text:
How to get the coordinates of β12 borophene and χ3 borophene?
Can someone tell me how to draw the β12 borophen and χ3 borophene in Materials Studio or any other software? I've just started my project work, and I'm a beginner in this field.
I have searched for the coordinates or a .cif file in the literature, but I could not find any. If anyone can provide me with those, it'll also be helpful for me.

If the references on Wikipedia (https://en.wikipedia.org/wiki/Borophene) don't offer a starting point, nor the ICSD database, what entries on materials project like https://materialsproject.org/materials/mp-160/#similar-structures ?

For obtaining optimized geometry, energies, and band structure, the Perdew−Burke−Ernzerhof [1] (PBE) functional in nonlocal corrected generalized gradient approximation [2] (GGA) was used. The OTFG ultrasoft pseudopotential was employed with a plane-wave basis with a kinetic energy cutoff of 520 eV in Monkhorst−Pack [3] k-mesh with a 0.04 Å separation in the two lattice directions for geometry optimization. The adjacent sheets were kept 20 Å away to avoid all possible interactions between two nanosheet layers. Individual atom position and lattice parameters have been simultaneously optimized with the chosen cutoff value for SCF tolerance (5.0 × 10-6 eV/ atom) and forces (0.01 eV/Å) to get the well-converged geometries of nanosheets.

All the calculations were performed using the Cambridge Ab initio Serial Total Energy Package (CASTEP) [4] program based on the above-mentioned density functional theory.

Perdew, J. P.; Burke, K.; Ernzerhof, M., Generalized Gradient Approximation Made Simple. Physical Review Letters 1996, 77 (18), 3865-3868. 10.1103/PhysRevLett.77.3865
Perdew, J. P.; Wang, Y., Accurate and simple analytic representation of the electron-gas correlation energy. Physical Review B 1992, 45 (23), 13244-13249. 10.1103/PhysRevB.45.13244
Ceperley, D. M.; Alder, B. J., Ground State of the Electron Gas by a Stochastic Method. Physical Review Letters 1980, 45 (7), 566-569. 10.1103/PhysRevLett.45.566
Clark, S. J.; Segall, M. D.; Pickard, C. J.; Hasnip, P. J.; Probert, M. I. J.; Refson, K.; Payne, M. C., First-principles methods using CASTEP. Zeitschrift für Kristallographie - Crystalline Materials 2005, 220 (5-6), 567-570. 10.1524/zkri.220.5.567.65075

Beta_12.cif file:
data_Beta_12_str
_audit_creation_date              2021-09-18
_audit_creation_method            'Materials Studio'
_symmetry_space_group_name_H-M    'PMM2'
_symmetry_Int_Tables_number       25
_symmetry_cell_setting            orthorhombic
loop_
_symmetry_equiv_pos_as_xyz
  x,y,z
  -x,-y,z
  x,-y,z
  -x,y,z
_cell_length_a                    2.5600
_cell_length_b                    20.0000
_cell_length_c                    4.2600
_cell_angle_alpha                 90.0000
_cell_angle_beta                  90.0000
_cell_angle_gamma                 90.0000
loop_
_atom_site_label
_atom_site_type_symbol
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
_atom_site_U_iso_or_equiv
_atom_site_adp_type
_atom_site_occupancy
B1     B     0.00000   0.00000   0.83833   0.00000  Uiso   1.00
B2     B     0.00000   0.00000   0.50500   0.00000  Uiso   1.00
B3     B     0.00000   0.00000   0.17167   0.00000  Uiso   1.00
B4     B    -0.50000   0.00000   0.68718   0.00000  Uiso   1.00
B5     B    -0.50000   0.00000   0.35385   0.00000  Uiso   1.00
loop_
_geom_bond_atom_site_label_1
_geom_bond_atom_site_label_2
_geom_bond_distance
_geom_bond_site_symmetry_2
_ccdc_geom_bond_type
B1     B2      1.420   .     S
B1     B4      1.433   .     S
B1     B4      1.433   1_655 S
B1     B3      1.420   1_556 S
B2     B3      1.420   .     S
B2     B4      1.497   .     S
B2     B4      1.497   1_655 S
B2     B5      1.433   .     S
B2     B5      1.433   1_655 S
B3     B5      1.497   .     S
B3     B5      1.497   1_655 S
B3     B1      1.420   1_554 S
B4     B1      1.433   1_455 S
B4     B5      1.420   .     S
B4     B2      1.497   1_455 S
B5     B3      1.497   1_455 S
B5     B2      1.433   1_455 S

Chi_3.cif file:
data_chi_3_str
_audit_creation_date              2021-09-18
_audit_creation_method            'Materials Studio'
_symmetry_space_group_name_H-M    'CMMM'
_symmetry_Int_Tables_number       65
_symmetry_cell_setting            orthorhombic
loop_
_symmetry_equiv_pos_as_xyz
  x,y,z
  -x,-y,z
  -x,y,-z
  x,-y,-z
  -x,-y,-z
  x,y,-z
  x,-y,z
  -x,y,z
  x+1/2,y+1/2,z
  -x+1/2,-y+1/2,z
  -x+1/2,y+1/2,-z
  x+1/2,-y+1/2,-z
  -x+1/2,-y+1/2,-z
  x+1/2,y+1/2,-z
  x+1/2,-y+1/2,z
  -x+1/2,y+1/2,z
_cell_length_a                    6.8829
_cell_length_b                    2.8510
_cell_length_c                    20.0000
_cell_angle_alpha                 90.0000
_cell_angle_beta                  90.0000
_cell_angle_gamma                 90.0000
loop_
_atom_site_label
_atom_site_type_symbol
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
_atom_site_U_iso_or_equiv
_atom_site_adp_type
_atom_site_occupancy
B1     B     0.30945   0.50000   0.00000   0.00000  Uiso   1.00
B2     B     0.10315   0.50000   0.00000   0.00000  Uiso   1.00
loop_
_geom_bond_atom_site_label_1
_geom_bond_atom_site_label_2
_geom_bond_distance
_geom_bond_site_symmetry_2
_ccdc_geom_bond_type
B1     B2      1.420   .     S
B1     B2      1.547   10_565 S
B1     B2      1.547   10    S
B1     B1      1.644   10_565 S
B1     B1      1.644   10    S
B2     B2      1.420   2_565 S
B2     B1      1.547   10_565 S
B2     B1      1.547   10    S

Jiang et al. 2018 (DOI: 10.1016/j.jechem.2018.01.026) may be relevant literature? Here's a quote:

These include three allotropes named borophene, beta-12(β12) and chi-3(χ3), which
have been successfully synthesized by physical methods [26,27].
[26] A. J. Mannix, X. F. Zhou, B. Kiraly, J. D. Wood, D. Alducin, B. D. Myers, X. L. Liu, B. L. Fisher, U. Santiago, J. R. Guest, M. J. Yacaman, A. Ponce, A. R. Oganv, M. C. Hersam, N. P. Guisinger, Science 350 (2015) 1513–1516.
[27] B. J. Feng. J. Zhang, Q. Zhong, W. B. Li, S. Li, H, Li, P. Cheng, S. Meng, L. Chen, K. H. Wu, Nat. Chem. 8 (2016) 563-568.

I also found this one:
Izadi Vishkayi, S., Bagheri Tagani, M. Edge-Dependent Electronic and Magnetic Characteristics of Freestanding β 12-Borophene Nanoribbons. Nano-Micro Lett. 10, 14 (2018). DOI: 10.1007/s40820-017-0167-z
Perhaps you may find the link to the .cif from all of the above?


text:
Does reduction of carbonyls using borohydride ever use NaOH in the reaction?
I was reading about borohydride reductions of aldehydes and ketones in one of the Clayden textbooks, and I saw that NaOH was used in the bottom reaction for some reason:

Is this really necessary? I know that usually we just use proton donors later on in the reaction to cap the alkoxide group but we have methanol and water for those. All the other examples seem to use either water or organic alcohols like methanol or ethanol, so why is NaOH being  used here?

Base helps to repress the slow reaction of NaBH4 (liberated BH3?) with alcohol, in particular methanol.

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sort: votes


text:
Do the αβ and βα states of homonuclear AX system have same energy?
I was studying nuclear Overhauser effect and came across this diagram$^1$:

This diagram shows the energy levels of four possible nuclear spin states of homonuclear AX system. The white/grey boxes refer to population deficit/excess, respectively, w.r.t the population at each of the two states at the centre (α$_\mathrm{A}$β$_\mathrm{X}$, β$_\mathrm{A}$α$_\mathrm{X}$ ). This is the population distribution at thermal equilibrium.

It is stated clearly in the text that α$_\mathrm{A}$β$_\mathrm{X}$ and
β$_\mathrm{A}$α$_\mathrm{X}$  have same energy, for homonuclear AX
system.
My question is, how can the α$_\mathrm{A}$β$_\mathrm{X}$ and
β$_\mathrm{A}$α$_\mathrm{X}$ have same energy?

The energy of each state is described by:
$$E(m_A, m_X) = -hv_Am_A -hv_Xm_X + hJm_Am_X $$
For the two states, third term at RHS are common, but the Larmor frequencies of A and X are not the same despite being homonuclear (already called 'AX', chemical shift very different, Larmor frequency also differ), so the states should have different energy. Did I understand something wrong?

Peter Atkins, Julio de Paula, James Keeler. Physical Chemistry (11th Edition). OUP. 2018. Page 516.

Or, 8th Edition: page 542; 9th Edition: Page 548

Yes it is confusing and rather strange, and does not make sense. The energy $\alpha\beta;; \beta\alpha$ levels are definitely different  both with and without spin-spin coupling and these are shown correctly in fig15.12. In specialist texts such as Levitt 'Spin Dynamics' they are shown correctly when NOE's are explained, see fig 16.36 in that book.

the Larmor frequencies of A and X are not the same despite being homonuclear (already called 'AX', chemical shift very different, Larmor frequency also differ), so the states should have different energy

Yes, that is precisely correct, so the two states $|\alpha\beta\rangle$ and $|\beta\alpha\rangle$ do not have the same energy. More formally, the Hamiltonian for the system (ignoring spin-spin coupling) is
$$\hat{H} = -\omega_I\hat{I}_{\!z} - \omega_S\hat{S}_z$$
where $\omega_I$ and $\omega_S$ are the Larmor frequencies of spin $I$ and $S$, and $\hat{I}_{\!z}$ and $\hat{S}_z$ are theie corresponding spin angular momentum operators, with the following properties:
\begin{align}
    \hat{I}_{\!z} |\alpha\rangle &= \frac{\hbar}{2}|\alpha\rangle \\
    \hat{I}_{\!z} |\beta\rangle &= -\frac{\hbar}{2}|\alpha\rangle
\end{align}
The energies of the states $|\alpha\alpha\rangle$, $|\alpha\beta\rangle$, $|\beta\alpha\rangle$, and $|\beta\beta\rangle$ can be evaluated fairly simply using
$$E = \langle \Psi | \hat{H} | \Psi \rangle$$
or in this case, since all of them are eigenstates of the Hamiltonian (without coupling), then
$$\hat{H}\Psi = E\Psi$$
and it suffices to read off the eigenvalue from the right-hand side. Proceeding in this manner, we have
\begin{align}
    \hat{H}|\alpha\beta\rangle &= -\omega_I \hat{I}_{\!z}|\alpha\beta\rangle - \omega_S \hat{S}_z|\alpha\beta\rangle \\
                               &= -\omega_I (\hbar/2)|\alpha\beta\rangle - \omega_S (-\hbar/2)|\alpha\beta\rangle \\
                               &= [\hbar(\omega_S - \omega_I)/2] |\alpha\beta\rangle \\
\Longrightarrow E_{\alpha\beta} &= \hbar(\omega_S - \omega_I)/2
\end{align}
Likewise,
\begin{align}
\hat{H}|\beta\alpha\rangle &= [\hbar(\omega_I - \omega_S)/2] |\beta\alpha\rangle \\
\Longrightarrow E_{\beta\alpha} &= \hbar(\omega_I - \omega_S)/2
\end{align}
and in general, since $\omega_I \neq \omega_S$, these energies are different. The introduction of spin–spin coupling does not affect these conclusions.
One question remains, which is: why does Atkins depict the states as having the same energy, despite them not being equal? One possible answer is laziness, but if you read their writings, Atkins (and Keeler) are quite particular about notation and technical correctness, so I feel like this is not the primary reason, or maybe I am more inclined to give the benefit of the doubt here.
A more convincing answer would be that the difference between the energies $E_{\alpha\beta}$ and $E_{\beta\alpha}$ is very small:
$$E_{\alpha\beta} - E_{\beta\alpha} = \hbar(\omega_S - \omega_I)$$
The difference in Larmor frequencies between spins, i.e. $\omega_S - \omega_I$, is generally on the order of Hz to kHz. On the other hand, the other energy differences on the graph are far larger:
$$E_{\beta\beta} - E_{\beta\alpha} = \hbar\omega_S$$
The Larmor frequency itself $\omega_S$ is on the order of MHz. So the energy difference between $\alpha\beta$ and $\beta\alpha$ is pretty negligible compared to the other energy differences shown in the diagram.

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text:
Help with two containers problem, Xenon
In two containers we have Xenon. The volume of the first container is 2780 L, temperature is 300 degree Celsius, and n = 200 mol. Volume of the second container is 4000 L, temperature is 1300 degrees Celsius and n = 100 mol. Temperature in both of the containers is constant. How much Xe do we have to transfer from one container to another so that the pressure in both containers is the same?
I have calculated pressure in container one and that is p1 = 342817.92 Pa, and in container two p2 = 326979.2275 Pa. Then I have substracted this two values which gives me 15838.6925 Pa. After that I have divided this value with two, to figure out how much do I have to decrease pressure in container 1, and that gives me 7919.34625 Pa.
After substracting this value from p1 I get the new pressure in container 1 which is 334898.5738 Pa which is the same as the pressure in container 2. From the new p1 I have calculated new n1 = 195.3798 mol, so I have to transfer 4.620 moles, or 606.593 grams to container 2.
My question is: When I take into the consideration the new n2 = 100 mol + 4.620 mol = 104.620 moles, and then I try to check if the p2 new is the same as p1 new I don'5 get the same result. p1 new = 334898.5738 Pa, and p2 new = 342086.14 Pa. Is my calculation wrong?

Using 10 valid digits is definitely wrong, if not backed up by the  accuracy of this value. Surely is there pressure  334898.5738 Pa ?   A methodical advice: Stay with symbolic algebra until you have the equation  providing the final result. After that, plug in the numbers.

For eventual writing and formatting of chemical formulas or equations, see how to use MathJax with mhchem extension

Why the votes to close as homework?? The poster gave the problem and explained how the wrong solution was obtained.

user33683 - I think the tone about significant figures was bad. If you put all the calculations into a formula, then the calculator obviously carries a ridiculous number of significant digits throughout the calculation. My take is that you do want to avoid rounding errors in intermediate calculations. So figure out how many significant figures the answer should have, then carry two extra digits in the intermediate calculations, if you use them. Any more is just wasted effort. I think 3 digits for this problem, so your answer should have been 4.62 moles, or 607 grams.

I have calculated pressure in container one and that is p1 = 342817.92 Pa, and in container two p2 = 326979.2275 Pa. Then I have substracted this two values which gives me 15838.6925 Pa. After that I have divided this value with two, to figure out how much do I have to decrease pressure in container 1, and that gives me 7919.34625 Pa.

You cannot generally just average both pressures. As $pV=nRT$, then ( for $\mathrm{T}$ and $\mathrm{V}$ constant ) :
$V \cdot \mathrm{d}p=\mathrm{d}n \cdot RT$
$\frac{\mathrm{d}p}{\mathrm{d}n}=\frac{RT}V$.
So pressure averaging would work only in the special case if $\frac{T_1}{V_1}=\frac{T_2}{V_2}$. Therefore, a weighted average must be used.

I suppose the scenario is vaguely defined. It can be the polytropic  expansion, or it may be the "locally isothermal" expansion. But for the former, it has like infinite number of possible solutions, depending on the rate of the transfer and of heat exchange.
I consider below the latter, i.e. isothermal conditions in each of containers.

We can consider the ideal behaviour, where
$$p=\frac{nRT}V$$
For equality of pressures, if $\Delta n$ is the moved molar amount of xenon from the container 2 to the container 1, the following equality must be true:
$$p=\frac{(n_1 + \Delta n)RT_1}{V_1}=\frac{(n_2 - \Delta n)RT_2}{V_2}$$
From that, we can easily calculate $\Delta n$:
$$(n_1 + \Delta n)RT_1V_2=(n_2 - \Delta n)RT_2V_1$$
$$\Delta n \cdot R(T_1V_2 + T_2V_1)=n_2RT_2V_1 - n_1RT_1V_2$$
$$\Delta n=\frac{n_2T_2V_1 - n_1T_1V_2}{T_1V_2 + T_2V_1}$$
Now you plug in the numbers here in the final formula, instead of working with literal numbers all the way:

In two containers we have Xenon. The volume of the first container is 2780 L, temperature is 300 degree Celsius, and n = 200 mol. Volume of the second container is 4000 L, temperature is 1300 degrees Celsius and n = 100 mol. Temperature in both of the containers is constant. How much Xe do we have to transfer from one container to another so that the pressure in both containers is the same?

$$\Delta n=\frac{(\pu{100 mol})(\pu{1573.15 K})(\pu{2.780 m3}) - (\pu{200 mol})(\pu{573.15 K})(\pu{4.000 m3})}{(\pu{573.15 K})(\pu{4.000 m3}) + (\pu{1573.15 K})(\pu{2.780 m3})}$$
In the sense below, if to be precise, we should round the temperature values for presented formula, because temperature was not known in 2 decimals accuracy ( otherwise there would be $\pu{300.00 ^{\circ}C}$, resp. $\pu{1300.00 ^{\circ}C}$), internally still computing with non rounded values.
$$\Delta n=\frac{(\pu{100 mol})(\pu{1573 K})(\pu{2.780 m3}) - (\pu{200 mol})(\pu{573 K})(\pu{4.000 m3})}{(\pu{573 K})(\pu{4.000 m3}) + (\pu{1573 K})(\pu{2.780 m3})} \approx \pu{-3.18 mol}$$
The negative value means the $\pu{3.18 mol}$ of xenon is to be moved from the first container to the second one, as we have formally assumed the opposite direction.

Internal full calculator precision of computation still makes good sense, to avoid artefacts of rounding errors.  But for presentation of final or intermediate values of quantities from the real world, the values have to reflect error propagation rules. As input data and constants have limited accuracy.

I still have the feeling the end of the point(1) If you want to answer a homework question, we much prefer that you give a full explanation of the method used to reach the answer  of hint-answers-revisited  does not mean you are expected to enumerate numbers in the formula to get literal result.

I don't think the differential equation $\frac{\mathrm{d}p}{\mathrm{d}n}=\frac{RT}V$ is the right one. $n$ is a variable too since gas is being transferred from one container to another. // Overall your point is correct. Don't expect averaging pressures to work.

@MaxW But that is the point. There is dn because of this transfer and RT/V is the rate of pressure drop dp/dn, because of this transfer. p and n are linearly proportional and RT/V is the proportionality constant.


text:
Do copper, copper(II) oxide, copper(II) hydroxide, copper(II) carbonate react with diluted sulfuric acid?

Which of the following will react with dilute sulfuric acid: copper, copper(II) oxide, copper(II) hydroxide, copper(II) carbonate?

Metals react with acids to give salt and hydrogen. So, copper will react with dilute sulfuric acid.
Metal oxides act as base. So, they neutralise acid to give salt and water. So, copper oxide will react with sulfuric acid.
Metal hydroxides are bases. So, they will neutralise acids. So, copper hydroxide will react with dilute sulfuric acid.
Carbonates react with acids to give salt, carbon dioxide and water. So, copper carbonate will react with dilute sulfuric acid.
So, my answer is all the given options will react with dilute sulfuric acid. Is this correct? The answers are not given in the book. I am afraid what if there are exceptions to the general reactions that I have written above and I am not aware of that.

Certainly true for Copper(II) oxide and copper(II) carbonate  https://edu.rsc.org/experiments/reacting-copperii-oxide-with-sulfuric-acid/1917.article   https://www.bbc.co.uk/bitesize/guides/zysdgdm/revision/3

Metallic copper does not react with dilute sulfuric acid. Its redox potential is too positive to do it.
All other copper compounds do react with sulfuric acid, as you had suspected.


text:
HPLC: Peak height calculation
I am trying to implement a small python script to detect peaks and calculate the corresponding area. The y-values are in mAU and x-values in minutes but there are some unit conversions in most HPLC softwares. For example in this image (found here):

The peak at 7.87 min has a height of 329451 but only approximately 20 V in the chromatogram. How can I calculate this height?

I suggest finding something much more substantive than what is at the link. Astonishingly, it actually says “What Is a Chromatogram? Now that you have understood everything about chromatography, let’s learn about chromatography and chromatographic analysis.”

Just took this example from their article because I got the same problem with our data. It is just about the unit conversion factors on the y-axis.

The table reads «329451» as area, e.g. after baseline correction (if there was one, but e.g., in powder diffraction crystallography [PXRD] one would look to perform a suitable one), followed by an area integration.  Thus, first digitize the plot (e.g., https://automeris.io/WebPlotDigitizer/).  For the subsequent processing, there already are solutions published which may serve as inspiration for you, e.g., https://fityk.nieto.pl/, PyMS, HappyTools. Area = half-peak width x peak height is one way.

@Buttonwood, the area calculation by  "half-peak width x peak height" gives really bad results in HPLC. He should rather use rectangular, Simpson's or trapezoidal method for Gaussian or exponentially modified Gaussians to get anywhere close to the true area.

@M.Farooq I agree, one way is not the best way, nor for all analyses encountered.  The formula mentioned is what I retained from Skoog and Leary's book about analytical chemistry as an easy approach.  Which may be inferior to the approaches mentioned by you.  And it doesn't matter if it still is mentioned (e.g., here) even if it is a step ahead from e.g., cutting and then weighting the paper printed peaks on a balance (some of my elder colleagues did this).

@Buttonwood, Interesting...I have done paper cutting once or twice...at that time I did not have a peak fitting software and it works exceptionally well. Modern students may not heard of the fact one could use a non-computer method to determine peak area :-)

Height in HPLC simply refers to the height of the peak (= maximum value of the peak) above the baseline. The point by z1273 is right in the sense that this voltage to height conversion factor is instrument specific but this height cannot be in milliabsorbance units- it is too large for absorbance. This height is apparently "intensity". Anyway, you do not need to worry about the "true height" for peak detection. It will be fine to use the height of 20 V for the first peak at 7.867 min. In older instruments, the detector output was displayed as Volts or mV and the manufacturer provided a conversion factor from volts to light absorbance or sometimes as intensity in the manual. See this discussion here (https://www.chromforum.org/viewtopic.php?t=6442). You should really look at modern HPLC data. The one you are showing here is from an ancient instrument.
I am assuming you are using derivatives to detect peaks? For any chromatography program, I would suggest that you do a simulation first, because all variables will be in your hand. Make a chromatogram as sum of large and small exponentially modified Gaussians or simple Gaussians of various heights and areas and add some noise to test your program.

The Height ($h$) in mAU is a function of the Signal A measured in volts at 254nm. The function should look like $h=16700\cdot A+1500$ or so, in this case. The calibration equation would be specific to a given installation.
I suppose the Height may be back-calculated from the area, but it looks circular reasoning to me.

Maurce, he used the term "circular reasoning" not circular height. Circular reasoning refers to the chicken and egg type problem. Who came first?

@Maurice Please take a moment to review the OP where mAU, Signal A at 254 nm and the context of the numbers are specified. As for the units, I suppose mAU/Volt and mAU may be appropriate for 16700 and 1500, respectively. What I meant by 'circular reasoning' is that deriving the height from an area that is itself calculated using the height appeared to me as a circular reasoning. To be sure, the link in the OP provides relevant specifics as to how they view the calculation of the area. Hopefully it is noted that I focused my answer to the OP's question 'How can I calculate this height?'


text:
Why does the ionization enthalpy of elements across a period not follow a regular pattern while the atomic size always decreases?
First of all, I would like to mention that I am only talking about elements that aren't from the d or f blocks.
In order to further elaborate on my question, I would like to take the third period as an example.

As we can see, the atomic radii decrease across a period constantly, whereas the ionization energies increase and decrease depending on the element. I just fail to understand how this can be. As far as my understanding goes, isn't work done against the electric field to remove the electron called ionization energy? And if this is true, then it must mean that the closer an electron is to the nucleus, the stronger the force of attraction experienced by it would be, which would in turn translate into a higher ionization energy. However, we can see that this does not seem to be the case. I would really appreciate it if someone could tell me where I'm going wrong and why this is the case.

The first thought that comes to mind that there's only one way to define ionization enthalpy for an atom. But there are multiple reasonable definitions of atomic radius.

@Zhe That makes a lot of sense. I just searched up the Van Der Waals radius of all of the period 3 elements and saw that the atomic radius isn't decreasing continuously. However, I could not see any definitive correlation between the Van Der Waals radius and the ionization enthalpy. Nevertheless, it seems convincing to me that the answer might lie in one of the ways of defining atomic radii. I sincerely thank you for helping me out.

https://chemistry.stackexchange.com/questions/5621/z-effective-charge-and-ionization-energy https://chemistry.stackexchange.com/questions/32363/anomalous-trends-in-ionization-energy

Put very roughly, we can boil this down to why aluminum and sulfur are depressed with respect to the average of the surrounding elements.
Aluminum is $\ce{[Ne]{3}s^2{3}p^1}$.  Its first ionization removes the $3p$ electron while leaving the more stable $3s$ orbital intact, whereas magnesium has to break its more stable $3s$ orbital.  So by comparison, the first ionization energy of aluminum drops off.
This does not mean aluminum will form ionic compounds more readily than magnesium.  Despite aluminum's subshell advantage, both elements have first ionization energies that are too high to be offset by the electron affinity of a nonmetallic element and electrostatic attraction between the resulting ions.  There is the possibility of retaining metal-metal bonding between +1 ions, which has become fairly well-known with magnesium, but ordinarily a reasonably low second or third ionization energy is needed to get a favorable energy balance with these elements.  So to form ionic compounds aluminum ultimately has to give up its $3s$ electrons, which are more tightly bound than magnesium's, and thereby the subshell difference is obscured.  Magnesium ultimately wins when it comes to ionic compound formation.
Turning now to sulfur, that element has the ground state configuration $\ce{[Ne]{3}s^2{3}p^4}$.  In this case loss of an electron gives a half-filled $3p$ subshell with all the electrons unpaired and having identical spins.  This stabilizes the half-subshell configuration through a quantum-mechanical exchange interaction between the electrons with like spin and total angular momentum.  Sulfur forms this favored configuration when it ionized once, while phosphorus has to break it, and so the first ionization energy of sulfur (but again, not second or higher ionization energies) is depressed as we saw with aluminum.  Note that the effect here is less than that involving full or empty subshells that we saw with aluminum versus magnesium.

Ah.  Surrounding the principal quantum numbers, with the right spelling for "principal", also seems to work.  Thanks @zhe.

"Principal" is a word I never spell right the first time. Only on proofreading...


text:
Combining solutions with multiple concentrations - Mixing NPK fertilizer ratios
I'm trying to mix my own fertilizer to grow some peppers.
Fertilizers are labeled with NPK ratios, so mixing two volumes and finding the result is easy: Just multiply the concentration of each component by the ratio of the solution over the total volume. For example, $\pu{1L}$ of a $3:4:6$ solution plus $\pu{1L}$ of a $5:2:6$ solution yields $\pu{2L}$ of an $4:3:6$ solution.
Something like:
$$ NPK_\mathrm{final} = \left(NPK_{s1} \times \frac{V_{s1}} {V_{s1} + V_{s2}} \right)+ \left(NPK_{s2} \times \frac{V_{s2}} {V_{s1} + V_{s2}}\right) $$
However, if I have a desired ratio in mind, how can I solve for how much of each of 2 solutions to add? I can plug in different values till I get close, but I'm curious if there's a general formula. I figured there's probably a way to get there with linear algebra (but I never took linear algebra!).

Note that you can chose just ratio of 2 from 3 elements and only in the range between these 2 source fertilizers. The ratio of the 3rd element is determined by your prior choice. You would need 3 fertilizers to mix to independently choose ratio of all 3 elements, again within the range of these sources.

Yes- mixing 3 fertilizers is the only way to alter the ratios of all 3 components. For simplicity's sake (and because I didn't want the example equation to be too long) I figure you could do a two step process. Mix two fertilizers, then mix the third into the resulting solution.

Yes, I was going to add exactly a note like this to my answer .

Let consider ratio of N/P.
Let fertilizer A has this ratio N/P=a.
Let fertilizer B has this ratio N/P=b.
Let you want the ratio c, where a < c < b.
Then you need to mix A : B in ratio $$\frac{ b - c}{c-a}$$
In the Central Europe with German scientific influence, it is called the "mixing cross rule".

Note that you can chose just ratio of 2 from 3 elements and only in the range between these 2 source fertilizers. The ratio of the 3rd element is determined by your prior choice for the other two elements.
You would need 3 fertilizers to mix to independently choose ratio of all 3 elements, again within the range of these sources.

A graphical chart method(*) may be more practical than a numerical method.
One can draw a equal-side triangular ternary chart. Each each corner of the triangular ternary plot represents formally a fertilizer with the solely N ( or P or K ) component.
If we take e.g. the N corner, than lines  parallel to the side opposite to N corner represent N relative content to the sum N + P + K. The same for other 2.
Within this triangle, one can draw 3 points, each representing 3 different particular fertilizers.
Then one can draw similar chart between these 3 points for these fertilizers.
Available ratios than fall into this smaller triangle.
Then one can on the big chart choose N:P:K ratios and than read from the small triangle chart the ratios of all 3 fertilizers.

There is small catch in all this. The amounts of fertilizers to be mixed should be normalized for their total NPK content.
So mixing fertilizers A, B, C would not affect the total NPK amount per volume/mass, only the component ratio.

(*) It is also called a ternary plot on Wikipedia. See also the listed referenced there:

an online tool to create ternary plots)
simple Excel template
Trip-Plot Excel package

Or you can used PDF chart template to be printed, provided by @MaxW.

Alternatively, I have prepared on Google docs the spreadsheet,

using linear algebra, which provides the mixing rations for given 3 fertilizer compositions, checking forbidden combinations.

I have correctly understood your cross mixing rule. Thank you. I also have been able to draw the small triangle, but not to use it. Sorry ! What is its use ?

@maurice It is projection of NPK ratio coordinates from big triangle to fertilizer ratio coordinates on small triangle. Usage is the same as of the big triangle, just it is not the same size triangle ( cannot recall the geometry term ).

@Maurice note that similar triangle is used for evaluation and tuning of mobile phases in reverse HPLC.

@Poutnik Something like this? link

@Poutnik Sorry. I don't understand. I will not dye of not knowing this point.

@beepy Exactly. For practical use, if the inner triangle is small, somewhat zoomed in bigger triangle cut would be more useful. The similar coordinate net would be drawn for the inner triangle. But it would not be NPK ratios, but fertilizer ratios. I.e. 100% of A would be solely usage of fertilizer A.

From the given NPK ratio you get the point location on the bigger triangle. And for the same point, you get the related fertilizer ratio from the small triangle. Obviously, if you are out of the small triangle, you are out of the game.

@Maurice - The gist is that you plot ($\frac{N}{N+P+K}, \frac{P}{N+P+K}, \frac{K}{N+P+K}$) on ternary scale for each type of fertilizer. With two types you can draw a line between them. You can make any mix along the line. With three different types you get a triangle when joining the three with line segments. You can only make mixes which would be inside the triangle.

@Maxw Unfortunately, I do not have any handy tool to provide an illustrative self-explaining chart. Wikipedia page offers a link to an Excel template for it, some dedicated software, also the link of the online tool www.ternaryplot.com looks promising.

@Poutnik - I'm old school. https://www.waterproofpaper.com/graph-paper/triangular-grid-graph-paper.pdf

@MaxW I cannot disagree - good paper beets them all. The gold good old time of specialized nomogram ( nomograph?) paper templates.

Well, When reporting three lines, all parallel to one side of the triangle, one for 10, one for 30, and one for 60, you do not obtain a triangle. You obtain one point !!

@Maurice One point for one fertilizer. another for another and the 3rd one for the 3rd one. You get a triangle, unless one fertilizer is a mixture of the other two. Then you would get 3 points on a line. Imagine fictituous fertilizers containing just one of each N, P, K elements. You would get then the 2nd triangle identical to the 1st triangle.

Alternatively, I have prepared on Google docs the spreadsheet, using linear algebra, which provides the mixing rations for given 3 fertilizer compositions, checking forbidden combinations.


text:
Is the polarisation of halogens of higher atomic number easier because the electrons are further away from its core?
This question concerns the dihalo addition reactions. I've learned that a halogen molecule (which, of course, consists of two atoms of the respective halogen) is more easily polarised by the double bond present in an alkene if it has a larger atomic number.
My question is this: Is this because the electrons of the halogens of higher atomic number are, on average, further away from its nucleus, so that they are easier to dislocate because the force that the protons exert on them is weaker?
A follow-up question would be: Is it, for the same reason, easier (in terms of energy requirement) to ionise elements of higher atomic number?

Welcome to Chemistry SE! In an attempt to increase the chances of a response, I would recommend adding a reputable reference.

The ionisation energy goes as $(Z_\mathrm{eff}/n)^2$, where $Z_\mathrm{eff}$ is effective nuclear charge and $n$ is principal quantum number. What you're describing is $Z_\mathrm{eff}$, but you can see here both play a role: smaller $Z_\mathrm{eff}$ and larger $n$ means smaller IE. For polarisability, your explanation makes sense conceptually, although I'm not sure if there's a similar formula.

Short answer: yes.
The Wikipedia article on Polarizability has a nice section on this, which I would like to supplement with further quotations from its cited source.$^1$
pg 24

Since electrons are charged particles they respond to electric fields.
...
The electron cloud is said to become polarized in response to the electric field. The ability of the electron cloud to distort in response to an external field is known as its polarizability.

pg 25

We define the polarizability of a molecule as the magnitude of the dipole induced by one unit of field gradient, which works out to be in units of volume. Often, the larger the volume occupied by the electrons, the more polarizable those electrons.

My understanding is that it's less that the electrons are farther away from the nucleus in higher elements, but rather that they occupy a larger volume and are therefore more greatly affected by the electric field.

There is probably a nice analytical expression for the relationship of the polarizability to the terms we're used to seeing in the Schrodinger equation, but I'm not aware of it. In my experience, it is typically calculated numerically using perturbation theory. I did find an example where they worked through the perturbation solution for a particle in a box being subjected to an electric field here, section 4.1.1.

$^1$ Anslyn, Eric V. and Dougherty, Dennis A. Modern Physical Organic Chemistry. University Science Books, California, 2006.

Well, a larger electric field does, of course, mean, that the electrons are more subjected to the shielding effect and less by the charge of the nucleus, because they are also more spaced out. I'm very grateful for the reference describing the relevant parts of perturbation theory, even though I'm as yet unfamiliar with Schrödinger's equation, whence I'll get back to it once I study the latter.


text:
Converting w/v to mole fraction
If I have 10 mL 0.9% w/v $\ce{NaCl}$ solution, I can deduce that the solution contains 0.09 g of $\ce{NaCl}$.
Imagine that this is standard laboratory conditions and 1 ml = 1 g is acceptable.
Is the mass of $\ce{H2O}$ simply $\pu{10 g} - \pu{0.09g}$?
This would lead to mole fractions of $\pu{76.43}$% $\ce{H2O}$ and $\pu{23.56}$% $\ce{NaCl}$.
Update
I realised I had made some errors in my original question. Do the above percentages look right and should I express them as unitless fractions?

How many moles of NaCl are there in 10 mL of 0.9% w/v NaCl? Did you figure this out? Once you do this step, you will see that 99.5 mol percent NaCl it is not possible. BTW, These percentage units are discouraged in analytical chemistry because of their ambiguity.

In a very strict sense, this problem is indeterminate because we do not know about the contribution of NaCl to the solution volume. Let us ignore that. But the numbers are still wrong. How many moles do you get for (10-0.09) g water and 0.09 g NaCl?

Thanks for your patience. 9.91 g H2O => 0.5501 moles, 0.09 g of NaCl => 0.00154 moles. Therefore H2O => 99.72% and NaCl => 0.28%?

Yes, this is what I get as well

You say that you have

10 mL 0.9% w/v NaCl solution

This is an unusual specification and as user M. Farooq pointed out it is ambiguous. So let's assume that this is a reagent that I am to add for some reaction. Is this:
$\quad$(1) 10 ml water (10 grams) and 0.09 grams of NaCl
$\quad$(2) 9.91 ml water (9.91 grams) and 0.09 grams NaCl
$\quad$(3) 10 ml aliquot of 99.1 ml water and 0.9 grams NaCl
Each of these would be slightly different. The gist is that (1) and (2) are different because the densities of the solutions will be slightly different as well as the volumes. (2) and (3) are different because the volumes will vary even though the densities are the same.
Pragmatically such reagent solutions are generally used in a large excess for whatever their use is, so the slight variations don't make any difference in the reaction. However it is just sloppy to give an ambiguous specification when a specific specification could and should be given.
Significant Figures
The other inconsistency with the question posed is with significant figures. The OP has specified 0.9% NaCl, but calculated "mole fractions of 76.43% H2O and 23.56% NaCl". You can't get four significant figures for the mole fraction when the NaCl concentration only has one significant figure.

In which sense the definition is ambiguous? It is uncommon but not ambiguous. It does require the knowledge of density or plugging the latter to be 1, as it would be with % m/m when operating with volumetric flask or cylinders. I read it as it is and not as in 1), 2), or 3). It seems to me that the answer to OP could have been a simple Yes (I did not check the values, by the way), accounting that s/he is aware of the assumption about density.

@Alchimista - So say you were to add this solution as a reagent for a reaction. Exactly what would you make? // As I said I doubt that pragmatically it would make any difference, but I gave three possible interpretations. Realistically I'd probably do some variant of my (3). Water and NaCl are cheap. It is a lot easier to weight out 0.9 grams than 90 milligrams. If I were going to dump 9/10th of a gold or platinum solution then that would be another matter. ;-)

I took it as an exercise. The only possible solution is to assume density to be 1. Of course pragmatically most likely I would have to use it for a work-up or something that probably I won't even worry about. The same happens all the time even if conc is expressed as m/m, that was my point. Point 3) is especially bizarre.

@Alchimista - I hadn't thought of (3) as bizarre, but I was an analytical chemist. I almost always did replicates when doing an analysis.

no doubt. Likely you had followed self-contained complete protocols. This is not.


text:
Why there is no sp2s hybridization in hydrocarbons?
If carbon's $\mathrm{sp^2}$ orbital is overlapped with hydrogen's $\mathrm{s},$ then why it is $\mathrm{sp^2}$ and why not $\mathrm{sp^2s}?$

Why it should be ?

Well, that's simply not a notation that exists.

You seem to have tried to combine the specification of two atomic orbitals to give a molecular bonding orbital. It doesn't work like that. For ethylene (see webpage) the two $\mathrm{sp^2}$ atomic orbitals of the carbon atoms overlap to give a C-C σ molecular bond. For each carbon hydrogen bond the overlap of the $\mathrm{sp^2}$ orbital of the carbon atom and the s orbital of the hydrogen atom give a C-H σ molecular bond.

The sp notation for hybridised orbitals applies only to the orbitals of single atoms. All orbitals that are hybridised must be centred on the same nucleus.
Once you move to molecular orbitals, i.e. bonding between atoms, the sp notation no longer makes sense and is not used. Instead, localised bonds are typically referred to as σ, π, δ (depending on their symmetry with respect to rotation around the bond axis) while orbitals across the entire molecule are typically referred to by their symmetry descriptor (e.g. $\mathrm{e_g}$ orbitals in an octahedral adduct).

Might be a good place to point out that these names are also abbreviations of abbreviations: $\mathrm{sp}^h \equiv \mathrm{s}^\frac{1}{1+h}\mathrm{p}^\frac{h}{1+h}$.

So you would like to hybridize $2s, 2p$ and $3s$ from the same atom, for example. Well ! Hybridization can only be done with orbitals having nearly the same energy. It is the case for $2s$ and $2p$. But it cannot be done with $1s$ and $2s$, or with $2s$ and $3s$. Their energies are too much different !

That is practically correct, but theoretically not. You can of course form a hybrid orbital from those components; physically it doesn't make any sense though. Since hybridisation is an entirely mathematical concept, it's still possible.


text:
Sodium and alcohol reaction
I wonder, what happens when sodium reacts with alcohol? What is the reaction mechanism?
I think that sodium is an electrophile that attacks the nucleophile oxygen atom in the -OH-group of the alcohol. Sodium will take one electron and bond with the oxygen molecule, while the bond between the oxygen and hydrogen will break. Am I correct?

Forms the sodium alkoxide and hydrogen gas. Its a standard way of making fresh sodium alkoxides with methanol/ethanol/propanols etc

@Waylander - Have you ever done the reaction? I'm guessing that you'd have to do it under a blanket of nitrogen or CO2 to keep the alcohol from catching fire.

@Waylander - Thanks, so the reaction mechanism that I guessed is correct?

@MaxW Many times. Depends on the scale, certainly under N2 for larger scale and with cooling as it is exothermic. Smaller scale in a f/c and just let the hydrogen bubble off.

@Csharpyikes  It is a moot point whether the nucleophilic alcohol is the attacking species as it is in solution and free to move around v the sodium is on the surface of a solid and immobile.

This is not an electrophile / nucleophile reaction. That would imply 2 electron processes, but sodium reduction is going to be a 1 electron process.

Sodium will not take one electron, as you propose. No, it is a redox reaction : the sodium atom will donate one electron to the Oxygen atom of the $\ce{ROH}$ alcohol, so as to repel one $\ce{H}$ atom and make an alkoxide ion. At the end, a $\ce{Na^+}$ cation and a $\ce{RO^-}$ alkoxide anion are produced in the solution. Of course the $\ce{H}$ atom will join the next $\ce{H}$ atom to produce $\ce{H2}$ bubbles that gets out of the solution. $$\ce{Na + ROH -> Na^+ + RO^- + \frac{1}{2}H_2}$$ The reaction is not dangerous. It will not catch fire if done in a sufficiently large amount of alcohol. I have done it plenty of times in the lab. No problem ! It is even a good way for getting rid of old residues of partially oxidized pieces of metallic sodium, in case you don't know what to do with them.


text:
Does hyperconjugation only arise from C–H bonds?
I have to an extent understood hyperconjugation but I don't understand why the donation can only occur from a $\mathrm{sp^3}$-hybridised orbital bonded with hydrogen? What about other bonding partners?

It doesn't have to be $\mathrm{sp^3}$ hybridised (the issue with $\mathrm{sp^2}$ and $\mathrm{sp}$ is not so much that they're not $\mathrm{sp^3}$, but rather because of the bond angles and restricted rotation in unsaturated systems which forbid correct overlap). It also doesn't have to be hydrogen, but hydrogen is the most common example.

Hyperconjugation do occur with C-F bond as well where the fluorine gains electron instead of losing(as in the case of hydrogen)

No.
Hyperconjugation is the donation of electrons from a one orbital to another, where at least one of those orbitals is a bonding orbital, and there is a net overlap between the orbitals.
A typical first example taught of hyperconjugation is that which occurs in the conformations of simple alkanes (e.g. ethane). When ethane is in the staggered conformation, electrons from the $\sigma_\ce{CH}$ bonding orbital are donated to the properly aligned $\sigma_\ce{CH}^*$ antibonding orbital of the adjacent carbon atom. The interaction of these two orbitals, the $\sigma_\ce{CH} \to \sigma_\ce{CH}^*$ lowers the energy of those electrons, and thus lowers the energy of the molecule.
Next up is to look at a longer alkane chain (e.g. butane), which will have $\sigma_\ce{CH} \to \sigma_\ce{CH}^*$ interactions like in ethane, but will also have additional interactions involving the $\sigma_\ce{CC}$ and $\sigma_\ce{CC}^*$ bonding & antibonding orbitals. Depending on which conformation of butane you are considering, it is possible to have $\sigma_\ce{CC} \to \sigma_\ce{CH}^*$, $\sigma_\ce{CC} \to \sigma_\ce{CC}^*$, and $\sigma_\ce{CH} \to \sigma_\ce{CC}^*$ interactions.
Continuing in a more general sense, these interactions are not limited to $\sigma$ bonding orbitals. In carbocations, which in principle have an empty p orbital ($\ce{C}_\mathrm{2p}$) on the electron-deficient carbon, there can be hyperconjugation in the form of $\sigma \to \ce{C}_\mathrm{2p}$. (Note that I didn't specify the atoms involved in the $\sigma$ bond; as long as the atoms form a stable $\sigma$ bond, this hyperconjugation interaction can occur, though how stabilizing it is will depend on the size of the orbitals involved.)
There are a variety of hyperconjugation interactions that can occur:

$\sigma \to \sigma^*$
$\sigma \to \pi^*$
$\sigma \to$ empty hybridized orbital
hybridized orbital $\to \sigma^*$
$\pi \to \sigma^*$

For example, in protonated formamide ($\ce{H-C(=O)-NH3+}$), there is hyperconjugation in the form of $\pi_\ce{CO} \to \sigma_\ce{NH}^*$ and $\sigma_\ce{NH} \to \pi_\ce{CO}^*$, as well as $\ce{O}_\mathrm{2p} \to \sigma_\ce{CN}^*$.

Nice answer! I might add that the $n \to \sigma^$ and $\pi \to \sigma^$ interactions are often called "negative hyperconjugation", cf. the IUPAC Gold Book.

Interesting, I hadn't heard of that naming before. (Also, thanks for cleaning up the post.)


text:
Symbolic form of equilibrium constant for ammonium carbamate dissociation

$$\ce{NH2COONH4(s) <=> 2 NH3(g) + CO2(g)} \qquad K = \left(\frac{p(\ce{NH3})}{p^\circ}\right)^2\times\frac{p(\ce{CO2})}{p^\circ}$$
Assuming that you start this experiment with pure solid $\ce{NH2COONH4}$ in an evacuated container and allow it to dissociate until equilibrium is reached, show that the equilibrium constant can be rewritten as
$$K = \frac{4}{27}\left(\frac{p_\mathrm{tot}}{p^\circ}\right)^3.$$

Let $n_0$ be the initial amount of ammonium carbamate and $\alpha$ be the factor describing the extent of dissociation:
$$
\begin{array}{cccccccc}
\ce{&NH2COONH4(s) &<=> &2 NH3(g) &+ &CO2(g)} &\quad &\text{Total}\\
& n_0 && 0 && 0 && n_0\\
& (1 - \alpha)n_0 && 2\alpha n_0 && \alpha n_0 && (1 + 2\alpha)n_0\\
\end{array}
$$
We can express mole fraction $x_i$ and partial pressures $p_i$ for $i$th component as follows:
$$x(\ce{NH2COONH4}) = \frac{n(\ce{NH2COONH4})}{n_\mathrm{tot}} = \frac{(1 - \alpha)n_0}{(1 + 2\alpha)n_0} = \frac{1 - \alpha}{1 + 2\alpha}\tag{1}$$
$$x(\ce{NH3}) =  \frac{n(\ce{NH3})}{n_\mathrm{tot}} = \frac{2\alpha}{1 + 2\alpha}\tag{2}$$
$$x(\ce{NH3}) =  \frac{n(\ce{CO2})}{n_\mathrm{tot}} = \frac{\alpha}{1 + 2\alpha}\tag{3}$$
$$p(\ce{NH3}) = x(\ce{NH3})\cdot p_\mathrm{tot} = \frac{2\alpha p_\mathrm{tot}}{1 + 2\alpha}\tag{4}$$
$$p(\ce{CO2}) = x(\ce{CO2})\cdot p_\mathrm{tot} = \frac{\alpha p_\mathrm{tot}}{1 + 2\alpha}\tag{5}$$
Finally, plugging in the partial pressures into the given expression for the equilibrium constant:
$$
\begin{align}
K &= \left(\frac{p(\ce{NH3})}{p^\circ}\right)^2 \times \frac{p(\ce{CO2})}{p^\circ} \\
  &= \left(\frac{2\alpha p_\mathrm{tot}}{(1 + 2\alpha)p^\circ}\right)^2 \times \frac{\alpha p_\mathrm{tot}}{(1 + 2\alpha)p^\circ} \\
  &= \frac{4\alpha^3p_\mathrm{tot}^3}{(1 + 2\alpha)^3(p^\circ)^3} \tag{6}
\end{align}
$$
This leads me to the right answer if I make an assumption that all of the ammonium carbamate dissociates. The question doesn’t say I could do this, hence why I’m not sure how to proceed. Any suggestions are greatly appreciated for deriving this.

Twice as many moles of $\ce{NH3}$ are formed as $\ce{CO2},$ so the partial pressure of $\ce{NH3}$ is $\frac{2}{3}p_\mathrm{tot}$ and the partial pressure of $\ce{CO2}$ is $\frac{1}{3}p_\mathrm{tot}$, where $p_\mathrm{tot}$ is the total pressure of the gases formed. Also, the carbamate is a solid species, so it is not involved in the equilibrium constant.

Let:
$C$ represent $\ce{NH3}$
$D$ represent $\ce{CO2}$
The expression of $K$ given can be set in terms of $K_P$:
$$K=P_C^2\;P_D\;\left(\frac{1}{P^o}\right)^{\Delta n}$$
$$K=K_P\;\left(\frac{1}{P^o}\right)^{\Delta n}$$
$K_P$ can be set in terms of $K_X$:
$$K_P=K_X\;\left(P\right)^{\Delta n}$$
Substituting above:
$$K=K_X\;\left(\frac{P}{P^o}\right)^{\Delta n}$$
For this reaction, $\Delta n=3$, so we have:
$$K=K_X\;\left(\frac{P}{P^o}\right)^3$$
Now, all we need to do is evaluate $K_X$.
Since total pressure $P$ only depends on products $C$ and $D$, and considering the fact that $C$ and $D$ are in a $2:1$ molar ratio:
$$X_C=\frac{c}{\Delta n}=\frac{2}{3}$$
$$X_D=\frac{d}{\Delta n}=\frac{1}{3}$$
We can now evaluate $K_X$:
$$K_X=X_C^2\;X_D=\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)=\frac{4}{27}$$
Finally, substituting this value for $K_X$ above, we get:
$$K=\frac{4}{27}\;\left(\frac{P}{P^o}\right)^3$$


text:
solubility and solubility product
I am a high school student and I am a little confused in solubility product.
My confusion is that when we talk of solubility of salts we say that suppose if any salt have solubility $s$ it means that whatever concentration of it we add on it will dissolve it will be dissolved by only $s$ concentration into the solution and other will precipitate.
This dissolved form will be present in ionized from in the solution, as degree of dissociation of salt is 100% so it will not be present as $\ce{AB(aq)}$ it will automatically dissociate.
But my question is that when we talk of solubility why do we treat weak bases like $\ce{Mg(OH)2}$ as sparingly soluble salts?
I.e they can be present in molecular form in aqueous solution without dissociating as well, as their  degree of dissociation is not 100%. But still, we treat them exactly like salts and apply all those things which I have discussed above.

Hm, try to read your question after a weak or a month - to simulate it is not your own question. Then ask yourself, if its formulation still seems OK for you. Like proper usage of diacritics, choice of the sentence lengths, paragraph separation of ideas/topics/questions. Even if you understand each word and the topic, it is difficult to get the idea unless you are highly focused. As a guidance, the recommended average sentence length of scientific texts is 20-25 words, with the advice to avoid sentences longer than 35 words.

Let assume the equlibrium chain
$$\ce{Mg(OH)2(s) <=>[K_\mathrm{sp}]Mg(OH)2 <=>[K_\mathrm{b1}] Mg(OH)+ + OH- <=>[K_\mathrm{b2}] Mg^2+ + 2 OH-}$$
.. and evaluate the related equilibrium equations:
$$\ce{[Mg(OH)2(sat)]=K_\mathrm{sp}}$$
$$K_\mathrm{b1} = \frac {\ce{[Mg(OH)+][OH-]}}{\ce{[Mg(OH)2]}}=\frac {\ce{[Mg(OH)+][OH-]}}{K_\mathrm{sp}}$$
$$K_\mathrm{b2} = \frac {\ce{[Mg^2+][OH-]}}{\ce{[Mg(OH)+]}}=\frac {\ce{[Mg^2+]{[OH-]}^2}}{K_\mathrm{b1} \cdot K_\mathrm{sp}}$$
$$K_\mathrm{b2} \cdot K_\mathrm{b1} \cdot K_\mathrm{sp} =K_\mathrm{sp}^{*} = \ce{[Mg^2+]{[OH-]}^2}$$
You can see that effectively, for practical purposes, $\ce{Mg(OH)2}$ acts like a salt, with minor glitches of occurence of $\ce{Mg(OH)2(aq)}$ + $\ce{Mg(OH)+(aq)}$


text:
Priority of the stabilising factors of resonating structures of Organic compounds
When finding the stability order of resonationg structures of certain molecules.. there is some confusion since some structures satisfy certain factors and others satisfy certain other factors. I want to know which factor has more weight to it i.e.. which factor matters the most. Can someone give me a priority order of the factors?

Here are some I know.. Please let me know if I missed any.

More covalent bonds
Octet config and less formal charge
Less charge seperation
Negative charge on electronegative atoms
More stable carbocations.

To know which resonance form contributes more you can follow the following priority
1)Contributing structures in which octet rule is exceeded for second row elements make no contribution.
2)As long as the octet rule  it is not exceeded for second row elements the contributing structure with greater number of covalent bonds contributes more to the resonance hybrid , maximizing the number of bonds and satisfying the octet rule normally go hand in hand.
3) when two or more structural satisfy the octet rule the major contributor is the one with the smallest separation of oppositely charged atoms.
4) among structural formulas that satisfy the the above rule and in which one or more atoms be a formal charge the major contributor is the one in which the negative charge precise on the most electronegative atom.

for stuctural formulas that satisfy 4) too , should we check for the stability of carbocation/anion?

@LOHO Yes, we can check for that further


text:
Freezing point depression and organic molecules
Do organic molecules reduce the freezing point of water in the same way that inorganic salts do?  I've recently been swimming in lakes with a high organic content (e.g. mostly tannins.  There's a lake in this area called Tea Lake).  I've got an infra-red thermometer and readings taken from the water at this time of year before total freeze up have been around -1 to -2 degrees C.  Are these reasonable (accepting of course possible inaccuracies and calibration issues with the thermometer)?
Thanks,
Phil

More likely there's some dissolved salt(s), though high concentrations of organic compounds certainly do depress freezing point, e.g. sugar-water sherbet or "ices".

Alcohol, an organic molecule, does just fine.

When using an IR thermometer you must be very careful when you choose the emissivity parameter that changes the measurement value a lot. Also water is a very reflective medium due to its high dielectric constant and you should be even more careful with your results as the heat your body emits can be reflected as well.

To get a freezing point depression of $\pu{−1.86 °C},$ the solution should be made by adding $\pu{1 mol}$ organic substance to $\pu{1 L}$ water. And $\pu{1 mol}$ per liter is much. For glucose, it is $\pu{180 g}$ per liter. It is nearly a syrup. I doubt your lake has such a high concentration whatever the nature of the solute.
For salts, it is less, because usual salts are made of two ions. For example, $\pu{29 g}$ $\ce{NaCl}$ in $\pu{1 L}$ water is enough to get a freezing point depression of $\pu{−1.86 °C}.$ Anyway, $\pu{29 g}$ salt added to one liter is not negligible. Are you sure of your measurements?

I'm not sure at all of the measurements (this IR thermometer is a crude tool and not designed for water) but I was just interested in the possibility that the lake water was lower than 0 degrees C before freezing.  I'll bring a better thermometer next time.   The main question was whether organic molecules reduce freezing point to the same extent that inorganic salts do.  From your answer I suspect my measurements are not accurate and that the water temperature was not that far below zero.  (My understanding of this is based upon high school chemistry from many years ago).

@Maurice Please format physical quantities appropriately. \pu{…} macro can help you with that. For example, instead of $180g$ $(180g),$ which implies $g$ is a variable (variables are typed in italics), use $\pu{180 g}$ $(\pu{180 g})$ for proper spacing and upright unit symbol. Also, try to avoid asking OP questions in the answer. If you need clarifications, sort it out in the comments.

@andselink I have still typography related doubts, if I should format short things like 180 g or 1 mol by pu{} in free text (as examples of many similar cases). In numerical formula it is clear I should, but in free text it often makes visual disturbance due different and bigger font. If left as it is, it looks more naturally.

@Poutnik You are correct, 180 g and −1.86 °C would make even more sense in this particular case (just make sure that a minus is used, not en dash), but since the original post already used MathJax for everything, so I used it as a basis for editing.


text:
What common observations in chemistry were unexplainable by an early twentieth century chemistry who had no knowledge of relativity
I was reading quite a well-known paper titled 'Relativistic Effects in Structural Chemistry' by PEKKA PYYKKO, and in it, he states that 'it was not until the 1970s that the full relevance of relativistic effects in heavy-element chemistry was discovered.'
I was just wondering if anyone could explain to me from the perspective of a chemist in the early twentieth century who would not have considered the importance of relativistic effects, what were some commonly observed phenomena that at the time chemists were unable to explain, but that today we know can easily explain using special relativity.
For instance I am aware that some chemical differences between silver and gold received a great deal of attention during the history of chemistry, although I am not quite sure why.

Pyykkö has a related review article, Relativistic Effects in Chemistry: More Common Than You Thought (https://www.annualreviews.org/doi/pdf/10.1146/annurev-physchem-032511-143755) in which he gives a comprehensive treatement of the subject, mentioning far more than gold and silver:  "Relativistic effects can strongly influence the chemical and physical properties of heavy elements and their compounds. This influence has been noted in inorganic chemistry textbooks for a couple of decades....[continued]

....This review provides both traditional and new examples of these effects, including the special properties of gold, lead-acid and mercury batteries, the shapes of gold and thallium clusters, heavy-atom shifts in NMR, topological insulators, and certain specific heats."   Why don't you see if you can get access to that paper (it's behind a paywall, but your local library can probably help), and read through it?  That would seem to be the best starting point.  Then if there's anything you don't understand, you could ask about it here.

Thanks for the paper it would be a great start for me!


text:
Why is the radial distribution at the nucleus 0 but not the radial wave function?
The modulus squared of the radial wave function gives the probability of finding an electron in an infinitesimal volume dv. On the other hand, the radial distribution gives the probability of finding an electron in a shell of radius r and thickness dr, or in other words, the probability of finding an electron at a radius r.
I know how the radial distribution is derived and it being 0 at the nucleus makes sense mathematically but not physically. The radial distribution being 0 at the nucleus means that an electron cannot be found in a shell of r=0 & thickness dr ( a sphere of radius dr), but we already know from the radial wave function that there is a non-zero probability of finding an electron in an infinitesimal volume at the nucleus. These results appear to contradict each other.
Is there something wrong in my physical interpretation of the 2 functions?

https://diego.assencio.com/?index=ec0bf1e2547df878846dd841b8d657b2

You state 'but we already know from the radial wave function that there is a non-zero probability of finding an electron in an infinitesimal volume at the nucleus.'
To add to Poutnik's answer, I think that this sentence states the misunderstanding. The radial part of the wavefunction is $R_{n\ell}(r)$ and when normalised
$$\int_0^\infty R_{n\ell}^*(r)R_{n\ell}(R)r^2 dr=\frac{4}{a_0^3}\int_0^\infty e^{-2r/a_0}r^2 dr=1$$
where the $r^2$ comes from the volume element needed to integrate over all angles which is $r^2\sin(\theta)drd\theta d\phi$. The probability from $r\to r+dr$ is then for the 1s orbital $\displaystyle p=\frac{4}{a_0^3}r^2e^{-2r/a_0}dr$ whereas the 1s wavefunction itself has the form $\sim e^{-r/a_0}$. For other orbitals, e.g $2p_{\pm 1}$ the wavefunction is a function with complex numbers so cannot represent anything physical.

For the wave function of $\mathrm{s}$ orbital $\psi(r)$, the radial probability is:
$P(r)=4\pi r^2  \cdot|\psi(r)|^2$.
So even if the differential probability density $|\psi(r)|^2$ is nonzero for $r=0$,  the integral probability over a spherical surface $P(r)$ in $r,r+\mathrm{d}r$ interval raises with $r^2$ and is zero for $r=0$.

i understand the explanation mathematically but I still dont know what that means physically. Can an electron be found at the nucleus or not?

The current wave function based model says it can. But considering the relative size of the region and the WF value, probability is very low.
Also note 2 important implied considerations in the current quantum atomic model:

a point-like character of a nucleus
as the center of spherical coordinates is taken the atom mass center, instead of nucleus itself. So $r=0$ is not at a nucleus, but typically 100 times farther away from a nucleus, compared to a typical nucleus size.

In classical analogy, the surface of a zero-radius sphere contains no adsorbed compound, even if surface concentration in $\pu{g/m3}$ is nonzero.

i understand the explanation mathematically but I still dont know what that means physically. Can an electron be found at the nucleus or not?

@OVERWOOTH You are confusing the radial wave function with something that has physical meaning in reality, which is the radial probability distribution. That function says that the probability of finding the electron at within some infinitesimal $dr$ of the nucleus is zero.

@OVERWOOTCH  ....the probability of finding the electron at within some infinitesimal dr of the nucleus is zero    .......which is true for any other infinitesimal space element, it is not limited to the one at r=0.(this may not be obvious at first glance from these functions of spherical coordinates )

so it is not possible to find an electron within some infinitismal dr of the nucleus, but it is possible to find it in a cube of infinitismal volume dv at the nucleus (as expressed by the non zero wavefunction at the nucleus)?? Don't these 2 statements contradict eachother?

@OVERWOOTCH No, as you misinterpret it.. "anything not infinite" times dV = 0, whatever coordinates is dV expressed with. Respectively, dP=psi^2.dV

I think I finally understand what you're saying. I was interpreting the probability density as the probability of finding an electron in an infinitismal volume when its actually p.density x dv, which is essentially 0. Would it, then, be correct to say that there is essentially no probability of finding an electron at the nucleus? And if yes, what does that say about the widely common statement that "the most probable location for a 1s electron is the nucleus"?

@OVERWOOTCH As nucleus is not really pointlike, so the probability is not zero. Their statement is inprecise, it just mean the probability density has at r-0 a maximum ( what is not exactly at nucleus as said earlier ). But this maximum multiplied by the nucleus volume is nowhere near 1.


text:
Change in enthalpy and entropy when sugar dissolves in water
We know that dissolution of sugar in water is a spontaneous process.
So, change in Gibbs free energy $(ΔG)$ must be negative for the overall process:
$$ΔG = ΔH - TΔS < 0$$
Hence either the enthalpy or entropy must drive the reaction.
Now, after searching for the values, most websites suggest the process is endothermic (i.e. $ΔH > 0).$
But Jan's answer has me convinced that the overall entropy of a system should decrease (i.e. increase in entropy of sugar molecules is outweighed by decrease in entropy of water molecules).
So, what's the real reason for reaction to be spontaneous if neither entropy and enthalpy are driving the reaction?

This doesn't answer your question, but serves as an important point of information:  You can have a reaction where the inherent enthalpy and entropy are both unfavorable, yet it will still proceed spontaneously if you are away from equilibrium, because of the entropy of mixing.  See:  https://chemistry.stackexchange.com/questions/129589/how-can-we-define-a-non-spontaneous-reaction/129628#129628

How can you say that neither energy nor enthalpy drive the reaction?

@ultralegend5385 For a reaction to be driven in a direction, ΔG must be negative in that direction. For ΔG to be negative, either ΔH must be -ve or ΔS must be +ve. (Neither of which is happening in this situation)

I think that no matter what, since the process is spontaneous, $\Delta G<0$ . As you said, $\Delta H>0$ is true for certain. So, that implies $\Delta S$ must be positive. I think that @Jan's reasoning has some flaws.  Read the comments and other answers in the link that you mentioned, they do point out that it is wrong.

What about the entropy change of mixing?

@ChetMiller can you please elaborate on that as an answer?

A lot less entropy to be gained does not imply that $\Delta S < 0$.

Since solubility of sugar in water increases with temperature, the change in enthalpy of the sugar-water system is positive. Since the sugar-water system is absorbing heat(at constant pressure, change in enthalpy is the heat absorbed) and since entropy change of an irreversible process is greater than that in an reversible process, the entropy of the sugar-water system will increase. So at constant temperature and pressure the enthalpy change is positive, entropy change is positive and the the change in Gibbs free energy is negative(before equilibrium is attained) . The process is spontaneous. By the time the system reaches equilibrium the enthalpy change will have been balanced by the change in entropy and so the change in Gibbs free energy is 0.

The entropy change for ideal mixing is $\Delta S=-R(x_A\ln{x_A}+x_B\ln{x_B})$

Here, 'x' refers to mole fraction, right?

Yes, that is correct.


text:
How is NaOH a base?
Firstly, how is $NaOH$ created? My teacher said that before it is $NaOH$, it is $NaO^-$.
$$NaO^- + H_2O = NaOH + OH^-$$
If so, that means that $NaO^-$ is a base, which would make $NaOH$ it's conjugate acid. However, $NaOH$ is a base. So, perhaps my teacher said wrong, and the fact is, the "prelude" to $NaOH$ is $NaOH_2^+$, i.e.:
$$NaOH_2^+ + H_2O = NaOH + H_3O^+$$
Meaning that $NaOH_2^+$ is the acid, and $NaOH$ is the conjugate base.
Okay, so assuming the latter is correct, I still don't see how NaOH is a base.
Bases, in their most general definition, are substances that easily accept proton(s). Now, this would make sense for $NaOH$ if the equation looked like this:
$NaOH + H_2O = NaOH_2^+ + OH^-$
This would make the solution more basic, due to the higher concentration of $OH-$, though considering how strong of a base $NaOH$ is, that means that $NaOH_2^+$ is a weak acid, and the solution would therefore revert back a lot (double arrow thingy).
However, I have never seen the equation like that. Instead, I've seen it like this:
$$NaOH + H_2O = Na^+ (aq) + OH^-(aq)$$
Now, that would increase the concentration of $OH-$, would would make the solution more basic, but there is no proton being accepted here. The so-called base that is $NaOH$ does not accept a proton, but rather releases an $OH^-$ group, which is quite the opposite of accepting a proton. Now, this would make sense if $NaOH$ was classified as "a basic salt", as in, a salt that makes a solution more basic. So, with such an explanation, $NaOH$ is just a salt, not a base, but it is a salt that makes the solution more basic.
Now, since it is a substance that makes a water solution more basic, it would by definition become an Arrhenius base. I can accept that, because the Arrhenius base definition only requires the substance to make the water solution more basic. $NaOH$ does that. However, I cannot see in any way how $NaOH$ fits the Brønsted-Lowry definiton of a base.
Please explain to me simply, but thoroughly, if you can. For some reason, the Acid-Base part of chemistry has evaded my comprehension for a while now.

Find other teacher. NaOH is a ionic compound, consisting of ions Na+ and OH-, that dissolve in water as Na+(aq) and OH-(aq).  OH- ions are scavenging H3O+ ions, forming water, according to the equilibrium $\ce{H3O+ + OH- <=>> 2 H2O}$.

Forget NaO- like a nightmare, forget NaOH2+, and read about the electrolytic dissociation instead.

Wow you have been severely misled my friend.

Btw, if you want to know how NaOH is actually made industrially, you can read up on chloralkali process. But I suggest you read up on acids and bases first. Wrap your head around the basics before diving deeper.

@Poutnik I am unfortunately stuck with this teacher. She relies solely on basic, limited rules, instead of explaining the underlying mechanisms, which only leads to confusion (for me at least). But, to be clear, are you saying that NaOH is not a base? Simply an ionic compound, making it a salt?

A ionic compound, releasing the base OH- as one of its ionic components. Base(1) is in the Broensted-Lawry theory context able to chemically bound H+ ( rather H3O+ or H3O+.nH2O). Base(2) is than substance to cause water solutions to be alkaline..

@Poutnik I see, and Base(2) is the Arrhenius theory, correct?

Not necessarily, it is from practical point of view. E.g.  NH3 is base(1) and base(2), but not base(Arrhenius), as it does not release OH-. It increaces OH- by capturing H+. NaOH(s) is base(Arrhenius), as it releases OH- when dissolved.

Let me analyze your sentences gradually, and try to correct the numerous mistakes.

Your teacher says that "before it is $\ce{NaOH}$, it was $\ce{NaO^-}$". No. First $\ce{NaO-}$ does not exist and cannot exist. Second NaOH has not been created out of an ion. Sodium hydroxide is usually produced from the sodium metal after reaction with water according to : $$\ce{2 Na + 2 H2O -> 2 NaOH + H2}$$ This equation may also be written in order to show the ions :
$$\ce{2 Na + 2 H2O -> 2 Na^+ + 2 OH^- + H2 }$$

Your teacher said that $\ce{NaOH}$ is the conjugate acid of the base $\ce{NaO^-}$. No ! it is impossible to find a structure for $\ce{NaO^-}$, as $\ce{Na}$ cannot be covalently bound to an oxygen atom. $\ce{Na}$ can only be bound to an oxygen atom by an ionic bond, and this bond will be immediately broken when dissolving it into water. Atoms included inside a polyatomic ion are always bound by covalent bonds, never by ionic bonds.

Your teacher said that "the prelude of NaOH is $\ce{NaOH2^+}$". First there are no "preludes" to $\ce{NaOH}$. Second $\ce{NaOH2^+}$ does not exist and cannot exist. If such an ion was imagined in any solution, it would immediately get decomposed into $\ce{Na+ + H2O}$

Your teacher believes that a reaction can produce simultaneously $\ce{NaOH}$ and $\ce{H3O+}$. This is wrong ! As soon as the smallest amount of $\ce{NaOH}$ and $\ce{H3O+}$ are in contact one another they would react producing $\ce{NaOH + H3O+ -> Na+ + 2 H2O}$

Your teacher said that $\ce{NaOH2^+}$ is a weak acid. No ! It is not a weak acid, as it doe snot exist.

You said that if NaOH is a base, it must have $\ce{NaOH2^+}$ as conjugated acid. No ! $\ce{NaOH}$ is not a base. It only contains the ion $\ce{OH^-}$ which is a base. $\ce{NaOH}$ is not a molecule. It is a juxtaposition of the two ions $\ce{Na+}$ and $\ce{OH-}$- When dissolved into water, this two ions are separated and can react indepentedly from one another.

It is right to say that $\ce{NaOH}$ behaves like a basic salt.

Apparently your teacher has never mentioned the fundamental property of $\ce{NaOH}$ which is its dissociation in water according to $$\ce{NaOH -> Na+ + OH-}$$ and this is why $\ce{NaOH}$ is basic in solution

Okay, so a fundamental misunderstanding here is, among others, that NaOH is even a base? Within the Arrhenius definition, it is a base, but not within the Brønsted-Lowry definition, right?

@A Kväle. You are right. According to Arrhenius, NaOH was a base, and it may still be for some contractors and technicians. But within Brønsted-Lowry definition, only OH- is a base.

However, I cannot see in any way how NaOH fits the Brønsted-Lowry definiton of a base.

Here are the two chemical equations that show you how the hydroxide anion present in an aqueous NaOH solution acts as a Brønsted-Lowry base:
$$\ce{OH-(aq) + H3O+(aq) <=> H2O(l)}\tag{1}$$
This shows that adding NaOH to water lowers the hydrogen ion concentration (i.e. raising the pH) by hydroxide reacting with hydronium ions to form water.
$$\ce{OH-(aq) + H2O(aq) <=> H2O + OH-(aq)}\tag{2}$$
This one is a bit strange, and the reason why NaOH is not your typical base (like ammonia or sodium acetate etc.). The hydroxide ion reacts with water to abstract a proton, turning into water. At the same time, water turns into hydroxide. The proton transfer is difficult to see because reactants and products are the same unless you somehow tag the two oxygen atoms involved.

My teacher said ...

As for your teacher, see Maurice's answer and the comments (which already contain the content of my answer but I think it helps to have it as an answer specifically about the question of Brønsted-Lowry bases).

I can see that the $OH^-$ is a Brønsted-Lowry base, but the hydroxide =! NaOH. The hydroxide is a base, obviously, but how does that make NaOH a base, by the Brønsted-Lowry definition? Just because NaOH can release a part of itself, said part being a base, how does that make it a Brønsted-Lowry base?


text:
Find the molarity and molality

We take 10mL of $H_2SO_4$ with a $96\%$ of mass and density $1.84g/cm^3$, and we add it into a flask that has $0.5L$ of capacity, filled up to half of the flask with distilled water. We shake the mixture and we add in more distilled water up to the level of $0.5L$. Find the molarity and molality.

$\mathbf{Attempt. }$ My problem comes up when trying to find molality, I've found molarity like this (and this meets with the book's solution):
Let $M$ be molarity, defined as $M=\frac{n_s}{V_{T}}$ where $n_s$ is the number of moles of the solute and $V_T$ the total volume of the mixture. As we know $V_s=0.01L$ and $d_s=1.84g/cm^3=1.84g/mL=1840g/L$. We can find the mass $m_s$: $$m_s=d_s\cdot V_s=18.4g$$ and then find the moles of the solute $n_s=18.4g\cdot\frac{1 mole\ H_2SO_4}{98g\ H_2SO_4}\approx 0.188 moles \ H_2SO_4$ and then the $V_T=0.5L$ so $$M=\frac{0.188moles}{0.5}\approx 0.38M$$ and this part agrees with my book now, the issue is with molality.
$\mathbf{Doubt/The\ part\ where\ I'm\ stuck.} $ With molality, which I'll denote $m$, defined as $m=\frac{n_s}{m_{solvent}(kg)}$, where $m_{solvent}$ is the mass of the solvent in kg. Okay well, I attempted this and it gave me $245.2m$ while the book's answer is $0.37m$. My attempt:
We know $n_s$, and $m_{solvent}=m_{mixture}-m_{solute}$, and from the percentage given of the solute $96\%$, we also know that $0.96m_{mixture}=m_{solute}\implies m_{mixture}=\frac{m_{solute}}{0.96}$, and we finally plug that in the previous equation: $m_{solvent}=\frac{m_{solute}}{0.96}-m_{solute}$, plugging $m_s=18.4g$ we get $m_{solvent}=0.767g$. Converting to kg, we get $m_{solvent}=0.000767kg$. And finally plugging into the molality: $$m=\frac{0.188moles}{0.000767kg}\approx 245.11m$$ where I have gone wrong so that it does not meet with my book's solution?

Your molarity calculation is somewhat incorrect. You forgot to take % wt/wt =96 into account. Your number are matching the book answer because of 96% but a similar calculation will fail for HCl which 37%.

oh! okay but then how should I take it into account? Is it because the volume $10mL$ and density $1.84g/cm^3$ is actually of the whole mixture? Because if so, we can find the mass of the mixture, and take into account the 96% to find the actual mass of the solute

You know the density and you know that mass. But that mass is only 96% sulfuric acid. Multiply the calculated mass by 96/100.

okay but that doesn't change a lot the molality, does it? I mean by multiplying 0.96*18.4 gives something less than 18 and it somewhat changes the molarity but not the molality and I'm still stuck on what did I do wrong on the molality

Okay, I will have a look at it.

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All in all this isn't a well stated problem. The molarity can be solved, but the molality can't really be solved.
The wording "with a 96% of mass" is odd. I'm not sure if this is your interpretation of the problem or if that is what is literately written in the book. I'll interpret the phrase to mean that the sulfuric acid was 96% sulfuric acid by weight.
For both problems the grams of sulfuric acid are:
$$\mathrm{grams\ }\ce{H2SO4} = 10 \times 1.84 \times 0.96 = 17.66$$
$$\mathrm{moles\ }\ce{H2SO4} = \dfrac{10 \times 1.84 \times 0.96}{98.079} = 0.1801 $$
Molarity
So the molarity, $M$, of the solution will be:
$$\mathrm{Molarity} = \dfrac{10 \times 1.84 \times 0.96}{98.079 \times 0.50} = 0.360 $$
Molality(1)
The problem with the molality calculation is that volumes are not additive. I'll admit that for two or three significant figures the calculation sort of works.
So for the 500 ml of solution, you must assume that 10 ml is sulfuric acid, the other 490 ml of volume is water. Water is assumed to be 1.00 kg/liter, so to figure out how many moles of sulfuric acid per kilogram of solvent we'll need to multiply the 0.360 moles per liter of solution by a correction factor of 1/0.980 which gives a value of 0.368 for the molality.
Molality(2)
As a second way to calculate the molality let's divide moles of $\ce{H2SO4}$ by the kilograms of water.
From above we know that the 500 ml of solution contained 0.1801 moles of $\ce{H2SO4}$.
Again we have to assume that the volumes are additive so that 10 ml of concentrated sulfuric acid and 490 ml of water yield 500 ml of solution. Given no temperature we can just assume that water is 1.00 g/ml, so 490 ml of water has a mass of 490 grams or 0.490 kilograms.
$$\mathrm{Molality} = \dfrac{0.1801}{0.490} = 0.368 $$

To perhaps beat a dead horse here let's back calculate the volumes for one liter of a similar solution. Steffen's Chemistry Pages give that a 4% solution of sulfuric acid by weight has a density of 1.0250 g/ml, 41.00 g sulfuric acid per liter and a molarity of 0.418 at 20°C.
To get 41.00 grams of sulfuric acid we'd need
$$\mathrm{ml\ conc\ sulfuric\ acid} = \dfrac{41.00}{1.0250\times 0.96} = \pu{41.7 ml}$$
But since water is 1 g/ml and the solution is 96% water
$$\mathrm{ml water} = 1000 \times 0.96 \times 1.0250 = \pu{984 ml}$$
$$41.7 + 984 = 1025.7$$

I'm still unsure where I've used that the volumes are additive on my attempt to find the molality. Could you please pin-point it out?

@Xetrez - You messed up the molality calculation so badly I don't have a clue as to what you were doing.

that doesn't help me much, I'd love to know what concept is wrong

@Xetrez - in your calculation is $$\mathrm{m_{solvent} = m_{mixture} − m_{solute}}$$ supposed to mean that $$\mathrm{molality(solvent) = molality(mixture) - molality(solute)}$$

no, it's true that I use the same letter for both, but I mean mass for that. I only mean m as molarity once when naming the formula to get it and to express the final result

I will elaborate on the problem a little bit in more detail. Your molarity part should have been solved. The problem is your original textbook/assignment question is that it does not tell us the total mass of the solvent for the molality part. If they had told us how much mass of water was added instead of "total" volume", the problem would have been better stated.

@Xetrez - How did you go from $m_{solvent}=m_{mixture}-m_{solute}$, to $0.96m_{mixture}=m_{solute}$ That doesn't make any sense.

that's not an step, I'm mentioning both equations that we know. And then I substitute one into the other one. The second equation can be obtained by the 96% mass thing


text:
Find the molality

Find the molality of a mixture formed by mixing $200cm^3$ of $HNO_3$ that has a $69\%$ of richness and density $1.41g/mL$, with $1L$ of the same acid with $1.2M$ and density $1.06g/mL$.

Attempt. The question asked the molarity too, and I think I've found it, but the issue comes with the molality. Pretty stuck here, I'm having troubles finding which is the solvent and which is the solute, is there any water going on here that it is assumed to be known? I know molality, let that be $m$, is defined as $m=\frac{n_s}{m_{solvent}}$, where $n_s$ and $m_{solvent}$ are the moles of the solute and the mass of the solvent respectively. But I want to be sure I know what the mixture is about. Are we mixing some mixture that has water and acid, with another mixture that has water and acid? Or is this just acid with acid, and the solvent is the acid with more proportion?
Continuation. Now that I know it is a mixture of 2 mixtures with both water and acid, let me post what I've tried:
I've used the letters to denote the first mixture, the letters with ' to denote the second mixture, and the letters with '' to denote the final mixure of both mixtures. $m''=\frac{n_s''}{m_{solvent}''}$, where $n_s''$ and $m_{solvent}''$ are the moles of the solute and the mass of the solvent of the new gas, denoted by the ''. I've found $n_s''$ by doing this: $$n''_s=n_s'+n_s$$, not sure if that thing above is true, but assuming it is, I've separated both mixtures with the information we are given about them: $$\text{Mixture }1: \begin{cases}m_{mixture}=V_{mixture}d_{mixture}\\ 0.69m_{mixture}=m_s\end{cases}$$ $$\text{Mixture }2: \begin{cases} 1.2=\frac{n_s'}{V_{mixture}'}\end{cases}$$, by simply just plugging the information we know and some unit conversions, with the equation above $$n''_s=n_s'+n_s$$ we get $$n_s''=1.2+0.69\cdot 0.2L\cdot 1410g/L\cdot \frac{1mole\ HNO_3}{63g\ HNO_3}\approx 4.289 moles$$ and we would have one part of the $m=\frac{n_s''}{m_{mixture}}$ done. Not sure on how to get the $m_{mixture}$ and not sure if what I've done is correct either.

Yes, we are mixing some mixture of water and acid with another mixture of water and acid.

okay, let me add in then what I've tried so far to find the molality

done, i'll be back in half an hour if you answer.

Get rid of the word "gas". The rest sounds about right. Now, what's that with $m_{mixture}$? You don't need that; you need $m''_{solvent}$. Well, go on, find it. The solvent is water. The mass of both solutions can easily be found. Whatever of them is not acid, is necessarily water.

yeah my bad I actually meant $m_{solvent}''$

For this problem too you'll need to make the assumption that the volumes are additive which isn't really true.

well, I'm guessing we have to assume so.

to find $m_{solvent}''$ I did this: $m_{solvent}''=m_{mixture}''-m_{solute}''$, and then as $m_{mixture}''=m_{mixture}'+m_{mixture}$, same stuff with the mass of the solute. We substitute in and get $$m_{solvent}''=(V_{mixture}'d_{mixture}'+V_{mixture}d_{mixture})-(\frac{1.2moles}{63g}+0.69V_{mixture}d_{mixture})$$, so far, without plugging numbers in, is this correct?

Ok, you correctly calculated the number of moles in the final solution with:
$$n''=n'+n$$
and
$$n''=\pu{1.2 M}\times \pu{1 L} + \dfrac{0.69\cdot \pu{0.2 L}\cdot \pu{1410 g/L}}{\pu{63.012 g/mol \ce{HNO_3}}} =\pu{4.2880  mol}$$
However I used 63.012 for the molecular mass of nitric acid instead of 63.
Now to calculat molarity you divide thus
$$\dfrac{4.2880}{\mathrm{liters\ of\ solution}} = \mathrm{molarity\ of\ \ce{HNO3}}$$
and to calculate molality you divide thus:
$$\dfrac{4.2880}{\mathrm{kg\ of\ solvent}} = \mathrm{molality\ of\ \ce{HNO3}}$$
At this point you have to assume that the volumes are additive, which isn't quite true..., so you end up with 1.2 L of solution. So for molarity:
$$\dfrac{4.2880}{\pu{1.2 L}} = \pu{3.57 M}\ \ce{HNO3}$$
Now the kg of solvent is a bit more convoluted. For the concentrated solution the mass of water is:
$$0.31\cdot \pu{0.2 L}\cdot \pu{1.410 kg/L} = \pu{0.08742 kg}$$
for the more dilute acid solution the mass of water is:
$$\pu{1.00 L} \cdot{(\pu{1.06 kg/L} - \pu{1.2 mol/L}\cdot \pu{0.063012 kg/mol} ) = \pu{0.98439 kg}}$$
so the total mass of water is
$$ \pu{0.08742 kg} + \pu{0.98439 kg} = \pu{1.07181 kg}$$
Now to calculate the molality
$$\dfrac{4.2880}{\mathrm{kg\ of\ solvent}} =  \dfrac{4.2880}{\pu{1.07181 kg}} =  4.00\ \mathrm{molal}$$


text:
How stable are surfaces functionalized with APTES?
I'm wondering specifically about magnetite micro particles functionalized/silanated with (3-aminopropyl)triethoxysaline (APTES), though I assume the stability would be the same for most surfaces.
By stability, I mean can the APTES "fall off" the surface over time or in extreme conditions (pH, temperature, pressure, etc.)?

Silanes do not bind very well with transition metal oxides. Many people tried covalently linking titanium oxide and zirconium oxide with silanes. The results were disappointing because it would wash off soon. Ignore such junkyard papers which say that the silane was covalently bonded metal oxide. Amino silane is no exception. If the silane is polymerized around the metal nanoparticles, then it is a different story.

@M. Farooq what would the stability of APTES polymerized around iron oxide particles be?

Stability is an ill-defined term. Chemists should always mention stability with to something. Polymerized APTES will disintegrate at high pH. What is your pH range?

Good point. My pH and temperature ranges are actually reasonable: pH 5.5-7 and 30-50 C. The issue is time. I need the functionalization to remain at least 50% in tact (relative to the starting functionalization measured as mg APTES per gram support) over several months with essentially continuous liquid flow over the surface.

I think you basically want amine groups on iron oxide particles? See if polymerization works. What is your procedure for "bonding" APTES to iron oxide?

Yes; protein immobilization is the end goal here. Sonication of the APTES and particles in methanol-water at 20 khz as described in this paper: https://www.researchgate.net/profile/Aidin_Bordbar-Khiabani/post/Simple-method-for-direct-coating-of-APTES-on-surface-of-Iron-oxide-Fe2O3/attachment/5cb10d443843b01b9b99a06b/AS%3A746987641114624%401555107140026/download/wang2016.pdf

They are using the word coating and apparently the nanoparticles are formed in the presence of APTES so worth trying to repeat their work. In the presence of ammonia and water the silane must have polymerized.

Interesting. I didn't know the silane would polymerize with ammonia and water. Does that very significantly increase the stability? Do you think it could ever remain functionalized for months, or is that a pipe dream?

I don't think so that ATPES will remain there for months. Silanes have a tendency to polymerize in the presence of water, acids or bases. My experience is limited to bonding chemistry related to silica particles. The funny thing is that silica bonded to propylamino silane hurts the silica.

Hmm... I will need to pursue a different route then. Thank you so much for your help! You've given me a lot of great information!

Good luck. Study the silica bonding literature first, search Langmuir journal as well. It did have a stability article.


text:
How to think about absorption and emission rates and intensity for stimulated emission?
I learned Fermi's Golden Rule applied to absorption and stimulated emission and encountered the following question:
Since absorption and emission rates are the same for stimulated emission, why can't we say that the incident beam is not attenuated by the sample?

Hence the need for a population inversion

@JonCuster I understand population inversion means there are more molecules at the excited than at the ground state. Can you please elaborate on how this idea is connected to the question?

@EdV I know that stimulated emission occurs when the excitation frequency matches the energy gap frequency. I don't see exactly how that can be useful for this question though.

You also have spontaneous emission, at least. So if absorption and stimulated rates were exactly equal, and there was no other de-excitation pathway, the medium would be transparent, i.e., have no net absorbance at the relevant wavelength. So you cannot get there with a passive medium by blasting it with light. With an active medium, like the Xe flashlamp-pumped ruby rod, you get population inversion, so incident light at 694.3 nm results in even more intense 694.3 nm light (from the stimulated emission, i.e., laser action) being emitted.

@EdV Is it correct to understand that, in the case that the ground and excited populations are the same, the beam is attenuated (incident and exiting intensities differ) because of other relaxation pathways such as fluorescence and non-radiative relaxation?

Ignoring degeneracies, you are not going to get them the same, unless some pumping mechanism is involved. If you simply illuminate a solution of a fluorescent substance, e.g., rhodamine 6G laser dye, some of the incident light is absorbed, so the exit intensity is lower than the incident intensity. And there is fluorescence and heat produced.

Doesn't the absorption and stimulated emission rates being equal imply that the ground and excited states populations are equal, since the rate is proportional to the population? Please correct me if I misunderstood: what I got from your previous comment is that, for a passive medium like a solution of molecules, if the absorption and emission rates were the same, the exit beam would always have lower intensity than the incoming beam because of the energy loss in non-radiative relaxation.

Both the rates (abs & stim. emission) are proportional to the light intensity and difference in population between the states which is why we get attenuation in passing light through a solution as upper level popl'n is so small. In a laser the upper population has to be greater than lower one is, and so we can get gain if a photon of correct wavelength passes through. To get a laser three or four levels are necessary as 2 levels cannot lase; transition up is always equal to transition down when pumped hard enough.

In my opinion you are over thinking at this. If you are stable at a situation in which the two rates equal, then in principle and in ideal condition a photon matching the levels should stimulate emission. So the beam isn't attenuated but enriched. If you want lasing you need inversion, so some kind of pumping.

My final comment: have a look at Einstein coefficients.


text:
Activated carbon + zeolite: Does relative humidity affect retention of VOCs?
I have a very common air purifier: HEPA filter with a heavy duty helping of activated carbon and zeolite.  My particular filter is designed for 5 years of operation under typical home conditions.  In reality, I expect a few years, if not 5. I found this to be the case in the past.
I put an air purifier in the bedroom, which I keep colder than the rest of the apartment unit.  After using a new filter for about 6 months, I noticed a sign that the filter was done: A sour smell.  Very odd, since another purifier was working in the living room and kitchen for 1.5 years with no such sign.
Strange also that the smell was greater when the thermostat was turned down for the night.  I would have expected higher temperature equals more energetic gases, and lesser binding with the filter material.  That points to less effective filtration with higher temperature, which was the opposite of what I observed.
The only other explanation I could fathom was relative humidity (RH).  Colder air has less vapour capacity, so RH goes up for the same amount of water in the air.  Maybe that was the cause (somehow).  I got some evidence that this was the case.  When I moved the dehumidifier into the bedroom, I could hardly notice the smell from the air purifier anymore.
The dehumidifier was dialed to 35%, but I actually had to make it operate continuously (unregulated) in order for it to run.  Without continuous operation, it would only regulate to down to 35% RH (plus/minus 3%).  With continuous operation, the readout showed 30%.  For all I know, this could be the lower limit of the sensor, but it does show that the RH is lower than under regulated operation.
I know that reliable RH measuring is tough, so I view the dehumidifier readout as a rough indicator only.  But it does seem to show that higher RH yields smellier filter output, possibly indicating that captured VOCs are being released.
I'm not a chemist, but was wondering whether those with expertise in the field can comment on whether retention of trapped "volatile organic compounds" (VOCs) actually lessens with higher RH?  Or perhaps there is a 3rd causal factor, leading to correlatino of RH with less filter retention?  Or maybe it's not even that -- maybe the colder temperature or higher RH simply makes one's sense of smell more sensitive.

For those who might have been wondering: https://en.wikipedia.org/wiki/Volatile_organic_compound

@Martin: Thanks.  I added that to my question.  I have had a year to observe the problem.  It seems that if I run the machine 24/7, the smell is much abated.  I know that it's not just me acclimatizing to the smell because I test the smell after coming into the dwelling, before I become acclimatized.  However, I can't run the machine 24/7 in the summer because I keep windows open.  Running the machine would be like trying to clean all of the outside air, and I live close to a major traffic intersection and a diesel bus transit hub.

I should add that it doesn't seem to make sense that running the unit 24/7 eliminates the smell most of the time and makes it barely noticeable the rest of the time. One explanation that was provided elsewhere was that the filter was contaminated with bacteria, which feeds off the impurities caught by the filter. I'm not necessarily dismissing this, but I would expect this to be worsened by that 24/7 operation.


text:
Is Molarity constant?
My friend and I were debating about if Molarity is constant in a disolution if this one changes volume due to evaporation.
There was this problem that stated that 125 [mL] of sodyum chloride 0.122 [M] evaporates over time and final volume becomes 113 [mL]
I say Molarity keeps being constant, he says no because volume changes.
What do you think?

$c_1=n/V_1;$ $c_2=n/V_2.$ What do you think the relation between $c_1$ and $c_2$ is when $V_1>V_2?$ Also, Wikipedia.

If you knew what exactly molarity is, there would be no debate. As you did not, why did you not rather search it up, instead of asking? It would be much faster.  //  Do people obtain salt by evaporation of seawater, or are they liers ? Does have saturated NaCl solution the same molarity as e.g. 0.1 M NaCl solution ?

Presumably this is an aqueous solution. So more fundamental than andselisk's question: (1) Does water evaporate from a sodium chloride solution?  (2) Does sodium chloride evaporate from a sodium chloride solution?

Interestingly, I was reading someone's thesis on corrosion yesterday and airborne salinity is indeed a problem near coastal areas, so in one sense sodium chloride can also be lost in the air too!

@M.Farooq But it s rather dynamic separation of solution into aerosol, eventually evaporated later. I have head you need on some island to work with tap water ventils/regulators every 2 weeks, otherwise they get "frozen" due salt and you cannot turn them anymore.

@Poutnik, You are right. The aerosol is formed first.

Do you have any idea of concentration in real life? This is a surprising question. If you read molarity M as moles number then nothing change. But it is not. What is constant is n. M = n / V

As molarity is defined as the ratio of molar amount of the dissolved solute and solution volume:
$$c=\frac nV$$
molarity increases when solvent ( like water ) evaporates.
Molarity can be considered constant only in context of later refilling the evaporated solvent and mixing the solution. Then molarity returns to the original value. It assumes the solute ( like sodium chloride ) content is constant, not undergoing any reaction.