text:
Kinetics of a simultaneous parallel radioactive decay
Question:

A radioactive isotope, A undergoes simultaneous decay to different nuclei as:
\begin{array}{cc}
\ce{A->P}&\,(t_{1/2}=9\ \mathrm h)\\
\ce{A->Q}&\,(t_{1/2}=4.5\ \mathrm h)
\end{array}
Assuming that initially neither P nor Q was present, after how many hours, amount of Q will be just double to the amount of A remaining?

My solving:
I got the $t_{1/2}$ for $A$ as $T=\dfrac{T_1T_2}{T_1+T_2}\ =\pu{ 3 h}$
Now from Rutherford-soddy law the number of atoms left after $n$ half lives are,
$$N=N_0\left(\frac{1}{2}\right)^n$$
and number of half lives $n=\dfrac{\text{Total time of Decay}}{\text{effective half life}}$
For isotope $A$, $\dfrac{N_A}{N_{0_A}}=\left(\dfrac{1}{2}\right)^{n}=\left(\dfrac{1}{2}\right)^{T/3}$ where $T$ is the common time of decay for $\mathrm A$ and $\mathrm Q$
For $\mathrm Q$ I had to assume that $\mathrm Q$ is also undergoing radioactive decay to form a relation between $\mathrm A$ and $\mathrm Q$
Now using the same formula as above $\dfrac{N_\mathrm Q}{N_{0_\mathrm Q}}=\left(\dfrac{1}{2}\right)^{T/4.5}=\left (\dfrac{1}{2}\right )^{2T/9}$
Now applying the condition in the question, $$\left (\frac{1}{2}\right)^{2T/9}=2\cdot \left (\dfrac{1}{2}\right)^{T/3} $$
Raising by power $9$ on both the sides,
$$\left (\dfrac{1}{2}\right )^{2T}=2^9\cdot \left (\frac{1}{2}\right )^{3T}$$
which implies,
$$ 2^{-2T}=2^{9-3T}$$
Therefore,
$$9-3T=-2T$$
$$
{\bbox[10px, border:2px solid red]{
 T=9\ \mathrm h.
}}
$$
Although I have got an answer and it satisfies the condition I still think it's wrong because of the assumption made for $\mathrm Q$ and something else needed to be done to make the relation between $\mathrm A$ and $\mathrm Q$ Moreover, it's given in the question that initially neither $\mathrm P$ nor $\mathrm Q$ was present and I guess the assumption violates that.

Parallel or side reactions of the first order: Concept
$$\require{cancel}\\
\ce{A ->[k_1] B} \ \ t=0\\
\ce{A ->[k_2] C} \ \ t=t$$
$$-\frac{\mathrm d[A]}{\mathrm dt}=k_1[A] + k_2[A] $$
$$-\frac{\mathrm d[A]}{\mathrm dt} = k_\text{eff} [A] \land k_\text{eff}=k_1+k_2$$
Effective order=1
$$\left(t_{1/2}\right)_\text{eff}=\frac {\ln 2}{k_\text{eff}} $$
$$\frac 1 {(t_{1/2})_\text{eff}}=\frac {1}{(t_{1/2})_{1}} + \frac {1} {(t_{1/2})_{2}} $$
$$A_\text{eff}\mathrm e^{-E_\mathrm a/(RT)}=(A_1+A_2)\mathrm e^{-E_\mathrm a/(RT)}$$
Differentiate with regards to $T$,
$${\frac{E_\mathrm a}{RT^2}}\cdot k_\text{eff}=\frac{(E_\mathrm a)_1 k_1}{RT^2}+\frac{(E_\mathrm a)_2 k_2}{RT^2}$$
$$(E_\mathrm a)_\text{eff}=\frac{(E_\mathrm a)_1 k_1 +(E_\mathrm a)_2 k_2}{k_\text{eff}}$$
$$[A]_\mathrm t=[A]_0\mathrm e^{-k_\text{eff}t}$$
$$a_t=a_0\mathrm e^{-(k_1+k_2)t}$$
$$\frac{\mathrm d[B]}{\mathrm dt}=k_1[A]=k_1a_0\mathrm e^{-(k_1+k_2)t}$$
$$\int\limits_{0}^{b_t}\mathrm d[B]=k_1 a_0 \int\limits_0^t\mathrm e^{-(k_1+k_2)t}\,\mathrm dt$$
$$b_t=\frac{k_1 a_0}{-(k_1+k_2)}[\mathrm e^{-(k_1+k_2)t}]_0^t$$
$$b_t=\frac{k_1 a_0}{k_1+k_2}(1-\mathrm e^{-(k_1+k_2)t}) $$
similarly,
$$c_t=\frac{k_2 a_0}{k_1+k_2}(1-\mathrm e^{-(k_1+k_2)t})$$
$$\frac{[B]}{[C]}=\frac{k_1}{k_2}$$

proportion of $B=\frac{[B]}{x}=\frac {k_1}{k_1+k_2}$ [times 100 for percentage]
proportion of $C=\frac{[C]}{x}=\frac {k_2}{k_1+k_2}$ [times 100 for percentage]

The actual problem
\begin{align}
&\ce{A->[\textit{k}_1]P}  &k_1 &= \frac{\ln 2}{t_{1/2}} = \frac{\ln 2}{9} \ \text{hr}^{-1} \\
&\ce{A->[\textit{k}_2]Q}  &k_2 &= \frac{\ln 2}{t_{1/2}} = \frac{2 \ln2}{9}\ \text{hr}^{-1}\\
\end{align}
$$Q_t=\frac{k_2a_0}{k_1+k_2}(1-\mathrm e^{-(k_1+k_2)t})=2A_t$$
$$\frac{k_2\cancel{a_0}}{k_1+k_2}\mathrm {(1-e^{-(k_1+k_2)t})}=2\cancel{a_0}\mathrm e^{-(k_1+k_2)t}$$
$$\frac{\cancel 2}{3}(1-\mathrm e^{-k_\text{eff}t})=\cancel 2\mathrm e^{-k_\text{eff}t}$$
$$\mathrm e^{-k_\text{eff}t} = \frac {1} {4}$$
$$\implies k_\text{eff}t = \ln 4 = \frac {3\ln 2}{9} t$$
$$\implies t= 6\mathrm h$$
So that gives the answer as 6 h.

For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details

Yashwini, though your solution is correct when applied to chemical reactions and the kinetics of radioactive decay, your reference to the temperature is problematic and may cause confusion and it is unnecesary.Even though you can model a nuclear reaction as a chemical reaction, the activation energy for these reactions are so high that temperature has no effect. I reccommend you to omit it from your explanation.

The question has already been solved by Yashwini and the answer given is correct.$^2$ A more intuitive and specific to question explanation would follow here.
Now, the two reactions given are:
\begin{array}{cc}
\require{cancel}
\ce{A -> P} &(t_{1/2} = 9\,\mathrm h) \\
\ce{A -> Q} &(t_{1/2} = 4.5\,\mathrm h) \\
\end{array}
Now using the rate law, we get,
\begin{align}
-\frac{\mathrm{d}[A]}{\mathrm{d}t}&=k_\mathrm P [A] \tag{1} \\
-\frac{\mathrm{d}[A]}{\mathrm{d}t}&=k_\mathrm Q [A] \tag{2} \\
\end{align}
The rate constant for a first order reaction having a half life of $t_{1/2}$ is defined as:
$$k=\frac{\ln 2}{t_{1/2}} \tag{3}$$
Now, substituting the given values of $t_{1/2}$ into the equations, we get $2k_\mathrm P = k_\mathrm Q$ (since $k\, \alpha \frac{1}{t_{1/2}})$
Now, intuitively since both reactions take place together, it would mean that for every one mole of P formed, two moles of Q forms. Therefore, for every mole of P formed, three moles of A react (since one mole is required for each mole of P and Q).
Now, we add the rate laws ($1$) and $(2)$, since the reactions take place simultaneously, to get:
$$-\frac{\mathrm{d}[A]}{\mathrm{d}t}=(k_\mathrm P +k_\mathrm Q) [A] \tag{4} $$
Now, since using the relation between $k_\mathrm{P}$ and $k_\mathrm{Q}$, we get $k_\mathrm{P} + k_\mathrm{Q} = 3k_\mathrm{P}$
Therefore using the integrated rate law for a first order reaction on equation $(4)$, we get:
$$A=A_0e^{-3k_\mathrm Pt} $$
Now, the amount of $A$ used here would be $A_0 -A$, and we get that value to be:
$$A_\text{used}=A_0\left(1-e^{-3k_\mathrm Pt}\right)$$
Now, as we have previously noted, for every three moles of A used, two moles  Q are formed. This means that the amount of Q now in the mixture would be two thirds of $A_\text{used}$. Therefore the amount of Q would be:
$$Q=\frac{2A_0\left(1-e^{-3k_\mathrm Pt}\right)}{3}$$
Now, we are given the condition, $Q = 2A$, substituting values of $Q$ and $A$ into the given relation we get:
$$\begin{align}
\frac{\cancel{2A_0}\left(1-e^{-3k_\mathrm Pt}\right)}{3} &= \cancel{2A_0}\left(e^{-3k_\mathrm Pt}\right) \\
\implies 1 -e^{-3k_\mathrm Pt} &= 3e^{-3k_\mathrm Pt} \\
\implies 4e^{-3k_\mathrm Pt} &= 1
\end{align}$$
Solving for $t$, we get:
\begin{align}
3k_\mathrm Pt&=2\ln 2 \\ \\
t&=\frac{2\ln 2}{3k_\mathrm P}\\
\end{align}
Now, using equation $(3)$, we get the rate constant $k_\mathrm P$ to be $\frac{\ln 2}{9}$. Substituting this value into the expression for time, we get:
$$t=\frac{2 \cancel{\ln 2}}{\cancel{3} \frac{\cancel{\ln 2}}{\cancelto{3}{9}}}$$
Therefore, time taken for this condition to happen is:
$$t=2\times 3 = 6\ \mathrm h$$

[2]: Re-formating said answer was an effort of about 90 minutes.


text:
Arrhenius equation and rate constant
I am a high-school student and I am currently taking a course on chemical kinetics. Towards, what seems, end of the course at highschool level, the Arrhenius equation came in. Now, I'm not sure if what I'm presented is just a general form of Arrhenius equation and there's more to it, or this is it. Well, anyways, one of the things I am being told to learn is the plot between graph of rate constant $k$ vs temperature $T$. $$k=A\mathrm e^{-E_\mathrm a/(RT)}$$
The graph is sort of looks like with increasing $T$, the value of $k$ is skyrocketing (kinda like $y=x^2$ in first quadrant) but if I'm not mistaken, the rate constant $k$ should equal $A$ as temperature tends to infinity. Now I believe there must be some other factor in the equation that is also a function of temperature and is causing the graph to be the way it is. Can someone help me understand the same?
(In case you didn't get, because I am unable to attach an image, the looks like $y=x^2$ instead of $x=y^2$, which basically interrupts my intuition of Temperature on $x$ axis ever approaching infinity. $k$ on $y$ and $T$ on $x$ axis.)

Imagine that you are small like an ant, only smaller, and looking at the initial portion of that graph. Now it will look convex all right.

This is how the graph between the rate constant and temperature looks like:

This might not seem intuitive looking at the Arrhenius equation which says:
$$k=Ae^{\frac{-E_a}{RT}}$$
After using some mathematical prowess someone might say that as $T\rightarrow \infty, ~k\rightarrow A$ and so the graph should look something more like:

Which on first glance seems fair but doesn't match the aforementioned curve.
To understand what is causing the problem we must check everything that might change during a change in temperature.
Upon examination we can find that only two things might cause this discrepancy, $E_a$ and the pre-exponential "constant", $A$.
The activation energy of a system is also independent of the temperature and thus the only possible problem point left is the pre-exponential "constant". Turns out that this constant isn't a constant in true sense, which isn't that disturbing of an idea if you know what that constant signifies.
The pre-exponential constant shows how many effective collisions can take place in a reacting mixture. It is defined as,
$$A=pZ$$
Where,
$A$ is the pre-exponential constant.
$p$ is the steric factor, which takes into account the steric effects during an effective collision.
$Z$ is the collision frequency.
Now as it might seem intuitive, the collision frequency is related to temperature and increases with increase in temperature according to the following relation(using equations for collisions between similar molecules to prevent unnecessary complexity):
$$Z=4nr\sqrt{\frac{\pi RT}{M}}$$
Where,
$Z$ is collision frequency.
$n$ is molecules per unit volume.
$r$ is the radius of molecule.
$R$ is the universal gas constant.
$T$ is temperature.
$M$ is the molar mass of the gas.
Thus the collision frequency and corresponding the pre-exponential constant tend to infinity when $T\rightarrow \infty$, which matches with our observation of $k\rightarrow \infty$ when $T\rightarrow \infty$.

Reference:
(1) The Effect of Temperature on Reaction Rates https://chem.libretexts.org/@go/page/24271 (accessed May 14, 2021).

As a consequence, Arrhenius law appears to be wrong. Don't you think it should be written without the "constant " A,  in such a way : $\pu{k = p 4nr \sqrt{\frac {\pi RT}{M} } e^{-\frac{E_a}{RT}}}$

@Maurice Yes, it's the correct form but the change in A is very small for changes in temperature which we encounter during chemical reactions and thus it behaves as a constant and thus also the name.

To understand the behavior described by the equation
$$\rho(T)=A\mathrm e^{-E_\mathrm a/(RT)}$$
a little better, consider a few  properties (where it is assumed that $A$ and $E_a$ are independent of $T$):

The value of $\rho(0)=0$ (as $T$ goes to zero $1/T\rightarrow \infty$ so that $\exp(-\infty)=1/\exp(\infty)=1/\infty=0$)

The value of $\mathrm d \rho (T)/\mathrm d T$ (the slope or derivative of the  rate function at $T=0$, how much it changes there with $T$). In the limit $T\rightarrow 0^+$ the derivative of $\rho$ is $0$ (this is a tad more difficult to show).

The value of $\rho(\infty) = A$ (since $\exp(1/\infty)=\exp(0)=1$).

The value of $T$ and $\rho(T)$ at the inflection point (when the second derivative is zero). This is discussed in another answer. The behavior of the function changes from monotonically increasing to decreasing at this point.

These four properties should allow you to generate a schematic plot of $T$ versus $\rho(T)$. What might be confusing is that the behavior described by points 2 and 4 might be hidden from view depending on the resolution ie which parts are emphasized by selecting a broad or narrow range of $T$,  as shown in the following plots (which refer to the same data):

Surely, as T tends to infinity k tends to A. The graph in this case will be exponential and not $y= x^2$ kind.
You can plot the graph by differentiating the equation (twice to be particular).

The graph looks like $y=x^2$ but only upto $T=\dfrac E{2R}$.
But don't misinterpret the same.

A is also temperature dependent, that's why k goes up as well

This is incorrect. The pre-exponential factor, $A$ is not independent of temperature. It is related the collision rate which is dependent of temperature. $A$ can be considered constant only for small temperature changes. As $T\rightarrow \infty ,~A\rightarrow \infty$ and thus the rate constant approaches infinity and not some finite value.

It looks to me more like $k=A/e$ at $T=E_a/R$

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text:
Will alkoxide ion in protic medium attack on double bond or carbonyl group?
I want to know what will happen next here-
I thought that the alkoxide ion would attack the aldehyde to form a secondary alcohol, but apparently, as the two pi-bonds are in conjugation, so, the ion attacks the carbon of the C-C pi-bond. Why is this so?

Your title should read "protic medium" as opposed to "acidic medium". Acidic medium would be EtOH/H+. Your reaction conditions are basic and protic. Nothing much will happen to the unsaturated aldehyde 1. 1.2-Addition leads to hemiacetal 2. Hemiacetals are unstable to base and reverse to the carbonyl compound. Adduct 3 would eliminate to form 1 in the presence of base. If you employed EtOH-d1 as the solvent, you would incorporate deuterium at the γ-position of 1-d2 via enolate 4. If heptanedial 5 were exposed to EtOK/EtOH, an aldol condensation would ensue to form unsaturated aldehyde 1. Under truly acidic conditions, the diethyl acetal (in the red box) of aldehyde 1 would form. Formation of the β, γ-isomer of 1 is unlikely because the α, β-form 1 is much more stable.

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text:
Standard electrode potential of copper disproportionation
I am having some difficulty with the below.
It says the standard potential for $\ce{2Cu+ (aq) -> Cu (s) + Cu^2+ (aq)}$ is $\pu{0.36 V}$.
The relevant half equations are:
$\ce{Cu+ + e- -> Cu}$ (potential $=  \pu{0.52 V}$)
and
$\ce{Cu^2+ + e- -> Cu+}$ (potential $=  \pu{0.16 V}$)
To get the overall equation, I flipped the second half equation and so reversed the sign of the potential also to get $\pu{-0.16 V}$.
The first half equation is therefore reduction and the second is oxidation.
If standard electrode potential is $\mathrm{RHS (reduction)} -\mathrm{LHS (oxidation)}$, would the answer not be $\pu{0.52 V} - (\pu{-0.16 V}) = \pu{0.68 V}$? Why is it $\pu{0.36 V}$?

It is RHS(reduction)-LHS(reduction)

how do I know which to put on which electrode if both are reduction? as in how do I know that the 0.52V corresponds to RHS? thanks!

Another Issue, emf of reactions are not additive; free energy is... read up on electrochemistry before you proceed.. the part undergoing oxidation is oxidation and part going reduction is reduction (seems redundant doesn't it)  ;-)

After Safdar, the standard potential of the reaction is 0.68 V/2 = 0.34 V (and not 0.36 V)

It should be: $\pu{0.52 V} - \pu{0.16 V} =  \pu{0.36 V} $.

Alas, the confusions related to signs in electrochemistry will never vanish. You mention that
$\ce{2Cu+ (aq) -> Cu (s) + Cu^2+ (aq)}$ is $\pu{0.36 V}$.
Let me start with a single equation, x-y = 10; There can be indefinite solutions if you can simultaneously change the value of x and y. However, the moment you fix the value of x, the value of y is fixed.
You stated that the overall cell potential is $\pu{+0.36 V}$. Electrochemically, this means that this reaction is spontaneous.
Now you also know that,
$$E_\text{cell} = E_\text{reduction} - E_\text{anode} \tag{1}$$
You are not supposed to change any sign of the half-cell from the electrode potential tables. People should stop teaching this nonsense to relatively innocent students. Suppose, I write
\begin{align}
&\ce{H2O (liquid) -> H2O (gas)} &T &= \pu{100 ^\circ C} \\
&\ce{H2O (gas) -> H2O (liquid)},    &T &=\pu{ -100 ^\circ C} ??
\end{align}
The equation (1), itself takes care of all sign flipping and all.
Your half-cell corresponding to the reduction is
$\ce{Cu+ + e- -> Cu}$ (potential $=  \pu{0.52 V}$)
And your half-cell potential for the oxidation is
$\ce{Cu^2+ + e- -> Cu+}$ (potential $=  \pu{0.16 V}$)
Using equation (1), what you do get (remember no sign flipping) = $\pu{+0.36 V}$


text:
ClO2 chlorine dioxide how to know if it is still active or not ? by color ? by smell ? by reaction?
my wife have ClO2 tablet form. she dilutes it to be put inside mister. each time she make the dilution:
package instruction:
1 tablet to 1 liter water => 100ppm ClO2
*we name this 100ppm as "original soln"
she takes 100ml from the "Original soln" to make 1 liter of solution to do household cleaning.
may i know how much ppm is the final solution that she used to do cleaning ?
i think should be 10ppm solution , right ?  10% mix.
thx

If you start with $\pu{100 mL}$ stock solution at $\pu{100 ppm}$ concentration, and dilute this by a factor of 10 to yield $\pu{1 L}$, then you are correct, the later one has a concentration of $\pu{10 ppm}$.  Keep both in a cool place.
As to check if the solution still is active, or not, and both stock and diluted solution at the expected concentration (which will decrease already if the solution is exposed to hot summer temperatures), you may use test strips you immerse dip into the solution, dry at room temperature, and compare against a scale.  An example were the following

(source)
Depending on the manufacturer, the scale of concentration to test may vary.

someone told me as long as the liquid still have greenish color, that means it is still active.. is that so ? ti saw a bottle of 10ppm ClO2 liquid under wash basin cupboard. it is still greenish color with little smell, no idea how long has she leave it there in room temperature. if soak my hand in 10ppm liquid, will i feel any thing ? i am in germany, i can't find this ClO2 test strip at all. are free chlorine test strip can be use? such as these https://sensafe.com/shop-all/?_bc_fsnf=1&Industry=Sanitizers+and+Disinfectants&gclid=EAIaIQobChMI2K7166yg6wIV1IbVCh0SgQDIEAMYAyAAEgLh7PD_BwE

«If it is still greenish in color», this depends too much on the light you use & the thickness of the solution you look across from wall to wall; thus I don't trust this approach. «Free chlorine» refers to Cl2 / chlorine gas, but this is not ClO2. Re test stripes, I found https://www.lovibond.com/en/PM/Water-Testing/Products/Reagents-Standards/Test-strips/Chlorine-Dioxide-HR-Test-Strips, https://chlordioxidteststreifen.de/, or on https://www.amazon.de with keyword «chlordioxid teststreifen». Some of the pharmacies still prepare medicine on demand, thus they might help you to get them, too.

Note ChemSE however has a policy to refrain from the provision of medical advice, already because most of us users here are not physicians.

your advice is good enough. even so, stackexchange policy is for caution. and some of ppl who are here are not stupid as well. lol . thanks for your advice , appreciate it.


text:
Converting SMILES string into valid sub-strings
I have a SMILES string as follows.
smi= 'CCOC(=O)C1=C(COCCN)NC(C)=C(C(=O)OC)C1c1ccccc1Cl'

Is there anyway that I convert this SMILES string into valid sub-strings which can be later processed in RDKIT.
For instance
smi_sub1= 'C1c1ccccc1Cl'
smi_sub2= 'CCOC(=O)C1=C(COCCN)NC(C)=C(C(=O)OC)'
smi_sub3= 'C1=C(COCCN)NC(C)=C(C(=O)OC)'

These sub-strings should be subset of the main SMILES string and should also be valid from the SMILES syntax point of view.
So that when I use the function MolFromSmiles, It get processed without any error or none in return?
Mol_sub1 = Chem.MolFromSmiles(smi_sub1)
Mol_sub2 = Chem.MolFromSmiles(smi_sub2)
Mol_sub3 = Chem.MolFromSmiles(smi_sub3)

Note that neither openbabel (version 3.1.0) accessed via its gui, nor the NIH pubchem sketcher (https://pubchem.ncbi.nlm.nih.gov/edit3/index.html) are capable to process your smi_sub1, smi_sub2, or smi_sub3; which may indicate an error in your processing.  While lacking clarity topically, your question may be interpreted like a scaffold identification.  The Murcko scaffold (e.g., https://www.rdkit.org/docs/source/rdkit.Chem.Scaffolds.MurckoScaffold.html) is one of them.

While this might work in some cases, I am not sure how well it will work in general. The problem here might be reflected in this statement:

These sub-strings should be subset of the main SMILES string

Not every substring will be a subgraph of the parent molecule. As mentioned in the comments, this seems essentially a fragmentation problem. Take this one of your examples:

smi_sub1= 'C1c1ccccc1Cl'

The reason this won't parse is that it starts with a 'C1' but has no corresponding 'C1' to finish the ring. So it should be 'Cc1ccccc1Cl', which is 1-chloro-2-methylbenzene.
Theoretically, you could take substrings, then post-process them to remove 'disconnected' ring numbers. However, it would make more sense to use proper fragmentation on the molecular graph, and then you would have correct subgraphs anyway, without the need to convert them from SMILES.

On a second thought, Daylight's SMILES tutorial (https://www.daylight.com/dayhtml_tutorials/languages/smiles/index.html) illustrates the theory (https://www.daylight.com/dayhtml/doc/theory/theory.smiles.html) behind the conventions used.  Equally note the related SMARTS useful for sub-structure searches (https://www.daylight.com/dayhtml/doc/theory/theory.smarts.html).


text:
Why does magnesium bromide transfers from a carbon atom to nitrogen in piperidine?

In this reaction, why does $\ce{MgBr}$ goes to the $\ce{>NH}$ group and form a bond with nitrogen by removing $\ce{-H}$.
Also where does the negative charge go?

To substantiate @Safdar's comment further, in the Evans pKa tables (e.g., https://www2.chem.wisc.edu/areas/reich/pkatable/index.htm), cyclopentane is estimated (section «-Benzyl») by a pKa(DMSO) of ~59, and pyrrolidine (section «-Amines») by a pKa(DMSO) of 44. Ammonia is listed by a pKa(DMSO) of 41. Thus, the differences are considerable.

Acid base reactions are more favoured here. N has an acidic hydrogen hence attack happens. So R-MgBr converts into R-H.

A better title would be, "Why does a hydrogen transfer from nitrogen to carbon in this reaction?"

There is no step 1 and step 2 in this reaction. Both are consecutive and second one is fast because it is an acid-base reaction:

Note that approximate $\mathrm{p}K_\mathrm{a}$ values of ring $\ce{C-H}$ and $\ce{N-H}$ are 50 and 40, respectively. Thus, as soon as Grignard reagent is formed, it exchanges relatively acidic protons of $\ce{N-H}$ fast (Remember, Grignard is also a strong base). To visualize the exchange process, I included schematic representation at the bottom of the image.

If the Grignard's reaction with the halogen followed by the exchange with the nitrogen-H, would it be reasonable to call the product of the first reaction the kinetic, the later one the thermodynamic product?

Buttonwood : I think the reaction rate depends on preparation of Grignard reagent in either way because it is the RDS. The proton exchange is much faster to my knowledge. Do you have any other thoughts?

Mathew is correct. There is no step 1 and step 2. There is just an acid/base reaction as the "Grignard reagent" is formed. To say there is a second step is to say, in the reaction of NaOH and HCl, the first step is the formation of HO-H. True. And the second step is the formation of NaCl. I think not.

@MathewMahindaratne I recognize now that my argumentation was (too much) influenced by a recent re-read about kinetic / thermodynamic deprotonations and metallations on halogenated pyridines. So it isn't sensible to this line of thought on saturated heterocyclic substrates as this one here.

How did you create the picture ? Can you tell me?

@Soham Chatterjee: I used ChemDraw.


text:
Detecting carbon dioxide concentration using mosquito neurons
At the Mauna Loa Observatory the concentration of atmospheric carbon dioxide is measured using the absorption of infra-red light. This is a statistical technique that requires constant careful calibration. Mosquitoes, drosophila, and other insects have specific cpA neurons that detect $\ce{CO2}$. These insects are orders of magnitude smaller and more portable than the equipment at Mauna Loa.
What work is being done on unraveling the chemical processes used by cpA neurons in order to create more accurate detection of $\ce{CO2}$ levels?

There is a great difference between sensitivity, stability and accuracy. Neurons have great sensitivity to stimuli, i.e., detecting minuscule amounts, but neurons are not very accurate, i.e., they don't repeatably show the same response to a given stimulus, and are subject to synaptic fatigue. Given a large number of neurons in a very tightly controlled environment, over a short time, the accuracy can be improved, but it is far simpler and less expensive to use existing sensors.
There are currently available electronic gas sensors that are of the same order of size as mosquito neurons, and are much easier to interface to devices and to maintain. For example, see the list of $\ce{CO2}$ sensors from DigiKey.
Mosquitoes do have an advantage in their ability to locate a source of $\ce{CO2}$, though they also rely on water vapor, heat and volatile organic chemicals to find prey. So, if you itch to try out interfacing with mosquito neurons, you might make a better directional tracker for sources of $\ce{CO2}$... or $\ce{NH3}$, $\ce{CH3COOH}$, etc. However, the Mauna Loa $\ce{CO2}$ measurements are to detect long-term changes in global $\ce{CO2}$, not to track down breathing prey. Mosquitoes, and their neurons, are a bit ephemeral for that.

I have not found any information about the use of telescopes for the detection of CO2 at Mauna Loa Observatory. I would be interested in any links that you could provide that describe this technique. As I understand it, the sensor for which you provide specifications uses the concentration of a range of VOCs to estimate the concentration of CO2, but does not measure CO2 directly.

@JamesNewton, the example I gave was invalid, and I rewrote the answer to be more specific to your question.


text:
What are the products formed in acidic hydrolysis of a tetrahydropyranyl ester?

Find the major product of the following reaction:

My  approach
From my knowledge of reagents, I figured out that that the acid will protonate the oxygen(s) in the reactant and after that $\ce{C-O \sigma}$ bond should break, but I am not able to decide which $\ce{C-O}$ bonds to break in the reaction after protonation by the aqueous acid.
Am I thinking in the right direction or are there some other concepts involved?

The original compound is dihydropyran. This group is a tetrahydropyranyl ether and is widely used as an alcohol protecting group  https://www.organic-chemistry.org/protectivegroups/hydroxyl/thp-ethers.htm

The actual mechanism
Waylander has given an accurate reaction since this is a THP ether

The actual mechanism of the de-protection using alcoholysis would be as follows:

The actual mechanism of the de-protection using acidic hydrolysis would be as follows:

A place to start from if you didn't know the mechanism
Seeing that the compound given resembles a carbonyl group that has been protected using ethylene glycol, a mechanism using a reaction similar to that of protection of carbonyl groups was used here to hypothesis what would happen(not knowing the original mechanism) and to decide the products formed.

Hence, taking a reversed approach here, the first question I ask is:

What could have formed this acetal in the first place?

One part of the acetal seems to be connected to the carbonyl group and is part of the main molecule. The other side, however, is a different molecule. So, this is what the acetal formation looks like in the mechanism. Here, I assume $\ce{R}-$ is just an inert part of the molecule and so is equivalent to $\ce{CH3-CH2}-$. This may not be the right mechanism but it would give us a good starting point.
Formation of the hemiacetal takes place as follows:

Formation of acetal using propanol can happen as follows:

Which is the THP ether in the question.

What are the final products?
Therefore, the products of acid catalysed hydrolysis of the acetal would be $\ce{R-CH2-OH}$ and 5-hydroxypentanal which exists in equilibrium with its cyclic form dominating in mildly acidic conditions.

The hydrolysis of the hydroxyderivative of THP would happen as follows. Both alcohols will get protonated but protonating the oxygen in $\ce{-OH}$ results in the same molecule being regenerated.

Therefore, the final products would give $\ce{R-CH2-OH}$ and a hydroxy derivative of THP.
However, if $\ce{HCl}$ or $\ce{NaHSO3}$ is used instead of a weak lewis acid, we see the dominating compound in the equilibrium being 5-hydroxypentanal as shown in this paper here:

[...] As a result of the acid-catalyzed hydration of 1, the 2-hydroxytetrahydropyran (2) was formed primarily which equilibrated with its open chain form, the 5-hydroxypentanal (3). The acid-catalyst (conc. hydrochloric acid), however, promotes not only the addition of water to the C-C double bond of 1, but also that of the OH groups in 1 and 2, thus resulting in the protected compounds 2-(2-tetrahydropyranyloxy) tetrahydropyran 0, and 5-(2-tetrahydropyranyloxy)pentanal (4). The ratio of 4 + 5 to 1 + 2: depends on the amount of acid used.[...]

Reference:

Peter Vinczer , Zoltan Juvancz , Lajos Novak & Csaba Szantay (1989) AN
IMPROVED SYNTHESIS OF 5-HYDROXYPENTANAL, Organic Preparations and Procedures International: The New Journal for Organic Synthesis, 21:3, 344-346, DOI: 10.1080/00304948909356391

The THP ether is usually formed from dihydropyran which is regenerated when the group is removed.

@user55119 It is SN1 (the reaction takes place via alcholysis). I'll modify that.  It's the same mechanism as aditya's answer, with the nucleophilic attack on the other side. I couldn't manually draw it for a generic compound since chemdoodle doesn't support it( currently do not have access to anything else), so I added the closest one.

@safdar: Back in pre-history we used the Fieser triangle.   https://www.sigmaaldrich.com/catalog/product/aldrich/z103772?lang=en&region=US
Alas, it has been discontinued.

5-hydroxypentan-1-al is the product.

That SN2 reaction is ugly as sin

Indeed, you made a bad shortcut here, as this C atom isn't nearly as electrophilic as the other one.

@Safdar  You are correct that under aq acid conditions the 5-hydroxypentanal is formed  https://en.wikipedia.org/wiki/3,4-Dihydropyran

@Waylander however, according to Greene's Protective Groups in Organic Synthesis, Fourth Edition, it depends on the conditions as you said. Using TsOH/MeOH yields a hydroxy derivative as shown in my answer (there are 30+ conditions giving different reactions).

@PCK: If the distasteful SN2 mechanism in the last step were operative, you couldn't form RCH2OTHP in the first place from THP and RCH2OH/H+?

@Safdar: In your response to Waylander are you saying that TsOH/MeOH forms tetrahydro-2H-pyran-2-ol and not 2-methoxytetrahydro-2H-pyran?

@user55119 It was an error on my part, I meant acid catalysed hydrolysis instead of TsOH/MeOH


text:
Latent heat of vaporization/evaporation
It's the amount of heat required for liquid ---> gas phase change of a substance at it's Boiling point.
But my question is, does providing energy equivalent to latent heat of vaporization to a liquid at a temperature below it's Boiling point (if you need, also assume above it's Freezing point) 'Evaporate' it?
eg. When water is kept inside a clay pot out in the Sun, we know a very small amount of water somehow seeps out through it's pores and 'evaporates' (if not vaporizes). Obviously the temperatures aren't even close to 373K(boiling point of water), but the water is spontaneously turning into steam. Because it's not boiling, but is simply evaporating which can take place any where below it's boiling point. But the point here is, 'how much' of the energy from the rest of water-pot system is entrapped in order to convert into a form that has got more energy than its predecessor.
Is it equivalent to only the amount of latent heat of vaporization or is it plus the amount of energy required to raise it's temperature upto it's boiling point.

Hint question: What is the relation of the latent heat of evaporation ( J/kg ) at 2 different temperatures and respective specific heats ( J/kg/K) of water and water vapour ? Can the total energy change going through the whole closed cycle ( both temperatures and phases ) be nonzero ?

@Poutnik I did think upon the hints but I didn't get you. Like, the first relationship you are talking about is something I didn't know, and even after acknowledging that I didn't have access to the 'papers' stating the relationship. Second, I didn't get which cycle you wanted to refer to. Kindly help.

@Poutnik You are asking? It's only =latent heat. But I don't have an intuition or even an understanding of it!

Latent heat of vaporization/evaporation:  It's the amount of heat required for liquid ---> gas phase change of a substance at it's Boiling point.

Not exactly. It's the amount of heat required for liquid ---> gas phase change of a substance at the particular temperature, there is usually and implicitly the substance boiling point. But generally, it is defined for any temperature.

Is it equivalent to only the amount of latent heat of vaporization or is it plus the amount of energy required to raise it's temperature upto it's boiling point.?

Neither, neither. It is between these 2 values.
The latent heat of water evaporation at temperature $T_2 \lt T_1$ (let we take $T_1$ as the boiling point ) is equal to the  latent heat of vaporization at temperature $T_1$
plus the heat required to warm up water  from $T_2$ to $T_1$
minus the heat released to cool down vapour from $T_1$ to $T_2$.
$$\Delta H_{\mathrm{evap},T_2} = \Delta H_{\mathrm{evap},T_1} + C_\mathrm{p,water}\cdot (T_1 - T_2) + C_\mathrm{p,vapour}\cdot (T_2 - T_1)$$
where $\Delta H$ are molar latent evaporation heats at constant pressure ( called molar enthalpies of evaporation) at respective temperatures,
$C_\mathrm{p}$ are respecive molar heat capacities of water and vapour ( for simplicity considered constant to avoid integrals).
It is application of the Hess' law, which is application of the energy conservation law. It says the energy change depends only on the initial and final state, not on the way how is the final state reached. If it were not so, we could easily construct perpetuum mobile by switching between 2 states by 2 different ways.
So getting A->B directly, or A->C->D->B is equivalent, where

A is cold water
B is cold vapour
C is hot water
D is hot vapour

Generally, the latent heat of evaporation increases with decreasing temperature, so you need more energy to evaporate water at lower temperature. At water critical temperature near $374 ^{\circ}\mathrm{C}$ the properties of liquid and gaseous water  converge to the same values and the latent heat of evaporation converges to zero.
If water at $50^{\circ}\mathrm{C}$ is provided by latent heat of evaporation at $50^{\circ}\mathrm{C}$, then it will evaporate. If water at $50^{\circ}\mathrm{C}$ is provided by latent heat of evaporation at $100^{\circ}\mathrm{C}$, then some water will not evaporate. Because the latent heat of evaporation is bigger at $50^{\circ}\mathrm{C}$, compared to $100^{\circ}\mathrm{C}$.
Technically, it is different question, as evaporation at ambient temperature takes energy from the environment, while you have to supply energy for the boiling.
Note that physicists/chemists prefer to talk about vapour, a gaseous phase of substance, that is not considered as a "permanent" gas. Steam is rather a technical/engineering/everydays term, covering pure water vapour, mixtures of vapour with gases like air, or the same with an aerosol of water or ice particles. See water phase diagram. Water vapour exists even at at mesosphere at altitude near 80 km at temperatures almost $-100^{\circ}\mathrm{C}$. Also, ice can sublimate directly into vapour, without melting.

You can calculate it, using heat capacities of vapour a water, as above.

This is an excellent answer. Just to clarify though, what do you mean when you say " At water critical temperature near 374∘C the properties of liquid and gaseous water converge to the same values and the latent heat of evaporation converges to zero." Surely at T=374 C, the latent heat of evaporation of water does not converge to zero but rather it converges to the latent heat of vaporization of water at T=374 C which is around $41 kJ/mol$, not zero?

@SalahTheGoat There is no evaporation at the critical temperature nor above, therefore the latent heat of evaporation is defined for T<Tc. The left limit of its value for T approaching Tc is 0 kJ/mol. Think about it: Why should there be any need of supplying energy to have exactly the same state as before ? Remember that state variables change near critical conditions highly non linearly

Aha okay I understand now. So by critical point you're referring to $T \approx 374 C$ and $P\approx 217 atm$, the point where liquid/vapour H20 becomes a supercritical fluid correct? Thanks again for the great response.

To answer your question, yes.

But my question is, does providing energy equivalent to latent heat of vaporization to a liquid at a temperature below it's Boiling point (if you need, also assume above it's Freezing point) 'Evaporate' it?

@Poutnik answered this fantastically, but with only theoretical examples. They did say though:

Generally, the latent heat of evaporation increases with decreasing temperature, so you need more energy to evaporate water at lower temperature.

…which is indeed what is stated in my College Physics textbook (Knight et. al, 2010, 2e). At standard atmospheric pressure, the (vaporisation) latent heat of water is:

24 × 105 J/kg, at 30°C
22.6 × 105 J/kg, at 100°C

Simply stated, you need 6% more energy to evaporate water at 30°C than you would at 100°C. Therefore, it can (and does) happen.

@Poutnik, you commented that you can calculate the latent heat of water vaporisation for other temperatures? Is this simple, like a linear relationship? Or more complex?

The linear relationship assumes the approximation the molar heat capacities are temperature independent.


text:
What is the minimum thickness for stainless steel crucible to melt aluminum
I am planning to make a stainless steel crucible (air-tight) to melt aluminum. Aluminum is known to oxidize in presence of air. I therefore want to create a custom crucible made with sheet metal stainless steel. What is the recommended thickness of sheetmetal in this case?
I shall be cladding the crucible with refractory material to ensure no heat losses from the stainless steel crucible. The crucible shall hold roughly 10-15kg of Aluminum. As per the comments I yet have to see now to have the crucible vacuumed to ensure no gasses are present or employing a one way valse, so that it doesn't become a dangerous to operate.

good question. keep in mind that you would need a pressure-regulator such as a check valve if you going to be making it air tight to ensure the gasses get vented out while heating (gas expands considerably) , otherwise it could be disastrous.

How large??? Are you melting 1 kg or 1,000? BTW, if it's covered, how is it heated? If electrical, what insulation?...

it is perfectly possible to melt aluminium in a normal, open crucible even in air. Search YouTube for gazillions of examples. The losses to oxidation are small and not worth trying to eliminate. Plus, extreme heat + sealed vessel is a bomb, not a cool chemistry experiment.

@DrMoishePippik 10kg of Aluminium, the stainless steel cruicible is covered but it is heated from the bottom, The insulation would be in the sides to mitigate the losses.

@matt_black, I am meting aluminum cans which are more prone to oxidation (even when crushed) compared to cast aluminum. I don't have any cast aluminium laying around to melt before adding cans.

@metron Even so, you don't need a sealed vessel. Which is really dangerous for other reasons.

When I search "crucible melt" on the net ,hundreds show up. Essentially all are silicon carbide  or ceramic . It may be simpler to do something the way everyone else does it. However ,professionally I would make the crucible of HK 40 about 1/2 " thick ( cast 310 SS with 0.4 % C).

@blacksmith37, that is the answer I was looking for. Ceramic won't last longer. I want to add actuators and strirrer so I want to custom make it so the pouring is done automatically. I don't have a helper or anyone who could assist me during casting so that's why I decided to make it with actuators with some added safety offcourse.


text:
Can a charcoal generating process be a low carbon footprint incinerator?
Noting that cellulose, hydrocarbons and carbohydrates are mostly C, H and O, it occurred to me that one might take low grade recyclables no longer welcome, and trash in general, and convert them to energy and carbon using a sophisticated incinerator that uses advanced sensors and computer monitoring, air and heating controls to burn the waste with little carbon footprint.
Heat generated could be recycled to start future  batches, and excess heat could drive power generation.
An interesting side discussion would be how to use the carbon without releasing it to the atmosphere.  Carbon fiber, filtration?
Another aspect is what sorts of other pollution is a problem, like sulfur compounds, carbon monoxide?

Yes, very simply. Add water and leave it in a pressure reactor vessel for some days. Zero sophistication, just watch the pressure and harvest the produced heat. I'm trying to remember the name of that process. The end result is solid carbon, i.e. charcoal.

https://en.wikipedia.org/wiki/Hydrothermal_carbonization 12h, 180°C, 10 bars, gives basically 90% conversion of the contained C to charcoal.

The trouble are those 180°C. That's basically waste heat, you cannot efficiently run a machine with that, e.g. a generator. My idea would be a https://en.wikipedia.org/wiki/Solar_updraft_tower that uses it as a night heat source, used for desalination and pumping, to water the agriculture/wood farming that produces the biomass.

Nice, I did not know about its high conversion efficiency and it was discovered in 1913! The only issue is that maintaining 180 C for 12 h. How to achieve this in a green "way"?

@M.Farooq The process produces a lot of heat. Once you have one reactor running, it can start five more. With biomass, microbial decay would start it automatically, but that takes a little more time.

When the process was "rediscovered" a few years ago I heard a talk by Antonietti about it, and my thought was that the whole thing is embarrassingly obvious. Any undergrad student should come up with charcoal if you asked him what happens if you keep some "sugar" with water at high temperature and pressure. ;-)

I guess the original experiment is done in a closed system, so you are basically in a steam bath at 180 C.

@M.Farooq Yea, it's always a closed system. Batch processing, which makes it not very economic.

Of course, converting these compounds to H2O and C will be net exothermic and so could drive a power producing process like a Stirling Cycle engine or IR solar cell.
What about other elements in the waste stream?
Can we get C with little or no CO2 or CO?
Is anyone currently incinerating with such a process?

Batch processing is not prohibited, or completely required, but you might need to process waste to ensure it flows through a streaming process without obstruction.  The most efficient pressure boundary is some sort of pump/turbine, for instance, but a restrictive orifice can hold back pressure and pass small chunks.  We're talking a long insulated stainless steel pipe for the process to complete during a slow flow.

Stirling engine, OK, a tried and true last resort engine, but IR solar cells? Have you got a reputable source for that? You get zero CO, the equillibrium is totally on the side of CO2, but you cannot get rid of the latter, see my answer below.

While a Stirling cycle engine can harness large amounts of low temp energy, it has moving parts.  While we know radiation energy drops with frequency and IR is below visible frequencies, it has no moving parts.  No excuse for intellectual laziness in the Google age: https://www.google.com/search?q=radiant+infrared+power+generating+cells  Last resort?  That sort of thinking gave us centralized coal and oil fired high temp power plants, huge power grids.  Prioritize by appropriateness and effectiveness, please.

If kept at 200°C with water under pressure, any organic molecules break down after a while, that's called hydrothermal carbonisation, basically how coal was formed geologically
With pure sugar (e.g. cellulose), the gross reaction is
$\ce{4CHO -> 3C + CO2 + 2H2O }$.
Lignin is approximately $\ce{C31H34O11}$ (varies a lot of course, but much lower in oxygen than sugar), so ideally
$\ce{C31H34O11 -> 31C + 6H2 + 11 H2O + 0 CO2}$
or
$\ce{C31H34O11 -> 28C + 3CH4 + 11 H2O + 0 CO2}$,
or other hydrocarbons, depending on the exact conditions.
You can also throw in finely shredded plastic, works the same.
The process obviously produces heat, and some $\ce{CO2}$, because once formed, it only very slowly reacts with hydrocarbon to carbon and water (you need higher temperatures and a potent catalyst for that).
Sulfur containing products can be handled by the same processes as in regular coal power plants, burn and wash out and turn into elementar sulfur.

I guess you need to introduce some oxygen to combine with the excess H but not so much it reacts with the carbon or other elements?

One applications would be on a vessel to convert ocean plastics into energy and carbon, which carbon can be discarded to sink, gravity speaking, and the ocean is a great cold sink, thermodynamically speaking, for a heat engine.

Having seen huge piles of un-recycled plastics in the third world and at home, we have a lot of high energy feed stock.  The general trash stream is a larger but lower energy stream, with lots of moisture, metals, other elements that might become toxic compounds and have to be dealt with carefully, responsibly.  Perhaps a galvanic process up front could pull out metals?  Some places pick over trash for recyclables, but some bits, like aluminum foil on materials like paper and plastic, are hard to recycle in current systems.  My town just recycles metal, paper, #1 and #2.

My nephew the Chem prof. suggest this would be lots of good college chem student projects: Trying this with different plastics and waste, different pressures, temperatures, duration, in batch or as a continuous flow.  Engineering, too, chemical and general.


text:
Calculating self-diffusion coefficient from MSD programmatically
According to the Einstein relationship, the diffusion coefficient $D$ is
$$\lim _{t\rightarrow \infty} \frac{\langle \left(\mathbf{r}(t)-\mathbf{r}(0) \right) ^2\rangle}{6t} = D$$
I have run a MD simulation with $N$ particles, and I have a file which has the location of each particle at every time step.
If I want to calculate $D$, I would have to do the following:
$$\langle (\mathbf{r}(t)-\mathbf{r}(0))^2\rangle = \frac{1}{N} \sum _{i=1}^N (\mathbf{r}_i(t)-\mathbf{r}_i(0))^2$$
Divide the above by $6t$, then take the limit as $t$ goes to infinity - which in my case is the final time step of my simulation.
My questions:

Is this the right way of calculating MSD?
If it is the right way of doing this, and my system reaches equilibrium and doesn't diverge, as $t\rightarrow \infty$, shouldn't my diffusion coefficient always be $0$, because of that infinity in the denominator?

In 1D diffusion along the $x$ axis by Brownian motion, let the mean jump lengths be $a$ each taking a time $\tau$ and this should be large compared to the time between successive collisions, which are $\approx 10$ fs, i.e. $a$ is the sum of all displacements during time $\tau$. Assuming this means that there is no correlation between collisions at time $t$ and $t+\tau$. From this the diffusion equation (Ficks Law)  can be derived and the diffusion coefficient defined by $D_x=\langle \Delta x^2\rangle/2\tau$ where $\langle \Delta x^2\rangle$ is the variance of the jump lengths $a$ or $\langle\Delta x^2\rangle = ( x-\langle x\rangle)^2$.  In 3D as there is no correlation between the motions on each axis,  $D=\langle \Delta x^2\rangle/6\tau$. This equation is subtly different from your equation and so you now need to connect these terms with what you have calculated. You should find that your diffusion coefficient settles down to a constant value (plus or minus statistical fluctuations) as the calculation proceeds.


text:
Hybridisation of similar energy orbitals
Hybridisation is the mixing of orbitals of similar energies, so when we perform hybridisation, why don't we prefer $4\mathrm s$ in place of the $3\mathrm d$ orbital for hybridisation, even though $4\mathrm s$ has less and more comparable energy to $3\mathrm p$ and $3\mathrm s$? Why can't we make an outer orbital complex like $\mathrm{sp^3s}$ instead of $\mathrm{sp^3d}$?

https://chemistry.stackexchange.com/a/76740/16683 This sp3d hybridisation thing is complete bogus anyway, so you're on the right track. I'd suggest unlearning these ideas of "outer orbital complex" etc. it is a very poor description of bonding in transition metal complexes. See linked answer for more info.

i didnt get my answer there

i need to know why cant we have an hybridisation of sp3s

There simply isn't a good rule to decide which orbital mixture is a "good" one. The concept works rather perfectly for $2s2p^{1,2,3}$ (covering all possible permutations there), and that has seduced some people to extend the concept into higher orbitals. Many other chemists think they fell into a little trap set up by nature with this extension. ;-)

https://chemistry.stackexchange.com/questions/82799/why-are-d-orbitals-used-in-the-hybridisation-of-pcl5-and-similar-compounds

Does this answer your question? Sulfur trioxide - vacant d-orbitals

M. Farooq is correct that hybridization is a concept that seems to lead students astray more often than not, but your intuition within this simple concept is entirely correct.
When s, p and d orbitals are all involved in bonding (I'll avoid the word "hybridization"), it is most often the orbitals that are close in energy, for example, 4s, 3d and 4p rather than 4d with 4s and 4p. This is why d orbital involvement with bonding is most often discussed with respect to transition metals (d-block elements).
For p-block elements whose only empty d orbitals are from the same shell as the p orbitals (eg 4d with 4p), d orbital participation in bonding is now known to be very limited because those d orbitals are so high in energy relative to the p. The same applies to the s orbitals that you propose to involve in bonding. Unfortunately, many textbooks and teachers still refer to this outdated model of hypervalent sulfur and phosphorus compounds, for example, as using those higher energy d orbitals for bonding, which is likely the source of your confusion.
Once you realize that d orbitals from the same shell that are much higher in energy are not generally involved in bonding, then your question becomes moot.

I would suggest that this entire factory of bogus hybridization stories should be eliminated from general and organic chemistry curricula. The word mixing is worse, as it is a jug where one can add different orbitals and run a blender and here is the mixture. This stuff is good for rote memorization only.
Only after a thorough QM course and mathematical foundation, these ideas should be introduced with an example along with historical development. Neither the textbook author, nor the teacher know what it is. The only real experimental observable is only electron densities via X-ray diffraction or other techniques.

Let us continue this discussion in chat.


text:
How to interpret BRICS fragments
I have a molecule with a SMILES string as given below.
smi='c1ccccc1OCCOC(=O)CC'

I want to obtain BRICS based fragments which I did as follows.
frag=BRICS.BRICSDecompose(Chem.MolFromSmiles('c1ccccc1OCCOC(=O)CC'))

I obtain the following fragments.
{'[1*]C(=O)CC', '[16*]c1ccccc1', '[3*]O[3*]', '[4*]CC[4*]'}

How to interpret [1*], [16*], [3*] ,[4*] ? What does these number with brackets and stars means with each fragment?

Cutting a molecule (expressed with SMILES strings) into fragments yields sub-structures.  Typically these fragments can't be considered as molecules for their open, non-bonding valencies.  Derived from SMILES, it is then the SMARTS notation which is used to represent them, or to search with this notation of sub-structures in databases, e.g. in bioinformatics.
Open Babel supports the display of SMARTS-encoded fragments with branching points marked by an asterisk. The phenyl group, [16*]c1ccccc1 thus is shown as:

Consequently, bidental fragments are represented with two asterisks, e.g., [4*]CC[4*]:

What does these numbers mean with the asterisk?

In the string of [16*]c1ccccc1, according to the Daylight documentation, the * is a placeholder / wild card for any atom. As for the integer, I do not know the answer.  The conversion of the SMARTS string into a .png, the program openbabel does no yield a visual difference in the output for using either [16*]c1ccccc1, or [5*]c1ccccc1, or even [*]c1ccccc1 (no integer at all); a weak evidence that it has have little for this type of processing the input. Perhaps Geoff Hutchison (https://chemistry.stackexchange.com/users/5017/geoff-hutchison) may help here.

Please see the original paper for the interpretations.
Briefly, the numbers correspond to the chemical environment of the BRICS fragments. e.g. the number 16 in [16*]c1ccccc1 indicates that the fragment is an all-carbon aromatic ring system. Below is the illustration of fragment prototypes and their numbers taken from the original paper by Degen et al.


text:
What does 1:4 t-BuOH/H2O mean?
I've got a - probably rather not-so-smart question regarding chemical reactions. In many publications, the authors say that they used e.g. t-BuOH/H2O (1:4) as solvent. What exactly does 1:4 mean? Is it one part of t-BuOH of 4 parts of solvent in total, i.e. $\pu{50 mL}$ t-BuOH and $\pu{150 mL}$ of H2O to give a total volume of $\pu{200 mL}$  (4*50), or is it one part of t-BuOH and four times as much water, i.e. $\pu{50 mL}$ t-BuOH and $\pu{200 mL}$ H2O to give a total volume of $\pu{250 mL}$?
Or otherwise said, do I add both numbers to get the total amount of parts, or is the higher number already the total number of parts? In my first example, we could say, that 4 is the total number of parts. In my second example, you'd have to add both numbers to get the total number of parts, i.e. 1 + 4 = 5 parts in total, one of which is t-BuOH and four of which are H2O.

one part of t-BuOH and four times as much water

In literature, when in experimental section, if it says 1:4 of solvent system but didn't verify, most of the time it means 1 part of one solvent and 4 part of the other solvent in volume units (total: 1+ 4, regardless they are additive or not). This is through specifically in chromatography. If it's not in volume, the authors would definitely verify whether it is $w/v$ or $v/w$ or $w/w$.

As @MaxW already pointed out, 1:4 means that you have to use four times as much water as t-BuOH. But I also want to explain why this is, because you'll probably see this more often.
In fact, 1:4 is a ratio, so what it represents is the ratio between t-BuOH and H2O:
$$
\begin{align}
\text{ratio} &= \frac{V(\ce{t-BuOH})}{V(\ce{H2O})}\\
&= \frac{50\text{ mL}}{200\text{ mL}}\\
&= \frac{1}{4}
\end{align}
$$
And that's why your ratio says 1:4. You could have just rearranged the equation:
$$
\begin{align}
1:4 = \frac{1}{4} = \frac{V(\ce{t-BuOH})}{V(\ce{H2O})}\\
\implies V(\ce{H2O}) = 4 \cdot V(\ce{t-BuOH})
\end{align}
$$
Here, both components, water and tert-butanol, are liquids.  Sometimes the implicit assumption the ratio refering to the ratio of volumes is expressed explicitly by an indication like (tBuOH-water, 1:4, v/v).  Conversely, a mass ratio sometimes may be expressed by m/m.
Side note: As discussed on ChemSE and here, or in the literature (example), be aware of the possibility of a volume contraction.  Thus, measure the volumes of the components independently from each other before pouring them into one container in common (e.g., for chromatography).

+1 for pointing out that exactly 50 ml of t-BuOH and 200 ml water will not yield exactly 250 ml of solution.


text:
Polarity of Methyl Radical
Is methyl free radical a polar molecule?
According to Advanced Organic Chemistry by Jerry March, it is given as non-polar.
But then wat about the moment of the single electron?

The methyl free radical is trigonal planar, hence the non-polarity.

What is the definition of polarity in chemistry? It does not mean anything specific despite its common use. You would rather ask, does methyl free radical have an electrical dipole moment?

I think you are getting confused see firstly methyl free radical is a planar with two hydrogens facing each other . The dipole moment vector of the single electron is away from the central carbon
Carbon has Electronegativity value 2.55 while for H it is 2.2 so all the hydrogens have a small positive charge and the carbon has a small negative charge . the direction of the dipole is from negative to positive so the dipole moment vector of the single electron egts cancelled by the resultant of that of the C-H bonds hence it is a non polar structure


text:
Why is N,N,2,6-tetramethylaniline less basic than N,N,2-trimethylaniline?
According to me, N,N,2,6-tetramethylaniline should be more basic than N,N,2-trimethylaniline due to more steric inhibition to resonance, but the answer is given opposite. What is the reason for that?

It increases basicity as well as acidity.That's the reason why 2-methyl-n,n-dimethylaniline is more basic than aniline

@Safdar SIR = Steric inhibition of resonance decreases lone-pair delocalization and should thereby increase basicity, but the opposite is happening here.

The lack of solvation might be a valid reason since both show SIR.  the more number of steric groups, the harder it is to solvate.[https://chemistry.stackexchange.com/a/8636/95133 ]

@Dylan Rodrigues: "It [SIR?[ increases basicity as well as the acidity...of the conjugate acid."

Steric inhibition of protonation is crucial here.


text:
Raoult's Law confusion
I have this confusion regarding the statement of Raoult's Law. It states that

The partial pressure of each component of an ideal mixture of liquids is equal to the vapour pressure of the pure component multiplied by its mole fraction in the mixture.

Does mole fraction in the mixture mean the combined liquid-vapour mixture or only the liquid which is left after evaporation on attaining equilibrium?
I had this problem in my textbook, one mole of two volatile substances each is mixed. Pure pressures of both of them are given. Should I just use mole fraction as one-half? Aren't some of the moles gonna go into vapour phase?

For the  liquid-vapour mixture to be at equilibrium, the concentrations in the vapor phase and the liquid phase would have to be stable.

Let me put it this way: does the vapour pressure of a mixture depend on how much amount (preferably in moles) of volatile components you put in a closed vessel, or on how much amount of that, which you put, is left in the liquid phase after equilibrium vapour pressure is attained.

You must have heard about the mass conservation law. Gas cannot contain more than you have put there.

It should be like:

The saturated partial pressure of each component of an ideal mixture of liquids is equal to the saturated vapour pressure of the pure component multiplied by its mole fraction in the liquid phase at equilibrium.

The Raoult law relates to an equilibrium state. Therefore saturated partial pressures of components in gaseous phase relate themselves to respective mole fractions of the equilibrium liquid phase composition.

So, how would you answer the question I mentioned I have in my textbook?

I intuitively thought that the mole fraction in liquid will be favoured for the component which is less volatile. Thus, the one-half mole fraction would be changed. But the solution uses the mole fraction of the amount initially put in, which confused me in the first place. This is the very reason I asked the question. Could you tell me if I'm correct in my intuition? Could you summarize it all in one sentence? (Solution = textbook answer, no pun intended)

It was Benzene and Toulene. How am I supposed to know if the gas phase volume is comparable to the liquid or not?

As a complete side note, if they volumes were comparable, how would I calculate the amount that evaporated?

The saturated partial pressure of each component of an ideal mixture of liquids is equal to the saturated vapour pressure of the pure component multiplied by its mole fraction in the liquid mixture. What is this mole fraction? The initial one or when liquid-vapour equilibrium is achieved

@Lllt How long have you been thinking about that? Few minutes  should be more than enough. I will let you extra time.

@Poutnik you say thinking, I have given a week to this question thinking , discussing, questioning but not explanation

See this https://chemistry.stackexchange.com/questions/44773/raoults-law-and-mole-fraction?rq=1

@Poutnik I think that it should be the one when liquid-vapour equilibrium is achieved but there are many questions which make me doubt my thought

@Lllt Became an advocate of both versions, collect evidence speaking for each case and accusations against them. The see which side wins.

@Poutnik Ok I will become advocate for both sides  but can you be the judge so that you could decide are my evidences true or false , accusations correct or incorrect and tell which side has more stronger claims :-)

@Lllt I will not. The whole sense in it is for you to get sure about the answer, not that you are told which one it is. As it is a trivial thought exercise. // Do phase compositions at equilibrium mutually relate to initial or current compositions ?

The equilibrium constant for the reaction gives the Concentrations or mole fractions at equilibrium in both the liquid and vapor phases

@jimchmst What reaction ?


text:
How to name aliphatic chain with carboxyl and amide terminal groups?
I know the name 5-carboxyl-2-methylethanoic acid is wrong for the compound $\ce{HOOC-CH(CH3)-CH2-CH2-CH(CH3)-C(=O)-NH2}$ drawn below. Can we just use 5-amido?

No, amido will not work here.
According to IUPAC 2013,

P-35.3 COMPOUND SUBSTITUENT PREFIXES

P-35.3.1 Names of compound prefixes derived from suffixes or functional parent compounds may be formed by substituting simple prefixes into other simple prefixes. When there is a choice, P-57.4 provides for the selection of a preferred compound prefix.
Examples:
–SSeH   selanylsulfanyl (preselected prefix)
–NH-Cl  chloroamino (preselected prefix)
–NH-CH3 methylamino (preferred prefix)
–PH-Cl  chlorophosphanyl (preselected prefix)

As we have a -CO-NH$_2$ on the 5th carbon (numbered from left to right (as carboxylic acid has higher priority)), it will be named aminocarbonyl because it is a carbonyl group (=O) attached to an amino group (-NH$_2$)
Therefore a systematic IUPAC name can be 5-aminocarbonyl-2,5-dimethylhexanoic acid.
Note that this is a systematic name and might not be the compound's Preffered IUPAC Name (PIN).

The suggested name "5-aminocarbonyl-2,5-dimethylhexanoic acid" $\ce{(C9H17NO3)}$ does not describe the compound $\ce{C8H15NO3}$ that is given in the question.

You have correctly identified the principal characteristic group as acid, which is expressed as suffix. The amide can be expressed as various prefixes.
The corresponding subsection in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-66.1.1.4.1.1 In presence of a characteristic group having priority for citation as suffix or if all carbamoyl groups cannot be included in the suffix, the $\ce{-CO-NH2}$ group is named in three different ways:

by using the two prefixes ‘amino’ and ‘oxo’ to denote such groups on terminal atoms of carbon chains having more than one carbon atom;
by using the acyl group name ‘carbamoyl’ (see P-65.2.1.5);
by using the prefix ‘aminocarbonyl’.

For generation of IUPAC preferred names, method (1) is preferred for chains and method (2) for rings and ring systems, heterogeneous chains, and on nonterminal atoms of carbon chains. Carbamoyl and amino groups may be substituted in the normal way.

The corresponding names for the compound that is given in the question are:

6-amino-2,5-dimethyl-6-oxohexanoic acid (PIN)
5-carbamoyl-2-methylhexanoic acid
5-aminocarbonyl-2-methylhexanoic acid


text:
Transition State search
I've been trying to find a  transition state for the reaction of Pd + F2 with no luck. Has anyone ever tried a similar reaction, have they managed to get the TS? The reaction proceeds from the reactants to a linear reactant complex forming a stable PdF2 product (104.4-degree angle).

Reaction of fluorine and metallic palladium produces palladium(II,IV) fluoride, $\ce{Pd[PdF6]}$ which can further be fluorinated to form palladium tetrafluoride, $\ce{PdF4}$

What search algorithms have you tried?

@jezzo I have started with a linear transit search and found the maximum then ran a TS reaction coordinate, which does not converge.


text:
Why isn't there a hydrogen bond present between the oxygen atom and hydrogen atom of adjacent alcohol groups?

*I know that there is inter-molecular hydrogen bonding between the molecules of glycerol.
*What makes the adjacent alcohol groups behave differently towards each other as compared to when they interact through separate molecules?
*(additional question) - How many Carbon-Carbon single bonds worth do hydrogen bonds persist?

The hydrogen and oxygen are not close enough. If they made a 6-membered ring with the h-bond they might be, but here the best they can do is a 5-member ring and that is too far.

@matt_black 5-membered rings are perfectly fine for h-bonds and OP is spawning nonsense here.

Do the adjacent "OH' have hydrogen bonds between each other?

@Mithoron Do you have an example? Given that a 'strong' hydrogen bond should have a ~180° angle around H, I wouldn't expect five membered rings to be anywhere large enough.

@Martin-マーチン What do you mean by " 'strong' hydrogen bond should have a ~180° angle around H"

@Martin-マーチン Well, I don't say they're strong, still https://www.sciencedirect.com/science/article/pii/S2210271X1000767X

https://chemistry.stackexchange.com/questions/31390/why-does-the-gauche-conformation-of-ethane-1-2-diol-show-hydrogen-bonding

@mithoron indeed, you're absolutely right. It is not strong, bit it would be wrong to disregard this interaction. // With that statement I meant that the hydrogen in the hydrogen bond should ideally be linearly coordinated, i.e. the bond angle $\angle\ce{O-H...O}\approx 180^\circ$.


text:
What gas law is involved when your ears "pop" when you are on a plane high in the air?
When you are up in the air, the air pressure decreases and the altitude increases. What gas law is involved in the situation wherein your ears "pop" because of the difference in pressure.

Related: https://physics.stackexchange.com/questions/187905/why-do-our-ears-pop

I’m voting to close this question because it's off topic. Physics at best, definitely no chemitry.

@Karl to be fair, such things are discussed in any introductory chemistry text.

Karl, I think you will agree that there is no distinct boundary between chemistry and physics.

The temperature and the volume of the inner ear are constant. When your ears pop during descent, air from the cabin goes into the ear, increasing the pressure. The law is the following:
$$n / P = const$$
You can derive this from the ideal gas law. It has no special name.


text:
How is resonance energy measured?
Is it the difference of energies of formation of the resonance hybrid and the average of the energy of formation of all the contributing structures? How would we even calculate the energy of a resonance contributer?

I think what you are asking is: how do we measure the energy of fictitious structures, which don't exist in real life but nevertheless are used to deal with the fact that our model doesn't properly delocalize electrons?

For contributing forms we are confined to calculations. For the rest we can imagine thermochemical cycle, the simplest of which should be that comparing benzene to what would be benzene with three localised double bounds. At least one answer exists here in Chemistry SE (by me) but a search should reveal many related Q & A.

I think this article will be helpful: https://mjcce.org.mk/index.php/MJCCE/article/view/274#:~:text=The%20topological%20resonance%20energy%20%28TRE%29%20of%20a%20catacondensed,%E2%80%930.223%2C%20C%20%3D%200.281%2C%20and%20D%20%3D%200.454.

@Zhe yes that is a nice way to put it

@Reihani thanks for your help but the article only states an approximate formula

I'm sure you have seen those tables in textbooks which list approximate energies of different bond types?

@Karl yes, I have

Then please edit your question and explain where the problem is.

Resonance energy is a fictitious value, it cannot be measured, it is always only estimated. The definition in the gold book[1] clearly states this:

The difference in potential energy between the actual molecular entity and the contributing structure of lowest potential energy. The resonance energy cannot be measured, but only estimated, since contributing structures are not observable molecular entities.

The arising question is how to obtain the energy of the hypothetical contributing structure. One way to do it is to use tabulated (averaged) values for 'normal' bonds to estimate the total energy of the hypothetical molecule.[2]
Another way to do it is to extrapolate from measured hydrogenation energies. This was famously done for benzene. For example, the hydrogenation energy of cyclohexene to cyclohexane is about $\pu{110 kJ mol-1}$. This is the energy to hydrogenate one double bond. In the hypothetical Kekulé structure of benzene you'd expect three of them, giving you a theoretical hydrogenation energy of about $\pu{330 kJ mol-1}$. The measured hydrogenation energy of benzene is about $\pu{200 kJ mol-1}$. This difference of $\pu{130 kJ mol-1}$ can now be taken as the stabilisation due to delocalisation, i.e. the resonance energy.[3]
There is no universal way of doing this, because none of the ways can be proven to be right; it's not real after all. Therefore you may find vastly different values for the resonance energies. It is better to think about resonance as a phenomenological, qualitative argument of mostly pedagogical value. If you are comparing values of hydrogenation (something that can be measured) against extrapolated values (or expectation values) then it is unnecessary to obscure this comparison by giving it another label.

Resonance Energy in: IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. DOI 10.1351/goldbook.R05333

You can find such values for example in this Chemistry LibreTexts article on Bond Energies.

Values are loosely converted to SI units from the article on Chemistry LibreTexts Structure and Resonance Energy of Benzene: A First Look at Aromaticity.


text:
Is this buffer possible?
Question
Can a buffer solution be obtained by mixing-
A) $\ce{NH4Cl}$ + $\ce{CH3COONa}$
B) $\ce{CH3COOH}$ + $\ce{NH4OH}$
My Thoughts
I know that a buffer solution can be obtained if we mix a acid and its salt of conjugate base or a base and its salt of conjugate acid.
But I was wondering whether a buffer solution can be prepared if we mix a weak acid and weak base or if we mix an acidic salt and basic salt?

To be precise, $\ce{NH4Cl}$ and $\ce{CH3COONa}$ are solids and $\ce{NH4OH}$ doesn't exist.

I meant there aqueous solutions. And you can assume $\ce{NH4OH}$ to be $\ce{NH3}$ dissolved in water.

Not really. NH4OH is kind of an immortal myth.

Both cases are equivalent in the case of    molar ratio 1:1. It would lead to a double buffer of $\ce{HAc/Ac-}$ and $\ce{NH4+/NH3}$. But the buffer capacity would be inferior, as the resulting $\mathrm{pH}$ is too far from the both respective $\mathrm{p}K_\mathrm{a}$.

That is correct but I was wondering since we are mixing an acidic and basic substance would they neutralise each other or not (as they are weak acid and base)

They would almost neutralise each other. For there would remain less than 1% of free acetic acid and less than 1% of ammonia, as both pKa are more than 2 units from pH 7 ( 4.75 and 9.25 ).

One more question : will it be a buffer if the molar ratio is not 1:1? If yes, then will it be a better buffer?

If it was a mixture acid : base  2 : 1, that it would be the ideal 1:1 acetic acid/acetate buffer of pH 4.75. If it was a mixture acid : base  1 : 2, that it would be the ideal 1:1 NH3/NH4+ ammonia buffer of pH 9.25, both at their maximum buffer capacity.


text:
Why for ideal gases , dU=nCvdT is always valid ?? I just know the formula but i didn't know the source from where this formula came or derived
So my question is:why ideal gas have same formula for every situation and how the formula is derive???

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic of the question by doing at least basic own topic review, writing what he/she has found and understood, and what is the stumble stone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed.

Question titles should be like book titles. They distinguish the book from
other books, give a hint about content, but do not tell the full story. The content should then elaborate the topic to full depth the author is able to do.

For pure substances and mixtures of constant composition, you will learn later that the more general equation for dU is $$dU=nC_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$For an ideal gas, the second term is exactly zero.


text:
Why do internal energy/heat and enthalphy differ? What is the physical significance of the $PV$ term for enthalphy of an ideal gas?
Let it be noted that this is not a duplicate of this downvoted-to-hell question, simply because I hope to ask it better!
I've been reading a thermo textbook and I've got a simple question.
The enthalpy of an ideal gas is $h = u + PV$. I understand that the specific internal energy, $u$, is the sum of all rotational, vibrational, translational, electronic/nuclear bonding, and lattice (ignored for ideal gasses, I presume) energies of the substance per unit mass. I further understand that $w + q =\Delta u$ for specific transferred  work and heat $w$ and $q$, implying that $\delta w + \delta q = du$ (forgive my deltas in place of a proper inexact differential symbol). Finally, I understand that since $h = u + PV$ and $PV = N\hat{R}T$, for a constant amount of ideal gas, $PV$ is solely a function of temperature. It can be demonstrated that for an ideal gas, $u$ is also a function solely of temperature, so that means $h$ is as well.
So what's the difference between $u$ and $h$? Clearly, it's that $PV$ term, but what does that physically represent? I don't grok it.
All incoming energy, whether mechanical or thermal, goes into a change of $u$--that makes sense. Add energy to a gas, it's going to vibrate/translate/rotate more & maybe change its bonding state. Say you adiabatically compress your gas. You do work on it and $u$ and $T$ rise, but $PV$ stays constant. Adiabatically unconfine the gas and the same thing occurs the other way. But $h$ still varies because $u$ does, it's not conserved over adiabatic actions.
So the "energy" of the $PV$ term represents something somehow separate to internal energy, inherent to the system, invariate to mechanical work done on it... unless that work ends up raising the temperature? What is this term physically represent? Where does it exist? Why bother with the enthalpy concept at all? I don't get it & I know I'm missing something fundamental.

Enthalpy is just a convenient function to work with in solving many kinds of thermo problems, particularly those that take place at constant pressure, so $P\Delta V$ is the work done.  My advice is not to drive yourself crazy trying to assign a deeper meaning to the PV.  As you get more experience solving thermo problems, you will see how enthalpy comes in.  Otherwise, I think you will be wasting your valuable time obsessing over this.

@chetmiller I couldn’t write a better answer.

Well, can be done in more straightforward way. U is total mass energy inherent to an object and H is mass energy of object that is compressed (or perhaps stretched).


text:
Why oxidation number is that important?
Oxidation number looks all but a convenient accounting tool for balancing chem equations. Or does it? Why would a molecule care about its oxidation state? Is a different oxidation state sort of a different chem specie with specific properties, reactions, etcetera... Does an oxidation state tend to maintain itself through "weak" exchange processes till some "strong" electron pushing/withdrawing agent gets into? Is it sort of dividing species into carnivores and herbivores?

Related: https://chemistry.stackexchange.com/questions/61040/what-are-oxidation-states-used-for

Thank you for the reference. It seems quite enough, indeed.


text:
How to calculate Hirshfeld charges from Gaussian?
I am new to Gaussian and want to calculate Hirshfeld charges from Gaussian 09 or 16.
I found this input file for Methanol:
%nproc=8
%mem=1600mb
# mPW1PW91/6-311+G(2df,2p) pop=Hirshfeld scf=tight
Methanol
0 1
8 -0.046153000 -0.751257000 0.000000000
6 -0.046153000 0.658469000 0.000000000
1 0.857779000 -1.062372000 0.000000000
1 -1.086394000 0.976180000 0.000000000
1 0.437376000 1.072716000 0.889654000
1 0.437376000 1.072716000 -0.889654000

but when I run it I get this error:
QPErr --- A syntax error was detected in the input line.
 p) Pop=Hirshfeld scf=tight Methanol 0 1
                            '
 Last state= "GCL"
 TCursr=      3523 LCursr=        51
 Error termination via Lnk1e in /opt/software/gaussian/g09.e01/l1.exe at Wed Aug 19 01:02:31 2020.
 Job cpu time:       0 days  0 hours  0 minutes  0.6 seconds.
 File lengths (MBytes):  RWF=      5 Int=      0 D2E=      0 Chk=      1 Scr=

Any suggestion which part is the problem?

%nproc=8
%mem=1600mb
# mPW1PW91/6-311+G(2df,2p) pop=Hirshfeld scf=tight

Methanol

0 1
8 -0.046153000 -0.751257000 0.000000000
6 -0.046153000 0.658469000 0.000000000
1 0.857779000 -1.062372000 0.000000000
1 -1.086394000 0.976180000 0.000000000
1 0.437376000 1.072716000 0.889654000
1 0.437376000 1.072716000 -0.889654000

will do. All I did was to introduce blank lines after Route section (the one starting with #) and Title section (Methanol). Also, make sure that there is a blank line after the Molecule specification part. Gaussian is usually a convenient entry point to computational chemistry and you can find a lot of advice for it. The Gaussian manual is not bad, either. Have a look at this to get an impression about the syntax of input files - this is something Gaussian isn't forgiving about.


text:
Find volume of ammonia gas required to prepare a solution of given pH

Calculate the volume of gaseous ammonia $(K_\mathrm{b} = \pu{1.8E-5})$ to dissolve (at NTP) in $\pu{100 mL}$ of water in order to obtain a solution with $\mathrm{pH} = 11.0.$ Consider volume variation negligible.

I started by evaluating molar concentration of $\ce{NH4+}$ and $\ce{OH-}$ knowing the $\mathrm{pH},$ then I used the $K_\mathrm{b}$ inverse formula to evaluate the molar concentration of ammonia. These are the results:
$$c(\ce{NH3}) = \pu{0.056 mol L-1}$$
$$c(\ce{NH4+}) = c(\ce{OH-}) = \pu{1E-3 mol L-1}$$
Then my idea was to work with water. Since $\ce{NH3}$ is a weak base it does not dissociate completely, the amount of water is
$$n(\ce{H2O}) = \frac{\pu{100 g}}{\pu{18.02 g mol-1}} = \pu{5.55 mol}$$
Then I run out of ideas. I tried to find the mamount of ammonia (but I think the result was not correct) by evaluating the degree of dissociation, but then I got lost.
The final result of the problem is $\pu{0.13 L}.$ How do I pursue the problem further?

I tried also with ideal gas equation. Probably I did some mistake in mole calculation of ammonia moles

Probably I solved it.
Molarity of ammonia = 0.056
Sinche the volume variation is negligible and 0.056 are the moles of ammonia required to form a liter of solution, the volume of the final solution will be 0.1L.
So, the number of moles of ammonia required to have a final volume of 0.1L are:
Molarity of ammonia * V final solution = 0.056*0.1 = 0.0056
Putting this number of moles in V=nRT/P the result is 0.1256
Is it correct?

Henry constant is not needed.

Is my comment correct?

If you cannot decide yourself, you do not understand what you have done yet. For numbers to be correct, your algebra must be correct. Show your algebra.( in the question ,not in comments ). Be aware SE Chemistry community is not much interested in validation of particular numerical results, but rather in the Q author understanding of the principles.

@Poutnik , why is henry constant not needed?

@Jayadithya We are not asked what ammonia partial pressure is needed to achieve the desired ammonia molarity. We are asked for the ammonia volume at STP. The Henry constant would still matter if the ammonia solubility were the limiting factor to verify the state is ever possible, i.e. what is the equilibrium ammonia molarity with STP ammonia. But this is not the case.

The $\mathrm{pH}$ of your final ammonia solution is $11.0$. When you dissolve ammonia in water it tends to ionize with water to give following equilibrium:
$$\ce{NH3 + H2O <=> NH4+ + OH-} \quad K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = 1.8 \times 10^{-5}$$
Suppose final concentration of $\ce{NH3}$ is $c$, and equlibrium concentrations of $\ce{NH4+}$ and $\ce{NH4+}$ are $\alpha$. Thus:
$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} = \frac{\alpha \cdot \alpha}{c - \alpha} = 1.8 \times 10^{-5}$$
$$\alpha^2 + 1.8 \times 10^{-5}\alpha - 1.8 \times 10^{-5}c =0 \tag1$$
Since, $\mathrm{pOH}$ of the solution is $3.0$ ($14 - \mathrm{pH}$), $[\ce{OH-}] = \alpha = 1.00 \times 10^{-3}$, from equation $(1)$:
$$(1.00 \times 10^{-3})^2 + 1.8 \times 10^{-5}\times 1.00 \times 10^{-3} = 1.8 \times 10^{-5}c $$
$$1.8 \times 10^{-5}c = 1.00 \times 10^{-6} + 1.8 \times 10^{-8} = 1.018 \times 10^{-6} $$
$$\therefore \ \ c = \frac{1.018 \times 10^{-6}}{1.8 \times 10^{-5}} = \pu{5.66 \times 10^{-2} molL-1} $$
This is exactly what you got without showing your work. However, you made only $\pu{100 mL}$ of solution. Thus, amount of ammonia dissolved in $\pu{100 mL}$ of water is:
$$$\pu{0.100 L} \times \pu{5.66 \times 10^{-2} molL-1} = \pu{5.66 \times 10^{-3} mol}$
Assuming gaseous ammonia is behaving like ideal gas, you can use $PV = nRT$ here to find the volume of $\pu{5.66 \times 10^{-3} mol}$ of ammonia at STP:
$$V= \frac{nRT}{P} = \frac{(\pu{5.66 \times 10^{-3} mol})(\pu{0.082 LatmK-1mol-1})(\pu{298 K})}{\pu{1 atm}} = \pu{0.138 L}$$


text:
Why are amino acids least soluble at isoelectric point?
At isoelectric point amino acids exist as zwitterions. As a molecule it is electrically neutral, but still has a positive and a negative charge. Then why are amino acids still least soluble at isoelectric point?
The positive and the negative charges of the amino acid can interact with the negative and the positive charges of water and can dissolve, right? Where am I going wrong?

They can and do dissolve. But at other conditions they dissolve even better.

It has also the other side, their mutual interaction within the solid lattice is even more stronger.

you are probably going wrong because you are not considering the pH of the solution

In order for molecules to precipitate out of solution, they need to aggregate together. Amino acids that have zero net charge can aggregate together much more easily than those that are charged.
Molecules that have net charge need counterions to aggregate with them to offset the charges or the electrostatic repulsion will be too great.
The concept of isoelectric point is more often applied to polypeptides and proteins than to individual amino acids. In those much larger molecules, the tendency to aggregate when there is no net charge can be quite high. This is especially true if there are any hydrophobic regions that can favorably associate with each other and reduce the interaction with water.
It is important to note, as mentioned in the comments, that this does not mean that all amino acids and proteins are insoluble at their isoelectric point, simply that they tend to be less soluble at that pH than at pH values where they carry net charge.

Good answer. Ions don’t form solids (they need to combine with counterions, which typically involves decreasing entropy).


text:
Pseudo chiral centre
I was reading about stereoisomers and I came across a term pseudo chiral centre.  I looked up on the net for its definition and found a weird formula for calculating no. of isomers if the compound has a pseudo chiral centre.
I can't understand how to deduce the formula and how to do analytically instead of using the formula in case of complex molecules say of more than 3 chiral carbon atoms.

Suppose you have the following compound,4-Methylhepta-2,5-diene.The central carbon atom-C 4 is  pseudo chiral because it is chiral only when the stereo centers at C2-C3 and C5-C6 are of different configuration other wise it is not chiral

See the gold book definition of pseudo-asymmetric carbon atom.

Yes I saw the definition but how to deduce the formula

It's something dependent on number of stereocentes

Carbon number 4 is a pseudo chiral center because its chirality depends on the chirality of the second and sixth carbons. For example, if the second carbon is S and the sixth carbon is R, then carbon number 4 can be r or s, but if the chirality of the second and sixth carbons are the same, then carbon number 4 is not the chiral center, because it doesn't have 4 different groups around itself.

When C2/C6 are of different configurations (R/S), C4 will be stereogenic and achirotopic (r/s). When  C2/C6 are of the same configurations (R/R or S/S), C4 will be chirotopic and non-stereogenic. There is no r/s assignment. See: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/RSrs.html

Pseudoassymetric center is the carbon that show chirality and achirality simultaneously.
The r/s (not R/S) nomenclature of carbon 4 depend on R/S configuration of directly attached carbon atom.
In the given molecule the carbon 4 is attached with constituitionaly same but configurationaly different group so it is a chiral center (note: R configuration  has priority over S configuration). if we imagine it has a plane of symmetry and that make it achiral.

What is the "given molecule" that you are referring to?

It is good that (interpretation reading your answer) that you discern chirality at the level of an atom, and its neighbours from chirality of the molecule.  The answer would improve if there were illustrations to support the argumentation. But then, if you focus on the lower level, i.e. environment around an atom, it is more than puzzling than atom should simultaneously experience an spatial arrangement both chiral and achiral when the leading «a» in «achiral» denotes the absence of chirality; the two are mutually exclusive.  Like e.g., is an information public .or. private.


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Why does branching increase stability?
The heat of combustion of neo- pentane is lower than n-pentane.This clearly means that neo-pentane is more stable than n-pentane.What is the reason for this?

You mean heat of formation! Neopentane can't be hydrogenated. In the future, you need to make an effort to address the issue. Try, https://chemistry.stackexchange.com/questions/123093/deciding-the-order-of-heat-of-combustion-of-isomeric-alkanes/130811#130811

Sorry,i meant heat of combustion

Deciding the order of heat of combustion of isomeric alkanes.

If the heat of combustion of neopentane is lower than that of n-pentane, then that is a fact, but it doesn't necessarily mean that one is more stable than the other in a general sense. It is only more stable with respect to that specific reaction.


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What is the correct expression of pressure-volume work?
My book defines work in pressure-volume work as:
$$\mathrm{d}w = - P_\mathrm{ext} \mathrm{d}V$$
However before doing so it mentioned the piston to be massless.
In physics, work is defined as
$$\mathrm{d}w = P_\mathrm{in} \mathrm{d}V$$

What is the correct definition of pressure-volume work in chemistry?
Does the definition of work, $$-P_\mathrm{ext} dV$$ change if the piston was not massless in chemistry?

P.S. My question was mainly pointed towards the mathematical expression of work in Chemistry,that is, 'What is the most general work expression in Chemistry in relation with the First Law'.
Some of the users have been suggesting me to check out answers on sign convention of work. I do appreciate it but my question is different and is perhaps best interpreted in @Chet Miller answer.

It the case $p_\mathrm{ext} \ne p_\mathrm{int}$ the work done on the piston is not the same as the work done on the system, as the piston itself is not considered as a part of the system.

Similarly, the work the system performs on the piston may not be the same as the work the piston performs on the neighbourgood. Imagine vacuum behind the piston. The work in TD context must be considered in relation to the change of the system internal energy. The massless piston simplifies such system-piston-neighbourhood quicks.

@Poutnik So the definition of work by external pressure is true only for ideal case?

If you think about it, you may answer yourself. There is no guarantee the energy $p_\mathrm{ext} \cdot \mathrm{d}V$ is passed to the system. It is the work done by the neighbourhood on the piston, but not necesserily the work done on the system by the piston.

At the inside piston face, the pressure exerted by the gas on the piston is equal to the pressure exerted by the piston on the gas. This just follows from Newton's 3rd law of motion.  But, in an irreversible process, the pressure exerted by the gas on the piston face cannot be obtained from the ideal gas equation (or other real-gas equation of state).  This is because the equation of state applies only at thermodynamic equilibrium.  Otherwise, the gas pressure on the piston involves viscous stresses which depend not only on the volume but also on the rate of change of gas volume.  For a reversible process, the gas pressure throughout the gas, including at the inside piston face can be determined using the equation of state, since a reversible process consists of a continuous sequence of thermodynamic equilibrium states.
In using these work relationships, it is important to specify what you are calling the system. If the system includes only the gas, and the piston is massless and frictionless, the pressure on the outside piston face is equal to the pressure on the inside piston face (and the gas pressure).  If the piston is included as part of the system, and the piston is not massless or frictionless, then the external pressure is the pressure on the outside piston face, but this is not equal to the gas pressure on the inside piston face (or, equivalently, the piston pressure on the gas).

Sir does that mean when we say work done on the system we mean 'Work done by Piston on the gas' keeping the system as gas and when we say work done by the gas we mean 'work done by the gas on the piston' ?Further both of these works are always equal and opposite following Newton's Third Law?

Yes, that's what it means.

You may notice that there are two conventions for defining the work $p\Delta V$. The same work is sometimes positive sometimes negative, for the same process. Let's be more precise.
Usually, the work is positive when it is given or added to the system (syringe), when the system is compressed. Work and heat are both positive when they are added to any system compressed by a piston. But, as $V$ decreases in such a compression, $\Delta V$ is < $0$ and the work $p\Delta V$ must be written with a minus sign to stay positive : $\delta w = - ~p\Delta V$. So the sum of all energies entering the system is the change of internal energy $U$ : $\Delta U = \delta w + \delta q = + ~ \delta q - p \Delta V$.
For some scientists, mainly engineers, the approach is different. The whole system is considered as a machine, where you put heat inside, so that it must produce as much work as possible to be useful ($\delta q > 0$). So the work done by the machine is positive if it works correctly, which happens if $\Delta V$ is positive. For engineers, the work is defined as : $\delta w = + ~p ~\Delta V$. As a consequence, the internal energy is the difference of the energy added as heat to the system, and the energy delivered b the machine. It is then $\Delta U = \delta q - \delta w = \delta q - p\Delta V$.
The final result is the same all over the world when speaking of change of internal energy, and of course of enthalpy. But the sign of the work is not always the same.


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Why is the test for sodium as dihydrogen antimonate so specific?
I have been attending lectures of salt analysis in my college. We are at group 6 cations or also known as group zero cations, namely $\ce{NH4^+}$, $\ce{K^+}$, $\ce{Na^+}$, $\ce{Mg^2+}$.
The lecturer wrote an equation:
$$\ce{NaCl + KH2SbO4 -> NaH2SbO4 (s) v  + KCl}$$
Now, I know that the electrochemical series follows the order $\ce{Li}$, $\ce{K}$, $\ce{Ba}$, $\ce{Sr}$, $\ce{Ca}$, $\ce{Na}$, $\dots$.
I would like to know why this reaction occurs so spontaneously that it is included as one of the lab tests. Sodium chloride is completely neutral whereas potassium dihydrogen antimonate is basic.

It has no (direct) relation to the electrochemical series nor to acido-basic properties.
It is not a true chemical reaction, rather just crystallization of the least soluble salt from the available ions.
There are no distinguished molecules $\ce{NaCl}$, $\ce{KH2SbO4}$, $\ce{NaH2SbO4}$, $\ce{KCl}$ in the solution, but mutually independent hydrated ions $\ce{Na+}$,  $\ce{Cl-}$, $\ce{K+}$, $\ce{H2SbO4-}$ instead (plus some minor hydrolysis products). Which salt precipitates first depends on their solubility products. $\ce{NaH2SbO4}$ has smaller solubility product than $\ce{KH2SbO4}$.
Similar things would happen if you mix solutions of potassium  chloride and sodium nitrate ( chlorate, perchlorate ). If original solutions are concentrated enough, and especially if you cool the mixture below ambient temperature, there will precipitate the respective potassium salt with the oxoanion.
Another case is the $\ce{NaCl}$ solution  with dissolved $\ce{NH3}$ And $\ce{ CO2}$, with ions $\ce{Na+, NH4+, Cl^-, HCO3-}$, leading to precipitation of $\ce{NaHCO3}$ and solution of $\ce{NH4Cl}$. It is used commercially in the Solvay process
producing sodium carbonate from sodium chloride.

Is there a way to  predict/ is there any general order for the $Ksp$?

There is much less general rules in chemistry than chemists would like. This is such a case of a missing rule. Just partial patterns.


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Cross-conjugation and aromaticity in pyrene
I have read about non-huckel double bonds and clar's rule, but what made me confuse is that huckel's perception gives me the feeling that central bonds in polycyclic aromatic hydrocarbons (ex: pyrene) is simply minimally affecting  aromaticity or even dropped off, while clar's rule shows different resonance forms which in total makes a complete sextet formed in each ring of pyrene.

So how could i resolve this issue?

what would happen to aromaticity if simply the central double bond was not in the pyrene?

is there any cross-conjugation involved? and how would cross-conjugation affect aromaticity?

have you seen https://chemistry.stackexchange.com/questions/16562/is-pyrene-aromatic-despite-failing-h%C3%BCckels-rule

@AniruddhaDeb Yes but the answers don't show the way both rules (clar and huckel) are applicable in the same molecule (pyrene) although they don't seem so. And they don't show the effect of central double bonds on stability as well as aromaticity and whether cross-conjugation is involved or not?