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Books showing colorful images of elementary particles....is this true?
Do electrons or protons or any other elementary particle have colors of their own or do they have smell?

No color and no smell.

But why do they dont have any color or smell?

Electrons can't just be "on their own".

Why should they have a color and a smell?!?

Isn't there a a sense of acidity associated with proton-donating substances? If I'm wrong, please clarify.

Books showing colorful images, using color to highlight elements in chemical formulas just want to make a visually noticeable difference between those elements, and (hopefully) don't intend on how those things look like. They don't have a color of their own (werrrrrl, mostly), unless  coupled with other tings.

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

The color is the interesting part here. They should react with photons to have any color. By molecules the chemical bonds can absorb some frequencies of the light, that why they have color. In theory a single electron or proton should be able to absorb photons too, so they might have color, but not necessary in the visual range of frequencies.

partial duplicate of: http://physics.stackexchange.com/questions/39076/what-color-would-a-proton-be-if-it-were-visible-to-the-human-eye

You can understand why they don't by understanding how smell and vision work
The mechanism by which we (or other animals) smell is not fully understood. But the key is that receptors in your nose are activated when certain molecules interact with them. The receptors consist of a wide variety of complex proteins with cavities of particular shapes. The cavities also have different patterns of surface structures which interact with the target molecules in a variety of ways (there might be some polar patches, patches that can hydrogen bond to another molecule and some that prefer non-polar interaction). When an appropriate target fits into the cavity, the cavity can send signals to nerves to signal that a target has been recognised. The targets can be very specific and there are a lot of different types.
Elementary particles don't (usually) exist alone but only as parts of other atoms or molecules. So they won't smell. We haven't evolved ways to detect free elementary particles. It is the 3D structure of molecules that creates smell.
The mechanism of sight is simpler. Molecules in the eye can absorb particular frequencies of light and then send a signal to the nervous system that they have detected some type of light. Light consists of photons. We see light when something emits photons, but only a narrow range of photons of about one octave. We can't see radio waves or X-rays as our eyes are not built to do detect potions of those energies. We see colour because photons of different energies can excite different receptors in the eye (human eyes have three different detectors tuned to red, green and blue light).
What creates coloured light is the either the emission of photons of the right energy or the selective absorption of light coming from another source. selective absorption happens when there is some way for the thing absorbing the light to capture the incoming photon's energy. This only happens when there is some part of the absorbing entity that can shift from one energy level to another and where that energy level matches the energy of the incoming photon. Some molecules have that property, but in many the energy levels don't match anything in visible light so they don't show colour.
Elementary particles don't have any energy levels in the appropriate range so won't be coloured. In molecules that do have energy levels absorption can happen, but this is a property of the whole molecule, not the atoms that make it up. It isn't the atoms that have colour, it's the molecule. Also atoms are about the same size as wavelengths of light, so even if you had a powerful microscope you could not see the individual components using light so the idea of them having colour makes no sense.
When presenting models of molecules, chemists often draw the atoms in different colours. But this is so they can tell them apart and in no way represents what an imaginary microscope would see. The colours are entirely arbitrary.

Electrons and protons do not have a smell because that means that they would have to reach our nose, which have molecule sized receptors, and cannot fit in the receptors.
Electrons and protons cannot have color since they sare smaller the visible light themselves! So how can they reflect something much bigger than themselves?

As far as I know, electrons are one of the most stable things in universe. And proton decay is still a matter of research.

@GyroGearloose electrons would be stable in pure vacuum for sure.

@_TanMath everything can be tackled if not in pure vacuum and attacked with some high energy particle.

So what do you reefer by "electrons and protons can be unstable by itself". I read "itself" as "not interacting with something else"

I think it is pretty clear that electrons per se have no color and are invisible . If you have a glass tube with a wire at each end and connect the ends to a high voltage power source, electrons do not flow. When you pump the air out of the tube until you have a pretty good vacuum, the electrons begin to flow and the tube emits light, due to collisions of the electrons with the atoms in the tube followed by decay of those excited atoms back to the ground state--the atoms are emitting the light, not the electrons themselves. (For example, if sodium atoms are present you get yellow light, and if mercury is present you get mostly ultraviolet light.) But - if you keep pumping on the tube until you have an almost perfect vacuum, the electrons still flow (you can measure that with an ammeter in the circuit) but there is no light. They are invisible.
If you dissolve sodium metal in liquid ammonia, you get a deep blue color. Some people would say that is the color of "solvated electrons", but they are not actually free electrons. It is believed that the result is ammonia-solvated sodium cations and ammonia-solvated ammonia radical anions. The blue color is due to absorption of visible light by the latter.

Who believes that the electron is associated with a specific ammonia molecules to form a radical anion?  Usually I see it in print as an electron sitting in a cavity between ammonia molecules.

According to Physicists find that an ultrahigh-energy proton looks like a black disk

What does a proton look like? The common answer to this question is that protons are much too small to scatter light, and since light is necessary for us to see things, protons do not “look” like anything. But in a new study, physicists have gathered sufficient evidence to show that, at least at very high energies, the proton is a black disk – sort of an elongated hockey puck.


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How does one determine the probability of H⁺ to H₂O in pure water?
The ratio of $\ce{H+}$ to $\ce{H2O}$ is something around $10^{-8}$. How did people determine that?
You cannot use the $K_\mathrm{w}$, $\mathrm{p}\ce{H}$, $\ce{[H+]}$ ion concentration value to answer because you get all that from the probability.                                                                   Is there any independent way to find the probability. Please elaborate on the steps.
I do not want to know why the ionic product is what it is. how was the ionic product found ?

I would guess that this number comes from conductivity experiments on ultra-pure water.

Technically, the probability in pure water would be zero...

@LewisMiller:  the conductivity of a solution also depends on electrical mobility, which I'm not sure that you can measure independently the conductivity itself.

You are allowed to use the conductivity values but please elaborate on the steps

Even pure water dissociates. Pure water has conductivity of the order of 10^-9

Possible duplicate of Why is the ionic product for water $10^{-14}$?, though this appears to be more of an experimental question rather than a ~philosophical one.

I disagree, that this is a duplicate. I think this question is really about the determination of the constant rather than the impact of a proposed change.

The answer to this question was surprisingly difficult to find because all modern ways to measure pH use reference solutions of known pH and thus rely on prior knowledge of the $H^+$ activity of said reference solutions.
Finally I found this on wikipedia:

it is possible to measure the concentration of hydrogen ions directly, if the electrode is calibrated in terms of hydrogen ion concentrations. One way to do this, which has been used extensively, is to titrate a solution of known concentration of a strong acid with a solution of known concentration of strong alkaline in the presence of a relatively high concentration of background electrolyte. Since the concentrations of acid and alkaline are known, it is easy to calculate the concentration of hydrogen ions so that the measured potential can be correlated with concentrations. The calibration is usually carried out using a Gran plot.[9] The calibration yields a value for the standard electrode potential, $E^0$, and a slope factor, f, so that the Nernst equation in the form
  $$E = E^0 + f\frac{2.303RT}{F} \log[\mbox{H}^+]$$
  can be used to derive hydrogen ion concentrations from experimental measurements of E. The slope factor, f, is usually slightly less than one. A slope factor of less than 0.95 indicates that the electrode is not functioning correctly. The presence of background electrolyte ensures that the hydrogen ion activity coefficient is effectively constant during the titration. As it is constant, its value can be set to one by defining the standard state as being the solution containing the background electrolyte. Thus, the effect of using this procedure is to make activity equal to the numerical value of concentration.

This means that the precise cocentration of $H^+$ is actually unknown what is known is it's chemical activity which has been defined as equal to the concentration for a certain standard solution.

You are, I believe, referring to the auto-dissociation reaction of water:
$$2\mathrm{H_2O}\text(l) \leftrightarrow \mathrm{H_3O^+}\text(aq)+\mathrm{OH^-}\text(aq)$$
This equilibrium reaction obeys the following equation (the square bracketed quantities are concentrations):
$$K_W=[\mathrm{H_3O^+}][\mathrm{OH^-}]=10^{-14}\:\mathrm{mol^2/L^2}$$
At $pH=7.00$ and in pure water this means:
$$[\mathrm{H_3O^+}]=[\mathrm{OH^-}]=10^{-7}\:\mathrm{mol/L}$$
$$pH = -\log[\mathrm{H_3O^+}]=7.00$$
The concentration of water in water is $55\:\mathrm{mol/L}$, so the ratio of $\mathrm{H_3O^+}$ to $\mathrm{H_2O}$ in pure and neutral water is about: $$1.8 \times 10^{-9}$$
Naked (non-solvated) protons $\mathrm{H^+}$ do basically not exist in water because $\mathrm{H_2O}$ is such a hard Lewis acid (nucleophile), which immediately bonds to any free protons.
The $pH$ of water is determined by means of a $\mathrm{H_3O^+}$-selective electrode, commonly known as a $pH$-meter.

The answer you have given is sort of circular. You cannot use the Kw, pH, [H+] ion concentration value to answer because you get all that from the probability.                                                                                Is there any independent way to find the probabitlity.

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Conductive Inks
I would like to know a few examples of Conductive (Electrical) Inks which can either easily be synthesised or bought online (Low Cost) and also happen to be transparent.
Thanks

By conductive I mean electrically conductive.

I think its a case of choose any two. Cheap and conductive is possible, but not transparent. Conductive and transparent has been researched, but not cheap. You'd also struggle to synthesise these things yourself without access to specialised lab facilities

If you don't need a rigorously transparent/conductive ink, PEDOT may work. It's often used mixed with polystyrenesulfonate (PSS) in an aqueous solution.

As @NotNicolau states, choices are limited. Indium-tin oxide is transparent, though it does absorb a bit of light and its high index of refraction causes some reflective loss. However, it is usually applied by vapor deposition, sputtering etc.
Nickel and silver conductive ink (paint) is readily available for window defroster grid or printed circuit board repair and has a resistance of a few ohms per square, but it's opaque.
Aquadag, a colloidal graphite paint, is inexpensive and has resistance ~ 1 kilohm per square, but is also is opaque.
That said, there is ongoing research on graphene-based conductive inks. In a thin (perhaps monomolecular) layer, graphene is close to transparent. You might buy or make graphene or graphene oxide, disperse it in ethanol, apply it as ink and test if its conductive enough for your purposes.
You can also buy In-Sn oxide coated PETE plastic.

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Enolate formation and tosylate displacement in cis/trans decalin
What will happen in the following reactions? This is Problem 3d from Chapter 6 of Zweifel and Nantz's Modern Organic Synthesis.

I thought that after the hydroxyl group is converted into a good leaving group ($\ce{OTs}$), an enolate would be formed which would attack $\ce{CH_2OTs}$, but I guess the fused cyclobutane thus obtained would be too strained. What else could happen?

There is a solution manual to the book which essentially fully supports the accepted answer.

I went looking for this question again today after talking to someone about a similar reaction, then realised I never got round to trying to actually answer it.
Leading reference: Tetrahedron 1969, 25 (21), 5267–5274.
In the first case, with the trans-fused ring system, we first get tosylation as expected. Using the bulky potassium tert-butoxide, we get enolate formation at the least hindered alpha position. This enolate is then setup, as can be seen in the conformational diagram below, to displace the tosylate forming the new ring. This is confirmed in the cited literature.

In the second case, with the cis-fused ring system, things are a little more complicated. Once again we get tosylation, however the the cis-fused product can be drawn in multiple 3D conformations, only one of which is productive (shown in orange below).

In this case, the steric argument for enolate formation doesn't hold, as the thermodynamic enolate is unproductive, and as such only the enolate formed at the most hindered position goes forward to form the product.
As you mentioned in your initial post, "the obtained structure would be too strained", which is what prevented me from thinking this pathway happened in the first place. However, looking at the original Tetrahedron paper, they do have evidence to back up the formation of these ring systems. One possible conclusion is that due to the conformational preferences of the 6,6-ring systems, the bonds are closer together than you'd maybe expect, meaning there isn't such a penalty to formation of the cyclobutane.

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What is the proper way to prepare a solution from hydrate?
I am following a protocol that requires me to prepare $10~ \mathrm{mL}$ $\mathrm{1~M}~ \ce{Ca(NO3)2}$. However, there is only $\ce{Ca(NO3)2*4H2O}$ in the lab.
Originally I thought that all I have to do to figure how much $\ce{Ca(NO3)2}$ to weigh is

and then just add  $10~ \mathrm{mL}$  water. However, that might dilute the solution and make the concentration less than $\mathrm{1~M}$ as there is already water in the hydrate.
So then I found a protocol and made new calculations as followed:

How can I verify that I got $\mathrm{1~M}$ $\ce{Ca(NO3)2}$ as opposed to $\mathrm{1~M}$ $\ce{Ca(NO3)2*4H2O}$?
Which protocol is correct?

The hydro-molecular equation is used to determine the volume of water contained within any given mass of a hydrated compound. This volume is then subtracted from the final volume of the solution.

Originally I thought that all I have to do to figure how much $\ce{Ca(NO3)2}$ to weigh is

and then just add  $10~ \mathrm{mL}$  water.

To properly make an aqueous solution of a specified concentration you should dissolve the desired amount of material in about $80\%$ ($8~ \mathrm{mL}$) the desired volume of water. Then dilute to the final solution volume (add water until the solution volume is $10~ \mathrm{mL}$).

How can I verify that I got $\mathrm{1~M}$ $\ce{Ca(NO3)2}$ as opposed to $\mathrm{1~M}$ $\ce{Ca(NO3)2*4H2O}$?

Hydrates are for the solid state.  You do not have "hydrates" in the aqueous phase.  Therefore the two solutions are identical and indistinguishable.

But the volumes of water added are different. How can they be the same solutions? I want it such that the concentration of Ca(NO3)2 is 1 M, no more, no less.

As AK says dilute solution TO 10ml total volume, not WITH 10ml.


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Under what conditions does MnO2 acts as an oxidizing agent?
It's very strong in acidic medium. Does it have low or nil oxidizing ability in neutral and basic media?
Does heating solid $\ce{MnO2}$ give it better oxidizing properties?

It certainly is  powerful oxidizer when heated! Thermite can be made with powdered aluminum and $\ce{MnO2}$, and is at least as spectacular as with $\ce{Fe2O3}$ as the oxidant.


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Non-destructive solvents for cyan-salts
What dissolves cyan-salts (NaCN and KCN) without destroying them the way water does via hydrolysis?
I have to make my own cyan-salts, so they're somewhat precious to me.  Every time I have to separate them (from metal cyanides as well as carbon and oxide impurities) using water I lose some: I can precipitate some from an aqueous solution with alcohol, but the rest hydrolyses significantly before it can be desiccated with the "amateur"-grade gear I have (I've tried both freeze-drying and vacuum drying).
What solvents can separate the cyan-salts from the other substances without hydrolysing, oxidizing, or otherwise destroying them?  (Ideally a solvent that is relatively volatile at room temperature so I can readily evaporate it to recover the salts.)

Would ammonia or hydroxylamine be doable with your resources/equipment?

@airhuff - I believe so, but in what form or solution?

What are the other salts/substances that you are separating out from the cyan-salts?

@airhuff: I smelt ferrocyanides, which produce NaCN and KCN mixed up with iron oxides and, inevitably, carbon.  (There are certainly some small portion of cyanates produced in the process as well, but those don't cause any significant problems for my purposes.)


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How can I name the products of the hydration of 3,5-dimethylhept-2-ene?
I know that the hydration of 3,5-dimethylhept-2-ene leads to several products, of which one is:

Is this called 3-oxo-3,5-dimethylheptane?

Is this called 3-oxo-3,5-dimethylheptane?

Not quite ;)
The prefix oxo is used if a compound has a $\ce{C=O}$ group and another group with a higher precedence that would give the suffix of the name.

If is quite obvious that there can't be an oxo substituent at the carbon atom that already bears the methyl group.
But you were pretty close!
The reaction with water under acidic conditions will be initiated by addition of a proton to the double bond. Of the two possible cations, the tertiary cation will be formed. When water adds as a nucleophile, you will obtain a 3-hydroxy compound.


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Is there a broader nomenclature convention to explain why the ‘a’ in ‘chlorophyll a’ is italicized?
By convention, the letter that specifies which side chain a chlorophyll molecule contains is italicized: for instance, chlorophyll a. Does this convention stem from any broader principle of chemical nomenclature, or is it more of a quirk unique to chlorophyll?

The trivial names, such as “chlorophyll a”, are listed in: Nomenclature of tetrapyrroles (Recommendations 1986). Pure Appl. Chem., 1987 59 (6), 779–832; Eur. J. Biochem., 1988, 178, 277–328. However, no rule is given for the italicized letters.

This is only a partial answer to your question.
The suffixes a and b to the name chlorophyll were most likely introduced by Richard Willstätter and Arthur Stoll. They are widely used in their book UNTERSUCHUNGEN ÜBER CHLOROPHYLL (Julius Springer, Berlin, 1913). Here, however, the suffixes are not yet italicized and appear as normal text.


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Interpreting notation format 1.64E-02 from particulate emission data
I'm doing a research project for school on air pollution and part of it is studying pollution caused by aviation and I have found a table of data for the emission indices of various pollutants but I'm not 100% certain, if I'm correctly reading the numbers.
The unit is $\mathrm{g\, kg^{-1}~of~fuel}$ and the numbers are like 1.64E-02 or 2.25E+00. I'm interpreting it as E being for Exa, so in short scale 1.64E would be 1.64 quintillion and that it means 1.64 quintillion particles per g per kg of fuel. Is that the correct meaning? I have no idea what the second bit of the numbers means so please could you explain for me?

@Martin-マーチン Added image

The format refers to a quite common scientific notation, where the $\mathrm{E}$ refers to exponent of $10$. It is often used when superscripted values become tedious to read and or write. It is widely known in computational sciences and was introduced for an easier display of huge or tiny numbers.
You can interpret the $\mathrm{E}$ as "times ten raised to the power of", hence the format can be interpreted like this: $$n\mathrm{E}m = n\times10^{m}$$
Therefore $1.64\mathrm{E}-02 = 1.64\times10^{-02}= 0.0164$ or $2.25\mathrm{E}+00=2.25\times10^0=2.25$.


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How to prove that the forward and reverse reactions have the same rate at equilibrium?
I was trying to answer a question from Zumdahl and Zumdahl's Chemistry textbook which asks me to show that the rate constant K is related to the forward and reverse reaction constants:
$$K=\frac{K_\mathrm{f}}{K_\mathrm{r}}.$$
In answering the question it is given that:
$$K_\mathrm{r}=Ae^{\frac{-(E_\mathrm{a}-\Delta G)}{RT}}$$
And something is lacking in my understanding here. As I understand it the activation energy of the reverse reaction is equal to $E_\mathrm{a}$ of the forward reaction plus the free energy. That takes you back to the transition state - but the transition state is defined as the point at which products always become reactants, so how are reactants being formed from the products?

Are $K_\mathrm{f/r}$ rate constants?

You can find your answer here, http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch22/react.html (take care with your second equation, above).

In the reverse reaction, the products and reactants and switched over from the forward reaction. Products are just what is made from a chemical reaction, not necessarily your desired or intended 'product'.


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Why carbon is so special?
One special branch in chemistry is allotted to compounds of only one element (carbon). Is it justified when there are more than 115 elements and their compounds but not with any special branches?

Most of the compounds that make up life on earth are based on carbon. That is why they are studied so thoroughly.

related http://chemistry.stackexchange.com/questions/43744/carbons-catenation

Carbon has a valency of four which helps in bonding with other atom or groups easier and adds to catenation. When one talks of Carbon's ability to bond, then Catenation property should come to your mind. Catenation is the linkage of atoms of the same element into longer chains. Catenation occurs most readily in carbon, which forms covalent bonds with other carbon atoms to form longer chains and structures. This is the reason for the presence of the vast number of organic compounds in nature. One such marvel is DNA! So, while studying chemistry, Carbon compounds form a class of peculiar natured compounds and the vast range of molecules and development of their own domain in their own scale might have made people feel that a separate category should be made for Carbon compounds because of their vastness and availability, also predominately manifestation of them in almost every thing, even life! In fact life became possible only because of Carbon, such is the importance. They have got their own nomenclature system in their domain.. Its really vast. The study of everything begins with Organic compounds first.. be it isomerism, medicinal reaction or anything. So, you may see, how important Carbon is. If you are a biology student, you must have learned about biomolecules.. DNA and gene pool, replication, transcription, translation, protein coding, defective proteins and some viral components and its lifecycle. Molecular basis of genetics is based on DNA and other organic compounds. That gives you thrilling experience to see how stuffs work and wonder about their importance and manifestation. Carbon is a wonder.


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Autoclaving certain plastics?
My professor is letting me use the autoclave in the future, but I am a novice. I asked him what materials were safe to autoclave for my notes, materials meaning glass or plastic-ware. He stated, "It depends on alot of factors, but some hard and opaque plastics are generally ok, also Teflon products."
I couldnt really ask him any more questions since it is still winter break and I guess he isnt responding to emails, so I decided to google the issue.
I found this source of an autoclave guide at: https://uwaterloo.ca/safety-office/sites/ca.safety-office/files/uploads/files/guideline-for-safe-use-of-autoclaves.pdf

Good Choices:
borosilicate glass (Pyrex) has very low thermo
expansion property and therefore resistant to breaking due to heating
polypropylene (PP) and polycarbonate (PC) are heat resistant plastics
stainless steel is a good heat conductor and thus facilitates
sterilization
Poor Choices:
polystyrene (PS), polylethylene (PE) and high density polyethylene (HDPE) do not resist heat well.
if there is a risk of material melting ensure they are placed in a
secondary container which is resistant to heat.

Would this be an accurate  summary of what plastics / glassware to use in an autoclave? I would ask about the validity of the website itself, but I assumed it was reliable because it was based off of a college.

The list looks accurate. For example, the melting point of polypropylene ranges from ~130 C to ~160 C, so it's likely safe to autoclave.  Commercially available polyethylene melts ~100 -110 C, so is unsuitable for autoclaving.
Another common plastic, not listed above, polyethylene terephthalate (PETE), used for packaging and soft-drink bottles, softens below 100 C and is also unsuited for an autoclave.
Electrical and optical items used in endoscopic procedures are often heat-sensitive and cannot be autoclaved, posing difficulties in removing contamination.
As suggested, if you're unsure but must autoclave an item, place it in an open or perforated container so as to reduce risk of damage to the sterilizer.


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Electrolysis of water for demonstration: ideal electrolyte for safety, stability, conductivity, and ease of maintenance
I'm developing a system for educational purposes where electrolysis of water takes place. The aim is to practically show how electrical energy can be converted to mechanical, by electrolyzing water, igniting the mixture of $\ce{H2}$ and $\ce{O2}$ produced, and accelerating a mass upwards.
Currently I'm using citric acid in distilled water ($4\%$ in weight, $18~\mathrm L$ water), platinum-coated mesh electrodes, $500~\mathrm W$ max power supply ($0\!-\!65~\mathrm{V_{DC}}$, $0\!-\!10~\mathrm A$). I'm wondering if there's any better option that meets the following requirements:

Not harmful to maintainers (that have to prepare and replace the liquid) and bystanders
Stable (not producing precipitates, and in general that can minimize the need for replacing the liquid)
Low/no interaction with the materials in the tank (Plexiglas, brass, platinum, silicone tubes, ABS)
Maximizes conductivity (the current setup at $20~^\circ \mathrm C$ has a conductance of $250\!-\!300~\mathrm{mS}$)
Not suitable for microorganism development (in the current setup algae develops in about two weeks)

If the last point can't be improved with just the electrolyte, I'd kindly ask for a suggestion for an additive that is not going to spoil the points above.

Sodium carbonate, maybe?

@IvanNeretin Signigificantly lowered conductance

Welcome to Chem.SE - great question!  I think something's wrong with your conductivity measurement, though. Per here, citric acid at 4% w/w should have a conductivity down around $4~\mathrm{mS \over cm}$.

Diluted sulfuric acid. High conductance (because hydrogen cations), definitely no electrolized itself (except persulfuric acid formation at low temperatures), nontoxic. Unfortunately, dangerous in high concentration, but that's workable.  Phosphoric acid may work as well.

@permeakra Phosphoric acid has terrible conductivity at anything under 10% w/w. Probably not safe enough.

I'm thinking white vinegar plus sodium bisulfate.  The big question at this point is the compatibility of the acrylic and the vinegar. Answer pending once I can get it fleshed out.

oxullo, are you using a brass cathode, or do you  have brass fittings or somesuch built into the system?

@Brian thanks for pointing me out the pdf. I believe conductivity is heavily affected by shape/surface and distance of the electrodes. Since they're a mesh, it's hard to provide a number and anyway they're quite close (5-6mm). Moreover, I calculate the conductance in DC with a steady 10A flow, definitely not a standard procedure.

The observed cell resistance definitely is affected by electrode shape/surface/separation.  Conductivity is an intrinsic property of a solution at a given composition and temperature. Going from the former to the latter in a system not specifically designed for the measurement is challenging. :-)

@Brian: only fittings. The electrodes are both platinum-coated. Currently the liquid gets tinted, turning from clear to yellow and slowly to green. I believe (I'm no expert) that brass has something to do with it.

@permeakra thanks for the suggestion, sulfuric acid was my first theoretical choice but I've never tried it. Can you suggest a concentration, please?

@oxullo I'd start with ~5%  (~20-times dilution) and work from here.

@permeakra The silicone tubing is the main concern, having unsatisfactory performance in $10%~\ce{H2SO4}$ (per Schweitzer in my answer to come shortly) and only "Fair" performance in $<10%$ per Cole-Parmer's compatibility guide.

@Brian Silicone tubes should not come into contact with the electrolite anyway.  I would be more concerned with plexiglas, I missed it. But I would recommend to move to glass tank anyway.

I suggest $1\%$ w/w $\ce{NaOH}$. Point-by-point:

$1\%$ w/w is $\sim\!10~\mathrm{g\over L}$, which for $\ce{NaOH}$ at $40~\mathrm{g\over mol}$ is $\sim\!0.25~\mathrm{M}$, corresponding to a $\mathrm{pH}$ of $13.4$. Standard personal protective equipment (coats, gloves, goggles) should suffice for observers and maintainers.
$\ce{NaOH}$ is extraordinarily soluble, so you will have no difficulties with precipitation.  You will want to keep the system covered when not in use, if you're not already, to minimize evaporation.  You may get some riming on the interior, but rinsing with DI water should take care of it.
According to Schweitzer$^\dagger$, the materials in your system have the following resistances to $10\%$ (not $1\%$) $\ce{NaOH}$:

Plexiglas (acrylic): Resistant, to $80~^\circ\mathrm F$
Brass: Good (less than 0.02" penetration per year), to $200~^\circ\mathrm F$
Silicone elastomer: Resistant, to $80~^\circ\mathrm F$
ABS: Resistant, to $140~^\circ\mathrm F$
Platinum: ..... darn well better hold up! (Though, if there are any gaps in the coating, the substrate might be exposed to attack.)

Per the first chart here (PDF link), the conductivity of $1\%~\ce{NaOH}$ should be about $50~\mathrm{mS\over cm}$, which is over ten times the $\sim\! 4.5~\mathrm{mS\over cm}$ of your current $4\%$ citric acid recipe.
I salute any common microorganisms able to grow in a $\mathrm{pH}~13+$ solution. (See, e.g., Table 3-5 here, where the maximum survivable $\mathrm{pH}$ for any species listed is $11$.)

Obviously, if you should switch, monitor the health of the system materials to be sure the above literature chemical resistance data actually holds in reality.  There's always the chance some small component of the system is of a material you haven't listed, too.
(To note, I re-thought the vinegar-plus-sodium-bisulfate approach after discovering that both brass and acrylic/Plexiglas are poorly resistant to vinegar, and that the bisulfate is not likely to provide as much of a conductivity boost as I was thinking.)
$^\dagger$ P.A. Schweitzer. "Corrosion Resistance Tables, Part C, P-Z." 4th Ed. New York: Marcel Dekker, 1995, pp. 2685-2688. (Amazon link)

Awesome suggestion, @Brian! I've just ordered a kg of pellets and I'll be able to perform a test early next week and feed the results back here.

Under test since yesterday, the solution stays clear, conductance is high (350mS at 10A, 20∘C) even though not 5-fold, electrodes look healthy (no fancy deposits), materials compatibility looks good (even though I can't really predict what happens on long term). I'll give it a try on one of our prototypes, crossing my fingers for a decent blast :) Thanks!

@oxullo Great! Hope it works out well for you. If you end up wanting more conductivity and the system appears to hold up well, you could probably bump the $\ce{NaOH}$ concentration up to $2$ or $3%$ without much worry to safety ($\mathrm{pH}$ still below $14$) or to attack on the materials. Caveat emptor, of course....


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Carbon's Catenation
I'm a ninth grade student and I don't understand how carbons catenation property allows the formation of many millions of organic compounds.My doubt is there is no need of linking with the same atoms to form many different compounds.

Could you say exactly what you're asking? Your second sentence suggests heteroatoms like oxygen, am I right? Then it's a good point - this way stable silicon polymers are obtained, instead of weak Si-Si bonds, they have strong Si-O bonds.

By 'catenation property' I'm guessing you mean the ability of carbon to form long chains? This is one of the reasons for the large number of carbon compounds, although it may not be the main reason (or it may, I just don't know).
Consider that it is possible to have molecules with 21 carbon atoms - the 'heneicosanes' apparently - and that just the number of possible tree-like molecules is 2,144,505 then we are already getting into the millions.
Now consider that we could have double or triple bonds anywhere in those molecules, along with atoms of other elements (O, N, S, etc) and you should see why there are so many possibilities. In contrast, you only get sulfur chains up to around 8 (?) atoms.
The other important property of carbon is having up to four neighbouring atoms. Although that's a bit harder to show, the number of possible arrangements is much larger than for elements that can only have two or three. It can be helpful to just try drawing out all possible structures for a small elemental formula (like C4H10).


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Explain how to filter Ca and Ba ions separately out of solution using insoluble compounds
I had a question on a test that said to use insoluble compounds to make precipitates to filter out Ca and Ba from solution separately, but I couldn't find anything that will react with one and not the other??

See part VI of  this. The process is explained quite well there. It is a downloadable pdf file

You could carry out the standard qualitative analysis procedure. Dissolve the mixture in  hot acetic acid solution.
Then add pottasium chromate ,barium is precipitated as its chromate salt(pale yellow ppt).
Filter it and you could then precipitate Calcium from the resultant filtrate by adding ammonium oxalate.(precipitated as calcium oxalate, white ppt)


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Why does fluorine stabilise a carbocation?
I have read that halogens have a stonger -I (inductive effect) than +M (mesomeric effect),then why do they stabilise a carbocation instead of destabilising it?
I asked this doubt to my teacher and he said that halogens like fluorine do not exert -I on a carbocation and hence they stabilise a carbocation through +M,is this true?

I had the same doubt. Check this: http://chemistry.stackexchange.com/questions/40076/stability-of-carbocations-cf3-vs-ch3

Fluorine belongs to the same period as carbon. Carbocation is electron deficient which has its two $\mathrm{2p}$ orbitals vacant. One electron of carbon in $\mathrm{2p}$ orbital goes into the formation of $\ce{C-F}$ σ bond. $\ce{F}$ has $\mathrm{2p}$ orbital containing a lone pair which can be donated to the $\mathrm{2p}$ vacant orbitals of carbocation. Same size $\mathrm{2p}$ orbitals of carbon and fluorine overlap effectively to form a π bond.

Carbocations are stabilized by neighboring atoms with lone pairs. The key stabilizing influence is that the neighboring atom can donate a pair of electrons to the electron-poor carbocation. In this way halogenes, nitrogen or oxygen can actually be an electron-donor group and stabilize the carbocation.

This is stated in the question. The actual question is about why the inductive effect does not dominate the mesomeric effect in carbocations, whereas it usually does for fluorine.

Most of the times the question that comes in the examinations is that why CH2F+ is more stable than CH2Br+ or CH2I+.
The reason behind it is that the p orbital of fluorine overlaps more effectively than that of Br or I with the carbon atom and therefore disperses the charge on it.
refer to K L Chugh for the details.

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


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What are the chemical similarities between gallium and barium?
I am analyzing nutrient concentrations in wood, and I have found that in all my samples the distribution of gallium and barium are closely related. As a related phenomenon, the presence of these elements also sometimes correlates to that of another pair of nutrients, calcium and strontium, but not always. For example, in samples from a humid site $\ce{Ba}$, $\ce{Ga}$, $\ce{Ca}$ and $\ce{Sr}$ are always closely related, but in samples from a dry site $\ce{Ba}$ and $\ce{Ga}$ remain correlated to each other, but are generally not strongly correlated with $\ce{Ca}$ and $\ce{Sr}$.
I would like to know what aspects of their respective chemistries might cause $\ce{Ga}$ and $\ce{Ba}$ to have this strongly correlated relationship in wood samples.  For example, is there some sort of similarity in how their valence electrons participate in bonding?

I would think we should explain why its probably not chemistry related: Metals are mainly biologically active as ions. Barium makes a large +2 ion while strontium makes a small +3 ion. This difference makes it unlikely that they would substitute for one another. In contrast, lithium treats bipolar disorder by interfering with the sodium systems in the brain; this works because they're both small +1 ions. This points to the idea that they do not serve similar roles. In addition to a biological system using both of them in separate roles, it may also be geologically caused.

See "Trace metals in soils, plants, and animals", ADVANCES IN AGRONOMY, Volume 24, especially at page 295, which suggests that solubility of barium and gallium are promoted by poor drainage conditions.  https://books.google.com/books?id=GUBABXKJiWYC&pg=PA295&lpg=PA295&dq=barium+gallium+plants&source=bl&ots=Rh7nPpsw3e&sig=-LDPkrlTXUwMXflPNYcpgak3Lew&hl=en&sa=X&ved=0ahUKEwjjrf3a9KnKAhWIKJQKHceSDjsQ6AEIQzAF#v=onepage&q=barium%20gallium%20plants&f=false

@ericksonla I agree, this aspect of the differences between the ions would be on-topic to Chem.SE. However, I think the question as-written is too broad. Darwin, if you're interested in the ion size aspects raised by ericksonla, by all means put together another question focused thereon and post it on Chem.SE.

@Brian since Ga, Ba, Ca and Sr are chemical elements (or nutrients) I posted my question here. I think my question is clear and it is related to observed chemical phenomenal in my measurements. Just I want to know some common characteristics between these elements. E.g. they use only the outermost shell orbitals in bonding or something like that. DavePhD gave a very good contribution, but you just want to put as off-topic my question with any contribution. Then, tell me how I can write to sound like a chemistry question?

@ericksonla - as a group II metal, strontium does not make  a +3 ion

authors: Νίκη Αθανασοπούλου


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Why plastic sheets contract when heated?
Why do thin plastic sheets contract when heated, contradictory to the behavior of most other materials ?
What are the things going on at the molecular level ?

what material are you considering?

Do you mean 3D/volume or 2D/area contraction ?

When plastic sheets are produced, they are rapidly cooled to keep the polymer chains oriented in a way that makes the sheets nice and flat. This is a relatively high-strain orientation since it is associated with the energy level of the molecules at the casting temperature.
Once the plastic is heated above its glass transition temperature, the polymer chains are no longer locked in that high strain orientation. They relax to a low energy orientation- curled and bending in a way that shrinks the bulk material.
As for the precise mechanism, I'm not exactly sure. I would guess that the shrunken conformation is entropically favorable because there are more arbitrary bends. This would decrease the Gibbs free energy, making it a more stable shape.
Alternatively, it could be that hydrogen bonding between chain elements makes the folded shape more enthalpically favorable.

Just for fun I heated up some yoghurt containers ... they all flattened out to the flat sheet of plastic they had been before they were heat-pressure deformed into cups.

It's the entropy for all polymers. Surface energy makes up a small additional term.

It's not glass transition, but melting of nanocystallites, which crosslink the sheet, stabilising it. The Tg of Polyethylene is below room temperature, and your saran wrap would be very brittle if it had to be used below its Tg.

Sheets are usually made by film blowing, or some other technique which involves uniaxial or biaxial stretching followed by rapid cooling.
The cooling (and, usually, crystallisation) of the material is usually so fast that the polymer chains have no chance to relax from their elongated state.
In the film or sheet, the chains are now frozen in in their oriented, stretched state, which strongly increases the toughness of the film. If you heat it again, so the microcrystallites melt, the film relaxes into its next best entropically favoured form, which is a wrinkly ball.

Different plastic types may behave differently, like thermoplastic and thermoset plastic. Thermoplastic sheet is made by melting the plastic. It further classifies into amorphous and semi crystalline type. Amorphous material doesn't have a sharp melting point. Semi crystalline material has a sharp melting point. Thermoset material like phenolic is not melted to form.
For thermoplastic amorphous material, like ABS, acrylic, PVC, polycarbonate, etc,
I think it depends on how you heat it up. If you heat it up from one side, the heat distribution is not even. One side is hotter and one side is cooler. One side then has a greater thermal expansion than another. The difference in thermal expansion creates an internal stress. When the temperature gets higher, the plastic goes softer. Until the plastic structure is weaken to a point that it cannot hold the stress, the plastic sheet bends.
Another scanerio is the internal stress already built up during manufacture. It highly depends on the manufacture process, the material use and the thickness. When you heat it up (evenly), the material is anyway soften. When the plastic sheet is so soft that it cannot hold the internal stress. It bends.
For thermoplastic semi crystalline material, like Polyethylene,
It itself is so soft having a low flexural modulus. It does not have much intenal stress otherwise it will contract itself.. When you heat it up closed to the melting point, it turns to a liquid state. The surface tension makes a tendency for it to turn to a sphere. Hence it contracts.
For thermoplastic sheet like phenolic, it behaves very different from thermoplastic when heated. It may bend a little but more often it burns.


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When dispensing liquid nitrogen, why does the surface get wet rather than frozen?
When I go to dispense liquid nitrogen after someone else has just used it (say in the last 30 minutes or so), the copper pipe used to transfer the liquid is covered in frost/ice.
This makes sense, as the pipe gets incredibly cold, condenses water out of the air which freezes extremely quickly due to the vast difference in temperature (77K for liquid nitrogen vs. water freezing at around 273K).
What I find confusing is that once I start dispensing the nitrogen, this frost/ice appears to 'melt' into a liquid, this seems less intuitive to me. My first thought is that so much new water is condensing out of the air that the frost/ice gets 'wet', however I'd expect this to freeze very quickly and the ice to eventually get so thick that this issue of wetting became irrelevant. Alternatively, one could imagine condensing something else out of the air which is indeed a liquid at these temperatures such as oxygen, but in this case I don't observe the characteristic blue colour.

An example of the type of dewar in question

Your "first thought" is basically right, with one clarification. The addition of "so much new water" to the existing ice results in the formation of liquid water due to the latent heat released upon condensation. This results in melting of the existing frost and thus the transitory existence of liquid water.
Also you can rule out oxygen condensation.  Firstly there is some insulation between the 77 K liquid nitrogen and the atmosphere, so the surface is not likely to cool all the way to the 90 K boiling point of oxygen.  More importantly, oxygen would not condense from the air at 77 K anyway.  You have to think in terms of saturation vapor pressures, and the vapor pressure of oxygen in our air is only about 0.21 atmospheres.
By analogy, albeit an extreme one, the temperature in the coldest regions of our atmosphere drops below the sublimation point of carbon dioxide.  Does that cause the formation of dry-ice clouds?  Not even close. The sublimation point of carbon dioxide in this example, like the boiling point of oxygen in your example, is only relevant at 1 atmosphere of pressure of that compound, and in neither case does our atmosphere contain 1 atmosphere of either of those compounds.


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How to compare the energies of sigma and pi bonds in molecular orbitals of diatomic heteronuclear species?
There is a rule for diatomic homo-nuclear species that says that if $Z<8$ then $$E(\sigma_{2p})>[E(\pi_{2p_y})=E(\pi_{2p_z})]$$ and the inverse if $Z\geq8$ and in both case we have energy of $$E(\sigma_{2s}^{*})>[E(\pi_{2p_y}^{*})=E(\pi_{2p_z}^{*})].$$
This is understandable but what if we have diatomic heteronuclear species like $\ce{NO}$ or $\ce{CO}$?
How can we determine which of the molecular orbitals $\sigma$ or $\pi$ is higher in energy? I.e. how should we know which to place below which in the molecular orbital diagram (sigma or pi orbitals)?

Actually, the gist idea is the difference of energy between the  orbitals $E_{2s(\mathrm{A})}-E_{2p(\mathrm{B})}$ and $E_{2s(\mathrm{B})}-E_{2p(\mathrm{A})}$.
If the difference is less than 10 eV in absolute value, there are interactions between the orbitals $2s$ and $2p$ and the diagram for diatomic homo-nuclear species with $Z<8$ is valid. (with a slight difference is that the levels of energy of the atomic orbitals of the  more electronegative atom are more negative than the those of the other one). This is the case for the molecules you propose, i.e. $\ce{CO}$ and $\ce{NO}$.

I am taking a general chemistry course and in any given question i am not given the values of the energies you mentioned.
Is there another method for solving this problem?

To the best of my knowledge, it is the only criterion. But, in general for the molecules of type AB, the molecular diagram of $\ce{N2}$ is valid (Z<8). Otherwise, the teacher has to report an indication or a hint in his question to use the other diagram (Z>8).


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Equilibrium saturation/depostion rate calculation
I found the Knudsen/Langmuir equation for calculating the sublimation rate of a solid. If I have a closed system, once the vapor reaches the saturation pressure, the solid will continue sublimating but will deposit at the same rate in a dynamic equilibrium.
Is there a relatively simply (ie, not-statistical mechanics modeling) way to calculate the dynamic rate of sublimation/deposition?


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Why does AsH5 not exist?
Why does $\ce{AsH5}$ not exist? Is it because the s-orbital of H doesn't and can't overlap with the d-orbital of As after As makes $\rm sp^{3}d$ hybridization?Theoretically the bond can occur because we can make $\rm sp^3d^2$ hybridization for As to give rise for 5 orbitals with 1 electron each and which can accept the  5 H atoms, but why is this not true in reality?

Related: Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't?

This is definitely not a duplicate question.  See especially page 15 of "Ligand Coupling Reactions with Heteroatomic Compounds"  https://books.google.com/books?id=gGNvHkXpKkIC&pg=PA15&dq=ash5&hl=en&sa=X&ved=0ahUKEwi72buEiazKAhUBy2MKHcR2D_wQ6AEIHDAA#v=onepage&q=ash5&f=false

@DavePhD You can convert your above comment into an answer.


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Liquid nitrogen superheating with a layer of oil?
edit: Thanks Ivan, also further research.
Pure water can be superheated to a few degrees above 100 degrees. However I remember reading reading that water can be superheated to even higher temperatures by covering it in a thin layer of oil that more effectively prevented the water from boiling. When disturbed the water can then suddenly boil and vaporise at a very fast rate. Apparently a microwave is an effective way to heat the water for this purpose.
However this can easily become very dangerous.
As liquid nitrogen boils at -196 degrees C, would using liquid nitrogen instead, with a thin layer of oil allow the same suppression of boiling, superheating, and sudden vaporisation to be observed but at much safer temperatures, as well as keeping the liquid nitrogen in it's liquid state at higher temperatures?
See also: physics.stackexchange.com/questions/29392/why-droplets-of-water-under-oil-explode-when-heated

Oil does not really matter much; you can pretty well superheat water without any oil. But in any case, you can superheat a liquid only by a few degrees, not hundreds. So superheated liquid nitrogen would look and behave pretty much like ordinary liquid nitrogen, which is not all that spectacular. Also, I wouldn't call it totally safe (though in most cases it is less harmful than boiling water).

I've just seen that this can be done with only pure water alone. I remember reading that using water and oil the temperature that the water can be superheated to is much higher than 100 degrees, although I don't remember where I saw this information and now cannot find it. I've also found the MythBusters episode on using only pure water, the explosion was a lot less vigorous than I remember reading this can be, there was still a lot of liquid water left and the microwave was fine. See physics.stackexchange.com/questions/29392/why-droplets-of-water-under-oil-explode-when-heated

I hope you don't mind if I incorporate this new information into the question.

According to this source, the highest you can metastably heat liquid water at ambient pressure before it boils spontaneously is about 300 ºC, which is actually much higher than I expected. For liquid nitrogen, the upper limit is -163 ºC, only 33 degrees higher than its normal boiling point.

Thank you Nicolau Saker Neto.
Superheating is possible up to 300°C for water, yet disappointingly only -163°C for liquid nitrogen.
Source: http://www.nist.gov/data/PDFfiles/jpcrd276.pdf


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Why would methanol and water be used for an aqueous layer in a separating funnel compared to just water?
I have a protocol for separating an inorganic component from an organic mixture, which asks for the mixture to be suspended in methanol and water and then doing an extraction with hexane. I understand that this allows separation of polar components from non-polar components. I am wondering what use methanol in water has in liquid-liquid extractions compared to just using water? My only ideas are: lowering the polarity of water, increasing the density, or somehow allowing some compounds to collect at the interface.

Lowering the surface tension, maybe?

@IvanNeretin what does a lower surface tension do?

Lower surface tension would make it easier to get good phase intermixing when agitating the mixture.  Droplets of water/MeOH in hexanes would be smaller, last (slightly) longer, and better facilitate the equilibrium partitioning of solutes between the two phases.

Another idea: the methanol is easier to remove than water, so after the extraction is done and the inorganic component is in the aqueous phase, the time & energy required to remove all the solvent is lower than if 100% water was used instead.

authors: 電訊巴打 - 電訊優惠比較平台


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How can borane exist?
Disclaimer: Sorry if this is a noob question, I am a high school chem student.
So I was reading my textbook in preparation for the midterm, and it talked about exceptions to the octet rule.
I understand that they exist, and what they are, but how are the chemicals structurally stable?
I also know that they are reactive, but that still doesn't help.
How do they exist?
I was thinking something like only under high pressure/heat, but I can't find anything to back that up.  Amazingly, google hasn't help.
Can somebody please explain?

Also, feel free to edit my tags.  I am new here.

Octet rule is less absolute than you seem to think. Sure, it would be nice to have a full octet, but what if the electrons are just not there? Then we'll have to do without, that's all.

Search for 'octet rule' here. There are a whole bunch of questions which will answer yours.

related http://chemistry.stackexchange.com/questions/228/what-makes-banana-bonds-possible-in-diborane

@IvanNeretin Okay.  I didn't know that, and assumed the atoms just wouldn't form a molecule in that case.


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O2 near magnets?
Now we know that $\ce{O2}$ is paramagnetic. I was just curious to know if we can find $\ce{O2}$ in gaseous state near magnets (of sufficiently strong intensity)? Or is it just that $\ce{O2}$ displays paramagnetism in liquid state only?

Indeed, gaseous $\ce{O2}$ is paramagnetic as well. As to your first question, we live on a planet with oxygen-rich atmosphere, so $\ce{O2}$ can be found pretty much everywhere even without magnets.

Yes - if you put a magnet in a room, theoretically, there would be more O2 near the magnet than further away from it. However, you need a pretty darn strong magnet to make the effect even barely noticeable, which is why we don't die when there's a magnet on the other side of the room from us.


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SN1 vs SN2 in low concentration of nucleophile
Why is SN1 mechanism favoured by low concentration of nucleophilic reagent? It is a first order reaction so shouldn't the reagent be in excess? Also higher the concentration of reagent faster the reaction right?

Let us assume that the competing reaction is a Sn2 based pathway.
Sn2=k[substrate][nucleophile]
Sn1=k'[substrate]
Relative rate of products formed=(ie.. Sn1 to Sn2 product)=k'/(k[nucleophile]
Sn1 Reaction doesnot depend on concentration of the nucleophile , so more the concentration of nucleophile, the relative rate of Sn1 product diminishes.
Hence ,when the competing reaction is a Sn2 pathway, lesser concentration of nucleophile would favour the Sn1 pathway.
Hope this helps.

Wow I never thought of it like that... You seem to be knowledgeable about organic chemistry. I am writing jee. Could you tell me how I should study organic chemistry and the books I should use?


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Why does phosphine reduce HgCl2 to Hg but LiAlH4 / LiH reduces it to Hg2Cl2?

Why does phosphine reduce $\ce{HgCl_2}$ to $\ce{Hg}$ but $\ce{LiAlH4}$
  or $\ce{LiH}$ reduces it to $\ce{Hg2Cl2}$ ?

It seems phosphine has more reducing tendency.But why?
Moreover can it be predicted whether a reducing agent will be reduce $\ce{HgCl_2}$ to $\ce{Hg}$ or $\ce{Hg2Cl2}$ ?

Actually $\ce{PH3}$ reacts with $\ce{HgCl2}$ to form a mercury(II) salt called  chloridomercury(II) phosphide. see this:- http://chemiday.com/en/reaction/3-1-0-9509


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How can sodium chloride melt ice or keep it frozen?
In European countries, they use $\ce{NaCl}$ or $\ce{KCl}$ to melt ice during the winter season.
In Asian Countries, they use $\ce{NaCl}$ to keep the ice without melting, for example in ice cream and beer boxes.
I asked my chemistry teacher about this and his answer was,

when you add ions to the ice, its melting point goes (i.e. freezing point of water) to -9 to -19 degree Celsius, depending on the amount of ions. The reason behind that is, when water becomes ice its water molecules re-arrange to specific shapes (see the link for the video at bottom). After we put ions to it, ions disturb that shape as they come in-between the water molecules. If water wants to be ice again, it should be like -9 degree Celsius because of these ions disturbing them to rearrange as a solid.

I also asked my Physics teacher about this and his answer was,

When we put salt into water, its temperature goes down to somewhere around $\pu{-6 ^\circ C}$, so that it takes more time to come to $\pu{0 ^\circ C}$ and begins melting

Link to the videos are as follows,

Salt melts ice (YouTube)
Salt reduces the temperature of ice (YouTube)

Could you give me a correct explanation for this question?

The apparent contradiction is resolved as follows: in both cases, addition of salt makes ice melt faster, which is important in the street and negligible in a beer box (or an old style cold bath used in the lab). But in doing so, it also causes the temperature of the ice and its surroundings to go well below $0^\circ\rm C$, which is important in the lab and negligible in the street.

also liquid has more surface area with the icebox than ice does, so adding the salt helps increase surface area by melting the solid water to liquid water

only the initial melting directly after adding salt will be faster but melting the bulk of the ice will be slower. @Ivan Neretin

Would it? Well, anyway, suppose there is no bulk, just a thin layer of ice all over the street, and you want to get rid of it. Then you may benefit from that initial faster melting.

The main difference is, that in order to melt the ice, you add the salt to the ice itself, where it lowers the melting point and thus (if the ambient temperature is only just below zero). Melts away the ice from the road.
On the other hand to keep ice cream cold, you add ice+salt around the ice cream that you want to prevent from melting.
This has three effects that help you cool your drink/ice cream faster to a lower temperature:

The melting point of the ice around your ice cream is below zero. This means that the temperature on the outside of your ice cream box cannot rise above zero until all the ice on the outside is molten and the temperature of your ice cream is kept below zero for that duration. The reason for this is, that it takes energy to break up the structure of solid ice. This is the same effect that keeps your coke at zero degrees until the last of your ice cubes has melted (and is the reason why adding more ice than necessary just reduces the effective volume of your container and does not help to keep your drink cool).
As your teacher correctly pointed out, when salt dissolves in water, the temperature of the water is lowered (because of the entropy vs. enthalpy of dissolving crystalline salt).
However, as pointed out in the comments, this effect is small compared to the others.
Your drink/ice cream is cooled faster because the very cold salty water (ideally with some ice cubes still swimming in it) around the bottle transports the heat away from your drink faster than some ice cubes touching the bottle would.

When salt ($\ce{NaCl}$) dissolves in water, the temperature stays pretty much the same, because the enthalpy of dissolution is all but negligible. And if you are talking about salts in general, there are examples of any sort: some would make water pretty hot, some pretty cold.

I did not specify how strong that effect was. I just looked it up and 1mole of NaCl cools down one litre of water by close to one degree Celsius. So with a solubility of 6 M, you could cool down water by about 6 degrees celsius. This is exactly as the teacher of the OP said and whether this is a lot or not depends on the use case. But considering that this could cool your drink well below freezing and thus make your tongue stick to the glass, I would say this should not be left out. And yes, I was talking about table salt since that was the question.

Well, yes, technically that's true. However, in any realistic setup some of that cold would be lost for cooling down the reaction vessel, and some would creep away to the surroundings, so you'll hardly be able to literally freeze anything at all. The enthalpy of melting of the ice is two orders of magnitude greater, so I think the word "negligible" (applied to the enthalpy of dissolution) is justified here.

Given enough salt and water, I am sure we could make an ice cube. However, this is completely besides the point. I was aiming to give a complete answer and since the op already mentioned this effect, I explained how it applies to his question. You are free to your opinion that it is negligible but please stop giving the impression that my answer was irrelevant.

By the way, in your own answer, you clearly mention that the decrease in temperature is very relevant in the lab, so I do not get why you are trying to argue that it is negligible now.

Why, I didn't mean that, your answer is totally relevant. However, I still would like to point out that any answer to any physics question is inherently "incomplete" in a literal sense (did we forget to consider the magnetic field of the Earth? or gravitational pull of the Moon?), so it may be worthwhile to single out the main reason and explicitly mark the rest as secondary. This is precisely the case with the decrease in temperature, which is for 99% caused by the enthalpy of melting and for 1% by the enthalpy of dissolution.

I am sorry, but where do you get these numbers? Dissolution of NaCl has an enthalpy of ~4 kJ/mol, melting ice requires 18kJ/mol so while melting ice cools more, I would not consider dissolution insignificant.

The first of those numbers is per mole of NaCl; the second is per mole of water. Suppose we have 1 kg of water (or ice) and 5 or 6 moles of NaCl, which is about as much as we need to get a saturated solution. That would make the difference more apparent. OK, maybe it's more like 97% vs 3%, but whatever.

As I said, the op specifically mentioned this, therefore I put it in my answer. But you are right the dissolution accounts for only a small amount of the total enthalpy (up to 6% to be precise) .

Great! We've come to a consensus (at least qualitatively). Let's leave it like this.

Dissolution happens only once and whatever it's effect was (although it is small) it will be diminished soon by the heat transfer, so it can safely be considered negligible in comparison with the continuous (but deceasing) effect of freezing point depression  due to the continuous salt(in water)-ice contact .

Neither of these answers solve my problem.

What you are saying is true. But it doesn't help my case here

@AVIE which part are you not getting. Could you try to rephrase the question?

Adding salt or any solute causes freezing point depression as one of the colligative properties of the material.
In the cold climate, when snow is left to accumulate it forms a slippery ice which is dangerous. Workers utilize the freezing point effect and add salt to snow to melt avoiding the danger of the slippery ice.
In the hot climate, when ice in the ice box is left it is going to melt. People utilize the freezing point (or melting point) depression effect of salt to ice to melt it at the depressed melting (or freezing) point (below the usual $\pu{0 ^\circ C}$). This can be described as a preemptive protective step to the ambient temperature effect, since the ambient temperature will melt the ice at a relatively high point (the usual $\pu{0 ^\circ C}$). This means a high rate of heat transfer and corresponding rapid melting to the remainder of the bulk of the ice, the salt-water will cover the ice with a layer of relatively cold water (below the usual $\pu{0 ^\circ C}$) which confers a slow rate of heat transfer and corresponding slow melting.

I don't believe this to be correct. See @Thawn's answer. Salt is added to the ice surrounding the ice box to increase surface area with the ice box + to lower the temp of the water (ice) surrounding it (b/c dissolving salt in water will decrease the temperature of the water)

Increasing the surface area by melting and lowering temperature by endothermic reaction are true effects but are of minor importance, the most important effect is melting the ice in new depressed melting(or freezing)point and keeping it in contact with salt-water to retard it's melting, by means of colder water contact and continuous effect of melting point depression but with gradually decreasing magnitude due to dilution of the salt by the melting ice.

would the salty water not just dissolve more salt in the surface of the solid water, melting that ice until all ice was liquid?

dissolution stops by the production of sufficient water by melting, it doesn't continue to the end of melting, except in case of salt has diffused in the ice or elsewhere.

Okay, interesting. good to know


text:
Calculate the heat of a reaction
Say I would like to calculate the heat produced by the following reaction:
$$\ce{2SO2 + O2\longrightarrow2SO3}$$
I have calculated the moles of the reagents. I can't seem to figure out how to calculate the heat of the reaction.
How can I calculate the heat of a reaction with moles of reagents?

in order to calculate the heat of reaction, you need the bond energies (which are measured in $\mathrm{kJ/mol}$ and usually provided in the question or in an appendix in the back of the chemistry book). The heat of reaction is equal to the energy of bonds broken minus the energy of the bonds created. It would be equal:
$[4*(\ce{S-O} $ bond energy$) ~+ ~1*(\ce{O=O} $ bond energy$)] - [6*(\ce{S-O} $ bond energy$)]$.
This will give a negative answer if the reaction is exothermic and vice versa. Since it is measured in $\mathrm{kJ/mol}$, multiply by moles to get the total energy released.


text:
NaOH to stabilize aqueous NaCN
This patent notes in passing:

It may be necessary to add $\ce{NaOH}$ to stabilize aqueous $\ce{NaCN}$. Anyone
  skilled in the art would know the amount of $\ce{NaOH}$ to add.

Obviously I'm not skilled in the art.  I assume the stabilization is against hydrolysis.  What is the mechanism by which $\ce{NaOH}$ would stabilize aqueous $\ce{NaCN}$, and how would one determine the correct amount of $\ce{NaOH}$ to add?
(I'm expecting it's some target pH level, but I'm too far along my "assumption" chain to know what the issues are.)

The mechanism is that the $\ce{OH^{-}}$ would drive those $\ce{HCN}$ (resulting from hydrolysis) all the way back to $\ce{CN^{-}}$. As for the amount... well, yes, looks mighty like it has to do with pH.

The pH is the tip of the ice berg. Many other safety protocols should be observed if working with cyanide solutions. What protocols would depend on the reaction, the nature of the setup (e.g. glassware or apparatus), and the amount of cyanide involved.

Hydrogen cyanide is an acid with a pKa of 9.2, meaning that at pH 9.2 in water you have about a 50-50 mix of cyanide anion and hydrogen cyanide. If you used sodium hydroxide to adjust the pH of a NaCN solution to, say, 12.2, you would inhibit the decomposition of NaCN by shifting the equilibrium to 99.9 percent cyanide anion and 0.1% hydrogen cyanide. Each pH unit higher gives you an additional factor of 10 lower concentration of HCN.
Another factor is temperature: since HCN boils at 28 C, it tends to evaporate from a room-temperature solution of NaCN, driving the reaction to the right. The solution would be more stable if kept cold.
Finally, it would be better to keep the solution in a tightly closed bottle, since carbon dioxide could be absorbed from the air, acidifying the solution and releasing hydrogen cyanide.

So (in a closed container) you can almost eliminate hydrolysis of a moderate acid using a strong base?


text:
General expression for entropy and free energy
The specific heat capacity of a solid at constant pressure and low temperatures is given by $$c_p=aT+bT^3$$ with $a$ and $b$ are constants.
How can I calculate a general expression for the entropy $S(T,p)$ and Gibbs free energy $G(T,p)$ as a function of $T$, assuming that $S$ and $G$ are zero at $T = 0~\mathrm{K}$?
I started with trying to calculate entropy by
$$\mathrm{d}S = \left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T = \frac {c_p}{T} \mathrm{d}T$$
and
$$S = \int_0^T \frac{c_p}{T}\mathrm{d}T = (aT + bT^3)\ln T$$

$c_p$ is itself a function of $T$, so $c_p/T = a + bT^2$ and $\int c_p\mathrm{d}T/T = aT + bT^3/3$ - you cannot simply take it out of the integral and integrate $\mathrm{d}T/T = \ln T$

Understood, but is that approach correct, can I do it that way? For the free energy we have $G(T,p)=H-TS $, but I dont know how to determine H from either its total differential nor $H=U+pV $...

The entropy part seems sensible to me, if there's a problem with it, I don't know. Gibbs free energy part, I suspect that you may need to use the equation $(\partial G/\partial T)_p = -S$, but I'm not entirely sure either. Sorry - I'm not that familiar with doing absolute thermodynamic quantities.

For $\mathrm{d}G=\left( \frac {\partial G}{\partial T}\right)_p \mathrm {d}T + \left(\frac {\partial G}{\partial p}\right)_T\mathrm {d}p = -S\mathrm{d}T+V\mathrm{d}p $ maybe the second part disappears because it's an isobaric process (because of the way the heat capacity is given) and because at $0$ K the free energy is zero ($\Delta G=G $) we would just have to integrate $S $ with respect to $T $?

Your equation is messed up but I get what you mean. I was thinking along those lines too - as with the process for $S$, it seems very reasonable to me.

Since G and S are taken to be zero at T = 0, H is also 0 at T = 0.  So you can calculate H by integrating the heat capacity from 0 to T:  $$H=\int_0^TC_pdT=\frac{a}{2}T^2+\frac{b}{4}T^4$$
Then you substitute H into the equation $$\frac{\partial (G/T)}{\partial T}=-\frac{H}{T^2}=-\frac{a}{2}-\frac{b}{4}T^2$$
and integrate between 0 and T.  This will give you G.  You can then solve for S from G = H - TS.  This gives you the same result as integrating directly to get S.

I calculated $G=-(aT)/2-(bT^3)/12$ and $S=a/2+(at)/2+(bT^2)/12-(bT^3)/4$, however I dont quite understand how you get to the expression $\frac{\partial (G/T)}{\partial T} $, how ddo you come up with $G/T $?

Your solution for G is correct, but your solution for S is not.  Probably an algebra error.  Try again.  The expression your are asking about comes from the equation for dG at constant pressure:  $$dG=-SdT=-\frac{(H-G)}{T}dT$$.  You can arrive at the final equation using the quotient rule for differentiation on the terms involving G.  This equation is very useful because you can get the temperature dependence of G without involving S, and only involving H (which is much more easily obtained).

If I differentiate $-\frac{(H-G)}{T}dT $ with respect to $T$ I get $\frac{(H-G)}{T^2} $, but I still dont understand how you get from there to partially differentiating $G/T $ with respect to $T $

$\frac{dG}{dT}-\frac{G}{T}=-\frac{H}{T}$ so $\frac{T\frac{dG}{dT}-G}{T^2}=\frac{d(G/T)}{dT}=-\frac{H}{T^2}$

Thats really neat. I calculated entropy to $S=-\frac {G-H}{T}=-a/2+bT^3/12-aT/2-bT^3/4$. Thanks man!

It looks like you have an algebra issue.  Please try again.


text:
Fluctuation in pH reading on pH meter
I was conducting a titration of a saturated $\ce{Ca(OH)_2}$ solution against $\ce{HCl}$ using a pH meter (the kind with a glass electrode). At the beginning of the titration, the pH reading would drop a bit every time I added a drop of $\ce{HCl}$. However, near the equivalence point, when I added a drop of $\ce{HCl}$, the pH reading would drop rapidly, but then it would rise again at a steady pace for a few minutes. Why does the pH rise again? My teacher said that the pH meter's membrane is somehow becoming "saturated" with the solution, after being left in the solution for awhile. But what's actually going on in terms of the ions and the membrane?

Your teacher gave you a poor answer.
You are observing three different things.
(1) The solution isn't instantly homogeneous when HCl is added. It takes a couple of seconds of stirring to mix it thoroughly.
(2) The calcium hydroxide solution initially has considerable buffer capacity. As the solution gets less basic then a small amount of HCl can change the pH by a relatively large amount. So a drop of HCl in 50 ml of 0.1 molar calcium hydroxide won't change the pH much, but a drop of HCl in a  neutral solution at pH 7 would change the pH by quite a bit.
(3) The electrode and meter have a response time. So if you could move the pH electrode instantly from an acid solution to a basic solution it would take the electrode and meter themselves a few seconds to equilibrate. (The pH electrode has a hydrated silica layer that forms the boundary on the surface which takes time to equilibrate. The pH meter itself will have some RC time constant so that electrical noise is minimized.)

I guess I wasn't clear about my question, but I'm mainly wondering what caused the pH to rise again.  And this rising happened for a few minutes after the drop was added (and it continued until I lost patience waiting for the pH to become constant and decided to add another drop of HCl). I'm pretty sure the solution was homogeneous by then.

The solution mixing and the meter response time would be a few seconds at most. For changes over minutes two other thing come to mind. First you could be getting carbon dioxide form the air dissolved in the solution. Particularity problematic if you're using vigorous stirring with a mechanical stirrer. Second the pH electrode could be damaged so that it takes much too long a time to equilibrate. Such electrodes get flaky if you let the electrode dry.  You should be able to test pH electrode response time by switching between buffer solutions. Expect a stable reading within 3 to 5 seconds.

I was using a magnetic stirrer, not a mechanical stirrer, so the stirring shouldn't have been too vigorous. Also, the pH meter did have a fast response time when I calibrated it. It was also working fine before and after the equivalence point, so I'm guessing it has something to do with the pH at the equivalence point dropping very fast?

If the pH was drifting down slowly at a pH of about 7, then I'd strongly suspect carbon dioxide going to bicarbonate. $$ \ce{CO2(g) -> CO2(aq) }$$ $$\ce{ CO2(aq) +H2O -> HCO3^{-} + H^{+}}$$

The pH was rising up steadily at a pH of about 7, not drifting down. Are there any side reactions that could have caused that?

The saturated $\ce{Ca(OH)_2}$ solution (or solid calcium hydroxide) absorbed carbon dioxide from the atmosphere before you started titrating. So you're seeing the reverse reaction. $$ \ce{HCO3^{-} + H^{+} -> CO2(aq) + H2O}$$ So the net result is that $\ce{CO2}$ is displaced from solution.


text:
Why copper sulphate is a catalyst in this reaction
Recently, we learnt about catalyst is a substance can increase the rate of reaction without any without itself undergoing any permanent chemical change.The experiment we did was the reaction between zinc and hydrochloric acid as the equation below
$$\ce{Zn + 2HCl -> ZnCl_2 + H_2}$$
However the thing I don't understand is why copper sulphate is a catalyst when the copper ion is reduced to copper atom as the equation below
$$\ce{Cu^2+ + Zn -> Cu + Zn^2+}$$
I mean how could it call as a catalyst when it is reduced from ion to atom

Where in the reaction is copper sulphate used? You just mentioned it as catalyst. The reaction you mentioned doesn't actually need to be catalyzed to occur at a reasonable rate

Anyway, in the particular ionic equation you provided copper sulphate could not in fact be called a catalyst, but maybe in some other step following this in the whole reaction copper sulphate is regenerated from copper metal

Is it the hydrochloric acid?

The hydrochloric acid is a reagent here. It's not a catalyst.

So your sulphate ion have to do something with copper atom after reduced

The reaction rate increases because the copper is deposited on particles of zinc, creating a galvanic couple causing galvanic (aka electrolytic) corrosion. [It could be argued that this is not catalysis, because the copper sulfate is entering into a reaction.]


text:
Nitration of Naphthalene Having Electron Withdrawing Group
In the following nitration which one out of 50 and 51 will be the major product.

In my book it is said that (51) will be major.But in this video it is said that (50) will be major.I'm confused.
Oh yes I saw this answer too Nitration in anthracene and naphthalene but did'nt help in my case.

If you look at p.450 onwards in Chemical Micro Process Engineering: Fundamentals, Modelling and Reactions by  Volker Hessel, Steffen Hardt and Holger Löwe, the major nitration product can be controlled to be mono. Conversely,  if you put this sort of system under forcing conditions,  you could favour a trinitro product. This is why these sort of set questions belong in the 19th century

@Beerhunter PLEASE read my question once again...

I think you cannot predict the outcome of 50:51 selectivity by theory and you need to look up the original paper (or try it yourself). However I wouldn't be surprised if 2 different labs report contradicting results since this kind of nitration reaction is not easy to control (violent and exothermic), and slight variations of the experimental conditions (scale, stirring, temperature control etc) can possibly lead to different results.

@K_P Ok ignoring this particular reaction.What if just mono nitro napthalene is nitrated using nitrating mixture? Even that can't be predicted?

Draw resonance structures of 48. Where are the positive charges?

@SanchayanDutta  I would expect as in your original question wrt to 50:51 selectivity that sterics would favour 50 but this would be on the assumption that there is no difference in electronics. But if your book is correct then this assumption is not right so maybe a more subtle / complex theoretic approach (computational chemistry? ) is required compared to the simpler explanation that works for 48:49 selectivity.

The reaction conditions are not plausible!  Dinitrogen trioxide $\ce{ON-NO2}$ has a "boiling" point around -3 °C, where it signficantly decomposes: $\ce{N2O3 <=> NO + NO2}$. I suspect that the reagent is in fact $\ce{N2O5}$.


text:
Internal energy change during phase change

Ice at $\mathrm{0\ ^\circ C}$ is converted to water at $\mathrm{0\ ^\circ C}$. If $\Delta H$ for the transition of ice to water is $\mathrm{1440~cal}$, calculate the change in internal energy.

Since internal energy $\Delta U=nC_V\Delta T$ and $\Delta T=0$, shouldn't $\Delta U=0$?
But if I use $
\Delta H=\Delta U +\Delta nRT$, I get $\Delta U=\Delta H\neq 0$.

The equation you wrote for $\Delta H$ applies only to an ideal gas (assuming that the nRT is all in parenthesis).  What is the general equation for $\Delta H$, since we are dealing here with a liquid and a solid?

Does the equation you wrote for $\Delta U$ when there is a change of phase?

@ChesterMiller I don't understand you.

I meant to say, does the equation you wrote apply...

@ChesterMiller so $\Delta U =nC_v\Delta T$ is only for expansion/contraction of gases?

For ideal gases, it describes much more than that.   But it doesn't describe $\Delta U$ for a change of phase.

You are considering the $\Delta H$ and the $\Delta U$ between the following two thermodynamic equilibrium states:
State 1:  1 mole of ice at 0 C and 1 atm.
State 2:  1 mole of liquid water at 0 C and 1 atm.
The relationship between $\Delta H$ and $\Delta U$ at constant pressure is:
$$\Delta H=\Delta U + p\Delta V$$where V is molar volume.  What is the molar volume of ice?  What is the molar volume of liquid water?  What is $\Delta V$?

So ∆H and∆U will denote the heat change and internal energy change between two equilibrium states at constant pressure?

what is the equation to be applied for phase change?

@the.eleventh.letter I have no idea what you are asking.  Please elaborate.

The question asked here was how to find ∆u when ∆h is given. However you ended up ur answer with questions 'what is molar volume of ice, water?' so I didn't quite understand that.

Did u mean, we jave to subtract molar volume of ice and water.. to find ∆v?

@the eleventh letter Yes.  That is exactly what I mean.


text:
What is the chemical composition of Turnbull's blue?
I was studying qualitative analysis from this book and  I found some tests for testing the presence of the cation $\ce{Fe^{++}}$.
There was a test in which we add $\ce{K4[Fe(CN)6]}$ solution to a solution of a ferrous salt which results in the Turnbull blue precipitate. Now according to Wikipedia, the Turnbull blue and Prussian blue are the same compound, ie, ferriferrocyanide $\left(\ce{Fe4[Fe(CN)6]3} \right)$.
But the book says that Turnbull blue is ferroferricyanide $\left(\ce{Fe3[Fe(CN)6]2}\right)$ which kinda makes sense because we are adding ferricyanide to ferrous solution.
I am preparing for an exam in which such questions can be asked SO I just want to settle this once and for all, which is wrong, Wikipedia or the book?

see also: Hansen, L. D.; Litchman, W. M.; Daub, G. H. Turnbull’s blue and Prussian blue: KFe(III)[Fe(II)(CN)6]. J. Chem. Educ., 1969, 46 (1), 46; http://dx.doi.org/10.1021/ed046p46

@Loong the link is good but what about user104014's answer? Can we say that the equilibrium hugely favours the formation of prussian blue?

Update: I have asked "What is the chemical composition of Ferrous Ferrocyanide?"

Though your textbook is right, Wikipedia isn't completely wrong as in aqueous solution, $\left(\ce{Fe3[Fe(CN)6]2} \right)$ gets converted and is in equilibrium with $\left(\ce{Fe4[Fe(CN)6]3} \right)$ (a redox change) and hence prussian blue and turnbull's blue are identical to look visually. But in a textual way, the naming is given as your textbook says, i.e. $\left(\ce{Fe4[Fe(CN)6]3} \right)$ is Prussian blue whereas $\left(\ce{Fe3[Fe(CN)6]2} \right)$ is Turnbull's blue.


text:
Why does ozone, being heavier than air, not settle down?
Is this because of some equilibrium between diffusion and gravity ? What exactly is responsible?

Not sure why this is considered a dupe. It asks about ozone where some of the issues are specific to the gas and not adequately answered in the other questions.

The absolute concentration of ozone in the ozone layer is very, very small.

While gravity does cause some separation of atmospheric constituents at high altitudes, the reason why ozone doesn't migrate to the ground has more to do with its chemistry
Ozone is pretty unstable and is formed from some chemical reactions and from the interaction of oxygen with UV light. But it is unstable and decays fairly quickly. This means that is only appears anywhere when the reaction forming it keeps happening.  This is true in the upper atmosphere which is why we have an ozone layer. Some hydrocarbons (not just from pollution from cars but also from volatile stuff released by trees) can react to generate ozone at ground-level. But we only see a significant concentration when the reaction forming it is happening.
The effect of gravity on the separation of the components of the atmosphere is small, especially in the lower atmosphere where turbulent circulation mix things up thoroughly. In the upper atmosphere the dominant process is diffusion and gravity can have more of an effect. This paper describes the process and has some details of the calculations about the size of the effect. It summarises the situation like this:

Turbulent mixing keeps the relative concentrations of gases nearly constant in the lowest 100 km. At higher altitudes, molecular diffusion controls the concentrations, with the lighter gases becoming relatively more abundant with increasing altitude.

It is a worthwhile read if you want the detail.
But the core issue is that, in the lower atmosphere, all gasses are well mixed so there is not separation by density. Ozone only appears there if there are some very specific reaction conditions that create it. The bulk of ozone in the ozone layer occurs because there are constant reactions there driven by UV light from the sun. Separation by density isn't relevant there either.

Hi Matt: I reckon only the second part of your post is the correct answer. Up to the turbopause/homopause turbulence causes a well mixed atmosphere. Gravity can't compete with eddy diffusion. How heavy a gas is (see question), does hence not play a role. However, while ozone does have a complicated photo chemistry, the answer does not have to do with that. The atmosphere is also well mixed for O2, N2, H2O, He, Ar, Ne, ... due to turbulence, and chemistry plays no role.

@tipavi The reason why ozone is found where it is is entirely to do with photochemistry and nothing to do with its relative mass. So this is relevant to the question. And I don't disagree with you on atmospheric mixing: it is certainly irrelevant here, though it does happen for some parts of the upper atmosphere.

I agree. But while its correct, that's not what he asked for. Many other correct things could be said, but the more concise and precise an answer is, the more helpful it will be. :)

@tipavi "... that's not what he asked for" That the OP accepted this is proof enough (in addition to the upvotes, one of which I cast myself) that the answer is better off without the omission you suggested. Thank you for sharing your opinion though :-)

@paracetamol I respectfully disagree. The question whether or not an answer is correct does not depend on a single opinion (e.g., the OP's) or a popular vote (e.g., 5 upvotes). Neither is "proof enough", see discussions about "alternative facts". While the photochemistry of ozone is interesting, it's not relevant in a question about mass ("being heavy than air"), gravity, and diffusion. If you turned off all the (photo)chemistry, ozone's altitude distribution would look different, but ozone would still not settle. Molecular diffusion and turbulence would keep it well mixed up to the turbopause

"If you turned off all the photochemistry, ozone's altitude distribution would look different, but ozone would still not settle" True, but in the real world there is photolytic decomposition going on, and to me it seems prudent to mention it. As I understand it, Matt didn't form his answer solely upon the photochemistry bit, rather, the photochemistry part was an insightful addendum, because it aims to justify why ozone is localized to the upper atmosphere. Matt then builds on this  "In the upper atmosphere the dominant process is diffusion and gravity can have more of an effect." O:)

In short: This answer isn't fixated over the photochemistry of ozone. ;-)

"But while its correct, that's not what he asked for" @tipavi though Matt's already posted an answer, that mustn't stop you from posting yours. If you're willing to write up a sufficiently well-tailored answer, by all means go ahead and post it :-) If the OP finds it helpful enough, he might accept it. Since you also understand this topic well-enough (although you might want to frame the answer different), in all likelihood, your answer will get my upvote as well O:)

@tipavi If you are going to complain that ozone photochemistry isn't what the OP asked for, you might want to also read the second sentence which is "What exactly is responsible?"

If ozone wasn't so reactive that it only exists in regions where it is produced, it would still never settle down, because no gases can unmix due to gravitational fields. Not even in the complete absence of winds and turbulence.
That is exept if the flight path of molecules is significantly influenced by an accelerating field on the lenght scale of the mean free path lenght. That's why ultracentrifuges can do it, and helium and hydrogen slowy escape out of our atmosphere.
Bulk gases of course have buoyancy in the surrounding atmosphere. That's a different issue. The concentration of ozone in the stratosphere is however below 10 ppm.


text:
Why perborate is greener bleaching agent?
All I know is, perborate degrades into supposedly less harmful compounds.  But what exactly happens that it is considered greener? Consider say perborate vs. hypochlorite.  Why perborate is "greener"?

According to Wikipedia:

Chlorination... can oxidize organic contaminants, producing chloroform and other trihalomethanes, which are carcinogenic, and many hundreds of [other]... by-products.
[W]hen household bleach and wastewater were mixed, 1–2% of the available chlorine was observed to form [halogenated] organic compounds."

Perborates, percarbonates and similar bleaching agents, while strong oxidizers, will not create these chlorocarbon chemicals (unless the chlorine is also supplied).
However, boron and perborates are not entirely "green". Boron compounds are somewhat toxic. If the prices were the same, I'd prefer percarbonate bleach to perborate.


text:
Possible approximations in complex equilibrium
If  I have the following equilibrium:
$$\ce{Cu^2+ + 4NH3 <=> Cu(NH3)4^2+}$$
Given an initial concentration of $\ce{Cu^2+}$ ($C_\ce{Cu}$) and an initial concentration of ammonia $C_{NH_3}$, the equilibrium constant $\beta$ can be expressed:
$$\beta=\frac{x}{(C_\ce{Cu}-x)(C_{NH_3}-x)^4}$$
My question is: is there any approximation I can use in order to solve this expression to know the value of $x$? $\beta$ is known.
Thanks

If you know that the equilibrium constant is small, then you could assume that [Cu]-x=[Cu] and/or [NH3]-x=[NH3], especially if one is in excess.

@Spontification The equilibrium constant for coordination is rather large though, around $\mathrm{2.1\times 10^{13}}$.

There is some chemistry behind this which you must understand to recognize the approximation. First the copper ammine complex doesn't really form in the manner indicated rather:
$$\ce{Cu^2+ + 4H2O <=>T[acidic solution] Cu(H2O)4^2+}$$
But in basic solution copper will form a gelatinous precipitate which can perhaps be represented as:
$$\ce{Cu(H2O)4^2+ + 2OH^{-} <=>T[basic solution] Cu(H2O)2(OH)2}$$
With ammonia in the aqueous solution various ammine complexes will be formed such that the soluble tetra-ammine complex is formed in a series of reactions something like:
$$\ce{Cu(H2O)2(OH)2(s) + NH3 <=> Cu(H2O)(OH)2(NH3)} $$
$$\ce{Cu(H2O)(OH)2(NH3) + NH3 <=> Cu(OH)2(NH3)2} $$
$$\ce{Cu(OH)2(NH3)2 + NH3 <=> Cu(OH)(NH3)3^+} $$
$$\ce{Cu(OH)(NH3)3^+ + NH3 <=> Cu(NH3)4^2+(aq)} $$
Ammonia is typically present in excess so the overall reaction is written "in shorthand" as the following equation (I highly doubt that the exact sequence of chemical reactions between the gelatinous precipitate and ammonia is fully understood):
$$\ce{Cu^2+ + 4NH3 <=> Cu(NH3)4^2+} \quad \text{K}_\text{eq}$$
and the equilibrium for this reaction is greatly favored to the right (the ammine complex form).
If you let $[\ce{Cu2+}]_i$ be the initial concentration of $\ce{Cu^{2+}}$, let $[\ce{NH3}]_i$ be the initial concentration of $\ce{NH3}$ and the final concentration of the ammine complex be $[\ce{Cu(NH3)4^2+}]_f$ then in excess ammonia a simplfying assumption would be:
$$ [\ce{Cu(NH3)4^2+}]_f \approx [\ce{Cu2+}]_i$$
$$[\ce{NH3}]_f = [\ce{NH3}]_i - 4[\ce{Cu(NH3)4^2+}]_f \approx [\ce{NH3}]_i - 4[\ce{Cu2+}]_i$$
the equations could of course be solved exactly, but you end up with a quadratic expression which really isn't a difficult mathematical problem to solve. It just takes a wee bit more calculating.


text:
What piezoelectric materials can be made at home?
I would like to know a few examples of transparent materials which exhibit piezoelectricity and can be purchased online or synthesized at home with ease.

The top of my list for exhibiting piezo activity and can be easily purchased and is also transparent..
Quartz
That's about it. It's used everywhere - as an oscillator for quartz crystal watches, quartz crystal microbalances, etc.
Most of the other piezoelectric options:

Lead Zirconium Titanate (PZT)
Barium Titanate ($\ce{BaTiO3}$)
Zinc Oxide ($\ce{ZnO}$)

Aren't transparent. They're white ceramics. You can easily purchase them online from most scientific supply companies, but they don't meet your transparency criteria.
Polyvinylidene difluoride (PVDF) is claimed to be transparent (e.g., here) but you would need to apply a high electric field to pole it, and if the formulation isn't crystalline enough, you may not get a high piezo response.
My group is designing some materials that would be good, but they're still in the research stage. Stay tuned.

Rochelle salt (potassium sodium tartrate tetrahydrate) exhibits piezoelectricity and forms transparent crystals.
It should be available from drug stores or online.


text:
What is the structure of [Co(NH3)5SO4]+?
What is the structure of $\ce{[Co(NH3)5SO4]+}$? If sulfate is a bidentate ligand, then the coordination number of cobalt would become 7. How is this possible?

Are you sure that the complex is uncharged?

I am sorry.  It has +1 charge. But then also how can the coordination number be 7?

It can, but for Co(III) it is unlikely. So, indeed, c.n. is six. what would be your solution for the problem?

There's a certain amount of misinformation here:

sulfate is a bidentate ligand

Sulfate is actually a flexidentate ligand, which is a term used to describe a polydentate ligand which is found to show different denticity in different complexes. Sulfate can act both as mono- or a bidentate ligand. Remember, just because it has two donor atoms doesn't make it a bidentate ligand.
The occurrence of complexes where sulfate acts as a bidentate ring would also be rare owing to the fact that four-membered rings are strained. This page lists $\ce{[Pt(SO4)(P(C6H5)3)2]}$ as one example of such an occurrence.

coordination number of cobalt would become 7. How is this possible?

If this implies that coordination numbers of 7 aren't possible, it isn't true. Coordination numbers of 3, 5, 7, 8-12, 14 are possible, albeit such compounds are quite rare. See this Wikipedia page for some examples.

Thus, the structure of the complex $\ce{[Co(NH3)5SO4]+}$ should be octahedral with cobalt showing a coordination number of 6.

Interestingly enough, it seems that crystal structure of $\ce{[Co(NH3)5(SO4)]+}$ has not yet been obtained. Probably it has something to do with trans-effect of $\ce{SO4^2-}$, which makes $\ce{NH3}$ ligand extremely labile. Also, C.N. 7 is also very common for cobalt and is often found in complexes like $\ce{[Co^{III}(Cp)L2]}$.
Anyway, to further illustrate Berry Holmes' great answer and show that sulfur oxoanions can not only be flexidentate, but also ambidentate (when denticity is 1), there are some examples how coordination abilities of various sulfur oxoanions deviate from each other in case of pentamminocobalt(III) complexes (only cationic parts are shown):

p-(Ammine-(sulfato-O)-tris(2-aminoethyl)amine-cobalt(III)) perchlorate $\ce{[Co(C6H21N5)(SO4)](ClO4)}$, CCDC #CULRIK [1]. $\angle(\ce{Co-O-S}) = 131.2^\circ$, $d(\ce{Co-O}) = \pu{1.958(3) Å}$.
$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#3050F8}{\Large\bullet}~\ce{N}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#FFFF30}{\Large\bullet}~\ce{S}$;
$\color{#F090A0}{\Large\bullet}~\ce{Co}$.

Pentaammine(sulfito-S)(IV)cobalt(III) chloride hydrate $\ce{[Co(NH3)5SO3](H2O)Cl}$, ICSD #37092 [2]. $\angle(\ce{Co-S-O}) = 110.0^\circ$, $d(\ce{Co-S}) = \pu{2.2138(8) Å}$.
$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#3050F8}{\Large\bullet}~\ce{N}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#FFFF30}{\Large\bullet}~\ce{S}$;
$\color{#F090A0}{\Large\bullet}~\ce{Co}$.

Pentaammine(thiosulfato-S)cobalt(III) chloride hydrate $\ce{[Co(NH3)5S2O3](H2O)Cl}$, ICSD #16144 [3]. $\angle(\ce{Co-S-S}) = 110.5^\circ$, $d(\ce{Co-S}) = \pu{2.286(6) Å}$.
$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#3050F8}{\Large\bullet}~\ce{N}$;
$\color{#FF0D0D}{\Large\bullet}~\ce{O}$;
$\color{#FFFF30}{\Large\bullet}~\ce{S}$;
$\color{#F090A0}{\Large\bullet}~\ce{Co}$.

References

Chun, H.; Jackson, W. G.; McKeon, J. A.; Somoza, F. B.; Bernal, I. Eur. J. Inorg. Chem. 2000, 2000 (1), 189–193 DOI: 10.1002/(SICI)1099-0682(200001)2000:1<189::AID-EJIC189>3.0.CO;2-V.
Elder, R. C.; Heeg, M. J.; Payne, M. D.; Trkula, M.; Deutsch, E. Inorg. Chem. 1978, 17 (2), 431–440 DOI: 10.1021/ic50180a048.
Baggio, S. J. Chem. Soc. A 1970, 0 (0), 2384–2387 DOI: 10.1039/J19700002384.