molecular system is described adequately by retaining the atomic orbitals (possibly hybridized) from each constituent atom, and overlapping them to form the needed description of bonding.  VB calculations are computationally much more complex: the AO basis is generally non-orthogonal, and the linear algebra and other techniques required to cope with this non-orthogonality are much more involved.
More broadly, the need to invoke resonance (superposition of multiple valence-bond structures) and other concepts highlights that, in fact, often this adherence to atomic orbitals and their valence-bond overlaps is not adequate for a proper description, at least not without extensive use of methods analogous to the configuration interaction of MO theory.  VB theory does hold certain advantages, however—for example, I believe the AO basis does sometimes allow for a more direct linkage to fundamental chemical concepts such as hybridization and Lewis electon-pair theory. To note, Sason Shaik is one of the strongest modern proponents of the theory I know of—he describes some of these strengths and their applications in his recent retrospective article (doi:10.1016/j.comptc.2017.02.011).
MO theory, at its core, casts away most assumptions regarding the mathematical structure of the electronic wavefunction, as long as properties like indistinguishability and the Pauli exclusion principle remain satisfied.  Calculations within MO theory can use a wide variety of mathematical structures to try to represent the wavefunction, ranging from gridded numerical quadrature approaches (see here and here, and also orbital-free DFT) to Slater orbital computations to the Gaussian-orbital based calculations used most commonly today.
One potential point of confusion is that MO theory in its most common implementations makes direct use of atomic orbitals in composing the molecular wavefunction (see LCAO theory). The distinction to be made here is that instead of attributing significance to the atomic orbitals as an intrinsically good descriptor of electronic structure, they are used merely because they afford a convenient mathematical form for efficient computation, in trying to construct the MOs.  Particular advantages are obtained from the use of atomic orbitals constructed from Gaussian functions, since the product of two Gaussians is another Gaussian (even in multiple dimensions) and numerical integration of n-dimensional Gaussian functions is straightforward.
It should be noted that, for as much of an improved description it affords, MO theory still is an approximation. For a given system, the electronic wavefunction just IS the highly-abstract mathematical object that satisfies the appropriate Schrodinger equation.  MO theory is used extensively because it provides one of the most computationally tractable paths to getting in the neighborhood of the "real answer."
Practical Aspects of MO theory
The "simplest version" of MO theory for a time-independent system under the Born-Oppenheimer Approximation is the Hartree-Fock method, which provides a remarkably good description considering the significant simplification represented by its mean-field approximation of the electron-electron interactions.  Hartree-Fock is in general not good enough for computations at the level of chemical accuracy (~$1\mathrm{\ kcal/mol}$), however, and numerous methods have been developed to refine it: the most often used techniques are Moller-Plesset perturbation theory, configuration interaction, and coupled cluster.
In-Depth Reading on MO Theory
For a deep dive into the underlying mathematics of MO computations, it's hard to beat the classic Szabo & Ostlund. As a more readable, lighter-weight introduction to MO theory and its implementations, I would recommend Frank Jensen's Introduction to Computational Chemistry.

A lovely answer, but are you then saying that the only real difference is due to the choice of basis sets and that one (MO) is thus  more adaptable (and computationally convenient) than the other?

@porphyrin I wrote this answer before I'd learned much about the modern implementations of VB theory. I now know differently, and have plans at some point to revise this. That said, yes, I think both points of your question are correct. However, the VB analogues of correlated post-HF MO methods are able to provide quite good wavefunctions in many cases, so modern VB theory is actually much more useful than this answer makes it sound. IIUC, though, one has to make a deeper dive into VB as compared to MO before obtaining practically useful results.

@porphyrin To note one key difference, whereas MO theory works intrinsically with orthogonal MOs, IIUC the "working orbitals" of VB theory are not constrained to be orthogonal.

thank you, very clear. So I suppose the non-orthogonality must make diagonalising matrices very time consuming vs orthogonal ones for the same level of accuracy?

@porphyrin I have to assume so -- I'm not too familiar with the mathematical particulars. I think it also makes the equation systems dramatically more complicated when you can't assume $\left<i|j\right>=0$ for $i\neq j$.

Comparing modern valence bond and electronic structure theories one can argue that the generalized valence bond (GVB) wave function can be regarded as a special form of the multi-configurational self-consistent field (MCSCF) wave function.1
Thus, for instance, for the hydrogen molecule, the GVB wave functions has the following form (ignoring normalization hereinafter for simplicity),
$$
  \Psi_\mathrm{GVB} = \Big( f(1)g(2) + g(1)f(2) \Big) \Big( α(1)β(2) - β(1)α(2) \Big) \, ,
$$
where $f = a + kb$ and $g = b + ka$ are non-orthogonal orbitals and $C$ is a normalization factor. This wave function is a linear combination
$$
  \Psi_\mathrm{GVB} = C_\mathrm{c} \Psi_\mathrm{c} + C_\mathrm{i} \Psi_\mathrm{i}
$$
of the covalent and ionic structures described by
$$
  \Psi_\mathrm{c} = \Big( a(1)b(2) + b(1)a(2) \Big) \Big( α(1)β(2) - β(1)α(2) \Big) \, , \\
  \Psi_\mathrm{i} = \Big( a(1)a(2) + b(1)b(2) \Big) \Big( α(1)β(2) - β(1)α(2) \Big) \, .
$$
The GVB orbitals $f$ and $g$ are determined variationally: $f$ and $g$ are expanded over a basis and expansion coefficients are varied to minimize the
energy of the wave function.
The simplest two configurational MCSCF wave function for the hydrogen molecule would also be a linear combination of two wave functions. But in MCSCF to avoid non-orthogonal functions (which complicates the matter) a bonding orbital $\phi_1$ and an anti-bonding orbital $\phi_2$ are introduced and the total wave function is written as a linear combination of two Slater determinants describing two different electron configurations, $\phi_1^2$ and $\phi_2^2$, i.e.,
$$
  \Psi_\mathrm{MSSCF} = C_1 \Phi_1 + C_2 \Phi_2 \, ,
$$
where
$$
  \Phi_1 = \left| \begin{matrix} \phi_1 & \bar{\phi}_1 \end{matrix} \right| = \phi_1(1) \phi_1(2) \Big( α(1) β(2) - β(1) α(2) \Big) \, , \\
  \Phi_2 = \left| \begin{matrix} \phi_2 & \bar{\phi}_2 \end{matrix} \right| = \phi_2(1) \phi_2(2) \Big( α(1) β(2) - β(1) α(2) \Big) \, .
$$
However different $\Psi_\mathrm{GVB}$ and $\Psi_\mathrm{MSSCF}$ appear to be, if the same basis set is used to expand the orthogonal MCSCF orbitals $\psi_1$ and $\psi_2$ as well as the non-orthogonal GVB orbitals $f$ and $g$,  it can be shown that these wave functions are identical. For details, see Exercise 4.9 in Ref. 1.

1) Szabo, A.; Ostlund, N.S. Modern Quantum Chemistry: Introduction to Advanced Electronic Stucture Theory; Dover: Mineola, NY, 1989 (revised in 1996). pp. 258-261.

Here, I am referring Valence Bond Theory by VBT and Molecular Orbital Theory by MOT.
1.In VBT, atomic orbitals of the combining atoms retain a large amount of their individual character. (For example, see Orbital Hybridization).
Whereas, in MOT atomic orbitals of the combining atoms lose their individual identity in the resulting Molecular Orbital.
2.Secondly, VBT fails to explain properties like para-magnetism of some compounds like O2, whereas MOT easily explains this.
3.Also, VBT fails to explain possibility and existence of ions like H2$^+$, whereas MOT is able to do this successful.
Why it is told that "MO theory provides a global, delocalized perspective on chemical bonding"?
Quoting from the Wikipedia page of MOT,

In MO theory, any electron in a molecule may be found anywhere in the molecule, since quantum conditions allow electrons to travel under the influence of an arbitrarily large number of nuclei, as long as they are in eigenstates permitted by certain quantum rules. Thus, when excited with the requisite amount of energy through high-frequency light or other means, electrons can transition to higher-energy molecular orbitals.

Why isn't such concept needed in MO theory?
See this question, which directly addresses this. However, I would recommend you to read How does molecular orbital theory deal with resonance?
Thank You!

I quoted the "MO theory provides a global, delocalized perspective on chemical bonding" from wiki; I couldn't understand the very lines of the wiki page that you have highlighted:(

In the words of a commentary in the Journal of Chemical Education:

At their theoretical limits, both VB and MO are equivalent; they simply approach that limit from different sides.

Source:https://doi.org/10.1021/acs.jchemed.1c00919
The authors argue that in introducing the VB method in General Chemistry, it is equated with Lewis structures, which does not fully represent modern VB theory.
In the article, they also describe the simplest forms of the two theories and how they differ:

As the names suggest, valence bond theory considers a molecule as a collection of bonds, whereas molecular orbital theory views the molecule as a whole. [...] VB is presented as a single configuration of electrons in hybrid orbitals, spin paired in bonds, while MO is presented as a single configuration of molecular orbitals arranged in order of increasing energy and filled with electrons in accordance with Hund’s rule.

One way of describing the different approaches is to look at the structure of methane. The symmetry of the atomic orbitals (the p-orbitals specifically) does not match the geometry of the hydrogen atom location. In VB, the carbon atom orbitals are mixed and matched (hybridized) to make localized bonds with hydrogen. In MO, the hydrogen s-orbitals are mixed and matched (linear combinations of AOs) to match the symmetry of the p-orbitals of carbon. At a higher level of theory, which recognizes that electrons are indistinguishable, you arrive at the same distribution of electrons in the molecule.


text:
Reaction of 1,1-diphenylethanol on Montmorillonite K10 under microwave radiation

What is the product of 1,1-diphenylethanol + Montmorillonite K10 Clay undergoing microwave radiation? What is the mechanism for it?

My effort:
1,1-diphenylethanol + Montmorillonite K10 Clay catalyst used in a microwave-assisted synthesis of pyrazoles from enaminones.
Used for:

Support for catalysts
Preparation of clay nanopaper composites of nacre-like structure

Catalyst for:

Conversion of triose sugars with alcohols to alkyl lactates
Deprotection of ketone dimethylhydrazones

Is my answer correct?

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

Montorillonite K10 is a solid acid catalyst and microwave is a source of heat, so... $$\ce{Ph2C(OH)CH3 ->[\ce{H+}][\Delta]}?$$

Is my answer correct?

No! Your "answer" is just a random collection of facts and does not address the question.
Actually, Ben Norris has already kindly provided all the necessary information in his comment:

Montmorillonite K10 is a solid, acidic catalyst.
Microwave irradiation means that you are applying heat to the reaction.

Under the given conditions, 1,1-diphenylethanol (1) is protonated, eliminates water (which might be driven out since heat is applied) and yields 1,1-diphenylethene (2).
Given that you're still applying heat in the presence of an acidic catalyst, the reaction will not stop here.
Protonation of 2 again results in the known bisbenzylic cation, which will add to a second molecule of 1,1-diphenylethene in Markovnikov fashion.

The resulting intermediate is will undergo an intramolecular Friedel-Crafts alkylation which furnishes the indane derivative 3.


text:
Oxidation numbers
An element have many oxidation numbers, then which one should we consider in the given question?
e.g. In the compound $\ce{HCN}$ as $\ce{C}$ {carbon} has +4,+2,-4 oxidation numbers and $\ce{N}$ {nitrogen} has -5,+4,+3,+2,+1,-3,-2 oxidation numbers

e.g. In the compound HCN as C {carbon} has +4,+2,-4 oxidation numbers and N {nitrogen} has -5,+4,+3,+2,+1,-3,-2 oxidation numbers

That is not correct. Each atom only has one oxidation number in a compound, although one element can have different oxidative states (i.e. sodium thiosulfate). Your statement should be:

C {carbon} has +4,+2,-4 oxidation numbers, and N {nitrogen} has -5,+4,+3,+2,+1,-3,-2 oxidation numbers. In the compound HCN, carbon has +2, hydrogen has +1, and nitrogen has -3.


text:
Byproducts in Indigo Synthesis?
I have been messing around with making Indigo dye while waiting for bacteria to grow. I have been following the Baeyer-Drewson method. Basically, you mix nitrobenzaldehyde with an acetone water mix and slowly add NaOH. The solution turns black and you get a dark sludge. I have been separating the sludge from the liquid by centrifugation. I have noticed that the supernatent has a dark reddish brown color.

I was wondering if anyone knew what the red byproduct was?
My indigo is black when dried, wikipedia shows nice bright blue chunks. What could be causing this color difference? Maybe the red byproduct?
Does indigo dissolve in anything without needing to be reduced first? I have had no luck with the organic solvents available in our lab.

I would love to know the answer to this question myself (and likely will provide a bounty). I do this synthesis with my students as a lab experiment, and the yields are dismal. Knowledge about what else is being formed may help me tweak the procedure to improve yields.

As a side note, I can make solutions that are dilute enough for spectroscopy by dispersing about 80 mg of indigo in the smallest amount of methanol possible and then diluting to 100 mL with acetone. Indigo is soluble in neither by itself, but this method produces (dilute) solutions.

My first guess is that some self-condensation of acetone occurs. It should give various conjugated polyenones, that should have yellow to red color mostly, up to black for longer chains. While acetone is usually condensed in acidic media, a proper base should work too. Phorone synthesis: http://www.youtube.com/watch?v=ll8FLbJpR14 . If this is the case, it should be possible to steam-distill away most of the colored byproducts, however the remaining black residues of aceton resin should be insoluble and undistillible.

For your question 3: I have no idea if it actually works, but based on structure and my own fevered imagination, I'd guess that a mixture of Dowtherm A (which is a eutectic mixture of biphenyl and diphenyl ether) and an amide such as N,N-dimethylacetamide  would be a good solvent for indigo.

@CurtF. - That is an interesting idea on the solubility. I wonder what the OP wants to do with the indigo. For my use (optical spectroscopy, the arenes in Dowtherm A are a deal breaker).

@BenNorris I just want to dye stuff. Right now I have dye in several containers and have been having trouble pooling it all together. I might try to disperse it in ethanol, combine them and centrifuge them. This might remove some of the byproduct as well.

@user137 You might not want to try Curt's suggestion of solvent either since  all three compounds have very low volatility. If you are interested in dyeing, why not just buy indigo, which is cheaper than 2-nitrobenzaldehyde?

@BenNorris Because then I'd be taking the easy route. Solubility probably wouldn't be good for dying anyway, since indigo only sticks to fabric by being so insoluble, a good solvent would carry the dye away from the fabric. If there was a good solvent that was also volatile, I would be interested because then I could easily combine all my reactions and remove the solvent.

@permeakra I tried to mix acetone, water, and NaOH to mimic conditions in the reaction and did not observe any color change. Could the nitrobenzaldehyde or indigo interacting with the acetone and promoting condensation?

@BenNorris I tried to suspend a very small of my indigo sludge in 500 uL of acetone or acetone-water mix. In the acetone water mix, the solvent turned kind of yellow and I could see tiny particles. In acetone, it turned green and I could not see particles. Do your solutions have a visible color? I suspect the green is caused by a more soluble yellow contaminant and the blue indigo.

@user137 Yellow (and to lesser extent red) color is typical for moderately large conjugated systems. For example, o-aminobenzaldehide is pale yellow.  (and, for that matter, dibenzilidenacetone and phorone) The conditions of the synthesis are typical for aldol condensation, so most likely we have some secondary condensation, but which one I have no idea.

After looking further into indigo production, I came across a related structure called indirubin.
Indirubin:

Indigo:

Indirubin is a byproduct of indigo synthesis and has a reddish color. It may also be more soluble than indigo, and indirubin derivatives have some anticancer activity. If you trust this supplier, you can buy indirubin for pretty cheap to compare with your lab's indigo reactions.

I also found a study done on old native american dyes that studied indigo and indirubin in fabric fibers, among other dyes. They provide a UV-VIS spectrum of both compounds, with Ind = indigo and Inr = indirubin. We can see that indigo has an absorption peak in the red region, so blue is reflected. Indirubin has a peak at around 540 nm, which is yellow-green, allowing some blue and red to reflect, but probably more red than blue. This paper also references a 1910 paper** studying indirubin levels in commercially available indigo samples, but I can't find that article. They suggest that higher levels of indirubin are produced when more indoxyl is converted to isatin. If you can find this 1910 article, it may provide some insight into reducing isatin and indirubin levels.

** William Popplewell Bloxam and A. G. Perkin, J. Chem. Soc., Trans., 1910, 97, 1460-1475.

Not sure if the labels are better this way, I wanted to resize them but imgur has troubles scaling png so they are just black squares... I included the link to the digital version of the linked article, It's behind a paywall for me so I can't read :(


text:
Why is 2-chloropropene achiral?
Could someone explain to me why some alkenes containing at least one couple of different substituents at the two double-bonded C atoms (that is, some of those with at least two different substituents for both C atoms) are considered achiral? For instance, for 2-chloropropene or 1-bromo-2-chloro-2-fluoro-1-iodoethene, where (and what) in the structural formula of each of them is the plane of symmetry that makes it achiral or, in other words, what do their superposable mirror images look like? Thanks.

Welcome to Chemistrty.SE If you have any questions about how this site works visit our [help] and take our [tour]. For your molecules, consider their  three-dimensional geometry. For example, what is the geometry around the central carbon atom in 2-chloropropene?

As your molecules  are planar. The plane of the molecule is the element of symmetry that renders the molecule achiral.

You can see the 3D geometry, e.g. 2-chloropropene and see that the molecule is flat.

Yes, I've checked it in the meantime and saw the plane of reflectional symmetry imposed by the C=C bond. It also explains why certain allenes are chiral (despite having no asymmetric C atom). Thanks!

You are welcome to contribute an answer to your own question @Adrian - that way, future readers can benefit from this knowledge.

@Adrian, once you have 15 rep you can Answer your own question, which is encouraged by the Stack Exchange model and by this community.

Any molecule that cannot be superimposed on its mirror image is described as being chiral. If the molecule is superimposable on its mirror image the object is called achiral.
For simple alkenes, the carbon atoms of the double bond and the four atoms bonded to them all lie in the same plane, which divides the molecule into two halves, which are mirror images of each other. This plane of symmetry ensures that the mirror image forms of the molecule are superimposable. Therefore, simple alkenes like 2-chloroprop-1-ene and 1-bromo-2-chloro-2-fluoro-1-iodoethene are achiral.
Nevertheless, owing to the barrier to rotation in alkenes, 1-bromo-2-chloro-2-fluoro-1-iodoethene exists as two (achiral) stereoisomers: (E)-1-bromo-2-chloro-2-fluoro-1-iodoethene and (Z)-1-bromo-2-chloro-2-fluoro-1-iodoethene.


text:
What is the voltage dependence of the thickness of oxide coating through anodising?
In titanium anodising, solid titanium reacts with the oxygen formed at it's surface (to form titanium(IV) oxide or $\ce{TiO2}$) through the electrolysis of water. Experimentation and research has shown that the thickness of the coating is dependent on voltage and not current. The potential difference obviously plays some part in the reaction then (besides the water electrolysis) due to the fact that merely placing titanium in an environment with more oxygen gas does not affect the thickness of the naturally occurring oxide, but I am struggling to find out how (through both thought and research) the voltage affects the depth of the coating formed.
I understand that the actual oxidation reaction can be written as:
\begin{align}
\ce{Ti &-> Ti^{4+} + 4e-}\\
\ce{O2 +4e- &-> 2O^{2-}}
\end{align}
Which are added together to get:
$$\ce{Ti + O2 -> TiO2}$$
I also understand that redox reactions usually include a reduction potential or $E^0$ value. Now this may or may not be relevant but it is the only related value I can find with a voltage associated. I have also thought that as the coating of $\ce{TiO2}$ builds up, the resistance of the circuit would increase. I just cannot understand how or why the voltage affects the depth (Is this the right word, as it is converting not depositing?) of the formed oxide layer.
So with that my question is more like a request for any insight/resources that could help with my understanding of the voltage-dependency of the depth of the formed oxide layer (specifically titanium, but I'm assuming it would be relatively easy to generalise from other metals such as aluminium or niobium (which I have also searched for)).


text:
Best Fuel for Potato Gun
I have been using deodorant to propel potatoes (potato gun :D), but I want to switch over to pure butane. However, in order to do this safely I have to know how much butane is typically in deodorant (percentage). If you do know, please also include a source.
Secondly, for a potato gun, is it wiser to use butane or propane?
And could a spark ignite butane? Or is a real flame needed? Tested this already, it does.

How much butane is typically in deodorant (percentage)

Here's a link to an MSDS for a deodorant; it claims the butane range is 30-60% (they don't state if this is by volume or weight). It might be cheaper to just go to a hardware store and buy a small butane cylinder, like the ones used for outdoor cooking, then you know exactly how much butane you are using.

For a potato gun, is it wiser to use butane or propane?

My first though was butane because it gives off a bit more heat per mole (or per unit volume) when it burns, than does propane.
$$\ce{C3H8 + 5 O2 -> 3 CO2 + 4 H2O + \pu{2219 kJ/mol}}$$
$$\ce{C4H10 + $\mathrm{6.5}$~ O2 -> 4 CO2 + 5 H2O + \pu{2878 kJ/mol}}$$
However, according to "Common Fuels for Combustion Spud Guns",

The key value for comparing two fuels based on their heats of combustion is not the actual heats of combustion. Instead, the "Heat per mole Oxygen" should be used since the amount of energy in the combustion chamber is limited by the amount of oxygen present in the chamber. Fuel is added to match that amount of oxygen. As you can see from the table, there is relatively little difference between the various fuels based on their "Heat per mole Oxygen" values.

Both propane and butane give off $\pu{105 kcal/mol}$ of oxygen, so they should perform equivalently in a spud gun. The article goes on to note that, of common fuels, only hydrogen ($\pu{119 kcal/mol}$ of oxygen) and acetylene ($\pu{120 kcal/mol}$ of oxygen) produce significantly more energy under these conditions than other fuels.
Now going back to your first question, if you know the volume of your chamber, you can back-calculate the correct amount of air (oxygen) and butane.


text:
Electrons jumping from 1 water molecule to another
Electrons can be anywhere in the universe and where they are is random.
But they are still discrete particles and thus it takes energy for electrons to move.
Is it possible for an electron to jump from 1 water molecule to another causing $\ce{H2O^+}$ and $\ce{H2O^-}$ to form?

Electrons can be anywhere in the universe

Kind of: There is a non-zero probability that an electron can be at any distance from some point of origin (for instance, the nucleus of an atom), but there is some distance which corresponds to the place where we're most likely to find it (the place where the probability amplitude is highest) most of the time.

and where they are is random.

Again, sort of: They are most likely located in certain regions of space for most of the time, but since it's a probability, it's not a certainty. That said, there is a common sense element that can be applied here. Yes, an electron from a hydrogen atom in a flask in your laboratory can be out near Jupiter, but realistically, it's not. And if it's on Jupiter, it's not really that hydrogen atom's electron anymore.

Is it possible for an electron to jump from 1 water molecule to another causing $\ce{H2O}^+$ and $\ce{H2O}^−$ to form?

Yes, it is possible (but probable? I don't know, but certainly there are conditions under which such a reaction will occur).

Yes, there certainly is incredibly small chance for tunnelling of electron from one water molecule to another.


text:
Is there an A value for the carbaldehyde and vinyl groups?
A methylene ($\ce{=CH_2}$) group ist sterically less hindered than a methyl ($\ce{CH_3}$) group.
I'm wondering if the carbonyl ($\ce{C=O}$) group is sterically more or less larger than a methylene ($\ce{=CH_2}$) group.

Bond length: ($\ce{C=O}$) is shorter than ($\ce{C=C}$) [1]
Diaxial i.a.: Me//H is larger than Me//n(O) (0.9 vs. 0.4 kcal/mol; no ref. for the 2nd value)

While the shorter distance makes the carbonyl more sterically, the lone pairs are less sterically than the Hydrogens of methylene. I think this effect is larger and hence a carbonyl to be more sterically hindered than a methylene group.
The A-values gives the energetical difference of a monosubstituted cyclohexane between the conformation in which the substituent is placed axial (1,3-diaxial interactions, hence higher in energy) versus equatorial.
To reformulate above problem of the sizes I have to set following question:
Comparing the axial conformations of vinylcyclohexane and cyclohexancarbaldehyde: Which molecule is more strained?

Why I am interested in above question?
If we don't assume protonation of a ketone's oxygen, I'd like to know if I have to place the carbonyl axial or equatorial when R=Me.
For R=long alkyl chain it makes sense to place R equatorial, but I'm not sure if a methyl group is also bigger than a carbonyl.
Because I think methylene and carbonyl are almost of the same size from above arguments, and I'm knowing that methyl is bigger than methylene, it sounds logical to me to place the methyl equatorial.

http://evans.rc.fas.harvard.edu/pdf/evans95.pdf

I prefer to use the table of "A values" from http://www.chem.wisc.edu/areas/reich/handouts/a-values/a-values.htm

Comparing the axial conformations of vinylcyclohexane and
  cyclohexancarbaldehyde: Which molecule is more strained?

As you point out, "A-values" can be used to get some idea of the relative "size" of a substituent in the cyclohexane system.  More specifically, the A-value reflects both the steric size of the substituent and the length of the bond connecting the substituent to the cyclohexane ring (the longer the bond, the further away the substituent and steric interactions are therefore reduced).
The A-value is just the free energy difference between the axial and equatorial conformers. Since destabilizing steric interactions can occur in the axial conformation, the larger the substituent, less axial and more equatorial conformer will be present.

(image source)
As a consequence, the "larger" the substituent the larger the A-value. The A-value for a methyl group is 1.8, using
$$\mathrm{\Delta G = -RT*ln(K)}$$
we can solve for K and find that the axial/equatorial ratio for methylcyclohexane is ~ 5/95 (as expected, the less sterically strained equatorial conformer predominates) at room temperature.

(image source)
Here is a link to a nice table of A-values.  The A-value for a vinyl group ($\ce{-HC=CH2}$) is 1.6, while it is only 0.7 for the aldehyde group ($\ce{-CHO}$). Hence the aldehyde group is the smaller of the two. Or said differently, axial vinyl-cyclohexane would be more strained than axial cyclohexane-carbaldehyde

I've seen now your added picture. I think this conformation of this axial methyl group is not related with the A-value, because I think this "bisected conformation" (name might be wrong) is higher in energy in my opinion than only one Hydrogen pointing in the middle of the ring and the two other hydrogens outwards.

Yes this conformation is higher in energy than the other conformation you describe, but the energy difference is very small. The barrier to methyl rotation is very low, therefore the barrier separating the various methyl rotamers is also very small.  It is a fair approximation to treat the methyl as a sphere which sterically comes close enough to H4 and H5 to produce electronic repulsion.


text:
Difference between ideal & ideal-dilute solution
These two terms has been frequently used in Physical Chemistry by Atkins. He describes Raoult's Law using ideal solution while Henry's law using ideal-dilute solution.
I thought, at first they were same but Atkins in order to contrast the two laws presented graphs in  which one curve was for ideal solution & another was for ideal-dilute solution. Though he wrote two different sections for these solutions, I really couldn't contrast what the difference actually is between ideal solution & ideal-dilute solution. I googled it but to vain! Can anyone please help what he is trying to say by these two terms?

Raoult's Law is for ideal solutions (all of it: solvent and solute):

Raoult's Law only works for ideal solutions. An ideal solution is defined as one which obeys Raoult's Law. "An ideal solution shows thermodynamic mixing characteristics identical to those of ideal gas mixtures [except] ideal solutions have intermolecular interactions equal to those of the pure components." (all quoted from above link)

As stated here concerning ideal-dilute solutions:

"Many liquid mixtures deviate from the ideal behaviour predicted by Raoult's law. Ideal-dilute solutions are those for which the solvent obeys Raoult's law and the solute Henry's law. The difference in behaviour is because the solvent molecules tend to be surrounded by other solvent molecules so that they behave in an almost ideal way. However, the solute molecules tend also to be surrounded by solvent molecules so that their environment and therefore their thermodynamic behaviour is very unlike that of the pure substance." (all quoted from above link)

In summary: Ideal solutions are those which obey Raoult's Law; ideal-dilute solutions are ones where the solvent obey's Raoult's Law and the solute obeys Henry's Law.

Doesn't in ideal solution, solute molecules get surrounded by  solvent molecules? Mean to say what is actually happening when the solute is in ideal solution compared to the solute in ideal-dilute solution? Both the solutions are dilute. So, why does the difference occur?

In an ideal solution, it is assumed that there are no interactions between solvent, solute, and solvent/solute molecules/atoms.  The difference is in the solute: if it's ideal, then the system is ideal; if the solute is not ideal, then it's ideal-dilute. A perhaps silly example is: Ideal = helium solvent, neon solute (very low concentration). Ideal-dilute: helium solvent, oxygen solute (very low concentration).

I have understood , sir. In ideal reaction the interaction between solvent-solvent & solvent-solute is same but not in the case of ideal-dilute solution, right? Can't I use Raoult's Law still for solvent in ideal-dilute solution?

Correct. And yes, you can use Raoult's Law for the solvent in the ideal-solute case.


text:
What are the fundamental parameter by which lubricants are graded?
I am into hobby machining using an old beaten CNC mill in my garage. Being novice my confidence is rather low and I have genuine fear of breaking something (even ancient CNC machines like mine cost a fortune), so when I got to the part of the users manual that listed recommended lubricants for the different parts of the machine I felt out of my league completely (especially since I just basically submerged the whole machine in a bath of general purpose lubricant spray).
So I would like to know, how are lubricants usually classified/graded?
Which classification/grade matters to which kind of use?
What are the risks of using the wrong lubricants?

Lubricants are usually graded for a host of different properties, but ultimately what you are looking for is it's thermal stability, and a viscosity index related to the temperature your mills parts run at. These properties are optimized for different parts of the machine because they run at different speeds and as a consequence temperatures. I could delve into more, but I would need more specifics about the machine itself. If you wish to pursue the problem further on your own I would suggest This website as a reference. Also take a look at this cautionary article regarding accurate measurements on viscosity index as it can be a bit tricky.
The consequence of using the wrong lubricants varies based on how wrong you are... But ultimately the lubricant will thin out to the point that is does not decrease friction and or cool the parts it is lubricating if it degrades at the temperature the part runs at.


text:
Can I make my own graphene at home in my kitchen?
Graphene is cool. Can I make some? What do I need? Is it at all feasible to make it in quantity?

This led me think: is there a way to check its strength at home?

There are several methods which you can use to make graphene at home.
There is the classic sticky tape method which was first used by researchers at the University of Manchester, UK and which won them the Nobel Prize for Physics in 2010.[1] Essentially you deposit some graphite onto a surface, such as paper, using a pencil and then use sticky tape to carefully remove a layer of graphite from the paper. Then use another piece of tape to remove a thinner layer of graphite from the first piece of tape. Repeat this until you end up with a layer of graphite only a few atoms (or hopefully only one) thick.
Another method, which was discovered more recently, is to put powdered graphite in a blender along with a surfactant to stop the graphite particles clumping together.[2] You can do this at home by getting a high speed kitchen blender and adding water, washing-up liquid and powdered graphite from a pencil lead. You want to keep blending until you end up with a suspension of very fine particles in the liquid. Unfortunately, the method isn't particularly accurate. The amount of washing-up liquid required depends on the quality of the graphite, which is difficult to assess without expensive analytical equipment and all you are left with at the end is a liquid full of tiny flakes of (hopefully) graphene and a bunch of larger particles which you didn't manage to blend all the way to single atom thickness.[3]
The most interesting way to make graphene at home, which albeit requires a little more effort than some of the others, involves using water and n-heptane to create a thin layer of graphene at the interface between the two solvents.[4] You can buy heptane fairly easily from chemical suppliers and distilled water is also advised to give a better result. The most expensive part is the ultrasound bath which is required to exfoliate the graphite (try buying one second hand online). Mix the two solvents in a 1:1 ratio in a hydrophobic container, such as a polyethylene vial, and then add powdered graphite. When the mixture is placed in a ultrasound bath and sonicated for a while, the graphite exfoliates and accumulates as flakes of graphene at the heptane-water interface. If you place a hydrophilic surface like a glass slide into the interface, the graphene climbs up the slide and forms a layer attached to the slide, which can then be removed, yielding a graphene coated glass slide.
References:
[1]http://www.nobelprize.org/nobel_prizes/physics/laureates/2010/press.html
[2]http://www.nature.com/nmat/journal/v13/n6/full/nmat3944.html
[3]https://www.newscientist.com/article/dn25442-make-graphene-in-your-kitchen-with-soap-and-a-blender/
[4]http://www.researchgate.net/publication/251567939_Conductive_Thin_Films_of_Pristine_Graphene_by_Solvent_Interface_Trapping

I Googled the phrase "How to make graphene" and got this result, which contains a section on how to make it at home.
The answer is, unequivocally, yes you can. Below are excerpts from the linked page.

(1) The first DIY method is to use a lead pencil to deposit a thick layer of graphite onto a paper. Then use ordinary sticky tape to peel off a layer of graphite from the paper. Use another piece of sticky tape to remove a layer of graphite from the first sticky tape. Then, use a third piece of unused sticky tape to remove a layer from the second piece of sticky tape, an so on. Eventually, the graphite layers will get thinner and thinner, and you will end up with graphene, which is single-layer graphite in the strict sense, or bi-layer or few-layer graphite (which acts almost like graphene in certain uses). Even though this way of making graphene is only a proof-of-concept, the sticky tape method works. It takes patience and time, but it's the DIY method which the Manchester group used in 2004. And remember, they actually won a Nobel prize for their work, so there's no messing with sticky tape!
(2) Shear exfoliation in liquids starts by pouring powdered crystal in a liquid, and then using a shear mixer to separate (exfoliate) layers of material from the crystal. The liquid that is used in such processes is chosen so that the small graphene particles don't clump back together, and the result is a liquid suspension of graphene. The suspension can then be dried to obtain graphene nanoflakes, or it could be directly used in later technological processes, such as making of graphene coatings and others.
  So, how do you make graphene at home using this method? Well, you can replace the shear mixer with a kitchen blender (since shear mixing is a fancy word for high-tech blending), and instead of special liquids you can use water and dish detergent. The detergent is added so that the particles don't clump, and acts as a surfactant. A great source of graphite powder are graphite pencils. Simply powder a few pencil leads and add the powder to the detergent solution, then blend for a while. Unless you have access to special microscopes and other equipment, you will probably be unable to confirm the existence of graphene in your detergent solution, but there's a quick rule of thumb which you can use to estimate the particle size. Generally speaking, if the graphene particles are too big, they will sink to the bottom of the vessel, while if they are smaller, they will float to the top. Nanoparticles are so small that they will be suspended in mid-water, and this is what you're aiming for if you want to make graphene. After you make enough graphene particles, you might want to filter the suspension and leave it to dry.
(3) If you pour a non-water-soluble liquid into water it will either float on top of the water, or the water will float on top of it, depending on whether the added liquid is heavier (more dense) or lighter (less dense) than water. A common example is oil floating on top of water. One group of scientists discovered that you can make graphene by pouring two solvents which do not mix together, such as heptane and water, into a glass and adding finely ground graphite powder and putting it in a sonication bath. You can buy heptane in most chemical supply stores, water is readily available (it's a good idea to use distilled water), and you can easily obtain graphite sticks in art shops. An ultrasound bath is needed, but these are not uncommon on ebay and the like. The resulting graphene is one to four layers thick, and is chemically stable. It can be extracted and deposited on a clean glass plate, or any other substrate. The idea is to mix the finely ground graphite with water and heptane (1:1 ratio of water and heptane) and sonicate it for a while. What happens next is that graphite flakes exfoliate at the interface between water and heptane and graphene actually starts "climbing" up the glass walls of the vial. If a hydrophilic substrate, such as a glass slide, is introduced through the interface, a thin film of graphene will climb the walls of the slide as well. The glass slide can then be extracted and after drying, you'll be left with graphene-coated glass. If you use a polyethylene vial, the resulting graphene will not climb the walls of the vial, but you can still extract it with a slide made of glass or other hydrophilic material. Graphene is deposited as a thin film on both sides of the glass plate, and its slight light-absorbing properties can be seen by looking through it (Graphene absorbs about 2.3% of visible light). The explanation of the process is that graphene self-reassembles from the small exfoliated flakes on the interface between water and heptane because the surface tension of graphene (54.8 mN/m) is nearly right between the surface tension of water (72.9 mN/m) and heptane (20.1mN/m). The thickness of the graphene film is limited by capillary forces on the interface and the excess graphene will simply fall to the bottom of the vial. The formation of thick aggregates is suppressed by the diffusion and the energy requirement needed to form a new layer. This is possibly the easiest way to make DIY graphene at home as an arbitrarily large sheet. Other methods only create graphene flakes, but these flakes could possibly be used instead of graphite flakes in this method as a refinement stage.

Another great reference is a video created by Dr. Derek Muller for his YouTube channel Veritasium. Check it out here.

@MatthewFox If you decide to describe the procedure outlined in the video in your answer, the post can easily be undeleated.


text:
Identification of solutions by solubility when mixed with each other
$\ce{Pb^{2+} (aq), Cl- (aq), Ag+ (aq), (SO4)^{2-} (aq)}$.  You number the four bottles, 1,2,3,4and while numbering them, you notice that bottles 1 and 3 have the word "cation" on their lids while bottles 2 and 4 have "anion" on their lids. You decide to test them by mixing them together in pairs, but you have only three empty beakers. When the solutions are mixed in pairs.
Three combination give white precipitates;
Bottle 1 + bottle 2. bottle 1 + bottle 4, bottle 2 + bottle 3. ive done each combination and theyre all insoluble? but my chart says there has to be 1 combination thats soluble and 3 that arent ?

Since It is a Homework Question I would Like to add a hint instead of giving full solution.
You might not be able to Watch the difference Carefully.
For Your information
$\ce{PbCl2}$ = insoluble
$\ce{PbSO4}$ = insoluble
$\ce{AgCl}$ = insoluble
$\ce{Ag2SO4}$ = slightly soluble
Try doing it again and watch the difference carefully afterall now you know which one is soluble.
Hope it helps.

i wrote that as the one that was soluble, (ag2so4) but then after looking through my solubility chart again got confused, thanks so much !


text:
Silicone vs Alcohol
The type of silicone sealant that you would buy at your local hardware store. Would it break down when exposed to 90% mixture of isopropyl alcohol?

According to this rather extensive chart, silicone can be exposed to isopropyl alcohol with no breakdown expected.

By breaking down, do you mean the destruction of the sealants ability to adhere or the destruction of the silicone polymer. Polar solvents will permeate and swell the polymer, separating the polymer from its filler, destroying the adhesive functionally but the polymer itself will remain intact. A strong base, such as NaOH will destroy the polymer, eating into and breaking the polymer chains, much the same way soda glass is much lower melting than quartz. Lower MW alcohols or water won't affect the sealant but higher alcohols or alkanes, acetates, ethers, aromatics, etc. will.

This is somewhat on the verge of a clarification-asking comment … but at the same time I realise the question itself was not great. Not sure what to do with this …


text:
Can ionic compounds can ever "dissolve" as molecular compounds do?
Here's an example. Say you take a 25˚C solution of NaCl, and lower the temperature suddenly. The equilibrium constant (solubility product) is immediately decreased, so ions start to join together into ionic compounds that float freely in the solvent before settling to the bottom. (In some cases they may be visible as precipitate.)

Can we refer to these free-floating ionic compounds as "dissolved" solute?
Does "XY(aq)" (the aqueous solution of an ionic compound) seem like a good description of this state?

Can we refer to these free-floating ionic compounds as "dissolved" solute?

No. Dissolved implies that the ions are solvated. If the ions are in a solid ionic lattice then they are not solvated by the solvent and so would not be described as part of the dissolved solute.

Does "XY(aq)" (the aqueous solution of an ionic compound) seem like a good description of this state?

Again no. $\ce{XY(aq)}$, in the context of ionic compounds, refers to $\ce{X^{n+}}$ and $\ce{Y^{n-}}$ ions solvated by water molecules. The dissociation is generally regarded as implicit. For example, $\ce{NaCl(aq)}$ implies an aqueous solution containing $\ce{Na+}$ and $\ce{Cl-}$ ions. A better description of the state you are thinking of would be $\ce{XY(s)}$, using state symbols, or even better you could describe it as a suspension.
From the IUPAC Gold Book:

Suspension:
  A liquid in which solid particles are dispersed.

This accurately describes your system.


text:
Calculate rate of reaction of a metal and an acid

At $10~^{\circ}\mathrm{C}$, a small piece of $\ce{Zn}$ reacts with $\mathrm{2.0~M}$ $\ce{HCl}$ to produce $\pu{15.0 mL}$ of $\ce{H2}$ gas in $\pu{25 s}$. What is the rate of the reaction?

My effort:
I know that the $\mathrm{2.0~M~} \ce{HCl}$ concentration needs to be used and started the following calculation:
$$\mathrm{2.0~M} \times \frac{1}{2} \times \frac{1}{\mathrm{25~s \times 0.015~ mL}} = \mathrm{0.00060~mol}~ \ce{H2}$$
$\mathrm{mol ~\ce{H2}} = \mathrm{mol~Zn} = \mathrm{0.0006~mol}$
However, the given answer is 0.060 mol of Zn.

They are asking for a rate, so I am unsure about the given answer - I would expect it to be a function of time. Can you check the problem statement to be sure the information given is correct?

one hint - So much much acid, so little hydrogen gas.

The problem has two answers that can be solved depending on the interpretation of a "small piece of zinc". Assuming that there is excess zinc (like a zinc disc), and the overall reaction is limited not by contact area, then one answer is obtained. If you assume that the zinc is a sphere and (a) the contact area is decreasing and (b) the zinc is totally consumes at 25 seconds then you get another answer. If you assume that the contact area changes and that some zinc is left at 25 seconds then the problem can't be solved with the information provided. // I over think dumb problems like this.

Assuming that there is excess zinc (like a zinc disc), and the overall reaction is limited not by contact area is the easiest way to solve the problem. Pick that one and state your assumptions in the solution.

The question as stated is asking for a rate, but the given answer is a mass of Zn. Voting to close as unclear, due to this inconsistency.


text:
How is Persian blue rock salt made?
How is Persian blue rock salt made and what does it contain that makes it look blue?

Firstly, there are two questions here which should be separated. I would regard the first one as off topic for chemistry.Se and better answered on earthscience.SE. The second one is probably on topic for both sites. I am trying to advise you so that you can get the best answer to your question.

To the matter if this is off-topic or not, I'll let the community decide. I personally have no idea what persian blue rock salt is, if it is a common mineral or a trivial name for a certain compound or something completely different. Finding that out is in my opinion not related to chemistry. If you research what this compound is, the colour aspect almost certainly is on topic on this site. The first part might be a fit on earthscience.se, but I am not certain about their scope.

While I googled a bit, I found this. You can check this. The color is not due to some mineral but rather an optical illusion!

Persian blue salt is formed by evaporating sea water, as are most if not all salt deposits. In the geosciences, we call salt deposits like this evaporites. Persian blue salt is sodium chloride (the mineral name is halite).
According to this description:

Blue salt is one of the rarest salts known to mankind, and is extracted from a salt mine in the northern province of Semnan in Persia.
The intriguing blue color occurs during the forming of the salt's crystalline structure, as intense pressure is exerted on the salt deposits. The individual crystals fracture the light in an unusual way and the resulting blue (which is caused by an optical illusion), becomes visible.

I'm quite surprised no one stated the obvious: formation of F-centers in the crystal lattice of $\ce{NaCl}$. Despite the fact that in nature blue rock salt is rare, it's quite easy to synthesize some in the lab by heating metallic sodium and sodium chloride in an enclosed vessel.
$\ce{Na}$ vapor is capable of dissolving in the crystal lattice formed by $\ce{Na+}$ and $\ce{Cl-}$ ions occupying vacancies around defects and providing extra $\ce{Na+ + e-}$ pairs. Liberated electrons occupy positions of $\ce{Cl-}$ anions forming F-centers. Once the halite is cooled down, F-centers tend to associate in crystal space, leading to very intensive color.
In nature blue halite is usually formed by radiation accompanying radioactive transformations.
Interestingly enough, once dissolved, blue rock salt often sold for over 1 \$/g forms transparent colorless solution with the composition indistinguishable from the table salt   costing 1 \$/kg.
There are also other factors, mostly impurities, that can lead to rock salt's crystals of virtually any color. Here is a part of the Table 1 from the review [1] regarding the shades of blue color:

Table 1. Common colors in rock salt.
$$
\begin{array}{ll}
\hline
\text{Color} & \text{Impurity} \\
\hline
\ldots & \ldots \\
\text{blue} & \text{lattice defects/colour centers} \\
  & \text{admixture of}~\ce{KBr + Au} \\
  & \text{in sylvinite zone} \\
  & \text{particles}~\pu{90-110 nm} \\
  & \ce{Pb, Cu, Ag, Au}~\text{as pigments} \\
  & \text{by adding metallic}~\ce{Na} \\
  & \text{in}~\ce{Na}\text{-vapor} \\
  & \ce{Cl-}~\text{removal by ionization} \\
  & \text{rapid growth} \\
\text{violet} & \text{slow groth} \\
  & \text{particles}~\pu{80-90 nm} \\
  & +0.5\%~\ce{KCl}~\text{or}~\ce{RbCl} \\
  & \text{exposure to radioactivity} \\
\text{purple} & \text{by cathode rays} \\
  & \text{in sylvinite zone} \\
\text{black to dark blue} & \ce{Na}\text{-vapour + pressure} \\
  & \text{large}~\ce{Fe2O3}~\text{platelets} \\
  & \text{by cathode rays} \\
\ldots & \ldots \\
\hline
\end{array}
$$

References

Sonnenfeld, P. The Color of Rock Salt—A Review. Sedimentary Geology 1995, 94 (3), 267–276. https://doi.org/10.1016/0037-0738(94)00093-A.