text:
Is an aqueous solution of potassium bicarbonate acidic or basic in nature?
The following is the definition of an acid salt according to my textbook:
The salts formed by the partial replacement of the replaceable hydrogen atoms (H+ ions) of an acid molecule by a metallic or ammonium ion are called acid salts. Acid salts ionise in water to give hydronium ions and therefore, their aqueous solutions show all the properties of an acid.
Going by this definition, potassium bicarbonate should be classified as an acid salt. Hence, an aqueous solution of potassium bicarbonate should be acidic in nature(as hydronium ions would be present in an aqueous solution of potassium bicarbonate).
However, my textbook also says, “The aqueous solutions of salts formed by a weak acid and strong base (as a result of a neutralisation reaction) are basic in nature.”
According to this, potassium bicarbonate solution should be classified as a solution which is basic in nature as potassium bicarbonate solution is formed as a result of a neutralisation reaction between potassium hydroxide, which is a strong base and concentrated carbonic acid, which is a weak acid even in its concentrated solution form.
So is potassium bicarbonate solution acidic or basic in nature?

I see the term acidic is in the question context used in ambiguous way. acidic(1) as being an acid, acidic(2) as forming solution with pH<7. While bicarbonate solution is acidic(1), as HCO3- is a weak acid, it is not acidic(2), as HCO3- is also a not so weak base.

Another example of an acid salt, that is not acidic(2) is Na2HPO4. The pH of an acid salt is approximately the average of 2 relevant neighbour pKas of respective acids.

You could simply calculate the $\mathrm{pH}$ of the solution. $K_\mathrm{a{_1}}$ of $\ce{H2CO3}$ is $\pu{4.30E-7}$ and $K_\mathrm{a{_2}}$ of $\ce{H2CO3}$ is $\pu{4.72E-11}$.The approximate $\mathrm{pH}$ of the solution can be given by:
$$\mathrm{pH}=\frac{(\mathrm{p}K_\mathrm{a{_1}} + \mathrm{p}K_\mathrm{a{_2}})}{2}$$
This yields $\mathrm{pH}= 8.347$, which indicates that the solution is mildly basic.

An aqueous solution of potassium bicarbonate is acidic ... towards a suitable base.  The existence of a dissociable proton in the anion guarantees that hydroxide ion could act as such a base, accepting  the proton from the bicarbonate ion; but says nothing about neutral water.  It turns out that neutral water is not so strong a base as to pull the proton off the bicarbonate ion, and the latter has some other atoms able to pull protons the other way ... so you end up with a net basic character.


text:
Is precision measured by significant digits in chemistry?
I was recently asked this question:

An experiment is done 3 times, with a different measuring tool used
for each time. The results of the experiment are 1.39, 0.9, and 1.0
(all centimeters). Which measurement is the most precise, and how do
you know?

On my paper, I put that the answer was 1.39 because it had the most digits past the decimal point and submitted. However, I've read online and found since then that my explanation actually referred to accuracy, not precision. My question is, would saying "1.39 is the most precise because it has the most significant digits" be a correct answer?

The question on the test is poorly worded and of the answer you should have written is: Nobody can answer this question with the given information. If your teacher, disagrees, explain with respect that his/her answers are not correct.

The single measured values provided by 3 different tools gives us just a hint about possible  accuracy or precision of the tools. The tool providing the value 1.39 is probably, but not necessarily more precise and accurate than the other two.
What these values really say is just the tool resolution.

There are 4 terms:

resolution
precision
accuracy(1), also defined by ISO as trueness
accuracy(2), defined by ISO as combined trueness and precision

Resolution relates to the least distinguishable difference of values provided by a tool. A tool able to tell ..1.38, 1.39, 1.40.... has resolution 0.01. It is not the same as the tick mark distance, as even the decimal fraction between the marks can be estimated for analogue output ( not digital one ).

Precision relates to the random errors during repeated measuring of the same value by the same tool or measurement method at the same conditions. It is usually expressed as a estimation of the standard deviation or confidence interval. It can be absolute, or relative wrt the measured value. It is usually worse than resolution, unless the latter is not fine enough. E.g. the resolution may be 0.001, but precission +- 0.02

Accuracy(trueness) relates to the systematic error of the tool or method aka the bias. It is the absolute or relative error of the mean measured value wrt the true or known value. The acccuracy(trueness) error is usually, but not necesserily bigger than precission error, e.g. the precission can be +- 0.05, but the accuracy(trueness) can have bias +0.1.

As formulated by Wikipedia in above links:

In measurement of a set, accuracy is closeness of the measurements to a specific value, while precision is the closeness of the measurements to each other.

Accuracy has two definitions:

More commonly, it is a description of systematic errors, a measure of statistical bias; low accuracy causes a difference between a result and a "true" value. ISO calls this trueness.
Alternatively, ISO defines accuracy as describing a combination of both types of observational error above (random and systematic), so high accuracy requires both high precision and high trueness.

Precision and accuracy(trueness) are mutually independent. The former is related to random errors, while the latter is related to the bias of the measuring tool a/o method. The accuracy(ISO) can be expressed as error $+0.1 \pm 0.05$.

source: Stack Exchange

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

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text:
How did Joseph Priestly discover carbon monoxide?
In my readings about the history of chemistry, I've come across the story of the incredible chemist and scientist Joseph Priestly. My question is just about one of his discoveries - the discovery of carbon monoxide.
Several places make mention of Priestly's discovery of carbon monoxide. However the best I can find is some vague language, like

There Priestley continued his research, isolating carbon monoxide
(which he called "heavy inflammable air") and founding the Unitarian
Church in the United States.

... but without a shred of detail!
I am wondering, can anyone provide some insight as to how Priestly went about doing this?
Where can I find some literature about his experiments and thought process?

As a Unitarian, I doubt Priestley was "Priestly".

Do a Google Book search with key words: "carbon monoxide" "inflammable air" and you will find tons of high quality results. https://www.google.com/search?tbm=bks&q=%22carbon+monoxide%22+%22inflammable+air%22

Converting my comment into an answer

It's true that Joseph Priestley is often credited to be the first one to synthesize carbon monoxide in 1772. Apparently when he synthesized it, he didn't realize it was a separate "air" (entity). So, he called it "combined fixed air". This was done by heating charcoal. For some time this was confused with "inflammable air" that is hydrogen, and also with methane since all three gases were flammable. Later other chemist refined the synthesis process and later it was identified to be a unique entity and was named carbon monoxide (but was erroneously called carbonic oxide at that time). You can find more details on how carbon monoxide was redefined here.
But point to note, there were several people often attempted to synthesize carbon monoxide before Priestley. There is a rich history of carbon monoxide mentioned here but since Priestley’s experimental success predominantly from his ability to design ingenious apparatuses and his skill in their manipulation, he is often credited to discover gases which no one has able to synthesize before. He gained particular renown for an improved pneumatic trough in which, by collecting gases over mercury instead of in water, he was able to isolate and examine gases that were soluble in water. Apparently, he is credited to discover 10 gases. For his work on gases, Priestley was awarded the Royal Society’s prestigious Copley Medal in 1773.
You can find more information from below articles:

Christopher P. Hopper, Paige N. Zambrana, Ulrich Goebel, Jakob Wollborn, A brief history of carbon monoxide and its therapeutic origins, Volumes 111–112, 2021, DOI: 10.1016/j.niox.2021.04.001.
McKIE, DOUGLAS. “JOSEPH PRIESTLEY (1733-1804), CHEMIST.” Science Progress (1933- ) 28, no. 109 (1933): 17–35. http://www.jstor.org/stable/43420726.
https://web.lemoyne.edu/~giunta/ea/PRIESTLEYann.HTML
https://www.britannica.com/biography/Joseph-Priestley
https://journals.physiology.org/doi/full/10.1152/ajplung.00310.2013

Here is a short description on how Joseph Priestly isolated carbon monoxide. He took an oxide of iron and charcoal, heated them together over a furnace in a gun barrel, than he passed the gases this reaction produced through limewater to remove the carbon dioxide. The remaining gas burned with a blue flame. He was able to weigh it and found it was slightly lighter than common air. He wrote about his findings in "Experiments And Observations Relating To Various Branches Of Natural Philosophy, With A Continuation Of The Observations On Air". This book can be bought on Amazon.

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


text:
Kinetics Vs Thermodynamics : Apparent contradiction in the definition of Enthalpy in the two disciplines
My book gives the following curve:

It gives the following relation :$$ΔH= E_{\mathrm{activation,forward}}-E_{\mathrm{activation,backward}} \tag{1}$$
But I suspect that $ E_{\mathrm{activation,forward}}-E_{\mathrm{activation,backward}}$ corresponds to change in Internal Energy of the system ΔU.
or $$ΔU= E_{\mathrm{activation,forward}}-E_{\mathrm{activation,backward}} \tag{2}$$
However,at a constant pressure, heat of reaction or Enthalpy change $$ΔH = ΔU + PΔV \tag{3}$$
Statement (3) contradicts statements (1) and (2) or is valid only for $ΔV=0$

This source contains the same graph : https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.04%3A_Potential_Energy_Diagrams

What is going on here?
In case, you feel that the graph is given only to cover a particular kind of reaction, please provide relevant comments.

Your book is sloppy?

@Zhe It does contain some misprints but overall it is a very good book for undergraduates. Moreover I have seen similar graphs in other books and on internet like https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.04%3A_Potential_Energy_Diagrams

Instead of using what appears to be the Arrhenius equation, trying looking at the Eyring equation instead: https://en.wikipedia.org/wiki/Eyring_equation

@Zhe Sir I am just an undergrad and Erying equation seems a little beyond my understanding. Please explain what are you trying to convey here through this equation.

The Eyring equation is from modern transition state theory. The Arrhenius equation is a fine starting point, and shares some similar features.

Related: https://chemistry.stackexchange.com/q/115923/72973

The reason might be that while drawing the reaction energy profile, we forget to mention what energy we are mentioning in the Y-axis. The following conventions are generally used:

If reaction conditions are constant NVT, energy in Y-axis should represent internal energy.
If reaction conditions are constant NPT, energy in Y-axis should represent enthalpy.
If reaction conditions are constant $\mu$VT, energy in Y-axis should represent Helmholtz free energy.
If reaction conditions are constant $\mu$PT, energy in Y-axis should represent Gibbs free energy.
If you are looking at single molecule, the energy will be the total energy of the molecule (kinetic energy + potential energy).

In case of reactions, the last three are generally used.
Hence, the "Energy" Y-axis changes based on reaction conditions. In the question, it seems you might have got mixed up somehow.

Summary of the terms used:

N: No of molecules
$\mu$: Chemical potential
V: Volume
P: Pressure
T: Temperature

Refer to NCERT Chemistry Class 12 Part 1 - Chemical Kinetics page 112 Graph 4.7 . It clearly mentions Potential Energy to Y-axis.

Even here they are deliberately using Potential Energy on the Y-axis along with same Enthalpy representation https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.04%3A_Potential_Energy_Diagrams

Please refrain from using a footnote that adds nothing to the answer. They can only reply in the comments and a greeting/footnote is unnecessary(adding very little value) in a Q&A forum.

So are you saying that in the Y-axis of my diagram,it should contain Enthalpy?

@TonyStark The Y axis is dependent on reaction conditions. You cannot expect to use Helmholtz free energy while studying reaction in $\mu$PT conditions.

Also, the Arrhenius activation energy is not equal to $\Delta^\neq H^\circ$. For a bimolecular reaction according to transition-state theory, $E_a =\Delta^\neq H^\circ +  2RT $, for a bimolecular reaction.


text:
Why isn't auto ionisation of water considered when we find the pH during salt hydrolysis?
$\ce{H+}$ concentration from the auto-ionisation of water is $10^{-7}$
If we have $\ce{HA}$ as weak acid and $\ce{BOH}$ as weak base having $\mathrm pK_\mathrm{a} = 3.2$ and $\mathrm pK_\mathrm{b} =3.4$ respectively, we are get a salt $\ce{AB}$.
When we find $\ce{H+}$ concentration for this salt using the formula,
$$\ce{H+} = \sqrt{\frac{K_\mathrm wK_\mathrm a}{K_\mathrm b}}$$
We get $\ce{H+}$ concentration in the solution to be $10^{-8}$.
In such a scenario, why don't we add $\ce{H+}$ concentration from the auto-ionisation of water to the $\ce{H+}$ concentration?

Nobody likes complex formulas. If all factors are considered, it can easily go out of hand. Formulation of few complications is quite easy.

What happens in salt hydrolysis is enhancement of water ionization with salts. Try asking yourself where do $\ce{H+}$ and $\ce{OH-}$ come from in the solution of $\ce{AB}$, and you'll find that they come from $\ce{H2O}$. Or equivalently, hydrolysis of $\ce{A-}$ can be written as
$$
\ce{A- + H+ <=> HA,H2O <=> H+ + OH-}\stackrel{\text{add}}{\Longrightarrow}\ce{A- + H2O <=> HA + OH-}.
$$
In a word, ionization has been taken into consideration in your calculation. There are multiple equilibria in the system, so you cannot simply add $10^{-7}$.

A more complex formula has to be considered, if the salt total/initial concentration is comparable with [H+] or [OH-]. In such cases, pH is not a simple function of just the respective pKx.


text:
How to identify cis and trans forms of cyclohexane
How can identify which one is cis and which one is trans . If I consider it as a planar molecule (though it is not) I understand that if the substituents are in same side it is cis else trans but this one confused me

At each carbon there is one substituent pointing in the general direction of up, and one pointing in the general direction of down, no?

@orthocresol shouldn't it be cis if botha re down and trans gor the other

Well, that's pretty much it, really. What exactly are you confused about?

@orthocresol in 1st both the substitients lie in perpendicular plane I.e in different planes so it should be trans whereas in 2nd both are in same plane isn't it?

Don't overthink it, there's no need for planes. Even in the chair, non-planar conformation that you've attached, you can already see that for every carbon, there is one substituent pointing upwards and one pointing downwards. Try building a model for yourself, if the 2D depictions on paper aren't sufficient.

Helpful tip: don't confuse cis- or trans- on the ring with axial or equatorial positions on the ring. The former refers to the relative positions of two groups. The latter refers to absolute positions of substituents on a fixed conformation.

Question: How to identify cis- and trans-forms of cyclohexane?

This is the title and main question raised in the OP's text. So I think that should be addressed before explaining the example. Following image shows the chair conformation of cyclohexane (in the box):

All $\color{red}{\text{red}}$ $\ce{C-H}$ bonds are in axial orientation (ax) and $\color{black}{\text{black}}$ $\ce{C-H}$ bonds are in equatorial orientation (eq). If you concentrate only on axial positions, you'd see 1,2-ax,ax-bonds are always trans (opposite direction). Therefore you can conclude followings:

Compounds with 1,2-ax,ax-substitutions are always trans.
Compounds with 1,2-eq,eq-substitutions are always trans as well.
Compounds with 1,2-ax,eq- or 1,2-eq,ax-bonds are always cis.

Similarly, since 1,3-ax,ax-bonds are always cis (same direction):

Compounds with 1,3-ax,ax-substitutions are always cis.
Compounds with 1,3-eq,eq-substitutions are always cis as well.
Compounds with 1,3-ax,eq- or 1,2-eq,ax-bonds are always trans.

Furthermore, since 1,4-ax,ax-bonds are always trans (opposite direction):

Compounds with 1,4-ax,ax-substitutions are always trans.
Compounds with 1,4-eq,eq-substitutions are always trans as well.
Compounds with 1,4-ax,eq- or 1,2-eq,ax-bonds are always cis.

For example, I have put structures of $\alpha$- ($\bf{1}$) and $\beta$-hexoses ($\bf{2}$) in the same image. Just look at orientations of 2,3-dihydroxy-groups in chair conformation of $\bf{1}$. They are in 1,2-ax,eq-orientation, thus are in cis-orientation (see point (3) above). You can confirm that by looking at planer version of the molecule.
In addition, if you look at the orientations of 2,4-dihydroxy-groups in the same chair conformation, you'd see they are in 1,3-ax,eq-orientation as well. Thus are in trans-orientation (see point (6) above). You can again confirm that by looking at planer version of the molecule.

According to Master Organic Chemistry-Geometric Isomers In Small Rings: Cis And Trans Cycloalkanes

$5$. Geometric  Isomerism: “cis-” And “trans-” Isomerism In Cycloalkane Rings
[...]For the case where the two groups are on the same side of the ring, we refer to it as ‘cis‘ (from the Latin, meaning, “same side of”.) For the case where the two groups are on the opposite side of the ring, we refer to it as “trans” (meaning “opposite side of'”).[...]

These are the possible structures of 1,4-dichlorocyclohexane (I've taken $\ce{-Cl}$ instead of $\ce{-CH3}$ to better highlight the difference using color as well)
Now, if you notice in the first structure, both chlorines are "above" the ring which means that they are the cis-isomer of 1,4-dichlorocyclohexane. (cis-1,4-dichlorocyclohexane)
In the second structure, one is "above" the ring and the other is "below" the ring. This means that they are 'trans' to each other and so this structure would be the trans-isomer of 1,4-dichlorocyclohexane. (trans-1,4-dichlorocyclohexane)

When reading the formulae, you need to translate these into 3D.  Imagine to see the cyclohexanes from the side for either the (cis)-1,4-substitution,

or the (trans)-1,4-substitution (here, intentionally the less stable conformer is chosen to highlight the difference),

From the side, the six carbon atoms of the cyclohexane are like an equator.  The substitutents are either on the same side of this (marked by the red line), or not.


text:
Reduction by Activated Carbon
Does reduction by carbon itself occur when filtration is performed using activated carbon? Is it part of the central mechanisms involved in activated carbon filtration?
One of the specificities of all activated carbon, per my understanding, is it's large surface area, which allows it to be treated in a way or another and maximize the surface area on which the respective reaction occurs. Now, just as all activated carbon has a high surface area (with some variations), can it be said that all activated carbon acts as a highly potent reducer on some ions?

Activated carbon is active mainly toward neutral and rather nonpolar substances, where reversible adsorption occurs based on van Der Waals force. It does not act as a ion reducer.

@Poutnik: Can't it accept oxygen to turn into CO2 or CO in an aqueous solution?

Carbon is quite inert wrt oxidation at ambient temperature, only strongest oxidation agents are able to partially oxidize e.g. graphite. But definitely not to CO2.

Activated carbon adsorbs unpolar substances. There is no chemical (re)activity.

@Hans Dispersed carbon may be used as a substrate  (think like an air craft carrier) for transition metals like platinum, rhodium, etc; then the metal decorated sites (think like its air planes ready to go) may reduce, e.g. $\ce{C=C}$ double bonds of alkenes / aromatic cycles into $\ce{C-C}$ single bonds.  Activated undecorated, metal free carbon alone typically is used to retain by adsorption (attraction on the surface) unpolar substances (e.g., cleaning prior to recrystallization, gas mask) and sometimes transition metal ions.  But these weak attractive forces are not very specific.

The only case I know of where activated carbon causes reduction at room temperature would be filtration of ozone, $\ce{O3}$. Even fluorine will not oxidize plain carbon:

Carbon is stable in a fluorine atmosphere up to about $\pu{400 ^\circ C}$ .

Oh, well perhaps dioxygen difluoride, $\ce{FOOF}$, could also oxidize carbon (or could be reduced by carbon, should you prefer a passive voice). Read Derek Lowe's Things I Won’t Work With: Dioxygen Difluoride, if you'd care to test that. Please let me know so that I could arrange to  be elsewhere.


text:
expression of apparent reaction quotient
question
When computing the Gibbs energy change of a reaction using Legendre-transformed $\Delta_fG'$ values for a given $\mathrm{pH}$, the formula of the $\Delta_rG'$ of the reaction is:

$\Delta_rG' = \Delta_rG'^\circ + RT \ln Q'$

How should $Q'$ be expressed in practice? Should it include $\ce{H^+}$? Based on the reasoning below, I think it shouldn't, but I would like to have confirmation.
example
Lets consider the following reaction;

$\ce{A^- + H^+ <=> AH}$

In conditions of constant temperature and pressure, the Gibbs energy change of the reaction can be formulated as

$dG_{T,P} = \mu_{\ce{A^-}} dn_{\ce{A^-}} + \mu_{\ce{H^+}} dn_{H^+} + \mu_{\ce{AH}} dn_{\ce{AH}}$

If we suppose that the reaction occurs in a buffered medium such that the activity of $\ce{H^+}$ is constant through the advancement of the reaction, a correct way to represent the system is as an open system exchanging $\{\ce{H^+}\}$ with its surrounding (c.f., Ref.1):

$dG_{T,P} = \mu_{\ce{A^-}sys} dn_{\ce{A^-}sys} + \mu_{\ce{H^+}sys} dn_{\ce{H^+}sys} + \mu_{\ce{AH}sys} dn_{\ce{AH}sys} + \mu_{\ce{H^+}buf} dn_{\ce{H^+}buf}$

where $sys$ denote the chemical system and $buf$ the buffer, which acts as the "surrounding" in the sense of the 2nd law of thermodynamics.
A commonly used way to enforce constant $\mathrm{pH}$ in this system is to apply a Legendre transform on $dG$ such that

$dG'_{T,P} = dG_{T,P} - d(n_{\ce{H^+}sys} \mu_{\ce{H^+}sys}) = \mu_{\ce{A^-}sys} dn_{\ce{A^-}sys} + \mu_{\ce{AH}sys} dn_{\ce{AH}sys} + \mu_{\ce{H^+}buf} dn_{\ce{H^+}buf} - n_{\ce{H^+}sys} d\mu_{\ce{H^+}sys}$

since $\mathrm{pH}$ is constant, $d\mu_{\ce{H^+}sys} = 0$ so

$dG'_{T,P,\mathrm{pH}} = \mu_{\ce{A^-}sys} dn_{\ce{A^-}sys} + \mu_{\ce{AH}sys} dn_{\ce{AH}sys} + \mu_{\ce{H^+}buf} dn_{\ce{H^+}buf}$

We can develop this formula by expressing $\mu_i = \mu_i^\circ + RT \nu_i \ln \frac{\{i\}}{c^\circ}$, with $\nu_i$ being the stoichiometric coefficient for chemical species $i$, $\{i\}$ being its activity and $c^\circ$ being its reference activity in standard state (i.e. $1$) to get to this familiar expression:

$dG'_{T,P,\mathrm{pH}} = \Delta_r \mu^\circ + RT \ln \frac{\{\ce{AH}\}}{\{\ce{A^-}\}} + RT \ln \frac{\{\ce{H^+}buf\}}{c^\circ}$

$RT \ln \frac{\{\ce{H^+}buf\}}{c^\circ}$ being constant, what is called $\Delta_r G'^\circ$ should be $\Delta_r \mu^\circ + RT \ln \frac{\{\ce{H^+}buf\}}{c^\circ}$, which means $Q' = \frac{\{\ce{AH}\}}{\{\ce{A^-}\}}$. The contribution of $\ce{H^+}$ is then not present in $Q'$ as it is in $\Delta_r G'^\circ$ already. Is that right?

Reference:

Lionel M. Raff, William R. Cannon, "On the Reunification of Chemical and Biochemical Thermodynamics: A Simple Example for Classroom Use," Journal of Chemical Education 2019, 92(2), 274–284 (https://doi.org/10.1021/acs.jchemed.8b00795).


text:
How to tell the difference between lysine and the N-terminal amino acid in DNP labeling?
The N-terminal amino acid of a protein can be identified by reacting the N-terminal amine with FDNB (1-fluoro-2,4-dinitrobenzene), or Dansyl chloride, or Dabsyl chloride, then digesting the protein with acid.
This question points out that the epsilon amino group of lysine can also react with the labeling reagents, creating a side-chain labeled lysine. Dinitrophenyl (DNP) derivatives of amino acids (specifically epsilon-DNP-lysine)
But if we're just looking for the labeled amino acid, how do we tell the difference between the N-terminal amino acid and lysine?
I suppose that if we ran the amino acid mixture on HPLC or TLC that the epsilon labeled lysine would have different retention times, and we could learn to ignore those peaks once we knew what lysine looked like.

Actually DNP-labeling of amino groups was first introduced in 1923 when 2,4-dinitrochlorobenzene (DNCB) was used for the identification of the terminal groups of a partial hydrolysate of silk fibroin (Ref.1). Then, Sanger has introduced 1-fluoro-2,4-dinitrobenzene (DNFB) in 1945, which bears his name, the Sanger’s reagent. DNFB was first used to detect free amino acids of insulin (Ref.2). In a nutshell, DNFB undergoes nucleophilic aromatic substitution with the $\ce{N}$-terminal amino group of a peptide or protein. DNP-group is stable to acid hydrolysis. After the DNP-labeled peptide or protein is hydrolyzed, the individual amino acids can be separated and only the DNP-labeled $\ce{N}$-terminal amino acid can be detected by a colorimetric detection at specific wavelength. Hence, DNFB can be used in protein sequencing to determine $\ce{N}$-terminal amino acids.
Not only the $\ce{N}$-terminal amino acid of the protein is labeled by Sanger’s reagent, but also the backbone amino acids with an active group to DNFB such as $\epsilon$-$\ce{NH2}$ in lysine, $\ce{SH}$ in cysteine, $\ce{OH}$ in tyrosine, and $\ce{NH}$ in histidine would be labeled as well. The labeled protein is then hydrolyzed (DNP derivatives are stable to acid hydrolysis), and extracted with ether. If lysine, cysteine, tyrosine, histidine are terminal amino acids, they should be di-DNP-labeled amino acids. If any of them are in the backbone (within $\alpha$-helix), they should have free amino group after hydrolysis. These mono-DNP-labelled amino acids with free amine group are soluble in acid phase and won't come to ether layer when extracted (Ref.2).
The concentrated ether layer is loaded on silica gel chromatography column and the elution profile is matched with the standard profile obtained from individual DNP derivative of all the amino acids. The reagents produce yellow colored derivatives which can be easily detected by absorbance (Ref.2). This method can be applied to any protein (best works with less than 50-70 amino acids). For larger proteins, before DNFB treatment, the protein should be hydrolyzed to small fragments by enzymes such as trypsin or pepsin, or by chemicals like cyanogen bromide.
Note: Extraction and separation methods are detaied in Ref.2, hence I attached the PDF file. For instance, in insulin (MW: ~$\pu{12000 Da}$), there are four terminals (two glycines and two phenylalanines), and two $\epsilon$-DNP-lysines.
References:

E. Abderhalden, W. Stix, Hoppe-Seyl. Z. 1923, 129, 143 (Reference 2 of Ref.2).
F. Sanger, "The free amino groups of insulin," The Biochemical journal 1945, 39(5), 507–515 (doi:10.1042/j0390507)(PDF).


text:
Is there any way we can arrive at a general principle of magnetic character based on the bond order (ie; fractional or integral)?
So, after going through two assignments on Chemical Bonding and the Molecular Orbital theory, I saw that in most of the problems where we were asked about the magnetic character of a molecule (paramagnetic or diamagnetic), I noticed that almost all of the paramagnetic molecules had fractional bond orders, whereas almost all diamagnetic ones had a whole number bond order. I highly doubt if it is a general rule, though. If so, what is the reasoning for this, and is there any other general pattern that relates bond order to magnetic character ? Sorry if the questions seems trivial.

A simple explanation for the paramagnetism of odd-electron species is you need two to make a pair.

An Oxygen molecule has two unpaired electrons and is paramagnetic and bond order 2.


text:
Identification of aqueous solutions of sodium hydroxide, sodium acetate and potassium thiocyanate using nickel(II)
I'm very new to chem — taking an intro to chemistry course where we are currently focused on substance identification. I have two substances, both aqueous solutions of either $\ce{NaOH},$ $\ce{KSCN},$ or $\ce{NaCH3COO}.$ Both substances are colorless with a pH higher than 7.
When mixed with $\ce{NiCl2},$ substance A forms a precipitate, and substance B does not. I know that you would expect a precipitate to form when $\ce{Ni^2+}$ ions are exposed to $\ce{OH-}$ ions, and I know that you would not expect a precipitate to form when $\ce{Ni^2+}$ ions are exposed to $\ce{CH3COO-}$ ions, so if $\ce{KSCN}$ was not a potential substance, I would know that substance A was $\ce{NaOH}$ and substance B was $\ce{NaCH3COO}.$
However, I am confused as to how $\ce{KSCN}$ fits into this. It's my understanding that since $\ce{SCN-}$ is the anion of thiocyanic acid, which is a weak acid, alkali metal compounds with $\ce{SCN-}$ anions are weak bases.
I have done some research and concluded that $\ce{NiCl2 + KSCN}$ will not form a precipitate. Is substance A $\ce{KSCN}$ or $\ce{NaOH}?$ Is there even a way to distinguish the two with this data?

Hint look up the solubilities of products, e.g. $\ce{Ni(OH)2},$ https://en.wikipedia.org/wiki/Nickel(II)hydroxide, or see https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules(Analytical_Chemistry)/Qualitative_Analysis/Characteristic_Reactions_of_Select_Metal_Ions/Characteristic_Reactions_of_Nickel_Ions_(Ni%C2%B2%E2%81%BA)

What is the color of the precipitate?

This is a homework question, not a lab, so this is all the data I have unfortunately.

Both nickel(II) acetate and thiocyanate are water soluble and gave green solutions in water. Therefore, to distinguish, you may need another solution, probably $\ce{Fe^2+}$ solution?

Probably $\ce{KSCN}$ is here just to be sure that the nickel salt is not contaminated by traces of $\ce{Fe}$ and $\ce{Co}$. Cobalt and iron are nickel neighbors in the periodic table, and $\ce{Fe^{3+}}$ and $\ce{Co^{2+}}$ ions are producing soluble but intensively colored complexes with thiocyanate ion $\ce{SCN-}$, dark red with $\ce{Fe^{3+}}$ and blue with $\ce{Co^{2+}}$. And nickel ions are not producing any colored complexes with thiocyanate.

I know that you would expect a precipitate to form when $\ce{Ni^2+}$ ions are
exposed to $\ce{OH−}$ ions

Yes, you are right, they will indeed form a green precipitate, nickel(II) hydroxide:
$$\ce{NiCl2 + 2NaOH -> Ni(OH)2 + 2NaCl}$$
Nickel(II) hydroxide does not dissolve in excess $\ce{NaOH}$.

I know that you would not expect a precipitate to form when $\ce{Ni^2+}$ ions
are exposed to $\ce{CH3COO−}$ ions

I haven't found any reaction like that to form nickel(II) acetate as it is generally formed from the reaction of nickel(II) carbonate and acetic acid but a little bit of searching gave me a site which shows the formation of nickel(II) acetate from various nickel(II) salts and acetic acid and various salts of acetic acid. The closest reactions I could find from that site is the reaction between acetic acid/lead acetate and nickel(II) chloride and reaction between sodium acetate and nickel(II) nitrate. Technically, the reaction can be written and balanced on pen-paper as it is a simple double displacement reaction but the reaction hasn't been performed practically. Also, I don't know the credibility of the site, so take that site as a reference in your discretion.

I have done some research and concluded that $\ce{NiCl2 + KSCN}$ will not form
a precipitate

Theoretically, it can be formed from a double displacement reaction, but practically it is found to be unfeasible as the end products get contaminated with unreacted reactants and byproducts to the point where yield is somewhat negligible. From a paper1:

Compounds containing nickel(II) thiocyanate are prepared mostly by metathesis between nickel(II) chloride or nitrate with alkali metal
thiocyanates. However, in such syntheses if precautions are not
taken, the final product may be contaminated with starting
materials. Methods known for preparing this compound include:
treatment of a dilute thiocyanic acid solution with nickel hydroxide
or carbonate, followed by evaporating the mixture at $\pu{150 ^\circ C}$; mixing and boiling aqueous solutions of ammonium thiocyanate and
barium hydroxide, followed by addition of an aqueous nickel sulphate.

Conclusion
A: $\ce{NaOH}$
B: $\ce{CH3COONa}$, as it is somewhat a weak base, pH = 8.9
Reference

Sarma, K.P., Poddar, R.K. A convenient method of preparing nickel(II) thiocyanate and its use in synthesis. Transition Met Chem 9, 135–138 (1984). DOI: 10.1007/BF00935928 (PDF)

I'm still confused as to how we know substance B is the sodium acetate, as opposed to KSCN. KSCN is also a weak base, no?

@Tornado547 KSCN has a pH in the range of 5.3-8.7, so, it varies more in the range of neutral pH.

And theoretically, all 3 compounds would form a precipitate with NiCl2. It's just that the latter two reactions has not been performed experimentally (at least I haven't seen them).


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Why can't methyl glycosides mutarotate?
I found this link trying to explain it: Why pyranosides do not mutarotate? however, I don't understand a step of the reasoning.
Why is it that acetalisation needs a stronger acid than for hemi-acetalisation? I was wondering why can't the mechanism be similar to the the one for the mutarotation of glucose as shown below:

Couldn't the $\ce{O}$ break the bond, and then instead of forming the aldehyde, rotate the $\ce{O-CH3}$ group and then reform the bond?

the second step, where the proton is removed by a base, woudn't work for a methyl group. Ether Cleavage needs a strong haloacid such as HI.

@AniruddhaDeb ok, but can't I skip that step and just rotate then whole thing and then reform that bond?

You can’t skip a step. You either move along the pathway or you hit a dead end/wall.


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fuel with the narrowest boiling temperature?
What commercially available fuel, such as kerosene or diesel, has the least variation in different hydrocarbon chains and thus the most uniform boiling point? I ask this because I'm searching for a fuel to use in a rocket engine, but unrefined fuels such as regular petroleum, which have a lot of variation in its constituents, will start to coke the engine plumbing when exposed to high temperatures. I understand that most commercially available fuels will do this, but are there any fuels which at the very least minimize this.

If I understand your question, you are looking for a pure hydrocarbon. Pure substances can be characterized by your requirements : 1) most uniform boiling point, and 2) least variation in hydrocarbon chain. So any pure hydrocarbon will meet your requirement.

Doesn't even have to be a hydrocarbon. Just any pure liquid that's got high energy content. Hydrazine is a classic one. There is. however, a specific grade of refined kerosene called RP-1 that has a narrower bp and is intended for rockets, though the bp range is still not what you'd get with a pure compound.

Compulsory reference to Ignition: an informal history of liquid rocket propellants by John D Clark. On Amazon here. he specifically talks about special grades of kerosine with these properties.


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entropy of vaporization vs entropy of fusion
Is the entropy of vaporization greater than entropy of fusion, for elements? And in which case they can be same, if it so happens?
I have asked this here because I couldn't find any satisfactory answers anywhere on the Internet and most of them were very limited in their description.

We define fusion here as transition from solid to liquid.

What is the issue with using latent heat, since thermodynamically dS is defined as dQ/T?

Yes, you are free to use latent heat, but my premise was that, can this dilemma only be solved by using latent heat? Let me edit this question. Now it looks better.

We expect the entropy of a solid to be less than that of its liquid to be less than that of its vapour. The change in values is also generally greater in that order, for example to melt ice $\Delta S = 22$ J/K/mol and to vaporise $109$ J/K/mol. Many liquids obey Troutons's rule of $\approx 85$ J/K/mol for vaporisation.

Of course, there is some temperature near the critical where the heat of vaporization at that temperature is equal to the heat of fusion at the fusion temperature.

@ChetMiller Say, at the melting point, will both of them be the same?

No they won't.  The heat of vaporization will be higher.

@ChetMiller OK, and what about the change in entropies?

Well, the temperature is the same for both at the melting point, but not the heat of fusion and heat of vaporization, so what does that tell you about the entropy changes?

@ChetMiller Yep. I get it. OK. So, a final question, if you know the melting point and the enthalpy of fusion for an element, could you calculate the change in entropy of vaporization?

@ShishirMaharana  The entropy of vaporization is the change in entropy associated with vaporization, i.e., it's $\Delta S_{vap}$.  By contrast, the "change in entropy of vaporization" means something else, since it means the change in the entropy change associated with vaporization, i.e., the change in $\Delta S_{vap}$. So unless you are actually interested how  $\Delta S_{vap}$ changes (which it doesn't seem you are), as opposed to what  $\Delta S_{vap}$ is, then you'll want to edit your post appropriately.  The same of course applies to the entropy of fusion as well.

Here I'm assuming you are interested in comparing the magnitude of $\Delta S^{\circ}_{fus}$ at $T_{fus}$ with that of $\Delta S^{\circ}_{vap}$ at $T_{vap}$, for the elements in their standard states (so, for instance, hydrogen would be $\ce{H_2}$ rather than $\ce{H}$).
Like you, I was unable to find a comparative tabulation of these values on the internet.  Fortunately, one can readily generate such a table using Wolfram Mathematica's chemical database.
For each element:
$$\Delta S^{\circ}_{fus} \text{ at } T_{fus} = \frac{\Delta H^{\circ}_{fus}}{T_{fus}}$$
$$\Delta S^{\circ}_{vap} \text{ at } T_{vap} = \frac{\Delta H^{\circ}_{vap}}{T_{vas}}$$
Wolfram has the above data for all of elements 1–93 (hydrogen through neptunium), except for helium (which can't be solidified at standard pressure, which is 1 bar), astatine, and francium.*
Here I've plotted $|\Delta S^{\circ}_{vap}|$ vs. $|\Delta S^{\circ}_{fus}|$* for these 90 elements, and added a y = x line. From the placement of the points relative to this line, you can see that all except one of the elements have $|\Delta S^{\circ}_{vap}| > |\Delta S^{\circ}_{fus}|$.
That single exception is hydrogen, for which:
$$|\Delta S^{\circ}_{fus}| \text{ at } T_{fus} = 39.8 \frac{J}{mol K}$$
$$|\Delta S^{\circ}_{vap}| \text{ at } T_{vap} = 22.3 \frac{J}{mol K}$$

*Note, however, this complication:  Most, but not all, of these measurements were done at standard pressure (1 bar). For instance:
"When heated at standard atmospheric pressure, arsenic changes directly from a solid to a gas, or sublimates, at a temperature of 887 K. In order to form liquid arsenic, the atmospheric pressure must be increased. At 28 times standard atmospheric pressure, arsenic melts at a temperature of 1090 K. If it were also measured at a pressure of 28 atmospheres, arsenic's boiling point would be higher than its melting point, as you would expect."
https://education.jlab.org/itselemental/ele033.html


text:
Formation of methane using Kolbe's electrolysis reaction
I have read in my book that methane cannot be formed using Kolbe's reaction (mechanism — Organic Chemistry Portal). And honestly, I kinda think I get where they are coming from: that two radicals are required to get a hydrocarbon which can in the least be two methyl radicals (of course when substrate is sodium acetate) which can combine to give ethane.
Why can't we use a mixture of sodium formate and sodium acetate and expect some methane to be released? Why shouldn't it work?

If you notice the actual mechanism that you have provided, you would notice that R-H is formed albeit in a very small amount.

Even though it deals with Wurtz reaction, tempted to call a dupe of https://chemistry.stackexchange.com/questions/65303/why-can-wurtz-reaction-be-not-used-for-preparation-of-alkanes-with-odd-number-of


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Statistical methods used for testing the homogeneity of the sample
So, if I have multiple FTIR spectra each taken on different location of the sample and averaged from ~20 spectra, how can I check if the sample is homogeneous, i.e., if there are statistically significant differences between the spectra based on location? For visuals, I tried to compute at each wavenumber confidence upper and lower limits, but I am not sure if this method is valid even for visualisation and it does not give anything exact or quantitative.

It is a statistical problem, which can be solved with maybe 95% degree of confidence, using the chi-squared test.

Without seeing the spectra it is hard to advise , but presumably you have a set of peaks that you expect to find. If you have a 'standard spectrum' then use this to compare peaks of others  and so remove poor samples. Plotting this spectrum on x vs each of the rest on y and making a contour plot may help to do this pairwise. As the spectra are likely to be so different due to extraneous (unimportant) lines a conventional stats method may not work.

@Maurice Here I tried to construct chi squared test in R, but got a very large value: https://stackoverflow.com/questions/63714065/chi-square-construction-in-r .

@ User. Well ! What can I say ? I don't know your numerical values. And even if I would have known them, I could not change the chi squared value...

@Maurice Ok, the problem was that I haven't divided everything by N.


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What is the size in m³ of a toxic benzene cloud when if 20.000 kg liquid benzene would vaporize?
Benzene is carcinogen, The short term exposure limit for airborne benzene is 5 ppm for 15 minutes see https://en.wikipedia.org/wiki/Benzene#Benzene_exposure_limits
I would like to know which volume of air could contain toxic benzene if an benzene ship with 2000 ton benzene should vaporise vaporization to the air. Benzene is heavier than air so it would be closed to the ground. How close is not a chemical subject but a physical phenomenon which is out of the scope of this question.
In this calculation I use a lot of assumptions, just because the lack of info. But at least it gives a idea what might happen. Hopefully, on a later moment, some assumptions could be replace by facts, making the calculation more accurate. For now it is meant to get an idea what would happen in the worst case scenario.
Assumption only 1% of the benzene vaporise. The question how much benzene could vaporise is out of the scope of this question.
Calculation adjusted with info of @maurice : Benzene is not a ideal gas. The vapor pressure of benzene is about 0.04 bar at 0°C.

Mol mass benzene: 78,11184 g/mol

Mol / kg: ~12,8 mol benzene (1000 g / 78,11184 g/mol)

0.022413969545014... m3/mol at 0 °C, see https://en.wikipedia.org/wiki/Molar_volume using Avogadro's law

The vapor pressure of benzene is at 0°C: 0.04 bar

1 kg benzene liquid: 0,01148 m³ benzene gas at 1 bar   (~12,8 mol benzene * 0.022413969545014... m3/mol * 0,04bar / 1 bar)

Benzene to vaporize: 1 %

Mass vaporized benzene: 20.000 kg ( 2000.000 kg * 1%)

Total 100% benzene gas: 230 m³ ( 20.000 kg liquid benzene * 0,01148 m³/kg)

Assumption height benzene cloud from the ground:

1 meter high: Surface: 230 m²
3 meter high: Surface: 77 m² (230 m² / 3m)

Assumption gas will spread as a circle:

Radius with height of 3 meter: 8,5 m  (230 m³ = 3,14 * r²)

Diameter cloud of 3 m high: 17 m

Which can move with the speed of wind which can be 0 m/s (0 Beaufort) up to 28 m/s (10 Beaufort)
Benzene will probably mix with air. But airborne benzene stay toxic up to 5 pmm.
So the maximum size of the toxic cloud would be:

The short term exposure limit for airborne benzene is 5 ppm
Maximum amount of gas which could be carcinogen: ~46 * 10⁶m³  (230 m³ / 5* 10-⁶ ppm)
Assumption height benzene cloud from the ground:

1 meter high: Surface: 46 * 10⁶m²
3 meter high: Surface: 9,2 * 10⁸m² ( 46 * 10⁶m³ / 3m)

Assumption gas will spread as a circle:
Radius with height of 3 meter: 3,8 km  (46 * 10⁶m² = 3,14 * r²)
Diameter cloud of 3 m high: 7,6 km

If the cloud gets larger it would be less than 5 pmm and is not toxic anymore. This will be a matter of windspeed and time, which both are beyond the scope of this question.
The key question for me is: Is the calculation for a vessel loaded with 2000 ton benzene where 1% leaks generate a cloud with carcinogen air between about 230 m³ and 46 * 10⁶m³ correct? Or do I miss important issues.

Benzene doesn't become toxic at a particular concentration. The rules about exposure define the acceptable concentration over a period of time to keep the risk low enough not to be a significant risk. And the big risks of a big cloud are not primarily toxicity.

@matt_black Is carcinogen in the category of toxic? If not, although it is not my question, what is than the main risk of a big cloud? I know it is said there should be no exposure to benzene at all. But is so no calculation can be made.

It is not clear if the limit 5 ppm for air is assumed to be V/V or w/w. I imply the former, but just guessing. The difference is almost 3 times.  1 ppm(V/V)  is approximately 78/29 ppm (w/w)

Your calculation uses Avogadro's law, which implies that benzene can form a perfect gas at $0$°C and $1$ bar. But benzene cannot form a gas with a pressure of $1$ bar at $0$°C. The vapor pressure of benzene is about $0.04$ bar at $0$°C. So all your calculations are about $25$ times too high.

@maurice I used your information to correct the calculation in my question

@Poutnik I used https://en.wikipedia.org/wiki/Benzene#Benzene_exposure_limits which does not give information about V/V or w/w So I can not tell. I am happy to adjust the calculation if needed.

@Bernard I got it, I have read it as well, that is why I have put there such a note. See my calculations in the answer.

@Bernard Contrasting to the initial answer given by @Poutnik, your argumentation is difficult to follow; in part because your lines of thought exclusively rely on bullet points.  This renders, e.g., a check of the dimensions difficult.  Beside formatting as an issue on itself (e.g. ChemSE as a website relating to the English speaking scientific community more likely uses the decimal point, than the comma; uses scientific notation), I notice the number of carried decimals once pretends an unlikely precision (molecular volume) when eventually aiming for a (reasonable) estimate.

@Bernard The toxicity of benzene is, indeed, primarily its carcinogenicity. But this is a long term chronic problem not a short term acute one. Until the 1960s benzene was uses as a standard lab solvent (people would even wash their hands with it to remove grease!) The short term risk is its flammability. The occupational risk levels are really about long term exposure in the workplace, not about short term risk.

@Buttonwood The answer of Poutnik is indeed much easier to follow than my calculation. And indeed the international scientific community uses a decimal point instead of a comma. And I also agree that the number of carried decimals seems to pretend an unlikely precision. But personally I feel some resistance to change my question after answers are given. If I would do so some comments might look weird in for a new reader. Poutnik gives, in my view a good answer in a good format.

@matt_black Indeed its carcinogenicity is a long term issue. But still an important one to consider. I am thinking about to make a calculation about its flammability. But that is outside the scope of this question.

The hypothetic benzene cloud will be tremendously large while it is still in the explosive range of a few percent, and surely find an ignition source. Poof.

@Karl Sounds logical and scary.

According to vapor tension online calculator, at $t=\pu{20 ^\circ C}$, the vapor pressure of benzene equates to $p=\pu{10.018 kPa \simeq 10 kPa = 0.1 atm = 100 000 ppmv}$
If we consider benzene vapor as an ideal gas, the saturated vapor density $\rho$ is the maximal concentration of benzene vapor at this given temperature:
$$\rho = \frac {pM}{RT} = \pu{\frac {10000 \cdot 0.078}{8.314 \cdot 293.15} kg m^-3}\simeq \pu{0.320 kg m^-3}$$
The volume of air, saturated by $m=\pu{20 000 kg}$ of benzene is:
$$V_\mathrm{saturated}=\frac m\rho  \simeq \pu{62500 m^3}$$
The air volume for the $\pu{15 min}$ safety threshold at $\pu{5 ppmv}$ is:
$$V_\mathrm{threshold}=V_\mathrm{saturated} \frac{100000}{5}\simeq  \pu{1.25e9 m^3}$$
Assuming an uniform, pie-shaped cloud of height $h=\pu{3 m}$, the cloud would have a diameter $D$ :
$$D=\sqrt{\frac{4 \cdot V_\mathrm{threshold}}{\pi h}}=\sqrt{\frac{4 \cdot \pu{1.25e9 m^3}}{3.14 \cdot \pu{3 m}}}\simeq \pu{23 000 m}$$
The diameter for the analogical cloud of the saturated air would be $\pu{163 m}$.
For obvious reasons, the real cloud shape and concentration distribution would be very different and hard to predict.
The 3D shape of the cloud very depends on the wind speed, the vertical air thermal gradient and the wind speed vertical gradient ( See the Richardson's number ). This determines the level of turbulent mixing in low hundreds meters of troposphere.
But the estimate above provides you a ballpark figure.

@Buttonword Thanks. Ehm, well, while your answer improvement is good, it is like a rewriting the answer. I see my English has weak times more often than I like. :-)

I agree my editing of the text went far; and while working in the text, wasn't recognized by mine as close to a rewrite as from your perspective. Regardless if these edits may be reverted, my future handling of urtexts is going to be more careful.

@Buttonword No hard feelings, I have left the changes there. I was smiling seeing all the green and pink fields. In fact, I appreciate your editor work. Some cases are stupid errors, some cases are language mistakes. I learn new fine features of English every now and then, like the rule for comma-separated adjectives or "ballpark figure".


text:
Is sulfuric acid necessary for Kevlar synthesis?
In the production of Kevlar, the 2 monomers needed for its creation are reacted in a solution. Due to it being water-insoluble, sulfuric acid is used. My question: Can other chemicals be used as well (for examplke salicyclic acid), or is sulfuric acid absolutely necessary? Is there maybe a benefit to using it, or is it simply the cheapest?

I think you need to tell us more about the process before we can answer this

It's not for synthesis, but for spinning out of solution.

The sulfuric acid isn't just required to make the reaction go: it is a solvent. And–even if you could do the reaction in a different solvent–the properties depend on the spinning process which is solvent dependent so you would get different bulk polymer properties.


text:
Safest way to conduct a reaction with bromine under reflux conditions
I am conducting an electrophilic aromatic bromination reaction according to literature. The substrate and intended substitution pattern do not matter, but the reported procedure requires the dropwise addition of bromine at room temperature followed by refluxing overnight. (The solvent’s boiling point is very close to bromine’s so it may be assumed that both compounds reflux.) The reaction is not conducted under inert atmosphere and as with every organic bromination I expect HBr to be a (gaseous) by-product.
Obviously, the reaction is being conducted in a fume hood whose front opening is closed as far as possible. Nonetheless, I want to limit the amount of bromine vapour that gets released into the surrounding atmosphere. I came up with the following setup:

Figure 1: A two-necked flask with a dripping funnel and reflux condenser. The glassware on top of the reflux condenser should be a drying tube (no accurate clip-art was found) containing damp $\ce{Na2S2O3}$ between cotton wool.
The dripping funnel is obviously used to add bromine dropwise to the reaction mixture and is sealed on top. The reflux condenser is open to the atmosphere through a drying tube; I filled the drying tube with sodium thiosulphate to quench any bromine vapour that makes it up there.
I did not attach the coolant flow to the reflux condenser until after the dropwise addition. This may have been a mistake, as bromine vapours had already diffused all the way up and penetrated the cotton wool and thiosulphate. It seems that these were far less effective in holding back any vapour than I had anticipated. Luckily, our coolant system is closed-circuit and the temperature adjustable. Thus, I set the cooling temperature to $\pu{0 ^\circ C}$ which seemed to reduce bromine’s vapour pressure sufficiently to clear the air in the drying tube. However, I believe there must be a better solution.
How can I run this reaction in reflux conditions while preventing bromine vapour escaping as much as possible? I had thought of sealing the entire reaction apparatus, but the reaction gives a gaseous side product (HBr) which could cause an undesired increase in pressure. I had also considered using a balloon to close off the reaction atmosphere, but I fear that both bromine and HBr will react with the rubber and destroy the balloon rapidly. What other methods exist that I may have missed?
Side note: the literature I am following was published in 1987. As expected of this era, they do not note specific precautions against bromine release.

Instead of the sodium thiosulfate filled dry tube you need a scrubber - a Dreschel bottle filled with thiosulfate soln would do it, though to be safe against suckback a train of 2 bottles with the first empty might be advisable as you're leaving it overnight.

You might look for another funnel or different flask, to make sure the bromine drops directly into the solvent. Add bromine slowly. That should reduce fuming significantly.

@Karl Simply swap over the addition funnel and condensor so that the addition funnel is vertical.

@Karl the picture is schematic. The actual flask I’m using is more pear-shaped and the dripping funnel has an outlet extending through the joint so that bromine did drop into the reaction solution. As there is quite a large amount of bromine compared to solvent, I am not sure whether more careful dripping will change any part of the general picture.

Based on your reaction solvent, you may dilute bromine 1:1 with solvent instead of using neat bromine. I was successful using $\ce{Br2/CHCl3}$ in aromatic bromination with 4-cyanophenol.

Hm, I think you should tell us what reaction exactly this is, and how much bromine, substrate and solvent you use.

Titanium metal will spontaneously ignite in room temperature bromine. I mention this because some corrosion resistant lab equipment is made of titanium , I see you have no exposed metal - but just in case.....

The biggest difference you could make would be to cool the condenser before starting with bromine addition. You already know this works from what you have observed later in the process.

Check if you may skip using elemental bromine as starting material altogether.  An alternative approach may be running the bromination in a heterogenic, biphasic setup with the couple $\ce{NaBrO3/NaHSO3}$ in water (e.g., 1998JOrgChem6023, 2000OrgProcRes Dev.30), or $\ce{NaBr}$ / sodium perborate (e.g., 1998SynthCommun925), or pyridinium tribomide (e.g.,2005JChemEduc306), or bromodimethylsulfonium bromide prepared by mixing aqueous $\ce{HBr}$ with DMSO (1997JOC4321, often at rt).  They share a slower release of bromine, and because of this, lower amount and concentration of reactive bromine and attenuate the hassle with corrosive gases.
If staying with liquid bromine:

Trapping unconsumed bromine with thiosulfate per se is a good idea, but the reaction of bromine gas with solid $\ce{Na2S2O3}$ in the drying tube is both incomplete, bears the potential that the drying tube gets congested, and tends to be messy to clean.  You could end up with a closed vessel, which isn't safe; and even dangerous as you generate HBr in stoichiometric amounts.  Thus passing the exhaust gas across scrubbers filled with aqueous thiosulfate solution has a much higher chance of win.  Don't forget the safety flask (different reaction, but same principle).

The the normal balloons from the toy shop does break under bromine, as well as HBr gas too quickly.

Beside @Mathew Mahindaratne's comment to dissolve bromine in chloroform, using acetic acid as solvent provides you a wider margin of temperature.  (If your substrate can withstand the acid.)


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How is elimination reaction (E1) possible without base and in presence of acid?
When learning about E1 we were taught that three things are important factors to judge if the reaction is E1 or E2

Substrate ( 3,2,1 favours E2 but only 3 favours E1 )
Base ( Stronger Base favours E2 , weaker Base disfavours E2 hence suggests E1 ) But E1 requires a base however weak
Leaving group ( E1 is more sensitive to LG because it initiates E1 ) Bad LG disfavours both , excellent LG favours E1 more

This was from David M Kleins Book " Organic Chemistry I as a Second Language "
But in Section 11.6 he says this

fact, the reverse path (starting from the alcohol and ending with the alkene) is a reaction that we have already studied. It is just an E1 reaction

Then he goes on to show this picture

Initial question : How do you have an elimination reaction without a base, that too in the presence of an acid ?
(Although I understand that E1 's rate doesnt depend on a Base but it is still needed)
Answer : there is a Base ( H20) which makes it look very much like E1 .
Edit : it behaves very much like an E1 but with an extra step
Step 0: converts the bad LG to an excellent LG
Step 1: LG leaves
Step 2 : Base abstracts proton to make alkene
New question:  Why does water behave like a base in an acidic mixture? And why does it not react with the H+ already in the solution ?

you are dehydrating an alcohol.. why not? Also, no, base is not needed for all E1 reactions..

then what are the necessary conditions for E1 reaction ?

so first of all, concentrated H2SO4 also has some amounts of water in it, so you have a base to work with. I think, "Why does water behave like a base in an acidic mixture? And why does it not react with the H+ already in the solution?" is indeed possible, but this reaction is in equilibrium forming some definite amounts of both. So, water is anyway abstract a proton.


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Can a metallic bond with a sea of electrons form inside the human body?
I’m interested if atoms or ions can for a metallic bond inside the human body?
I’m talking about a metallic bond as described here: https://www.britannica.com/science/metallic-bond
So I’m thinking of a metallic bond where the valence electrons of the atoms or ions are not associated with specific atoms or ions of the metallic bond. My understanding is that this is described as a sea of electrons.
If such a metallic bond can not form inside the human body, I would be very interested in the reason why. Thank you very much.

Umm... pierced earrings? Spinal fusion hardware? Gold dusted food? https://observer.com/2017/12/gold-plated-most-expensive-food-and-drink-in-the-world/

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3232296/

If you zoom in on a metallic bond, you’ll realise that it is ultimately identical to a covalent bond at atomic levels. That said, the question is how do you define form? Do metallic solids get created de novo within a body? Likely not – due to the absence of starting materials and reaction pathways. Can metallic solids exist within a body? Absolutely!

We may not produce nails from our bodies, but certain enzymes and proteins do involve metal clusters, especially clusters involving iron [1].  In the natural biomolecules iron-iron and iron-nickel compounds serve as functional groups for electron transfer.  Also discussed are iron-sulfur clusters, which have characteristics similar to all-metal clusters despite sulfur usually being considered a nonmetal.  Chemists have sought to mimic these natural cluster compounds with artificial molecules, not only using iron and nickel but also moving on to less common metals such as molybdenum and ruthenium.
Reference
1.
C. Fehl C and B.G. Davis, "Proteins as templates for complex synthetic metalloclusters: towards biologically
programmed heterogeneous catalysis.", Proc. R.
Soc. A 472: 20160078. (2016)
Link


text:
Why is it rational that salt-bridges (in proteins) are described as a combination of H-bond and ionic bond?
This is a fundamental question that came to me as I was trying to draw the electrostatic attraction between Aspartate (D) and lysine (K). I have always considered the charge on nitrogen, in the lysine side-chain (-NH3+), as a real charge. This is because it has always been described to me as an ionizable amino-acid, that can form salt-bridges to carboxylic anions e.g. aspartate (D) or glutamate (E).
The general definition of a salt-bridge is:  a strong electrostatic attraction, a combination of H-bond and an ionic-bond. This is simple enough to memorize and makes sense when one is studying enzyme structure. However, as I started to draw this salt-bridge attraction between lysine (K) and aspartate (D) - it started to seem counter intuitive that a salt-bridge is a type of ionic-bond, as the hydrogens on nitrogen seem to be in the way? I also started to wonder when the charge on nitrogen is real, or simply a formal charge.

Questions on the the definition of salt-bridges in proteins:

Why is a salt-bridge not just simply a hydrogen bond (i.e. not partially ionic)?
If I accept the notion that this is (partially) an ionic bond, how can I tell that there is a real charge on the nitrogen (an not just simply a formal charge)? And furthermore, why won’t the hydrogens (protons) on the ammonium group ‘disrupt’ the ionic attraction?
Is there, perhaps, a defining bond energy that would classify it as ionic (and it happens to be just within this?) otherwise we would simply call it a hydrogen bond?

I also noted that the question is very similar to this , but it does not really highlight how salt-bridges are a combination of both H-bonds and ionic-bond.

"It also assumes salt-bridges as purely ionic" - no, not really, just poorly worded. If you read answers properly, it may clear it up a bit. Also your wordiness here obfuscates points of questions.

point taken, ill rephrase. My point here was that the linked question (not the answers)  is not focusing on 'the combination of H-bond and ionic-bond' confusion. I dont understand how this could be called an ionic bond.

Actually that's perhaps more of an ionic bond then what is commonly ascribed as such - even in something like KF covalent component is significant and here, it tenuous. If not hydrogen bond, it would be almost pure ionic interaction.

Another thing is that this picture is overdone, hurts eyes a bit. While bifurcated h-bonds are of course possible, it's not one of these cases. There would be just one, generic bond, two at most.

For two bonds, asp:arg is a good pair


text:
'Ab initio' couplings for DNA used in charge transport
I came across a publication by Senthilkumar et al. [1] in order to compute charge transfer through the DNA. Table 3, for instance, gives the coupling between bases along the leading (or retarded) strand of DNA:

Table 3. Charge-Transfer Integrals, $J$ (in eV), Overlap Matrix Elements, S, and Generalized Charge-Transfer Integrals, $J′$ (in eV), for Nucleobases Stacked at a Distance of 3.38 Å with a Twist Angle of 36°

For concreteness I will talk about only on columns 3 and 6 of the aforementioned table. I have two questions:

For instance, is it 5'-A-C-3' physically different from 5'-C-A-3'? If I 'erase' the direction 5' to 3', then A-C or C-A should be coupled by the same number (for instance, 0.042). However, I think that 5'-A-C-3' is different from 5'-C-A-3', because in the first example, a phosphate is "attached" to A from the left and a sugar is attached to C to the right; in the second case, C is attached to the phosphate to the left and A is attached to the sugar to the right. Is this interpretation correct?

5'-A-C-3' the coupling says it is 0.042, for 5'-C-A-3' it is −0.002. If I regard the previous argument as true, then both DNA systems are different, but not too different since they contain the same basis A and C. Does it make sense that by just changing the position of phosphates and sugars, one is approx. 40 times bigger than the other?

Reference

Senthilkumar, K.; Grozema, F. C.; Guerra, C. F.; Bickelhaupt, F. M.; Lewis, F. D.; Berlin, Y. A.; Ratner, M. A.; Siebbeles, L. D. A. Absolute Rates of Hole Transfer in DNA. J. Am. Chem. Soc. 2005, 127 (42), 14894–14903. DOI: 10.1021/ja054257e.


text:
why heavier transition metals can stabilise higher oxidations states
I believe similar questions have been asked but this is different and the answers I have looked at don't answer this question. I have read that oxidation states of heavier transition elements (Ru, Os etc) are more stable than first row transition metals.
I understand how they can get to the oxidation state (which is the usual answer to this question) but im asking why they are more stable then the oxidation state is reached. And although an element like Os is bigger meaning the ionisation energy should be small for the first few electrons doesn't the lanthanide contraction mean the valence electrons have a stronger effective nuclear attraction.
By more stable I mean why is RuO4 explosive but OsO4 on the other hand is stable and FeO4 doesn't exist.

And see the attached links therein.

Basically saying " they can get to the oxidation state (which......" is basically the answer. There is one answer whatever worded, as there is one question.

@Alchimista but doesn't the lanthanoid contraction lead to a stronger effective nuclear attraction between outer electrons and the nucleus. So it shouldn't be that easy to ionise an element like Os. Or is the contraction not big enough to increase the ionisation energy by that much?

@NilayGhosh to sum up the answer it seems to say the weak effective nuclear attraction causes ionisation energies to be lower and let the heavy metal to get to the higher oxidation state. but for Os doesn't the lanthanoid contraction lead to a stronger effective nuclear attraction between outer electrons and the nucleus?

@muhammadhaider at the moment I have no idea about particular elements. I commented on your passage. How they can get to an higher ox state and stay in that high ox state is the same question/answer.


text:
Behavior of ionic salts in solution
The electrons of an isolated sodium chloride bound pair in vacuum reside at a semi-classical level in their ground state, so that the Born-Oppenheimer approximation applies, and the 'molecule' acts like a sort of rigid rotor, whose relevant degrees of freedom are translational momentum and orbital angular momentum. In solution, however, one might expect things to become a bit more complicated. Electrons and holes from water, H+ and OH- and other metastable arrangements would tend to perturb the electronic structure of the NaCl molecule, and it seems that this could greatly impact its effective behavior (after integrating out the water molecules, say.) It does not seem totally unreasonable that the NaCl pair, once in solution, would spend most of its time partially dissociated, and that the constitutive ions would tend to be far enough from each other to be described reasonably well by a 'semi-classical' Langevin dynamical model. However, if one simulates overdamped Langevin dynamics of two charged ions separated by distances that are within the screening length, one finds that the ions tend to approach each other arbitrarily closely. Because of this, it would seem that the ions must also spend an appreciable time in some sort of quantum mechanical bound state. My question is, to what extent is the structure of this bound state known (beyond that it probably resembles the bound state in vacuum), and if so, what are the relevant 'soft' degrees of freedom?

Sorry ! The molecule NaCl does not exist. The salt NaCl is made of an arrangement of Na+ and Cl- ions. But there is never a real NaCl molecule. Dissolution into water has only one effect, that is to separate the ions from one another.

@Maurice https://chemistry.stackexchange.com/questions/14174/what-is-sodium-chloride-like-in-gas-state

Still OP here seems to like jargon way too much, still not enough to hide things like "Electrons and holes from water" - seriously? Not to mention "weave themselves into the electronic structure" isn't exactly proper description of single no so lofty word "hydration" that the post lacks.

Rather all the time fully dissociated, ions hydrated and moving independently. Even in gaseous state, it is rather a ion pair than a molecule.

In the vapour phase NaCl is a molecule, albeit with a lot of ionic nature. The vibrational frequency $v=0$ is $\approx 380$ wavenumbers. In solution, if you repeat the calculation with a large dielectric constant (80 for water) you should see that the ions will separate as the attraction is comparable or smaller than thermal energy.

@porphyrin so it would seem that the bound state is unstable (at least in water-like solvents.) Still, it is not obvious to me that the configuration would be completely irrelevant (as long as the associated binding energy is not immensely smaller than the thermal energy.)

yes not bound in water or similar high dielectric solvents. At the sum of the ionic radii $\approx 0.3 nm$ the interaction energy is only $\approx 2.3, k_BT$ the thermal energy at room temperature. In dilute solution the ions once they become separated are unlikely to diffuse together v often, (just too much space to diffuse into, entropy wins ) the thermal energy is only an average and energy fluctuations will ensure that the ions will always separate.

From the comments, it would seem that NaCl, while existant, is a not-especially-stable molecule that in solution tends to dissociate rapidly, with bare charges screened either with readily available (if intermittent and short-lived) water-based ions or even more readily available aggregate dipole distributions. This would allow the ions to be treated effectively as simple uncoupled Brownian walkers at large scales. I am still not entirely convinced, however, that this picture is adequate when the ionic concentration is low enough that the water is only weakly conductive. Presumably, the 'hydrolysis'(?) of NaCl triggers a cascade of similar dissociations until the free charge recombines elsewhere. If the medium isn't fully conductive, then hydrolysis must produce an effective dipole pair or aggregate/collective dipole moment somewhere around the original source (or, in non-idealized conditions, at least drive the recombination of an existing dipole nearby.)
It might be more sensible to recast the question in terms of the 'dissolution' of (dilute) salt in a more stable solvent (e.g. oil or liquid carbon dioxide or what have you.) Presumably there is a somewhat subtle interplay or trade-off between the electronic structure of the solvent near the salt and the Coulombic attraction between oppositely charged ions. If the solvent is sufficiently stable, then the screening length could be quite large, and it isn't obvious to me that the semi-classical coupled diffusive behavior discussed in the original post would be physically irrelevant.
One could investigate this question in detail using density functional theory, but at a heuristic level it might be enough that at very small length scales (and sufficiently low temperatures, near the melting point), the 'core' degrees of freedom of the solvent would, in its quantum bound state, tend to exhibit gel-like behavior, and that the ionic 'molecule'/pair would act as a sort of inclusion or notch (or inclusion/notch 'pair') in the gel, perhaps with a 'trail' that would depend on how the ions approach the bound state, and which might gradually dissipate as the solvent molecules adjust to the neutral inclusion (and at a faster rate at higher temperatures.) Dissociation of the ions might then be driven by 'micro-quakes' in the solvent matrix (and/or cosmic rays), as ambient oscillations concentrate stress at the inclusion. In any event, vibrational modes are probably more relevant in solution than in vacuum, and the 'bound state' is probably kinetically frustrated or arrested by solid-like behavior of the 'solvent matrix' at small length scales.

Note that the solid NaCl does not contain NaCl molecules either, but is a collection of alternating Na+ and Cl- ions, where each ion is surrounded by 6 equivalent opposite ions. When dissolved, is interacts mainly with neutral water molecules, as H+ and OH- ions are far from abundant, having concentration $\pu{10^-7 mol/L}$. Each ions is then dissolved separately by being hydrated by water molecules.

It might be more sensible to recast the question in terms of the 'dissolution' of (dilute) salt in a more stable solvent (e.g. oil or liquid carbon dioxide or what have you.)  Ionic compounds are not generally soluble in nonpolar solvents, unless ions are rather big organic ones.

Ionic compounds are not generally soluble in nonpolar solvents in bulk, but I'd expect at least a few NaCl pairs floating around from the marginal entropy gain if nothing else. Also, "abundant H+ and OH- ions" should perhaps be interpreted as "readily available H+ and OH- ions" (which may be obtained from any of the adjacent water molecules.)

H+ and OH- ions do not play any significant role in NaCl dissolution. And I would be very curious about the way how oil molecules are tearing ion pairs from NaCl ion crystals. Rather residual NaCl dust.

Does Na+ not bind to OH-, and Cl- to H+, in order for them to form neutral pairs that can meander freely?

Of course they are not. NaOH as another ionic compound dissolves as Na+ and OH- and HCl as highly polar molecule completely dossiociate as H+ ( rather H3O+, or H3O+ . n H2O ) and Cl-.

Is it obvious how Na+ and Cl- interact with water in order to unbind from each other?

Let us continue this discussion in chat.


text:
IUPAC salt definition makes bases be classified as salts?
The IUPAC salt definition is: "A chemical compound consisting of an assembly of cations and anions" (Compendium of Chemical Terminology Gold Book, Version 2.2.3, 2014)
So it means that a base like NaOH, that's also an ionic crystal, is a salt? I'm quite certain it's not but it seems the definition says so.
Then, a if a base is not a salt, what would be a better definition for salt?

There is no good definition of salt, or even "ionic" compound.

$\ce{NaH2PO4}$ acts as an acid when dissolved in pure water, yet it is a salt, no?

https://chemistry.stackexchange.com/questions/117914/is-an-acid-a-salt-or-not https://chemistry.stackexchange.com/questions/96090/what-is-the-difference-between-an-ionic-compound-and-a-salt

NaOH can be formally considered as a salt of the very weak acid H2O, making it a very strong conjugate base of the H2O acid.


text:
Can carbon dioxide be reduced to carbon monoxide and oxygen to produce energy?
There are lots of questions about reducing or burning CO2 to carbon and oxygen to solve climate change, but of course that wouldn't work because it takes a lot of energy. But carbon monoxide is more stable than the dioxide, so could CO2 be split into CO and oxygen to create more heat? Of course this is a bad idea since it'd produce a toxic gas but is it at least theoretically possible?
Apologies if I'm misusing chemistry terms or concepts; please inform me in the comments section.

It will take a lot of energy too.

CO could be captured more easily than CO2

It will take energy to break that C=O bond. I can't imagine that there's an easy way to do this on an industrial scale. Cool idea, though, I'm sure the chemists of the future will be working on something like this!

Related reading: Syngas is used as a fuel, and consists mainly of H2 and CO - this suggests that the thermodynamics are unfavourable (as confirmed by the answers below).  The article suggests some ways that related reactions might be useful

Forget carbon monoxide. Why not take it all the way back to gasoline?

Perhaps one could use hemoglobin or similar protein structures to capture carbon monoxide as it has a higher affinity for CO than CO2 and CO2 than O2

Unfortunately, the question as stated is thermodynamically impossible. Let's look at the proposed reaction:
$$\ce{CO2(g) -> CO(g) + O(g)}$$
This reaction is simply a bond dissociation (specifically, a carbon-oxygen covalent double bond is broken). We can look up the enthalpy change associated with it. From a table of values on Wikipedia, we find in the row for carbon dioxide that this reaction has an enthalpy change of $\mathrm{+532\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$. The proposed reaction is therefore heavily endothermic. That is to say, it must absorb energy. Interestingly, it's true that the extreme strength of the bond in carbon monoxide has a measurable effect, making this process more favourable than expected. However, it is still overall extremely unfavourable, and therefore requires a large input of energy. I stress that this is unavoidable, no matter how fancy your machine - if the end result is the reaction stated above, then you must pay the energy cost somehow.
Part of the problem, though, is that we have monoatomic oxygen as a product, which is a very reactive, high energy species - it doesn't actually exist except in special conditions. A simple adjustment therefore is to have molecular dioxygen, $\ce{O2}$ (the kind in the atmosphere that you breathe). The reaction then becomes:
$$\ce{2 CO2(g) -> 2 CO(g) + O2(g)}$$
So what's the enthalpy change associated with this reaction? Looking up another table, this turns out to be $\mathrm{+283\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$. Again, this reaction is endothermic, though much less so than the first one. Regardless, once more this reaction is an energy sink.
If you want more visceral confirmation of this fact, consider the following. It is well known that pure carbon monoxide burns in an oxygen atmosphere. The reaction is self-sustaining and releases considerable heat. If you pay close attention, the reaction in the video is the exact inverse of the second equation. By chemical thermodynamics, if the combustion of $\ce{CO}$ to $\ce{CO2}$ releases heat, then it is necessarily true that cleaving $\ce{CO2}$ to form $\ce{CO}$ and $\ce{O2}$ will consume energy.
As a last point, there are ways to make the production of $\ce{CO}$ from $\ce{CO2}$ feasible, but it requires changing the products. For example, if hydrogen gas is used as a reagent, the following becomes possible:
$$\ce{CO2(g) + H2(g) -> CO(g) + H2O(g)}$$
The enthalpy change for this reaction is $\mathrm{+41\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$ , which is still endothermic, but approaching the break-even point. This is not too surprising, as hydrogen gas can behave as a reductant, and the bonds in water molecules are strong, pushing the reaction forwards. Let us make one last tiny modification:
$$\ce{CO2(g) + H2(g) -> CO(g) + H2O\color{red}{(l)}}$$
By assuming the water produced is in the liquid state as opposed to a gas, the reaction surrenders a little more energy, and the calculated reaction enthalpy becomes $\mathrm{-3\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$. This reaction is very mildly exothermic, which is to say it releases heat (admittedly, so little that it's within margin of error, and slightly different conditions could make the reaction overall endothermic).
If you're not dead-set on having carbon monoxide as a product, then there are further options still. For example, here is the complete reduction of $\ce{CO2}$ to methane ($\ce{CH4}$), a considerably exothermic process with a reaction enthalpy of $\mathrm{-253\ kJ\ mol^{-1}}$ at $\mathrm{298\ K}$:
$$\ce{CO2(g) + 4H2(g) -> CH4(g) + 2 H2O(l)}$$
Methane is not an ideal product, as it too is a greenhouse gas and is a low value chemical feedstock due to its abundance and relative lack of useful chemistry. There is much more interest in conversion of $\ce{CO2}$ to compounds such as methanol $\ce{CH3OH}$ and formic acid $\ce{HCOOH}$. These two particular reactions are also exothermic.
There are several issues with using hydrogen reduction of $\ce{CO2}$ as a carbon capture strategy to combat climate change, but perhaps the main one is a real-world factor: most hydrogen we currently produce is derived from fossil fuels, notably from the partial combustion of fossil methane (natural gas) with water at high temperatures, known as steam reforming. Therefore, while alternative sources of hydrogen gas using renewable low-carbon intensity energy are not available, this is a poor strategy to remove anthropogenic $\ce{CO2}$ from the atmosphere.

All true, good answer. But if we had readily available hydrogen gaz, I don't think we be burning fossil fuels and producing CO2 in the first place. So, we can assume we don't have hydrogen available.

And let's not forget that capturing the carbon dioxide is one of the biggest hurdles in the first place. Sure, it's accumulating in the atmosphere, but in the bulk, it's still just a couple hundred ppm - you need to process ridiculous amounts of air to extract the carbon dioxide, and that adds massive costs to anything you do with the carbon dioxide afterwards. Of course, there were interesting alternative proposals - e.g. feeding the exhaust from thermal power plants directly through vats of algae, recovering much of the fuel with the help of sunlight; not without its own problems, though.

@Luaan: That would be more than a "interesting alternative proposal". The Dutch commonly run greenhouses on the heat and CO2 of small power plants. It accounts for 10% of the total Dutch electricity production. It's a bit less impressive when you account for the fact that these greenhouses are also illuminated at night, which uses that same electricity. In effect, these greenhouses turn natural gas into food - the other byproduct of burning natural gas is water, which like CO2 is also consumed by the plants.

@MSalters That's pretty much what I meant, yeah. We're very far from being anywhere near to a closed loop, but there's certainly at least some cases where it's profitable (money and ecology likewise). Even the fact that they're illuminated at night isn't as bad as it sounds, given that night electricity is still a bit of a waste product - the electricity would have been produced in much the same quantity anyway.

You’re looking at bond dissociation energies. They, however, do not give a good picture. A better place to start looking is the standard enthalpy of formation. The linked Wikipedia article provides an extensive list of compounds but only two matter:

$\displaystyle\Delta_\mathrm fH^0 (\ce{CO}) = \pu{-110.525 kJ/mol}$
$\displaystyle\Delta_\mathrm fH^0 (\ce{CO2}) = \pu{-393.509 kJ/mol}$

Next would be invoking Hess’ law which is based on the fact that enthalpy is a state function and thus the route getting from A to be in the reaction $\ce{A->B}$ does not matter, only the enthalpies of A and B. In mathematical terms, this is usually summed up as:
$$\Delta H^0_\text{reaction} = \sum \Delta_\mathrm f H^0_\text{products} - \sum \Delta_\mathrm f H^0_\text{reactants}\tag{1}$$
Plugging the values of a hypothetical reaction that goes only from carbon dioxide to carbon monoxide (and hitherto ignoring other reactants or products) gives us:
$$\ce{CO2 -> CO}\tag{A}$$
$$\begin{align}\Delta H^0_\text{(A)} &= \pu{-110.525kJ/mol} - (\pu{-393.509kJ/mol})\\
&= \pu{+282.984kJ/mol}\end{align}\tag{2}$$
This is an endothermic reaction and quite greatly so.
‘But wait!’, I hear you exclaim. ‘I’m producing oxygen too, aren’t I?’
Indeed you are. In simplest terms, you would be producing oxygen gas ($\ce{O2}$). And this does not help you, as the standard enthalpy of formation of elements is, by definition, zero. (At this point, we also nicely see that the reverse reaction – combustion of carbon monoxide to give carbon dioxide – is energetically favourable.)
That is not to say this reaction is impossible. You would need something else to put in – the reaction (A) as is isn’t balanced anyway. A real reaction would have to look something like this:
$$\ce{CO2 + X -> CO + Y}\tag{B}$$
Obviously, Y will have to be related to X but contain more oxygen. X and Y would then need to be carefully balanced to give an overall negative reaction enthalpy. One such example would be the creation of (liquid) water whose enthalpy of formation is $\pu{-285.8 kJ/mol}$ according to the same Wikipedia article linked above. As water would be produced by adding hydrogen gas to the reaction, the reactant side only gains another zero.
More complicated systems are evidently possible, the reactions and equations will just get more and more complicated. The key takeaway is, however, that the reaction as proposed is endothermic and you need some sort of driving force (something more exothermic) to make it happen.
Note that my calculations all centre around enthalpy. To determine whether a reaction will actually occur under a given set of conditions, the important value to check is Gibbs Free Energy. However, enthalpies usually provide a very good starting point.

Your values regarding the hydrogen reduction of CO2 made me look around a bit more carefully. According to ACS Energy Lett. 2018, 3, 1938−1966, equation (1) on page 1943 shows a value of $\mathrm{+41.2\ kJ\ mol^{-1}}$, which is corroborated by Top Curr Chem (Z) 2018, 375, 23. It seems I had the right numerical value, but swapped the sign. I'll have to partially revise my answer.

Also after rewriting my answer, I believe all our values are in agreement (to within fractional errors) - water being produced in the liquid state drives the reaction enthalpy to almost zero, which agrees with your calculations. You really can trust Hess's law!

@NicolauSakerNeto Ah, wonderful, I was scared where this minor discrepancy might have come from but glad to see we agree even numerically in the end.

To get heat, you need to go the other way (add $O_2$ to $CO$). In steel mills, $CO$ is a waste byproduct of the blast furnaces that gets used as fuel. It gets burned in boilers and makes high pressure steam. The steam spins turbines and makes megawatts of electricity. It's not a great fuel compared to methane, but you can't beat the price.


text:
Place cooling bath apparatus in freezer to speed up the process?
As part of an at-home lab, my chemistry class is creating cooling baths to measure the freezing point of different solutions. We have to determine the molar mass of the solutes in these solutions by observing the freezing point. The cooling bath is constructed from ice and salt inside a 500 mL beaker, into which are placed test tubes of solutions (one at a time). A thermometer sticking out of the test tube hypothetically measures the freezing point. The problem is that the ice bath is taking hours to freeze the solution.
To speed up this process, can I just stick the whole apparatus in the freezer and check it every 10 minutes or so to see the temperature at which the solutions freeze?
Will the freezing point of salt water and calcium chloride water (the solutions in question) be radically different if frozen in an ice bath vs. frozen inside a kitchen freezer?

The freezing point remains the same, but its measured value may depend on the way of measurement.
In case the solution freezing point is too close to the ice/salt 3:1 bath temperatures, it may be the reason for too long freezing time. The bath ice/$\ce{CaCl2 . 6 H2O}$ bath 1:2 could help, if applicable.
Note that fast freezing may lead to biased value, especially, if supercooling occurs. E.g. a PET bottle with water can get supercooled in a freezer deep below the freezing point.


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Does exchange energy affect completely filled orbitals?
In my textbook, it's written that exchange energy stabilises half filled and completely filled orbitals. But my teacher said electrons can't be exchanged in full filled orbitals because of pairing up of electrons. So which explanation is correct? My textbook has an explanation for half filled electron orbitals, by it had no explanation given for completely filled orbitals.

It seems like you are just being introduced to quantum mechanics and is confused with the concept of exchange energy. The concept can be confusing at first, since it neither involves exchanging electrons as classical particles nor has a classical analogue. I will try explaining the problem with a simple two electron system.

Before we discuss, here is a short introduction. In quantum mechanics, one solves the Schrodinger's equation and tries to estimate the wave function of the system. We are talking about time-independent case here, so we solve time independent Schrodinger's equation to obtain the wavefunction.
$$\hat{H}\psi=E\psi$$
Here $\hat{H}$ is the Hamiltonian (energy operator), $\psi$ is the wavefunction and $E$ is the energy. This equation is an eigenvalue equation, which essentially means any valid wavefunction should have a fixed energy irrespective of coordinates. In the equation, the $\hat{H}$ is the only known quantity (operator), $\psi$ and $E$ has to be estimated based on boundary conditions.

Let us consider a system with two electrons, with wavefunction $\psi_1(r)$ and $\psi_2 (r)$, where $r$ denotes the four dimensional space spin coordinates. Now, we know that electrons are indistinguishable and two electrons (fermions) cannot have same space-spin coordinates (Pauli's exclusion principle). Two electrons not having same space-spin coordinates is same as saying that the electrons should not have all same quantum numbers.
So, in order to consider the indistinguishability  of the electrons, we can have two cases of the total wave function: (1) $\psi_1(r_1)\psi_2(r_2)$ or (2) $\psi_1(r_2)\psi_2(r_1)$. On a side note, these are known as Hartree products and are valid for bosons. However, in case we wish to include the Pauli's exclusion principle, we write the system as a Slater determinant, i.e.
$$\psi(r_1,r_2)=\frac{1}{\sqrt{2}}\begin{vmatrix} \psi_1(r_1) & \psi_2(r_1)\\\psi_1(r_2) & \psi_2(r_2)\end{vmatrix}=\frac{1}{\sqrt{2}}\left[\psi_1(r_1)\psi_2(r_2)-\psi_1(r_2)\psi_2(r_1)\right]$$

Now let us talk about the Hamiltonian (energy operator). The Hamiltonian for an atom/molecule can be divided into three parts:

Constants: This include any constant terms, for example, nuclear-nuclear repulsion for poly-atomic molecules.
One electron operator: This part contains terms dependent on single electrons, for example, kinetic energy and nuclear-electron attraction.
Two electron operator: This part contains terms dependent on two electrons, and usually contains only electron-electron repulsion terms. This part gives rise to the exchange energy. We shall expand this part.

Before we go into details, if one is provided with a operator $\hat{A}$ and a wavefunction $\psi$, the expectation (average) value of the observable corresponding to the operator is given as $\langle\psi|\hat{A}|\psi\rangle$, where $\langle\cdot\rangle$ implies integral over all coordinates.

In atomic coordinates, the electron-electron repulsion is given as $1/r_{12}$, where $r_{12}$ is the distance between the electronic coordinates. Now, if we use the wavefunction in the Slater determinant, we get the total two electron interactions as:
$$\left\langle\psi(r_1,r_2)\left|\frac{1}{r_{12}}\right|\psi(r_1,r_2)\right\rangle \\= \frac{1}{2} \left[ \left\langle\psi_1(r_1)\psi_2(r_2)\left|\frac{1}{r_{12}}\right|\psi_1(r_1)\psi_2(r_2)\right\rangle + \left\langle\psi_2(r_1)\psi_1(r_2)\left|\frac{1}{r_{12}}\right|\psi_2(r_1)\psi_1(r_2)\right\rangle - \left\langle\psi_1(r_1)\psi_2(r_2)\left|\frac{1}{r_{12}}\right|\psi_2(r_1)\psi_1(r_2)\right\rangle - \left\langle\psi_2(r_1)\psi_1(r_2)\left|\frac{1}{r_{12}}\right|\psi_1(r_1)\psi_2(r_2)\right\rangle \right]$$
Now, in this cases the first two terms are called Coulomb terms and are easy to imagine classically, i.e. what is the repulsion of electron 1 if present in coordinate $r_1$ with electron 2 if present in coordinate $r_2$ for the first term and vice-versa for the second term. These are repulsive in nature.
The third and fourth terms are attractive terms, and are known as exchange terms. There is no classical analogue of these, and appears purely from the indistinguishability of electrons and Pauli's exclusion principle.

Now remember we discussed that $r_1$ and $r_2$ are space-spin operators, and $r_{12}$ depends only on space? This essentially means that we can separate the space and spin parts while calculating the terms. So the exchange terms vanish while the two electrons have different spins, and remains only when electrons have same spins. This is the origin of exchange energy and IT DOES NOT INVOLVE PHYSICALLY EXCHANGING THE ELECTRONS.

I hope I could explain you in brief the origin of exchange energy. In reality, it is very difficult to calculate the exact energy of a system with N-electrons and say half filled/full filled systems are more stable due to "exchange interactions" simply because:

We do not exactly know how the wavefunctions look. We assuming hydrogenic wave functions but that may not work out always.
Optimising the wavefunctions based on Hamiltonian is carried out self consistently and is costly.
Let us say one resolves the first two issues somehow, still total energy has other terms like kinetic energy, electron-nuclear attraction, Coulomb terms, and even nuclear-nuclear repulsion for polyatomic molecules. So it is very difficult to assert only "exchange energy" is responsible for the extra stability.

Whatever I have tried to explain in short takes up at least two quantum chemistry courses in Bachelor's degree and Master's degree. In case you have some doubts, please feel free to ask in comments. In case you wish to study more, I will suggest the following books (in the order given):

D. J. Griffiths, Introduction to Quantum Mechanics
A. Szabo and N. Ostlund, Modern Quantum Mechanics
I. Levine, Quantum Chemistry

Thanks a lot for the answer, actually I'm new to quantum mechanics, so I didn't get some points. But thanks a lot for the book refrences.


text:
Aromatisation of a ring which includes nitrogen
Question

Complete the following reaction sequence:

Answer

My solution
I was able to complete all the steps except for the last one. This is what I ended up with:

What I'm not getting is how the $\ce{-OH}$ groups are removed and the ring is aromatised. If anyone could provide me with a mechanism for the same, it would be really helpful.

Hint: An aldimine forms which reacts with the ketone to form the desired compound.

Does the aldimine form first and then attack the ketone( which means my final product may be incorrect)

That's would be rather easy part - simple elimination. Perhaps NH3 misled you about pH, as it's acid catalysed?

@Mithoron yes, i would think that was the problem.. I got it now, though!

This would be the mechanism I propose for the reaction:

I have taken the $\ce{pH}$ of the medium to be $4.5$. That should explain the protonation of the $\ce{-OH}$ groups.

Your final product is missing a methyl group. And I wouldn't draw both at the same time (but that's nitpicking). By the way, you should use the N lone pair to kick out the water next to it, then deprotonate.

Thanks for pointing it out @orthocresol! I shall make the changes..

@orthocresol would it be correct now?

Yup, that’s pretty much what I’d have done, I think.

Also the last step should be removing of $\ce{H3O+}$ and not $\ce{H2O}$ but I think that was understood..

Well, two oxygens shouldn't be protonated at once, otherwise looks reasonable.

@Mithoron I would happily correct it, but what worries me is the continuous bumping of this question to the top of the front page...

Your q. and answer were edited only once - there are post that were edited much more, also as long as it was still a front page, that would be just moving it a bit. If it was a few dozen different questions edited, that would be an issue.


text:
How to Choose if a reaction will go elimination of substitution?
I want to know how to choose if a given reaction will prefer going substitution mechanism or elimination mechanism.Both elimination and substitution reactions have types like Sn1 ,Sn2 ,SnNGP,Sni and E1 ,E2 ,E1CB
I know how to decide (if given that reaction is substitution reaction or elimination reactiin) which type among these a given reaction will go but I am always confused on choosing that the reaction will be substitution or elimination.
Any help/contribution appreciated.Thanks!

It mainly depends on the reagent. If the attacking reagent has more nucleophilic character ie. tendency to act as a nucleophile, Substitution reactions will be preferred. On the other hand, if the reagent has more tendency to act as a base, elimination reactions will occur. In cases where a reagent falls under both categories, you can go into additional details like whether the compound has a Beta-Hydrogen to eliminate, leaving group tendencies etc.