text:
Can a colorant be mixed into zirconia powder?
I’m wanting to color zirconium dioxide into beads after it is fired but I would like to get more colors than just white and the high temperature black oxide. Can I mix other metal oxides to zirconium powder to change its color once fired?

Setting aside commercial sources whose manufacturers are unlikely to reveal their secrets, Chen et al.[1] describe the coloration of zirconia with oxides of cerium(IV), iron(III), and bismuth(III):

The coloration of zirconia by the three kinds of metal oxide was determined. With the addition of $\ce{CeO2}$, the lightness of the zirconia was about 84–87, a* decreased slightly, and the colour shifted to a more yellow-green hue. With an increasing content of $\ce{Fe2O3}$, the chroma of the ceramic became more intense, the lightness decreased, and the colour changed to a red-yellow hue. With an increasing concentration of $\ce{Bi2O3}$, the colour of the materials shifted to a light saffron yellow. X-ray diffraction spectra revealed only tetragonal phase. Results show that zirconia ceramic can be colourised using $\ce{Fe2O3}$, $\ce{CeO2}$, and $\ce{Bi2O3}$.

Reference

Chen, J., Zhang, Y., Dong, S., Zhao, L., Gu, B., Liao, N. and Wen, N. (2015), "Effect of coloration with various metal oxides on zirconia". Coloration Technol, 131: 27-31. https://doi.org/10.1111/cote.12118


text:
Color of coordination compounds
My book(Page $259$ section $9.5.5$) says that the color mainly comes from absorption of light in the coordination compound:

The color is complementary of the wavelength it absorbs , the wavelength left over is the color of the complex.

But there are two instances of these coordination complexes:
$\ce{[Co(NH3)5(H2O)]^{3+}}$ which absorbs blue green light of wavelength $\pu{500nm}$ which appears to be red in color and
$\ce{[Ti(H2O)6]^{3+}}$ which absorbs blue green light of wavelength $\pu{498nm}$ which appears to be violet in color.
Question: If a color of a complex is determined with what color it absorbs then why is it that two complex's have the same absorption region yet appear to be in different colours

As you can see here, wavelengths of the maximum absorption says nothing about the shape and width of very broad absorption bands, typical for condensed phases. Thin absorption lines are seen (almost(?)) exclusively for atomic absorption spectra in gaseous phase only.

When we see the colour, the way we perceive it is very complicated, it depends somehow on how much is absorbed vs reflected at many wavelengths. But, in practise to characterise molecules we use a spectrophotometer and measure exactly what the absorption is vs. wavelength. This enables us to work out where the energy levels in the molecule are.

@porphyrin oh okay , then we can say titanium complex is having stronger reflection of violet light and strong absorption of blue green and , for the cobalt complex strong absorption of blue green and stronger reflection of red ?

@Poutnik thank you , it seems the materials absorb a range of wavelengths which makes them recieve their characteristic colour

As I mentioned the way we perceive is complex, a solid of the same substance may appear quiet different to its dilute solution hence the use of instruments to characterise. I have seen a  'color wheel' that helps one to understand absorption/reflection and what you actually see, you could search or possibly someone else has a reference.

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text:
What exactly are the stages of the reaction between an oxygen ion and a water molecule?
I found information on the website water.lsbu.ac.uk - Oxygen and water:

Between altitudes of $80-\pu{100 km}$, $\ce{O2^{\small{\bullet}+}}$ and $\ce{NO+}$ ions are the dominant cationic species. $\ce{O2^{\small{\bullet}+}}$ ions react with water molecules to form  $\ce{H2O.O2^{\small{\bullet}+}}$   species (see figure left) [4020].. At lower altitudes, this further reacts with more water molecules to form hydrogen ions.

that as a result of the reaction of an oxygen ion with water, a hydronium ion can be obtained. If I understand correctly, the oxygen ion in interaction with water dissociates the water molecule, but how does the formation of the hydronium ion occur?
I hope that there is someone here who would be able to write down the stages of the reaction for me. Thank you.

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text:
Understanding the Role of Gibbs Free energy and Enthalpy as the energy required to drive a reaction
I was looking at the enthalpy change for water-splitting reaction:
$$ \Delta H^o_R =  [\Delta H^0_{H_2(g)} +\frac{1}{2}\Delta H^0_{O_2(g)}]-\Delta H^0_{H_2O(l)} = \pu{285.83  kJ/mol}$$
According to the book "Thermal physics"by Schroeder; at constant T and P; if there are no other forms of work on the system besides compression/expansion, then $\Delta H^o_R = Q$. However, when there are other forms of work being done we then have $\Delta Gº_R  \leq W_{other}$ and $\Delta H^o_R = Q + W_{other}$. The value for the Gibbs free energy in this case is $ \Delta Gº_R =  \pu{237 kJ/mol} $. We can relate $\Delta G$ and $\Delta H$ by $\Delta G =\Delta H -T\Delta S$ .
My confusion arises first from reading that the Gibbs free energy is the work we need to drive the reaction, say electrical work. However, the enthalpy change shows that the energy required could be done via heat and/or another form of work is higher than that of the Gibbs free energy? My guess so far is that we can take some energy from the environment for "free", but what happens when we drive this reaction only via heat, such that $\Delta H = Q$,  would $\Delta G = 0 $ ?

To get the properly formatted $\pu{285.83 kJ/mol}$ you can write $\pu{285.83 kJ/mol}$ or  $\pu{285.83 kJ mol-1}$ for $\pu{285.83 kJ mol-1}$.

Be aware that for the given conditions  -ΔG/T = ΔS_total. You need to convert some nonvolume work to heat to ensure ΔS_total >= 0

First of all WE are doing nothing except possibly mixing chemicals and perhaps lighting the match. The energies involved reside in the internal energies in the reactants, products and the environs.
Enthalpy is U + PV. It is not necessarily heat because heat change is not a state function. The utility of enthalpy is that it is a state function and can be measured under certain conditions. Those conditions are at constant pressure and no work except PV work. Under these conditions Q, heat exchange, becomes a state function. Q can be measured by the art of calorimetry. Standard values are calculated from heat capacities and latent heats involved using Hess's Law. H, U, [even S] are based on a total amount of reaction; all the reactants become products. This causes a problem because all reactions do not go to completion.
We pchem students are not the only ones that struggle with this; imagine how the pioneers had to struggle: no textbooks, lousy equipment, no standards, fighting established beliefs, not even knowing what matter and energy really were [i am not saying that we do know now]. They accomplished wonders in developing the Laws of Thermodynamics, laws that seem almost infallible. One big question remained reactions did not go to completion, the question of equilibrium.
Gibbs, Helmholz and many others realized that there were two competing or complementary processes in literally everything: the release of energy and the increase of entropy. The increase in entropy happened because of the release of energy. When the energy release slowed the entropy increase slowed until at equilibrium they were equal. At equilibrium forward and reverse reactions have equal rates. There are some reactants that will not react. This effectively means that this heat from the reaction cannot be removed from the reaction. This energy is quantified by TdS. The maximum amount of heat that can be turned into work is A = U - TS at constant volume or G = H -TS at constant pressure. The First Law establishes that work and heat are equivalent. The Second Law states that some of the heat must remain at equilibrium and cannot be converted into work [without changing the conditions].

[OP] what happens when we drive this reaction only via heat, such that ΔH=Q, would ΔG=0?

The Gibbs energy is a state function, so no matter how we run the reaction, the Gibbs energy of reaction stays the same. It is still positive, and that means you still need non-PV work to make the reaction go forward. As you know, water is stable at room temperature (it does not split into the elements while cooling the surrounding). So that hypothetical path is not available (in that direction). In the other direction, you have a choice. You can let the gas mixture explode, or feed it into a fuel cell to generate electricity.

[OP] Understanding the Role of Gibbs Free energy and Enthalpy as the energy required to drive a reaction

Enthalpy and Gibbs free energy are both state functions. So these are not parameters you can play with to drive a reaction. What you can do to drive a reaction are things like putting in work or removing a product (or - similarly - couple the reaction with a second one). Finally, you have no control over heat flow in a quasi-isothermal situation (you can't magically "push" heat from the solvent into the reactants).

[OP] My guess so far is that we can take some energy from the environment for "free"

Yes, that is true.

Thank you very much for your explanations! I checked with the book again, and they sort of mention, that a reaction with $\Delta H > 0$ will only take a certain amount of heat:   ($T\Delta S$), and the rest will need to be done as non-PV work. I am a bit thrown off by considering a "limit" to the heat we can use to drive the reaction (I thought Q could supply all $\Delta H^0$). Or is my understanding going astray somewhere? Kind regards.


text:
What is the major product on mono chlorination of 2-methypentane?
I understand that mono chlorination of 2-methylpentane are :

My Question is : How can you determine the major product from those results ?
The probability of monochlorination of :
Amount of 1-chloro-2-methylpentane: 6 (number of 1°hydrogens) × 1.0 (relative reactivity) = 6.0
Amount of 1-chloro-4-methylpentane: 3 (number of 1°hydrogens) × 1.0 (relative reactivity) = 3.0
Amount of 2-chloro-4-methylpentane: 2 (number of 2°hydrogens) × 3.8 (relative reactivity) = 7.6
Amount of 3-chloro-2-methylpentane: 2 (number of 2°hydrogens) × 3.8 (relative reactivity) = 7.6
Amount of 2-chloro-2-methylpentane: 1 (number of 3°hydrogens) × 5.0 (relative reactivity) = 5.0
From the calculation above, the major product will be 2-chloro-4-methylpentane and 3-chloro-2-methylpentane with the same yield. But we should choose one from the two. I am confused. Do you have any idea ?


text:
Formation of cocrystals troubleshootin
In cocrystal formation what would the possible reason be for the precipitation of an amorphus oily precipitate not suitable for PXRD? Solvent was DMF and the reagent were L-thyroxine and isonicotinamide
I had the same problem of not being able to form cocrystals with other combinations with L-thyroxine and at this point i dont know what im doing wrong. The other used solvents were water and methanol and the other reagents were all organic like 3-hidroxyipiride and nikotiamide for an example. Pls help

For readers: PXRD stands for Powder X-Ray Diffraction (also seen as PXD) // For Luka: It is a good habit to explain in text less used acronyms. Valuable answers may be provided even by those not familiar with them. It is not very considerate to make multiple readers to google it if you can explain it in a single action. Overall, the question is written in a way like if you were catching a bus. Low effort in question formulation seldom attracts good answers.

Not obtaining crystals is the typical result. Even if you have published results describing how someone else made crystals, not obtaining crystals the first time you try is the typical result. There are a lot of parameters, and you don’t have full control over them.
This is not a problem you can solve remotely. If you can talk to a local expert, your chances of obtaining crystals increase.
Some crystals rely on a badly designed incubator, some allegedly rely on smoking a certain brand of cigarettes, and some rely on letting the samples age so that they decompose slightly, but not too much.


text:
What will be the product if ethane reacts with SO2?
At first I thought it was Reed's reaction but then I looked up the reaction and the reactant was actually $\ce{SO2Cl2}$ (sulfuryl chloride).
I couldn't find the product which would be produced when any alkane reacts with $\ce{SO2}$. Does it react and what would be the product?

AFAIK there is no reaction between SO2 and alkanes in the absence of other reagents

At what temperature? At room temp, likely nothing happens. At 1,000°C, likely there would be multiple products, as the molecules decompose and recombine.

Electronically excited SO2 do reacts with alkanes: https://pubs.acs.org/doi/abs/10.1021/acs.jpca.8b04643

Methane is oxidized by SO2 at high T Other hydrocarbons might be also. https://ase.tufts.edu/nano-cel/publications/directReductionSO2.pdf#:~:text=Methane%20is%20an%20attractive%20reductant%2C%20due%20to%20abundant,the%20various%20elemental%20sulfur%20forms%20%28S2%2C%20S6%2C%20S8%29.

Alkanes can react with $\ce{SO_2}$ if the reaction is photoinitiated via via a hydrogen abstraction to form $\ce{HOSO^.}$ and organic radical $\ce{(R^.)}$ species and that reactivity is correlated with the energy required to break a C–H bond and the length of the alkane chain. you can find references here

For future reference: for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations.


text:
State functions and change in state functions
So, I have been studying thermodynamics, in it I read that state functions are those functions which depend only on one state of the system and are independent of the path taken, which is pretty easy to understand. And this clears why properties like enthalpy, pressure, volume etc are state functions.
Then there was the statement,

Change in a state function is not a state function.

which took me some time to understand, but I got it once I referred to the original definition of a state function. But now when I read about Hess law, which basically states that change in a state function is independent of the path taken, I'm confused as doesn't this imply that change in state function is infact a state function. Can someone please clarify this.

Does this answer your question? Is delta H a state function?

@Poutnik i had seen this answer to clarify my doubt regarding why change in state function is not a state function but this doesn't clarify my doubt regarding hess law

If you do not comment you have seen it, it is like if you have  not seen it. // The Hess law is about different paths between 2 given states, leading to path independent change. If one of the states changes, the change value changes, not being function solely of a single given state. The single state is the key, not the change being independent on the path.

Imagine hiking scenario and the elevation as the state function of your position.
The elevation value for given location does not depend on the way how you have got there nor on any parameter not defined by the state, therefore it is a state function.
OTOH, your elevation change when getting there depends on the initial point and does not depend solely on the destination. Therefore the state function (elevation) change is not state function.
If, like in the Hess law , both the initial and final states are given, then the difference in the respective state function values (elevation change) does not depend on the path taken from the initial to the final point.
As state function change depends on two states, not on the current state only, it is not a state function. Multiple initial states, leading to the common final state, would mean multiple state function value changes for the same state.

Thanks for the amazing explanation. So to be classified as a state function, the function must be only dependent on the state of the system, while the condition that it is independent of the path taken may not be enough to say that the function is a state function. Is this correct?

State function = function of state, not function of state + other factors.


text:
Can species with diffrent number of electrons be isoelectronic?
According to IUPAC gold book-

Does this meam that 2 species with diffrent electron number, but same valence electrons, be considered isoelectronic
Say a molecule XY and ZY where X and Z belong to the same group

Does IUPAC say/assume all electrons are valence electrons?

I don't think so, "valence electrons" leads me to oxidation state and valence leads me to "the maximum number of univalent atoms that can combine with the element"

(Consider the above as a rhetorical question.)

That depends, often only valence electrons are considered, so the concept is of more use.

The wikipedia page for isoelectronicity shows an example image of Serine, Cysteine, and Selenocysteine.
These three molecules can all be described by chemical formula "HOOC−CH(−NH2)−CH2−XH" where X represents Oxygen, Sulfur, or Selenium.

Thanks!, I was also  does this mean that one can describe the atoms O, S and Se as isoelectronic?

@ThatApollo777 The isoelectronic concept is about comparing the overall structure of one molecule to another; for some big complex molecules it may make sense to consider some substructures as being isoelectronic substructures... but it seems like a stretch to make this kind of comparison all the way down to single-atom 'substructures'.

@ThatApollo777 Furthermore, even if that particular O in Serine is analogous to the Se in Selenocysteine it is probably bad practice to say things like "O is isoelectronic to Se" because that's erasing the entire context of where the comparison applies and makes it potentially sound like it happens for ANY/ALL contexts (which will not always be the case).


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Formula derivation for adiabatic process

$\pu{1 mol}$ of an ideal gas at $\pu{400 K}$ and $\pu{10 atm}$ is allowed to expand adiabatically against $\pu{2 atm}$ external pressure. Find the final temperature of the gas if it is diatomic.

While looking for the solution of the problem above, I came across the formula for adiabatic compression in a reference book, and I was wondering how I could derive it:

$$T_2 = T_ 1 \left(\frac {C_V + P_\text{ex}\frac{R}{P_1} } { C_V + R}  \right)$$

First I tried using the enthalpy equation
$$∆H = ∆W + ∆nRT,$$
but it proved to be futile. Then I tried using the definition of an adiabatic process, $PV^\gamma$, but that too didn't yield the desired result.
Could someone kindly help me solve this question?

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The definition of adiabatic process is "no thermal exchange with environment" and not what you said. $PV^\gamma$ is derived from that. So you are not against memorizing formulae when they are nice. (Same with me.) Now to the point: what is $P_{ex}$?

Sorry for the confusion. I forgot to give the context in which the term P (ex) is used.

Imagine that you gave me a problem on calculating a potential energy of an apple. Now imagine that I come up with an answer that involves the mass of the apple, the gravity of Earth, the distance to the Moon, and the pressure at the bottom of the Mariana trench. Without looking at the actual expression, do you think it may be correct? Why or why not?

I apologise for the confusion, but I am as clueless as you are. I checked the solution for that question, and the author has directly used the above mentioned formula. My main goal is not just to prove the formula, but to arrive at the answer to the given problem.

@IvanNeretin, I do understand why you may feel that the formula is incorrect. However, I'm sure that the formula used is exactly correct. Here $P_{ex}$ is the final pressure after expansion.

What's $P_2$, then?

I'm extremely sorry, I should have noticed that before. For this question, $P_2$, as you said, is nothing but the final pressure after expansion, i.e., $P_{ex}$.

This is an irreversible adiabatic change there is already an answer here https://chemistry.stackexchange.com/questions/97677/irreversible-adiabatic-process-against-constant-external-pressure

Thank you so much, @porphyrin!


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Explanation for Raoults Law
The vapour pressure of pure liquids is independent of quantity of substance.
But when in a solution, the vapour pressure of the components are given by
$$P_A=P^°_AX_A$$$$P_B=P^°_BX_B$$
Since $A-A$, $A-B$ and $B-B$ interactions are assumed to be equal for an ideal solution:

In case of liquid-liquid solutions how does it affect the vapour pressure(like whats is mechanism at molecular level). I presume it has something to do with volumes of both liquids being added as a result of which it is kind of like each liquid occupying a larger volume than permitted so making is somewhat similar to effect on gaseous equilibrium upon adding inert gas at constant pressure But I  hope someone might provide a better,  more intuitive explanation.
In case of non-volatile solid solutes, I cannot think of any explanation. Since solids occupy intermolecular spaces of liquids, there is no change in volume of solution, so it should essentially have no effect on its vapour pressure similar to adding inert gas at constant volume to gaseous equilibrium.

Please explain with intuition why vapour pressure is not independent of concentration.

Starting with the easy answer, a solid can attract and absorb a liquid. For example, $\ce{CaCl2}$ is hygroscopic, and is sold packaged to reduce humidity in enclosed locations. Simply think, water molecules are more attracted to $\ce{CaCl2}$ than to the space above.
For liquids in liquids, consider those same intermolecular forces. For example, 100 mL of ethanol ($\ce{C2H5OH}$) + 100 mL of water ($\ce{H2O}$) combine to take up ~4% less volume. Again, thin of it as ethanol  being more "attracted" to water than to air.
BTW, surface tension is similar, in that one measures cohesion of a liquid compared to adhesion to an interface.

The lower molar fraction of the volatile liquid A, regardless of if the other component B is volatile or not, means the lower rate of the liquid A evaporation.
It can be easily imagined that there is a lower number of molecules on the surface unit and therefore lower molecular evaporation rate
\begin{align}
\frac{\text{d}N_\text{A}}{\text{d}t \cdot \text{d}A} &= k_\text{A,f} \cdot N_\text{A} &- k_\text{A,b} \cdot p_\text{A}\\
&= k_\text{A,f} \cdot N^\circ_\text{A} \cdot x_\text{A}  &-k_\text{A,b} \cdot p_\text{A}
\end{align}
At equilibrium, the rate of evaporation and condensation is equal.
\begin{align}
k_\text{A,f} \cdot N^\circ_\text{A} \cdot x_\text{A}   = k_\text{A,b} \cdot p_\text{A}\\
k_\text{A,f} \cdot N^\circ_\text{A}   = k_\text{A,b} \cdot p^\circ_\text{A}
\end{align}

Legend:

$N_\text{A}$ is molecular surface density of the component A.
$N^\circ_\text{A}$ is molecular surface density of the purecomponent A.
$t$ is time.
$A$ is surface area.
$k_\text{A,f}$ is the evaporation kinetic constant for given temperature.
$k_\text{A,b}$ is the condensation kinetic constant for given temperature.
$p_\text{A}$ is the partial pressure of the component A.
$x_\text{A}$ is molar fraction of the component A.


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Why is it okay for the central carbon to lose a bond when forming a resonance structure for acetone?
As the title reads, I'm confused as to why it's okay for the central carbon in acetone to have a +1 formal charge when we push the electrons from one of the bonds with oxygen to form a lone pair and give the oxygen a -1 formal charge.

Try to read this question after a month, pretending it was written by someone else, if it is still clear to you after a single look.

Resonance structures are a model, and as such, it may or may not reflect reality. It's only as good as the predictions are. Does drawing resonance structures in this way for acetone produce any useful conclusions?

You can push the electrons around in many different ways when creating a resonance structure, but some of the resulting structures will be "good" and others will be "bad." The "good" structures will contribute more to the actual properties of the molecule than the "bad" structures will.
As always, when talking about resonance structures, it's important to remember that both the name and the notation are extremely misleading. There is no vibration, oscillation, or transition between the two structures. Rather, the molecule has one structure that is a weighted average of the different structures, with the "good" structures having higher weight in the average.
The resonance structure of acetone with a formal negative charge on the oxygen and a formal positive charge on the carbon is a relatively "bad" structure (because of the formal charges) compared to the "good" one with a double bond and no formal charges. Therefore, the properties of acetone are mostly described by the "good" double-bond structure, with a much smaller contribution from the "bad" structure with formal charges.
There is actually a third possible resonance structure for acetone, where you push the pi electrons the other way. leaving the carbon negative and the oxygen positive. But this is an even more "bad" resonance structure (because oxygen is electronegative and "prefers" a negative charge) and makes an even smaller contribution to the actual properties of the molecule.
Here is another explanation (including all three possible structures for acetone) to the same effect, from Dr. Ian Hunt at Calgary: https://www.chem.ucalgary.ca/courses/353/Carey5th/Ch01/ch1-7.html


text:
Clarifications on enthalpy and change in heat energy
From my understanding, an exothermic reaction has a negative change in enthalpy meaning that energy is released to the surroundings. However, is it correct to say that during this exothermic process, the enthalpy change of the surroundings is positive? Does this mean that whenever a system has a negative change in enthalpy, another system must have a positive change in enthalpy?
Also, I was wondering what the distinction between change in enthalpy and change in heat energy is. I was told that the change in heat energy has an opposite sign to that of enthalpy during a process. Hence, does heat energy refer to the surroundings??

Review the guide How to ask and Asking FAQs to prevent need of clarification, objections, down-voting or closure.

To your first question:
Consider a huge, isolated system $\text{A}$, consisting of two other systems:

A closed system $\text{B}$,
A huge closed system $\text{C}$, fully surrounding the system $\text{B}$, acting as a heat bath with constant temperature.

As $\require{cancel} \text{A}$ is isolated, its enthalpy $H_\text{A}$ is constant and  $\text{d}H_\text{A}=0$
If an exothermic reaction occured in $\text{B}$,  and thermodynamic equilibrium is reached in $\ce{A}$, then:
\begin{align}
ΔH_\text{A} &= ΔH_\text{B} + ΔH_\text{C} = 0\\
ΔH_\text{C}&= - ΔH_\text{B}\\
ΔH_\text{B} &\lt 0 \implies ΔH_\text{C} \gt 0
\end{align}

For isothermic and isobaric conditions, the enthalpy change of a system is the sum of the heat and nonvolume work (e.g. electricity during battery charging) provided to the system.
By definition:
$$ΔH = ΔU + Δ(pV)$$
For isothermic, isobaric conditions:
\begin{align}
ΔH &= W_\text{volume} + W_\text{nonvolume} + Q  + pΔV\\
ΔH &= \cancel{-pΔV } + W_\text{nonvolume} + Q  + \cancel{pΔV}\\
ΔH &= W_\text{nonvolume} + Q
\end{align}

The heat of reaction means that, for a closed system operated at constant pressure, the amount of heat that must be added to the system in order to also hold it at constant temperature is $Q=\Delta H$.  For an exothermic reaction, heat must be removed from the system to hold it at constant temperature.  So, for a chemically reacting system held at constant temperature and pressure, the heat of reaction is equal to $\Delta H$.


text:
Determining the molar mass of a substance by the osmotic pressure as a function of a mass concentration
Let's say that I have a solution of some substance in some solvent. The density of the solute is $\rho_\text{solute}$ and the density of the solvent is $\rho_\text{solvent}$. Suppose that the osmotic pressure of this solution is $$\Pi(\gamma) = k\gamma$$ where $k$ is some constant. We know that osmotic pressure is generally given by $$\Pi = icRT.$$ If we express the molar concentration as $c = \gamma / M$ and rearrange the equation, we get $$\Pi(\gamma) = \frac{iRT}{M}\gamma.$$ That means that $$k = \frac{iRT}{M}.$$ Now, if I know the constant $k$, but not the temperature $T$, that means that I can't calculate the molar mass $M$? Is there maybe another way to calculate it? Perhaps using the given densities? Thank you in advance.

Well, the temperature can be determined from the density of the solvent. Or estimated.

By that, do you mean looking at charts or something else? @Poutnik

Charts or using empirical formulas. But estimation of T could be good enough.

If you have been able to calculate the constant $k$ in an osmosis experiment, it means that you know the formula and the molar mass of the solute (and the temperature).

I was given the value of $k$ and need to find out the molar mass.


text:
The contact process
In the contact process, concentrated sulphuric acid instead of water is used to absorb the sulphur trioxide gas produced.One of the reasons suggested online is that the reaction between sulphur trioxide and water is highly exothermic.
But I have these concerns:

Both the reaction between sulphur trioxide with concentrated sulphuric acid and that between oleum and water are exothermic
If the reaction is too exothermic, a coolant can be applied in the industrial process

So, I propose two possible reasons.

The two previously mentioned are less exothermic
More concentrated sulphuric acid can be produced by the reaction between concentrated sulphuric acid and sulphur trioxide, hence the process can become more economical

What should be the best explanation for why water is not used to absorb sulphur trioxide?

The explanation is kinetics. $\ce{SO3}$ reacts with water producing fumes of sulfuric acid. On the contrary it is dissolved into sulfuric acid without fumes, producing oleum. This is why sulfuric acid is used in practice instead of water for absorbing $\ce{SO3}$ produced in the contact process.

AFAIK, SO3 reacts with water rather violently and forming thick fumes, while it reacts with conc. H2SO4 relatively peacefully, forming oleum,  a mixture of mainly H2SO4 and H2S2O7. // Two step production of conc. H2SO4 is similar to preferred two step dilution of H2SO4: Mixing remaining diluted H2SO4 with water and after cooling,  mixing conc. H2SO4 with the prior mixture. The exothermic dilution is then divided in 2 steps, easier to manage.


text:
Concentration term PPM
What volume do we assume when the concentration is written as mass by volume ppm?
I mean when mass by volume percentage is say 20%, we can say that 20g solute is present in 100 ml solution.
So if it is written that concentration of given solution is say 20ppm (mass by volume) how can I understand this concentration term like I did in above example?


text:
What is that substance coming from tractor tire?
I have old tractor with Bridgestone tires. When I press a clean white tissue against the tire, a black spot show on the tissue. Is that substance carbon black and it’s harmful to soil?

I suppose it is applicable to all weathered black tyres on all vehicles. They progressively releases rubber and carbon particles. What exactly does "harmful to soil" mean?

I mean those tires are in powered tiller used for small garden.  This black substance will be released to soil effecting plants and humans(if its harmful).

Compressed air intended to clean analytical devices in the lab / medical use (e.g., scuba diving) has to pass filters to remove small droplets of oil (an aeorsol) which may originate from the lubrication of the compressor, and water.
I speculate the compressors at the gas / petrol station need not to fulfill quality standards this high and over time (years) these droplets accumulate in the tires with every refill of compressed air. By smaller diameter of the corresponding molecules and lesser surface tension, it is easier for air to pass through small defects of the tire mantle, bypass the neck of the valve, than it is for air.  (I back this up with the observation that I have to refill the tires of my bicycle at least monthly, despite the presta valves itself as well as their inner screw and the cap appear to be tightly closed.)  As for the composition, beside water (by humidity of the air compressed) and lubricant, it were plausible if this blend equally contains some soot/carbon black and rubber from the vulcanization, as well as debris because of the mechanical stress the polymer has to sustain (youtube hosts some videos of gopro cameras mounted inside a tire in used, example).


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How does polarity increase when electron density in oxygen decreases when comparing acidity
My book says the following statement:

Due to the higher electronegativity of $sp^{2}$ hybridized carbon of phenol , the electron density in oxygen decreases. This increases polarity of O-H bond and results in ionization of phenol

My question is , since the electron density in oxygen atom decreases , the overall partial negative charge that oxygen has should decrease which makes the O-H  bond less polar.
What Have I misunderstood here ?

Generally, higher electronegativity of $\ce{X}$ in $\ce{H-O-X}$ leads to higher acidity of the hydrogen atom.
If the oxygen atom in a phenol molecule can take less electron density from the bond with $\ce{C_{\mathrm{sp^2}}}$ (which is more electronegative than $\ce{C_{\mathrm{sp^3}}}$), it takes more electron density for $\ce{O-H}$ bond, polarizing it more, making the hydrogen more acidic.
Another factor, decreasing phenolate basicity and therefore increasing phenol acidity, is delocalization of the charge on the oxygen atom, involving the aromatic ring.
Both effects can be seen in perchloric acid $\ce{H-O-ClO3}$ too, where electronegativity of $\mathrm{Cl}$ is boosted by three oxygen atoms and the charge of perchlorate anion is delocalized among all four oxygen atoms.

So the electron density belonging to oxygen  we are speaking is of two types in the example , oxygen with carbon since carbon has relatively higher electronegativity it allows the second electron density o-h to be more polar(since csp^2 pulls majority of electron density towards itself , this allows more attraction and charge seperation)

Please correct me if I am wrong

Not two types, rather two contribution sources.  Consider a family analogy:  A boy (O) is asking for his monthly allowance at his parents. He is not greedy, he wants just the amount that was promised.  If the parent A (C)  gives less money than expected, the more money the boy (O) asks at the parent B (H). // Rather, O pulls less of the electron density from C_sp2 than from C_sp3, as the electronegativity order is O >> C_sp2 > C_sp3

Ah ! That is a very good analogy thank you for clearing this for me


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Production of town gas
Town gas contains quite a lot of chemicals.
Based on my research, the order of abundance of its constitutes is hydrogen, carbon monoxide, nitrogen, carbon dioxide, gaseous naphtha and oxygen, if arranged in descending order).
I do not understand why the abundance of carbon monoxide is greater than that of carbon dioxide since carbon monoxide can be converted into carbon dioxide due to the reaction $\ce{2CO + O2 -> 2CO2}$. So, in my opinion, the abundance of carbon dioxide will certainly be greater.
I propose several possibilities:

The problem boils down to inaccurate or incorrect data of my research
The reaction $\ce{2CO + O2 -> 2CO2}$ is hard to take place

Are there any other possibilities, or what should be an answer to my question?

The border between "there is" and "there isn't" is not where you think it is. There isn't any significant amount of oxygen there, hence CO stays CO.

For future reference:  for the body of questions, answers, and comments, chemistry.se offers to use mhchem as a comfortable method to add chemical equations.

Reactions like the oxidation of CO to CO2 don't just happen. The production process of Town gas is  designed to minimise the proportion of CO2 and to maximise the proportion of CO (CO2 is a useless product, CO is a useful fuel). H2 and CO are the major useful components of the gas.

Starting from carbon (which coal is not), the two most common oxidations yield carbon monoxide (incomplete combustion, $\ce{2C + O2 -> 2CO}$), or carbon dioxide (complete combustion, $\ce{C + O2 -> CO2}$).  However, there equally is a temperature dependent chemical endothermic equilibrium between carbon, carbon monoxide, and carbon dioxide ($\ce{C + CO2 <=> 2CO}$) equally known as Boudouard reaction which is of significance e.g., in blast furnaces to generate iron:

(image credit to openstax.org)
A high concentration of $\ce{CO2}$ in town gas is not desirable because contrasting to $\ce{CO}$ it does not oxidize further (which eventually is the source of heat why town gas principally is/was used for).  Instead (similar to nitrogen), $\ce{CO2}$ picks up heat which otherwise would be available for a combustion of pure $\ce{CO}$ with pure $\ce{O2}$, and dilutes the reagents; both lower the maximal temperature of the flame.  Hence the generation of town gas runs within reaction parameters (e.g., temperature, pressure, residential time of the reagents in the reactor) where the yield of carbon monoxide is maximized.
While it is possible to burn pure $\ce{CO}$ with pure $\ce{O2}$, the required purification of town gas to remove nitrogen; carbon dioxide, and other gases at scale (and flux) of consumption by cities would require large additional technical installations.  There is too little (economical) incentive to offset these investments.


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What's the reason of curvature in Tanabe-Sugano diagrams?
Analysing Tanabe-Sugano diagrams it's evident that some of the lines are pretty parallels to that of the ground field whilst others are inclinated or have some curvature.
I know what's the effect on the line width of the spectra, but I'd like to have an explanation on why we have so different behaviours.
The fact that in Orgel diagrams some lines are straight and others have curvature is related to this?

For readers:  Wikipedia: Tanabe-Sugano diagrams and Orgel diagram // See also Libretexts   here and here


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How are Fermi resonances between normal modes possible?
As I understand, Fermi resonances occur when two (nearly) resonant vibrational states—most commonly a fundamental and overtone or combination mode—that are of the same symmetry interact, which, in the usual way, causes mixing between the two states and an increase in the energy difference between the two (a Rabi splitting, essentially).
But it puzzles me that this is even possible if I assume that both of the two states that comprise the resonance are actually normal modes. The resonance/interaction relies on the matrix element between the two states, $\langle \psi_1 | H | \psi_2 \rangle$, being nonzero. But by the definition of normal modes, these states should be eigenstates of $H$ and orthogonal, shouldn't they? Then the matrix element is necessarily zero, so is the resonance only possible in the presence of some additional perturbation or did I misunderstand something?

I think you have it backwards: the matrix element in this case is zero because you are imposing the constraint via normal modes.

@ToddMinehardt After reading and thinking a bit more, I think my reasoning is correct but irrelevant, since the concept of Fermi resonances is (apparently?) only applied to local modes, which are generally not the normal modes. And then the matrix element is non-zero and everything works out. Do you agree?

Your reasoning is correct - local modes are not the normal modes unless an approximation has been made. I was glues to a copy of Wilson, Decius, and Cross for 4 years. R = LQ = B$\delta$, where local modes R are related to normal modes Q by the L-matrix, which is sparse but not just on the diagonal. And the expectation value of the transition dipole moment operator - sometimes taken as the vector r - is then calculated with the force field parameters which include but are not limited to Morse functions plus assorted higher-order terms.


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How does tBu2Si(OTf)2 react in uridine in DMF at 0°C
I have to write the mechanism of a four steps reaction that aims to transform uridine into a deoxythymidine by following these steps :

(tBu)2Si(OTf)2, DMF, 0°C
ClCSOPh, DMAP, 20°C
AIBN, Bu3SnH, 80°C, toluen
TBAF, THF

I am stuck in the first step. I know that DMF promotes a SN2, I found that the Si atom is electrophilic but in the uridine, I couldn't find the most nucleophilic atom that can react in Si. Could you please help me to find it?

here is the reaction I wrote after the explanation :

Think about the process here. The object of the sequence is to remove one of the oxygens of the N-heterocycle so you do not want the oxygens on the sugar reacting with the ClCSOPh and the last step (TBAF) removes silyl protecting groups.

@Waylander thank you for your reply I want to show you what I did in order to be sure that I got all what you said

Looks correct to me


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If air can be dissolved in water, is water vapour dissolved in air?
If O2 dissolved in water acts like a liquid even though it is a high temperature, then is water vapour in the air "dissolved" as it acts like a gas at less than 100C? Or is there a different word for it (suspension)?

Mixtures of gases are never called solutions.

Water vapour is just part of the air. You can call air "gaseous solution" if you want, IUPAC won't sue you. Heck, air is even a suspension if there are liquid droplets, or dust particles.

While gas mixtures may be formally called solutions as homogenous mixtures, there is important different to conventional solutions. Water vapor as gaseous form of water is not being dissolved by air. Vapor would be there even if there would be no air.


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Does dissolution and crystallization always occur together?
Does dissolution and crystallization always take place simultaneously?
Also does crystallization also occur in unsaturated solution but its rate is less than rate of dissolution or is rate of crystallization 0?

Getting to, and staying at, thermodynamic equilibrium is a dynamic process.

For your first question:
Yes, crystallisation and dissolution takes place simultaneously.
When a solid solute is added to the solvent, some solute dissolves and its concentration increases in solution. This process is known as dissolution.
Some solute particles in solution collide with the solid solute particles and get separated out of solution. This process is known as crystallisation.
For your next question:
Crystallization rate is NOT 0 even in unsaturated solution.
As I have already mentioned crystallisation occur simultaneously when solute is getting dissolved.
But the rate of crystallisation is less than rate of dissolution that's why amount of solute being dissolved is more that amount of solute being crystallised.

Are you really saying that in an unsaturated solution there actually is crystallisation happening? That does not make any sense to me.

@Martin-マーチン I am taking reference to my book according to which some dissolved atoms collide with solid solute particles and get separated out. But the rate of crystallisation is lower than rate of dissolution so we don't see actual crystals separating out. As more and more solute keeps on dissolving the rate of crystallisation and dissolution come in equilibrium and no more solute dissolve. My book : NCERT Chemistry Part 1 Class 12

An unsaturated solution is a homogeneous phase, there is no solid. What you state only holds for a dissolution process, or saturated solutions. It doesn't even hold for super-saturated solutions as you need a nucleation site.


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Separation of methyl acrylate from ethanol
I have liters of ethanol contaminated with methyl acrylate (MA) from cleaning UV resin 3D prints. Is there an easy way to purify this ethanol by separating MA (as unwanted product) from it? Distillation is not a good option, because MA has boiling point of 80°C, less than 2° higher than ethanol.

Do you want to recover the methyl acrylate or just want rid of it?

Waylander: no, I do not want MA, it is useless for me, I just want to recycle/regenerate ethanol for another use as cleaner. May be it is not problem if it contains MA, but it has very bad odor, much worse that ethanol itself.

Could you then add sodium hydroxide to hydrolyse the MA to sodium acrylate + Methanol and then distill to get your ethanol?

Yes, that seems easy. Sodium acrylate will precipitate? What about reaction of ethanol with sodium hydroxide, is there any? [link] https://chemistry.stackexchange.com/questions/90281/will-naoh-react-with-ethanol

As the ethanol solution concentrates then it will eventually precipitate

@Waylander: Thanks, I will try it and write here about the result.

Probably need to heat the EtOH solution to get complete reaction of the MA

OK, I can confirm that it works. I have tried it in small scale, about 20 ml of contaminated ethanol, after adding NaOH and heating to 60°C and stirring, cloudy precipitate has formed and odor is now "alcoholic", stronger than ethanol itself, but I think it is thanks to methanol formed. Thank you for advice.


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What meaning does Gibbs energy acquire when applied to solutions?
I have been picturing functions like enthalpy and Gibbs free energy as mathematical tools, which were defined in such a way as to be useful for some specific cases. I understood enthalpy as a useful tool for calculating heat when the only work being done was P-V and the system was closed and at constant pressure. Enthalpy being a state function can be represented as a function of intensive properties, this makes it useful to calculate heat, heat being a path function it is difficult to tabulate experimental values to calculate heat in other cases (different temperature ranges, different substances, etc), but for enthalpy it is not, it is easy to tabulate experimental values that do not depend on mass or path.
For Gibbs energy I did the same, i thought it was a useful function to determine if a process would be spontaneous or not, or in an equilibrium with quantities easier to measure and to control in experimental settings, those being temperature and pressure.
But now that I've been studying solutions all the meaning I had given to Gibbs energy has been thrown out the window. My text book just grabs Gibbs energy to study solutions without a particular justification to do so
If $nG=f(T,P,n_1,..,n_j)$ the total differential is
$$d(nG) = nVdp - nSdT + \sum_i \mu_i dn_i$$
Where a new a new property is introduced, the chemical potential. I can follow the math and everything, but what does Gibbs free energy mean here? Why it gets picked up to study solutions? Why not enthalpy or enthropy? Physically now what does $G$ mean? It's not work or heat, nor a criteria to determine spontaneity, I don't know what it is and neither why does it work. Not resolving these doubts makes it difficult to follow what happens in the rest of the chapter. Everything is build upon this function, but what is this function? A result of $G=2 \frac{J}{mol}$ in this context what would it mean? How does $G$ manage to keep everything together?
Another thing, this is an easier question to answer, why do thermodynamics books always use $\equiv$ to define such functions? Why not just $=$?

Note that G,H,U,V,S are generally extensive properties. Then there are various intensive molar versions with explicit indexes.

This Enthalpy being a state function can only be represented as a function of intensive properties is wrong.

you're right, what I wrote wasn't quite right, I'll fix it

And about $G,H,U,V,S$, I know they can be extensive properties or intensive properties, they have to be devided by mass in order to be intensive, but in general they are computed via intensive properties (at least on my experience) making them intensive for a "brief time" before multiplying by the systems mass

Well, if you measure reaction enthalpy, you measure extensive enthalpy, dividing it by amount of substance to get the molar formation or reaction enthalpy. The point is the state variable symbols U,H,S,G,A with implicit indexes are implicitly considered as extensive variables with energy units. // Ask rather, why is equality operator often incorrectly used as the definition or assignment operator.

You may want to review search results for site:chemistry.stackexchange.com OR site:libretexts.org OR inurl:hyperphysics Gibbs free energy meaning

Gibbs energy (also called free enthalpy) and it's change in context of isothermal and isobaric processes have two major meanings:

It is the hard limit for maximum non-expansion work the system (or it state change) can provide without violating the second law of thermodynamics. By layman terms, it is available energy usable for other things than transformation to heat. It does not say anything about if this limit can be practically reached nor the way how it could be done.
The system entropy change is the parameter determining process spontaneity in isolated systems. The same role is played by Gibbs energy for closed, isothermal and isobaric systems. As

$$- \frac{\mathrm{d}G}{T} = \mathrm{d}S_\mathrm{sys} + \mathrm{d}S_\mathrm{surr} = \mathrm{d}S_\mathrm{tot}$$
The negative change of the system Gibbs energy means the positive change of the universe entropy, so the process is thermodynamically spontaneous (still can be kinetically frozen}.
If the Gibbs energy change is negative, the process is spontaneous, even if it is endothermic (dissolving salts leading to cooling down the solution) or if it decreases system entropy (vapor condensation).


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Are any two elements indistinguishable by a single measure?
Elements can be investigated using a range of different techniques to determine their fundamental properties.

Some properties of an element can be observed only in a collection of atoms or molecules of the element. These properties include color, density, melting point, boiling point, and thermal and electrical conductivity.
Infoplease: Properties of the Elements

Are there any two stable(ish) elements that are completely indistinguishable in terms of their fundamental properties by a single measure?
That is to say, if I investigated one element to the absolute limit of human technology in that one specific measure, I wouldn't be able to tell one element from another?

Closest I can think off the top of my head is iridium and osmium, whose densities in ambient conditions differ by 0.13%. This is within measurement error for even rather accurate analysis methods, but obviously still the realm of possibility. Similarly, there have been attempts to make fake gold bars with tungsten, whose densities differ by around 0.25%, and you can believe there was significant economic pressure to tell the difference.

Nd and Pr are so similar that it is very difficult to separate them: https://www.sciencedirect.com/science/article/abs/pii/S1002072118309918. Zr and Hf are another example.

Can you explain better what do you mean by "single measure"? Do you mean if there are elements with identical physical properties? (Same identical density?) because a lot will depend on the accuracy of your measurements (see Nicolau answer.. 0.13% is not a lot but it is still a difference)

@Jojostack - That's exactly it. Are there any elements with (for example) completely the same density? That if you measured their density and density alone, to the maximal limit of human science, that you still couldn't tell one from the other.

@Jojostack -"Fakeium and Notrealium have similar melting points. So similar in fact, that modern science cannot tell them apart..." That sort of thing.

@Valorum Every measurement comes with errors, which either can be of systematic (relating to accuracy), or random origin (relating to precision; for a discern of the two, see e.g. Wikipedia's illustrations here). It hence depends on the method deployed and distribution of individual measurements about sample A and sample B if the two are significantly different from each other to distinguish the samples as about different elements (e.g., Student $t$-test).

This question seems to be attracting a bunch of downvotes but without explanation. Anyone care to suggest scope for improvement?

If I make the 'fundamental property' to be only crystal structure at STP, then there are many fcc, bcc, or hcp elements. If instead it is crystal structure and lattice parameters at STP, then the answer is no.

The absolute limit of human technology, huh? While it's technically impossible to prove a negative, the answer is almost certainly no, at least in spirit. The problem is that metrology is just too powerful!
Let's do a very crude analysis. We know 118 elements, and say there are 100 "reasonable" experimental properties for each. Nowadays, metrological studies can determine a wide variety of physical parameters with 5 or more significant digits. Let's say that the measurements are perfectly random within that range. That means you have 118 attempts at pulling 100 balls with replacement out of a box with 100 000 balls, and you hope to pull the same ball out twice. That probability is about 11%, which isn't very high.
If there is a way around it, you have to stack the odds in your favor. Not every physical quantity can be measured with the same accuracy in different contexts. For example, while something like LIGO can implicitly measure changes in distance up to ~22 significant digits, it turns out that precisely measuring volumes of macroscopic objects (which are fundamentally distances cubed) is tricky due to shape variability. What that means is certain volume-dependent properties, most obviously density, may be comparatively difficult to measure precisely.
And it turns out that there are a few elements which happen to have similar densities. Presumably for over 100 years, it was known that either osmium or iridium was the densest of the stable elements, but it was hard to definitely say which. It took arguably until 1989 for the matter to be settled by a combination of x-ray crystallography measurements, which determined the density of osmium to be about 0.12% larger than iridium (22.588 ± 0.015 g/cm³ and 22.562 ± 0.009 g/cm³, respectively). There are a few other pairs of elements whose densities differ on the order of 1%, and I don't know to what accuracy their densities have been established, so you could say there is still room for overlap confusion.
But even then, this is only true when comparing the data we have currently. In reality, though, the vast majority of properties measured for elements has not been performed with the "absolute limit of human technology" backing them, because that's very expensive. Usually it's just a few labs around the world working semi-independently with standard amounts of funding. However, when there is a strong coordinated international effort to determine such quantities, then it may be possible to tack on as many as 3-4 additional significant figures on demand.
I mentioned how density measurements are typically constrained by volume errors. Well, a few years ago we had the Avogadro project as part of a drive to change the definition of the kilogram. This eventually resulted in the manufacture of 99.9999999% pure, monocrystalline, 99.9999%-perfect spheres of silicon-28 (99.99% isotopic purity) and ultimately allowed determination of the macroscopic density of the silicon-28 isotope to 7 significant digits (approximately 2.320071 ± 0.000002 g/cm³). The same technology could be applied to silicon with a natural isotopic abundance for slightly less precision, to determine a true density for elemental silicon. Other elements would require modifications to the techniques, but it's conceivable that similar precision could be obtained, and it would be enough to distinguish every element with virtual certainty. There is no reason to assume similar disambiguation cannot be performed for every other reasonable physical property.


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Do diatomic molecules undergo hybridization
For the steric number to be 1,the molecule must be diatomic or linear in geometry and shape.
First of all, do they undergo hybridization?
Secondly, Just like the hybridization for steric number 2 is sp and for steric number 3 is sp2, what is the hybridization for steric number 1.

Hybridization is a technique that has been invented by Linus Pauling for explaining the angles and the geometric shape of the polyatomic molecules. As diatomic molecules are always linear, without any angle, hybridization is not necessary for explaining their geometry.

I agree with that. Sometimes it is helpful though. Like in oxygen, the mixing of s and p orbitals to create a bond that is lower in energy then the pi bonds is easier to understand  with hybridisation. But for the general case: yes, don't use it where you don't need it. You'll get an up-goat from me.

As a typo that's almost as good as cow-orkers. Will try to use it in future.

@Ian the pun was indeed intended. But it is a very old and brief one (and I didn't even spell it right). https://chat.stackexchange.com/search?room=3229&q=upgoat

It might not be necessary to explain geometry but it helps explain some molecular properties. AOs are fine but MOs are different.


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Is there any equation to predict the salting out concentration of a polymer?
I have to roughly calculate the salting out concentration of a polymer with respect to NaOH.
Is there any method to calculate?

There is a method, using  diagrams and/or empirical equations. It is universal, applicable to all  systems. It is not in favour, as it has unknown parameter values, particular for particular systems, that have to be gathered experimentally.


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Correct structure of compound tetraiodine nonoxide
I have a confusion about the correct structure of tetraiodine nonoxide.
I came to know about it while reading the equation :
$$\ce{6 I2 (dry) + 11 O3 -> 3I4O9 + 3O2 }$$
Now again I saw the equation :
$$\ce{ I4O9 <=>I^3+(IO3-)3}$$
Now this equation and this is structures in chemistry stack exchange make me to understand that ($\ce{I4O9}$) has the following structure:

But when I saw the following structure (shown below ) on Wikipedia I have a doubt about that which structure is correct:

On seeing all the structures, according to me, due to the small size of oxygen to iodine (large size) are unstable on it, therefore the structure 3 is not possible and structure 1 and 2 which are same structure are the correct structures of $\ce{I4O9}$.

Again after thinking I came across the result that the structure on Wikipedia has more chance to be correct because Wikipedia has made this page after my search but the place where I saw the structure 1 and 2 the writer has doubt on itself and this structure is a assumption by him.

The 2019 edition of Holleman Wiberg talks about oxidation states, not structure: https://i.stack.imgur.com/x08bv.png. Other sources mention a polymeric ionic structure. The sources mentioned by the OP say that their structure suggestions are hypothetical.

The Wikipedia structure for $\ce{I4O9}$ is plausible, but there is a twist — literally.
The simpler oxide $\ce{I2O5, O2I–O–IO2}$ indicates that a bridging oxygen between two iodine atoms is indeed possible. From Wikipedia:

I2O5 is bent with an I–O–I angle of 139.2°, but the molecule has no mirror plane so its symmetry is C2 rather than C2v. The terminal I–O distances are around 1.80 Å and the bridging I–O distances are around 1.95 Å.[1]

The angle at the central oxygen is much larger than that found in molecules where oxygen is bonded to small atoms or groups (cf.
104.5° in water, 112° in dimethyl ether, from Libretexts). Also the lack of mirror-image symmetry (which makes $\ce{I2O5}$ chiral) implies that the small size of the atom does have a steric effect, forcing the rest of the molecule to twist in exchange for the oxygen serving as a bridge.  We may expect the same thing in $\ce{I4O9}$.
Cited reference

Selte, K.; Kjekshus, A. (1970). "Iodine Oxides: Part III. The Crystal Structure of I2O5" (PDF). Acta Chemica Scandinavica. 24 (6): 1912–1924. doi:10.3891/acta.chem.scand.24-1912

Two questions. (1) Can you try again? (2) Is there a stylus for Android phones?

Molecuke?! More like molecute... Anyway, I don't find the bent surprising at all. Iodate is pyramidal in shape. And that is an electronic effect. The twist might well be due to "steric repulsion", but that's also just an electronic effect. Think about biphenyls, even a proton causes a twist. Steric effects are generally explained wrong though.

Determination of the absorption cross-sections of higher order iodine oxides at 355 nm and 532 nm - Scientific Figure on ResearchGate. Available from: https://www.researchgate.net/figure/Molecular-geometries-of-I2O3-I2O4-I3O6-cyclic-I3O6-linear-I3O7-cyclic-I3O7_fig3_341682472 [accessed 10 Sep, 2023]

Why use Android phone — which is intended for content consumption — for content production and complain about the inconveniences it causes when someone points out typos and lack of proper formatting? Why not just get a laptop and stop torturing yourself (it's evident you are trying your best, but the platform fights you back) while doing tech writing? This problem of yours persist for at least three years, maybe it's the time for a change?


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Solubility of silver chromate in ammonia
Does anyone know where I can find data on the solubility of silver chromate in an aqueous ammonia solution (100 mL) of concentration 2 mol/L at 25 C? Values for different concentration and temperature would also be helpful. I am solving problems in chemistry, so I am interested in whether the obtained value makes sense. Specifically, for the mass of silver chromate that can be dissolved in 100 g of an aqueous ammonia solution with a concentration of 2 mol/L at 25 C, I got about 6.7 g. I have no idea if this value is too high, too low, or possible. Thank you for your answers.

Did you calculate it or measure? It's not 1:1 in moles - a quarter of that at best.

The Handbook of Chemistry and Physics states that $\ce{Ag2CrO4}$ is soluble in aqueous ammonia solutions.

@Mithoron I calculated it. However, I found a mistake in my calculations and now I got 15 g instead of 6.7 g.

Not much of a progress - it was already at least 4 times too much. BTW it can't be calculated - just an upper bound based on how the reaction goes.


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LC-MS/MS analysis of rosuvastatin in rat plasma two peaks
I am trying to analyse samples of Rosuvastatin in rat plasma using LC-MS/MS and ran into the problem that I get two peaks for Rosuvastatin, one after a retention time of around 7 mins and one around 10. Both peaks are existent if only the plasma that does not contain any Rosuvastatin is measured (blankplasma; see the first Figure) and further regardless of concentration, however the higher the concentration, the smaller the relative size of the second peak (example in the second Figure).
If I measure samples containing Rosuvastatin solved in water instead of Plasma, no second peak is present.

I redid the analysis with newly created samples but the second peak persists.
Since the samples containing rosuvastatin solved in only water ran fine after the plasma based samples, I assume that the LC-MS is clean, and that every component works as intended.
My thoughts were that there is a compound inside of rat plasma that has the same m/z value as Rosuvastatin, but my search has not yet yielded any results.
Does anyone have an idea what the reasons could be or how to interpret this result?
Also, I am a newbie in this forum, if this is more fit for the BiologyExchange feel free to tell me.
Thanks in advance.
Edit below: Clarification of experimental details.
I'm showing multiple reaction monitoring on a Waters instrument (Micromass Quattro Ultima PT & a Waters 2695 HPLC).
In the shown MRM method the sample gets analyzed for the mass transition of rosuvastatin (482 > 258) and the internal standard carbamazepin (237>194).
The chromatograms shown are results of the processing with QuanLynx for quantification.
In the QuanLynx method the retention time of rosuvastatin is set to ca. 7 minutes, such that the shaded peak is recognized as rosuvastatin and therefore integrated by QuanLynx.
The non shaded-peak is the unknown peak which appears when measuring with rosuvastatin spiked plasma for the calibration curve, rat plasma samples after rosuvastatin injection in the rat, and in the blank plasma that did not contain any Rosuvastatin.
When I inject only acetontrile however, I only see a small peak at 7.18 min rosuvastatin trace without the unknown second peak.

I think SE Chemistry is a good home for this question. Are you showing multiple reaction monitoring on a Waters instrument? A bit more experimental detail would be helpful. Also, what is shaded vs not shaded on the chromatogram?

@Karsten I have added the additional information regarding MRM, as well as the devices and quantification method used to the post. Thanks for your feedback.

Can you explain better this "Both peaks are existent if only the plasma that does not contain any Rosuvastatin is measured (blankplasma; see the first Figure)" so you mean that the blank has two peaks, but there is presence of Rosuvastatin you should only see one of them?

@Jojostack I analyzed plasma containing no rosuvastatin as a "blank" to get an idea what is wrong with my measurements.
I performed (as with every sample) a protein precipitation with an acetonitril-carbamazepin-soultion and therefore would only expect the carbamazepin peak but expect none of the two peaks of the rosuvastatin trace which can be observed in the first figure.

I think you can easily find some other MRM transitions that are specific for rosuvastatin, and then see if the mysterious peaks from the blank plasma show up at that time.  For example, the MS2 spectrum of rosuvastatin leads me to believe you could try 482 -> 446 or 482 -> 376.  If you don't see peaks with those MRMs instead of your original one, then you've identified that the problem is an unrelated contaminant that just happens to be active on your chosen MRM.

Analysis of biomolecules chromatography can be a challenging exercise because these biomolecules can stick to any fluidic path of the chromatograph if they want to. Just to make sure, is every connection (autosampler needle, connection tubings etc) biocompatible? Now you have three separate problems:

Seeing two peaks for a single analyte "rosuvastatin": In a chromatographic experiment a single molecular species will not have two retention times. Two peaks at different retention times means two different molecular species, perhaps same molecule but a different shape. A different conformation of the same molecule, if it is formed for whatever reason might have a different retention time. It could have formed due to the solvent conditions/pH. Check the quality control sheet that came with the compound. Usually they will mention related impurities which might undergo the same transitions in MS. For example, N-desmethyl rosuvastatin might be present.

Both peaks are observed even when measuring plasma that does not contain any Rosuvastatin, referred to as "blank plasma."

Seeing peaks in a blank plasma means "carry over" from previous injections. Even high-end instruments have carry over. Do you have an autosampler needle wash protocol? Before injection a blank, make large volume injections (three times the volume you are injecting for the sample) of several wash solvents in which rosuvastatin is highly soluble. Does the peak intensity go down by repeating this protocol? It should.

As the concentration of Rosuvastatin increases, the relative size of the second peak decreases. This behavior is exemplified in the second figure.

Again it indicates, either a very closely related impurity or conformation change of a certain fraction of your analyte under your solvent system.


text:
Explain estimating intercellular fluxes
Formula that I am using to calculate intercellular fluxes:
$$ J = \frac{4D}{\pi d} \big([C]_1 - [C]_2\big), $$
where the flux through a hole is $d$ of a solute with a diffusion coefficient $D$ that is present in the signaling cell at concentration $[C]_1$ and in the receiving cell at $[C]_2$.
($d = \pu{4nm}$ , $D = \pu{10^{-5} cm^2/s}$, $[C]_1 - [C]_2 = \pu{100 \mu M}$)
Step 1: Standardize the units of $d$ and $D$:
$$D = \pu{10^{-5} cm^2//s} = \pu{10^9 nm^2//s}$$
Step 2: Substitute values in the formula
\begin{align}
J &= \frac{4 \cdot \pu{10^9 nm^2}}{3.14 \cdot \pu{4 nm s}} \left(\pu{100 \mu M}\right)\\
J &= \frac{\pu{10^9 nm}}{\pu{3.14 s}}\left(\pu{10^2 \mu M}\right)
\end{align}
You can compute it even further but my answer is not correct as the correct answer is $2.4  \times 10^5$ molecules per pore per second.
This problem is an example from a presentation. Here is the slide:

Kindly explain what I am doing wrong.

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

Welcome to Chemistry StackExchange. There are at least twi problems here. First, you do not give us the area of the hole. Second, if you want to get results in molecules instead of moles, you have to multiply with Avogadro's constant. See also this worked example.

I believe the equation has mistakes: Why is the diameter of the hole ($d$) in the denominator? This means that the larger the hole, the smaller the flux!?
Let me try to give the whole calculation. The flux $J$ through a hole of thickness $l$ of a solute with a diffusion constant $D$ and the concentration difference $\Delta c$ is given by the Fick's law:
$$J = D\frac{\Delta c}{l}.$$
Note that the flux is a quantity that tells us what is the solute current per unit area, namely $J = \Phi/A$.
Let's assume the thickness of the pore is $l=\ce{4 nm}$, then given your values of $D=\ce{10^{-5} cm^2 s^{-1}}$ and $\Delta c=\ce{100 \mu M}$ we obtain:
$$J = \ce{2.5 \cdot 10^{-2}  mol m^{-2} s^{-1}}.$$
We can convert this to a number flux by multiplying with Avogadro's constant:
$$J_N = J N_{\rm A} = \ce{ 1.5 \cdot 10^{22} m^{-2}  s^{-1}}.$$
To calculate the ion current through one pore, we multiply flux with the area of the pore, which is $A=\pi (d/2)^2$ for a circular hole with diameter $d$:
$$\Phi_N = J_N A = J_N \pi (d/2)^2 = \underline{\ce{1.5\cdot10^5 s^{-1}}}.$$

I don't see the diameter $d$ in that equation.

@Martin-マーチン, ah, I've introduced $d$ prematurely in the text. I'll fix it.


text:
Voltammogram during calibration is not smooth
During calibration for differential pulse voltammetry, the voltammogram obtained has two distinct peaks. I am not sure why I am getting two peaks in the range from $\pu{-0.6 V}$ to $\pu{-0.9 V}.$ I have used three electrode setup. I used screen printed carbon working electrode, platinum counter electrode and saturated calomel electrode as reference electrode. Mercuric chloride was used for deposition of mercury over the carbon working electrode.
The $y$ axis shows current, and the $x$ axis shows potential. This is just one example, similar types of voltammograms were obtained. The peak on the far right in the range of $\pu{0.09 V}$ might be of mercury. This is the voltammogram obtained for $\pu{0.1 ppm}$ cadmium nitrate standard solution.

Do you mean in your question that you find extra peaks you are not able to assign?

@Jojostack Not related to this question, but your frequent title formatting: Chemistry SE  (in contrary to some other SE sites) strongly recommends plain text question titles without MathJax for index/search reasons and due possible displaying issues in question title lists.

@jojostack I am very new to voltammetry. I am using DPV to quantify cadmium in my sample. I am required to make a calibration curve of concentration vs peak current. However, i am getting double peaks. Which peak current value should i take?

Also, on increasing the concentration of cadmium, the smaller peak goes on increasing and the larger peak becomes smaller. Its strange. On increasing the concentration, the peak current should increase, shouldn't it ?

Cd peaks is the one near -0.745 E/V (as far as I know) is this what you were asking?

Okay, so the smaller peak. Alright, but what is the cause of peak splitting, given that i have used a standard solution of cadmium nitrate?


text:
If diborane has 3c-2e bonds, then why are both boron atoms still electron deficit?
This is an image of the structure of diborane.

It has a 3-center-2-electron bridge bond.
As we can see in the bridge bond, the electron in the hydrogen atom and one electron in either boron atom is shared between the hydrogen and boron atoms two times, making two $\ce{B-H-B}$ bridges. Does this mean that both boron atoms as well as the hydrogen atom share those 2 electrons? If this is true, then both borons' octet should be satisfied since the 2 electrons are shared equally between all 3 atoms. So why are both boron atoms still electron deficit?
These are the sources confirming that boron has an electron deficit.

Chemistry LibreTexts: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_13%3A_The_Boron_Family/Z005_Chemistry_of_Boron_(Z5)/Boranes_and_Borohydrides

Coord. Chem. Rev. 2002, 226 (1-2), 61–69

Wikipedia: https://en.wikipedia.org/wiki/Diborane

In this arrangement, Boron atom is not electron deficient. But the bonds are not typical covalent bonds as they are in $\ce{H2}$ molecules, for example.

@Maurice According to many online sources, diborane is indeed electron deficient. There are not enough valence electrons to form the expected number of covalent bonds. My question was asking why it as such.

Provide references and reasoning of such sources. If there are so many of them, they surely also say why or why not are boron atoms still electron deficient even in B2H6.  As depending on a point of view, they are and at the same time they are not electron deficient. ( They are not as they do deal with the full electron octet. They are, otherwise they would not need to achieve that by 3c-2e bond.)

If a 3c-2e bond was as good as two normal, you'd see them all the time.

In all of these sites, it mentions that the boron atoms in diborane are supposedly electron deficient. I am asking why it is so, since the 2 electrons in the bridge bond are shared equally between all atoms, boron's octet should be satisfied.

3c-2e bond is electron deficient by its nature.

Thank you, could you please elaborate a little more on this, and post it as an answer to the question?

Quoting the wikipedia article to which you link: "Traditionally, diborane has often been described as electron-deficient, because the 12 valence electrons can only form 6 conventional 2-centre 2-electron bonds, which are insufficient to join all 8 atoms.[28][29] However, the more correct description using 3-centre bonds shows that diborane is really electron-precise, since there are just enough valence electrons to fill the 6 bonding molecular orbitals.[30] Nevertheless, some leading textbooks still use the term "electron-deficient". One of the references is to Bill Lipscomb's Nobel lecture.

If you don't know who Professor Lipscomb was see https://www.nobelprize.org/prizes/chemistry/1976/lipscomb/facts/ - if anybody knew about boranes it was him. From his Nobel Prize Lecture: "One of the simple consequences of these studies was that electron deficient molecules, defined as having more valence orbitals than electrons, are not really electron deficient. I mean by this non-sequitur that the three-center two-electron bonds make possible a simple description of these molecules and ions as filled orbital species." https://www.nobelprize.org/uploads/2018/06/lipscomb-lecture.pdf

@IanBush that might as well be an answer. I don't think you would get to that conclusion in a better way.

Maybe later. I'm not sure I really know what "electron deficient" means. I'm not sure anybody does.

Well at least according to Professor William Lipscomb, who won a Nobel Prize "for his studies on the structure of boranes illuminating problems of chemical bonding", they are not electron deficient. Quoting from his Nobel Prize lecture

"One of the simple consequences of these studies was that electron
deficient molecules, defined as having more valence orbitals than
electrons, are not really electron deficient. I mean by this
non-sequitur that the three-center two-electron bonds make possible a
simple description of these molecules and ions as filled orbital
species."

However as Wikipedia notes some leading text books still use this description. Personally I'm not sure I really know what is meant by "electron deficient", it being a term that seems to mean less and less with the more chemistry you know (c.f. hybridization). So I will bow to Professor Lipscomb's authority.

I think if it is or is not electron deficient depends on what is meant by that by the context. It may be both at the same time. In sense of using less valence electrons than the double of the bond count, it is electron deficient  In the sense you have described, it is not.

To be honest nowadays I would just solve for the orbitals and aufbau them with electrons. Let the maths do the talking, there is (usually) no ambiguity then

What I have meant that sometimes a term is not ambiguous, it just have more meanings and is good to explicitly say which one is assumed.

I just don't think it an especially useful term. Just marvel in the extraordinary variety of ways that atoms can combine. That, for me, is the wonder of Chemistry, Then get down to the maths.

I agree. Scientists often think nature does things this or that way, what is often true for a long time until they are shown the cases it does not.


text:
Why don't P graphs contradict Raoult's law graphs?
Our teacher gave us the following question:

1 mole each of liquids A, B are mixed in a piston-type vessel and piston is moved slowly and isothermally. (PA0 = 75 torr and PB0 = 25 Torr)
Find:
(a)  Find pressure when first bubble is formed. And its composition.
(b) Find pressure when last drop of the liquid is remaining.
Ans:  (a) 37.5 Torr     (b) 50 Torr

Subsequently, he drew a graph to introduce us to bubble-point and dew-point curves:

And made us write,
"At a given temperature, a pure liquid has a fixed/definite vapor pressure, but an ideal solution doesn't have a fixed vapor pressure."
Now the doubt which irks me is that
(i) When Raoul's law's straight-line graphs were drawn, each composition corresponded to only one value of vapor pressure. But here (through the attached graph) we can see multiple values of vapor pressure at a particular composition, (since I'm assuming that external pressure and vapor pressure would be the in this case). Don't these 2 graphs contradict each other? Since one shows a single value of vapor-pressure while the other (the attached one) shows no-fixed value?
(ii) How do the terms "pure liquid" and "ideal solution" differ in this context?

The same compositions of liquid and gaseous phase belong to different pressures and temperatures. OTOH, horizontal isobaric lines determines composition of both phases at the given pressure.

At $x_B=1/2$ (1:1 mixture , fixed ) and at v. low pressure, only vapour is present,  then increasing pressure at $37.5$ torr the vapour first condenses, then at $50$ torr it is all liquid. You can calculate the total pressure $P=(1-x_B)p_A^* + x_Bp_B^$ where indicates pure A or B, and the dew point (lower) curve as $x_B(g) = x_Bp_B^*/P$ where $x$ is a mole fraction.

There is important to consider that the gaseous phase will have at equilibrium always higher relative content of the more volatile liquid $\ce{A}$, compared to the equilibrium composition of the liquid phase.
For the $1:1$ molar mixture of liquids:

The higher pressure $\ce{50 torr}$ on the straight line crossing 1:1 ratio belongs to the equilibrium pressure for the liquid phase mixture 1:1. The gaseous phase has at that point roughly $80\ \%$ of liquid A vapor (a horizontal line crossing the curve).

The lower pressure $\ce{37.5 torr}$ on the curve, crossing 1:1 ratio belongs to the equilibrium pressure for the gaseous phase mixture 1:1. The liquid phase mixture has at that point roughly $15\ \%$ of liquid A.(a horizontal line crossing the straight line)

For pure liquids, both the straight line and the curve converge to the same point. Composition of both gaseous and liquid phases is identical and there is single pressure value for both bubble and dew point.

The two curves in your diagram have different meanings. More specifically, the x-axis means different things for the two curves.
The curve labeled "bubble point" shows the relationship between pressure and composition of the liquid. According to Raoult's law, this curve should be a straight line. (Raoult's law is often not true, or even close to true, but an ideal solution is a mixture where it is true, and that curve actually is a straight line.)
The curve labeled "dew point" shows the relationship between pressure and composition of the gas. Raoult's law has nothing to say about this curve, because Raoult's law only talks about the pressure of a vapor above a liquid of known composition.
That is why there is no contradiction.


text:
Why is m-fluorophenol less acidic than m-chlorophenol?
Why is m-Fluorophenol less acidic than all the other m-Halophenols, even though it has the most inductive effect

The data is from :
A guidebook to Mechanism in organic chemistry for the JEE, 6th edition, page no. 3.12 from the chapter, the strength of acids and bases

What are these values in the table?

What is the origin of the data?  For one, it is to provide an easy access of the data (for future readers of the question).  For two, how are the data obtained?  Isolated, i.e. gas phase acidities, or in a solvent?  The experimental recordings about meta chloro- or fluorophenol dissolved in DMSO for example, the Bordwell p$K_a$ table assigns both 15.8. reference.

@Martin-マーチン, updated the book name, these are pka of the acids

@Buttonwood I think it's in aqueous medium like the most of data but not sure because they haven't mentioned anything specific

The trends are similar to Hammett sigma constants https://en.wikipedia.org/wiki/Hammett_equation  The halogens have opposing effects electron withdrawing by induction and electron donating by resonance. Fluorine by its size seems to be more effective in the electron donation than Cl, Br I.

@jimchmst but at meta position the donating effects cancel right?

Mesomeric effects are lower at the meta position.  Canonical forms are not the total answer. Check the differences between meta and para sigma constants  for the halogens, fluorine is interesting.


text:
Bond order of carbon monoxide and nitrosyl cation
The following question was asked in JEE Mains exam in 2022, a competitive exam for engineering in India.
The difference between bond order of CO and NO$^+$ is $\frac{x}{2}$ where $x$ = ____. (Round off to the nearest integer).
Here, I have taken the bond order of CO as 2.6 (which is cited as the actual bond order of CO) as per:

This Stack Exchange answer
Vocation India
Kayson Education
Research Gate

and many other sources, the reason being the anti-bonding character in the non - bonding orbital in the Molecular Orbital diagram.
Upon solving we get:
$$3 - 2.6 = \frac{x}{2}\\
0.4 = \frac{x}{2}\\
x = 0.8\\
x = 1$$
But the answer given for said question is 0. They have taken the bond order of CO and NO$^+$, both as 3. So their solution is:
$$ 3 - 3 = \frac{x}{2}\\
0 = \frac{x}{2}\\
x = 0$$
On a technical level, the bond order of CO should be ~2.6, so my answer should be correct. Is the solution given wrong? The views of others on this topic would be welcome.

At this level they both have 3. If you want to nitpick it, then what makes you think NO cation would have effective bond order of 3 while CO doesn't?

Well, the difference between $\ce{CO}$ and $\ce{NO+}$ is about a proton in the nucleus of nitrogen. They are isoelectric and at that level of approximation, that means they are basically the same.

I think at this level you are supposed to make the simplifying assumption that bond order is $(1/2)×$(the number of electrons in bonding orbitals minus the number of electrons in antibonding orbitals); the $1/2$ coming from needing two electrons to make a bond. Ignore the "effective" bond order formalism on which the $2.6$ value for carbon monoxide is based (if we were to assess the "effective bond order" in the nitrosyl cation the same way that probably would also be less than $3.0$).
Here, you should verify that the molecular orbital structures for both species give eight bonding electrons and two antibonding ones, so a bond order of $(8-2)/2=3$ for both and that's where they get $x=0$.


text:
Is the gas produced by boiling bioethanol safe?
As part of a small experiment, I'd like to boil bioethanol (I believe it should boil at 80 degrees).
Would the resultant gas be safe (to humans)?
Presumably it would be quite flammable?
In small amounts, would it safe to burn the gas as it is being emitted?

Bioethanol is the same thing as ethanol and its behaviour as the major compound is identical. There would be differences in minor or trace impurities or denaturating additives (these depend on the source and country). Note that bioethanol for fuel purposes is implicitly assumed to be denaturated as well.
Define safe. As it could be very safe and very dangerous, depending on scenario.
Ethanol boils near $\pu{78 ^\circ C}$ , forming ethanol vapor as the gaseous phase of ethanol. Its inhalation has much stronger effect with much faster raise, compared to drinking the same amount. As it gets through large lung surface directly to tissues, body liquids, blood and brain. It can be very dangerous due difficult control of the level of exposure.
Yes, it is very flammable, forming explosive mixture with air in range $\pu{3.3 - 19} \text{ vol %}$ of vapor. In countries where personal alcohol distillation is prohibited, badly maintained  illegal distilleries sometimes ended in violent explosive destruction of the room (or the house).
Note that ethanol vapor is used in ethanol bases campsite/outdoor burners where ethanol is being evaporated and gaseous ethanol serves as the gaseous fuel in conventional gas burner. It is simpler/cheaper than petrol/spirit burners, has less issues than LPG in air traffic, but has less heating power than both petrol or LPG ones.

Thanks! I should have clarified, I would be boiling very small amounts, say 10 or 20ml. I certainly wouldn't be looking to inhale it. If you had a naked flame near the source of the rising gas, would that safely burn it off (minus 'explosion')? Nothing would be contained. By safe I'm just trying to avoid any vapors that would irritate/poison people (or get them drunk) or that would trigger a large fire/explosion. I never suspected that would be the case when dealing with small volumes in an open container, I just wanted to be sure...

As the general precaution, no open flame nearby boiling or hot ethanol, unless the ethanol burning is the primary reason.  If you remember school demonstations of explosion of bubbles,  filled with hydrogen and air/oxygen, you can have rough idea of explosion intensity of comparable volume of ethanol/air mixture.

Thanks. I should have clarified - a 'small' explosion is indeed the desired outcome! I will start with VERY small volumes and build up from there until I deem the effect reasonable enough to be entertaining, while still within safety margins! Thanks for taking the time to help!

So, Ben Logan,  what is your question?

I think it has been answered - ambiguity around safe, obviously, but agreement that in small amounts and in a well ventilated space, the gas resulting from boiling bioethanol is safe to humans and flammable, but not to the extent that you couldn't burn a small amount of it safely.


text:
Possible technique for measuring pH in solutions
I would like to develop a device that measures pH in aqueous solutions. I'm not a chemist, but I have some basic knowledge about electrochemistry. So, I thought I could use two electrodes quite close together; the first (-) is ideally isolated from the solution under test and within another reference solution with 0V reduction potential. The other electrode (+) is located inside the solution under examination and both are connected by a salt bridge. Then, using a microcontroller I will increase the voltage until I see that current starts to flow. Knowing the applied voltage, I can ideally measure the concentration of H3O+ ions in the solution under test. Do you think it could work or is it a bad idea?

Measuring pH via dynamic processes involving significant current is a bad idea. Aside of the need of knowledge of the relation pH as function of electrode potential, there are various potential differences/drops due kinetic reasons and there are overpotentials for evolving gaseous hydrogen and mainly oxygen. It would need precise calibration and even then, various solutions with the same pH could have different electrode potential. Aside of the glass electrode, there is also older technique of quinhydrone electrode using a Pt wire.

Correction - the QH electrode seems to use antimony - I falsely remembered Pt from my memory. It has also limitation for pH<=8 and absence of strong oxidation or reduction compounds.

Measuring pH with current flowing is not an appropriate idea. You would like to measure something without chemically disturbing the system. The moment you pass current in a solution, you start to change the local pH near the electrode. There is a reason why the glass electrode is popular despite the fact that its working is not fully understood is that
(i) No electrolysis is involved
(ii) The thin glass membrane has a very high resistance
(iii) It is highly hydrogen ion selective (in a limited pH range). At pH extremes acid or alkaline errors begin to show up.
Now, not to completely disregard your idea, there are contactless conductivity measurments which work pretty well. Think along those lines and see if you can find anything useful for pH measurements.
The key idea for electrode based sensor is that they are selective to the ion being measured. If you just random pick an electrode, and start electrolysis, it will be responding to all ions in solution not just the hydrogen ion.


text:
Would having a greater H-bond donor count make a substance more readily enter a twig?
Yes, weird plant scientist asking weird plant questions...
Background:
I am looking at how the twigs of trees absorb water, a poorly understood means by which trees alleviate water stress. Of 4 fluorescent tracers in an aqueous solution, experimentally applied to twig surfaces, 2 enter the twigs and 2 do not. The 2 that are absorbed have a hydrogen bond donor count that is 6-8X greater than the 2 that don't. Of all the chemical properties, this is by far the most different between the pairs. Oddly, it is the largest and smallest molecules that are absorbed, so I am thinking it is not molecular weight. All 4 compounds have a formal charge of zero.
The solutions were applied directly to the intact surface of detached (but live) twigs, on the twig epidermis (not leaves), during the day. Transport was detected throughout, both symplastically (inside the cells so a membrane was crossed) and apoplastically (outside the membrane, so a layer of suberin was crossed).
The apoplastic tracers were:
Sulforhodamine B sodium salt (not absorbed) and
Calcofluor white (absorbed)
The symplastic tracers were:
HPTS (not absorbed) and
5(6)-Carboxyfluorescein (absorbed)
Question:
What does it mean? Can an H-bond donor count positively influence absorption, particularly across suberin/other waxes or membranes (the 2 possible entry points)? Or could the smaller HBD:HBA ratio impact absorption? How?
Thanks for any insight! :)

Unfortunately I cannot answer your question, but I think it would be helpful if you could indicate the nature of the fluorescent probes you are using in the experiments.

Well, if someone had started the biology class with that, I would have been very curious to know more. Anyway, to the point, and without knowing anything about how twigs gobble up water, I could imagine any kind of hydrogen bonding will positively enhance permeability. But I guess that's hardly helpful here. Still think it's an interesting question. +1

All else being constant, more H bond donors/acceptors will correlate with lower hydrophobicity. Greater hydrophobicity is expected to lower aqueous solubility. This is a well known principle within drug design as it influences the compound's ADME properties (adsorption, distribution, metabolism and excretion). See for instance Lipinski's rule of 5 or RO5. Note that in the case of RO5 too many HA or HD implies poor ADME due to low lipophilicity and therefore poor membrane permeability. Too high lipophilicity (indicated by the octanol/water partition coefficient) also correlates with poor ADME. The "sweet spot" is somewhere in between.
Similar concepts have been developed for plants, see Ref. 1. You don't explain exactly how the solutions were applied (nighttime or daytime, for instance), or where the compounds were detected, and there may be differences in the transport of the compounds in different parts of the plant (within a leaf as opposed to in sap transported by conducting cells within the phloem). It is expected that transport of the larger compound by diffusion alone is slower (see the Stokes-Einstein equation for instance). Diffusion across a membrane without participation of a transporter would by the same principle also be expected to occur more slowly for larger or heavier compounds (for organic compounds molecular weight is a reasonable proxy of size). In the case of the RO5, some violations occur due to active translocation.
References

Tomer Malchi, Sara Eyal, Henryk Czosnek, Moshe Shenker & Benny Chefetz (2022) Plant pharmacology: Insights into in-planta kinetic and dynamic processes of xenobiotics, Critical Reviews in Environmental Science and Technology, 52:19, 3525-3546, DOI: 10.1080/10643389.2021.1946360


text:
Determining weight to within 0.1mg?
For a recent laboratory session we were asked to weight a 20mL beaker three times on an analytical scale. Each time, we were asked to determine the weight to within +/- 0.1mg. I am just a little confused about what this means as I have never been asked to round this way.
The three values are: 13.5448g, 13.5436g, and 13.5435g.
What would these three measurements actually be when applying this uncertainty?
EDIT: Please let it be known that I understand the math and that is not what I am referring to. I was specifically wondering about the semantics since I know balances have uncertainty but was unsure what this specific wording meant.
EDIT2: I have corrected first value

For broader context, you may want also to search for meanings of terms resolution, precision, trueness, accuracy.

Weighing something to the nearest 0.1 mg is a common wording/terminology in analytical chemistry. In simplest terms, it is an instruction to use a certain type of balance- a balance which can weigh upto 0.1 mg or 0.0001 g. One should not round off the values you got if the instruction was to record to the weight to 0.1 mg. Actually, given the variations in your observations about the weight of the same beaker three times, shows the problem of precision of the balance user (oils from fingers, air drift etc.) or balance issues or even electrostatics if the atmosphere were too dry.
Alternatively, another interpretation is that if someone tells you to measure an exact amount of a certain material to a precision of 0.1 mg. This may mean that one has to be extra careful during weighing this substance and the variation in your mass should not be +/- 0.1 mg from the suggested amount.
In short the instruction depends on the context.

this makes sense thank you! this is one of my first analytical chemistry classes and we have not worked very closely with the balances or their uncertainty before. i appreciate you being kind :)

A dry beaker should show constant mass, or maybe 0.1 mg or 0.2 mg variation when measured with the same analytical balance in sequence. Typical user errors are leaving the doors open when taring, changing the angle of the balance between measurements (by moving the balance or leaning on the table/bench), or working with a balance that is not clean. Oils from fingers are more a 0.01 mg problem.

Your masses are already given to within $\pu{0.1 mg}$. $\pu{0.1 mg}$ is equal to $\pu{0.0001 g}$. You used a balance (I assume) that gives four decimals beyond the decimal point. The difference between - say - $\pu{13.5448 g}$ and $\pu{13.5449 g}$ is in fact $\pu{0.0001 g}$ or $\pu{0.1 mg}$.
Had you dropped the last digit (but there is no reason to do that), your mass would be to within $\pu{1 mg}$. Had you used a scale made for larger masses (like a doctor's scale), the masses might be given to within $\pu{100 g}$. For the small masses in your example, that measurement would be meaningless.
EDIT: corrected values