text:
Buffer Solution Problem (Acid/Conjugate Base)
To create a buffer mixture, I understand that mixing an equal concentration of a weak acid and it's salt (conjugate base) will form an acidic buffer.
However, why is it that when $\pu{25 cm^3, 9.100 M}$ $\ce{CH3COOH}$ and $\pu{50 cm^3, 0.100 M}$ $\ce{CH3COONa}$ can also form a buffer solution?
Is there any specific rule for the creation of buffer solution from weak acid and it's salt?

You need excess weak acid. Not an equal amount of both of them.

Welcome to ChemistrySE. Please feel free to take a [tour] of our site to know how our site works and what's on-topic here. Hope you get a satisfactory answer!

There is no such rule that you need equal concentrations of weak acid and salt of conjugate base to make a buffer. (In this case acidic buffer)
A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it.
In this case, if the solution contained equal molar concentrations of both the acid and the salt, it would have a $\mathrm{pH}$ of $\mathrm{4.76}$ because $\mathrm{pK_a}$ of acetic acid is $\mathrm{4.76}$. It doesn’t matter what the concentrations were, as long as they were the same, the $\mathrm{pH}$ remains at $\mathrm{4.76}$.
You can change the pH of the buffer solution by changing the ratio of salt to acid, or by taking a different acid and one of its salts.
Note: The $\mathrm{pH}$ of a buffer is calculated using Henderson-Haselbalch equation, which is,
$$\mathrm{pH = pK_a + log\frac{[Conjugate Base]}{[Acid]} }$$
Also, as indicated by @VicNeal in the comments, the buffer solution has highest buffer capacity at equal concentrations of the constituents, meaning its more effective at resisting the change in $\mathrm{pH}$.

The relationship between $\mathrm{pK_a}$ and $\mathrm{pH}$ of a weak acid, $\ce {HA}$, is mathematically represented by Henderson-Hasselbach equation where $\ce {A-}$ is its conjugate base (there is a equation for a weak base as well).
$$\mathrm{pH = pK_a + log\frac{[\ce {A-}]}{[\ce {HA}]}}$$
Thus, the $\mathrm{pH}$ of a buffer of weak acid can be calculated using above  equation, which gives close approximate value. As @MollyCooL pointed out, the highest buffer capacity is in effect when $\mathrm{pH}$ of the buffer is equal to $\mathrm{pK_a}$ of the acid, meaning,
 $\mathrm{log\frac{[\ce {A-}]}{[\ce {HA}]}= pH - pK_a =0}$. Hence, $\mathrm{\frac{[\ce {A-}]}{[\ce {HA}]}= 1}$ or $\mathrm{[\ce {A-}]=[\ce {HA}]}$. At this $\mathrm{pH}$, the effective buffer capacity is $\mathrm{pK_a \pm 1}$. Because of this reason, it is common practice with biochemists that they made bufer solutions at or about $\mathrm{pK_a}$ of the relavent acid or base. For example, if they need buffer of $\mathrm{pH}\space 4.5$, they use acetic acid/sodium acetate (acetic acid: $\mathrm{pK_a} = 4.76$), but use $\ce {KH2PO4/K2HPO4}$ system to make $\mathrm{pH}\space 7.0$ buffer (phosphoric acid: $\mathrm{pK_{a2}} = 7.2$). To understand this concept, see Plot 1 below:

This plot illustrates various protonation states of acetic acid as the pH changes. When $\mathrm{pH}$ is below $\mathrm{pK_a}$ of the acid ($\mathrm{pH \lt 3.5}$), most of the acid has been protonated. When $\mathrm{pH}$ is significantly above $\mathrm{pK_a}$ ($\mathrm{pH \gt 6.0}$), most of the acid is deprotonated. If the $\mathrm{pH}$ close of equal to the $\mathrm{pK_a}$, the acid is about $\mathrm 50\%$ protonated and $\mathrm 50\%$ deprotonated. That means, acetic acid is not suitable for the situations where sought $\mathrm{pH}$ is below $\mathrm 3.5$ or above $\mathrm 6.0$.
Now, let's do little quick and dirty calculations for your buffer solution:
$$\pu {Final acetic acid concentration = 9.100 \space M\times\frac{0.025 \space L}{0.075 \space L} = 3.033 \space M}$$
$$\pu {Final acetate concentration = 0.100 \space M\times\frac{0.050 \space L}{0.075 \space L} = 0.067 \space M}$$
Now, apply these two values to Henderson-Hasselbach equation:
$$\mathrm{pH = 4.76 + log\frac{[\ce {0.067}]}{[\ce {3.033}]}= 4.76-1.66=3.1}$$
Thus, the buffer system selected is not good choice for $\mathrm {pH} \space 3.1$ buffer. ascorbic acid (ascorbic acid: $\mathrm{pK_{a1}} = 4.10$) or phosphoric acid (phosphoric acid: $\mathrm{pK_{a1}} = 2.15$) would have been a better choice. Nonetheless, citric acid (citric acid: $\mathrm{pK_{a1}} = 3.15$) is the ideal one for this $\mathrm {pH}$.
The plot is from: https://bio.libretexts.org/LibreTexts/University_of_California_Davis/BIS_2A%3A_Introductory_Biology_(Easlon)/Readings/06.1%3A_pKa

Your image wasn't working. Is this the plot you were referring to . Images are recommend to be uploaded to stack.imgur since many times the image is deleted on the source site which renders your answer/question obsolete.


text:
Why does the E1cb mechanism proceed in the specific way it does?
We have been learning the E1cB mechanism in organic chemistry, but we have not been delving into the motivation as to why this mechanism occurs instead of others. Consider the following mechanism:

I am confused as to why we push the electrons to the left, creating an anion on the oxygen, instead of simply performing an E2 reaction on the right hand side. What is the functional reason for drawing this intermediate? Why does it arise?

In terms of the difference between the 3 main types of elimination (E) reactions, this earlier answer may be helpful.

-OH is a poor leaving group and so E1 and E2 are not favourable. In the case of E1, the formation of a primary or secondary carbocation is not stable either.
For E1cb, the intermediate enolate is more stable and since the removal of the -OH is not in the rate determining step, it is more favourable than E1.

In the two mechanisms, two bonds in the substrate are broken. In the E1 of t-butyl bromide, the C-Br bond breaks first to leave a tertiary carbocation. Secondly, a C-H  bond breaks to form isobutylene. In the E1cB one bond is also broken before the other one. In your example, a C-H bond breaks to form a carbanion, then the C-OH bond is broken afterward. In the E1 example, the C-Br bond is more labile. In the E1cB, it is the C-H bond.

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text:
Does Grignard reagent react with oxygen?
I thought that Grignard reagent might take a proton from water and just form ethane. Does the following reaction take place?
$$\ce{C2H5MgCl + O2 ->[H2O] C2H5OH}$$

This paper says it does and notes that while the yield of alcohols from aliphatic Grignards is 60-90%, formation of phenols is much less satisfactory. Other methods exist for the formation of phenols from aryl organometallics with much better yields.

To add on to the answer by Waylander (edited by Nilay Ghosh):
Waylander's reference [1] does not actually propose reacting the Grignard reagent with oxygen in the presence of water. Rather, the reaction with oxygen is carried out in "an ethereal solution", and the water is used to work up the product after the Grignard reagent is oxidized.
The reaction to produce an alcohol is actually not very practical (there are cheaper and more efficient ways to get alcohols), but by reacting under conditions where the oxygen is in excess (saturating the ether with oxygen and adding the Grignard reagent slowly), the reaction can be directed to form a hydroperoxide product.
Reference

Cheves Walling and Sheldon A. Buckler (1955). "The Reaction of Oxygen with Organometallic Compounds. A New Synthesis of Hydroperoxides". J. Am. Chem. Soc. 77, 22, 6032–6038. https://doi.org/10.1021/ja01627a069.

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text:
Do I really need to wear protective clothing while using isopropyl alcohol?
I recently purchased a 500ml bottle of 99% isopropyl alcohol (aka isopropanol aka rubbing alcohol) for everyday use at home. The label for the bottle worryingly states:

Wear protective gloves/protective clothing/eye protection/face protection

I'm surprised by this particular warning because I've always known of isopropanol being used for such everyday household appliances as cleaning and sterilising wounds, and have never known even the experts to wear or recommend wearing any sort of protective clothing while doing so.
The most I expect to handle the isopropanol is when pouring the solution from the 500ml bottle into smaller dropper bottles, and after that either using the 500ml bottle to wet cleaning cloths, or the dropper bottle pipettes to apply the solution directly to electronics.
Do I need to wear protective clothing for such applications? If not, at what volumes/for what applications would it be necessary to start wearing protective clothing?

Well, that's rather opinion based, and as far as my opinion goes, if you won't drink it instead, you should be OK even if you won't use any protective gear. Label needs to warn, otherwise producer could be sued.

You don’t want it in your eyes, that is for absolute sure. It will irritate your skin. Read the MSDS and act accordingly.

Actually, rubbing alcohol is 10% (maybe up to 20%) isopropanol, not 100%. Diluted isopropanol is less harmful (unless ingested) than pure isopropanol. If you are careful, one can even work with concentrated sulfuric acid without protective clothing (although I would not advise it).

@JonCuster Regarding the MSDS, I came across this answer here: https://chemistry.stackexchange.com/a/4607/61951, which is partly why I asked the question and included the volume of alcohol that I'm working with.

@LDC3 You're right that 99% isopropanol technically isn't rubbing alcohol, but it's far higher than 10% - Wikipedia says 70%. Although 99% is still commonly used in the electronics and computing industries as well as for sterilising wounds.

This is quite usual with warning and material safety data sheets. A non chemist or a non professional handler can overestimate them. But that is by purpose according to a principle of high caution. A lab coat is anyway always fast removed as compared to pullover or shirts, when it comes to fire hazard. Think about Tabasco sauce. If it would be sold as a chemical it would surely have warnings on the label.  Disclaimer: follow the bottle recommendations ;-)

You'd be surprised at what things you might not think of as "chemicals" can do.  One of my co-workers got a spritz of  oil in his eye while peeling an orange, and now he is afraid when I tell him that I eat orange rinds because they are rich in vitamin C.  The moral of the story: your eyes are so sensitive that at the very least, eye protection is a must with just about anything in the lab, even oils.  I second Jon Custer's suggestion of reading the MSDS.


text:
Determine the structure for C6H13Cl using the following data.
The H NMR for C6H13Cl states that there are 3 different types of protons:
6H, triplet
4H, quartet
3H, singlet
What will the structure look like?

Welcome to ChemistrySE. What are your own efforts to solve this question?

The molecule you're looking at is 3-chloro-3-methylpentane.
Symmetry causes H NMR to integrate 2x for the 6H and 4H signals.

Rats!  My cover is blown! +1 for noting the symmetry.

I was about to draw out this structure but I was overthinking it. Thanks so much :)

Where was Samantha's effort to solve the NMR problem?

Here's a useful hint:  when you have a triplet and a quartet with the area ratio given as above, it indicates the presence of ethyl groups.  Since the total area under those peaks is 10 H, there are two of these  groups equivalent to each other.  Only one isomer with the given formula does that.

Hoping you could have taken it the rest of the way, but my cover was blown.


text:
Describe the electrons defined by the following quantum numbers?
n=3, l=1, ml=2; and
n=4, l=2, ml=-1
I know that the first one is in the third energy shell and the second one is in the fourth energy shell because n=3 and n=4 respectively. I also know that when l=1 it should be in the p orbital and l=2 should be in the d orbital. The next part I am a bit confused about!
In the third energy shell, my textbook shows there is no ml=2. Am I missing something?
In the fourth energy shell, where ml=-1 in which of the three p orbitals should the electron be, that is the x, y, or z one?

the $m_l$ quantum number range from $\pm l$ in steps of $1$ so there seems to be a typo in the first question.


text:
What is required to achieve 999.9999 or finer gold?
After doing some research on gold purity, I found that the current record of gold fineness is at an astounding 999.999 fine achieved by the Perth Mint, Australia 1957. However, I could not find any other attempts to break this record.
I have also read the other related answers like this one: Obtaining 100% pure metal
I was hoping that perhaps some insight could be given as to how the current fineness record was achieved, and what refinement or fineness verification barriers may need to be overcome to achieve finer purities.

And to what purpose?

Ever heard of zone-refining?  http://chemistry-desk.blogspot.co.uk/2012/08/zone-refining.html

My ion accelerator makes pure gold all the time - if an ion makes it down the beam line into the end station, it has mass 197 and charge 1. Now, getting a macroscopic amount takes something on the order of the Y-12 plant...

There are many approaches that can be used to purify an element such as gold. Gold has a different melting point than any likely contaminant so zone refining can be used.
Imagine a bar or rod of gold with some contaminant with a lower melting point than gold, melt one end of the bar then move the heated and molten zone the length of the bar, the atoms of the contaminant will be more likely to remain molten and not solidify, thus the contaminants with a lower melting point will move slightly towards the end that was molten last. If the procedure is repeated, then the gold in the bar will become more and more pure.
Once all the lower melting point impurities have been reduced in concentration to an acceptable level, then the process is continued to move the gold away from the higher melting impurities and a section of the bar is selected to optimise the reduction in impurities.
Zone refining was first used commercially for semiconductor manufacturing; germanium was purified to 999.999999 purity, using the notation in the original question. I'm unsure of the utility of purifying gold to the level described above, apart from the 'we can do this' factor. You could possibly use a centrifuge and purify gold (III) fluoride in the gaseous state, although materials of construction would be a problem; it is very corrosive. There would also be hydrometallurgical techniques, but I don't know any chelating agents specific for gold.

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text:
Can a zeroth order reaction be reversible?
After recently studying about chemical equilibrium, I was convinced that the forward and backward reaction rates meet each other at equilibrium.
However thinking about zeroth order reaction annoyed me a little, as I could not figure out how the rates would become equal, if at all, in cases where the rate remains static throughout. Here's what I guessed:

The backward reaction of a zeroth order reaction must not be a zeroth order reaction in itself, so that it may reach the reaction rate equivalence point at equilibrium.But then this generalization is difficult to digest (for me).

or

A zeroth order reaction can not attain equilibrium (highly unlikely).

Well, some of these reactions can't attain equilibrium, and some do it according to your first argument.

A zeroth order reaction does not achieve equilibrium as a zeroth order reaction.  Instead the kinetics will go away from zeroth order when we get close to the equilibrium condition.
Zeroth order kinetics is an approximation we get to when some aspect of the reaction approaches a saturation condition, for example the reaction mechanism may involve adsorption on a surface and there is enough reactant to saturate all the adsorption sites.  Such conditions are perforce relatively far from equilibrium; "saturation" and "reversibility" do not come about simultaneously.

If you ever want a nice, intuitive proof that zeroth-order reactions can only be a (concentration-dependent!) approximation, try imagining the reaction rate when the reagent concentration is zero!


text:
What is the Lewis dot diagram for molybdenum and cerium?
How do I draw a Lewis diagram for $\ce{Mo}$ and $\ce{Ce}$?
 I am a bit confused since they are transition metals.
Attempt
The outer electron configuration for $\ce{Mo}$ is $\ce{5s^1 4d^5}$. The number of valence electrons would be 6 if I am correct. Wouldn't the Lewis dot structure look like the one for group 6A elements with six dots?
EDIT: I spoke to my lecturer, he said I have to do research and draw the Lewis dot diagram of Mo and Ce. My textbook says that transition metals have incompletely filled inner shells, and that generally we cannot write simple Lewis dot symbols for them. Yet, my lecturer said it would be similar to other elements. I am very confused!

From what I gather from the definition of Lewis dot structures from Wikipedia they exist for bonded atoms or compounds. Not for single atoms. Do you mean the electronic configuration instead?

I spoke to my lecturer, he said I have to do research and draw the Lewis dot diagram of Mo and Ce. My textbook says that transition metals have incompletely filled inner shells, and that generally we cannot write simple Lewis dot symbols for them. Yet, my lecturer said it would be similar to other elements. I am very confused!


text:
Will the electrolyte surface become charged if one electrode is raised out of solution?
What would the potential distribution be if I were to immerse a single inert (graphite) electrode in an aqueous solution with good ionic conductivity and hold it at a high (ca. $\pu{24V}$) potential vs. an electrode that would remain in the air above the solution's surface?
I don't expect anything to happen at the immersed electrode interface because the solution should remain equipotential.
But, what about the air-water interface? Would the electric field induce a surface charge?

But, what about the air-water interface? Would the electric field induce a surface charge?

Yes.  Basically, by locating one of your electrodes out of the solution, you've assembled an air-gap capacitor, though a fairly inefficient/low-quality one.
Per that above link, typical metal-plate "[a]ir capacitors have a small capacitance which usually lies between $\pu{100pF}$ and $\pu{1nF}$". High-level, the definition of capacitance is:
$$
C = {q\over V} \quad\quad\Longrightarrow\quad\quad q = CV
$$
Assuming that this water/electrode air capacitor has similar properties to the above-referenced metal-plate air-gap capacitor, the charge accumulated at the electrolyte surface under a $\pu{24V}$ applied potential should be between $2.4$ and $\pu{24nC}$. A similar amount of charge, but of opposite sign, should of course accumulate at the surface of the elevated electrode.

I don't expect anything to happen at the immersed electrode interface because the solution should remain equipotential.

This will be true at steady-state, as the entirety of the potential drop is going to be supported by the air gap.
However, you will have some minuscule amount of transient electrochemical activity at your submerged electrode right when you turn on your power supply, as the system adjusts from the $\pu{0V}$ to the $\pu{24V}$ applied voltage.

Thank you! Seeing it as an air capacitor makes complete sense.


text:
Is there any book as Landau's for Physics but for Chemistry?
Landau's series of books are a collection of books that describe briefly but very deeply the core part of the main fields of Physics. He wrote them rigorously and they are very very abstract. He gave Physics new approaches in those books. If you want to understand them, your starting point should be knowing all those main fields of Physics previously.
Is there any book like it but in Chemistry?
Thank you all!

Groups Theory by Cotton, Szabo for Q mechanics, I don't know a nice book for statistical thermodynamics...maybe Terrell Hill intro to statist thermodynamics, although I don't like it..

Related: https://chemistry.stackexchange.com/q/37303/31775

Feynman's introduction to his "Lectures on Physics" gives a hint why there cannot be such a book for chemistry:

‘This is one of the most fantastic pieces of detective work that has ever been done – organic chemistry… The physicist could never quite believe that the chemist knew what he was talking about when he described the arrangement of the atoms. For about twenty years it has been possible, in some cases, to look at such molecules by a physical method, and it has been possible to locate every atom, not by looking at colors, but by measuring where they are. And lo and behold!, the chemists are almost always correct.’

Possible duplicate of Resources for learning Chemistry

Chemistry is as much an artisanry, a handicraft as it is science. Pure rigour get's you nowhere, except for the development of fundamental understanding in physical chemistry (statistical mechanics, QM, etc.).


text:
Where i am wrong using these reagents of lowrys protien estimation?
Reagent A : 2g Na2CO3 and 0.1N NaoH
B :1% copper sulphate
C : 2% sodium potassium tartarate
Mixing them into 98:1:1 ratio.

I'm sure you found this method somewhere in the literature but it hasn't give full description of reagents used in Lowry Protein Assay, modified or not. For example, your reagent A should have been $\mathrm 2\%$ $\ce {Na2CO3}%$ in $\pu {0.10 N} \space \ce {NaOH}$ solution, according to Lowry's original paper (Reference [1]). Thus, for the benefits of future readers, I'd include here two updated protein assay methods suggested by RUF at Rice University, Houston, TX (Reference [2]). Their recommendations are as follows:
Considerations for use of Hartree-Lowry or Modified Lowry Protein Assays
The Lowry assay (see Reference [1]) is an often-cited and generally used protein assay since 1951 (claimed to be "The Most Highly Cited Paper in Publishing History"). For sometimes, it was the method of choice for accurate protein determination for cell fractions, chromatography fractions, enzyme preparations, etc. until the introduction of the Bicinchoninic Acid (BCA) Assay or "Smith Assay" in 1985 (Reference [3]). BCA Assay is based on the same principle, but can be done in one step, therefore it has been suggested  that the 2-step Lowry method is outdated (Reference [4]). However, the Lowry method has now been modified to be done entirely at room temperature. The Hartree version of the Lowry assay, a more recent modification that uses fewer reagents, improves the sensitivity with some proteins, and is less likely to be incompatible with some salt solutions. It provides a more linear response, and is also less likely to become saturated.
Standard Procedure for Hartree-Lowry Protein Assays ([Reference [5])
Reagents: Reagent A consists of $\pu {2 g}$ of sodium potassium tartrate tetrahydrate, $\pu {100 g}$ of $\ce {Na2CO3}$, $\pu {500 mL}$ of $\mathrm {1N}$  $\ce {NaOH}$, diluted to one liter with deionized water (final concentrations: $\mathrm {7mM}$ sodium potassium tartrate, $\mathrm {0.81M}$  $\ce {Na2CO3}$, & $\mathrm {0.5N}$  $\ce {NaOH}$). This can keep for 2 to 3 months.
Reagent B consists of $\pu {2 g}$ of sodium potassium tartrate tetrahydrate, $\pu {1 g}$ of $\ce {CuSO4.5H2O}$, $\pu {90 mL}$ of $\ce {H2O}$, $\pu {10 mL}$ of $\mathrm {1N}$ $\ce {NaOH}$ (final concentrations: $\mathrm {70mM}$ sodium potassium tartrate, $\mathrm {40mM}$ $\ce {CuSO4}$, & $\mathrm {0.1N}$  $\ce {NaOH}$). This can keep up to 2 to 3 months as well.
Reagent C consists of $\mathrm 1$ volume of Folin-Ciocalteau reagent diluted with $\mathrm 15$ volumes of ultrapure water.
Assay: First, as standards, prepare a series of dilutions of $\pu {0.3 mg/mL}$ bovine serum albumin (BSA) using the same buffer, which has been used to prepare solutions with protein-containing unknowns. Do the dilution to give concentrations of $\pu {30}$ to $\pu {150 \mu g/mL}$ ($\pu {0.03}$ to $\pu {0.15 mg/mL}$) of BSA in the buffer.
Add $\pu {0.90 mL}$ of Reagent A to each test tube containing $\pu {1.0 mL}$ of each of your series of standards, your protein-containing unknowns, and buffer solution (for the reference). Then, vortex all test tubes to mix. Incubate the tubes for $\pu {10 min}$ in a $\pu {50~^oC}$ bath, then cool to room temperature.
Add $\pu {0.1 mL}$ Reagent B to each tube, vortex to mix, and incubate for $\pu {10 min}$ at room temperature.
Rapidly add $\pu {3 mL}$ of Reagent C to each tube, vortex to mix, incubate again for $\pu {10 min}$ in the $\pu {50~^oC}$ bath, and cool to room temperature. Final assay volume should be $\pu {5 mL}$. Now, measure absorbance at $\pu {650 nm}$ in glass or polystyrene cuvettes (pathlength: $\pu {1cm}$).
Analysis: Prepare a standard (calibration) curve of absorbance versus $\pu {\mu g}$s of protein (or vice versa) using BSA readings, and determine protein amounts in your unknowns from the curve. Then, you can determine concentrations of your original samples from the amount of protein determined, by applying all the factors including volume/sample, dilution, and other factors, if any.
Procedure for Modified Lowry Protein Assays (room temperature) ([Reference [6])
Reagents: Dissolve $\pu {20 g}$ of $\ce {Na2CO3}$ in $\pu {260 mL}$ of $\ce {H2O}$, $\pu {0.4 g}$ of $\ce {CuSO4.5H2O}$ in $\pu {20 mL}$ of $\ce {H2O}$, and $\pu {0.2 g}$ of sodium potassium tartrate in $\pu {20 mL}$ of $\ce {H2O}$. Mix all three solutions to prepare the copper reagent.
Prepare $\pu {100 mL}$ of a $\mathrm 1\%$ solution ($\pu {1 g}$ in $\pu {100 mL}$) of sodium dodecyl sulfate (SDS).
Also prepare a $\mathrm {1M}$ solution of $\ce {NaOH}$ ($\pu {4 g}$ of $\ce {NaOH}$ in $\pu {100 mL}$ of $\ce {H2O}$).
To make the $\mathrm {2\times}$ Lowry concentrate, mix $\mathrm {3}$ parts copper reagent with $\mathrm {1}$ part SDS and $\mathrm {1}$ part $\mathrm {1M}$ $\ce {NaOH}$. This final solution is stable for 2-3 weeks. Warm the solution to $\pu {50~^oC}$ if a white precipitate forms, but discard it if there is a black precipitate. It is better to keep the three stock solutions separately, and mix them just before use (recommended).
Prepare $\mathrm {0.2N}$ Folin reagent by mixing $\pu {10 mL}$ of $\mathrm {2N}$ Folin reagent with $\pu {90 mL}$ of ultrapure water. Kept $\mathrm {0.2N}$ Folin reagent in an amber bottle (the dilution is stable for several months).
Assay: Dilute protein-containing unknown samples with buffer to give concentrations from $\pu {0.025}$ to $\pu {0.25 mg/ml}$ if it can be estimated. If the concentration can't be estimated it is advisable to prepare a range of 2-3 dilutions spanning an order of magnitude. Prepare $\pu {200 \mu L}$ each dilution. Duplicate or triplicate samples are recommended.
Prepare a reference of $\pu {400 \mu L}$ buffer. Prepare standards for calibration curve using $\pu {0.25 mg/ml}$ bovine serum albumin (BSA) solution. Prepare concentration series of standards by adding BSA solution to $\pu{13 \times 100 mm}$ tubes (BSA volumes ranging from $\pu {40}$ to $\pu {400 \mu L}$) and bring up volume of each tube to $\pu {400 \mu L}$ by adding the appropriate amounts of buffer solution.
Now, add $\pu {400 \mu L}$ of $\mathrm {2\times}$ Lowry concentrate to each and every test tube including reference buffer, mix thoroughly, incubate at room temperature for $\pu {10 min}$.
Add $\pu {200 \mu L}$ of $\mathrm {0.2N}$ Folin reagent very quickly, and vortex immediately. Complete mixing of the reagent must be accomplished quickly to avoid decomposition of the reagent before it reacts with protein. Incubate for $\pu {30 min}$ more at room temperature (Final volume: $\pu {1.0 mL}$).
Use glass or polystyrene cuvettes (pathlength: $\pu {1cm}$) to read the absorbance at $\pu {750 nm}$. If the absorbance is too high, they may be read at $\pu {500 nm}$.
Comments: Recording of absorbance needs only to be done within $\pu {10 min}$ of each other for this modified procedure whereas the original Lowry required precise timing of readings due to color instability. This modification is less sensitive to interfering agents and is more sensitive to protein than the original. As with most assays, the modified Lowry assay can be scaled up for larger cuvette sizes, however more protein will be consumed. Be aware that proteins with an abnormally high or low percentage of tyrosine, tryptophan, or cysteine residues will give high or low errors, respectively.
References:
[1] Protein measurements with the Folin phenol reagent: Oliver H. Lowry,  Nira J. Rosebrough, A. Lewis Farr, Rose J. Randall, Journal of Biological Chemistry, 1951, 193, 265-275 (http://www.jbc.org/content/193/1/265.full.pdf).
[2] Hartree-Lowry and Modified Lowry Protein Assays by Reformed University Fellowship (http://www.ruf.rice.edu/~bioslabs/methods/protein/lowry.html)
[3] Measurement of protein using bicinchoninic acid: P. K. Smith, R. I. Krohn, G. T. Hermanson, A. K. Mallia, F. H. Gartner, M. D. Provenzano, E. K. Fujimoto, N. M. Goeke, B. J. Olson, D. C. Klenk,  Analytical Biochem., 1985, 150(1), 76–85 (doi:10.1016/0003-2697(85)90442-7).
[4] Chapter 6. Quantitation of Protein: Christa M. Stoscheck, In Methods in Enzymology; Murray P. Deutscher, Ed., 1990, 182, 50-68 (https://doi.org/10.1016/0076-6879(90)82008-P).
[5] Determination of protein: A modification of the lowry method that gives a linear photometric response: E. F. Hartree, Analytical Biochem., 1972, 48(2), 442–427 (https://doi.org/10.1016/0003-2697(72)90094-2).
[6] Optimization of Folin-Ciocalteu reagent concentration in an automated Lowry protein assay: G. M. Oosta, , N. S. Mathewson, and G. N. Catravas, Analytical Biochem., 1978, 89(1), 31–34 (https://doi.org/10.1016/0003-2697(78)90723-6).


text:
Why does my sodium acetate look like metalic magma?
I tried to make sodium acetate.
Ingredients:

Store bought vinegar - 8% acetic acid, water, E150C  (caramel)
Baking soda - $\ce{NaHCO_3}$

Steps:

I first emptied sacks of baking soda to a container
I kept adding vinegar until the resulting fuming was mild
I started heating the solution slowly, I kept adding more vinegar as long as I could see bubbly reaction
I increased heat until all water evaporated. I was left with white-brown chunks of (presumably) sodium acetate and the impurities from vinegar
I kept heating this until it melted into opaque black liquid
I poured the hot molten stuff into ceramic bowl

This is my result:

I might upload better picture tomorrow, when there's more light available.
So, this does not really look as sodium acetate at all. It looks like a volcano. Clearly there's more volcano-ish experiments with vinegar than I thought.
How do I purify this? Is there even any sodium acetate? Could I burn (decompose) it or something?

You don't want to purify this.

Could you elaborate?

"E150C (caramel)" made it failure from the start. You'd need to distil it to have a chance.

And maybe remove this picture, my poor eyes! ;)

Can anyone explain the downvote so that I can improve the question?

Question isn't bad, but you had pretty much no idea what happened. TL:DR You used vinegar like it was chemical grade not food grade. Caramel made your product a black unholy abomination of science ;)

Yeah, I had no idea, that's why I asked question, hoping to figure out what went wrong exactly. But yeah I understand now - it's not the first downvote I got for having absolutely no clue about the topic I'm asking about.

I wouldn't bother trying to purify it. As Mithoron said in his legendary comment "Caramel made your product a black unholy abomination of science". I would have distilled the vinegar and diluted it (because concentrated acetic acid is a bit dangerous if you don't know what you're doing) and start again. If you want pure product, make sure your starting materials aren't filled with "unholy abomination of science" causing impurities. If you really want to purify it, maybe recrystallise it in hot water? Or a solvent that sodium acetate can become saturated in.

I did dissolve it and filter it and it looks better. The thing is I made sodium acetate with the purpose of making concentrated acetic acid through redox reactions. So if I just distill the vinegar, it kinda ruins the purpose of the experiment - I could just as well buy the acid from some e-shop.

Good point. Has the caramel colour disappeared?

It was reduced, but when I dried it again, it is still pretty black.

Have you tried dissolving the sodium acetate in alcohol?

That's a great idea! Will try it!


text:
Dissociation energy of benzene dimer
It is known that the benzene dimer has two stable structures, T-shaped ($\ce{C_{2v}}$) or parallel-displaced ($\ce{C_{2h}}$) structure.
Now the question is: does the dissociation energy ($\ce{D_0}$) $$\ce{(C_6H_6)_2 -> C_6H_6 + C_6H_6}$$ depends on the structures?
Or it is independent of the structures?

Of course it has to depend, at least a bit...

Of the three configurations mentioned by @rbw  the parallel-displaced is most likely to be the most stable.; see https://chemistry.stackexchange.com/questions/60055/pi-electron-stacking-how-does-it-work/60095#60095 and you can calculate these for yourself.

In fact you have three configurations to consider: t-shaped, parallel-displaced and sandwich. The dissociation energy depends on the configurations of the structures since they may interact differently. As the pi-pi interaction is directional, the different structures of the dimer interact differently.

I think it would best if you elaborate a bit. Don't leave answers that require clarification from the asker.


text:
The structure of sodium cuprous thiosulphate
While studying about copper sulphate, I came across a reaction involving sodium thiosulphate (Hypo) and copper sulphate.
Although I don't know the utility of this reaction, I felt that it was important to understand the structure of $\ce{Na4[Cu6(S2O3)5]}$, which is formed as a the major product of the overall reaction.
The full set of reactions is given as follows:
$$
\begin{align}
\ce{CuSO4 + Na2S2O3 &-> CuS2O3 + Na2SO4}\\
\ce{2CuS2O3 + Na2S2O3 &-> Cu2S2O3 + Na2S4O6}\\
\ce{3CuS2O3 + 2Na2S2O3 &-> Na4[Cu6(S2O3)5]}
\end{align}
$$

What is the structure (3D, not Lewis structure) of $\ce{Na4[Cu6(S2O3)5]}$ (or the ion $\ce{[Cu6(S2O3)5]^{-4}}$)?

Reference: Allen JEE-Mains+Advanced Chemistry, Unit 10, Chapter: Transition Elements:

https://image.isu.pub/111221092114-db19be06b02a46f5ad076d7beb294282/jpg/page_46.jpg and https://books.google.co.in/books?id=6yhNuZqLF5AC&pg=SA3-PA111&lpg=SA3-PA111&dq=na4%5Bcu6(s2o3)5%5D&source=bl&ots=-J8xxtssa9&sig=dLzuGQQKMC2eu2CE3QQoftz4XRM&hl=en&sa=X&ved=2ahUKEwis9pGtubHaAhWLvo8KHbhCAUsQ6AEwBHoECAMQAQ#v=onepage&q=na4%5Bcu6(s2o3)5%5D&f=false  Here's something else I found...

Pretty sure it is a polymeric structure with no discreet complex anionsm kinda like many silicates.

For knowing these three dimensional complex structures, which are hard to predict theoretically, we have to use X-Ray diffraction or spectroscopic techniques (IR, NMR, or, mass). I don't have any such facilities to verify the complex structure, but I may logically predict a  logical (possibly incorrect) structure.
In the complex ion $\ce{[Cu_6(S_2O_3)_5]^4-}$, all the $\ce{Cu}$  is in +1 oxidation state, and thus can form a tetrahedral geometry around it with the help of other $\ce{Cu^+}$ ions and $\ce{S_2O_3^2-}$ ions. On the other hand,  thiosulfate can act as both a bidentate and a monodentate ligand.
So, in the predicted structure, all the $\ce{Cu^+}$ are in tetrahedral geometry. Three $\ce{S_2O_3^2-}$ act as bidentate ligands, whereas the other two $\ce{S_2O_3^2-}$ ions act as monodentate ligands.
Note that the donor atom of the thiosulfate anion can be either all $\ce{S}$, all $\ce{O}$, or a combination of $\ce{S}$ and $\ce{O}$. It is indeed hard to predict (because in some complexes, $\ce{Cu}$ binds with $\ce{O}$ regardless of being soft acid to preferably bind with $\ce{S}$ - as in $\ce{CuCl_2.4dmso}$) and thus is best to determine through IR spectroscopy by looking at $\ce{S-S}$ stretching frequency. Thus, the donor atoms may be different from what is shown in the possible structure.
The possible structure is:

In that case, two $Cu^+$  will be tri-coordinated, which is very unlikely to happen. and , more Importantly, This is a possible Structure.

Has no spectroscopy or X-RAY diffraction ever been done for this compound? Is it really that uncommon?

Spent quite a number of hours spread over days, to end up with no reference of the complex
Could someone please refer to any related primary or secondary reference of the complex...


text:
When to disconnect tubing in vacuum filtration?
The UCLA site says in step 6 to disconnect the vacuum tubing before turning off the vacuum and the second link from Leeds says in step 11 to turn off the vacuum source first and then disconnect the tubing.
I've been taught both ways and am confused about the difference in methods.
When should one actually disconnect the tubing and what is the rationale for each method?

First disconnect the tubing else the vacuum in the erlenmayer flask can suck in water (if you use a water pump). This can be tedious even if your filtered through phase has to be discharged (e.g. solvent wastes separation). If your target is the liquid, than is detrimental.

What about a regular vacuum pump?

I would do the same but with a normal setup nothing shall happen. It depends on pump,  what was used before.... With a line or a pump there should be a trap in between vacuum and erlenmayer. ...

If you were to turn off the pump while leaving the object attached then if you have an oil filled pump or a water pump then you will have a suck back which will either damage the pump or contaminate the thing you are filtering, or if you are very unlucky both.
If you have a house vacuum system then you should close the valve / tap between your filtration and the vacuum manifold before you remove the hose to your flask. If you do it the other way around then you will interfere with the use of the house vacuum by others.
I think you need to know if you have your own vacuum pump or if you are using a house vacuum system where a centralised pump serves many people.


text:
What is the relation between absolute temperature, and rates of diffusion and effusion of a gas?
According to Graham's diffusion law, rate of diffusion is directly proportional to square root of temperature but rate of effusion is inversely proportional to square root of temperature.
Why is this so? I would assume that rate of effusion would also increase with increase in temperature due to increased speed of the gas.

Who said it is inversely proportional to $\sqrt T$?

@IvanNeretin Actually, it is inversely proportional to $\sqrt{T}$. See here: https://en.m.wikipedia.org/wiki/Effusion under measures of flow rate.

@AbhigyanC That depends on your point of view. Say, you measured it in a system of two connected constant-volume vessels. Then you increase T. Pressure in both vessels will increase proportionally to T, and so will $\Delta P$. Also, the denominator will grow as $\sqrt T$. Guess what will happen to the overall expression?

The self diffusion coefficient in a gas is given by $\displaystyle D=\frac{\lambda}{2}\bar c$ and the rate of effusion by $\displaystyle \frac{\rho}{4}\bar c$ where $\bar c$ is the average velocity of the gas molecules of mass $m$, and $\displaystyle \bar c= \sqrt{\frac{8k_BT}{\pi m}}  $, $\lambda$ the mean free path and $\rho$ the gas density. The diffusion constant is $\displaystyle D=\frac{1}{\pi \sigma^2 n}\sqrt{\frac{k_BT}{\pi m}}$ where $\sigma$ is the molecule's diameter, and the rate of effusion $\displaystyle \rho \sqrt{\frac{k_BT}{2\pi m}}$. The diffusion constant can also be written as $\displaystyle D=\frac{m}{\pi \sigma^2 \rho }\sqrt{\frac{k_BT}{\pi m}}$ thus the dependence on temperature is the same for diffusion and effusion if the gas density is held constant.
[ The rate of effusion is calculated by  considering the gas bouncing around in a container and then suddenly opening a tiny trap door of area $dA$ in the wall so that some minute proportion of the gas streams out in time $dt$. The trap door diameter should be no more than 1/10 of the mean free path.   If each molecule travels with average speed $\bar c$ the number leaving would be $n\bar c dA dt$ if there are $n$ molecules /unit volume. The efflux in terms of the mass/unit area/unit time is then $mn\bar c =\rho \bar c$.
However, half the molecules are moving away from the trap door at this instant so the rate is divided by 2. Additionally molecules come out at different angles so it is necessary to integrate over the volume element $\sin(\theta)d\theta d\phi$ and this leads to a further factor of 1/2.]

I just have a simple answer for this.
Rate of effusion is the rate at which a gas is allowed to escape from a given cross sectional area(orifice) due to pressure difference.
While diffusion is the tendency of gases to mix with each other spontaneously and form a homogeneous mixture.
In case of diffusion, increasing temperature increases the rate of diffusion because it increases the kinetic energy of molecules which increases the mixing ability of the gases and thus, diffusion.
In effusion the same thing happens, kinetic energy of molecules increases, which increases the speed of the molecules. But Increase in speed of the molecules decreases the probability of the molecules to escape through the given orifice/hole in a given time.
Therefore, rate of effusion decreases with increase in temperature.
Thanks:)


text:
What to say 'one mole of C atoms or one mole of C molecules?
What to say 'one mole of C atoms or one mole of C molecules in CO? Please point out if I am wrong here
As we know,
one molecule of CO = one mole of CO molecules = 6.223 X 10^23 number of CO molecules
If I split one molecule of CO, we get one C and one O. Here are my questions, What to say these ''one C and one O" one mole of C atoms and one mole of O atoms or ''one mole of C molecules and one mole of O molecules'' or just like above like ''one C and one O''?
Could you clarify it, please? I am confused

C molecules is not a thing at all.

What you need to remember when using terminology involving moles is that moles are just a way of counting things but that the terminology only makes sense if you unambiguously specify the thing you are counting. It is entirely possible to use moles to count non chemical things (for an amusing example of the consequences of having a mole of moles (the small burrowing mammals) see this.) Most confusion arises because the way the thing is specified is ambiguous.
Some examples:

a mole of CO molecules is perfectly correct (the units being counted are single molecules of CO)
A mole of C atoms is perfectly correct (and, if you are counting C atoms in CO molecules that is also fine and the answer will be the same)
a mole of C molecules in CO isn't correct but only because the molecules is CO which consists of atoms of C and O.
if you were counting the atoms in CO2, then it would be correct to say there are twice as many moles of oxygen atoms than of carbon atoms or carbon dioxide molecules

The use of moles to count doesn't imply anything about what you are counting, you have to specify that as clearly as possible. When talking about the parts of a molecule we usually talk about the atoms that make up the molecule. CO2 doesn't consist of molecules of oxygen and carbon, but atoms of oxygen and carbon. But this is independent of the use of moles as a way of counting those components.

One non-standard use of mole that I find convenient is that (according to some estimates), the number of stars in the visible universe is about $\frac{1}{10} mole$.  http://www.skyandtelescope.com/astronomy-resources/how-many-stars-are-there/


text:
How to compare standard entropy of diatomic gases?
Say, for example, I wanted to compare the standard entropies of $\ce{Cl2(g)}$ and $\ce{F2(g)}$. They both are equimolar.
I understand that entropy is a measure of how many states a particle can occupy (loosely speaking), so I tend to think the larger chlorine molecule has higher standard entropies. This is backed by experimental data, but why?

You can calculate the entropy exactly using the partition function from statistical mechanics for rotations, vibration and any electronic component plus the Sakur-Tetrode equation for the translational entropy $S_T$.  This depends on the mass and the partition functions depends on vibrational frequency and moment of inertial. If $Z$ is the partition function  then S=R(\ln(Z)+Td\ln(Z)/dT). The partition function is $Z=\sum_i g_iE^{E_i/k_BT}$ for levels with energy $E_i$ and degeneracy $g$. Phys Chem text book give formulas for calculating $Z$. The total entropy $=S_{vib}+S_{rot}+S_{elec}+S_T$


text:
Apparent contradiction in the type of deviation of an ethanol-water mixture
For a mixture of $\ce{EtOH}$ and $\ce{H2O}$: $\Delta H_{\text{mix}} < 0$ (exothermic) and $\Delta V_{\text{mix}} < 0$ (volume contraction). This means that the unlike interactions are stronger than the like interactions, because the mixture exhibits a lower enthalpy and volume than that would be expected for ideal mixing.
Also, $\ce{EtOH}$ and $\ce{H2O}$ form a minimum boiling azeotrope and the $T-\chi$ graph is below the Raoult's law $T-\chi$ plot. This means that the pair exhibit positive deviation from Raoult's law. This implies that the mixture is more volatile than the pure components. Thus, like interactions should be stronger than unlike interactions.
I am not able to resolve this contradiction. Does the mixture actually exhibit positive deviation? Which interactions are stronger - unlike or like?

Look for the T dependence of Gmix. It should be positive at about  azeotrope boiling point.

Water-ethanol is a v complicated system. The partial molar volume of ethanol decreases then increases as ethanol mole fraction increases. That of water does the opposite in accordance with Gibbs-Duhem eqn. Entropy changes are also likely to be v important as hydrogen bonding plays such a large part in these mixtures.  For example EtOH/water mixtures of the same viscosity but different composition have very different properties.

But how can it exhibit positive deviation, but still have Vmix and Hmix negative? It is contradictory.

Those are all T dependent. Check for that as it seems the only explanation.

As suggested in the comments, there is an entropic contribution to the excess free energy which should not be overlooked, as it is the excess Gibbs free energy that determines whether the deviation from Raoult's law is positive or negative. All of the statements made by the OP are correct, except that negative $\Delta H$ (exothermicity) and $\Delta V$ (volume contraction) do not necessarily imply a reduced volatility (negative deviation from Raoult's law), that is they are not sufficient conditions for a negative deviation, a negative excess Gibbs free energy instead being the sufficient condition. Again, as alluded to in the comments, a simple analysis based entirely on $\Delta H$ is not appropriate here, largely because of the hydrogen bonding (associative) nature of the liquids. In fact, the negative mixing volume and enthalpy might partly explain the negative excess entropy (and vice-verse: they are correlated). The excess entropy of mixing of water and ethanol is highly negative over the entire range of composition at room temperature, leading to a net positive excess Gibbs free energy.

Does the mixture actually exhibit positive deviation? Which interactions are stronger - unlike or like?

Therefore the mixture does exhibit a positive deviation from Raoult's law, even as interaction enthalpies between unlike molecules are stronger.

I know I'm really late to the party, but I was interested in this as well.  Essentially, it appears that water and ethanol form a solution that has some structure to it, not unlike micellar aggregation.  The OH end of the ethanol molecules aggregate with each other and the polar water, which creates a packing structure that has negative excess volume.  However, these aggregated structures still have a bunch of hydrophobic tails sticking out that interact unfavorably with water, resulting in the increased fugacity (effective vapor pressure) associated with positive deviation from Raoult's Law.


text:
Magnetic non-equivalence of acetal protons
I have been using bromoacetaldehyde diethyl acetal recently and noticed something interesting about the 1H NMR. The methylene protons of the ethyl acetal are magnetically non equivalent. You see two quartets integrating for two protons with different shifts.
You can see this in the spectra here.
I have found a short description, however it is a little unclear:

Diethyl acetals such as (1 6) and triglycerides such as triacetin (1 7) have a molecular symmetry plane and consequently are optically
  inactive. However, this plane does not bisect the connecting line of
  the geminal protons, which are magnetically non equivalent. This is
  manifested by the relative complexity of the AA'BB'X type proton
  spectrum of the glyceryl moiety of triacetin (Figure). Similar
  magnetic non-equivalence can be induced across an ester bond; e.g.,
  the two methyl groups of the isopropyl ester (18) area non-equivalent.

I don't really understand as there is no restriction of rotation. And does this mean that this would disappear using high temperature NMR?

This issue has been addressed. The methylene hydrogens on the acetal ethyl groups are diastereotopic, i.e., different. https://chemistry.stackexchange.com/questions/86205/ch2-group-in-alkyl-glucoside-gives-two-separate-hnmr-signals/86451#86451

Sorry! Didn't see this when I searched

See ADDENDUM in my answer.

This link on SE explains why the methylene hydrogens of the ethyl groups are different or diastereotopic. For your specific case, if a deuterium is substituted for Ha on the left side of the structure, it leads to an R-configuration while substitution of deuterium at Hb leads to an S-configuration. Both of these hydrogen sites "see" the same handedness at the methine carbon. Thus, the two hydrogens, Ha and Hb, are in different environments, they have different chemical shifts and they are coupled to each other and the vicinal methyl group. The same argument applies to the ethyl group on the right side of the structure. This difference in chemical shift is more apparent at high field (e.g., 300 MHz spectrum) rather than at low field (60 MHz).
There is no restricted rotation and temperature will not make a difference. Don't feel bad, these arguments were made for a monoalkyl derivative of diethyl malonate by a chemistry Nobelist!
ADDENDUM: Here is an additional link to this answer. Here is a related spectrum: Intermediate Level #12 at this link.


text:
Deriving heat capacity in terms of internal energy U and natural variables S & V
My lecture notes have the following equation regarding heat capacities:
$$
C_p = C_v + T\left(\frac{\partial p}{\partial T}\right)_v\left(\frac{\partial V}{\partial T}\right)_p
$$
(just to be clear, all terms after the brackets in this question indicate a constant process of that term)
It then states that by using internal energy $U$, volume $V$ and entropy $S$, you can express heat capacity at constant pressure, $C_p$, just in terms of these 3 variables:
$$
C_p = \frac {\left(\frac{\partial U}{\partial S}\right)_v\left(\frac{\partial ^2 U}{\partial V^2}\right)_s }{\left(\frac{\partial ^2 U}{\partial V^2}\right)_s\left(\frac{\partial ^2 U}{\partial S^2}\right)_v - \left(\frac{\partial ^2 U}{\partial S \cdot\partial V}\right)^2}
$$
with no inbetween steps given.
From a previous derivation of $C_v$ I know that:
$$
C_v = \frac{\left(\frac{\partial U}{\partial S}\right)_v}{\left(\frac{\partial ^2 U}{\partial S^2}\right)_v}
$$
and I think I understand at the least how to express the $\left(\frac{\partial p}{\partial T}\right)_v$ term, because you can express $p$ and $T$ in terms of $U, S, V$ like so:
$$
\begin{align*}
-p &= \left(\frac{\partial U}{\partial V}\right)_s \\
T &= \left(\frac{\partial U}{\partial S}\right)_v
\end{align*}
$$
The real killer is the second term $\left(\frac{\partial V}{\partial T}\right)_p$, because the only thing I know I can do with this equation is use the Maxwell relation:
$$
-\left(\frac{\partial V}{\partial T}\right)_p = \left(\frac{\partial S}{\partial p}\right)_T
$$
but this still leaves a constant temperature term $T$ which is not present in the final equation. So, any help with this would be really appreciated.

Suggested reading: Physical Chemistry, Peter Atkins and Julio de Paula.

Read this Ref.1 and Ref.2

In the first equation you quoted, the natural variables are either $T$ and $P$ or $T$ and $V$. On the other hand, in the equation you are trying to derive, the natural variables are $S$ and $V$. It should in principle be possible to start from the former and arrive at the latter through a series of manipulations, but it should be much easier to simply start from scratch with the natural variables $S$ and $V$.

Usually this part of thermodynamics is not presented in the most efficient way. To derive the expression for, e.g., $C_P - C_V$, one goes through a series of steps involving Maxwell's relations and the triple product rule. Figuring out what to do in each step is done mostly by guesswork.
However, all relations of these sorts really come from just a single mathematical relation that more generally holds. That is to say,

Consider a twice-differentiable,  two-variable function $f(x,y)$ and its Legendre transform with respect to $y$, defined by $g(x,f_y) := f - f_y y$. 1  Here, $f_y$ denotes the derivative of $f$ with respect of $y$ with other independent variables held fixed. Then, the following identity is satisfied:
  \begin{equation}
\boxed{g_{xx}  - f_{xx}  = - \frac{f_{xy}^2}{f_{yy}} = \frac{g_{x f_y}^2}{g_{f_y f_y}}}.
\end{equation}
  Proof. We have
  \begin{equation}
\bigg(\frac{\partial f}{\partial x} \bigg)_{f_y} = f_x + f_y \bigg(\frac{\partial y}{\partial x}\bigg)_{f_y} = f_x - f_y \frac{f_{xy}}{f_{yy}}.
\end{equation}
  On the above, the first equality holds because of the chain rule [for the change of variables $(x,y)$ $\to$ $(x, f_y)$], and the second equality because of the triple product rule. Then,
  \begin{equation}
g_x = \bigg(\frac{\partial f}{\partial x} \bigg)_{f_y} - f_y\bigg(\frac{\partial y}{\partial x}\bigg)_{f_y}   =f_x,
\end{equation}
  and
  \begin{equation}
g_{xx} = \bigg(\frac{\partial f_x}{\partial x} \bigg)_{f_y}= f_{xx} + f_{xy}\bigg(\frac{\partial y}{\partial x}\bigg)_{f_y} = f_{xx} - \frac{f_{xy}^2}{f_{yy}}.
\end{equation}
  Also, using the chain rule for the change of variables $(x,y)$ $\to$ $(x, f_y)$, we get
  \begin{equation}
g_{f_y} = \bigg(\frac{\partial f}{\partial f_y} \bigg)_{x} -\bigg(\frac{\partial y}{\partial f_y} \bigg)_{x}f_y - y = -y.
\end{equation}
  Therefore, Legendre transform of $g$ with respect to $f_y$ is simply $g - g_{f_y} f_y = f$. Then, performing a similar analysis as above for $g$ gives
  \begin{equation}
f_{xx} = g_{xx} - \frac{g_{x f_y}^2}{g_{f_y f_y}}.
\end{equation}

Now, all that remains to be done is to choose the variables and functions $x$, $y$, $f$ and $g$. Below are some examples.
(1) $x = S$, $y = V$, $f = U$, $g = H$, and $f_y = -P$:
\begin{equation}
H_{SS} - U_{SS} = -\frac{U_{SV}^2}{U_{VV}}.
\end{equation}
Note that
\begin{equation}
U_{SS} = \bigg(\frac{\partial^2 U}{\partial S^2}\bigg)_V = \bigg(\frac{\partial T}{\partial S}\bigg)_V = \frac{T}{T(\partial S/\partial T)_V} = \frac{T}{C_V},
\end{equation}
and that, similarly,
\begin{equation}
H_{SS} = \bigg(\frac{\partial^2 H}{\partial S^2}\bigg)_P = \bigg(\frac{\partial T}{\partial S}\bigg)_P = \frac{T}{T(\partial S/\partial T)_P} = \frac{T}{C_P}.
\end{equation}
Hence, we have
\begin{equation}
\boxed{\frac{1}{C_P} - \frac{1}{C_V} = -\frac{U_{SV}^2}{T U_{VV}} = -\frac{\Big(\frac{\partial^2 U}{\partial S \partial V}\Big)^2}{\Big(\frac{\partial U}{\partial S}\Big)_V \Big(\frac{\partial^2 U}{\partial V^2}\Big)_S}}
\end{equation}
which is equivalent to the relation OP wanted to derive.
(2) $x = T$, $y = V$, $f = A$, $g = G$, $f_y = -P$, and noting that $f_{-y} = -f_y$,
\begin{equation}
G_{TT} - A_{TT} = \frac{G_{T,-P}^2}{G_{-P,-P}} = \frac{G_{TP}^2}{G_{PP}}.
\end{equation}
Here,
\begin{equation}
A_{TT} = \bigg(\frac{\partial^2 A}{\partial T^2}\bigg)_V = -\bigg(\frac{\partial S}{\partial T}\bigg)_V = - \frac{1}{T}C_V,
\end{equation}
\begin{equation}
G_{TT} = \bigg(\frac{\partial^2 G}{\partial T^2}\bigg)_P = -\bigg(\frac{\partial S}{\partial T}\bigg)_P = - \frac{1}{T}C_P,
\end{equation}
\begin{equation}
G_{TP} = \frac{\partial^2 G}{\partial T \partial P} = \bigg(\frac{\partial V}{\partial T}\bigg)_P = V \alpha,
\end{equation}
where $\alpha := \frac{1}{V}\Big(\frac{\partial V}{\partial T}\Big)_P$ is the volume expansion coefficient,
\begin{equation}
G_{PP} = \bigg(\frac{\partial^2 G}{\partial P^2}\bigg)_T = \bigg(\frac{\partial V}{\partial P}\bigg)_T = - V\kappa_T,
\end{equation}
where $\kappa_T := -\frac{1}{V}\Big(\frac{\partial V}{\partial P}\Big)_T$ is the isothermal compressibility. Then, it follows that
\begin{equation}
\boxed{C_p - C_V = \frac{TV\alpha^2}{\kappa_T}}.
\end{equation}
(3) $x = V$, $y = S$, $f = H$, $g = G$, and $f_y = T$:
\begin{equation}
G_{PP} - H_{PP} = \frac{G_{TP}^2}{G_{TT}}.
\end{equation}
We have
\begin{equation}
H_{PP} = \bigg(\frac{\partial^2 H}{\partial P^2}\bigg)_S = \bigg(\frac{\partial V}{\partial P}\bigg)_S = - V\kappa_S,
\end{equation}
where $\kappa_S := -\frac{1}{V}\Big(\frac{\partial V}{\partial P}\Big)_S$ is the adiabatic compressibility. We have already shown that
\begin{equation}
G_{PP} = -V\kappa_T, \quad G_{TP} = V \alpha, \quad G_{TT} = - \frac{C_P}{T},
\end{equation}
whence we obtain
\begin{equation}
\boxed{\kappa_T - \kappa_S = \frac{TV\alpha^2}{C_P}}.
\end{equation}

1 To be more precise, one should first write down $f(x,y) - y f_y$ and then express $y$ as a function of $x$ and $f_y$. Also, for the Legendre transform to exist, $f(x,y)$ should be a convex or concave function of $y$.


text:
Diluting Stock Solution with NaOH
So I know that if you want to dilute a stock solution to a smaller concentration you simply use the $M_1V_1 = M_2V_2$ equation where the $M_1$ and $V_1$ are the molar concentrations and volumes of the stock solution and $M_2$ and $V_2$ are the molar concentrations and volumes of the resulting dilute solution. This is assuming if you are diluting with water.
My question is, if you have a stock solution of say para-nitrophenol and you wanted to dilute it with $\pu{0.100 M}$ $\ce{NaOH}$ solution, what modifications would you use to calculate how much of the stock solution you would need to get a dilute para-nitrophenol solution?
What I'm thinking is that the density of the $\ce{NaOH}$ solution would come into play here but I don't know how I would go about it.

First, you didn't tell us the whole story and what you have done so far. Your sought compound, p-nitrophenol is a solid. Thus, I assume you already have a solution of it (your stock solution). Is it in $\mathrm{0.100M}$ $\ce{NaOH}$ as well? Or, in some other solvent? In any solvent, what is the concentration of your stock solution?  You want to the find final concentration of $\ce{NaOH}$ or sodium p-nitrophenolate? (Remember,  p-nitrophenol  reacts with $\ce{NaOH}$)

Ok, so I really asked this as a general question as in if I were to dilute a solution with something other than water . But I'm doing a lab using a 96-well plate to do spectroscopy to create a standard curve for pNP. I have a 100 micro-molar solution of stock pNP (paranitrophenol, given to us so I assume its dissolved in water). The stock solution is probably colorless and will change after I dilute it because the instructions say that the solution will turn yellow after diluting with NaOH. So the current goal is to create dilute solutions of 10, 20, 30, 40, 50, 60, and 70 micro-molars.

What is your expected final volume in each well? $\pu{300 \mu L}$?

The final volume for each standard solution should be 1.00 mL.

Since you'd like to use the equation, $\mathrm{M_1V_1 = M_2V_2}$, I'll show you here, how to do dilution for your extreme concentrations. You can follow same method to do the rest.
Suppose, your final volume of each of dilute solutions is $\pu {V \mu L}$. Then, for all calculations, $\mathrm{V_2 = V}$ and $\mathrm{M_1 = 100\space\mu M}$ (concentration of stock solution). Note that, concentration of $\ce{NaOH}$ does not play a role here, since: (1) it is $1000\times$ more than that of the stock solution (pNP) and hence, dialution of itself is minimal; and (2) it acts just as any other solvent, except for reacting with pNP to give the desired color.
Now, let the volume of $\ce{NaOH}$ solution needed to make your highest concentrated pNP solution is $\pu {V_{1} \mu L}$. Then, using the equation $\mathrm{M_1V_1 = M_2V_2}$, you get:
$$\mathrm  {100\space\mu M \times V_1 = 70\space\mu L \times V\space\mu L}$$
$$\mathrm  {V_1 = \frac {70\space\mu M \times V\space\mu L}{100\space\mu M} = \frac {7}{10}V \space\mu L}$$
Thus, you dilute $\pu {0.7V \mu L}$ stock solution with  $\pu {0.3V \mu L}$ $\ce{NaOH}$ solution to make $\pu {70 \mu M}$  pNP solution.
To see if there are enough  $\ce{NaOH}$ in $\pu {0.3V \mu L}$ portion, calculate $\pu {\mu mol}$ of $\ce{NaOH}$:
$$\mathrm  {100\times 1000\space\mu M \times 0.3V\space\mu L = 3\times10^{-2}\times V\space\mu mol}$$
Similarly, you can calculate $\pu {\mu mol}$ of pNP in final solution:
$$\mathrm  {100\space\mu M \times 0.7V\space\mu L = 7\times10^{-5}\times V\space\mu mol}$$
Thus, all p-nitrophenol will be converted to sodium p-nitrophenolate in the highest concentrated solution. Therefore, it is safe to assume that rest of the solutions would undergo the same conversion.
By similar calculations, we can show you must dilute $\pu {0.1V \mu L}$ stock solution with  $\pu {0.9V \mu L}$ $\ce{NaOH}$ solution to make $\pu {10 \mu M}$  pNP solution. Since you know  value of $\pu {V \mu L}$, you can get numerical values.

Density changes when mixing liquids is not only applicable to NaOH solutions. Even when you dilute a stock solution with water you may find that the volume is not additive. The factors that matter are the starting concentration and the concentration change.
At the concentrations you're using, you can probably safely ignore density changes and assume volumes are additive, unless you need extreme precision (but then you could use volumetric glassware and do the dilutions properly, instead of just adding volumes).
You are correct: when you dilute your $100 \ \mu M$  pNP solution with $0.1 \ M$ NaOH, it will turn yellow, because you form the pNP anion, which is intensely yellow.
However, for the spectroscopy part to work correctly, you must make sure you always have at least the stoichiometric amount of NaOH, i.e. you must convert all pNP to its anion.
This will be the case in all your diluted solutions, simply because the concentration of NaOH is 1000 times larger than the pNP concentration ($0.1 \ M = 100 \ mM = 100000 \ \mu M$), so whenever the volume of NaOH solution you add to the pNP solution is greater than 1/1000  times the volume of the latter, you're already in excess of NaOH.
For the rest, the formula you mentioned works correctly. For instance, to make the $70 \ \mu M$ solution of pNP anion: $V_2 = \frac {100 \cdot V_1} {70} \approx 1.43 \cdot V_1$, so you could add $0.43 \ mL$ of $0.1 \ M$ NaOH to $1 \ mL$ of $100 \ \mu M$  pNP. You can see that you are in huge excess of NaOH, so there is no risk to leave any non-ionised pNP.
You can achieve the same result by a 'quick' subtraction method, which also works when both solutions contain the substance of interest.
In this case, your stock is $100 \ \mu M$ in the substance of interest (pNP). You dilute it with a solution that contains no pNP, so it's $0 \ \mu M$. You want to obtain a $70 \ \mu M$ solution.
Subtract the final desired concentration from the concentration of each solution you're mixing, take the absolute values and swap the results.
So, pNP stock minus final desired solution is $|100 - 70| = 30$. NaOH minus final desired solution is $|0 - 70| = 70$. Swap the two: 30 applies to NaOH and 70 applies to pNP.
This means that, to obtain the desired final solution, you need to mix 30 'parts' of NaOH with 70 'parts' of pNP. You can easily verify that indeed $100/70-1 = 30/70$.
This may sound more complicated than the other formula, but in fact if you write the concentrations of the two 'starting' solutions on the top left and bottom left corner of an ideal square, respectively, and the final concentration in the centre of the square, and then do the subtractions 'across the diagonals', you will get in the right corners of the square the 'parts' to mix for each starting solution.
E.g. if you had a $10 \ \mu M$ solution (A) and a $90 \ \mu M$ solution (B) and wanted to obtain a $40 \ \mu M$ solution by mixing them, this method would quickly tell you that you need to mix 50 'parts' of A and 30 'parts' of B.
Obviously any mixing-based dilution will work only if the final desired concentration is 'bracketed' by the concentrations of the solutions being mixed. In these examples, $0<70<100$ and $10<40<90$.

Thanks. I actually just asked my professor and she just wanted a percentage of pNP to NaOH. So like 100 uL of pNP and 900 uL NaOH for the 10 uM solution. This was just theoretical and the actual concentrations should be close to these values. I kind of over thought about this :)


text:
Structure of benzene dimer in an vibrationally excited state
It is known that the benzene dimer has few stable structures, T-shaped, parallel-displaced, and sandwich structure.
Now if we excite the dimer such that it goes to a higher vibrational level (which is close to the dissociation limit but the dimer is still bound).
What happened to the structures then?
Can we still have different structures or it will be some kind of mixture?

Presumably you mean the inter benzene vibrational frequency. This will have a very low frequency indeed and several vibrational (and rotational)  levels would be excited  at room temperature  so exciting another level should have a small effect. ($k_BT \approx 210$ cm $^{-1}$ at room temp.)

The binding energy of the benzene dimer is really quite small. Ref. [1] reports the parallel-displaced and T-shaped dimers to have a binding energy at about 2.6 kcal/mol. This is about 900 cm$^{-1}$.
So, it is quite likely that if any vibrational mode with more energy than this is excited that upon redistribution of the energy, the dimer will fall apart.
On the other hand, excitation of vibrations less than the binding energy will likely lead to a very rapid interconversion between the T-shaped and parallel-displaced (or maybe sandwich) isomers. I have read that it is believed this happens basically no matter what.
Also, the sandwich isomer has a binding energy closer to 1.5 kcal/mol, so it is potentially possible to excite some low frequency mode which would cause the sandwich isomer to dissociate but not the T-shaped or parallel-displaced isomers.
Note that because this is such a weakly bound dimer, experimental measurement can only be made at very low temperatures. Ref. [2] describes one measurement of the conformers using supersonic jet expansion to form the dimers. This means temperatures close to milliKelvin.

References:
[1]:  Janowski, T., & Pulay, P. (2007). High accuracy benchmark calculations on the benzene dimer potential energy surface. Chemical Physics Letters, 447(1-3), 27-32.
[2]: Scherzer, W., Krätzschmar, O., Selzle, H. L., & Schlag, E. W. (1992). Structural isomers of the benzene dimer from mass selective hole-burning spectroscopy. Zeitschrift für Naturforschung A, 47(12), 1248-1252.

Thank you very much for the answer and nice explanation.


text:
What is the difference between the mean radial distance and the most probable radial distance of an orbital?
I was reading JD Lee's Concise Inorganic Chemistry and it has given some values of mean radial distance which is increasing from s to p to d orbital. I had earlier read that the most probable radial distance of an orbital has the order s>p>d>f. This makes me wonder what is the difference between the mean radial distance and most probable radial distance. Also, why is the order of mean distance reverse from that of most probable radial distance?
Also can you please explain why is the energy of an orbital proportional to mean radial distance as given in the book?

Image Source: Concise Inorganic Chemistry by JD Lee

If $\psi$ is the radial part of the wavefunction the mean or average radial distance is $\langle r \rangle=\int r\psi^\psi dr/\int\psi^\psi dr$ where the denominator is the normalisation. The most probable is (presumably) the value of $r$ at the maximum of $\psi^*\psi$. ( The $\psi$ is an equation involving $r$ and  the * indicates the complex conjugate. If $\psi$   is real is just the same as $\psi$ and if not replace $i$ with $-i$ where $i=\sqrt{-1}$).  These values are not the same as $\psi$ is not 'symmetrical' as $r$ increases.

Do you have a source for the most probable radial distance ordering? http://slideplayer.com/6234381/20/images/36/Overlay+of+Radial+Distribution+Functions+4pr2R%28r%292+for+the+hydrogen+atom.jpg

@Tyberius As it can be seen in the image you provided, the most probable radius of 3s>3p>3d.

In multi-electron atoms the energy of the s, p and d orbitals for a given $n$ is determined primarily by shielding by inner electrons and penetration of the orbital towards the nucleus. As s orbitals have some probability of being at the nucleus their energy is lowest, next is p then d orbitals in increasing energy. Also there is spin-orbit coupling which splits sub-levels of p and d orbitals or any orbital which has angular momentum so this does not effect s orbitals. This effect is much smaller in d than p orbitals and generally smaller than the separation of s to p to d.
The statements in the book extract you give seem contradictory. The mean radial value of an H type orbital is calculated using $\displaystyle \int_0^\infty r^3R_{n,l}^*R_{n,l} dr $ where $R_{n,l}$ is the (normalised) radial part of the wavefunction for an atom with quantum numbers $n,\;l$. The $r^3$ arises because the volume element in the integration is $r^2\sin(\theta)d\phi d\theta$ in spherical coordinates. The other $r$ is how the average is defined, i.e. $\int r\psi^*\psi d\tau$ for some normalised wavefunction $\psi$.
The radial parts of the wavefunction for the 3s,3p and 3d are define with $q=Zr/a_0$ ($a_0$ is the Bohr radius) as
$$\begin{align}
R_{3,0}&=N(27-18q+2q^2)exp^{-q/3}\\
R_{3,1}&=N(6q-q^2)e^{-q/3}\\
R_{3,2}&=Nq^2e^{-q/3}
\end{align}$$
and $N$ is the normalisation which is different for each wavefunction. The integrals are not hard to evaluate but very tedious as there are many terms. However, there is a general formula for the average of H atom type orbitals and is
$$\langle r\rangle =\frac{n^2a_0}{Z}\left( 1+\frac{1}{2} \left( 1-\frac{l(l+1)}{n^2}  \right) \right ) $$
The average values are
$$3s \;\langle r \rangle = 0.476 \text{ nm,}\quad 3p \;\langle r \rangle = 0.441\text{ nm,}\quad 3d\; \langle r \rangle = 0.370\text{ nm}$$
The s orbital value is the same as in your question but the others differ both in value and in order of size. Perhaps there are typos in the text you give or some other unstated factors producing these values.
The maximum values have to be calculated by making the derivative (of $q^2R_{n,l}^2$) zero and solving the equation. Alternatively the value can be taken with sufficient accuracy  from a plot of the function which is far easier. The values are
$$3s \;r_{max} = 0.46 \text{ nm,}\quad 3p \;r_{max} = 0.43\text{ nm,}\quad 3d\; r_{max} = 0.32\text{ nm}$$
the plot below shows the radial functions $q^2R_{n,l}^2$  each normalised to the same maximum height. The mean is greater than the maximum in each case as expected as the main part of the radial probability is 'lopsided' towards larger distances.

(In evaluating the integrals $\displaystyle \int_0^\infty r^ne^{-\beta r}= \frac{n!}{\beta^{n+1}}$ is useful.)

Thanks for the answer! I think the average radius value that you calculated differs from the book because the book has given mean radius for a phosphorus atom and not hydrogen. Now I don't know why the s orbital value matched because intuitively, it feels like they should differ.

Also, according to you since the mean radius value for H atom is in the order 3d<3p<3s, the statement that energy of an orbital is proportional to the mean radius is false, right?

I used Z=15 for phosphorus so all values should fit. It seems that that statement is not correct or at least needs to be explained. The picture above seems to contradict it. But the figure is for H type atoms not one with multiple electrons but for an atom I don't see why this would have a big effect on the radius such a changing the ordering of the average sizes.

First of all,in the radial plot of probablity distribution of  $3s$ and $3p$ subshells have  more than one peaks. So, if we take the highest peak as the most probable  radius, then we have to solve ,  $1$. the maxima for $\ce{P(r_{3s}) }$ which is proportional to $\ce{|$\Psi$_{3s}|^2r^2}$. So, actually, we have to solve $$\ce{\frac{d}{d$\rho$}($\rho$^2(27 - 18$\rho$ + 2$\rho$^2)^2e^{-\frac{2\rho}{3}}) = 0, where $\rho$ =\frac{Zr}{a_0}}$$Solving this and the global maxima of $\rho$ will come at approximately $13$. $2$. Similarly, for $\ce{P(r_{3p})}$ which is proportional to $\ce{|$\Psi$_{3p}|^2r^2}$ (note, here we are concerned aboutr radial part which is same for all $p_x,p_y, p_z$ orbitals). So, here we have to solve,$$\ce{\frac{d}{d$\rho$}($\rho$^2(6 - $\rho$)^2e^{-\frac{2\rho}{3}}) = 0, where $\rho$ =\frac{Zr}{a_0}}$$ Solving this and the global maxima of $\rho$ will come at exactly $12$ $3$. and finally, for $3d$ orbitals, the maxima will be a single peak and that will come from the equation, $$\ce{\frac{d}{d$\rho$}($\rho$^4e^{-\frac{2\rho}{3}}) = 0, where, $\rho$ = \frac{Zr}{a_0}} $$ This will give the maxima at $\rho$=$6$.  So, it is clear that the most probable radius follows the order, $r_{mp}(3d) < r_{mp}(3p) < r_{mp}(3s) $ But when you come to the mean radius, we just cannot bring same analogy as the most probable radius. Because the mean radius is actually an overall effect of radial probability distribution throughout the entire space. So, unless we calculate the integral,$$<\Psi|r|\Psi> =\int_{0}^{\infty} r|\Psi|^2  r^2drsin\theta d\theta d\phi$$ it is difficult to say, which will have the larger mean radius, and which one will have smaller. It totally depends on the integral. So, we actually have to evaluate the integral to tell the answer (which can be done but very much calculative. You can use the idea of gamma functions to evaluate this kinds of integral $\int_{0}^{\infty}r^n e^{-\alpha r }dr$ easily which will appear lot in this mean radius calculation).  Coming to the energy, the statement is said in a loose way i.e, As an orbital becomes farther from the nucleus, its energy becomes lesser negative and increases towards zero.

Mean radius is actually no more calculative than what you just did. Factor away the angular part, and you'll be left with a relatively manageable integral.

Also, you might want to use \langle...\rangle for $\langle\Psi|r|\Psi\rangle$.

Thanks! So it is not necessary that the mean radius of a d orbital will be greater than that of p and that of s, am I correct in saying this?

So when we look at the energy of an orbital, do we look at the mean radius? If we were to at the most probable distance from the nucleus, then the reverse trend of energy would be predicted.


text:
Are phenols good antioxidants?
Are phenols good antioxidants? If so what makes them good? I think it is because they oxidize easily but I'm not sure how that helps.

https://chemistry.stackexchange.com/questions/85486/how-do-antioxidants-work https://chemistry.stackexchange.com/questions/62729/what-makes-a-good-antioxidant

https://chemistry.stackexchange.com/questions/16088/what-has-stronger-antioxidant-potential-bht-or-bha

Phenolic compounds, or in general aromatics actually, has electrons which many oxidants can easily abstract or bind to molecule.
http://www.springer.com/cda/content/document/cda_downloaddocument/9781441978257-c1.pdf?SGWID=0-0-45-1106848-p174056970
From the above link in page 17,

as you see the highest rates, order of magnitude or even more higher ones come from aromatics, or at least pi-bond between carbons. It can reasonably assumed that if you have these compounds in a mixture, the possibility of any oxidant introduced reacts with them is higher than the other ones. In short, they are good anti-oxidants in that way.


text:
Is endpoint pH concentration dependent? Choosing an acid/base indicator
We use titrations to determine analyte concentrations with a standardized titrant and an acid/base indicator, BUT how do we choose the indicator? It is my understanding that indicators are supposed to be matched to the endpoint pH of your titration, but unless you already know the concentration of the analyte in solution, how do we determine the endpoint pH to match the indicator to? And if we already know the concentration of our analyte, doesn't that defeat the purpose of the titration in the first place?
I must be missing something...

related https://chemistry.stackexchange.com/questions/8048/shouldnt-the-ph-at-the-equivalence-point-always-be-7 https://chemistry.stackexchange.com/questions/8653/the-significance-of-the-dissociation-constant-of-an-indicator

https://en.wikipedia.org/wiki/PH_indicator


text:
When is it necessary to include solvent effects?
When modelling a process like keto-enol tautomerization or SN2 ring opening reaction, I am unsure as to what factors need to be considered when deciding whether solvent effects need to be accounted for. Because if they are important, a solvation approach, implicit or explicit is chosen to model the process.
For example, in a keto-enol tautomerization, solvent effects are important as the equilibrium is maintained by the solvent.
What I was thinking is that using the explicit water model, as few as  1 to 2 water molecules can be used to model the tautomerization. But many studies are showing that continuum solvent models (implicit) are the benchmark for keto-enol tautomerizations. I don't understand why and what factors determine this?

Generally you only need explicit solvent if there are important effects due to the actual finite size of water and its geometry. For instance, if hydrogen bonding is quite important then to be quantitative an explicit solvent is probably needed. Note, however, that an implicit solvent can capture basically all of the field effects which the water introduces. So if all that matters is how the reaction changes when it is in the presence of an electric field that can shift to accommodate charges, there is no reason to do explicit solvent.

Well, first of all, explicit is more expensive than implicit. So, most people do not go for the former if they can help it. So, in my experience, you want to use explicit solvent only when you have to, like in the formation of $\ce{H2SO4}$ from $\ce{SO3}$ and $\ce{H2O}$. Here, extra water molecules act as catalyzers [1]. As for the case keto-enol tautomerization, I guess the results are in good agreement with the experiments using implicit solvent.

The bifunctional catalytic role of water clusters in the formation of acid rain. Chem. Commun. 2017, 53, 3516-3519.


text:
Would the Formation of Benzaldehyde Be Possible Under Specific Conditions?

Our process:
We cure an epoxy resin using SLA. The triarylsulfonium hexafluoroantimonate salts in propylene carbonate act as the photoinitiator for the oligomeric reactions between 4,4 isopropylidenedicyclohexanol and 1 chloro 2,3 epoxypropane to crosslink the resin. We then rinse the part in propylene carbonate multiple times, rinse in water once, and finally end with time in the post-curing apparatus. We were finding that the piece was not biocompatible so we turned to further testing.
We sent a cured and cleaned sample in for 24 hour extraction at 37C and tested the extract with GCMS (results on table). The uncured resin itself was also tested and found positive for propylene carbonate and cyclopentanol.
My question is specifically about the benzaldehyde that is present. Would benzaldehyde be able to form using the listed chemicals (and exposure to UV light) or would it likely be introduced from an outside contaminant?
Thanks


text:
Electron Affinity negative?
Why is electron affinity negative? An electron carries energy so if we had an electron doesnt that mean we add energy? Please describe in a non-PhD lingo that a first semester chem student can comprehend.

Because the potential energy of the system decreases

The electron affinity is the energy released when an electron attaches to an atom in gas phase. The sign of the energy depends on the convention used in the textbook. For instance, in Atkins' Physical Chemistry textbook it is defined as a positive quantity which means that the process is exothermic. However, if your point of view is the potential of the system as mention in the comments by Avnish Kabaj then you will have your negative electron affinity since the reference energy is zero when the electron is far from the gas phase atom and when it is brought close to the atom the final potential energy is negative.


text:
Reduction of Carbonyl Compounds with LiAlH4
The following appeared in the October/November CIE Camebridge A level chemistry paper 4.

Two isometric aromatic compounds, V and W are shown below.
Draw the structures of the two organic products from the reaction of V and W with $\ce{LiAlH_4}$.

The answer key simply removes the oxygens from the structures. However, I was under the impression that they would be replaced by hydroxyl groups since reducing carbonyl compounds gives one alcohols. Does anyone know why this is the case?

Though, the carbonyls are reduced to alcohols after the treatment with $\ce{LiAlH_4}$,here the $\ce{C=O}$ double bond is part of the Amide($\ce{-CONH_2)}$group, and this group has definitely different chemical functionality than pure carbonyls.So, though both the aldehyde(or, keto) and amides have same $\ce{C=O}$ double bonds, but thir property is different due to extra presence of adjacent of $\ce{NH_2}$ group in amides.  In Aldehydes or ketones, the $\ce{C=O}$ has more double bond character. So, when it is treated with $\ce{LiAlH_4}$, due to the strong bonding between $C$ and $O$, only cleavage of the $\pi$-bond is possible and thus it can possibly form alcohols, because, Nucleophillic attack by $\ce{H^-}$ of $\ce{LiAlH_4}$, can form $C-H$ bonds, and energetically it is not so favourable that it will break two $C-O$ bonds for that.So, that's why, the end products are alcohols in reduction of aldehydes or ketones.  But in Amides, the $\ce{C-O }$, bond has a very less double bond character due to the resonance of carbonyl carbon with the $\ce{-NH_2}$ group, and therefore, the bond in this case becomes very weak, and easy to break.  More Over, if you see the mechanism, for this reduction of amides, in a transition state, that carbonyl carbon forms double bonds with the $\ce{-NH_2}$ group ( $\ce{-C=NH}$), and this becomes energeticaly more favourable ( As, the difference in energy of the $2p$ orbitals of $C$ ans $N$ is very less thaan that of the $O$, so the $\pi$-overlap can form strongly )and thus the $C-O$ bond becomes easier to break totally,  which in turn give the amines as the end product in reduction of amides.

(-1) this has little if anything to do with bond strength in the starting material. It is simply that the mechanism allows for loss of OH in only one case.


text:
Make an eco friendly cleaning product
I am trying to make an eco-friendly cleaning product and want some advice on what the best chemicals are to use.
I plan to make a laundry detergent and all purpose cleaner. I want both to be free from toxins and harmful chemicals. I would also like to use essential oils. Even better and kinder on the skin a honey or milk product included. Can this be done? I cannot find any cleaning products so far with milk or honey in.

I think you need to tell us more about this proposed cleaning product. What do you want to clean?

Making your own soap from oils/fats and sodium or potassium hydroxide is something well described and within the compass of a kitchen chemist. Running an internet search on "making your own soap" produces pages of hits such as this one  https://www.diynatural.com/how-to-make-soap-2/
Laundry detergents are a different matter as the Wikipedia entry makes clear https://en.wikipedia.org/wiki/Laundry_detergent
Soaps are less effective for this and particularly struggle with hard water. Modern laundry detergents contain multiple ingredients such as alkylbenzenesulfonates which are beyond the kitchen chemist. They also contain bleaches based on salts of peracids  (typically perborate and percarbonate) which are not readily available and, I think, you would consider toxic.


text:
Number of water molecules in a box to simulate liquid water
I have seen that depending on the simulation approach, different number of water molecules are employed for studying liquid water properties.
32 in early DFT calculations
The electronic structure of liquid water within density-functional theory
The Journal of Chemical Physics 123, 014501 (2005); https://doi.org/10.1063/1.1940612
216 in early classical molecular dynamic simulations
Molecular Dynamics Study of Liquid Water The Journal of Chemical Physics 55, 3336 (1971); https://doi.org/10.1063/1.1676585
256 I'm not sure if this number of water molecules is just for computational reasons $2^8$= 256
So I'm wondering, What is the reasoning behind them?,
Edit:
In other words, If I want to obtain the ambient density of the system and I don't know anything about it (number of  molecules or the size of the simulation box, assuming cubic), what should I do?
As a starting point, the only thing I know is that the size of the simulation box should be chosen considering a number density that corresponds to the experimental density of the system.

Different simulation methods require different amounts of computational horsepower. You simulate as many particles as you can afford to...

Generally speaking, you use as many molecules in a simulation as you possibly can. There are a couple of caveats. For instance, if you wish to do a simulation using an expensive method (relative to a classical potential) like DFT, you will need to use fewer molecules. This has the obvious drawback that when you use periodic boundary conditions, each molecule will interact with its own reflections much more than we would like (which is not at all). Nowadays, we can certainly do simulations with more than 32 water molecules using DFT.
Also, there is no reason to just keep using larger and larger numbers because it is known that interactions between molecules in the liquid die off reasonably quickly with distance, so at some point you stop gaining anything by adding more molecules. All of the long range effects can be handled by using periodic boundary conditions, which works very well if each molecule is far enough from its own image that a molecule does not have short-range interactions with itself.
Another consideration is that the number of molecules you use depends a lot on what property you are interested in studying. Conventional wisdom about the number of molecules needed is usually associated with the number of molecules that give converged average energies. If you are interested in something like diffusion coefficients, however, you may need quite a few more molecules before convergence is achieved. I have read that up to 512, or even 768, can be needed to get good diffusion coefficients around room temperature for water.
It is true that sometimes a specific number of molecules is chosen to match a specific density you want to do simulations at. This, however, isn't usually too big of a deal as you can always change your box size to match the density you are trying to achieve.
There is no physical reason that people choose powers of 2 for the number of molecules other than that computational scientists tend to use powers of 2. The number 216 is probably used because some seminal paper used this number and got good results, so people tend to just stick with the trend.


text:
Experimental measurment of thiosulphuric acid decomposition kinetics
Problem
I am trying to do a kinetics analysis on the following reaction:
$$\ce{Na2S2O3 {(aq)} + 2 HCl {(aq)} -> S {(s)} + 2 NaCl {(aq)} + SO2 {(g)} + H2O {(l)}}$$
Suggested Method
However, most practicals online suggest I use an 'X' Mark and let the sulphur precipitate cloud the mark (Here's an Example) . To measure the rate of the reaction, I would begin timing as soon as I mixed the two solutions and end as soon as the 'X' was completely obscured.
Alternatives
However this form of time measurement is somewhat qualitative.
I considered looking at:

pH or temperature:       Those readings would not provide substantial data.
Change in volume or mass:   For the gas released, but $\ce{SO2_{(g)}}$ is dissolvable in water.
Conductance:         Although this is viable, I was told there is an alternative.
Colour (Spectrometry):    This probably won't work because the spectra observed and                  measured will suddenly reach its maximum as solids are                becoming present that have complete absorption.

I was told that the last option (colour or spectrometry) was "on the right direction".
Question
So, with all this in mind...
Are there any quantitative means of measuring the rate of the above reaction?

You could try arresting the reaction and then measure the important quantities.

Hmm, this is all rather too simple methods to get much insight in the reaction. It's much more complicated then it may seem and mechanism is highly non-trivial. Potentiometry  may be better, but more sophisticated methods would be needed to study it in depth.

Measuring conductivity over time might do the trick (7 ions (8 charge equivalents) in, 4 ions/charge equivalents out).
Regarding spectrometry: you could deliberately make use of the light scattering caused by the precipitating sulphur. Should be somewhere in the 600+ nm range. It's a standard method in biological sciences to follow the growth of bacteria.

The consumption of [S2O3]2- over time could be monitored using in situ FT-IR spectroscopy (eg. using a react IR instrument). This way the measurement is done in reflectance mode rather than transmission mode which solves the issue of interference from the solids formed such as sulfur. This will provide the reactions profile at different concentrations of [S2O3]2-. Then there are various options for processing the data with the best ones being either initial rates or even better reaction progress kinetic analysis. For mechanistically simple reactions you could use integrated rate equations as well.


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What is causing a blue colour when sodium bicarbonate, vinegar and aluminium are reacted together?
My 10 year old has been experimenting in my kitchen!
The other day he wanted to know why when he reacted bicarb soda and boiling vinegar in a container with some aluminium foil, the solution turned blue.
I have no idea!
Your help is much appreciated.

Search "home chemistry for kids". This looks a good one  https://mommypoppins.com/kids/50-easy-science-experiments-for-kids-fun-educational-activities-using-household-stuff

Would be better to know which kind of vinegar. Wine apple or just acetic acid and caramel might made a difference. Or there was some cupper around

Aluminum foil could contain copper as an impurity.  This "shouldn't" react with the nonoxidizing acid but could be oxidized by air.

Thanks everyone for your comments. It was standard kitchen white vinegar that he used. Cheers.

Hi all. Even though I saw the blue result at the time, I thought I'd better let you know that I tried replicating the experiment exactly the way he said he did it just now, same porcelain mug, same ingredients and didn't get any blue - very mysterious! He promises there's no food colouring tricks in there (and the blue did seem too light to be a drop of food colouring).


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Volume of gas in containers
If gases occupy all the volume in a container then how can we define 'volume percent' as all different gases will have same volume?

Most likely it's an open container with a piston i.e fixed pressure not volume

They had different volumes before we've mixed them.

@IvanNeretin it's not like that, as you can see that it's often stated  'oxygen is 21% by volume in air'.

@ShivamAgarwal say you have 2 mol of oxygen and 8 mol of nitrogen. Then the oxygen is 20% by volume (and 20% by mols if we assume they are all ideal gases)

@ShivamAgarwal This may be useful for answering your question if you are familiar with the concept of partial pressure. https://en.wikipedia.org/wiki/Partial_pressure#Partial_volume_(Amagat's_law_of_additive_volume)

I think this is a fair question. There isn't really much to add, although an example could help. In many high-school questions, the composition of the mixture if specified as x% gas X, y% gas Y, etc. Yet, when applying the ideal gas law, for each gas, the volume term $V$ is specified as the volume of the container, which implies they do occupy the whole volume.||Even if the answer to this question ends up being a single line, it does not - imho - does not justify closing the question. Thanks

I know it doesn't make much sense, but the term volume percent  typically has the meaning "mole percent."

Imagine nitrogen and oxygen in a cylinder separated by an infinitely thin barrier between the two gases. The height of the oxygen layer will be 21% of the total height and nitrogen 79%. Because we assume ideal gas behavior that means that if the barrier is removed, then after mixing a sample removed from the mixture will be 21% oxygen molecules and 79% nitrogen molecules.

We can and do use volume fractions to express the level of a gas of vapour in air, it is volume ppm.  For example air to a first approximation is 20 % oxygen and 80 % nitrogen.
This means 20 % of the molecules in air are oxygen while 80 % are nitrogen. If we treat the gases as ideal gases then air is by volume 20 % oxygen.

The point you have to remember about ideal gases (and gases at atmospheric pressure a a good approximation of ideal gases) is that the particles of the gas don't interact. This applies to mixtures as well as pure gases.
So, in one sense, if you fill a vessel with a mixture of gases, each does occupy the whole vessel. But the gases are also perfectly mixed as they don't separate out. If you calculated the pressure for a fixed amount of each gas separately in a vessel of a given size, then the answer you would get would be the partial pressure of the gas in a mixture containing the same amount of the specific gas.
As an alternative, if you kept the pressure fixed and calculated the volume occupied by the gas, the volume would be proportional to the amount of the gas. The volume of the components would add up the total volume of the vessel containing the mixture at the same pressure. This means that the volume % is proportional to the amount of gas molecules in the mixture and to the partial pressures. Volume and partial pressure are proportional for ideal gases so volume %, partial pressure and mole % are the same.


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Do aldehydes give bromine water test?
If they give bromine water test, what do they form?
I think that aldehydes will get oxidised and so decolourise bromine water but I don't know if that happens in normal conditions.

Think about how readily aldehydes enolise

Are you saying that they are quite reactive and hence might bromine water?

Possible duplicate of Does Br2/H2O oxidize all aldehydes?

The answer is yes. Aldehydes will decolourise bromine water.
The oxidation of aldehydes to acids is discussed in this answer Does Br2/H2O oxidize all aldehydes?
In addition alpha-bromination of the aldehyde via the acid catalysed bromination of the enol is a well-documented process. https://chem.libretexts.org/LibreTexts/Athabasca_University/Chemistry_360%3A_Organic_Chemistry_II/Chapter_22%3A_Carbonyl_Alpha-Substitution_Reactions/22.03_Alpha_Halogenation_of_Aldehydes_and_Ketones
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch18/ch18-3-1.html


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How does something like elemental oxygen oxidize anything if each atom's valence shell is full?
Been learning in biology about $\ce{O2}$'s role as an oxidizing agent, but I don't really understand how $\ce{O2}$ can take electrons from anything if it's not reactive....

Also, see this: https://chemistry.stackexchange.com/questions/90773/why-is-oxygen-an-oxidizing-agent

And before I forget, welcome to Chemistry.SE! Take the [tour] to familiarize yourself with the site! We hope you enjoy your stay!

O2 is pretty reactive, even though each atom's valence shell is full.

@Mithoron But that question doesn't have an answer that metaphorically explains everything as domestic abuse.

Most compounds consist of atoms that have "filled" their valence shells. It is basically what defines "reasonably" stable molecules. That does not mean that they cannot react with other molecules. Nature always seeks to find the lowest energies possible for its molecules, and the oxygen molecule has a very high potential energy.
The oxygen molecule is like two bullies who live together. They'd rather find a pair of victims that are weaker than themselves, split up and live an abusive but for them good life alongside their weaker partners. There are worse compounds out there, like the homicidal maniac Chlorine Trifluoride who will rip apart the stablest of marriages to abominable and atrocious results.

Lolz, nice vivid answer. I take it that's because oxygen is so electronegative. Thanks.
Edit: but wouldn't it take something more electronegative than oxygen (ie fluorine) to break the bond of an oxygen to another oxygen?

@JurpBurper Victims and bullies. Or bullies and homicidal maniacs. As long as there is difference between them and the resulting molecule(s) are stable  reaction can happen. If a molecule that is highly electronegative find a molecule that is much less so, and the ambient temperature is high enough for it to happen, the reaction will happen and the result is one or more averagely electronegative ones. It is like hot and cold, nature prefers lukewarm.