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What happens when you mix Hypochlorous Acid (HClO) and Hydrogen Peroxide (H2O2) in water (H2O)
I currently use Hypochlorous Acid (HClO) in my plant reservoir. I really want to use Hydrogen Peroxide (H2O2) also. Is there any reason why not to use Hypochlorous Acid (HClO) and Hydrogen Peroxide (H2O2) together in water (H2O).

Could you edit your question to include the concentrations that you are talking about, and to explain what a plant reservoir is? Otherwise, the question might be closed as too broad. Also, can you edit your question to indicate what the adverse consequences are that you want to avoid (change in pH, explosion, exothermic reaction, one reagent not stable in the presence of the other)? It might be unclear what you are asking otherwise.

This is a famous reaction to generate singlet oxygen. I am not how do you generate HOCl. If these two components (hypochlorite ion, and peroxide ion) are reacted under certain alkaline conditions, oxygen is immediately evolved with a red glow. One is left with chloride ions and water. Video link here
https://www.youtube.com/watch?v=vMi8kKXgshk


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Is xylene an acid?
What is the pH of xylene, and is it considered to be a corrosive substance?
Specifically I am interested in how xylene might interact with electrical conductors and components, which contain copper, aluminum, and steel as well as plastic.
I can't seem to find a pH reading online (checked Wolfram Alpha, Wikipedia, and SDS for xylene and ethylbenzene). The SDS I found did not contain the symbol for corrosive materials, but could it be even lightly corrosive?

Xylene is not acid and is likely to attack the insulation on electrical components, not the metal conductor.

Interesting, thank you. I was so preoccupied with the conductor I didn't think about possible interactions with the insulator.

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What is the activation energy barrier of uncatalyzed nitrogen fixation?
I'm interested in the activation energy ($E_\mathrm a$) of uncatalyzed aqueous nitrogen fixation:
$$\ce{3 H2 + N2 -> 2 NH3}$$
Biology performs this reaction with the nitrogenase enzyme at room temperature. Humans have come up with ways of doing this reaction with a metal catalyst like rubidium at ~300 °C. I am not interested in the activation energy of nitrogen fixation when catalyzed by biology or by industrial chemists. I want to know the activation energy of the uncatalyzed reaction.
The most luck I have had is with a book titled "Ammonia: Principles and Industrial Practice", which states, "According to estimates[99], the homogenous reaction of nitrogen with hydrogen to form ammonia in the gas phase requires an activation energy of 230–420 kJ/mol." Reference [99] is cited as "W. G. Frankenburger, Catalysis 1954-1960 3 (1955) 171" which I haven't found a copy of yet. I would prefer to have a measured (not estimated) value that is a bit more precise.
Any help would be much appreciated!

For an activation energy, you first need a sensible mechanism under some defined conditions. I don't think there is one. The thermal decomposition is sure to take another route. You can of course make assumptions, but I strongly doubt the number has any scientific value. Sorry. I'd be glad to get convinced of the opposite.

Side note: 400 kJ/mol is equivalent to a temperature of about 50 kK. That's right, kilo Kelvin.

@Karl I understand what you mean that a specific mechanism is needed to define an activation energy. I'm not sure what the mechanism would be. I would imagine though that at ~2000 K it would still be thermodynamically favorable for nitrogen and hydrogen to make a fair bit of ammonia, and some small fraction of the molecules that have enough energy would react via some mechanism. I am interested in the activation energy of that scenario.

I think the point is you won't get any ammonia at all. You'd need a three- or four-particle reaction, and if those 400kJ/mol are anywhere near a realistic value, at 2000K you will get just nothing. I'm sure you could do it at extreme pressures, but that's likely a different reaction then.

What are you actually interested in, i.e. why are you asking this? The number itself seems not so terribly interesting, perhaps we can string up sth. else?

@Karl I'm interested in the rate enhancement of the nitrogenase enzyme complex, and a bunch of other other enzymes. There are cool papers like this one which show the rate enhancement of a bunch of enzymes. The paper linked doesn't describe the rate enhancement of some of the oldest enzymes like nitrogenase or the enzymes in methanogens or CO consumers. I would really like to know how effective these old enzymes are at speeding up chemical reactions, and what kinetic barriers they had to overcome.

Sorry, but that number is just pointless. The uncatalysed reaction does not take place, period. Comparing different catalysts, or different mechanical pathways is one thing. What you want to do however, is, citing E. Rutherford, stamp collecting. ;-)

@Karl haha, well maybe we can agree to disagree. I think understanding how life has been able to overcome certain kinetic barriers is interesting, and not just stamp collecting. By "that number" do you mean rate enhancement? and in your earlier comment you said "at 2000K you will get just nothing". what you mean by "nothing?"

Agree to disagree is a good solution to many discussions. ;-) I mean no ammonia. You cited those 400 kJ/mol. That is equivalent to a temperature of around* fiftythousand* Kelvin. http://www.colby.edu/chemistry/PChem/Hartree.html 400 kJ/mol, or 4 electronvolts, are just not available in chemistry. And once you have reached that transition state, you'd need to get rid of this huge amount of energy again. And you just can't, says I.

know how effective these old enzymes are at speeding up chemical reactions, there's the problem. The reaction with the enzyme is a totally different reaction than the hypothetic uncatalysed one.

You say that 400 kJ.mol is equivalent to a temperature of around 5e4 kelvin. This doesn't make sense. Temperature is a measure of the distribution of kinetic energy in a material. In this image link I plotted the Maxwell-Boltzmann distribution for 2000K and 50,000 kelvin (so we are talking about an ideal gas). Your are right in saying that the average kinetic energy of a mole of N2 at 50,000 K is ~400 kJ/mol but the average kinetic energy isn't the most crucial bit of information...

Even at 2000K, there will be a 1e-10 chance (integrate 2000k curve from dotted line to $\infty$) that a N2 molecule will have sufficient energy to overcome the 400 kJ/mol activation energy barrier. 1e-10 chance means out of a mol of molecules ~6e13 molecules will have enough energy to overcome the activation energy barrier. So this activation energy is surmountable at "lower" temperatures.

@Karl I understand what you mean about the enzyme doing a completely different reaction than the uncatalyzed reaction. It may feel like comparing apples to oranges.

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How to calculate the aqueous solubility of CO2 vs Pressure
I would like to make Na2CO3 react with HCl in a closed fixed volume vessel, and accordingly I'd like to understand how ensuing pressure will affect the equilibrium of my reaction (and to which extent my reaction will affect the level of pressure when equilibrium is reached).
I guess that the link between pressure and the reaction mostly has to do with CO2 retention into the solution, which leads me to want to understand how I can calculate CO2 solubility in water vs pressure.
Would anyone care to explain me how I could calculate CO2 solubility vs pressure? And tell me if there's anything big I am missing other than CO2 solubility?

The acidity of HCl is so great compared to H2CO3 that you have no equilibrium between Na2CO3 and HCl; the equilibrium is only between CO2 and water and the vapor space. H2CO3 is one of the forms of CO2 in water, but it is unnecessary to delve into this.
After your reaction, you will have a solution of NaCl in which some of your CO2 is dissolved; the rest of the CO2 will be in the vapor space. We will assume that the solubility of CO2 in this solution is the same as in pure water. Data from CRC Handbook, per 100 cc H2O at 1 atm pressure: 171 cm^3 at 0C, 90 cm^3 at 20C. In grams: 0.348 at 0C, 0.145 at 25C, 0.097 at 40C, 0.058 at 60C.
That's the chemistry. Now the mathematics is more complicated, and I'll leave that up to you. But a few points to help you get started: if you have a container with a volume of 100 cc and put 100 mL of water into it, you could also dissolve 171 cm^3 (0.348 grams) of CO2 into that same volume at 1 atm pressure (OK, maybe a slight volume increase - ignore it for simplicity). If you put only 50 mL H2O in that 100 cc volume and added 171 cm^3 of CO2, the pressure in the vessel would be greater than 1 atm (but less than 1.71 atm, which is the pressure you would have with no water in it at all).

This gives me clear structure of how to think about the issue and what I will need to learn and calculate. That's an awesome start for me as it is exactly what I need! Thank you James!


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Why is there only little difference between molecular dipole moments in CHX3 compounds (X = Cl, Br, I)?
Why is there only little difference between molecular dipole moments in $\ce{CHX3}$ compounds ($\ce{X}$ = $\ce{Cl}$, $\ce{Br}$, $\ce{I}$)?

As we can see there is a huge difference in dipole moment of methyl halides $\ce{F}$, $\ce{Cl}$, $\ce{Br}$, $\ce{I}$ excluding $\ce{CHX3}$.

Please explain the reason why is it so only in $\ce{CHX3}$ but not in other methyl halides?

Monohalides: electronegative element just on one corner. dihalides: electronegative elements on one side. haloforms: electronegative elements on three of four sides, and they have to divide the electrons of carbon and just one hydrogen between them. Doesn't leave much for each.

Plus: Cl, Br and I are so bulky that the molecule is no longer nearly tetragonal. I don't find a web source for the actual bond angles now. The gist is that there are several effects that give haloforms a low dipole moment. That the resulting number is practically identical for three of them is an "accident". And you notice that except for CH2F2, already the dihalides have a significantly lower dipole moment, esp. methylene iodide.

Data here: According to the Computational Chemistry Comparison and Benchmark DataBase, the experimental H-C-X angles are 110.4 deg for chloroform, and 107.2 deg for bromoform
Refs: 1. https://doi.org/10.1063/1.1743153  2. Landolt-Bornstein: Group II: Atomic and Molecular Physics Volume 15: Structure Data of Free Polyatomic Molecules

If I had to guess, I would look into the definition of the dipole moment. The dipole moment for two point charges depends both on the charge difference and on the distance.
If I had to give a rationalization for the results, I would rely more on the VDW radius rather than the bond angle: Cl, Br and I have decreasing electronegativity, but increasing VDW radius. Probably, in haloforms the two effects (luckily) "balance" themselves, and this results in a small difference in dipole moments.

Every sum of vectors does. It is why falling is faster than sliding. Even treating electronegativity as not influenced by what goes on the rest of the bonds. That beyond a legitimate but rudimentary analysis I guess it is not true.


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Thermodynamics of Acid Base Neutralisation
I was attempting an exam in which they questioned us about the thermodynamics of acid-base reaction.
The question was this:

In neutralisation reaction of a strong acid and a strong base in a solution, the following statements are correct

and then they gave these four options

∆G = negative
∆H = negative
∆S = negative
∆U = negative

We can easily see that Option 1 and 2 are correct, but in the key, it has been specified that Option 3 as well as 4 are correct, too.
Can anyone please explain this to me.

Sorry, but can you please elaborate upon your statement

As $\Delta G =\Delta H - T\Delta S$ option (1) is always correct. As $\Delta H = \Delta U +p\Delta V$ and as volume change is negligible  $\Delta U \equiv \Delta H$. As long as $\Delta G $ is negative $\Delta H $ and $\Delta S$ can be positive or negative.

@porphyrin Ok, So ∆U is negative too but how would you explain ∆S.

Suppose that the reaction is $\ce{Na+OH- + H+Cl- \to H2O + Na+Cl-}$ then you may argue that entropy is reduced as initially there are 4 species and finally only 3 since all are fully dissociated except water which is only partially so. If this is true then $\Delta H $ must be more negative than $T\Delta S$ is positive to make $\Delta G$ negative.

@porphyrin Ok, I understood your Reasoning. Thanks


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Calculating distance between adjacent planes (hkl) in a simple cubic crystal
My textbook says the following in a section on Miller indices:

Adjacent planes $(hkl)$ in a simple cubic crystal are spaced a distance $d_{hkl}$ from each other, with $d_{hkl}$ given by:
$$d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \tag{2.12}$$
where $a$ is the lattice constant. Equation $2.12$ provides the magnitude of $d_{hkl}$ and follows from simple analytical geometry. To generalize this expression, notice that for a plane $(hkl)$ with $hx + ky + lz = a$, the distance from any point $(x_1, y_1, z_1)$ to this plane is:
$$d_{hkl} = \left| \frac{hx_1 + ky_1 + lz_1 - a}{(h^2 + k^2 + l^2)^{1/2}} \right| \tag{2.13}$$
Hence when that point is at origin $(0, 0, 0)$, we find Equation $2.12$ back.
Example 2.1: With $a = \pu{5 Å}$, we find $d = a = \pu{5 Å}$ for $(100)$ planes and $d = \frac{a}{\sqrt{2}} = \pu{3.535 Å}$ for $(110)$ planes.

I'm confused as to how the author calculated sample 2.1. For instance, if we take $a = \pu{5 Å}$ for $(100)$ planes, then we get
$$d_{hkl} = \left| \frac{x_1 - \pu{5 Å}}{(1 )^{1/2}} \right| = \left| x_1 - \pu{5 Å} \right|$$
So how did the author get $d = a = \pu{5 Å}$?
I would greatly appreciate it if people could please take the time to clarify this.

He/she must have calculated w.r.t the origin so that $$(x_1,y_1,z_1) \equiv (0,0,0)$$. Then $$d_{100} = \bigg|\frac{0-a}{\sqrt1}\bigg| = a$$ and $$d_{110} =  \bigg|\frac{0-a}{\sqrt{1+1}}\bigg| = \frac{a}{\sqrt2} = \frac{5}{\sqrt2} \approx 3.535$$


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Is this reaction possible?
When mixing HCl with K2CO3, the reaction that is usually given is the following:
K2CO3 + 2HCl → 2KCl + CO2 + H2O
I am wondering, however, if I could produce KHCO3 from K2CO3, by using half as many moles HCl. The reaction I imagine could happen would be expressed as follows:
K2CO3 + HCl → KHCO3 + KCl
I've gathered the solubility data for this reaction hereafter:

K2CO3: 112 g/100 mL  (20°C)
HCl: 72,5 g/100ml (20°C)
KHCO3: 22.4 g/100 mL (20°C)
KCl: 34.2 g/100ml (20°C)

From solubility data I see that it wouldn't be extremely practical to conduct as both KHCO3 and KCl are fairly soluble, but I'm wondering if the reaction would occur, or if it would just produce half the amount of KCl + CO2 + H2O through the first reaction (which is generally attributed to these reagents).
If the reaction occurs, can I assume that the first reaction is strongly dominant until I get close to have added as many moles of HCl as the initial moles of K2CO3?
And if it doesn't occur, why does it not?
Note that this question had been asked already, but had received no satisfying answer.

You obviously have to dissolve the carbonate in water first. Look at the ions you have in solution then, and imagine slowing adding HCl, and remember the fastest reaction (very good rule of thumb) is always $\ce{H3O+ + OH- -> 2 H2O}$

It is interesting to consider that either outcome is possible, depending on the order of addition.
If a solution containing 0.5 moles of HCl is added to a solution of 1 mole of K2CO3, slowly, with mild stirring, cold, the products will be 0.5 KCl + 0.5 KHCO3. Nothing is lost to the atmosphere.
However, if a solution of 1 mole of K2CO3 is added to a solution of 0.5 moles of HCl, it would be possible to get 0.5 KCl + 0.5 K2CO3 with slow addition, rapid stirring, warm/hot. The difference here is that the initial K2CO3 is completely overwhelmed by the HCl and produces KCl + H2CO3, which can be degraded to H2O + CO2. By the time half of the K2CO3 is added, the products are 0.5 KCl + CO2, and the CO2 could be mostly evolved as gas and disappears. The last half of the K2CO3 just comes in and sits there.

Thanks for your answer! You meant to write 2KCl + CO2 + H2O at beginning of the last paragraph right?

Oops! What I meant was that as you add the K2CO3 (1 mole in solution) to the HCl solution, at the halfway point, you can get reaction to KCl + 0.5 H2CO3 (which goes to H2O + CO2 and has the CO2 forced out of solution before it gets reacted with K2CO3 to form KHCO3). Thanks for noticing that!


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How to generate Density Functional Tight Binding parameters (Slater-Koster files)?
What is the simplest way (or least complicated way) to generate parameter files (.skf) to be used with DFTB+, for example?
Assuming I want to be able to reproduce geometries for a certain type of molecules, how would I go about it?

There is no reliable black-box method for SK parameters. There are some useful tools, but even with those a reliable parameter set requires lot of work.

Do you care to elaborate on the useful tools and the kind of work required?

You can check out Hotbit (https://github.com/pekkosk/hotbit) and corresponding documentation and code examples, or automatic ("automatic"?) parametrization by SKOPT (https://bitbucket.org/stanmarkov/skopt/src/master/) or adpt_core (https://bitbucket.org/solccp/adpt_core/src/master/).


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How much ethylacetate is produced?

1 mol of acetic acid and 1 mol of ethanol was mixed to react at a constant pressure to reach the equilibrium, and 0.75 mol of water was found in the product.
A. Calculate the equilibrium constant of this reaction at the above temperature
B. At the same temperature, 1 mol acetic acid, 1 mol ethanol, and 4 mol of water were initially mixed to react. Calculate the amount of ethylacetate product after the mixture has reached equilibrium.

A. The equilibrium constant = 9
B. I thought that 4 mol of water reacted with previous product of acetic acid and ethanol so there are 2 mol of acetic acid and 2 mol of ethanol now. I get ethylacetate is 1.5 mol.  The answer is 0.5 mol
How is the reaction actually react?

0.5 mol is correct. Now that you have Keq = 9, and water is x+4 and ethyl acetate is x, now solve the quadratic equation. Remember the quadratic has two roots.

How much acetic acid and ethanol at initial?

The first line of the question states the initial concentrations. Then determine how much of each is at equilibrium. If you solved part A for Keq, then you should know how to set up an equation to express equilibrium conditions.

Acetic acid left and ethanol left are 0.25 mol. There are 0.75 mol of ethyl acetate and water each. Then added 1 mol of acetic acid and ethanol each. 4 mol water. So. There are 1.25 mol of acetic acid and ethanol now?

How can there be more acetic acid and ethanol than you started with?  9 = [x][x+4]/[1-x][1-x]. Solve for x! Only one of the two roots is physically possible. x>0 and x<1.

I get $8x^2 -14x + 9$

Watch your signs for the middle term. $8x^2 - 22x + 9 = 0$

Ok thanks. You don't want to answer in section. So i can give you point

I didn't want to solve it with an ANSWER. I wanted to direct you in the right direction.


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Is adding the word "solution" necessary when listing the materials used in a synthesis?
I am editing a chemistry-related paper and the synthesis procedure described is as follows:

Then, 10 mL of the intermediate were mixed with 10 mL of 1% (w/v)
  VCPL and 2 mL of 0.1% (w/v) KPS.

I was wondering if one should add the word "solution" after the name of the ingredients as follows:

Then, 10 mL of the intermediate were mixed with 15 mL of a 2% (w/v)
  VCPL solution and 3 mL of a 0.2% (w/v) KPT solution.

I think that indicating a volume and a % (w/v) gives it away that you are referring to a solution containing the active ingredient, but should the word "solution" still be added?
Thanks.

Perhaps the name of the solvent should be mentioned

No, the repeated addtion of the word "solution" is not needed.
All you want here is to provide a measured quantity (e.g., volume), concentration of your reagent in the solvent, and inferred quantity (e.g., mmol).  This may be reported in the following pattern, too:

"1-Cinnamyl-4-methylbenzene (4a): The compound 4a was synthesized from
  1-methoxy-4-{[(2 E)-3-phenylprop-2-en-1-yl]oxy}benzene 2b ($\pu{240.9 mg}$,
  $\pu{1.00 mmol}$) with p-tolylmagnesium bromide 3a ($\pu{1.4 mL}$, $\pu{1.1 M}$ in THF, $\pu{1.5 mmol}$) in the presence of 1b ($\pu{4.1 mg}$, $\pu{1 mol\%}$) in $\ce{Et2O}$ ($\pu{5 mL}$) at room temperature for $\pu{24 h}$ ($\pu{198.0 mg}$, $\pu{95\%}$ yield, colorless liquid)."

(source: Hashimoto et al. in Molecules 2019, 24, 2296 (doi.org/10.3390/molecules24122296), section 3.2; open access publication.)
This pattern is seen for inorganic reagents like $\ce{HCl}$ dissolved in water, yet equally available in methanol, 1,4-dioxane, etc. (example); and organometallic reagents, e.g. butyllithium, methylmagnesium bromide.


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Deriving the equations for two-electron integrals in Hartree-Fock theory
In Szabo and Ostlund's Modern Quantum Chemistry, the matrix elements for the hamiltonian of a two-electron system, with an operator $\hat O$, are written on pages 64-66 as
$$\langle \Psi_0 | \hat O | \Psi_0 \rangle=\int dx_1 dx_2 [2^{-\frac{1}{2}} (\chi_1(x_1)\chi_2(x_2)-\chi_2(x_1)\chi_1(x_2))]^* \cdot \hat O [2^{-\frac{1}{2}} (\chi_1(x_1)\chi_2(x_2)-\chi_2(x_1)\chi_1(x_2))]$$
for a spin orbital $\chi$ of electron x defined as the product of a spacial orbital $\psi(r)$ and spin $\alpha(\omega)$ or $\beta(\omega)$, e.g., $\chi(x)=\psi(r)\alpha(\omega)$.
This is multiplied out into
$$\langle \Psi_0 | \hat O | \Psi_0 \rangle=\frac{1}{2} \int dx_1 dx_2 \{\chi_1^*(x_1)\chi_2^*(x_2)\hat O \chi_1(x_1)\chi_2(x_2)+\chi_2^*(x_1)\chi_1^*(x_2)\hat O \chi_2(x_1)\chi_1(x_2)-\chi_1^*(x_1)\chi_2^*(x_2)\hat O \chi_1(x_1)\chi_2(x_2)-\chi_2^*(x_1)\chi_1^*(x_2)\hat O \chi_2(x_1)\chi_1(x_2)\}$$
This expanded equation is what I will base my question on.
For the core-hamiltonian (one-electron integrals),
$$\hat O=h(r_1)+h(r_2)$$
and for the electron repulsion (two-electron integrals),
$$\hat O=\frac{1}{r_{12}}$$
For the core hamiltonian, you can integrate out the electron which is not being acted upon by the operator to reduce the expanded equation to a one-electron integral.
For the electron repulsion, on page 66 the two-electron integral is simplified by stating the first two terms of the expanded equation are equal. This means:
$$\int \chi_1^*(x_1)\chi_2^*(x_2)\frac{1}{r_{12}} \chi_1(x_1)\chi_2(x_2)= \int \chi_2^*(x_1)\chi_1^*(x_2)\frac{1}{r_{12}} \chi_2(x_1)\chi_1(x_2)$$
This is what I do not follow. How is this derived? I can permute the electrons to achieve the same form, but that would introduce a negative sign.
Furthermore, for the equality to be true, does it not imply one of two scenarios: that either the spatial functions are degenerate, i.e.,
$$\chi_1^*(x_1)=\chi_2^*(x_1)$$
$$\chi_2^*(x_2)=\chi_1^*(x_2)$$
$$\chi_1(x_1)=\chi_2(x_1)$$
$$\chi_2(x_2)=\chi_1(x_2)$$
or otherwise that the electrons have the same spin and coordinate such that $x_1=x_2$
$$\chi_1^*(x_1)=\chi_1^*(x_2)$$
$$\chi_2^*(x_1)=\chi_2^*(x_2)$$
$$\chi_1(x_1)=\chi_1(x_2)$$
$$\chi_2(x_1)=\chi_2(x_2)$$
neither of which would make sense.

I believe your statement

I can permute the electrons to achieve the same form, but that would introduce a negative sign.

is mistaken. The exchange integrals arise from permuting one pair of integration variable instances, not two. The latter is the case necessary here. Szabo and Ostlund state on the same page:

Since $r_{12} = r_{21}$, we can interchange the dummy variables of integration in the second term [...].

where the second term appears to be you are troubled with.

I wasn't sure on what they meant by the dummy variables of integration, or why $r_{12}=r_{21}$ is a relevant property for this interchange. What precisely is being permuted here? I only have the one pair here, $x_1$ and $x_2$, to interchange, correct?

Basically, one is interested in the distance between the electrons. (That is a directionless quantity, thus $r_{12} = r_{21}$.) For the integral, it is then without consequence how the integration variables are chosen or ordered. (That's why Szabo and Ostlund call them dummy variables.)


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Explain why can't zinc be plated out from a Zn(II) solution using standard reduction potentials
I am currently learning about electrochemistry using Zumdahl's Chemistry [1, pp. 709–710]. One of the critical thinking questions is

Why can $\ce{Zn}$ not be plated out from an aqueous solution of $\ce{Zn^2+}$ using the choices in Table 18.1?

I feel like it would be plated out when with a half-reaction with a lower potential such as
$$\ce{Al^3+ + 3e- -> Al}$$
Combined,
$$\ce{3 Zn^2+(aq) + 2 Al(s) -> 3 Zn(s) + 2 Al^3+(aq)}$$
Also, looking at the reactivity series, it seems that there are plenty of options that should be able to plate out zinc. It would great if someone could explain to me what's wrong with my thinking.
References

Zumdahl, S. S.; Zumdahl, S. A.; DeCoste, D. J. Chemistry, 10th ed.; Cengage: Boston, MA, 2017. ISBN 978-1-305-95740-4.

Hint: the crucial point is aqueous solution.

Thanks for the hint. I think I get it now. Just to be sure, since the zinc is in aqueous solution, it is more electrically favored for aluminum to reduce the H+ in water to H2 than Zn2+ to Zn. But then H+ isn't reduced because it needs external potential to occur. Anything I left out?

Please note that it's preferred to 1. use self-sufficient and non-clickbaity title (e.g. an equivalent to "click here to see more"); 2. add complete citations, including publisher and page number(s); 3. preserve quoted text as close to the source as possible. These tech details aside, this is a good question. I also took a liberty to re-upload the table for better readability.

Zinc can be purified via the electrowinning process: https://en.wikipedia.org/wiki/Zinc_smelting . From the article, Zn is deposited on Al cathodes, oxygen gas is evolved at the anode and the electrolyte is sulfuric acid (possibly with other stuff, but I don't know). This would be an interesting thing to test: a beaker with zinc sulfate and sulfuric acid, an Al cathode, a Cu anode and a couple of batteries or DC power supply. The standard reduction potentials do not take things like overpotential into account, and that apparently gives some opportunities. Good question (+1).

What external potential? H2 against Al, that's a difference of 1.7V in standard potentials. As long as you have a blank Al surface, you get hydrogen gas. I don't think the overpotential of H2 on zinc is larger than 1.7V. (de.wp says it's 0.77 https://de.wikipedia.org/wiki/%C3%9Cberspannung_(Elektrochemie))


text:
Electrolysis of a calcium salt in polar aprotic solvent
If i found a calcium salt that is soluble in acetone or ethyl acetate, which are polar aprotic solvents, and attempted electrolysis, what would happen? Since polar aprotic solvents are used in metal-air batteries I don't think there would be a reaction between the newly formed metal and solvent. At the same time, no one uses such method industrially. Maybe it's just inefficient, or maybe it doesn't work?

https://www.nature.com/articles/183816a0

Hmm, there was a very similar q. recently... Why you guys think about solvents that can be reduced?


text:
How does de Broglie actually prove Bohr's postulates?
We know Bohr said that the angular momentum of an electron is an integral multiple of $nh/(2π).$ And in de Broglie's wave equation, he said the circumference of the path of the electron traveling as a way is $n$ times the wavelength, which is equal to $2πr.$
But in Bohr's model $n$ is the principal quantum number, while the $n$ in de Broglie's equation is the number of wavelengths in the one revolution. I don't understand how they are equal.

What does n determine in Bohr's theory? Larger the n, larger the circle's radius  and larger the energy of the electron revolving around that the proton. You can see the connection by noticing that for a circle of a given radius, only a certain number of integral wavelengths can be fitted on its circumference.

Tough question to answer considering that the Bohr model is a step in the right direction, but ultimately, wrong in several regards.

Does it? I agree with @Zhe, the Bohr model is pretty rudimentary, its postulates are not "proven".

Note also that in the Bohr model, the lowest quantized value of the angular momentum of an electron in a hydrogen atom is $\hbar$, but via the Schrödinger equation, it is zero.

de Brogile explains why orbitals are quantised
Strictly speaking de Brogile doesn't prove Bohr's postulates which are mostly wrong. But he did provide an explanation for the most important of Bohr's ideas: electron orbitals are quantised.
Bohr's whole model starts with the classical idea that electrons "orbit" a nucleus. But this has several problems not least of which is why don't they just radiate away their energy and fall into the nucleus (which is what accelerating charges should do in classical electromagnetism). Bohr made the extra assumption that the electrons around a nucleus could only "orbit" in a small number of fixed paths or orbitals which explained the actual observations. He didn't have any good reason why this was true.
de Brogile provided a reason. According to his postulates things like electrons are not just classical particles but have some of the characteristics of waves. and the thing about waves is that they interfere with each other. Electrons orbiting a nucleus can only do so if their wave-like property fits into a specific orbit (the way the vibrations in a guitar string produce only specific notes made from standing waves of integer divisions of the string length). Imagine integer numbers of electron waves around a nucleus and you are a third of the way of getting the picture (electrons orbit in 3D so the compatible vibrations are spherical harmonics in 3D space not the easier to picture vibrations of a 1D guitar string).
de Brogile, in short, produced a justification for the Bohr assumption that electron orbits had to be quantised.
Those pictures you see of electron "orbitals" are, in essence, pictures of the possible "waves" electrons can form when orbiting a nucleus.
All the above simplifies things a bit but the key ideas were from de Brogile and were more fully mathematicised by Dirac and Schroedinger giving us the modern quantum picture of how electrons behave.


text:
Calculating bond energy given the enthalpy value of burning reaction

The burning enthalpy (exothermic) of $\ce{NO +O2 ->NO2}$ is $\pu{57 kJ mol^-1}$ and $\ce{O=O}$ bond energy is $\ce{494 kJ mol^{-1}}$, then what is the bond energy of $\ce{N-O}$ ?

So I balanced the equation
$$\ce{2NO + O2 ->2NO2}$$
wrote down Lewis structures and got
$$(2\Delta H(\ce{N=O}) + \Delta H(\ce{O=O}) - 2(\Delta H(\ce{N-O}) + \Delta H(\ce{N=O})) =  \pu{-57 kJ mol^-1}$$
($\pu{-57 kJ mol^-1}$ since it is an exothermic reaction), so we are left only with
$$\Delta H(\ce{O=O}) - 2\Delta H(\ce{N-O}) = \pu{-57 kJ mol^-1}$$
thus
$$\Delta H(\ce{N-O}) = \frac{494+57}{2}\pu{kJ mol^-1} = \pu{275.5 kJ mol^-1}$$
However the right answer is
$$\frac{494-2\cdot 57}{2} \pu{kJ mol^-1}= \pu{190 kJ mol^-1}$$
and I really don't get what am I doing wrong?

Have a look at a very similar question asked recently: Calculating bond dissociation enthalpy of F-F bond. Also, I've never seen square brackets used to denote bond energies and I'm wondering what textbook does that (in chemistry "[" and "]" are commonly used to denote either a coordination complex, or equilibrium concentration). It also appears that there is something off with the unit symbols or lack thereof. I already corrected Kj to kJ, but it looks like those should be per mole (?).

@andselisk Thanks, I edited the units and the notation. For the given link - I used the same method and got the correct answer, which make me think there is a catch in this exercise which I don't get.

Consider the reaction enthalpy of $$\ce{2NO + O2 ->2NO2}$$ and $$\ce{20 NO + 10 O2 ->20 NO2}.$$ They are different, and also different from the reaction enthalpy of burning one mole of NO.

@KarstenTheis How can I calculate the enthalpy for those balanced reaction given the data of $\ce{NO + O2 -> NO2}$ enthalpy?

How can I calculate the enthalpy for those balanced reaction given the data of $\ce{NO +O2 -> NO2}$ enthalpy?

The way the enthalpy data is presented is sloppy (is this the original text of the exercise?). For some types of enthalpy, it is unnecessary to specify the chemical reaction equation because there is a convention. For example, for the enthalpy of formation, it is understood that the substance in question is the only product, the stoichiometry coefficient is one, and that the reactants are elements with coefficients so that the equation is balanced. I am not aware of any such convention for the enthalpy of combustion.
What is the enthalpy of $\ce{2NO(g) + O2(g) -> 2NO2(g)}$?
We can look up the enthalpies of formation for the nitrogen-containining compounds:
$$\Delta H_f(\ce{NO}) = \pu {90.25 kJ/mol}$$
$$\Delta H_f(\ce{NO2}) = \pu {31.18 kJ/mol}$$
From this, we can calculate the reaction enthalpy:
$$\Delta H_r = 2 \cdot \pu {31.18 kJ/mol} - 2 \cdot \pu {90.25 kJ/mol} =  \pu {-118.14 kJ/mol}$$
Which combustion reaction has an enthalpy of -57 kJ/mol?
If we divide the balanced chemical equation by 2, we get:
$$\ce{NO + 1/2 O2 -> NO2}$$
The reaction enthalpy is one half of that of the original reaction, -59.07 kJ/mol. So for the reaction describing the combustion of one mole of NO, the reaction enthalpy is about -57 kJ/mol.
What about the bond dissociation energy?
Everything else in the OP's calculation is correct. If you plug in the correct combustion enthalpy, the calculated bond energy will be correct.


text:
Why is it necessary to correct the pressure of a gas that is collected over water?
Basically, what are some mistakes that you can make while using the downward displacement of water method and how to fix them? This is actually a homework question and I'm not too sure what it means either. The original question:

State and explain Dalton’s Law of Partial Pressures. Why is it necessary to correct the pressure of a gas that is collected over water? {Background: Gases formed in a chemical reaction are collected using a technique called the downward displacement of water, feel free to research this technique before you respond to this question}

Could you please add more context? It's not quite clear how (and why) one would "correct" the pressure in the first place. Posting an example of an experimental setup or citing a textbook where you encountered this info would help to understand your question better.

Here's the full question:
State and explain Dalton’s Law of Partial Pressures.  Why is it necessary to correct the pressure of a gas that is collected over water?  {Background: Gases formed in a chemical reaction are collected using a technique called the downward displacement of water, feel free to research this technique before you respond to this question}

This comment should be an [edit] to your question. Since it's a homework question, you are expected to demonstrate your effort in solving the problem (literature/internet research also counts).

He never actually taught us this so I was hoping to find some help somewhere else. But the question also says that we can research before actually answering.

see eg https://www.quora.com/What-is-downward-displacement-of-water

You have to correct for the vapor pressure of water.

Generally, gases which are insoluble in water are collected by putting them into an inverted vessel via a tube, so that the gas rises to the top and 'displaces' the water downward. Take a look at the following picture—

Now, the problem is that water has a certain vapour pressure at any temperature. In other words, whenever water is exposed to a gaseous phase, it will start to evaporate until the partial pressure of the water vapour in the gas phase reaches a certain value (this value depends on the temperature). This value is known as the 'vapou pressure of water'.
In this case, the gas you collected by displacing the water will be contaminated by water vapour. Therefore, the pressure of the gas which you measure will be the sum of the partial pressure of the product gas, and partial pressure of water vapour.
$$\begin{align*}
Measured\;pressure &= Partial\;pressure\;of\;product\;+\;partial\;pressure\;of\;water\;vapour \\
&=Actual\;pressure + Vapour\;Pressure
\end{align*}$$ This means that you must correct the pressure, by subtracting the vapour pressure of water from the measured value to get the actual pressure of the produced gas.
Note: There are standard tables which list the value of vapour pressure of water at different temperatures. These tables are generally used for the correction.


text:
Molecular Orbitals (only pi system) of Allyl system using group theory
I have just started reading group theory and their applications in the MO theory. I tried to get the equations of the pi molecular orbitals of allyl system using group theory, but I am stuck at the last part.
This is what I did so far—

Determined the point group of the allyl system to be $\ce{C_{2v}}$.
Used the 3 p orbitals as basis.
Found out the reducible representation. Reduction gives $\ce{A2 + 2B1}$

I am stuck at this point. I named the three p orbitals of C atoms as $\ce{\psi1, \psi2, \psi3}$. However, when I try to use the 'projection operator method' using the $\ce{\psi1}$ as generation vector, it gives me $\ce{\psi1 +\;\psi3}$ and $\ce{\psi1 - \psi3}$ (unnormalized). I don't know how to proceed.
 [image copied from Master Organic Chemistry webpage ]

The projection operator method only relates symmetry-equivalent orbitals, so if you ‘start’ from ψ1, you can’t account for ψ2. Symmetry alone cannot tell you how much of ψ2 goes with ψ1, because any linear combination of ψ2 and (ψ1 + ψ3) will still transform as B1.

@orthocresol Then what do I need to do to get the component of $\ce{\psi2}$?

The coefficients of the AOs in the bonding and antibonding MOs will depend on the energies of the constituent AOs. Hückel theory is the simplest (also the most simplistic) way of getting the coefficients.

@orthocresol, Ok, I will try reading about Huckel theory. But can I get the same result by applying group theory?

Symmetry, or group theory, can only tell you whether a coefficient is positive, zero, or negative. It cannot give you an actual number.

The calculation below is not exactly the same as yours but I had it to hand and it is very similar. The bit you specifically want is towards the end. In your case you add will vectors along the $\pi$ orbitals.
What projection operators  operators do is to extract the symmetry adapted functions from the basis functions used to form them and so produce a linear combination L of the basis functions. Many basis functions can be chosen, but the easiest to use are usually vectors pointing along bonds or those representing a p$\pi$ orbital. To form a linear combination, a 'victim' vector is chosen and operated on with each symmetry operation in the point group, $C_2\,, \sigma(xz)$, etc. to find what vector this turns into. This vector's name (the basis function) is then multiplied with the character from the point group corresponding to that symmetry operation and symmetry species. Finally, the names are added together by moving from one symmetry operation to the next along the top row of the point group. We expect the result to be the sum or difference of the displacement vectors for a molecular vibration, such as $2v_1 + v_2 - v_3 \cdots$ and so forth. The same formula or algorithm is used if the combinations of atomic orbitals that make up molecular orbitals are sought.
The equation to make a linear combination is
$$ L_M=\frac{d}{h}\sum_{j=1\cdots h}c_jS_j(v)$$
where $M$ is the symmetry species label, $A_g, B_{3g}$, and so forth, $h$ is the order of the group, $d$ the dimension of the irreducible representation, and the sum is over all the classes. Because the resulting vector will be normalized d/h can be ignored. One further point is important. In the $C_{3v}$ point group, for example, there is a heading $2C_3$ as there are two members in this class, and this must be split in the summation into $C_3^+$ and $C_3^-$. This is because moving a vector $120$ degrees to the right, say, will turn it into a different vector than turning 120 degrees to the left, similarly, any mirror planes must be separated out. The character for symmetry species $M$ is $c_j$ and $S_j(t)$ is the effect that operator $S$ in column $j$ of the point group has on our victim vector $v$. The result of this operation is to produce a vector whose name is recorded. For example, if $S$ is the identity then the result is $E_1(v) = v$, other operations may leave $v$ unchanged or change it into another vector. Finally, note that if the reduced representation of the vibrations or molecular orbitals contains two or more symmetry species of the same type, e.g. $2A_1$, then two or more different victim vectors will have to be chosen to obtain all the linear combinations.
In H$_2$O, SO$_2$%, and other triatomics with $C_{2v}$ symmetry, it is clear that two of the vibrational modes stretch the bonds and one changes the bond angle. To work out the normal mode vectors, some representative vectors are placed along the bonds as in the figure. You can choose where to put these but physically realistic choices will usually make the calculation simpler.

In $C_{2v}$, the normal mode vibrations have been found to comprise the symmetry species $2A_1$ and $B_1$. The $L_M$ formula is shown as a table with $v_1$ as the victim vector. The table also shows the characters for each symmetry species.
$$\begin {array}{l|llll|l}
C_{2v}  & E & C_2 & \sigma(x,z) & \sigma'(y,z) & L_m \;(d=4,h=4)\\
\hline
A_1& 1&1&1&1\\
L_{A1} & v_1 & v_2& v_1& v_2&  2(v_1+v_2)\\
\hline
A_2& 1&1&-1&-1\\
L_{A2} & v_1 & v_2& -v_1& -v_2&  0\\
\hline
B_1& 1&-1&1&-1\\
L_{B1} & v_1 & -v_2& v_1& -v_2&  2(v_1-v_2)\\
\hline
B_2& 1&-1&-1&1\\
L_{B2} & v_1 & -v_2& -v_1& v_2&  0\\
\hline
\end{array}$$
Notice that what the vector changes into is (obviously) the same for each symmetry species as this is determined by the symmetry operations. Only the value of the character changes in front of each term The final $L_M$ is not multiplied by $d/h$ because the resulting vectors are instead normalized giving, $L_{A1} = (v_1 + v_2)/ \sqrt{2}$ and $L_{B1} = (v_1 - v_2)/\sqrt{2}$. Note also that $v_3$ and $v_4$ do not enter into this table, they are at $90$ degrees  to $v_1$ and $v_2$ and no operation in this point group can inter-convert them.
There are two $A_1$ species produced from the reducible representation but only one has been found so far. This is because the vectors $v_1$ and $v_2$ cannot produce a bend, which is the other normal mode. Using vector $v_3$ as the victim this species appears. As this mode has $A_1$ symmetry all the characters $c_j$ in the $L_M$ formula are $1$, making the calculation easy and the combination is shown below.
$$\begin {array}{l|llll|l}
C_{2v}  & E & C_2 & \sigma(x,z) & \sigma'(y,z) & L_m \\
\hline
L_{A1} & v_3 & v_4 & v_3 & v_4 &  2(v_3+v_4) \\
\hline
\end{array}$$
which describes the bending mode shown in the figure in terms of its unit vectors. When normalized this mode is $(v_3 + v_4)/\sqrt{2}$. (As the centre of mass cannot move during a vibration, there being no force to make it do so, therefore the O atom has to move a small distance away from the H atoms. Vectors
could be added onto the O atom to show this and $L_M$ terms re-calculated.
In the case of degenerate symmetry species as occurs in the $C_{3v}$ and most other point groups the results of the summation $L_M$ is to produce two or more vector sums that are not orthogonal. Either these have to be made orthogonal, by choosing another victim for the degenerate term and then using the Gram-Schmidt method to make the two vectors orthogonal. R. Carter 'Molecular Symmetry and Group Theory' and A. Vincent 'Molecular Symmetry and Group Theory' both give method by which to do this.

But how do I apply this to the MO problem? By using the $\psi_1$ and $\psi_3$ as generating vectors, I have got $\psi_1 + \psi_3$ and $\psi_1 -\psi_3$. However, the orbitals of allyl system as we can see from the diagram, should be $\psi_1+\psi_2 + \psi_3$, $\psi_1-\psi_2+\psi_3$ and $\psi_1 -\psi_3$. So, I am getting at least one orbital wrong, even if I use another set of generating vectors.


text:
Facing a problem in enthalpy calculation

If 1 mole of gaseous carbon atoms are converted into diamonds, calculate enthalpy change of process given that bond enthalpy of $\ce{C-C}$ bond is $\pu{400 kJ mol-1}.$

According to me, the answer should be that since C(diamond) has 4 bonds and C(gas) doesn't have any bonds, 4 bonds are being formed so enthalpy change should be
$$ΔH = 4\cdot (\pu{-400 kJ mol-1})$$
But in the correct solution it says each bond is counted twice, so answer will be
$$ΔH = \frac{4\cdot (\pu{-400 kJ mol-1})}{2}$$
I don't get why are we doing this. In other questions we don't consider the double counting of bonds. For example, another question I did had 1 mole phosphorus gas converting into $\ce{P4}$ molecule, and even though $\ce{P4}$ has 6 bonds, we did not consider what we did with diamond, so what is the reason for that?

Consider the simpler question: If 1 mole of hydrogen atoms (H) form dihydrogen (H-H), how many bonds are formed? Using your logic (each hydrogen makes one bond), there should be one mole bonds. But only 1/2 mole of dihydrogen was made. That would be 1 mole of single bonds in 1/2 mole of dihydrogen, not possible.

$\ce{P4}$ forms a molecular solid, whereas diamond is a covalent-network solid (see eg definitions here).
In a molecular solid, molecules containing a discrete number of covalently bonded atoms are arranged in a lattice stabilized by weak interactions such as van der Waals forces or hydrogen bonds between the molecules. In a covalent-network solid, the "molecules" are extended with no finite stoichiometry definining a molecular size.
When estimating the enthalpy change, you consider only covalent bonds, not weak intermolecular forces. Therefore, for $\ce{P4}$ you consider only the 6 P-P bonds, whereas in diamond you consider covalent bonds to nearest neighbours, and correct for the fact that when you rupture a bond to a neighbour, that neighbour also loses a bond, therefore the division by 2.

Thanks ! , but if in future questions , I am unable to identify is the given substance if a covalent network solid or not , can we say that any molecule of the form Xy (like O2) where y is any number will not be a covalent-network solid ?

@bscripts If you can envision the molecule existing on its own in solution or in the gas phase, then it is likely to be a molecular solid. I can't think of exceptions of the bat, but there might be some, particularly if you change pressure or temperature to the right values.

I saw a question where nitrogen atoms where present on all the corners of a truncated octahedron  and the energy required to break the octahedron was asked , and something similar was done to what was done with diamond above , would you call nitrogen a covalent network solid ?

@bscripts Apparently there are exotic phases of nitrogen, see https://en.wikipedia.org/wiki/Solid_nitrogen. I would not call solid phases of molecular nitrogen $\ce{N2}$ covalent network solids.


text:
Finding the pH of the solution
In this video, at 3:30 min professor solves two equations
$$\ce{H2O —> H+ + OH-}$$
and
$$ \ce{HCl —> H+ + Cl-}$$
and deduces $[\ce{H+}] = [\ce{Cl-}] +[\ce{OH-}]$. How come this happen from those two equations?
Can anyone elaborate and explain?

Left sides are neutral. If nothing else is going on, for every H+ made, either an OH- or a Cl- are made. It keeps the solution net neutral, which it should be.

For future reference, ChemSE benefits from mhchem's markup which eases input and display of chemistry-related content in questions, answers and comments -- just have a look at https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here.

Let's take the degree of dissociation of water be $\alpha$ and the $[\ce{H2O}] = c$, then at equilibrium the $[\ce{H^+}] = [\ce{OH^-}] = c\alpha$. And since $\ce{HCl}$ is a strong acid it will completely dissociate. So  $\ce{HCl}$ is a then the $[\ce{H^+}] = [\ce{Cl^-}] = a$. So $[\ce{H^+}]_{\textrm{net}} = a+ c\alpha$, which is equal to $[\ce{OH^-}] + [\ce{Cl^-}]$.

What is the meaning of the second alpha? And what is the concentration of water? Can you add subscripts to the two [H+] terms. They seem to have different values.

For a quick general answer, this equation represents the charge balance. In a solution, the charges of all ions should add up to zero. You charge balance reads:
$$\ce{[H+] = [Cl-] + [OH-]}$$
If you had divalent ions, you would have to correct for that, e.g. if the solution also contained magnesium ions:
$$\ce{[H+] + 2 [Mg^2+] = [Cl-] + [OH-]}$$

How come this happen from those two equations?

$$\ce{H2O —> H+ + OH- }\tag{1}$$
$$\ce{HCl —> H+ + Cl- }\tag{2}$$
Both equations are balanced, which includes charge balance. For every $\ce{H+}$ made by dissociation of water, an $\ce{OH-}$ is also made. For every $\ce{H+}$ made by dissociation of hydrogen chloride, an $\ce{Cl-}$ is also made. So the total $\ce{H+}$ concentration is equal to the sum of the $\ce{OH-}$ and $\ce{Cl-}$ concentrations.

[OP's comment] Won't it be HCl+H2O =2 [H+]+ [OH-] + [Cl-] ?

This is a bit a step in the wrong direction, but I'll explain it anyway.
It looks like you are trying to add the two reactions. This would be appropriate if they were coupled, i.e. if a nanomachine grabs one molecule of hydrogen chloride and one molecule of water, and spews out two protons, a hydroxid and a chloride. The overall equation of that would be:
$$\ce{HCl +H2O ->2 H+ + OH- + Cl-}$$
For example, one mole HCl and one mole water would make two moles $\ce{H+}$, one mole hydroxide and one mole chloride. This would maintain the charge balance.
However, these two reactions are typically not coupled but happen independently. In fact, usually when reaction (2) increases the concentration of $\ce{H+}$, reaction (1) would go in reverse, lowering the hydroxide concentration (and the $\ce{H+}$ concentration).
Deriving the charge balance from the two balanced equations
The two reactions are independent in the sense that there is no set stoichiometric factor linking them. So let's say reaction (1) makes a certain amount $x$ of $\ce{H+}$, and reaction (2) makes a certain amount $y$. If we start out with no ions at all, and ions are only produced by these two reactions, we can write down all the amounts:
$$n_{\ce{H+},\mathrm{total}} = x + y$$
$$n_\ce{OH-} = x$$
$$n_\ce{Cl-} = y$$
Adding up the anions:
$$  n_\ce{OH-} + n_\ce{Cl-} = x + y $$
So
$$n_{\ce{H+},\mathrm{total}} = n_\ce{OH-} + n_\ce{Cl-}$$
To turn this into a statement about concentrations, divide the amounts by the volume they are in. Because it is the same volume for all (they are all in the same solution), we get a similar equation for the concentrations:
$$[\ce{H+}]_\mathrm{total} = [\ce{OH-}] + [\ce{Cl-}]$$
Example with numbers
For example, if reaction (1) made $\pu{e-10 M}$ of $\ce{H+}$, and reaction (1) made $\pu{e-4 M}$ of $\ce{H+}$, there would also be  $\pu{e-10 M}$ of $\ce{OH-}$ and $\pu{e-4 M}$ of $\ce{Cl-}$.


text:
Which has maximum bond energy? F2, N2 or O2?
I think that it should be F2 because F is smaller than N and O thus having a small bond length. And small bond length means greater energy (my teacher told me).
But the answer is N2.

Your textbook probably contains the data in a table such as this:
$$
\begin{array}{|c|c|c|}
\hline
\mathrm{bond} &\mathrm{length(pm)}  & \mathrm{energy(kJ/mol)}  \\
\hline
\ce{N-N} &   145 &  170 \\
\hline
\ce{O-O} &   148 &  145 \\
\hline
\ce{F-F} &   142 &  158 \\
\hline
\hline
\ce{N#N} &  110 &  945 \\
\hline
\ce{O=O}&  121 &  498 \\
\hline
\end{array}
$$
So if you consider that dinitrogen has a triple bond and dioxygen a double bond, you see that breaking dinitrogen into atoms requires a much higher bond dissociation energy. Also, you see that the bond length in dinitrogen is the shortest, so here your teacher's rule of thumb (shorter bond, stronger bond) works nicely.

F is smaller than N and O thus having a small bond length

That rule does not work, even if you compare just the single bonds. It probably works if you stay in the same group (comparing F, Cl, Br, I), but in the end, you have to look at the experimental data or, if there is no data available, measure it yourself.

And small bond length means greater energy (my teacher told me)

That rule often works, but again not for the single bonds N-N, O-O and F-F. The differences in bond length and energy are subtle, and that's when these rules often break down.


text:
What is the approximate cost per unit length to synthesize carbon nanotubes free of defects?
As implied near the end of this article, carbon nanotube fibers have been synthesized in chunks on the order of about a centimeter in length (Fei Wei at Tsinghua University in 2018). What is the cost per unit length for making these fibers? Does it seem feasible to do so in the next 5 years, 10 years, 20 years?

OK, that's better, but what do you mean with "Does it seem feasible to do so in the next 5 years, 10 years, 20 years?"

Zhang et al. reported the creation of CNT fibers with growing rate 5 mm/s to up to final length of 550 mm (surpassing half of a metre), in ACS Nano 2013, 7,  6156-6161 (doi 10.1021/nn401995z). They used motorized equipment (e.g., a furnace on rails) which might point into "carbo nanotube spinning".

I mean, assuming that cost is the primary factor, are there technological advances on the horizon which will bring the cost down in the near future, or does it seem more like we'll have to wait 50 or 100 years?

You should check if the authors have spun off a start-up or have licensed their process (if patented), and see if they are selling the tubes. Maybe they are in talks with manufacturers and resellers who can scale up the process and distribution.

@BuckThorn Apparently the experiment was run without a bobbin.  But from "normal scale world" we know cotton may be collected on a bobbin, as copper wires, too, provided the radius of the bobbin is adjusted.  The article describes many of the parameters, including ratio of $\ce{H2}/\ce{CH4}$, temperatures, catalyst and water loading, analytical techniques of "quality control"; which closes a bit the window of novelty patent offices tend to require.


text:
How would I find the molarity/concentration of a gas using pV = nRT?
The problem I'm working on is:

At $\pu{1100 K}$ and a total pressure of $\pu{1 atm}$, $\ce{SO3}$ is $75.0\%$ dissociated.
$$\ce{2SO3(g) <=> 2SO2(g) + O2(g)}$$
Assume $\pu{1.00 mol}$ of $\ce{SO3}$ was present before dissociation.
Calculate the molarities of $\ce{SO3},$ $\ce{SO2},$ and $\ce{O2}$ at equilibrium.

I've already found the moles at equilibrium to be $\pu{0.25 mol}$ for $\ce{SO3},$ $\pu{0.75 mol}$ for $\ce{SO2},$ and $\pu{0.375 mol}$ for $\ce{O2}.$ I'm just not sure how to go about finding their molarities.

Molarity is the number of moles of a paticular solute divided by the volume of the solution (in litres). You can find out the volumes of each using the ideal gas law. Then you get the total volume from that and you can calculate the molarities.

Supposing the dissociation process is isochoric, the total volume $V$ required for calculation of molarity at equilibrium of $i$th component $c_{\mathrm{eq},i}$
$$c_{\mathrm{eq},i} = \frac{n_i}{V}$$
can indeed be found from the ideal gas law using initial amount of undissociated sulfur trioxide:
$$V = \frac{n_0(\ce{SO3})RT}{p} = \frac{\pu{1.00 mol}\cdot\pu{8.21e-2 L atm K-1 mol-1}\cdot\pu{1000 K}}{\pu{1 atm}} \approx \pu{82.1 L}$$
You have found amounts of substances at equilibrium correctly, the rest is just math:
$$c_\mathrm{eq}(\ce{SO3}) = \frac{\pu{0.25 mol}}{\pu{82.1 L}} \approx \pu{3.0e-3 mol L-1}$$
$$c_\mathrm{eq}(\ce{SO2}) = \frac{\pu{0.75 mol}}{\pu{82.1 L}} \approx \pu{9.1e-3 mol L-1}$$
$$c_\mathrm{eq}(\ce{O2}) = \frac{\pu{0.375 mol}}{\pu{82.1 L}} \approx \pu{4.6e-3 mol L-1}$$

Thank you! I realize I had been inputting the wrong values for the equation. I had been inputting the specific equilibrium moles instead of the delta n. Just to be clear, when doing this, n is (#moles of gaseous products - #moles of gaseous reactants)?

@user81232 Yes, this is what you achieve by applying reaction's stoichiometry (e.g. when building up an ICE table). Also, please note that "number of moles" or "# of moles" isn't quite a correct term (see this Meta post) – amount of substance is.

It seems to me the problem statement implies that the system is isobaric, not isochoric.


text:
Where to find experimental spin-orbit coupling constants?
I am searching for experimental spin-orbit coupling constants, particularly for main group elements and Cr(0) to calculate correlation consistent Composite Approach (ccCA) energetics. This method requires the application of experimental spin-orbit coupling constants (Table 1, [1]). I have found and read several other papers that use experimental spin orbit coupling constants, but none have a corresponding reference. I have also searched other places like NIST, etc. but have not found anything. Where can I find these experimental values for spin-orbit coupling constants?
References:

Peterson, C.; Penchoff, D.; Wilson, A. Prediction of Thermochemical Properties Across the Periodic Table. Annual Reports in Computational Chemistry 2016, 3–45 DOI: 10.1016/bs.arcc.2016.04.001.

This thread might be helpful. https://www.researchgate.net/post/How_do_I_determine_spin-orbit_coupling_experimentally_or_the_strength_of_the_spin-orbit_coupling

Also, check this out: https://pubs.acs.org/doi/10.1021/acs.jpca.8b09218

I believe you want this website: https://srdata.nist.gov/xps/EngElmSrchQuery.aspx?EType=DS&CSOpt=Retri_ex_dat&Elm=Cr
This is the NIST database for XPS experiments and you can search by element and find the doublet separation. When the data is available, you can choose between splitting in different orbitals. Each value contains the reference as well.

In addition to the comments above, I'd recommend the following reference:
This looks like a good reference:"Spin-orbit coupling in the first and second transition series" BY T. M. DUNN. 10.1039/TF9615701441


text:
Crystallographic terminology associated with the honeycomb structure
I'm trying to quickly learn basic crystallographic principles in 2D under pressure (it's me whose under pressure, not the lattice) so I'm checking many sources to find those that "speak to me" the best (including this excellent answer). Many sources (books, papers, researcher's websites) cite the "honeycomb lattice" as an example of a lattice which is non-Bravais in order to reinforce their explanation of what a Bravais lattice is.

some online examples where honeycomb is a "lattice" but not Bravais: slide 3 and here and slide 7
some online examples where the "lattice" designator is avoided: here and slide 20
ambiguous or noncommittal as to being a "lattice" slide 6

Briefly (I don't have the various texts with me right now to transcribe the paragraphs) it seems to me that a 2D Bravais lattice any set of points $m \vec a + n \vec b$ for all integers $m, n$.
The vertices of a honeycomb don't satisfy this because if you try to use that expression, 1/3 of the points are missing. If you want to use a Bravais lattice, you have to define a unit cell with two "atoms" in it.
Question: Can the vertices of a honeycomb or "honeycomb lattice" really be called a proper lattice (but just not a Bravais lattice), or should we really call it either a net† or a regular complex apeirogons or something else, but not refer to it as a lattice?

From http://www-personal.umich.edu/~sunkai/teaching/Winter_2018/01302018.pdf (click for full size)

†The Five nets
From Elizabeth A. Wood's classic Vocabulary of Surface Crystallography Journal of Applied Physics 35, 1306 (1964); https://doi.org/10.1063/1.1713610

In triperiodic structures the equivalent points form
  a three-dimensional lattice in which the space units are
  unit cells. In diperiodic structures the equivalent points
  form a two-dimensional net in which the area units are
  unit meshes. The fourteen Bravais space lattices are
  replaced, in the diperiodic case, by five nets, described
  in Table I and diagrammed in Fig. 2.

After some thought, I think this is mostly a question of what you need to convey by using a particular term. If 'net' is commonly used in the community you are trying to engage, go for it. But I'm not sure there is a definitive answer to be had - use what somebody else will understand.

@JonCuster sage advice for sure! Yes, the more sources I read, the more I begin to wonder if the term "lattice" doesn't have a universally accepted definition that includes or excludes the honeycomb vertices.

Well, Bravais lattice is well defined. And the honeycomb is a lattice, just a Bravais lattice with a basis. Like diamond cubic...

the concept of lattice is very (too!) often misinterpreted outside crystallography. By definition, the lattice points are all identical, that is each lattice point has identical surroundings to all others. In a honeycomb pattern you can distinguish 2 different sets of points, the red and the blue ones in the figure.
So honeycomb pattern is actually a 2-dimensional structure (not a lattice), resulting from the convolution of a basis (in this casis constituted of one blu + one red point) with a 2D lattice (the hexagonal one in the figure representing the five nets)

Thank you for your excellent answer, it's great when a new user picks up an old question and answers it nicely (i.e. clearly, completely and concisely). Yes this makes complete sense.

To the longer form of the question statement in the body of the question post "Question: Can the vertices of a honeycomb or "honeycomb lattice" really be called a proper lattice (but just not a Bravais lattice), or should we really call it either a net† or a regular complex apeirogons or something else, but not refer to it as a lattice?" (bumpy wording because I was grappling with concepts back then) I think the answer is "NO if 'vertices' = atoms or honeycomb points" and "YES if 'vertices' = ring centers or either the red or blue atoms as indicated by the unit cell vectors in the diagram"

I still don't fully understand the usage of the term "net" here or in general. In the block quoted passage it looks like the author uses "lattice" for 3D lattices and "net" for what I would call a 2D lattice. Perhaps that's a historical artifact. To me I would call the honeycomb pattern a net, and the hexagonal lattice that underpins it a lattice, but I really don't know if net has a specific meaning anymore or not.

I agree with you. you can find reliable definitions of net here: [Journal of Solid State Chemistry 152, 3 (2000)]; [Journal of Solid State Chemistry 178, 2480 (2005)]


text:
Solvated Electrons
We all know that the alkali metals give solvated electrons in liquid ammonia which is responsible for the blue color of the solution.
It is stated, "the blue color of the solution is due to the ammoniated electron which absorbs energy in visible region of light."
But i couldn't understand why solvated electron is absorbing energy?

Interesting question! These are different, but related questions: What is happening in this video of solvated electrons donated from sodium in ammonia? and also How to think of solvated electrons?

A solvated electron is not a free electron in a vacuum. It is associated with the solvent as the name indicates.

@M.Farooq Now I want to ask if an electron in a dense plasma can be thought of as solvated ;-) Something like "Are dense, partially ionized plasmas also solutions?"

Are we discussing solvated electrons or are we discussing plasma based reactions.The comment and your question are different.

Basically the cage of solvent and the electron are bound. This entity has its own energy levels and, at least at a basic level, you shouldn't wonder it has specific transitions

@uhoh, Plasma physics is not my area. However the "Are dense, partially ionized plasmas also solutions? is purely a semantics issue. As Shakespeare said "What's in a name? that which we call a rose. By any other name would smell as sweet". Just remember that in a typical plasma only small fraction consists free electrons and ions. The emission spectrum of plasma is a continuum with superimposed lines of the element.

Is it that the solvated electron appears to go around absorbing energy from the environment? Rather, consider that the solvated electron just sits there (stably, in the dark), but when white light illuminates it, it can jump to a higher energy state by absorbing some of the red light from the white, leaving blue light to be seen by the observer.

A solvated electron sufficiently disturbs the solvent to form a cavity. Hence, the electron is confined in a relatively small space. Anything as light as an electron which is confined will have quantized energy levels as quantization is typically due to boundary conditions. From this it is clear the electron can absorb energy. Whether or not the first excited state is below the top of the well will depend on temperature, pressure, and the solvent. Also, this is in no way a plasma.


text:
Gamma spectrometry
I want to calculate minimum detectable activity (MDA) at 200 and 300 keV energy of a gamma spectrometry by HPGe detector. 662 keV Photopeak is present for 137Cs.
Assume 2 scenarios:

There is a 'peak' at 200 keV. Let's say net counts is 'b' unit. And gross count is 'a+b' unit.
There is no peak at 300 keV. Gross count is 'c' unit

Using Curie's formula, we can calculate MDA.
What background value $B$ should I use in the above cases?
Pl see the image.

With such Gamma ray spectroscopy you'd be dealing with counts not area. The detection limit would depend on what else in the sample was emitting gamma rays and hence raising the background counts (noise). Another gamma ray near 662 keV would result in peak overlap and substantially increase the background noise.

@ggs Can you take a minute and explain this a little more clearly? I've added some formatting to your question but I think you can explain more systematically. Also, what is Curie's formula? There is Curie's law but that's a horse of a different color.

The Currie algorithm* for the calculation of the minimum detectable activity (MDA) is generally considered obsolete. Today, detection limits are calculated according to ISO 11929 and related standards (for example, see also https://chemistry.meta.stackexchange.com/a/3283/7951)
Nevertheless, when using the Currie MDA algorithm, the detection limit $L_\mathrm D$ is calculated using
$$L_\mathrm D=k^2+2L_\mathrm C$$
which can be translated into MDA.
In calculating the detection limit, the continuum is based on data on each side of the peak region for peaks that were found. It is based on the equivalent of the peak region itself for peaks that were not found.
For example, for a found peak with the linear continuum model in effect (ypou could also use a step continuum model instead), the continuum is calculated as
$$B=\left(\frac N{2n}\right)\left(B_1+B_2\right)$$
Where
$N$ is the number of channels in the peak region,
$n$ is the number of channels on each side of the peak region used for the determination of the continuum counts,
$B_1$ is the sum of counts in the $n$ channels to the left of the peak region, and
$B_2$ is the sum of counts in the $n$ channels to the right of the peak region.
For peaks that had a background interference component subtracted, the equation for $L_\mathrm C$ becomes
$$L_\mathrm C=k\sqrt{B+\left(\frac{T_\mathrm s}{T_\mathrm b}I_\mathrm b\right)+\left(\frac N{2n}\right)^2\left(B_1+B_2\right)+\left(\frac{T_\mathrm s}{T_\mathrm b}\right)^2\sigma_{I_\mathrm b}^2}$$
Where
$B$ is the value of the continuum subtracted,
$I_\mathrm b$ is net peak area of the background measurement,
$T_\mathrm s$ is the live time of the sample measurement, and
$T_\mathrm b$ is the live time of the separate background measurement.
This equation is used as a critical level test criterion, for example at $95\ \%$ confidence level. If the net peak area is less than $L_\mathrm C$, then a “not detected” decision is made. If the net peak area is greater than $L_\mathrm C$, then a “detected” decision is made.
* Currie, L. A. Limits for qualitative detection and quantitative determination. Application to radiochemistry. Anal. Chem. 1968, 40, 586–594.


text:
Use IUPAC system to name the structural diagram below
Can someone break down the steps for naming $\ce{CH3-C(Cl)=C(Cl)-CH2-CH3}$?

I redrawn your structure using free LaTeX package chemfig based on your comment and previous attempt to format it using source code markup tool. There are also numerous free GUI tools such as MarvinSketch to draw molecules if you cannot reach the wanted representation with ASCII graphics (which is understandable).

Also, for the reference, see How to ask and answer nomenclature questions?. I think there is a very similar question asked before, but I'm unable to find it for now; probably other users could help.

(E)-2,3-Dichloropent-2-ene
Explanation:
The longest carbon chain is a 5 membered chain with a carbon-carbon double bond. Since double bond must be given the least locant, we number it from the leftmost carbon as in your image. The substituents are named accordingly by indicating their positions.
For the stereochemistry aspect of the compound, we observe the following:
The highest priority group on the second carbon atom from the left, as per the Cahn–Ingold–Prelog priority rule is the chlorine atom, and on the third carbon atom from the left it is again chlorine. Clearly, both the higher priority groups (here chlorine atom) are opposite, and so the configuration is E. To learn about E-Z notation click here.

(E)-2,3-dichloropent-2-ene?

It would be best to [edit] your answer to actually contain the correct answer. Please also see our guidelines on how to ask and answer nomenclature questions.

@Martin-マーチン, Could you please tell whether the answer is fine now. I have improved the answer to my best. This was my first attempt in answering.

As per the rule:

Longest carbon chain is to be considered.
Numbering of carbon should start from the side which has lowest set of locants i.e from right side, locants are 3,4 and from left side, locants are 2,3.

So answer is 2-3-dichloro pent-2-ene.

I'm afraid without a the stereodescriptor this answer is incomplete. It would also benefit from a better formatting, i.e. correct spelling and punctuation, as well as an correctly rotated image.


text:
Is Ammonia (NH3) base?
In my text book, it is written that $\ce{NH3}$ is a base. They mentioned the reason for this is that it produces $\ce{OH-}$ ions in water. They stated this reaction:

$\ce{NH3 + H2O -> NH4OH}$

So That's why, they said that it is a base.
But in a neutralization reaction, mentioned by our teacher,
Acid + Base = Salt + water
So, if $\ce{NH3}$ is a base then,
it should be,
Acid + $\ce{NH3}$ = Salt + Water
But it does not happen. When reacting with acid, it doesn't produce water.
For example,

$\ce{NH3 + HCl -> NH4Cl}$
--  Reference

Again,

$\ce{NH3 + H2SO4 -> (NH4)2SO4}$
--  Reference

So it is not producing water in the neutralization reaction. Hence, I'm confused whether it is base or not.

You can still make it work like a base just by changing the stoichiometric coefficients of the reaction.  $\ce{2NH_3 + 3 H_2O -> 2NH_4 OH + H_2 O}$

Your problems are caused by using the Arrhenius theory, which is based on electrolytic dissociation.
According to the Arrhenius definition, acids are compounds that dissociate and release hydrogen ions $(\ce{H+})$ into the solution:
$$\ce{HCl -> H+ + Cl-}$$
Bases are defined as compounds that dissociate and release hydroxide ions $(\ce{OH-})$ into the solution:
$$\ce{NaOH -> Na+ + OH-}$$
The products of a neutralization reaction of an acid with a base are a salt and water:
$$\ce{HCl + NaOH -> NaCl + H2O}$$
This definition works for metal hydroxides such as $\ce{NaOH}$, which can exist as pure compounds. However, it doesn’t work for bases such as $\ce{NH3}$.
In order to solve this problem for $\ce{NH3}$, the fictitious compound “ammonium hydroxide” $\ce{NH4OH}$ was proposed (see also this related answer). Assuming this formula, the Arrhenius definitions of bases and neutralization reactions work again:
$$\ce{NH4OH -> NH4+ + OH-}$$
$$\ce{HCl + NH4OH -> NH4Cl + H2O}$$
However, ammonium hydroxide cannot be isolated; the fictitious solid compound does not exist. Nevertheless, the misleading traditional name “ammonium hydroxide” is still widely used for solutions of ammonia in water.
In order to really solve your problems with $\ce{NH3}$, you need to switch from the definition of bases according to the Arrhenius theory to the definition of bases according to the Brønsted–Lowry theory.

Well, you can explain it as a lewis base. In ammonia, nitrogen forms 3 single covalent bonds with 3 atoms of hydrogen. However it has 5 electrons in its valence shell. Which means that there are two electrons left a.k.a lone pair of electrons. Hence, it acts as a lone pair electron doner making it a lewis base.

"Acid + base gives salt and water" is only possible when acid and base both can dissociate into two ions. Let's take an example
$\ce{HCl}$ and $\ce{NaOH}$ gives $\ce{NaCl}$ and $\ce{H2O}$
So, in that case, $\ce{HCl}$ and $\ce{NaOH}$ both can dissociate.
But in case of $\ce{NH3}$ and $\ce{H2O}$ only $\ce{H2O}$ dissociate into $\ce{H+}$ and $\ce{OH-}$ but $\ce{NH3}$ remains the same.
And reaction proceeds with the help of lone pair of $\ce{NH3}$.
$$\ce{NH3 + H+ + OH- ->NH4+ +OH-}$$ and in later stages both combine together due to electrostatic force.


text:
Does this electronic configuration violate the Hund's rule of maximum multiplicity?

I know this configuration violates the AufBau principal but does it also violate Hund's rule of maximum multiplicity?

No, I don't think it does. Why do you think it violates the maximum multiplicity rule?

Because Hund's rule says that all unpaired electrons must have the same spin, but I don't know if unpaired electrons in one orbital can have one spin and unpaired electrons in another orbital can have the opposite spin without breaking the rule

Have a look at https://en.wikipedia.org/wiki/Hund's_rule_of_maximum_multiplicity and https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configurations/Hund%27s_Rules

You got the definition slightly wrong, Hund's rule basically states that degenerate (i.e. equal in energy) levels must be filled up before pairing of electrons starts, and  those unpaired electrons must have the same spin.

So, $\ce{3p_x}$, $\ce{3p_y}$, $\ce{3p_z}$ are degenerate, because they have the same energy. So, these levels will each have one electron, all with same spin before the pairing of electron begins.

But the same rule does not apply to 3p and 3d electrons because their energies are different. So your diagram does not violate Hund's rule (but it is wrong due to Aufbau principle, as you have noticed)


text:
Are there examples of crystals with random but static mixtures of two unit cell configurations?
The image in this question turned out to probably be wrong, but it still makes me wonder. Are there examples of crystals where the atoms within a unit cell can have two (or more) different configurations or orientations who's energies are so close that a sample crystal at room temperature could have a random but static mixture of the two?
With sufficiently large organic molecules (e.g. proteins) this could be true and possibly even common, but I'd like to know about a simple example with a small number of atoms in the unit cell. It might be produced by rapid cooling or crystallization from solution in unusual conditions, and it might be detectable (at least in some cases) by a slight broadening in high resolution X-ray diffraction peaks. (sort of a solid-state version of Raleigh scattering perhaps?)

Exact position of each and every atom in different unit cell would slightly deviate from each other due to thermal motion — that's why ORTEP diagrams are preferred by crystallographers. Also, there are modulated crystal structures.

@andselisk the modulated crystal structures seems along the line of what I was thinking of; I've updated to "random but static" to exclude thermal broadening. Thanks for these!

A crystal with lots of stacking faults might qualify. At room temperature under no stress it will be pretty static.

@uhoh Do you refer to something like in the «answer»?

Partial disorder is a pretty common thing both in organics and inorganics.

@IvanNeretin While I agree with you the question of «randomness» was not clear for me. If random in space but fixed, or mobile (pedal motions), both may of course contribute to diffuse scattering.  At least on first look neither of the two picks revealed such.  Their site occupations was refined in the models.

True, static and dynamic disorder produce more or less the same pattern and can be pretty hard to tell apart. Still, both are ubiquitous.

@IvanNeretin I've added comments here.

If you see an "ALTLOC" record in a PDB or CIF file, you know the electron density has been modeled with two sets of atoms at alternate locations. Typically, this is static disorder, especially for cryo-methods. I did not find a small molecule example, but I don't have access to the Cambridge structural database.

Possibly retracted soon, because I'm not aware how to include a picture in a comment to clarity your question.  Do you mean instances like the following, where sites seemingly are occupied by two atoms simultanously -- as in analogues of Prussian blue -- just because the model deposit in the .cif is the average of the unit cells' content?

(Chen et al. in Angewandte Chemie,  129, June 2017, 9151-9156, doi 10.1002/ange.201702955)

No problem as far as I'm concerned; if done with care, answer posts like this can  evolve quickly into the correct answer. I don't think I mean this. I am asking about a real crystal fixed rather than software artifact. One hypothetical example would be a unit cell which is almost but not quite symmetric under a 180 degree rotation. So it can fit on the lattice in two ways; right-side-up and up-side-down, with only a tiny difference in energy, say of order $k_B T$. This way maybe 1/3 of them are up-side-down, randomly. But once the crystal is built, they are stable ("static") and won't flip.

That's not the only way it could happen, it's just meant as an illustrative example. btw it's late where I am, so I might be off line for 6-8 hours...


text:
How to control pH in a fermenter based on the acid produced by mircoorganism
I have got a bioreactor with an aqueous medium which produces biomass (yeast; S.cerevisiae) after inoculation.
I am observing and collecting data over the whole process via probes/etc. on different process parameters such as pH, temperature, offgas of $\ce{O_2}$ and $\ce{CO_2}$, base uptake ($\ce{NaOH}$), weight of bioreactor and some little more like stirrer speed and so on.
How can one calculate or make a model on adding base to control pH based on the produced acid  (for now I've got no information on which one that is exactly) of the yeast or other factors?
(Is there a mechanistic or only a deterministic solution?)
Edit:
I can indirectly calculate how much acid is produced because that corresponds to the biomass whose growth data I collect. I don't know the exact acid that is produced, f.ex. $\ce{HCl}$ or so, but I think that the molar ratio between my used base and the produced acid is either 1:1 or 2:1, based on my assumption that yeasts will not produce acids with 3 or more protons in their metabolism.
The medium used for growing:
I am currently using glucose as the main substrate, the fermentation itself lasts only about 5-7 hours which means the $\ce{EtOH}$ that is produced can be "neglected" as second substrate. Furthermore I use $\ce{(NH4)2SO4}$ and $\ce{(NH4)H2PO4}$ as N-source and P-source respectively.
[Also some trace elements but those are in minor concentrations/ negligible for my pH model.]
My question is if there is a way to make a model on the acid production to control the base feed to stabilize the pH.

So let me get this straight: you don't currently have a way of measuring how much acid the yeast produces, but wonder if, knowing this, you can predict how much base you'd need to add to keep the pH at a given value?

No, I can indirectly calculate how much acid is produced because that corresponds to the biomass whose growth data I collect. I meant that I don't know the exact acid that is produced, f.ex. $\ce{HCl}$ or so. But I think that the molar ratio between my used base and the produced acid is either 1:1 or 2:1, based on my assumption that yeasts will not produce acids with 3 or more protons in their metabolism.
My question is if there is a way to make a model on the acid production to control the base feed to stabilize the pH.

What is the medium on which the yeast grows? Does it contain sucrose or starch or something else?

I am currently using glucose as the main substrate, the fermentation itself lasts only about 5-7 hours which means the $\ce{EtOH}$ that is produced can be "neglected" as second substrate. Furthermore I use $\ce{(NH4)2SO4}$ and $\ce{(NH4)H2PO4}$ as N-source and P-source respectively. [Also some trace elements but those are in minor concentrations/ negligible for my pH model.]

You should always [edit] in the additional information, not just include it in the comment sections.

I agreed with Martin - マーチン♦, because these comments may disaapear with time. Besides, information in your multiple comments is important to answer the question.

It is a mix of acids and could include citric, malic, lactic, acetic. To which pH do you adjust? Are there any bacteria present as well? Did you test the concentration of titratable acid at the beginning and end of the fermentation? See https://beerandbrewing.com/understanding-ph-and-titratable-acidity-in-sour-beer-tools-for-brewers-and/

Do you batch feed the glucose or is it continuous feed? If you're starting with high glucose, the yeast will still ferment ethanol primarily, so the acid is carbonic acid produced 1:1 with ethanol. If you use glucose limitation to get complete oxidation, carbonic acid is again the main acid product, but the ATP yield per carbonic acid (and therefore per biomass) is much higher. You can estimate the ratio of acid to biomass based on the known ATP yields if you know whether or not you are producing ethanol.

That said, simple feedback control connected to a continuous pH probe is much more reliable and is standard practice for well controlled fermentation.


text:
How can I determine the electrode potential of ethanoic acid redox reaction?
I'm conducting a research paper on the effect of surface area ratio from anode to cathode and the electrode material on the rate of electrolysis of ethanoic acid by measuring the volume of carbon dioxide collected. In order to determine the material that I am using (so that it is kept constant), I need to determine the electrode potential of ethanoic acid.
If anyone knows of a standard electrode potential for ethanoic acid, that would be great so that I can avoid testing for it experimentally.

Perhaps this link might help: http://ceae.colorado.edu/~silverst/cven5534/REDOX%20HALF%20REACTIONS.pdf.

Do you know the chemical equation of the reaction, or at least of the half reaction?


text:
Why is detergent better at cleaning than soap?
Soaps and detergent are both molecules that have a polar head and a long organic tail. In Soaps, the head is often a carboxylate ion $\ce{COO-}$ where as a detergent has a benzosulphonate $\ce{C6H11SO3-}$(or other) head:

I have often seen that detergents are 'stronger' than soaps. Why is this? It is given as the reason why we don't wash our skin with detergents. The only answer I could find is that detergents don't form insoluble salts with Mg and Ca (scum) but doesn't seem like it would have a large impact. So why are detergents stronger than soaps?

Actually one of the advantage of detergents is right that they are less aggressive (pH) to the skin than soaps. Bath shower and hairs formulations are detergents. Still, all the pros that comes to my mind are not due to cleansing power. If there are regular differences (the process has multiple steps or facets) likely someone will answer.

hm, if you wouldn't mind, could you describe the pros? I haven't heard of many except being 'stronger' which apparently isn't even true!

The pH of anionic detergents is less alkaline than that of soap, so they are better for skin and especially hairs. This is because benzensulphonic ac. is stronger than fatty acids. Those detergents should be in standard body shower formulation. Tough we are speaking of formulations that might contain more or have changed. The others are as you and especially the answer so mentions are linked to water hardness and mass production.


text:
Temperature dependence of reaction rates
Is it correct to say that "Increasing the temperature will always increase the rate of any chemical reaction"?
Many articles on the internet and textbooks say yes but what about biochemical reactions that take place in our body?
To my knowledge  at higher temperatures some enzymes (which are catalysts in chemical reactions in the body) become less effective which in turn slows down the reaction rate.
So is it correct to say "any chemical reaction" in the statement or do I have to think of the temperature dependence of the biochemical reactions without the enzymes?

Many enzymes stop working at higher temperature because the enzymes are proteins, and the structure of the protein breaks down above a certain temperature. The reaction slows down due to the loss of catalyst. Generally, the statement that 'reactions become faster at higher temperatures' indicates the ability of more molecules to overcome the activation barrier at high temperatures, rather than any catalysis. Look into Arrhenius equation.

Also you can switch from spontaneous occurring to non spontaneous. Look at Gibbs free energy You can reasonably say what you refer to at least for on going single reactions.

@Alchimista - Spontaneous or non-spontaneous is not an issue when discussing the rate of a defined reaction step.

@Karsten.  Not clear what you mean. Beside the fact that I did not refer to any step but reaction as something going from reactants to products of interest without being coupled to another reaction, again so defined.

My meaning was you can stop a reaction, beside enzymes denaturatiom and pot melting :) Good point is arisen by the answer of DivMit.

Why are none of you upvoting the question (or the answers)? Is this a bad question according to Chemistry.SE standards?

Well mostly it does but there can be exceptions.
Eg Consider the reaction $$\ce{2NO + O_2 -> 2NO_2}$$
It has a negative temperature dependence. The rate decrease in increasing the temperature.
There is a simple reason to why this happens.
We have a RDS with an intermediate species. The rate determining step must have a positive temperature dependence.  But the preceding equilibrium that produces the intermediate species could have a negative temperature dependence , that affects the overall reaction.
The Mechanism
$\ce{NO + O2 <=> ONOO}$ Fast but reversible
$\ce{ONOO + NO -> 2NO2}$ Rate determining step
If Step 1 is exothermic, as a bond is being formed between the reactants.
This has a greater impact on the overall reaction and by le Chatlier's principle the equilibrium shift backward . We get a negative temperature dependence!

Good point, I weakened the statement in my answer accordingly.

Well, all the elementary steps get faster. Because the back reaction gets faster by a larger degree, the overall rate is lower. Nice example!

It is correct to say: "If nothing else changes, an increase in temperature will increase the rate of an elementary step of a chemical reaction".
All kinds of things could go wrong. If the temperature is so high that your reaction vessel melts, spilling a liquid reactant and removing it from a solid reactant, the rate will not increase. If the temperature is so high that one of the reactants undergoes a different reaction (maybe a decomposition), the rate will not increase. If the temperature is so high that your solvent turns into a plasma, the rate will not increase.
What I am saying is that enzymes are not the only aspect of a reaction that might not withstand higher temperature. On the other hand, within the range of temperatures where the enzyme has the proper conformation, the rates do go up with increasing temperature.

I think this is the perfect answer!

Since you mention it, can you elaborate on the plasma case? I would think that the dependence is inverse here - a lower temperature leads to less dissociation, so the effective rate of molecule formation increases for low plasma temperatures. In fact I know this to be true in chemical equilibrium, but I've always wondered what the model for the reaction rates is.

@PoorYorick Not the creation of the plasma slows down, the reaction slows down because it needs solvent but now the solvent is gone. I know nothing about plasma reactions, and I used it as an out of this world example, but I saw your question on plasma reaction, alas.

The statement is basically true, but only if you take into account that any reaction has two directions, and goes towards equillibrium.
Two issues still arise

temperature changes the point of equillibrium
in equillibrium, the reaction rate is, by the very definition of equillibrium, zero

Seeing that, it's better to throw away that overgeneralistic statement. The rate of elementar reaction steps indeed increases with temperature. As long as it can  still work, of course, and e.g. your enzyme is still alive.


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How do computational chemistry programs determine the irreducible representation of a molecular orbital?
I was coding a simple RHF code to test my understanding of computational chemistry when I stumbled upon this question. How exactly do computer programs like Gaussian, GAMESS-US etc. know which irreducible representation to assign to the molecular orbitals generated?

A molecule has a point group; a molecular orbital (or a group of MOs) has an irreducible representation.

The software constructs a symmetry adapted linear combinations (SALC) of the AOs. The IRREPs of the MOs follow directly from that: each MO is formed from SALCs of the same IRREP only.

Thanks. How exactly does the software construct these SALCs? Projection Operator/Schmidt orthonormalisation?


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Modification of Janus nanoparticles
I would like to synthesis gold/polyacrylic acid (PAA) Janus nanoparticles in solution.
Any ideas how I can functionalize gold with a photosensitizer and PAA with an antibody so that there would be no other unwanted interactions?


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Does aminosugars give positive response in alkaloid tests?
Does aminosugars give positive response to test for alkaloids such as tests with Dragendorff's reagent, mayer's reagent, Hager's reagent, Wagner's reagent etc?

Dragendorff's reagent works for tertiary amines (and some secondary), amino sugars contain primary amines so there is no reaction nor do N-acetylaminosugars

I saw "aminosugars" but read it as "aminosugaris" and was thinking I don't recognize that dinosaur name, and what's it have to do with alkaloids?  Too much with the kids I guess :)

@CrossRoads I dont know about any such Aminosaurus :P :D

Most alkaloids are tertiary amines. Almost all tests for alkaloids, reagents of which are  are solutions of the salts of heavy metals, are precipitation tests. For example, all of tests with Dragendorff's reagent, Mayer's reagent, Hager's reagent, and Wagner's reagent give color precipitates: The heavy metal atom in the reagent with the nitrogen in the alkaloid would form ion pairs. These ion pairs form an insoluble precipitate. The color of the precipitate depends on reagent, plant species, etc. (Ref.1). Most amino sugars are either primary amines or acetamides, thus, I doubt they'd give precipitates with sought reagents. However, some reagents give false-positive tests for non-nitrogenous extracts indicating the presence of alkaloids. The best example is Dragendorff's spray reagent (Ref.2).
Mean time, some reagents are giving negative results even with alkaloid extracts, e.g., Mayer's reagent (Ref.3). Therefore, my answer is purely opinion based.
References:

Felix G. Coe, Gregory J. Anderson, "Screening of medicinal plants used by the Garífuna of Eastern Nicaragua for bioactive compounds," Journal of Ethnopharmacology 1996, 53(1), 29-50 (https://doi.org/10.1016/0378-8741(96)01424-9).
Abdel‐Azim M. Habib, "False‐positive alkaloid reactions," Journal of Pharmaceutical Sciences 1980, 69(1), 37-43 (https://doi.org/10.1002/jps.2600690111).
J. W. Field, M. Kandiah, "A note on the use of mayer's reagent for the detection of quinine in alkaline urine," Transactions of the Royal Society of Tropical Medicine and Hygiene 1935, 28(4), 385-390 (https://doi.org/10.1016/S0035-9203(35)90133-X).

If I get scope to do this simple lab tests I'll publish it as answer. Currently aminosugars are not available to me. We have chitin flakes though, that makes technically a different question because chitin is polysaccharide.


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How a titration curve is affected when a poorly soluble salt is formed?
Suppose that I have $\pu{100 mL}$ solution of strong acid, $\ce{HA}$, at $\pu{0.10 mol L^{-1}}$.
This solution is titrated with a strong base solution, $\ce{BOH}$, at $\pu{0.10 mol L^{-1}}$.
Suppose yet that a poorly/sparingly soluble salt, $\ce{BA}$ $(K_\ce{sp}=\pu{2.30×10^{-3}})$, is formed during titration.
The reactions involved are:

ionization of strong acid: $$\ce{HA + H2O -> H3O+ + A-}$$
dissociation of strong base: $$\ce{BOH -> B+ + OH-}$$
self-ionization of water: $$\ce{2 H2O <=> H3O+ + OH-}\qquad K_{\ce{w}}=\ce{[H3O+][OH-]}$$
Equilibrium of poorly soluble salt: $$\ce{BA <=> B+ + A-}\qquad K_{\ce{sp}}=\ce{[B+][A-]}$$

Charge balance of the solution is given by the equation: $$\ce{[H3O+] + [B+] = [OH-] + [A-]}$$
Once that $\ce{B+}$ and $\ce{A-}$ are spectator ions, are come from strong substances, and disregarding the ionic strength effect, I think that titration curve don't be affected. I'm right?
If not, how to write a mass balance expression to take account into the precipitation of $\ce{BA}$?

The definition of a strong acid and strong base is that they dissociate completely in an aqueous solution. // $\ce{B+}$ and $\ce{A-}$ are spectator ions with reference to the acid base reaction, but obviously not in respect to the ppt reaction.

How a titration curve is affected when a poorly soluble salt is formed?

If by "I think that titration curve don't be affected" you mean that the graph of pH vs volume of strong base will not be affected by the precipitations, then you are right. This is because the species that form the precipitate are not part of any other equilibrium reaction.

how to write a mass balance expression to take account into the precipitation of BA?

In general terms, the mass balance (really the amount balance) can be expressed in concentrations if the system is homogeneous (all in solution). For example, it could look like this:
$$[E] + [F] + [G] = \mathrm{constant}$$
If there is a solid species $H$ that you have to consider, use an expression based on amount of substance:
$$n_E + n_F + n_G + n_H= \mathrm{another\ constant}$$
If you want to, you can express parts of this as concentration again:
$$([E] + [F] + [G]) \cdot V_\mathrm{solution} + n_H= \mathrm{another\ constant}$$