text:
SN2 with thionyl chlorides

Grignards don't couple with alkyl halides. I know that thionyl chloride reacts with alcohols and other $\ce{-OH}$ group containing molecules to replace the $\ce{-OH}$ with a $\ce{-Cl}$. So we probably aren't making an alkyl halide from the alcohol with the thionyl chloride.
The book says that the starting product, which is an alcohol, attacks the sulfur in the $\ce{S=O}$ bond of $\ce{SOCl2}$. I got that the starting product should be an alcohol, but I did not expect the alcohol to attack the thionyl chloride.
Nevertheless, this generates an $\ce{R-O-SOCl}$ molecule, and the resonance-stabilized $\ce{O-SOCl-}$ ion is a good leaving group. This is the substrate that the Grignard can perform an SN2 type reaction on.

My question is: how common is it that alcohols attack thionyl chloride in this pattern? I'm guessing "not very common" because this would make thionyl chloride an undesirable reagent for making alkyl halides from alcohols. Is this reaction synthetically useful?

Maybe not Sn2, maybe addition-elimination instead..  2.  Grignard reagents do react with alykl halides and can couple, but yields are low and elimination byproducts are common.  It certainly isn't the answer in this case.  3. ~SO2Cl is a decent leaving group, but it will decompose into SO2 and ~Cl.  SO2 gas being liberated will drive equilibrium. 4.  It's a great reaction for alcohol conversion.  5.  I do not like this question from the pedagogical perspective.

This generates an R−O−SOClR−O−SOCl molecule, and the resonance-stabilized O−SOCl−O−SOCl− ion is a good leaving group. This is the substrate that the Grignard can perform an SN2 type reaction on.

I could be wrong about this, because I always doubt everything that I do, but I don't think that's quite right.
Here is a good discussion about thionyl chloride works.  It should turn your alcohol into an alkyl chloride, and now that chloride is your leaving group for the SN2 reaction.
Knowing that, we can work retrosynthetically.
Your product is octane.  You get to the octane by reacting something with ethyl magnesium bromide (a great nucleophile), so that's a strong indication that you did a substitution reaction.  If you add on an ethyl group and get octane, then your substrate was an ethyl group shorter, so it would have contained a hexyl chain.  You got to that hexyl species by treating an alcohol (as it was described in the initial question) with SOCl2, and you know that SOCl2 turns alcohols into chlorides, so your hexyl species that you treated with EtMgBr was 1-chlorohexane, which you obtained by treating hexan-1-ol with SOCl2.

The use of Thionyl Chloride is a very good way to prepare Alkyl Halides, especially because it is a "clean" reaction where the byproduct $\ce{SO2}$ is a gas so purification of your product is much easier.
You shouldn't put in the Grignard without working-up the reaction because the HCl and/or any alcohol that remains in solution will destroy the Grignard. However, working up the reaction also means that your R-OSOCl intermediate will not survive and will become R-Cl. The original reaction you show is also clear about this given how it drew two arrows instead of just one.
And finally, is this type of reaction synthetically useful? Why yes it is! This type of reaction allows you to control the stereochemistry of the product. This particular sequence (Thionyl Chloride followed by Grignard) allows you to invert the stereocenter of the final product, whereas using thionyl chloride with pyridine or $\ce{PCl5}$ first will retain the stereocenter.

@K_P yes, but if coupled with the next step which is SN2 then you get inversion. I guess that part wasn't clear and I'll edit the answer now!

There are so many better ways to invert stereocenter than this lousy sequence, that would give you low yields and lots of byproducts. You can use Mitsunobu reaction or easily mesylate the alcohol and then attack with some good nucleophile/weak base like alkyne, cyanide, malonate, etc. Stereospecifity is usually very good and there are rarely any elimination byproducts.
Tosylates and mesylates
Mitsunobu reaction


text:
Find molar concentration of sodium carbonate dissolved in water
The question is

A solution of $\ce{Na2CO3}$ is made by dissolving $5.3~\mathrm{g}$ of the powder in enough water to make $250~\mathrm{mL}$ of solution. Find the molar concentration.

I have done the following but I don't know why it's not working.

Molar mass: $106~\mathrm{g}$
$5.3~\mathrm{g} / 106 = 0.05\ \mathrm{mol}$
$0.05\ \mathrm{mol} / 250\ \mathrm{mL} = 0.0002\ \mathrm{mL}$

However, the answer is $0.2\ \mathrm{mol/L}$ but I got $0.0002$. It does become $0.2$ when I multiply by $1000$ but the answer is already in $\mathrm{mL}$. How do I arrive at the correct answer?

Please fix your units!

Use units! Consistently, properly, all the time!
You have units all over the place sometimes fitting the quantity you are calculating, most often not. If you approached the problem with correct use of units, there would be no question.
First, the molar mass.
$$M(\ce{Na2CO3}) = 105.99~\mathbf{\frac{g}{mol}}$$
Grams are not equal to grams per mole. Then, let’s do the rest without losing or inventing units as we go along. (I bolded the units for you.)
$$n(\ce{Na2CO3}) = \frac{m(\ce{Na2CO3})}{M(\ce{Na2CO3})} = \frac{5.3~\mathbf{g}}{105.99~\mathbf{\frac{g}{mol}}} = 0.05~\mathbf{mol}$$
$$c(\ce{Na2CO3}) = \frac{n(\ce{Na2CO3})}{V} = \frac{0.05~\mathbf{mol}}{250~\mathbf{ml}} = \frac{0.05~\mathbf{mol}}{0.250~\mathbf{l}} = 0.2~\mathbf{\frac{mol}{l}}$$
Different versions of jumbling around the powers of 10 (especially in the millilitre part) will get you the same result if done correctly. It really bears repeating.
Use units! Consistently, properly, all the time!

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text:
Can hydrogen gas reduce copper(II) ion?
I have a solution of copper(II) sulfate, and I would love to precipitate the copper metal. From looking at the standard reduction potential table, it appears that I can do this by bubbling hydrogen gas through the solution. However, the potential is not very large, so I am unsure of how well this approach would work. I was also wondering if sodium metabisulfite could be used for this purpose. I know I can precipitate copper carbonate with sodium carbonate and reduce with carbon, but this will take more time, and probably result in a large loss of product. I would also like to avoid precipitating the metal with another metal, because I am demonstrating how metals can be isolated so it would be silly to use a metal that is even harder to reduce.

With proper catalyst (platinum) this should work even at room temperature. Without it, I don't think so.

http://chemistry.stackexchange.com/questions/52091/can-hydrogen-gas-reduce-copperii-ion?rq=1

Copper reduced from solution is readily oxidized by air.
When I cared, I made a moderately fine copper powder from citrate. Threat copper salt solution with slight molar excess of sodium (potassium) citrate solution (can be made on site from corresponding hydroxide and easily available citric acid), shake and let the solution stay for a while. It immediately turns deep blue, but the precipitate may appear several minutes later. Filter off the precipitate and then heat on a sand bath at ~200-300 Celsium in a vial plugged with cotton wool. I didn't check if evolved gases are toxic, so use a fume hood or good ventilation. The citrate goes through several changes in color, but eventually produces metallic copper powder. Cool it with wool in. The powder seems to be mildly (inconsistently) pirophoric, so store in a small air-tight container.
Hydrogen on it's own certainly can reduce copper, but the process is neither fast, nor pragmatic.

Use electrolysis, it is much easier and faster to set up, without high temperatures or poisonous gases ($\ce{SO4^2-}$ is very hard to oxidise). So copper will plate out on the cathode and oxygen gas will bubble out on the anode.

Yes as Copper is less reactive than Hydrogen(or Hydrogen is more reactive than Copper) in the reactivity series of metals with respect to it

Please convert it into a comment ASAP.

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Why do coefficients become exponents in the speed equation when dealing with elementary reactions
In school, while studying chemical kinetics, we were learning about the speed law and equation for different chemical reactions. We learned that in general, the exponents of the concentration of reactants in the speed equation must be determined experimentally. However, earlier this week, we learned that if we are talking about basic elementary reactions, the coefficients of the reactants become the exponents in The equation. This puzzled me. Can anyone explain why the coefficients would become exponents? Thanks in advance

A reaction occurs when its reactives are in the same region of the space. If you have an elmental reaction of the form $a A + b B \rightarrow $, for the reaction takes place it is needed that $a$ molecules of $A$ and $b$ molecules of $B$ collide. How probably is this event? Well, if you only need one molecule of $A$ ($a=1$), the probability that an $A$ molecule be in this specific place (at some time) is proportional to the $A$ concentration. If $(a=2)$ we need another $A$ molecule, as the probability for a second molecule being in that space region (at some time, in particular at the same time) is independent, the probability of finding two $A$ molecules in the same space region (at the same time) is the product of the two individual probabilities (is like throwing a dice two times). In general, for the mixture the probability of the encounter is proportional to $[A]^a [B]^b$.
The probability of encounter is proportional to the number of encounters in some time interval. And the later to the reaction velocity.

This sort of made sense but you made it seem as though the concentration is the probabillity of finding a particule at that soace. Is this true?

It is not the probability, but it is proportional to the probability. Proportionality factor depends on temperature and volition radii. Also, once the volition takes place, there is another probability for the collision be reactive

Volition = collision after spellchecker

Ok, so say the coefficient of one of the reactants is 3 and at a particular moment in the reaction its concentration is 2 moles/ liter. I would think that since the reaction requires 3 of these molecules to collide at the same time, it would be less likely than if it only needed 2 and thus the reaction would take longer. However, the value of 2 cubed is larger than 2 squared wich leads to think the reaction would take less time, where is my logic flawed?

In the omission of proportionally coefficient, if it is small enough to make the product between it and concentration smaller than 1, then the cube will be smaller that the square


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Copper acetate synthesis without hydrogen peroxide
Usually, when people synthesize copper acetate, they use a mixture of acetic acid and hydrogen peroxide. I assume this is because the acetic acid actually reacts with the various oxides on the copper, rather than with the copper itself. However, assuming I don't have access to hydrogen peroxide, can I do a similar thing (but slower), just by waiting until fresh copper surface is exposed and then taking it out. Then, I will heat it up (or just wait) until oxide layer forms again, and keep repeating until copper is corroded away.

You can electrolyze  a conc. aqueous solution of calcium acetate with copper electrodes. see wikipedia for the description of the process.

Related: Copper Doesn't React with Hydrochloric acid

I placed 1tbs salt, 1L vinegar, and 200g solid copper in a jar. After 3 months in a dark place the solution was extremely dark blue - so blue I couldn't see through it. This is what happened (bottom answer).
How may copper acetate ligands be manipulated to change colors?
Heat may be able to substitute for time and quicken the formation of copper acetate...I really haven't tried. Peroxide is dirt cheap. I would have used it if I had known.

I think this at least tries to answer the question. It would be better with more explanation, but I see no reason to delete it.

See my rapid prep 'Experimental Variants of Bleach Battery to Obtain Metal Oxides (like CuO)' where from Bleach (NaOCl), a piece of copper metal, and a noble metal cathode (nickel alloy coin), one can in a few minutes with microwave assisted heating, create a cloud of black CuO particles.
Add acetic acid to the CuO to prepare your Copper Acetate.

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Does conductivity change in a solution due to rust?
I plan on conducting an experiment in which I place an iron nail in water and analyze its rate of corrosion into rust with different concentrations of NaCl.
My original plan was to cover the water in a beaker with oil then use a dissolved oxygen probe to measure the drop in oxygen levels as the reaction went on. However, I was thinking that maybe conductivity would change as the reaction went on?
In short, in a solution containing iron, salt, oxygen, and water, will conductivity change as the iron rusts?

This is an interesting question. As iron rusts, you will have a couple of reactions. First, as you expect is the oxidation of iron:
$\ce{Fe ->Fe^2+ + 2e^-}$ or $\ce{Fe ->Fe^3+ + 3e^-}$
Then, either one of these may precipitate as their oxide, or hydroxide or their chloride.
On the flip side, somewhere else on the same piece of nail, there has to be a reduction. This may be one of:
$\ce {2H+ +2e- -> H2} $ or $\ce{2O2 + 4e- +4H+ ->2H2O}$
The overall reaction will probably be a combination of the above depending on the various rates of reaction. A priori, I would say it is very hard to predict which mechanism would be operative and therefore it is hard to predict which way conductivity would go.
Another point is even if the overall mechanism was such that there was some imbalance that would affect the conductivity, that would be very hard to measure. The changes would be a very small fraction of the total conductivity which would make it a hard measurement.
The typical way this is measured is "weight loss", which involves cleaning the corrosion products off the metal specimen and measuring the weight after given time periods. You can find lots of cleaning procedures and details of weight loss measurements over the net.

Assuming starting out with distilled water, both reduction half cell reactions consume $\ce{H+}$. Thus the solution would get more basic. $\ce{O2}$ is uncharged and therefore not conductive. Since  $\ce{H+}$ has a greater conductivity than  $\ce{OH-}$, the conductivity of the solution must go down.


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Alternative to Hanging Drop Crystallization
Forgive my ignorance if I am way off, but I was having a look at the crystallization of a number of different substances (e.g. proteins) through the hanging drop vapour diffusion method.
The examples given use a precipitating agent (e.g. $10~\%\ \ce{NaCl}$) at the bottom of the well and then you put the liquid protein you are looking at upside down on the top cover and the water moves from the protein into the precipitating agent, slowly crystallizing the protein.
So my question is, for simplicity would it be possible to use a divided container (e.g. petri dish) which has two compartments that share the same atmosphere to achieve the same thing, that is have a container that has a salt solution on one side, and the solution I wish to crystallize on the other?
Would this likely work?

Yeah, that might work.

The process you are describing is essentially the sitting drop crystallisation.
Yes, it is used. Frequently, sitting drop is used in robotic setups, because robots have a much harder time creating hanging drops. With a sitting drop, it becomes simple; just pipet a drop into the sitting drop reservoir.
Gerhard mentioned that the hanging drop method is the preferred one for manual handling since it becomes easier to handle the crystals. Possibly because they sit at the bottom of the hanging drop rather than stick to the ground of a sitting drop’s reservoir.

I think sitting drop is used much more than hanging drop, also when manually setting up crystallization screens. One of the additional reasons to use a hanging drop setup is that (usually) you use a bigger drop, so you can get bigger crystals.

@vonBeche: actually,  it is the other way round: hanging do experiments are limited to around 5 ul (otherwise the drop won't stay on the cover slide any longer), whereas there is no upper limit to sitting drop.  When sitting drop is used for screening, lower volumes can be used through robotics to minimize sample consumption - but this has nothing to do with the limits of the method.

The main reason to use hanging drop to my mind is that they are easier to work with for crystal manipulation and that crystals tend to stick less to the surface, but drop to the surface of the drop.

@Gerhard I hope you don’t mind that I shamelessly copied that into my answer ;) I only had a very short introductory lab course into protein crystallisation, so I’ll admit not being an expert … and it was rather long ago …

@Jan: no worries...


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Reaction of Phenolic acids
I'm puzzled as to what can happen to phenolic acids when they are heated. (Chemistry, not health or food science) How do they dehydrate? Do they form anhydrides? Also, is it possible for the carboxylic acid part to form an anhydride with an ortho phenol group (given that the COOH group is on a sufficiently long alkyl chain).

That would be an ester, not an anhydride.


text:
Ozone depletion distribution
I am working with ozone measurements from two different sites (one rural and one urban). I am looking at the distribution of hourly averaged concentrations in air. The first figure shows the ozone distribution from the rural site and the second one from the urban site.

The first one (rural) looks normally (or lognormally) distributed, whereas the second one (urban) looks exponentially distributed.
My hypothesis is that it is caused by the fact that there is more pollution in an urban area, therefore more ozone depleting substances and this is why we have mostly small values, creating an exponential distribution. I am wondering if there is a more precise explanation that these distributions of the same molecules are so different.

Why are your scales so vastly different?

It isn't even obvious from the pictures you show that they are very different. Your scales on both axes are wildly inconsistent and make it almost impossible to do a simple comparison.


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Reaction between manganese(IV) oxide and sulfuric acid
$$\ce{2MnO2 + 2H2SO4 -> 2MnSO4 + O2 ^ + 2H2O}$$
How does this reaction occur? We get the oxygen moiety of the manganese dioxide suddenly oxidized to gaseous oxygen, while the manganese gets reduced from +4 to +2.
The sulfuric acid is apparently unperturbed: its proton $\ce{H+}$ and its $\ce{SO4^{2-}}$ parts remain the same in terms of their oxidation states. What role does the sulfuric acid play here at all?

Mechanisms in inorganic chemistry are often not well-studied and not well-characterised. There are thousands of possible compounds that you can form with varying amounts of oxygen and varying charges and you can bet that the reaction produces a lot of them. They do not involve "arrow pushing" like in organic chemistry and often you don't get much chemical insight from them. This is an example of what I am talking about: http://i.stack.imgur.com/euf6M.png I don't think there is a definitive answer to this question.

I edited your question, as it isn't really about exact mechanism

I am afraid I cannot provide the exact mechanism. Nevertheless, here is what I know.
The reaction is most certainly more complex than the equation
$$\ce{2 MnO2_{(s)} + 2 H2SO4_{(aq)} -> 2 MnSO4_{(aq?)} + 3 O2_{(g)} + H2O_{(l,g)}}$$
suggests. One might say it is even misleading, but it is definitely not the whole story.

A purple smoke

After adding concentrated sulfuric acid to manganese(IV) oxide, a purplish smoke manifests. This suggests to the species $\ce{MnO4-}$. Yet, what is the balancing particle? Is it $\ce{Mn^2+}$, or rather $\ce{H}$? I have no clue.
$$\ce{2 Mn_{(aq)}^2+ + 5 MnO2_{(s)} + 6 H3O+_{(aq)} -> 2 HMnO4_{(aq)} + 5 Mn^2+_{(aq)} + 8 H2O_{(l,g)}}$$

A green to brown mixture

The initial mixture is green. This points to either $\ce{MnO4^2-}$ or $\ce{Mn2O7}$, even $\ce{MnO4^3-}$ could be it. Again, which one is it? Maybe it is all of the above? Upon standing, we observe a brown precipitate. The manganese(IV) oxide appears to reform … Speculation (more in the next chapter):
$$\ce{MnO4_{(aq)}^2- + H2SO4_{(aq)} -> H2MnO4_{(aq)} + SO4_{(aq)}^2-}\\\ce{3 H2MnO4_{(aq)} ->  2 HMnO4_{(aq)} + MnO2_{(s)} + 2 H2O_{(l,g)}}$$
$$\ce{3 MnO4^2- + 2 H2O_{(l)} -> 2 MnO4- + MnO2_{(s)} + OH_{(aq)}-}$$

Oxygen, where's that from?

I did the classic match experiment; it did indeed burn more vigorously near the reaction mixture.
$$\ce{HMnO4_{(aq)} -> MnO2_{(s)} + 3 O2_{(g)} + 2 H2O_{(l,g)}}$$
$$\ce{2 MnO4^2- -> MnO4^3- + MnO2_{(s)} + O2_{(g)}}$$
$$\ce{2 Mn2O7_{(l)} -> 4 MnO2_{(s)} + 3 O2_{(g)}}$$

I think you get the idea

Sorry I could not be of more assistance. The mixture is indeed complex with multiple processes happening in parallel. (This is actually the case with every reaction.) It should at least point you in some direction.

Extra

I was a bit loosey–goosey with the ionic equations; you should not do that ;). In my opinion, it made the process more clear.
One might also wish to consider $\ce{Mn2MnO4}$ or $\ce{Mn3O4}$ which, I believe, are also brown.

Also, as @orthocresol mentioned, there are too many combinations, generally, which is why there aren't many arrow-pushing mechanisms in inorganic chemistry. Glance at the link Orthocresol provided.


text:
Educational practice problems/systems/reactions in computational chemistry
I am not sure if this question is suitable for Chemistry SE, but I think fairly objective answers can be given, and not just I will benefit from them

I have almost completed the first year of a MSc program in theoretical (and computational) chemistry, and have gone from knowing absolutely nothing about computational chemistry, to having heard of some of the stuff I need to master. I just took the step of installing GAMESS on my personal computer, and I intend to start doing test calculations on my own.
While courses provide theoretical foundations, my experiences are that actual computations are quite limited. I took the initiative to engage in a student research project which started February this year, and during this time I have gained a lot of practical experience with calculations (exploring the PES, finding and optimizing transition state structures, calculating MEP's, looking at molecular orbitals). And to me this kind of experience is extremely valuable.
So I would like to practice with well-studied systems/reactions/whatever. In short, I want a diverse set of systems, some of which may require "manual labour" (e.g. CASSCF). With well-known systems I can find experimental and theoretical values to compare with, and likely also discussions about how certain methods perform for each system. These systems should be small enough so that they can be comfortably run on a Macbook Pro (2,9 GHz Intel Core i7, 8GB RAM, SSD) without taking too long.
So question basically is this: What is a good set of educational systems/reactions, suitable for an interested student in computational chemistry?

http://www.amazon.com/Exploring-Chemistry-Electronic-Structure-Methods/dp/1935522035/ref=sr_1_1?s=books&ie=UTF8&qid=1464884867&sr=1-1&keywords=Exploring+Chemistry+With+Electronic+Structure+Methods

I was looking for something similar, but have not yet been able to try the suggestions. http://chemistry.stackexchange.com/q/20308/4945 (And I would not recommend the Gaussian book, it is very much just a tutorial for g09.)

What you are asking is kind of vague. What area of computational chemistry are you interested in? Reaction path modeling? NMR calculations? Excited states? Biological systems? Molecular interactions? Spectra?

There's a free software quantum chemistry package called Psi4. If you're a GNU/Linux user, under Ubuntu or any Debian-based distro, you can install it with a simple terminal command.
sudo apt-get install psi4

They have a well documented site, including a section dedicated to education with simple projects and friendly message boards. As of today, they list the following projects in their education section:

What is the radius of an atom?
What makes a molecule polar or non-polar?
How do we calculate the most accurate energies for a boron atom?
Can we visually predict binding energies?
When does water become $\ce{OH + H}$? A study in molecular orbitals.
What happens to the orbitals as a one-dimensional box gets longer? A study in molecular orbitals.
Is $\ce{C3H+}$ present in the Horsehead nebula?
What is symmetry in chemistry?

The version available in Ubuntu repository is functional, but quite outdated, 1.1.5, as of Mar 2020. If one is serious about learning it, my advice is to download it directlly from their site. For me the only issue with it is that Avogadro, my output viewer of choice, seemingly doesn't provide native support to view Psi4 output files. But I count that as a missing feature in Avogadro, not Psi4.


text:
What will happen if I put this chemical into my water tank?
There is an 8000 liter water tank on top of my house which provides water for the house.
Bacteria, fungi, algae, etc often start developing in the tank, giving unpleasant smells, etc.
This chemical seems to be chlorine based. According to the bottle, the composition is:

Sodium hypochlorite 5% max.
Sodium hydroxide 1% max.
Amine oxide 1% max.

I'm thinking of maybe putting 50 ml into the tank about once a week.
Nobody is drinking or cooking with the water. It's used for stuff like laundry, showers, washing dishes, etc.
I know it seems like a strange idea, but I live in the countryside in a developing country and it's not easy to find stuff here.
The water is ground water that we pump up there with our own pump.
What's gonna happen? Should I decrease or increase the amount? Or drop the idea entirely? (And maybe do something else?)
It's very hot here, maybe 40 celsius in the shade. This tank is in the baking sunlight.

Once a week is maybe too often. You don't use those 8000 l in a week, do you?

@IvanNeretin Nope, we use maybe 800 liters per day. We top it up maybe once or twice a week.  BTW, won't chlorine evaporate into the air gradually?

Is your tank sunlight proof ?

@ankit7540 Yes, it's made of 10 cm thick concrete and is covered.

Could a foam fractionator and ozonation be an alternative for you?

@aventurin Haven't heard of the former. but I have great experiences with the latter for disinfecting vegetables. water for dishwashing, etc. How big of an ozonizer would i need for a 8000 liter tank?

I'm not an expert, and its difficult to say since it depends on the amount of biological and chemical material that must be oxidized as well as on the technique used to bring the ozone into the water. Usually this is done by bubbling air that contains ozone through water (therefore the combination with a foam fractionator). The residue of ozone should be less than 0.05 mg/L (50ppb) for drinking water. Also for drinking water bromine and trihalogen methanes might be a problem.
https://en.wikipedia.org/wiki/Water_chlorination#Ozonation

@aventurin Interesting. I have RO, Ozonizer and UV inside the house. I take the water from this tank and run it through it to create drinking water.

The product which you have mentioned in your post may have other chemicals like fragrance etc... which may make the water not suitable for washing dishes.

You can purchase sodium hypochlorite (bleach) which is sold separately as powder or solution and add to the water tank. The concentration added to water tank is crucial.
You can use unscented / fragrance free laundry bleach solution also. This will be easier to find.
If you get solution (6% solution of bleach in water) then you can add 40mL to the 8000L full tank first time and mix.
Then, say after a week when your tank has 2400 Liters full (using 800L per day ); you can add 28-30mL again and then fill up tank to full (8000L). (Approximate ratio is 1-1.5mL per 200 Liters of water).
Preventing algal growth:

To prevent the growth of algae it is always recommended to prevent light exposure of stored water. Dark colored/black tanks do well. You can cover the manholes/ crevices through which light may enter the tank.
Additionally, warm water aids in algal growth, so you can try to cover the tank with some tarpaulin etc... to prevent heating.
Regular cleaning is important. Algae already growing on the inner wall of the tank is difficult to to deal with. Proper regular cleaning will help to remove such growth.

Thank you for this answer. Unfortunately, I have no idea how to get a hold of sodium hypochlorite here. But I understand the point about the other chemicals.

@Fiksdal : In India it is not that difficult to get bleach powder or solution.

But the product I mentioned in OP is bleach solution, no?

Yes, the RinAla product is bleach solution but it also has some chemicals (specifically which I do not know), which gives fragrance. These chemicals may make the water not be suitable for washing dishes.

How can we know whether a particular solution has extra chemicals in it?

Unscented/fragrance free product will have no such chemicals and in the composition on the product bottle you can see that. Usually the ones which are used laundry have some fragrance (to make clothes smell better).

Let us continue this discussion in chat.


text:
Finding pH using the -log, why does my calculation differ from the book's answer?
An aqueous solution containing 0.0020 mol/L barium hydroxide, what is the pH? I do the following: 14-(-log(0.0020))=11.3pH. The answer is 11.6.
2.0*10^-3 mol/L sodium hydroxide solution, what is the pH? I do the following:
14-(-log(0.002))=11.3pH. The answer is 9.3
When I solve for pH of solutions with acidic solutes, I get the accurate answer, such as in the question: 2.0*10^3 mol/L nitric acid solution, where I do:
-log(0.002)=2.7pH, which matches the answer, 2.7.
My question is, why am I getting the discrepancy between the calculated and the given answer when calculating pH for solutions with base solute? Or am I not doing anything wrong?

Because barium hydroxide is $\ce{Ba(OH)2}$
The book is assuming each mole of barium hydroxide release 2 moles of hydroxide ions, while you are assuming only 1 mole is released.

Please disambiguate.

@ArthurAlexKarapetov OK, I added to the answer.

I see. But what about sodium hydroxide (NaOH)? There is only 1 hydroxide ion released there. In my question, it doesn't work out either. Please help understand why?

Your answer is right for 0.002M NaOH


text:
Unable to find sources for uncommon specific heat
I'm calculating the energy, temperature, and gas volume outputs of gunpowder. To solve this I need to find the specific heat of Potassium sulfate (K2SO4) and Potassium carbonate (K2CO3). If anyone can refer me to any directory that can list the values for these it would be very helpful.
Thank you for your time.

For potassium sulfate see here.
For potassium carbonate see here.
Generally, search in the NIST Chemistry WebBook and when you find the compound of interest, either click on "gas phase thermochemistry data" or "condensed phase thermochemistry data" as appropriate.


text:
Why does Cu(OH)2, which is generally insoluble due to low Ksp become soluble in excess of a base, such as NaOH?
Where does the "excess" criteria come into the picture? Also, can the equilibrium
$$\ce{Cu^2+ + 6 H2O <=> [Cu(H2O)6]^2+}$$
participate in the solubility process in any case? So there are 2 questions: one regarding the solubility of $\ce{Cu(OH)2}$ and other regarding the equilibrium mentioned above?

see http://chemistry.stackexchange.com/questions/51793/which-is-more-amphoteric-zinc-or-copper-hydroxide?s=1|0.3711

How does that 'question' answer my question?!

It doesn't but is related.

Also, can the equilibrium $$\ce{Cu^{2+} + 6H2O <=> [Cu(H2O)6]^{2+}}$$ participate in the solubility process in any case?

This equilibrium does contribute to the solubility of copper(II), but probably not in the way it appears you are envisioning it to.  When copper(II) is fully dissolved in moderately acidic solution, as long as the anion of the acid is a weak ligand for copper (e.g., $\ce{ClO4-}$ or $\ce{NO3-}$), the copper ions are dissolved as the aquo complex, $\ce{[Cu(H2O)6]^{2+}}$.
As the $\mathrm{pH}$ is adjusted upward with, e.g., $\ce{NaOH}$, though, other reactions come into play (see below) and the hexaaquo species contributes negligibly to the overall solubility behavior.

Where does the "excess" criteria come into the picture?

To answer this question, one really must first understand what's actually going on when solvated metal ions undergo acid/base reactions.  As I indicated above, in acidic solution copper(II) exists as an aquo complex, typically written as the hexaaquo species, $\ce{[Cu(H2O)6]^{2+}}$.  As the $\mathrm{pH}$ rises, the following series of acid-base ("hydrolysis") reactions occurs$^\ddagger$:
$$
\ce{[Cu(H2O)6]^2+ + OH- <=> [Cu(H2O)5(OH)]+ + H2O}\tag{1}
$$
$$
\ce{[Cu(H2O)5(OH)]+ + OH- <=> [Cu(H2O)4(OH)2]^0 + H2O}\tag{2}
$$
$$
\ce{[Cu(H2O)4(OH)2]^0 + OH- <=> [Cu(H2O)3(OH)3]- + H2O}\tag{3}
$$
$$
\ce{[Cu(H2O)3(OH)3]- + OH- <=> [Cu(H2O)2(OH)4]^2- + H2O}\tag{4}
$$
(For simplicity, I have omitted the $\ce{M2L2}$ hydrolysis complex reported in Vuceta$^\ddagger$, and have written the above in terms of addition of alkali so as to highlight the similarities among the series.)
So, to some extent, the copper can be treated much like a polyprotic acid: as hydroxide ion is added, the copper is able to "absorb" it, forming water/hydroxide complexes with increasing numbers of hydroxides attached.
Per the low $K_\mathrm{sp}$ of $\ce{Cu(OH)2}$, however, the following (fictitious) solubility relation must also be considered:
$$
\ce{ [Cu(H2O)4(OH)2]^0 (aq) <=> Cu(OH)2 (s) + 4H2O} \tag{5}
$$
Eq. $(5)$ is a re-interpretation of the typical solubility expression:
$$
\ce{[Cu^2+][OH^{-}]^2} \leq K_\mathrm{sp} \tag{6}
$$
Qualitatively, Eq. $(5)$/Eq. $(6)$ indicates that the copper existing as $\ce{[Cu(H2O)4(OH)2]^0}$ is limited to a certain concentration in solution, depending on the conditions of the system.
Thus, as base is added to an acidic solution of copper and the equilibria are swung to a range that favors $\ce{[Cu(H2O)4(OH)2]^0}$/$\ce{Cu(OH)2}$, much of the copper precipitates out.  But, as more base is added, eventually there's enough hydroxide present that the equilibria can start to shift toward $\ce{[Cu(H2O)3(OH)3]-}$ and $\ce{[Cu(H2O)2(OH)4]^2-}$, and the copper will start to redissolve.

This, finally, leads back to question: what does "excess" mean in this context?
To help answer this question, I input the stability constants from Vuceta$^\ddagger$ into HySS, a program that is designed to simulate these sorts of equilibria and that is able to handle precipitates like $\ce{Cu(OH)2 (s)}$.  The HySS file used to generate the below figures can be downloaded from Google Drive here.  The below figure shows the simulated speciation diagram for $10~\mu\mathrm M$ total copper over a $\mathrm{pH}$ range of $4$ to $14$ (all ligating water molecules are omitted from the labels; click the image for a larger version):

As expected, the copper is fully dissolved at an acidic $\mathrm{pH}$ of $4$, but starts to precipitate at around $\mathrm{pH}~7$.  Coincidentally, the concentration of the dissolved $\ce{[Cu(H2O)4(OH)2]^0}$ is predicted to be equal to the "concentration" of the precipitated $\ce{Cu(OH)2 (s)}$ (mass of solid divided by the liquid volume) over much of the insoluble range. Starting at about $\mathrm{pH}~11.5$, the base is in sufficient excess (~$316$-fold) over the copper to begin to re-dissolve it.  Only around $\mathrm{pH}~13$ or so does all of the solid finally re-dissolve, at a ~$10,000$-fold excess of base.
At the higher copper concentration of $10~\mathrm{mM}$, re-dissolution has hardly begun to happen at all by $\mathrm{pH~14}$ (click image for larger version):

This is perfectly sensible, as the base is only in excess of the copper by ~$100$-fold at that $\mathrm{pH}$.  In order to observe re-dissolution at this copper concentration, one would presumably have to use extraordinarily concentrated alkali in order to achieve a sufficient excess of base.

$^\ddagger$ Vuceta & Morgan, Limnology and Oceanography 22(4): 742, 1977, doi:10.4319/lo.1977.22.4.0742


text:
Coagulation with Ferric Chloride
I'm having a hard time understanding the use of coagulation wastewater treatment. I have read in literature that ferric chloride is used as a coagulant to remove COD, TSS and ammonia from landfill leakage. Their dose was $\pu{10 g/L}$ at a $\mathrm{pH}$ of 8. Since there is so much ferric chloride, it will neutralize all of the alkalinity in the leakage and then start decreasing the $\mathrm{pH}$. To maintain the current $\mathrm{pH}$ of 8 lime, $\ce{Ca(OH)2}$, is added so that the ferric chloride will neutralize that instead.
$$\ce{2 FeCl3 + 3 Ca(OH)2 -> 2 Fe(OH)3 + 3 CaCl2}$$
What I don't understand is what exactly is going on in this reaction? What do the products do in the leakage? What will happen to the reaction at higher or lower $\mathrm{pH}$?

According to this Wikipedia article:

Iron(III) chloride is used in sewage treatment and drinking water production.$\mathrm{^{[1]}}$ In this application, $\ce{FeCl3}$ in slightly basic water reacts with the hydroxide ion to form a floc of iron(III) hydroxide, or more precisely formulated as $\ce{FeO(OH)−}$, that can remove suspended materials:
$$\ce{[Fe(H2O)6]^3+ + 4 HO− → [Fe(H2O)2(HO)4]− + 4 H2O → [Fe(H2O)O(HO)2]− + 6 H2O}$$

1) Water Treatment Chemicals, Akzo Nobel Base Chemicals. 2007.

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


text:
Salt crystal growing inside a mold inside a jar
I have recently found an experiment where you can grow a salt crystal on a jar. The experiment works like this: we take a jar, fill it with hot/warm water and start dissolving salt in it until it saturates (add food coloring - optional) and then leave it for a couple days after which on the bottom of the jar there will be a salt platform with crystals growing out of it. We take this formation out of the water and break it into pieces and take the nicest crystal, tie it with a string and then suspend it back in the water and over time (1-2 months) our crystal will grow in size.
My question is, if I create a solid plastic mold in which the water from the jar can infiltrate, if I put this mold in the saturated water and the crystal suspended right at the very top part of the mold and make sure the water is saturated from time to time, will the crystal grow perfectly or almost perfect all the way to the feet of the mold? I mean I want to start making salt figurines using a 3d printed mold and the salt experiment is it going to work?Thank you for you help I really appreciate your work.

I was wondering too
https://www.amazon.com/Thames-Kosmos-643522-Crystal-Growing/dp/B007WDGZUM
you could make something similar?
:)
Pam

If you are still interested in this I'd be glad to give my thoughts on the best way to go about it.  But as hard as you tried to describe it I'm not sure I'm getting it. A picture is worth way more than a thousand works in my opinion ;)  If you could post a sketch of what you are proposing that would be tremendously helpful!


text:
In a condensation reaction, which of the reactants loses their OH group?
You have Propanoic Acid ($\ce{CH3CH2CO-O-H}$) so an $\ce{OH}$ on the end
You have Ethanol ($\ce{CH3CH2-OH}$) so an $\ce{OH}$ on the end
When you combine the two for a condensation reaction you have the $\ce{OH}$ group from 1 reactant and the $\ce H$ group from the other that joins to release a water molecule correct?
So which of these groups loses its $\ce{OH}$ group and which group loses its $\ce H$ group?
How can i determine this is im guessing it may have something to do with strength of bonding and shielding but can somebody please explain i cant find the answer

related http://chemistry.stackexchange.com/questions/18378/are-all-esterification-reactions-reversible-what-is-their-general-mechanism

Acid loses its OH and alcohol loses its H, that's the way it is.

Long story short, oxygen from alcohol's $\ce{-OH}$ attacks the carbon from the acid's $\ce{-COOH}$. That carbon, being linked to two oxygens, carries significant positive charge, so this is much more likely than the other way around.
Also, it has been proven rigorously with the use of isotope-labeled compounds. Say, you take some acid with "normal" oxygen ($^{16}\ce O$) and alcohol with $^{18}\ce O$, and behold, the ester comes out with one $^{16}\ce O$ and one $^{18}\ce O$, so the latter must have come from alcohol, so it is the oxygen from acid that was lost in the form of water.


text:
Why does changing the dilution of an aqueous equilibrium shift the equilibrium?
Say we have the equilibrium $$\ce{A_{(aq)} + 2B_{(aq)} <=> C_{(aq)} + D_{(aq)}}$$ meaning the formula for the equilibrium constant is $$\ce{K_{c} = \frac{[C][D]}{[A][B]^2}}$$ My understanding is that adding more water would shift the equilibrium to the left (in the same way that decreasing the pressure would shift the equilibrium to the left if it was a gaseous equilibrium).
Why is this the case? What mathematically (using the equation) and physically (in terms of kinetics?) means that $\ce{[A]}$ and $\ce{[B]}$ must increase?
And why does the equilibrium constant remain the same?
Apologies if this is a duplicate, but I couldn't find a specific answer to this question.
Many thanks!

Mathematically, say, if you dilute everything with equal amount of water, all concentrations become 2 times less; guess what would happen to $\ce{\frac{[A][B]^2}{[C][D]}}$?

Sorry, I wrote the $K_c$ equation wrong. But halving the values of $\ce{[A], [B], [C], and [D]}$ would increase the value of $K_c$ ($\ce{\frac{[C][D]}{[A][B]^2}}$), which I know isn't the case.

That's right; now what else do you have to change in order fo $K_c$ to actually stay constant?

The concentrations of A, B, C, and D have to change. But why does Kc have to remain the same?

That's right, too, and you seem to know which way they have to change. As for why the equilibrium constant is the way it is, see http://chemistry.stackexchange.com/questions/6924/why-does-the-reaction-quotient-use-the-products-multiplications-of-reactants-a or http://chemistry.stackexchange.com/questions/14836/why-is-the-equilbrium-constant-expressed-the-way-it-is


text:
Is this Boiling point apparatus?
I have some chemistry equipment but I am not sure what it is for; I think it may be apparatus for measuring elevated boiling points. I have included a drawing of how I think it should go together. Basically, it consists of a conical flask connected to a graduated boiling tube (D) (for volume); the conical flask has a two-holed bung: one aperture contains a delivery tube (B) connected to the boiling tube (there are small holes in the delivery tube within the boiling tube, and the other has a glass cylinder passing vertically upwards, which is angled at the bottom (A; this might be to do with maintaining atmospheric pressure?). The boiling tube has a bulb on the upper portion. The boiling tube is contained within a larger glass container (C). There is an aperture in the boiling tube opening into the larger container (E). The larger container has a spout at the bottom.
So the liquid in the conical flask is heated and passes through E and condenses in the boiling tube (D). I am not sure if the delivery part, E, has to be lower, so that as the liquid condenses the delivery tube becomes submerged in the liquid. Then I am guessing the liquid that condenses in the boiling tube will at some point be hot enough to pass from the boiling tube, through aperture E into the larger container (C). Quite what is happening, I am not sure...
If anyone has seen this equipment and knows how it works, that would be much appreciated.

To me this looks like an apparatus to measure water in fat, butter or oil. A weighed sample of butter is placed in the conical flask. The tube A goes below the level of the melted butter or oil. It's purpose is to maintain atmospheric pressure or in case of a bump in the comical flask or back pressure or suction from the boiling tube D or the condensation flask C. Flask C should have a hole open to the atmosphere through the bung. The hole at the bottom of C is to remove water from the condensed ice. The conical flask is heated. The butter (oil) has a much higher boiling point than the water in it. So the water boils out. The tube from the conical flask will go well down into the boiling tube D. More than you have indicated in your diagram. Flask C should be full of an ice water salt mixture at −4 °C surrounding the boiling tube up to below the ventilation hole in D. It must not cover it or spill over into D as this would give you a wrong result. The thermometer measures the temperature in the region of D where the water condenses. If the thermometer indicates a rise in temperature above a set limit (I do not now have access to data where I could look up this temperature. However, I think it should be in the region of 5–10 °C) then the rate of heating show be reduced as you might overload the condenser and loose moisture. If the test has been running for some time it may be necessary to add ice as the ice melts to water the cooling efficiency of your condenser would be compromised. To do this some water should be run off from the tube at the bottom of flask C. This would be done Using the glass spout which you have shown. What you have not shown is the piece of rubber tubing (now long perished) and the clip attached to the spout at the bottom of C which would allow you to remove excess water from C. If you do not remove the excess water from C you could cause water to spill over into flask D as the ice is added from the top. If water were to get into D from C then your result would be destroyed and after cleaning everything out you would have to start over again. Remember it is necessary to have some water in C so that the ice can make better contact with D and act as a much more efficient condenser. The volume of water in the butter can be read on the side of flask D. It works well on low fat spreads which contain lots of water. If you want to use do not send for me. Good luck.


text:
Bond strength of different isomers of an organic compound
Out of the following isomers of C7H7Cl containing benzene ring, i.e, out of
o-Chlorotoulene, m-Chlorotoulene, p-Chlorotoulene and Benzyl chloride, benzyl chloride has weakest C-Cl bond.
Why is it so?

Welcome to Chemistry! This seems like a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed.

Because the corresponding cation (or radical, for that matter) is the most stable of all.

I believe it is a result of resonance structures. When a chemical entity (ion, radical etc.) has a lot resonance structures it is considered stable beacuase it allows the electrons to move around realtivily freely. If you remove the Cl from the benzyl chloride you can draw resonance structures by transfering the positive charge in the benzyl ring. Thus the positive charge can be delocalized making this structure more stable than the others that would be created from the chlorotoluenes.

Electrons do not move in resonance.

Other than that, the explanation is about right.

By electrons moving I meant that the charge can be delocalized. Is this incorrest?

The charge is delocalized all right, but not because electrons are moving. Instead, it is customary to say that electrons sit on the molecular orbitals, and thus each electron by itself is delocalized in such-and-such manner.

authors: Αντώνιος Κελεσίδης


text:
What is the structure of trifluoromethanesulfonic acid in the gas phase?
During an interesting discussion in our group’s seminar, the question arose what actually causes trifluoroacetic acid to be a much better nucleophile than trifluoromethanesulfonic acid. Part of the discussion was whether the oxygens of trifluoromethanesulfonic acid were equal or not.
Traditionally, $\ce{TfOH}$ is drawn as shown in the structure below on the left. The structure would assume the molecule having point group $C_\mathrm{s}$ due to the non-equality of the oxygen atoms only one of which bears a hydrogen. The triflate anion $\ce{TfO-}$ in solution, on the other hand, has three homotopic oxygens as can be seen in the structure on the right, leading to the more symmetric point group $C_\mathrm{3v}$.

One of the opinions voiced in the discussion was that even the gas-phase molecule $\ce{TfOH}$ adopted a $C_\mathrm{3v}$ point group, the hydrogen being in the same axis as carbon and sulfur and somewhat centred between the three oxygens. This makes perfect sense when thinking about it. Therefore, my questions are:

Has the structure of trifluoromethanesulfonic acid or a structurally strongly related acid been studied in the gas phase? (Please provide a reference)

Does the experiment suggest equivalent or non-equivalent oxygens?

How well can that idea be extended? e.g. should one consider the structure of formic acid to be $C_\mathrm{2v}$ symmetric rather than $C_\mathrm{s}$?

Do (solid-state) crystal structures or other experiments exist that confirm a different structure for solid $\ce{TfOH}$?

If solid and gas phase structures differ substantially, what about the structure of the molecule in liquid phase/solution?

I can't see any good reason why it should be symmetric. It would be symmetric bifurcated H-bond there. I wonder if they even exist.

I agree with @Mithoron. In the more compact $\ce{-COOH}$ group the angle does not even get close to allowing a substantial attractive interaction $\ce{-OH\bond{~}O\bond{-}}$. See this question and my answer to it. It's an interesting thought provoking question nevertheless.

Also, dear compchem people, if you want to contribute with calculations, you are hereby warmly encouraged. That is, if no reference exists already.

Solid state crystal structures do exist, but everything I've seen indicate hydrates, making the question of symmetry a bit complicated.

related: http://chemistry.stackexchange.com/questions/47346/do-symmetric-hydrogen-bonds-in-neutral-molecules-exist?rq=1

Has the structure of trifluoromethanesulfonic acid or a structurally strongly related acid been studied in the gas phase? (Please provide a reference)

Yes. See Gas Phase Structure of Methyl Trifluoromethanesulfonate, CH3OSO2CF3, and Conformational Properties of Covalent Sulfonates Inorg. Chem., 1999, 38 (13), pp 3051–3055 and references cited therein, especially Electron Diffraction Investigation of the Molecular Structure of Trifluoromethanesulfonic acid (triflic acid) Z. Naturforsch. Teil A 1981, 36, 917

Does the experiment suggest equivalent or non-equivalent oxygens?

Non-equivalent.
The 1981 paper finds
S=O 141.7 ( 1) pm
S-O 155.7 ( 2) pm

How well can that idea be extended? e.g. should one consider the structure of formic acid to be C2v symmetric rather than Cs?

From the following gas phase study, I don't think that gas phase formic acid monomer is C2v.
http://actachemscand.dk/pdf/acta_vol_23_p2848-2864.pdf

Do (solid-state) crystal structures or other experiments exist that confirm a different structure for solid TfOH?

Quoting from Trifluoromethyl Chlorosulfonate, CF3OSO2Cl: Gas Phase and Crystal
Structure, Conformation and Vibrational Analysis Studied by
Experimental and Theoretical Methods Inorg. Chem. 2003, 42, 7297-7303
"The structure of triflic acid, HOSO2CF3 (trifluoromethansulfuric acid), was also studied by GED, and an anti orientation
was assumed in the analysis of the diffraction pattern. [reference 6: the 1981 reference above] In
contrast, an X-ray investigation of 2HOSO2CF3 [dot] H2O [ie. the hemihydrate] resulted
in gauche orientation of the O-H bond,[reference 7] in accordance with
theoretical calculations. [references 8 and 9]"


text:
How can a conductivity experiment have higher voltage than battery?
I did an experiment where I tested the conductivity of different solutions and some substances.
We connected a 9 bolt battery, some aluminum foil sticks, and a multimeter that reads the voltage.
It connects like  liquid --> aluminum --> multimeter --> battery --> aluminum --> liquid.
My understanding is that if a liquid contains more ions, it conducts electricity better.
Here's my question. We tested the battery just by itself and it read about 9 volts. When we tested with baking soda, it reads about 9.2 volts.
How can it be higher than the energy source? Does baking soda produce its own electricity?
Also, when we tested deionized water, it came out to be about 8.5. It seems close to 9, but it was one of the lowest of the tested solutions.
Why is it like this? If it's deionzed, shouldn't it be nearly zero? since it doesn't conduct electricity at all? (although there can be impurities so it can have some, but I think 8.5 is still too much).
I don't really understand. Can someone explain why the results came out like this?

A voltmeter is a zero current(i.e. very high resistance) device. Therefore in your scheme, the readings you are getting are not reliable. The way to make that measurement is by either switching your multimeter into an ammeter in the configuration you have and measure the current, or adding a resistor to the circuit in place of where you have the multimeter now and connect the multimeter in parallel to that resistor. In either case, you will be measuring the current going through the solution, which will tell you about the conductivity.


text:
Lithium metal not reacting with water
To start off with I want everyone to know that I'm not doing this for nay kind if drugs or anything illegal, just having some fun. So I know that lithium( Li) is the least alkaline metal in group one and that it will react with water like Na ,k, Fr, Rb and Cesium (Cs). So I decided to take apart some rechargeable D cell Li batteries to gather some lithium, the first time I tried it didn't react with the water at all. The second time I did all my work covering everything in mineral oil to make sure non if it reacted with the moister in the air ,then dried it. But when I placed that in water it still did nothing after 10 minutes. I even tried hot water but still nothing. Can someone give me an explanation if what's wrong with my lithium?
The batteries came out of a cordless drill that no longer took charge in case that has anything to do with it. Thank you ( I'm sorry for any spelling mistakes or chemical symbol errors, been quite a few years since I have been out of chemistry)

Some good answers below, but I'll add that if somehow you had lithium, you would know it. It not only reacts with water, but floats on the surface due to its lower density. The rate of reaction, far less reactive than sodium or potassium. Hydrogen is evolved in the reaction, but usually not enough heat to ignite it into flame.

The lithium-ion batteries do not contain pure metallic lithium, nor do they contain any metallic alloy of lithium. They contain various intercalation compounds which are relatively harmless by themselves.
When a charged battery is short-circuited (by throwing it into the water or otherwise), it may react quite violently, but that's another story. Any battery would do the same, more or less. It's about physics, not chemistry (I mean, chemical reactions would occur as well, but they don't explain the explosive part; they would occur during the normal operation all the same).

@Ivan I would expect in the charged state that the anode would be quite reactive.  Typically, it is some form of carbon intercalated with lithium.  In the uncharged state, the lithium would be in ionic form at the cathode and completely harmless.

@Qubit1028 True, it would be reactive, but still far less so than metallic lithium.

Inside batteries it isn't pure lithium metal used. I believe that it is an alloy of lithium and cobalt or lithium and iron. This would mean that the metal strip that you put into the water won't react as vigorously as you would expect lithium in it's element form to.

Well I did not know that to just assumed it was mostly pure lithium. I have seen on the internet tho that the lithium battery's react violent in water. That's y I am going with it has something to do with the charger or the age


text:
Physical reason for why delocalisation leads to stability?
Why does delocalization of electrons generally make compounds more stable (e.g. in carboxylate anions, where the lone pair on the negatively charged oxygen is delocalised into the C=O π* orbital)? Is there any proper explanation involving quantum mechanics?

Electrons are always delocalised across the entire molecule according to quantum mechanics. It is just that the Lewis depiction of localised bonds is so much simpler to teach that we use it, and then we need resonance to overcome its weaknesses.

A simple QM explanation I can think of would invoke Hückel theory. It would go this way:
Suppose you have a two-state system (analogous to having two orbitals) with eigenvalues of the Hamiltonian $\lambda_1$ and $\lambda_2$. These correspond to the orbital energies. So just for argument's sake, let's say that
$$\begin{equation}
\mathbf{H} = \begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix}
\end{equation}$$
then obviously you have $\lambda_1 = 1, \lambda_2 = 2$. Let's say this system has two electrons. They would both go into the orbital with the lower energy, i.e. $\lambda_1$. In the context of organic chemistry, this would typically mean that the first orbital would be bonding in nature (hence filled), and the second orbital would be antibonding (hence empty), although that's not always the case.
Now, you perturb the system by making the non-diagonal elements of the Hamiltonian negative (analogous to allowing the orbitals to overlap, i.e. allowing delocalisation of electron density between the two). Just for argument's sake, let's change $\mathbf{H}_{12}$ and $\mathbf{H}_{21}$ to both be $-0.1$. If you recalculate the eigenvalues you get
$$\lambda_1 = 0.99, \lambda_2 = 2.01$$
The new eigenvectors will be linear combinations of the old eigenvectors – one in-phase and one out-of-phase combination. The "bonding" or in-phase combination of the two orbitals will have decreased in energy, whereas the out-of-phase combination of the two orbitals will have an increased energy. If you put both electrons into the bonding orbital, you get a net stabilisation.

In general, interactions between filled and unfilled orbitals are stabilising in nature, since both electrons occupy a lower-energy bonding orbital.

On the other hand, interactions between two filled orbitals are destabilising, because you would have four electrons and you would have to fill both of the new orbitals. If you add up the two new eigenvalues, the total energy is greater than it was before when $\mathbf{H}$ was unperturbed.
This is not evident in my example above. The reason is that I made the assumption that $\langle i | j \rangle = \mathbf{S}_{ij} = \delta_{ij}$. In extended Huckel theory you would solve the generalised eigenvalue equation, $\mathbf{Hc} = E\mathbf{Sc}$. However with my assumption, $\mathbf{S}$ is simply the identity matrix and this reduces to $\mathbf{Hc} = E\mathbf{c}$. A theorem from linear algebra states that the sum of the eigenvalues of $\mathbf{H}$ is simply equal to the trace of $\mathbf{H}$. Therefore, as long as we assume $\mathbf{S}_{ij} = \delta_{ij}$, merely changing the off-diagonal values will not change the sum of the orbital energies.
In general, the basis vectors are not orthogonal. If you solved the full equation, $\mathbf{Hc} = E\mathbf{Sc}$ then you would find that the sum of the new eigenvalues is greater than the sum of the old, unperturbed eigenvalues.
Obviously this is ignoring other important aspects, e.g. electron-electron repulsions (that is a problem intrinsic to Huckel theory), but it's a start.

Great explanation and easily understood! Thanks!

Even simpler:  Particle in a box.  The longer the box, the lower the energy.  Delocalization equals a longer box.  However, this trite model is not quite as good as orthocresol's explanation.

Actually, is it always true that delocalisation leads to stability? Consider the case of various antiaromatic compounds. In particular, let us look at cyclobutadiene. It exists more stably in its "rectangular form" with localised double bonds, rather than with electronic conjugation across the whole molecule.

@TanYongBoon Would one describe an antiaromatic system as well delocalized? In cyclobutadiene, two bonds have constructive pi intereference, while two other bonds have destructive pi interference. I'm not an expert in it, but perhaps the instability is due in part precisely to less delocalization being available.

@Blaise I would like to disagree. In the "fully-antiaromatic" cyclobutadiene, the 4 $\ce {C-C}$ bond lengths tends towards being equivalent, favouring maximum delocalisation. The degree of delocalisation cannot be predicted by just looking at the MOs in the MO diagram. I think your description is also slightly incorrect. In the MOs of cyclobutadiene, two are nonbonding, one is bonding while one is antibonding. However, the degree of delocalisation cannot be inferred from the diagram.


text:
What really is an equation of state (EOS) and how are they developed?
Are EOSs merely correlations that mathematically represent (in a better or worse way) experimental data or do they have a theoretical background? How are they developed (suppose one wishes to develop his/her own EOS -- how would he/she start and which script should he/she follow?).

related http://chemistry.stackexchange.com/questions/549/what-is-a-rigorous-definition-of-gas-volume-and-how-is-the-van-der-waals-equati

At least some EOSes may be derived from first principle using statistical physics approach.  However, originally they were developed from general considerations and trial and error.

They are merely brief and approximated ways to describe observed relations among state parameters. They do not need to have a theoretical background, but many times they have. It helps to create them and provides some reliability. The theory used to guide the construction of a EOS is anyone that turns to be useful.
how would he/she start and which script should he/she follow?
It is a very broad question. In thermodynamics there are few state parameters used normally. Thermodynamic theory is a macroscopic theory of somewhat general applicability, but EOS can be very specific, and this specificity can not be provided by thermodynamics, so, it turns to be a wide search.
In cases where simple general relations exist (mostly already discovered), it can be found just by fitting a data to a function.

Statistical thermodynamics provides the methodology for deriving the virial equation of state based on first principles.  It starts out by looking at the potential energy interactions between a pair of molecules of the gas (to get the first virial coefficient).


text:
Liquid for cooling beverages
I am looking for any kind of fluid which has a freezing point below the temperature of a regular freezer  (about -20 C). My plan is to keep this liquid in my freezer for fast cooling of beverages, so the liquid needs to be harmless to drink in case some is left on the can.
Suggestions? Or maybe the liquid will cool down the beverage too fast making it impossible to get a proper even drinking temperature without freezing the beverage?

Ethanol, liquid nitrogen or ether jump to mind.

But they aren't that harmless to drink ;)

@Mith Wait, a Polish person saying ethanol is not harmless? =O

If brine is either unpalatable or has too high a freezing point for you, other options include a solution of glycerol in water. Glycerol is perfectly safe to ingest in modest quantities -- it's used as a food additive and a medicine -- although it tastes a little weird (a bit like extra-sour sweet and sour sauce).
Although it's mostly been replaced by ethylene glycol now, glycerol used to see a lot of use as an antifreeze, and is still used in domestic applications like ice cream makers. Apparently it's best used in a 60-70% solution, when it can get a freezing point below -40°C.
The paper linked below puts the optimum point at -46.5°C with two-thirds glycerol, one-third water by weight.

"Freezing Points of Glycerol and Its Aqueous Solutions", Leonard B. Lane, Ind. Eng. Chem., 1925, 17 (9), pp 924–924, DOI: 10.1021/ie50189a017

Aim for a brine solution: mixture of water and NaCl. The lowest freezing point for a brine solution is about -20$^\circ$C, obtained with approximately 23 mass % NaCl, as is seen in the phase diagram below (you want to stay within the "brine" part of the diagram). This is not much below the freezing point of a regular freezer, but you may get away with some brine + ice. Or set your freezer to -18$^\circ$C.
Concentrated salt solutions is toxic, so make sure to rinse and wipe off any residues. In comparison, sea water is around 3,5 % salt, and we all known we should not drink that!

(Ref: http://people.ucalgary.ca/~kmuldrew/cryo_course/cryo_chap6_1.html)


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What happens to gasoline when it lands on your hand?
I was filling the mower with gasoline and some gas splashed on my hand. Not even 2 seconds later it was gone...
1) Does gasoline turn into vapor that fast? And if it does, why doesn't it evaporate in the gasoline can?
or
2) Does it seep into your skin and into your blood?
Also.. after it had 'dried' there was a white residue leftover, what is it?

Yes, gas vaporizes quite easily which is a big help in the engine. That is why there is a lid on the gas cap to limit the escape of vapor.

It evaporates, but it will first take some hydrocarbon compounds in your skin
Gasoline is a mixture of a variety of volatile hydrocarbons, some branched chain alkanes and some aromatics. The warmth of your hand is enough to cause small quantities to evaporate. But it will also dissolve some of the hydrocarbon-soluble oils from your skin. Then when it evaporates, they will be left but they will form a microcrystalline powder rather than returning to the parts of the skin they were extracted from: this is the white residue. This is a common when volatile non-polar solvents are spilled on the skin (don't do it too much or your skin will be deprived of the oils that keep your skin flexible and impervious to outside contamination). Small quantities are nothing to worry about, just don't get into the habit or it will hurt your skin.

You might want to add that some (admittedly, very small) portion of it actually does seep into through your skin. Back in the days of leaded gasoline it was quite a thing to worry about.

Wow, this is great information! Thanks a lot for your sharing of knowledge ;)
and, p.s. don't worry; I don't intentionally pour the gas on my hand - it just falls on accident when trying to refuel :P

Gasoline is a mix of light hydrocarbons, and it is meant to vaporize quickly. In countries where winters are harsh, gasoline is even lighter than the one fabricated in warmer places; that is done to prevent the formation of hydrocarbon crystals in the gasoline (the temperature in which those crystals begin to form is called "cloud point"). Small amounts of gasoline evaporate fast on your hand because of your body's warmth and the large superficial area of the spill, but if you pour some amount of gasoline into a glass and leave it in contact with the air, there will be no more gasoline left after some time.
The "white residue" is your dry skin, because gasoline washes out your skin's natural oils.
You shouldn't be concerned about occasional gasoline spills on your skin as it doesn't penetrate too deeply into your body, but constant contact with the product may irritate your skin.

why doesn't it evaporate in the gasoline can?
It does.  But the air in the can quickly gets saturated with gasoline vapors, and -- just like on a day with 100% humidity -- no more evaporates.


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Why are axial bonds are longer than equatorial bond in sp3d hybridized atoms?
Why are axial bonds are longer than equatorial bond in case of $\mathrm{sp^3d}$ hybridization? I have done some research but I can't seem to find the answer.

more repulsion in axial bond I guess (one confront three equatorial in 90 degree)?

I would love to answer, however I don’t know any pentacoordinated species that has $\mathrm{sp^3d}$ ‘hybridisation’. =C

$\ce{sp^{3}d}$ hybridization is not involved here. See this earlier answer for an explanation as to the hybridization\bonding involved  and why the axial bonds are longer.

You are asking for $\mathrm{sp^3d}$ hybridisation, but I do not know of a case where $\mathrm{sp^3d}$ hybridisation actually happens. Either it does not make sense to discuss hybridisation at all (the iron in pentacarbonyliron) or the hybridisation is actually not $\mathrm{sp^3d}$ but $\mathrm{sp^2 + p}$ (the phosphorus in $\ce{PCl5}$). I shall discuss the latter, because I believe it could be the one you are on about.
Phosphorus pentachloride is often drawn like in figure 1 below for simplicity. This depiction implies five identical covalent 2-electron-2-centre bonds. However, that does not agree with the octet rule.

Figure 1: Simplified structure of $\ce{PCl5}$.
Instead, one can draw a set of two mesomeric structures, each one conforming with the octet rule — see figure 2.

Figure 2: Mesomeric structures of $\ce{PCl5}$ conforming to the octet rule.
This already hints us towards the answer: Rather than assuming two bonds which are equal to the other three bonds we need to consider three ‘classical’ 2-electron-2-centre bonds to the equatorial chlorines and one 4-electron-3-centre bond to the two axial chlorines. This bond’s order is $0.5$ rather than $1$. Typically, the lower the bond order the weaker a bond and therefore the greater the bond length is — the experiment is in fine agreement with theory.
But no bonding discussion is truly complete without an orbital consideration. Check out the three collinear p-orbitals of phosphorus and the two chlorines, that together form three molecular orbitals labelled $\Psi_1$ to $\Psi_3$ in figure 3.

Figure 3: Representation of the three molecular orbitals that form the 4-electron-3-centre bond.
As you can see, the lowest MO $\Psi_1$ is bonding with respect to both $\ce{P-Cl}$ bonds. The highest MO $\Psi_3$ is antibonding with respect to both. And the middle one is bonding with respect to $\ce{Cl-Cl}$ but nonbonding if we add the phosphorus atom. We need to fill in four electrons into these three orbitals ($\ce{PCl3}$ has one lone pair, and we are effectively using the bond of a $\ce{Cl-Cl}$ molecule as our second electron pair). Thus, the bonding and the nonbonding orbitals are filled. Bond order can now be calculated by:
$$\text{bond order} = \frac{(\text{electrons in bonding orbitals}) - (\text{electrons in antibonding orbitals})}{\text{number of bonds}\times 2}\\= \frac{2-0}{2 \times 2} = 0.5$$
Again, a lower bond order typically correlates with a greater bond length.

Please, consider partial charge interactions in bond order determination.


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What is the logic behind the ion electron method of balancing redox reactions?
The ion electron method is a.k.a the half reaction method.
How are we just allowed to add electrons and protons and water molecules? Why does it work anyway?

What are you asking really? Why stoichiometry and oxidation numbers work?

Why can't we just add protons and water molecules? According to the law of conservation, we can neither create nor destroy matter. Hence, to balance the atoms of each respective element on either side, we add protons and water molecules. Also, I think the method is based on the fact the a certain amount of reduction would entail that an equal amount of oxidation would take place.
If one atom of a P4 molecule underwent a change in oxidation state of +5 (say forming PCl5), then the total change would be +20 and hence you would need an equivalent amount of reduction (i.e. -20) of other species participating in the reaction.


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Anions produce flame colour?
I know that flame tests can be used to distinguish between some metal ions, and that the colours come from excited electrons returning to the ground state.  My question is, why don't non-metal anions also produce colours?  Are their emission spectra in the non-visible range?  Is there a simple reason why?

Anions also can produce flame color. E.g. boron, tellurium, arsenic, antimony.

The OP specified "non-metal anions" so that would seem to exclude anion complexes of tellurium, arsenic, and antimony.

There are two reasons, depending on the anion of interest:
1) For the flame emission of any species, anion or cation, metal or not, to be observable, it must emit light in the visible wavelength range, and most non-metal anions do not do this very strongly.
2) Even if there is some emission from the non-metal anion, in many cases their flame excitation efficiency is very low.
The answer then is the combination of these two things, and the fact that many metals have the opposite tendencies (e.g. high excitation efficiency, strong emission in the visible region of the spectrum), which overrides any weak visible emission from the non-metal anions.
As noted in a comment, this is not always the case, but it is the explanation for this strong trend.


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Pinacol pinacolone rearrangement
I was reading the topic alcohol in that I saw a reaction Pinacol pinacolone rearrangement .
But I am not able to find the sterochemistry of the reaction .
Can anyone please tell me the stereochemistry

In that example, there is no stereocentre.

see this:-http://chemistry.stackexchange.com/questions/43858/why-do-we-have-an-alkyl-shift-rather-than-epoxidation-in-a-pinacol-reaction/43859#43859

@Martin-マーチン if there is a stereocentre in any other example then what would be the stereochemistry

@NilayGhosh in that question about stereochemistry is not mentioned .

This is perhaps the best explanation available- https://chemistry.stackexchange.com/questions/64459/ring-contraction-in-pinacol-rearrangement-of-cyclohexane-1-2-diol

The migrating group attacks trans to the leaving OH group:

This answer implies a concerted mechanism, while the mechanism is usually written stepwise. Can you point us to papers that investigated the mechanism and came to this conclusion? As it stands, your answer seems claiming something without proving it.

Why in second mechanism it becomes five membered ring ?


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Mechanism of acetylene formation from iodoform and silver
My book mentions a synthetic route for turning iodoform into propyne.
That conversion was somewhat as follows:

$$\ce{CHI3->[Ag] HC#CH ->[NaNH2] NaC#CH ->[CH3Br] CH3C#CH}$$

The first step is the conversion of iodoform into acetylene with silver. I am unsure how that reaction works. Could someone provide a mechanism?

What happens when "chloroform reacts with silver powder". Putting this in here so that future google searches redirect here, since this a popular question.

What happens is that the C-I bond being weak, the iodide group is attacked by the silver atom (similar to what Na does in Wurtz reaction). As, there are 3 I-atoms and 2 molecules of iodoform usually react, 6 silver atoms are used up in the reaction and they precipitate out 6 molecules of AgI. The 2 carbon atoms form a triple bond to form acetylene.
EDIT- $$\ce{2CHI3 + 6Ag-> C2H2 + 6AgI}$$
$$\ce{C-I + 2Ag +I-C->I\bond{...}Ag + Ag\bond{...}I + C\bond{...}C}$$
As there are 3 C-I bonds, 6 Ag atoms form bonds with I-atoms, Hence, triple bond is formed between the 2 C-atoms. I hope that explains it.

How does C-C bond formation occur and how do you end up with a triple bond? Can you provide a mechanism?

What @bon said and additionally can you expand on how "the iodide group is attacked by the silver atom"?

While this still isn't what I'd call mechanism, the reaction is redox on surface of Ag powder, so exact mechanism may be elusive.

Perhaps something related to the Ullmann-reaction? Forming an alkyl-silver-iodide, which reacts with a second silver to form silver iodide and alkyl-silver. Another iodoform enters the reaction, the c-c-bond is formed and silver iodide leaves. That would create something like $I_2HC-CHI_2$ and perhaps this further eliminates iodides with the help of silver in a similar fashion?

Here the iodine atoms of iodoform are the leaving groups, which creates a precipitate of silver iodide by reacting with silver. But, there are no nucleophile to occupy the vacant places of the leaving groups. So it creates a triple bond with other molecule. I think that is the process.

It might be undergoing the same mechanism as the Wurtz reaction. Which is through free radical mechanism (Some say that Wurtz reaction also takes place through the ionic reactions).
Firstly the $\ce{C-I}$ bond breaks from both the iodoforms to form $\ce{CHI2}$ free radical
which causes them to fuse forming $\ce{I2HC-CHI2}$. The same step repeats causing further breakng of $\ce{C-I}$ bond leading to formation of alkenes and alkynes later.

It is always best to have literature backing up your mechanisms. Opinion-based answers aren't very helpful, especially when you give no rationale behind why that mechanism might be considered correct.


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Phenol or ethanol as a nucleophile for ether formation?
If we react phenol with $\ce{C2H5I}$ in $\ce{C2H5O- Na+}$ (excess), $\ce{C2H5OH}$ (anhydrous) solutions .
In this reaction why is diethyl ether not formed and the actual product is phenyl ethyl ether?

You are adding excess sodium ethanolate so there is the possibility of deprotonation of phenol: $$\ce{C6H5OH + C2H5O- <=> C6H5O- + C2H5OH}$$
If we look at the $\text{p}K_\text{a}$ of ethanol and phenol we have 16 and 9.95 respectively and so the equilibrium above will strongly favour the right hand side. Therefore, the phenoxide ion will dominate as a nucleophile in the mixture, and the major product will be ethyl phenyl ether.

Yes, that's what I thought too, until I saw that the EtO- is apparently "in excess"... Ethoxide is definitely more nucleophilic than phenoxide and if both are present I would not be surprised if lots of Et2O was formed.

@orthocresol Hmm. You are probably right.

So, you mean to say that phenoxide is not a better nucleophile than ethoxide (on the basis of resonating structures), but only happens to be the dominant nucleophile in this case just because it is present in more amount than ethoxide?

@GaurangTandon Yes that is what I am saying.


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Is there wavefunction interference between alternate reaction pathways?
I'll give a specific example to clarify my question. If we look at the approach of chloromethylenecarbene's electron pair to cis-2-butene in the reaction of creating 1-chloro-2,3-dimethylcyclopropane, there's a symmetry in the choice of where the nucleophilic attack can happen.
These two alternatives can be represented by wavefunctions and so I would expect there to be an interference effect between them that would make the reaction different than if one of the spots was somehow unavailable for reaction (perhaps by the existence of some electron donating/withdrawing groups or other hindering groups attached on the periphery of the butene's double bonds).
Has reaction pathway alternatives interference been observed?

I'm not getting your point.

I've already asked the question as best as I think I could.


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How do Shaving Creams, Foams, Soaps ensure cut Hair Fragments that stick & clump between Safety Razor blades are easily washed away?
This is my first post on Chemistry SE and I figure it's pretty chemical oriented.
On Sustainability SE we were trying to find "organic & natural" alternatives which are okay for lubrication & moisturizing (don't really care about foaming etc), but so far we have one key problem:
Hair fragments clumping and getting stuck in between 3-4 blades of a safety razor and not easily washing out using water, which is not as much an issue with commercial shave products.
Is it some (one or more) active ingredients/chemicals that make it happen or some other chemical phenomenon by which this works?

The same clumping between the blades still happens with commercial shaving creams , I find that it always happens with cheap razors. Also depends how long the hair is too.

It washes out easier. Should I rephrase question and add that?

@Joel rephrased the question to point out that it's easily washed away whereas with natural products tried it wasn't so. So what're the active ingredients & phenomenon at play

Active chemicals are called surfactants. It is well explained in colloid chemistry books and kind of in wikipedia (https://en.wikipedia.org/wiki/Surfactant)
And it is possible to find alternatives for such chemicals in nature, but they won't be as effective as artificial.

Could you improve/ detail more in your answer; list some of these inside Shave products and maybe some natural ones?


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Oxidation of polycyclic aromatic compounds using KMnO4
How is tetracene oxidised by $\ce{KMnO4}?$ It appears that the product is a white solid with a benzene ring inside.
What is the reaction mechanism? Is it radical based? Or does the $\ce{MnO4-}$ ion react with tetracene in a similar manner as alkenes are hydroxylated to diols?
Do I have to consider the inherent aromaticity of tetracene to deduce which "double bond" (because there's no "true" double bond) the $\ce{MnO4-}$ species reacts with?

My money is on either phthalic acid or the 1,2,4,5-tetracarboxylic acid.

I'm guessing quinone will be at least intermediate.

@Jan Sounds reasonable, since naphthalene is oxidized to phthalic acid under these conditions.


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Why is the oxime tautomer more stable than its corresponding nitroso isomer?
We know, generally, that the keto tautomer is more stable than the corresponding enol tautomer. I would appreciate an explanation that goes beyond the thermodynamic reasoning, too.
In the nitroso-oxime tautomerism, the nitroso form is analogous to keto isomer and the oxime is analogous to the enol form.
Then why is the oxime more stable than the nitroso isomer?

I think the more suitable reason is that nitro compounds are stable when there is no alpha hydrogen.

In keto-enol tautomerism generally the keto tautomer is more stable than its enol counterpart because in the former a significant ionic character exists which stabilises it. However, there are certain special cases. For instance if the enol form is aromatic(phenol), or very bulky mesityl groups are attached, etc.
In oxime-nitroso tautomerism however, the electronegativity difference comes into play.
In oxime, nitrogen is attached to carbon by a double bond;
and in nitroso, nitrogen is attached to oxygen by a double bond.
Now it is known that greater the difference in the electronegativity values, stronger is the bond. This property makes the nitrogen carbon double bond stronger hence making oximes more stable than nitroso.
(electronegativity values: N=3.04, O=3.44, C=2.55)

sorry but I can not at all agree with the above answer! The greater the difference in electronegativity between 2 atoms, the weaker is the bond and not the reverse, when we compare atoms of the same period (C, N, O, F). For example H-F is the weakest (most acidic) linkage i.e., F-H <O-H <N-H <C-H. The C-H bond is the strongest and therefore the most difficult to break.

@lemos The farther away the electronegativity values of 2 atoms, the stronger the bond generally is. Cesium has the lowest, and Fluorine has the highest and the make the strongest ionic bond. Be nice to people.

For your acidic strength thingy: http://chemistry.stackexchange.com/questions/39711/electronegativity-and-how-it-relates-to-bond-strength

"it is known that greater the difference in the electronegativity values..." -- a reference would be good here, even a titorial type reference that has primary sources documented.

@lemos The H-F bond is the strongest of the hydrogen-halogen bonds. Look in any bond enthalpy table.

I wonder if the $\alpha$ effect is in play here. Two hetero atoms with lone pairs bonded to each other have a weaker bond because of lone pair repulsion. This probably destabilizes the corresponding $\pi$ bond as well. That would favor the oxime.


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Theoretical Concentration (in PPM) of CO2 production after 1 minute
I recently conducted an experiment to determine ideal ratio of baking soda and citric acid that would produce the most amount of CO2 in water. We used 250ml flasks filled with 120ml water. The samples themselves all weighed a total of 6.5 grams contained various ratios of Sodium bicarbonate and Citric Acid.
The mixtures were dropped in the water, and stoppered with a CO2 meter which measured the production over a minute. We recorded the PPM of CO2 in the container after a minute for each sample. Overall, the rank of the results supported the hypthesis. However, I am having trouble determining the theoretical value to determine the accuracy.
For example, a 50:50 ratio of 3.25g Sodium Bicarbonate and 3.25g citric acid was recorded to have 102,946 PPM of CO2 in the container after 1 minute. How should I go about calculating the theoretical PPM of CO2 after a minute?
To recap:

250mL container
120mL water in the container
Mixture Dropped in water
Stoppered with CO2 meter, which measures PPM of CO2 in the air inside
the container
PPM recorded after 1 minute

How can I determine the theoretical value?
Any help or guidance will be greatly appreciated.


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How is NaOH able to clear a solution of Paraformaldehyde?
While making a solution of $4~\%$ para-formaldehyde for immunohistochemistry (to fix the tissue), we add para-formaldehyde into water and stir. It looks like a chalky supersaturated solution even after stirring with a magnetic stirrer. And as per protocol, you are supposed to add 2 drops of $2~\mathrm{N}\ \ce{NaOH}$ to clear the solution. It surprisingly turned crystal clear!
I was amazed at how just 2 drops could make such a huge difference. Does it react with para-formaldehyde? Does changing pH change solubility? But para-formaldehyde is a polymer, how could it suddenly become soluble? Does adding $\ce{NaOH}$ make it a monomer?

para-formaldehyde consists of long chains of the following type:
$$\ce{HO-CH2-O-[CH2-O]_{$n$}-CH2-OH}$$
When being dissoluted in water, this needs to be broken down into formaldehyde monomers (and then further into formalin):
$$\ce{(CH2O)_{$n$} + H2O <=> $n$ CH2O + H2O <=>> CH2(OH)2}$$
Acids and bases are both able to speed up the first step. Bases by deprotonating one end and causing a domino effect of bonds being broken and formed with a hydroxide released on the other end. Acids by protonating one end and then the domino occuring in the other direction.
$$\ce{HO-[CH2-O]_{$n$}-CH2-O-CH2-O-CH2-OH ->[\ce{OH-}] \\HO-[CH2-O]_{$n$}-CH2-O-CH2-O-CH2-O- ->\\ HO-[CH2-O]_{$n$}-CH2-O-CH2-O- + H2C=O -> \\HO-[CH2-O]_{$n$}-CH2-O- + 2 H2C=O ->}\\ \dots\\\ce{HO- + ($n$+3) H2C=O}$$

so ultimately every PF solution has to form a formalin solution?

Yeah, it’s not going to stay para-formaldehyde. Indeed, any time you are using the stuff in chemistry, you require it to break down and actually want to use ortho-formaldehyde. But gases are hard to weigh @Polisetty ;)


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Phenolphthalein as a indicator for the titration of benzoic acid

Question: Justify the use of phenolphthalein $(\mathrm{p}K_\ce{in}=9.4)$ as a indicator for the titration of benzoic
  acid $(K_\mathrm{a}=6.3\times 10^{-5})$ with $\ce{NaOH}$.

My attempt:
I can calculate the $K_\mathrm{b}$ of $\ce{NaOH}$ from the above data.
And I also know the formula
$$\mathrm{pH} = \mathrm{p}K_\mathrm{b} + \log \frac{\ce{In}}{\ce{HIn}}$$
I know that this is the titration of a strong base and a weak acid. So the equivalent point will be at a $\mathrm{pH}>7$ and hence phenolphthalein is a good choice.
But how can I conclude mathematically that this is a correct choice?

I am slightly confused. Your Henderson-Hasselbalch equation should include the most relevant species which are benzoic acid and benzoate. Also, what you did was calculate and thus it is a mathematical proof, isn’t it?

Geez simply compare pKb of acid with pKind. If it's similar then OK.

As far as I have understood, I have to compare the $\mathrm{p}K_\mathrm{b}$ of sodium benzoate with the $\mathrm{p}K_\mathrm{ind}$ of phenolphthalein.
Thus what I get is:
$K_\mathrm{b}$ of sodium benzoate $=\frac{K_\mathrm{w}}{K_\mathrm{a}}=1.59\times 10^{-10}$
$\mathrm{p}K_\mathrm{b}$ of sodium benzoate $=9.79$
$\mathrm{p}K_\mathrm{ind}$ of phenolphthalein $=9.4$
So, $\mathrm{p}K_\mathrm{ind}$ lies in between the values $\mathrm{p}K_\mathrm{b} \pm 1$.
Hence phenolphthalein is a good indicator for the titration.
Thanks all for helping.

This reasoning is totally wrong. You want a difference between the pKa's of the acid and the indicator. Using the pKb (14 -pKa) of the acid  is just wrong. Consider if the pKa of the acid was 2. Then this reasoning says that the pH of the final solution should be +/- 1 pH unit of pH 12. That would introduce a significant blank value which would then need to be corrected for.

You have to take two reactions into account.
First the protolysis of benzoic acid
$$\ce{HA + H2O <=> H3O+ + A-}$$
$$k_A = \frac{[\mathrm{H_3O^+}] [\mathrm{A^-}]}{[\mathrm{HA}]} \tag{1}$$
and second the (simplified) protolysis of phenolphthalein
$$\ce{HInd + H2O <=> H3O+ + Ind-}$$
$$k_{Ind} = \frac{[\mathrm{H_3O^+}] [\mathrm{Ind^-}]}{[\mathrm{HInd}]} \tag{2}$$
The color change is around the point where the concentrations $[\mathrm{HInd}]$ (colorless) and $[\mathrm{Ind^-}]$ (colored) are equal, i.e. $$k_{Ind} = [\mathrm{H_3O^+}] \tag{3}$$
If you substitute (3) into (1) and take the negative logarithm to base 10 you get
$$\mathrm{pK_A} - \mathrm{pK_{Ind}} = -\log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} = 4.4 - 9.4 = -5.2$$
what shows that the benzoic acid has been completely titrated at this point.

One needs to compare pKb with pKind here

This is what the difference $\mathrm{pK_A} - \mathrm{pK_{Ind}}$ does.

The answers given by SchrodingersCat and aventurin are both bogus.
The gist is that for an titration with sodium hydroxide we'd like;

A blank to use essentially zero volume of the titrant.
The acid to be "completely" neutralized

We can assume a fairly standard situation for the analysis. We can assume 0.1000 molar NaOH as the titrant and want to use about 30 ml of a 50 ml burette. That would mean weighing out about 3.0 millimoles of benzoic acid (0.37 grams) which would be dissolved in 25 ml of distilled water.
Significant Figures
All calculations done in chemistry involve errors. We can assume that at best a 50 ml burette must deliver at 0.01 ml to be detectable. We'd normally use about 30 ml of titrant. Thus any error or bias less than 0.33 parts per thousand is insignificant.
To give ourselves a bit of leeway, we can assume a desired precision of 0.1% overall, or about 1 part per thousand.

You have to be careful here since 10 factors with -0.1 parts per thousand bias each would introduce a -1.0 ppt error overall which would be significant.

Preparation of Phenolphthalein Indicator (Acid/Base Indicator)
Use a fairly standard "recipe" for this.

Weigh out 0.5g of phenolphthalein.
Prepare 50% ethyl alcohol solution contained of 50mL ethanol and
50mL water.
Now, dissolve the phenolphthalein in the 50% ethyl alcohol solution.
Store in dropper bottle for use.

Let's say that 2 drops of the indicator would be used for a titration. Then that would be:
moles indicator = $\dfrac{2}{20*100}\dfrac{0.50\ \mathrm{grams}}{318.328\ \mathrm{g/mol}} = 1.6\times10^{-6}$
Blank Value
Now for the blank we'd just titrate 25 ml of distilled water with two drops of the indicator. The notion here is that if 0.01 ml of titrant would change the indicator then we can ignore the necessity of using a blank. (Think of the blank as "noise" in a signal-to-noise measurement. No noise means we're only measuring signal.)
Now 0.01 ml of 0.1 molar NaOH is $1\times10^{-6}$ moles of base ($\ce{OH^-}$). In 55 ml of solution that yields a pH of 9.3. The pKa of phenolphthalein is 9.4 so almost half of the phenolphthalein would converted to the colored form so the color change should be very detectable at that point.

Here we have ignored how much base that would react with the indicator itself. This is mitigated because we only have $1.6\times10^{-6}$ moles of indicator in solution. If we assume half of the base reacts with the indicator we still have a pH of about 9.0 for the solution.

Assuming that half the indicator must change color to be detectable puts us on the ragged edge of needing a blank correction. Since we used $3\times10^{-3}$ moles of acid, but $1.6\times10^{-6}$ of indicator that gives a ratio of about 0.3 parts per thousand. In reality the color change should be detectable at a pH of about 8.5, so much less than half of the indicator will react to make the colored form.
Thus all in all no blank should be needed.
"Total" Reaction of the Acid
For any acid , $\ce{A}$, we can assume the ionization reaction to be:
$\ce{HA + H2O <=> H3O+ + A-}$
and
$K_a = \dfrac{[\ce{H_3O^+}] [\ce{A^-}]}{[\ce{HA}]}$
Now at the pKa we know that $\ce{[A^-] = [HA]}$. We can't ever remove all of the $\ce{HA}$ form since we can always calculate the relative amounts of  $\ce{HA}$ and $\ce{A^-}$ based on the pH. So let's calculate pH at which:
$\ce{[HA] = 1\times10^{-3}[A^-]}$
That turns out to be at:
$\ce{[H^+] = 1\times10^{-3}K_a}$
or at $\text{pH(99.9% conversion)} = \mathrm{p}K_a + 3.00$
Given that benzoic acid has a pKa value of 4.20 that means that at a pH of 7.20 99.9% will be in the unprotonated form.
We can assume that the amount of acid necessary to take the solution from a pH of 7.00 to a pH of 7.20, or even about 8.6 is negligible from our blank calculations.


text:
Is the reaction of silver nitrate and copper endothermic or exothermic?

Is the following reaction endothermic or exothermic and why?
$$\ce{2 AgNO3(aq) + Cu(s) -> Cu(NO3)2(aq) + 2 Ag(s)}$$

I have a feeling that it is exothermic due to the fact that it is spontaneous and will end in around 30 minutes (which I am guessing is because it will run out of energy to burn) and no source of energy is there for it to gain energy from, but is this right?
I found on the internet, that I was right about the reaction being exothermic as $\Delta_rH^\circ= \pu{-147.39 kJ}$, but my real question is why? Is its spontaneity the only means of support for this answer?

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You're basically saying that it's exothermic because it happens which of course isn't true.

Many endothermic reactions are spontaneous as well.

You may already know that the world is not about enthalpy but about Gibbs free energy. A reaction is sponanteous if the change Gibbs free energy is negative.
$$\Delta G = \Delta H - T \cdot \Delta S$$
As you see, enthalpy and entropy play into this. Looking at your reaction, we note that we are taking two ions out of a solution to release one into solution. Compare the following, slightly different description:
$$\ce{2 Ag+ (aq) + Cu (s) <=>> 2 Ag (s) + Cu^2+ (aq)}$$
From an entropic point of view, this is bad. Rather than having two dissolved ions, we have one on the products’ side. Therefore, we can assume $\Delta S < 0$, therefore $T \cdot \Delta S < 0$. Negating it gives a positive value.
But the reaction is spontaneous, so $\Delta G < 0$. This can only be true if $\Delta H$ is sufficiently negative to counteract the loss of entropy. Thus we conclude that $\Delta H < 0$ and the reaction is exothermic.

I have carried out the experiment myself and from observation, the reaction is exothermic (releases heat) even though I didn't quantitatively determine the value of heat change occurred.

If you didn’t quantitatively determine the heat released, how can you be sure that the reaction is exothermic? As it stands, this answer needs additional references …


text:
Why do some acids have pH 8?
Is it necessary for an acid to have pH less than 7? If no, please elaborate with examples.

Yes, the pH will be less than 7.
Remember that the pH of 7 in neutral water comes from its autodissociation.  That autoionization diminishes when we add an acid or a base, but it does not disappear immediately.  If the acid or base is very weak, you may still have much of the solvated proton/hydroxide ion concentration comes from water rather than acid/base dissociation.
If you have an acid that appears to be generating less than $10^{-6}$ mol/l solvated protons from the acid, or a base  generating less than $10^{-6}$ mol/l hydroxide ions, then water autodissociation remains appreciable.  You then need to look at both water dissociation and acid or base dissociation.  When you do, you properly get pH below 7 for an acid, or above 7 for a base.

Neutrality doesn't necessarily mean that pH = 7. Neutral water at 25 degrees Celsius has a pH of 7, yes. The pH of neutral water will vary with temperature. The only requirement for neutrality is that [H+] = [OH-].

I know.  But usually these problems are posed in the context of ambient conditions.  Regardless of the temperature or exact dissociation product, the point that we may need to be ready to consider solvent autodissociation as well as acjd/base dissociation stands.  A practical example is the very slight acid hydrolysis of magnesium salts and maybe even calcium salts of strong acids.