text:
Glutaraldehyde or Glutardialdehyde
Recently I was searching for a chemical compound to meet my research needs, i.e. glutaraldehyde, and later I found that Merck has been released different name of compound: (1) glutaraldehyde and (2) glutardialdehyde, with the same CAS number, i.e. 111-30-8:

Many articles I've read, just mention glutaraldehyde, not glutardialdehyde.
Does glutardialdehyde and glutaraldehyde are the same compound?

if the Merck says they are the same and they have the same CAS number, they are the same. I have never heard it referred to as glutardialdehyde and I worked for the business unit that made it. Merck is just trying to be "inclusive".

IMO, glutaraldehtyde follows from glutaric acid, which is already difunctional. Glutardialdehyde appears to be redundant.

Many articles I've read, just mention glutaraldehyde, not glutardialdehyde.

Not only Merck, but also Sigma-Aldrich listed glutardialdehyde (25% solution) and glutaraldehyde (25% solution) under the same CAS # (111-30-8). Surprisingly, glutaraldehyde is more expensive (USD 48.70/100 mL) than glutardialdehyde (USD 34.10/100 mL)! :-)

Does glutardialdehyde and glutaraldehyde are the same compound?

Answer to this question is yes. However, the preferred name for the compound glutardialdehyde or glutaraldehyde (CAS #: 111-30-8) with the formula $\ce{OHC-(CH2)3-CHO}$ is pentanedial (Ref.1). Other names for the same compound included glutaraldehyde, glutardialdehyde, glutaric acid dialdehyde, glutaric aldehyde, glutaric dialdehyde, and 1,5-pentanedial (Wikipedia).
References:

Nomenclature of Organic Chemistry : IUPAC Recommendations and Preferred Names 2013 (Blue Book); Prepared by Henri A. Favre and Warren H. Powell; The Royal Society of Chemistry: Cambridge, United Kingdom, 2014, p. 907 (ISBN: 978-0-85404-182-4).


text:
Effect of Solvent - Nucleophilicity of RS- and RO-
While reading on order of nucleophilicity of $\ce{RS-}$ and $\ce{RO-}$, I came across this:
Source: Page 355, Organic Chemistry by Jonathan Clayden, Nick Greeves, and Stuart Warren
However, my teacher has said that in polar protic medium, nucleophilicity of $\ce{RS-}$ dominates over $\ce{RO-}$ due to the heavy solvation of oxygen and the relatively low solvation of sulfur. And, in polar aprotic medium, due to the high potential energy of the $\ce{RO-}$ anion, it becomes a better nucleophile than $\ce{RS-}$ anion.  (Just like the order of nucleophilicity halide anions).
Although they have given data for reaction in polar protic ($\ce{EtOH}$) medium in the book, I feel like nucleophilicity of $\ce{RS-}$ should still be higher than $\ce{RO-}$, independent of the medium, because of the reason of HOMO-LUMO interaction mentioned in the book (which should be independent of the solvent)

Also, electrostatic interactions must be a less dominating factor than HOMO-LUMO interactions for $S_N 2$ reactions as the C-X bond is not that polar.
In the above data of reaction with $\ce{MeBr}$, the difference in rate is in the order of $10^4$. I feel like that must account for something more than the effect of solvent.
However, I could not find data for the rate of $S_N 2$ in polar aprotic medium. So, in polar aprotic medium, which of $\ce{RO-}$ or $\ce{RS-}$ is more nucleophilic and why?

The statement in the textbook may not be all that general. And nucleophilicities may reverse for polar aprotic and, especially, nonpolar aprotic solvents, more so at lower concentrations (trying my best to simulate gas-phase like conditions here).
Although I was not able to find a study performed in polar aprotic or nonpolar aprotic solvents, I did find one performed in gas phase where $\ce{HO-}$ is a much better nucleophile than $\ce{HS-}$.$^\text{1}$ I have attached some values of bimolecular rate coefficients for the following reaction:
$$
\ce{RX(g) + Nu-(g) ->RNu(g) + X-(g)}
$$

Alkyl Halide
$\ce{HO-}$
$\ce{HS-}$

$\ce{CH3Cl}$
20
0.12

$\ce{CH3Br}$
22
3.2

$\ce{C2H5Cl}$
25
<0.001

$\ce{C2H5Br}$
26
2.0

I know this doesn't exactly answer your question; however, it helps to know that sulfur is not plainly a better nucleophile than oxygen for saturated carbon.
Reference

DePuy C. H., Gronert S., Mullin A., and Bierbaum V. M. (1990). Gas-phase SN2 and E2 reactions of alkyl halides. J. Am. Chem. Soc., 112, 8650-8655.

That does seem a bit convincing, atleast to know that in gas phase the order reverses, even if by a slight amount. Thanks for the answer! I will keep this question open if someone does turn up with data for polar aprotic medium.

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text:
If Qc> Kc how can forward reaction can take place
If Qc>Kc then there will be a net backward reaction but still forward reaction will take place. My problem is if Qc>Kc forward reaction will become non spontaneous right. So how can that happen, does it like the gibbs free energy which released during backward reaction absorbs by products for forward reaction?

Please define Qc and Kc

Molecules do not know Qc/Kc ratio. nothing tells them "Now you can react." nor "Now you cannot." // Trolleybuses do not spontaneously climb uphill. Yet 2 or 3 ones going downhill can create by recuperation enough energy for one going uphill.

A Graph
I haven't seen this graph anywhere, so I thought it would be interesting to draw it. Consider a simple reversible reaction:
$$
\ce{A <=>[k_\text{f}][k_\text{b}] B}
$$

Time
$\ce{[A]}$
$\ce{[B]}$

$0$
$1$
$0$

$t$
$1-\xi$
$\xi$

$t_\text{eq}$
$\dfrac{k_\text{b}}{k_\text{f}+k_\text{b}}$
$\dfrac{k_\text{f}}{k_\text{f}+k_\text{b}}$

Reaction quotient at time $t$ is defined as
$$
Q_\text{c} = \dfrac{\ce{[B]}}{\ce{[A]}}=\dfrac{\xi}{1-\xi} \tag{1}
$$
Forward and backward rates at time $t$ are given by:
$$
R_\text{f} = k_\text{f}\ce{[A]} = k_\text{f}(1-\xi); R_\text{b} = k_\text{b}\xi \tag{2}
$$
From Equations (1) and (2):
$$
R_\text{f} = \dfrac{k_\text{f}}{Q_\text{c}+1}; R_\text{b} = \dfrac{k_\text{b}Q_\text{c}}{Q_\text{c}+1}
$$

You can play around with this interactive plot.
Some More Clarification
There is a reason we draw the double, forward-backward arrow in $\ce{A <=>[k_\text{f}][k_\text{b}] B}$: it shows that the reaction proceeds in both directions simultaneously.

When $Q_\text{c}<K_\text{c}$, forward reaction rate is greater than backward reaction rate:

$$
\ce{A <=>>[k_\text{f}][k_\text{b}] B}
$$

When $Q_\text{c}>K_\text{c}$, backward reaction rate is greater than forward reaction rate:

$$
\ce{A <<=>[k_\text{f}][k_\text{b}] B}
$$
Equilibrium Conditions
Equilibrium is achieved when forward and backward rates are equal:
$$
R_\text{f} = R_\text{b}\implies k_\text{f}\ce{[A]_\text{eq}} = k_\text{b}\ce{[B]_\text{eq}}\\
K_\text{c} = \dfrac{k_\text{f}}{k_\text{b}} = \dfrac{\ce{[B]_\text{eq}}}{\ce{[A]_\text{eq}}}
$$
Free Energy
Since both processes are spontaneous, $\Delta G$ for both processes, from nonequilibrium state to the equilibriums state, is negative. Free energy is minimum at equilibrium.

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text:
What are the dissociation constants of hydronium and hydroxide ions?
Context
The answer to How to calculate Ka for hydronium and Kb for hydroxide?, water was regarded as solvent once and base (or equivalently acid) once while citing Reference 1 (1986) and made no comments on the dissociation constants of $\ce{H3O+}$ and $\ce{HO-}$ in water.
$$
K_{\text{a,} \ce{H2O}} = K_{\text{b,} \ce{H2O}} = \dfrac{K_\text{w}}{\ce{[H2O]}} \approx \dfrac{10^{-14}}{\ce{[H2O]}} = ; \text{room temperature}
$$
The question was marked duplicate to What is the pKa of the hydronium, or oxonium, ion (H3O+)?, and the answer disregarded auto-protonation of water as a thermodynamic process while citing Reference 2 (1998). However, since the reference did not consider it, the answer also did not contain any information on the dissociation constant of $\ce{HO-}$.
I don't think these questions are duplicate since these have been answered differently, and neither provide dissociation constants for both $\ce{H3O+}$ and $\ce{HO-}$. Moreover, neither question or answers have considered the relation:
$$
K_\text{a}K_\text{b} = K_\text{w}
$$
I think there is room for further discussion here.

Question
Are there any more recent developments on this topic? Has a definitive answer been provided somewhere?
References

Starkey, R., Norman, J., Hintze, M. (1986). Ronald Starkey, Jack Norman, and Mark Hintze. J. Chem. Educ., 63(6), 473. 10.1021/ed063p473
Burgot, J. (1998) New point of view on the meaning and on the values of Ka○(H3O+, H2O) and Kb○(H2O, OH–) pairs in water. Analyst,123, 409-410. 10.1039/A705491B

Housecroft and Sharpe$^\text{3}$
Formulation of $K_\text{a}$ and $K_\text{b}$:$^\text{3}$

This seems to side towards Reference 1 (1986) and not Reference 2 (1998):
For general reactions $\ce{HA + H2O -> H3O+ + A-}$ and $\ce{B- + H2O -> HB + HO-}$, $K_\text{a}$ and $K_\text{b}$ are given by:
$$
\dfrac{\ce{[H3O+][A-]}}{\ce{[HA]}} \equiv \dfrac{[\text{Conjugate Acid}][\text{Conjugate Base}]}{[\text{Acid}]} \tag{1}
$$
and
$$
\dfrac{\ce{[HB][HO-]}}{\ce{[B-]}}\equiv \dfrac{[\text{Conjugate Acid}][\text{Conjugate Base}]}{[\text{Base}]} \text{,} \tag{2}
$$
respectively.
Self-Ionization of Water
Consider the following:
$$
\ce{\underset{\text{acid}}{H2O} + \underset{\text{base}}{H2O} -> \underset{\text{conjugate acid}}{H3O+} + \underset{\text{conjugate base}}{HO-}}
$$
$K_{\text{a,} \ce{H2O}}$, from Equation (1), is given by:
$$
K_{\text{a,} \ce{H2O}} = \dfrac{\ce{[H3O+][HO-]}}{\ce{[H2O]}} = \dfrac{K_\text{w}}{\ce{[H2O]}}
$$
$K_{\text{b,} \ce{H2O}}$, from Equation (2) is given by:
$$
K_{\text{b,} \ce{H2O}} = \dfrac{K_\text{w}}{\ce{[H2O]}}
$$
An Example
Consider this example:

Calculation for $\ce{H3O+}$
If we consider our acid as $\ce{H3O+}$:
$$
\ce{H3O+ + H2O -> H3O+ + H2O}\implies
K_{\text{a,}\ce{H3O+}} = \dfrac{\ce{[H3O+][H2O]}}{\ce{[H3O+]}} = \ce{[H2O]}\\
$$
This is in agreement with $K_\text{a}K_\text{b} = K_\text{w}$
$$
\implies K_{\text{a,}\ce{H3O+}}K_{\text{b,} \ce{H2O}} = \dfrac{K_\text{w}}{\ce{[H2O]}}\ce{[H2O]} = K_\text{w}
$$
Calculation for $\ce{HO-}$
If we consider our base as $\ce{HO-}$:
$$
\ce{HO- + H2O -> H2O + HO-}\implies K_{\text{b,}\ce{HO-}} = \dfrac{\ce{[HO-][H2O]}}{\ce{[HO-]}} = \ce{[H2O]}
$$
This is in agreement with $K_\text{a}K_\text{b} = K_\text{w}$
$$
\implies K_{\text{a,}\ce{H2O}}K_{\text{b,} \ce{HO-}} = \dfrac{K_\text{w}}{\ce{[H2O]}}\ce{[H2O]} = K_\text{w}
$$
Remarks
I am siding with Housecroft and Sharpe and Reference 1 (1986) and not Reference 2 (1998) because of the agreement with $K_\text{a}K_\text{b} = K_\text{w}$. However, this would also mean:
$$
\boxed{\mathrm{p}K_{\text{a,} \ce{H3O+}} = \mathrm{p}K_{\text{b,} \ce{HO-}} \approx 3.16 \times 10^{-56}}
$$
as stated in Reference 1.
References

Starkey, R., Norman, J., Hintze, M. (1986). Ronald Starkey, Jack Norman, and Mark Hintze. J. Chem. Educ., 63(6), 473. 10.1021/ed063p473
Burgot, J. (1998) New point of view on the meaning and on the values of Ka○(H3O+, H2O) and Kb○(H2O, OH–) pairs in water. Analyst,123, 409-410. 10.1039/A705491B
Housecroft C. E., Sharpe A.G. (2008). Inorganic Chemistry, 3$^\text{rd}$ ed. Pearson  Education Ltd.

Ananta, You appear to be a very keen student of science which is good. This is not for criticism but what I have learned over time. Science, chemists, textbook authors do not know everything. However, sometimes these debates become debates of questions like, "How many angels can land on a pinhead"? Water is one of the miracles of creation. Its bulk structure is still full of active research in physical , and there is perhaps nothing such as $\ce{H3O+}$ in water. Chemists do have a tendency to talk about a lot of stuff without having any decent knowledge about it.

@AChem I never knew about non-existence of $\ce{H3O+}$, this would be a very interesting topic to discover. Could you provide some citation that could guide me to it?

It is certainly an interesting topic but out of my area of expertise but I have attended several advanced talks on the structure of water which perplexes best of the best. What I am trying to teach is to be skeptical. By that I mean don't take any "answer" as final. I meant that $\ce{H3O+}$ does not exist independently, it is surrounded by many many water molecules, in a dynamic structure. See solvation of hydronium here: https://en.wikipedia.org/wiki/Hydronium. One may discuss hydronium in non-aqueous systems, then it makes more sense. See https://pubs.acs.org/doi/abs/10.1021/ja00508a040

@AChem thank you. Although I have provided an answer, i don't consider it as the answer. As I stated in the question, I think there is room for further discussion here. I am as skeptical as a sparrow going to pick a grain next to a snake's lair on this one.

For a serious research, which may be too advanced for you, one may start physical chemistry journals of ACS and other journals. Hydronium ion is being discussed for more than hundred years. So I feel, it is meaningless to even worry about its pKa for most purposes, even if some textbook casually mentioned it. Also learn to ignore the down voters which may pour in randomly for no rhyme and reason.

Many chemists consider the hydratation of $\ce{H+}$ goes much further than $\ce{H3O+}$, so that $\ce{H3O+}$ must be replaced by $\ce{H5O2^+}$ (with a proton between two $\ce{H2O}$ molecules), or even more complicated structures like $\ce{H13O6^+}$ (with one proton between six $\ce{H2O}$ molecules). Serious arguments may be given in favor of these complicated structures. But for the ordinary analyst, they are not fundamental. For titration purposes, it is sufficient to know that the acidic proton is hydrolyzed, whatever the number of water molecules attached to it.


text:
Is it possible for a reaction to change its order mid-reaction?
I encountered a problem which stated that the order of the reaction remained zero, until the concentration was halved. Then it became first order. Already, zero ordered reactions don't actually exist (We can create conditions though, such as the ones in the Haber process for ammonia). Changing of a zero order into first is a bit perplexing.

Autocatalytic reactions could well do this, but I can't provide a definitive example

Changing the order of the reaction with change in concentration; Can a first order reaction change to a second order reaction under specific temperatures?

"Saturated" photolytic reaction with all active light absorbed ( and the volume being mixed) may be the zeroth order, not depending on concentration. But when the reactant is disappearing and the active light partly  passes through, the photolysis rate would become depending on the reactant concentration.

An example might be the Lindemann scheme for gas phase 'unimolecular' reactions such as the isomerisation $\ce{CH2NC\to CH3CN}$. The scheme is
$$\displaystyle A+M\overset{k_2}{\underset{k_{-2}}{\rightleftharpoons}} A^* +M; \qquad A^*\overset{k_1}\to P$$
where A is the reactive molecule and M a collision partner. A$^*$ is an energised molecule that undergoes reaction and P is product. Using the steady state approximation the rate of formation of product can be found but we do not necessarily need to do this.
If we assume that the rate of reaction of A$^*$ to form product is slow ($k_1$ small) relative to the equilibrium rate constants (left hand above which is forming A$^*$), this equilibrium can be satisfied and the reaction to form the product will be first order because there is always a sufficient amount of A$^*$ to react without disturbing the equilibrium.
At low pressures, on the other hand,  the equilibrium cannot be maintained i.e.  A$^*$ is formed very slowly w.r.t. its reaction to form product, and now the reaction of A$^*$ depends on how quickly it is formed, not on how quickly it reacts to form product, and so the reaction is bimolecular.


text:
Why is Li easiest to oxidise among the alkali metals despite having the greatest ionization energy 1 in the group?
If we use the magnitude of the IE1 as an indicator of ease of oxidation, Li is the most difficult of the alkali metals to oxidise. On the other hand, if E0redn (standard reduction electrode potential) is used as the criterion, Li is the easiest to oxidise. How can we explain this discrepancy?

You should include in questions a short conclusion from prior info searching, so others do not write things you already know, or should know if searched about it. If nothing was found, write at least what you have searched for.  Aside of general keyword search, search also with the search modifier site:stackexchange.com OR site:libretexts.org to find similar topics on the respective site.

First ioanization energy describes the energy needed to take electron from an isolated atom in its gaseous state. However, standard reduction potential describes the tendency of losing or gaining electrons typically in aqueous solution. So in case of the reduction potential you also need to consider 3 different steps that contribute to the calculation of reduction potential:

heat of vaporization of the metal (separation of atom in gaseous state from bulk metal in solid)
the ionization energy of the atom (removing electron from the atom to create ions)
the enthalpy of hydration of the metal ions (turning into aqueous ion by water molecules)

As you pointed out, ionization energy of sodium (495 kJ/mol) is less than lithium (520 kJ/mol) meaning that you would spend less energy to ionize sodium. However, hydration of enthalpy of lithium is -515 kJ/mol while for sodium it is -405 kJ/mol as charge density of lithium ion is higher and its ion size is smaller (that helps with making bond with lone pair of electrons on oxygen atom in water molecule, thus a stronger bond and more energy is released). You can notice that the difference between the hydration entalphies of lithium and sodium is much higher than the difference in ionization energy. So in comparison to lithium it is easier to ionize sodium atoms, however it is much harder to hydrate the sodium ions compared to the lithium ions. That is why aqueous/hydrated lithium ions have less reduction or higher oxidation potentials than aqueous/hydrated sodium ions. In summary, you are "spending" less energy in general to remove electron from bulk lithium metal based on the steps I mentioned before. Hope this helps!

You compare a single formal step (atom ionization) with a multi-step process leading from solid bulk metal to hydrated metal ions.
The higher ionization energy means the higher energy barrier as a kinetic obstacle, so metallic lithium has in general the lower kinetic reactivity in the group I.
But in context of electrochemical processes in water,

formal metal atomization
atom ionization
electron capture by bulk metal
metal ion hydration

lithium has due high ion hydration enthalpy the most negative change of Gibbs energy. Therefore, it has the most negative reduction potential and the highest thermodynamic reactivity.


text:
What factors determine the distribution of chemical energy between work and heat in a chemical reaction?
In the context of a chemical reaction, what factors determine the allocation of chemical energy between performing pressure–volume work and generating heat? Are there any key parameters or variables that dictate the proportion of energy converted into useful work versus dissipated as heat?

How long have you been thinking about this question before posting it? What are results of your thinking? What is your idea how it could be? Posts consisting just of questions are not received well on Chemistry SE site.

I have been thinking about this question for a while but need some direction. I am a high school student and my chemistry book says that energy is converted into either pressure-volume work or heat. I am really curious as to what determines the proportions

It is a question of experimental situation. If the experiment is done at constant volume, no work is produced, so that the heat produced is equal to the change of internal energy of the substances in reaction. If the experiment is done in contact with the atmospheric pressure, the volume may change a bit. But the heat produced is equal to the change of enthalpy of the substances in reaction.

First Law of Thermodynamics
$$
\Delta U = q + W
$$
Work and Heat Depend on Process
Suppose we fix the initial and final states, then $\Delta U$ is fixed and constant, or:
$\Delta U = q + W = c \tag{1}$
Let's carry out severable reversible processes for a $PVT$ system, which means there is no nonmechanical work. For such processes the work is given by $W = -\int_{V_1}^{V_2}P\mathrm{d}V$, which is the area under the $P$-$V$ graph.

Since area under process 3 is the highest, and process 1 is the lowest:
$$
W_3>W_2>W_1
$$
Sum of $W$ and $q$ is fixed, $\Delta U = c$, so,
$$
q_3<q_2<q_1
$$
Work and Heat Depend on Reversibility
It is known as the maximum work theorem:$^\text{1}$

...for all processes leading from the (a) specified initial state to the (a) specified final state of the primary system, the delivery of work is maximum (and the delivery of heat minimum) for a reversible process.

Reference

Called H. B. (2005). Thermodynamics and an Introduction to Thermostatics. 2$^\text{nd}$ ed. John Wiley & Sons Inc.


text:
What are the best conditions for EDC coupling of 2 small molecules?
I want to couple 2 small molecules, one is a DNA oligo with an amine group and the other one a small molecule with with a carboxyl group.
I found protocols for EDC (1-Ethyl-3-(3-dimethylaminopropyl)carbodiimide) coupling of amines and carboxylic acids, but they always refer to proteins. My molecules are stable in a wide range of pH, what buffer conditions would work best for my coupling reaction?

I wouldn't count DNA oligomers as small, but with biomolecules perspective is different.  ;)

For this reaction the optimal buffer conditions are a pH of 4.5-5.5, which is slightly acidic. The reaction mixture should be stirred at room temperature for several hours to overnight. Common buffers used for this reaction include MES and MOPS. It is important to use an excess of EDC and the appropriate stoichiometric amount of the other reagents to ensure a high yield of the desired product.

What about the amines on the nucleobases? Wouldn’t they react as well, giving you an unintended product mixture?

Good Question. I would consider using an amine-blocking reagent on the nucleobases, so it will selectively block the amine on the nucleobases while leaving the amine on the DNA oligo's end.


text:
How do alpha particles contribute to pressure of nuclear decay products?

A closed vessel with rigid walls contains 1 mol of $_{92}\ce{U}^{238}$ and 1 mol of air at 298K. Considering complete decay of uranium to $_{82}\ce{Pb}^{206}$, the ratio of final pressure to initial pressure of system at 298K is?

The solution of this question is that during this decay, 8 moles of alpha particles will be emitted, so total moles of gas will be 9. At constant temperature and volume, pressure is proportional to moles of gas. So the answer is 9.
My question is: are helium nuclei considered gas molecules, and should we count them while finding $n$ in the ideal gas equation? My thought process for this question was to find out the number of $\beta^{-}$ particles (electrons) emitted (which is 6 here) and then figure out that they would combine with 3 helium nuclei to form 3 helium atoms. Then I would use the ideal gas equation and the answer would come out to be 7.

The net decay is neutral, with two electrons left behind around the decaying nucleus, so the alphas find electrons eventually as they slow down.

You seem to be in a grave misconception about the nature of radioactive decay. Sure, helium nuclei are most emphatically not considered gas molecules. But then again, helium nuclei do not exist. Let that sink in. They do not exist. Not for any significant amount of time, that is. They all instantly turn into helium atoms, and then you seem to know what to do.

A multi-MeV alpha particle most certainly does not ‘instantly’ turn into a neutral atom. It must slow down to where the possibility of grabbing an electron with reasonable probability can happen. One can happily propagate an MeV He++ beam for many cm in air without them becoming neutral (while creating many ions in the air along the track).

Ahh, relativity of terms! Multi-MeV alpha particle would pass these centimetres in less than nanosecond - much shorter time than duration of many "instant" processes.

When it is an alpha particle, it is not a molecule(IUPAC), that is by definition neutral.

An electrically neutral entity consisting of more than one atom (n>1). Rigorously, a molecule, in which n>1 must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state.

In contrary, a molecular entity(IUPAC) covers ions too, so an alpha particle is a molecular entity:

Any constitutionally or isotopically distinct atom, molecule, ion, ion pair
, radical, radical ion, complex, conformer etc., identifiable as a separately distinguishable entity.

When an alpha particle becomes a molecule, it is not an alpha particle anymore, as an alpha particle is a ion, a nucleus of helium-4. When they slow down enough, they are able to take electrons from literally any atom but helium. Well, in the first step even from helium, as
$$\ce{He^2+ + He -> 2 He+}$$

In the idealized case when no high energy particle leaves the idealized contained, the total charge balance remains neutral. There would be neutral atoms, molecules, ionic pair, ions and electrons that would eventually recombine to neutral $\pu{1 mol}$ of lead-206, $\pu{8 mol}$ helium-4 of , together with $\ce{1 mol}$ of air ($\ce{1 mol}$ of any aerial molecules).
Note that until all uranium decays (what would take much longer than the age of universe), lead would be already oxidized by oxygen.

I suspect the question doesn't expect a full analysis of all the decay products (including electrons) and assumes that alpha particles will rapidly abstract electrons from somewhere. 8 moles of doubly charged ions would a hell of a driving force for electron abstraction even if the decay itself were not charge neutral.

Neither I suspect that. I also noted recombinations.

I disagree with the sentence "Note that until all uranium decays (what would take much longer than the age of universe), lead would be already oxidized by oxygen."The decay of uranium has a time constant T and at t >5T all the uranium atoms have practically been decayed

@Volpina Well, lets see: The halftime for uranium-238 is about 4.5 billions years. Universe age is assumed (I guess) about 13.8 billion years.  After 10 halftimes, i,e about 45 billion years, roughly 3 time universe age, there would be still 1/1024 mol of uranium-238, almost  1/4 g.

@volpina In 5 half lives about 3% of the uranium would remain (that's twice the age of the universe). In 10 half lives it would be about 0.1% or the original amount. It would take about 80 half lives to get to just a single uranium atom starting with 1 mol (though statistical noise would kick-in much earlier).  ;-)

@Poutnik Dammit, you beat me to that! ;-)

@matt_black Waiting for that last atom would be funny.  :-)

@Poutnik But decay would take infinite time if you were a mathematician and ignored the particulate nature of atoms. This generated a long, frustrating thread here a few days ago on a related topic.

@matt_black thats why sometimes you must be able to evaluate the result produced by mathematics you cant just blindly follow it.In practice the last uranium atom will decay into finite (but very big)time.The error comes from how we do integration.

@matt_black  As Lacedaimoneans could have said: "If ..."

The error comes before doing integration.

We assume that for a infinitesmall change of a function f(x) in dx the value of f(x) doesnt change  but in reality this isnt really true but it is a really good approximation as we take the limit of $dx\rightarrow 0$.However we should stop right here because this is a seperate question for Math SE and isnt related to chemistry anymore.

It is blind extrapolation on a mathematical model of reality far beyond its applicability limit.


text:
In a flame test, what is being excited?
I have done a flame test with $\ce{NaCl}$ and got the yellow characteristic flame of sodium. I am trying to understand exactly how this yellow color is being produced and when I searched for it, I got conflicting answers.
I understand that it is being produced by excitted electrons that fall back and produce photons. I also understand that different elements have different energy levels for their electrons so when the electrons fall back, they will produce a different wavelength of photon and therefore a different color.
What I was trying to find out is whether the sodium that gets excited is still bound to the chlorine or if it is dissociated, becoming a free ion and then excited.
Most sources said the sodium ion is dissociated and then excited, but some sources argued the yellow flame is apparently produced by electrons getting excited from the 3s orbital to 3p and the sodium ion doesn't have the electron in the 3s orbital which is being excited, therefore it is actually a neutral sodium atom which is being excited. Would a neutral sodium atom, a sodium ion and a sodium ion bound to chlorine produce different colors when excited? And if the sodium ion is dissociated and then excited, then why does the anion still sometimes affect the color of the flame?

https://chemistry.stackexchange.com/questions/141382/wavelengths-of-colors-emitted-by-alkali-metals-during-flame-test https://chemistry.stackexchange.com/questions/9857/flame-test-is-the-metal-atom-or-the-metal-ion-responsible-for-the-flame-colour

There's no ions in gas - you need plasma for that.

https://chemistry.stackexchange.com/questions/147950/flame-test-of-alkali-metals-and-alkaline-earth-metals https://chemistry.stackexchange.com/questions/32686/is-it-possible-to-have-a-diatomic-molecule-of-sodium-in-gaseous-state

@Mithoron Thank you! It confirms some of my assumptions and answers part of my question, but I am not completely satisfied with the answers on the other questions. What I am mainly wondering about is if the sodium dissociates from the chlorine and if it is a free sodium ion which gets excited. The answer mentions that the sodium gets ionized, but the answer also mention again that the yellow color is caused by electrons from the 3s to 3p orbital, but I still don't understand how that is possible in a sodium ion.

@Mithoron also if there are no ions in a gas, but the sodium gets dissociated from the chlorine what is it then? I would think it is a free ion like a gas molecule, but is it plasma then?

I found a related article which I am satisfied with. Will still read answers tho.

The sodium emission is shown here: https://chemistry.stackexchange.com/a/164168/79678. The emission is from excited sodium atoms, not sodium ions. In the flame, there are various species, including excited sodium atoms, chlorine atoms, some NaCl molecules and some ions, though not many in an ordinary flame such as I show in my linked answer. The flame is electrically neutral, as expected.


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When adding a weak acid to a strong base in an acid-base titration, why does multiplying the pKa of the acid by two give the pH at neutralisation?
I'm not sure if the premise of the question is even correct, but when adding ethanoic acid to sodium hydroxide solution, I was told that the pH at the point of neutralisation should be around 2x the pKa of ethanoic acid (so around 9.52), and the data I collected showed this as well. Why is this the case? I know that at the half equivalence point pH = pKa when adding a strong base to a weak acid, so what changed in this scenario?

You will find your answer searching for "neutralization of a weak acid with a strong base".

pH=7 + 0.5(pKa + log c) (simplified but approximately valid for most cases)


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What words can I use to differentiate between cases where two atoms are of the same element versus when two atoms are of different elements?
I am writing my thesis, and usually I'm quite good with picking the right words for specific phenomena, yet I am having trouble finding the right words for this specific circumstance. I am trying to differentiate between halogen•••halogen contacts of the form R–X•••X'–R' where X and X' are the same element versus where X and X' are different elements. X in this case is a halogen, so a word for different halogens is okay too (I'd still like to know the general word, assuming there is one). I know that "congener" means two elements of the same group, but I'm unsure if that implies they are distinct elements. If it doesn't, I imagine the answer is "homo____" and "hetero____" but what could that ______ be?

Could you just write a sentence and what you don't like in there?

You may be able to use "homonuclear" and "heteronuclear" depending on the exact wording

"X•••X contacts may be represented with the general structure R–X•••X'–R, where X and X' are either [same element word] or [different element word]." I want to concisely emphasize that X and X' might be different elements, but they might not be either. To go from the @Andrew example, "X•••X contacts can be represented with the general structure R–X•••X'–R, where X and X' may either be homonuclear or heteronuclear." I like those, but I don't want to accidentally imply covalence between them or imply anything about the R group.

Could you please send us an example of a phenomena where the "interaction X-X and/or X-X' " occurs, and where you are reluctant to imply covalence, or to imply anything about the R group ? We might better understand your question.

These bonds are explicitly non-covalent (well, they are intermolecular—it is possible they adopt slight covalent character occasionally, but they are mostly non-covalent). I don't want to imply that the R group has to be any particular group, because while it is often an organic molecule, it can be another halogen, a metal, all sorts of things. These interactions come up in crystal formation/engineering a bunch and can be used in things like OLEDs.

Contact
Name
Alternatively

$\ce{RX$\cdots$XR'}$
homonuclear contact
homohalide contact

$\ce{RX$\cdots$X'R'}$
heteronuclear contact
heterohalide contact

+1 Interesting to see this terminology...homohalide occurs only 22 times in Google Scholar (out of millions of articles and mainly non-English speaking authors) which means that this is not a popular term.

"X•••X contacts may be represented with the general structure
R–X•••X'–R, where X and X' are either [same element word] or
[different element word]." I want to concisely emphasize that X and X'
might be different elements, but they might not be either.

In scientific writing, we don't need to worry about Pulitzer Prize winning prose. The main question is whether others can understand your idea unambiguously. Your sentence is as clear as it can be. You can perhaps modify to
"The halogen atom to halogen atom contacts may be represented with the general structure R–X•••X'–R, where X and X' can be the same or different halogens."
Congener will be too broad here and perhaps incorrect too.


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How to interpret gas chromatography results?
I am having trouble interpreting gas chromatography (GC) results for the production of ethyl acetate from acetic acid and ethanol. I am trying to find the reaction order from these results, as well as the proportionality constant $k_i = A_i/c_i$ for the relation between the $i$th peak area $A_i$ and corresponding concentration $c_i.$
I have the peak areas for each substance. I assume the corresponding concentrations are the arguments on the $y$-axis. With that I can get the proportionality constant $k_i$ for each analyte, but I'm failing to see how to calculate the reaction order from this.
Can I just ignore the proportionality constant $k_i$ and plot the three linear cases for the reaction orders to determine the order: $c$ vs $t$, $1/c$ vs $t$, and $\ln c$ vs $t$?

The problem has nothing to do with gas chromatography. It has to do with the kinetics of a chemical reaction.
Whatever the method used to get the concentrations, when you get the concentrations of the reactants (A, B, or C), you report them versus the time. The concentration of a reagent must decrease.
To analyze these results and get the order, you plot the logarithm of the concentration of A versus the time on a diagram. Look at the diagram ! Are the obtained points alined? If they are, the order of the reaction is $1$ with respect to A : the slope of the line joining the points is the 1st order rate constant. If the points are not alined, you should start again by plotting the inverse of the concentration of A versus the time. Here again the points should be alined. If they are, the reaction is $2$nd order with respect to A. If they are not alined, the order is not known. It is different from $1$ and from $ 2$. But it is not known yet.

I see, thank you!! If I may, I only have one more question just to make sure I get it, is the y-axis on the graph the concentration? Or do I need to do something else to get the concentration of each reactant? Thank you in advance

The y-axis on the chromatogram is proportional to the concentration of the corresponding stuff.


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Why I am unable to see any electrical conductivity in Permalloy nano powders?
We recently purchased some permalloy nano powders (Ni80Fe70Mo3) I checked the electrical conductivity of these powders using a multimeter which can measure resistance to the order of giga ohm's. I was able to see electrical conductivity in carbonyl iron powders. While we were not expecting to see high electrically conductivity we expected to see some conductivity. I am wondering if I am missing something here or are these powders the issue.

Nanoparticles are often pyrophoric, and may be coated to prevent spontaneous combustion. The particles' surface might be nitrided, or oxide coated, or coated with an organic chemical. Permalloy itself is quite conductive, but the coatings are insulators.
One doesn't think of a lump of iron as flammable -- but sprinkle iron filings in a flame to make a "sparkler". Steel wool is effective kindling.
Check with the source of your permalloy powder to see if they're coated, and what that coating is.

Thermite: iron oxide powder + aluminum powder + heat = iron + aluminum oxide + lots of heat.

Also, uranium.  It's pyrophoricity + self-sharpening attributes are part of what makes DU soooo effective in cannons.


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Different types of bond representations
What do the following different kind of bonds mean?

Although this answer to the question "What are the meanings of dotted and wavy lines in structural formulas?" does describes some of the bonds, it does not provide a complete description and some bonds are not mentioned. Also, the question has a dotted-line bond, which maybe misleading.

While your effort in aggregating info about graphical representations of chemical bonds is appreciated, but I don't think posting a different question with a subset (which is also far from complete) of notations from What are the meanings of dotted and wavy lines in structural formulas? is justified. Posting textual description as an image pretty much nullifies the distinction as this question won't be discovered on its own. Maybe the previous question's answer should be a wiki. Also, IMO an MD table would be more clear and searchable.

@andselisk the previous answer did not cover all the different plain/stereobonds, which is why i decided on posting this. Thank you for the suggestion, I have formatted the description of bonds as a table.

@NilayGhosh hope the latest edit addresses your comment.

Then make a new answer in that post addressing other type of bonds initially not mentioned. I also agree with andselisk's suggestion.

@NilayGhosh an accurate description of most commonly used bonds may be helpful to the community. I have incorporated andselisk's suggestions. As I mentioned, the other question is misleading. This question is more thorough, and the answer more accurate. I am not claiming to describe the entirety of stereochemistry in a single page, there is a reason several books have been written on it; and as mentioned, this is a topic of ongoing discussion and developments.

Disclaimer: This is topic of ongoing discussion, and different authors use different conventions. This answer is, mostly, based on IUPAC recommendations (2006). This is not a complete list, as there are a vast number of representations for stereochemistry of a compound, Fischer and Newmann projections to name two. Furthermore, in biochemistry, the representations are different.
This page covers the topic in more detail with lots of examples.

Name
Representation
Meaning

Dashed bond

Partial bond in plain of paper or of unspecified stereochemistry. Also represents hydrogen bonds. Also represents bonds with known relative and unknown absolute configuration in an enantiopuric product; similar to hashed wedged bond.

Solid bond

Bond in plain of paper or unspecified stereochemistry.

Hashed wedged bond

Narrow end: towards reader; wide end: away from reader.

Hashed unwedged bond

Both atoms are below the plain of the paper. Also represents configuration of racemates.

Solid wedged bond

Narrow end: away from reader; wide end: towards reader.

Solid unwedged bond

Both atoms are above the plain of the paper. Also represents configuration of racemates.

Hollow unwedged bond

Bonds with known relative and unknown absolute configuration in an enantiopuric product; similar to solid wedged bond

Wavy bond

Bond with unknown stereochemsitry. Also represents an unknown moiety.

Arrow bond

Dative bond in plain of paper or with unknown stereochemsitry.

Controversy Surrounding Hashed Wedged Bond
There is a bit of controversy surrounding the hashed bond.$^\text{1}$

Unfortunately, bonds below the plane of the drawing have historically been represented in many different ways. Each of those representations has involved a bond drawn with small line segments either coincident with or perpendicular to the main axis of the bond. The two schools of thought that prefer the use of a hashed wedge bond assign it two directly opposite interpretations—one school says that the atom at the narrow end should be considered in the plane of the paper, while the other says that the atom at the wide end should be so considered. Additionally, a dashed line often represents a partial bond, delocalization, or a hydrogen bond. The biggest problem is that there is no way to intuit an author’s desired meaning from a chemical structure drawing alone.

Hashed and Solid Wedged Bonds
In the article,$^\text{1}$ IUPAC recommends using the hashed wedge bond in a similar way to that of the solid wedged bond:

In cases where such below-the-plane bonds are required, there is no option that will please everyone. Earlier recommendations 12 proposed the use of an unwedged hashed bond, but today such bonds are in fact encountered in the literature least frequently of all options. This document now recommends the use of a hashed wedged bond interpreted in a sense similar to the solid wedged bond (starting from an atom in the plane of the drawing at the narrow end of the wedge). This recommendation is made primarily because the hashed wedge is easier to analyze visually than the unwedged type, particularly when it is used in a sense similar to the wedged bold bond.

The unwedged hashed or solid bonds are used specifically to indicate that both atoms are below or above the plane, respectively:$^\text{1}$

Strictly speaking, unwedged bold and hashed lines show that both atoms are above or below the plane of the drawing (as is used in Haworth drawings of carbohydrates).

Hashed and Solid Unwedged Bonds
The hollow wedged bond represents bonds with known relative but unknown absolute configuration in enantiopuric compounds. And unwedged hashed and solid bonds are also represents recemates:$^\text{2}$

Wiggly Lines
Although it is not recommended by IUPAC, some authors (e.g., Clayden J., Greeves N., and Warren S. who call it wiggly lines) do use the wavy bond to show an unknown moiety: $^\text{3}$

The Dotted-Line Bond
Note: a dotted-line bond has not been recommended by IUPAC. I have seen it represent an ionic interaction, such as between carboxylate and sodium ions.
References

Brecher, J. (2006). Graphical Representation Of Stereochemical Configuration. Pure Appl. Chem., 78(10), 1897–1970. 10.1351/pac200678101897
Maehr, H. (2002). Graphic Representation of configuration in Two-Dimensional Space. Current Conventions, Clarifications, and Proposed Extensions. J. Chem. Inf. Comput. Sci, 42, 894-902. 10.1021/ci025518w
Clayden J., Greeves N., and Warren S. (2012). Organic Chemistry, $\textit{2}^\textit{nd}$ ed. Oxford University Press Inc., New York


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Prepration of standards for analysis of chromium using AAS
I have conducted experiments to study specific adsorption of chromium in various soils and I made my all solutions of cr with using 0.01M NaNO3 as background electrolyte. (I made solutions of 5 different ppm concentrations)
So now for calibration of the AAS instrument should I prepare standards using only distilled water or my background solution( electrolyte).

How are you preparing your Cr standard and instrument do you have? What is the concentration range? Write more details. Cr analysis on AAS requires more special conditions.

IMHO, the rate and efficiency of metal atomization may strongly depend on the dissolved matrix..... I would try to make as many variables constant as possible.

What do you mean when stating that you "made solutions of 5 different ppm concentrations" ? Five different concentrations of what ? Of chromium or of NaNO3 ? And are the ppm concentrations expressed in weight per liter or in mole per liter ? Weight or mole of what ?

@Maurice I made 5 different solutions of cr (e.g 1 2 4 8 ppm.chromium and made up the volume with 0.01M sodium nitrate.)

@Poutnik okay thanks I am thinking about making the standards also with sodium nitrate. I mean Im thinking about diluting my stock solution using the NaNO3 solution

What is the reason for the sodium nitrate? Does it have an effect on the Cr atom concentration in the flame? If YES it must also be in the standards. If NO you best rethink your method. Remember AA analysis requires that the element must be converted quantitatively into atoms under the flame conditions for best results.

@jimchmst okay I think Calibration solutions and my samples should have similar chemical composition so I thought I should make standards also with sodium nitrate. I am not sure of effect of chromium concentration I don't think it will impact the concentration in any way.


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Does heating 3-phenylpropionic acid decarboxylate it?

Here, when P is reacted with $\ce{SOCl2}$ and then NaCN to give 3-phenylpropionitrile and then hydrolysed to give 3-phenylpropionic acid. Why shouldn't it further decarboxylate and dehydrogenate to give styrene instead of continuing with the reaction scheme ahead, as heating is also mentioned?
Answer key given says that (B) is the correct answer.

No, 3-phenylpropionic acid does not readily decarboxylate. The sulfuric acid protonates the carboxylic acid which loses water creating the R-C=O+ cation which elecrophilically adds to the aromatic ring.

My teacher told me that it would decarboxylate at 112°-113°C thus i asked the question

Your teacher is in error. Wikipedia quotes a boiling point of 280C for it  https://en.wikipedia.org/wiki/Phenylpropanoic_acid

Thanks for clarification!


text:
Does Grignard reagent on reaction with PbCl2 give PbR4 and not PbR2?
I was told by my teacher that $\ce{RMgX}$ forms $\ce{PbR4}$ when it reacts with either $\ce{PbCl2/PbCl4}$ the later case is clear to me but why does $\ce{PbCl2}$ forms a compound of the form $\ce{PbR4}$.

What the teacher said when you had asked him/her? If one says something it need not to be true, even if it is said by a teacher.

PbCl4 "is best stored under pure sulfuric acid at -80 °C in the dark." (Wikipedia)

Note with very bulky side groups Pb(II) compounds can be isolated, e.g. https://pubs.acs.org/doi/10.1021/om990614g https://pubs.acs.org/doi/10.1021/om9708337

@IanBush "The compound is in effect the first dialkylead(II) species to be structurally characterized". Cool! It took way longer than it should have been given the first organolead compound was synthesized in 1858 ;)

More is involved here than the simple nucleophilic exchange normally associated with Grignard reagents. Essentially, the proposed dialkyl lead disproportionates with displacement of metallic lead. According to bionity com, the main reactions with both organolithium and Grignard reagents bring about this disproportionation:

Lead(II) chloride is the main precursor for organometallic derivatives of lead. The usual alkylating agents are employed, including Grignard reagents and organolithium compounds[1]:
$\ce{2 PbCl2 + 4 RLi -> R4Pb + 4 LiCl + Pb}$
$\ce{2 PbCl2 + 4 RMgBr -> R4Pb + Pb + 4 MgBrCl}$
$\ce{3 PbCl2 + 6 RMgBr → R3Pb-PbR3 + Pb + 6 MgBrCl}$
Notice how these reactions produce derivatives that are more similar to organosilicon compounds, i.e. that Pb(II) tends to disproportionate upon alkylation.

The "inert pair" effect in heavy $p$-block elements becomes less prominent when the heavy atom is attached to less electronegative ligands that favor covalent instead of ionic bonding (in this case, by exchanging chlorine for alkyl groups), so the higher oxidation states ($+4$ in tetraalkyl lead $+3$ in hexaalkyl dilead) become more favorable in such situations.
Cited Reference

Housecrift, Catherine and Sharpe, A.G.. Inorganic Chemistry (2nd edition). Pearson College Div (January 1, 2005), p. 524.

Note that the dialkyl/diaryl lead(II) derivatives are still an intermediate products before getting disproportionated.

Simple dialkyl- and diaryl-lead(II) derivatives cannot be isolated because of their facile disproportionation to elemental lead and hexaorganodilead compounds. Nevertheless, they are presumably the initial products of the reaction of lead(II) chloride with organolithium or Grignard reagents. A deep brown–black reaction mixture which exhibits reactions attributable to the presence of diphenyllead can be obtained from the reaction of lead(II) chloride with exactly two moles of phenylmagnesium bromide in ether at −20 °C (the reported isolation of diphenyllead as a red solid by Krause and Reissaus is most probably in error). Thus addition of water to the freshly prepared "diphenyllead" solution produces lead(II) oxide and benzene, whilst iodine affords lead(II) iodide and iodobenzene together with a small amount of diphenyllead diiodide. Reaction with phenyllithium gives triphenyllead lithium. Disproportionation occurs on allowing the solution to warm, and hexaphenyldilead and lead metal result.

Source: P.G. Harrison, Comprehensive Organometallic Chemistry, 1982 (Sciencedirect Link)


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Oxidation Number of the middle carbon in $\ce{C3O2}$ (Carbon Suboxide)
By taking the oxidation state of the terminal oxygen atoms as -2, and oxidation state of the carbon atoms adjacent to these oxygen atoms as +2, we are left with the oxidation state of the middle carbon atom as 0.
However I researched a bit into the structure of $\ce{C3O2}$ and found out that it is easy to bend (and can be non-planar?), would this affect the oxidation state of the central carbon atom or would it remain as 0.
My thought process initially:

The oxygen atoms being more electronegative will pull the electron clouds towards themselves leaving the adjacent carbon atoms with a positive charge.
I then thought that this positive charge will be distributed among the carbon atoms leaving the middle carbon atom with a non zero oxidation state. (On reading the comment by Poutnik, I think this thought is wrong since it is not taking total bond breaking operation into account)

Why do you think it could affect the oxidation number of the central C? Remember that ON is the formal charge, obtained after undergoing formal total bond breaking operations under specific conditions. (Elaborate your reasoning in the question.)

@OscarLanzi No, they aren't? It's irrelevant if bonds are single or double, as long as they are between carbons. This compound was even modelled as carbido complex of two CO molecules - what you mention, but on both sides and it's still $0$ ox. state in the middle.

A different molecular structure can surely lead to a different chemical environment and consequently different charges. However, if the rules of oxidation state are correctly employed, the middle carbon will always be zero. Iirc the candidate partial charge also is close to zero. And the bonding pattern fits better to C(CO)2.

The $\ce{C(CO)2}$ complex model actually looks a lot like the combination of structures you would draw using valence bond theory. The central carbon in the complex would have its pi-symmetry orbitals filled, and then they overlap with the pi acceptors in the $\ce{CO}$ ligands so pi electrons are delocaluzed the same way they would be with the ordinary valence bond theory. Bond lengths in $\ce{C3O2}$ suggest stringer multiple bonds than those in a nonconjugated alkene or ketone.

First off, carbon suboxide is much closer to being linear than formally isoelectronic species such as $\ce{N5^+}$. The energy difference between linear and bent conformations is so small it gets smeared out by quantum-mechanical zero-point energy, and chemists call the structure "quasilinear".
If we work out the effective hybridization of $s$ and $p$ orbitals as a function of bond angle, a 160° bond angle which is considered the most stable configuration (Wikipedia, citing Reference 1), gives $sp^{1.06}$ (the number is the negative of the secant of the bond angle). So even if the premise of a linear geometry is ambiguous, the $sp$-hybridized electronic structure is an accurate electronic model and on that basis, the middle carbon may be assigned oxidation state $0$ (and the other carbons $+2$).
Some additional examples of carbon(0) bonded to only two other atoms is given by degruyter.com.
Note that the middle carbon comes out with a significantly lower oxidation state than the other two. This dovetails with the middle carbon being nucleophilic (accepting protons when the carbon suboxide reacts with water or hydrogen chloride) while its neighbors are electrophilic.
Cited Reference

Brown RD (1993). "Structural Information on Large Amplitude Motions". In Laane J, Dakkouri M, Veken Bv, et al. (eds.). Structures and Conformations of Non-Rigid Molecules. NATO ASI Series. Vol. 410. Springer Netherlands. pp. 99–112. doi:10.1007/978-94-011-2074-6_5. ISBN 9789401049207.


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Why hot iron turns from red to white and then blue?
While reading structure of atom chapter no. 2 from ncert class 11th chemistry part 1 book, I got a doubt. In the book it says on page 40 that:
Hot objects emit electromagnetic radiations over a wide range of wavelengths. At high temperatures, an appreciable proportion of radiation is in the visible region of the spectrum. As the temperature is raised, a higher proportion of short wavelength (blue light) is generated. For example, when an Iron rod is heated in a furnace, It first turns to dull red and then progressively becomes more and more red as the temperature increases. As this is heated further, the radiation emitted becomes white and then becomes blue as the temperature becomes very high. This means that red radiation is most intense at a particular temperature and the blue radiation is more intense at another temperature. This means intensities of radiations of different wavelengths emitted by hot body depend upon its temperature.
So if on heating there is higher proportion and intensity of shorter wavelengths, then why red hot iron turns to white before blue as white light is consisting all wavelengths of visible colours?
Here is the link to download book:
https://ncert.nic.in/textbook/pdf/kech1dd.zip

See results of googling site:stackexchange.com OR site:libretexts.org OR inurl:hyperphysics Planck law

Iron sounds very patriotic.

White is not a color. It is the signal sent by the eye to the brain when the light contains the same proportion of red, orange, yellow, green, blue and violet as in the sun light.
If the light entering the eye contains more blue than other colors, the eye sends the "blue" information to the brain.
So consider what light is emitted when the temperature of any body is increasing. Before 500°C, hot bodies do not emit any visible light. Maybe they emit some infra-red, but no visible light.
When hot objects are at around 600°C, they begin emitting a little light, but with more red light than any other color. At around 1500°C, they are brighter. They still emit all colors, but the light contains more orange rays than any other color. They look orange, but overall they look very bright. It is necessary to wear spectacles to look at such a hot body.
At still hotter temperatures, near 6000°C, the hot bodies look extremely bright, but they look yellow. Well ! This yellow is nearly similar to the sun light. So the light looks white, like the Sun, which is a little yellow.
Blue light emitting bodies are not easy to discover on Earth, because we are unable to produce the high enough temperatures. At these high enough temperatures, all ovens will melt and be vaporized. But blue lights can be observed on some stars, which are much brighter and thousand times hotter than the sun. But they emit a huge proportion of ultra-violet light.
Iron can never be hot enough to produce light perceived as blue. Hot iron may produce red light. Orange is the maximum.

Rather, iron can produce blue light (430-470 nm), but cannot produce light perceived as blue (having majority of emission in short visible wavelengths)

Note that the Sun, observed from orbit, is like white. It becomes yellow observed from surface, after Rayleigh dispersion of large portion of violet/blue light in atmosphere.

OK Poutnik. All you say is correct. But I have simplified the description of the phenomena. I am not sure our correspondant would have been happy by reading  an answer containing the change of the black body radiation spectrum  when the temperature is increasing.


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Reaction Order from Percent Yield?
Referencing this paper, I am trying to approximate the order of reaction for cyclohexane (4a). Trials 1-3 provide the relevant information of mmol changes and corresponding percent yield but I'm struggling to see how I can use this to get the reaction order. Am I missing something? Thank you, I appreciate any help!

That's impossible. Yield describes, if anything to that matter, thermodynamics. The order of the reaction comes from kinetics.

It's probably pseudo first order looking at the equation and type of reaction. I'll leave the rest to your imagination and searching skills.

Thank you, that is what I was thinking along the lines of. It feels dubious regardless but I'm glad to have a second opinion to help reassure my conclusion now.

We don't know why the yield is lower than 100%. It could be because the blue LED was used for a limited time. It could be because the reactants turned into a side product (the GC data might show that). It could be because of product inhibition. Or it could be some combination of these and other factors.
Without at least two data points taken at different times, or prior knowledge of why the yield is less than 100%, you should not make any statements about reaction order.
To be fair, there are intro lab experiments that pretend to get a reaction order without measuring a time trace (use different concentrations of reactants and measure how long it takes to use up a reactant), but they say in the introduction what the underlying assumptions are, and that those have been tested already. Also, they give a range of "allowable" concentrations to make sure you get the "right" answer.
In this case, with the complexity of the reaction mixture including the requirement for photo-catalysis, it seems you can't pull this off. If you naively plot concentration of cyclohexane against yield after 24 h (and assume zero yield for leaving it out altogether), you don't get a curve that fits zero, first or second order.


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Is potassium Sulfate a good electrolyte for the production of oxy-hydrogen?
I am currently conducting an electrolysis experiment using a 30 amp power supply with a voltage ranging from 10 to 16 volts.I am thinking of using potassium sulfate (K2SO4) dissolved in water as an electrolyte. I am using graphite rods as the electrodes. My primary goal is to produce hydrogen and oxygen gases through the electrolysis process. However, before proceeding further, I would like to clarify a few points:
-Is potassium sulfate a suitable pH-neutral electrolyte for my intended purpose of producing only hydrogen and oxygen gases?
-Will the electrolysis of potassium sulfate generate any harmful byproducts that I should be aware of?
-Are there any specific considerations or precautions I should keep in mind while using graphite rods as the electrodes with potassium sulfate electrolyte?
I appreciate any insights, suggestions, or relevant information that can help me optimize my experiment and ensure the safe and efficient production of hydrogen and oxygen gases. Thank you!

Potassium sulfate is a good electrolyte for electrolytic purposes. The only risk is the production of persulfates as secondary product at the anode if the current is too high. The intensity of the current depends on the surface of the anode. Apart from this, nothing disagreeable will occur.
Graphite electrode can do the job of course, but you should know that they will progressively be corroded by the electrolysis and graphite powder will slowly be produced to fall down the solution.
The only parameter to check is the temperature. With a 30 Am power supply, the temperature of the solution will quickly increase. It should not make it boil.

As peroxosulfates/peroxodisulfates ($\ce{SO5^2-}$, $\ce{S2O8^2-}$) are assumed to be created roughly at $U \ge \pu{2.4 V}$, do not raise the applied voltage above $2.0 - 2.1 \pu{V}$.
To increase the passing current during the experiment design stage, do not increase the voltage but improve the geometry of the electrolytic cell to minimize internal resistance. Increase electrode surfaces, decrease the electrode distance(s), increase (reasonably) electrolyte concentration.
Note that, aside of unwanted side reactions, applying too high voltage leads to excessive Ohmic heat loses and overall low energy efficiency, if that is important parameter for your case.
And last but not least, evaluate all aspects of possible danger and danger precautions, while manipulating/storing hydrogen. At no circumstances mix it with air nor even oxygen. The only exception is small volume of few dozen milliliters max for explosion demonstration purposes, avoiding brittle materials like glass or ceramics.

Aren't those values you listed for sulfate oxidation products versus SHE (it's always best to relate the voltage to a reference in these cases because it is very system dependent otherwise)? The cell voltage  is not a good measure to determine the potential at the anode side. There will be losses on the cathode and throughout the solutions (the latter will probably be the biggest contributing factor). I do agree that this experiment is dangerous, definitely not something someone should just do in their backyard when they have to pose the question about the electrolyte.

Well, the SP of H2/H2O at pH 7 is about -0.4 V wrt SHE, for peroxodisulphate is it near + 2.0 V. Sure Ohmic loses and eventual overvoltages add. The suggested limit of applied voltage was just ballpark number guidance. For one time experiemnt, it is not that much important nor dangerous to have some peroxodisulphate present.

10-16 V is too high. There is needed much lower voltage with high enough current supply.

Any can be used, but what you want may not be the same as what you get. About 8 V will be wasted on heat, so about 4/5 of the power. You can use it for boiling eggs if left ON for long enough.

So if I use 10 volt it will not create byproducts and only waste it’s energy as heat?

It will produce byproducts, oxidizing things that normally would not be. Like if water level is too high, water flows away even by paths that were not intended for it.

Since my power supply can not go lower than 10 volt and I don’t really want it to create byproducts, is there any other electrolyte possible that I can run at high voltages and not create harmful byproducts? (I don’t having byproducts as long as they aren’t dangerous.)

Why not to put several cells in series? Probably five. The same current would then produce five times more gases.

Use a screw thats conductive with alot of surface area.. if you have some change to spare.. this is what I would do.. buy a 1 once bar of silver and coat the screw in it.. silver is the most conductive element on the periodic table.. once done you cant really improve unless you wanna pay a fortune for precision machine work to improve on a fine threaded screw.Electrolyte wise, Sodium sulfate would be good, or if you got the resources or know how silver sulfate.. or sodium acetate, tho many just use salt lol.. dont know why you chose the one you did. but.. I assume availability and price, but graphite rods dont seem like a cheaper option lol unless you just were aware of surface area of the electrodes being a bottle neck. a rod does not have good surface area at all.. at all.. lol it just lasts longer. Next thing I would do tho im not aware of its efficiency increase is I would lower the temperature of your electrolyte solution. Lower temperatures equal less resistance in metals.. Im not positive but assume the same is true universally to a point. So you could go about that several ways. Interestingly this is not the part of the reaction where I had questions when I was gathering an compressing oxy-hydrogen, My questions were on the compression side. An the release as fuel. Just remember going forward.. silver then copper for all things electrical. Silver isnt too ludicrous.. its much more expensive then 8 years ago when i loaded up on pounds of it.. but.. still.. its underpriced.. the market is rigged and flooded with paper silver.. dont invest in silver you dont earn is another pointer.. bc eventually your gonna find out.. you dont own silver, you own an IOU lol.

There are plenty of errors in your post. Silver is a really bad catalyst for HER, for the OER I'm not sure, but Nickel is a commonly used, better and cheaper metal (although not stable in acidic solutions). Table salt will generate chloride which is dangerous. Higher temperatures are typically more efficient as the rates of all processes but the electronic conductivity rise with temperature and the latter is almost never a bottle neck. In general, your response is lacking a lot of structure and is wrong in many ways. I advise you to research more about the topic if you're interested of course!


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What nucleophilic reaction takes place in case of bad leaving group as well as 1* alpha carbon?
If we add HI to a substrate Ph-C-C-OH what will be the major product and what reaction mechn. will be followed?
My attempt: as -OH is bad Leaving Group, we cannot just simply proceed with Sn2 and remove the OH.
So first H2O (due to the H+ and lone pair on O) will form and then it will leave the substrate and then 1-2 hydride shift in Ph-C-C+ --> Ph-(C+)-C and we will add I- at the carbocation.
My dilemma is how can we even form a 1* carbocation (even if it can be stabilised by rearrangement )
Is my attempt valid? If not what would be the real mechanism and why?

This reaction has been studied and yields 2-phenylethyliodide (Ph-CH2-CH2-I) in around 90% yield. The authors postulate a symmetrical non-classical carbocation with involvement of the aromatic system.

Yes thank you.....(SNGP will take place)

Another possibility: a hydride ion half-shifts to float over the carbon-carbon bond of what is originally the $\ce{CH2CH2OH}$ group. This would also displace the water molecule and distribute the positive charge over both carbons.


text:
Is there any solvent I can mix with DMSO to lower its boiling point?
Is there a way, or a solvent I can mix with DMSO to lower its boiling point?

Using Cosolvent
Any miscible solvent with a lower boiling point than DMSO, especially ones that show positive deviation from Raoult's law, will lower the boiling point of the solution. According to Reference 1, the following may be utilized:

Alkyl-benzenes from benzene (unsubstituted) to tert-butyl benzene.
Haloalkanes such as tetrachloromethane, 1-chlrohexane/pentane/butane,
and 1,1,2,2-tetrachloroethane.

If you are trying to do a reaction, haloalkanes maybe somewhat reactive.
Do try to achieve an azeotropic solution so the concentration and boiling point does not vary while you are evaporating the solvent.
Lowering the Pressure
By definition, boiling point is the temperature at which vapor pressure of liquid becomes equal to the external (atmospheric) pressure. Thus, if you reduce the external pressure, it will reduce the boiling point of DMSO.
Reference

Marongiu, B., Piras, A., Porcedda, S., Tuveri, E. (2007). A comparitive study of Thermodynamic Properties of Binary Mixtures Containing Dimethylsulfoxide. Journal of Thermal Analysis and Calorimetry, 90, 909–922. 10.1007/s10973-007-7885-2.


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Stark and Zeeman effects and their causes
In high school we are taught that Zeeman and Stark effect directly lead to the findings of further divisions of an electron's shell in an atom.
What is exactly the reason that causes the spectral lines of an atom to split when an external magnetic or electric field is applied ?
And in what ways does this explain the existence of subshells?

It's not a "reaction", but rather a breaking of symmetry. Consider a sphere - a vibration viewed from any direction appears the same. Squish the sphere a bit, and there would be new modes of vibration that could be differentiated from each other.

See http://hyperphysics.phy-astr.gsu.edu/hbase/Atomic/stark.html and http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/zeeman.html

These effects split the energy of already degenerate energy levels, say the px, py, pz levels in an atom. This lowers the overall energy,  1 level does down, one remains unchanged, and the last one goes up.  As these levels are split this leads to new transitions and so new spectral lines. (Btw the Zeeman effect splits the nuclear spin levels in, for example, H atoms and so leads to NMR spectroscopy)


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Ionsiation enthalpy of Zinc Cadmium and Mercury
Why is ionisation enthalpy of Zinc, Cadmium and Hg in the increasing order, when ideally, all periodic groups have decreasing order of enthalpy downward?
An argument could be that F and D orbitals show less screening effect, but this is true for all the transition metal groups.
Another argument about fully filled shells being more stable also arises,... but even 3rd, 4th and 5th groups have fully filled S shells. When it comes to D orbitals, how does it being fully filled affect the ionisation enthalpy? It is not the valence shell afterall.


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Reaction of Heliox and seawater on commercial diving vessel
I work on a commercial diving vessel in the Persian Gulf. We have equipment that regularly goes sub sea. On occasion this equipment springs a gas leak, the gas leaking is a mixture of oxygen and helium (Heliox), usually somewhere between 8% to 18% oxygen. When a small leak occurs and goes unnoticed for some time there is a white buildup aroud the leak area. These buildups can grow quite significant in size compared to the leak and, in the odd case, self seal the leak!
The best way I can describe it is a calcification, it is brittle and breaks off quite easily into granule not unlike sand. The metal components involved are 316 stainless steel and Tungum, which is a brass alloy with aluminium in it, but the buildup occurs anywhere there is a leak, so I don't think the metals are relevant.
I am hoping someone here could explain this phenomenon to me, if this is not the right place for a question like this I apologies and ask if you could point me in the right direction.

Is that deposit everywhere, or only below water? If under water, likely some marine organisms are taking advantage of a steady O2 supply. Examine the deposit under a microscope. If the ship is engaged in petroleum exploration, there might be a petroleum geologist aboard who would be familiar with foraminifera, likely colonizers.
For a citation, "Species of benthic foraminifera presumably colonized the bacteria- and diatom-rich biofilms on the metallic surfaces of... petroleum platforms soon after their emplacement."
The heliox mix should not leave a residue, by itself -- one wouldn't want to breathe anything that left deposits like that!

Hi Dr Pippik, thank you for the reply. Yes the equipment is subsea for prolonged periods of time so marine growth does make sense. Although we are in the oil and gas industry we do not have a geologist or a microscope onboard. That would have been interesting to see. The deposit (or growth) is not negatively effecting any operations, in fact it makes leaks easy to identify and correct, just interested in causes. Thank you again for your reply.

@Dalmain, it might be worthwhile using a pair of magnifying glasses, or an inexpensive USB microscope, to take a look. Some of these critters shells are incredible: https://paleonerdish.wordpress.com/2013/06/17/an-introduction-to-foraminifera/

Could you comment the image displayed in the first question ? What is the red light ? What is the white stuff near the junction  ? Is it the result of corrosion? Why is there a junction ? Has water entered the gas tubing ? Why don't we see any bubbles around ?


text:
Hydration of alkenes

In the image I've provided, I offer two alkenes. The first is in conjugation while the second isn't. Additionally, both of the C=C groups are symmetrical, in the sense that they both have one hydrogen at each carbon.
My question is: what would happen when you hydrate each of these (acid catalysed hydration) with heat? Markovnikov's rule doesn't help because both carbons of the double bonds have one H each.  Where does the H go and where does the OH go?
I'm also interested to know if the conjugation (first molecule) has a different result from this reaction?

These two examples are unrelated - compound structures only look similar.

@Mithoron I am aware they are two different molecules.  My question is which carbon of each alkene group gets the H and which gets the OH in acid catalysed hydration, since Markovnikov's rule is not a consideration.
I have offered the two molecules in case conjugation produces a different result (e.g. possibly one product is more favoured when conjugation is present).

Two is such cases tends to be one too much. Another thing is you should analyse a bit the nature of intermediate carbocations.

In the 1st compound when you add water the intermediate carbocation is formed at the carbon nearer to the benzene ring as benzene ring being an electron withdrawing group delocalises the positive charge. So automatically $OH$ group gets attached in that position. But do not do anything to the benzene ring as if you do so, it loses its aromaticity.
For the 2nd one there is a hydride shift from the carbon lying at the intrsection of the ring and chain to the double bond position which forms more stable carbocation intermediate as positive charge resides on 3° carbon. So rather the $OH$ gets bonded to the $C-1$.
The products are as follows


text:
MS Waste Products
I'm looking for info about what happens to chemical species after undergoing MS. As I understand the process:

We have an high-energy electron beam to ionize our sample. This often leads to fragmentation, yielding things like radical species and cations. At this point, the neutral radicals don't get affected by the EM field and just sort of hang around until they're sucked out by the vacuum. What happens to these radicals after they're no longer in a vacuum environment? I would assume they react with something eventually, but this seems potentially dangerous (admittedly in small quantities, but still).
The ionized species then go through to be detected and then my understanding is that they're vented out of the lab, presumably for safe disposal somehow. What exactly happens to these ionized species once they're out? Again, the species seem like they'd be quite reactive (depending on the particular sample). What happens to these ionized species after we're done analyzing them with the mass spec?

I'm sure it depends on the precise sample, but presumably (hopefully) the waste systems are setup for a wide variety of waste particles, so it would be helpful to have a general answer. Thanks!

Well, depending on the instrument, molecules you detect may be destroyed.  In instruments that don't detect ring current, detection requires the ion of interest slamming into a hard surface and stimulating the emission of photons or electrons and detecting those resultant photons or electrons.  But in most mass spec, the molecules you detect are a small fraction of the molecules that go into the ion source.

If you are doing MS on potentially toxic substances, you need to treat what comes out of the pump (e.g. use a burn box). And be careful changing your pump oil.

The good news is that there is no shortage of electrons in the world. Gas phase radicals at ambient pressure do not survive long (perhaps on the order of femto to nanoseconds), so there is no concern of free radicals leaking from a mass spectrometer into the environment. Also recall, that we are constantly bathed in cosmic rays as well, that can balance any random electron lost from air. Secondly, the actual amount injected in the MS is so small (nanograms/micrograms), that it is not a major issue of environment pollution as well.
There are so many times of MS and ionization strategies but your query indicates standard electron impact MS. It is true that waste from electrospray ionization sources (liquids) is more serious and proper venting is needed (outside to open air).
In general waste from an MS instruments and vacuum pump exhaust should be vented outside the labs. The oil from vacuum pumps has to be replaced periodically to avoid build up of waste products. Finally, the internal components of MS do get dirty with time, and  they have to be cleaned periodically by qualified engineers.


text:
Do we understand why Dimethyl Mercury is so toxic?
I have read extensively about Prof. Wetterhahn and in the Wikipedia Wetterhahn mercury poisoning article it said her blood mercury (I do not know if the test measures mercury or the compound) was 4000 millionths of a gram per liter. I realize all the mercury was not in her blood but this does not sound like a lot of mercury, about a fiftieth of a gram. And the amount spilled was a few drops -- which however are very heavy and maybe those few drops were a few grams which maybe explains a lot?
Naively, I would think that poisons would tend to be bound up with the cells they damage. But perhap dimethyl mercury does damage and yet keeps circulating? Or is it something like in the brain a very small amount of some vital compound(s) is somehow destroyed or can't be made and therefore dimethyl Hg might not work by directly killing neurons?

Yes, have a look: https://chemistry.stackexchange.com/questions/88995/what-makes-dimethyl-compounds-so-extreme

Also see: https://chemistry.stackexchange.com/questions/103663/what-makes-dimethylmercury-ch32hg-so-penetrative-through-latex-and-human-ski and https://chemistry.stackexchange.com/questions/125999/accidental-dimethyl-mercury?noredirect=1&lq=1

As related in the book titled “Ignition!”, by John Drury Clark, it was once proposed to fire up a test rocket engine using 100 pounds of dimethyl mercury reacting with red fuming nitric acid. Fortunately, Eastman Kodak declined to produce the dimethyl mercury. The person who proposed using dimethyl mercury as rocket fuel thought is was “harmless as mother’s milk”.

@edv: story i heard was american scientists were attempting to brain damage or kill the soviet rocket wizards by getting them to try to use dimethyl mercury in the own space program. hydrazine is a very toxic fuel component that is actually used but i think dimethyl Hg would be far worse, almost no way anyone involved would not get sick. i bet the soviets would have realized the dangers.

@EdV: I noticed in chem in HS and after was sort of casual about toxic stuff. In university my pre-med friends would joke about coughing on fumes but by the mid 1980s or, they had begun to eliminate some stuff from chem labs. The chem teacher I had repeated the same stories over and over -- i have to say this guy probably addled himself.

@EdV: this grad student may be in real trouble. why did he do this stupid thing?

But at least the grad student survived, unlike Jack Parsons: https://en.wikipedia.org/wiki/Jack_Parsons_%28rocket_engineer%29.


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How do polyurethane varnishes cure after only the solvent evaporates
How do polyurethane varnishes cure after only the solvent evaporates?
I believe that polyurethane varnishes have as components isocyanates & polyols that react to form the polyurethane portion of the cured varnish. How is it that this doesn't happen spontaneously while the varnish is still in the can & only after it's been left for the solvent (water or organic solvent) to evaporate? What's the chemistry there?
Does the polyol only become a polyol after the solvent have evaporated? If so, again, what's the chemistry there?

Don't tell me the moisture is the polyol & it only gets to the isocyanate after the hydrocarbon solvent evaporates. Because that's only true for water repelling, hydrocarbon-based solvent formulas. What about water-based formulas, which is what most polyurethane varnishes are?

A lot of polyurethane resins start with beads of pre-made polymers dispersed in water with very little polymerisation happening on curing (or limited controlled cross linking of existing long chains using other types of cross-link).

What stops these pre-made polymers from polymerizing completely? The fact that they need to get settle down & get close to each other as the solvent evaporates?

A good answer might require a whole book to explain. Perhaps this one . But one highly simplified explanation is that the cross linking of dispersed droplets or resin is different to the reactions creating the oligomers and is inhibited in solution by the surfactants creating a stable dispersion.


text:
Confusion in n factor in the reaction
What will be the n-factor of $\ce{O2}$ in the reaction:
$$\ce{C4H10 + \frac{13}{2} O2 -> 4 CO2 + 5 H2O}$$
My tutor said it will be 26 but I am confused because he used stiochiometric coefficients in calculation too. Isn't n-factor for a mole or formula unit ?

Does this answer your question? The definition of n-factor, what was it when it was introduced?

No it doesn't.I am confused in this specific reaction.

Isn't n factor for 1 mol or a formula unit.My teacher taught it in equivalent  concept which main purpose is to calculate without using st coefficients.Here using the st coefficients just don't make sense to me.

Some teachers do not accept equations containing non integers. They say that it is impossible to take half a molecule . So the real equation should be : $$\ce{2 C4H_{10} + 13 O2 -> 8 CO2 + 10 H2O}$$ Here the $n$ factory is $26$
Edit: In this reaction, it  may be admitted that the $\ce{C}$ and $\ce{H}$ atoms from the butane molecule $\ce{C4H10}$ are all at the oxidation number equal to zero. In the oxidation of butane, each $\ce{C}$ atom offers $4$ electrons to $\ce{O}$ atoms. So, in total, it makes $16$ electrons for the $4$ $\ce{C}$ atoms. Each H atom offers $1$ electron, in total $10$ electrons for $10$ $\ce{H}$ atoms. So one molecule $\ce{C4H10}$ offers $16 + 10 = 26$ electrons to $\ce{O}$ atoms. As a consequence, the factor $n$ is equal to $26$ per molecule $\ce{C4H10}$.

But isn't n factor (or valency factor as my teacher calls it) defined for 1 mol or formula unit only??

In this reaction, one can admit that $\ce{C4H10}$ contains Carbon atoms and Hydrogen atoms both at the oxidation number zero. Each carbon atom offers $4$ electrons, in total $16$ electrons for all carbon atoms. Each H atom offers $1$ electron, in total $10$ electrons for all hydrogen atoms. One molecule $\ce{C4H10}$ offers $16 + 10 = 26$ electrons to oxygen atoms. The factor $n$ is equal to $26$ per molecule $\ce{C4H10}$

I would strongly advise you to go through this answer before reading through the rest of this post.

In the reaction
$$\ce{C4H10 + \frac{13}{2} O2 -> 4 CO2 + 5 H2O}$$
Let's see from the perspective of $\ce{O2}$
$$\ce{\frac{13}{2} O2 -> 4 CO2 + 5 H2O}$$
Initially the oxygen atom has oxidation state of zero, and in the water and carbon dioxide molecules the oxidation state is -2. So for each oxygen atom the number of electrons involved is 2.
n factor is not defined only for 1 mol only, it is defined based on the way the reaction is written and in this case it is written as $$\ce{\frac{13}{2} O2}$$ so 13 oxygen atoms are involved, giving the final answer as 26.

Difficult to understand how $13/2$ $\ce{O2}$ can produce $5$ $\ce{H2O}$

fixed the error, thanks for pointing it out

Ah ok now I think I am beginning to understand.I was taught n factor in equivalent concept where we weren't using st coeffs.Thanks.


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How much water is needed or released in the process of burning 1000 kcalories from fat reserves?
Recently, I have started working on my weight and, having a curious mind, I looked deeper into this whole exercise/metabolism/energy production thing. And I became curious how much water is needed to burn 1000 calories while doing aerobic exercise.
At first I looked into glucose oxidation and it looked pretty straightforward - fully oxidizing 1 glucose molecule generates so much energy and releases 6 water molecules. So unless a lot of water is involved in breaking down glycogen to individual glucose molecules, it's a no-brainer.
Then however I realized that glycogen is used in moderate-to-intensive exercise, while I am curious about slow energy burning exercise like walking. And apparently in this case, especially after some time walking, the body starts burning fat. This is where I am completely lost, as it is way above my high school level of biochemistry, most of which I have forgotten.
So my question is, is water consumed or released while burning fat? And how much would it be in order to generate 1000 kcalories?

As biochemical, metabolic and tissue structure processes are involved, it is a huge simplification to reduce it to direct chemical fat–water equivalence. // It is known fact that as fat has lower density than water and in initial stages is being replaced by the same water volume, so one can even gain weight. This does not related to the chemical equivalence.

There are 9 (dietary) calories in a gram of fat, as a rule of thumb. So for 1000 kcal (or 1000 dietary calories), you would metabolize about 110 g of fat. To estimate how much water is made, we assume the fat is made exclusively from oleic acid. The net reaction for complete oxidation is:
$$\ce{C18H34O2 + 51/2 O2 -> 17 H2O + 18 CO2}$$
To get the mass of water, we take the mass of fat and multiply it by the ratio of stoichiometric coefficients and the ratio of molar masses:
$$m_\mathrm{water} = m_\mathrm{fat} \cdot \frac{\pu{18 g mol-1}}{\pu{282.47 g mol-1}} \cdot \frac{17}{1} = \pu{119 g}$$
This is just the core reaction. There are lots of other things that happen, including the hydrolysis of the triglyceride (which uses up one water for every fatty acid released). As the fatty acid is oxidized, you might be making ATP, which has an effect of the water balance as well. You would have to specify exactly what you mean with "burning" fat to give an idea what you want to consider as the net effect. The net reaction I used looks like a combustion reaction, but the set of reactions going on in the body is more elaborate and more controlled.

How do you produce ATP from fatty acid oxidation?

This is exactly what I wanted, including the bit about triglyceride hydrolysis. Thank you, Karsten. I am aware of subsequent ATP hydrolysis to release the stored energy, but the reverse also happens in order to make ATP from ADP. What I meant by "burning" was the ultimate use of released energy by muscles. I guess I have one follow-up question - I read that energy released in the process of oxidation is used to make ATP molecules. My question is can the released energy be directly used in muscles right away, or is the ATP creation a necessary step?

@Renat The creation of ATP is a necessary step in the way metabolism currently works. There might be some extinct "primitive" species with a more direct path, or some future species. ATP is genius because you can take different chemical reactions that all make ATP, and you have a choice what to do with the ATP, so it is a very flexible system. Just like having money instead of 1:1 bartering where you trade your chemistry skills against some reputation points but really you need some music instruction.

@JiaoCtagon It is a three-step process, and some students spend an entire semester on learning it. Briefly, fatty acids are oxidized while reducing NAD+ to NADH. NADH is oxidized with O2 while pumping protons across the mitochondrial matrix. Protons flowing back through ATPsynthase is coupled with turning ADP and phosphate into ATP and water. Water, again. Here is a textbook chapter (also Ch. 19).


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Question about vapor pressure in solids and liquids
I've been reading about vapor pressure and it's a bit confusing. For example when water is in equilibrium between the solid and liquid phase, my textbook says that the vapor pressure is equal between the two.
Now I know this might sound stupid for someone who knows much more than me, but why is there vapor pressure in the equilibrium between a solid and a liquid?
Do the molecules with high kinetic energy go directly from the solid to the vapor form?( same thing with liquid). If not I don't know what that vapor pressure means in this case.
I only understand it in the case of equilibrium between liquid and gas.
Also note that I'm talking of the equilibrium between solid and liquids in water so the temperature would be increasing from 0 celsius up until it's fully a liquid

the T maintains zero Celsius until the ice phase vanishes.

First, your assumption about molecules going from solid directly to gas (vapor) is correct. This is called sublimation. Though many substances melt before there is significant sublimation, some sublime noticeably before melting, particularly at low pressure. Some well known examples are "dry ice" ($\ce{CO2}$), iodine, camphor, carbon or tungsten incandescent lamp filaments blackening the bulb, and ordinary water ice crystals growing from repeated sublimation and condensation.
Second, consider that if all three phases of water at the triple point are in contact, they must be in equilibrium, or at least one of the phases would disappear at the expanse of others. At the molecular level, there is constant interchange, with some fast-moving molecules in the solid becoming liquid or subliming to gas, while some less energetic molecules in the gas are condensing to liquid or solid. At a macroscopic level, one can observe a piece of ice slowly change shape as parts melt and refreeze elsewhere. However, in a closed system, the total amount in each phase should stay the same.
[Trivia: gallium is useful in high-temp thermometers because very little sublimes or evaporates. Ga has the widest range between melting point and boiling point, and a low vapor pressure until quite hot.]

In the case of gallium the triple point pressure is so low we can't consistently measure it as nonzero.

There is vapor pressure above liquids and there is vapor pressure above solids.
If their values would not be the same, the respective solid and liquid could be in equilibrium. As the phase with higher pressure would manifest net evaporation and the other net condensation.
It would be like if you put an open jar with water and another with sulfuric acid to a closed container. The volume of liquid water would decrease with time and the volume of sulfuric acid would increase. As water vapor pressure above water is greater than above sulfuric acid.
If you look at phase diagrams, you can see the direct solid–gas transition is common.

Is it because of Maxwell-Boltzmann distribution that the solid has vapor pressure? Can sublimation of ice happen at $0$ celcius?

Look at the diagram in the link. What do you see? Conclusions you make yourself have bigger value in several aspects, compared to if you are told so.

I know it's better but I also know my conclusions might not be correct.

The biggest mistake is being afraid of making mistakes. One learns from mistakes more than from successes.

My conclusion is that because 0C is the triple point of water, then all three phases coexist, so that is where the vapor pressure comes from.

Points on 2D charts have 2 coordinates....

At the same temperature solid and/or liquid and vapor, at equilibrium have the same Average Kinetic Energies but they do not have the same velocities because of the added forces of intermolecular attraction and, on Earth etc., of gravity. The differences manifest in the latent heats of melting and vaporization. [It seems that the place to study this is in microgravity, freefall, in a vacuum; or at the critical point.] In a pure water system [A one component system]: ice, water and vapor coexist only at the triple point. At the triple point all 3 phases have the SAME CHEMICAL POTENTIAL. That means the solid and liquid have the same vapor pressure. If temperature change and head space remains either vapor and liquid, or vapor and solid are present [subject to physical constraints]. No head space results in liquid-solid equilibrium or a single phase.
If the headspace is an inert gas such as air, water becomes a 2-component system. You will have to read about the Gibbs Phase Rule in a PChem text for the consequences. Each inert gas pressure changes the vapor pressure of both liquid and vapor differently depending on their molar volumes. This means that the melting point changes and that the melting point at each pressure is a triple point where all three phases are in equilibrium, ie. the liquid and solid have the same vapor pressure. If the temperate changes from the triple [aka freezing] point either the solid or liquid phase vanishes. Each pressure of inert gas has a distinct triple point and set of equilibrium curves.
I like to think of it as the 4th dimension in a 3phase brane.


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Dipole moments of trans and cis nitrous acid
I have a paper due about the isomers of nitrous acid, using various calculations methods to find molecular properties like their dipole moment. I've found the dipole moment of trans isomer to be 1.987 and the dipole moment of cis to be 1.458. I've also found that ALL bond angles are bigger for cis, and most bond lengths are longer for cis, other than N-O (single bond) which is longer for trans. Also if that helps, the HOMO and LUMO orbitals of trans have a bigger overlap in the N-O bond.
I know that the longer the bond, the bigger the charge separation, thus a bigger dipole moment, which doesn't seem to match my data and the literature, as the bonds are longer in cis (also checked the distance between the oxygens in both isomers). Does it have anything to do with the N-O bond being longer for trans, as the dipole moment for trans is bigger?
I'm matching my data to "Microwave Spectrum, Structure, Dipole Moment and
Quadrupole Coupling Constants of cis and trans Nitrous Acids" by A. Peter Cox, Alan H. Brittain and David J. Finnigan from 1970 (The article is quite old but it's the first one I had found, and it is very similar to more recent ones).
Here are the calculated isomers:

Thanks!

I do not know much about theoretical modelling but just keep in mind that dipole moment units may be different in older papers of microwave spectroscopy. Are you sure your units and their units in the calculations are identical and perhaps there are no scaling factors?

Also microwave spectroscopy can measure conformers.

 The dipole moments are aligned in trans-nitrous acid, adding up yield a higher dipole moment than that of cis-nitrous acid.

Dipole Moment ($\mu$)
Charge and Distance

I know that the longer the bond, the bigger the charge separation, thus a bigger dipole moment.

$\mu$ constitutes two terms, charge and distance:
$$
\mu = q\times d
$$
You need to account for both factors, even though the bond may be longer, the charge separation may be lesser reducing $\mu$. As an example, consider hydrogen halides:

$\ce{HX}$
Bond Length ($\pu{pm}$)
$\mu$
Source

$\ce{HF}$
94
$ \pu {1.51 D}$
1

$\ce{HCl}$
127
$ \pu {1.76 D}$
2

$\ce{HBr}$
147
$ \pu {1.73 D}$
3

$\ce{HI}$
168
$ \pu {1.55 D}$
4

Notice that even though the bond lengths are increasing, $\mu$s are not following the same trend, this is because of increasing covalent character of the bond (lesser charge separation) which is becoming prominent going down the group.
Vectorial Nature
Also note that $\mu$ is actually a vector quantity. Thus, the correct formula would be:
$$
\vec{\mu} = q\vec{d}
$$
Although this is a simplification, see note at the end, if you have multiple bonds with individual $\mu_i$s:
$$
\vec{\mu} = \vec{\mu_1} + \vec{\mu_2} + \vec{\mu_3} + \cdots \tag{1}
$$
When these vectors are aligned, the dipole moments add up vectorially. A classic example would be the $\ce{CF4}$, where all bonds are polar, yet $\mu$ is close to zero because of the symmetry of the molecule.
In case of cis- and trans-nitrous acid, $\mu$s are aligned for the trans isomer, adding up vectorially, and vice-versa for the cis isomer.

Structure
Polarity
$\mu$
Source

cis-nitrous acid

$\pu{1.46 D}$
5

trans-nitrous acid

$\pu{2.56 D}$
6

Note
In a molecule, charges are not localized on the atoms or within bonds, and thus, Equation 1 is a simplification. You can read more about the quantum mechanical nature of $\mu$ for a better understanding.