text:
Molarity of pure water
The molarity of pure water is 55.5 M at which temprature is obtained this value? Further, you will come across that 1000 mL = 1000 g (for water only, but at which temprature?)

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Water is very unique liquid, because it has a higher density in the liquid state than the solid state. The maximum density of water is found at $\pu{4 ^\circ C}$, which is reported as $\pu{999.9720 kg\:m^{-3}}$ (Temperature Effects on Density). Therefore, molarity of water at $\pu{4 ^\circ C}$ can be calculated as $\frac{\pu{999.9720 g\:L^{-1}}}{\pu{18.015 g\:mol^{-1}}} = \pu{55.51 mol\:L^{-1}}$.
Following chart shows accepted densities of pure water at range of temperatures:
$$\text{Table of Densities of Pure Water at Various temperatures under }\pu{1.0 atm}$$
$$
\begin{array}{ccc}
\hline
\text{Temperature }(\pu{^\circ C }) &\text{Density }(\pu{g\:cm^{-3} }) & \text{Temperature }(\pu{^\circ C }) &\text{Density }(\pu{g\:cm^{-3} }) \\\hline
-30 & 0.98385 & +30 & 0.99565 \\
-20 & 0.99355 & +35 & 0.99406 \\
-10 & 0.99812 & +40 & 0.99225 \\
-5 & 0.99930 & +45 & 0.99025 \\
0 & 0.999840 & +50 & 0.98807 \\
+4 & 0.99997 & +60 & 0.98313 \\
+10 & 0.99970 & +70 & 0.97763 \\
+15 & 0.99910 & +80 & 0.97180 \\
+20 & 0.99821 & +90 & 0.96506 \\
+25 & 0.99705 & +100 & 0.95838 \\\hline
\end{array}
$$
So you may calculate the molarity of pure water at each temperature using the chart. An example for step by step calculations to find the molarity of pure water at particular temperature ($\pu{20 ^\circ C}$):
$$\text{Density of pure water at }\pu{20 ^\circ C } = \pu{0.99821 g\:cm^{-3}}$$
Since molarity $= \pu{mol\:L^{-1} }$, let's consider $\pu{1.0 L}$ of water.
$$\text{Mass of } \pu{1.0 L} \text{ of pure water at }\pu{20 ^\circ C} = \pu{1.0 L}\times \frac{\pu{1000 cm^{-3}}}{\pu{1.0 L}} \times \pu{0.99821 g\:cm^{-3}}= \pu{998.21 g}$$
$$\text{Molar mass of pure water at any temperature} = \pu{18.015 g\:mol^{-1}}$$
$$\therefore \text{ # of moles in } \pu{1.0 L} \text{ of pure water at }\pu{20 ^\circ C} = \frac{\pu{998.21 g}}{\pu{18.015 g\:mol^{-1}}} = \pu{55.41 mol}$$
$$\therefore \text{Molarity of pure water at } \pu{20 ^\circ C} = \frac{\pu{55.41 mol}}{\pu{1.0 L}} = \pu{55.41 mol\:L^{-1}}$$

Problem: undefined control sequence\pu...this statement is repeated 4 times in your answer, what to do?

Please don’t post text and equations as a picture. The content cannot be found by search engines and screen readers. It makes it difficult for other users to edit your answer.
Especially since this answer is directed at beginners, you might want to consider using proper equations and typography. Please do not use names of quantities and names of units in equations. And please use correct units symbols.

@yaseenwazir Your soft can't parse this mathjax package, dunno why M.M. used pu instead of mathrm.

@yaseenwazir Please refer to Undefined control sequence \pu in official SE Android app. Note that mobile Stack Exchange app is using outdated MathJax and hasn't been updated for a long time. I strongly recommend to use a mobile web browser instead.

@Mithoron Mathew is doing it right, or at least according to the way it should be done on Chemistry.SE, see How to write physical units.

@Loong: Thanks for the valuable point. I apologize the community for been lazy. I found this post in a website in PEG format, but so sleepy to write it up and decided to attach as a picture. Won't happen again.


text:
What should be the order of specific conductance (κ) of floride, chloride and bromide ions?
The equivalent conductance order is $\ce{Br-} > \ce{Cl-} > \ce{F-}$.
$$Λ = \frac{κ}{c},$$
so $Λ$ is proportional to specific conductance?
But in ion chromatography we get peak area where $\ce{F-} > \ce{Cl-} > \ce{Br-}$ (specific conductance, μS/cm) for the same concentration, let us say 1 ppm.
Why?

Are they of the same concentration? As without that knowledge, the order of the conductivity ( specific conductance ) is undefined.

yes sir. lets say 1ppm

ppm is ambiguous quantity, it can be v/v, w/v,w/w,n/n.
By other words, all 1 ppm very probably is not the same molar concentration, unless it is n/n ppm, relating itself to the molar amount of the solution.
$\Lambda = \kappa /c$ is correct, but $c$ must be the molar concentration.
$$\Lambda[\pu{Sm^2mol^-1}]=\frac{M[\pu{g/mol}]}{c[\pu{ppm as g/m^3}]}\cdot \kappa[\pu{S/m}]$$

Thank u sir. I missed the conc. unit.

@ggs Please don't post a "thank you" as a comment. Instead, upvote and/or accept answers that were useful to you.

source: Stack Exchange

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

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text:
Cause of continuous spectral lines
We know different types of atoms have different spectral lines. But what exactly are the spectral lines representing? What causes white light to have a continuous spectra while other atoms show discontinuous spectral lines?

As a corollary to your question and answers : white light is not necessarily a continuous spectrum, the brain only needs stimuli of the three different receptor cells in a pattern to "see white". So white light is rarely a continuous spectrum, but it is broad enough to stimulate all the receptors.

The spectral lines represent transitions between pairs of discrete energy levels in the atom; an electron is excited by absorbing the energy of a photon and a transition from one of these levels to another occurs as the photon is destroyed. Emission lines occur when the reverse process happens.
If enough energy is absorbed the atom can be ionised and an electron with un-quantised energy can be liberated and moves to a position of complete separation where attractive forces between electron and nucleus is insignificant. As the liberated electron's energy is un-quantised the absorption giving rise to ionisation is continuous. The absorption is truly continuous, no matter how high a spectral resolution is used discrete absorption lines will not be observed.
Black body radiation is continuous emission from a hot body. The analysis of this emission lead Plank in 1901 to propose, for the first time, the idea that the photons were quantised, thus avoiding the 'ultra-violet catastrophe'. He did this by allowing ultra-violet photons to have more energy than visible ones which in turn had more energy that infra-red ones. Nowadays we know that he energy is directly proportional to frequency via  the Planck constant, $E=hv$.
In the black body, all the numerous energy levels in the body are highly excited and transitions between them continuously occur as the body is being continuously heated. These transition energies overlap one another in energy so that the spectrum is continuous.
(In molecules there are also discrete energy levels, in the ground state these are due to the vibrational and rotational motion of the molecule.  A molecule may also have electronically excited states and these can also have discrete vibrational and rotational energy levels. Furthermore there are dissociative excited states that have continuous absorption spectra.)

just as a side comment (it might be of interest for the OP), the continuum observed in electrons' energy when ionization occurs is the photoelectric effect; all energy excess increases electron's speed.

yes; a good point

Natural white light had continuous spectrum, as the source - black radiator - has continuous emission in all wavelengths.
Atoms emit or absorb only at the wavelengths, that are corresponding to energy differences between discreet energy levels of electrons in atoms.

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text:
How to convert wedge–dash structure into Fischer projection?
I am very confused when a compound with more than one chiral centre is given. How to find whether the two compounds are enantiomers or diasteromers?
Please also explain how to convert more than one chiral compound with wedge–dash structure to Fischer projection.

Where should i start?What are thoughts?what did you attempt?

Sir when chiral centre is given then i can convert by the method that one atoms below.or above the plane of paper are lie on horizontal line and other lie on same plane on a vertical but when it come to more than one chiral centre i am confuse .when question asks predict wether pair entaiomers or something

Could you edit your question and give an example that you attempted? Also, you are asking two questions, so you should probably post two questions. Otherwise, the title does not match the questions, and it would be difficult to find your second question.

Method 1

Method 2
Quoting from
http://oscar.iitb.ac.in/OSCARPP/Chemistry/UploadedStoryboards/SCH023_3P0431Wedge-dash%20to%20Fischer.ppt
The following is conversion of molecules with two chiral centres from wedge to fischer projection yhrough a series of operations as shown below.

Reference :
http://oscar.iitb.ac.in/OSCARPP/Chemistry/UploadedStoryboards/SCH023_3P0431Wedge-dash%20to%20Fischer.ppt

How to find whether the two compounds are enantiomers or diasteromers?
Diastereomers are stereoisomers that are not mirror images of one another and are non-superimposable on one another.$\ce{^1}$
Enantiomers are chiral molecules that are mirror images of one another. Furthermore, the molecules are non-superimposable on one another. This means that the molecules cannot be placed on top of one another and give the same molecule.$\ce{^2}$
The figure below will explain.
Further in diastereomers only part of the molecule is a mirror reflection.(circled in the image) ,while the other stereo-center is identical.
NOTE:Geometrical isomers are also diasetreisomers , but achiral.In the examples given below Diastereomers are chiral.

References:

http://www.chemeddl.org/resources/stereochem/definitions17.htm

2.http://www.chemeddl.org/resources/stereochem/definitions16.htm


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Confusion regarding calculating the pH of a salt of weak acid and weak base
I need to calculate the $\mathrm{pH}$ of a salt of weak acid and weak base in an aqueous solution.
Method I used for calculation
For a salt of a weak acid and a weak base (assuming complete dissociation of salt) both the cations and the anions in the salt undergo hydrolysis. When the anions undergo hydrolysis, since they are the conjugate base of the weak acid, they grab $\ce{H+}$ from the water and release $\ce{OH-}$ ions. Similarly, when the cations undergo hydrolysis they release $\ce{H3O+}$ ions.
These $\ce{H+}$ ions and $\ce{OH-}$ ions may react to give water. If these ions have the same concentration, the solution will have $\mathrm{pH}$ equal to $7$. If there is any excess of any of these ions after reaction, then the $\mathrm{pH}$ of the solution may be greater than or less than $7$. But my answer didn't match up with what is given in my text.
Textbook answer
Formula given in my textbook is
$$\mathrm{pH} = \frac{1}{2}\left(\log K_\mathrm{b} - \log K_\mathrm{a} - \log K_\mathrm{w}\right)$$
where $K_\mathrm{b}$ is dissociation constant of base, $K_\mathrm{a}$ is the dissociation constant of acid, and $K_\mathrm{w}$ is the ionic product of water.
When I try to derive this formula I need to make these assumptions:

The degree of hydrolysis of both the anionic and the cationic parts of the salt is equal.
The $\ce{H+}$ in the solution is contributed by the dissociation of weak acid.

But how can this be possible? The dissociation of the weak acid formed from salt hydrolysis has already been suppressed by the available concentration of its conjugate base (anion part of salt).

Stephen Fraser has a derivation on his site that uses arguments of major species and reactions to avoid solving multiple equations simultaneously, see http://www.chem1.com/acad/webtext/acid2/acid2-3.html#SEC2

There is calculation here also https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566 and here https://chemistry.stackexchange.com/questions/100346/calculate-ph-of-a-mixture-of-a-strong-base-and-acid-knowing-only-the-ph-wt-v/100355#100355

This simplified calculation is direct analogy to pH of acidic salt of a diprotic acid, where pH=(pKa1+pKa2)/2, where in this case pKa2=pKw-pKb. In the former, there is requirement -log c <<pKa1, here it would be -log c << min(pKa,pKb)

https://chemistry.stackexchange.com/questions/75047/ph-of-ammonium-acetate-solution/99988?sfb=2#99988

@user79504"when anion part undergoes hydrolysis they release $\ce{H3O+}$ ions." I think they release $\ce{OH-}$

Another approach, I believe, is this:
Definitions
\begin{align}
\mathrm{pH}
  &= \frac{1}{2}\left(\log K_\mathrm{b}
     - \log K_\mathrm{a} - \log K_\mathrm{w}\right)\\
2\,\mathrm{pH}
  &= \log\left(\frac{K_\mathrm{b}}{K_\mathrm{a}\,K_\mathrm{w}}\right)
\end{align}
It suggests equations' sum, like this:
\begin{align}
  \ce{HA &<=> A- + H+} &
  K_\mathrm{a} &= \frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}\\
  \ce{BOH &<=> B+ + HO-} &
  K_\mathrm{b} &= \frac{[\ce{HO-}][\ce{B+}]}{[\ce{BOH}]}\\
  \ce{A- + B+ + H2O &<=> AH + BOH} &
  K &= \frac{[\ce{AH}][\ce{BOH}]}{[\ce{A-}][\ce{B+}]}
     = \frac{K_\mathrm{w}}{K_\mathrm{a}\,K_\mathrm{b}}
\end{align}
Finding $\ce{[H+]}$ in relation to $K$
If the salt concentration is $C_\mathrm{s}$, and the mass balance:
$$
C_\mathrm{s} = [\ce{AH}] + [\ce{A-}] = [\ce{BOH}] + [\ce{B+}]
\tag{Mass balance}
$$
Dividing by $C_\mathrm{s}$ we define $\alpha$ and $\beta$ such that
$$
\frac{C_\mathrm{s}}{C_\mathrm{s}} = \alpha + (1-\alpha)
  = \beta + (1-\beta) \tag{Mass balance}
$$
And if we admit that $\alpha$, $\beta <<1$ which also implies $\alpha\approx\beta$ (this is possibly a weakness of the answer) then $$\alpha^2 \approx \frac{K_\mathrm{w}}{K_\mathrm{a}\,K_\mathrm{b}},$$
and
$$
K_\mathrm{a}
  = \frac{(1-\alpha)[\ce{H+}]}{\alpha}
  \approx\frac{[\ce{H+}]}{\alpha},
$$
then
\begin{align}
  (\frac{[\ce{H^+}]}{K_\mathrm{a}})^{2}
    &\approx \frac{K_\mathrm{w}}{K_\mathrm{a}\,K_\mathrm{b}}\\
  [\ce{H^+}]^{2}
    &\approx \frac{K_\mathrm{w}\,K_\mathrm{a}}{K_\mathrm{b}},
\end{align}
and we get the equation.

As an example, $\ce{pH}$ for ammonium acetate salt is exactly 7 because $pH=\frac{14}{2}$


text:
How would a temperature profile in wall change if the rate of heat transfer increases

Given the system below, how would the temperature profile in the wall change if the heat transfer in Fluid II was very fast?

If the heat transfer in Fluid II is very fast, then the T-profile for $\ce{T_{2∞}}$ would experience less resistance and hence have a more linear slope. However, I don't understand how this would affect the T-profile inside the wall? Wouldn't Fluid II have the same temperature and interact with the interface at the same point, meaning nothing changes within the wall?

Do you mean ‘wall is good thermal conductor’?

If the heat transfer in fluid 2 is fast, the thickness of the boundary layer wall - fluid 2 is very small (the 'not so linear' part of the temperature profile on that side)
One can then approximate that the temperature of fluid 2 directly at the wall is equivalent to the bulk temperature.
So the temperature profile through the wall would be steeper, indicating a larger overall heatflux - see it as a self fullfilling prophecy: fast heat transfer will allow you to transfer heat faster


text:
Why does this reaction produce an alcohol in one case and an alkene in the other?

Does it have something to do with the fact that the chlorine atom is bonded with the primary carbon, or with higher temperature in the upper reaction?

Try to use multiple tags to increase your audience.

Please also see Why substitution and elimination reactions are favored respectively at lower and higher temperature?
The reaction conditions are the clue you are looking for.
The first thing to recognize is that in any situation where there is competition between substitution and elimination, you will always get a mixture of products.  The question is, which one will dominate.
The effect of temperature is subtle, and sometimes misleading. In practice, organic chemists heat many (maybe even most) reactions, in order to get them to occur in a reasonable timeframe.  When we are teaching about the effect of temperature on the competition between substitution and elimination, we often indicate heating ($\Delta$) is present for the elimination and not the substitution.  You should take this to mean that higher temperature favours elimination.
Now, subtleties aside, let's look at why heating favours elimination.

One difference between the two situations is the relative changes in entropy.  We can see that in the case of elimination, the overall entropy of the system is increased after the reaction (there are more different particles) while in the case of substitution, the entropy change is negligible.
We can consider the competition between the two reaction pathways like this:
$\Delta G_{S_N2} = \Delta H_{S_N2} - T\Delta S_{S_N2}$
and
$\Delta G_{E2} = \Delta H_{E2} - T\Delta S_{E2}$
At low temperature, the enthalpy term ($\Delta H$) dominates, while at higher temperature, the entropy term ($\Delta S$) dominates.
Therefore, the increase in temperature favours the reaction with the larger entropy contribution.


text:
Axial Equatorial NMR graph difference
I wanted to ask a question about NMR for Axial and Equatorial molecules.
I was asked to describe how to separate the following two product molecules from this reaction:

and the answer that my colleague mentioned was NMR.
He stated that for the product on the left, the $\ce{P}$ atoms were in the same environment (axial) so they would show only $1$ peak but on the right hand side product, the $\ce{P}$ atoms were in different environments (equatorial and axial) hence two different peaks would be shown.
I had a previous introductory course to NMR where coupling constants were discussed, and as the maximum bond coupling "length" was considered to be $3$, this made perfect sense.
My question is, how are axial and equatorial substituents (if they are the same substituent as in this case) in such different environments? Is it to do with the surrounding substituents deshielding by varying amounts? What is the case?
I failed to find a definitive answer on Google or StackExchange, but only statistical interpretations were given, no actual explanations.

Not an answer because I don't have a reference, but it is likely the spacing from the other ligands. It's important to consider what is axial/equatorial relative to the ligand you are interested in. For the left structure, the $\ce{PR_3}$ have the same ligands equatorial to them and each other axial. For the right structure, one of them is axial to CO and the other is axial to O, so they feel a different amount of interaction from those ligands. @vik1245

When discussing about same environment and different environment, not only the different position should be noted (axial/equatorial), but also the different electronic effects implied.
In the first case:

In short: the molecule has a plane of symmetry, and the substituents are "mirrored" through the plane.
"Long" explanation: This means that they "see" the same electronic environment: the
equatorial substituents have the same electronic effects on the two
axial substituents. This means that their resonance frequency, is the same.

In the second case:

Short answer: no symmetry exists between the substituents
"Long" answer: every group neighboring the substituents is different. One of the two substituents feels the effects of $\ce{CO}$, the other one feels the effect of -$\ce{O}$-. Except in some unlikely and unlucky cases, this means two different signals.

Note to the reader: shielding, resonance frequencies and NMR behavior arise from (way more) complex interactions: I did not mention, above, any possible effect of non-neighbouring substituents, for instance, neither I talked about the shielding or deshielding effects of near substituents.
Nonetheless, a perception of the "symmetry" of the involved groups might give you a "dirt cheap" and operational insight on what could be the outcome of an NMR experiment, in cases as simple as this.


text:
Rate of reaction
For a reaction say
$$\ce{nA->B}$$ with $$\text{rate} =k[A]^n$$
So do we write rate as $ \frac {-\mathrm d[A]}{\mathrm dt}$  or $\frac{1}{n} \frac {-\mathrm d[A]}{\mathrm dt}$ We have only been taught for $n = 1$, and I couldn't thin which one should be correct. It will be really helpful if you can also include the reason why.

$1/n$ https://goldbook.iupac.org/html/R/R05156.html

$$\text{Rate of disappearance}~A= \frac {-\mathrm d[A]}{\mathrm dt}$$
$$\text{Standard rate} =\frac{1}{n} \frac {-\mathrm d[A]}{\mathrm dt}=\frac {\mathrm d[B]}{\mathrm dt}$$
The vlaue of $k$ depends on how you define the rate.


text:
Can anybody recommend me a Calculus Based General Chemistry textbook?
I'm looking for a Calculus Based General Chemistry textbook. I checked Atkins Chemical Principals but I didn't liked it that much. I'd be awesome if I had some more options
Thanks!

"Basic Chemistry" doesn't use much calculus. Calculus is more applicable to PChem. Specifically such topics as thermodynamics and kinetics.

Most general chemistry courses include a chemical kinetics unit, where calculus would be appropriate (but is often avoided). Outside of that, there's not many places in general chemistry that calculus would be useful or appropriate (except of the nuclear chemistry section, which is mathematically equivalent to first order chemical kinetics).


text:
Why must polar molecules be asymetrical?
As I understand, polarity in molecules comes from the difference in electronegativity or the ability of the atoms to attract eletrons so the electrons spend more time or are more probable to gravitate more around the atom with stronger electronegativity right? So I wonder why polar molecules must necessarily have an asymetrical structure? (as I've read on multiple sites). Shouldn't a symetrical atom with strong electronegativity with 2 atoms bonded linearly 180° on both sides which are significantly less electronegative, be polar since the electrons would gravitate more towards the center and the sides would then be more positively charged? What am I missing here?

Why is the first molecule (1) not possible? Is it becuase of the less polar atom on the left side repells electrons just as much as the one on the right? But then the electrons would still orbit the middle atom more because its more elctronegative, right?

I meant gravitate not orbit in the last sentence

You ain't missing anything. The first molecule is surely possible, and indeed does exist (think CO2), and is quite definitely very polar. Some people would claim it isn't polar because it has zero dipole moment. They are irrelevant.

So symetrical molecules can be polar and the sources that say otherwise are wrong? Thank you.

Or https://chemistry.stackexchange.com/questions/14641/why-is-carbon-dioxide-nonpolar

Zero net dipole moment means zero net dipole, just that. Molecules shouldn't be call polar, it only leads to misunderstandings. Solvents can and should be, but still with some measure of caution.

@IvanNeretin I have a feeling OP might just not get how dipoles work.

@IvanNeretin and @ Mithoron, I think you are touching on something important.  Most courses that introduce the concept of polarity/dipoles in the context of molecular structure use vectors to describe the dipoles of the individual bonds, and then the sum of those vectors to determine the dipole moment (at least qualitatively) of the molecule.  In that case, a molecule like CO2 would be non-polar.  Perhaps you should elaborate for the OP and other readers.

@MichaelLautman We did, on a number of occasions.

@IvanNeretin, I only saw the link that was posted after my comment.  I can retract my comment if you like.

Thank you for answers and for this link whoever posted it https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_(Zumdahl_and_Decoste)/08%3A_Bonding_General_Concepts/13.03_Bond_Polarity_and_Dipole_Moments  I thought the dipole came from the individual parts of the molecule now I understand that its the whole molecule's +/- that is Directionaly not negated or that the symetrical polar parts of the atom cancel eachother out directionaly so the molecule doesn't seek equilibrium in others because it alredy has it

@MichaelLautman No, leave it, because it is still useful.

Zan what you say is certainly true as far you look at parts of the molecule. Those are polar bonds. The various electrical moments can sum up to cancel each other and this requires a symmetrical arrangement.

If you start with the wikipedia definition of chemical polarity, which is standard (if not IUPAC) usage:

In chemistry, polarity is a separation of electric charge leading to a molecule or its chemical groups having an electric dipole moment, with a negatively charged end and a positively charged end.

then it's not hard to understand why a molecule such as $\ce{CO2}$ containing a center of inversion lacks a dipole moment and is therefore "nonpolar", as are, by this definition, other molecules with high symmetry properties and lacking a net dipole moment:

The only groups compatible with a dipole moment are Cn, Cnv and Cs. In molecules belonging to Cn or Cnv the dipole must lie along the axis of rotation.

It's not " bound to be canonical". IUPAC def. only applies polarity to solvents,  which this article is missing almost entirely.

@Mithoron I used "canonical" to mean "standard" in this case, but you have a  point, I was a bit loose with language.


text:
Where does the energy come from to lower the temperature of a brine solution?
Suppose I have a densely packed bucket of ice chips / cubes made from pure $\ce{H2O}$. It has been at room temperature for days, so the interior temp of the ice is likely close to 0 °C / 32 °F. (I have a working model of this in a clear ice - ice maker in a kitchen that does not have refrigeration; the ice is kept cold solely by the continuous addition of more ice.)
To a measured quantity of that ice (think about an ice bucket to cool wine) if I add:

23% of the weight of the ice in $\ce{NaCl}$ (table salt);
Some water to make a brine solution;
Stir.

The temperature of the resulting solution declines to roughly -22 °C / -8 °F.
I know that the $\ce{NaCl}$ depressed the freezing point of the water and that is why it can be liquid at this temperature. I suspect that $\ce{NaCl}$ is forming ions in the solution.
My question is: Where does the thermodynamic energy come from to lower a mass of ice and water to a significantly lower temperature?

You don't need energy to cool things down. Quite the opposite, in fact.

All three answers were great. Thanks all.

The minimum temperature reached should be −21 °C at 29 % NaCl. Initially suppose that the ice,  water and water vapour are at 0 °C in a thermally insulated container. Say point A in the figure and some salt is now added.
However, the mixture is now not at equilibrium, and some ice melts and dissolves some salt. The resulting solution is now too concentrated to be in equilibrium with the ice so more ice melts, diluting the solution and so more ice melts. At the same time the temperature of the whole system spontaneously falls until after sufficient salt is added a temperature of −21 °C is reached.
This is the eutetic point (point E) and equilibrium has been reached. The temperature falls because for a phase change to occur the ice uses the heat of fusion it needs to melt by drawing that much heat from its surroundings.
Many other salts can be used instead of NaCl, such as ammonium chloride and calcium chloride.

Figure 1. Crude phase diagram of the ice/water/salt mixture.

In contrary to mixing/dissolving salts with/in water, doing the same with ice instead works with much less total thermal energy.
Note that melting ice needs as much energy as heating the same mass of water by $\pu{ 80 ^{\circ}C}$
Water molecules in saturated solution at ice/salt contact have lower chemical potential than in ice, so ice is dissolving with expanse of lowering molecule kinetic energy and therefore temperature.

My question is: Where does the thermodynamic energy come from to lower a mass of ice and water to a significantly lower temperature?

In order to decrease the temperature of a sample, you don't need any energy. It is the opposite, you have to lower the kinetic energy. So the question becomes where does this energy go.
The answer is given by Poutnik: Some of the ice melts, and that takes energy.
What drives this process?
In ice, water is in a pure state. In the liquid state, water is either pure or mixed with sodium and chloride ions.
In the absence of salt, the ice and the water would be at equilibrium at zero degrees Celsius (the ice only melts slowly because there is heat transfer from the warmer surroundings, keeping the temperature at zero degrees Celsius).
After adding salt, some of the ice melts quickly, driven by the entropy of mixing. Once the temperature has dropped sufficiently (and the salt concentration has decreased because of the additional liquid water), the system is at equilibrium again. At that point, the ice slowly melts and the temperature slowly increases as the concentration of salt decreases.


text:
Synthetic Solution: How much nitric acid to add?
I am making a synthetic solution to imitate a real life situation. I am adding calcium nitrate to the solution to meet a target calcium level of $\pu{334 ppm}$. However, using excel calculations the amount of calcium nitrate that is needed imparts a total of $\pu{522 ppm}$ calcium. I am also adding $\pu{100 ppm}$ calcium carbonate which will increase the levels of calcium. My supervisor said I need to add nitric acid to lower the calcium imparted by the calcium nitrate. How and why does that work? and what calculation do I need to do to figure out how much nitric acid I need to add?

Hi, and welcome to Chemistry SE.  In order for us to help you, it would be useful for us to see your any work you have done and where you are having trouble.

I don't think nitric acid is the solution. I think your supervisor meant oxalic acid. In that way, excess $\ce{Ca^2+}$ ions may precipitate as calcium oxalate. Phosphoric acid may work as well.

Could you list all the concentrations you would like to achieve in this solution? Right now it is not clear what is "forcing" you to add more calcium than you need.


text:
How do we know the taste of dangerous compounds?
For example,   this site tells that the taste of Copper sulfate, (which is very dangerous for consumption) is of "metallic, nauseating taste" and as many dangerous chemical compounds often have taste descriptions, does this mean people eat them to taste test?

Yes, perhaps many years ago... and some of those rather cavalier experimentalists are no longer with us. Read https://www.acsh.org/news/2016/02/14/fluorine-element-hell or https://blogs.sciencemag.org/pipeline/archives/2010/02/23/things_i_wont_work_with_dioxygen_difluoride, for some other examples.

Related: How do people know HCN smells like almonds?; How do we know a nerve agent is tasteless?; Do all salts taste salty?

Newton in his older age was experimenting with testing of chemicals, including mercury compounds what is considered as reason for his related neurological problems.

Also, while working, you get to taste and smell a lot of things. Knowledge and regulations protect us, but I know the smells of hundreds compounds that I didn't deliberately smell or want to smell. If you would apply copper sulphate to trees or grape, you will get a feeling of its taste. (If not nowadays, surely it was at our grandparents time). Think of gluing something, too.

@poutnik did you mean testing or tasting?

Ah, tasting. I guess tasting was more frequent in time of alchemy than it is now.

@poutnik why not revive the practise? of course you can start small to build “immunity” however that works

Not all abandoned practices were good. Currently, if a substance is not considered dangerous, eventual tasting includes just few grains put on the tongue, then spilling out. Definitely not eating.

For the sake of curiosity, the history of modern and past artificial sweeteners is mostly a funny series of accidental tasting


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Difference between acetocarmine and feulgen stain
We did an experiment on observing mitosis in onion root tip cells, and I have observed that there are two ways of staining it, using acetocarmine and feulgen stain. What are the advantages of using each of them and the disadvantages, if there is one? What would be the difference, after all? Thank you in advance for your kind response.


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


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Visualizing rotation about a double bond

So I'm having trouble visualizing rotation about a double bond without using a model kit. The 2nd image has a red dot where I rotated it to make the molecule look exactly like the second image with my head and then my model kit. This is also consistent with my answer key that says these molecules are identical.
Now, the next two molecules:

I visualize rotating the left molecule at the red dot carbon and in my head, it matches up with the right molecule. However, when I pull out my model kit, it does not rotate like that and does not match up. This is also consistent with my answer key that says these molecules are diastereomers. So my question is, are there any rules that I could use or a way to visualize that will help me realize if a bond can be rotated to match a molecule without using a model kit.
Also, when people talk about it is not possible to rotate a double bond without breaking it, what exactly are they talking about that makes it different from this?

In your first example, you are not rotating about the double bond but rather the C3-C4 single bond. Both structures are identical. The pair of chlorides are diastereomers, Z and E, respectively. If in either one you change the geometry of the double bond, which is a structural change and not a rotation of a bond, then you will have a pair of enantiomers. To "rotate about a double bond" requires the breaking of a pi-bond, an unlikely event.

If your rotate about the C3-C4 bond, the atoms attached to C3 (the H and CH3O) would change from wedge to dash, not making the structures identical. I meant just rotating C4 (the red dot).

Rotating just the red dot does mean anything. At most is an handle for rotation around the near bond, when you have a model in your hands.

There is a difference between rotating the entire molecule along an axis (no change in conformation and energy) and rotating about a bond (leave one part of the molecule where it is and rotate the other part - change in conformation).

The red dot atom is C3. C4 is the carbon bearing the CH3O group. C4 and C5 and everything attached to them stays as is in the Fischer projection. A 180 deg. rotation about the C3-C4 bond gives the two views you have shown in your structures. I hope this issue is clear now. Make a molecular model!

Adding one more element to help you long-term: I think the underlying reason you are having a problem with this is that the visualization you are using is a little misleading.
You're selecting an atom with your red dot.  You can't rotate around an atom. You can rotate around the bonds between the atoms, though, assuming that the constraints of that bond allow it.
So in the first example, there are two possible carbon-carbon bonds you can rotate around that would both correspond to that red dot. Your dot should be on the sigma bond, based on your diagram.
In the second, you have three carbon-carbon bonds connected to your red dot. The one you are rotating is the double bond, not either of the bonds in the ring.  So the dot should instead be on that bond.
If you change the way you're marking the connection in your head, it could help you think about the nature of that connection, and that could help you with the rest of the information everyone is providing.

You can't freely rotate double bonds
Double bonds don't have free rotation. This is "built in" to the nature of the bonding in them: in effect you would have to break a bond to allow the molecule to rotate around a double bond.
In the first picture you drew, you didn't rotate around the double bond but actually rated around the single bond attached to the red dot. The second molecule has no single bond that can rotate (the single bond attached to the red dot is part of a 3-membered ring which locks to bond so it can't rotate). The "rotation" you show is not possible: the apparent result is a structurally different compound that can only be reached by breaking some bonds and not from free rotation.
The results you see with a model kit are right. And the rule you need to remember is that you cannot rotate around double bonds.

It is not possible to rotate a double bond without breaking the double bond .
Below 300$\ce{^o}$ C, double bonds are configurationally stable; that is, cis stays cis and trans stays trans.$\ce{^1}$
In alkenes, there is no free rotation about the double bond, unlike in alkanes. Consequently, groups attached to a double bonded carbon cannot exchange places. Therefore, the cis and trans isomers are separable compounds because they cannot interconvert.

source :ORGANIC CHEMISTRY , PETER VOLLHARDT &NEIL SCHORE pg439
In the first molecule ,red dot is on carbon bonded to a double and a single bond .If rotated along an axis passing through the red dot same molecule is formed.
However,in the second molecule,rotation along the red dot is not possible since the bonds are locked.
Your molecular model kit is correct.


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Can I make a mirror on PET plastic using Tollen's reaction?
I want to make a very shiny surface, like a mirror. I tried to use chrome paint but it's barely shiny. It's more like gray even though I bought one of the most expensive chrome looking paint.
I know mirrors can be produced using Tollen's reagent on glass (seen some demonstrations on YouTube).
But glass is quite expensive for what I want - I would like to make the mirror on plastic - more specifically on PET - the same material that a normal Coca Cola bottle is made from. I have big sheets of this PET material and I intend to transform them into mirrors.
Will silver deposit on the PET as it does on glass?
I need to order the ingredients online, then wait for it a couple of days - also because they are quite large quantities, it will end up being quite expensive. So, before making a mistake, I thought to ask here.
Could I use a Tollen's reaction to make a mirror on plastic?

As I remember from the book "Surely you are joking, Mr. Feynman!",stable silver coating of plastics is very difficult to achieve. It may depend on plastics and may require mechanical a/o chemicsl preprocessing of surface.

It will be endangered by plastics eventual flexing and would need fixing, otherwise it would be damaged soon by scraping and falling off. I suppose you cannot avoid experimental testing.

Definitely worth a test, if nobody knows of course.

See this for alternative metalization techniques: https://www.thomasnet.com/articles/custom-manufacturing-fabricating/how-to-metalize-plastic

The real barrier to doing this is not finding chemistry that will work: it is finding conditions that yield a good, stable, unblemished coating. In the early days of CD production, many CDs got their mirror coating chemically (later that was mostly replaced by sputtering). But this, even in well designed factories, often had a high defect rate as many CDs ended up with holes and imperfect coatings. The chemistry worked but it proved very hard to get surfaces clean enough to give good coatings.

I found a video on YouTube making a mirror on PET bottles.
Just that it consumes quite a lot of substance, and is not using Tollens reagent. My question is specifically about Tollens reagent.
What she uses is some "secret mixture" called RA-205.
It just proves that Ag can precipitate on PET bottles, which further supports the fact that will work with Tollens reagent. I'm will edit this answer once I execute my experiment, so this finding may only be somewhat helpful.

Even though this is very interesting and relevant information, I think currently it looks like this is better suited as an edit to your question and not as an answer. However, do keep in mind that answering your own question is a good practice:)

@andselisk yes indeed, started with an edit that was quite long, then i changed my mind and made it an answer. I intent to continue it, i will provide the results of the experiments in an edit on this answer. I need the package with ingredients delivered first. People can post answers in the mean time, and i'm gonna choose those as the best answer if they answer the question. Nice to see your effort of keeping this community high quality though :)

The silver deposition itself is much smaller problem them mirror layer durability.

Worth noting that whatever method you choose, you can test it without committing to buying large quantities of reagents. Try it first on small pieces of the PET film and only buy large quantities once you have refined the method.


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How much salt (or any other substance one can find in a kitchen) do I need to add to make water boil at 104 °C?
I've seen some formulas around in other questions and Google searches, but my chemistry is pretty much dead so I have no clue where to find the relevant values to calculate it myself. I just need to make an oil bath at 104 °C, but I would prefer to use a water solution instead since it is cheaper and easier to clean up. Doesn't need to be salt, can be anything commonly available at your average household.

According to [1, pp. 281–282], solution of sodium chloride $\ce{NaCl}$ prepared by dissolving 25 g of salt in 100 g of water has boiling point of 104.6 °C.
Additional data is available in the following table for the aqueous solutions of common salts and bases.
English transcription; column 1: Compound; columns 2–6: Concentration, g/100 g water — boiling points $(t_\mathrm{b.p.},~\pu{°C})$.

Note, however, that there is a significant drawback of using boiling salt solution due to the shift of boiling point upwards as the water boils off.
You either have to add water to the mark from time to time, or use a bath with another heat carrier and a thermal sensor.
References

V.A. Rabinovich and Z.Y. Havin. (Eds.) "Kratkii khimicheskii spravochnik" (Brief chemical handbook), 2nd ed., Khimiya, Leningrad, 1978. (in Russian)

Consider the altitude at which you will run your operation, too. @Louis Victor.

Wow, lithium chloride is powerful stuff!

Boiling point elevation of water by adding salt can be predicted with the formula:
$$\Delta T_b= K_b\, b_{solute} \, i$$
We look for $\Delta T_b=4$, using water $K_b=0.512$, and $i=2$ for salt $\ce{NaCl}$
$b$ is the molality, and an estimate $b=\pu{4 mol/kg}$ as the number $\ce{NaCl}$ moles per kilogram of solution.
Then
$$ \begin{align} \frac{g_{salt}}{100 g_{H2O}} &= \frac{b\times100}{(1000-b\times PM_{salt})}\, PM_{salt} \\ \frac{g_{NaCl}}{100 g_{H2O}} &=\frac{4\times100}{(1000-4\times 58.5)} \times 58.5\\ &=30.55 \end{align} $$ (results).

$i$  of Different salts can be found (some examples here).

A pressure cooker will allow boiling temperature to increase to 115 degrees or so by pressuring the cooking space.  Its not quite salt, but you could find one in a kitchen.
Commonly set to HIGH or LOW using a weight on the vent
Setting LOW would be 0.4~0.55 bar and HIGH would be 0.9~1 bar so double atmospheric pressure.  Readings are on top of the 1 bar normal atmospheric pressure of the room.
By comparison, a car tyre would be about 2.2 bar, so just over triple normal room pressure air.
0.5 bar would be a boiling point of 112 degrees C
1.0 bar would boil at 120 degrees C
2.6 bar would be 140 degrees C, but good luck getting such pressures in the kitchen.
Disclaimer, even 0.5 bar is a lot of pressure, and any rupture will cause significant damage.

Maybe you should mention that these values refer to gauge pressure, and not to absolute pressure which is normally used in chemistry.


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If carbon dioxide is non polar, why does it react with water?
I’ve read in many places that carbon dioxide forms carbonic acid with water since the carbon is partially positive and thus the oxygen bonds with it. But isn’t carbon dioxide in its entirety non polar since the 180 angle removes any dipole moment, so how would oxygen be attracted to the carbon?
Is it that individually each atom has partial charge but not collectively? Or am I wrong in some way?

Fluorine gas is about as non-polar as you can get, but you don't want to mix it with water (or anything else...). Meanwhile, caesium fluoride is extremely polar, but happily dissolves in water with no reaction, and can be completely recovered by drying.

Carbon dioxide is pretty polar. True, it has zero dipole moment, but that's irrelevant.

There is need to distinguish polarity of the whole molecule and polarity of its parts. CO2 as a molecule us not polar ( but is polarizable), but CO bonds are polar.

https://chemistry.stackexchange.com/questions/67543/intermolecular-forces-between-carbon-dioxide-and-water https://chemistry.stackexchange.com/questions/67073/why-is-carbon-dioxide-considered-a-lewis-acid

Related: Why is carbon dioxide nonpolar?

@IvanNeretin Co2 is non-polar, cant agree with you saying Co2 is polar

@DJphy Too bad.

@DJphy If two people pull a box in opposite directions through ropes, can you say that as net force is zero the ropes will be loose and have zero tension?

I can only repeat myself here: Polarity is an ill-defined concept that has a nice potential for confusion.
In most cases, when specifying a molecule as polar, one is colloquially referring to the presence of a dipole moment, i.e. one actually categorises the molecule as dipolar. As described by ron in "Why is carbon dioxide nonpolar?", $\ce{CO2}$ has no dipole moment, it is therefore not dipolar, or colloquially it is not polar.
However, $\ce{CO2}$ has two very dipolar bonds, and a significant quadrupole moment. If one were to extend the nomenclature, one would say the molecule is quadrupolar. However, this may lead to complications down the line.
On the other hand, as I have written in the linked question, toluene is often considered an unpolar/non-polar solvent, which is not really true considering it has a small dipole moment.
There are a couple of things one can predict with the concept of polarity, and fortunately, the more complex the molecules become, the better the approximation becomes. It is small highly symmetric molecules, which break these approximations.
Related reading:

Some more extended reading by others and myself can be found in this question: What are dipole moments in a molecule supposed to act upon?

And to further the confusion: Is the triiodide ion polar?

Where to draw the line? What is practical and what not? Why is CO practically nonpolar? What is the dipole moment direction in the nitrosonium ion?

Carbon dioxide actually is polar.  It is not di polar, but it has a quadrupole -- a combination of two opposing dipoles.  Quadrupoles interact only weakly at a distance; the electrostatic interaction energy with an external charge falls off as $1/r^3$.  But as with a dipole, a close-up external charge or dipole can interact selectively with one of the component charges by drawing close to the favored component, like the hydrogen atoms of a water molecule drawing close to one of the negatively charged oxygen atoms in the carbon dioxide quadruple.
There is also something of a "bonus".  Water is best known as a dipole, but it too has a quadrupole along the line through the two hydrogen atoms (perpendicular to the dipole).  The water quadrupole has its negative charge in the middle whereas carbon dioxide has a positive charge in the middle; and the two molecules are similar in size.  We therefore have a "hand in glove" situation where the water has its quadrupole aligned with the carbon dioxide quadrupole but the charges oppositely distributed, enabling every atom of the water molecule to simultaneously attract an appropriate atom from the carbon dioxide.  This extra solvation interaction not only improves solubility, it gets the molecules properly arranged to combine and form (a trace of) carbonic acid.

I would like to suggest this answer, since this also explains the carbonic acid formation.

A molecule is said to be "non-polar" as a whole and carbon dioxide falls into this category. However, it does not mean that parts of the molecule aren't indeed polar, even if their effects cancel out when considering the whole molecule.
EDIT: So the CO2 molecule is indeed very polar and, more especially, the central carbon atom is really electrophilic. That is why it reacts with water.

A molecule of carbon dioxide has a slight negative charge near the oxygen and a slight positive charge near the carbon. CO2 is soluble because water molecules are attracted to these polar areas. The bond between carbon and oxygen is not as polar as the bond between hydrogen and oxygen, but it is polar enough that carbon dioxide can dissolve in water.


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Argon vs nitrogen for preserving wine
I have a question regarding whether argon or nitrogen is to prefer when it comes to preserving wine.
Specifically, I am interested in knowing if there is a difference in preservation periods.
Also, if there is a risk of them reacting with the wine, which might affect taste and smell?

I don't believe there is any measurable difference between the two.

Nitrogen is a lot cheaper.

Point taken. Yes, I guess that counts.

https://www.google.com/url?sa=t&source=web&rct=j&url=https://physics.stackexchange.com/questions/213903/behavior-of-two-gases-in-a-closed-container&ved=2ahUKEwjPqp6M9c3iAhUlmeAKHZ-RBWIQFjADegQIBhAB&usg=AOvVaw2YsvCQwvzCN89yiWoimoxJ Related discussion on physics SE.

Thank you for all the answers! It helped a lot.

In chemistry, some chemical species can still react with nitrogen so argon is usually favored, but, in the case of wines, none of the two should react whatsoever.
In both cases, however, the high density of argon is an advantage because it will always fall down to be in contact with the liquid in order to avoid contact with oxygen.
By contrast, nitrogen would just dilute oxygen in the remaining air and just slow down oxidation, because air is already composed of 80% nitrogen.


text:
Having a contradiction while deriving Gay-Lussac's law using Boyle's law and Charles' law
From Boyle's law:
$$P_1V_1 = P_2V_2 \quad\implies\quad V_2 = \frac{P_1V_1}{P_2} \label{eqn:boyle}\tag{1}$$
From Charles' law:
$$\frac{V_1}{T_1} = \frac{V_2}{T_2} \quad\implies\quad V_2 = \frac{V_1T_2}{T_1} \label{eqn:charles}\tag{2}$$
Equating \eqref{eqn:boyle} and \eqref{eqn:charles} gives:
$$\frac{P_1V_1}{P_2} = \frac{V_1T_2}{T_1},$$
which upon solving will give:
$$P_1T_1 = P_2T_2,$$
which is exact opposite of Gay-Lussac's law. I can't find the flaw in my logic. Help!

(1) is true for isothermal processes, (2) for isobaric, and it would be prudent not to expect anything good when you mix the two.

There is no flaw. Or rather, if there is a flaw it is to think that your equation holds for any T1/T2 and P1/P2 pair when in fact (as Ivan explains) you have implicitly imposed constant T and P conditions during the derivation.

@Ivan Neretin But my textbook says you can prove the Gay Lussac's law by Boyle's law and and Charles' law. If mixing those two wouldn't get anything meaningful how would you prove Gay Lussac's law through Charles' law and Boyle's law ?

As was pointed out in the comments, the problem arises when you assume that the two starting points are equivalent.  @IvanNeretin pointed out that Boyle's Law is only useful for situations where the temperature doesn't change (isothermal) and Charles' Law is only useful in situations where the pressure doesn't change (isobaric).
This means that even though we can mathematically set the equality, it doesn't have any physical significance.
This may be a bit clearer if we consider that both Boyle's Law and Charles' Law can be seen as simplified versions of the ideal gas law.
$PV = nRT$
I'm adding the subscripts B (Boyle's Law) and C (Charle's Law) to highlight the differences.
The only way for Boyle's Law to hold true is if the $nRT$ portion of the equation is identical in both cases.
$P_{1B}V_{1B} = nRT$ and $P_{2B}V_{2B} = nRT$
$P_{1B}V_{1B} = nRT = P_{2B}V_{2B}$
For this to be true, both the number of moles in the system n and the temperature must be held constant between the two systems.
We can do the same analysis for Charles' Law, in this case holding the pressure (and the number of moles) between the two systems constant.
$PV_{1C} = nRT_{1C}$ and $PV_{2C} = nRT_{2C}$
$\dfrac{V_{1C}}{T_{1C}} = \dfrac{nR}{P} =\dfrac{V_{2C}}{T_{2C}}$
So although we can establish two equations that are equal to $V_2$
$V_{2B} = \dfrac{P_{1B}V_{1B}}{P_{2B}} = \dfrac{nRT}{P_{2B}}$
and
$V_{2C} = \dfrac{V_{1C}T_{2C}}{T_{1C}} = \dfrac{nRT_{2C}}{P}$
It should be clear that the $V_{2B} \neq V_{2C}$, and therefore the premise of the rest of your derivation isn't valid.

The derivation is not that hard. If my memory is correct this how we have done that:
From Boyle's Law:
$$P_1V_1 = P_2V_2 \label{eqn:1}\tag{1}$$
From Charles' Law:
$$\frac{V_1}{T_1} = \frac{V_2}{T_2} \label{eqn:2}\tag{2}$$
Multiplying \eqref{eqn:1} and \eqref{eqn:2} gives:
$$\frac{P_1V_1^2}{T_1} = \frac{P_2V_2^2}{T_2} \label{eqn:3}\tag{3}$$
If $V_1=V_2$ (as Gay-Lussac's law states), \eqref{eqn:3} can be simplified to give:
$$\frac{P_1}{T_1} = \frac{P_2}{T_2} \tag{4}$$

How can you multiply 1 and 2?

The two law apply for the same amount of gas.

But Temperature is constant for (1) and pressure is constant for (2).

They still cancel out if that condition applies. The key is same amounts.


text:
Guide to DFT functionals
Having recently begun studying Quantum and computational chemistry in depth... I am looking for a manual, text, or guide of any form that could explain the basic formulations and potential of the Functionals used in DFT.
My focus lies primarily on their physical significance, preferably derived mathematically...
Could someone kindly help me out with this...

I would suggest a book entitled "A Chemist's Guide to Density Functional Theory" as a gentle introduction. This has everything you cited.

And I just noticed that the aforementioned book is not in that list, however, there are plenty of other books that may do the job as well. Somewhat related: DFT Functional Selection Criteria


text:
Measure low DOM absorbance values in a microplate reader. Is there a suitable deep well microplate with flat bottoms available?
I would like to use a microplate reader to measure the absorbance of dissolved organic matter (DOM), but the measured values are close to the lower detection limit. To overcome this, I thought of deep well plates as the pathlength is longer then. Nevertheless, I have not found any deep well plates with flat bottoms, which might be best for absorbance measurements. I want to avoid measuring in a photometer with 5 cm pathlength because the available amount of sample is limited and the microplate would need much less.
Does anybody have experience with similar problems?
Can anyone suggest a suitable microplate?
What is your opinion on the influence of the material?

I am not aware of suitable microplates for this application.  Here are some alternative approaches.
1. Low-volume quartz cuvettes.
There are quartz cuvettes in "micro" formats that may be suitable for you.  Mostly commonly they work with 1 cm pathlength.  The volume of the cells is controlled by the width of the non-optical dimension.  This can be made extremely small.
One example I found lists cuvettes that still have path lengths of 1 cm with total fill volumes as low as 15 μL.
2. Sample concentration
Details depend heavily on the details of your DOM and application, but you could concentrate your samples and then use existing microplates.
3. Longer measurement times.
Most plate readers (and cuvette-based spectrophotometers) let you control the number of "shots" or of the total amount of time a measurement of light intensity is integrated over.  You could crank this number up to get better sensitivity.


text:
Is dihydrogen the only example of overlap between two s orbitals?
I heard that $\ce{H2}$ is the only example of s–s orbital overlap. Can anyone give an example which contradicts this statement?

Metallic Li will do.

Are you suggesting that 1A and 2A mettalic bonds are all masses of covalent compound?

What's 1A and 2A? Also, no, a metal is definitely not a covalent compound. But there still is certain orbital overlap in it.

@IvanNeretin  I meant the groups in modern periodic table.

Oh, of course. Well, they all are as good as Li in this regard.

You don't need particularly exotic things to get a counterexample. Metal–metal bonding in mercury(I) compounds is quite prevalent, where the electronic configuration is $\ce{Hg+}$: $\ce{[Xe](4f^14)(5d^10)(6s^1)}$, and the bonding involves 6s–6s overlap. The prototypical example is the $\ce{Hg2^2+}$ cation, which features in mercury(I) chloride $\ce{(Hg2^2+)(Cl^-)2}$; this cation is valence isoelectronic to $\ce{H2}$.
Fundamentally, it is a relativistic contraction of the 6s orbital which makes such bonding common in Hg(I): see also Why does mercury form polycations?. The bonding in alkali metal dimers is qualitatively similar, but significantly weaker.
If we play the "every bond has orbital overlap" card (i.e. not just covalent bonds, which I think is perfectly fair), then the delocalised bonding in Group 1 and 2 metals arises from overlap of ns orbitals on each atom to form bands, as has already been alluded to in the comments. Even simpler would be lithium hydride, which likely has some degree of 1s–2s orbital overlap, even though the bonding may be primarily ionic.

Hydrogen forms, in addition to $\ce{H2}$, ion clusters of varying size.  Following are two prominent examples.
The trihydrogen cation, $\ce{H3^+}$, is found in the interstellar medium and in the atmospheres of Jovian planets.  The three-center, two-electron bond that binds this ion together simultaneously covers all three atoms and all three atom-atom linkages, giving a strongly stabilized structure.
The hexahydrogen ion, $\ce{H6^+}$, mentioned in the first reference above, has a turnstile structure in which three dihydrogen units are linked together with three mutually orthogonal orientations; alternatively it may be viewed as two tri-hydrogen triangles joined vertex to vertex.  This species can be formed in solid hydrogen, where it migrates through an exchange mechanism similar to the Grotthus mechanism of proton conduction in water and other protic solvents.


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Reason behind hybridisation
I have been taught that orbitals can overlap only when all of them are in the same energy state. That is the reason why the orbitals hybridise in a kind of 'mixing up' and give rise to orbitals having same energy state.
But my question is: Why do the orbitals need to be in the same enrgy state to overlap with orbitals of other atoms?

Hybridization is not a real physical process.

@Greg — Well could you be a bit more explicit?

:/ If you have 4 identical bonds in product and non-identical orbitals in substrate... - you confusing cause and result.

Related: How does localized bonding theory and hybridisation work?

I wouldn't normally vote to close as duplicate when the questions don't match, but considering this is already getting closed as unclear and you're already very likely to get your answer by reading the answers to that question, and the fact that this question isn't worded importantly search-wise . . .


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How to explain disagreement between Le Châtelier's principle and the simplified Gibbs free energy equation?
For example, for the dissolution of a salt in water that is exothermic, heating the solution would drive the reaction towards the solid form of the salt according to Le Châtelier's principle.
However, according to the simplified Gibbs equation $(\mathrm dG = \mathrm dH - T\mathrm dS),$ the reaction would be spontaneous towards the products due to the increase in entropy and the negative enthalpy.
To me, these seem to contradict each other. How is this reconciled? Am I making a mistake somewhere?

To see how $\Delta G^\circ$ or $K_{eq}$ behave with T you should use the van't Hoff equation.

The dissolution of salt in water is endothermic.

I think the use of the word "salt" in the question is generic; it refers to any ionic compound. So yes, enthalpy of solution of sodium chloride is (slightly) endothermic, but many salts have exothermic enthalpies of solution.

The primary flaw in your reasoning is assuming that $K$ is proportional to $-\Delta G^\circ$, so that a reaction with $\Delta S^\circ >0$ and $\Delta G^\circ<0$ must have a larger $K$ at a higher temperature because $\Delta G^\circ$ is more negative. If that were true, we would have a relationship of the form $\Delta G^\circ = -cK$, where $c$ is a constant. Instead, the key relationship is
$$\Delta G^\circ = -RT\ln K.$$
So $\Delta G^\circ$ is proportional to $-T\ln K$. In the case above where $\Delta G^\circ <0$ and $\Delta S^\circ>0$, as T increases, $\Delta G^\circ$ increases in magnitude (becomes more negative), but so does $-RT$, so we don't necessarily need to have a larger $K$ to satisfy the equation. To figure out the temperature dependence of $K$, we need to substitute $\Delta G^\circ$ with $\Delta H^\circ - T\Delta S^\circ$ and then rearrange things:
$$\Delta H^\circ - T\Delta S^\circ=-RT\ln K$$
$$\frac{\Delta H^\circ}{T}-\Delta S^\circ=-R\ln K$$
From that equation, hopefully it is clear that if $T$ increases (which reduces the magnitude of the $\frac{\Delta H^\circ}{T}$ term), $K$ will only increase if $\Delta H^\circ > 0$. If $\Delta H^\circ < 0$, K will have to decrease with increasing $T$ to maintain the equality. $\Delta S^\circ$ is a constant term that does not affect the change in $K$. Thus, our result is completely consistent with both Le Chatelier's principle and with the van't Hoff equation analysis.

You gave a good first-order explanation, which should be sufficient for the OP.  Note, however, that neither delta_H nor delta_S are constant terms; they are both temperature-dependent.  The reason why the van't Hoff equation has the form it has (i.e, the reason why delta S doesn't appear in it, and why delta_H appears as a constant, in spite of the T-dependencies of both terms) is that the T-dependency of delta_S is exact T times the temperature-dependency of delta_H.  Hence (since the terms are subtracted) the two temperature-dependencies exactly cancel out.  This can be proven analytically.

Yes, I was trying to match the level of the question, and treating $\Delta H^\circ$ and $\Delta S^\circ$ as constants with respect to T is a standard simplification in introductory level chemistry.

The dependence of an equilibrium constant on temperature is given by the van't Hoff equation:
$$\left(\frac{\partial{\log(K)}}{\partial{T}}\right)_p=\frac{\Delta H^\circ}{RT^2}$$
Therefore for an exothermic reaction ($\Delta H^\circ<0$) you expect a decrease in $K$ with temperature, as Le Châteliers principle would predict.
To give an example, the following figure displays solubilities for LiCl at different T, obtained via PubChem, and the associated values of $\Delta H^\circ$ obtained from the slope using the van't Hoff equation. Note that the increasing positive slope indicating an increasing solubility is also consistent with an increasing value of the positive (endothermic) enthalpy.

Using the values of $\Delta H$ derived from an analysis based on the van't Hoff equation is a form of self-fulfilling prophecy: the van't Hoff expression can't fail, because it's a way to derive $\Delta H$ for the process from the temperature dependence of $K$. Of course $\Delta H$ has an independent meaning, and values can be provided by calorimetry, but the point is that van't Hoff's expression will make correct predictions provided you use the right value of the enthalpy. van't Hoff's equation will not predict the behavior expected according to Le Chatelier's principle if you use the wrong $\Delta H$. In particular, you should not use the value for the limiting (infinitely dilute solution) heat of solvation if attempting to predict how temperature will affect solubility. Rather you should use the value associated with transferring a solute molecule into a saturated solution.

Update
Andrew gives the right explanation on how to interpret the standard expression for the Gibbs free energy in terms of enthalpy and entropy (multiply it by $-1/T$).
I think I misunderstood the original question, and Andrew nailed the answer because he saw where the confusion lay (it has nothing to do with van't Hoff's equation or the particular value of $\Delta H$).
I leave the following comment which I included in my original answer but it is no longer valuable as a clarification on how to reconcile the apparent contradiction.
The expression for $dG$ you provide is derived from the definition of the Gibbs free energy  $$G=H-TS$$ by imposing the condition of constant temperature, and so is applicable only to a process carried out at constant T. This leads to
$$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$
with the values of $\Delta H^\circ$ and $\Delta S^\circ$ specific to the specified T. Sometimes these values are weak functions of T, but not necessarily.

Does this type of analysis apply to dissolution, which is nominally a physical process.  The van't Hoff equation postulates an inverse relationship between K and T for exothermic reactions.  Empirically, however, we know that for the vast majority of ionic compounds, solubility increases with increasing temperature.  The van't Hoff equation as written excludes entropy, which is the bit can account for the increased solubility.  In these cases both the OP's Le Châtelier analysis and the use of the van't Hoff equation are incorrect.  Am I wrong?

@BuckThorn, unfortunately that just isn't the case.  As an example, LiCl.  Enthalpy of solution = -37 kJ/mol.  Solubility in water: 0C = 68 g/100mL, 25C = 84 g/100mL, 100C = 123 g/100mL.Source

@BuckThorn Huh... I hadn't thought of it that way.  You make a very valid point.  That makes my initial comment seem a little silly.  I still think that the OP's mistake in the qualitative Le Châtelier's analysis is the exclusion of entropy.  I'm going to see if I can find some enthalpy of solution values at different temperatures.

@MichaelLautman I turned your comment into a question, here: https://chemistry.stackexchange.com/questions/116365/why-does-the-solubility-of-licl-increase-with-temperature-if-the-dissolution-is

@MichaelLautman my earlier comment "if solubility increases with increasing T I would predict that the dissolution is endothermic" was clearly wrong (in light also of my answer to a later question). This should have been "if solubility increases with increasing T I would predict that the process of formation of a saturated solution is endothermic".

@BuckThorn, this whole discussion has been very enlightening for me.  I hadn't thought deeply about this topic previously.  I did a major edit on my answer here in light of what you and Andrew have pointed out.

Note: This answer has been heavily edited, because the original answer I posted contained some assumptions and was very limited (and incorrect) in the analysis.  I would like to highlight the patience of @Buck Thorn, @Andrew and @Karsten Theis in pointing out my errors, expanding the discussion and going into a lot of detail.

The trouble is not with any inconsistencies in your analysis.  The difficulty is actually in finding consistencies in the behaviour of ionic compounds when they dissolve.
For compounds with low solubilities (and endothermic enthalpies of solution) like silver(I) chloride, using either the van't Hoff equation or Le Châtelier's principle are consistent with observations.  That is to say, that solubility (and the associated equilibrium constant) increases with temperature.
The equilibrium for the dissolution is:
$\ce{AgCl(s) -> Ag^+(aq) + Cl^-(aq)}$
We can write the following equilibrium:
$\ce{K_{sp} = [Ag^+][Cl^-]}$
At 25$^oC$, $\ce{K_{sp} = 1.77*10^{-10}}$
At 50$^oC$, $\ce{K_{sp} = 1.32*10^7}$
This corresponds to a solubility of 1.9 mg/L at 25$^oC$ and 5.2 mg/L at 50$^oC$.
Gibb's free energy and equilibrium are related by
$\Delta G = \Delta H - T\Delta S = -RTlnK \quad(1)$
The trouble starts when we try to apply this type of analysis to compounds that are very soluble in water.

Source
We can see that different ionic compounds exhibit a wide variety of changes in solubility in response to increases in temperature.  The bizarre behaviour of sodium sulfate was addressed in this question.  $\ce{NaCl}$, which has a slightly endothermic enthalpy of solution, shows almost no variation in its solubility as the temperature increases.  The cerium salt shows a decrease in solubility as the temperature increases.
As pointed out in this question and this question there are often multiple factors at play when an ionic compound dissolves in water. With large solubilities, the values of the $K_{sp}$ are considered unreliable, which might explain discrepancies between experimental and calculated results.
So, the answer to the question is that it depends. Some soluble salts will obey Le Châtelier's principle and the van't Hoff equation accurately predicts their solubility behaviour, but others do not (sometimes quite spectacularly). In those cases a closer look at the specific circumstances surrounding the dissolution of that compound in order to understand what is happening.

You provide a lot of useful data to compare, which I'll have to look at more closely. I think as far as the posted question is concerned, the answer is that it's sort of a self-fulfilling prophecy: the van't Hoff expression can't fail, because it's a way to define $\Delta H$ for a process from the temperature dependence of K. Of course $\Delta H$ has an independent meaning, and values can be provided by calorimetry, but the point is that van't Hoff's expression is always right, provided you use the right value of $\Delta H$.

I think I misunderstood the original question, and Andrew nailed the answer because he saw where the confusion lay (it has nothing to do with van't Hoff's equation or the particular value of $\Delta H$).

Finally, van't Hoff's equation will not predict the expected Le Chatelier's behavior if you use the wrong $\Delta H$. In particular, you should not use the value for the limiting (infinitely dilute solution) heat of solvation. Rather you should use the value associated with transferring a solute molecule into a saturated solution.


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When 2-pentene reacts with HBr, what will be the major product?
If we consider the first step (formation of carbocation by the attack of the proton).
A couple of textbooks I referred say that when the substituents on the $\ce{C=C}$ are different, the direction of shift of the electron is decided by the inductive effect of the substituents. For example, in this image:
$$
\begin{align}
\ce{\underset{1}{C}H3-\underset{2}{C}H=\underset{3}{C}H-\underset{4}{C}H2-\underset{5}{C}H3 &-> \underset{1}{C}H3-\underset{2}{\overset{\large\ominus}{C}}H-\underset{3}{\overset{\large\oplus}{C}}H-\underset{4}{C}H2-\underset{5}{C}H3} \label{rxn:a}\tag{a} \\
\ce{\underset{1}{C}H3-\underset{2}{C}H=\underset{3}{C}H-\underset{4}{C}H2-\underset{5}{C}H3 &-> \underset{1}{C}H3-\underset{2}{\overset{\large\oplus}{C}}H-\underset{3}{\overset{\large\ominus}{C}}H-\underset{4}{C}H2-\underset{5}{C}H3} \label{rxn:b}\tag{b}
\end{align}
$$
Assuming that a proton $\ce{H+}$ is approaching, both the textbooks say that the electronic shift happens as in the case \eqref{rxn:a} because \eqref{rxn:a} has 2 ethyl groups to stabilize the carbocation by inductive effect. But isn't hyperconjugation more important than inductive effect in determining the stability?
By hyperconjugation, case \eqref{rxn:b} should be the preferred pathway as there are 5 alpha-hydrogens (including the $\ce{H}$ that is attacking and will join at the $\ce{C-}$) and case \eqref{rxn:a} has only 4? By this logic 2-bromopentane should dominate. Am I missing anything?

This sort of reaction isn't investigated seriously anymore. But 80 years ago, Kharasch et al. (J. Am. Chem. Soc. 1939, 61 (6), 1559–1564) wrote that

Both results agreed within the limit of error (3%) and indicated an equimolar mixture of 2- and 3-bromopentanes [...]

So, for all intents and purposes, there is no major product. It is not always possible to get a definitive answer based on simplified "rules" and concepts. All we can genuinely say in this case is that both intermediates are very similar in stability, and so both products will be formed in significant amounts. In fact, the carbocations may even interconvert rapidly by means of hydride shifts, because when pentan-2-ol or pentan-3-ol is treated with HBr, you get a mixture of the bromides (J. Am. Chem. Soc. 1930, 52 (6), 2440–2451).
This lack of selectivity is precisely why nobody investigates these seriously anymore. Maybe with modern analytical methods we could accurately measure it and find that the product ratio is 52:48, but what's the point? That's just as useless as a 50:50 mixture.

Indeed, and another thing that precise values would vary depending on multitude of factors.

Your argument of inductive effect and hyperconjugation, made me to tell you a story about theory behind these two effects. I think after reading that you may able to understand how far we have come from them:
Wagner and Saytzeff (Ref.1; in German) prepared 2-pentene by dehydrohalogenation of 3-iodopentane. Addition of hydrogen iodide to this pentene yielded a product, which they characterized as the 2-iodopentane. From these results, they formulated the rule that the negative portion adds to the carbon atom bearing the shorter carbon chain.
Assuming that carbon compounds are polar in nature and using the work of Wagner and Saytzeff as experimental proof, Cuy (Ref.2) in 1920 proposed the hypothesis of alternatively charged carbon atoms. Accordingly, if hydrogen bromide (or other unsymmetrical reagent) is added to 2-pentene, the reaction would follow to give 2-bromopentane as the predominant product. It is a well-known fact (at the time as well) that primary propyl bromide in the presence of catalysts such as aluminum bromide, goes over to the secondary propyl bromide. Cuy had proposed that such isomeric rearrangements of alkyl halides can also be readily accounted for, on the basis of his hypothesis.
Lucas and Jameson (Ref.3) disagreed with Cuy’s theory of alternate polarization of carbon atoms, based on the fact that Wagner and Saytzeff’s reaction of hydrogen iodide and 2-pentene has given a mixture of isomeric iodides as products (Ref.1), not 2-iodopentane alone as Cuy assumed. Thus, they had advanced their own theory of electron displacement.
Lewis (Ref.4) was the first to show the effect of substituents upon the strength of organic acids, and to show that this effect extends throughout the entire carbon chain. Lucas and Jameson applied the work of Lewis to propene, to acrylic acid, and to dimethyl allene. They found that the addition of hydrogen halides could be explained by the theory of electronic displacement. The work of Lewis suggests that the alkyl group is more positive than hydrogen, while the carboxyl and halogen groups are more negative. Because of the positive character of the alkyl group as compared to hydrogen, the structural formula of propene can be written with the electrons constituting the double bond ($\pi$-bond) being closer to the terminal carbon atom. Thus, addition of an unsymmetrical reagent would result in the more negative reagent attaching to the number two carbon. This theory also explains the rearrangement of 1-bromobutane to 2-bromobutane. Application of this theory to the addition of hydrogen bromide to 2-pentene would lead to a predominance of 3-bromopentane (Ref.5).
Lucas and Moyse (Ref.5) prepared the 2-pentene, and added hydrogen bromide to the olefin using glacial acetic acid as solvent. Using the refractive index method of analysis, they found 78% of the 3-bromo isomer and 22% of the 2-bromo isomer (74% overall yield). Lucas then considered the theory of electronic displacement confirmed (Lucas claimed the results obtained are in harmony with the hypothesis of electron displacement, but not with that of alternately polarized carbon atoms).
In l935, Baker and Nathan (Ref.6) advanced the theory of hyperconjugation in order to explain certain abnormal reactions in the halogenation of alkyl-substituted benzenes. They first postulated that the accelerating effects of alkyl groups must be related to their capacity for electron-release. The relative magnitude of such electron release by alkyl groups increases in the order $\ce{C(CH3)3 < CH(CH3)2 < CH2CH3 < CH3}$. This order is exactly the reverse of that anticipated on the basis of the general inductive effect ($+I$) of an alkyl groups. They called this newfound electron releasing mechanism of the alkyl group attached to the necessary system is a type of tautomeric effect, which is often referred to as the Baker-Nathan effect in English literature for some extended time until they coined the word, hyperconjugation.
Further proof of this theory and beyond from a physical-chemical basis has been reviewed by Deasy (Ref.7) including quantum mechanics point of view. Basically, according to the theory of hyperconjugation, the addition of hydrogen bromide to 2-pentene would produce a predominane of 2-bromopentane, since it is possible to write three hyperoonjugative structures involving the three -hydrogens of the methyl group, but only two forms involving the two hydrogens of the ethyl group.
Based on above literature, the products outcome of this type of reaction is also depending on the conditions used, workup procedure, etc.
Now, it is well accepted that polarization can not be predicted, but product outcome can be determined by various factors, including hyperconjugation, 1,2-hydride shift, condition used, etc. (Ref.8, ref.9).

References:

G. Wagner, A. Scytzeff, “Ueber amylenbromür und Amylglycol aus Diäthylcarbinol,” Justus Liebigs Annalen der Chemie 1875, 179(3), 302–313 (https://doi.org/10.1002/jlac.18751790305).
E. J. Cuy, “The electronic constitution of normal carbon chain compounds, saturated and unsaturated,” J. Am. Chem. Soc. 1920, 42(3), 503–514 (https://doi.org/10.1021/ja01448a016).
H. J. Lucas, A. Y. Jameson, “Electron Displacement in Carbon Compounds I. Electron Displacement Versus Alternate Polarity in Aliphatic Compounds,” J. Am. Chem. Soc. 1924, 46(11), 2475–2482 (https://doi.org/10.1021/ja01676a018).
G. N. Lewis, “The Atom and the Molecule,” J. Am. Chem. Soc. 1916, 38(4), 762–785 (https://doi.org/10.1021/ja02261a002).
H. J. Lucas, H. W. Moyse, “Electron Displacement in Carbon Compounds II. Hydrogen Bromide and 2-Pentene,” J. Am. Chem. Soc. 1925, 47(5), 1459–1461 (https://doi.org/10.1021/ja01682a037).
J. W. Baker, W. S. Nathan, “429. The mechanism of aromatic side-chain reactions with special reference to the polar effects of substituents. Part V. The polar effects of alkyl groups,” J. Chem. Soc. 1935, 1844 –1847 (DOI:10.1039/JR9350001844).
C. L. Deasy, “Hyperconjugation,” Chem. Rev. 1945, 36(2), 145 –155 (https://doi.org/10.1021/cr60114a001).
B. A. Howell, R. E. Kohrman, “Preparation of 2-bromopentane,” J. Chem. Educ. 1984, 61(10), 932–934 (https://doi.org/10.1021/ed061p932).
H. Pines, A. Rudin, V. N. Ipatieff, “Investigation of the Preparation of Bromides from 1-, 2- and 3-Pentanol. Synthesis of Pure Bromopentanes,” J. Am. Chem. Soc. 1952, 74(16), 4063–4067 (https://doi.org/10.1021/ja01136a027).


text:
Copper Thin Film Oxidation?
So I am a clueless undergraduate student trying to get a glass substrate sputtered with a 100 to 200 nanometer copper thin film from one lab to another. I need to know how much time I have before the copper begins to oxidize, or at least how I could figure out how long it would take in room temperature.

Could you edit your question to add some detail. What were the conditions of sputtering? Is this vacuum sputtering? Could you transport it under nitrogen? Why do you have to transport it at room temperature?

And how much oxidation is acceptable?

Why don't you keep it in an inert atmosphere, like argon or nitrogen if oxidation is to be minimized.


text:
butane extraction (microbials)
I was wanting to know using a butane extraction method for hemp flower (cannabis sativa) would also extract the microbial DNA present on the flower as well as deriving the cannabis oil (oleoresin)? This extraction process is used to obtain oleoresin(hash) from flower and I want to know if microbial DNA (or living microbes) can also be extracted? The goal of the extraction is to obviously have no microbes present in the end (this is for human consumption). If someone could explain the chemistry as to why it does or doesn't work I would appreciate it.
Thanks,
Kelsey

Are you asking whether a DNA extraction method would yield microbial DNA as well as plant DNA? Could you edit your question and give more detail on the goal of the extraction, and the method of extraction?

@KarstenTheis yes that's exactly what I am wanting to know. I edited it in my question if that helps? Thanks!!

DNA contamination
DNA is water-soluble and non-volatile (large molecules). You would not expect it to be present in a butane extract. However, traces of DNA are difficult to remove. If you look for DNA (using PCR, for example), you will probably find it. After all, it is a forensic molecule of choice.
Contamination with living microbes
I don't think living cell would survive the extraction. Fungal spores might, though.
Human consumption
There are lots of legal foods where the presence of microbes is desired - yoghurt, beer, cheese are just some examples. All fresh food has its share of microbes, and healthy humans have them too - on the skin and in our digestive tract. Many of these microbes have a role in keeping us healthy, but some make us sick.
As the laws concerning your project differ with location, and the opinions on benefits and risks of your product range widely, I won't comment on those except to say that residual butane would concern me more than traces of microbial DNA.


text:
Why does the solubility of LiCl increase with temperature if the dissolution is exothermic?
In an answer to another question, the following data on the solubility of $\ce{LiCl}$ at different temperatures is presented.

This suggests that dissolving LiCl is endothermic (according to the van't Hoff relationship).  However, the enthalpy of dissolution has been measured and extrapolated to infinite dilution, and it is negative (the value given in one of the comments is $\Delta H = \pu{-37 kJ/mol}$, reference).
What is the reason for this discrepancy? Is it because $\ce{LiCl}$ is so soluble that there is some non-ideality effect? Maybe dissolving the first couple of molecules of $\ce{LiCl}$ is endothermic, but then when you dissolve more it becomes exothermic? Or is the activity of $\ce{LiCl}$ at increasing temperatures actually decreasing even though the concentration is increasing? Or is the van't Hoff relationship not a general one?
There is another question ($\ce{NaOH}$ solubility with temperature) which claims that Le Chatelier's principle sometimes does not apply, but it does not mention van't Hoff (and the answer might be incomplete).

Reference for enthalpy of solution of LiCl Enthalpy of solution of lithium chloride and of lithium chloride monohydrate in water

The key to the answer lies in this comment in the answer to the linked OP:

the dissolution of NaOH is exothermic, yet it's solubility increases with temperature.

So in short it's the opposite of #2: dissolving the first couple of molecules of $\ce{LiCl}$ is exothermic, but then if you dissolve more it become endothermic. The heat of dissolution to form an infinitely dilute solution is not the same as the heat associated with formation of a saturated solution.
Alternately you might say that the heat of dilution of $\ce{LiCl}$ is also exothermic (see e.g. Reference 1).

Reference

YC Wu, TF Young, Journal of Research of the National Bureau of Standards Vol. 85, No.1, January- February 1980, p. 11-17.

For NaOH, that is definitely the case. As the concentration increases, the pH increases and the exothermic $\ce{H+ + HO- -> H2O}$ reaction ceases, leaving only the endothermic $\ce{NaOH(s) -> Na+(aq) + HO-(aq)}$ reaction. Anyone who has made a concentrated NaOH solution can tell you that it gets quite cold if not continually heated. I don't know the details of the LiCl dissolution.


text:
All possible determinants forms a complete basis
How do we mathematically prove that the set of all possible excited determinants forms a complete basis for expanding the wavefunction (as in full configuration interaction)?

I think this is an interesting question, but it would benefit from your thoughts on the problem or what you have tried to arrive at a solution. @HBHSU

I have no idea how to prove that FCI is exact, but I am curious and cannot find the answer in my books or online.

A proof is given in section 2.2 of Szabo and Ostlund's Modern Quantum Chemistry. if you can't easily obtain a copy, I can summarize their argument here when I get a chance. @HBHSU


text:
Can a low concentration of an acid disolve solids just as well?
I'm in the process of removing some hard water stains with vinegar.  It's not working so well.
The stains are in a large fish tank, which I have filled with water and vinegar.  I would expect a lower concentration of acetic acid to cause the acid to dissolve the calcified material slower, but it doesn't seem to be having any effect.
Is there a threshold-like effect on an acid's ability to dissolve such solids, such that a lower concentration would simply be unable to dissolve them (rather than merely dissolving them slower)?

The rate of dissolution depends on $\mathrm{pH}$ and $\mathrm{pH}$ dependent equilibrium reaction chain :
$$\begin{align}
\ce{2 CH3COOH + CaCO3 v &<=> (CH3COO)2Ca + Ca(HCO3)2}\\
\ce{2 CH3COOH + Ca(HCO3)2 &<=> (CH3COO)2Ca + 2 H2CO3}\\
\ce{H2CO3 &<=>   H2O + CO2(aq)}\\
\ce{CO2(aq) &<=>   CO2(g)}\\
\end{align}$$
what can be written in the equivalent ionic form:
$$\begin{align}
\ce{CaCO3 v &<=> Ca^2+ + CO3^2-}\\
\ce{CH3COOH &<=> CH3COO-  + H+}\\
\ce{H+ + CO3^2- &<=> HCO3-}\\
\ce{HCO3-  + H+ &<=> H2CO3}\\
\ce{H2CO3 &<=>   H2O + CO2(aq)}\\
\ce{CO2(aq) &<=>   CO2(g)}\\
\end{align}$$
6% vinegar (1M) acetic acid has $\mathrm{pH=p}K_{\mathrm{a}}/2=2.37$.
Vinegar with 1/11 acid  spent has pH 3.75, with activity of $\ce{H2O+}$  24 times lower, so does is lower the reaction speed.
Vinegar with the acid half spent has pH 4.75, with activity of $\ce{H2O+}$ 240 times lower.
Vinegar with 10/11 acid  spent has pH 5.75, with activity of $\ce{H2O+}$  2400 times lower.
If pH raises enough, equilibrium concentration of $\ce{CO2(aq)}$ does not cross the gas solubility limit anymore and reaction approaches equilibrium, as production of gaseous carbon dioxide ceases..
If initial concentration crosses some minimal threshold, $\mathrm{pH}$ raises to level not able to produce gaseous carbon dioxide before water gets saturated by $\ce{CO2}$.
Some carbonate gets dissolved as bicarbonate, but with low initial acetic acid concentration, this dissolution is slow.
See also carbonic acid Wikipedia article.
Another effect is diffusion limitation. Without stirring, the acid near walls may get spent, while the acid abundance in free volume is intact, without access to limescale. Rate of acid diffusion to walls limits rate of dissolution.
Personally I would suggest rather citric acid, that have much higher acidity.
And/or, instead of filling the whole volume by diluted solution , rather using wall washing by concentrated solution.

Depending on your stain, you might want to consider a soap as well. As for your more theoretical question, yes there is a threshold. If the concentration of acetic acid is high enough, it will be less effective. However, it would have to be very high. This is described by Le Chatelier's principle