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How to calculate the concentration H3O+ in a solution with pH=6.99?
What is the correct way to calculate the concentration $\ce{H3O+}$ in a solution with $\ce{pH}=6.99$?
Attempt 1.

pH<7, therefore there are only $\ce{H3O+}$ particles in the solution. $[\ce{H3O+}] = 10^{-\ce{pH}} = 10^{-6.99} = 1.02 \cdot 10^{-7}$

Attempt 2.

We have $[\ce{H3O+}] = 10^{-\ce{pH}} = 10^{-6.99} = 1.02 \cdot 10^{-7}$ and $[\ce{OH-}] = 10^{-\ce{pOH}} = 10^{-7.01} = 9.77 \cdot 10^{-8}$.   Because of $\ce{H3O+ + OH- -> 2 H2O}$ we are left with $[\ce{H3O+}] = 1.02 \cdot 10^{-7}- 9.77 \cdot 10^{-8} = 4.6 \cdot 10^{-9}$

When the pH is smaller than 6 or greater than 8, one will not notice the difference, but here it is logarithmically speaking very large. So I wonder what the correct way is?

Let's put it bluntly. What is the concentration of $\ce{H3O+}$ in a solution with pH=7.00? Try calculating it using your first way. And your second way, too. Where is the truth now?

@IvanNeretin I believe the second. So it should be always the second way. However, someone with a degree in chemistry claimed that chemists agreed one should use the first way, because using the second way would be superfluous work and the difference is small anyway. I did not believe it, hence my question.

The second attempt is wrong. There exists an equilibrium between the ions. The ions don't combine to form water molecules (they actually do but the rate at which they combine is equal to the rate at which water molecules dissociate to produce the ions at equilibrium, hence no net change). Your first attempt is correct.

@wythagoras OK, let's try the other way around. At pH=7, using your second way (which is wrong, in case nobody told that before) you would get the concentration of $\ce{H3O+}$ as 0. But wait; what is pH? How it is defined?

Here's what's wrong with the second method. When you subtract $[\ce{OH^{-}}]$ from $[\ce{H3O+}]$ to get the "excess" $[\ce{H3O+}]$, you are implicitly assigning an equilibrium constant of $+\infty$ to the neutralisation reaction. This is not true; the equilibrium constant is high ($\mathrm{k_{w}^{-1}=10^{14}}$) but it is not infinite. At these very low concentrations, you cannot perform such a subtraction, and must take the finite value of the equilibrium constant into account.

If you take a sample of pure water, there will be few hydroxide and hydronium ions. Of course, they can combine to form water and yes they do combine but there will be few water molecules which break/combine to form the ions again. Hence, there exists a dynamic equilibrium between concentration of ions and water molecules.
$\textrm{pH}$ by definition is the negative logarithm of hydronium ion concentration.
$$\textrm{pH} = -\log [\ce{H^+}] = -\log [\ce{H3O^+}]$$
You can obtain the concentration of H+ ions by substituting the value of pH in the following formula,
$$[\ce{H3O^+}] = 10^{\mathrm{-pH}}.$$
Your attempt 2 is flawed because your assumption that all the ions combine to form water molecules is incorrect. There will always be some concentrations of the ions and all of them needn't combine to produce water molecules. Your attempt 1 is correct.
It appears like you are not aware of the concept of equilibrium and self ionization of water, I have picked few good materials which you might(should) want to refer to,
Chemical Equilibrium
Self Ionization of water
The concept of chemical equilibrium is very important and you will come across it frequently in chemistry, so you must learn it. Also, self-ionization of water along with chemical equilibrium are central concepts for learning acids and bases.

I think you are confusing two different concepts. If you want to know how much acid you need to add to get to a pH of 6.99, it is important to take account of the fact that water is slightly dissociated. But that was not the question. The question was simply

what is the concentration of H3O+

And that follows directly from the definition of p in pH:
$$\rm{pH = -\log_{10}([H_3O^+])}$$
A simple mathematical rearrangement gives you
$$\rm{[H_3O^+] = 10^{-6.99}}$$
Don't confuse yourself with random bits of science that don't belong in the answer... it just makes it harder than it needs to be.

Please discard earlier answer as there was a small misunderstanding.
Here self ionization of water will also take place which will increase H+ concentration and will reduce OH-  concentration. Also [H+ ] from water will not be equal to 10-7 due to common ion effect. Net [H+]=10-pH
Also [H+ ] = [H3O+ ] because a single H+  combines with a single water molecule to give H3O+ without involving OH-  as you did in attempt 2.

In earlier attempt by mistake I was considering that conc. Of HCl is given and pH is to be calculated

The pH is close to 7.
So the hydronium ion concentration of water can't be neglected.
[H3O+fro water + H3O+ from acid][OH-]=10^-14
Please note that H2O dissociates partially to form H3O+ and OH- and that this process reaches equilibrium with finally the ionic product:
[H+][OH-]=10^-14
If an acid is added to water.
H+ increases and hence by the Law of Mass Action the equilibrium is pushed to the left and the concentration of OH- decreases. This is how concentration of H+ becomes greater than the concentration of OH-.
So you can in fact take H+ concentration as 10^(-ph) which gives the total concentration of H+ due to both acid and water.
Your attempt 2 is conceptually erroneous as you have taken the difference of H+ and OH- and not found PH itself.
I think the point that you have forgotten is that both H+ (rather H3O+) and OH- exist together in solution although one might be in excess of the other.
So your first approach is more suitable.
pH is by definition the negative of the common logarithm of total H+ concentration in the solution.

It is true that the $\ce{H+}$ from water shouldn't be neglected here. But it shouldn't be considered either. We know the pH already, so we don't care where those protons came from.

What do you mean by "considered either"? We have to consider most of the H+ as coming from the water and not from the acid. Only a small excess amount is contributed by the acid and it is this which lowers the pH to 6.99.

And it's true that now we know the pH we don't care about where the protons came from.

Well, I mean just that: since we know the pH, we don't have to make any calculations involving the water self-ionization constant.

Oh. In that sense. I simply explained about the ionization constant because the question had some confusion regarding that.

PH = - log 10 [H3O+]
[H3O+] = -antilog 10 (PH)
[H3O+] = - 10^6.99
Because antilog b (x) = b^x
Therefore , [H3O+] = 9772372.21

You've reached an impossibly high concentration because you've incorrectly used the properties of logarithms/exponentiation and misplaced the minus sign.

This post is highly corrosive! Beware!


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What is the relation between chemical thermodynamics and chemical kinetics?
Can one be understood on the basis of the other or are they not interrelated at all?
The first thing my kinetics textbook demonstrated was how thermodynamics ignores time taken for a process whereas kinetics considers the time and rate of the process and hence kinetics is a more effective way to view change.
Also, I saw examples of highly thermodynamically feasible reactions which were not kinetically favourable in the sense that they were so slow that the entire process could be neglected as not happening.
Can one understand the thermodynamics of a reaction by studying it kinetics?
Or can one figure out the kinetics of a reaction by thermodynamical consideration?
Also, would it be a wise observation to say that thermodynamics is a largely theoretical subject while kinetics is a more experimental field?

http://chemistry.stackexchange.com/questions/25077/kinetics-vs-thermodynamics

No. Both I think are equally theoritical or experimental. You must study both seperately. Thermodynamics doesn't speak about rates. While kinetics doesn't tell you stability.
If a reaction produces more than one product, thermodynamics will tell you which is more stable, while that doesn't necessarily mean it is the major product. It may happen that rate of formation of other product is more under certain conditionds. Read about thermodynamic vs kinetic product and you will get it.

But why is it that the two are so (at least apparently) non-interrelated. Knowledge of either yields nothing of the other right? Why is that?

This is because both emphasise two different things. If you look at developement of thermodynamics it was developed for handling energy in form of heat. Thermodynamics deals with  macroscopic systems. So it doesn't tell about rates.

Consider for example a vessel in which a rxn is taking place. Thermodynamics being macroscopic science will measure heat released, enthalpy, entropy etc. It will tell you equilibrium constant. While kinetics being microscopic science will focus on a single molecule. Collision theory is an example. It will tell you rate constant and rate equation. So a reaction will only be fully studied when both aspects are studied.

But what is the scientific causes(or probable causes) for the disparity between what the two say?

That's what I am saying. The motive behind their developement itself separates them. For knowing rate microscopic study is required. While for handling heat macroscopic study is required. And hence they are not directly related.

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What makes a glowing splinter glow and a burning splinter burn
This question bugs me for years since I started learning the oxygen, carbon dioxide and hydrogen tests. I know you can make a glowing splinter burn when doing the oxygen test due to enough oxygen. But why won't it burn in ordinary air?
So, my question is what factor differentiate a glowing splinter and a burning splinter

I'm guessing you are defining a burning splinter as having visible smoke, while a glowing splinter does not. Is this correct? If so then both splinters are burning.  The burning in the glowing splinter is complete.

Ordinary air is made up of only approximately $21~\%~\ce{O2}$, the primary component being nitrogen. According to the partial pressure rule, at standard laboratory temperature and pressure, this is a partial pressure $p(\ce{O2}) = 21~\mathrm{kPa}$. In a test tube filled with pure oxygen, e.g. from water electrolysis, the partial pressure of oxygen is $p(\ce{O2}) = 100~\mathrm{kPa}$ — five times as much.
According to the rate law, the combustion of wood ($\ce{C_{$n$}H_{$m$}O_{$l$}}$) with oxygen will be dependent of oxygen’s partial pressure somewhat. Thus, we can assume at least a five-fold increase in reaction rate. A five-fold rate increase also means a five-fold increase of enthalpy per time-frame which means five times the heat is liberated which means that the glowing splinter will get hotter much more quickly and finally ignite.
Technically of course, glowing is just a weaker form of burning and oftentimes only requires more oxygen to reignite. Hence why sometimes a fire can be reignited by blowing onto the embers.

A splinter can undergo multiple types of thermal reactions depending on both the maximum temperature and amount of oxygen present.  The answer to the question seems to be in the difference between total and partial inceneration.
In total inceneration, which is very likely in the total oxygen atmosphere at high temperatures, all of the organic carbon is converted to carbon dioxide. Carbon dioxide is colorless so the emission of electrons, i.e. the glowing, is observable.
In partial inceneration, which occurs when oxygen is present but deficient, the organic carbon is converted into both carbon dioxide and a variety of volatile compounds observable as smoke. Smoke forms at lower temperatures than thermal emissions. Smoke can also obscure thermal emissions.

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What is the formula of manganese oxohydroxide: MnOOH and MnO(OH)2?
I found various formulae of manganese oxohydroxide. Some site says it is $\ce{MnOOH}$, other say it is, $\ce{MnO(OH)2}$. So, which one is correct?

The mineral manganite is considered manganese oxide-hydroxide $\ce{MnOOH}$, but yahoo answer claims that it is $\ce{MnO(OH)2}$.
In Winkler titration, $\ce{MnO(OH)2}$ is formed:
$$\ce{2 MnSO4(s) + O2(aq) → 2 MnO(OH)2(s)}$$

There is some uncertainty about whether the oxidised manganese is tetravalent or trivalent. Some sources claim that $\ce{Mn(OH)3}$ is the brown precipitate, but hydrated $\ce{MnO2}$ may also give the brown colour. So, is $\ce{MnO(OH)2}$ hydrated $\ce{MnO2}$?
So which one is manganese oxohydroxide- $\ce{MnOOH}$ or $\ce{MnO(OH)2}$?

The difference is that one contains $\ce{Mn^{3+}}$ and the other $\ce{Mn^{4+}}$, and they coexist together perfectly fine.

@IvanNeretin Are you sure they contains Manganese ions? Simon's answer says that they are bonded covalently......

That's not important. All ionic compounds are in fact partly ionic and partly covalent. You may interpret those +3 and +4 as just the oxidation states.

First, the difference the substances is the oxidation number of manganese. In $\ce{MnOOH}$, the oxidation number of manganese is 3+. While in $\ce{MnO(OH)_2}$, the oxidation number for manganese is 4+.
However this doesn't mean $\ce{MnOOH}$ is manganese (III) oxyhydroxide and $\ce{MnO(OH)_2}$ is manganese (IV) oxyhydroxide, not for $\ce{MnO(OH)_2}$ at least. If you move the element around, you will get $\ce{MnO(OH)_2}$ is actually $\ce{H_2MnO_3}$. $\ce{H_2MnO_3}$ is dihydroxy(oxo)manganese which is an acidic substance.
For the question regarding the structure for both substances, I'm sorry that I don't haven't the image but I can describe it (Hope that would do). For $\ce{MnOOH}$, manganese is single bonded with $\ce{OH}$ and manganese is double bonded with oxygen. For $\ce{MnO(OH)_2}$, manganese have two single bonds with both of the $\ce{OH}$ and manganese is double bonded with oxygen.
In conclusion, $\ce{MnOOH}$ is manganese (III) oxyhydroxide and $\ce{MnO(OH)_2}$ is dihydroxy(oxo)manganese. In structure wise, manganese is single bonded with $\ce{OH}$ and manganese is double bonded with oxygen for $\ce{MnOOH}$. For $\ce{MnO(OH)_2}$, it likes $\ce{MnOOH}$ but have one more single bond with $\ce{OH}$.

I understood your explanation but a picture would have been better. Actually, I need a visualization.

I will find it for you or draw it

Are the compounds covalently bonded? Ivan (see comment) says it contains manganese ions..... Also in this site , the empirical formula mentioned is Mn3+OOH

What I assume is that is the manganese oxidation state instead of ion. But I will hope there is more people more expert in this topic may help you.

dihydroxy(oxo)manganese or manganous acid is actually a hypothetical compound.

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Why are chelate complexes more stable than other complexes with unidentate ligands?
I read that complex compounds with chelating ligands are thermodynamically more stable than complexes without them. Why would that be so?

also http://chemistry.stackexchange.com/questions/31005/whats-so-special-about-chelation


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Color of Anhydrous Transition Metal Salt
Why are some anhydrous transition metal salts colored?
Here are some examples I found.

Anhydrous cobalt(II) chloride ($\ce{CoCl2}$), with sky-blue color.
Anhydrous chromium(III) chloride ($\ce{CrCl3}$), with purple color.
Anhydrous copper(II) chloride ($\ce{CuCl2}$), with yellow-brown color.

I already know that complex ion could undergo d-d transition from its unfilled d orbitals, and that's what makes them pretty-colored. (The wavelength of absorbed light corresponds to the gap in energies between d orbitals.)
But in anhydrous salt, which I think isn't a coordination compound, the d orbitals are degenerate. Thus, no energy gap and excitation could occur, no light is absorbed, and the compound would be colorless / white-colored as it reflects all wavelength.
I summarize my understanding in this problem as below.

Solution of complex ion could give color (with exception of $\ce{[Zn(H2O)6]^2+}$ and some other complex ions). e.g. $\ce{CuSO4 (aq)}$, which form $\ce{[Cu(H2O)6]^2+}$ complex ion with blue color.
Hydrated salt is indeed a coordination compound (found related question here), so it could give color. e.g. $\ce{CuSO4.5H2O (s)}$ with blue color.
Anhydrous salt, because it doesn't have any ligands coordinating with the metal ion, it's not a coordination compound. It's a mere ionic salt instead. Shouldn't it be colorless?

Is there any explanation to this case?
Note that in all compounds mentioned above, the metal ions are transition metal. We don't consider s-block metal ions such as $\ce{NaCl}$ and $\ce{MgSO4}$. (Additional question: They should be colorless, am I right?)

They are coordination compounds.

The real question is why copper(II) sulphate is colourless in its water-free form.

@Mithoron I thought ionic salt isn't called coordination compound by definition?

Most so called salts are more covalent then ionic

A ‘mere ionic salt’, when inspected properly, turns out to be nothing different from a coordination compound. One side (the cation) is electron deficient while the other side (the anion) has spare electrons, so there will be some kind of coordination happening from one to the other. Indeed, all salts adopt crystal structures so that the favourable coordination interactions are maximised.
For anhydrous cobalt(II) chloride, I would immediately expect a closest packing with cobalt ions sitting in tetrahedral voids, occuping ¼ of them to give $\ce{[CoCl4]}$ tetrahedrons where each chloride coordinates two neighbouring cobalt ions. Analogously for chromium(III) chloride, I would expect chromium to occupy one-third of the anion structure’s octahedral voids, giving rise to $\ce{[CrCl6]}$ octahedrons where again each chloride coordinates two chromiums. Both are thus clearly coordination compounds. And going by what I know about $\ce{Cu^{II}}$, that salt will probably be a set of square-planar $\ce{[CuCl4]}$ fragments, again where each chloride coordinates two coppers.
Note that all of these are rough guesses (I wouldn’t even call them educated) since you can only predict solid-state (ionic) structures so far a priori. Oftentimes you can arrive at multiple possibilities and even more times you are surprised at what is actually adopted but then retrospectively you can usually explain the structure somehow with coordination.

Well, thanks for answering! I quite grasp what you're telling. Since they're kind of 'coordination' compounds, I'd suspect them to have some kind of excitation involving electrons from either the 'ligand' (anion) or the metal ion, and thus, their colors.

Crystallography indicates that anhydrous cobalt chloride has cobalt in an octahedral environment with chloride anions as ligands. As the chloride is a very weak liagnd field ligand this accounts for the fact that cobalt chloride is blue and not red like hydrated cobalt chloride.

@NuclearChemist But $[CoCl_6]^{4-}$ is faintly pink.?


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Corey-Schmidt oxidation of primary alcohols
Why does pyridinium dichromate (PDC) in dimethylformamide (DMF) oxidize alcohols to acids while in methylene chloride the oxidation stops at the aldehyde level?

As stated in the previously posted answer:

This oxidation into
  carboxylic acids succeeds when the intermediate aldehyde equilibrates with a liberal proportion of hydrate; that is, when the intermediate aldehyde belongs to the aliphatic kind and is not highly hindered. The water necessary for the formation of the intermediate aldehyde hydrate may proceed from the decomposition of PDC.$^{[1]}$

Though the text does state that water may form from decomposition of the dichromate anion, since DMF is so hygroscopic as a result of its high polarity ($\epsilon_\mathrm{r}=38^{[2]}$), I would not be surprised to learn that little to no over-oxidation occurs in completely dry DMF under an inert atmosphere.
This is only speculation though, and the dichromate anion may very well be decomposing to produce water in the solution. As such, I will propose a reaction mechanism for this decomposition. The first step can only occur when pyridinium dichromate is well solvated and protons are free to be transferred between the species, which occurs in DMF but not in DCM:
$\hspace{.6cm}$
This produces chromic acid, another chromium(VI) oxidizing agent. Unlike dichromate oxidation, however, chromic acid oxidation actually produces water as a byproduct, which is then free to hydrate the resulting aldehydes:
$\hspace{.8cm}$

$[1]$ Tojo, G.; Fernández Marcos. Oxidation of Alcohols to Aldehydes and Ketones: A Guide to Current Common Practice; Springer: New York, NY, 2006; p 28.
$[2]$ Wikipedia: Solvent

I attempt to provide an reasonable answer to this question.
The Corey-Schmidt oxidation uses PDC (pyridinium dichromate) to primarily oxidise alcohols to carbonyl groups. PDC exists in the form of stable bright-orange crystals that remain unaltered by manipulation in the open air. Its lack of hydrophylicity facilitate its handling and allows the selective oxidation of alcohols in the presence of very sensitive functional groups.

The medium generally used is $\ce{CH2Cl2}$, ethyl acetate, benzene, methyl cyanide or chloroform. PDC is far more reactive in more polar solvents like water.
Now, coming to your question the interesting thing is that DMF, which is very efficient in dissolving PDC,causes the over-oxidation
of certain alcohols into acids, something that may happen even
in the absence of added water.

This oxidation into carboxylic acids succeeds when the intermediate aldehyde equilibrates with a sizeable proportion of hydrate (when the intermediate aldehyde belongs to the aliphatic kind and is not highly hindered). The water necessary for the formation of the intermediate aldehyde hydrate may proceed from the decomposition of PDC.

So given that, it may be concluded that DMF causes PDC to over-oxidise due to following reasons (possible):

Decomposition of PDC in DMF to form water molecules, so that hydrate formation of intermediate aldehyde occurs and then conversion into carboxylic acid.
This may be aided by the solvation (solubilising) of PDC in DMF.

Reference:
G. Tojo, M. Fernández Oxidation of Alcohols to Aldehydes and Ketones:
A Guide to Current Common Practice; Springer Science+Business Media: New York, 2006.

I would love to see a mechanism of hydrate formation in the absence of water — especially if you postulate it occuring via the dichromate anion.

With the mechanism I could accept this.

A reading of the Corey-Schmidt1 paper on the oxidation of non-conjugated primary alcohols and aldehydes to the corresponding carboxylic acids with dipyridinium dichromate (PDC) in dimethylformamide (DMF) makes no explicit mention of drying DMF but flame-dried glassware is employed. Perhaps this implies that the solvent was dried. @Ringo has speculated (Ringo's admission) how water may be generated from PDC in situ to then form the aldehyde hydrate necessary for oxidation. Given the stoichiometry of the reaction (3.5 equiv., PDC/RCH2OH), sufficient but limited water can be produced in this manner. Nonetheless, this pathway is viable. An equally speculative method of "hydration" invokes DMF as an alternative to water. continued

A species akin to hemiacetal 3 (it and other intermediates may be protonated) may add to chromium trioxide 4, or PDC itself, which forms 5 and collapses to a Cr[IV]-species 6 and 7, which is nothing more than a reactive, masked anhydride. The Corey-Schmidt paper experimental section calls for the addition of the reaction mixture to water.
1) E. J. Corey and G. Schmidt, Tetrahedron Letters, 1979, 5, 399.

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Charge of an amino acid sequence
I have the following problem:

A polypeptide has the following sequence (amino end on the left)
  $$\mathbf{lysine - aspartic~acid-histidine-arginine-serine}$$
Using the $pK_a$ information that follows, what would be the approximate charge on this molecule at pH = 3.3?
$pK_a$ values for the various groups in the molecule are as follows:
\begin{array}{|cc|}\hline
\rm Alpha~carboxyl & 2.2 \\ \rm Alpha~amino & 9.5 \\ \rm R-group~amino & 11.7 \\ \rm R-group~carboxyl & 4.5 \\\hline\end{array}

After my calculations I got +2 and my answer was wrong. How to solve it correctly?
My effort:
$\ce{NH3}$ has $+1$ when on and $0$ when off. $\ce{NH3}$ are Lysine, Histadine, and arginine.
$\ce{COOH}$ has $0$ when on and $-1$ when off. $\ce{COOH}$ are Serine and aspartic acid.
Based on the pH and $pK_a$ values the answer should be 2.
Lysine=+1 Aps=0 His=+1 Arg=+1 Serine=-1

This is 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. Start by further explaining what you did. Just telling us that you got a 2 as an answer isn't helping us help you.

updated the post

I find your current wording a bit confusing. What do you mean by "ammonia are lysine, histadine, and arginine". Do you mean they are the compounds with an amine group?

Serine isn't a carboxylic acid.

When the pH is lower than the pKa, the group will be protonated, and when the pH is higher than the pKa, the group will be deprotonated. Your explanation is correct that basic groups will be positively charged when protonated and neutral when deprotonated and that acidic groups will be neutral when protonated and negatively charged when deprotonated. Shown below is the compound in it's fully protonated state with the given pKa's.*

The given pH is 3.3 which means that:

N-terminus: pH < pKa, protonated basic group, +1
lys-sidechain: pH < pKa, protonated basic group, +1
asp-sidechain: pH < pKa, protonated acidic group, 0
his-sidechain: pH < pKa, protonated basic group, +1
arg-sidechain: pH < pKa, protonated basic group, +1
C-terminus: pH > pKa, deprotonated acidic group, -1

The overall charge is +3.
*An aside, the pKa's for the side chains are a gross approximation here. It turns out they are sufficient to answer the question, but given a higher pH, the answer wouldn't be correct.


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Kinetic Barrier of ATP
I do a fair bit of biochemistry, and one of the things I've taken on faith up until now is that ATP is stable because it has a high kinetic barrier for falling apart despite the spontaneity of its decomposition. I understand that it does, but what about the transition state of ATP hydrolysis in an aqueous solution is so high energy? For that matter, what is the transition state of spontaneous ATP hydrolysis?

This is not exactly an answer making strong use of references etc. but it does give a good hint. The most ‘energetic’ bonds in ATP are the phosphoric acid anhydride ones i.e. the $\ce{P-O-P}$, which, to hydrolyse ATP, must be attacked by water. The most logical path of attack is one according to the $\mathrm{S_N2}$ mechanism, where a lone pair of water attacks the rear side of a phosphorus-oxygen bond and liberates some phosphate species. ($\mathrm{S_N1}$ can be ruled out, since it would result in a phosphocation which is surrounded by three electron-withdrawing oxygens making the cation highly unstable.)
The problem in this mechanism is physiological pH. It is safe to say that the entire ATP molecule is at least a dianion, if not even a tri- or tetraanion. Now water’s lone pair, i.e. its partially most negatively charged area must approach this already highly negative molecule — but that would result in the repulsion of same charges leading to a high activation barrier.
This would be circumvented by protonation, and indeed even at equilibrium there is a certain probability that a higher number of protons attaches to ATP, making it less negative. This probably helps to account for the slow but steady decay of ATP.

facepalms Should have realized this. XP Thank you!


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Hydrogen Bonding (with oxygen) - restricted to oxygen singly bonded to H or not?
I understand the following definition of a hydrogen bond from Merriam Webster:

an electrostatic attraction between a hydrogen atom in one polar
  molecule (as of water) and a small electronegative atom (as of oxygen,
  nitrogen, or fluorine) in usually another molecule of the same or a
  different polar substance

However, I wasn't sure if this applied to F, N, or O atoms that were not covalently bonded to H in the first place. A question regarding hydrogen-bonding came up on an exam for which the marks scheme read that both element X (with an O-H bond) and element Y (with an O in an ether group). However my teacher doesn't not agree with the marks scheme and, "[believes that the marks-scheme is] incorrect because Y doesn't contain O-H bonds so the O will not carry a large enough δ- to create hydrogen bonds with water."
Is she correct, or is the marks scheme?

Note: I thought the same as my teacher up until I saw the MS.

The answer sheet is correct. Your teacher is mistaken (polite way to say flat out wrong).

Indeed hydrogen bonds are not just restricted to fluorine, oxygen and nitrogen either. They may be found to other elements in some cases. Additionally, they are not just electrostatic interactions - they contain a significant covalent component.

Hydrogen-bonds require a $\delta^+$ hydrogen atom attached to an electronegative element, i.e. O, F, N, or Cl, and also an "active" lone pair on another electronegative atom.  An electrostatic attraction between these two is an extremely strong kind of intermolecular dipole-dipole interaction.
Compounds which have O-H or N-H groups are able to both form hydrogen-bonds through the $\ce \delta^+$H atom and accept hydrogen-bonds through the lone pair on the oxygen or nitrogen.  These compounds tend to have higher then expected melting and boiling points due to the hydrogen-bonding.  Compounds are even able to form internal hydrogen-bonds with other O or N atoms in the structure.
However hydrogen bonding can occur between substances that cannot form the bonds with their own molecules.  Because compound Y is an ether and can only accept hydrogen-bonds there are fewer bonds made overall between the mixture of substances, so the effect on the physical properties isn't so great.  Nonetheless water can form hydrogen-bonds through its proton and functional groups such as ethers, esters, amines, amides, alcohols and carboxylic acids can all accept those bonds through the relevant lone pairs.
Another example would be a tertiary amine such as $\ce{N(CH3)3}$, which is miscible with water because of the hydrogen bonds it accepts from the water molecules even though it has no N-H bonds itself.  Even $\ce{N(C2H5)3}$ is miscible with water at room temperature, and other tertiary amines continue to be soluble in water for the same reason: they are hydrogen-bond acceptors.
So the mark scheme is correct, and the teacher has shown a poor understanding of intermolecular interactions.


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Where do we report yields in a research report?
I'm in an integrated laboratory course at university that combines all the disciplines of chemistry that we have here. I was instructed to place the yield in the methods section of the report by my professor, but found that to be odd.
Is reporting the yield in the methods section a standard protocol in some paper formats?

Your professor is right! Giving the yield in the method section, that is, together with detailed information on synthesis, workup and characterization of the products obtained, is standard in scientific journal I know or I have published in.


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Does Friedel-Crafts reaction fail with benzoic acid?
I know that Friedel-Crafts reaction  (either alkylation or acylation) fails with phenol and aniline, due to formation of a coordinate bond between Lewis acid catalyst and the lone pair of the respective functional group.
But I have read that even benzoic acid fails to undergo this reaction. I couldn't think of a reason for this. Can anyone explain why is it so?

Still the same reason.

I personally performed Friedle-Crafts acilation on 4-hydroxy-[2,2]-paraciclphane  (a kind of phenol) using TiCl4 and AcCl (or TiCl4 and (COCl)2). TiCl4, however, is required in slight excess to equimole amount in this reaction.

@Mithoron I did fell like that but is benzoic acid 'basic' enough to get coordinated?

Is quite nucleophilic

@Mithoron Through which 'O' atom does it get coordinated to the catalyst.

Hi, does phenol really behaves as aniline when treated with Lewis acid (Friedel-Crafts reactions), and when treated with Bronsted acid (e.g. during nitration and sulphonation)? Many websites just say only aniline acts as Lewis & Bronsted base

It could be that $\ce{-COOH}$ group is electron withdrawing hence there is an increased electron density in $\ce{-COOH}$ group due to -R resonating effect of electron from benzene ring.
Since friedel craft catalyst is a Lewis acid (electron deficient) the catalyst (anhydrous $\ce{AlCl3}$) reacts with electron rich COOH group of benzoic acid and gets bonded.

Hi, abhishek. Did you reach a conclusion in your answer? You have stated two possibilities - a deactivated ring due to $-R$, and a bond formed by $\ce{-COOH}$ with the Lewis acid - for FC reaction to not happen. Which one of these is your answer exactly? Thanks!

I think Friedel Craft reaction with Benzoic acid fails because the Benzene ring is deactivated. Here electrophile is Carbocation  which is a weak electrophile. So the necessity for Friedel-Craft reaction is activated Benzene ring.

I don't get it when you say "C+ is a weak electrophile". A carbocation? I don't think that's a weak electrophile. Anyway, do you mean that all mono-substituted benzene rings with deactivating groups (i.e., atleast as strongly deactivated as benzoic acid) fail to give Friedel- Crafts reactions.

I think yes. And also I am doubtful whether carbocation is a strong electrophile.

@AadityaJoshi Carbocations are pretty good electrophiles. The more important factor here is the coordination of the lewis acid to the carboxylic acid group.

@bon Yes, now I think it is so but through which of the two oxygen atoms does benzoic acid coordinate to lewis acid?

@user104014 Probably the carbonyl oxygen but I'm not entirely sure.


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Acidity of hydrogen chloride in acetic acid
A question asked to find the $\mathrm pK_\mathrm a$ of $\ce{HCl}$ in acetic acid. The equation of $\ce{HCl}$ disassociating in water is: $$\ce{HCl + CH3COOH \leftrightharpoons CH3COOH2+ + Cl-}$$ I predicted that the $\mathrm pK_\mathrm a$ should be higher  as acetic acid is a weaker base than water so it would be less willing to accept a proton from a $\ce{HCl}$.
However when I finally calculated the $\mathrm pK_\mathrm a$, I got 8.56. This was much higher than I expected. Also doesn't this imply that $\ce{Cl-}$ is stronger as a base than $\ce{HCl}$ is as an acid in acetic acid, which doesn't sound right.
So I am asking if this value totally wrong or is it actually quite reasonable?

Properly, "Acidity of hydrogen chloride in acetic acid". "Hydrochloric acid" is defined as a solution of hydrogen chloride in water.

So basically I am asking, is this value totally wrong or is it actually quite reasonable?

It's quite reasonable. Compare to page 32 of this reference.

This was much higher than I expected. Also doesn't this imply that
  Cl− is stronger as a base than HCl is as an acid in acetic acid, which doesn't sound right.

To answer whether Cl- is a stronger base than HCl is an acid, you need to know the self-dissociation constant of acetic acid, the equivalent of Kw for water. It is about $10^{-10}$. So whereas the middle of the pH scale is thought of as 7, the middle for acetic acid would be 5. So, yes, Cl- is a stronger base than HCl is an acid in acetic acid.

thanks for you answer and for the reference. Also, what I meant to ask is 'is $\ce{Cl-}$ a stronger base than $\ce{HCl}$ as an acid' if you get what I mean. I will edit my question. Sorry for my badly worded question.

@Nanoputian To answer whether Cl- is a stronger base than HCL is an acid, you need to know the self-dissociation constant of acetic acid, the equivalent of Kw for water.  It is about 10^(-10).  So whereas the middle of the pH scale is thought of as 7, the middle for acetic acid would be 5.  So, yes, Cl- is a stronger base than HCl is an acid in acetic acid.

Glacial acetic acid is a powerful solvent, but not as polar as water. Whereas HCl has a pKa of -6.3 in water, its pKa in acetic acid will be less acidic (greater numerically).
It helps to think of the three hydrohalic acids as covalent molecules (especially HCl, since that is what we are discussing). They are ionic only when hydrogen bonding provides a lower energy state for solvated ions, especially H+, but also the anion. Cl- is poorly solvated by acetic acid compared to Br- and I-, and the protonation of acetic acid is not especially energetic.
Acetic acid is called a differentiating solvent for the hydrohalic acids: HI is strongest, HBr less so and HCl is least strong.


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Energy levels transitions of an electron?
I have an example that tells me that the energy transition between $n=1$ and $n=3$ is $12.1\ \mathrm{eV}$ using the following formula:
$$E= -RH (\frac{1}{m^2} - \frac{1}{n^2})$$
and that the conversion from $12.1\ \mathrm{eV}$ gives a wavelength of $102.6~\mathrm{nm}$.
I understand that $-RH$ is the Rydberg's constant that corresponds to $13.60\ \mathrm{eV}$ and $m$ is the inner orbit and $n$ the outer orbit (but I might be wrong with the latter). So,
$$E= -13.60\ \mathrm{eV}\ (\frac{1}{1^2} - \frac{1}{3^2})$$
$$E= -13.60\ \mathrm{eV} \cdot 0.83$$
$$E= -11.288$$
I don't understand how they get the first value for the electron Volts and then convert these into wavelength. Additionally, my task is to find the energy level for the 4th orbit in the Lyman's series.

If I remember correctly, you can use the equation:
$$E = \frac{hc}{\lambda}$$
and
$$\nu = \frac{c}{\lambda}$$
to solve for the frequency, wavelength, or energy of the photon. Note that $E$ is energy, $h$ is Planck's constant, $c$ is the speed of light, $\lambda$ is the wavelength, and $\nu$ is frequency.
The energy, (actually 11.97eV not 11.2eV)converted into joules (2.719 $*10^{-18}$ J) and then put into the first equation:
1.918 $×10^{-18}$ J = $\frac{6.626× 10^{-34} × 2.998 × 10^7 m×J}{\lambda}$
then lambda can be solved for:
$\lambda$ = $\frac{6.626× 10^{-34} × 2.998 × 10^7 m×J}{1.918×10^{-18} J}$
$\lambda$ = $1.036×10^{-7} m $
= 103.6 nanometers.
This solves for your wavelength like you asked for.


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Which temperature to subtract from which when calculating a difference?
Is $\Delta T$ solved by using: $T(\mathrm{final})-T(\mathrm{initial})$?  Or $T(\mathrm{initial})-T(\mathrm{final})$?
I am getting some conflicting info from class and online.

$\Delta x$ is always defined as $x(\mathrm{final}) - x(\mathrm{initial})$.

The delta symbol ($\Delta$) means change, so $\Delta T$ means the change in temperature.  To find the change in temperature the correct formula is $T(\mathrm{final})−T(\mathrm{initial})$.
For example, if the temperature of a sample changes from $70~\mathrm{^\circ C}$ to $80~\mathrm{^\circ C}$ then $\Delta T = 80~\mathrm{^\circ C} - 70~\mathrm{^\circ C} = 10~\mathrm{^\circ C}$. The positive number indicates the change was an increase in temperature.
An example of a decrease in temperature would be if a sample changes from $80~\mathrm{^\circ C}$ (our initial temperature) to $60~\mathrm{^\circ C}$ (our final temperature) then $\Delta T = 60~\mathrm{^\circ C} - 80~\mathrm{^\circ C} = -20~\mathrm{^\circ C}$. The negative number indicates the change was a decrease in temperature.


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What is the lowest mass measurable using a 4 decimal point analytical balance?
I think it is 1 mg, but I've heard people saying 0.1 mg. As 0.1 mg = 0.0001 g, which is the uncertain digit, I think such a measurement is not reliable at all.
Any opinions ? Thanks.

An odd point is that for significant figures one would assume that the true measurement of such a balance would be ± 0.00005 g.

Keep in mind that for many analytical instruments, you cannot assume that the uncertainty is ± the last digit.
It's usually necessary to look at the specifications in the manual or on a sticker to find the measured uncertainty.
For instance, one of our balances, a Sartorius CPA225D (manual), is readable to 0.01 mg (i.e. the display can display increments of 0.01 mg), but the specifications list the repeatability as only ±0.05 mg (±1 s.d.).
This means that repeated measurements of the same mass will be (ideally) normally distributed with a standard deviation of 0.05 mg and that even though the instrument has the resolution to report in 0.01 mg increments, the uncertainty in a measurement is higher.

This precisely!

I think that by 4-digit analytical balance; you mean 4 decimal places. in which case the error would be ±0.0001 g.
If you were correct in that it was actually a 4-digit balance then the uncertainty would be ±0.001 g. However, this would mean that the balance only weighs up to 9.999g and balances usually allow more than that by my understanding. Hence why I think it would be 4 decimcal places and correct to ±0.0001 g.

EDIT

The scales I use have a repeatability of ±0.0001 g. My understanding is that if I have 5.0006 ±0.0001 g then there is doubt about the last digit. For every reading we take there will always be a measure of doubt. In this case it is important to highlight how big that doubt is, so it cannot simply be ignored. we have an uncertainty of ±0.0001 g meaning the true value could read 5.0005-5.0007 g. If you thought the uncertainty was 0.001g of the balance reading then our uncertainty would become 5.000-5.001 g. But the balance is certain that it doesn't weigh 5.000 g.

Sorry about that. I meant 4 decimal points. I've corrected the question.

i'll edit my answer about the reliability of the measurement

@jan I have seen analytical balances with 7 decimal places in the pharmaceutical industry

@jan ah I think I understand you. are we agreeing that the last value is the uncertainty?

@LiamH If the balance readout contains 4 decimal points, doesn't that mean the 4th decimal to be the uncertain digit ?

@LiamH Please explain "if I have 5.0006 ±0.0001 g then the last digit has some reliability because I know i don't have 5.007 g".


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pH, conc vs activity
After studying a somewhat overview of "gamma" activity coefficient (it is not actually in our course but may come in exam as a part of unknown paragraph in ionic equilibrium) I can understand most things , but can somebody please just explain in layman's terms (so i can understand it logically, not experimentally) how ionic strength can increase the coefficient and how is distance of approach (lost here) comes into play.
{I am familiar with laplace equations but by only using fourier tricks etc, so, i cant solve any general question with my knowledge}

Are there any more very well known effects only to be understood by activity like
ex- When a 0.1 M hydrochloric acid solution is added to 5 M solution of magnesium chloride, pH decreases — when in fact the acid has been diluted.


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Naming amitriptyline by IUPAC
Can someone guide me through step by step how amitriptyline is named by IUPAC?
I know the name per se, but it is very confusing to me, and I'd like someone to explain it to me. The systematic name is

3-(10,11-Dihydro-5H-dibenzo[a,d]cycloheptene-5-ylidene)-N,N-dimethylpropan-1-amine.

Please note that the fusion nomenclature of the name “3-(10,11-dihydro-5H-dibenzo[a,d]cycloheptene-5-ylidene)-N,N-dimethylpropan-1-amine” corresponds to obsolete IUPAC recommendations from 1979 and is no longer encouraged since 1993. According to the current IUPAC recommendations from 2013, the preferred name is “3-(10,11-dihydro-5H-dibenzo[a,d][7]annulen-5-ylidene)-N,N-dimethylpropan-1-amine”.

@Loong [7]annulen sounds a lot better than cycloheptene, because it doesn’t make me think incorrectly that there be only one double bond in cycloheptene.

Note: As stated in a comment to the question by Loong, the preferred IUPAC name according to the 2013 nomenclature recommendations is:

3-(10,11-dihydro-5H-dibenzo[a,d][7]annulen-5-ylidene)-N,N-dimethylpropan-1-amine

The only change this makes to the analysis it that we can ignore the ‘incorrect’ cycloheptene, because [7]annulen is automatically a cycloheptatriene-type molecule.

First, let’s separate the name into its fragments. We have:

3-something-propane-1-amine
N,N-dimethyl
(10,11-dihydro-something)
(something-5H-dibenzo[a,d]cycloheptene-something)
(something-5-ylidene)

I’m going to start with item 4, since that is the main bicycle. Cycloheptene sounds like this molecule:

But the [a,d] and the dibenzo is telling us that we have a fused polycyclic unsaturated hydrocarbon, so we will need to add two more double bonds. (Skipping a picture of this step.) Dibenzo means that there are two fused benzene rings, and the letters in square brackets symbolise the attachment points. a stands for 1,2; d for 4,5 (analogously for other letters). With that, we get a tricyclic compound with many double bonds. However, it is still ambiguous how exactly to place the double bonds, we need the information 5H — it tells us that the ‘additional’ hydrogen must be at the 5-position. Beware: The 5-position does not reference the initial numbering we used for fusing the rings; the atoms are renumbered according to the rules for aromatic polycycles: Starting at the top right and going clockwise, ignoring all bridgehead atoms. So we get 5H-dibenzo[a,d]cycloheptene which looks like this:

Next up, item 3: 10,11-dihydro. This means that the double bond in the seven-membered ring is hydrated, i.e. there is a single bond instead.

And finally, item 5: 5-ylidene. We are attaching something to carbon 5 via a double bond. We’ll get back to that.
Item 1: 3-something-propane-1-amine. That’s nice and simple, we’ll first consider propane-1-amine:

N,N-dimethyl (item 2) tells us that there are two methyl groups attached to the nitrogen, giving us the following:

And now we need to add the entire tricyclo-blurb to the propane’s 3-position using a double bond:

And voilà, there is amitriptylene!

Wonderful! Very detailed and accurate, and quick response. I'm very thankful :)


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How can collission theory of chemical reaction and Law of mass action be connected?
From Wikipedia,

Collision theory is a theory ... that qualitatively explains how chemical reactions occur and why reaction rates differ for different reactions. The collision theory states that when suitable particles of the reactant hit each other, only a certain percentage of the collisions cause any noticeable or significant chemical change; these successful changes are called successful collisions. The successful collisions must have enough energy, also known as activation energy, at the moment of impact to break the preexisting bonds and form all new bonds.

From Britannnica,

Law of mass action, law stating that the rate of any chemical reaction is proportional to the product of the masses of the reacting substances, with each mass raised to a power equal to the coefficient that occurs in the chemical equation.

How can both explain and imply each other?
My theory: Due to more availability of reacting species (due to more amount), rate of reaction of increases. But it does not explain the reason of exponentiation of stoichiometric coefficients.

For exponentiation, we can say that it is actually repeated multiplication. Comparing two cases:

$$\ce{A + B -> AB}$$
Here, the rate expression is $r=K[\ce{A}][\ce{B}]$. This is because one molecule of $\ce{A}$ and one molecule of $\ce{B}$ is needed for the product to form.
$$\ce{2A + B -> A2B}$$
First, write it in this way: $$\ce{A + A + B -> A2B}$$
Now, rate expression is $r=K\ce{[A][A][B]}$.
That's it - $r=K[\ce{A}]^2[\ce{B}]$

An intuitive starting point for me to understand why concentrations are multiplied is probably to connect it with multiplication theorem in probability (for independent events).


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The naming of benzene
Why is benzene not called as a cyclo triene? Wikipedia says benzene is a IUPAC name, but isn't it a common name?

Benzene is not a triene, the bonds are delocalized.

@ron I think that entirely exceeds the scope of the question. (Also note that the cyclopentadienyl anion is generally called that, although the bonds and charge are also delocalised.)

@Jan Well, just my opinion, nothing more. Note that derivatives of 1,3,5-cyclohexatriene have been prepared; their properties are noticeably different from those of benzene.

@ron I would love to see relevant papers, could you poke me towards them?

@Jan  See Peter Vollhardt's work on the synthesis and characterization of 1,3,5-triscyclopropabenzene and 1,3,5-triscyclobutabenzene.

how about phenyl hydride?

@LiamH Phenyl is a neutral entity, hydride is the hydrogen anion, making phenyl hydride $\ce{[C6H6]-}$. Hydrogen phenyl might do the trick though.

Cyclohexa-1,3,5-triene does not exist.
It is just a hypothetically possible structure for the formula $\ce{C6H6}$.
The whole discussion is intended to explain that if it were to exist it would have such and such properties but it does not exist.
Instead we have benzene.

Perhaps you could give a brief explanation of why it doesn't exist. At the moment this seems a bit unsatisfactory as an answer.

Not to say that this answer is probably a mile away from what the question is actually asking.

[citation needed] Ron’s evidence seem to show that derivates of cyclohexatriene exist.

It would be wrong to say cyclotriene, rather cyclohexatriene would be an appropriate name. And it would also be a highly systematic IUPAC name for said compound, following the rules of e.g. the cyclopentadienyl anion, the cycloheptatrienyl cation, cyclooctatetrene and more.‡
Let’s look at benzene’s history. It was discovered as early as 1825 by Michael Faraday who called it bicarburet of hydrogen. In 1833, Eilhard Mitscherlich isolated it and named it benzin. In 1836 it was isolated by Auguste Laurent who called it phène. The name benzene also dates back to the first half of the 19th century and, like benzin, derives from the gum benzoin. Note that this already predates the founding of IUPAC by about 80 years. The first IUPAC nomenclature was not established until 1921.
At the time when it was discovered, chemists had no clue as to what structure it had. It wasn’t until 1865 when Friedrich Kekulé published his ground-breaking paper[1] on the structure of benzene which in the end turned out to be correct (-ish). At that time, nomenclature of chemicals was still the Wild West, since there was hardly any way for structure elucidation and thus most chemicals acquired trivial names — some of which survive up to date (acetic acid being a prime example).
When IUPAC finally derived comprehensive nomenclatures for organic (and inorganic) compounds, and when structure elucidation was finally good enough to prove structures, the names benzene, benzin and possibly also phène had been in use for many decades, much like many other trivial names. Had IUPAC decided to obsolete them all altogether, they could have just as well trashed their own recommendations, because nobody would have followed them — even today, many chemists are too traditional to take up newer nomenclature schemes even if they have been around for two decades.
Probably to help gain acceptance, IUPAC has always allowed a certain set of retained names to persist under official nomenclature. While the list has gotten smaller over the years, it is extremely unlikely that benzene, phenol, phenyl etc. will ever get removed from it, since calling those compounds that way is so prevalent — and short.
Similar things happened to carboxylic acid. Hardly anybody says methanoic acid or butanoic diacid in place of formic acid and succinic acid. And sometimes, trivial names even become part of the official nomenclature, as was the case with fullerene because it provided a much shorter way to label a certain class of compounds.
So the short answer to your question: IUPAC has retained a common name. The reasons are tradition, prevalence and simplicity.

Note:
‡: Yes, as a chemist I know that the structure of benzene is incorrectly represented by drawing three pure double bonds and three pure single bonds, that in reality the molecule has a very high $D_\mathrm{6h}$ symmetry rendering all bonds equal and the electrons in the π system are delocalised all across the molecule. However, that is not what the question is asking about, and not something IUPAC nomenclature cares about.
Reference:
[1]: F. A. Kekulé Bull. Soc. Chim. Paris 1865, 3, 98.

I disagree, cyclohexatriene is an incorrect name for benzene.  As I mentioned in my comments under the OP's question, derivatives of 1,3,5 cyclohexatriene have been prepared and characterized by Vollhardt's team.  These 1,3,5-cyclohexatrienes behave quite differently from benzene - the names benzene and cyclohexatriene describe different compounds.

@ron In this and in many other cases concerning nomenclature, IUPAC looks at the compound to be named in an entirely theoretic manner and decides solely by theoretic considerations what to call it. IUPAC’s aim in nomenclature is to be able to take a compound of which you only know a skeletal formula and turn it into a name. Therefore, IUPAC nomenclature would never distinguish between a resonance-stabilised an a non-resonating cyclohexatriene.

Sounds like a serious shortcoming to me, as 3 double bonds do not exist in benzene.  Since the properties of 1,3,5-cyclohexatriene and benzene differ, it just seems wrong to me to call them both by the same name - but understand, that's just my opinion.

I second @ron's concern, but following that logic we would have to give the cyclopentadienyl-anion a new name too, because there are certainly no localised double bonds in there, as well as no localised negative charge. In that regard I support the decision to ignore delocalised systems completely and name such species according to it's resonance structures.

great, I was always wondering about phenyl etymology

@ron When I type "Peter Vollhardt synthesis and characterization of 1,3,5-trichloro benzene" in Google Scholar, I get zero relevant results. Is it possible that the PDF has been moved or is it behind a non-indexed paywall?

@GaurangTandon It is not trichloro, see these two papers and references therein; ref_1, ref_2

@ron If you disagree with Jan's answer, could you please post a new answer of your own? Thanks!


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Which tautomer of hexane-2,4-dione is more stable?

Which tautomer will be more stable?
Both 1 and 2 show intramolecular hydrogen bonding and form 6 membered cyclic ring, thus increasing the stability.
I think I need to compare the effect of the two alkyl groups ,ie, methyl and ethyl  but how?

You need to consider the substituent effects on both the carbon-carbon and carbonyl double bonds.  In (1) you have a methyl group stabilizing the carbon-carbon double bond and an ethyl group stabilizing the carbonyl double bond, whereas in (2) you have an ethyl group stabilizing the  carbon-carbon double bond and a methyl group stabilizing the carbonyl double bond. That doesn't seem like much of a difference; therefore I'd expect roughly equal amounts of (1) and (2).

Stability here is roughly the same, like what ron said. But by stability if you mean less susceptible to attacks by other species, I would go with (1), since the electron deficiency on the carbonyl carbon is slighly less due to the greater +inductive effect of ethyl group, compared to the methyl group.

TL;DR
The question you are asking is not worth asking. Systems like these are hard to judge and the equilibrium between these structures is extremely fast. Considering any structure as more stable than another is certainly a mistake, because it completely ignores that the molecule is not static.
The double minimum of the tautomers of acetylacetone
 Long time it was believed, that for the equilibrium of acetylacetone (AcAc) the enol tautomer (4-hydroxypent-3-en-2-one) has a symmetric hydrogen bond and shows therefore only one minimum, like the following scheme by DMacks from Wikipedia suggests:

A quite recent publication by Howard et. al. nicely explores the potential energy surface of this molecule and manages to explain the transitional barrier.[1] They find the barrier to be around 10.5 kJ/mol (2.5 kcal/mol) which translates to a rate of 1011 s-1, which is ultrafast. At room temperature you are not able to distinguish between those structures.[2,3]
The equilibrium of hexane-2,4-dione
When we move away from the symmetrically substituted case, we can expect things to be similar, but a bit more complicated. First of all, we would expect a transition barrier to be roughly around the same size, so the transition from one form to the other is still ultrafast.[4]

I did a few, very rudimentary calculations on the subject and I have to admit, that they were far more complicated than I have anticipated. For reasons I will explain in the notes section, I choose to only discuss the electronic energies on the DF-BP86/def2-SVP level of theory.[5]
First of all it should be noted, that the dione open form 3 is more than 35 kJ/mol higher in energy and will play an insignificant role in the whole mixture. However, I chose this to be the reference state.
The (4Z)-5-hydroxyhex-4-en-3-one 1 is therefore 39.3 kJ/mol lower in energy, while (3Z)-4-hydroxyhex-3-en-2-one 2 is 37.2 kJ/mol lower in energy. This separates the two states by just 2.1 kJ/mol. At room temperature it is impossible to really distinguish those two. It is an incredibly shallow potential energy surface. The transition state 4 is at -36.5 kJ/mol, which makes the activation barrier less than one kJ/mol. Now we should not trust these values absolute, but they give us a rough idea about the whole process. The virtual mode in the transition state is depicted below.

The same mode in 1 looks like this:

And in 2 it looks like this:

You can see how these modes are all essentially the same. Again you will probably have to cool your system to below 50 K to actually see a separation of those.
One remarkable feature is that against the predictions of judging the +I effect, this level of theory would actually predict 1 being more stable than 2. This is not even surprising since the ethylene chain is also rotating out of plane in this structure.[6]
Conclusion
The tautomers of hexane-2,4-one are in a very fast equilibrium with each other. Based on simple models, like +I effect, it is impossible to judge which tautomer would be more stable, i.e. which tautomer has the lowest absolute energy. Not even higher level approximations like density functional theory are reliable enough to predict this.
Notes and References

Daryl L. Howard, Henrik G. Kjaergaard, Jing Huang, and Markus Meuwly, J. Phys. Chem. A 2015, 119 (29), 7980–7990.
Assuming that every transition leads to the formation of the associated structure, i.e. $\kappa =1$, which is reasonable for this small a barrier, we can approximate via transition state theory the rate for a reaction with a barrier of around 10 kJ/mol:
$$ k = \kappa\frac{k_\mathrm{B}T}{h}e^{\frac{- \Delta G^{\ddagger }}{RT}}\approx 1\times10^{11}~\mathrm{\frac{1}{s}}.$$
You might need to cool the system to about 50 K to actually see the formation of two minima. Of course for the symmetric case these two would be identical, so you would still need to mark one of the sides with an heavier isotope. This of course will influence the measurement, but I think you can get where this is going to.
If you are as confused as me about the correct IUPAC nomenclature of the compound see: How to name the tautomers of hexane-2,4-dione correctly according to IUPAC rules?
This system is incredibly hard to treat with DFT. It might be also very hard with more elaborate quantum chemical methods. The transition state is easily obtained, but a caluculation following the intrinsic reaction coordinate seems to be impossible. The reason for this is that the transition state itself is in very close proximity to the structure of 2. The position of the proton does differ by less than 0.2 Å. See a figure of the superimposed structures.
Accordingly all calculations involving frequencies will fail, since the harmonic treatment is simply not enough for this system. This is actually an interesting case, where the structures are in very close proximity, so that adding thermal contributions leads to completely nonsensical results. Here it is, that the activation Gibbs energy between the tautomers is negative.
Fast equilibria like these cannot any longer be described accurately in the Born-Oppenheimer approximation, hence molecular dynamics need to be involved, or at least high level geometry optimisations on the coupled-cluster level with a triple zeta basis set. (The last time I tried that, it already took forever.)
Reasons for this to happen are plenty. It is partially discussed in Why is the eclipsed conformer more stable than the bisected conformer in 1-butene? Also related to this is Why do sp2 carbons tend to prefer eclipsed conformations whereas sp3 carbons prefer staggered?

(I forgot to mention, that we also completely neglect the tunnelling probability of a single proton.)

The inductive effect: +I(Et) > +I(Me) so O near Et is more attactive for the H, therefore the equilibrium between isomer1 and isomer2 is in favor of the second.
But if you want to be sure, a simple IR-spectrum should confirm the real composition of the mixture, there are several extern parameters who can influence the ratio between 1 and 2, eg the polarity of the solvents used.

The ethyl group has a stronger positive inductive effect than the methyl group. So 2 should be more stable than 1.


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Difference in the change in oxidation states of transition elements and p-block elements
In transition elements the oxidation states differ by one (+2 to +7 in Mn). However, in the p-block elements, the oxidation states differ by two (-1,+1,+3,+5 in the halogen group). Why is this so?

This is not true of the whole p-block. For example, carbon -1 (ethyne) -2 (ethene) -3 (ethane). It is also not true of the halogens as they can be 0 ($\ce{Cl2}$).

it's not entirely correct, for example almost all oxidation states from -1 to +7 are known for chlorine. HCl (-1), Cl2 (0), ClF (+1),  ClF3 (+3), ClO2 (+4), ClO2F (+5), ClO3 (+6), HClO4 (+7). All oxidation states from -2 to +6 are known for sulfur. It is more complicated than that.

But in general isn't it this way? @permeakra

@IndoUbt Nope. Actually, s/p block typically demonstrates that (covalency+formal charge) usually differs by two except some special cases. This goes to stability of free radicals, which are more stable when the orbital holding unpaired electron is either hidden behind other orbitals (case of 1st d-row), or  is very large (case of neutral nitrogen and chlorine oxides, superoxide anion and so on). This doesn't restrict oxidation state all that much.

In d block, variable valency occurs due to small differences in successive ionization enthalpy. Hence, it shows valency with difference of 1
While in p block, variable valency occurs due to inert pair effect. Due to this 2 electrons remain paired and become inert for bonding. Hence, in p block the difference of 2 occurs.


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Nitrogen molecule vs carbon-carbon triple bond stability
Why does the triple bond in the nitrogen molecule convey stability, whereas a triple bond between carbon atoms conveys instability?

Hi and welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Note that — like everywhere on the internet — all-caps words are perceived as shouting. Rather, use markdown such as italics \*like this\* (like this) to emphasise.

What do you mean by stability? In what way are alkynes unstable?

Why, alkynes are unstable in many ways. Consider positive $\rm{\Delta_fH}$, for starters. Acetylene is explosive, and diacetylene even more so. That being said, though, the $\rm{C{\equiv}C}$ triple bond is quite strong, almost as strong as $\rm{N{\equiv}N}$.

I guess reactivity would be better word here.

You might already know that elements like to follow the octet rule, which means that atoms tend to combine in a way that each atom can have eight valence electrons. Nitrogen can exist diatomically because through the triple bonds, each nitrogen will have one lone pair and three shared pairs. As seen in the picture, each nitrogen satisfies the octet rule.

Carbon, on the other hand, cannot satisfy the octet rule diatomically. Since carbon has one less valence electron than nitrogen, a triple bond would not work because the carbon does not have enough electrons to bond and satisfy the octet rule. It would leave each carbon atom with three bonded pairs of electrons and only one lone electron, giving each atom a configuration of only seven electrons per atom, not eight. Sorry I don't have a picture of a triple carbon bond. Hope this answers your question though!

It doesn’t really, because there are a high number of $\ce{C#C}$ bonds known, e.g. in acetylene $\ce{H-C#C-H}$.

that's true, but in that case,  the hydrogen bonds with the lone electron in carbon, to give carbon a full valence shell.  C2H4 like you mentioned above a stable molecule. @Jan

If that's your response to my comment you missed the question and I must give you $-1$.

authors: Александр Рябоненко


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Baking Powder instead of soda error
Bit of a random one but I had an oven fire yesterday due to grease.  I decided to clean my over as a result but stupidly used baking POWDER rather than soda and wiped it off with vinegar.
Now I am really worried as I have a gas oven, and some of the baking powder has seeped through the holes at the bottom where the naked flame.  Do you have any idea whether this is dangerous?  Will the baking powder ignite or set fire? Or will it just burn off? I'm very worried about what will happen to the oven now.

For what it’s worth, baking powder = soda in a household environment.

Baking powder is baking soda with an acid component added.  In your case it should make little difference.  After wetting the acid and carbonate have reacted into a more stable salt. https://en.wikipedia.org/wiki/Baking_powder

Baking powder is baking soda ($\ce{NaHCO3}$) plus a mild acid salt or dry acid, such as $\ce{Ca(H2PO4)2}$ or tartaric acid. These ingredients should pose no hazard in a stove unless there is remaining powder blocking a vent hole or gas nozzle. In that case, just use a paper clip or other tool to open the vent.
Tartaric acid is theoretically flammable, but the amount you might have left is so small as to be innocuous. You've probably cleaned the stove about as well as baking soda would, once you remove all obvious deposits of the white powder. Bon appétit!


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Is styrofoam soluble in gasoline or paint thinner?
Why doesn't gasoline or paint thinner melt through Styrofoam when I try it yet there are videos of others melting Styrofoam in gasoline and paint thinner? Is there something I am doing wrong? This is the link to the paint thinner, I used.
http://www.kleanstrip.com/uploads/products/Paint-Thinner-gallon-metal-GKPT94002-900.jpg
For gasoline, I used gasoline from a jerrycan which was sitting for a couple of months.
Please help me out by pointing out where and how I might have made a mistake. No excessive chemistry terminologies please. I am merely a fourth grader ( guardian ) trying to work this out.
Thank you,
Guardian of fourth grader

The styrofoam experiment works well with acetone or toluene. Gasoline is no good; as for paint thinner, I guess it comes in various sorts and I don't know which one do you have (the picture doesn't help either).

They add a good deal of petroleum to many paint-thinner products which can have varying strength as a styrofoam solvent. Pure organic solvents are ideal here, e.g. 100% acetone. The main ingredient of paint thinner that would dissolve Styrofoam, I think, is methanol. Methanol can be extracted by distillation but you probably don't have the equipment for that. Toluene likewise but the 'safer' products have methanol.

Can you use this to seals leaks on plastic, roof, or any oil leaks?

Unless you're trying to make napalm, don't use gasoline.
Use acetone. It dissolves egg cartons within just a few minutes, and you can then take it out and reshape it (use gloves!). Set it out to dry and once the acetone has evaporated, it should be a (crude) solid.
Just ensure whatever container you're using to hold the acetone isn't polystyrene. Check to make sure recycling code 6 isn't on the container. I had a pretty good laugh at my teacher one day who poured acetone into a red solo cup and it went running all over the table.

Polystyrene foam can be dissolved by gasoline, which contains toluene and its derivatives. Either the foam you have is not polystyrene, or the gasoline you're using is poor in toluene. Check for the code found on the styrofoam you've got, it should look like this:

If your foam is indeed polystyrene, the problem is with your solvent. Check if the fuel you've got is indeed gasoline, and not diesel or bioethanol. Using pure toluene is guaranteed to work.


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What chemicals react with alcohol?
I am curious if there is any chemicals that react with alcohol to either produce a visible colour change or an odour someone could smell.

As it stands this question is rather broad. What application are you looking for here? For example, potassium dichromate will react with alcohol to produce a colour change, but it will also react with other things to produce the same colour change.

For contamination purposes I wanted to be able to see if there is alcohol in a substance by either sight or odour.

How can I tell if a liquid contains alcohol with the addition of a chemicals with the cheapest and least toxic option

When you say "alcohol", I am not sure if you are referring to ethanol (drinking alcohol) or alcohols in general so in my example I will use ethanol. When an alcohol is combined with a carbocylic acid and a catalyst, they will produce esters (fragrances). For example, ethanol and ethanoic acid would react to produce ethyl ethanoate (a fruity smelling substance) and water.

There are a plethora of different reactions from all different types of alcohols and carbocylic acids that produce countless different smells. The fragrances also come in a variety of colors as well. If you want to find out more, I would suggest looking up ester synthesis. Hope my answer helps.

carboxylic acid not carbocyclic acid.


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Tungsten 4+ ion magnetism?
I just had a quiz and struggled on a particular question:

Tungsten forms a $+4$ ion. It is observed to be diamagnetic. The electron configuration for this ion may not be what you would expect. Write the electron configuration for tungsten(IV) that is diamagnetic.

So if tungsten has a configuration of $\ce{[Xe] 6s^1 4f^{14} 5d^5}$ and it loses 4 electrons wouldn't it be $\ce{[Xe] 6s^0 4f^1^4 5d^2}$? And this would be paramagnetic since there are 2 unpaired electrons in the 5d subshell. How could it be diamagnetic?

Now here’s a hint: ‘The electronic configuration for this ion may not be what you would expect.’

Maybe the energy difference between 6s and 5d orbitals is very low, leading to a more stable configuration 6s2 (occupied sub-shell).

Oddly the NIST database for atomic spectra suggests that the $\ce{W^{4+}}$ ion has the configuration $\mathrm{Xe^{14}(5d)^2 }$ and a triplet ground state, which would make it paramagnetic after all.

The electron configuration for Tungsten is [Xe] 6s2 5d4 4f14. If it were to lose its electrons, and the answer is that it is diamagnetic, then it would lose its 4 d-electrons.
References

https://www.quora.com/What-are-the-differences-between-paramagnetic-and-diamagnetic-elements
https://www.tabulka.cz/english/elements/configuration.asp?id=74
https://socratic.org/questions/what-are-columbic-and-exchange-energy-and-how-it-it-determined

Hey Aaron, welcome to Chem.SE! Could you please cite the actual resources, instead of linking to other forums? Thanks!

You have a different electron configuration for the neutral atom than the OP. It would be nice to point that out and comment on it. Also, you should summarize the argument (with citations) rather than just posting links.


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Hair Gel Derivative
I need a water-soluble, extremely fast drying adhesive gel.  My first thought is hair gel.
pros:

Water Soluble (easily removed)*** most important
High viscosity in liquid form ~60 cps
Easily dispensed.

Cons:

Drying is slow when dispensed in a thick bead. Slow is bad.
Looses quite a bit of volume when fully hardened.  Decreased coverage.

I'm not planning to use this on my body.  It would be used for masking off a portion of a metal production part for non-destructive testing.  It is temporary and must be removed easily.
Does anyone know of a water soluble fast drying gel that doesn't lose much volume when hardened?  Something similar to hair gel.

Why does it have to be water-soluble? Art-gum (rubber cement) can be removed by rubbing. Other products can be dissolved by mineral turps or acetone.
Kaolin and talcum powder can be used as thickening agents, although I haven't tried them with art gum.

I need to integrate it into an automated system.  It's hard to reliably remove a material by rubbing it off using a machine.  I can't use volatile liquids to remove it because of the risks.


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Minimum resolution to acquire exact mass on a spectra from m/z peak
I'm having a little difficulty with finding what the minimum resolution required to obtain the exact mass of a 400 $m/z$ peak in a mass spectrum. I would appreciate any help.

Depends what you mean by "exact mass"—exact mass generally refers to the calculated mass of a monoisotopic species, so if you mean what resolution do you need to unambiguously identify the elemental composition of that ion, it depends on the circumstances. If you know the ion only contains carbon, hydrogen, and oxygen, you may not need as much resolution or if the ion is multiply charged, you may need far more.

What is the formula used in this case: mass (400) over what delta m?  The composition and the conditions are not specified. What could be a minimum resolution to obtain the exact, yes the monoisotopic mass of the compound appearing at 400 m/z, being not multiply charged.

It's hard to say exactly, it depends on what other compounds are possible near the mass you're looking for. As is, this is a poorly posed question. If we consider only C,H,N, and O this has a figure that shows you the accuracy needed for unambiguous identification for different m/z and formulas for calculating the resolution you need.

If you're looking at a single peak I doubt that any mass spec has enough resolution to uniquely identify the formula of the ion. Part of mass spec analysis is the analysis of the isotopic pattern. For example sulfur and chlorine both have multiple isotopes. For instance the difference between $\ce{^{32}S^{37}Cl}$ and $\ce{^{34}S^{35}Cl}$ is only about 0.001 amu. I'm sure that there are other isotopic combinations that would be even closer.

In mass spectrometry, resolution and mass accuracy are distinct yet related concepts.  Eliding the difference between the two often causes confusion.

I'm having a little difficulty with finding what the minimum resolution required to obtain the exact mass of a 400 m/z peak on a mass spectrometry spectra is.

Resolution is a concept that always involves at least two chemical species that differ in $m/z$.  It doesn't make sense to talk about resolution if you are talking about a single ion.
Mass accuracy is a concept that more or less assumes there is a single chemical species at a single $m/z$.  It's very easy to define: how different is a measured $m/z$ from the correct $m/z$?
Now, due to the presence of multiple stable isotopes for most elements commonly seen in organic mass spectrometry, almost no compound will give rise to a single chemical species with a single $m/z$ value.  If your compound ionizes to give singly charged ions ($z=1$), these isotopologues will be separated by approximately 1 $m/z$.
A nice coincidence among commonly occurring elements in organic chemistry ($\ce{CHNOPS}$) is that the lightest stable isotope is by far the most abundant.  (This isn't true of $\ce{Cl}$, $\ce{Br}$, $\ce{Fe}$ though.)  For molecules as small as 400 Daltons, this means that the most abundant isotopologue will be (a) the lightest of all possible isotopologues, and (b) monoisotopic.
Thus, to get an accurate mass (really accurate $m/z$) measurement of an compound you'll need a resolution that is better than unit mass resolution.
However, you won't necessarily need a much greater resolution than that.  A lot depends on your sample.  Can you assume that there is only one chemical compound present that gives an $m/z$ of 400 Da?  If so, then even "low resolution" mass spectrometers can give very accurate masses.  However, if your sample is complex or contains other compounds that give ions very close to and $m/z$ of 400, then you will need more resolving power to separate the contaminating ions from your peak of interest.  In that case, it's impossible to say how much resolving power you need without specifying what the contaminants are.
For example, say your compound is $\ce{C29H37N}$, for example, maybe it's N-(3,3,3-triphenylpropyl)octan-1-amine.  In electrospray mass spectrometry it ionizes as $\ce{(C29H37N + H)+}$ at an $m/z$ of 400.3010.  If that was the only compound in your sample, then even old TOF instruments with resolutions as low as 2000 would suffice to obtain an accurate mass.  However, if there was also some $\ce{C26H41NS}$ in your sample, say 1,6-dimethyl-4-pentadecylquinoline-2-thione, its $\ce{(C26H41NS + H)+}$ ion would have an $m/z$ of 400.3043.  That would mean you would need a resolution of at least 122,000 to tell those two peaks apart, generally available only on Orbitrap or FT-ICR instruments.  But if there was no such contaminant in your sample, you wouldn't need that degree of resolution.


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Why do Pop Rocks fizz longer in milk than in Sprite?
After doing an experiment with Pop rocks in various liquids, it was noted that they fizzed the longest in milk. My daughter compared Sprite, apple juice, water and milk, and milk fizzed the longest. Why?

How much longer are we talking about here? If you can provide times from the experiment, that would be helpful.

There are a number of experimental factors which you haven't detailed. Did your daughter use same amount of pop rocks for volume of liquid? Was the temperature of the liquids the same? What was the experimental error in timing for each liquid? I assume no stirring. But I'm also unsure about the size distribution of the pop rocks. Could an abnormally large pop rock been with the milk? // Off the cuff I'd guess the milk was taken from the fridge and was much colder than the other fluids.

Likely the milk was a colder temperature than the other liquids, which will give a slower reaction hence lasting longest. Also milk contains a lot of stuff mixed in it, fats and proteins, so that may or may not cause the candy to be less soluble compared to water.


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why does this reaction place the nitro group specifically there?

the amide up top is activating and the sulfonic acid below is deactivating. I'm just wondering how we know that the nitro group is placed ortho to the amide or meta to the sulfonic acid.

Bad example, there's no such problem in this case.

If you draw resonance structures for ortho and meta attacks with respect to amide group,
Resonance structures in ortho attack will exhibit carbocation directly attached to amide group, which is resonance stabilised.
Resonance structure of meta attack will exhibit carbocation directly attached to SO3H group. Which is destabilised. So ortho attack is predominant.
You can check the image I have attached as an answer.


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How to calculate the mass of a single atom?
If we know the molar mass of a certain element and Avogadro's constant, how can we calculate the mass of a single atom? Do we need to multiply the molar mass with Avogadro's constant?

For most atoms it's around Ryan's answer.
E.g. Carbon-12:
$$\frac{\ce{12 g}~\ce{C}}{\pu{1 mol}~\ce{C}} \times
\frac{\pu{1 mol}~\ce{C}}{\pu{6.022E23 atoms}} =
\pu{1.993E-23 g//atom} = \pu{1.993E-27 kg//atom}.$$
That was the molar mass $M$ multiplied by $1/N_\mathrm{A}$, where $N_\mathrm{A}$ is Avagadro's constant.
Thus $M/N_\mathrm{A}$ gives you a calculation for mass of an atom for the specific element.

The average mass of a single element is its atomic mass on the periodic table (measured in $\pu{u}$, the unified atomic mass unit). One $\pu{u}$ equals about $\pu{1.661E-27 kg}$.


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Method to synthesize KAu(CN)2 (or KCN)?
Every reasonably effective method of electroplating $\ce{Au}$ onto another metal's surface (e.g. $\ce{Cu}$,$\ce{Ni}$,$\ce{Ag}$, etc.) involves use of $\ce{KAu(CN)2}$ or in some cases $\ce{KAu(CN)4}$ in acidic, neutral, or (in almost all cases less effective) basic solution. Ideally it should be $60$ to $80~\mathrm{^\circ C}$ with a certain current density applied (c. $0.5~\mathrm{A/dm^2}$). In the past I have tried a direct route of dissolving the gold in warm $\ce{HCl / HNO3}$ as the deposition solution to little effect (a brownish/blackish tint is left on the receiving surface with very little tint of gold). The short story is that I have found the use of cyanide unavoidable.
$\ce{KAu(CN)2}$ runs $157/g from Sigma-Aldrich so I would like to synthesize it, if possible.
I have:

1g solid $\ce{Au}$
Excess $\ce{KNO3}$, $\ce{KI}$
Various $\ce{C-N}$ organic compounds, e.g. urea, uric acid, etc.

I think my best route is indirect — to first synthesize $\ce{KCN}$ and then form the product:
$$\ce{4 Au + 8 KCN + O2 + 2 H2O → 4 K[Au(CN)2] + 4 KOH}$$
...and follow by vacuum filtration on the solid precipitate. Which begs the question of the best method for safe synthesis of $\ce{KCN}$.

Much of my information comes from this difficult-to-find and expensive book, "Properties of Electrodeposited Metals and Alloys: A Handbook" by Safranek. It's got quite a lot of proprietary data. Also Kohl's chapter on gold electrodeposition.

please clarify how your workplace looks like. It is a kind of pro-lab, a DIY garage, something in-between ?

It's a makeshift lab at home. I have a good amount of equipment but any synthesis of CN I'd be doing outside. Wearing PPE of course. I'm a computational chemist at university but do experimental work as a hobby at home

I kindly suggest to order it by mail than. As a last resort you may find a buddy with access to a proper fume hood and ask him. I actively discourage producing potassium cianide at home. Even on open air. Hydrogen cyanide is an intermediate and you definitely do not want to deal with it outside a fume hood.  This said, potassium cyanide is sometimes employed in organic synthesis, so you may find some there.

Related: How to prepare potassium cyanide from potassium ferrocyanide

I read that you are a computational chemist at university, so I assume you went through chemical training. This is a good thing. But please remember that cyanides are potent poisons and that protonated cyanide ($\mathrm{p}K_\mathrm{a}(\ce{HCN}) \approx 8$) aka hydrogen cyanide is a poisonous gas.
If you are not a trained chemist (or if you are one but do not feel $100~\%$ safe about using cyanides), stick away from them!
With that in mind this leaves two methods of acquiring $\ce{K[Au(CN)2]}$ — either buying it directly from YourSuppliers™ or by synthesising it. My preferred route of choice would definitely be to buy it from YourSuppliers™, as that is the safest way possible — and always always ensure that it remains in an alkaline solution.
I have no idea about the experimental procedure to synthesise $\ce{K[Au(CN)2]}$ according to the reaction you propose, but it didn’t escape my notice that you are generating $\ce{4KOH}$. This means, I’ll have to assume that the reaction runs under acidic conditions. If that is the case, do not use that method unless you have access to a proper and tested fume hood! If there are other procedures out there that employ alkaline reaction conditions you may (as a trained chemist!) decide to use them at your own risk.
Assuming you did find a method that employs alkaline conditions, the question remains where to get the cyanide from. Well, SigmaAldrich gives me a price of 38 € for $100~\mathrm{g}$ which means that the chemical is very cheap. Note that acquisition of potassium cyanide may well be regulated in your jurisdiction — and rightly so!
Whatever you do, do not attempt to synthesise $\ce{KCN}$ (or any other cyanide for that matter) at home!
Remember that a certain percentage of the population can sense $\ce{HCN}$ which they perceive as an almond-like smell. Don’t attempt to test which part you belong to. If the concentration of $\ce{HCN}$ is such that you can smell it, you are in severe danger!
This bears repeating for future visitors:
Do not do this unless you are a trained chemist and have access to a proper fume hood! Never, ever try this at home! Never work with cyanide in the open air or in any other place where unsuspecting passers-by may be endangered!

It seems the wise answer here is a focus on matters of practicality and safety and not chemical procedure. I think buying KCN and borrowing a hood is indeed best. Going to do further research on options

Sodium cyanide can be very easily made by fusing sodium ferrocyanide with sodium carbonate. The guy in the video is still alive, but I would recommend wearing gloves at least.
Video including Au dissolution


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What is the difference between D and L configuration, and + and −?
What is the difference between D and L configuration, and + and −?
My textbook says they are two different things.
It also says that the correct way to name glucose is D(+)-glucose.
Could someone please explain what D, L and +, − represent, and why they are different?

The D-L system corresponds to the configuration of the molecule: spatial arrangement of its atoms around the chirality center.
While (+) and (-) notation corresponds to the optical activity of the substance, whether it rotates the plane of polarized light clockwise (+) or counterclockwise (-).
D-L system tells us about the relative configuration of the molecule, compared to the enantiomers of glyceraldehyde as the standard compound. Compounds with the same relative configuration as (+)-glyceraldehyde are assigned the D prefix, and those with the relative configuration of (-)-glyceraldehyde are given the L prefix.
It's kind of another way to tell the configuration of molecules beside the Cahn–Ingold–Prelog convention (R/S system), with little difference. (D-L system labels the whole molecule, while R/S system labels the absolute configuration of each chirality center.)
In short, the D-L system doesn't have direct connection to (+)/(-) notation. It only relates the stereochemistry of the compound with that of glyceraldehyde, but says nothing about its optical activity. We may have compound with same relative configuration as (+)-glyceraldehyde (thus, it's given the D prefix), yet it rotates the polarized light counterclockwise (-), such as D-(-)-ribose.
And also, don't confuse the D-L system with d- and l- naming. d- and l- is the exact same with (+) and (-) notation.

Additional explanation
D-L system (also called Fischer–Rosanoff convention) is mainly used for naming α-amino acids and sugars. It compares the relative configurations of molecules to the enantiomers of glyceraldehyde. This convention is still in common use today.
Rosanoff in 1906 selected the enantiomeric glyceraldehydes as the point of reference[1]; any sugar derivable by chain lengthening from what is now known as (+)-glyceraldehyde (or named D-glyceraldehyde) belongs to the D series. In other words, we used a D to designate the sugars that degrade to (+)-glyceraldehyde and an L for those that degrade to (-)-glyceraldehyde.
In assigning the D and L configurations of sugars, we could direcly look for the OH group of the bottom asymmetric carbon in the Fischer projection. If it's located on the right, we designate it with D, and vice versa, since they would have the same relative configurations with glyceraldehyde for the bottom asymmetric carbon.
Reference
[1]: IUPAC and IUBMB. Joint Commission On Biochemical Nomenclature. Nomenclature of Carbohydrates. 1996, 7.

That was very clear and informative. Thank you. Is it right to infer from your explanation that there is no way to find out whether a molecule turns light left or right just by looking at it?

@user93868 Yes, it is. The direction which it rotates the polarized light, (+) or (-), is something that we measure in the polarimeter. It's directly the physical property of that compound. We can't easily tell the direction just by looking at it.

does the lower case d- and l-  that you mention refer to dextrorotatory and levorotatory?

@LiamH Yes, they are, respectively.

While d- and l- are synonymous with (+) and (-), respectively, as the signs of optical rotation in modern terminology, such was not always the case in Fischer's day (~1891). The designation (d)-glucose in many instances was equivalent to today's (D)-glucose while (+)-glucose referred to the sign of dextrorotation. Rosanoff (J. Am. Chem. Soc., 1906, 28, 114) revealed inconsistencies in Fischer's assignment of the "D-" and "L-" families of carbohydrates. Rosanoff introduced the δ-(delta) and λ-(lambda) series that have evolved into today's D- and L-series, respectively.


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Does turning propane or butane gas tank upside down help to use the remaining gas?
Some think that when gas (propane-butane mixture for cooking) in a gas tank is almost used (only remainings are left inside and the flame is hardly noticeable) and you turn it upside down it will improve gas burning. Is it true?

Are you asking about propane or butane? IIRC, tanks tend to contain one or the other, rather than a mixture of both, and they have sufficiently different properties to change the answer to a certain extent.

Question updated.

Can you hear any liquid inside?

Yes, and if I leave the tank long enough without using I can burn flame for a few seconds in normal position.

I was using a propane fed heater to dry drywall mud. I noticed that the tank was starting to get low, so I turned the tank upside down. Immediately the flame went back up to high. After a few minutes, I noticed a bulge form in the hose. I turned the tank upright and shut it off. The heater burned for 10 more minutes after I turned it off. I imagine that I was probably seconds away from the hose blowing up and burning down the house. Don't do it. It scared the crap out of me and I consider myself lucky.

In principle it won't make any difference, but in practice it sometimes can.
Since what you want to emerge from the tank is gas, in principle nothing you do will make more gas come out. The liquid inside will vaporise until there is no gas left and the pressure won't change while there is still liquid in the vessel (both gases liquefy easily under pressure and the pressure will be relatively constant while there is still liquid.) All this is true at equilibrium and constant temperature.
However, in practice the assumption that the inside of the vessel is in equilibrium and constant temperature is not always true. If you use a lot of gas, the vessel will cool as the liquid evaporates (and you sometimes see a ring of frost on the outside of the vessel corresponding to the level of the liquid). If this happens then shaking the vessel may have the effect of warming the liquid and increasing the pressure of the gas. So shaking it might help a little.


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Could any constituent of normal human urine produce a false positive in a Kastle-Meyer test for presence of blood?
From my understanding, one forensic test for the presence of traces of blood at a crime scene is (or was) the Kastle-Meyer test. Doing some research, I understand that in the test, hemoglobin reacts with hydrogen peroxide which oxidizes phenolphthalin into phenolphthalein which then changes it's colour to bright pink.
My question is, are there any constituents of normal urine that could cause a false positive in this test?
To be clear, my interest is academic, not personal.