text:
Is it feasible to measure the pH of solution consisting of hydrofluoric acid, methanol and water using redox potential method?
Is it feasible to measure the pH of $\ce{HF}:\ce{H3COH}:\ce{H2O}$ solution using redox potential method?
Can I use this method to calculate the $\ce{H+}$ ions in this solution and specifically, can I measure the changes in pH simultaneously as auto ionization is taking place?


text:
Which compound should I avoid in deodorants to avoid "sticky, unwashable" feeling?
I tried some personal deodorants to avoid the smell of sweat, both in stick, no gas, gas sprays. Some can be washed away easily with body soap on the next day, some require either hair shampoo (because it is significantly harsher than body soap and washes much better) or to wait for two days for the effect to disappear by itself (whether the compounds break down or the compounds get transferred to the shirt is untested).
I don't like to have on me a product I cannot wash away, so I would like to know which compounds in the deo's I should check to avoid the sticky/unwashable effect.
The answer should list or indicate the compounds or rules in a way that people with little knowledge of chemistry can understand, like "the name comprises this string" or something like that. Generic classes which include compounds with a significantly different name is regarded as too complex.
Trying each product may work of course, but then quite some flacons would end up wasted or I would need to use products I don't like for a long time (one flacon lasts months, usually). Also, the composition may change.

Who voted this question down? It's a good question.

Especially without explaining the reason

Regrettably, most deodorants work as antiperspirants,  by clogging the pores of apocrine glands. These include salts of aluminum and zirconium, which hydrolyze (break up and combine with water). One example is aluminum zirconium tetrachlorohydrex glycosine complex, which works by forming a sticky, colloidal gel with water (and sweat). It's like the tale of the scorpion and the frog -- its their nature to act that way.
A bad alternative is deodorants with antibacterial compounds such as triclosan, which don't block sweating, but kill some odor-causing bacteria. Some of these antibacterials are endocrine disruptors, and all promote a population of antibacterial-resistant germs on the skin.
Traditionally, personal deodorants were just masking scents, and in 16th century Europe, flowers were worn to cover the stench... personal and environmental.

Interestingly, the sticky effect is something I don't remember before about 5 years ago. Maybe the market has changed. I will test whether "48 hours" products present more or less of the effect. I will edit the answer in a couple of days to post a known "sticky" composition and a known "non-sticky" one.

If those two solutions are the only solutions used in deodorants, I guess it's much better to stop using them.

@FarO, there are also masking scents, which are stronger than B.O.

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text:
Why does the O atom of N-phenylnitrosamine donate its lone pair towards H+ when O is more electronegative than N?

Nitrogen has lone pair as well which it can donate towards the proton but instead we see the oxygen being protonated despite being more electronegative. Also, this is not like the next part where $\ce{H3O+}$ is formed which is a good leaving group.

The N can protonate, but this is reversible and there is no reaction pathway forward from the protonated form, so it is of no relevence to the mechanism of diazo formation.

If you look at complete mechanism, this wasn't hard to understand why only oxygen get protonated instead of readily available nitrogen, which is supposedly more nucleophilic than oxygen. OP has given N-phenylnitrosamine as the direct product from reaction between aniline and nitrosonium ion, avoiding the first intermediate I (see following reaction mechanism of formation of diazonium ion). As Waylander correctly pointed out in his comment that all of these steps are reversible until you get diazonium ion (counterion $\ce{X-}$ is chloride ($\ce{Cl-}$) in this case):

As I emphasized in the Blue box of the depicted diagram, nitrosamine can be protonated on either $\ce{N}$ (see in blue box) or $\ce{O}$ (see in the total mechanism). As shown in the total mechanism, protonation is slightly favored towards $\ce{O}$ because charged intermediate is resonance stabilized. Nonetheless, this stabilized intermediate is the one directing to the expected product, diazonium salt (see also Waylander comment), making it more favorable. Also note that protonation on $\ce{N}$ make the intermediate going back to original first intermediate I, which cannot give the any other product.
Note: Also keep in mind that direct conversion from intermediate I to resonance stabilized, $\ce{O}$-protonated intermediacte can be achieved by $\ce{N}$ to $\ce{O}$ proton transfer, which is highly likely.

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text:
Why does cobalt(II) chloride (CoCl2) crystal melts?
I made $\ce{CoCl2}$ crystals by making a concentrated solution. After the crystals formed I took them out of the solution for drying.
 After a time I noticed that they were still wet so I dried off the water with a cloth and left them. Again they got wet and it seemed that the crystals just melted enough to get wet. No matter how many times I wipe them they again get wet.
Currently I am keeping them in an airtight container inside refrigerator.
So why does it happen and is there a way to preserve $\ce{CoCl2}$ crystals?

My explanation: I think that it is possible that $\ce{CoCl2}$ crystals that I made exist as $\ce{CoCl2.2H2O}$ (which is purple, as are my crystals) and it reacts with humidity to form $\ce{CoCl2.6H2O}$ (pink in color, as was the wet melted crystal)
Though I may be wrong.

What color was your product initially and afterwards? You likely isolated crystals of a hexahydrate which is highly deliquescent.

I have added in the question, anyway the salt was dark purple-pink, the crystals are very dark purple and the melted crystal is dark pink. So how do I preserve deliquescent crystals ?

Well, your explanation makes sense, the salt is hygroscopic and forms higher crystallohydrates gradually which are all deliquescent. There is no melting to speak of, the salt is getting dissolved in water vapors from the air, that's all. Are you asking why this is happening in this particular case or how to prevent it?

Did you flush the airtight container with dry N2 before sealing it? That's a good way to keep things dry. Also put something in the container that will absorb any remaining moisture.

@andselisk I was asking why this was happening and how to prevent it. I guess keeping in airtight container and removing moisture is the only way? Can I try coating the crystals with something like transparent nail polish ?

@NilakshSingh Sure you can use coating if you keep crystals for the demonstrative purpose only; otherwise you need to either store it with a stronger (than $\ce{CoCl2}$) desiccant or, as Andrew suggested, in the dry/inert atmosphere.


text:
What is wrong in this argument that dG must always be zero?
Under constant temperature and pressure, the change of the Gibbs free energy can be written as
$$\mathrm dG_\textrm{sys} = \mathrm dH_\mathrm{sys} - T\,\mathrm dS_\mathrm{sys}$$
And in the textbook that I have (Theodore L. Brown, Pearson), when deriving this equation, they use the equation
$$\mathrm dS_\mathrm{univ} = \mathrm dS_\mathrm{sys} - \mathrm dH_\mathrm{sys}/T$$
I can understand to this step but the question occurs. Can't we write $\mathrm dS_\mathrm{sys}$ as $\mathrm dH_\mathrm{sys}/T$ also? If this T and the T that is used in the equation
$$\mathrm dS_\mathrm{univ} = \mathrm dS_\mathrm{sys} - \mathrm dH_\mathrm{sys}/T$$
are the same, than shouldn't $\mathrm dG$ always be zero? At which point am I wrong?

Ah, I've figured out what the problem is: $\mathrm dS_\mathrm{sys}$ is only equal to $\mathrm dH_\mathrm{sys}/T$ for a reversible process, in which case $\mathrm dG_\mathrm{sys}$ is indeed zero.

Your textbook's derivation is done under the assumption of constant $T$, which means $T_{sys} = T_{surr} =T$.  However, this does not mean $dG_{sys}$ is always zero.  Let's start with the following:
$$dS_{univ}=dS_{sys}+dS_{surr}= \frac{\text{đ}q_{rev, sys}}{T_{sys}}+\frac{\text{đ}q_{rev, surr}}{T_{surr}}$$
Since heat flow always affects the surroundings reversibly (the surroundings are in the limit of being infinitely large, and thus heat flow affects them only infinitesimally), the reversible heat flow into the surrondings equals the actual heat flow, which is the negative of the actual heat flow into the system.  And since $T_{sys} = T_{surr} =T$, we have:
$$\frac{\text{đ}q_{rev, surr}}{T}=-\frac{\text{đ}q_{sys}}{T}$$
Which gives:
$$dS_{univ}=dS_{sys}+dS_{surr}= \frac{\text{đ}q_{rev, sys}}{T}+\frac{\text{đ}q_{rev, surr}}{T}=\frac{\text{đ}q_{rev, sys}}{T} -\frac{\text{đ}q_{sys}}{T}$$
Under the restrictions of constant $p$, and no non-$pV$ work,
$$\text{đ}q_{sys}=dH_{sys}$$
Hence:
$$dS_{univ}=\frac{\text{đ}q_{rev, sys}}{T} -\frac{dH_{sys}}{T} = dS_{sys} - \frac{dH_{sys}}{T},$$
which is what your textbook provides.
Note, however, that
$$\frac{dH_{sys}}{T} = \frac{\text{đ}q_{sys}}{T} \ne \frac{\text{đ}q_{rev,sys}}{T}=dS_{sys},$$
unless the process is reversible, i.e., non-spontaneous, in which case we could replace the "$\ne$" with "=", and $dG_{sys}$ would equal $0$.
This connects directly to the great beauty of using $\Delta G_{sys}$ at constant $T \text{ and } p$, when there is no non-$pV$ work: Under those conditions, $\Delta G_{sys} = -T \Delta S_{univ}$.  I.e., under those conditions, we can use $\Delta G_{sys}$ (something we can calculate) as a surrogate $\Delta S_{univ}$, whose value is the criterion for spontaneity (and is something we typically can't calculate directly).
Here's how we get there:
If the system is at constant $T$:
$$\Delta G_{sys} = \Delta H_{sys} - T \Delta S_{sys}$$
That's because: $G=H-TS => dG = dH - d(TS) = dH -TdS -SdT$; at const $T, dG= dH -TdS $; note that a constant-$p$ restriction is not required to write this.
As we mentioned above, if we add the restrictions of constant $p$, and no non-$pV$ work, we obtain:
$$\Delta H_{sys} = -T \Delta S_{surr}$$
This in turn gives:
$$\Delta G_{sys} = -T \Delta S_{surr} - T \Delta S_{sys} = -T \Delta S_{univ}$$

For a microscopic step at constant $T$ and $p$ $$\mathrm dG=0\tag{constant $T$ and $p$}$$ implies:

reversibility (equilibrium)
$\mathrm dS_\mathrm{univ} = 0$
$\mathrm dH_\mathrm{sys} = T\,\mathrm dS_\mathrm{sys}$ since $\mathrm dG = \mathrm dH_\mathrm{sys} - T\,\mathrm dS_\mathrm{sys} \tag{constant $T$ and $p$}$

The derivation you suggest seems strange.
A derivation that shows why the Gibbs free energy serves as a condition of spontaneity under constant $T$ and $p$ could start from the expression
$$\begin{align}
\mathrm dS_\mathrm{univ} &= \mathrm dS_\mathrm{sys} + \mathrm dS_\mathrm{surr} \\
&= \mathrm dS_\mathrm{sys} - \frac{\mathrm dq}{T} \\
&\ge 0
\end{align}$$
Combining this result with the 1st Law leads to the condition
$$\mathrm dG\le 0 $$


text:
Do the procedure used to find the empirical formula from combustion analysis be extended in other cases when is not specifically mentioned?
The situation is as follows:

A sample of $\pu{19.40 g}$ from a certain compound consisting only in $\ce{K, Cr}$ and $\ce{O}$, receives a treatment. This gives $\pu{7.6 g}$ of $\ce{Cr2O3}$ and $\pu{9.70 g}$ of the compound is transformed into $\pu{13.85 g}$ of $\ce{KClO4}$. Using this information find the empirical formula.

In this particular situation there isn't given any specification on what sort of treatment the sample received. Given such situation, can the steps used in finding the empirical formula during combustion analysis be used as well?.
Typically in the mentioned scenario (combustion analysis) a mass of the sample is given (as in the problem).
After that the sample is burned in oxygen. From this we determine the masses of the combustion products $\ce{CO2}$, $\ce{H2O}$, $\ce{N2}$, $\ce{SO2}$ and using this mass we'll find the number of moles of each combustion product and use the atomic masses to calculate the masses of the elements other than oxygen in the original sample.
Such mass is found by subtracting the mass of the sample to that of the combined mass of the other combustion products. Later, percentages can be used to a $100\,g$ sample to find the number of moles of $\ce{C}$, $\ce{H}$, $\ce{N}$ and $\ce{S}$. Then all is left to divide the moles of each element by the moles of elements present in the smallest ammount. In the last step only is needed a multiplication by non integral ratios as necessary to give small whole numbers.
Using the same logic I've attempted to solve this riddle, but would that be also applicable to any sort of problem involving finding the empirical formula?.
First I did found the grams of Chromium in the given $\pu{7.6 g}$.
Assuming FW for $\ce{Cr2O3} = 2(52)+3(16)=\pu{152 gmol-1}$
$\pu{7.6 g}\,\ce{Cr2O3} \times \frac{2\times \pu{52 g}\,\ce{Cr}}{\pu{152 g}\,\ce{Cr2O3}}=\pu{5.2 g}\,\ce{Cr}$
FW for $\ce{KClO4} = 39+35.5+4(16)= \pu{138.5 g mol-1}$
Then:
$\pu{13.85 g}\,\ce{KClO4} \times \frac{1\times \pu{39 g}\,\ce{K}}{\pu{138.5 g}\,\ce{KClO4}}=\pu{3.9 g}\,\ce{K}$
Then the rest of the mass should be the oxygen, subtracted from the sample:
$19.40-(3.9+5.2)=10.3$
Therefore the moles for each would be as follows:
$\pu{5.2 g}\,\ce{Cr}\times \frac{\pu{1 mol}\,\ce{Cr}}{\pu{52 g}\,\ce{Cr}}=\pu{0.1 mol}\,\ce{Cr}$
$\pu{3.9 g}\,\ce{K}\times \frac{\pu{1 mol}\,\ce{K}}{\pu{39 g}\,\ce{K}}=\pu{0.1 mol}\,\ce{K}$
$\pu{10.3 g}\,\ce{O}\times \frac{\pu{1 mol}\,\ce{O}}{\pu{16 g}\,\ce{O}}=\pu{6.4375 mol}\,\ce{O}$
Dividing between the smallest yields:
$\frac{\pu{0.1 mol}\,\ce{Cr}}{0.1}=1$
$\frac{\pu{0.1 mol}\,\ce{K}}{0.1}=1$
$\frac{\pu{6.4375 mol}\,\ce{O}}{0.1}=6.4375$
Which would give:
$\ce{K1Cr1O_{6.4375}}$
However this isn't very convincing to me. However since the chromium and potassium are in the same proportion it looks very suspiciously to be potassium dichromate. But does it exist a justification for this. Can somebody help me to clear out the basis for the right calculation?. The most importantly what should be subtracted and from where?.

The wording is ambiguous and potentially misleading. Confirm two things with the source: 1) That the compound contains only K, Cr, O or some other atoms might be there and 2) should you read that 19.40 g produce 7.6 g of Cr2O3 and 13.85 g KClO4 or that 19.40 g produce 7.6 g of Cr2O3 and 9.7 g of the compound produce 13.85 g of KClO4 .

@permeakra The source implies that such compound has only K, Cr and O. I became confused at how exactly should I understand those $9.7,g$ of the compound.

I'm voting to close this question as off-topic because the exercise you are discussing is no sufficiently well-defined to give an unambiguous answer. It would help to cite the source of the exercise.

@KarstenTheis This question comes from a collection of riddles with no author so I dont know exactly how I can met the criteria you are asking. The original source is not in english and I had to translate it the closest possible to the original text, but I will review it to update this question.

@KarstenTheis I've already updated the question and corrected any misleading clue in the question and added what was being asked. As it stands now the question is well defined with what is requested in this community.

The "riddle" as translated by the OP reads:

A sample of 19.40 g of a certain compound containing K,Cr and O, receives a treatment. This gives 7.60 g of Cr2O3 and 9.70 g of the compound is transformed into 13.85 g of KClO4.

In the answer by Mathew Mahindaratne, this is interpreted as (I paraphrase):

19.40 g of a compound containing K, Cr and O exclusively reacts to form 7.60 g of $\ce{Cr2O3}$. In this reaction, the amount of chromium in reactant and product are equal (i.e. there is no other reactant or product containing chromium, and the yield is 100%).
9.70 g of the same starting material reacts to form 13.85 g of $\ce{KClO4}$. Likewise, the amount of potassium in reactant and product are equal (i.e. there is no other reactant or product containing chromium, and the yield is 100%).

Without giving the numerical answer, I will sketch how you could solve this:

Find the mass percentages of chromium and potassium in the original compound. First, calculate the mass of these elements in the product from mass of the product using the known composition. Then, divide by the mass of the reactant to get the mass percentage.
Calculate the missing mass percentage of oxygen (percentages have to add up to 100%). Then, proceed as you would usually to get the empirical formula from the mass percentages.

For the numerical answer, work through the correct answer by Mathew Mahindaratne.

I made a different approach. I transform the second experiment in such a way that the same amount of the initial compound $\ce{K_xCr_yO_z}$ is used in both experiments. As the second experiment uses exactly one half of what was used in the first, I can simplify both data in saying that 19.4 g of the compound react with something containing at least Cl to produce 7.6 g $\ce{Cr2O3}$ and 27.718 g $\ce{KClO4}$. In moles, it produces 0.05 mole $\ce{Cr2O3}$ and 0.2 mole $\ce{KClO4}$. 0.2 is 4 times 0.05. So the equation must be something like :
...  +  ... -> $\ce{Cr2O3}$  + 4  $\ce{KClO4}$
On the other hand 19.4 g is exactly 0.1 mole of $\ce{K2CrO4}$. Furthermore, 0.1 mol $\ce{K2CrO4}$ produces exactly 0.05 mol $\ce{Cr2O3}$ whatever its reaction. So the equation is probably :
2 $\ce{K2CrO4}$  +  ....   ->  $\ce{Cr2O3}$  +  4 $\ce{KClO4}$
By subtracting atoms in the equation, one finds that the mysterious missing substance must contain 4 Chlorine atom plus (19 - 8 =) 11 atoms Oxygen. It might be $\ce{Cl4O11}$. But, in my opinion, this does not seem reasonable.
Another way of handling the problem is to weigh for both experiment the amount of the mysterious substance. It is :  7.6 + 27.7 - 19.4 = 15.918 g. Surprisingly enough, this mass is exactly equal to 0.05 mol $\ce{Cl4O11}$. So the equation is :
2 $\ce{K2CrO4}$ + $\ce{Cl4O11}$  --> $\ce{Cr2O3}$ +  4 $\ce{KClO4}$
There must be a mistake somewhere, because I seriously doubt that $\ce{Cl4O11}$  exist. Never heard of this strange molecule !! Or maybe it is a mixture of existing molecules like $\ce{Cl2O7}$  +  2 $\ce{ClO2}$ .. Who knows ?
I would be pleased to hear any sensible reaction to my odd calculations.

You should probably try K2Cr2O7, or perhaps don't wate time on such vague question. What is "treatment" there? I doubt if values even need to add up.

K2Cr2O7 does not work. It yields non integers as coefficients

@Maurice Interesting observation. How do you come to the conclusion that potassium dichromate does not yield integers can you show this?.

$$\ce{4K2CrO4 + 4Cl2 + 11O2 -> 2Cr2O3 + 8KClO4}$$ would work.

OP's question:

A sample of $\pu{19.40 g}$ of a certain compound containing $\ce{K, Cr}$ and $\ce{O}$, receives a treatment. This gives $\pu{7.60 g}$ of $\ce{Cr2O3}$ and $\pu{9.70 g}$ of the compound is transformed into $\pu{13.85 g}$ of $\ce{KClO4}$.

And then, OP has wonder, given only above information where no other information is given regarding how to solve it, which would be the most adequate way to find the empirical formula of such compound. And somehow, OP has taken a correct approach to solve the problem, but lost in close to the end of it. I'd put an answer using OP's approach and showing where OP assume wrong:
I assumed the molecular formula of original compound is $\ce{K_xCr_yO_z}$. First, OP found the amount of $\ce{Cr}$ in the given $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$ using the first given information (OP has worded it wrong but done the correct calculation). The first information given is: When $\pu{19.40 g}$ of a certain compound containing $\ce{K, Cr}$ and $\ce{O}$ is treated in certain way, it would give $\pu{7.60 g}$ of $\ce{Cr2O3}$. Actually, $\ce{Cr}$ contains in this $\pu{7.60 g}$ of $\ce{Cr2O3}$ is equal to $\ce{Cr}$ contains in $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$, if the conversion is $100\%$.
Also, OP should know that formula weight of $\ce{Cr2O3}$ is $\pu{151.99 gmol-1}$ (This is a fact so does not have to assume it). Hence:
$$\pu{7.60 g}\text{ of }\ce{Cr2O3} \times \frac{2\times \pu{52.0 g}\text{ of } \ce{Cr}}{\pu{151.99 g}\text{ of }\ce{Cr2O3}}=\pu{5.20 g}\text{ of }\ce{Cr}\\ (\text{this is the amount of $\ce{Cr}$ in $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$})$$
This can also be used to find the mass of $\ce{Cr}$ in $\pu{1.00 g}$ of $\ce{K_xCr_yO_z}$ at this point if needed:
$$\frac{\pu{5.20 g}\text{ of }\ce{Cr}}{\pu{19.40 g} \text{ of } \ce{K_xCr_yO_z}} = \pu{0.268 g} \text{ of }\ce{Cr} \text{ in } \pu{1.00 g}\text{ of }\ce{K_xCr_yO_z}$$
Keep in mind that the usefulness of this unit value is evidence in the calculations of the next step.
Second, OP correctly found the amount of $\ce{K}$ in $\pu{9.70 g}$ of $\ce{K_xCr_yO_z}$, using the second given information. The second information given is: When treated, $\pu{9.70 g}$ of original compound ($\ce{K_xCr_yO_z}$) gives $\pu{13.85 g}$ of $\ce{KClO4}$. However, it seems, OP has wrongly assumed that the amount of $\ce{KClO4}$ given here is for $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$ used in first conversion. Actually, amount of $\ce{K}$ contains in this $\pu{13.85 g}$ of $\ce{KClO4}$ is equal to $\ce{K}$ contains in only $\pu{9.70 g}$ of $\ce{K_xCr_yO_z}$, assuming the conversion is $100\%$ again.
Again, the formula weight of $\ce{KClO4}$ can be found in online easily, which is $\pu{138.55 g mol-1}$. Now, we can plug in the values as follows:
$$\pu{13.85 g}\text{ of }\ce{KClO4} \times \frac{1\times \pu{39 g}\text{ of }\ce{K}}{\pu{138.55 g}\text{ of }\ce{KClO4}}=\pu{3.90 g}\text{ of }\ce{K} \\ (\text{this is the amount of $\ce{K}$ in $\pu{9.70 g}$ of $\ce{K_xCr_yO_z}$})$$
For convenience, let's use this value to find the mass of $\ce{K}$ in $\pu{1.00 g}$ of $\ce{K_xCr_yO_z}$:
$$\frac{\pu{3.90 g}\text{ of }\ce{K}}{\pu{9.70 g} \text{ of } \ce{K_xCr_yO_z}} = \pu{0.402 g} \text{ of }\ce{K} \text{ in } \pu{1.00 g}\text{ of }\ce{K_xCr_yO_z}$$
At this point, thinking above calculated ${\pu{3.90 g}\text{ of }\ce{K}}$ is from $\pu{19.40 g} $ of $ \ce{K_xCr_yO_z}$, OP has wrongfully assumed that the rest of the mass in $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$ is oxygen. Simply, what OP should have done is make a simple conversion to find the correct amount of $\ce{K}$ in $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$:
$$\text{The amount of $\ce{K}$ in $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$}\\=\pu{3.90 g}\text{ of }\ce{K} \times \frac{\pu{19.40 g} \text{ of }\ce{K_xCr_yO_z}}{\pu{9.70 g} \text{ of }\ce{K_xCr_yO_z}}\\=\pu{7.80 g}\text{ of }\text{of } \ce{K} $$
Or, you can simply multiply above unit value ($\pu{0.402 g} \text{ of }\ce{K} \text{ in } \pu{1.00 g} \text{ of }\ce{K_xCr_yO_z}$) by $\pu{19.40 g}$: $0.402 \times 19.40 = 7.8$.
Now, it is safe to assume rest of $\pu{19.40 g}$ is $\ce{O}$, since the compound contains only $\ce{K, Cr}$, and $\ce{O}$. Thus, you can subtract to find the amount of $\ce{O}$ in $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$ as follows:
$\text{The amount of $\ce{O}$ in $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$} =\pu{19.40 g}- \pu{(7.80 + 5.20) g} = \pu{6.40 g}$
Now, the easiest way to calculate formula is as follows:
$$\text{The mass ratio of $\ce{K:Cr:O}$ in $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$} = \pu{7.80 g}: \pu{5.20 g}:\pu{6.40 g}$$
$$\text{The mole ratio of $\ce{K:Cr:O}$ in $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$ }(x:y:z)\\ = \frac{\pu{7.80 g}}{\pu{39.10 gmol-1}}: \frac{\pu{5.20 g}}{\pu{52.00 gmol-1}}:\frac{\pu{6.40 g}}{\pu{16.00 gmol-1}} \\=\pu{0.2 mol}:\pu{0.1 mol}:\pu{0.4 mol} \\=\pu{2 mol}:\pu{1 mol}:\pu{4 mol}$$
This ratio would give the molecular formula: $\ce{K2CrO4}$.

I'm really sorry for any word which has not been used accurately and may had caused any misconception, but the original source did not helped much neither. Btw, when I mentioned assuming I intended to say that I'm rounding the masses for the atoms $K$, $Cl$ and $O$ to their nearest integers or least decimals. I'm well aware that this is a fact. I'm a bit lost about the conversion factor. How exactly should I understand $\frac{\textrm{19.40 g},\textrm{K}_x\textrm{Cr}_y\textrm{O}_z}{\textrm{9.70 g},\textrm{K}_x\textrm{Cr}_y\textrm{O}_z}$?

Does it imply $\textrm{19.40 g},\textrm{K}_x\textrm{Cr}_y\textrm{O}_z= \textrm{9.70 g},\textrm{K}x\textrm{Cr}y\textrm{O}z$? Why the $5.2,g,\textrm{Cr}$ is the same in quantity that in the sample and a conversion factor must be used for the $3.9,g,\textrm{K}$. Aren't those grams the same what is in the sample?. I am getting the idea that in the passage is stressed that $19.40,g$ of the compound is transformed totally into $Cr{2}O{3}$? But when it mentions $9.70,g$ of the compound is transformed into $\textrm{KClO}{4}$.

Doesn't mean that $19.40-9.70=9.70$ is transformed in $\textrm{Cr}{2}\textrm{O}{3}$? and a conversion factor must be used for $\textrm{Cr}$ as well?. Can you explain this part to me please?

It is a mathematical ratio, you learned from algebra: If $2x$ has $4y$s, then how many $y$s in $20x$? To solve it, find how many $y$s in $1x$: $\frac{4y}{2x}=2y$ per $x$. This is the unit. Then multifly it by $20x$ to find the answer, which is $40$ $y$s.

Yes I know it is a ratio. I think the part where I got confused is why should be considered a ratio. Perhaps is due this is meant to concentration and this is preserved constantly to the whole substance. Therefore the ratio can be used.

In, real problem, $\pu{9.7 g}$ of $\ce{K_xCr_yO_z}$ gives $\pu{3.9 g}$ of $\ce{K}$ so to find how many $\pu{g}$ of  $\ce{K}$ in $\pu{1.0 g}$ of $\ce{K_xCr_yO_z}$, you divide: $\frac{3.9}{9.7} = 0.402$ (this is the unit means per gram of your compound).  Now, to find $\pu{g}$ of  $\ce{K}$ in $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$, you multify: $0.402 \times 19.40 = 7.8$.

Can you upgrade your last comment as part of your answer?. I think it is very relevant as means of clearing doubts of the meaning in conversion units. There's an unanswered doubt which is regarding why subtraction must be made to find the oxygen?. It doesn't say specifically that it is a combustion analysis. I'm assuming that since oxygen is a gas it is measured indirectly as it is not collected as a solid given the salt and the oxide mentioned in the problem. Am I right with this assumption?.

My answer is updated as you requested.

Formatting note: please avoid using "textual formulas" where unit symbols are followed by text of chemical formulas. This is incorrect, makes parsing harder and leads to strange results such as $\pu{3.90 g K}$ which reads as 3.90 gram-kelvin.

@andselisk: Corrected all possible ones as suggested.

I also meant there should be no prepositions as well, it's better to keep textual explanation to the preceding or following text paragraph and use symbols and labels for the formulas. For example, instead of $$\pu{7.60 g}\text{ of }\ce{Cr2O3} \times \frac{2\times \pu{52.0 g}\text{ of } \ce{Cr}}{\pu{151.99 g}\text{ of }\ce{Cr2O3}}=\pu{5.20 g}\text{ of }\ce{Cr}\ (\text{this is the amount of $\ce{Cr}$ in $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$})$$ in my opinion it would be more correct to use the following:

Mass of chromium $m(\ce{Cr})$ in $\pu{19.40 g}$ of $\ce{K_xCr_yO_z}$:
$$m(\ce{Cr}) = m(\ce{Cr2O3})×\frac{2 × m(\ce{Cr})}{m(\ce{Cr2O3})} = \pu{7.60 g} × \frac{2 × \pu{52.0 g}}{\pu{151.99 g}} = \pu{5.20 g}$$


text:
Relation of temperature and intensive property
I came across a question that inquired about whether there exists any intensive property of a system, that is independent of temperature and then seeks proof for the presence/absence of such properties, from a quasi-theoretical standpoint. The question mentions that the system shall be assumed to be non-ideal (ideal gas laws don't hold true et al).
How to proceed, for the latter? Does there exist any such property, in the first place?

Make a list of intensive properties from here https://en.wikipedia.org/wiki/Intensive_and_extensive_properties#Examples, I can think of concentration as one example.

Concentration depends on temperature via temperature dependent density.

@Poutnik, yes but what does density have to do with concentration? If we have 1 M NaCl in a closed container in water, what will temperature do to concentration, as long as there is no phase change?

Liquid thermal dilation decreases concentration.

@M. Farooq If you heat up 1 M NaCl to 100 deg C, the concentration will be several percent lower, as the volume will be several percent bigger.  It affects also accurate preparation of titration stock solution and volumetry vessel volume calibration.

After looking at the list in the Wikipedia article linked in the comment by M.Farooq, I suspect there are few properties that are guaranteed to be independent of temperature, an exception being molality and mole fraction in a closed system and absence of reactions. Other concentrations that are functions of volume are bound to be functions of temperature. Even in the presence of reactions, the atomic mole fraction is bound to be a constant (absent nuclear reactions, that is).
Standard molar quantities and some other properties are defined at fixed temperatures and are therefore strictly temperature independent, but that makes for a circular argument, and presumably this is not the type of property the question is about.

The atomic/molecular/formula weight (choose whichever is appropriate) of a substance is an intensive property (it is inherent to the substance itself, and is independent of how much of the substance you have), and is independent of temperature (if you raise the temperature high enough to cause a chemical or nuclear reaction, it's no longer that substance).
If you wanted to apply this to a system, it's straightforward if the system consists of only a pure substance.  Otherwise you could say that, if the system is closed, in the absence of chemical and nuclear reactions, the average atomic/molecular/formula weight is independent of temperature.


text:
Can hydrogen peroxide be considered an alkali?
Hydrogen peroxide is miscible with water, can neutralize acids, and produces H2O, but I've never seen it labeled as an alkali, could we still consider hydrogen peroxide an alkali?

No, and it doesn't neutralize acids. Even if it did it wouldn't be enough. And what gave you such idea?

Interestingly, H2O2 has been considered an acid because its dissociation constant is smaller than that of H2O:
"The available literature indicate that hydrogen peroxide is a weak inorganic acid with a dissociation constant between 2.24 and 2.4E-12 at 25 °C, which corresponds to a value for pKa ranging from 11.62 to 11.65.
Hydrogen peroxide - Registration Dossier - ECHA
echa.europa.eu/registration-dossier/-/registered-dossier/15701/4/22"
However, it would hardly be useful in neutralizing an alkali, but strong bases would generate OOH- ions.


text:
Reduction of amino acids to corresponding amino alcohols
I've stumbled upon an interesting article about reduction of amino acids and from there to reduction of L-valine to L-valinol using $\ce{LiAlH4}$ in THF.
Article says that because of the cost of $\ce{LiAlH4}$, better method using $\ce{NaBH4}$ with $\ce{I2}$ in THF has been developed.
But I haven't found anything about $\ce{HI}$. Is it possible to reduce amino acids all the way down to their corresponding amines with concentrated $\ce{HI}$ solution?

For alternative process: http://www.orgsyn.org/demo.aspx?prep=CV7P0530.

No, all you would get is the HI salt of the amino acid. The I2/NaBH4 system works because it produces BH3 in situ. HI would not do that.

And what would you propose to reduce the hydroxyl group? The best option would be without need to protect the amino group. So, HI/P reduction might work here?

I would use the BH3/Me2S procedure from the orgsyn reference above. I would not use HI/P for anything

Doesn't BH3/Me2S reduce only to alkanoamines? I meant all the way to amine. And what is the problem with HIP method?

If you want to reduce the alcohol to the methyl group I would tosylate, displace with Me or PhS- then reduce with Raney Ni. You'll need the NH protected

I fail to follow your reasoning in this question. HI doesn’t exactly strike me as a commonly used and/or cheap reducing agent.

Chiral amino alcohols such as L-valinol are generally prepared by the reduction of corresponding α-amino acids and suitable reducing reagent. Most commonly used reducing agent is $\ce{LiAlH4}$ since most common reducing reagent, $\ce{NaBH4}$ does not reduce carboxylic acid to corresponding alcohols. However, use of $\ce{LiAlH4}$ has notable disadvantages such as being expensive and highly flammable. To avoid these disadvantages on high scale syntheses, some novel methods have been developed. One such method is reduction of carboxylic acid using $\ce{NaBH4/I2}$ reducing sysrem (Ref.1). The most remarkable feature of this reagent is its selectivity over other known strong reducing agents. For example, if $\ce{NaBH4/I2}$ is used to reduce bicarboxylic acid, it would reduce both carboxylic groups to $100\%$. However, if one carboxylic acid i
 group was protected (say, ester-carboxylic acid combo), $\ce{NaBH4/I2}$ would reduce only carboxylic acid group leaving ester group alone (Ref.1).
Two years after this important discovery, McKennon, et al. (1993) found out that this same method can be used to reduce N-protected or naked amino acids to their corresponding N-protected amino alcohol or amino alcohol, respectively (Ref.2). A typical procedure is as follows:

Preparation of (S)-tert-Leucinol: A dry 5-L three-necked round-bottom flask equipped with a mechanical stirrer and a 250-mL addition funnel was charged with sodium borohydride ($\pu{55.6 g}, \pu{1.47 mol}, \pu{2.41 equiv}$) and $\pu{1.6 L}$ of anhydrous THF. The solution was stirred while (S)-tert-leucine ($\pu{80.0 g}, \pu{610 mmol}$) was added in one portion. The flask was cooled to $\pu{0 ^\circ C}$ and fitted with a reflux condenser. The addition funnel was charged with a solution of $\ce{I2}$ ($\pu{155.0 g}, \pu{610 mmol}, \pu{1 equiv}$) in $\pu{200 mL}$ of THF, which was added dropwise to the flask over a $\pu{1.5 h}$ period with considerable gas evolution. The solution was allowed to warm to room temperature. When the brown color had dissipated to give a cloudy white solution, the reaction was brought to reflux for $\pu{19 h}$. The cloudy white suspension was cooled to room temperature with the aid of a water bath. The addition funnel was charged with $\pu{150 mL}$ of $\ce{MeOH}$, which was added drop-wise with rapid stirring. Vigorous gas evolution was observed. Small aliquots of $\ce{MeOH}$ were then added until all of the solid white material had dissolved. The solution was concentrated by rotary evaporation to give a white pasty oil that was dissolved in $\pu{1.20 L}$ of $20\% (w/w)$ aqueous $\ce{KOH}$ and mechanically stirred for $\pu{6 h}$ at room temperature. The light green solution was extracted with $\ce{CH2Cl2}$ ($\pu{3 \times 1.50 L}, \pu{1 \times 600 mL}$). To minimize the emulsion, $\pu{200 mL}$ of brine was added to the aqueous layer after the first extraction. The combined organic extracts were dried over $\ce{Na2SO4}$, filtered through glass wool, and concentrated in vacuo to give $\pu{74.1 g}$ of a white crystalline solid in a thick, slightly yellow oil. Distillation of (S)-tert-leucinol, bp $70$-$\pu{73 ^\circ C}$/$\pu{2 mmHg}$ (Lit. $117$-$\pu{120 ^\circ C}$/$\pu{57 mmHg}$) from a $\pu{30.0 g}$ reduction reaction performed in the above manner yielded $\pu{21.3 g}$ of product ($79\%$ yield) as a clear oil, which solidified upon cooling to room temperature.

References:

J. V. Bhaskar Kanth, Mariappan Periasamy, “Selective reduction of carboxylic acids into alcohols using sodium borohydride and iodine,” J. Org. Chem. 1991, 56(20), 5964-5965 (https://doi.org/10.1021/jo00020a052).
Marc J. McKennon, A. I. Meyers, Karlheinz Drauz, Michael Schwarm, “A convenient reduction of amino acids and their derivatives,” J. Org. Chem. 1993, 58(13), 3568-3571 (https://doi.org/10.1021/jo00065a020).

Matthew, thank you for your answer. However, the main question was about HI reduction to amines. Nontheless, thank you for your contribution with many valuable information.

I did not address about reduction by $\ce{HI}$ since it wasn't clear whether you are mainly asking about $\ce{NaBH4/I2}$ system. However, to my knowledge, any $\ce{HI}$ system reducing carboxylic acid is not known. Only one existing is discussed here: https://chemistry.stackexchange.com/questions/15586/hi-p-reduction-mechanism.

I thought so as well, but then I managed to find some reference on sciencemadness forums. It's a paper discussing reduction of ketone to hydrocarbon with special section to HI here  : http://www.sciencemadness.org/talk/files.php?pid=161627&aid=8575

Thanks for sharing. It's also $\ce{HI/P}$ system as one in my previous comment. But it is interesting chemistry, which reduces aromatic aldehydes and carboxylic acids all the way to toluene (note that not to benzyl alcohol). I wonder it would do the same to aliphatic counterpart.


text:
Do tertiary ethers tolerate oxidative workup of an ozonolysis reaction?
I am attempting to cleave a double bond into a carboxylic acid and carbon dioxide using an ozonolysis reaction followed by an oxidative workup. I am concerned with the presence of a tertiary ether in my reactant, however. I am planning on using standard ozonolysis procedures as described in Vogel's Textbook of Practical Organic Chemistry on page 592 using an oxidative workup with either potassium permanganate or hydrogen peroxide. I am worried that the oxidative conditions will also affect the tertiary ether. I cannot find any literature detailing the compatibility of tertiary ethers with peroxide or permanganate. Is there another protocol that would be milder on my substrate, or does it seem as though this reaction would not affect other functional groups? Here is my scheme:

I don't think this will be a problem. Tertiary ethers are pretty robust.

Oxidation of ethers is normally an issue of there is a C-H bond directly adjacent to the oxygen. Your molecule has neither.

Since oxidation is not an issue, would using the permanganate workup be more effective due to the possibility of the formation of tertiary and benzyl radicals using the peroxide?

But you can't form any tertiary or benzyl radicals. There are no tertiary C-H or benzylic H's to abstract.

As an alternative, consider RuO4/NaIO4. https://en.wikipedia.org/wiki/Ruthenium_tetroxide

Ruthenium tetroxide would definitely provide for a more gentle oxidative cleavage of the double bond, and I've read about it along with cleavage using osmium tetroxide; however, because I will be performing a gram-scale synthesis, I think my best option would be ozonolysis followed by oxidative work up with silver oxide.

Osmium Tetroxide is available encapsulated which is much safer than using the neat liquid. The Johnson-Lemieux procedure as modifed here http://aether.cmi.ua.ac.be/artikels/MB_09623.pdf is very reliable.


text:
Nucleophillic addition reaction in the presence of an acid catalyst on a molecule with an alkene
This is a solution for a question in my textbook.

In the presence of an acid catalyst, am I wrong in my understanding that the acid will protonate the alcohol, and then do an electrophillic addition reaction onto the alkene (in addition to the nucleophillic addition reaction from the alcohol)? If so, was it omitted from the solution because it is not a major product (if it's not a major product, why)?

Alcohols are not nucleophilic enough to do Michel addition. If it is a thioalcohol, then it is a different story.

The acid will protonate the O of the carbonyl carbon, so a positive charge is generated. The oxygen of the alcohol with the free pair of electrons will act as nucleophile. So the alcohol is added to the cycle and a pair of electrons of the double bond of the carbonyl will compensate the positive charge generated. As the medium is acid, in the second step the hydroxyl group is protonated, so it is generated a water molecule, the alcohol attacks again at the same carbon and the water is released to the medium.
This is why this process is used to protect carbonyl groups. The reaction is reversible so if you add water, the reaction direction is reversed and you get the keto form again.


text:
Degree of dissociation of weak electrolyte at infinite dilution
Why the degree of dissociation of weak electrolyte is 1 at infinite dilution and how do we get the result? Like I know that at infinite dilution the concentration of solution is 0 but what does that have to do with degree of dissociation, secondly in oswald's law of dilution we already assume that dissociation is almost negligible so we can't obtain the result from that also

There are a number of ways to explain this. In words, the entropic gain with dissociation overwhelms any attraction between counterions as the electrolyte is diluted.
Mathematically: for a m:n electrolyte undergoing dissociation according to the reaction
$$\ce{A_mB_n -> mA^{\nu_A -} + nB^{\nu_B +}}$$
if the initial concentration of electrolyte $\ce{A_mB_n}$ is $c_{T}$, define the degree of dissociation as
$$\alpha = \frac{c_{AB}}{c_{T}}$$
where $c_{AB}$ is the concentration of dissociated $\ce{A_mB_n}$. Assuming ideality and unit molarity as standard concentration, the equation for the equilibrium constant can be written as $$K=\frac{c_A^m c_B^n}{c_{AB}}$$ and solved in terms of the degree of dissociation. For a 1:1 electrolyte this leads to
$$\alpha = \frac{\beta}{2}\left[\left(1 + \frac{4}{\beta} \right)^{1/2} -1\right] \tag{1}$$
where $\beta = K/c_T$. As $c_T \rightarrow 0$ it follows from this last expression that $\alpha  \rightarrow 1$ #QED
PS The following equation (from Wikipedia's page on Ostwald's law of dilution) $$\alpha=\sqrt{\frac{K}{c_T}} \tag{2}$$ does not hold over the entire concentration range $c_T$. It can be obtained from Eq. (1) by assuming $K<<1$ and therefore $\beta<<1$. For increasingly dilute solution Eq. (2) fails and  $\alpha  \rightarrow 1$.

Not always does $\alpha\to 1$.  Other answers give counterexamples.

@OscarLanzi Good point, I didn't even think about water autodissociation since I was considering simple electrolytes and not about other interfering equilibria.

For infinite dilution, water acts as a buffer with $\mathrm{pH}=7$.
Therefore, $\alpha=f(\mathrm{p}K_\mathrm{a})$.
For $\mathrm{p}K_\mathrm{a}=9$ is $\alpha=\frac{1}{101}\overset{cca}=0.01$.
For $\mathrm{p}K_\mathrm{a}=8$ is $\alpha=\frac1{11}$.
For $\mathrm{p}K_\mathrm{a}=7$ is $\alpha=0.5$, as only a half of the acid is dissociated.
For $\mathrm{p}K_\mathrm{a}=6$ is $\alpha=\frac{10}{11}$.
For $\mathrm{p}K_\mathrm{a}=5$ is $\alpha=\frac{100}{101}\overset{cca}=0.99$.
The same applies for conjugate acids of respective weak bases.

If the weak electrolyte is a weak acid, we have to reckon with the autoionization of water.  With the autoionization, the solvated ions formed from the acid do not reach infinite dilution and therefore the dissociation is limited to below 100%.
Let's look at acetic acid, for which (if I remember and can type correctly) $K_a=1.8×10^{-5}$.  Thereby*
$\dfrac{[\ce{H^+}][\ce{C2H3O2^-}]}{[\ce{HC2H3O2}]}=1.8×10^{-5}$
*Ion formulas omit water of solvation.
In the limiting case of infinite dilution of the acid we will have $[\ce{H^+}]\to10^{-7}$ forcing
$\dfrac{[\ce{C2H3O2^-}]}{[\ce{HC2H3O2}]}\to180$
and since the amount of acetate ion equals the amount of dissociated acetic acid molecules the degree of dissociation is limited to
$\dfrac{[\ce{C2H3O2^-}]}{[\ce{C2H3O2^-}]+[\ce{HC2H3O2}]}\to\dfrac{180}{181}=99.4\text{ per cent}$
The general formula for the dissociation limit would be
$\dfrac{K_a}{K_a + 10^{-7}}$
In the case of acetic acid the autoionization places very little limitation on the dissociation because the acid is comparatively strong versus the maximum autoionization from the water.  If we had chosen a weaker acid with $K_a<10^{-7}$ then the (relative) flooding of solvated hydrogen ions from autoionization would prevent the acid from ever reaching majority dissociation.  For example, $\ce{HCN}$ in the absence of anything that complexes cyanide ions would remain predominantly molecular in water solution, no matter how much it's diluted.


text:
Naming a coordination compound
I was asked to determine the coordination compound which has the name:
$$\text{triamminetrinitritocobalt(I)}$$
I thought it must be: $$\ce{[Co(NH3)3(ONO)3]^2-}$$
The $-2$ charge can be deduced by adding up the oxidation numbers.
However, in this case, shouldn't the name be written as cobaltate instead of cobalt? Where is my mistake and what did I miss here?

-ate is used for if metal has a latin name

In coordination compunds nomenclature, -ate is used when the complex has a net negative charge.

Yes, you are correct: an -ate prefix is used when the complex has a net negative charge and the name should be triamminetrinitritocobaltate(I) ion
Reference:

If the overall coordination complex is an anion, the ending "-ate" is attached to the metal center. Some metals also change to their Latin names in this situation. The rest of the metals simply have -ate added to the end (cobaltate, nickelate, zincate, osmate, cadmate, platinate, mercurate, etc. Note that the -ate tends to replace -um or -ium, if present).

Chemistry Libretexts - Nomenclature of Coordination compounds


text:
Identifying molecule with three adjacent six-membered rings from blurry image
I have found this isolated image on one of my old hard drives and I would like to know what molecule does the image represent:

Hint: It's a pothead tattoo

It appears to be Δ⁹-tetrahydrocannabinol, also known as Δ⁹-THC:

(Image source: 2D structure of tetrahydrocannabidiol (THC) by Harbin (Wikimedia))
The comment by The_Vinz about “pothead tattoo” was very … insightful.


text:
Neither Raman nor IR Active vibrational modes of water
How can we find vibration modes of water that are not in IR and Raman spectroscopy? I thought these modes creat when a transition happens between 2 modes i.e from mode to another mode.

Do you have a specific example of a vibration which neither changes a dipole moment (then spotted in IR spectroscopy), nor the polarizability of the electron cloud (Raman)?  Of course, there are vibrations spot with both techniques, too.


text:
What is J in the rigid rotor model?
In my class notes regarding the rigid rotor model, J is described as simply the angular momentum (replacing roe or p) and is also referred to as the angular momentum quantum number.
Are these two thing synonymous? If so in what way? If not, what do each mean and when is it appropriate to use which one?

TL;DR It depends on context and is unfortunately something that isn't going to change in the near future. However, the more you understand angular momentum, the easier it becomes to figure out the meaning from the context.

Quantum mechanical angular momentum is a tricky topic and it is not helped by some books which use confusing nomenclature. In QM, angular momentum is a vector quantity (much like in classical mechanics).* This is denoted with various letters;$^\dagger$ we will stick with J for the time being:
$$\text{angular momentum: }\mathbf{J} = (J_x, J_y, J_z)$$
Alternatively, you may also see vectors denoted with an arrow sign: $\vec{J}$.
All three components of the (QM) angular momentum are observable quantities. They are more commonly referred to as projections of the angular momentum onto an axis, which means the same thing. In any case, each one is associated with a Hermitian operator :
$$\text{ang. mom. projection operators: }\hat{J}_x, \hat{J}_y, \hat{J}_z$$
Apart from the $x$-, $y$-, and $z$-components of the angular momentum, the magnitude of the angular momentum is also an observable. This is similar to the length of the vector $\mathbf{J}$, and classically we would write $|\mathbf{J}|^2 = J_x^2 + J_y^2 + J_z^2$. The corresponding quantum mechanical operator is usually written without the boldface as
$$\text{ang. mom. magnitude operator: }\hat{J}{}^2 = \hat{J}{}_x^2 + \hat{J}{}_y^2 + \hat{J}{}_z^2$$
So, now, we have four unique operators: $\hat{J}{}^2, \hat{J}_x, \hat{J}_y, \hat{J}_z$. Because they do not all commute with one another, we usually focus on two of them, which are (conventionally) $\hat{J}{}^2$ and $\hat{J}_z$, and ignore the other two.
The quantum mechanical states of interest are those which are simultaneous eigenstates of these two operators, namely $\hat{J}{}^2$ and $\hat{J}_z$. It turns out that these eigenstates have the properties
\begin{align}
\hat{J}{}^2|\psi\rangle &= s(s+1)\hbar^2 |\psi\rangle \\
\hat{J}_z|\psi\rangle &= t\hbar |\psi\rangle \\
\end{align}
where $s$ is constrained to be a non-negative integer $0, 1, 2, \ldots$ and $t$ is constrained to be $-s, \ldots, -1, 0, 1, \ldots, s$. Different eigenstates will have different values of $s$ and $t$, which immediately suggests that we should label the eigenstates $|\psi\rangle$ by the values of $s$ and $t$ that apply to them. Constants such as $s$ and $t$ which can only take on certain values are called quantum numbers.
We can therefore associate the eigenstate $|\psi_{s,t}\rangle$ with the angular momentum $\sqrt{s(s+1)}\hbar$ (because the angular momentum squared was $s(s+1)\hbar^2$). We could write this as:
$$J_s = \sqrt{s(s+1)}\hbar$$
which says that the (magnitude of the) angular momentum $J$ of a state with quantum number $s$ is equal to $\sqrt{s(s+1)}\hbar$. In the rigid rotor system, we ignore $\hat{J}_z$ and instead look at another observable, which is the energy. The energy of a state with quantum number $s$ is:
$$E_s = \frac{\hbar^2}{2I}s(s+1)$$
where $I$ is the moment of inertia. So far, so good. But confusingly, instead of using innocuous labels like $s$, the convention is to use $J$ instead. And $t$ is replaced with $M$ (sometimes $M_J$), which at least is slightly less confusing.
\begin{align}
J_J &= \sqrt{J(J+1)}\hbar \\
E_J &= \frac{\hbar^2}{2I}J(J+1)
\end{align}
And therefore we have a situation where $J$ can mean several things, and it completely depends on context. To be extra clear, I'll colour the $J$ which means "the value of angular momentum" in red, and the $J$'s that refer to the quantum number in blue:
\begin{align}
\color{red}{J}_\color{blue}{J} &= \sqrt{\color{blue}{J}(\color{blue}{J}+1)}\hbar \\
E_\color{blue}{J} &= \frac{\hbar^2}{2I}\color{blue}{J}(\color{blue}{J}+1)
\end{align}
and the corresponding eigenvalue equations (recall that the operator for energy is always denoted $\hat{H}$, for the Hamiltonian):
\begin{align}
\hat{\color{green}{J}}{}^2|\color{blue}{J}\rangle &= \color{blue}{J}(\color{blue}{J}+1)\hbar^2|\color{blue}{J}\rangle \\
\hat{H}|\color{blue}{J}\rangle &= \frac{\hbar^2}{2I}\color{blue}{J}(\color{blue}{J}+1)|\color{blue}{J}\rangle
\end{align}
Here the green colour indicates an operator; the red in the previous equation was not an operator, but rather the quantity of the observable. Some texts will help you out by putting a hat on operators, but often it is omitted (either because authors are lazy or because it gets really cluttered after a while), so be careful!

* Technically a pseudovector.
$^\dagger$ $J$ is usually used for the total angular momentum of a system. In classical mechanics, there is only really one type of angular momentum, but in quantum mechanics there is orbital angular momentum and spin angular momentum.
Orbital angular momentum is the one which is much more closely related to the classical idea of angular momentum (for example, the orbital angular momentum operators are directly derived from $\mathbf{L} = \mathbf{r} \times \mathbf{p}$), and is typically denoted with $L$, not $J$. The quantum number is usually (helpfully) denoted $l$, although some books will also use $L$.
Spin angular momentum, on the other hand, is denoted with $S$. It has no classical analogue and has even weirder behaviour in that the associated quantum number $s$ can take on half-integer values, i.e. $0, \frac 1 2, 1, \frac 3 2, \ldots$
In quantum mechanics, the total angular momentum $J$ is the sum of both these two contributions. In the rigid rotor model we are not concerned with the spin of the particle(s), so $J$ is simply equal to the orbital angular momentum $L$, and there is no ambiguity about which is which. But if you start thinking about particles like nuclei or electrons, then it becomes important to distinguish between the two.

when you use s, that is not the spin state variable right?

What do you mean by "variable"?


text:
Finding the wavelength of maximum absorption and Crystal Field Splitting of [Ni(H2O)6]2+
(I haven't been here for a long time, so tell me if my formatting is fine)
Given

The $\ce{[Ni(H2O)6]^2+}$ complex appears green in solution from the transmitted light. Estimate the wavelength of maximum absorption, and also the crystal field splitting $Δ_0$.

I got 1.59ev for the CFS, but I want to verify if I'm doing this right.
I wasn't sure if I also need to know the wavelengths of light and spin pairing energy for this problem. So I just looked up the average wavelength of red light (650nm) and plugged it in (and ignored spin pairing energy since my prof never talked about it)
My Method
Since H20 isn't charged, so Ni is 2+ to make this complex 2+.
Since $\ce{Ni^2+}$ is $[Ar]3d^8$, and water makes a weak ligand field, I basically did Crystal Field method and filled the bottom row (t2g) with 6 electrons and put the remaining 2 electrons in eg.
I then used stablization energy: $ΔE = -6(2/5)Δ_0+2(3/5)Δ_0=-6/5Δ_0$
And then $ΔE = hv = hc/λ= 6/5Δ_0$ (I think I can exclude that negative sign since I'm just finding the splitting)
to get $Δ_0 = 5hc/(6(650nm)) = 1.59ev $

Absorption of a photon (loosely speaking) causes the transition $\mathrm{t_{2g} \to \mathrm{e_g}}$. The energy needed for this is (by definition) exactly $\Delta_O$, which doesn't actually have anything to do with the CFSE or the electron configuration, so you've gone one step too far.


text:
Why can a wave-function be written as a product of 1-electron wave-functions?
Why can a wavefunction be written as a product of 1-electron wave-functions when making approximations to solve atoms other than hydrogen? (ie. helium)
We used this when solving multielectron systems such as that of helium. Why is this assumption valid or at least reasonable?

Related: Trying to understand the necessity of orbital approximation or the independent electron model.

I still don't quite understand

You may need to edit your question to address what about Ortho's link is still unclear.

To start with your first sentence: "Why can a wavefunction be written as a product of 1-electron wave-functions"? Well, you can write whatever you want to approximate a wavefunction; physics police is not going to come and take you away. But the two important questions are: How useful will the results from your approach be, and how hard is it to work with mathematically? These criteria obviously reduce your number of options quite a bit.
A wavefunction that is constructed as a product of one-electron orbitals definitely has that second point going for it. As you have already experienced yourself, it is straightforward to write down and calculate with. Arguably the most important implementation of this approach is Hartree-Fock theory where the electrons never see each other as distinct point charges in space, only as fuzzy charge distributions that each one has to wade through individually. Hartree-Fock gives you fairly straightforward equations to work out the optimal way to build the independent-particle wavefunction; and while it cannot accurately capture the energetics of interactions among the electrons "out of the box", it's a clean and simple starting point to describe the quantum mechanics of your molecule. And it's often not all too bad, either - for typical well-behaved cases, Hartree-Fock and its one-electron orbitals can already land you within a few percent of the full energy value. The remainder is still way too large for it to be practically useful, but you can reuse this initial wavefunction in more refined algorithms to reduce the error further.


text:
Calculate the percent composition of a sodium carbonate/sodium bicarbonate mixture

In a lab, we titrated a solid mixture of sodium carbonate and sodium bicarbonate with $\pu{0.1 M}\ \ce{ HCl}$. We are asked to calculate the percent composition by mass of sodium carbonate and sodium bicarbonate.
Data:

Unknown mass used: $\pu{0.107 g}$
Volume of $\ce{HCl}$ used: $\pu{16.6 mL}$

Here was my approach:
1) I determined the moles of carbonate by doing the following:
$$(\pu{16.6 mL} \ \ce{HCl}) \times \frac{\pu{1L}}{\pu{1000 mL}} \times  \frac{\pu{0.1 mol}\ \ce{HCl}}{\pu{1 L}\ \ce{HCl}} \times  \frac{\pu{1 mol}\ \text{carbonate}}{\pu{1 mol}\ \ce{HCl}} = \pu{0.00166 mol}\ \text{of carbonate}$$
2) I calculated the molar mass of sodium carbonate and sodium bicarbonate:
MM of sodium carbonate $= \pu{105.99 g/mol}$
MM of sodium bicarbonate $= \pu{84.01 g/mol}$
3) This is the part I'm unsure about. I said that
$$\frac{x}{\pu{105.99 g/mol}} + \frac{(0.107 - x)}{\pu{84.01 g/mol}} = \pu{0.00166 mol}\ \text{of carbonate}$$
Doing this math, I determined that the compound was 52% sodium carbonate and 48% sodium bicarbonate but I'm unsure if this is correct.
The instructor seemed to want us to solve it in terms of H+ using the following ratios:

$\pu{1 mol}$ of carbonate: $\pu{2 mol}$ of $\ce{H+}$

$\pu{1 mol}$ of bicarbonate: $\pu{1 mol}$ of $\ce{H+}$

but I'm not sure how you would use this information to answer the question. Any help would be greatly appreciated!

In your work in step 3, you have treated both carbonate and bicarbonate equally. That contradicts what your teacher has said.

Thanks for your response. I'm not sure how to write an equation that treats them "unequally". Would saying that 2 moles of carbonate and 1 mol of bicarbonate = 1 mol H+ be correct?

Not correct. Each mole of carbonate counts for two moles of $\ce{HCl}$.

Ah I see. So                                                                                1 mol Na2CO3: 2 mol HCl
and                                                                                1 mol NaHCO3: 1 mol HCl are the correct ratios?                                             From there I can calculate the number of moles of each, convert to grams, and then say that 0.107 grams = x(grams of Na2CO3) + (1-x)(grams of NaHCO3)?

Your modified approach is fine but your teacher is teaching this analytical method the wrong way. No wonder students are confused. What happens in reality is that you use $two$ indicators in steps. One indicator shows the end-point when all carbonate is converted to NaHCO3. Then we add another indicator to titrate all the bicarbonate. Google "Lab 2: Determination of Carbonate/Bicarbonate." The assumption  0.107 grams = x(grams of Na2CO3) + (1-x)(grams of NaHCO3) should never be used in any real analysis, because we are assuming that the sample is pure i.e. made of only two components.

Thank you both for the help and resource suggestions!


text:
Changes in entropy of the universe in a reversible reaction
Does entropy of the universe always increase in one of the forward or backward reactions of a reversible reaction and decrease in the other?
If no then how do you explain the entropy of the universe increasing in both directions or decreasing in both directions?!?
If yes, do we get more of the reaction that increases the entropy of the universe? I would presume yes as all real processes increase the entropy of the universe.

I need to think more, but usually you define a system and surrounding, and these may make up your universe. The entropy of the universe will always increase. The system may increase or decrease, but the surrounding will more than compensate this to ensure the entropy of the universe increases.( I have been teaching thermo to mechanical engineers for too long and they don't do reactions... I will get back to you)

Like I said in my comment in your other thread, there is a terminology difference between a reversible reaction and a reversible process.  It is possible to devise a reversible process that takes a reversible reaction mixture from its initial state to its final state without increasing the entropy of the universe.

First, a defintion of terms:  The meaning of reversible in "reversible reaction" is different from the thermodynamic meaning of reversible.  A reversible reaction is one that is kinetically reversible.  I.e., one where you can adjust the relative concentration of products and reactants back and forth as you change conditions.   By contrast, an example of an irreversible reaction would be cooking an egg, which causes the proteins to denature.  The proteins can't be readily undenatured, even if you change the conditions to thermodynamically favor denaturation because, kinetically, the proteins can't find their way back to their undenatured state.
Anyways, let's suppose we have a "reversible reaction", $$\ce{A <=> B},$$  in which the equilbrium constant is, under certain conditions (e.g., at a certain temperature and pressure):
$$K_{eq} = \frac{[B]}{[A]} = 10,$$ i.e., at equilibrium, we will have 10x as much B as A.
This represents the state at which the entropy of the universe is a maximum under those conditions.  Thus if, under those conditions, $\frac{[B]}{[A]} <10$, the reaction will proceed in the forward direction until it reaches equilibrium, increasing the entropy of the universe in the process.  Likewise if, under those condtions, $\frac{[B]}{[A]} > 10$, the reaction will proceed in the reverse direction until it reaches equilibrium, also increasing the entropy of the universe in the process.
So it is not the case that if, say, $\frac{[B]}{[A]} = 9$, then the entropy of the universe would increase in both directions. Rather, it would increase in the forward direction (until it reached equilibrium), and (theoretically) decrease in the reverse direction.  If we instead had $\frac{[B]}{[A]} = 11$, the opposite would happen: decreased entropy of the universe in the forward direction, increased entropy of the universe in the reverse direction (again, until it reached equilibrium).

If I understand you correctly, a 'reversible reaction' is one that is kinetically reversible. I.e., one where you can adjust the relative concentration of products and reactants back and forth as you change conditions. But what is your definition of 'thermodynamic reversibility'?

@ETS That's an entirely separate question, and there are entire answers on Chemistry SE devoted to this (I'd recommend a search). But, briefly: a thermodynamically reversible process is an idealized, infinitely slow process (consisting of an infinite no. of infinitesimal steps) in which the system is in continous equilibrium with the surroundings, and for which $\Delta S_{universe} = 0$.  [continued]

....It is called reversible because each step creates only an infinitesimal change, and thus one can do a backwards infinitesimal step and restore both the system and the surrondings  to their original states.  Since all real processes require $\Delta S_{universe} > 0$, a reversible process is a limiting idealization that is not possible in the real world.

will these do as definitions: an irreversible reaction is one where activation energy in one direction can be achieved without constant human intervention and where activation energy in the other direction cannot be achieved without constant human intervention; a reversible reaction is one where the activation energy in both directions can be achieved without constant human intervention?

Another thing. Why is it that at equilibrium of a reversible reaction, entropy of the universe is maximum? Surely the side of the reaction with lower Gibbs Free Energy yields lower entropy of the universe?

No, you're referring to just the Gibbs free energy of the pure reactants and products, not the total Gibbs free energy of the system, which includes the Gibbs free energy of mixing, which arises mostly from the entropy of mixing. That's why reactions don't go all the way to the side with the lowest pure Gibbs free energy, instead finding an equilibrium somewhere away from pure reactants or pure products: because by having both reactants and products, we increase the entropy of mixing.  At const T and p, with no non-pV work, the total Gibbs free energy of the system is minimized at equilibrium.

thank you so much for the clarification! You explain these complex notions well. I think that you might have missed my previous comment. Are my definitions correct?

Thanks.  I'll add that the logarithmic functional form of the equilibrium constant actually comes from the entropy of mixing.  As to your question, I don't think so, since if you have a cooked egg, the reason it can't be directly reversed back to undenatured protein is not because of insufficient activation energy. It's kinetically stuck because the proteins can't find their way back to their original, complex folded conformations.

If you really want to explore your activation energy view further you should re-read mine and other's accepted defintions of reversible reactions, and then, if you still want to pursue it, post it as a separate question; we've over-extended the amount of discussion that's supposed to take place in comments.

Related: https://chemistry.stackexchange.com/a/124602/72973

@BuckThorn   Yes, that's a good point --- I don't think his argument quite passes muster now, but if it were reframed from activation energy to Gibbs energy of activation, which would include an entropic prefactor to account for the difficulty of refolding the proteins, then it might work.  But this is why I suggested he give it more thought and then, if he wished, post it as a separate question.   Regarding your last sentence, since we're talking about kinetic reversibility now, how you get between the states of course does matter.


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What is the purpose of atomisation in atomic absorption spectroscopy?
What is the purpose of the atomisation process in atomic absorption spectroscopy?
I understand that solution is evaporated by the flame and then the molecules decompose into atoms.
But why must the molecules become individual atoms before the light is passed through them?

To complement the previous answer:
Atomic absorption lines are very sharp, with high extinction coefficient, low detection limit and are specific for the given element.
Molecular absorption is band-wise, diffuse, with much lower ext. coefficient, higher detection limit and last but not least, the selectivity is lost.

Atomic absorption spectroscopy is a quantitative measure of the elements in the analyte.
We atomise the analyte so that the absorption of light corresponds to the characteristics of the elements.
Otherwise how we can predict the fundamental elements when these are still part of some molecules?
We measure the absorbance and refer to the Beer-Lambert law.


text:
What will be the name of CH2=CH-O-CO-CH3?
We have this question in the textbook but there is no answer given and I have no idea how to name this compound.
Edit: I tried naming the compound.
Correct me if I am wrong here
1-Methylpropenoate

Most people would call it vinyl acetate. If you want the systematic name, then it is ethenyl ethanoate.
It is formally derived from "vinyl alcohol" and acetic acid, from which the name follows easily. See here (https://en.wikipedia.org/wiki/Ester#IUPAC_nomenclature), for example.
The compound is https://en.wikipedia.org/wiki/Vinyl_acetate.


text:
Does H3CO3+ ion exist and what is its structure?
Why doesn't $\ce{H3CO3+}$ ion exist and why doesn't the carbonyl group in carbonic acid $(\ce{H2CO3})$ seem reactive?

A protonated carbonic acid (trihydroxymethyl cation) $\ce{H3CO3+}$ does exist, and is in fact incredibly stable in the solutions of superacids [1] up to $\pu{0 °C}.$ One way of obtaining $\ce{H3CO3+}$ is dissolving inorganic carbonates and hydrogen carbonates in magic acid at $\pu{-80 °C}$ [2]:

$$\ce{CO3^2- or HCO3- ->[FSO3H-SbF5/SO2][\pu{-80 °C}] H3CO3+ ->[\pu{-10 - 0 °C}] CO2 + H3O+}$$

As for the structure, the $C_\mathrm{3h}$-symmetrical molecule (triskelion shape, structure 7 below) was found to be the global minimum [2] (reference numbers updated):

We also carried out ab initio and IGLO calculations […] on bicarbonate as well as free and mono-, and diprotonated carbonic acids to rationalize the observed experimental results. Initially the possible geometries were optimized at the HF/6-31G* level. The most stable isomers were subsequently determined from
  relative energies at the MP2/6-31G*//HF/6-31G* level. Finally, the lowest energy structures were further optimized at the MP2/6-31G* level. MP2/6-31G* optimized structures were used to calculate $\ce{^{13}C}$ NMR chemical shifts.
[…]
Protonated Carbonic Acid. The symmetrical $C_\mathrm{3h}$ structure 7
  was found to be the global minimum. This structure was suggested for the observed species in superacid solution by Olah et al. [1] on the basis of $\ce{^1H}$ and $\ce{^{13}C}$ NMR spectroscopy. The ion at $\pu{-80 °C}$ shows single sharp peaks at $δ(\ce{^1H})~12.05$ and $δ(\ce{^{13}C})~165.4$ in the proton and $\ce{^{13}C}$ NMR spectra, respectively.
Protonation of carbonic acid on the hydroxyl oxygen atom (8) was shown to be unfavorable by $\pu{23.4 kcal/mol}$ over protonation on the carbonyl oxygen (7). Frequency calculation at the HF/6-31G*//HF/6-31G* level showed that the structure 8 is not a minimum, as it contains two imaginary frequencies.

$\ce{H3CO3+}$ shares structural similarities with its triaza-analog, the guanidinium ion, as both are possessing resonance stabilization via their onium forms [3, p. 60].
References

Olah, G. A.; White, A. M. Stable Carbonium Ions. LXIV. Protonated Carbonic Acid (Trihydroxycarbonium Ion) and Protonated Alkyl (Aryl) Carbonates and Hydrogen Carbonates, and Their Cleavage to Protonated Carbonic Acid and Carbonium Ions. The Possible Role of Protonated Carbonic Acid in Biological Carboxylation Processes. J. Am. Chem. Soc. 1968, 90 (7), 1884–1889. https://doi.org/10.1021/ja01009a036.
Rasul, G.; Reddy, V. P.; Zdunek, L. Z.; Prakash, G. K. S.; Olah, G. A. Chemistry in Superacids. 12. Carbonic Acid and Its Mono- and Diprotonation: NMR, Ab Initio, and IGLO Investigation. J. Am. Chem. Soc. 1993, 115 (6), 2236–2238. https://doi.org/10.1021/ja00059a020.
Dewar, M. J. S., Hafner, K., Heilbronner, E., Itô, S., Lehn, J.-M., Niedenzu, K., Rees, C. W., Schäfer, K., Wittig, G., Boschke, F. L., Series Eds.; Topics in Current Chemistry; Springer Berlin Heidelberg: Berlin, Heidelberg, 1979; Vol. 80.

I think the OP was looking for H3CO3 as in,without any formal charge,while you have talked about H3CO3+ . Still upvoted for a great piece of information though

@YusufHasan True, this is a stretch, but the question sort of implies a cation, otherwise it is indeed wrapping one's head around the justification for the existence of a neutral molecule like this.

@andselisk amazing answer and it gave me new insights to inorganic chemistry

I can't look it up, so I'm just commenting this: the level of theory HF/6-31G*//HF/6-31G* doesn't make a lot of sense; it would translate to a single point calculation on the same level of theory as the structure optimisation. There must be a difference between the first and the last part, that might be interesting to know. On the other hand, that level is not the most reliable to begin with...

@Martin-マーチン I added extra details from the R&D section, probably this could help to rationalize the authors' methodology. I agree with your comment, but since I'm far from being an expert in QC, I'd rather avoid commenting on this on my own.

That makes some more sense now. The way I read it, they have conducted a screening on the HF/6-31G* level of theory and calculated relative energies as single points om MP2/6-31G*, which is commonly denoted as MP2/6-31G*//HF/6-31G*. In the term HF/6-31G*//HF/6-31G* they are probably referring to the frequency calculations, meaning they have done them on the same level of theory (which is not surprising, but still not that common to write it that way). In any case, thank you for the update.

@YusufHasan actually I am asking about the reactivity of the oxygen in carbonyl group of carbonate ion and I used H3CO3+ just as an example

@mohamed In this particular case, yes.

@andselisk I just want to ask about the reactivity of carbonyl oxygen in the carbonic acid and what I understood from your answer is that hydrogenating the carbonyl oxygen is more favourable than hydrogenating the hydroxyl one if we are to form an H2CO3 .Is it correct?

$\ce{H3CO3}$ does not exist. Why should it exist ? Where does this formula come from ? Of course you are allowed to write any formula you want. It is not forbidden to write $\ce{HC4O3}$ $\ce{H3CO5}$ $\ce{H3C2O3}$ or any other combination of C, H and O atoms that you may think of. Just for fun ! But Nature ignores your fantasy. Nobody has ever been able to make these substances. So these substances do not exist.
The second part of your question may be explained by the fact that $\ce{H2CO3}$ is extremely instable, and is always largely decomposed into $\ce{CO2}$ and $\ce{H2O}$. $\ce{H2CO3}$ exists only in very dilute solution in water, and in equilibrium with $\ce{CO2}$ and $\ce{H2O}$. If you try to make a reaction with the carbonyl group of $\ce{H2CO3}$, it will first be decomposed before any other reaction. The only exception is its reaction with the OH- ion. But both $\ce{H2CO3}$ and $\ce{CO2}$ can react with NaOH, producing the same $\ce{CO3^2-}$ ion. So you cannot be sure about the nature of the molecule that reacts with NaOH. It may be $\ce{H2CO3}$ or $\ce{CO2}$.

I was always taught that $\ce{H2CO3}$ is very stable, just not in protic solvents.

Your answer starts with the statement $\ce{H2CO3}$ does not exist.  The second paragraph states that it does exist.  It were better to state that the majority of $\ce{CO2}$ in water is dissolved (physically), and a small portion only actually constitutes $\ce{H2CO3}$.  And for this tiny portion of «true» $\ce{H2CO3}$, I think the acidity experienced in water is low, rather than darn tiny to be insignificant and negligible.


text:
Properties of starch in packaged pudding
I had a mug of packaged pudding. The top layer of the pudding was of typical pudding consistency, while the bottom was more runny. What could have caused this?

The top layer is usually a solid elastic film. You mean the top half, or third or so, right?

yes I mean the top third @Karl

Starch makes a glue or a paste with water. Humidity from the air may be sufficient to transform your starch into a glue. Water produces strong hydrogen bridges between starch molecules. This transforms a pile of independent starch molecules into a strong block of interconnected starch molecules.

What you have witnessed is gelation coupled to phase separation. It depends on the composition of your pudding.
The process of preparing a starch gel begins with starch granules swelling and eventually dissolving with uptake of water. This initial step requires disrupting the structure of both amylopectin (more amorphous) and amylose (more crystalline) domains to form an aqueous solution of extended polymer chains. Melting of the granules occurs with heating. When the solution is cooled structure begins to reform, involving cross-linking (crystallization and entanglement) of the mixed chains, a process known as retrogradation. As structure reforms water is pushed out of the gel, resulting in water-rich and water-depleted regions (phase separation). The less dense gel settles on top of the more dense polymer-depleted water-rich region.
For more information try a search for "phase separation gelation". The literature on the subject is vast and there are various mechanisms that can explain the phenomenon depending on the composition of the gel and other parameters such as temperature.


text:
Why do atomic radii decrease across a period?
It's said that atomic size/radius decreases across a period, in spite of addition of electrons. But how does this actually work?

Atomic radius decreases across a period because  electrons are being added to the same energy level and at the same time the nucleus is increasing in protons. The increase in nuclear charge $$Z_{eff}$$ attracts the electrons more strongly, (addition of one proton has a greater effect than addition of 1 electron) thus pulling  electron closer to the nucleus and making atomic radius decrease.


text:
Benzannulation of 18-crown-6
I know crown ethers can be used to complex metal ions.
Question: What important properties do 18-crown-6 change during the benzannulation from 18-crown-6 to DB18C6 (dibenzo-18-crown-6)? How do these change?
(Disclaimer: I know that DB18C6 does not necessary to be synthesised from 18C6)
Motivation: I saw that some complexations of metal ions use 18C6, but some of the other use DB18C6.
Thoughts and Research: As crown ethers are used to make size-specific ions complexes, the sizes of the hole in the ethers are important. But the benzannulation does not change the size, so I suggest that the process changes the equilibrium constant of $\ce{M(ether)X_n}$ or the rate of complexation. Investigated by this article, the distribution constant between polar phase and non-polar phase increases significantly and the equilibrium constant in non-polar phase increases. I understand the former because of the property of benzene ring. But how can the latter one happen? Are there another reasons why DB18C6 is chosen?

Of course does the geometry of the crown ether change! Four sp3 carbon atoms are now sp2!


text:
What is the difference between dynamical and non-dynamical correlation and which one is accounted for in Hartree-Fock theory?
It is my understanding that dynamical is related to the movement of individual electrons. I don't fully understand what this means though.
Non-dynamical correlation is still very unclear from my notes and how it relates to HF theory is very confusing. Any clarification would be appreciated.

Related: https://chemistry.stackexchange.com/q/50804/41556

@Harley, this is a good question, you might find it useful to support the proposal for a Stack Exchange dedicated specifically to the "expert-level" modeling side of things: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=Y2JlNjkxZWYyMWEyNzdkZmMzM2UwMWIwMzhlYTIwMTI1OGMzNWY4ZTVjMTAyZDU3MWFkNjEwZjIzMzgxZDVhZdnlQ-Z1y5YVnxsjGvH9jFkvFuyZGzROg66XTw-wSrQL0


text:
Why does table salt (NaCl) dissolve branded food?
To dissolve branded food from e. g. a baking tray, there is the advice to pour table salt on it and heat it up in a baking oven or on a hotplate until the salt turns brown. From practice I confirm that it works.
But I'd like to know why it works. Can somebody please explain it in layman's terms (my chemistry knowledge is at a-grade)? What reaction(s) happen(s) here?

As far as I understand it, the salt (due to its large surface) takes up whatever becomes liquid in the pan, removing it from the hot surface. That way the iron becomes hotter faster. Thermally destroying old oil (or evaporating it) needs a lot of energy and takes time. Im interested to hear about other explanations!

Monovalent ions generally tend to dissolve things because they "sneak" into structures. I am curiou about a more scientific reasoning.

@GyroGearloose Hm, it sounds not unreasonable that single ions can diffuse into a polar solid, but I would expect that this hardens the structure instead of softening it.


text:
Thermal decomposition of steam to hydrogen and oxygen; at what temperature is half decomposed and how to calculate?
In comments below this answer in Space Exploration SE I've said:

(Wikipedia) says that dissociating water costs about 4.4 eV per bond. If = 4.4 eV and $k_B$ is 8.617E-05 eV/K then that corresponds to a characteristic temperature of something like 50,000 K. Does a nuclear rocket's exhaust get close to that? (I have no idea)

and the reply was:

but en.wikipedia.org/wiki/Water_splitting says 3000 K. Can anyone clarify?

That Wikipedia article actually says:

At the very high temperature of 3000 °C more than half of the water molecules are decomposed, but at ambient temperatures only one molecule in 100 trillion dissociates by the effect of heat.

Question: If it takes 4.4 eV to break one hydrogen-oxygen bond, how could half of the water be dissociated at only 3273 K where $k_B T$ only about 0.28 eV?
$$\exp \left( -\frac{0.28 \ \text{eV}}{4.4 \ \text{eV}} \right) \approx 0.94$$
so I'd naively expect only about 6% of the water to be dissociated at 3000 °C.

update: I've just realized that in my naivety I've forgotten to consider that forming H2 and O2 from free atomic H and O is probably exothermic, so my 4.4 eV is an overestimate of the energy difference between initial and final states. But I don't know how to take this into account.

https://chemistry.stackexchange.com/questions/37562/difference-between-dissociation-energy-and-dissociation-temperature-for-water

@Mithoron Thanks for the link! Yep, answers there certainly looks helpful and consistent with my update to the question, but an answer to this question addressing "...and how to calculate?" would need to go further than "For example, $\ce{H2O <=> H + OH}$ and perhaps even $\ce{H2O <=> H2 + O}$ or $\ce{H2O <=> 2H + \frac{1}{2}O2}$ and related equilibria may be happening."


text:
Combustion of carbon with hydrogen
Can coal (carbon) be "burned" in a hydrogen atmosphere? I know it would not match the definition of "combustion".
But is the reaction $$\ce{C +2H2 ->CH4}$$ exothermic?
(I'm fairly convinced this is a duplicate, but I didn't find any question like that.)

To find the enthalpy change associated with the reaction you can find heats of formation and use Hess' law. The heats of formation you need can probably be easily found online or in a textbook.

@orthocresol Thank you, but I need some more help. I have no textbook, I'm out of school for some decades. A link would be useful.

Google is your friend! https://en.wikipedia.org/wiki/Methane_(data_page)

Please mention the state of reactants so that there is no confusion while calculating the enthalpy change. I suppose that you asked it out of curiosity, in that case yes it is exothermic.

@PRITIPRIYADASBEHERA I did not know that curiosity contributed to making a reaction exothermic...

Yes, I have seen that I was thinking too complex, I didn't expect an easy solution, but just adding up entropy enough. Will promptly delete that silly question.

The formation reaction of CH4 is exothermic. The heat of formation is : -75 kJ/mol

In my opinion, citing a reputable source for the data would make it a better answer. Also, probably a brief note on terminology (something along the lines why "combustion" isn't the best term here) wouldn't hurt either, I think.


text:
Overlap matrix for finite sized square lattice of hydrogen atoms
So I'm trying to use the LCAO method to calculate molecular orbitals on a square lattice of hydrogen atoms. To this end, I need to compute the Hamiltonian matrix. With the nearest neighbor interaction approximation, the matrix should be pentadiagonal, analogous to the Laplacian matrix.
Now the problem is that I don't want to assume orthogonality of nearest neighbor atomic orbitals so there will be an overlap matrix. I believe it's also pentadiagonal but when I solved for molecular orbitals and their energies, the result is wrong. So I wonder if anyone has done similar things and could provide me some insight? Thank you!
If needed, I can post my Matlab codes and the computed Hamiltonian and overlap matrix here.
PS: The LCAO approach I described here can be found in Simon’s Oxford Solid State Basics textbook, Chapter 6, exercise (6.5)
PPS: I'm not constructing Bloch functions because I don't want to impose boundary periodicity constraints. I'm trying to calculate band structures of finite-sized nanoparticles but not bulk material in my project.
My code:
N = 10; %Size of the lattice is N^2
S = eye(N) + diag(zeros(N-1,1)+s12,1) + diag(zeros(N-1,1)+s12,-1); % Initialize the overlap block
Hop = 4*(eye(N)) + diag(zeros(N-1,1)+v12,1) + diag(zeros(N-1,1)+v12,-1); % Initialize the hopping block

B0 = S*e_atom + Hop;   % block matrix for interactions between same row of atoms
B1 = eye(N)*(e_atom*s12+v12);  % block matrix for interactions between adjacent row of atoms
Stmp = kron(eye(N),S)+ kron(diag(1+zeros(N-1,1),1),eye(N)*s12) + kron(diag(1+zeros(N-1,1),-1),eye(N)*s12); % Pentadiagonal Overlap matrix

H = kron(eye(N),B0) + kron(diag(1+zeros(N-1,1),1),B1) + kron(diag(1+zeros(N-1,1),-1),B1); %Hamiltonian matrix, justification see my notebook

[EVec, EVal] = eigs(H,Stmp, N^2,'sm'); %EVec contains eigenvectors, EVal contains eigenenergy on diagonal
E = real(diag(EVal));
[E, ind] = sort(E);
tmp = EVec(:,ind);
EVec = tmp;

For two adjacent hydrogen atoms, denote their $1s$ orbital by $|1\rangle$ and $|2\rangle$. $s_{12}=\langle1|2\rangle$, $v_{12}=\langle1|V_1|2\rangle$, $v_{22}=\langle2|V_1|2\rangle$ where $V_1$ is the Coulombic potential centered on the first hydrogen nuclei, $e_{atom}=-13.6eV$ is the energy of electron in an isolated $1s$ orbital.

It probably would help to see your code. At this point, what you are doing sounds right in principle, so it is likely some type of coding error. If that's the case, the question maybe better on the scicomp SE

@Tyberius Hi Tyberius, I've attached relevant codes to my problem. Perhaps I'm making wrong assumptions somewhere?

What are you comparing your calculation against? Or I guess phrased another way, how do you know your result is wrong?

@Tyberius Hi Tiberius, sorry for the late reply! I'm checking my density of state distribution with that in the textbook Orbital Interactions in Chemistry by Albright. I'm not sure if the DOS plot presented in that book is for wavevectors in the full first Brillouin zone or just along the specific path through high symmetry points.


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Simple sandbox software for an interactive molecular dynamics simulation
I'm not sure if this is the right site to ask a such question.  But I'm getting to learn the world of the small.
I know there is software out there like NAMD, and LAMMPS.  But these tools are very non-intuitive, having complex interfaces that seem to hide the underlying mechanics.
I'm looking for something that can display live molecular simulations in 3D, with some kind of inbuilt editor to simply place, or remove atoms.
Again, sorry if this is the wrong place for a such question, I'm sure from what you can tell, I am extraordinarily new to this area of knowledge.  I would greatly appreciate any feedback.  I fear although, I will have to create my own intuitive frontend to NAMD or equivalents to have essentially a 'quantum sandbox'

Albeit not as complete as other packages can be, Ascalph is what you describe http://www.biomolecular-modeling.com/Ascalaph/index.html

Also: I have not tried it myself, but it should be possible to set up VMD so that it "streams" the output of MD packages like LAMMPS

I strongly recommend openmm. It is the easiest of the packages to interact with, followed by LAMMPS. I believe Macsimus also has video options, but I have never used it

Avogadro has a live minimizer when editing - pulling, adding, deleting atoms, changing bond orders, etc. The next version (v2) will support full MD.

Since this question has been up for a couple weeks and still doesn't have an answer, perhaps it might be something that a dedicated computational chemical physics community would be interested in. There is a proposal for a Stack Exchange for Materials Modeling for example: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=ODVjZmM2YzJmOTc0MzBkZDVkMTZmMTIzN2EyNmZmNTQyOWYyY2Y0YWFmNGYxNTQ2OTQzNmMwOWIyYjhhOTZjMhwblBomlTMtPu8EgXeAKtp-8dFsxghojkD0vBSe1wZl0. Perhaps you might consider supporting it if you think it would be useful for you! there's already MD questions there.

@The_Vinz that's if you can get VMD to compile for you. With Catalina, Apple decided to up the VMD compilation difficulty level to boss-mode.


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Does magnesium carbonate exist? what makes magnesium bicarbonate different from calcium bicarbonate
In my book, I found the following reactions:
$\ce{Ca(HCO3)2 ->[\Delta] CaCO3 + … }$
$\ce{Mg(HCO3)2 ->[\Delta] Mg(OH)2 +...}$
What makes these two reactants different? Why is one forming hydroxide while the other carbonate?

See https://en.wikipedia.org/wiki/Magnesite

Both of these reactions are thermal decomposition reactions, but the difference basically occurs in the polarizing power of the cations $\ce{Ca^2+}$ and $\ce{Mg^2+}$ .
The thermal decomposition of $\ce{Ca(HCO3)2}$ is just the reverse reaction of the lime water test for the detection of $\ce{CO2}$ , where the carbonate combines with $\ce{CO2}$ and $\ce{H2O}$ to give the bicarbonate, so it stands to reason that the bicarbonate will break down to give the reactants we started with.
We can therefore, imagine that a similar decomposition will occur with $\ce{Mg(HCO3)2}$ to yield $\ce{CO2}$, $\ce{H2O}$ and $\ce{MgCO3}$. And in fact, a similar reaction does occur on drying the bicarbonate solution.
However, a small amount of $\ce{Mg(OH)2}$ is also obtained as a side product.
The reason is that, $\ce{CO3^2-}$ is relatively quite a large-sized anion.
Also, if you recall the diagonal relationship between $\ce{Li}$ and $\ce{Mg}$, then sizes of both $\ce{Li+}$ and $\ce{Mg^2+}$ will end up being similarly quite small relative to other members of their respective groups. Thereby,Fajan's rule will become important to consider here, and the carbonates of both these cations will experience similar internal polarization, and tendency to decompose relatively.
Hence, since we know that $\ce{Li2CO3}$ is the only Group 1 carbonate to decompose at laboratory temperatures to give $\ce{Li2O}$ and $\ce{CO2}$ ,(see this) we can expect $\ce{MgCO3}$ to also  break down into $\ce{MgO}$ and $\ce{CO2}$ on prolonged heating.
$\ce{CO2}$ being a gas will escape, and since Group 2 bicarbonates exist only in aqueous form, so the magnesium oxide produced will hydrolyse within the solution to produce the hydroxide, that is, $\ce{Mg(OH)2}$.
Note: Decomposition of $\ce{Li2CO3}$ is only "easy" relative to other members of Group 1. It still has a pretty high decomposition temperature of 1300 °C approximately
This value of 1300 °C might still not be accurate, see the green Note under 'Heating the carbonates' for more insight.

MgCO3 gives CO2 not O2.  There is a magnesium peroxide, MgO2, which is metastable and gives off O2 on heating or by reaction with water, but this is not in scope.

@OscarLanzi Oh damn,there isn't any O2 being given off as well,I wrote it by mistake, wasn't paying attention. Thanks for pointing it out!

Rubidium carbonate decomposes at only 900C.  Can't say lithium carbonate decomposes more easily than heavier congeners.

@OscarLanzi Well, the entry you are pointing to says it's the boiling point at 900C, and maybe that's also the decomposition they are talking about in brackets? The wiki page hasn't clearly distinguished b/w the thermal decomposition and boiling point.Anyhow, as per HSAB logic, the carbonates of group 1 should become more thermally stable as we go down the group as said here

Usually indicating (decomposes) with a boiling point indicates the "boiling" is a decomposition.  Could be seeing vapors made of Rb2O and CO2 instead of Rb2CO3 molecules.

@OscarLanzi I found another link (under the part 'Heating the carbonates") he discusses in the green comments about why the data I found in Wikipedia is not entirely correct, and lithium carbonate can maybe decompose at as low as 790 C. Do see this part


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Confused about identifying delocalized electron pairs in Isoniazid
I have to identify delocalized electron pairs in Isoniazid (pairs not shown in the image below):

I know the nitrogen in the ring has a localized electron pair, since it already forms a pi bond. I'm confused about the other two nitrogen atoms. When we draw one of the resonance structures, we can see the amide nitrogen has a delocalized electron pair:

And if we push the lone pair from the amine nitrogen, we get another resonance structure (even if it is insignificant, due to all of the formal charges, it still seems to be a valid resonance structure):

Because these two lone pairs engage in resonance, I thought that they were both delocalized. But my textbook (Klein's Organic Chemistry, section 2.12, 2nd Ed.) says only one pair is delocalized:
Therefore, if the electron pairs on both the amine and amide nitrogens in Isoniazid seem to engage in resonance, why is only one pair delocalized?

The reason why the amine lone pair is delocalized is that to make the last conformer you proposed, the nitrogen closest to the oxygen would have to form five bonds. Nitrogen is a second-row element that cannot readily possess 10 electrons from those 5 bonds, even if nitrogen has to possess those 10 electrons for the very short time required for the electrons to switch conformations between different resonance conformers.

Molecules do not switch between resonance structures. All resonance structures exist simultaneously.

@Andrew can you write a answer for this query by OP , i think i am having same doubt as OP


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Clarifying question about chloride ions, water softening & corrosion
I have a question regarding the use of NaCl as a water softener.
Recently i was asked to comment on a series of test results from a water softening device were a lot of corrosion had occurred post softening. Just to be sure we are on the same page, NaCl was added to the water to dissolve* the CaCO3, Ca(CO3)2 & CaSO4.
*Ion exchange is not my field of expertise, so I'm not quite sure if this is the correct wording
Now, according to the folks who ran the test, there were no signs of ‘Chlorides’ in the water sample, what they mean by this is still unclear. I was also informed that no precipitate was collected. Now to the actual question: they wanted to know why the pipes used, post softening, had crevice corrosion inside the pipe. My first intuition was that there were still $Cl^-$ ions in the water which is causing the steel to corrode. However, the folks running the test assured me that the chloride had bonded with the calcium ions.
I’ve been trying to figure out what in the world could cause the pipes to corrode, assuming that they are correct, and the chloride have formed CaCl2.
Have you guys got any ideas? Have I misted something, or am i right to assume that the folks who ran the test don't know what they are talking about.
If you need any more information, feel free to ask. As stated earlier, this is not my field of expertise so if I'm wrong about something please correct me.

A traditional softener has a zeolite resin that "holds" positive ions. When it is charged the Na ions fill the resin. Then in service , the positive ions - Ca , Mg , etc , exchange places with the Na ions, putting Na in the water . In theory the back flush of the softener after charging, removes the Cl ions. So that in service the only Cl present would be what is in your raw water as the Cl is not affected by the softening. Water corrosion results from conductivity ( any ions) and oxygen; Although Cl  ions are especially active. Generally "waters" leave a CaCO3 type scale that forma a protective layer ( scaling ) that  reduces corrosion. Serious evaluation of water corrosion looks at the water "scaling index" to get a initial idea of how corrosive it may be. Softening water reduces the scaling index and promotes corrosion . I removed a softener from one house I owned for that reason; However it was Lake Michigan water which has a very good scale index - enough to be protective but not enough to restrict water flow. I also had to replace a couple sections of copper pipe that had corroded through in the 20+ years the softener had been in service. On the other hand , softeners are useful in very hard natural waters.

I'll try and ask a few more question of the folks who ran the test based on your answer. So far I've learned that the pipes were made of 'en 1.4404 stainless steel' and that the hardness of the water is around 28 (dGH/°dH). I don't know if they are using zeolites as a medium, the pictures I've seen so far does not seem to indicate that they do.
I will return with more info when I have it.

Okay, I've acquired the product data sheet from the company who operates the equipment. Rough translation of the description: "Salt-pellets for use in chemical industry as well as water treatment and food production". I believe this means that they aren't in fact using not using zeolites with NaCl added, and simply ads a bunch of NaCl to the solution to soften the water.

I am not a water specialist , however, hardness is similar to scale index but not quite the same  , but I never used dH. I expect the stainless is in the softener , not the house piping. The softener has resin ( maybe another name ) in it.