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Where to buy/How to build a Devanathan-Stachurski cell?
I'm preparing to do some hydrogen permeation tests to calculate the diffusivity of some iron samples. I would like to do this using the Devanathan-Stachurski cell, which consists of two cells, one where the hydrogen is charged and the other where it is oxidized (each containing an electrolyte solution), with the sample in the middle. The current produced by the diffusing hydrogen can be used to calculate diffusivity.
Does anyone have experience using this cell and know where to purchase one? So far I've only been able to find one company that sells it, but I would like to know about more options. If anyone has built it from scratch, can you provide some specifications on dimensions and materials used and how it worked for you?

I have searched this item, and I have found on: https://www.zimmerpeacocktech.com/products/accessories/devanathan-cell/ I have found with professor Cecílio SADAO Fugivara (UNESP) (email: sadao.fugivara@unesp.br) in Brazil too, that makes it on a considerable low price than the first option. About 400 dolars.

I found this website. They make Devanathan-Stachurski cells.
https://www.landtinst.com/product/devanathan-stachurski-cell-h-cell/ The price is around USD600.00. Looks like that they could do a lot of customization. Go check with it.

redox.me has so far the best Devanathan-Stachurski Permeation Cell available on the market.

I would recommend you to do a research to find scientific articles on using DS cells. In this case if you have any specific questions you can try to contact authors directly, hopefully they will reply and be able to help you.
You can start from this article (Mike J. Danielson, Corrosion Science 2002 44 (4), pp. 829-840) http://www.sciencedirect.com/science/article/pii/S0010938X01001032.
The author made his own DS cells as described follow

Welcome on the Chemistry SE! Note, we prefer text copy-paste against screenshots. Screenshots are bad on multiple reasons, not only they are ugly but they are also far lesser indexed (if et al) and are practically not searchable/fixable/editable.

I worked with Tyler, and something to add that I've recently found is doing a search for "H-type ion exchange membrane" cells or similar. An alternative would be to get 4" diameter recycled teflon rods from McMaster and machine them to suit your purposes as the main housing of each cell. A spherical ball joint clamp or threaded nylon rods could be used to secure the two cells together. The taller you make the cells and liquid height, the tighter the seal will need to be to prevent leaking.
Update 2020-05-28
I've been able to successfully 3D print and operate Devanathan cells. Mainly the chemical compatibility needs to be checked (e.g. by putting a cube of the 3D-printed material in the electrolyte and testing weight change after a day or so). For example, FormLab's Clear resin (e.g. Clear V2) is compatible with NaOH (I use 0.1 M NaOH) and seems to work just fine during the electrochemical tests. 3M has some teflon 3D printing services (not sure about cost), and other services like Xometry (not affiliated with either company) have some inexpensive chemically compatible materials that can be printed with (e.g. Ultem).

(I wrote the question) - For those who still have this problem, I've found that the best way to prepare for this experiment is to talk with a glass blower. Most people who have done the experiment are vague in their details on how they made the cell, and the paper referenced above is the most detailed I have found. Many of the authors are also hard to get ahold of. I've found one supplier of the entire set-up, but it is a little pricey and would still have considerable lead-time for them to make (However if you'd like their info you can ask me separately). The more affordable method is to find a glassblower who can make it. They can make a hole/flange to turn a normal cell into an H-cell, and you need 2 of these for the experiment (or you can find 2 H-cells that work for your application if you can). Then you need O-rings to hold/seal your specimen in between and a way to clamp the flanges together. You also need to make sure you have enough entry ports for your electrodes and sparging if you choose to do that.
Short answer: This is not something that companies have in stock, and whether you make it yourself or hire someone else to make it, it requires considerable time and research to make sure you have all the details right.


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Why is more than one spectral line created on the de ecxitiation of a hydrogen atom
When an electron is de excited from a higher energy level to a lower one only one lone fits the bill then why is it that so many lines are created example n=4 to n=1

Not sure what you mean here. Can you provide an example?

Potentially related, depending on what he's asking - https://chemistry.stackexchange.com/questions/5513/why-are-there-multiple-lines-in-the-hydrogen-line-spectrum?rq=1

If you're talking about what I think you are (and if you're not I will delete this answer) but what you mean by "one lone fits the bill" is only 1 energy value represents the drop from n=4 to n=1. However, Hydrogen's electron has many jumps that it can make. It can drop from n=4 to n=3 or n=2 or n=1. It doesn't even have to start at n=4 it can start at n=3 and drop to n=2 and n=1, and it can start at n=2 and drop to n=1. While it is true each movement of an electron (from one energy level to another) will only produce a certain amount of energy for that jump, there are many jumps Hydrogen can make, resulting in the multiple lines you see on an absorption/emission spectrum.
Also note this question may also answer your question

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Where can I find ranges of isotope abundances?
I'd like to know how much stable isotopes like $\ce{^{13}C}$, $\ce{^2H}$, $\ce{^{17}O}$ and $\ce{^{18}O}$ deviate from their natural abundances in the earth's crust among all possible compounds we may come across, especially organic molecules. I'd like to know a rough minimum and maximum percentage of such isotopes that a given compound is likely to contain, and if possible how likely it is to contain such a percentage. It might be really good to have a histogram showing how common compounds with different percentages of such isotopes are. Does anyone know where I could find such data?

I am aware of this range diagram about $\ce{^{81}Br}$ that appeared in the recurrent (open access) IUPAC recommendations of atomic weights in Pure and Applied Chemistry:

(source: doi 10.1351/PAC-REP-13-03-02)
which cites data of a publication in Environmental Chemistry extending a little to $\ce{^{37}Cl}$ (doi 10.1071/EN10090, open access), too.
Even if the elder (2000) IUPAC compilation in the same journal includes a graphical representation of the range in occurrence of $\ce{^2H}$ as known at this time:

(source: doi 10.1351/pac200375060683, open access)
it may be worth to read more in detail, as -- in contrast to the 2011 edition (32 pages) -- the 2000 edition (120 pages) includes a dedicated section "Isotope fractionation from natural processes" and "Element-by-element review of the standard atomic weights" for a detailed view.

Otherwise, if you were "just" interested in an overall overview of isotopes, their masses and their percentage of occurrence, NIST has a searchable dedicated database here.

After looking through the sources from the gratefully acknowledged Buttonwood's answer, I also came across CIAAW, which seems to have some of the best available data for the natural isotope abundance ranges of elements like $\ce{C}$, $\ce{H}$, $\ce{N}$, $\ce{O}$, and $\ce{Cl}$. It is based on IUPAC Technical Report 2013 [1].
References

Meija, J.; Coplen, T. B.; Berglund, M.; Brand, W. A.; De, B. P.; Gröning, M.; Holden, N. E.; Irrgeher, J.; Loss, R. D.; Walczyk, T.; et al. Isotopic Compositions of the Elements 2013 (IUPAC Technical Report). Pure and Applied Chemistry 2016, 88 (3), 293–306. https://doi.org/10.1515/pac-2015-0503.

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Propylene Glycol Degradation due to Sunlight
I had a translucent container of propylene glycol exposed to sunlight.  When I went to use it, I noticed it has a strong "acetone" like odor.  Someone said it was a photo-oxidation reaction of the glycol into ketones or aldehydes.  What happened and will the odor get worse even though it is is now removed from sunlight?  Do I need to worry about these ketones or aldehydes in the glycol?

The thermal degradation of propylene glycol and its uses in e-cigarettes has been a discussed in a previous question here on Chem.SE.

Upon complete oxidation and degradation, propylene glycol will fully
  decompose to carbon dioxide and oxygen. However, if there is partial
  oxidation, you will get some intermediate species - formaldehyde or
  acetaldehyde.(1st answer)
Propylene glycol does not decompose on heating, it simply boils. In
  theory, under certain conditions (pressure/encapsulation/catalysts) it
  could dehydrate to either acetone or Propionaldehyde, probably the
  former.(2nd answer)

Also, this link shows the decomposition products of propylene glycol which are all acid based:

The main degradation product of propylene glycol was lactic acid (2-hydroxypropionic
  acid), with formic, acetic, and oxalic acids as minor products.

So, mainly degradation of propylene glycol in sunlight would yield a bunch of acids and small amounts of acetone. Worrying about the odors is a personal medical question and is off-topic in Chem.SE.


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Why is there an increasing solubility in water for chlorides, chlorates, perchlorates in that order?
According to Wikipedia $\pu{100 mL}$ of water dissolve at $\pu{25 ^\circ{}C}$ about $\pu{35 g}$ of $\ce{NaCl}$ (1), but about $\pu{79 g}$ of $\ce{NaClO3}$ (2), and around $\pu{210 g}$ for $\ce{NaClO4}$ (3).
This looks to me roughly like its doubling solubility for every single higher level of oxidation.
Earlier I was under the impression that the % of $\ce{Cl}$ or % of $\ce{Na}$ in the water would still be the same, and the amount of additional oxygen is what could be causing the increase in solubility.... ie. now each molecule has 3 oxygen atoms as opposed to none in case of $\ce{NaCl}$. But this doesn't hold true.
I would also like to know what is the effect of multiple solutes sharing same atoms on each other.  For example does dissolving some chlorate cause the solvent to hold more chloride ?
As you may have guess Chemistry isn't my strong suite, I know basic chemistry, but now when I am looking at it again.. the whole thing has changed atleast w.r.t what I was taught during my school days. (right from how we determine the number of electrons)

There's no simple numerical dependence for solubility of different salts. That's why solubility tables are useful.

Science is all about discovering hidden regularities, patterns, and laws. In that sense your bold generalization is good, but will it hold if we check just one more data point? Let's see: $\ce{KCl}$ has solubility of 25g/100ml, $\ce{KClO3}$ has 8, and $\ce{KClO4}$ has 1.5. Er, well...
What we have here is an interplay between solvatation energy and lattice energy, the latter being determined by the crystal structure, which is a tricky thing, and hence so is solubility. Don't expect it to be predictable from qualitative considerations. You don't know it until you measure it. Chemistry is an experimental science, after all.

The solubility of a salt depends on 2 factors:

Force of attraction between H2O molecules and the ions of the solid. This force tends to bring ions into solution. If this is the
  predominant factor, then the compound may be highly soluble in water.
Force of attraction between oppositely charged ions.This force tends to keep the ions in the solid state. When it is a major factor,
  then water solubility may be very low.

According to point 2, if the force of attraction between the oppositely charged ions is more, more energy is needed to break them and hence lattice enthalpy of that salt is more.
First, we compare sodium chloride and sodium chlorate. Force between sodium cation and chloride ion is way stronger than that with chlorate ion. So lattice enthalpy of sodium chlorate is less than sodium chloride. So, sodium chlorate is more soluble in water.
Force between sodium cation and perchlorate anion is way less than that with chlorate ion.
So lattice enthalpy of sodium perchlorate is less than sodium chlorate. So, sodium perchlorate is way more soluble in water than sodium chlorate.
Arranging the above points, we get solubility order:
Sodium perchlorate > sodium chlorate > sodium chloride.


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fatty acid vs polyethylene
Both fatty acid and polyethylene are made (mostly) of carbons with hydrogens with a very similar configuration, yet, their properties are very different. What am I missing?

Not that similar structure and not that different properties.

Would you care to elaborate?

Well, you should elaborate what you consider different.

For start, from the top of my head: state in room temp, digestion by body, abundance in nature

Well, there are lots of solid ones and  "abundance in nature" isn't exactly property. Short answer may be that fatty acids have chains short enough that presence of COOH group is still very important.

Still, not as solid as polyethylene, and doesn't explain why the body can't digest polyethylene

@KobiT try this: https://en.wikipedia.org/wiki/Fatty_acid_metabolism

@KobiT Fatty acid metabolism depends on the $\ce{-COOH}$ functional groups that makes a fatty acid, well an acid. Polyethylene is a polymerized ethylene units, and do not contain $\ce{-COOH}$ groups, and do not respond to biological metabolism.

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

I agree with @Mithoron's earlier comment that even if these two types of compounds are somewhat similar in structure (e.g., they both consist of long aliphatic chains) they are not exactly alike. If you perhaps reformulate your question it may allow us to answer it more precisely. I will make an attempt below.
A fatty acid contains a carboxylic acid group attached to an aliphatic chain. This chain can be either saturated or unsaturated and still be referred as a fatty acid. Most naturally occurring fatty acids can have somewhere between 4 up to 28 carbons in their chain.

Polyethylene (also named polyethene or poly(methylene) is a polymer with a variation on the chemical formula $\ce{(C2H4)_n}$ lacking a carboxyl acid group. Some important similarities between the two compounds include that the melting profile is dependent on the chain lengths. As there are greater intermolecular forces between long chains compared with smaller simple molecules.

Polyethylene is therefore a wholly non-polar compounds, while the overall polarity of fatty acids depends on which functional group is the most dominant, the head carboxyl acid group or the alkyl chain group.
In the table below I have summarized the melting temperatures of some fatty acids and polyethylenes (which are equivalent to small straight-chain alkanes) to show the difference. Observe how the melting point changes drastically by the introduction of a double bond in the saturated versus unsaturated fatty acids of the same chain length.

A characteristic property of very long chained polyethylenes (with molecular weights numbering in the millions) is that they are thermoplastics, meaning they can be heated to their melting point, cooled, and reheated again without significant degradation. Fatty acids can be derivatized to other compounds, including polymers, to gain similar properties, but are not thermoplastic in their underivatized form.
I hope this extended answer somewhat helps you to find and answer to your question.

I disagree with the other answers that polarity is the most important.  I think the degree of polymerization is the most important difference in common fatty acids and common polyethyelenes.
The DP of commercial polyethylenes is much higher than the fatty acids you are considering, north of 100,000 in some cases.  For example, stearic acid is a C16 molecule.  A polyethylene with the same number of carbons would be hexadecane and would have a DP of 8, only about 10,000-fold less than commercial polyethylene.  Hexadecane's physical properties are reasonably similar to stearic acid in my book.
Or conversely, the fatty acid $\ce{CH3(CH2)100000COOH}$ probably has properties that are indistinguishable from the corresponding polyethylene.

The main difference is the polarity. Alcanes and polyethylene are nearly not polar, because both $\ce{C}$ and $\ce{H}$ have nearly the same electronegativity. Fatty acids contains the group $\ce{-COOH}$. As the oxygen atom electronegativity is rather different from Carbon or Hydrogen, the molecules of fatty acid are polar and behave quite differently form the alcanes, whatever the length of the chain.

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Is electric current always necessary to make cathodic protection possible?
Can a cathodic protection system still work if there is no current between the anode and cathode?

Yes I think it can be done but problem is that the process would become slow and  less effective so we need a current also to overcome driving forces.

By definition, cathodic protection relies on the electric current to prevent (or redeposit) the protected metal. For example, galvanized iron is protected by the dissolution of zinc and the current that generates.

To clarify the question: some cathodic protection systems work by impressing a specific (small) current using a power source connected to an inert electrode. Other cathodic systems use a sacrificial anode which (in theory) need not dissolve and produce a current, but just develops a protective voltage that prevents corrosion at the protected metal. Of course, the protected metal always corrodes a bit, and so the sacrificial anode corrodes a bit to prevent major corrosion.


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Can I anodize an aluminum coating on steel to have better corrosion inhibition for the steel?
I am planning to coat aluminum on steel and I want to give the steel its maximum corrosion protection. Is anodizing the coating necessary?

In my opinion I think it's not necessary, since aluminum will kinda anodize itself when exposed to air. But that's ny opinion and don't hold by it.

Would attaching the traditional aluminum anodes to steel work ? Certainly it  would be easier to do.

Does this answer your question? Can I claim that a hot dip Aluminum coating in steel provides cathodic protection to a substrate since Al can act as sacrificial anode?

The aluminum oxide layer formed by anodizing will have substantially higher corrosion and abrasion resistance than non-anodized aluminum.  For most applications, simple Type II sulfuric acid anodizing is sufficient.  For maximum protection, Type III hard anodizing will yield a much thicker coating with substantially higher corrosion and abrasion resistance. If you require a thinner film, the older Type I chromic acid anodizing process may be used.
If you plan to paint or dye the part, you will most likely need to stick to Type II sulfuric acid anodizing or Type I chromic acid anodizing.  Type III hard anodizing, depending upon the alloy, will result in a dark grey to an almost black finish, so dyes will not show. Paint adhesion is also an issue with Type III.
See MIL-A-8625 for more complete information about the various types of aluminum anodizing.


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Activation energy of chlorine trifluoride and hydrazine
Chlorine trifluoride was once tested in a rocket engine as an oxidizer for hydrazine. I am looking to simulate in software a rocket engine using this propellent combination. Activation energy is one of the inputs to the simulation software.
What is the activation energy of a chlorine trifluoride and hydrazine reaction?
$\ce{3 N2H4 + 4 ClF3 -> 3 N2 + 12 HF + 2Cl2}$
Is there a online chemistry source or book which might contain this value? If it has not been measured is there a way to calculate this value based on the properties of the chemicals involved? How to get the value is less important than a number I can enter into a text file.
Searching for papers on the subject mostly results in unrelated chemistry homework examples. The closest thing thing I found to an answer was looking up chemicals in the National Institute of Standards and Technology (NIST) Chemical Reactions Database. The database includes data for hydrazine reactions and chlorine trifluoride reactions separately, but not with each other.
$\ce{ClF3 -> *F + ClF2}$ reaction has $E_a = 118 \frac{kJ}{mol}$. Is it reasonable to assume this might be the first step if it is a radical reaction?

Welcome to Chemistry.SE! Take the tour to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution, otherwise it might get closed.

Do you really think someone was crazy enough to measure this?

In case it is a high order/concerted reaction the most realistic way to get it is from the literature. In case it is not it could be a radical reaction then the rate determining step most likely is a homolytic Cl-F cleavage. That one might be calculated with some effort to even realistic accuracy.

Chlorine trifluoride is potentially the most dangerous substance ever. Please do not use it at home, or without EXTREME safety, especially when simulating a rocket engine. See the chemical spill many years ago, where 50 tons of this substance was dropped, and ate through more than a meter of concrete.

@Equinox And on that note, nor is hydrazine any good.


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Polarising Power and Covalent Character

Higher the positive oxidation state of the central atom, more will be its polarising power which, in turn, increases the covalent character of bond formed between the central atom and the other atom.

This seems counter-intuitive to me. When the polarising power of the central atom increases, shouldn't the bond be more ionic instead of covalent? If not, can you tell me where I am wrong?

Have a look at the Wikipedia entry for Fajan's rule: "https://en.m.wikipedia.org/wiki/Fajans'_rules "

I believe it is about perspective. Note that "covalent molecule with ionic character" and "ionic compound with covalent character" are different.
If we are looking at a covalent molecule and assessing its ionic character, we cannot talk about polarising power, because in a covalent molecule, there would be no cation and anion to polarise in the first place. Thus, we would go by the rule: the bigger the electronegativity difference, the higher the ionic character and therefore, the more polarised the bond.
If we are looking at a ionic compound and assessing its covalent character, then we will go by the rule: the more polarising the central atom, the higher the covalent character.
Now, from the way you phrased your question, I believe that you are conflating the concepts "big electronegativity difference" and "polarising power". They are distinct concepts and should be thought of separately.
Although bond polarity is associated with electronegativity difference, "polarising power" is about the polarisation of the electron cloud of the anion by the cation and not the polarisation of the bond. This is also why "polarising power" is a concept associated with ionic compounds with covalent character.
Hope this clarifies your doubt! If you are still unclear and require some examples for clearer understanding, please leave a comment on my post.

Fajans' Rules help to explain the covalent character of ionic compounds. According to these rules, as charge on ion increases, the covalent character of the bond increases.
For example, between $\ce{NaF, MgF_2, AlF_3}$ and $\ce{SiF_4}$, order of covalent nature of bond would be as follows:
$\ce{NaF < MgF_2< AlF_3 < SiF_4}$

Kindly note, we are talking about polarising power of 'cation'. This means there already exists an ionic bond and electron transfer has already taken place and the ions are formed. Now that cation again attracts the electron cloud of anion.
So more polarising power means that the among ions of ionic bond, the cation is attracting electron cloud of anion and hence inducing a 'covalent character'.

No one mentioned anything about a cation in the question or any of the answers, as far  as I can see.


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Solubility of potassium nitrate and potassium chloride
Why more water is needed to dissolve potassium chloride than potassium nitrate of equal amount? Is it to do with the reaction that occurs when the compounds dissolve or the properties of the compounds?
This is for an experiment on the effect of temperature on solubility in various chemicals.

Welcome to Chem SE. Take a [tour] of this site. Visit the [help] for more info.

Recently, I answered a similar type of question. The reasoning for that question is also applicable for this answer also.
Electrostatic force between potassium cation and chloride ion is way stronger than that with nitrate ion. So, more energy is needed to break the  bond between potassium chloride than potassium nitrate. Hence, lattice enthalpy is more in case of potassium chloride and thus more water is needed to dissociate the bonds. So, potassium chloride is less soluble in water.
You can compare the solubilities of potassium nitrate and potassium chloride by observing the graph taken from this site:

You can observe that solubility of both potassium chloride and potassium nitrate is same at approximately 23℃ but then as temperature is increased, solubility of potassium chloride increases steadily but solubility of potassium nitrate increases drastically.
So solubility order:
Potassium nitrate > potassium chloride


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Is propanone responsible for photochemical smog?
Many volatile organic compounds (VOC) are responsible for photochemical smog. Propanone is a VOC. It's produced in a great amount for industrial purposes. So, propanone can be a potential contributor to photochemical smog.
I have 2 questions:

Is propanone really responsible for photochemical smog?
If it is, then what kind of photochemical reactions happen with propanone? (Please describe the mechanism also)

Most of the information is taken from Wikipedia or is mentioned therein

First some definations.

What is a VOC?

Volatile organic compounds (VOCs) are organic chemicals that have a high vapor pressure at ordinary room temperature. Their high vapor pressure results from a low boiling point, which causes large numbers of molecules to evaporate or sublimate from the liquid or solid form of the compound and enter the surrounding air, a trait known as volatility. For example, formaldehyde, which evaporates from paint and releases from materials has a boiling point of only –19 °C (–2 °F).

What is a photochemical smog?

Noxious haze of minute chemical droplets and ozone, resulting from sunlight induced chemical reaction between industrial and automobile pollutants (mainly nitrogen dioxide, hydrocarbons and aldehydes).

Indeed, propanone(acetone) is a VOC. Majority of acetone is released from industries origins. It indeed evaporates rapidly, from water and soil and has a 22-day half-life and degraded by UV light via photolysis primarily into methane and ethane or other alkyl, alkoxy or acyl radical.
As said, the main components of photochemical smog is nitrogen dioxide, hydrocarbons and aldehydes. The reaction include breakdown of hydrocarbons and aldehydes into alkyl and alkoxy radicals. These radicals react with nitrogen dioxide to form PAN(peroxyacetyl nitrate) which is the main cause of photochemical smog. The reaction mechanism is given here:

Another mechanism is given here:

What is observed is that the active component is acyl radical which is necessary in the PAN formation. Both aldehyde and acetone necessitate in acyl radical through their decomposition. From here:

The primary photochemical decomposition of aldehydes produces organic
  free radicals. For ex: acetaldehyde decomposes into methyl and formyl
  radicals.
$$\ce{CH3CHO + hv → •CH3 + HCO•}$$
Similarly, acetone decomposes into methyl and acetyl radicals.
$$\ce{(CH3)2CO + hv → •CH3 + CH3CO•}$$

But acetone contribution to photochemical smog is negligible as acetone is quite stable to irradiation and photolysis. From here:

[...]Acetone has shown little reactivity to irradiation. Its
  contribution to produce significant amount of PAN or ozone in ambient
  air is negligible.


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Why can higher voltages lead to less efficient electroplating?
I was reading up on electroplating and I came across some things which really confused me:
Faraday's First Law says that the amount of metal plated onto the cathode is proportionate to the current, so it should be that a higher current= more metal plated onto the cathode (I assume).
And from Ohm's Law, a higher current= higher voltage so (by syllogistic reasoning) a higher voltage should give you more metal plated onto the cathode.
However this isn't the case, because I did an experiment recently where the highest voltage didn't produce the highest amount of metal plated onto the cathode. I did some reading on why, and I read (on the internet) that it's because when you increase the current by a lot, the ions formed at the anode can't keep up with the rate of electron flow and electrons just end up combining with other things in solution and no metal forms.
I have some issues with this though; I used a solution with ions already in it, so shouldn't there be no problem for me? Even if the ions cannot be formed at the anode fast enough, I have ions already in solution, so I thought it would be fine. Plus, the experiment was run for only 25 minutes, so I doubt they would've all been used up or anything.
Also, even if it were true that the current was just too high for the ions to keep up; wouldn't the amount of metal formed be at least equal to or greater than the amount formed for lower voltages? It's just that you're forming more metal in less time, ie it gets faster (I think); so you should get the same amount at say 5 V that you do that 3 V, even if towards the end of the experiment in 5 V it gets inefficient and you stop plating. Because the amount of ions or anything doesn't change.
Sorry for the huge amount of writing, I just thought it'd be easier if I explained my thought process. I'd appreciate it a lot if anyone could help me! Thanks so much. :)

I added some clarification to my answer, is it good enough for you?

When you apply the required voltage for the desired cell, then your desired product is being produced, but if you provide an overpotential then you forced the other reaction to also occur(the undesired one).
For eg. If in a reaction two species can be oxidized then, of the required potential is applied then you will get the required product but if an over potential is applied then the other species will also get oxidized and you will have two oxidation products.

There are two types of control of what reaction may occur: thermodynamic control and kinetic control.
Thermodynamic control dominates in near-equilibrium conditions, which in case of electroplating means small voltages and relatively high concentrations.
When applying high voltage, metal ions in near-electrode space are quickly exhausted and we move from equilibrium conditions to kinetic conditions when the first reaction that may happen occurs. In water this is oxidation/reduction of water.
The main criteria for choice of the voltage in electroplating is so that diffusion of metal ions from solution happened fast enough, but electrolysis of water and/or oxidation of anions didn't happen. This means low voltages, preferably below 1 V when possible. Electrolysis of water may happen if voltage applied is above 1.23 eV, so exceeding this value is allowed only if the procedure and experimental setup is proven to work.
Why electrolysis of water in this setup bad? Leaving aside waste of energy, it creates bubbles of gas stick to the electrodes, resulting (at the very least) in inhomogeneous plating.  Also, mind that the transport of ions in water is a diffusion process and electrode potential does not have that much effect on it.

I still dont understand why people describe blackened product when using higher voltages (say above 3V).  Electrolysis of water should just produce hydrogen and oxygen gas.

Also with an excess of the cupric sulfate we are driving the reaction forward by mass action so it must not be in equilibrium.  Interesting about the kinetic aspect though is it as if we are raising the temperature higher than what would be possible in water?

@user391339 most metal electrodes oxidize before water, so instead of oxygen we get metals entering the solution of producing oxides. If the electrodes contain impurities that oxidize harder than the main electrode components, the impurities form a precipitate.

@user391339 A lot of times when people describe blackened product it's when they're using tap or well water, which generally contain a ton of other ions (especially sulfates and chlorides, and often tap water has a fair amount of fluorides too) that react readily with the electrode material and/or interfere with the final result. Distilled water gives a much cleaner finish.


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What is this formula representing?

What is the formula on this cup representing?

It's caffeine, what else you'd expect?

You may "get it" even as a pendant (e.g., https://www.etsy.com/market/caffeine_pendant), albeit some representations appear, chemically speaking, questionable.

It's a coffee mug, and coffee is (in)famous for one particular substance: Caffeine.
The 2D picture on Wikipedia is the same as on the mug, except that it's rotated 180 degrees. The 'sticks' are methyl ($\ce{CH3}$) groups.


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stability and energy gap
We know that the lower the energy gap between bonding and antibonding orbital the easier it is to break the bond making it unstable. However, in a conjugated double bond, the overlapping of wavefunctions cause the energy gap between the bonding and antibonding orbital go down. Then why conjugated double bonded molecules are more stable??


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Which gas has the greatest volume?
I was given $4$  gases, which gases had the same temperature ,pressure and mass and was told to determine the gas with the greatest volume.
The gases were $\ce{CH4}$, $\ce{NO}$, $\ce{C2H2}$ and $\ce{CO2}$.
So I assumed the mass was $1$ gram and used the formula $n=\frac{\text{mass}}{\text{molar mass}}$  and found that $\ce{NO}$ had the greatest volume but the correct answer given to me was $\ce{CH4}$.
I cannot find out where I went wrong?

Check your molar masses and calculations.

First of all, if in a certain condition it is said that mass (or any other parameter) is constant, you cannot just assume ${1gm}$ (or any other number). Mass being constant means $\Delta m =0$, not that it is ${1gm}$. You have no way of knowing what the mass is (unless stated in the question, of course). The correct approach is to assume the mass to be $'m'\space{units}$. Now, this mass $'m'$ can assume any value, according to the condition.
Moving on, you have made a mistake while comparing molar mass and volume.
The ideal gas equation gives us the relation,
$$PV=\frac{\text{Mass}}{\text{Molar Mass}}RT$$
For your question,
$$V=\frac{1}{\text{Molar Mass}}.\frac{(\text{Mass})RT}{P}$$
$$V\propto \frac{1}{\text{Molar Mass}}$$
Recheck your calculations.


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What research methods can I use in identifying Aluminum in a solid sample?
I do have a cube of solid aluminum. But I need to verify that it is really Al to validate my research. What simple scientific techniques do you recommend?

Do you want to verify purity? Or the question is that if it is aluminum, and not eg iron? In the later case simple density measurement would show you a lot.

Would density measurement be valid/acceptable in research studies? I just want to clarify if a sample is mostly made up of Al.

Acceptable or not, it depends on the field, the accuracy you need, the probability of alternative materials, your exact question... Most studies already know what is their standard sample, or else they wouldn't use it.

If you want a simple one, then you might want to just perform qualitative analysis (QA) using aqueous sodium hydroxide and aqueous ammonia. Firstly, dissolve a portion of the cube using a small volume hydrochloric acid. Then, add a small volume of sodium hydroxide to a small volume the metal chloride solution. To another small volume of the metal chloride solution, add a small volume of aqueous ammonia.
For the first test, aluminium ions would first give a white precipitate but will dissolve in excess. For the second test, aluminium ions would first give a white precipitate but will not dissolve in excess.
Lastly to confirm, add aqueous potassium iodide to the sample. If no precipitation occurs, the sample is indeed aluminium chloride solution and the cube does indeed contain aluminium.
Alternatively, this paper here also provides some insights on verifying if something is indeed aluminium:
http://www.iaea.org/inis/collection/NCLCollectionStore/_Public/08/346/8346019.pdf
Unfortunately, the method I have mentioned and the methods in the paper are destructive as they require you to chip of a piece of metal from the cube for analysis. So... if you can't afford to do that then they cannot be used.

Can you give the source of your information in the method that involves sodium hydroxide and aqueous ammonia?

I learnt it in chemistry class. Online sources include: http://www.rsc.org/learn-chemistry/content/filerepository/CMP/00/000/534/cce-80.pdf (Pg 207) http://chemistryforsg-olevel.blogspot.sg/2012/01/qualitative-analysis-testing-for_02.html

Although RSC claims that it is a "colourless precipitate", from my recollection during the lab sessions and the lab demonstrations I have watched online, it should be a white precipitate.

Are those 2 test methods that uses aqueous sodium hydroxide and aqueous ammonia different/separate from each other?

@J.C.Hernandez They are separate tests as in they are conducted separately. However, the results of the tests corroborate with each other to give you the conclusion (i.e. confirmation that it is aluminium).

Can I just used the NaOH only and then confirm using KI? I don't have supply of ammonia.

@J.C.Hernandez Unfortunately, I don't think you can confirm it, with this method without ammonia.


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How do foundries prevent zinc from boiling away when alloyed with Aluminum?
How do foundries prevent lower boiling point metals such as zinc from boiling away when alloyed in a furnace with higher boiling point metals such as aluminum?

When alloys are made by mixing molten metals (actually an alloy only need contain one metal and at least one other compound, metal or not) the metals only need to be heated to their melting point, not all the way to their boiling point. In the example you've given, the melting point of aluminum is $\pu{660^oC}$, which is $\pu{247^oC}$ below the boiling point of zinc, so the volatilization of zinc is negligible under these conditions.
However, the issue you bring up does present problems in other cases. For example this article states the following:

One difficulty in making alloys is that metals have different melting points. Thus copper melts at $\pu{1,083^oC}$, while zinc melts at $\pu{419^oC}$ and boils at $\pu{907^oC}$. So, in making brass, if we just put pieces of copper and zinc in a crucible and heated them above $\pu{1,083^oC}$, both the metals would certainly melt. But at that high temperature the liquid zinc would also boil away and the vapour would oxidize in the air. The method adopted in this case is to heat first the metal having the higher melting point, namely the copper. When this is molten, the solid zinc is added and is quickly dissolved in the liquid copper before very much zinc has boiled away. Even so, in the making of brass, allowance has to be made for unavoidable zinc loss which amounts to about one part in twenty of the zinc. Consequently, in weighing out the metals previous to alloying, an extra quantity of zinc has to be added.

Summary, TL;DR:
In your example of aluminum and zinc, each metal melts well below either of their boiling points so that loss via volatilization is not a problem. There are cases however, such as alloying copper and zinc, where the boiling point of one metal is lower than the melting point of the other. One way to minimize (but not eliminate) the loss of the more volatile metal is to quickly dissolve it in the high-melting metal and then cool the solution. Although this does not eliminate losses due to volatilization, it can greatly reduce the problem. And actually, since alloys are frequently composed of predominantly one metal, it is not uncommon to dissolve the lesser components into the primary component as a matter of practice anyway.
I hope the example I gave addresses your question. Please don't hesitate to ask for any clarifications in the comments below.

I assume you then have to process the (e.g.) brass alloy melt relatively quickly, to avoid continuous further loss of zinc?

@hBy2Py , yes, I'm actually impressed that they keep the losses of Zn so low (~5%, see above). I assume the addition of Zn would be about the last step of the process before cooling it down. There is probably some decrease in the vapor pressure of Zn due to solvation, but don't quote me on that.

, I also would assume the Zn vapor pressure is lower in Cu solution... but it's by no means certain!

In the real world zinc losses from brass are kept very minor by keeping temperature below the boiling point of Zn and fluxes added to make a covering of slag. When a founder screws-up and over heats a furnace it makes a " zinc boil" , it produces an exceptionally dense white "fog" of zinc oxide. I will admit to having done it once. In steel the Zn, Pb, etc. are deliberately vaporized (and collected in bag houses)  to clean the steel.

This reminds me of recycling  , at one time ( and maybe today) ; Die cast scrape was melted on one furnace hearth . Liquid would run down to a lower furnace through an air trap ( like a plumbers "P" trap ). The second furnace is enclosed and hotter, above the boiling point for zinc. Relatively pure zine is distilled and condensed .  On about a monthly basis the  residual from the second furnace is drained; it was lead, zinc , copper, etc.   and could be used at a brass/bronze smelter.


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Please help regarding finding the correct approach {solid states}

In Diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of unit cell is 356 pm   then diameter of carbon atom is ________?

I got the answer but is it with correct method? Anything wrong here (in the attachment first one is purely wrong as no answer but why)
Is it wrong attempt in Second attempt? So in order to find answer I thought this as assuming in octahedral void which has  distance 2(radius){assumption} and as the atoms are in alternate Tetrahedral voids so again +2(radius) and as at corner (both) (radius) + (radius)
and this equals √3(a)  {a = unit cell edge length }
Answer is 154.07

Any help here please

It is true that the answer is √3/8 * a. But you already knew that. I am not smart enough to understand how you arrived at that. I agree that not only is 1+1 = 2 but that 2r + 4r + 2r = 8r. But I hope you already knew that as well. Restating a√3/8 = r as  a√3 = 8r is also correct. After that, you have lost me completely. I never re-visit questions, so if you are sure that you can correctly justify your bald assertions that 2r + (2r)*2 + 2r = √3*a, then your solution is one of the correct ways to arrive at the answer. As for me, I had to recall enough trigonometry to determine that if the edge was "a" then the distance to any face's center from a vertex is a½√2 and that the tetrahedral angle was Θ = arc-cosine(-1/3). From there we use the three points: a vertex, a face center, and a tetrahedral occupancy to form a triangle with sides D, D and a½√2 as well as the angle between sides D and D of Θ. Simple decomposition of the triangles into right triangles gives D = a√2/(4*sin(Θ/2)) which is indeed a√3/4.

Digonal of a cube is a* sqrt{3} so i applied that and imagined it as containg atoms in digonal


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Reference for assignment of symmetry (term symbols) of electronic states
I am looking for a textbook and/or internet resource where I can find the reference for assignment of symmetry of electronic states systematically.
For example, $$ \ce{^1\Delta^{}_{g}}, \ce{^3\Sigma^{}_{g}}, \ce{^1\Sigma^{}_{g}} \hspace{1em} \text{for ground state} \hspace{1em} \ce{O_{2}}$$

Pretty much any containing an introduction on electronic structure theory. I prefer $Carl J. Ballhausen:$ "Introduction to Ligand Field Theory".

Ballhausen is a pretty tough book to start learning with;  try Carter 'Molecular Symmetry and Group Theory' and an older book Bishop 'Group Theory and Chemistry'. Easier books are Molloy 'Group Theory for Chemist ' and for beginners very good tutorial type book  Vincent ' Group Theory for Chemists'. Most phys chem textbook also cover these topics but usually briefly. You could also look at specialist spectroscopy books of which there are many.


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Is Heisenberg uncertainty principle really a demerit of Bohr model?
My teacher taught that Heisenberg's uncertainty principle is a demerit of Bohr atomic model.
But as the below paragraph from the article What Is Heisenberg’s Uncertainty Principle? says and as per my thinking Heisenberg's uncertainty principle tells about the uncertainty, if we try to do some practical observation; while what Bohr told us by his formulas was almost fully theoretically derived. The paragraph basically tells that when photon strikes any subatomic particle then change in velocity occurs and uncertainty in position is incurred:

One way to think about the uncertainty principle is as an extension of how we see and measure things in the everyday world. You can read these words because particles of light, photons, have bounced off the screen or paper and reached your eyes. Each photon on that path carries with it some information about the surface it has bounced from, at the speed of light. Seeing a subatomic particle, such as an electron, is not so simple. You might similarly bounce a photon off it and then hope to detect that photon with an instrument. But chances are that the photon will impart some momentum to the electron as it hits it and change the path of the particle you are trying to measure. Or else, given that quantum particles often move so fast, the electron may no longer be in the place it was when the photon originally bounced off it. Either way, your observation of either position or momentum will be inaccurate and, more important, the act of observation affects the particle being observed.

Is Heisenberg's uncertainty principle really a demerit of Bohr model?
Reference

Jha, A. What Is Heisenberg’s Uncertainty Principle? The Guardian. November 10, 2013.

...and it's still not rewritten...

The text is actually describing the observer effect, which is very commonly confused with the Heisenberg uncertainty principle. The latter is a far more fundamental property of the Universe, and has strong theoretical backing (the non-commutativity of certain mathematical operators in quantum mechanics), regardless of experimental setups. The uncertainty relations thus establish a fundamental "fuzz" to any object, precluding its complete and precise characterization even in principle.
The lack of the uncertainty principle is indeed a severe conceptual limitation to the Bohr model (or rather it is better to say that the lack of the groundwork which results in the Heisenberg uncertainty principle is a limitation); whereas the Bohr model still makes predictions in terms of the classical concept of orbits, a more fully-fledged theory of quantum mechanics applied to an atom necessarily predicts the formation of orbitals, where the position and momentum of the electrons is smeared out.
Indeed, the uncertainty principle is ultimately why electrons have stable configurations around atomic nuclei that stay on average around 100 picometers away from the nucleus. The nucleus itself is only around 1-2 femtometers (~0.001 picometers) in diameter, so electrons could have their potential energy greatly reduced if they closely hugged the nucleus at a similar distance. So what's stopping them? If the electrons did scrunch up so tightly against the nucleus, they would essentially be localizing themselves in space very accurately, and thus the fundamental uncertainty in their positions would be very small. However, because of the Heisenberg uncertainty principle, this would cause a very large increase in the uncertainty of the momentum, which in turns causes a very large increase of the average momentum and from there a large increase in their kinetic energy, so that the electrons would ultimately need a higher total energy to stay so close to the nucleus. Again, this behaviour is completely independent of whether the system is being measured or not.
Furthermore, Bohr's model is actually very empirical. In particular, one of its most important considerations is that the angular momentum of the electrons around an atom obey the formula $m_evr=n\hbar$, where $n$ is the positive integer responsible for introducing quantization to his model. However, this is just a postulated equation, with no rigorous derivation. It was chosen because it happened to describe the emission spectrum of hydrogen exceedingly well, so he knew it was right in at least some regime, but had no proof for it.

Minor nitpick: you mean the average magnitude of the momentum in the second-to-last sentence of the second paragraph.

Actually we cannot actually call it a demerit. It is more of an opposing principle. It is one of the reason why bohr's model got disregarded. According_to bohr electrons revolve in specific orbit with radius= n^2x0.529A. But heisenburg's uncertainty principle says otherwise.


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Does the emf of an electrode potential change over time?
In an electrode potential with a salt bridge and high resistant voltmeter, does the emf change over time as the reactions occur?

Yes. Remember that the standard electrode potentials are for 298K at 1 atm of pressure and 1N concentration of reagents. As the concentration of ions to be reduced at the cathode decreases the emf will fall as well and is reflected in the Nernst equation.$$E_{cell} = E^\circ_{cell} - \frac{RT}{nF}ln(Q)$$

what about the mercury button cells their emf is independent of concentration of ions so theoritically its constant 1.35V but varies 1 % over shelf life.

@Physicsapproval The mercury button cells have solid products and reactants, therefore, Q is a constant 1 and emf does not change.


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What happens if hydrogen peroxide is added to silver oxides?
I have some powders that contain $\ce{Ag2O}$ and $\ce{AgO}$.
What happens if I add $\ce{H2O2}$ and some water to these silver oxides?

What else do your powders contain?

AgO (silver(I, III) oxide) is unstable and decomposes to produce $\ce{O2}$ in aqueous solutions. Hydrogen peroxide is thermodynamically unstable too and slowly decomposes to form water and oxygen. Decomposition of hydrogen peroxide can be catalyzed by different compounds, including transition metals (such as Ag) and their compounds. Probably, the silver (I) oxide will directly/inderectly catalyze the decomposition of hydrogen peroxide according to the reaction:$$\ce{Ag2O + H2O2 → 2Ag + H2O + O2}$$
Given that the majority of silver compounds/salts are photosensitive, they will be decomposed when exposed to regular light, especially to the ultraviolet light. Even if you have other substances/compounds as impurities in hydrogen peroxide solution which can react with silver oxide you still will probably end up with elemental silver ($\ce{Ag}$).

and some water?

Sounds like a joke. The available hydrogen peroxide solutions are approx. 3% (drug store) or 30% (most chemical labs), thus, the solutions already have a plenty of water, at least 70%. Do you really need more water?

You said AgO decompose to produce O2. Does AgO decompose to Ag2O?

When hydrogen peroxide is added to silver(I) oxide, it reduces to silver(here)
$$\ce{Ag2O + H2O2 → 2Ag + H2O + O2}$$
Silver(II)oxide is unreactive towards hydrogen peroxide and hence it is a good method to separate a powder containing silver(I) oxide and silver(II) oxide.


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When you hydrolyze asparatame what do you get as a product
On my Organic Chemistry 1 class final, the instructor threw us off with the last question. It asked:

What products would be obtained if asparatame were hydrolyzed completely in an aqueous solution of HCl?
Bonus: Write a proposed mechanism.

I couldn't figure out how to solve this problem because I didn't know which oxygen attacked the hydrogen from HCl.
 (Source: Wikipedia - Aspartame)

Hint: The molecule contains a methyl ester, which will yield (after neutralisation) a carbonic acid, and methanol.  At the same time, the molecule contains an amid; its hydrolysis will yield (after neutralisation) a carbonic acid, and an amine.  In the special case here, both carbonic acids happens to be $\alpha{}$-amino acids.

Since aspartame is made up of phenylalanine and aspartic acid, upon hydrolysis it will break into corresponding amino acids. But in basic or acidic medium, it generates methanol. According to Wikipedia article of aspartame:

Aspartame is a methyl ester of the dipeptide of the natural amino
  acids L-aspartic acid and L-phenylalanine. Under
  strongly acidic or alkaline conditions, aspartame may
  generate methanol by hydrolysis. Under more severe conditions,
  the peptide bonds are also hydrolyzed, resulting in free amino acids

Since this is more or less a HW question, I am not going to answer the 2nd part of question. Type - "peptide bond hydrolysis mechanism" in Google and you will get your answer.

Firstly, find the types of functional groups present in the given compound and then use the theory of hydrolysis of the functional groups. Here functional groups are ester and amide. Finally, use the special characteristic of alpha amino acids.


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Direct connection in a zinc copper battery
Why can't you just hook up a piece of wire between a piece of zinc and  copper and get an electric current? What is it about the electrolyte that makes zinc give up electrons?  Also, why will the lack of a salt bridge stop the electric current from moving?   Why  does balancing out the positivity and negativity at the cathode and anode matter?

A close circuit is needed for electrons to flow. The electrolyte and salt bridge serves to complete the circuit. Additionally, the salt bridge also serves to function of maintain electrical neutrality in each half-cell, preventing the accumulation of charges, aiding the flow of electrons.
If charges were to accumulate in each half-cell, the potential difference between the two half-cells, which is the key driver of electrons, would decrease over time to 0 V and no electrons would flow.
There have been many discussions on this site. To cite a few:
What is the real purpose of the salt bridge in an electrochemical cell? (look at the comments)
Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?
electrochemical (Galvanic) cell solution purpose


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Is this an aluminum screw dislocation?
Is this an aluminum screw dislocation?
https://youtu.be/zbUPjHHml1E?t=55s
I have heard of zinc pest which looks similar.

No, what is occurring is the mercury is forming a eutectic liquid (also called an amalgam) with the aluminum. Aluminum is quite reactive with air/water, but typically aluminum can form an oxide coating that protects it from a complete reaction with air/water (passivation). When the mercury is added and the surface of the aluminum is damaged (by the drill) the two are free to mix per the phase diagram below now the aluminum in the liquid may react with the air/water, but now on the liquid the oxide formed is easily disrupted, and thus ineffective in protecting the aluminum from oxidation. This allows more aluminum to oxidize and the mercury to dissolve more aluminum as aluminum is depleted from the liquid. This creates a steady state condition where aluminum is continually dissolved and oxidized until the aluminum is completely consumed.

A screw dislocation is a microscopic feature only visible with atomic scale imaging such as with a tunneling electron microscope (TEM).

Please explain this: https://en.wikipedia.org/wiki/Whisker_(metallurgy)#/media/File:Zinc_whiskers.jpg

Those are zinc whiskers. That is not what is happening to the aluminum.


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Will unused spare razor blades rust if stored for a long time?
I am considering purchasing 16 razor blade spares for a reduced price, but am unsure if they will be good as new if I were to use them several months from now.
If left alone in the packaging (they come in plastic cartridge cases with one side exposed), will the quality degrade by the time I use the last several (if the blades were replaced monthly)?

If you store it properly, and if the blades are of good quality... there will be no appreciable rusting by the time you use them ;) [ "Store properly" = Minimize contact with moisture-laden air]

@paracetamol. Would keeping them in a drawer in the cabinet attached to the bathroom sink suffice? Would that be enough to store properly?

Aside from storing a in dry location, you might use a vapor corrosion inhibitor (VCI) in a container with the blades. Amazon offers a similar product for ~US$11.
My experience has been that VCI is effective and does not leave a noticeable film that needs to be removed. Whether this is cost-effective for your intended use is another question (perhaps you can use the remaining VCI on other objects or share with ten other people).
And then, beards are even more cost-effective ;-)


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Decarboxylation mechanism and stability of α,β-unsaturated and α-hydroxy acids
I was wondering why α,β-unsaturated and α-hydroxy acids undergo decarboxylation. Also, I'm aware of neither the products nor mechanism. All I know is that β-keto acids and α-nitro compounds have resonance stabilized anions and have stable transition states.
But α,β-unsaturated have none of those and it isn't that electronegative as its sp2. Someone please help me out; I can't find the answer anywhere else.

i cant find the answer anywhere else Strongly disagree: http://pubs.acs.org/doi/abs/10.1021/ed064p591#/doi/abs/10.1021/ed064p591

Im sorry I cant understand it, Im not an undergraduate student can you please explain it to me? Without the oxidative cleavage or free radical hunsdicker reaction. Just how on heating aplha beta unstaurated acids undergo decarboxylation?

Decarboxylation mechanism involves creating a carbanion at the carbon attached to $\ce{-COOH}$ group. Does this help you decide why?

Related: https://chemistry.stackexchange.com/questions/53879/why-are-beta-ketoacids-better-at-decarboxylation-than-gamma-or-delta

But isnt a vinyllic carbanion unstable?

I know why beto keto acids and beta gamma acids undergo decarboxylation.

@LostGuardiano It's not vinylic, it would possibly be allylic.

@PrittBalagopal I noticed you'd linked to a J. Chem. Ed. paper in the comments (DOI: 10.1021/ed064p591). I suspect that the OP doesn't have journal access, it might be more productive to paraphrase what that paper actually says about the reaction in question

@PrittBalagopal the mechanism is concerted. Nowhere does carbanion forms


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Is it possible to perform ROMP2 numerical optimizations in Gaussian09?
I wish to perform an restricted open-shell MP2 (ROMP2) geometry optimization.
Gaussian09 currently only provides energies at ROMP2 level of theory.
It should be possible to perform an optimization by displacement of the structure followed by single-point-energy evaluations at the displaced geometries to obtain a gradient. In other quantum-chemical software packages this (slower) numerical optimization technique is readily available. I cannot seem to get it to work in the case of ROMP2 in Gaussian09. I have already tried the usual options, such as #p opt roMP2/cc-pVDZ scf=tight and so on. All of my approaches failed and yielded the following error:
 NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-NEF-
 NUMERICAL EIGENVECTOR FOLLOWING MINIMUM SEARCH
 INITIALIZATION PASS

 ************************************************
 ** ERROR IN INITNF. NUMBER OF VARIABLES (  0) **
 **   INCORRECT (SHOULD BE BETWEEN 1 AND 50)   **
 ************************************************

I am aware of this website: http://www.somewhereville.com/?p=1571
and I have tried the solution outlined therein, but it has failed in the same manner as well.
Any help or pointers towards other software that is capable of ROMP2 optimizations would be appreciated.

Could you post a sample geometry that this fails for?

For numerical optimization you have to provide a limited number of parameters you want to optimize. Gaussian doesn't find those in your input.

Is there a reason, apart from computational cost, that you are not interested in unrestricted MP?

@Vic Lineal: Severe spin contamination and symmetry broken orbitals. The system has multi reference character, therefore all single determinant methods will yield more or less non sensical data. I still want to benchmark to which extend cheaper methods yield accurate results.

I've seen this error when I want to combine an optimization with an orbital deletion (NBODel) in Gaussian. It cannot handle an optimization with more than 50 degrees of freedom, so you need to freeze as much as you can, or pre-optimize your molecule with a more traditional method and then freeze whatever you can and do the ROMP2 optimization on the parameters of interest.

I do not think this answers the question. Gaussian explicitly says that it'll only compute energies on the RO-MP levels of theory. The lack of analytical gradients would have to be compensated by numerical ones, which simply do not seem to exist in G09 for RO-MP. No gradients, no optimisation, and I am equally unaware of a random walk optimiser.


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What is the correct way to transfer large volumes of chemicals?
I am an engineer starting a film developing and processing lab in my house for personal use. I am currently using a bathroom, and am concerned about contaminating my chemical process and the bathroom using what seem to be accepted practices in developing labs.
Many developing chemicals (e.g. Kodak D-76) come in exact quantities to mix 1 gallon of stock solution. I will mostly be using quantities 500 mL or less, though certain chemicals will be larger arbitrary volumes.
What methods/equipment are used for transferring large volumes?

what kind of engineer?

drum pumps! If you want to get large volumes of unreactive liquid out of a barrel or other large container you can do this easily with a drum pump and transfer it into measuring cyclinders like @NotEvans posted.

@A.K. Electrical

Given that the chemicals in question aren't air sensitive (i.e. theres no need for an inert atmosphere), a measuring cylinder would suit your needs.
They're available in glass and various plastics, in sizes upto around 3000 mL. The glass ones are generally better as theres less risk of the chemicals reacting with the glass compared to with plastic, but it really does depend on what you're using.
Once you're done with the measuring cylinder, just rinse it with water (into a suitable toxic waste container if the chemicals cannot go down the drain, which I suspect they can't), and leave it to dry somewhere (else put it in a warm oven for an hour or so).

What is the correct way to get liquid into the cylinders if the bottles are too full to pour?

@JimmyStrates: A large pipette would do (but not overly convenient for large volumes), some kind of siphon would be best!


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An ionic compound dissociates into 3 ions of 3 different elements?
When an ionic compound dissolves in water and becomes ions.
1) Typically we see:
$\ce{AB ->}$ Ions $\ce{A+ + B-}$
$\ce{ABC ->}$ Ions $\ce{AB+ + C-}$
2) But how about the one below, is there any case that will fit the situation
$\ce{ABC2 -> A+ + B+ + 2C-}$
or $\ce{AB2C-> A+ + 2B+ + C^3-}$

Another is the de-icing compound calcium magnesium acetate, $\ce{CaMg2(C2H3O2)6}$ (https://en.m.wikipedia.org/wiki/Calcium_magnesium_acetate). Like all such "double cation salts", this compound has a crystal structure with two different sites for cations; in this case one kind of site has the calcium ions and the other has the magnesium ions.  This combination enables a reproducible stoichiometry, but when the crystal structure is broken down by dissolving the salt you just have "ion soup".

Yup, they are definitely possible. Such salts are called as mixed salts. They furnish more than two ions when dissolved in water. A popular example is Mohr's salt, popularly used as a standard titrant to measure concentrations of oxidizing agents. It's (anhydrous) formula is:
$$\ce{Fe^{II}(NH4)2(SO4)2}$$
This salt on dissolving in water would split as:
$$\ce{Fe^{II}(NH4)2(SO4)2 <=>> Fe^2+ + 2NH4+ + 2SO4^2-}$$
This is one such example. There are many others, notably alums. Alums are a general type of mixed salts, of the form:
$$\ce{A^{I}M^{III}(SO4)2}$$
where $\ce{A^{I}}$ is a monovalent metal ion, and $\ce{M^{III}}$ is a trivalent metal ion. Clearly, these are also mixed salts.
These are just a few practical examples. There's literally quite nothing that can stop imagination of course.


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What is the effect of the surface area of an electrode on the emf of a cell?
I've read that a platinum electrode has no effect in a hydrogen fuel cell - why is this?
Also, what is the effect of the surface area of an electrode such as zinc in a zinc copper cell?

Area? No - it only increases the maximum Current, However surface curvature, crystallographic texturing, and crystallographic strain can, which often confounds results and leads people to believe that area does.

To amend my comment, Imagine you have a second identical electrochemical cell. Does the emf increase? No, the EMF is the same even though you have doubled the surface area in this case.

I think you Should split the is question into two independent questions, as they are independent of each other. This would make it easier to provide an appropriate answer.


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Is free expansion of an ideal gas necessarily adiabatic?
I wanted to know if the term "free expansion" automatically implies an  adiabatic  vessel.Is it possible that there can be a transfer of heat in the free expansion of an ideal gas (as the term seems to imply to me that only external pressure is zero)?
Can we have something like Q=∆U

Of course there can and even must be transfer, process can be only approximately adiabatic.


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Why does chlorine replace a hydrogen in propene instead of adding to the double bond?
I read about the following reaction in my book:
$$\ce{CH3-CH=CH2->[Cl2][\pu{773 K}] (Cl)CH2-CH=CH2}$$
It surprises me that the $\ce{Cl}$ doesn't attack the $\pi$ bond, rather it attacks the $\sigma$ bond.
It seems to me that the reaction follows free radical mechanism, for the high temperature of the reaction system. That's why, the $\ce{Cl}$ attacks the $\sigma$ bond.
But, I am not sure, why won't free radical $\ce{Cl}$ attack the $\pi$ bond.

It's not that the chlorine radical doesn't attack the $\pi$-bond, it just doesn't do it as fast. Alkyl radicals are much less stable than allyl radicals, so hydrogen abstraction will occur much faster than radical addition ($k_1 >> k_2$).

Isn't the question rather why the chlorine abstracts the hydrogen in the first place instead of directly adding to the double bond?


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Synthesis Golf II: Erythronolide B
A full FAQ post has been written on meta.chem.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this if you haven't already.

This round of synthesis golf concerns the Corey synthesis of Erythronolide B (J. Am. Chem. Soc. 1978, 100 (14), 4620–4622).

In order to narrow the scope of the challenge, only a small fragment of the natural product will be considered. Fragment 2 was an intermediate on the Corey route, and provided C10-C13 of Erythronolide B.
Crucially, 2 was needed as a single enantiomer and diastereomer. In the original Corey report, 13 steps were needed to accomplish the synthesis of the fragment, a length which prevented significant amounts of material from being made.
The challenge is to propose a synthesis of 2 which sets up the required 1,2-stereochemistry with absolute configuration. You may start from anything commercially available in the Sigma–Aldrich catalogue, including chiral building blocks.

What is the iodine set up for? Could a triflate suffice here?

I'd like to see future synthesis golf questions where a biosynthetic answer would fit in.  There are viable engineered biosynthetic routes to erythronolide B, but since this question is about some random synthetic intermediate that is not biological and is of no use to anyone except synthetic organic chemists, I guess biosynthesis is off-topic for this particular synthesis golf.

@CurtF. The intermediate was chosen since some felt that anything larger than a small fragment was too broad (and therefore likely to be flagged/closed).  If you'd care to expand, theres a post on meta.chem asking for input!

As an aside, some bio-transformations could still be used to set up the stereochemistry here.  There is presumably only one true biosynthesis, which isnt really any individuals work rather than natures.

I was thinking of using a VMAR reaction (vinylogous Mukaiyama aldol reaction) then defunctionalising the ketone and adding iodine. But I’m not really finding the sources I would be looking for. Oh well, maybe join next time since I really should be doing something else rn =D

And yes, that’s just to add a not-quite-Evans aldol to the mix, because try as I might I keep falling back onto Evans as the natural reaction to use here …

@Beerhunter the next step is the formation of the vinyl Grignard, followed by addition to a thioester

I think we should switch to another problem. One month period is too long for a such a narrow question. Ten days have passed and there are no new ideas. Either a shortening of the period between two questions or posting questions with a wider scope would in my opinion improve the challenge.

@Marko, the current question hasn't really received a lot of attention. If we ignore my answer, only two people have actually participated

@NotEvans As I said I think it is because of the narrowness of the question. Even our approaches are very similar.

I have established the stereochemistry via Evans aldol addition:
Then, after protection of the hydroxyl group, the chiral auxiliary can be transformed to a Weinreb amide which will be attacked by Grignard reagent:

The methyl ketone is then transformed to an alkyne which is methylated:
Finally, hydroboration (should be very regioselective) of the alkyne will provide the trans alkenylborane which on treatment with NaOH and iodine will provide target 12:

Length (total steps and longest linear sequence): 7 steps

Nice! Not free to answer this time round, but I was thinking of something along this line too.

@Ben Norris I counted 8 steps. Are one-pot reactions considered one step?

@Marko, for things in the same pot (such as the aldol), its usually written as Bu2BOTf; RCHO, with the semicolon indiciating then. Different numbers do imply (to a bystander at least), different reactions.

@NotEvans Okay, I should have mentioned that I use number designations for different steps in one pot reactions and reactions in different vessels are designated by different arrows. Now I see that hydroboration/iodination is done one pot, too therefore there are 7 steps in total.

Plan:
The key things that need setting up are the 1,2-syn stereochemistry and the vinyl iodide.

The 1,2-syn stereochemistry is easiest setup using a syn aldol, via a Z enolate. The Evans' (or variant) is the obvious way to gain control of absolute stereochemistry as well as relative stereochemistry.
The vinyl iodide can be made in many ways, but hydrozirconation of a methyl alkyne is a good way of ensuring double bond geometry and regioselectivity for the iodide ending up where it does.

Forward synthesis:
1 → 3: Acylation of the Evans' auxiliary to append the propionyl unit followed by a modified Evans' aldol (Crimmins' type chemistry, avoiding the boron triflate reagents), these conditions give the Evans' syn product.1
3 → 4: TBS protection, fairly hindered alcohol so using TBSOTf along with a lutidine base.
4 → 5: In order to introduce the alkyne, need to first convert the Evans' auxiliary to something functional, an aldehyde would be a good starting point and could be obtained in multiple ways. The two most obvious (both with the same step count) are either direct reduction of the auxiliary to the alcohol followed by oxidation (Parikh Doering often useful in these systems), or formation of the Weinreb amide and DIBAL reduction directly to the aldehyde. In both cases, the aldehyde is chiral at the alpha-position, so must use immediately to avoid possible racemisation of the methyl stereo centre.
5 → 7: An Ohira-Bestmann reaction installs a terminal alkyne. Could also get the same result in a two-step procedure using a Corey-Fuchs (possibly more reliable, but precedence is sound for the Ohira-Bestmann).
7 → 12: Introduction of the methyl group at the end of the alkyne via deprotonation (terminal alkynes have fairly low pKa) alkylation sequence. The alkylated alkyne may then undergo hydrozirconation/trapping to install the vinyl iodide, furnishing 12, as required.

Summary:

8 steps overall (longest linear and total steps), with all starting materials and reagents being commercially available
Control of absolute and relative stereochemistry via the use of an Evans'-type aldol reaction
Control of the vinyl iodide geometry using a selective hydrozirconation

References:
1: DOI: 10.1055/s-2004-825626

Good call for Ohira-Bestmann reaction.

Before my answer, I'd like to inform you that I'm an undergrad student with some knowledge of organic chemistry. If there are any errors in my answers, I'm sure you'll tip me off!
Step 1: (Z)-pent-2-ene reacts with $\ce{HOCl}$ to give the (threo)-2-chloropentan-3-ol.

Step 2: (Threo)-2-chloropentan-3-ol reacts with 1-propyn-1-yllithium

Step 3: Reaction with $\ce{TBDMS-Cl}$

Step 4: Reaction with $\ce{9-BBN}$

Step 5: Reaction with 3 eq. $\ce{NaOH}$ (3M aqueous solution) and 2 eq. $\ce{I2}$ in $\ce{THF}$ at $\ce{23^\circ C}$ for 1 hr.
Reference: C. Wang, T. Tobrman, Z. Xu, E.-i. Negishi, Org. Lett., 2009, 11, 4092-4095

And finally, for simplicity,

I realise that I have represented the incorrect (from the requirement point of view) enantiomer, but seeing that my first step will produce both enantiomers, we should get the required product as 50% of the yield.
Note: I later realized that I have missed a methyl group on the protecting group. Please consider it is there.

(1) You did a racemic synthesis, but the stipulation was to set absolute stereochemistry, though. (2) Curious as to how regioselective the first step is?

@orthocresol I was thinking the first step would be an electrophilic addition via a chloronium ion intermediate, which gave the anti-diastereoselectivity.

I was also thinking that the first step could be modified to be enantioselective using a bulky chiral catalyst, and using $\ce{TBDMSO-Cl}$, if that works, but I don't think the reagent is available. Further, maybe the substrate is too small (?) for steric interactions to work effectively to prefer one enantiomer over the other.

Regio, not diastereoselectivity. i.e. How sure are you that the chlorine will end up attached to C-2 and not C-3

@orthocresol The chloronium ion is a non-classical carbocation, so the sterically more hindered (and better +I) side of the three-membered chloronium would be attacked by the $\ce{HO^-}$. After the attack, we should end up with $\ce{-OH}$ on C3 and $\ce{-Cl}$ on C2 in the anti- form.

For someone unversed in organic...with an otherwise fair chemistry (plus physics/coding/math/etc) foundation... this discussion makes my head spin! Usually we phys scientists, in subjects we don't know, can hold our own, or can talk about it at least. I see a benzene in one of the pictures, that's about it for me!  If any introductory chem student ever thinks "oh chem must be simple, it's a increasing combinations of these elements, these kinds of bonds", just point them here and watch a head explode? You've won the web's most complex comfortable-sounding comment thread in all history.

It's an ethyl vs methyl group - I really don't know how this turns out in the lab, so take what I say with a tablespoon of salt, but on paper it doesn't seem like there is a significant difference in their inductive effects, such that the formation of one regioisomer is strongly favoured.

@orthocresol That's what my concern was - yes...

A synthesis using Sharpless epoxidation for stereoinduction.

MeLi will destroy your ester before it opens your epoxide. I’ve seen some chelation-controlled epoxide openings with hydride sources before, but not sure if there’s an analogous one with carbon nucleophiles (surely it exists?).

@orthocresol good point, might have to use something like THP instead, which would still be able to direct the MeLi hopefully.


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Synthesis of 2-phenoxybutanal from but-1-yne

How can I convert 1-butyne to 2-phenoxybutanal as shown above? I thought of using a hydroboration-oxidation sequence to introduce the carbonyl group, but I do not know how to add the phenoxy substituent.

Do you have any ideas about how you might accomplish the transformation, you could [edit] your post to include these so that people may help you more easily (if they know what exactly you're getting stuck at).  Also, welcome!  You might find the [tour] useful to learn more about chem.SE and how the stack exchange format works.

@AshleyAustin Consider using phenol for your addition instead of benzene. Also your product is 2-phenoxy-butanone (not phenyl).

How would I remove the H of the phenol group then?

So adding say h2o and phenol and the br- would work?

Phenol is a pretty decent nucleophile

Do you think you know how to do this reaction if it were an alkene instead?

Water, BH3, H2O2, NaOH

This reaction, an hydroboration, yelds butanal

Br2, H2SO4

This reaction, carried without light, is an alpha carbon bromination, and yelds 2-bromobutanal

Sodium phenoxide

This is a simple Williamson Ether Synthesis, which yelds the final product

The edited question was wrong, and was not the OP's original question. The target compound was an aldehyde and not a ketone. I have changed it back to what it was supposed to be. Of course, this simply means that you just need to take out steps 2 and 3. (However it would also be appreciated if you could add some elaboration as to why you chose this sequence of events.)

I'm sorry, the question I answered explicited the request of reagents, without pointing out the reactions (the review " edited Jun 12 at 4:24 by Pritt Balagopal"), and the molecule in that review was a ketone.

Something like this would do:

I made the alkene symmetrical so that it would lead to fewer regioselective problems. However, I would advise starting with another starting material as this synthesis might take too long in practice.

The edited question was wrong, and was not the OP's original question. The target compound was an aldehyde and not a ketone. I have changed it back to what it was supposed to be.

I'd strongly recommend updating your answer to reflect that, or else it's technically not an answer.


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Extimating the repartition of water between hygroscopic compounds
Sulfuric acid is sometimes used as a drying agent.
According to literature, one possible application is, for instance, in the drying of bromine.
Some questions came to my mind: which concentration of sulfuric acid would be suitable for such a task? Under which concentration would the acid stop absorbing water from bromine, and there would then be an "equilibrium" between the water bound to bromine and that bound to sulfuric acid?
To put the problem in a general and less specific way: is there a way to estimate the capability of a compound to "catch" water from another compound, and in case of liquid compounds or of hemihydrated molecules, to estimate the maximum amount of water present before an equal repartition of water between the two substances occurs?
I was thinking about the enthalpies of hydration, but I honestly didn't come up with an idea for using them in such calculations.

It's characterised by equilibrium water vapor pressure over agent.


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Is the surface of the planet Mercury made of copper minerals?
I have been looking at photographs of the planet Mercury, and I notices that in most of the pictures I have seen, the rocks on the surface seem to be covered in blue in green. From this, is it correct to say that the surface of Mercury is mostly made up of copper deposits?

The planet Mercury? No, the photos are very much grey. The surface is most likely covered mainly in silicates, i.e. rock.

https://en.wikipedia.org/wiki/Mercury_(planet)#Surface_geology

If you're referring to the picture @PrittBalagopal posted, these are not real colors but a color-coding for a height-representation.

@Fl.pf. Thanks for the insight but I already know that (I would need a heck of an imagination to think mercury is colored blue and orange!). I shared the link for the OP to search if he can find out about the surface of mercury minerals.

@PrittBalagopal I wasn't saying that for you, I was guessing OP saw the picture you posted and thought it was real-color so I wanted to clear that up for him, not you.

I think you probably fell into a trap concerning color-coding.
In reality, mercury is dark grey and mostly covered in silicate rocks/dust.
From a scientific point of view, the "real color" of an object like mercury is of little interest. What's more interesting, is it topography etc.
To make it easier to interpret data, man often uses color-coding (since our color-vision is quite ok) to communicate different information about such things.
Or ist just about beauty:
When you see pictures of nebulae like the pillars of creation you notice they are in a lot of different colors. This is because a lot of the radiation is not in the visible spectrum, and you can "color-code" it to make it interesting to look at.


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Krazy Glue Color Change
A type of Krazy Glue changes from purple to clear when it dries. What is the chemistry behind the color change? Is it due to a particular compound only found in this particular Glue?
A description of glue can be found on their website.


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What is the general mechanism for a lactam ring reacting with water and an acid?
For example, let us say we have 2-Pyrrolidone and react it with water and $\ce{HCl}$. I do know that the product is a salt, $\ce{HO(C=O)CH2CH2CH2NH3Cl}$.
I've tried coming up with a mechanism but fell short. I do know that at the first step we must have protonation of the carbonyl-oxygen which gives us a resonance stabilized carbocation intermediate. Nitrogen can better handle a positive charge.
For the second step, we use the addition of water to the carbonyl-carbon followed by a proton transfer from the oxygen to the nitrogen. But when we do this we get $\ce{NH2-}$ which is not favorable. This is the step that I'm stuck on. After this step it would be easy to see the collapse of the tertiary carbocation intermediate.

You're right that -NH2 is a poor leaving group, even worse than another common "bad" leaving group, -OH. What do we do to an alcohol to get the hydroxyl to leave?

We protonate it

Exactly, the simplest and most common way to activate leaving groups. Could that work with -NH2?

@electronpusher has nailed it. The hydrolysis mechanism is quite similar to the mechanism of ester hydrolysis. Does that help you? Btw, if you get your answer, do consider posting it as an answer here, that would help future visitors to learn :)

I have the solution. I can't post it here because I'm using the ship's computer (a military computer) and we don't have USB ports, so I can't transfer an image from my mobile phone. There is also no webcam utility for screenshots


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Dependence of EMF of a cell on stoichiometry
In the reaction $\ce{Zn + 2Ag+ -> Zn^2+ + 2Ag}$ , my book says that the EMF is the difference between the standard reduction potential of $\ce{Ag}$ and $\ce{Zn}$.
Shouldn't it be the difference between twice the potential of $\ce{Ag}$ and $\ce{Zn}$ since two moles of $\ce{Ag}$ are getting reduced per mole of $\ce{Zn?}$

The electromotive force of a redox reaction is not dependent on the coefficients in the balanced equation as it is an intensive property (i.e. a property that does not depend on the size of the system). Thus, your reasoning is incorrect.
You can read more about electrochemistry here: https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Basics_of_Electrochemistry
Also, I am wondering if your textbook is actually correct because $\ce{Ag2+ + e- -> 2Ag}$ and $\ce{Ag+ + e- -> Ag}$ are definitely different. Thus, the standard reduction potential of silver (I) should not be used. Instead, the standard reduction potential of $\ce{Ag2+}$ should be used. Someone who is more knowledgeable in this area please correct me if I am wrong.

The standard cell potential is an intensive property and therefore doesn't depend upon the stoichiometry rather you can use the relation ∆G=-nFE(standard potential) to prove the result for the standard cell potential because ∆G is an extensive property.


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What is the use of silver ion?
I was going through the mechanism of Arndt-Eistert synthesis and got stuck in the step where silver ion is used in the rearrangement step.
If anyone can clarify the use of silver ion?
It would be a great help!

The silver oxide is used as a catalyst. So it lowers the activation energy of the reaction, but comes out unchanged.

So why only silver ion?

Or can I use some other ion also? If yes, then can you provide an example!

So, this means that the use of silver ion is determined experimentally.

Yes probably. They tested which catalyst is the most effective. Probably was silver.

I did some catalysis research early in my studies. Often you don't really have a way to be certain which metal would be the best to catalyze a reaction so you test a couple...

Let us continue this discussion in chat.

The step in question here is the so-called Wolff Rearrangement.
In this rearrangement, silver in the form of for example $\ce{Ag2O}$ is used as a catalyst. A catalyst facilitates a reaction (lowering the activation energy), but it comes out of the reaction unchanged (which doesn't mean it's not changed during the reaction, it's just changed back at the end ;))
So why silver? Well, transition metals are good catalysts most of the time for different reasons:

They have vacant d-orbitals which leads to
An easy tendency to exhibit variable oxidation states.
They can form reaction intermediates with reactants.
The presence of defects in their crystal lattices.

(I conveniently copied this list from a quora post, because im lazy)
What this essentially means is that due to their electronic structure, transition metals are often able to bind reactants to their surface and activate them for example. There are quite some catalyst reactions that we don't even understand today, but sometimes it's as easy as this

And why silver? Well probably Wolff just tried a couple of transition metal catalysts and chose the one with the best effect/cost ratio! Theoretical expectations don't always meet reality so often you just have to try and find the right reactants/catalysts.

can whoever downvoted explain why so I can rectify my mistake?


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reaction between Green tea and sodium bicarbonate
There is a tea in northern Pakistan called noon chai or Kashmiri chai. It is distinct between it is a pink colour.
The method involves brewing green tea leaves in cold water and adding sodium bicarbonate.
What is the chemistry here?  How does this reaction work and why does the green tea become pink when sodium bicarbonate is added?

https://www.google.co.in/url?sa=t&source=web&rct=j&url=/amp/s/amp.reddit.com/r/tea/comments/tzexm/baking_soda_to_tea_water/&ved=0ahUKEwjwjN7h5LfUAhVLs48KHZFpDQwQFghbMA0&usg=AFQjCNHeRGLsykGeNr6ag2fr5QCQzIB6OA&sig2=F_X1IF51tfRojOdbj_oAIQ


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Balance equation for trilead tetra oxide -> lead monoxide + oxygen
Please give me the answer for this question.
I am having difficulty in this question.

I'm voting to close this question as off-topic because there is no effort in the question.

Homework questions like this are often better when you show an effort to solve the problem yourself, and then come to us for any particular issues you are having.

Your question become victim of downvote.

I won't do your homework for you, but I can show you how to balance equations. Balancing equations is pretty simple once you get the hang of it. First, let's look at an unbalanced equation:
$$\ce{Al2O3 -> Al + O2}$$
The number of atoms for each element on either side of the reaction must match. On the left side there are 2 aluminum atoms and 3 oxygen atoms. On the right side there is 1 aluminum atom and 2 oxygen atoms. Keep in mind that oxygen is one of the diatomic gases that must be bonded to itself when alone as $\ce{O2}$. Now that we have counted the atoms on each side, we can balance them out, by changing the coefficients or number of each molecule in the reaction. There needs to be the same number of atoms on each side, yielding:
$$\ce{Al2O3 -> 2 Al + 1.5 O2}$$
While the number of atoms is now balanced, there is a problem. You can't have a fraction of a molecule in an equation, so we have to double all of the coefficients to change the 1.5 into 3.
$$\ce{2 Al2O3 -> 4 Al + 3 O2}$$
We are finally done. As you can see there are 4 aluminum atoms and 6 oxygen atoms on the right side and the left side. That is how you balance an equation and I hope it helps you solve your equation! Feel free to ask any questions.


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Determining the chemicals in potash and crystallizing them
I have made potash by leaching wood ash in water. The potash in solution has a high pH indicating it is alkaline, and reacts with acetic and citric acid. Is there an easy way to determine the different chemicals in solution? The neutralization reaction creates an odorless gas. I believe it is carbon dioxide which would make the following equation work:
$$\ce{K2CO3 + 2 CH3COOH -> H2CO3 + 2 KCH3OO}$$
The carbonic acid ($\ce{H2CO3}$) would then decompose into water and carbon dioxide:
$$\ce{H2CO3 -> H2O + CO2}$$
Which leads me to believe the potash contains potassium carbonate (from the reaction) and potassium hydroxide (from the pH.) Is there any flaw in my reasoning and how can I determine and separate the other components? Potash can contain potassium hydroxide, potassium carbonate, and potassium sulfate, among other things.
Once I determine the chemicals, is there a way I can easily crystallize certain compounds out of solution. Thanks.

@NilayGhosh Your formula is incorrect. The potassium cation has a single positive positive charge and the acetate anion has a single negative charge.

I think it would be quite difficult to separate all of the different compounds dissolved in the sample into their solids because they are all in the same phase. I do not think any method exists for this kind of separation.
However, determination is absolutely possible.
Firstly, regarding your deductions...
Although carbonates give carbon dioxide gas, which is colourless and odourless, it is nevertheless important to test that the gas is indeed carbon dioxide by bubbling it into limewater and looking out for white precipitate. If you have not conducted that test, then you cannot claim that the gas produced is carbon dioxide. Otherwise, you can conclude that there is carbonate in the sample.
The high pH is indeed indicative of high hydroxide ion concentration in the sample. That is for certain.
However, you have not proved that the cation is indeed potassium. Perhaps, you want to conduct a flame test or do analysis using atomic absorption spectroscopy (AAS). Procedure for flame test can be found here: https://www.soinc.org/sites/default/files/uploaded_files/flametest.pdf
If you would like to find out what other ions are present in the sample, carry out a simple qualitative analysis using some common lab reagents. Procedure can be found here: http://www.wiredchemist.com/chemistry/instructional/laboratory-tutorials/qualitative-analysis
By boiling off the water, you would likely get some ionic solids but I believe that it should contain a lot of impurities.