text:
Best regression model for a sigmoidal pattern
Students submit the data for an acid–base titration to my app.
The abscissa axis corresponds to the volume.
Along the ordinate axis are the pH values.
The curve I’m looking for must work automatically for any experimental data I receive.
The logistic regression model fits to the data points poorly:

It seems I’m looking for a sigmoid function of the form
$$y = \frac{L}{1 + \mathrm e^{k(x - x_0)}},$$
where $k$ is the steepness of the curve, and $x_0$ is the midpoint of the curve.
I guess I need something closer to a spline, but I don’t know how to compute a regression model based on a spline.
All I can find are interpolation models.
Also, I would like to be able to compute the derivative of the regression.
With a spline interpolation, I don't know how to compute the derivative over $x$ so that it appears as a function of a single variable.

There are several function that you could use,  look at the 'sigmoid' function in Wikipedia, and then look at the examples section. You will probably need to add a multiplier to fit the size of your data, and a parameter to make the curve sharper or shallower.,e.g a/(1+exp(-b*x)$ and vary a & b.

Why not to use the same function the curve theoretically follows?

Does this answer your question? Titration curve equation/function

https://chemistry.stackexchange.com/questions/117984/how-to-derive-the-conductivity-titration-curve-which-accounts-for-salt-formation

@porphyrin yes, a sigmoid function is exactly what I need. But what I'm having trouble with is finding the parameters a & b. I have to find a way to find these parameters automatically.

@Poutnik I can't because those points are experimental results that students have sent to my app.

@Mithoron thanks but not really, this question ask for a theorical equation whereas I'm looking for a regression curve that fit experimental data

You can as the function type is not dependent on experimental data, it depends only on the type of titration. It is a task for finding the minimum of the user function of sum of squares, changing function parameters.

@Poutnik the curve I'm looking for must fit the experimental points of the students, even if they've sent points that are falsy. The goal is to show the regression curve they did obtain with their data, even if it's a wrong one

That is what I advice. The least square method, not by some not related regression function, but with the true function with unknown yet parameters.

You will see from the answers below that you can fit the data to various arbitrary functions given enough parameters, but this is meaningless. What you need is a model of your experiment and fit using this equation and so extract some parameters relating the the processes involved. You will need to use non-linear least squares with your own function, available in many computer packages such as Origin or Igor.  (btw. What the arbitrary fitting functions do is, in effect, to smooth the data).

@porphyrin One of the statistical tests in crystallography is Hamilton's $R$-factor ratio test (1964ActaCryst502). Put simply, passing from a higher space group symmetry to a lower (which is always possible), you model describing a fixed number of observations uses more and more parameters. Hamilton's test tells if the larger set of parameters offers a significantly improved model (i.e., better than just by mere addition of parameters). Is there a reliable test («Is model X for a small set of 20 points better than Y?») than $r^2$ hunting alone?

@Buttonwood quite so, it is true that if you have two or more models to test against the data then you can statistically determine which is best and is the normal process, but my point is that just fitting an arbitrary function misses the point and generates no understanding from the data.

The evolving consensus appears to be that the primary issue is that of fitting titration curve data to an arbitrary sigmoidal curve rather than to the relevant titration curve equations. An entirely secondary issue is that of over-fitting in whatever curve fitting procedure is employed. The most recent comment by @porphyrin nails it: “just fitting an arbitrary function misses the point and generates no understanding from the data”. (This question, unfortunately, appears to have been the X-Y problem.)

Ideally, the regression function would need only 2 parameters for strong-strong titration and 3 parameters for strong-weak titration. And 1 is already given, so 1 or 2 parameters. This way it would not be just math for math, but parameters would have direct chemical meaning.

I'm sorry but I'll have to down vote this question: it, and all given answers are terribly misleading and show absolutely no respect for the chemistry involved. It you want to fit a couple of random points to a sigmoid function, then maths.se will be your friend. If you want to provide an actual service with your app, try to understand the chemistry involved and develop a model based on that. Polished brass isn't gold just because it shines similar in the right light. Whatever the purpose of this fit is, it looks right, while it is wrong.

It looks like your model is not complete. The current shape of your model makes the function go to zero for small values of $x$. There is nothing to indicate that the function should go to zero, instead you should expect the function to go to some constant pH. You can incorporate this by adding one more parameter $y_0$:
$$y=y_0+\frac{L}{1+\exp(-k(x-x_0))}$$
As to how to do this automatically in software: this can be easily done in a lot  of scientifically oriented programming languages, like Python, Matlab or Mathematica. Python is also free. You should be able to get this working even if you don't have a lot of programming experience. Here is an example using python (using the packages numpy and scipy, which are both used widely)
    import numpy as np
    import scipy

    def sigmoid(x, L, k, y0, x0):
        return y0 + L/( 1 + np.exp( -k*(x - x0)) )

    x = [ 0. ,  2. ,  4. ,  6. ,  8. , 10. , 12. , 14. , 16.1, 18. , 19. ,
           19.8, 20.4, 21. , 21.5, 22.8, 24. , 26.2, 27.9, 29.9]

    y = [ 2.02,  2.1 ,  2.18,  2.38,  2.38,  2.49,  2.57,  2.65,  2.97,
            3.3 ,  3.82,  4.84,  7.02,  8.85,  9.9 , 10.92, 10.92, 11.24,
           11.07, 11.15]

    fit, _ = scipy.optimize.curve_fit(
            sigmoid,
            x, y,
            p0=[10, 1, 1, 15]
            )

and here is how to plot the results
    import matplotlib.pyplot as plt
    plt.figure()
    # convert x to numpy array so you can insert it in function
    x_plot = np.linspace(0, 30)
    y_plot = sigmoid(x_plot, L=fit[0], k=fit[1], y0=fit[2], x0=fit[3])

    plt.plot(x_plot, y_plot)
    plt.scatter(x, y, color='r')

Note: the code I wrote above is not ideal, but I tried to make it as accessible as possible for someone with little programming experience.

(+1) Welcome to the chemistry stack exchange and your first upvote here for posting a helpful answer!

Thanks to Buttonwood for the data!

While the scripting is nice, the image also impressively shows that this function is a terrible fit for the data.

Some of the examples on the page EdV's comment points to are programmatic (as in: knowledge of a scripting/programming language is beneficial/a requirement) in addition to statistics.
If you are more inclined in using a GUI, fityk can be an alternative for you.  It is a freely available open source program (GitHub), running cross platform, validated against NIST's statistical reference set, and can be moderated by a script.

A coarse recuperation of your raw data with WebPlotDigitizer yielded the .csv
0.000, 2.020
1.984, 2.099
3.967, 2.178
5.991, 2.378
8.015, 2.375
9.999, 2.494
12.023, 2.573
14.006, 2.652
16.071, 2.974
17.973, 3.296
19.026, 3.822
19.755, 4.835
20.362, 7.025
21.009, 8.850
21.455, 9.904
22.750, 10.916
24.005, 10.915
26.151, 11.237
27.932, 11.072
29.875, 11.151

then imported (data -> load file, or Ctrl + O) into fityk.  Then Functions -> Functions Type -> Sigmodal to combine be points (somewhat) followed by a curve fit (Fit -> Method -> e.g. least squares Marquardt) and a Ctrl + R to launch the fit:

As you can see (final screen image, right hand side), the parameters of the fit are available to you. With its background in crystallography, the green trace provides you a help to visualize the difference between the data of your experiment vs the model's prediction.  The program is very useful for data of other origin (e.g., spectroscopy, chromatography), too.
Addition: a comment by the OP suggests the question in context of teaching.  It then can be instructive to optionally fix some/all of the fitting parameters (padlock symbol), or to manually alter them.  Departing from the same raw data and still assuming a sigmoid function:

Wojdyr, M. Fityk : A General-Purpose Peak Fitting Program. J. Appl. Crystallogr. 2010, 43, 1126–1128. doi 10.1107/S0021889810030499. (link to the author's preprint).

@EdV The answer as amended because there are occasions where the manual definition of a parameter with some constraints is more reasonable, than a totally free «refinement» of a model (borrowing some vocabulary from X-ray crystallography).

My initial comment to the OP was when I did not know the titration curve fitting context for the experimental data the OP wants to fit. So I am a bit puzzled, as apparently is @Poutnik, why the data are to be fitted with an arbitrary sigmoidal curve instead of the relevant titration curve. Even my simple answer here shows the curve for a simple acid base titration. Anyway, your answer is good and I will go back to working on a physics SE answer that is getting no love.

Thanks for your answer, this software looks great. Unfortunately I haven't been clear enough in my question. I'm building a software myself, and what I'm looking for is the way to find the equation of the curve given any set of data in an automatic way.

@krirkrirk The additional information (to identify the function type [not only the parameters of a function elected] which fits best the recorded data) changes the direction, indeed. A potential pitfall is looking on «the typical numbers» only (e.g., $r^2$) which (Anscombe's quartet is one of the better known examples) can be misleading.

I am answering to the comment that you made to the @Buttonwood's great answer.
Clearly the $y$ values are the $\text{pH}$ of the solution and the $x$ values the volume of base added. If you don't mind, I will use $y$ as the dependent variable, and $t$ as the independent variable.

1. Set of equations Our objective is to  minimize the error given by
\begin{align}
  E(\mathbf{P}) &= \sum_{i = 1}^N [\hat{y}(t_i) - y(t_i;\mathbf{P})]^2
  \tag{1} \\
\end{align}
where $i$ refers to the $i$-th experimental data, $N$ the number of points you have, $\hat{y}(t_i)$ the experimental value at time $t_i$, and $y_i(t;\mathbf{P})$ the function that tries to minimize the error evaluated at time $t_i$. The locations $t_i$'s are known to us, but we want to find the vector $\mathbf{P}$ that globally minimizes Eq. (1).
It $E$ has a minimum, then the gradient of $E$ is zero at $\mathbf{P} = (P_1, P_2, ...,P_k,..., P_m)$. Hence
\begin{align}
  \require{cancel}
  \frac{\partial E}{\partial P_k}(\mathbf{P})
  &= \sum_{i = 1}^N 2[\hat{y}(t_i) - y(t_i;\mathbf{P})]\frac{\partial [\hat{y}(t_i) - y(t_i;\mathbf{P})]}{\partial P_k} \\
  &= \sum_{i = 1}^N \cancel{2}[\hat{y}(t_i) - y(t_i;\mathbf{P})]\cancel{(-1)}
  \frac{\partial [y(t_i;\mathbf{P})]}{\partial P_k} =  0 \\
  &= \sum_{i = 1}^N [\hat{y}(t_i) - y(t_i;\mathbf{P})]
  \frac{\partial [y(t_i;\mathbf{P})]}{\partial P_k} =  0 \tag{2} \\
\end{align}
Eq. (2) can be written for the $m$ unknowns that form $\mathbf{P} \in \mathbf{R}^m$, that form a set of $m$ nonlinear equations.

2. Function proposed For the equation you proposed we can make two modifications regarding the horizontal asymptotes:

We would like that as $x \to \infty$ then $y \to L$. Thus, we add a minus sign in the exponential function.
We would like that as $x \to 0 $ then $y \to \beta \approx 2$. Thus, we add a constant term $\beta$ to the function.

\begin{equation}
 y(t; \mathbf{P}) =
 \frac{L}{1 + \exp[\mathbf{-}k(t - t_0)]} + \beta \tag{3}
\end{equation}
Eq. (3) has three parameters, $L$, $t_0$, and $k$. But I will follow your list and suppose that it has two parameters. From a practical point of view, it is clear that $L \approx 12$. This is a good decision because it diminishes by one the number of equations to be solved.
The partial derivatives of $y$ with respect to these two are immediate
\begin{align}
  \frac{\partial y(t; \mathbf{P})}{\partial k} &=
  -\frac{L}{\{1 + \exp[-k(t - t_0)]\}^2}\exp[-k(t - t_0)](-1)(t - t_0) \\
  &= \frac{L\exp[-k(t - t_0)](t - t_0)}{\{1 + \exp[-k(t - t_0)]\}^2} \tag{4} \\
  \frac{\partial y(t; \mathbf{P})}{\partial t_0} &=
  -\frac{L}{\{1 + \exp[-k(t - t_0)]\}^2}\exp[-k(t - t_0)](-k)(-1) \\
  &= -\frac{kL\exp[-k(t - t_0)]}{\{1 + \exp[-k(t - t_0)]\}^2} \tag{5}
\end{align}

3. Combining Now we combine Eq. (2) with Eq. (4) to get the first equation, and then Eq. (2) with Eq. (5) to get the second equation
\begin{align}
  \sum_{i = 1}^N \left\{\hat{y}(t_i) - \frac{L}
  {1 + \exp[-\color{blue}{k}(t_i - \color{blue}{t_0})]} - \beta\right\}
  \frac{L\exp[-\color{blue}{k}(t_i - \color{blue}{t_0})](t_i - \color{blue}{t_0})} {\{1 + \exp[-\color{blue}{k}(t_i - t_0)]\}^2} &= 0 \tag{6} \\
  \sum_{i = 1}^N \left\{\hat{y}(t_i) - \frac{L}
  {1 + \exp[-\color{blue}{k}(t_i - \color{blue}{t_0})]} - \beta\right\}
  (-1)\frac{\color{blue}{k}L\exp[-\color{blue}{k}(t_i - \color{blue}{t_0})]}
           {\{1 + \exp[-\color{blue}{k}(t_i - \color{blue}{t_0})]\}^2} &= 0 \tag{7}
\end{align}

4. Solve Eqs. (6-7) are a set of 2 non-linear equations with two unknowns, emphasized in color blue. I count $20$ experimental points in the geogebra plot you posted, so each equation has $20$ terms. This is the starting point. Although they look fearsome, one immediate way is to use Microsoft Excel. Put aside two cells, one with an initial guess of $k$ and other with $t_0$, and the next $5$ columns:

The first one has the volume of the base added.
The second one has the $\text{pH}$ values.
The third one has the function given by Eq. (3) evaluated at the rows of column 1.
The fourth one has the derivative with respect to $k$ evaluated at the rows of column 1.
The fifth one has the derivative with respect to $t_0$ evaluated at the rows of column 1.

Points 3-5 also will need the $k$ and $t_0$ you put in the cells apart. Combine these values with operations to have two final cells, each one is the value of Eqs. (6) and (7). Finally, use the solver tool to get the values of the unknowns. You set the objective function as Eq. (6) and add the restriction that Eq. (7) yields zero. The solve will vary its values to find a (hopefully) global minimum.
If you want further math just add in the comments, there are additional methods, like Gauss-Newton which can be programmed by yourself.

Looking at the plot it seems that as x approaches minus infinity y converges to around 2 and not to zero as in your example function. This can be easily adjusted by taking your general form (3) and adding a constant term. One would expect the optimal solution to give a value of around 2 for the constant term and consequently around 10 for L so that the limit for x approaching plus infinity still works. From the plot t_0 should come to around 20.5.

@quarague Wow, good catch... I had to look left and right. Thanks for the comment! I added that term as $\beta$. It is true that we could also guess $t_0$ and reduce even more the non-linear equations to be solved. I sticked to the parameters the OP listed in the bullet points, so I retained it.

The best regression after (automatically) surveying your data is Pearson IV with five parameters (a, b, c, d, e). Thanks to Buttonwood for digitizing your data.
$$
y=a \frac{\left[1+\frac{\left(x-\frac{c e}{2 d}-b\right)^2}{c^2}\right]^{-d} \exp \left[-e\left(\tan ^{-1}\left(\frac{x-\frac{c e}{2 d}-b}{c}\right)+\tan ^{-1}\left(\frac{e}{2 d}\right)\right]\right]}{\left(1+\frac{e^2}{4 d^2}\right)^{-d}}
$$
The $r^2$ value is 0.99909 and fits the curve very nicely! Now note that the finding a perfectly fitting equation does not reflect reality. It is just a function that fits given data the best within this range of x-y values that you provided.
Systat already sells such automatic curve and peak fitting software.

You looked through a large number of possible curves and then picked a 5-parameter family to model 20 points. Unless you can find a good reason from the context or source of the data that explains why this is an appropriate model this looks like a textbook example of overfitting.

Well, read my disclaimer in the answer-fitting does not reflect reality or a real model. I have been working with the developer of this software for other types of peaks and I have asked this question several times about overfitting. His answer is that  one should always look at the statistical significance of the fit parameters.

The generalised logistic function might be a good fit for your data. You will probably have to test the effects of each parameter to determine whether you need them to model your data.
At first glance, I would propose:
$$ y = \dfrac{k_2}{\left(1+\exp{\left(-\dfrac{k_3}{k_1}\left(H - k_4\right)\right)}\right)^{\dfrac{1}{k_5}}} + k_1 x + k_0 $$

$k_0$ is the offset at $x = 0$
$k_1$ is the slope far away from the transition
$k_2$ is how high the curve jumps
$k_3$ is the slope during the transition
$k_4$ is where the slope is
$k_5$ is the symmetry of the curve as it goes into and leaves the transition

A value $>1$ will make the end of the transition sharper than the beginning and vice versa.
Beware! This makes the other parameters non-linear in their behavior, it gets worse the farther this value is away from 1.

I had a very similar data set, which formed of a hystersis loop. Some minor changes to the generalised form gave me an analytical expression which fit the data well. For reference, I have included my model and an image indicating the parameters influence below.
$$ B_{\genfrac{}{}{0pt}{2}{\mathrm{lower}}{\mathrm{upper}}} = \dfrac{\pm 2 k_1}{\left(1+\exp{\left(-\dfrac{k_3}{k_1}\left(\pm H - k_4\right)\right)}\right)^{\dfrac{1}{k_6}}} \mp k_1 + \mu_0 H $$

The derivative is a bit disgusting, but it serves its purpose.
$$ \dot{B}_{\genfrac{}{}{0pt}{2}{\mathrm{lower}}{\mathrm{upper}}} = \left(\dfrac{2k_3\cdot\exp{\left(-\dfrac{k_3}{k_1}\left(\pm H - k_4\right)\right)}}{k_6\left(1+\exp{\left(-\dfrac{k_3}{k_1}\left(\pm H - k_4\right)\right)}\right)^{\dfrac{k_6+1}{k_6}}}+\mu_0\right)\cdot\dot{H} $$

Use the R drc package which was created to solve your problem exactly: sigmoid regression; its acronym stands for Dose Response Curve. It is a well-established package backed up by scientific publications.
The package includes many forms of sigmoid curves as detailed in the documentation (drc package documentation with examples).

To prevent ambiguity: do you refer to the R package of said name (source)? If so, name the source, add a link.  With a manual of 149 pages, what is the page (or better: what is the specific function within this package) you have in mind to address the problem? The addition of the name can serve as an anchor.

You are right. Information added.


text:
Is preferential oxidation/reduction at electrodes only dependent on the reduction potential or is it also dependent on kinetics?
It is commonly taught that in electrochemical cells where there are 2 different chemical species that can be oxidised/reduced, species with more positive actual reduction potentials (concentration being considered as well as standard reduction potential) are preferentially reduced at cathodes, and species with more negative actual reduction potentials are preferentially oxidised at anodes.
It appears to me that, theoretically speaking, this need not always be the case, since reduction potentials predict thermodynamics, not kinetics. Therefore, it seems plausible that a case could arise where: 2 species are present at the anode of an electrolytic cell; one species has a more negative reduction potential, but for some reason (charge, hydration shell etc) has much slower oxidation kinetics compared to the other species, which has a more positive reduction potential but has much more rapid kinetics, causing the species with a more positive reduction potential to be oxidised preferentially.
Is this reasoning flawed, and if not, is there such a case in real life experiments?

You more or less reinvented the nature of polarography. If the rate of reaction kinetics or diffusion for the "first choice" is slow enough, it allows electrode potential to be externally raised or lowered enough to start the less preferred choice.

@Poutnik Meh, looks like he only got next to overpotential.

BTW OP should make clear it's about kinetics already in the title, not end of the body of question.

That's my bad, will edit, thanks

The reasoning is fine, there are huge amounts of examples. For anodic reactions one can think about how a lot of metals actually form chlorine rather than oxygen in aqueous sodium chloride solutions. For reduction reactions there are a lot of examples that are currently hot topic, like CO2 reduction to a lot of species which some are thermodynamically preferred over hydrogen gas formation. Same with ammonia formation from nitrogen etc.

Thanks everyone, will read up on the mentioned concepts

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sort: votes


text:
Speed of substitution reactions
What is the order of speed of SN1 and SN2 for the following molecules?

My idea:
For SN1 I think it is C>A>B>D
C - the carbocation has resonance
A - I believe the double bond acts as an electron donating group and helps stabilize the positive charge on the carbocation.
B - just a secondary carbocation
D - vinylic carbocation are highly unstable
For SN2 I think it is the same order C>A>B>D
C - the allyl group is completely planer and as a result it allows the nucleophile an easier approach to the alpha carbon
Honesty I'm not confident between B and A
A - while the CH2=CHCH2- group isnt planer I think the double bond allows for a more linear geometry of the group and a more condensed electron cloud which results in a smaller steric interference and an easier approach for the nucleophile.
B - all carbons in the propyl group are sp³ hybridization therefore the electron cloud is less linear and condensed resulting in a larger steric interference than compound A
D - vinylic carbons dont react in substitution since the nucleophile has to go through the plane of the electron dense double bond to attack the alpha carbon
What do you guys think is this correct?

Compare the acidity of 3-Butenoic acid, (E)-2-Butenoic acid, and Butanoic acid. What do you think?

I think your rankings are all correct other than for SN2 I would say B>A and most of the rationale is as well.
In both cases C is definitely fastest, both because of the reasons that you gave but also due to the SN1’ and SN2’ mechanisms where the nucleophile attacks the double directly and the double bond shifts to either kick off the bromine or donate to the carbocation.
D definitely slowest in both cases for the reasons you gave.
For the SN2 I would have B>A just because the weaker rotation barriers will minimise steric interactions. When one considers the transition state in the SN2 this is reasonably considerable as the A transition state will be appreciably higher in energy

Thanks a lot for answering!.

Actually there is something I don't understand, I thought the transitional state in SN2 mechanism has a trigonal bypyramid geometry which is was what I based my explanation upon. Can you try giving me a better insight on the mechanism and nucleophile attack since I'm not quite sure I understand how the double bond on C helps pushing the leaving group. Also another thing,  you say A is only faster than B only in SN1 mechanism because in SN2 the double bond limits the movement of the substitutes which results in more steric interference? This is all very interesting I really appreciate the help!

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text:
A silly doubt on the concept of volume percent and concentration
I am trying to solve the following elementary problem of volume percent, however, I am not quite sure where my mistake is:
Ethanol has a volume-by-volume concentration of 93.3%. Calculate the volume of water in $\pu{10 L}$ of this alcohol.
My Attempts:
By definition,
$$\text{volume concentration} = \frac{\text{volume of solute}}{\text{volume of solution}} \times 100\%$$
From the exercise, we know that $\text{volume concentration} =93.3\%$ and $\text{volume of solute} = \pu{10 L}$. By substituting those values in the equation above, we then have
$$93.3\%=\frac{\pu{10 L}} {\text{volume of solution}} \times 100\%$$
Working forward from the last equation and solving it for the volume of the solution, we find
$\text{volume of solution} = \pu{10.71 L}$. Therefore, the volume of water or the volume of solvent becomes
$$\text{volume of solution} = \text{volume of solute} + \text{volume of solvent}$$
$$ \pu{10.71 L} = \pu{10 L} + \text{volume of solvent}$$
Therefore, $\text{volume of solvent} = \pu{0.71 L}$.
However, the actual answer to this problem is $\pu{667 g}$.
Based on the above, what is my mistake here?

For future reference: for the body of questions, answers and comments, chemistry.se allows to use mhchem.  In addition to assistance typing chemical formulae, the \pu{} command allows to type values with units including a non-breakable space (and units typically are upright, too).

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

First of all it is a stupid problem.

Volumes don't add. So 90.00 ml of ethanol and 10.00 ml of water doesn't yield 100.00 ml of solution.
The problem messes up significant figures. 10 L and 93.3% ethanol in the problem statement suddenly become 10.000 liters and 93.33% in the problem solution.
The problem statement is given in terms of volumes, but the answer for water is given in grams.

So the problem assumes that volumes do add and that the final solution (a mixture of ethanol and water) has a volume of 10.000 liters. Since the solution is 93.3% ethanol, 6.7% is water. So about 670 ml of water. Now further assuming that water is 1 gram per ml (since no temperature is given) that means that there is about 670 grams of water.

Your mistake - In your second equation the 10 L should be in the denominator not the numerator.

Well, does "10 L of the alcohol" mean 10 L of the impure liquid, or the volume of liquid that contains 10 L of pure alcohol after separation? I agree with "stupid problem".

@Karsten - LOL In order to get the "right" answer it must be 10 L of the ethanol/water mixture. // Getting crotchety in my old age. I think problems out to adhere to the real science. The problems are supposed to be teaching chemistry too.

Yes, they ought to.


text:
Is the liquid in this question water? How can one know for sure?
I have been studying and there is a question in my book asking:

A faulty thermometer reads freezing and boiling point of a liquid as -5°C and 95°C respectively on the Celsius scale. What is the correct value of temperature when it reads 60°C on the faulty thermometer?
(a) 60°C (b) 65°C (c) 64°C (d) 62°C

The solution on back of the textbook is as follows:

(b) 65°C

For all temperature scales,
(X - LFP)/(UFP - LFP) = constant
where LFP is Lower fixed point, and UFP is Upper fixed point
[X - (-5)]/[95 - (-5)] = (C - 0)/(100 - 0)
=> (60 + 5)/(95 + 5) = C/100
=> C = 65°C

In this, I do not understand from where the C - 0 and 100 - 0 came in step 2
EDIT: Fixed typo from -95°C to 95°C

It seems it is not only a faulty termometer, but also faulty assignement. there should be -5  deg C and +95  deg C, with the correction of adding +5 deg C, probably due accidental shifting of the scale bar. Therefore 60->65, as the liquid is obviously  water. C-0 and 100-0 comes from mapping function  -5..+95 to 0..100.

I think you're getting "the lamest title of the day" prize... but the day's still young here ;)

Please change the title of question to reflect what is being asked in the body of the post.

I’m voting to close this question because it contains typos that should be fixed (or if the textbook problem had the typo, include the source). In fact, always include the source if you cite something.

@Poutnik I'm sorry, i accidently wrote -95 instead of 95 degrees.

@Karsten can you please point out the typos? I can't find them. If you mean the formula, I've double checked it and that's exactly what the textbook said

Also, if not chemistry stack exchange, which site can i post this question to?

@InfoDaneMent You fixed the typo. Did your version of the question state that the sample is water? Could you add which textbook you found this in? You can find similar problems and answers here, here, and there.

@Karsten but how can it be assumed that the correct readings are 100 and 0 degree Celsius?

@Karsten no, it isn't written in the question anywhere that the liquid is water.

@InfoDaneMent please edit the question to state    where the problem set comes from.

@Karsten you mean the textbook name and version?

Indeed a faulty question and data as pointed out by Poutnik. Instead of "a" liquid, it should be water. Don't use plug and chug formulae. When it is known that the liquid is water, recognize the fact that the freezing point is -5 °C, and the +95 °C for the boiling point on this thermometer. This statement is to show that the systematic error is constant in this temperature range. What do you have to do to make -5 °C to 0 °C and +95 °C to 100 °C? Add +5 °C. Make this correction with this understanding in observed 60 °C and you get the textbook answer.

The question is nonsense, Chemically to assume a liquid is water is foolish, dangerous, and potentially fatal. A typo is the responsibility of the typist not the reader. The scale is inverted so a 60 degree reading is colder than the FP T; no listed answer can be correct.  All that is needed to fix the question is one defined Kelvin temperature measurement  [the BP of water was -95 at ~600Torr. =368 K or the ice point 273K less variable] That with absolute zero  defines the absolute scale, a translation defines the new scale. The student who figures it out learns something.

@jimchmst Other cases than constant offset are highly improbable, so take it as educated guess.

But the question does not mention that the liquid is water

@InfoDaneMent, That is why we said that it was a poorly designed question by the question author (not you!).


text:
Interpreting NMR: what does it mean when there are two signals when there should be (and looks like) only one, and integration where there is no peak?
A couple of basic questions.
So in this spectrum, there seem to be two peaks at 2.38-2.347, but this signal should be a singlet and correspond to the methyl group. So does this mean anything?
Another question is, there is an integration of 1.02 at around 2.00-1.50, but there is no peak. Could this mean impurities, or do I just ignore?

I rolled back the edit because the question (and the answer) makes little sense without being able to see the spectrum. It would be nice to specify what instrument was used (e.g. 300 MHz) and what the solvent was, if any.

It's impossible to see from your spectrum whether there's one or two peaks. You need to zoom in.
But I get the sense you're referring to the 'peaks' at the top (i.e. the numbers) which are chosen by automatic peak picking, I wouldn't worry about them: the peak picking algorithms can often be over-sensitive and report many 'peaks' that aren't really peaks.
As for the broad peak at around 1.6 ppm, that's probably just water, nothing to be concerned about. You can find a very thorough list of common impurities and their NMR chemical shifts at: Organometallics 2010, 29 (9), 2176–2179. (By the way, just because the peak picking algorithm didn't tell you there was a peak there, doesn't mean that there isn't one. Look at the spectrum, not the numbers on top; there are certainly peaks in that region.)
There are impurities in your product, for sure — you can see extra peaks at the baseline — but nothing too significant.

The only other thing it's perhaps worth mentioning is that while predicted peak locations from a structure given by e.g. chemdraw are usually "about right", they are decidedly not completely right, particularly where labile protons are involved...


text:
Relative magnitude of the crystal field and spin-orbit coupling
In the EPR part of the book 'Physical methods in Inorganic Chemistry' by S. Drago i found that relative magnitudes of crystal field and spin-orbit coupling affect the g values.
I understand that the chemical environment of a metal complex partly removes the orbital degeneracy, but i don't find the way to imagine a sustaining effect which implies that if an electron has an orbital angular momentum this is manteined by coupling to the spin angular momentum. I have read that consequently a net orbital magnetic moment results, giving a g value different from the free electron.
I also can't imagine how when the crystal field increases the LS coupling is broken down completely.


text:
Reactivity towards electrophilic addition reaction among unsaturated cyclohexane derivatives
I am a class 8th student preparing for chemistry olympiad. I am having slight confusion regarding how to decide the reactivity towards electrophilic addition reactions. What should we give more priority to? To the stability of carbocations formed, or the electron density on the substrate?

In this question if we finally add an electrophile and check the stability of cation using hyperconjugation effect.
b = d > a > c seems to be the answer, but the solution says a > b due to the electron density because of the hyperconjugation effect on both sides of the π-bond which increases the electron density.

I am still slightly confused by this question. It would be great so see the actual wording and what is asked. Could you cite the source of it, please. I'd guess it's part of a preparation kit of some kind.

The reaction rate depends on the energy difference between the state of the substrate and the transition state. The energy of the transition state is close to that of the carbocation but not quite so there can be subtle surprises.
My first blush predictions are: a, d, b, c
Reasoning: a. extra methyl stabilizes both ground and transition states but should have greater effect on transition state. Also entropy favored 2 sites d. Same intermediate as b; the transition state has lower ring interactions. Exo double bond is of higher energy [I think; I did not cheat and look it up]. Models might help here. b. tertiary cation.  c. secondary cation higher energy.
Both ground state and transition state must be considered. The classic example of this is benzene that introduced the new concept of resonance stabilizing the ground state.

The reputable source here is the original data giving the rates etc. of these compounds under SN1 conditions., not the answer on an answer sheet to a MC question.

I agree with the poster above. A, D, B, C seems to be the correct order. The logic behind this answer is that A gets more inductive donation from the methyl groups across the double bond. For D, the double bond is less substituted, so more unstable and more prone to reactivity. For B, inductive donation from the methyl group. For C, it does not receive any inductive donation at all, and it's quite substituted, so it is stable as is.

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center.

It is the responsibility of the framer of the test question to support the answer with data and citations. The real question about considering ground state or transition state or intermediate energies is much deeper than the "corr

@jimchmst I guess you wanted to write some more.


text:
Is it possible to have multiple nitrogen atoms bonded to one carbon atom?
So I was wondering, if we replaced two of the hydrogens in methane with amine groups, would it be a stable compound? What about for R groups of varying size?
Similarly, if I tried to make a compound with a terminal acetal except I now replace the hydrogen at the terminal carbon with another -OR group, would that be stable? Or is it too much steric hindrance?

Addressing the title: guanidine NH=C(NH2)2 and urea CO(NH2)2

Uhh, tetranitromethane?

also amidines are a thing

also ortho-esters are a thing too, I think that answers your question about a carbon bonded to three oxygens.  also maybe carbonate esters.

Title should be rather something like "Stability of aminals"...

A carbon atom can be bonded to as many as four nitrogen atoms, with those nitrogen atoms bonded to additional alkyl groups. Tetrakis(dimethylamino)methane, $\ce{C(N(CH3)2)4}$, has been known since 1966 [1]. The compound is stable enough to allow finding a reported melting point of 122.5-124.5°C.
Alkylating the nitrogen atoms, at least with methyl groups as above, actually gives a more stable compound than would be obtained with $\ce{N–H}$ bonds despite the steric appearances. With the alkylation, possible decomposition by labile exchange of protons (liberating ammonia) would be excluded.
Reference
Harold Weingarten and William A. White (1966). "Preparation of Tetrakis(dimethylamino)methane". J. Am. Chem. Soc. 88, 12, 2885.
https://doi.org/10.1021/ja00964a071


text:
Specific question about the total synthesis of B-12
In the process of synthesizing B-12, Woodward & Co. produce an intermediate thioether they refer to as Type I. "Type I because we found early in working with these substances that they were very labile, and undergo quite a plethora of changes. [Type I is prone] to an isomerization in which the C-N double bond is transformed into a C-C double bond to give thioether Type II." I may be missing something basic here, but what's behind the ready conversion of Type I into Type II? (I haven't taken o-chem in a while and am reading out of curiosity.) Also, does he mean that as soon as he produces Type I it starts converting into Type II, almost as if, practically, Type I doesn't exist?

Generally imines are a bit more favorable then enamines. It's technically thioimidoester, (imine+thioether at one carbon), though, and apparently tautomerisation equilibrium is somewhat shifted.


text:
Why is iron(III) more stable than iron(II)?
What I have seen, and what has been answered before (Why Fe 3+ is more common than Fe 2+), is that it is due to the half filled d orbital of $\ce{Fe^3+}$ compared to $\ce{Fe^2+}$. I don't understand why exactly this half-filled d-orbital is more stable.
The most obvious thing to me would be to minimise electron pair repulsion, but this both seems both too simple an answer, and also doesn't explain why elements like copper would actively promote 4s electrons to 3d orbital (which suggests some extra stability besides negating repulsion).
Are the reasons why copper promotes electrons to 3d and the reason iron prefers to be $\ce{Fe^3+}$ due to different reasons, or due to the same effect that I haven't properly understood?

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

$\ce{Fe^3+}$ is more vulnerable to hydrolysis than $\ce{Fe^2+}$ and has oxidative properties, e.g. oxidates $\ce{Cu}$ to $\ce{Cu^2+}$

The reason iron III is more common than iron II is because the atmosphere contains sufficient O2 and there is sufficient liquid water present to oxidize ferrous to ferric. The reason the potentials are just right resides in the electronic structure of the elements and compounds.

The reactions with copper are interesting. Ferric oxidizes copper; Ferrous does not. This means in a similar situation ferric is less stable than ferrous as indicated by electrode potential etc.  Stability evaluation requires a complete equation. Ferrous and ferric ions are each perfectly stable until one electron is involved.

When considering stability, you first think in terms of electron orbitals.
The “d” orbitals have five degenerate orbitals and can accommodate up to 10 electrons.
Fe3+ has 23 electrons and Fe2+ has 24electrons.
When electrons are accommodated in d orbitals, up to five orbitals are empty, so electrons are placed into separate orbitals one by one according to the Hund's rule, but after the sixth orbital, electrons can only be placed into the orbital that already contains an electron.
When two electrons are placed in one orbital, the electrostatic repulsion between electrons is greater than when electrons are distributed in separate orbitals, so the stability of the divalent iron ion is lower than that of the trivalent iron ion.
This is Hund’s rules in wikipedia English.
https://en.m.wikipedia.org/wiki/Hund%27s_rules

Each progressive loss of an electron requires more energy.


text:
Does this reaction that make an Alum happen?
One day I asked a question in high school.
The problem is “Show the reaction that occurs when potassium permanganate is added to dilute sulfuric ferrous sulfateⅡ.
The answer is $\ce{2KMnO4 + 8H2SO4 + 10FeSO4
-> 2MnSO4 + 8H2O + 5Fe2(SO4)3 + K2SO4}$
I know this reaction because I have studied oxidation reaction.
But I have a question.
The reaction that  $$\ce{KMnO4 + 4H2O + 5FeSO4 + 8H2O -> MnSO4 + FeK(SO4)2・12H2O + 2Fe2(SO4)3}$$ (like a Alum) is happen?
The reason this question arose is that I can think of combinations that are alum when each becomes an ion after iron oxidizes.
Would you mind solving my question?

How can you distress a question? If we find out maybe we can scare off some of them ;)

More to the point, all products are dissolved. If you tried to crystallise some of the stuff, then we could talk.

Thanks to see my question.The textbook only written the upper  reaction equation and not written the equation that I think to produce alum(FeK(SO4)2) that I thought of.
I wondered if alum(FeK(SO4)2) can be made.

Yes, I know that all material is ion in water.
But in the reaction equation, wouldn't you write it without it being an ion?
For example, although Na+ Cl- in water, HCl+NaOH->NaCl+H2O in a neut.
I was thought that if there is an ion, it is a half-reactive equation.

BTW, alum is a specific example of a double salt, https://en.wikipedia.org/wiki/Double_salt . Since no aluminum was involved, you might use the general term.

Thanks for looking DrMoishe Pippik.
My textbook said that anything that can be represented as M1M2SO4 is called alum.
And, the Double salt article on Wikipedia first says that alum can be represented by MIMIII(SO4)2-12H2O.
However, the alum article says it is a salt of aluminum, so it would have been better to write it as a double salt.
Thank you for correcting.

Also, I appreciate you DrMoishe Pippik for your help in correcting the reaction equation.

Thank Michael Lautman for the formula edit. As for alum vs. double salt, see https://en.wikipedia.org/wiki/Alum .

Your 2nd reaction is disbalanced, 5 sulfates on left, 9 on right.

sorry Poutnik, I mistakenly write 4H2O where 4H2SO4 was.

Note that the above reactions do not happen in the presented molecular form, but in ionic form. Dissolved anions and cations are separated and independent (safe minor ion pair formation) until the solubility product of the least soluble salt is achieved. Then the respective ions mutually combine, eventually with crystal water and form the respective least soluble solid salt.
The oxidation reaction goes like this:
$$\ce{MnO4-(aq) + 5 Fe^2+(aq) + 8 H+(aq) -> Mn^2+(aq) + 5 Fe^3+(aq) + 4 H2O(l)}$$
The remaining ions $\ce{K+(aq)}$ and $\ce{SO4^2-(aq)}$ are bystanders not taking part in the above reaction. They are therefore not explicitly added to the reaction equation.
As the next step, e.g. via evaporation, there can happen eventual crystalization of respective salts, according to their solubility products:
\begin{align}
\ce{K+(aq) + Fe^3+(aq) + 2 SO4^2-(aq) + 12 H2O(l) &<=> KFe^{III}(SO4)2 . 12 H2O(s)}\\
\ce{2 K+(aq) + SO4^2-(aq) &<=> K2SO4(s)}\\
\ce{Mn^2+(aq) + SO4^2-(aq) + $n$ H2O(l) &<=> MnSO4(s) . $n$ H2O(s)}\\
\ce{2 Fe^3+(aq) + 3 SO4^2-(aq) + $n$ H2O(l) &<=> Fe2(SO4)3 . $n$ H2O(s)}
\end{align}
The Wikipedia: Solubility table can be used for respective salts as the guidance, but it is valid for pure salts with stoichiometric rations of respective ions. An excess of one of ions (aka common ion effect), e.g. sulfate from sulfuric acid, decreases the solubility of the affected salt.

What can be gleaned from the discussion is that the most efficacious method to effect a synthesis involves the simplest combination of reactants. To make FeK[SO4]2 the reaction with permanganate could be eliminated.  The complete "molecular" equation does not necessarily depict the actual products isolated.

@jimchmst But the task is not the simplest method to synthetize the alum, but to resolve outcome of the redox reaction, regarding possible isolation of products.

My second sentence. The reaction to give the mixed salt is suspect because permanganate requires an acidic reaction to be reduced to Mn++.  The second equation fails to describe a possible reality.

I have made no claim about the validity of his 2nd reaction, which is disbalanced.

Thanks to answer my question Poutnik.So, when each salt crystallizes, there are four possible reaction equations.I understand.

I assume, but not sure, that there could be formed also Mn equivalent of the Mohr salt, i.e. K2Mn(SO4)2. 6 H2O.


text:
Enthalpy change in an isothermal experiment at constant pressure
I read, in a passage regarding isothermal and adiabatic calorimetry for determining heat of solution, that

In a truly isothermal experiment at constant pressure, $\Delta H$ of the system in the calorimeter differs from zero only because of changes in numbers of moles of various substances in the system.

I am having difficulty understanding this. Under constant pressure,
$$\Delta H = \Delta U + p \Delta V$$
However, in an isothermal process, $\Delta U = 0$. Thus, wouldn't it be the case in the example in the passage that $\Delta H = p \Delta V$?

According to the passage, $\Delta H$ is not equal to zero only when there is a change in the number of moles. Doesn't this then mean that there has to be a change in number of moles so that there can still be a change in volume, allowing for expansion work to be done? Could anyone please point out what I am misunderstanding?

Why do you think ΔU=0 for isothermal process when generally Q<>0 and W<>0 ? 1 kg of ice at 0 deg C has definitely different U than 1 kg of water at the same temperature.

That is true, but what threw me off was the statement that the change in enthalpy only differs from zero due to the change in the number of moles. I did not know how to interpret this otherwise.

It is true for ideal gases.

This assumes that you have a change of phase or a chemical reaction where U and H depend not only on temperature (in the case of ideal gas mixture, say), but also depend on the amounts of the various chemical species present.

About ice water chemical potential energy is involve not just heat transfer.

As suggested by previous comments, for an ideal gas $p \Delta V = RT \Delta n$.

@ChetMiller could you provide some further reading on this?

See Smith and van Ness, Introduction to Chemical Engineering Thermodynamics


text:
Problem adding non-polar hydrogens whilst converting from pdbqt to pdb
I'm using OpenBabel 3.1.1 and the Python to C++ bindings to load pdbqt files and convert them to pdb with explicit hydrogens. Unfortunately, I can't add nonpolar hydrogens using OpenBabel. I can do this using Autodock's PMV GUI as a test and comparison, so my concern is not the underlying molecule.
In the python bindings, I've tried combinations of AddHydrogens(), AddNewHydrogens(), AddPolarHydrogens(), AddNonPolarHydrogens(), checking the OBMol.NumAtoms() each time, and nothing changes.
In the OpenBabel GUI I've loaded the pdbqt files and tried combinations of "Add hydrogens (make explicit)", "Add hydrogens to nonpolar atoms only", and "Adds hydrogens to polar atoms only". Nothing changes the pdb output. Note: There is one polar-hydrogen present which can be removed and added back.
Ultimately, I must determine why the Python bindings aren't adding hydrogens. Using the GUI was a sanity check.
I've included for you below a copy & paste of a pdbqt file.
Many thanks for your help.
REMARK VINA RESULT:    -8.009      0.000      0.000
REMARK INTER + INTRA:         -11.923
REMARK INTER:                 -10.819
REMARK INTRA:                  -1.105
REMARK UNBOUND:                -1.105
REMARK  Name = ZINC000004125379
REMARK                            x       y       z     vdW  Elec       q    Type
REMARK                         _______ _______ _______ _____ _____    ______ ____
ROOT
ATOM      1  C   LIG    1       -7.551 101.695  50.079  0.00  0.00    -0.030 A
ATOM      2  C   LIG    1       -7.837 101.486  48.730  0.00  0.00    +0.090 A
ATOM      3  C   LIG    1       -9.147 101.402  48.306  0.00  0.00    +0.030 A
ATOM      4  C   LIG    1      -10.180 101.524  49.216  0.00  0.00    -0.050 A
ATOM      5  C   LIG    1       -9.907 101.732  50.562  0.00  0.00    +0.130 A
ATOM      6  C   LIG    1      -11.610 101.431  48.746  0.00  0.00    +0.100 C
ATOM      7  C   LIG    1       -8.592 101.813  50.998  0.00  0.00    +0.120 A
ENDROOT
BRANCH   1  13
ATOM      8  C   LIG    1       -3.446 102.767  52.833  0.00  0.00    +0.240 A
ATOM      9  N   LIG    1       -2.257 102.405  52.445  0.00  0.00    -0.250 NA
ATOM     10  N   LIG    1       -2.330 101.778  51.323  0.00  0.00    -0.230 NA
ATOM     11  C   LIG    1       -3.581 101.712  50.944  0.00  0.00    +0.080 A
ATOM     12  S   LIG    1       -4.712 101.147  49.709  0.00  0.00    +0.190 S
ATOM     13  C   LIG    1       -6.145 101.786  50.535  0.00  0.00    +0.170 A
ATOM     14  N   LIG    1       -5.709 102.349  51.619  0.00  0.00    -0.250 NA
ATOM     15  N   LIG    1       -4.348 102.331  51.887  0.00  0.00    -0.340 N
BRANCH   8  17
ATOM     16  C   LIG    1       -3.372 102.687  55.311  0.00  0.00    +0.060 C
ATOM     17  C   LIG    1       -3.777 103.522  54.095  0.00  0.00    +0.160 C
ENDBRANCH   8  17
ENDBRANCH   1  13
BRANCH   5  18
ATOM     18  N   LIG    1      -10.956 101.853  51.480  0.00  0.00    -0.620 N
ATOM     19  H   LIG    1      -11.561 102.610  51.427  0.00  0.00    +0.410 HD
BRANCH  18  20
ATOM     20  C   LIG    1      -11.124 100.921  52.438  0.00  0.00    +0.510 C
ATOM     21  S   LIG    1      -10.962  99.261  52.056  0.00  0.00    -0.470 S
BRANCH  20  22
ATOM     22  N   LIG    1      -11.419 101.297  53.699  0.00  0.00    -0.640 N
ATOM     23  H   LIG    1      -12.344 101.330  53.991  0.00  0.00    +0.410 HD
ATOM     24  C   LIG    1      -10.431 101.629  54.553  0.00  0.00    +0.530 C
ATOM     25  O   LIG    1       -9.272 101.589  54.186  0.00  0.00    -0.380 OA
BRANCH  24  26
ATOM     26  C   LIG    1      -10.754 102.041  55.932  0.00  0.00    -0.180 A
ATOM     27  C   LIG    1      -10.619 103.376  56.316  0.00  0.00    +0.070 A
ATOM     28  C   LIG    1      -10.922 103.754  57.607  0.00  0.00    +0.000 A
ATOM     29  C   LIG    1      -11.361 102.814  58.524  0.00  0.00    +0.080 A
ATOM     30  C   LIG    1      -11.500 101.489  58.153  0.00  0.00    -0.010 A
ATOM     31  C   LIG    1      -11.205 101.095  56.861  0.00  0.00    +0.170 A
ATOM     32  F   LIG    1      -11.341  99.801  56.499  0.00  0.00    -0.110 F
ENDBRANCH  24  26
ENDBRANCH  20  22
ENDBRANCH  18  20
ENDBRANCH   5  18
TORSDOF 6

EDIT #1
I suspect I do not understand the underlying mechanism behind adding hydrogens in OpenBabel. I created a benzene ring in Avogadro and then removed the hydrogens. Avogadro and Autodock PMV can add hydrogens back, but OpenBabel will not. This is the test benzen.pdb:
COMPND    UNNAMED
AUTHOR    GENERATED BY OPEN BABEL 2.3.90
HETATM    1  C   UNL     1       0.371  -1.202   1.209  1.00  0.00           C1-
HETATM    2  C   UNL     1      -1.019  -1.145   1.197  1.00  0.00           C1-
HETATM    3  C   UNL     1      -1.701  -1.052  -0.011  1.00  0.00           C1-
HETATM    4  C   UNL     1      -0.994  -1.017  -1.209  1.00  0.00           C1-
HETATM    5  C   UNL     1       0.396  -1.074  -1.197  1.00  0.00           C1-
HETATM    6  C   UNL     1       1.078  -1.167   0.011  1.00  0.00           C1-
CONECT    1    2    6
CONECT    2    1    3
CONECT    3    2    4
CONECT    4    3    5
CONECT    5    4    6
CONECT    6    1    5
MASTER        0    0    0    0    0    0    0    0    6    0    6    0
END

Like a .cif (see here), your .pdbqt does not store the bond order between atoms in your model.  So OpenBabel (or other programs) have to guess by presence of other atoms already present if the addition of hydrogens is sensible by valence rules, or not.  There equally are checks for interatomic distances and geometry before hydrogens can be add on idealized positions.  If the conformer presents a case outside the thresholds, the addition of hydrogens on idealized / riding positions will fail.
You can run a forcefield optimization followed by an addition of hydrogens all over the molecule
obabel example.pdbqt --gen3d -O intermediate.pdb
obabel intermediate.pdb -h -O output.pdb

however on expense to loose this conformer to obtain one OpenBabel* assumes good enough (depends on the force field selected).  It is not obvious if this is acceptable for your in silico docking.
The problems become (interactively) visible if you load the structure e.g., Jmol and subsequently request an addition of hydrogens.  On the program's console
load "example.pdbqt";
connect aromatic modify;
calculate aromatic;
calculate hydrogens;

yields a visualization.  Note the nitrogen bound hydrogens, the methyl/methylene group (pointing to the west), and the ethyl/ethylene group (to south east):

Here, you need additional information about the molecule initially submitted to your docking experiment.  For example, was there a $\ce{-CH3}$ methyl group and is there evidence it still is there?  What happened to the (presumed) carbonyl $\ce{C=O}$ which appears now elongated like an alcohol?  The unit of thiourea?
The structure exported, by an additional write ex2.pdb;:
ATOM      1  C   LIG     1      -7.551 101.695  50.079  0.00  0.00           C
ATOM      2  C   LIG     1      -7.837 101.486  48.730  0.00  0.00           C
ATOM      3  C   LIG     1      -9.147 101.402  48.306  0.00  0.00           C
ATOM      4  C   LIG     1     -10.180 101.524  49.216  0.00  0.00           C
ATOM      5  C   LIG     1      -9.907 101.732  50.562  0.00  0.00           C
ATOM      6  C   LIG     1     -11.610 101.431  48.746  0.00  0.00           C
ATOM      7  C   LIG     1      -8.592 101.813  50.998  0.00  0.00           C
ATOM      8  C   LIG     1      -3.446 102.767  52.833  0.00  0.00           C
ATOM      9  N   LIG     1      -2.257 102.405  52.445  0.00  0.00           N
ATOM     10  N   LIG     1      -2.330 101.778  51.323  0.00  0.00           N
ATOM     11  C   LIG     1      -3.581 101.712  50.944  0.00  0.00           C
ATOM     12  S   LIG     1      -4.712 101.147  49.709  0.00  0.00           S
ATOM     13  C   LIG     1      -6.145 101.786  50.535  0.00  0.00           C
ATOM     14  N   LIG     1      -5.709 102.349  51.619  0.00  0.00           N
ATOM     15  N   LIG     1      -4.348 102.331  51.887  0.00  0.00           N
ATOM     16  C   LIG     1      -3.372 102.687  55.311  0.00  0.00           C
ATOM     17  C   LIG     1      -3.777 103.522  54.095  0.00  0.00           C
ATOM     18  N   LIG     1     -10.956 101.853  51.480  0.00  0.00           N
ATOM     19  H   LIG     1     -11.561 102.610  51.427  0.00  0.00           H
ATOM     20  C   LIG     1     -11.124 100.921  52.438  0.00  0.00           C
ATOM     21  S   LIG     1     -10.962  99.261  52.056  0.00  0.00           S
ATOM     22  N   LIG     1     -11.419 101.297  53.699  0.00  0.00           N
ATOM     23  H   LIG     1     -12.344 101.330  53.991  0.00  0.00           H
ATOM     24  C   LIG     1     -10.431 101.629  54.553  0.00  0.00           C
ATOM     25  O   LIG     1      -9.272 101.589  54.186  0.00  0.00           O
ATOM     26  C   LIG     1     -10.754 102.041  55.932  0.00  0.00           C
ATOM     27  C   LIG     1     -10.619 103.376  56.316  0.00  0.00           C
ATOM     28  C   LIG     1     -10.922 103.754  57.607  0.00  0.00           C
ATOM     29  C   LIG     1     -11.361 102.814  58.524  0.00  0.00           C
ATOM     30  C   LIG     1     -11.500 101.489  58.153  0.00  0.00           C
ATOM     31  C   LIG     1     -11.205 101.095  56.861  0.00  0.00           C
ATOM     32  F   LIG     1     -11.341  99.801  56.499  0.00  0.00           F
ATOM     33  H33 LIG     1      -7.017 101.388  48.003  1.00  0.00           H
ATOM     34  H34 LIG     1      -9.370 101.238  47.241  1.00  0.00           H
ATOM     35  H35 LIG     1     -12.434 101.528  49.468  1.00  0.00           H
ATOM     36  H36 LIG     1     -11.829 101.266  47.681  1.00  0.00           H
ATOM     37  H37 LIG     1      -8.372 101.970  52.064  1.00  0.00           H
ATOM     38  H38 LIG     1      -1.411 102.587  52.945  1.00  0.00           H
ATOM     39  H39 LIG     1      -1.753 102.201  50.624  1.00  0.00           H
ATOM     40  H40 LIG     1      -6.345 102.790  52.253  1.00  0.00           H
ATOM     41  H41 LIG     1      -2.291 102.489  55.276  1.00  0.00           H
ATOM     42  H42 LIG     1      -3.920 101.733  55.300  1.00  0.00           H
ATOM     43  H43 LIG     1      -3.614 103.238  56.232  1.00  0.00           H
ATOM     44  H44 LIG     1      -4.239 104.520  54.135  1.00  0.00           H
ATOM     45  H45 LIG     1     -10.678  99.132  50.794  1.00  0.00           H
ATOM     46  H46 LIG     1      -9.234 101.292  53.232  1.00  0.00           H
ATOM     47  H47 LIG     1     -10.271 104.126  55.590  1.00  0.00           H
ATOM     48  H48 LIG     1     -10.815 104.806  57.909  1.00  0.00           H
ATOM     49  H49 LIG     1     -11.600 103.123  59.552  1.00  0.00           H
ATOM     50  H50 LIG     1     -11.846 100.747  58.888  1.00  0.00           H
CONECT    1   13   13
CONECT    2    3    3
CONECT    4    6    6
CONECT    5    7    7
CONECT    8   17   17
CONECT   26   31   31
CONECT   27   28   28
CONECT   29   30   30

* Note, you seem to use an old version of OpenBabel.  It currently is shipped as version 3.1.1 (source of the releases, and the most recent documentation).

Thank you, I suspected, but I wasn't sure this was the case. You raise an excellent point concerning losing the conformation. The incoming drugs are from an Autodock Vina virtual screen, and therefore I can't mess with the heavy atom conformation.   Autodoc Vina uses a united atom force field; the only hydrogens are polar hydrogens.
Your answer had helped, and I've been left with food for thought. Thanks


text:
Hartree-Fock calculations of the states of Nitrogen molecule ion
I was reading the third chapter of "Modern Quantum Chemistry" by A. Szabo and N. Ostlund about HF calculations, and in the end of the chapter it shows as an example a calculation on $\ce{N2}$ MOs. It's said that HF predicts that the $1\pi_{u}$ MO is higher in energy than the $3\sigma_{g}$, what we know to be wrong. There are two hypotheses to explain it: the negligence of the electronic correlation or of the relaxation effects. To test the effects of relaxation, UHF calculations are done for the ion $\ce{N2^+}$. The energies of the states $^{2}\Sigma_{g}$ and $^{2}\Pi_{u}$ of $\ce{N2^+}$ are computed and subtracted from the energy of $\ce{N2}$ ground state to compute the ionization potentials.
My question is about the determination of the energy of these two states of $\ce{N2^+}$ in practice using a program like Gaussian 09. The program computes the energy of the ground state, but how can I compute the energy of different states of a molecule? I know that there is a relation between the total energy and the energies of the MOs, so I suspect that I should use the MO energies to do this, but I don't know a way to do it using the output of Gaussian for example.

To calculate electronic excitations you need to use a non-variational scheme like CIS, CISD, TDDFT, EOMCCSD, etc. These are generally quite expensive and don’t even do a great job, but it can be quite illustrative. Gaussian will be perfectly happy to give you excited states with CISD if you ask for them.

Some programs, and I do not know whether Gaussian is among them, can restrict the symmetry of the molecule and thus force certain states by ruling out lower-lying states of a different symmetry. Note that this can also happen accidentally when the initial guess if of such a symmetry and the converging algorithm does not check for instabilities: basically, a saddle point instead of a minimum is found. The instability check is non-trivial and expensive.

Please keep in mind that depending on the actual revision of Gaussian, it is at least a 7 year old program with no support whatsoever. You should only be using it for the sake of reproducing old data and comparing it to new data. Symmetry isn't a key feature of Gaussian, I think it is limited to D2h. However, it should still be possible. Since this question is less about the chemistry, or theory, and more about the execution with a specific software, I suggest transferring it to [mattermodeling.se].


text:
Can the ionisation energy of hydrogen be calculated using Coloumb’s Law?
Using Coloumb’s Law: $$F=k\frac{qQ}{r^2}$$
and the potential energy function: $$U(r)=\int_\infty^r F dr$$
I get:
$$U(r)=k\frac{qQ}{r}$$
Where k is the Coloumb constant and q and Q are the values of negative and positive charge respectively.
Inputting the values of charge for one electron and one proton for a hydrogen atom, the value of k and the Bohr radius I get:
$$U=8.99\times10^9 \frac{-1.60\times10^{-19} \times1.60\times10^{-19}}{5.29\times10^{-11}}$$
Which works out as:
$$U=4.35\times10^{-18}J$$
Multiplying this by the Avogadro number, dividing by 1000 and rounding up gives:
$$U=2600 kJ mol^{-1}$$
This value is approximately double that of the actual ionisation energy of hydrogen and closer to that of helium but I can’t see the reason for it.

Hello! Good try to come up with a new idea to calculate something. Unfortunately, Bohr did it like a $100$ years ago, combining Newton's second law with Coulomb's law, and also got a value of the ionization energy. You may want to read the Bohr's model...

You didn't consider the energy electron already has.

Yes, the ionization energy of hydrogen can be calculated using Coulomb's law for the potential energy, but you must also take into account the kinetic energy. In Bohr's model, the electron in the ground state is spinning around the nucleus at around 1% of the speed of light. As you bring this electron to infinity (=ionization) it progressively loses this speed, since overall angular momentum $mvr$ is conserved.
You can avoid actually calculating that kinetic energy by using a very convenient theorem called the Virial Theorem (https://en.wikipedia.org/wiki/Virial_theorem) which states that the kinetic energy is (minus) half of the potential energy. Therefore, the ionization energy you are seeking is simply half the value you've calculated above.
If you do want to calculate the kinetic energy explicitly, you'll need to know the speed of the electron, which follows from Bohr's model that you'll have to look up elsewhere. I'll just give the result here: the speed of the electron is $\alpha$ times the speed of light, where $\alpha$ is the so-called fine-structure constant that is approximately 1/137. You can then state the kinetic energy in the ground state as :
$$E = \frac12\cdot m\cdot v^2$$
Entering the electron mass, the speed of light and the fine structure constant gives $2.18 \times 10^{-18} J$, which is indeed half of what you calculated for the potential energy above.


text:
Principal quantum number and 'good' quantum numbers
When we discuss about configurations we specify n, l, m quantum numbers for the individual electrons.
My question is: why when we pass from configurations to atomic terms in order to use the total angular momentum we don't use anymore the n quantum number? Does exist an 'equivalent' number for the terms or is conceptual wrong to think about it?
I know the formal procedure for the change of basis set from uncoupled to coupled basis and i understand why this is important in terms of commutation properties of hamiltonian, but i simply don't see anymore this n quantum number, so i'm a bit confused about it.

The quantum number n does not contribute to angular momentum, in contrary to the others.

As Poutnik points out in the comments, the angular momentum of an electron does not depend on its quantum number $n$; it only depends on $l$ and $m_l$.
Thus, for example, $\mathrm{1s}^1$ and $\mathrm{2s}^1$ configurations (which might be, for example, the ground and first excited state of hydrogen) both correspond to the same term symbol $^2\!S_{1/2}$.
Clearly, this is fine if all the states you care about have different angular momentum properties. But there are cases where this isn't fine, and you need to distinguish. I don't know if there is 'official' notation specified out there, but what people usually do is to specify the orbital configuration along with the term symbol, or $n$ along with the term symbol (if that provides enough information).
Both of these possibilities are mentioned in the following extract from Hollas' Modern Spectroscopy, 4th ed. (p  246):

In the sodium atom pairs of $^2\!P_{1/2}$, $^2\!P_{3/2}$ states result from the promotion of the 3s valence electron to any $np$ orbital with $n > 2$. It is convenient to label the states with this value of $n$, as $n{}^2\!P_{1/2}$ and $n{}^2\!P_{3/2}$, the n label being helpful for states that arise when only one electron is promoted and the unpromoted electrons are either in filled orbitals or in an $s$ orbital. The $n$ label can be used, therefore, for hydrogen, the alkali metals, helium and the alkaline earths. In other atoms it is usual to precede the state symbols by the configuration of the electrons in unfilled orbitals, as in the $2p3p~{}^1\!S_0$ state of carbon.

this is what i was searching for. Let me be sure to have well understand: so if i wanna put the energies in increasing order i write down the configurations in increasing order and for each configuration i write its terms in increasing order?

@Chemistry. Yup, pretty much.


text:
Nephelauxetic effect and covalency of metal-ligand bonds
As I was reading about nephelauxetic effects, which is the effect that atomic d-orbitals are bigger in a complex than in gaseous metal ions. The Racah interelectronic repulsion parameter gets smaller, indicating that doubly occupied d-orbitals are bigger in complexes than in an atom (why only doubly occupied though, shouldn't there also be less repulsion between, lets say singly occupied d-orbitals?). I wondered what reasons there could be for that. The two most cited reasons are:

decreased effective nuclear charge, partially neutralized by the ligand sphere
formation of covalent bond character with the ligand

The first option seems fine enough for me, but 2) makes little sense for me. The ligand field splitting arises because the partially filled d-orbitals want to avoid the double occupied ligand orbitals, therefore assuming a position where the electronic repulsion is smallest.
Why would they form covalent bonds then, or is that just a consequence of MO theory that we have to acccept?

There's practically no such thing as gaseous metal ions - you need plasma for any significant amount of "free" metal ions (a nitpick).

The Racah repulsion parameters are smaller in a transition metal complex than in a free atomic ion because

Decreased effective nuclear charge: As the effective charge localized at the metal decreases, the orbitals increase in size (they're pulled in less tightly). This larger size means the electron density is more diffuse so electrons feel less repulsion from each other

Ligand covalency: Mixing between the d orbitals and the ligand orbitals causes the wavefunction to extend further. This delocalization also increases the general interelectron distance, leading to less repulsion.

Your point about ligands repelling d orbitals comes from a crystal field understanding of metal–ligand interactions rather than a ligand field understanding. Crystal field theory tries to explain all of coordination chemistry using repulsion, and it turns out it's just too simplistic. The fact that crystal field theory ignores all metal–ligand bonding (covalency) is one of the major failings of crystal field theory (it's the reason it can't justify the spectroscopic series or the nephelauxetic series). This failure is the reason chemists gradually moved to ligand field theory.
Ligand field theory explains d orbital splitting using orbital overlap and covalent mixing of orbitals — not using repulsion. This is very reminiscent of diatomic MO diagrams, where localized orbitals overlap in constructive (bonding) and destructive (antibonding) patterns. Repulsion is treated separately through Racah parameters (just like crystal field theory). An advantage of this approach is that you can start to look at the interplay of covalency, symmetry (sigma/pi), 10Dq, B, and C.
So I suppose the short answer is that: you can't ignore MO theory like crystal field theory does. If you take into account the molecular orbitals, then bullet (2) makes sense.


text:
Determination of the chloride concentration by the modified Volhardt method
Would someone help me solve a task:

The concentration of chloride ions in aqueous solutions can be
determined by the Volhardt method. An excess of silver nitrate is
added to the sample containing the chlorides, and the extremely
insoluble silver chloride is precipitated. Titration determines the
excess of silver ions, from which the concentration of chloride ions
in the initial sample can be determined. However, silver nitrate is a
rather expensive reagent. One alternative that could be used in the
same method is some soluble lead(II) salt, eg lead(II) nitrate.
However, lead(II) chloride is slightly more soluble than silver
chloride, so it is necessary to calculate whether the use of lead(II)
salts will affect the precision of the method.
If a solution of lead(II) nitrate (7.45 g in 100 mL of water) is added
to 200 mL of a sodium chloride solution with a concentration of 0.1023
mol/L, how much lead(II) chloride will precipitate? What is the
concentration of lead(II) ions in the solution after precipitation?
What is the equilibrium concentration of chloride ions remaining in
the solution after percipitation? Calculate the error in the
determination of chloride ions. The solubility product of lead(II) chloride is $1.7 \times 10^{-5}\,\text{mol}^3\,\text{L}^{-3}$.

Solution:
First, I determined the limiting reagent, which in this case is sodium chloride. This helped me to calculate the mass of precipitated lead(II) chloride which is 2.84 g. I then calculated the mass of excess lead(II) ions and determined their concentration in the final solution (300 mL) to be 0.041 mol/L. I'm not entirely sure about the chloride ion concentration calculation -- I calculated it from the expression for the solubility product $$[\ce{Cl^-}] = \sqrt{\frac{K_\text{sp}}{[\ce{Pb^2+}]}}$$ and by including the calculated concentration of lead(II) ions present in the solution I got 0.02039 mol/L. I am struggling with determining the error of the chloride ion concentration since I only calculated one chloride concentration.
I would be very grateful if someone could check $[\ce{Cl^-}]$ and help me calculating the error.

Why do you want to use Pb and introduce an error and work with Pb? You  seem to be working with large amounts of reagents! rewrite your method to use 10mL samples, use a simple Mohr titration and a 10mL low actinic buret and use as little silver nitrate as possible. Another consideration is to buy a chloride sensing electrode. [or possibly a conductivity meter]. PbCl2 is too soluble for quant analysis.

crvenikupus - What you are really looking for in this situation is the negative bias in the analysis. You assume that the analysis should yield 100% of the chloride, then you calculate how much chloride is really left in solution. The chloride left in the solution is the negative bias.

@MaxW hmm, okay. do you think that I calculated the equilibrium concentration of chloride correctly?

Ok, your formula for calculating the chloride concentration:
$$[\ce{Cl^-}] = \sqrt{\frac{K_\text{sp}}{[\ce{Pb^2+}]}}$$
is fine as far as it goes. The equation can give a ballpark estimate of the method's bias. However the solubility product of lead(II) chloride,  $1.7×10^{−5}$ is only known to two significant figures, and the solutions are concentrated enough that activities should be used instead of concentrations.
For the lead
$$\ce{(7.45 g)/(331.2 g/mol) = 0.0225 mol}$$
For the chloride
$$\ce{(0.2 L) x (0.1023 mol/L) = 0.0205 }$$
No since each lead cation precipitates two chloride anions the moles of excess $\ce{Pb}$ is:
$$0.0225 - (0.0205/2) = 0.0123$$
In 300 ml of solution this gives a lead molar concentration of
$$ 0.0123/0.300 = 0.041$$
So:
$$[\ce{Cl^-}] = \sqrt{\frac{K_\text{sp}}{[\ce{Pb^2+}]}}=\sqrt{\frac{1.7×10^{−5}}{0.041}} = 0.0204$$
Now for 300 ml of solution the moles of chloride is:
$$0.3 * 0.0204 = 0.00611$$
which means that there is a loss of
$$ 0.00611/0.0205 *100\% = 29.8\%$$
of the chloride. So the method with lead will have a very significant negative bias.


text:
Vapor pressure of a weak acid for a solution of its salt
I need to calculate the vapor pressure of an acid in equilibrium with an aqueous solution of its salt. For example, say I have a solution of 40% potassium formate by mass in water at 50°C. How do I find the vapor pressure of formic acid in equilibrium with that liquid?
My best guess is that I first calculate the concentration of formate ions via
pKb = -log10([HB+][OH-]/[B]).
Then I can calculate the concentration of formic acid using pKa=log10([HA]/[A-][H+]).
Then I can use Raoult's law to estimate the partial pressure.
This is the best I can come up with, but it doesn't feel quite right to me. Does anyone recommend any corrections to this process?

Sounds fine altho why calculate formate? Just dissolve it calculate formic acid, convert to mole fractions and apply Raoult's law! do the calculations, a sniff test and head space analysis. an iterative calculation might be needed to account for the self- ionization of water. Report back the completed calculations and the analysis information.

Another point here is a 40% solution of  potassium formate in water by weight is very concentrated. You'd really need to use activities not concentrations in the calculations.

Note that I assume strong deviation from Raoult's law for formic acid/water or  formic acid/formate water solution  mixture, forming maximum azeotrope boiling at 107.2° C. and containing 22.5 wt. % water.

The part that probably doesn't feel right is that the two equilibria are coupled. However, if the amount shifting to the gas phase is small compared to solution it will hardly affect the solution concentrations, making your approach reasonable. You can check whether if it is so by using an iterative approach.


text:
Regression equation for peak fronting/tailing in chromatography
Background
I would like to have a sum of functions $$\sum_{i=1}^k f_i(x)$$ which approximates my chromatogram. The variable $x$ would be my retention time, although I don't mind if answers change it to $t$, $t_r$ or anything else they prefer.
One issue that arises is that the peaks are not always symmetric. This is called tailing or fronting:

Troubleshooting Basics, Part IV: Peak Shape Problems

Effort
Fiddling around in Desmos I came up with a sort of "split-variance Gaussian function" to model a single peak. Or at  least that's the name I made up writing this post.

(Sorry for using an image. The $\LaTeX$ wasn't rendering correctly.. Feel free to edit: $$f(x; a, m, s_1, s_2) = \begin{cases}a\exp\left(-\frac{1}{2}\left(\frac{x-m}{s_1}\right)^{2}\right) & x \leq m \\ a\exp\left(-\frac{1}{2}\left(\frac{x-m}{s_2}\right)^{2}\right) & x > m \end{cases}$$)
This seems to

have relatively few parameters per peak
give fronting or tailing depending on the pair $(s_1, s_2)$
$\lim_{x \rightarrow m^-} f(x) = \lim_{x \rightarrow m^+} f(x) = f(m) = a$
$\arg\max_x f = m$

But I know that just b/c I have an attempt that superficially serves the purpose doesn't mean that someone hasn't come up with something better. I'm looking for what a chromatographer would use (rather than my mathematical guesswork).
Question
Are there conventional mathematical models for asymmetric peaks on chromatograms? If not, would my split-variance Gaussian function be alright from a chemist's point of view?

It depends on the regression method too.

@ananta Oh NUTS that is a good point.

You also need the tops of the peaks to touch. So the two equations need different a's, i.e. $a_1 \ne a_2$.

You could try this equation and vary $N$ and $p$. The shape of a chromatogram is closely related to the Poisson distribution
$\displaystyle P(n) = p\frac{(np)^{(N-1)}}{(N-1)!}e^{-np} $ where $N$ is the number of theoretical plates, $p$ the probability related to the partition constant $K$ the total volume of the mobile phase $V_m$ and of $V_s$ the stationary phase,$\displaystyle p=\left(1+K\frac{V_s}{V_m}\right)^{-1}$(chapter 4 in chemistry-maths-book.com)

Maybe this: https://en.wikipedia.org/wiki/Log-normal_distribution

@MaxW I don't think that is necessary. The function is continuous and at its maximum when the case switches.

@porphyrin Theoretical plates is not necessarily an integer  (although it does appear to be non-negative) so putting it in a factorial is undefined except in rather special cases. But maybe extending to the non-negative real numbers using the gamma function might be fine. Would have to check that the normalization is fine under this generalization.

Actually, the normalization isn't important in this case.


text:
Besides diamonds, are there any stones without oxygen in composition?
In my inpression, the most of stones are oxides and hydroxides, correct me if I am wrong.

Review the guides Asking and How to ask to prevent misunderstanding,  need of clarification, objections, downvoting or closure.

related: https://chemistry.stackexchange.com/questions/75478/aside-from-carbon-what-other-substances-can-be-made-superhard

By stones you mean minerals? https://en.wikipedia.org/wiki/Rock_(geology)

Define «stone».  Just something «hard» like a rock? A mineral like rock salt (halit/NaCl)? An iron meteorite?

@Buttonwood I think, stone is something hard, brittle, non-alloy-like (insulator, no metallic luster), not volatile, crystalline (not polymeric), not molecular, not organic, non-easily solvable, not flammable, non-fibrous (non asbestos-like). Basically, a gem.

There are minerals of every kind you'd like, and you're not entitled to make some artificial definitions of words.

So biblical stoning was de facto making a heap of gems on persons being gifted.

Stone: hard solid nonmetallic mineral matter of which rock is made, especially as a building material. (Oxford dictionary)

LOL - Let's not make this complicated. For all practical purposes stones are composed of minerals. A stone may be one mineral or many. // Now the OP needs to clarify if the question is over all minerals, or the relative fraction of the minerals that compose the earth.

@Karstenyes. And rock cannot be made of volatile, easily solvable or polymeric materials.

Silicates are polymeric minerals. And rock can be soluble.

To your credit "mineral" is not an unambiguous term. However it does have a common use where it refers to compounds with a solid structure. There are other uses that might refer to liquid substances (mineral acid, or your reference to petroleum as a mineral). In that case "mineral" is an adjective that distinguishes the origin of the substance, contrasting it to an organic or botanical origin. I had a look at the IUPAC gold book which compiles chemical terminology and there is no entry for "mineral". Assume then the definition in the Wikipedia or a good dictionary.

In general please use standard chemical nomenclature. You can ask what is standard use of course. Your question is a bit diffuse because "stone" is a common usage term, not a technical term. It refers to rocks of a certain size, and rocks are pure minerals or solid mixtures of various minerals. A second point of confusion is whether you are interested in compounds with a single element or only those excluding oxygen in their composition.

It's always interesting how people downvote questions, yet the top users are all replying it. Hypocrites lol.

@Minsky this question was closed twice: closed, reopened, closed by mods.

Mods get fed up of their boring job, and become convinced that they have the authority to shame on others (even if they do not understand at all what the post is about), so most SE sites have a growing number of questions asked by non-verified accounts. This behaviour is then normalised and copied by new users. (In the beginning though, when these sites wanted users, it was quite friendly. Now business matters more.) The simplest solution is to just ask with random accounts, currently.

Most stones are silicates and aluminosilicates (which can be vaguely described as combinations of oxides), and thus do indeed contain oxygen. Among the rest, many are just oxides. Your impression was correct. Oxygen is ubiquitous. One doesn't get called the most abundant element in Earth's crust for nothing.
As for the minerals containing no oxygen: why, there are some. Pyrite is the first that comes to mind. (Source: Wikipedia)
Despite its majestic look and the fact that you don't find it often among roadside stones, it is nowhere as rare as diamond. In fact it is quite cheap, hence the name "fool's gold".
So it goes.

I wonder if the OP meant elemental minerals: https://en.wikipedia.org/wiki/Native_element_mineral

I think not. After all, it would be hard to expect any of those to contain oxygen.

@Mithoron I think this will be closed soon "of natural causes".

@Mithoron I object. The question is not too broad: it does have a meaningful answer. Maybe it was asked by a person who would not pass your intelligence test. Maybe. Actually, I like the idea of giving such test to anyone whom I converse with during the day. It would make a lot of things easier. But before such practice is instituted, I'd rather stick to my usual routine.

Ivan, the thing is this could have a hundred answers, each one telling about a different mineral - that's what makes it too broad. It's a typical HNQ material. The answer is there's loads upon loads of such minerals, that's pretty much that, but we could easily have half of mineralogy put here. But, oh well, maybe I worry too much...

This thing on the photo does not look like a stone. Looks more like an alloy.

@Anixx Please address the questions you have been asked in order to reopen the post. Pyrite is a mineral, iron sulfide.

Yes, there are plenty:
-Sulfide (e.g. Pyrite)
-Silicide (e.g. Suessite). Here is a good link: https://encyclopedia.pub/entry/18800
-Nitride (Siderazot ($\ce{Fe3N_{1.33}}$) is the only terrestrial nitride mineral)
-Carbide
-Phosphide (e.g. Schreibersite)
-Selenide
-Halide (There are some non-oxygen ones)

But are they stones rather than volatiles, alloy-likes, salts or organic compounds?

They are minerals.

Petroleum is also mineral but not stone.

@Anixx Define "stone" in your question. Accordingly, answer will be modified.


text:
Can an electrode run out of the active species?
I have a background in solid state physics but am new to electrochemistry. I have been following the example of a physical chemistry (Mortimer) book that looks an electrochemical cell with a hydrogen electrode and a silver-silver choloride electrode.
The two half reactions are:
\begin{align}
\ce{H2(g) &-> 2 H+(aq) + 2 e-(aq)}\\
\ce{AgCl(s) + e^- &-> Ag(s) + Cl^-(aq)}
\end{align}
The silver-silver chloride electrode consists of a silver wire coated with a silver-chloride solid.
If we connect the two electrodes and allow current to flow then there will be a flow of electrons from left to right.
The half-reaction on the silver-silver chloride reaction says that the chloride will come off of the solid and go into solution in the electrolyte.
The hydrogen electrode has hydrogen gas being constantly flowed into the system. However, the silver-silver chloride electrode, for me, will eventually run out of silver-chloride if the cell is run continuously. I can't seem to find any texts that treat this.
I found this question here, but it doesn't quite address the same issue.
This seems to be a key point in the cell-type reactions that the reactants are supplied such that the activities/concentrations do not change. Thus there is a stable current. It also seems to bring up a few ideas on chemical equilibrium. In my intuition as long as there is an electrical current the reaction is ongoing, thus the reactants and products are not in thermodynamic equilibrium...
Excuse the rambling question, it is not all quite straight in my head yet.
Thanks

Note that reference electrodes as SHE or AgCl ones are NOT assumed to be used under any significant current load. Rather measured in stationary state with very high input impedance of digital voltmeters. When there is ongoing significant current activity  in electrolytic or in galvanic mode, there are used "working electrodes" and the reference electrode is used as the 3rd, reference point in "currentless" mode.  Note that AgCl electrode will be kinetically very limited due very low Ag+ ion concentration.

Note you can write $$\ce{H2(g) -> 2 H+(aq) + 2 e-(aq)}$$ to get  $$\ce{H2(g) -> 2 H+(aq) + 2 e-(aq)}$$ as Chemistry SE site uses by default the optional mhchem MathJax package (usable on other SE sites too via \require{mhchem} )

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

Yes, it will absolutely run out. To put it simply, this is a battery, and it will run out of juice like any other battery.
You're also correct about which reactant would run out first. The hydrogen can presumably be pumped in indefinitely, but there is only so much $Ag^+$ available. So in this cell, the oxidizer would run out first.
As Poutnik noted in the comments, these are reference electrodes. They are not intended to supply a singificant amount of power. In other words, as a battery, this setup is very poor, and would never actually be used. But if you were trying to design a silver-hydrogen battery, you would want to cram in as much silver chloride as possible, perhaps by having it on the surface of something like a mesh or sponge. That way you could get the reaction to continue for the longest possible time.


text:
Doubt on vibronic transitions
When we justify the presence of electronic transitions forbidden by Laporte selection rule, the coupling between electronic and vibrational states help us. Why we say that in this way we are overcoming the Born-Oppenheimer approximation? Possibly I'm searching for an intuitive explanation and/or a theoretical one.

The BO approximation defines any potential energy surface, and so also the vibrational/rotational levels it has. The selection rules are thus always within this approximation.

In the BO approximation you decouple electronic and nuclear motion. To obtain potential energy curves, you typically solve the SE as function of nuclear coordinates. Vibronic couplings break down this approximation because the hierarchy of the motions (electronic >> vibration >> Rotation) is no longer valid.

@porphyrin so, can i say that when i consider vibronic transition i overcome Franck-Condon principle and why? Are Franck-Condon and Born-Oppenheimer related? Thanks for the patience

There are two effects, vibronic transitions and vibronic coupling.
In the BO approximation the wavefunction is a product of electronic $\psi$ and vibrational $\chi$ wavefunctions, for the ground state and excited state
$$\displaystyle \Psi^g_v=\psi^g(r,R)\chi^g_v(R)\\
\Psi^e_u=\psi^e(r,R)\chi^e_u(R)$$
with vibrational quantum numbers $v,u$, electron coordinates $r$ and nuclear coordinates $R$. The electronic wavefunction depends parametrically on $R$, i.e. the electronic energy is calculated separately at each $R$, and the vibrational ones $\chi$ only on the potential energy in their respective states at position $R$.
In vibronic transitions the transition moment for an electric dipole transition is the integral over coordinates $r$ and $R$
$$\displaystyle \mu^{ge}_{vu}=\int\int \psi^g(r,R)\chi^g_v(R)\; \hat\mu \; \psi^e(r,R)\chi^e_u(R)\,dr dR$$
and where $\hat\mu$ is the dipole operator, a vector. In symmetry terms this transforms as $x,y$ or $z$ and has the form of 'charge times displacement' with displacement vectors $\hat r, \hat R$,
$$\displaystyle \hat \mu=-e\sum_i \hat r_i+e\sum_jZ_j\hat R_j=\hat\mu_{elec}+\hat\mu_{nucl}$$
If this operator is put into the transition dipole equation the nuclear term $\hat\mu_{nucl}$, which depends on position $R$ of stationary nuclei, factors out the integral in electronic coordinates $r$ which becomes zero because the ground and excited electronic states are orthogonal, i.e. $\int \psi^g(r,R)\psi^e(r,R)=0$ thus we only need to evaluate $\hat\mu_{elec}$. The transition dipole therefore becomes
$$\displaystyle \mu^{ge}_{vu}=\int\chi^g_v(R)\,\hat\mu^{ge}\, \chi^e_u(R)\, dR$$
where $\displaystyle \mu^{ge}=\int \psi_g(r,R)\hat\mu_{elec}\psi_e(r,R)dr$
and this depends on $R$ through the electronic wavefunction. To overcome this difficulty the Condon Approximation is now made, which means evaluating the electronic transition moment at the equilibrium internuclear position, (or its average value) so that it no longer depends on $R$ and so is separated from the integral
$$\displaystyle \mu^{ge}_{vu}=\hat\mu^{ge}_0\int\chi^g_v(R) \chi^e_u(R)\, dR$$
and the integral is now the overlap of the vibrational wavefunctions and produces the Franck-Condon factors as the square of the absolute value, $F=|\langle v|u\rangle|^2$ where the bra-ket indicates the integral.
The vibronic coupling case is a breakdown of the BO approximation which can often be treated as a perturbation. One particular example is the the case where a non-totally symmetric vibration distorts the electronic state and changes its point group and hence the transition dipole is changed. Benzene is a commonly discusses example where an $e_{2g}$ vibration make the lowest transition weakly allowed by borrowing intensity from a nearby allowed transition. This is sometimes called Herzberg-Teller Coupling and /or Intensity Stealing.


text:
Density of unknown metal

An empty flask weighs $\pu{115.713 g}$. When filled water at $\pu{20 °C}$ it now weighs $\pu{365.263 g}$. The flask is then emptied and a $\pu{15.000 g}$ sample of unknown metal is added to it. It is then filled with water at $\pu{20 °C}$. The combined mass is $\pu{378.426 g}$. What is the density of the unknown metal? The density of water at $\pu{20 °C}$ is $\pu{0.9982 g/ml}$.

Here is my calculation:
Weight of the water: $\pu{249.552 g}$
The volume occupy by the water: $\pu{249.551 g}$ divide by $\pu{0.9982 g/mL}$ yield $V = \pu{250 mL}$.
Then, I don't know what to do I think I should divide the volume of the water by the mass of the unknown metal but I don't understand why I should to that.

For future reference: chemistry.se offers to use mhchem for the body of questions, answers, and comments.  Do not use it for the headers of questions, because it is not guaranteed every browser processes this equally well.  It equally provides the \pu{} command for numerical results, which keeps units typed upright; and the space between the figure and unit in input yields a non-breakable space in the output which mostly is fine (except for the description of angles, and percentage/permille sign in English typography).

\begin{array}{|l|l|l|l|}
\hline \text{Parameter} & \text{Symbol} & \text{Relation} & \text{Value} \\
\hline
\text{Flask mass} & m_0 & &\pu{115.713 g} \\
\text{Flask mass with water} & m_1 & & \pu{365.263 g} \\
\text{Water mass} & m_2 & m_2=m_1 - m_0& \pu{249.55 g} \\
\text{Water density} & \rho_\ce{H2O} & &\pu{0.9982 g cm-3} \\
\text{Flask volume} & V_0 & V_0 = \frac{m_2 }{ \rho_\ce{H2O}} &\pu{250.00 ml} \\
\text{Metal mass} & m_3 & & \pu{15.000 g} \\
\text{Mass of flask+metal+water} & m_4 & & \pu{378.426 g} \\
\text{Mass of water only} & m_5 & m_5 = m_4 - m_0 - m_3& \pu{247.713 g} \\
\text{Volume of water only} & V_1 & V_1 = \frac{m_5 }{ \rho_\ce{H2O}} & \pu{248.160 ml} \\
\text{Volume of metal} & V_2 & V_2 = V_0 - V_1 & \pu{1.840 ml} \\
\text{Density of metal} & \rho_\ce{m} & \rho_\ce{m} = \frac{m_3 }{ V_2} & \pu{8.151 g cm-3} \\
\hline
\end{array}


text:
Residual current in electrolysis
On this site: https://water.lsbu.ac.uk/water/electrolysis.html there is a graph showing current-voltage characteristic for electrolysis of water:

I am wondering, if I apply voltage lower than required to decompose water (lower than 1.23 V, like for example 0.8 V), will there still be residual current flow (called "background current" in the picture) and if yes, will the ions, that normally participate in chemical reactions at the electrodes, be charge carriers? Also, if there is current flow even with applied voltage below 1.23 V, how high is typical current (like in the range of μA or mA)?

According to the footnote a in the page you provided, the specific resistivity of pure water, for example, is $\zeta = \rm 18;M\Omega\cdot cm$.  The background current is therefore $I = U/R = UA/(\zeta \ell)$ with $A$ being the area of electrodes, $\ell$ distance between them, and $U$ applied voltage.

Also, as you can see in the picture, 1,23 is hardly enough - you need at least some 0.4-0.5 V of overpotential on top of that, to have significant reaction.


text:
Reaction of silver with hydrogen sulfide
Usually, when we talk about the tarnishing of silver in air, we think of it as the reaction of silver with hydrogen sulfide in the air.
However, when looking at the standard potentials, silver should not reduce hydrogen sulfide to hydrogen gas and silver sulfide. In this answer (https://chemistry.stackexchange.com/q/99812) it is said that the precipitation of silver sulfide drives the reaction to completion, however, silver is also a solid, and hydrogen also a gas, so there shouldn't be a reasion why silver should completely decompose in a hydrogen sulfide stream, rather reach a small equilibrium value.
However, since the Nernst equation breaks down (as there is no dissolved Ag+), I don't know how to calculate it. How to do this?

And how is your question not a duplicate of the one you linked?

The reaction needs oxygen. Pure H2S does not react with silver. But it does slowly in the presence of oxygen

See: https://chemistry.stackexchange.com/questions/141365/silver-tarnish-and-reactivity-series

Usually, when we talk about the tarnishing of silver in air, we think of it as the reaction of silver with hydrogen sulfide in the air.

According to this account by Tim Burstein from Cambridge University, the main two reactions in dry air are:
$$\ce{4Ag(s) + O2(g) -> 2 Ag2O(s)}$$
and
$$\ce{Ag2O(s) + H2S(g) -> Ag2S(s) + H2O(g)}$$
The oxide layer is passivizing, so it remains thin and invisible. In contrast, the sulfide layer grows over time, giving rise to interesting changes in coloration.

[OP] However, when looking at the standard potentials, silver should not reduce hydrogen sulfide to hydrogen gas and silver sulfide.

The paper has a list of reaction steps with Gibbs energies of reaction. It makes the comment that standard potentials (electrochemical series) most often involve the metal ions in aqueous solution rather than metals salts. $\ce{Ag2S}$ is very insoluble, driving the process.

[OP] In this answer (https://chemistry.stackexchange.com/q/99812) it is said that the precipitation of silver sulfide drives the reaction to completion, however, silver is also a solid, and hydrogen also a gas, so there shouldn't be a reasion why silver should completely decompose in a hydrogen sulfide stream, rather reach a small equilibrium value.

If the reaction steps given above are a good representation of what happens, then there is no evolution of elemental hydrogen, and the argument in the answer to the related question is moot.

[OP title] Reaction of silver with hydrogen sulfide

I'm by no means an expert in this field, and am relying on the source I gave. If I understood the document, you should change your title to "Reaction of silver with dioxygen and hydrogen sulfide" or "Reaction of silver oxide with hydrogen sulfide" to reflect what might be happening in silver tarnishing.

[OP] However, since the Nernst equation breaks down (as there is no dissolved Ag+), I don't know how to calculate it. How to do this?

In the presence of humidity, tarnishing is much faster. In the presence of water on the surface, the process becomes an electrochemical one, allowing oxidation and reduction to happen at a distance.

Yeah that makes sense, at neutral pH the oxygen reduction should be favored electronically also by about 0.1 V (using the electrochemical series).