text:
Could this solution be resolved by titration?
I have a mixture of $\ce{NH_4NO_3}$ and $\ce{NH_3}$ at $\ce{pH}$ of about 13.
It is possible to determine the concentration of both on them by acid-base titration?
I was told it's possible, and that you need first titrate with $\ce{HCl}$ and then with $\ce{NaOH}$.
Thanks.
Edit: In the lab were I was working, there were an auto titrator, that had a program that could tell the concentracion of $\ce{NH_4NO_3}$ and $\ce{NH_3}$ in a solution, just by titrating it and controlling the $\ce{pH}$ with a pH meter. I simply don't understand how it does it, because depending of the $\ce{pH}$, $\ce{NH_4+}$ and $\ce{NH_3}$ would be in the same form and I don't know how you could tell them apart.

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.
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Note that at pH=13, NH4+ ions are practically all converted to NH3 by OH- ions. Notice the pKb constant for ammonia and what NH4+/NH3 concentration ratio exists at pH 13.

The ammonium nitrate dissociate in water:
$$\ce{NH4NO3 -> NH4+ + NO3-}$$
The $\ce{NH4+}$ and $\ce{NH3}$ form together the ammonia $\mathrm{pH}$ buffer with $\ce{NH4+}$ acidity constant $\mathrm{p}K_\mathrm{a}=9.25$:
$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+ \log { \frac {c_{\ce{NH3}}}{c_{\ce{NH4+}}}}$$
The titration curve as $\mathrm{pH}=f(V)$ is the most flat at $\mathrm{pH}=\mathrm{p}K_\mathrm{a}=9.25$, where the buffer has the biggest capacity.
For a given solution, containing $\ce{NH4+}$ and $\ce{NH3}$, titration with $\ce{HCl}$ or $\ce{NaOH}$ causes  respective acid-base reactions:
$$\begin{align}
\ce{NH3 + H+ &-> NH4+ \\
NH4+ + OH- &-> NH3 + H2O} \\
\end{align}$$
The ammonia buffer causes   $\mathrm{pH}$ being quite stable during titration, until one of the 2 ammonia forms is practically spent.
At that point, the $\mathrm{pH}$ titration curve gets very progressively bent up or down, respectively and the titrator stops the titration.
The content of $\ce{NH4NO3}$ and  $\ce{NH3}$ is then obtained by obvious routine calculation.
$$\begin{align}
m_{\ce{NH3}}&= M_{\ce{NH3}}\cdot c_{\ce{HCl}}\cdot V_{\ce{HCl}}\\
m_{\ce{NH4NO3}}&= M_{\ce{NH4NO3}}\cdot c_{\ce{NaOH}}\cdot V_{\ce{NaOH}}\\
\end{align}$$
Note that at $\mathrm{pH}=13$, $\ce{NH4+}$ ions practically do not exist, as $ \frac {c_{\ce{NH3}}}{c_{\ce{NH4+}}}=5620$
If we consider $\mathrm{pH}$ of ammonia solution itself, it is
$$\begin{align}
\mathrm{pH}=14 - 0.5 \cdot ( \mathrm{p}K_\mathrm{b}-\log c) \\
13=14 - 0.5 \cdot ( 4.75 -\log c) \\
\end{align}$$
$c_{\ce{NH3}}$ would be 563 mol/L, what is nonsense. Such $\mathrm{pH}$ is not achievable by ammonia alone, but a mineral hydroxide would be present.
If there was initially any ammonium nitrate, it would be converted to ammonia:
$$\ce{NH4NO3 + NaOH -> NaNO3 + NH3 + H2O}$$
In such a case, determination of both compounds by titration is not possible, as there are no ammonium ions and ammonia would be titrated together with hydroxide.

Ok, I get it, thanks. I don't know why, but such a simple thing and I was completely stuck. And sorry, $\mathrm{pH = 13}$ was just an example, but now I see that it would make no sense with ammonia alone.
So the titrator would first titrate with $\ce{HCl}$ and then with $\ce{NaOH}$, having to titrate first the $\ce{HCl}$ that it used.


text:
Calculating pH of weak base and strong acid: Why is K(a) not required?
Question: 35mL of 1.5M $\ce{HCN}$, a weak acid(K(a)= $6.2*10^{-10}$) is mixed with 25mL of 2.5M $\ce{KOH}$. Calculate the pH of final solution.
The solution given in my textbook is as follows:-

$\ce{KOH}$ completely dissociates to give 0.062 mol $\ce{OH-}$(The $\ce{K+}$ becomes irrelevant)
0.052 mol of $\ce{HCN}$

3.The equation is $\ce{HCN + OH- ⇌ CN- + H2O}$

So 0.062-0.052 mol=0.01 mol of $\ce{OH-}$ is left over
so 0.01 mol/(0.035l+0.025l)=0.17M of $\ce{OH-}$

6.pOH=0.76 so pH=13.24
7.Also, $\ce{CN-}$ is a weak base relative to $\ce{OH-}$ so we consider it negligible in the calculation.

Finally, it says that K(a) was not required and it was just to distract.

However, I think K(a) is actually required because $\ce{HCN + OH- ⇌ CN- + H2O}$ is a reversible reaction and I don't understand how all of the 0.052 mol of $\ce{HCN}$ can react with $\ce{OH-}$(and in turn, 0.01 mol of $\ce{OH-}$ is left) when there is also a backward reaction producing $\ce{OH-}$ and $\ce{HCN}$ back.
Please help.

You can directly calculate, what is HCN/CN- ratio in 0.17M NaOH, And What would be pH of NaCN alone Consider more thorough elaboration of your effort to avoid closing future questions.

Your doubt is legitimate because you are assuming that $K_{(a)}$ is a parameter that you can plug in every reaction. Actually $K_{(a)}$ is calculated using this equation in a solution of the acid with water:
$$
\ce{K_{(a)} = \frac{[H+][CN-]}{[HCN]}}
$$
That is derived from this ideal equilibrium:
$$\ce{HCN ⇌ CN- + H+}$$
Even though it reflects the strength of that specific acid in a pure water solution it's practical usage is limited when the equilibrium is not perturbated by others factors.
Anything that will interfere with this equilibrium can cause problems to your assumptions.
In your specific case, the conditions are different there is an excess of $\ce{OH-}$ in the solution due to the strong base, so every $\ce{H+}$ produced will react with the $\ce{OH-}$. Due to the Le Chatelier principle the equilibrium will respond shifting to the right: dissociating more and more $\ce{HCN}$ from producing more $\ce{H+}$.
In simple words, if one reaction is favourable the solution doesn't care about your $K_{(a)}$ it will go following the most thermodynamic favourable path to the equilibrium.
Possible practical learning aim of this question?
So if you want to have high products amount (i.e. $\ce{CN-}$) and you have cheap reactant (i.e. $\ce{KOH}$) you can use an excess of reactant to shift the equilibrium where you want.
Bonus question: what can you do if you want to have HCN?

Really good answer, your explanation of the equilibrium shifting to the right in a continuous cycle until no more HCN is left clarified my doubt in an instant. Thanks for your help.

source: Stack Exchange

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sort: votes


text:
Reaction of borax with NaOH
How does borax  react with NaOH?
My textbook says that sodium peroxoborate is formed. I couldn't find the reaction anywhere else.
Can anyone verify as well as explain it?

This is nonsense.

The following reaction schemes are adapted from [1, pp. 76–77].
Amorphous borax $\ce{Na2B4O7}$ forms tetrahydroxoborate in concentrated sodium hydroxide solution:
$$\ce{Na2B4O7 + 7 H2O + 2 NaOH → 4 Na[B(OH)4]}$$
Upon fusion sodium metaborate is formed:
$$\ce{Na2B4O7 + 2 NaOH →[\pu{700-750 °C}] 4 NaBO2 + H2O}$$
In order to form peroxoborate, a peroxide should be introduced in the first place, e.g. by using hydrogen peroxide (see also Preparation of Sodium Peroxoborate):
$$\ce{Na2B4O7 + 2 NaOH + 4 H2O2 + 11 H2O → 2 Na2[B2(O2)2(OH)4] * 6H2O}$$
References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)

Andselisk has given a more satisfactory answer for this question, which has been accepted. Yet, I'd like to give some interesting facts and view about borax, if some readers are interested.
The name borax is derived from the Persian word borak, meaning white. IUPAC name of borax is sodium tetraborate decahydrate ($\ce{Na2B4O7.10H2O}$). Also note that, after analyzing its crystal structure, borax is more properly formulated as $\ce{Na2[B4O5(OH)4].8 H2O}$ and given its systematic name 2,4,6,8,9-pentaoxa-1,3,5,7-tetraborabicyclo[3.3.1]nonane-3,7-diolate, sodium salt, hydrate (1:2:10), accordingly.
When dissolved in water, borax is subject to completely dissociate, and then,  hydrolyzes to orthoboric acid ($\ce{B(OH)3}$) and $\ce{ OH-}$ anions, yielding a $\mathrm{pH}$ of about 9.13 according to the equation:
$$\ce{ Na2[B4O5(OH)4].8 H2O (s) -> 2Na+ (aq) + B4O5(OH)4^{2-} (aq) + 8H2O (l)}$$
$$\ce{ 2Na+ (aq) + B4O5(OH)4^{2-} (aq) + 5H2O (l) <=>  2Na+ (aq) + 4B(OH)3 (aq) + 2 OH- (aq)}$$
Therefore, borax reacts with strong acid solutions. For example:
$$\ce{ 2Na+ (aq) + B4O5(OH)4^{2-} (aq) + 2 HCl (aq) + 5H2O (l) -> 4B(OH)3 (aq)  + 2Na+ (aq) + Cl- (aq)}$$
The liberated $\ce{B(OH)3 (aq)}$ acts as a Lewis acid on $\ce{ OH-}$ ions from autoionization of water, and hence, in equilibrium with water:
$$\ce{ B(OH)3 (aq) + 2H2O (l) <=> B(OH)4- (aq) + H3O+ (aq)}$$
Thus, it can also be back-titrated with alkali solutions such as $\ce{NaOH}$. However, note that the boric acid is a weak monoprotic acid ($K_\mathrm{a} \approx 6.4 \times 10^{-10}$). Therefore, it cannot be directly titrated with standard alkali for analytical purposes. Yet, by the addition of an organic polyhydroxy compound such as 1,2- or 1,3-diol, it is converted to a much stronger acid, which can be titrated using phenolphthalein. This is due to the complex formation between hydrated borate ion and either 1,2- or 1,3-diol (see illustration in the diagram):

It is noteworthy that the combination of these two reactions is commonly used by some analytical chemists to standardize their acidic/basic solutions.
Now, consider the following two reactions:
$$\ce{ 2Na+ (aq) + B4O5(OH)4^{2-} (aq) + 5H2O (l) -> 2Na+ (aq) + 4B(OH)3 (aq) + 2 OH- (aq)}$$
$$\ce{ 4B(OH)3 (aq) + 8H2O (l) <=> 4 B(OH)4- (aq) + 4 H3O+ (aq)}$$
If you add these together, you’d get:
$$\ce{ B4O5(OH)4^{2-} (aq) + 9 H2O (l) -> 4B(OH)4- (aq) + 2 H3O+ (aq)}$$
To cancel $\ce{2H3O+ (aq)}$, add $\ce{2NaOH (aq)}$ to both side. Now, you’d get:
$$\ce{ 2Na+ (aq) + 2OH- (aq) + B4O5(OH)4^{2-} (aq) + 5 H2O (l) -> 2Na+ (aq) + 4B(OH)4- (aq) }$$
Now, compare the equation Andselisk wrote in his answer for the borax reaction with strong alkali solutions. For example, $\ce{2NaOH (aq)}$:
$$\ce{ Na2B4O7 + 7H2O + 2 NaOH -> 4NaB(OH)4 }$$
Both are acceptable ionic and condensed formations, respectively.
Sodium peroxoborate ($\ce{ Na2[(OH)2B(O-O)2B(OH)2]}$), on the other hand, can be prepare by the oxidation of sodium metaborate ($\ce{NaBO2}$) using hydrogen peroxide ($\ce{H2O2}$):
$$\ce{2NaBO2 + 2H2O2 -> Na2[(OH)2B(O-O)2B(OH)2]}$$
Sodium peroxoborate can be precipitated out of water as hexahydrate: $\ce{ Na2[(OH)2B(O-O)2B(OH)2].6H2O}$. Sodium metaborate ($\ce{NaBO2}$) can be prepared by heating borax and $\ce{NaOH}$ together:
$$\ce{ 2NaOH + Na2B4O5(OH)4.8H2O  -> 4 NaB(OH)4 + 3H2O -> 4 NaBO2 + 11 H2O}$$

Did you write thus answer based in memory or from a source? I would like to read the source if possible, to study more(if possible).

It was not one source. There are few of them. I made a Google search under few key words including the title of the question. Trick is you need to be in a campus computer to be able to see all the references. Sorry on that part.

What does a campus computer change? I have never used one.

They have campus' subscribed journal access, of which you can't have with your personal computer at home. That's all.

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text:
Why does Azo coupling of β-naphthol takes place at alpha position and not at gamma position

I have searched for coupling reaction of $\beta$-naphthol with benzene diazonium salt at quite a few places. But everywhere the coupling has been shown at alpha position. Why doesn't it take place at gamma position to minimise steric hindrance.
Chemistry LibreTexts

Chemguide

My book

It is already addressed here https://chemistry.stackexchange.com/questions/97067/mechanism-of-formation-of-2-naphthol-red-dye-aka-sudan-1

Coupling reaction of β-naphthol with benzene diazonium is an example of electrophilic aromatic substitution.
If the electrophile attacks at alpha position ,then two resonance structures 1 and 2 , with aromatic rings are possible.

If the electrophile attacks at gamma position ,only one resonance structures 3 , with aromatic ring is possible and 4 is not aromatic.

Therefore attack at alpha position is the major product.
References:
https://pubs.rsc.org/en/content/articlehtml/2015/gc/c4gc02381a
https://en.wikipedia.org/wiki/Azo_coupling
Mechanism of formation of 2-naphthol red dye (aka Sudan 1)


text:
Why does internal rotation of ethane give an energy spectrum asymptotically approaching a rotation based on $\nu=1$ rather than $\nu=0$?
I'm a physicist with very little knowledge of chemistry. Recently a chem colleague left a molecular model in my classroom, so I brought it back to her, and as I was walking down the hall with it, a hydroxy group started spontaneously swinging around in circles. This made me curious about whether such motions could actually be observed. After asking around and doing some web searching, my understanding is that this sort of thing is called an internal rotation, and usually there is a barrier to rotation, which is modeled by a potential $V(\phi)=(1/2)\sum V_n(1-\cos n\phi)$.
If I'm understanding the history correctly, then originally people thought that the spectrum would be that of a free rotor, but that produced incorrect thermodynamic predictions. Then in 1937 Pitzer published a famous paper, "Thermodynamics of Gaseous Hydrocarbons," in which he showed that you could get correct thermodynamic predictions if you put in a barrier to rotation.
The classic textbook example seems to be ethane. I found this explanation  of the physics (di Lauro, "Floppy Molecules," in Rotational Structure in Molecular Infrared Spectra, 2013), which I only partly understand. He gives the following theoretical spectrum of energy states:

So at low energies we have roughly a set of vibrational energy levels with energies $\hbar\omega (\nu+1/2)$, which makes sense to me because I guess we're sitting in a minimum of the potential where the geometry is staggered. He says, "The high levels, on the contrary, correlate and eventually almost coincide with the free internal rotor eigenfunctions, whose energy increases parabolically as $AK_i^2$, where $K_i$ is the eigenvalue of $J_γ$, in ℏ units." This also sort of makes sense to me, because if you put in a lot of energy, you shouldn't see the effect of the little hills and valleys in the potential $V(\phi)$, so it should act more like a free rotor.
Do the units of inverse cm mean $1/\lambda$, or $2\pi/\lambda$? If the former, then $1/\lambda= 200\ \text{cm}^{-1}$ corresponds to have $\hbar\omega_0/k\approx 300$ kelvin, so it makes sense that the barrier to rotation would have significant thermodynamic effects at temperatures on the order of room temperature or below, since the rotational-vibrational modes are not available. Am I getting this right?
But why is it necessary to place this fictitious rotational band on top of the $\nu=1$ state, rather than $\nu=0$? I would think that the fictitious band should be build from the ground state, it wouldn't match the actual energies at low energies, and then at high energies they would match up.

Wavenumber units are cycles-per-cm, which I think corresponds to your $1/\lambda$. So, if you're using $\omega_0$ to represent the wavenumber value, you would need to use $h\omega_0c\over k$ to calculate that representative temperature. By my math $200\mathrm{cyc\over cm} \rightarrow 288\ \mathrm K$.

The rotor potential is cyclic which means that levels in each potential well interact with one another and become split. In addition there is quantised angular momentum both about the C-C axis and perpendicular to this which leads to further sets of energy levels. The unique axis is defined by quantum number $K$.  Look for spectroscopy of symmetric  top molecules. To me it is unclear what the levels in the figure refer to. ($ k_BT \approx 208, \mathrm{cm^{-1}}$ at 300 K)

@porphyrin: To me it is unclear what the levels in the figure refer to. (kBT≈208cm−1 at 300 K) I think that rotational band is supposed to be a fictitious one that he is saying gives the right energy levels in the limit of high spin.


text:
Nice textbook illustration of microwave spectrum for a nearly free internal rotation?
Is it possible to find a nice textbookish example of experimental data for an internal rotation of a molecule, in which the rotation is nearly free? For example, 1-butene is supposed to have a nearly free rotation about its middle bond.
It would be fun, for example, to be able to show a microwave spectrum for this molecule that demonstrates the evenly spaced photon energies of a rotor.

I did find some data on NIST. Perhaps you could find more there: https://webbook.nist.gov/cgi/cbook.cgi?ID=C106989&Mask=80. Look at "Data at other public NIST sites." I was personally disappointed by the apparent lack of a nice visual spec for it, but perhaps I'm looking in the wrong place.


text:
What is the significance of antinode at r=0 in the plot of radial wave functions of s-orbitals?

Why should there be an antinode at r=0?
I can comprehend the presence of node at r=0 for p,d,f orbitals; but I don't understand that of s-orbital.
Is there any intuitive approach?

Welcome to Chemistry.SE! We have a site policy for homework questions. Please edit your question to include your attempt at the problem, where you got stuck, and let us know where you're finding difficulty so we may best help you.

possible duplicate of : 2s orbital wavefunction has non-zero probability at r=0?(duplicate) from another site on the network (physics.SE). Interestingly, that question is also a duplicate.

@ANBENZENE please do research (at least a little!) on google before asking a query, it really helps.

A nice little meta.SE post for other users in doubt

Ask yourself: what do ordinate and abscissa in the plots represent? Then follow the logical implications of a value y=0.


text:
Which of the following is the most basic compound?

Which of the following is the most basic compound?

$\ce{HS-}$
$\ce{H3Si-}$
$\ce{H2P-}$
$\ce{Br-}$

I can't figure out which of these is the strongest base. Would it be $\ce{H3Si-}$, since $\ce{Si}$ is the least electronegative of all these central elements?

The size of the atom carrying the charge is generally more important than the electronegativity.

I agree with Doctor Insult's argument that "the stronger the base, weaker is the conjugate acid." In that sense, higher the $\mathrm{p}K_\mathrm{a}$ of acid, weaker is the acid, and hence, stronger is the conjugate base. The $\mathrm{p}K_\mathrm{a1}$ values of $\ce{HBr}$, $\ce{H2S}$, $\ce{H3P}$, and $\ce{H4Si}$ are -8.7, 6.89, 27, and ~35, respectively (Ref.1). Therefore, the order of the strength of their conjugate bases should be:
$$\ce{H3Si- \gt H2P- \gt HS- \gt Br-}$$
Hence, strongest base is $\ce{H3Si-}$.
Note: Ironically, this is the same decreasing order of electronegativity of center atoms $\ce{Si, P, S, Br}$: $1.90 \lt 2.19 \lt 2.58 \lt 2.96$, respectively.
References:

William L. Jolly, In Modern Inorganic Chemistry; 1st Edn.; McGraw-Hill, Inc.: New York, NY, 1984, p 177.

Basic strength of a base is inversely proportional to the electronic negative charge density on the charged atom, because it will more readily to try to stabilize it charge by abstracting a proton. Size varies as Br>Si>S>P, so $\ce{H2P-}$ will be the strongest base. Even from the ${K_a}$ data, I concluded that $\ce{H2P-}$ will be the strongest base. Look up the ionisation constants of their conjugate acids. The stronger the base, weaker is the conjugate acid. Hope that helps.

In this case, the charged atoms are all similar in size, so the electronegativity is the dominant effect.


text:
How to remove wash anthracene oil/creostae oil from stainless steel tanker?
I once loaded anthracene oil/creosate Oil in my stainless steel tanker and I had to transport it to some factory. The material was inside the tanker for throughout the journey that was of around 8 days.
The material got unloaded at the said factory. However I had another order to load Formaldehyde from the same city. I took my truck to the usual tanker service station to wash the tank with Detergents/water/diesel/scrubbing and everything which is quiet a normal procedure after and before any other liquid chemical.
However I have never had trouble with washing any other material as much as I am having with anthracene oil/creosote oil. It just doesn't seem to come off , the servicing person had severe respiratory problem when he went inside the tank.
Even after washing with several detergents, fast water streams and other things... the chemical is just not leaving the surface of my tank. One more thing, I think the service would have been much easier if the fumes/gas inside the tanker weren't so strong and scorching to the skin of the service person. He straight away told us to never come back with the same material.
Kindly let me know how to make sure to remove the essence of the gas/fumes left in the tank due to 8 days of material in the tank, how to remove the material from the surface of the tank and how to make this process well by using any other chemicals.

Do you have any information on what the components of the creosote oil are? What is its origin?

No I don't . I am not the manufacturer. Just a transporter. Do you think mixing water with caustic soda powder and let it remain inside the tank for few hours can help?

It depends on the constituents; wood-derived creosote oil contains a lot of phenols which will dissolve in sodium hydroxide solution. Coal-derived creosote oil is mainly polyaromatics which will not, it would be more likely to dissolve in toluene or gasoline. I would take sample of it and see what it does with sodium hydroxide or toluene/gasoline.

It's coal derived. My main concern is to remove the gas from the tank that gets entrapped in the tank because of several days of material inside the tanker during transit time..what could be done for that?

Safest to flush with inert gas, nitrogen or carbon dioxide.

That is a too costly procedure. Not feasible.

@Kushgra Gupta two statements   "the fumes/gas inside the tanker weren't so strong and scorching to the skin of the service person"   and  " he servicing person had severe respiratory problem when he went inside the tank " make me wondrous ,are you using chemical protection suits with face masks while cleaning.

Anthracene oil (CAS registry number: [90640-80-5]): Both European Chemical agency (ECHA) and US Toxicology Date Network (TOXNET) classify anthracene oil as:

A complex combination of polycyclic aromatic hydrocarbons obtained from coal tar having an approximate distillation range of $\pu{300 ^{\circ}C}$ to $\pu{400 ^{\circ}C}$ ($\pu{572 ^{\circ}F}$ to $\pu{752 ^{\circ}F}$). Composed primarily of phenanthrene, anthracene, and carbazole.

Anthracene oil  is also classified as a carcinogen the risks to human health and to the environment as well (though the relevant symbol is not issued).

Creosote oil (CAS registry number: [8001-58-9]): Creosote oil is a synonym for coal tar creosotes. It is a distillation product of coal tar. Creosote oil is an oily liquid consistency and range of color from yellowish-dark green to brown.  It is a mixed compound, composed primarily of aromatic hydrocarbons, anthracene, naphthalene, and phenanthrene derivatives. At least 75% of the coal tar creosote mixture is polyaromatic hydrocarbons.
Creosote oil is heavier than water (specific gravity: 1.07-1.08) and has a continuous boiling range beginning at about $\pu{200 ^{\circ}C}$  (range: $\pu{194\!-\! 400 ^{\circ}C}$).  They are slightly soluble in water, but miscible with alcohol and ether (Ref.1).
Koppers' Safety Data Sheet has following warnings for creosote oil:

Hazard Statements:

Causes eye irritation.
May cause cancer.
May cause respiratory irritation (No wonder your servicing person had severe respiratory problem when he went inside the tank).
Very toxic to aquatic life.
Toxic to aquatic life with long lasting effects.

It also listed following precautionary statements for prevention:

Avoid breathing dust/fume/gas/mist/vapors/spray.
Wash thoroughly after handling.
Wear protective gloves/protective clothing/eye protection/face protection.
Obtain special instructions before use.
Do not handle until all safety precautions have been read and understood.
Use only outdoors or in a well-ventilated area.
Avoid release to the environment.

I put all this effort to show that how dangerous your situation is and how ignorant you have been. Your first mistake is not receiving all special instructions for handling these kind of chemicals from responsible people (see above bold-face precaution). In your case, I assumed the manufacturer of the chemicals should provide these details. What you should have done now is go back to the manufacturer, and request the instruction for clean-up procedure safely. And be aware that whoever the person following that instructions, should wear with all personal protective equipment (PPE) required, specially suitable breathing equipment.
Alternatively, since the both oils are viscous, you may slant your tanks backward to make them drain out to safety container (for long time) before starting to wash them (assuming drainage is in the back). Once drained, a person wearing PPE can go inside and wipe out inner walls with cloths soaked with alcohol several times, since both oils are miscible in alcohol. Also, kerosene or gasoline or even coconut oil may work as well (please dispose those cloths accordingly). Finally, you should use appropriate detergent (or soap) with water to remove trace of remaining hazard chemicals (if there any) and washing organic material used.
References:

B. F Hibbs, J. Georges (Agency for Toxic Substances & Disease Registry), Toxicological Profile for Creosote (updates):  Wood Creosote, Coal Tar Cresote, Coal Tar, Coal Tar Pitch, and Coal Tar Pitch Volatiles; Forwarded by Dr. D. Satcher; Prepared for US Department of Health & Human Services, 1996.


text:
Distribution of polymer lengths
Suppose we have a chemical reaction where a linear chain can form (reversibly) from building blocks (e.g. atoms or molecules):
$$\ce{X_$i$ + X_$k$ <=>[$k_\mathrm{fwd}$][$k_\mathrm{rev}$] X_$i$X_$k$}$$
where $k_\mathrm{fwd}$ and $k_\mathrm{rev}$ are rate constants.
What is the distribution of polymer lengths (at equilibrium)? What are some canonical references that derive these distributions under variations of the above polymerization process (e.g. step versus chain growth)?

I allowed myself to edit your question a bit by merging both forward and reverse reactions into one (since you are speaking of equilibrium) and I also changed notations for the rate constants to the standardized ones $k_\mathrm{fwd}$ and $k_\mathrm{rev}$ to avoid ambiguity.

When this exercise come from?

As the expression is written at present you have numerous  pairs of species in equilibrium. There is no way for them to grow. You should look for references to kinetics of radical and of condensation polymerisation. Many chemical kinetics texts cover these topics.

Concentrations are a geometric series
First, let's focus on the reactions where one of the reactants is a monomer $\ce{X_1}$:
$$\ce{X_1 + X_n <=> X_{n+1}}$$
and call the equilibrium constant for this reaction $K_{\mathrm{extend}}$.
From this, we can figure out the equilibrium constant for making a polymer from monomers:
$$\ce{n X_1 <=> X_n}$$
$$K_n = (K_{\mathrm{extend}})^{n-1} = \frac{[X_n]}{[X_1]^n}$$
So we can express the concentration of a polymer $X_n$ dependent on the monomer concentration:
$$[X_n] =  (K_{\mathrm{extend}})^{n-1} [X_1]^n = \frac{1}{K_{\mathrm{extend}}}(K_{\mathrm{extend}} \cdot [X_1])^n$$
This is a geometric series (if you take the concentration of polymers as a series, the ratio of neighbors in the series is constant). Because the amount of matter is finite, $K_{\mathrm{extend}}\cdot [X_1] < 1$ for an equilibrium scenario (as opposed to ending up with a single chain containing all the monomers). So the concentration of $[X_1]$ is the highest, and concentrations drop with increase of $n$.
Expression for $[X_1]$
Let $[X]_\mathrm{tot}$ be the total concentration of monomer, no matter whether in a polymer or not.
$$ [X]_\mathrm{tot} = \sum_{j=1}^{\infty} j[X_j]$$ $$ = \frac{1}{K_{\mathrm{extend}}} \sum_{j=1}^{\infty} j (K_{\mathrm{extend}}\cdot [X_1])^j $$
Using $$\sum_{j=1}^{\infty} j \beta^j = \frac{\beta}{(\beta - 1)^2}$$
we get
$$ [X]_\mathrm{tot} = \frac{[X_1]}{(K_{\mathrm{extend}}\cdot [X_1] - 1)^2}$$
which you can solve for the monomer concentration.
Relationship to real polymerization reactions
The polymerization reactions in chemistry and biochemistry that I am aware of are kinetically controlled. There are fibers (such as actin) in biochemistry that are dynamic (i.e. there is polymerization and depolymerization), but the processes are affected by nucleotide binding, complicating matters. I would be interested to hear whether the OP has an example that behaves like this.

here there is polymerization/depolymerization too because the end of the polymer can come off. can you say what properties of actin fibers change the dynamics you outlined?

Can you point to a book or article which contains this derivation?

I did the derivation from scratch, but here is a more comprehensive treatment that has the result as the simplest 1D case: https://hal.archives-ouvertes.fr/hal-01565995/document. @TonyRuth

@TonyRuth Here is another reference that considers real systems: https://pubs.acs.org/doi/10.1021/jp981592z


text:
Can Acids have pH of more than 7?
Consider a solution of $\text{HCl}$ with a concentration of $10^{-10}$ $\text{M}$. Now, if I find it's pH:
$$\begin{aligned}\text{pH} &= -\log([\text{H}^+])\\
\text{pH} &= -\log(10^{-10}) = 10\end{aligned}$$
At room temperature, $\text{pH} + \text{pOH} = 14 \implies \text{pOH}$ of the given acid is $4$. This means that the $[\text{OH}^-]$ ions is $10^{-4}$ $\text{M}$. But from where do these $\text{OH}^-$ ions come?

At this dilution rates you can no longer get away with using concentration instead of activity, and you have to calculate $\mathrm{pH}$ as $\mathrm{pH} = -\log a(\ce{H+}).$ Note that $\mathrm{pH}$ also heavily relies on solvent/medium.

I object. At dilutions this great, you can more than ever  rely on using concentration instead of activity. Now to the point. What is the pH of pure water?

@andselisk right answer, but to another question ;-)

In one word: Autoprotolyis

^ That said, I think this is a useful duplicate for search, concisely worded.

In acid base reactions in water there is always the reaction $\ce{H2O <=> H^+ + OH^-}$ occurring and this cannot be ignored. The equilibrium constant $K_w=10^{14}$ so the $\ce{[H^+]}=10^{-7}$ from this reaction far exceeds that from $10^{-10} $ molar acid. see answer https://chemistry.stackexchange.com/questions/100346/calculate-ph-of-a-mixture-of-a-strong-base-and-acid-knowing-only-the-ph-wt-v/100355#100355


text:
How can the potential energy at the infinite energy level in an atom be zero?
I have recently learnt about atomic structure and the Bohr model of the atom and have observed a discrepancy between it and my previous knowledge based on physics.
For simplicity, assume the atom is of hydrogen.
Total energy of electron $$E= -13.6 \left(\frac{z^2}{n^2}\right) \pu{eV/atom}$$ where $z$ is atomic number, $n$ is energy level.
Thus, at $n=\infty$, $E=0$.
Now, by the relation Total Energy = potential energy/2 (from $TE=-KE=PE/2$),
$$PE=0/2$$ so $$PE=0$$
Now, the infinite energy level ($n=\infty$) occurs at a finite distance from nucleus (as the difference in distance between the energy levels decreases at higher energy levels).
Thus, $PE=0$ at a finite distance from the nucleus.
But, I have previously studied that $PE$ of a system of a negative charge (electron) and positive charge(nucleus) is $=0$ only at an infinite distance.
Thus, the discrepancy.
Please let me know where I have gone wrong.

You incorrectly assume that when $n=\infty$ the electron and proton are at a finite separation.

Let's forget for a moment that the Bohr model is pretty rudimentary. Why do you think that the orbital radius for n-> infinite is finite? Can you explain this a little more so we can answer?

Please see the following image:  https://www.google.com/search?q=orbit+of+n%3Dinfinity&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjKsc7e7sfiAhUe4XMBHQx5CNYQ_AUIECgB&biw=1280&bih=578#imgrc=LUbuupHc87igmM:

Let me know if it makes my question clear...

Yes your question is clear, and clearly based on a wrong premise. You know the formula for energy; you plug $n=\infty$ and correctly deduce that $E=0$. But you don't know the formula for radius and can't plug $n$ there, so you have to rely on obscure googled pictures instead. That's the root of all evil.

See page 10 in https://users.physics.ox.ac.uk/~smithb/website/coursenotes/qi/QILectureNotes2.pdf

The picture is not obscure, it is a great representation of the relationship between n and E in the H atom. However, it has nothing to do with radius. The distances shown between the bars are not spatial distances, they represent the energy differences. The vertical axis is energy.

@electronpusher i think you have the best point...i have got a solution...thanks


text:
Derivation of time probability distributions for single molecule kinetics
If we have a protein $\ce{A}$ that can bind to protein $\ce{B}$:
$$\ce{A + B <=>[$k_\mathrm{fwd,1}$][$k_\mathrm{rev,1}$] AB}$$
and the bound complex $\ce{AB}$ can be converted into $\ce{C}$:
$$\ce{AB <=>[$k_\mathrm{fwd,2}$][$k_\mathrm{rev,2}$] C}$$
Assume we have a volume of 1 and some initial amount of $\ce{A}$ and $\ce{B}$ (which are $A_0,$ $B_0$) and that $C_0 = 0.$ We then measure the amount of time elapsed before the first molecule of $C$ is formed. The experiment is repeated many times, and the frequency of occurence of first $C$ during each time interval 0 to $t_1$, $t_1$ to $t_2$, $t_2$ to $t_3$, etc. is plotted, so that the result is a probability distribution of the time elapsed before the first molecule of $C$ is formed.
I would like to know what is the mathematical function that describes this distribution and how is it derived?
UPDATE: this is not a homework question and the point is not a specific answer. I would like references to literature/textbooks that derives these distributions analytically for simple examples like the above would be helpful. Have not seen a setup where distribution of duration is derived. Commenters say it's standard. Please give references.

This is a very standard set-up for a kinetics problem. The typical way involves making the steady-state approximation which says that the change in concentration of AB with time is zero (essentially because it is eliminated immediately). If one can't make an assumption like this, there typically won't be a closed-form solution. One can always solve these types of differential equations numerically though. Basically I'm saying you probably need to give more info for this to be answerable.

It would help if you could at least give the equilibrium constants for the two steps to define the distribution of species at equilibrium. "The first C" is a strange thing to ask about - the standard kinetics are about bulk properties.

@jheindel happy to assume AB doesn't change with time. do you have a link to an example where durations are computed for a problem of this sort?

not a homework question. happy with references instead of answer. closing it as off topic is silly

Here's a more basic write-up of the derivation for a unimolecular first-order reaction. Extending it to the more complex system is a matter of a bit more math using the same concepts. http://makarov.cm.utexas.edu/resources/Lecture-notes,-tutorials-etc./single-vs-bulk.pdf

@user248237 - I edited your question to make it (hopefully) more clear, so that it might get reopened. I hope you don't mind. Please correct any errors I might have introduced. It's a good question, and I don't see any reason why it should be closed.

@EdV: thank you for the clarification. is there an example derivation of the distribution of times at which some fixed percentage of moiety C was formed? it is not a standard textbook problem

To see how this calculation is done, let us consider the simpler case of just the second step shown above, ie $\ce{AB ->[k_{f2}] C}$. Since we are interested in the situation where no $C$ has yet formed, the reverse reaction can be ignored. I will assume that this reaction is first order in $AB$, so $k_{f2}$ has units of $s^{-1}$.
In this case, the rate constant $k_{f2}$ represents the average frequency of the reaction.
Now let's consider that we establish an infinite number of evenly spaced time intervals of length $\Delta t$. When we consider the probability distribution, we need to consider two factors:
1) The probability of reaction occurring in an interval $t_n + \Delta t$ and
2) The probability that the reaction has not yet occurred at $t_n$
The first value is simply the product of the rate constant $k_{f2}$ and the length of the time interval: $k_{f2}\Delta t$.
The second probability is slightly more complicated, since it is the chance of the reaction $not$ occurring in each of the $n$ intervals preceding $t_n$. That probability is given by $1-(k_{f2}\Delta t)^n$. Conveniently, as our time interval $\Delta t$ goes to $0$, this probability becomes a continuous function $e^{-k_{f2}t}$.
To get our final probability, we simply multiply these two probabilities:
$[\text{total probability of reaction occurring in interval }t+dt] = [\text{probability of occurring in interval }t+dt] \times [\text{probability that reaction has not occurred by time }t]=k_{f2}dt\times e^{-k_{f2}t}$
The probability as a continuous function of $t$ therefore can be written as
$p_{\ce{AB->C}}(t)=k_{f2}e^{-k_{f2}t}$.
As expected, integration from $t=0$ to $t=\infty$ gives an overall probability of $1$.
Reference: Adapted from the lecture notes published by Dmitrii Makarov here: http://makarov.cm.utexas.edu/resources/Lecture-notes,-tutorials-etc./single-vs-bulk.pdf
For the two-step process, if $\ce{A+B <=> AB}$ establishes equilibrium very fast relative to the $\ce{AB->C}$ reaction, then the above is a reasonable approximation (after adjustment of $k_{f2}$ to reflect the equilibrium concentration of $AB$). If the equilibrium is not established quickly, then you would use the same reasoning to determine the probability first of $AB$ having formed by time $t$ and then the probability of it reacting based on how long it has been in existence.
I have not read Makarov's textbook on single molecule kinetics, but I assume it covers these kinetics in much greater detail and might be a good resource:
Single Molecule Science: Physical Principles and Models
1st Edition by Dmitrii E. Makarov; CRC Press 2015; ISBN 9781466559516

So with the combination of the two reactions as given in the OPs question, would there be a lag time (for AB to build up) and then exponential decay?

Yes, instead of a y-intercept of $k_{f2}$, the probability would start at 0 and increase with the concentration of AB.

@Andrew: thank for un-closing my q and for very helpful explanation

@Andrew: quick clarification: shouldn't you take into account not only the rate $k_{f2}$ but also the amount of $AB$ you have formed thus far in time? the formation of $C$ should depend on the rate and on that abundance

Yes, the k should be scaled by a factor [AB]/[1 M] for concentrations other than 1 M. I left that out for simplicity, but I should have noted it.

This is an alternative derivation to that given by @Andrew in his answer for the same first order reaction.
We suppose that the species has a probability $p$ of reacting in time interval $\Delta t$ and that this is independent of any past history, i.e. it depends only on the length of $\Delta t$ and for a sufficiently short time is proportional to this time, i.e. $p=k\Delta t$. The probability of not reacting in the time interval is $1-p=1-k\Delta t$.
If there is no reaction in this time interval  and also in the next, the chance of surviving both is $(1-k\Delta t)^2$; and, by induction, for $n$ such intervals is $(1-k\Delta t)^n$.
If the total time $t$ is defined as $t=n\Delta t$ then the survival probability is $(1-kt/n)^n$. The chance of surviving unreacted at time $t$ is just this value when $\Delta t$ is made infinitesimally small, that is in the limit of $n \to \infty$ and this produces $e^{-kt}$ for the limiting value, when we recall the definition $e^x= \lim_{n\to \infty} (1+x/n)^n$.
As we do not consider just one molecule but many (initially $n_0$) the fraction remaining unreacted ($n$) at time $t$ is $\displaystyle \frac{n}{n_0}=e^{-kt}$.
We can also calculate the probability $q(t)$ of a molecule lasting a time interval that has a value between $t \to t+dt$. This is calculated as the product of not reacting between $0\to t$ and of reacting from time $t\to t+dt$. The first of these is $e^{-n_0kt}$ and the second, the probability that any of the $n_0$ molecules react in time $dt$ is $n_okdt$, thus the total probability is $q(t)dt=n_0ke^{-n_0kt}dt$. To test this experimentally a large number $s$ of time intervals of short length $\Delta t$ are needed then  the number $t\to t+\Delta t$ should be $sn_0ke^{-n_0kt}\Delta t$. A plot of the log of the number of such intervals vs time should be linear.
Notes:
The chance of not reacting $0\to t$ can be found using the Binomial Distribution. This describes the very general case that given a large set of objects in which the chance of having an object of type $w$ is $p$, and then $n$ objects are withdrawn then the chance that exactly $r$ of these are of type $w$ is $\displaystyle W(r)=\frac{n!}{(n-r)!r!}p^r(1-p)^{n-r}$.
In our example we will have $m$ reactions in time $t$ from $n_0$ initial molecules. We make the connections $n_0\equiv n$ the number chosen or withdrawn, and $m\equiv r$ where the $m$ have the property that these many react in time $t$, thus  $\displaystyle W(r)=\frac{n_0!}{(n_0-m)!m!}p^m(1-p)^{n_0-m}$. Furthermore, the chance of not reacting in time $t$ is  $(1-p)$ and this is equal to the number that survive unreacted to the total number which is $n/n_0$, and is derived above, thus $1-p=e^{-kt}$.
Thus the chance of $m$ reacting is $\displaystyle W(m)=\frac{n_0!}{n_0!0!} (1-e^{-kt} )^m (e^{-kt})^{n_0-m}$.
In the case of no reaction $0\to t$ the probability is  $W(0)=e^{-n_0kt}$


text:
Can Rydberg constant be in joules?
In my textbook (Chemistry Part - I for Class XI published by NCERT), there is an equation for the energy of an electron in an energy state: $$E_n = -R_\mathrm H\left(\frac{1}{n^2}\right)$$ and there is a paragraph below it with the following text:

where $R_\mathrm H$ is called Rydberg constant and its value is $2.18\times10^{-18}\ \text{J}$.

There is another section with the expression for the wavenumber ($\overline{\nu}$):
$$\overline{\nu}=109\,677 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\ \text{cm}^{-1}$$
with a paragraph with the following text:

The value $109\,677 \space\text{cm}^{-1}$ is called the Rydberg constant for hydrogen.

I checked online and found that in most (all) websites (incl. Wikipedia), the value of Rydberg constant is $109\,677 \space\text{cm}^{-1}$. But when I searched for its value in joules, I found this website with the value of Rydberg constant $= 2.18\times10^{-18}\ \text{J}$.
How can Rydberg constant be written in joules?

How can a price of a cheeseburger be written in ¥? Well, just like that. Rydberg constant is energy, and Joule is energy, so what's the problem?

Look at the two different equations. Consider the dimensions in each case, and why that means in one case the Rydberg constant must have dimensions of Energy, and in the other it must have dimensions of wave number. Now what relations do you know relating Energy and wave number?

Although wavenumbers are not strictly energy you will see from the value in joules that it is much more convenient to use wavenumbers. The conversion is $ 1 ,\mathrm{cm^{-1}}\equiv 1.986\cdot 10^{-23}$ J.

Authors may be sloppy about notation in this matter. I recommend considering $R_\ce{H} \approx \pu{10973 cm-1}$ and $Ry \approx \pu{2.18e-18 J}$, noting $Ry = hc \cdot R_\ce{H}$. Units of wavenumbers $(\pu{cm-1})$ and energy are often considered interchangeable in practice because they are proportional to each other by the constant value $hc$.
In my notes, I would always be sure to write $R_\ce{H}$ or $Ry$ to explicitly remind myself "which" Rydberg constant I was using (in fact I merged the R and y into a single symbol because I didn't like the suggestion of multiplication.)
Note also that there is a unit of energy known as a Rydberg, with $\pu{1 Ry} = Ry = hc \cdot R_\ce{H}$.

Check those numbers, maybe a digit is missing in $R_H$?

Rydberg constant $R_∞$ is usually given in reciprocal length units historically and because it's determined from hydrogen and deuterium transition frequencies [1].
Current value (in $\pu{m-1}$) is listed at NIST [2] website (accessed 2019-06-05):
$$R_∞ = \pu{10973731.568160(21) m-1}$$
Since it's an energy unit, one can convert it to SI rather trivially via multiplying the value in reciprocal length units by $hc$ ($h$ is the Planck constant; $c$ is the speed of light in vacuum):
$$E = hν = \frac{hc}{λ} \quad\text{or}\quad R_∞[\pu{J}] = hc\cdot R_∞[\pu{m-1}]$$
resulting in the following value:
$$R_∞ = \pu{2.1798723611035(42)e-18 J}$$
References

Mohr, P. J.; Newell, D. B.; Taylor, B. N. CODATA Recommended Values of the Fundamental Physical Constants: 2014. Reviews of Modern Physics 2016, 88 (3). https://doi.org/10.1103/RevModPhys.88.035009.
Tiesinga E.; Mohr P. J.; Newell D. B.; Taylor, B. N. "The 2018 CODATA Recommended Values of the Fundamental Physical Constants" (Web Version 8.0). Database developed by J. Baker, M. Douma, and S. Kotochigova. Available at http://physics.nist.gov/constants, National Institute of Standards and Technology, Gaithersburg, MD 20899. 2019.

In spectroscopy and related fields it is common to measure energy levels in units of reciprocal centimeters (e.g., IR and Raman spectroscopy). Strictly speaking, these units ($\pu{cm^{−1}}$) are not energy units, but units proportional to energies, with $hc$ being the proportionality constant (Wikipedia). In general, $hc$ can be attributed to the value $\pu{1.986E-23 J cm}$. Hence:
$$R_∞ = \pu{109677 cm^{−1}}$$
$$\pu{1 Ry} = \pu{109677 cm^{−1}} \times hc = \pu{109677 cm^{−1}} \times \pu{1.986E-23 J cm} = \pu{2.178E-18 J}$$


text:
What is a molecule’s smallest set of smallest rings?
It is been a while since I've studied chemistry. Now, I am reading the documentation of RDKit, where they mention the expression "smallest set of smallest rings" (SSSRs). I understand that a ring is a cycle in a molecule. I also understand that the SSSRs is, as the name suggest, the smallest set of smallest rings, but how do you determine what the smallest sets and rings are?  What is a molecule's smallest set of smallest rings? What is a SSSRs used for?

This is a tool to classify molecules with multiple rings. Here is an example from a PhD thesis:

So even though you can find six-membered and eight-membered rings in this structure, these are made up of multiple four-membered rings, and are thus not parts of the set of rings. As the picture illustrates, it is sufficient to choose five rings to show the connectivity of the atoms, so even though there are six four-membered rings, the SSSR includes only five of them. In this case, which five is ambiguous.

how do you determine what the smallest sets and rings are?

It turns out that this is ambiguous, so different software might get different results. In contrast, giving the size of all rings is unambiguous, but - according to the thesis I quoted - computationally more expensive and chemically less intuitive.

What is a SSSRs used for?

To classify cyclic molecules in cheminformatics. For a given software, two molecules are different if they have different SSSRs. Conversely, two identical molecules will have the same SSSR. This makes it easier to classify, sort, and search this class of molecules.

I think you mean "identify" instead of "classify". But I don't get this. If the SSSR is ambigous, that makes it rather useless for identification.

@Karl Yes, rather useless or worse: https://www.ics.uci.edu/~dock/manuals/oechem/pyprog/node123.html


text:
What is kekulization (in RDKit)?
It's been a while since I've studied chemistry. Now, I am reading the documentation of RDKit. At a certain point, the term "kekulization" is mentioned. What is kekulization (in RDKit, if this is not a standard chemical term)?

This is a method to generate alternate Lewis structures of the same molecule. Here are two examples from the cumulative dissertation by Sascha Urbaczek[1]:

If you were to search for the left molecule in panel (1) using an image search or a SMILES string, you might miss the right molecule in that panel.
According to the RDkit document cited in the question, the software routinely generates the alternate position of double bonds, and then (in a second step they call "aromatization") labels the ring as aromatic. In panel (2), there are three possible Lewis structures contributing to the actual structure (i.e. there is resonance), so the software would have to generate all three to be able to search for identical structures.

Urbaczek, Sascha. A consistent cheminformatics framework for automated virtual screening. Ph.D. Thesis, Universität Hamburg, August 2014. URL: http://ediss.sub.uni-hamburg.de/volltexte/2015/7349/; URN: urn:nbn:de:gbv:18-73491; PDF via Semantic Scholar

I do not really get the connection that led to the naming, but fwiw i suppose this is named after Kekulé? https://en.wikipedia.org/wiki/Kekul%C3%A9_Program

@bukwyrm Kekule of benzene structure fame. Alternate positions for double bonds, or delocalising the double bond electrons into hybrid orbitals.

SMILES also exist in kekulized and non-kekulized forms. The kekulized form makes the double bonds in aromatic rings explicit. Instead, the normal (canonical) SMILES omits the double bond symbol = in favor of lowercase characters to denote aromaticity. For example:
Canonical SMILES:
CC(C)(C)c1cc(O)ccc1O

Kekulized:
CC(C)(C)C1=CC(O)=CC=C1O


text:
Building an average model from MD snapshots
I am performing a molecular dynamics simulation, and I am taking "snapshots" of the system at regular intervals.
The aim is that of finding the preferred geometry of a molecule in solution.
Are you aware of any software capable of automating to some extent the superposition of different geometries, and maybe of building an "average" structure?

I think what you are asking is a fairly difficult and likely not automatable procedure as the details will vary a lot depending on what you are interested in and the system you are studying. The simplest (but not necessarily simple) thing to do is just define a set of internal coordinates and transform your cartesians to these and average over your snapshots.

I have also been a part of a project where we were interested in averaging over a particular coordinate that is sensitive to low frequency vibrations which themselves are sensitive to thermal fluctuations. So, we first performed a Fourier filtering over these low frequency vibrations and then averaged the ones we were actually interested in. I'm just trying to get across that doing this in a reliable manner can be highly non-trivial and is certainly not unique as the systems get bigger and which coordinates are good ones becomes less obvious.

Indeed, averaging the coordinates was my idea as well, but I hoped in a less time-consuming procedure :)

Ya, unfortunately defining coordinates is the bane of many computational chemists' existence. It is time-consuming and often has to be done by hand if you want to be able to easily interpret the chemical relevance of the coordinates. The actual transformation from one set to another isn't too bad, but you have to write the code to do it at least once. I don't know of any well-established code to do this, but maybe there's a github repo somewhere...

If you want to get really modern you could try to use some kind of unsupervised learning to pick out important motifs and then average over these somehow. See: Ceriotti, M. (2019). Unsupervised machine learning in atomistic simulations, between predictions and understanding. The Journal of chemical physics, 150(15), 150901.

For people interested, I have not tried it yet, but it looks like there is a script for PyMol (free for academic purpose). http://muralab.org/~cmura/PyMOL/pymol_mainFrame.html PyMol also allows alignment of different stuctures

GROmaps may be an option for you too GROmaps

This is possible with the Python package MDTraj. Assuming you have your trajectory saved as trajectory.dcd and topology in topology.pdb:
import mdtraj as md
from mdtraj.geometry.alignment import compute_average_structure

traj = md.load('trajectory.dcd', top='topology.pdb')
average = compute_average_structure(traj.xyz)

I believe it will RMSD align the trajectory first, but you may need to specify that explicitly first.
You might also want to check out MDAnalysis as well, it may have similar functionality. I've used both packages and both are still actively maintained.


text:
Shouldn't it take more energy to break CO2 compared to CO?
I read in a book that bond dissociation energy of carbon monoxide $(\ce{CO})$ is $\pu{749 kJ/mol}$ and carbon dioxide $(\ce{CO2})$ is $\pu{532 kJ/mol}.$ Shouldn't it take twice as much energy to break $\ce{CO2}?$

532 kJ/mol times two...

I'm voting to close this question as off-topic because your question lacks context. This question needs revision before it is ready for a great answer. Please edit to include how this question came up and how you tried to answer it. This will help writing an answer that is useful for you and for others.

The bonds are not equivalent, therefore they have different bond energy.

Carbon monoxide consists of one carbon atom and one oxygen atom, connected by a triple bond that consists of two covalent bonds as well as one dative covalent bond.It is the simplest oxocarbon and is isoelectronic with other triply-bonded diatomic molecules having ten valence electrons, including the cyanide anion, the nitrosonium cation and molecular nitrogen. In coordination complexes the carbon monoxide ligand is called carbonyl.

Source: Wikipedia page on Carbon Monoxide

The bond dissociation refers to each bond.  Therefore, the number you cited for $\ce{CO2}$ is for only one of the C=O bonds.

Look at it simply. Bond strength is proportional to the number of bonds present. CO has a triple bond which is consisted by 1 sigma and 2 pi bonds. This kind of covalent bond is relatively hard to break compared to a single and double bond.


text:
Things to consider to determine stability of a compound to another?
Some Compounds with higher bond dissociation energy seem to be more reactive towards certain compounds then ones with lower bond dissociation energy , so what are the things you need to factor in to know if a compound will show reactivity with another?

BDE's refer to homolytic bond strength and not heterolytic bond strength. While the BDE of the C-H bond of acetylene is much higher than that of ethane, the pKa of the acetylene C-H bond (~25 kcal/mol) is more acidic than that of ethane (~50 kcal/mol).

First, you need to determine the relative stability of the compounds. Usually, this can be known by comparing the preferred state of the element or compound to its current state. For example, metals commonly want to donate electron so seeing a negatively charged metal should give a hint that the molecule/atom you're dealing with is unstable.
Second, the spontaneity of the reaction is a big help in determining if the two molecules would atleast react (the spontaneity is given by how negative the change in gibbs energy is).
Third, compounds with variety of resonance structures prefer to be themselves because the electron density is delocalized in the molecule.
Lastly, reactivity of the reactants is accompanied by energy. Presence of heat, current, work, etc. usually pushes the equilibrium of the reaction to favor the formation of the product. However, not all reaction have this kind of characteristics especially exothermic ones.


text:
Product of this Michael addition
I'm having problems when trying to know the outcome of this reaction.

The first step must be a Michael addition of the 1,3-dicarbonyl compound to the  α,β-unsaturated carbonyl compound, followed by a acid hydrolisis and a decarboxylation of the ester.

LDA must deprotone the least hindered hydrogen, that must be the $\alpha$-$\ce{CH3}$ to the non cyclic ketone. But at this point, I don't know how this reaction can continue to give one of the above products.

Intramolecular aldol reaction?

@user55119 So it was as easy as a Robinson annulation, but the LDA step was driving me crazy, it makes sense now, thanks.

I don't think you need LDA to finish an aldol condensation. The first step probably effect the Robinson annulation with the ester group intact. Acid hydrolysis would lead to CO2, methanol and compound C because the precursor is a vinylogous beta-ketoester. Probably a made-up problem. –

Although the work of Wilds and Werth1 does not align perfectly with the conditions of the OP's question, it does illustrate that Robinson annulation and decarbomethoxylation can occur under alkaline conditions. Catalytic base was found to be optimal to effect the Michael addition with methyl vinyl ketone (MVK). The aldol condensation was conducted in aqueous KOH at reflux.
Whether decarbomethoxylation occurs prior to aldolization (2-->3-->5) or subsequently (2-->4-->5) was not addressed. α'α-Dialkylated β-ketoesters generally undergo attack at the keto group rather the carboxylate by base. This generality may be tempered by the aromatic nature of the keto group in 2. Initial aldolization of 2 to 4 affords a structure that is a vinylogous β-ketoester; it is susceptible to base initiated decarboxylation.

1) A. L. Wilds and R. G. Werth, J. Org. Chem. 1952, 17, 1154.

The OP's original question has some flaws. Since he  is asking the most possible product among the given four, I'd give reasonable mechanism to achieve that product, which would be C. Following is the suggested mechanism:

Late addition: Evidently, lithium diisopropylamide (LDA) has been used to selectively abstract proton from the least steric position among three available acidic hydrogens (irreversible generation of the kinetic enolate). Such a non-nucleophilic, sterically-demanding, strong base will always abstract a proton from the least hindered side. The resultant lithium enolates will avoide Proton transfer at low temperatures in ethereal solvents, so that addition of a second carbonyl group will produce the desired aldol product (Organic Chemistry Portal). For an example for use of lithium dialkyl amide in aldol condensation, see the given reference:
Atsushi Seki, Fusae Ishiwata, Youichi Takizawa, Masatoshi Asami, "Crossed aldol reaction using cross-linked polymer-bound lithium dialkylamide," Tetrahedron 2004, 60(23), 5001-5011 (https://doi.org/10.1016/j.tet.2004.04.026).

Mathew Mahindaratne: Do you have a reference for the use of LDA in the context of your example? If so, could you post it in your answer?  Conceptually, it works! Practically, does it? I'm curious. PS: In your fourth structure, the keto form is undoubtedly better than the enol form.

@user55119: I have added a paragraph and a reference for your request.


text:
Enthalpy of combustion of 1-butanol
1-butanol $\ce{C4H10O}$
So I have absolutely no idea how to find the enthalpy and I did some looking on the internet with no use since I don't understand.
I came here hoping someone generous could do the sample calculations so I can understand and do the rest my calculations and test for my experiment.

That's the data for one the mixture of alcohol.
How would I figure out enthalpy for test 1?
(maybe I could cut out test 2, 3 and average in the picture since its irrelevant in figuring out the enthalpy of test 1, and causes confusion)
For temperature before and after. The solution was used to heat up 200 ml of water. By burning the mixture, igniting it.

Welcome to Chemistry Stack Exchange. Can you clarify two things about your question? You want to figure out "enthalpy", but determining the absolute enthalpy is challenging. I assume you are trying to calculate the enthalpy change. Can you verify that is correct? Also, can you tell us a little bit more about this experiment? What procedure are you following? That will help us provide a better answer that is more specific to your situation.

Yes, I did 5 different mixture of alcohol and water. and we are seeing if the enthalpy changes. That is all. there is 4 more table like this for different mixtures and that is it. I will compare the enthalpy of each solution and talk about it.

45 ml of butan-1-ol + 15 ml of water do not have a mass of 156.3–156.5 g. Does the mass include the container? (It also cannot include the additional 200 ml of water that are heated during the experiment.)

Answering your question as it stands is impossible, as we readers do not have enough information about your actual experiment. What do you mean with 'enthalpy of 1-butanol'? You can only measure changes in enthalpy, so you need to specifiy what 'change' you're looking at.
You could mean the enthalpy of evaporation or the reaction enthalpy of some chemical reaction or hundreds of other possibilities. It's essential to clarify what process you're referring to, so one can associate the corresponding starting (e.g. liquid) and end state (-> gas) to this energy change.
However, you stated that 200 ml of solution were heated up by whatever your process was. At constant pressure, the enthalpy required to change the temperature by $\Delta T$ is
$$\Delta H = m c_p \Delta T$$
where $c_p$ is the isobaric heat capacity. Knowing the value of $c_p$, you can calculate this energy change. Now where did that energy come from? Presumably, from your process. At this point, assume that there are no energy losses (f.e. neglect the heat lost to the surrounding air). Then by conservation of energy, the enthalpy increase measured by the temperature increase must be equal in size to the energy released by the process.
Finally, note that by convention we say if a process releases energy, it's enthalpy change is negative (we could've just done it the other way around; think of it as the same idea as defining a sea level).

I ignited the solution made a fire that heated up the water. the mass of the fuel was before and after.

Yes so what you are trying to measure is a enthalpy of reaction (combustion energy). Calculate the energy change as outlined, and then divide that by the mass of fuel that has burnt and you obtain a value of energy released per gram burnt. Best practice would be to calculate this for each test seperately, then average only the final values for the energy content.


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Can we tell if a compound is explosive by just looking at its chemical structure?
If I showed you a compound's chemical structure will you be able to tell if it can be used as an explosive or we can only know that by conducting chemical experiments?

Some materials you wouldn't think of as explosive (depending on your background) are very explosive under the "right" (or very wrong) circumstances.

As @BuckThorn pointed, it depends on the conditions in which the compound has been put and for unknown or new ones, you will have to experiment. There is no known (none to me atleast) way of telling whether something's explosive.

https://chemistry.stackexchange.com/questions/7587/what-makes-substances-explosive?noredirect=1&lq=1

Looks like more precise dupe of https://chemistry.stackexchange.com/questions/117195/can-we-tell-the-chemical-properties-of-a-compound-with-its-chemical-structure

It's not like "no way"; no, there is a way, but it is far from absolute.

Surely looking at cubanes or tnt at least one suspects something.  Experiments are nevertheless necessary, also because something that might be explosive enough to be potentially dangerous doesn't necessarily qualify as an explosive in a technical sense (mining, weaponry, demolition).

A compound, a mixture may explode under different conditions.  As an example, for some of them it is heat, other require electric discharge or mecanical impact to decompose.  There are classes of compounds with known examples of explosives (azides, picrates, fulminates) so one may infer new compounds within these classes equally might be explosives, too.
Among modern explosives, increasing the number of nitrogroups or / and increase of nitrogen content over other elements present is a theme, like octanitrocubanes.  See the Klapötke group in Munich / Germany working on 5,5’-azotetrazolate salts, too.

(dihydrazinium 5,5’-azotetrazolate dihydrazinate, source)

In short, things are explosive if they contain lots of nitrogen. For example, C2N14 (1-diazidocarbamoyl 5-azidotetrazole) is highly explosive because of its high nitrogen content. This is because nitrogen gas is very stable, so when compounds release it, they also release a lot of energy. The nitro group is often found in explosives because of this and because it contains oxygen, which can react and oxidize the rest of the molecule.
We can also look at bond strength. For example, in peroxides the O-O bond is weak and can easily break, releasing reactive radicals. Many peroxides, such as ascaridole, are explosive.
Also, there are steric effects, which occur when parts of a molecule are close together. Nitrogen triiodide is explosive because its three iodine atoms are very close together and are under lots of strain.
There are many exceptions to these rules. For example, urea, despite containing nitrogen, is not explosive.

Being explosive is ill defined as context matters
Part of the issue with what is explosive is the context of the situation where the substance explodes. Flour (the sort used for making bread) is not normally thought of as an explosive, but serious explosions in four mills are known and easy to recreate in the lab (finely divided flour dispersed in air is an explosive mixture).
This illustrates that "being explosive" is not an entirely simple idea.
So, even if we stick to the narrower question of whether a single pure compound is explosive, it isn't always entirely clear what we mean.
Some classes of compounds are usually explosive. Azides and fulminates often are. Highly nitrated compounds often are, but then again, circumstances matter. Guncotton, for example, is highly nitrated cellulose. It has been used as a propellant in firearms (where it deflagrates–which is an explosion travelling slower than the speed of sound in the substance). early billiard balls made from it occasionally exploded with little provocation during normal play. But as a loosely aggregated ball of cotton or a thin film of paper, it doesn't explode, it just burns quickly.
The military explosive C-4 clearly is an explosive: armies use it as a demolition charge. But they have also been known to just burn it in the field to heat rations. It will explode but needs a good detonator to trigger that behaviour.
The point it that the situation and context matters for what will explode.
So, while there are many classes of compounds that are likely to be explosive in some contexts, it is a little hard to generalise. Highly nitrated compounds are often explosive but some will explode at the slightest provocation or when there is a full moon (like nitroglycerin); others need the right context (like TNT).
Many highly nitrogen rich compounds are explosive and highly nitrated, nitrogen rich compounds are in a class of their own And Derek Lowe explains in his "Things I Won't Work With" series talking about Hexanitrohexaazaisowurtzitane:

Hexanitro? Say what? I'd call for all the chemists who've ever worked with a hexanitro compound to raise their hands, but that might be assuming too much about the limb-to-chemist ratio. Nitro groups, as even people who've never taken a chemistry class know, can lead to firey booms, and putting six of them on one molecule can only lead to such.

So, while there are classes of compounds that are very likely to be dangerous and very likely to explode given their structure, there is a big grey area where compounds can be explosive in some contexts but not others and some where it is almost inevitable that they will be dangerous.
Thank goodness there are some people who are willing to do the experiments on this rather than you.


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Reduction of conjugated double bond by sodium borohydride
Some of my teachers were debating about whether a double bond in conjugation with a ketone is reduced or not by $\ce{NaBH4}.$ One of my teachers presented the evidence from Clayden which had an example where a cyclic ketone was reduced alongside the conjugated double bond.
However, there was no example for a straight chain compound. Also, the book said that steric hindrance may affect the reduction process and only favour the reduction of ketone.
I want to know that is there any authentic source which has an example of reduction of a conjugated double bond and ketone via $\ce{NaBH4}.$

Scifinder shows lots of reactions such as you describe. Often (but not always) copper is used to help favour the 1,4-addition (alkene reduction) process.

The simplest example I can see is the reduction of 3-penten-2-one[[2]] - when treated with NaBH4 the majority (89%) is reduced at the ketone only, while 11% of the doubly-reduced product is formed:

Obviously different substrates would give different product ratios, but certainly this is a plausible (and precedented) reaction.

Thanks a lot. But why does it have such a low yield in the example you gave. I don't see much steric hindrance. Do you know the reason?

Related question: https://chemistry.stackexchange.com/questions/87138/why-does-lialh4-reduce-the-double-bond-of-cinnamaldehyde-whereas-nabh4-does-not/87195#87195 As a rule, NaBH4 reduces 1,2.


text:
Can you use a photon's torque to change a molecule?
Can you use the torque of a photon to change bond angles, etc.? 11-Cis-Retinal to the all trans version, in  the eye, is a good illustration [1].
Note: This phenomenon can be observed by closing blinds, allowing the sun to shine through, you get light beams. If you blow a smoky material past it, you can observe this phenomenon of light.
References

Rego, L.; Dorney, K. M.; Brooks, N. J.; Nguyen, Q. L.; Liao, C.-T.; Román, J. S.; Couch, D. E.; Liu, A.; Pisanty, E.; Lewenstein, M.; et al. Generation of Extreme-Ultraviolet Beams with Time-Varying Orbital Angular Momentum. Science 2019, 364 (6447), eaaw9486. https://doi.org/10/gf4f6h.

Molecules are solid, they don't change. Except the rare cases when they do. Can you change a piece of metal with your bare hand?  Normally no, but sometimes yes. Same thing here.

Probably not, normally the total energy of a photon is needed to produce an excited state, (by being absorbed by the molecule), and from which isomerisation/bond breaking  can occur. This is the case in your example where 2 to 3 eV are needed, and the torque is not likely to be of sufficient energy. You should ask a question in Physics StackExchange to better understand what this torque actually is, as opposed to circular polarisation, for example.

Hey Ivan, thanks for the information!

Porphyrin- I am not trying to get the photon to be absorbed by an electron. I am wondering if bombarding a specific bond with photons with torque could change the orientation of a bond. i only provided the study to give more information. Thank you! Just trying to figure things out.

What makes you think that the reference you've provided is relevant to 11-Cis-Retinal?

The torque could be used to align whole molecules, see the info at CFEL in Hamburg https://www.controlled-molecule-imaging.org/, and corkscrew -laser technique

@porphyrin At technique uses optical spin angular momentum (which is tied to the polarization), not the orbital angular momentum (tied to the spatial dependence) which is at play in the referenced publication, i.e. the two have very little to do with each other. They're similar, but they operate at vastly different length scales - OAM will only couple to objects at the size of the wavelength.

@Emilio Pisanty Thanks for clarification.

Hey Guys, i only added the study to show photons can have a torque associated with them. That is really all. The same answer basically for the question about 11-cis-retinal. I only included it to give an example. Thanks guys!

What relevance does this have to spin?? "This phenomenon can be observed by closing blinds, allowing the sun to shine through, you get light beams. If you blow a smoky material past it, you can observe this phenomenon of light."

I think it is cool to observe natural phenomena that surrounds us. I would say with 95% confidence that the beam of light created by the method i described has photons with a torque. I think it confirms the experiment i linked to since you can actually see the phenomena as the study i linked originally mentioned states when interacting with matter. I can literally see the movement.
https://www.nature.com/news/people-can-sense-single-photons-1.20282

@kevindelahoussaye If your core question is whether optical forces are sufficient to directly twist a molecule, then your reference acts in a way to make it directly harder for your question to be answered. (This is because the reference you provided is highly suboptimal to addressing molecular structures compared to e.g. optical centrifuges.) If you're asking about the applicability of that specific paper to twisting molecules, then ask explicitly. (Full disclosure - I'm an author in the reference you cited.)

Hey Emilio, Thanks for the response! As mentioned above the only reason i included your paper was to show photons can have a torque. I believe your paper does that. I am not asking about the applicability of your paper to twisting molecules; only to show they have a torque. Which i then wondered if you could use the torque on photons, that exist in plenty in the world, to change a bond. I do not know of any techniques to produce photons with a certain torque. That is also another reason. I think it is probable that photons can change orientation given it seems to happen in our bodies. Thanks!

Sorry if my reference skills are not up to par. I was not even aware of optical centrifuges. Thank you for your contribution! Please pass along my warmest regards.


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Activators for non-sulphide compounds in froth floatation
I was recently studying about Metallurgy.
Concise Inorganic Chemistry by J. D. Lee claims the process of froth floatation can also be applied to non-sulfide ores such as Malachite ($\ce{CuCO3.Cu(OH)2}$) and Anglesite ($\ce{PbSO4}$) by using $\ce{Na2S}$ as an activator.
The coatings formed are $\ce{CuS}$ and $\ce{PbS}$ respectively. However, in a previous example, they used $\ce{CuSO4}$ as the activator over a $\ce{ZnS}$ surface and a coating of $\ce{CuS}$ was formed.
Therefore, going by the same co-relation in the case of Malachite and Anglesite, shouldn't the coating be a sodium compound instead?

Partly related: Froth floatation process - why CuSO4 is activating floating character of ZnS?

Sodium compounds are soluble, they won't stay.

A previous comment noted that sodium compounds are generally soluble and would not be strongly adsorbed on malachite and anglesite.
However, the other compound formed in the reaction between Na2S and a copper or lead salt would be CuS or PbS, both of which are very insoluble. Three cases can be visualized: 1) a sodium ion approaches the Pb or Cu salt and this aquated sodium ion forces off a lattice-bound dication, or 2) a sulfide ion approaches the Pb or Cu salt and either deposits on the crystal surface, forcing a nearby OH-, CO3-- or SO4-- to leave in order to maintain electrostatic neutrality, or 3) the low (but not zero) solubility of the Cu and Pb salts allows precipitation of PbS or CuS as tiny crystals which more frequently than not become bound to the original salt by epitaxial considerations.
In the case of CuSO4 applied to ZnS, ZnS is about 20 times more soluble than CuS, so an exchange of cations will occur, with the copper going into the insoluble phase.
Case 1 seems far-fetched. Cases 2 and 3 are similar, and may occur simultaneously. The key is that the insoluble product is what sticks to the material that you want to float. And the surfactant that forms the bubbles sticks to the activator (CuS or PbS).
It's interesting that ZnS is not easily flotable. ZnS in water may acquire an aquated layer, that is, it is activated by water, which is not selected by the surfactant, so it sinks. Both the copper carbonate and lead sulfate would seem to have surfaces more compatible with water than their sulfides, too, so converting the surface is required.


text:
How to calculate the number of atoms in a gas molecule?

$\pu{20 mL}$ of sulfur vapour at $\pu{1000 °C}$ reacts with $\pu{40 mL}$ of oxygen gas to form $\pu{40 mL}$ of sulfur dioxide gas. Determine the number of atoms in the molecule of sulfur in the vapour state at this temperature.

My working out is as follows:
$$\ce{\underset{\pu{20 mL}}{S(g)} + \underset{\pu{40 mL}}{O2(g)} -> \underset{\pu{40 mL}}{SO2(g)}}$$
I then used the formula $n = m/M$ to find the amount of substance.
$$n = m/M = 20/32.07 = x$$
Then, to find the number of atoms, I used the formula $n = N/N_\mathrm{A}$.
$$N = nN_\mathrm{A} = x\cdot\pu{6.022e23} = \pu{3.80e23}~\text{(atoms)}$$
However, I am uncertain as to whether this is correct as I did not utilize any of the gas laws. For this question, which of the gas laws should I have used?
Any answers would be appreciated.

Welcome to our site! BTW I noticed that you used the volume $(20\pu{mL})$ of $\ce{S}$ in the place of mass "$m$". Are you sure you want it that way?

@WilliamR.Ebenezer Should I be using one of the gas law equations to solve for "n" and then use the formula N=N/Na to solve for the number of atoms?

Instead of using any 'gas laws', why don't you just remember the equation $pV = nRT$?

I'm afraid I misread the question, further fueled by seeing your usage of Avogadro's number.

I have rewritten the answer, sorry for the confusion caused earlier.

@WilliamR.Ebenezer That's alright.

The question isn't about counting all the atoms of sulphur but the number of atoms of sulphur in a single molecule of sulphur vapour. This can be done with simple gas-law reasoning.

Reason for rewrite: refer to edit summary.

From Wikipedia on Disulfur ($\ce{S2}$)

This violet gas is commonly generated by heating sulfur above 720 °C

Since the sulfur is well above $720^\circ\pu{C}$, it would most likely be in the diatomic state, more like an analogous structure to dioxygen ($\ce{O2}$).
We can also judge this from the molar ratios too, as rose22 has pointed out. Since it is given that $40~\pu{mL}$ of $\ce{SO2}$ is formed from $20~\pu{mL}$ of $\ce{S_x}$ (assuming the atomicity of $\ce{S}$ to be $x$ at this temperature) and $20~\pu{mL}$ of $\ce{O2}$, they must come to terms with a simple whole number ratio.
$$\ce{\underset{20~\pu{mL}}{S_x}+\underset{40~\pu{mL}}{xO2}->\underset{40~\pu{mL}}{xSO2}}$$
Statement 1: Since the ratio of volumes of gases at a particular temperature and pressure is equal to the ratio of amount of substance in moles, it is clear that:
$$1:x:x=20:40:40$$
That is to say, $x=2$, so the atomicity of sulfur here is $2$.
In the question, the OP said:

However, I am uncertain as to whether this is correct as I did not utilize any of the gas laws. For this question, which of the gas laws should I have used?

Statement 1 is a  result from the universal gas law.
$$PV=n\text{R}T$$
When $P$ and $T$ are constant,
$$\frac{V_1}{V_2}=\frac{n_1}{n_2}$$
Some may argue that this relationship is called Avogadro law, but again, this law too is a gas law, evidently.

@KarstenTheis If you're referring to the line "dead wrong  to say that there are no gas laws involved", I was replying to question itself, which was(and is still) asking if gas laws were required or not. Not referring to the answer, sire. Anyway, I respect you for explaining your downvote.

I did not realize until now that the question and the other answer were by the same person. You answer makes more sense to me know. I also appreciate the bit about properties of $\ce{S2}$ you included.

The question is (my emphasis):

20 mL  of sulfur vapour at $\pu{1000 ^\circ C}$ reacts with 40 mL of oxygen gas to form 40 mL of sulfur dioxide gas. Determine the number of atoms in the molecule of sulfur in the vapour state at this temperature.

For this question, we must understand that the number of sulfur atoms in the sulfur molecule (the subscript in its chemical formula) is actually an integer ($x$). The volume ratios determine the mole ratios and thus, are able to determine the stoichiometry of the equation. From this stochiometry, the value of ($x$) can be determined.
The working out is as follows:
Let $\ce{S_x}$ be the formula of the molecule making up the sulfur vapour.
Volume ratios (Sx:O2:SO2) = 20:40:40 = 1:2:2 = mole ratios, so the chemical equation with stoichiometric coefficients is:
$$\ce{S_x + 2O2 -> 2SO2}$$
Thus $x = 2$ (to have a balanced equation).
Therefore, the molecule making up the sulfur vapour is $\ce{S2}$.
To conclude, there are 2 atoms of sulfur in this particular molecule of sulfur.

I agree with the corrected statements made in your answer - however, what I meant was that no gas laws equations are required to solve this problem. Also, thank you very much for your additional information.

@rose22 Without using gas laws you wouldn't be able to use direct proportionality between gas volumes and the amounts of substances. It doesn't matter whether you write those down or not — they are the basis of your assumption. Also, note that this doesn't really answers your the question (where is the answer for the number of atoms?) and is more justified as an edit or a shortened comment.

@andselisk Okay, thank you for your input. I'll ensure that the answer is more coherent and distinguishable next time.

I am up-voting because the answer is correct, and the incorrect statement about not using the gas laws have been removed.

@andselisk There was some misunderstanding about the wording of the question, which asks for the number of sulfur atoms in one sulfur molecule, if I understand correctly. I hope folks who downvoted for that reason (or for the now removed statement about not needing any gas laws) reconsider their downvote.

@KarstenTheis I noticed that, but since OP initially attempted to find the total number of sulfur atoms, I thought there might be some incorrect translation of the question. Note that the first revision of William's answer also addresses this number, using an assumption about the composition of sulfur molecule (which, as it's turned out, should be the answer) as an aid to solve the problem.

Also, this solution only makes sense if the question would explicitly say that the reaction was complete, which it doesn't. I think the question is still badly formulated (not OP's fault, I agree).


text:
Conditional kinetic equations
From a biological system, I got the following reaction:
$$A+B+D\rightarrow C+C_1D$$ ($C_1D$ is a certain complex)
However, if at any given time there are no $D$s to react, and only then, instead the following reaction does happen (until a new $D$ appears, at which point the first reaction switches in instead):
$$A+B\rightarrow C+C_2$$
So, the second reaction happens if and only if there are no $D$s.
Therefore, this system is like a "conditional" system, where under existence of $D$ the first reaction happens, until all $D$s are used, and then . At any given time, more $D$s can appear and then the whole reaction system acts again as described above.
In the above system, what are the kinetic equations?
(Note: of course, this is not the complete reaction scheme, but the minimum one to extract equations for. However, it should be noted that I consider $[D]=g(t)$, i.e. not constant equal to the initial concentration of $D$. Of course $g(t)$ also depends on the aforementioned reactions too, but on others also)

Without a mechanism you can't get the rate equation (i.e. rate vs concentration of species). And the mechanism will be complicated because the way you set it up, reaction 2 is shut off completely as soon there is a little bit of D. That is very unusual.

You will only be able to solve these equations numerically if D varies in time externally to the scheme given, ie. as some arbitrary g(t). If you want exclusive reactions (or nearly so ) then the rate constant for reaction 1 should be vast compared to that for 2 such that when D is present  1 is favoured, but you will always have some of reaction 2.


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Does aminosugars give positive response in alkaloid tests?
Does aminosugars give positive response to test for alkaloids such as tests with Dragendorff's reagent, mayer's reagent, Hager's reagent, Wagner's reagent etc?

Dragendorff's reagent works for tertiary amines (and some secondary), amino sugars contain primary amines so there is no reaction nor do N-acetylaminosugars

I saw "aminosugars" but read it as "aminosugaris" and was thinking I don't recognize that dinosaur name, and what's it have to do with alkaloids?  Too much with the kids I guess :)

@CrossRoads I dont know about any such Aminosaurus :P :D

Most alkaloids are tertiary amines. Almost all tests for alkaloids, reagents of which are  are solutions of the salts of heavy metals, are precipitation tests. For example, all of tests with Dragendorff's reagent, Mayer's reagent, Hager's reagent, and Wagner's reagent give color precipitates: The heavy metal atom in the reagent with the nitrogen in the alkaloid would form ion pairs. These ion pairs form an insoluble precipitate. The color of the precipitate depends on reagent, plant species, etc. (Ref.1). Most amino sugars are either primary amines or acetamides, thus, I doubt they'd give precipitates with sought reagents. However, some reagents give false-positive tests for non-nitrogenous extracts indicating the presence of alkaloids. The best example is Dragendorff's spray reagent (Ref.2).
Mean time, some reagents are giving negative results even with alkaloid extracts, e.g., Mayer's reagent (Ref.3). Therefore, my answer is purely opinion based.
References:

Felix G. Coe, Gregory J. Anderson, "Screening of medicinal plants used by the Garífuna of Eastern Nicaragua for bioactive compounds," Journal of Ethnopharmacology 1996, 53(1), 29-50 (https://doi.org/10.1016/0378-8741(96)01424-9).
Abdel‐Azim M. Habib, "False‐positive alkaloid reactions," Journal of Pharmaceutical Sciences 1980, 69(1), 37-43 (https://doi.org/10.1002/jps.2600690111).
J. W. Field, M. Kandiah, "A note on the use of mayer's reagent for the detection of quinine in alkaline urine," Transactions of the Royal Society of Tropical Medicine and Hygiene 1935, 28(4), 385-390 (https://doi.org/10.1016/S0035-9203(35)90133-X).

If I get scope to do this simple lab tests I'll publish it as answer. Currently aminosugars are not available to me. We have chitin flakes though, that makes technically a different question because chitin is polysaccharide.


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Does equality of molarity or molality determine constancy of boiling (or freezing) points of different solutions?
I encountered a question among papers for JEE (2005) that demands completing a statement:

Equimolar solutions in the same solvent have _______

There are four options and the correct answers are

same boiling point

and

same freezing point

I don't think this is correct as the change in boiling and freezing point depends on molality and not molarity. Hence, according to the options, the one saying different boiling and freezing points seems to be correct to me.
However, such exams very rarely have mistakes which are not corrected later. Please clarify.

Is this in a question block about ideal solutions? Because at higher concentrations, both answers would even be wrong if you ignored the "lality"/"larity" mistake.

@Karl it's about ideal solutions. I forgot to mention it

Equimolar solutions are also equimolal !

@Maurice not at all! Molarity is a volumetric property, therefore depends on density.

@Buck Thorn. Equimolar solutions do not have the same molality. But compared to one another, their molalities are equal.

@Maurice I don't understand your statement, the second sentence seems inconsistent with the first. In any case, for dilute solutions (within the scope of the problem) the units are interchangeable.

At low concentrations for which the ideal bp elevation/fp depression expressions usually apply, molarity is linearly proportional to molality, therefore the statements are equivalent.
This webpage explains nicely why low concentrations are important:

Raoult's law only works for low concentration solutions. Why? Well, in order for our approximation to work, the interactions between the solute and solvent molecules must be nearly identical. If the interactions are stronger, then the heat of vaporization of the solvent will change and thus our whole approximation falls apart. Since we know that intermolecular forces vary greatly between molecules, the affect of those forces must be kept at a minimum by keeping the solute concentration real low.

In any case (as explained by additional statements in the abovementioned website, which discusses dissociation of ionic solutes), the statement made in the answer to your exam would not be true generally even if molality instead of molarity was mentioned.
The point of the question seems to be that a student should associate a universal response of the solvent that is independent of solute with colligative properties as described using Raoult's law, which is behavior most often approximated by dilute solutions (the statement that Raoult's law only applies to dilute solutions happens to be false, btw), and that in this realm of application molarity is approx linearly proportional to molality.

Hm, the question never says that the cocentration is low.

@Karl No, some details seem implicit - or at the very least, the point seems to be that a student should associate a universal response that is independent of solute with colligative properties as described using Raoult's law, which as my answer explains, is behavior most often approximated by dilute solutions (the statement that Raoult's law only applies to dilute solutions happens to be false, but then the point is not to write an essay on solution chemistry, just to clarify the confusion), and that in this realm of application molarity is approx linearly proportional to molality.

@BuckThorn Very well said


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Which compounds react positively in the aniline dye test?
I read that to test for the presence of phenol groups in an organic compound the aniline dye test can be performed. If the solution has phenol present it gives a red or orange precipitate to confirm the presence, however as a side note it was mentioned that this test is applicable for the detection of small quantities of phenols, which have a free para position. Why is it so?

Umm, why don't you just read up about synthesis of azo compounds? That should make things clear.

First, the aniline is converted to a diazonium halide salt in the presence of acid and sodium nitrite, per this reference.
Then, the diazonium salt reacts with the phenol to give an azo compound, per this reference (see the section entitled Coupling reactions of diazonium ions
Note that the diazonium ion is shown reacting with the para position of the phenol. That is because the -OH group in phenol is an ortho,para-director (reference), meaning that it enhances the nucleophilicity of the ortho and para positions on the phenol ring. However, the para position is significantly more nucleophilic because of the relative lack of steric hindrance in comparison to the ortho positions, which are hindered by the proximity of the -OH group.
So, having a free para position on the phenol means that the phenol is especially reactive towards electrophiles, including the diazonium salt in this example, meaning that it will react faster and in higher yields - increasing the sensitivity of the aniline dye test.


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Doubts Regarding Ceric Ammonium Nitrate Test for Alcohols
I read about Ceric Ammonium Nitrate Test for Alcohols and a red colour is produced if the solution has an alcohol functional group however as a side note it was mentioned that this test is only useful for compounds having less than 10 Carbon atoms. Why is it so?

I have a strong feeling you could have put those two questions together - while you could think that reasons would be different, that's virtually impossible - such difference in length of carbon chain simply doesn't affect the reactivity of OH group, so solubility is the thing here.

@Mithoron ok I've been told in the past to make separate questions so to avoid that I broke it into 2

Ah, it may be problematic to decide, but these should be  at least linked: https://chemistry.stackexchange.com/questions/119157/why-is-the-lucas-test-not-recommended-to-differentiate-higher-alcohols

The question does not mention in what the ammonium cerium nitrate is dissolved in, but I guess it is water. If so, one expects that sufficiently large compounds that one wishes to test are not sufficiently soluble in water, such as decanol. Obviously, this is not entirely correct because disaccharides such as lactose do are soluble in water.
I expect a sufficient solubility to be important for the test such that it finishes in a timely manner and does not become inconclusive due to prolonged reactions with air, water etc.

Is there any reference you can provide?


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Why is the Lucas test not recommended to differentiate higher alcohols?
I read about the Lucas test for alcohols in which a cloudy precipitate is produced with a tertiary or secondary alcohol. However, on a side note it was mentioned that this test is useful only for compounds having less than six carbon atoms. Why is this so?

This "degre" nomenclature for alcohols is terribly outdated. Profs teaching that today are really old and have learned it from teachers which were already oldfashioned then.

The problem is that the alcohols themselves might be insoluble. Ethanol is soluble, yes, but as the alkyl chain length increases, the hydrophobic nature due to the chain starts dominating over whatever hydrophilicity the $\ce{-OH}$ group had been providing for lower homologues.
From source [1]:

.. any alcohol that is insoluble in the reagent might appear to be giving a positive 3° test.

The paper goes on to properly quantify several things:

Based on the results of this study, using 1 drop of alcohol added to 10 drops of reagent, it can be said that, at the very least, all saturated acyclic monofunctional alcohols having six or fewer carbons are soluble in the Lucas reagent.

It is clear that the relative amounts of the sample and the reagent become important in such borderline cases (~6 carbon long chains). If too much alcohol is added, it may cause an illusion of positive lucas test.
From source [1]: (emphasis mine)

That statement may not be true when 3 or 4 drops are used
  per 10 drops of reagent. In fact, sometimes when 10 drops of
  reagent are added to 3 or 4 drops of some of the alcohols, an
  interesting event occurs after a delay of a couple of minutes
  or so. Apparently, some undissolved alcohol that has been
  clinging to the bottom of the test tube suddenly breaks loose
  and, by streaming to the surface of the very dense Lucas
  reagent, gives a momentary illusion of a positive test. The
  real cloudiness, alkyl chloride formation, appears later—all
  the more reason to use a high reagent/alcohol ratio.

References:
[1] "A study of the Lucas test"
R. A. Kjonaas and B. A. Riedford
J. Chem. Ed. 1991 68 (8), 704
DOI: 10.1021/ed068p704

To get a cloudy "precipitate", you need to dissolve the alcohol in the acid solution. Alcohols get less soluble in water as their chain length increases. By the time you pass 6 carbons, the alcohol itself is more or less insoluble. If it reacts along the interface, the chloride product would just mix into the alcohol layer, with no visible change.

Please share the reference you've used

I had no specific reference in mind, just general knowledge. The other answer is more complete than anything I had in mind.

no problem still I appreciate your efforts


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Electrolytic cell - Potentials

I'm an electronic engineer student and I'm trying to understand the basics of electrolytic cells and practical values' meaning. Based on my figure and electrode potencial values (Vce, Vre and Vwe), can you help me how to describe the following concepts:

Standard cell potencial
Equilibrium potential
Overpotential
Open circuit potential
Reference voltage (electrochemical perspective)

EDITED:
Actually I'm asking about direct relations between this definitions and circuit voltages if there are any
For example, is OCP = Vre when it's not applied external potential?
Thank you for your time!

I didn't get your point, I know how the potentiostat works. I'm asking about electrochemical concepts and relation between them and potentiostat voltages

Also, it's unclear what the question is. "Can you help me how to describe the following concepts" is a vague request suggesting certain amount of tutoring, which is a poor fit for a Q&A site. @EdV made a good point that preliminary literature/internet research might give you a better overview of the topics you've listed. Refer to the Help Center, especially How do I ask a good question?

Ok, maybe I didn't explain it right. Actually I'm asking about direct relations between this definitions and circuit voltages if there are any. For example, I think OCP is Vre when its not applied external potencials, is it right?

Ok, I need to review my question I get it. Btw I'm not asking about electronics. Thank you anyway