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Low viscosity, low surface tension liquid for volume measurement
General Case
This question concerns the general problem of accurate volume measurement of a given object. Among other techniques, that of liquid displacement is sometimes used for this purpose.

In case of a solid material, the latter may be immersed in liquid, and its volume calculated by measuring the amount of fluid displaced in a graduated cylinder.

In case of a hollow object, the latter may be filled in with a liquid which amount is accurately determined, either while filling in, or by emptying the object in a graduated container a posteriori (or - maybe more accurately - by weighing).

In both cases, the choice of the liquid is of crucial importance for the accuracy of the measurement.
Specific Issue
I have a CO2 gas sensor with a complex hollow embodiment, also containing electronics, and since I wish to perform response time measurements, I would like to precisely know its inner volume. I came up with the idea of filling it with a liquid, as stated above. I do not have and cannot have accurate drawings of the inner sensor.
Questions
1. Is there other methods to perform such a volume measurement which may be preferable?
Indeed, I came up with this idea because I find acceptable to sacrifice a sensor to perform such volume measurements. Alternatively, I also thought of placing the sensor in a closed, accurately known volume, along with some amount of chemical reagents that would produce a given amount of CO2. Then, by dilution and measurement of the CO2 percentage inside the enclosure, I may know the volume occupied by the sensor. However, I believe that there are far too many sources of errors in such procedure for it to yield accurate results. I may be wrong, though, and other variations of the latter idea may work well...
2. Which liquid may be the best for such measurement?
This is the main issue here. I would tend to say that such a liquid should have both a low surface tension and a low viscosity. Additionally, low volatility would be a plus, since it would prevent the amount of poured liquid to decrease with time. I find methanol or ethanol to be good canditates despite their high-volatility. I cannot use acetone because of the risk of plastic dissolution and the resulting leakage of the sensor's embodiment. Would you know of a more suitable liquid, possibly readily available in the lab?
3. Can surfactant be added to ethanol or methanol? If so, do they further decrease surface tension? Which are they?
A quick search in the literature revealed that methanol or ethanol are themselves considered as water surfactant. Their mixture with other surfactants is often analysed in emulsions of oil, or other chemicals, but not alone.
Related topics:

Non-wetting fluids for precise volume measurements
Looking for a non-volatile liquid less viscous than water

Edit:

As pointed out by Poutnik, low volatility and low viscosity are mutually incompatible with each other. Indeed, low viscosity tends to be positively correlated with high volatility, both of them being the consequence of low intermolecular forces. More resources on the topic are available here (saved version). The requirement on low volatility was not mandatory though, as it was just to ensure that not too much of the liquid evaporates during the filling of the sensor...

Low viscosity and high volatility are two correlated attributes of low intermolecular forces, so these requirements go strongly against each other. You may found a liquid with good one value and acceptable other value, but it may be tough.

@Poutnik, Indeed, I should have thought about that in the first place. Actually, the constraint on volatility is not that important. It was just to ensure that not too much of the liquid evaporates during the filling of the sensor... I edited my question accordingly, thanks :)

Air would be a convenient fluid. Fill your sensor with air at a known (atmospheric) pressure - well, it is probably filled with air already.
Then connect it to a vacuum chamber tube of known volume with a pressure sensor, i.e., a tube with two valves and a pressure sensor. Attach a vacuum pump to one valve and evacuate the tube; attach the $CO_2$ sensor to the other valve and then open it and measure the new pressure in the closed system. All the air will have been supplied by the sensor volume, and the new system volume will be the sensor volume plus the (known) tube volume. Doing it this way is likely to be less messy and more accurate than using liquids, and it won't sacrifice a sensor - and you can check every sensor!
A problem with liquids in a blind experiment (you cannot see the internals) is that there could be blind alleys that would be hard to detect and hard to fill with liquid (or which would not empty easily).

1 - Your idea may be good with a solid sensor with an inlet and outlet that may be easily connected with pipes but it is not the case (it is a sensor for free air measurement, with no in / out ports, rather a thin mesh exposed to ambient air). We could however place it inside a known volume enclosure and apply your method.

2 - Its main drawback, in my opinion, is that - while you are certain that no "dead space" remains (as could have been the case with liquid filling, with a hard-to-fill volume) - the accurate measurement of the volume of your pipe and valve is quite impossible. In other word, there are too many little uncertainties (is my pipe pushed enough on the nozzle, what is the gas volume at the entrance of the valve, etc.) that makes the global measurement inaccurate. It may work for big enough sensors though, given these inaccuracies are negligible compared to the inner sensor volume...

The internal volume is secondary, if you want to find the response time. When you open the system and allow CO2 in, the output of your sensor will reach equilibrium after some time. Why do you need to know the volume? If holdup of gas in the sensor extends the response time by limiting diffusion, would putting the sensor into different CO2 concentrations establish a time to reach equilibrium easier than calculating from a volume?

I would like to place the sensor against a medium which will produce (exhale) CO2. What I really want to measure in the end is the rate of production of CO2 by this medium at t=0. To do so, my idea was to place (air-tightly) the sensor against this medium, wait a few minutes and deduce the diffusion rate from the slope at t=0 which will be the quotient of the diffusion rate by the sensor's volume. Hence my need to know the sensor's volume...

@mranvick: How about making "synthetic media": fill a balloon with CO2, fill another with air. Place your sensor against the one with CO2, plot response vs time, then against the one with air. The rate of CO2 production from a balloon will depend on how big you blow it up (pressure, thickness of membrane); it should be constant for several minutes. You can get the actual rate of CO2 production per cm^2 by weighing the balloon and calculating its total area. One liter of CO2 (a small balloon) weighs about 2 grams plus the balloon; the pressure inside is only about 10-20 mm Hg above atmospheric.


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find minimum value of p1+p2+p3+p4 where pi represents position of substituent and functional group in iupac name p1-bromo-p2-chloro-pent-p3-en-p4-one
I have tried all I could for this question the answer I reached is 7 as in the attached pic but the book says 6

Please rotate your image.

Voting to close. Needs more details and the structure in image is not the same as that in title.

You can put Br and Cl in the same carbon. In 1-bromo-1-chloro-pent-1-en-3-one , p1+p2+p3+p4 is equal to 6. I think this is the smallest p1+p2+p3+p4 which can be reached according to IUPAC rules of nomenclature.

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Is HCl a Lewis base?
Cl in HCl has three lone pairs of electrons, so it can donate a lone pair and thus becomes a Lewis base. Just like water is called a Lewis base because oxygen has two lone pairs and can donate a lone pair. Am I right?

It can, but most commonly considered Lewis acids do not interact well with HCl. Still, PtCl4 dissolves in HCl(aq) forming H2[PtCl6].

Please don't believe that all molecules having atoms with lone pairs of electrons are Lewis base. It is a necessary condition but it is not sufficient. $\ce{O2, CO2, N2}$ are not Lewis bases.

@Maurice Hmm, a naked proton is quite a very strong Lewis acid, able to react with many very weak Lewis bases.  "Dioxidanylium  ($\ce{HO2+}$) is the conjugate acid of dioxygen. The proton affinity of dioxygen (O2) is 4.4 eV."

@Maurice They are very weak bases, but all of them can coordinate to form complexes, or even be protonated with carborane superacid.

The strength of Lewis acids relative to your Lewis base has been commented on. A fuller answer to your question needs to include the Lewis acid you are considering.
The original Bronsted-Lowry theory of comparing acidity and basicity, or acids and bases, involved $\ce{H^+}$ and $\ce{OH^-}$ and the transfer of a proton.
The broader Lewis theory allowed consideration of other cations besides the proton as acids, and other nucleophiles than hydroxide ion to be bases. So water can be considered a Lewis acid relative to chloride ion because $\ce{H2O}$ hydrates chloride ions by hydrogen bonding. And water is a Lewis base relative to metal cations because water hydrates the metal ions through electron donation from oxygen.
So the answer to the question is yes and no. Yes, $\ce{HCl}$ can be considered a Lewis base - relative to a very strong acid, as mentioned in the comments - but in the ordinary world, I would not first think of $\ce{HCl}$ as an alkali. Stretching the word "base" by adding "Lewis" in front of it needs more, like adding the reference acid, to make the comment clear rather than confusing.

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Working out energies of hydrogen bonds by comparing chemical potential of vapour and condensed phase - equation derivation
I want to ask a question about working out the energy of a hydrogen bond between two water molecules, $w_{AA}$ using the chemical potentials of vapour and condensed phases.
I was reading K. Dill, Molecular Driving Forces, 2nd Ed, 2011, p. 257 which lines out the framework for the Clapeyron equation and I was confused on the following:
The book states that you can calculate $w_{AA}$ using the following formula, which is derived by calculating the chemical potentials of the vapour $\mu_v$ and $\mu_c$ respectively:
$$\mu_v = kT\ln\left(\frac{p}{p_{int}^{*}}\right) $$
where $\mu_v$ is the chemical potential of a a pure substance in the liquid phase, $p$ is the partial pressure of the vapour phase and $p_{int}^*$ is the standard vapour pressure
The chemical potential of the condensed phase, $\mu_c$ is achieved by differentiation of the Helmholtz equation, with the knowledge that the $TS$ term of $U-TS$ is set to $0$ since a pure liquid's molecules are indistinguishable upon interchange.
$$\mu_c = \left(\frac{\partial F}{\partial N}\right)_{T,V} = \frac{zw_{AA}}{2}$$
where $z$ is the number of neighbours of each molecule in the liquid phase, $w_{AA}$ is the bonding interaction between two molecules, and divided by 2 to account for any double counting.
Using $\mu_A = \mu_B$, both terms can be equated:
$$kT\ln\left(\frac{p}{p_{int}^{*}}\right) = \frac{zw_{AA}}{2}$$
and more simply written as
$$\ln\left(\frac{p}{p_{int}^{*}}\right) = \frac{zw_{AA}}{2kT}$$
and for those interested, this can be expressed as:
$$p=p_{int}^*e^{\frac{zw_{AA}}{2kT}}$$
Now, I was presented with the following question from the book:

Given that water’s vapor pressure is $1 \ atm$ at $T = 100^{\circ}$ and $0.03 \ atm$ at $T = 25^{\circ}C$, find the value of $zw_{AA}$.

This all makes sense.
However, Dill states the following

Take the logarithm of the equation to get the boiling pressure $p_1$ at temperature $T_1$
as $$\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}$$
and the boiling pressure $p_2$ at temperature $T_2$ as $$\ln\left(\frac{p_2}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_2}$$  Subtract the second equation from the first to get $$\ln\left(\frac{p_2}{p_1}\right) = \frac{zw_{AA}}{2R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$ neglecting the small temperature dependence on $p_{int}^*$.

However, I'm not sure whether this derivation is correct. Below shows my working out from the second equation stated from the start of the body:
First of all, they transformed
$$\ln\left(\frac{p}{p_{int}^{*}}\right) = \frac{zw_{AA}}{2kT}$$
to
$$\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}$$
and somehow stated the relationship between the gas constant $R$ and $k$ is $R = 2k$ which doesn't make sense.
Also, the denominator term containing the $2$ disappears in both of their initial statements for $T_1$ and $T_2$ e.g. $\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}$, yet it returns in the final equation $\ln\left(\frac{p_2}{p_1}\right) = \frac{zw_{AA}}{2R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$ and I can't see that from the following working that I did to prove this equation does not work (taking $R = 2k$ to be true):
$$\ln\left(\frac{p_1}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_1}, \ln\left(\frac{p_2}{p_{int}^{*}}\right) = \frac{zw_{AA}}{RT_2}$$
Subtracting the second equation from the first:
$$\ln\frac{P_1}{P_2} = \frac{zw_{AA}}{RT_1} - \frac{zw_{AA}}{RT_2} = \frac{zw_{AA}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$
which does not compare with the derived result.
My question is therefore two-fold:

How has Dill interchanged $R$ and $k$ in his solution?
Is there an error in his derivation for his final expression.

The solution (as transcribed above) is also attached as a screenshot for illustration (for illustrative purposes only):

Switching between k and R is just a matter of changing units from J to J/mol (or energy to molar energy units). The factor of two should be retained and its disappearance from some of the expressions is  was probably a typo.

Going between R and k is more than just a dimensional conversion. You also need to divide or multiply by $N_A$, since R is J/K/mol, while k is J/K/particle:  $R = N_A k$.

@theorist you are strictly right of course. You encounter this type of manipulation so often that you ignore the finer details. Forgot to note: upon conversion from k to R, $w_{AA}$ must be expressed as a molar quantity.

@BuckThorn yeah that's what I thought - all quantites relevant would then have to be expressed explicitly as molar quantites - shame that this book doesn't choose to mention it properly. Many thanks for your suggestions! I think I've solved the matter!


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Work done on a Gas?
I'm extremely confused about work done on a gas. So if you look at the system pictured,

From what I've read online, the net work done, which is equal to the work done by the objects, is equal to the pressure times the change in volume, $W=p\,\Delta V$.
But I'm confused on is how can this be true. Why wouldn't the net work done on the gas be equal to the work done on both the gas and the atmosphere, which is also applying a force on the gas? It doesn't make sense that the net work being done on the gas is equal to only the work being done by the objects.
Thanks, I hope I explained my issue correctly here.

The pressure doing the work is the sum of the atmospheric pressure plus the effect of the additional masses present on the piston.

When you do a force balance on an object A that is acted upon by a force from an object B, do you include both the force that B exerts on A and the equal and opposite force that A exerts on B, or just the force that B exerts on A?

You have differential equations.
Differential equations are ALL about the initial conditions, and A PROCESS.

So, what’s the initial condition?

You have a piston with a given volume of gas.

What’s happening then?

I apply an additional force with an object. (The atmospheric pressure already was there)
And what if you want to know the work made by the atmosphere?
No problem. You have to start with a very large volume and start to add the gravitational force. Performing a PdV :-)
Do what it works best to you. (Nerd Joke)

Yeah, I actually agree with this answer. Your states can only be assigned at equilibrium. Your initial state must be such that p_gas = p_atm (or else not at mechanical equilibrium, states not assigned) . Then you apply external pressure, and it is that extra pressure that does work. If p_atm is to be taken into account, just leaving the gas there doing nothing will have non-zero work done which does not make sense.

The total work ( and its value for the constant force case )
for pressure $$p=p_\mathrm{force} + p_\mathrm{atm}$$
$$W_\mathrm{tot} = - \int_{V1}^{V2}{p \cdot  \mathrm{d}V} $$
is shared between the source of the explicit mechanical force acting on piston and atmosphere.  Atmosphere would do work
$$W_\mathrm{atm} = - \int_{V1}^{V2}{p_\mathrm{atm} \cdot  \mathrm{d}V} $$
and the force acting on the piston
$$W_\mathrm{force} = - \int_{V1}^{V2}{p_\mathrm{force} \cdot  \mathrm{d} V} $$
If we consider a simple case of constant forces, expressions are simple:
$$W_\mathrm{tot} =  -p \cdot  \Delta V$$
$$W_\mathrm{atm} =  -p_\mathrm{atm} \cdot  \Delta V$$
$$W_\mathrm{force} =  - p_\mathrm{force} \cdot  \Delta V =  -\frac FA \cdot  \Delta V $$

Why did you subtract the  $p-p_{atm}$ in the final expression?

Is not it obvious ? Part of the work does the atmosphere and part is done by the local pressure on the piston. Without proper thinking, only the latter seems to happen

Why wouldn't the net work done on the gas be equal to the work done on both the gas and the atmosphere, which is also applying a force on the gas?

Your phrasing is not quite correct:  The work done on the gas in the cylinder is equal to the work done by the atmosphere (if any), plus the work done by the lowering of the weights (if any).
I.e., the gas is part of the system, and both the weights and the atmosphere are part of the surroundings.  The work done on the system is done by the surroundings.  Hence the work done on the gas in the cylinder is the sum of the work done by the weights and the atmosphere.
[It is not, as your wrote, the work done on both the gas and the atmosphere, because the gas is part of the system, and the atmosphere (like the weights) is part of the surroundings.]
Thus:
$$\mathrm{w} = - \int_{V_{i}}^{V_{f}}{p_\text{ ext} \cdot  \mathrm{d}V},$$
where
$$p_\text{ ext} = p_{\text{ from atmosphere}} + p_\text{ due to weights}.$$


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Applicability of Enthalpy-Heat relation for a non-isobaric process
Consider a process in which state variables changes from $(p_1, V_1, T_1)$ to $(p_2, V_2, T_3)$ . I've learned that to calculate the change in enthalpy for any process, we consider a hypothetical isobaric process (Since enthalpy is a state function, it doesn't depend on the path).
Should the hypothetical process have the same initial and final state variable or just the same initial and final temperature? Because if we consider the same initial and final state variables, then we can't form a hypothetical isobaric process since the pressure changes in the actual process and the formula $\Delta H = q_p = n \cdot c_p \cdot \Delta T $ fails here.

For an ideal gas, H is a function only of T.  So it doesn’t matter if P and V change.

Of course, unless P is constant, q is not equal to delta H.

Could you clarify if you were asking about ideal gas or general?

Disclaimer: The below discussion only holds to be completely true for reversible processes and for ideal gases.
I think the problem is you're thinking of it the wrong way, that is, you first think that $ q= nC_p \Delta T$ and for constant pressure process $q_p= \Delta H$ , and hence, $ \Delta H = nC_p \Delta T$
In actuality, it is the other way around. Similar to how internal energy is always $nC_v \Delta T$, the enthalpy is always $nC_p \Delta T$ for an ideal gas. A simple proof is shown below:
$$ \Delta H = \Delta U + \Delta (PV)$$
For $n$ fixed moles of an ideal gas,
$$ U = n C_v \Delta T$$
and,
$$ \Delta (PV) = nR \Delta T$$
Hence,
$$ \Delta H =n\Delta T [  R + C_v]$$
Now, from Mayers relation,
$$ C_p  = C_v + R$$
And hence,
$$ \Delta H = n \Delta T C_p$$
So, as seen above, if you can accept that $\Delta U= nC_v \Delta T$ for an ideal gas always then you must be able to accept a similar statement for enthalpy

Regarding state variables:
Yes, there are indeed many states for an ideal gas which correspond to the same enthalpy since enthalpy is a function of the only temperature. The state variables of pressure and volume are irrelevant for enthalpy calculations unless you don't have the final and initial temperatures. In such a case, you can use the ideal gas law to find expression for the temperature at different states.

Note:
A discussion of proving $q_p =\Delta H$ can be found in this post
You may think the above proof is circular reasoning since regularly Mayers relation is proven by using $ \Delta H = nC_p \Delta T$. However, there exists another way to prove it by considering path functions. Have a look at the first twenty minutes of this lecture by MIT OCW.

If you suddenly drop the external pressure on the gas to a new lower value and then hold it constant throughout the process, would you consider that a constant pressure expansion?

Didn't I ask you this question on physics stack exchange? xD I'm not sure if the constant variable process means constant variable for state or constant variable for external. However, for reversible it's simple because both are at equilibrium for each point in the process.

Is that Joule-Thomson expansion?

No, not necessarily.

If your process is not isobaric, there is no way you can consider it as only one hypothetical isobaric step. So you need to 'decompose' your process into more than one steps, and among those several steps, there can be isobaric step.
For example, your process of interest is
$$(p_1, V_1, T_1) {\rightarrow} (p_2, V_2, T_2) $$
where $p_1 ≠ p_2, V_1 ≠ V_2, T_1 ≠ T_2$.
You can consider it as two steps:
$$(p_1, V_1, T_1) {\rightarrow} (p_1, V_a, T_2) {\rightarrow} (p_2, V_2, T_2) $$
where your first step is isobaric heating/ cooling and your second step is isothermal expansion/ compression.
Your formula applies only to the first step which is isobaric. If your system made up of perfect gas, $ΔH$ for second step is zero as commented by Sir Chet Miller so you can apply that formula to get final $ΔH$ directly. Else, you will need to find $ΔH$ for second step and sum up $ΔH$ of both steps to get $ΔH$ of process of interest.
In short, when comparing process of interest and hypothetical steps, the initial states for both must be the same, and the final states for both must be the same. If your process is not isobaric, and you want to 'decompose' it into only one step which is isobaric as illustrated by your question, the first sentence of this paragraph will not be true, this cannot be done!


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How to make krypton gas glow
So hypothetically, if one was to buy krypton and wanted to make it glow, would a little tesla coil be enough to make it glow? If not, what would be enough? And would one be able to make it glow multiple times or just once, like will making it glow wear down the electrons in the noble gas or somethin’?

Great question! I don't believe it'll wear out in the short term. Light is produced when the electron is excited to a higher orbital and drops back down; this is completely reversible. Neon tubes work on the same principle and last for years!
Eventually (probably after many years) a whole bunch of effects will probably kick in; if there's some impurities in the krypton, they might react with the plasma; the walls of the container will degrade and become opaque, etc.
Whether a certain voltage will cause the krypton to glow seems to be a pretty complicated problem! I believe that would depend on the frequency of the tesla coil (V = dB/dT), how it's coupled to the gas (is there a wire in the tube?), and how large the volume of gas that you want to light.

Likely would need a pressure less than atmosphere in the tube for best results. A Tesla coil is overkill - just a simple rf coil will do.

Your question reminds me to Nixie tubes.  Well, these typically deploy Ne and Ar (e.g. in the Czech republic, Dalibor Farny, documentation video). But there are Penning mixtures for Krypton, too, as examined e.g. by Heylen in 1985 in the International Journal of Electronics (doi 10.1080/00207218508939078) with propylene:

(loc. cit.)


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Mercury thiocyanate synthesis
My friend and I are trying to synthesize mercury thiocyanate, however the only reagents we have access to are mercuric acetate and ammonium thiocyanate.
We figured that we could possibly force a substitution reaction with these reagents in alcohol over heat with stirring.
It worked, but after a few moments the solution turned yellow and after diluting with water (since mercury thiocyanate is insoluble in water and ammonium acetate is the opposite) the precipitate was yellow.
Was there a possible side reaction that occurred? If so what was it? And why is it yellow? I am accrediting that to the sulfur, but I could be wrong as I am a novice chemist.
I don’t have many peers to ask and would appreciate some input!

Mercury salts are toxic. Do not attempt to synthesize if you don't know what is happening.

We used lab safety and are looking for answers thank you

2 questions: 1) why use alcohol; 2) why use heat? Mercuric acetate, ammonium thiocyanate and ammonium acetate are all very soluble in water. Mix, let stand, filter, dry (gently), ignite, get Pharaoh's Serpent! (Remember to clean up and dispose PROPERLY.)


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Find the Ionisation Potential and Electron affinity of X
$N_0/2$ atoms of $X_{(g)}$ are converted into $\ce{X^+_{(g)}}$ by energy $E_1$. $N_0/2$ atoms of $X_{(g)}$ are converted into $\ce{X^-_{(g)}}$ by energy $E_2$.
Hence ionisation potential and electron affinity of $\ce{X_{(g)}}$ per atom are:
I tried solving this by unitary method, which gave ionisation potential and electron affinity as, ${2E_1/N_0}$, ${2E_2/N_0}$ respectively.
But the answer given is  ${2E_1/N_0}$, ${2(E_2-E_1)/N_0}$
Can anyone please tell me where I went wrong and help me with the solution. Thanks in advance!

The Ionisation Energy of an atom is defined as the minimum amount of energy required to remove the most loosely bound electron of an isolated neutral gaseous atom or molecule. In this case, since $E_1$ amount of energy is used to ionise $N_0/2$ atoms, the ionisation energy of each atom is $\frac{E_1}{N_0/2} = \frac{2E_1}{N_0}$.
The Electron Affinity is defined as the amount of energy released when an electron is attached to a neutral atom or molecule in the gaseous state to form a negative ion. In this case, $E_2$ amount of energy is released when $N_0/2$ atoms accept electrons to form negative ions, hence electron affinity of each atom is $\frac{E_2}{N_0/2} = \frac{2E_2}{N_0}$.
Your answers are correct. The textbook must have made a typo in the second question. For the textbook's answer to be correct, the second question should read

$N_0/2$  atoms of $\ce{X+_{(g)}}$ are converted into $\ce{X−_{(g)}}$ by energy $E_2$

This gives the answer as $\frac{E_2 - E_1}{N_0/2} = \frac{2(E_2 - E_1)}{N_0}$ for electron affinity.


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Does mixing two different foaming soap make it ineffective?
To save soap during this pandemic, I mixed foaming soap of two different brands and then the solution became cloudy (both originally transparent) . It seems foaming okay. But does this make the soap less efficient in washing away germs/virus? Does the foaming show that the mixture is still working as a regular soap?
I know that soap is both hydrophilic and hydrophobia which makes it capable of washing away bacterias. Is it possible for the ingredients of two soap make it non polar?
Thanks.

According to the answers, more details are added. The ingredients of the two foaming soaps are given below. According to the ingredients, can we tell if the effectiveness of soap is reduced by mixing them?

About 1/3 is method foaming hand soap with ingredients--WATER (AQUA), SODIUM LAURYL SULFATE*, COCAMIDOPROPYL BETAINE*, GLYCERIN*, ALOE BARBADENSIS EXTRACT*, TOCOPHERYL ACETATE (VITAMIN E), CITRIC ACID, SODIUM CHLORIDE*, SODIUM CITRATE*, METHYLISOTHIAZOLINONE, METHYLCHLOROISOTHIAZOLINONE, FRAGRANCE (PARFUM: LINALOOL, LYRAL), EXTERNAL VIOLET 2 (CI 60730), BLUE 1 (CI 42090).

2/3 is equate anti-bacteria foaming soap with ingredients in the picture.

The answer to your last question is that that's not really possible. If it foams then the washing properties of the soap should be ok. Also, stability (or instability as you observed) depends on the particulars of composition, and not knowing what it is you mixed it's not possible to say what exactly happened.

To the level of accuracy required by the question certainly your mix is still working and I agree with the comment above.

Thank you @BuckThorn. I added the ingredients list of two foaming soap I mixed.

Thank you @Alchimista. I added the ingredients list of two foaming soap I mixed.

The point is that you most likely ruin the optimized formulations but the mix will certainly have washing and antibacterial properties. However from a chemical point of view you didn't mix two soaps. See the answer by James Gaidis. Say that the next time you might well use each single product to the last drop. - Let's stay safe everybody.

@Alchimista, thank you! I am relieved as long as the mixture still functions as regular soap.

The extra information from the ingredient list helps suggest some reactions that may be occurring.
The antibacterial foaming soap contains four detergents (not salts of fatty acids), probably totaling 10-15% by weight of the product. The antibacterial action is provided by the benzalkonium chloride at 0.13%. The label tells you to avoid contact with the eyes, but it is used as a preservative in eye drops (up to 0.01%). So this product contains foamers (considered inactive ingredients) and the very dilute, but aggressive, cationic antimicrobial surfactant.
The amine oxides have positive polarity on the nitrogen, but the attached oxygen, which is negatively charged, creates a dipole, unlike the unambiguous cation in benzalkonium chloride. A betaine has its anion chained into the molecule already, so its availability to connect with another anion (or cation) is minimal. So the antibacterial foaming soap is internally compatible: the benzalkonium chloride does not develop a tight bond to any other of the fatty molecules in this product. Everything is soluble, and there is plenty of surfactant to give good foaming, without an excess of benzalkonium chloride, which at higher concentrations would be more likely to cause skin irritation.
There are two primary surfactants in the foaming hand soap. Neither is a fatty acid soap. The cocamidopropyl betaine is present in both products, so should not cause an incompatibility. The issue of sodium lauryl sulfate is not clear to me (apparently not clear to the mixture of solutions either). On the one hand, the negative charge on the oxygens of the sulfate group could be so delocalized that it does not bind tightly with an amine cation. On the other hand, perhaps there is some interaction, but so little benzalkonium that the combination, a precipitate (which should float) may be emulsified by the remaining surfactants, of which there are plenty.
The extra cloudiness suggests that the benzalkonium chloride has reacted with the sodium lauryl sulfate. The combination might separate and float, or might just be incorporated into an emulsion. The surfactant concentration is so great that it could easily emulsify ~0.13% of scum. However, that suggests that the benzalkonium ion is not available for antibacterial action, even if there is still plenty of surfactant available for foaming!
The bottom line is that using the products separately is more certain to do what they claim to do. You can't combine the claims.

Thank you @JamesGaidis. Based on your analysis, is it correct to conclude that the claimed 'anti-bacteria' action is reduced but the regular soap's amphiphile and ability to wash away virus/bacteria isn't so negatively affected?

It would seem that way. Theoretically, the antibacterial action is neutralized, but there's plenty of foaming and washing capability remaining, so it's not bad, just not as good as using the separate soaps separately.

The soap solutions were originally transparent. An interesting experiment, if you still have some of the original soaps, is to shine a light thru the solutions and look for a Tyndall effect:

The light source is on the far right, out of the picture. The beaker on the right shows light scattering from particles comparable in size to the wavelength of light due to the Tyndall effect; the beaker on the left shows no scattering, even tho the light beam is passing thru it. Its contents are unknown, but there seem to be no particles of sufficiently different index of refraction and size to scatter light.
The tiny particles which scatter the light (just like in a smoky room) are micelles, aggregates of several hundred or thousand soap molecules. In the original soap solutions, this scattering is small, due to selection of manufacturing conditions (exact formulation, concentration, pH, other ingredients, etc., etc.), so the solutions appear transparent. What is optimal for each soap separately may not be optimal for the combination: perhaps the mixing produced larger micelles which scattered more, so the mixture scattered more light and became cloudy to the eye. I'm going to suggest that both soap solutions show a Tyndall effect, and the mixed solution shows a much larger Tyndall effect. It may change with time.
Now the question becomes: is the combination as effective as either solution alone. Well, it is different (cloudy). It could be worse; it could be better! It might not be observably different.
The foaming test could cast some light on the effectiveness: If you can test the foaming (height of foam and time stability) of the original solutions and of the mixture, and if they are all the same (+/- ~10-20%), the combination is probably OK.
There is a case where this foaming test will indicate immediately that the combination is not good. The molecules in soaps and detergents have a hydrophilic end and a hydrophobic end. Soaps are salts of fatty acids, and have a negative (hydrophilic) carboxyl group at one end and a (hydrophobic) hydrocarbon at the other end, and a soluble alkali metal cation. There are nonionic detergents which have no charge; they are compatible with anionic or cationic soaps and detergents. Cationic detergents (usually amine salts) have a hydrophilic positively charged nitrogen at one end and a hydrophobic hydrocarbon at the other, and some anion, usually chloride. If you should mix solutions of a cationic detergent with an anionic soap, the positive and negative charges associate to produce a neutralized ion-pair with two hydrophobic hydrocarbon groups - together this looks like a big fatty molecule with a little polarity in the middle. Overall, this molecule is not very soluble in water and precipitates out (but it floats like scum). And since it is not soluble in water, it won't foam.

This is so detailed. Thank you @JamesGaidis. I added the ingredient list and hopefully that can provide more information to answer the question. Do I need any tool to test the foaming? Thank you.


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Realistic conditions for fuel container explosion
First off, I'm not actually trying to blow anything up. Also, my apologies if this isn't the right SE to ask. If so, I'd appreciate it if you could point me to the right SE.
I'm writing something where at a point I need a big explosion to happen. I've done some digging and I found out that the old bullet-fired-at-gas-tank trope is essentially just a trope, and that it alone is unlikely to cause an explosion.
The explosion needs to happen near a big mansion on an otherwise uninhabited island. The mansion is powered by an outdoor generator that runs on some kind of fuel, for example propane, though it can be anything else that makes sense and can more easily get me the explosion I need. There are fuel containers
to refill the generator sitting somewhere outside the mansion, and they need to be accidentally blown up during an action sequence.
The ideal explosion would be strong enough to blow up the closest wall of the mansion, so that the fire can spread inside the mansion, and this leads to my first issue: I have no idea if it's realistic to expect such a dramatic effect or not—for all I know, it might take far more fuel than it makes sense to keep in stock.
Secondly, as said just shooting one of the containers won't trigger an explosion. I understand that you need the right mixture of oxygen and fuel, and the containers would have to be exposed to sufficient heat for some time before they actually blow up. I thought that, maybe, if some containers are shot and start leaking, and then the leaked fuel somehow catches fire during the sequence, this might trigger an explosion, because air would start leaking into the containers that got a hole punched in them.
To sum up, my questions are:

Can a fuel explosion realistically blow up a nearby wall and spread a cloud of fire into the mansion so that at least the nearest room is almost completely on fire very quickly, without unrealistically huge amounts of fuel involved?

Can the explosion be caused by a bunch of fuel cans that were hit by bullets, started leaking, and were heated up by the flames that followed when the leaked fuel somehow caught fire? Would this take too long to happen (eg, ten minutes), or could it happen within a minute or two?

If your generator runs off propane and is in a sturdy shed (maybe brick-built) attached to the wall of the mansion then a propane leak could easily set up an explosive fuel/air mixture in the shed. Flicking the lightswitch could easily be enough to set off the explosion.

Lookup bleve explosion

Hint: near-empty gasoline containers are much more likely to explode, but then calculate the amount of fuel at an assumed 14:1 ratio with air, and the energy available in that fuel.

The topic of explosives in movie production is a specialty application as one does not generally employ classic explosives!
In movies, the aim is NOT to kill/injure anyone (especially, your famous lead actor). As such, one can select even a flameless explosive, which, for example, could include Ammonium nitrite ($\ce{NH4NO2}$). The latter is primarily a pressure explosive which, per a chain reaction, quickly releases a massive amount of gas. Such explosives find applications, for example, in blasting in coal mines. Apparently, mines can develop a potentially explosive mixture of methane and air, where a flame/spark could result in a massive secondary explosion!
Also, distinct from a battlefield setting, consider low brisance explosives, which do not produce lethal shockwaves.
In essence, your objective should differ significantly from the design/use of military-grade explosives. Instead, aim at much flash and dust, but less lethal. The sound level of the explosion can be addressed in the post-editing of the film.
To accomplish this, you need more research and the opinion of someone who actually specializes in this area.


text:
Does #4 LDPE Plastic Bottle Leach into Water?
I understand that Water Bottles expire due only to the fact that the plastic can degrade over time and leach into the water making the water's taste not as desirable as the manufacturer might intend.
Does anyone happen to know the degradation rate/expiration for #4 LDPE? I'm using a plastic water storage vessel for long term water storage in m emergency preparedness supplies and would like to make sure the plastic doesn't leach. Please let me know! Thanks!

For long term storage, the plastic vessel is not the only factor, but also water itself, catching smells and growing life. Note that plastic leaching is not directly related to plastic degradation.Degradation is rather related to releasing plastic microparticles, while leaching is diffusion of minor components from plastics to water ( like phthalates used as plasticisers ).

The key is regular water replacement and occasional container cleaning. Durability of PE is not a concern, aside of manipulation weariness and optical unusability to serve any more.


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Synthesis of Pyrene from Benzene
I know of this ring expansion reaction that can fuse a benzene ring with another benzene ring:

Performing this reaction twice gives us phenanthrene

How would one go about synthesizing pyrene from phenanthrene or benzene? I couldn't find any literature online, as most pyrene is obtained via fractionating and cracking petroleum. Synthesizing it from phenanthrene also turns up a dead end, as the 9 and 10 positions on it are most reactive, and I would not obtain pyrene in that way. Diels-alder reaction of phenanthrene is also not possible, as highlighted in this answer.

I would imagine that you could do it from the commercially available 2,2'-dibromobiphenyl by e.g bis-allylation with allyl tributylltin, ozonolysis to the bis- aldehyde and acid catalysed ring closure.

Coupling   of 2-bromostyrene followed by treatment with strong lewis acid and then elemental sulfur at about 200-300 celsium might work

Petroleum cracking and incomplete combusion of organic compounds may give pyrene in a complicated mixture as discussed in else where. However, if you looking at a method to prepare pure pyrene or its derivative you may need to go through [2.2]metacyclophane procedure (For review: Ref.1):

As seen in the above image, [2,2]metacyclophane can undergo transannular ring closure to yield products with the 4,5,9,10-tetrahydropyrene (THPy) skeleton (e.g., Ref.2), which may then be dehydrogenated to pyrene (Ref.3).
The synthesis of [2,2]metacyclophane(s) can be achieved from m-xylene. This can be in two steps: First, the bromination of two methyl group using NBS to achieve 1,2-bis(bromomethyl)benzene and then subjected the brominated m-xylene to Wurtz reaction (e.g., Ref.4). Recent approach of some substituted [2,2]metacyclophanes is described in Reference 5 starting with substituted m-xylenes (The electronic supplementary material is attached for readers benefit).
The synthesis of highly substituted pyrenes via cyclisation of bis-alkynylbiphenyls has been also achieved (e.g., Refs.6-8). It is worth noting that these starting biphenyls were made using simple benzene derivatives (Ref.1):

References:

Juan M. Casas-Solvas, Joshua D. Howgego, Anthony P. Davis, “Synthesis of substituted pyrenes by indirect methods,” Org. Biomol. Chem. 2014, 12, 212-232 (DOI: 10.1039/C3OB41993B)(PDF).
Fritz Vögtle, Peter Neumann, “Stereochemistry of [2.2]Metacyclophanes,” Angew. Chem., Int. Ed. Engl. 1972, 11(2), 73–83 (https://doi.org/10.1002/anie.197200731).
Peter P. Fu, Ronald G. Harvey, “Dehydrogenation of polycyclic hydroaromatic compounds,” Chem. Rev. 1978, 78(4), 317–361 (https://doi.org/10.1021/cr60314a001).
Virgil Boekelheide, Rodger W. Griffin Jr., “Synthesis of [2.2]metacyclophanes,” J. Org. Chem. 1969, 34(6), 1960–1961 (https://doi.org/10.1021/jo01258a095).
Marco Blangetti, Helge Müller-Bunza, Donal F. O'Shea, “First asymmetric synthesis of planar chiral [2.2]metacyclophanes,” Chem. Commun. 2013, 49(55), 6125-6127 (https://doi.org/10.1039/C3CC42275E)(Electronic Supplementary Material).
D. Barney Walker, Joshua Howgego, Anthony P. Davis, “Synthesis of Regioselectively Functionalized Pyrenes via Transition-Metal-Catalyzed Electrocyclization,” Synthesis 2010, (21), 3686-3692 (DOI: 10.1055/s-0030-1258238).
Matsuda Takanori, Moriya Taisaku, Goya Tsuyoshi, Murakami Masahiro, “Synthesis of Pyrenes by Twofold Hydroarylation of 2,6-Dialkynylbiphenyls,” Chemistry Letters 2011, 40(1), 40-41 (https://doi.org/10.1246/cl.2011.40).
Mareike M. Machuy, Christian Würtele, Peter R. Schreiner, “2,6-Bis(phenylethynyl)biphenyls and Their Cyclization to Pyrenes,” Synthesis 2012, 44(9), 1405-1409 (DOI: 10.1055/s-0031-1290754).

A quick search on Reaxys(tm) report only 7 synthesis that involves pyrene and benzene (and they can include also 13 multi-step reaction). See ref. 1 (Physical Chemistry Chemical Physics) for more details. The abstract of the reference is as follows:

Measurements of concentrations of high molecular hydrocarbons in laminar flat low pressure flames with various hydrocarbons under sooting conditions are reported. The results show that for all fuels investigated in that region of the flames where the bulk of soot is formed besides the main combustion products some characteristic groups of species are to be found. Within these groups of species only few fuel specific characteristics are stated. The main fuel specific influence lies in the relative concentration for the species. A reaction scheme for the qualitative explanation of the results found is discussed.

In the full text article (that I cannot report for copyright issues), there is a reaction scheme for the formation of high molecular hydrocarbons. It's also for sure a good starting point.
References:

H. Bockhorn,  F. Fetting,  H. W. Wenz, "Investigation of the Formation of High Molecular Hydrocarbons and Soot in Premixed Hydrocarbon‐Oxygen Flames," Berichte der Bunsengesellschaft für physikalische Chemie (Physical Chemistry Chemical Physics) 1983, 87(11), 1067-1073 (https://doi.org/10.1002/bbpc.19830871121).


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O-formyl benzoic acid or p- formyl benzoic acid. Which is more acidic?

As far as I can think, the ortho effect comes into play which makes o-formyl benzoic acid more acidic.
But according to some, due to Hydrogen bonding between aldehyde and carboxylic group, hydrogen atom will be tightly bonded to it's two neighbouring oxygen atoms decreasing acidic character of o-formyl benzoic acid.
These two assertions are confusing and hence I seek for an experimentally verified result along with a satisfying explaination.

You can't compare them at all.  If the ortho form dissociates at all then the carboxylate ion combines with the aldehyde function, forming a lactone ring.  See Wikipedia.
This question invokes a comparison of the similarly structured nitrobenzoic acids.  There the ortho compound is found to be strongest, beating its isomers by about one $pK_a$ unit and apparently indicating that the ortho effect is stronger than intramolecular hydrogen bonding.
Upon further review, we find that it wasn't a fair fight.  Count the number of atoms that would be formed in a hydrogen-bonded ring: the hydroxyl portion of the carboxylic group, the carboxyl carbon, two aromatic ring carbons and two of the atoms from the nitro group.  We generally do not get good intramolecular hydrogen bonding with a seven-atom ring, the arrangement just does not work sterically.  Intramolecular hydrogen bonds are more common with five- or six-atom rings.
So in the case of 2-nitrobenzoic acid versus 4-nitrobenzoic acid the real answer is that hydrogen bonding is not in the game.  Nor would it be for the formyl benzoic acids being described here, if the ortho compound were not prone to tautomerization; here too such a hydrogen bond would unfavorably form a seven-atom ring.  We might say that the  ortho compound, failing to form an intramolecular hydrogen bond, is thereby made prone to the strong dissociation and subsequent tautomerization noted above.

Wasn't clear before the edit. I deleted my comment now.

Then which of the two dissociates more in aqueous solution? Ortho or para?

Again we can't tell experimentally when the ortho form launches off and becomes a lactone (or lactol).

Can you try it theoretically?

"Can you try it theoretically?" Maybe I could, definitely I won't. Chemistry is still an experimental science; theory is intended to suggest possible experiments and outcomes.


text:
Do some particles get ionised to 2+ or higher in a mass spectrometer or do they only get ionised to 1+?
In a mass spectrometer the sample is vapourised, then ionised, then accelerated.
My textbook says that the high energy electrons from the electron gun knock electrons off the sample molecules and hence ionise the particles by 1+.
But do some particles get 2 or more electrons knocked out or is it always just one ?
I suppose once it has been ionised by 1+ it gets accelerated through the tube so it won't hang around for long but I would think that there is still a small window of time for further ionisations.

Don't forget that the abscissa in mass spectroscopy is $m/z$, i.e. the ratio of mass divided by charge.  An example where double-charged ions play an important role, published as «open access» is «Identification of impurities in nafamostat mesylate using HPLC-IT-TOF/MS: A series of double-charged ions» by Zhang et al. in the Journal of Pharmaceutical Analysis 10 (2020) 346-350, https://doi.org/10.1016/j.jpha.2020.03.002.

@M.Farooq your comment would make an excellent answer...

@CurtF., Changed it to an answer.

In electron impact, you would see mostly +1 charged ions. Electron impact is a "harsh" way of ionization. However, +1 charge is not a universal rule in MS. The state of ionization depends on the ionization source. Later in your courses you might encounter electrospray ionization. There you might have +14 charges, etc.

Your downvote will not change the facts! Sorry.

Interesting , but in what sense is it ''harsh'' ?

EI is considered as a hard ionization technique because it breaks the molecule into many fragments. You rarely see the molecular ion-peak in an EI mass spectrum.


text:
How does an acid dissociate in water
I read somewhere that hydrogen cation combines with water to form hydronium ion. I also read it combines with various water molecules. Can someone please explain?

Have you tried searching and reading e.g. about hydronium ?

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected the author elaborates the question by at least basic textbook/online search and writes down what he/she found/understood and what is the eventual stumble stone.
The "quick questions" without explicitly expressed particular effort are not very welcome, and may be closed. OTOH, explicitly written effort significantly raises probability of receiving appreciated answers.

https://chemistry.stackexchange.com/a/14119/9961 https://chemistry.stackexchange.com/a/72892/9961

Usually a pure acidic molecule like HCl gives one proton $\ce{H+}$ to a base like NH3 or $\ce{CH3COO-}$, and the proton once fixed on the base gives a new species which is $\ce{NH4^+}$ or $\ce{CH3COOH}$. And that's all what's happening. With water, it is a bit different. The $\ce{HCl}$ molecule starts by giving one proton $\ce{H+}$ to one $\ce{H2O}$. This produces an ion $\ce{H3O+}$ and usually chemists consider that it is the final result. It is not, because it is possible to show that the proton may attract a second water molecule in order to make a heavier ion $\ce{H5O2^+}$ having a charged H atom in a central bridge $\ce{H2O-H^+-OH2}$. Thermodynamic measurements are even able to show that still more $\ce{H2O}$ molecules are attracted around this $\ce{H5O2^+}$ ion, to produce an ion $\ce{H_{2n+1}O_n^+}$. The trouble is that the exact value of the parameter $n$ is not known with precision. It should be between $6$ and $10$. That is why many chemists forget about this difficulty and have decided that the only species in water is $\ce{H3O^+}$. This simplifies also the determination of dissociation constants of acids in water.


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Concentration of Amino acid calculation?
I found the $\mathrm{pH}$ of alanine solution, and I need its concentration, I'm given the formula below:
$$\mathrm{pH}=\frac 12 (\mathrm{p}K_\mathrm{a}- \log c)$$
My try:
I have calculated $\mathrm{pH}$, to find $c$ I need $\mathrm{p}K_\mathrm{a}$. Can someone explain that formula, what is $\mathrm{p}K_\mathrm{a}$?
I know that this amino acid has $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$, which do I use?

If a question is asked on Chemistry SE site, the site policy expects authors to elaborate the question, e.g. by searching in textbook+online resources, writing what has been found, understood or tried and what is the stumble stone. The quick questions without explicitly expressed solving effort are not very welcome, and may be voted to be closed. OTOH, explicitly written effort to answer the question raises the probability to get a satisfying answer.

Search for the term acid dissociation constant .

@Poutnik I dont understand, I know the term pKa... I just wonder which one to use? of the acid correct? since there are 2 pkas for each amino acid

OK, but your question is unclear in that.

To clarify:
$\mathrm{p}K_\mathrm{a1}: \ce{NH3^{+}-R-COOH <=> NH3^{+}-R-COO- + H+}$
$\mathrm{p}K_\mathrm{a2}: \ce{NH3^{+}-R-COO- <=> NH2-R-COO- + H+}$
For aminoacids, the the pH calculation is the same as for acidic salts of diprotic weak acids $\ce{MHA}$.
Unfortunately for you, $\mathrm{pH}=\dfrac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$ as the very good approximation, so it is almost independent on the concentration, so you cannot determine the concentration from $\mathrm{pH}$.
More exactly. you could, but by a much more complex formula as you listed above, as it would be very inacurate because of high $\dfrac{\mathrm{d}c}{\mathrm{d(pH)}}$
By other words, aminoacid solutions act as low capacity pH buffers ( because of quite a big difference of their $\mathrm{p}K_\mathrm{a1}$ and $\mathrm{p}K_\mathrm{a2}$.


text:
What is the significance of enthalpy?
I an getting confused between enthalpy and internal energy. My textbook says that change in enthalpy is the sum of change in internal energy and the amount of work done in expansion or compression but isn't it the same as the amount of heat lost or given. So why do we need enthalpy?

In your chemistry lab, what container do you perform most of the reactions in? A test tube. Since most of the reactions are carried out at constant pressure not volume, we defined Enthaply function. It is more useful for most of the reactions as it is defined as the heat transferred in an isobaric process

Have you tried reading enthalpy article on Wikipedia ?

Does this answer your question? Need help understanding Enthalpy

The internal energy $U$ is the sum of all the energies stored in the chemical bonds. If a chemical reaction happens, the bonds are losing or getting energy. Heat is getting in or out of the system, and this heat $\Delta Q$ can be measured. $\Delta U$ =$\Delta Q$, if the volume is constant.  But if the transformation is carried out in contact with the atmosphere, at constant pressure, the volume may change. A fraction of this energy is converted into work $\Delta w$ to repel the atmosphere, because the volume of the system may change. Usually this volume change is small (except if a gas is produced or consumed). And this volume change may be difficult to know with precision. But it must be taken into account to calculate $\Delta U$. So if this small amount of work is neglected, a correction must be added to the internal energy $U$, that depends on pressure and volume. Neglecting this volume correction gives a sort of “apparent internal energy”, which is called enthalpy $H$, with $H = U + PV$. And then, in all transformations made at ordinary pressure, the following expression is valid : $\Delta H$ = $\Delta Q_p $.
Enthalpy is a sort of "apparent internal energy". There is a similar difference between the weight and the "apparent weight", which is measured in air. The weight of an object should be measured in a vacuum. The weight, as measured with a dynamometer in the usual atmosphere, is a little bit smaller than in a vacuum, because of Archimedes.


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Difference between Helmholtz free energy and normal free energy
I know that free energy is defined as the amount of work that a system can do at constant temperature and and its formula is F=U-TS and that by definition Helmholtz free energy is the amount of work a system can do at constant temperature and constant volume but the formula is still A=U-TS. Why is that , why is there no difference in the formula or some kind of correction factor to account for the different conditions (like the PV in Gibbs free energy ) ?

Please rewrite your question ! You get two different parameters (F and A) for the same definition. What is the correct definition ? Is it F = U-TS or A = U-TS ?

Well that is the thing , they have the same formula but one is said to happen at constant T only while the other is at constant T and constant V

That is false interpretation. They do not happen and they do say identical thing. A/F nor U definitions are NOT limited to constant V,  same as G or H definitions are not limited to constant p. But their usage at the respective constant p or V is with advantage.

BTW, the pV term is rather in enthalpy than In Gibbs energy, that is derived from enthalpy.

By "normal free energy", you must mean Gibbs free energy $G$, i.e. free energy at constant pressure ( = the same start and final pressure, as state functions do not depend on the process path.)
Both symbols $A$ and $F$ are eventually used for the Helmholtz free energy ( free energy at constant volume ) for historical reasons, so the particular symbol usage is source dependent. They mean the same thing.  The fact each of them was used in the context of your question in different cases does not change that.
Note that $F/A$ nor $U$ state function are NOT limited to be used at constant volume ( see e.g. adiabatic expansion ), same as $G$ nor $H$ are not limited to be used at constant pressure. But their usage at the respective constant state variable is with an advantage, as you can see below.
To refresh the relations of $F(A), G, U, H$:
$$H=U+p \cdot V$$
$$G=H-T \cdot S=U + p \cdot V - T \cdot S=A(F) + p \cdot V$$
$$A(F)=U-T \cdot S = H - p \cdot V - T \cdot S = G -  p \cdot V$$
For closed, not isolated systems with zero non volume work:
(Non-volume work is e.g. a system charging a battery. )
For constant volume: $$\mathrm{d}U=\text{đ}Q$$
For variable volume: $$\mathrm{d}U=\text{đ} Q
 - p \cdot \mathrm{d}V$$
For constant pressure and zero non volume work, $$\mathrm{d}H=\mathrm{d}U + \mathrm{d}(pV) = \mathrm{d}U + p \cdot \mathrm{d}V= \text{đ}Q$$
For variable pressure, it would be $$\mathrm{d}H=\mathrm{d}U + \mathrm{d}(pV) = \mathrm{d}U + p \cdot \mathrm{d}V + V \cdot \mathrm{d}p= \text{đ}Q + V \cdot \mathrm{d}p$$
Now involving free energies:
$$\mathrm{d}F = \mathrm{d}U - \mathrm{d}(T \cdot S ) = \mathrm{d}U - T \cdot \mathrm{d}S  - S \cdot \mathrm{d}T$$
$$\mathrm{d}G = \mathrm{d}H - \mathrm{d}(T \cdot S ) = \mathrm{d}H - T \cdot \mathrm{d}S  - S \cdot \mathrm{d}T$$
Isothermal cases:
For $\mathrm{d}F$ and $\mathrm{d}G$, there would be just the additional term $-T \cdot \mathrm{d}S$
$$\mathrm{d}F= \text{đ}Q
 - p \cdot \mathrm{d}V  - T \cdot \mathrm{d}S$$
$$\mathrm{d}G=\mathrm{d}F + \mathrm{d}(pV)= \text{đ}Q + V \cdot \mathrm{d}p - T \cdot \mathrm{d}S$$
Non-isothermal cases:
There appears another term: $-S \cdot \mathrm{d}T$
$$\mathrm{d}F= \text{đ}Q
 - p \cdot \mathrm{d}V  - T \cdot \mathrm{d}S - S \cdot \mathrm{d}T  $$
$$\mathrm{d}G=\mathrm{d}F + \mathrm{d}(pV)= \text{đ}Q + V \cdot \mathrm{d}p - T \cdot \mathrm{d}S - S \cdot \mathrm{d}T$$
Then for reversible processes, as $dS = \text{đ}Q_\mathrm{rev}/T$:
$$\mathrm{d}F=
 - p \cdot \mathrm{d}V  - S \cdot \mathrm{d}T  $$
$$\mathrm{d}G=  V \cdot \mathrm{d}p  - S \cdot \mathrm{d}T$$
You can see now, that for zero non-volume work  and reversible processes, $\mathrm{d}F=0$ for constant $T$,$V$ and $\mathrm{d}G=0$ for constant $T$,$p$.
For irreversible processes, as $\mathrm{d}S_\mathrm{irr} \gt \text{đ}Q/T$, $\mathrm{d}F \lt 0$ for constant $T$,$V$ and $\mathrm{d}G \lt 0$ for constant $T$,$p$.

@ Poutnik. I repeat the question given by David Sterlinsky ; " Why is there no difference in the formula or some kind of correction factor to account for the different conditions ? ". Well ! Don't you think the answer might be : There is no difference because A is sometimes called F ?

@Maurice Sure, it can be. But the question formulation is quite unclear about the question scope, so I took answering in a broader way as well.

I am sorry for the unclear question , what I meant was that if one is said to be happening at constant temperature only (P and V are able to vary) but the other one (Helmholtz) has T and V constant (as in the case of Gibbs energy would be constant P and constant T ) why don't we have some sort of factor to account for the difference in conditions (in one case V is able to change as is P) and in the other (only P is variable) ?

Such question clarifications by the OP should be expressed by modification of the original question, rather than by a comment, where should me rather just mentioning of the question update. It is helpful to read the question after its writing, if it is clear to you, impersonating yourself to an eventual reader.

I have updated the answer to address your clarification


text:
Does cellulase catalyze the hydrolysis (or break down) of soluble starch?
I’m attempting to perform an Iodometric redox titration to evaluate the vitamin c concentration in a particular solution. These solution have dissolved within them up to 2% cellulase enzyme.
In an iodometric redox titration, a 0.5% starch indicator solution would be used to inform when endpoint occurs via it’s reaction with excess, un-neutralized iodine indicator solution.
Will the cellulase in solution interfere with this titration process by perhaps hydrolyzing the starch indicator solution?
If so, should I try and denature the cellulase in solution before performing the redox titration?


text:
Using the method of differential absorption to determine the ethanol content of an unknown sample
I have completed a whole set of FITR-ATR spectra analysing 6 solutions of ethanol-water mixtures, all at different concentrations of each.
I have now been instructed to use the method of differential absorption to determine the concentration of ethanol. I am subtracting the absorbance at the base of the C-H stretch of ethanol from the maximum absorbance of the same stretch so now I have a difference in absorbance value. At this stage, I am a little confused as to how I use this value to obtain the concentration of ethanol. All I know is that the Beer-Lambert law is involved somehow.
Any help with this would be appreciated. Thanks.

The absorbance is proportional to the concentration of ethanol.

See Beer–Lambert law

@Dave, Can you add more details? Are you correcting the baseline?

A starter: https://www.researchgate.net/post/How_can_i_use_FT-IR_spectroscopy_in_quantitative_analysis


text:
What does the equilibrium constant mean for reactions with very small concentrations?
When you calculate the equilibrium constant for a reaction with a higher concentration of reactants, the answer is fairly intuitive. For example, for the reaction:
$$\ce{A + B <=> C}$$
when we have $\pu{0.10 M}$ of A, $\pu{0.30 M}$ of B and $\pu{0.45 M}$ of C the equation for the equilibrium constant is:
$$ K = \frac{[\pu{0.45 M}]}{[\pu{0.10 M}][\pu{0.30 M}]} = \pu{15M^{-1}}$$
This is relatively intuitive, we can see that the concentration of C is larger than the product of A and B and so saying the products are favoured makes sense.
However, an issue comes about when you consider really small amounts of all of them. For example, for the same reaction
$$\ce{A + B <=> C}$$
If we have $\pu{2.4E-4 M} $ of A, $\pu{2.4E-4 M} $ of B and $\pu{7.5E-6 M} $ of C we have the equation:
$$ K = \frac{[\pu{7.5E-6 M} ]}{[\pu{2.4E-4 M} ][\pu{2.4E-4 M} ]} = \pu{130 M^{-1}}$$
In this case, the equilibrium constant suggests the products are favoured but if you look at how much product we have compared to reactants there's so much more reactants than products. The constant makes sense since multiplying two numbers less than 1 will get a smaller number but it's confusing when you consider the implications that has on the equilibrium. My immediate assumption is that I am misunderstanding the relationship between the equilibrium constant and the concentrations of species in the reaction and if this is the case is there a more intuitive way of looking at this?

You have it the wrong way round. At a given temperature for any equilibrium $K$ has a fixed value, i.e. is constant. The concentrations must always match this so you cannot just guess arbitrary values. The extent of dissociation changes to ensure  concentration ratios  agree with $K$.

@ Nathan. I think you simply forget the units. The equilibrium constant has a unit. For your system, the equilibrium constant is not $130$, as you state. It is $130~ L/mol$

"In this case, the equilibrium constant suggests the products are favoured but if you look at how much product we have compared to reactants there's so much more reactants than products."  This is what I take as the heart of your question.  To that point:
I believe you need to compare the concentration of all the reactantS, not just one of the products.  In this case, the concentration of all of the reactants is 5.76 X 10 -8 which is a lot less than the 2.4 X 10 -4 which you might have been comparing to the product in the equation.

if you want to find the concentrations, you should add them, why did you multiply? As @porphyrin said in the comments to the question, concentrations of chemicals will change to satisfy the value of $K_{eq}$, not the opposite


text:
Am I understanding this IR spectrum correctly?

Is it safe to say that this IR spectrum belongs to a carboxylic acid because of the 3100-2700 stretch indicating OH group, and the sharp peak at 1728 showing a carbonyl group?

@ M. Farooq: The IR is not a spectrum of carboxylic acid derivative. It is just a carbonyl compound. I'd like to let OP to express their thoughts before giving any hint.

@Mohamed:  Don't link to images outside of ChemSE, but embed them (e.g., as a .png file).  This keeps the illustration accessible even if the link to a third-party is invalidated (e.g., change of the address) and allows the users of ChemSE to connect quickly your question to the spectrum in question.


text:
Resulting ions concentration after mixing solutions

$\pu{1 g}$ of $\ce{NaHCO3}$ and $\pu{2 g}$ of $\ce{Na2CO3}$ were dissolved in $\pu{50 mL}$ of water known as solution A. Solution A was mixed with $\pu{450 mL}$ of $\pu{0.154 M}$ of $\ce{NaCl (aq)}$ known as solution C. What is the resulting concentration of $\ce{Cl-}$ ions and $\ce{Na+}$ ions in solution C in $\pu{mM}?$

I started of by finding the amounts of $\ce{NaHCO3}$ and $\ce{Na2CO3}$ which were $1/84$ and $1/53,$ respectively, but I do not know what to do with these numbers.

General advice: never omit units and solve the problem algebraically first, plug in the physical quantities second.

@Sarahbaker  Using algebraic approach does not have only the aspect of advantage of keeping track of proper approach.  The another aspect is the community one. The Questions/Answers are here not just for the OP(i.e. you). Many people search solutions for similar problems all the time. It is of much greater value, if the problem is evaluated generally first in algebraic form. If they have to work just with literal numbers of a particular task, it is much less useful, as it is more difficult to orient in the task and generalize to apply elsewhere

@Poutnik It's neither. From Wikipedia: "A physical quantity can be expressed as the combination of a numerical value and a unit. For example, the physical quantity mass can be quantified as n kg, where n is the numerical value and kg is the unit. A physical quantity possesses at least two characteristics in common, one is numerical magnitude and other is the unit in which it is measured."

@andselisk Sure . But I had in mind rather symbols versus literal instead of [symbols/values] with units or without them. In case of algebraic expresions units are often implicit, as they can vary and can be implied from symbols. Literals require units  explicitly as a unit cannot be implied from a number.

You have 1/84 + 2/106 moles (twice as many sodium ions in a mole of carbonate as in bicarbonate) of sodium ions from the carbonates. Add that to the (0.45 x 0.154) moles of sodium ions from the NaCl. This is your total moles of sodium ions. The total volume of solution is (0.05 + 0.45) L so divide your total moles of sodium ions by 0.5 to get the concentration of sodium ions.
The concentration of Cl- ions is the moles of chloride ions in solution C (0.45 x 0.154) divided by the total volume of solution (0.5L)

Hey there thanks for replying but did you mean 1/53 rather than 2/53. And for calculating the total moles of sodium, did you mean adding 1/84 + 1/53 + (0.0693). I did not realise you could add moles together

Edited to correct

Calling combined amounts of substance "total moles" is as illiterate as calling the combined masses "total grams". Also, current policy suggests the answers should be self-contained and provide straightforward solution. Hints, guidelines and suggestions are better suited as comments.


text:
What concentration of HCl is safe to smell?
I want to give students the chance to gently waft and smell HCl.
What concentration would be considered safe for a once-off single waft and smell ?

Why would you want your students to smell HCl?

See this and this answer. Bottom line: Don't smell chemical compounds. They are not perfumes.

That's not really answerable, thing is highly dependent on how you do it and what this "safe" means. BTW @NilayGhosh It's not like you can smell anything but chemical compounds ;>

@Mithoron It is late at night. I should catch some zzzs.

That will naturally occur if they are supposed to be trained in practical chemistry. If they don't, there are better things to smell.

Concentrated HCl solution is about 35% - 36%. Its odour is strongly corrosive and should not be smelled. In my school we consider that the aqueous mixture 1+1 has no odor. It is about 18% HCl.

I would rather let them see it, under a hood in a graduated cylinder.  Pour a small amount of the concentrated acid (37% works well) into the cylinder.  When the acid mixture evaporates the HCl molecules attract water vapor from the air and form a solution with reduced vapor pressure, so you see condensation.

I think it is enough to watch the Safety Data Sheet to understand. I found this one browsing on the internet:
Safety data sheet of a Chloridric acid manufacturer
On section 11 you can find: ATE US (gases): 1560.000 ppmV/4h. ATE means "Acute Toxicity Estimates".

As a teacher, you should lead by example how to engage with chemicals respectfully, as a professional. Which neither is to be fearful, nor to be sloppy and negligent for their potential hazards.  This includes to plan ahead, to limit exposure to dangerous goods (like concentrated HCl is), to use protective gear, safe manipulation of the chemicals in response to their possible action, and to have a rescue plan should something go wrong.
The students may already know the smell of acids from cuisine (e.g., balsamic vinegar, kimchi stew, choucroute garnie, etc.), but teach them early the right way chemists do:

(screen photo from here; it is not concentrated HCl)
Or would you trust a dentist as professional if forceps were not both clean and functional?


text:
How much grams oxygen is in 7 gram magnesiumnitrate? Mg(NO3)2
I calculated this:
molar mass (M) of Mg(NO3)2 is 148,32 (24,3 + 214,01 + 616)
so when there is 7 grams Mg(NO3)2 there is 0,0472 mol (7/148,32)
But how do i find out how much of that is oxygen? Do i just divide by 8 and multiply by 6? (because 8 atoms of with 6 are oxygen)

No. Each mole of the compound contains 6 moles of oxygen atoms, and has a mass of about 6 * 16 grams. You don’t have a mole, so use proportional reasoning to determine the actual mass in the sample.
To help to understand why it is simply a factor 6 and does not depend on the total number of atoms in the compound, perhaps imagine a chemical reaction that atomizes the compound:
$$\ce{Mg(NO3)2 ->[atomizer] Mg + 2N + 6O}$$
Irrespective of the number of other atoms in the compound, there are six moles of oxygen atoms for every mole of magnesium nitrate.

Do i just divide by 8 and multiply by 6? (because 8 atoms of with 6 are oxygen)

No, that would not work because the other atoms are heavier or lighter than oxygen. You already incorporated the information about the other atoms in the compound when you divided by the molar mass of the compound.

I did in this way:
Molecular masses:
Mg: 24.305;
N:14.007;
O: 15.999;
Total mass of Mg(NO3)2:
24.305 + (14.0072) + (15.9953*2) = 148.289 g/mol
Then the proportion:
Mass of oxigen in the molecule / total mass * 7 = (15.99532)/148.289 *7 =4.53 g

Oh wow, thank you, i finally understand :)

Generally the right idea but numerous mistakes. Molecular masses: Mg: 24.305; N:14.007; O: 15.999 - these are not molecular masses but rather atomic masses // Total mass of Mg(NO3)2: 24.305 + (14.0072) + (15.9953*2) = 148.289 g/mol the equation is wrong. It should be $24.305 + 2*(14.0072) + 6*(15.9953) = 148.291$g/mol

(cont.) Finally the mass of oxygen is wrong. It should be $$\dfrac{7}{148.291} \times 6 \times 15.9953 = 4.53$$ // not sure about how many significant figures should be in the answer. I'll assume 7.00 grams of the magnesium nitrate.


text:
Silver Tarnish and Reactivity Series
I read that silver in the presence of hydrogen sulfide corrodes to form silver sulfide and hydrogen.
$$\ce{Ag + H2S -> Ag2S + H2}$$
But in the reactivity series silver is placed much below hydrogen.
So, how is this displacement reaction taking place at all?
Reference:
Silver to Black - and Back. JCE
Single Replacement Reactions. Chemistry LibreTexts.

Are you aware the reactivity serie applies to hydrated ions ?

I don't quite believe this happens. Oxygen is a part of the process too.

@Shub Please note ChemSE uses mhchem to ease writing chemical equations / formulae.  It functions well in questions, answers, comments; but because it is something special, you should not use it to format the title of a question.  Its scope is described on https://mhchem.github.io/MathJax-mhchem/.

Though this is looked at as a displacement reaction, this reaction here is more of an redox reaction. Let me elaborate..
(Ag) is less reactive than (H) in reactivity series meaning (Ag) can't displace (H). But when you look at it as redox reaction,
2Ag + H2S → Ag2S + H2
Mechanism:

Ag atoms reacts with H2S gas

Ag atoms undergo oxidation (lose electrons) and becomes Silver ions (Ag+).

The electrons lost by Ag are gained (reduced) by Sulphur atoms in H2S and becomes Sulphide ions (S2-).

The Ag+ ions and S2- ions react to form Ag2S, which is deposited as a dark layer on the surface. This is the process of tarnishing.

So, in this reaction, Ag does not displace H. Instead, Ag+ ions react with S2- ions to form Ag2S, leading to the tarnishing of silver.

Consider formatting guides for texts and formulas/equations/expressions, including mhchem  \ce{} specifics.

thank you sir  .

It is rather $\ce{4 Ag + 2 H2S + O2 -> 2 Ag2S + 2 H2O}$, similarly as $\ce{2 Cu + 4 CH3COOH + O2 -> 2 (CH3COO)2Cu + 2 H2O}$

Cu and Ag will react with H2S in an anaerobic setting because of the extreme insolubility of the sulfides. H2S reacts with the copper in Monel metal rather than the nickel.

@jimchmst that preference may be a kinetic one. Copper assumes an oxidation state of +1 in the sulfide versus nickel +2, and the lower oxidation state compound tends to form faster due to lower charge ions being less tightly held and diffusing faster.

We do indeed have an exothermic reaction between hydrogen sulfide and silver, which goes against the usual EMF series. The EMF series applies specifically to species in water solution, so we should not be surprised that it does not hold when we go to gas-phase reactions (even with steam, let alone with hydrogen sulfide).
We compare the standard enthalpy changes associated with oxidation of silver by water vapor and by hydrogen sulfide (technically, we should be using free energies but the enthalpies make the point and are easier to come by). Using the data from this table:
$\ce{2Ag + H2O -> Ag2O + H2}; \Delta H  =+210.7\text{ kJ/mol}$
$\ce{2Ag + H2S -> Ag2S + H2}; \Delta H  =-11.2\text{ kJ/mol}$
The reaction with water comes nowhere near being thermodynamically favorable, whereas the reaction with hydrogen sulfide releases energy and thus may be favored.
This selective reaction of silver with hydrogen sulfide may be explained in tetms of molecular-orbital structure. Sulfur atoms are large with valence orbitals having multiple nodes, which leads to poor covalent overlap and thus weak bonding with hydrogen. Therefore some metals that cannot break the hydrogen-oxygen bonds of water, or do so only with difficulty, may break the hydrogen-sulfur bonds of hydrogen sulfide. Silver is one of several late transition and post-transition metals with this selectivity; these are the chalcophile metals that often occur in nature as sulfides.

At least 9 out of 10 downvotes come with no alternatives or suggestions for improvements. Here's your chance to stand out from the crowd.


text:
When MnO2 acts as a catalyst to speed up the decomposition of H2O2 what does it actually do?
Everywhere I look it just states that $\ce{MnO2}$ acts as a catalyst but I have yet to find a simple explanation for what it actually does. What is the mechanism involved in simple terms?

Decomposition of H2O2 is catalyzed by pretty much everything.

It's heterogeneous catalyst - surface chemistry isn't well understood.

related https://chemistry.stackexchange.com/questions/131904/what-is-the-rate-expression-for-decomposition-of-ceh2o2-in-the-presence-of


text:
What is the change in enthalpy for an adiabatic expansion?
I am just wondering if ∆H can be equal to 2∆U for an adiabatic expansion as q=0 in adiabatic process so ∆U would be equal to work done for changing volume and ∆H=∆U+work done in expansion.

First of all, in adiabatic expansion, $\Delta U=-\ work$ (according to your sign convention).  Secondly, $\Delta H=\Delta U+\Delta (PV)$, not $\Delta H=\Delta U+\ work$.  But, Work = $\Delta (PV)$ only if P is constant (equal to the starting gas pressure), which can't be the case because you have to drop the pressure on the gas to get to expand.


text:
Detection limit for log-log calibration curves
The limit of detection ($LOD$) for an analytical method can be derived from its calibration curve using the following equation:
$$ LOD = {\dfrac{(3.3 \cdot {\sigma_
\mathrm{resids}})} {m}} \tag{1} $$ [see reference at bottom]
where:

$ \sigma_\mathrm{resids} $ is the standard deviation of the residuals of the regression fit
$m$ is the slope of the regression curve.

Would the above equation still apply if I were to apply it to a calibration curve generated through linear regression of $\ln(y)$ on $\ln(x)$?
Here's an example data set that I generated (artifically - hence apparent rounding errors) to reflect what would be seen in some electrospray ionisation-mass spectrometry data (non-linear and heteroscedastic):
$$\begin{array}{|c|c|c|c|}
\hline
\text{Calibrant} & x \ (\pu{mg/L}) & y \text{ (peak area, AU)} & \ln(x) & \ln(y) \\ \hline
\text{S1} & 1 & 550 & 0 & 6.309918 \\ \hline
\text{S2} & 1 & 500 & 0 & 6.214608 \\ \hline
\text{S3} & 1 & 450 & 0 & 6.109248 \\ \hline
\text{S4} & 2 & 1100 & 0.693147 & 7.003065 \\ \hline
\text{S5} & 2 & 1200 & 0.693147 & 7.090077 \\ \hline
\text{S6} & 2 & 1300 & 0.693147 & 7.17012 \\ \hline
\text{S7} & 10 & 6500 & 2.302585 & 8.779557 \\ \hline
\text{S8} & 10 & 7000 & 2.302585 & 8.853665 \\ \hline
\text{S9} & 10 & 7500 & 2.302585 & 8.922658 \\ \hline
\text{S10} & 50 & 40000 & 3.912023 & 10.59663 \\ \hline
\text{S11} & 50 & 45000 & 3.912023 & 10.71442 \\ \hline
\text{S12} & 50 & 50000 & 3.912023 & 10.81978 \\ \hline
\text{S13} & 125 & 140000 & 4.828314 & 11.8494 \\ \hline
\text{S14} & 125 & 150000 & 4.828314 & 11.91839 \\ \hline
\text{S15} & 125 & 160000 & 4.828314 & 11.98293 \\ \hline
\text{S16} & 250 & 400000 & 5.521461 & 12.89922 \\ \hline
\text{S17} & 250 & 420000 & 5.521461 & 12.94801 \\ \hline
\text{S18} & 250 & 440000 & 5.521461 & 12.99453 \\ \hline
\text{S19} & 500 & 1000000 & 6.214608 & 13.81551 \\ \hline
\text{S20} & 500 & 1200000 & 6.214608 & 13.99783 \\ \hline
\text{S21} & 500 & 1400000 & 6.214608 & 14.15198 \\ \hline
\end{array}$$

Applying equation (1) and exponentiating the result, the $LOD$ would come out as:
$$ LOD = \exp {\left(\dfrac{3.3 \times 0.177}{1.2265} \right) } = \pu{1.61 mg/L} $$
The strange thing for me is that if I multiply the original concentrations ($x$) by 1000, say to convert to from $\pu{mg/L}$ to $\pu{\mu g/L}$,and then generate the calibration curve of $\ln(y)$ on $\ln(x)$, the $LOD$ comes out at exactly the same value, i.e. 1.61. This is because the slope and the standard deviation of the regression fit are unchanged by multiplying the concentration by 1000 - only the intercept value changes. Note, given that concentration axis was in $\pu{\mu g/L}$ before $\ln$-transformation, this would presumably mean an $LOD$ of $\pu{1.61 \mu g/L}$?
My suspicion is that I've missed a crucial step in calculating $LOD$ from the $\ln$-transformed calibration curve model.
Thanks for any pointers you can provide!
Reference: https://www.ema.europa.eu/en/documents/scientific-guideline/ich-q-2-r1-validation-analytical-procedures-text-methodology-step-5_en.pdf

Very interesting question. I do not think the LOD equation is valid anymore after log-log transformation. The other issue by multiplication by 1000 seems like a coincidence. What instrument are you using?

Hi M. Farooq, I'm using an LC-MS instrument that has a quadratic-type response and shows heteroscedasticity in the residuals of the regression when plotted in 'raw' form. By multiplying my concentrations by 1000, before ln-transformation, I see no change in the gradient of the slope. If I did, this approach would mostly likely work okay.

Your data at low values of x is fairly linear with y , so you could find a (probably quite good)  approximation to your limit from just this data, or better still, if it is possible, would be to add more data points in this concentration region and fit to a normal linear plot.

@MRJ, What type of LCMS is this? ESI-MS, Why do you think it is fundamentally quadratic?

The ln() transformation functionally changes the weighting of the error at each point. In the linear regression, you assume that the magnitudes of the errors (ie the residuals) are independent of the true value of y. That is, a deviation of 0.1 from a true value of 0.1 is the same as a deviation of 0.1 from a true value of 1. With the ln() function, this is no longer true, since an 0.1 deviation from 0.1 results in a larger deviation of the ln function than an 0.1 deviation from 1. In some cases, this might in fact be a more accurate representation of the error of your machine.

I can add some points for the OP which are too long to serve as comments. I have talked to a couple of leading persons who work with a lot a of LODs. The first lesson, which is important, that there is no "the limit of a detection" of an analytical procedure. Second, some analytical chemists also state that it should be abolished because of the same reason- there is no the limit of detection. Limit of detection varies from method to method of calculation. You will find a couple of common things in legally accepted methods for LODs as the German DIN-32645, and ISO-BS 11843-2:2000 (Incorporating corrigendum October 2007). They all deal with linear calibration curves and no one want to mess around with quadratic or log transformed calibration curves. In the literature, in my opinion, LOD is the most incorrectly reported number and the experts agree with this opinion in personal conversations.
Now quadratic calibration curves are completely kosher. They are being used in atomic absorption spectrophotometers for ages (~ 40 years). Similarly, another HPLC detector based on light scattering does a small manipulation that makes the calibration curve non-linear. These detectors are used in pharmaceutical industry all the time. So, if your sole purpose is to do quantification, within 1 to 500 mg/L, this calibration curve should serve you well but do not calculate the LODs using the formula 3.3xsigma/slope. The problem is the determination of sigma, and the multiplying factor. When Kaiser developed these formulae (mostly in German papers), he never bothered to say anything about nonlinearity because nobody wanted to work with non-linear calibration curves before the computers.
The log transformation as suggested in the comments changes the noise behavior. If you really wish to determine a number called LOD for this instrument you can do the following (for the satisfaction of a novice reviewer of your paper or another higher up who wants a LOD number from you).
P.S. Do not round off concentrations and areas, the instrument readings look simulated to me because they are only changing in units of 50 and 100. Round off is a bad idea, it should only be done at the end.
a) Check the linearity range of your method. I plotted the data up to 50 mg/L. It is linear.
b) Obtain the readings in the linear range in triplicate or quadruplicate. You already have this data.
c) Look at the standard deviation of peak area vs. concentration curve. As you can see it is not a flat line (slope of 0). Hence the Kaiser's formula fails right here. You cannot apply 3.3*sigma/slope formula. The noise is heteroscedastic. Kaiser's formula is valid for homoscedastic noise.
Suggestions to obtain an LOD of your method.
d) Make about 7-8 standards in the range 0.1 mg/L to 20 mg/L. Repeat the process a to c. What we need is a standard near the estimated LOD.
e) If you still see a non-zero slope for the peak area standard deviation vs. concentration, apply a weighted least squares method. ISO-BS 11843-2:2000 has all the solved examples.

Your answer is very interesting. But it makes me wonder if you understand the meaning of a LOD, or if you just question its utility? To me the meaning is simple. Too small a signal (here <3.3 sigmas) can't be distinguished from random noise. Assuming a linear fit is appropriate and noise is uniform (no heterosc), do you see an inherent problem with application of the equation in the OP? To me it seems simple and intuitive. It seems to me the biggest problem the OP is facing is the nonlinear nature of his problem, but for application under ideal conditions the LOD computation seems just fine.

So determining LOD is simple: determine a function for the noise, establish if it depends on the concentration of sample, and determine a function for the signal.

Buck Thorn, Well, I wish it were that easy. You will have to talk to experts to fully understand the intricacies of correct LOD and the philosophy behind it. I have had detailed discussions with people who have written an entire book on the limit of detection or written at least 20-30 articles on this topic. The deeper you are exposed...the lesser you feel to know. I feel I now know very little about LODs. However, I now refuse the accept the typical undergraduate spam taught in standard analytical textbooks.

The problem is the behavior of noise at very low concentrations. Second question is how do you define noise? Is it the noise in the baseline (the way chromatographers accept it) or is it the noise in the signal (the way spectroscopists define it).

I can understand that things can get quite complicated but I think that questioning the meaningfulness of LOD is unnecessary (if interesting).

It's an extensive field, but a quick dip might be useful: https://www.eflm.eu/files/efcc/Zagreb-Theodorsson_2.pdf

LOD is one of the most hotly debated topic in analytical chemistry. The debate will never end. I will quote a sentence from a speaker in a conference, that analytical chemists do not know enough statistics and statisticians do not know enough analytical chemistry. My contacts who are cited couple of times in your linked pdf agree with that the way undergraduate LOD concept is taught is garbage.

@M.Farooq I have edited the question to clarify the instrument type and the fact the data were artificially generated (hence apparent round errors).

@M.Farooq In response to your answer/comments: 1) I'm not against quadratic models and/or weighted least squares regressions. Rather, I'm simply interested in how one could use log-transformation as an alternative approach and still derive an LOD by equation from the model. 2) Thank you for your LOD calculation suggestions. Option one (d) is laborious/time consuming, but I understand that it would allow us to hone in on a 'real world' LOD.
More generally, it seems the lack of consensus among standards (ISOs/regulatory documents) for LOD calculation is what renders the value so contentious.

@MRJ, I was also interested in your original question long time ago, I can promise that dealing with quadratic equations and log transformations will be horribly complex. The Kaiser's formula cannot be used after log transformation.

TL;DR
No, you can't use your LOD equation to estimate LOD for a log-log regression.
Dissecting the LOD equation and it's meaning via dimensional analysis
The version you give is $\textrm{LOD} = \frac{3.3 \sigma}{m}$.  Here, the $m$ parameter is in units of "peak area units per concentration unit", or (AU-L / mg).  The $\sigma$ parameter has units of $y$, or peak area units, or "AU".  Thus, the $m$ parameter in this equation is converting the $\sigma$ parameter from AU units to mg/L units.
A log-log regression and its dimensional analysis
In a log-log regression, this isn't true.  Although the statistical details and handling of the uncertainties can be very different, a log-log regression is effectively modeling the response as
$$y = m x^b$$
Take the log of both sides to see why:
$$\log(y) = \log(m) + b \log(x)$$
This is is a linear regression between $\log(y)$ and $\log(x)$ with slope $b$ and intercept $\log(m)$.  However, if you look at the original model $y = m x^b$, you will see that $b$ is a nonlinearity parameter that must have no units, and $m$ is the "unit-conversion" parameter that converts from mg/L to peak area units (AU).
So in "logspace", the $\log(m)$ parameter, or the intercept of your regression, would be more relevant to use in the LOD equation.  It converts the units of uncertainty from $\log(\textrm{AU})$ to $\log(\textrm{mg/L})$.  If you do this, the "weird effect" you noted where changing the unit doesn't change the numerical result will disappear.


text:
3' terminal phosphate (p at the end of RNA sequence)
I am reading the paper:
Petersheim, M. and Turner, D.H. (1983) Base-stacking and base-pairing contributions to helix stability: thermodynamics of double-helix formation with CCGG, CCGGp, CCGGAp, ACCGGp, CCGGUp, and ACCGGUp. Biochemistry, 22, 256-263.
The p at the end of the sequence apparently stands for 3'-terminal phosphate.
What is the difference between RNA sequences with a 3' terminal phosphate and a "normal" RNA sequence? Why would they consider RNA with a 3' terminal phosphate?

This is just a guess, but maybe it's related to how oligonucleotides are made? That first phosphoramidite would have a 3' phosphorous attached, which would become a 3' phosphate upon cleavage from the support. This could be removed after synthesis, or it could be left attached.


text:
Wavelengths of colors emitted by alkali metals during flame test
While reading my NCERT chemistry book (Page number 300), I came across a table that mentions colors emitted by alkali metals and corresponding wavelengths.  It mentions that Potassium emits a violet color (as far as I know it's more close to lilac) but the corresponding wavelength written is 766.5 nm (I mean this is closer to the red region), which does not even lie in visible spectrum range. Does that mean that infrared rays are emitted and the book suggests that or is it wrong to say so?

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The NCERT textbook is partially wrong for elements after sodium. The flame colors and wavelengths do not match. Keep in mind that for alkali metals, the Bunsen burner flame shows multiple wavelengths for potassium, rubidium and cesium.  It is very easy to excite their outermost electron in such a low temperature flame to higher energy levels. My point is that the table should mention multiple lines for K, Rb and Cs.
What you textbook is showing is showing the wavelength of the strongest intensity for all elements except Cs.
The strongest lines or the so-called resonance lines for the alkali metals are
Li = 670 nm (falls in the red region)
Na = 589 nm (orange-yellow)
K  = 769 nm (deep deep red)
Rb = 780 nm and 794 nm  (still deep red)
Cs = 852 and 894 nm (near infrared, not visible to human eye)
Now you are right, K flame appears violet to us and Cs is indeed blue. As stated above, potassium salts in Bunsen flame also show
K = 404 nm (deep violet)
Now violet and red wavelengths appear somewhat lilac to us.
The same goes for Cs, the strongest emission is near infrared, but the visible spectrum has a faint line of 455 nm - this is bluish. Hence the flame appears blue. The name Cesium also comes from bluish gray. Rubidium comes from red from its red lines in the spectrum.
The source of wavelengths is a very old book by Grotrian.
The Bunsen flame spectrum of Lithium and Sodium is pure Li and Na spectrum. The sodium line is a doublet. I could not find the experimentally measured emission spectrum of K, Rb, and Cs.

You're absolutely right: the visual appearance of the potassium spectrum is a moderate purplish or lilac, but it has a very strong peak in near IR, which we don't see (but would likely show as very bright viewed with a digital camera, particularly with internal IR filter removed). See the spectra in this paper.
BTW, a digital camera clearly shows the light from IR LED remote controls, which you can sometimes, barely, see in a totally dark location. So potassium would make a good flare for IR photography.


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Comparison between Molecular Orbital Theory and Valence Bond Theory
I have been reading about introductory Molecular Orbital Theory lately. Till now, I have used Valence Bond Theory to evaluate bonding. I have a few questions about it:-

In VBT, we used the idea that bonds form when orbitals overlap. In Oxygen molecule, two bonds forms in VBT as the other orbitals were far enough to not overlap . But in MOT we say that all the orbitals overlap to form new MOs. Aren't the two statements contradictory?

In Coordination Compounds, when we study about non-classical $\pi$-acceptor ligands,we say that Bond Order of the ligand decreases when it receives electron pair from transition metal in its Anti-Bonding orbital. However, if I try to use VBT for the same, I feel that Bond Order should increase or at least remain the same as any kind of overlapping leads to stability in VBT. Where am I wrong?

Please try to explain the matter without using complex mathematics.

Roughly speaking, the two theories are all but unrelated. You don't have orbitals in VBT. Also, you don't really have bond orders in MOT. There is simply no area where they could possibly contradict each other.

The two theories necessarily describe the same sum total of electron density at any point in space, but break that sum into different components. Just as 5 and 7 are different numbers from 9 and 3, but 5+7 and 9+3 are both equal to 12.

The statement that orbitals do not overlap is wrong. They might not overlap much.Also, the idea of sharing electron density using anti-bonding orbitals is bound to confuse, I agree. It's best to think of involvement of anti-bonding orbitals resulting in high-energy MOs which, as the name suggests, reduce sharing of electron density.


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Alkalinity measured by titration after degassing of CO2
I recently received some lab results from a sample which contained a lot of CO2, which degassed before the total alkalinity was measured by titration to a pH 4.5 end point. The best analogue for this sample would be a carbonated (sparkling) mineral water, which was measured 5 min after opening the bottle.
Is the total alkalinity, which was reported from these conditions correct? From my understanding, it is not, as degassing of the CO2 leads to a higher pH, which leads to a speciation shift from CO2 to HCO3-. This actually means, that more acid is required for the titration in order to reach the end point.
I have the feeling that there is a flaw in my argumentation as this would actually mean that for many samples, the actual total alkalinity cannot be correctly determined, or that you would have to measure the amount of degassed CO2 and correct alkalinity with a speciation calculation, which is not typically done.


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Does the lamellar structure in say a copper-silver phase diagram have any unique mechanical properties?
For example, at lower amounts of silver the alloy gets stronger due to precipitation hardening, so in the composition as the wt% of silver increases how do the mechanical properties change? Both of them are FCC structure, so does the ductility remain the same as well?

If there is precipitate hardening, what happens to the ductility?

Oh, by both of them I was referring to the silver and copper respectively, not for lower amounts. Is there any precipitate hardening in a lamellar structure? I am trying to compare the mechanical properties of say 2% and the a alloy with a eutectic composition (Although now that you mention it I assume the lower amount would also have potentially lower ductility)


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Point group specific definition
Does the term only refer to molecules that have the same symmetry elements or exactly the same symmetry operations. Online I see that its the same symmetry operation however my professor's slides show only symmetry elements.
Or is it depend on the writer; Dnh refers only to symmetry elements but D3h refers to symmetry operations but they are both point groups?

Each element corresponds to each operation, so doesn't same elements imply same operations?

More precisely, each element corresponds to a set of operations. For example one $C_3$ symmetry axis implies three operations $(C_3)^0 \equiv E$, $(C_3)^1$, and $(C_3)^2$ (notation?). But yes, same symmetry elements corresponds to same symmetry operations.

Have a look at this answer https://chemistry.stackexchange.com/questions/58609/understanding-group-theory-easily-and-quickly/58613#58613. which should explain things