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Molar Absorptivity of Copper(II) Sulfate
What is the molar absorptivity of copper sulfate? I am trying to find the molarity of copper sulfate solution by absorption in a spectrophotometer with a cuvette.
However, I can't find $\varepsilon$, the molar absorptivity coefficient of $\ce{CuSO4}$. I searched it up online, and I found experimental data saying that is $\pu{2.81 L mol-1 cm-1}$, but another source says that it is $0.91$.
I have searched on various databases including NIST's, but I couldn't find anything. Is there a definite molar absorptivity for everything, or is it all experimental?

The value will depend on wavelength so both the values above could be correct; it would be worth checking.  Also if you can, measure absorbance vs. concentration. The eqn. is $A=\epsilon [C]l$ where l is the cell path length and $\epsilon $ the extinction coefficient in $\pu{dm^2mol^{-1}cm^{-1}}$

Right, I searched up 635nm, but I forgot to make sure that it was actually that. Thanks for reminding me

"I searched it up online, and I found experimental data saying that is 2.81 Lmol−1cm−1, but another source says that it is 0.91." Please give your sources as clickable links (and the use of web archive links is encouraged) so that other can check it as well.

From a purely mathematical point if view, the absorptivuty at 635 nm where you searched appears to be about one-third the max value in Ed V's plot. Your different numbers may reflect that ratio.

These are the absorbance spectra of copper sulfate in water (open markers) and heavy water (filled markers) (1):

The absorbance in water is maximum at 780 nm, as stated in the reference. From the spectrum, the molal absorption coefficient, at 780 nm, appears to be approximately 12.5 $\mathrm{(mol/55.51 mol \ of \ water)^{-1} cm^{-1}}$ with aquamolality equal to 0.5.
Since only ordinary water, i.e., $\ce{H2^16O}$, is of interest in the OP's question, the aquamolality is the same as the molality, so the solution is simply 0.5 molal. The measurements were reported to have been made at room temperature, with no specific temperature given. Hence, the solution molarity is approximately 0.5 M.
Reference

G. Jancso, "Effect of D and $^{18}O$ isotope substitution on the absorption spectra of aqueous copper sulfate solutions", Radiation Physics and Chemistry, 74 (2005) 168-171. DOI: 10.1016/j.radphyschem.2005.04.011

TL;DR: the values that you may find online will most likely only be accurate up to a few percents. I would strongly encourage the use of a reference solution of known concentration that you can further use for calibration on you own hardware (cuvette and spectrophotometer).

Your question is no trivial one. Indeed, absorption spectra may be influenced by a variety of factors, a phenomenon referred to as chromism. Among these factors are temperature, pressure (mechanical forces), solvent (including the use of heavy water instead of water, as mentioned in Ed V's answer), for instance. As an example, see below the absorbance spectra of a given molecule (termed Conmpound 1 in [1]) in different solvents.

Additionally, even when measuring the same solution, prepared in the exact same way, in identical cuvettes, on the same day, and with the same spectrophotometer, there can be a significant dispersion in the so-obtained results. For instance, below are raw spectra of diluted haemolysed blood equilibrated with CO$_2$, as described in [2] (personal measurements).

As you can see, there is a significant dispersion of the results, even if care was taken to reduce the external influences to a minimum. This kind of deviations explains why—even for a molecule as simple and ubiquitous as water—considerable amount of variation may exist in the literature between reported absorbance values coming from different papers. See below an example with Segelstein[3] and Kou[4] data.

"Is there a definite molar absorptivity for everything, or is it all experimental?"
The answer is thus mitigated: in theory, every molecule has a given absorbance spectrum in a fixed set of settings (solvent, temperature, EM-field, pressure, etc.). However, since it is often exceedingly hard to reproduce these settings or compensate for measurement artefacts (id est influence of the cuvettes, spectrophotometer, etc., even with blanking), this knowledge is in practice often experimental.
Of note, there exists some databases of absorbance and fluorescence spectra, which—alas—are often very patchy. You mentioned NIST database, below are a few more that I compiled (sorry about it being a picture and the missing references).

What does this mean in practice?
It means that even by comparing different papers measuring the same substance, the molar extinction coefficients that you can find may vary from one source to the other. This in turn makes quantitative measurements based on these coefficients unfeasible or inaccurate in practice.
What can you do?
Common practice to achieve what you intend to do is to prepare a standard solution, id est a solution of known concentration, for instance by diluting a known mass of CuSO$_4$ into a known volume of water. You can then compare the absorbance of this standard solution at a given wavelength (usually chosen at point of null derivative such as a maximum, so that a slight error in wavelength does not result in a huge absorbance change) with that of the solution which you want to characterise. Beer-Lambert law may be useful here...

References
[1] Synthesis of New 2-(2´-Hydroxyaryl)benzotriazoles and Evaluation of Their Photochemical Behavior as Potential UV-Filters, Farkas et al. (2010), DOI: 10.3390/molecules15096205
[2] Measuring hemoglobin spectra: searching for carbamino-hemoglobin, Dervieux et al. (2020), DOI: 10.1117/1.JBO.25.10.105001
[3] The complex refractive index of water, Segelstein (1981), URL: http://hdl.handle.net/10355/11599
[4] Refractive indices of water and ice in the 0.65- to 2.5-µm spectral range, Kou et al. (1993), DOI: 10.1364/ao.32.003531

The molar absorptivity of copper sulfate in water at 810 nm (the position of the absorbance maximum) is about 12.3 L/mol/cm

Could you cite your source?


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What is meant by 'penetration of atomic orbitals'?
While studying atomic structure, I have faced this term very often. I assume this is a very detailed type question!
But if anyone can please provide just the preliminary idea about this, it will be helpful.

Hey there! That's a good question, however, the answer to question that user @Pritt linked carries a brief explanation of "penetration of atomic orbitals" and is presumably what you want. So, assuming you require nothing else, we'll be closing this question as a "duplicate" (we do this to reduce clutter on the site). However, if that answer does not have what you're looking for, then by all means go ahead and edit your question to include specifics and it (may) be re-opened. We look forward to your future participation on Chemistry.SE. Cheers!

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How can I separate sodium ions from brine or molten NaCl?
Is possible to separate sodium ions from an aqueous solution of NaCl? If so, how do I go about it?

separate for what purpose? To produce sodium or to produce pure water?

@matt_black To produce sodium ions(Na+)

You can't produce isolated sodium ions in any volume. You can electrolyse molten NaCl but this gives sodium metal (you are basically adding electrons to neutralise the ions).

Research the Contact Process. The products of the chlor-alkali industry are Chlorine gas, hydrogen gas and aqueous sodium hydroxide. A titanium anode is used and a steel cathode is used.

Brine ($\ce{NaCl (aq)}$) and molten ($\ce{NaCl (l)}$) are very different things...

You can't simply produce isolated sodium ions, you also need to have an equivalent amount of counter ions of the opposite charge. So, in this case, you are going to get sodium chloride unless you introduce some other negatively charged species.
So you could just evaporate water which will give an increasingly concentrated solution of sodium chloride and, eventually, solid sodium chloride (which will contain all your sodium ions alongside an equal number of chloride ions.
This sort of separation is commonly used for a different reason: to produce pure water. Sea water is too salty to drink but there is far more of it than fresh water. So distillation (retaining the fresh water) or reverse osmosis (which produces more concentrates salt on one side and fresh water on the other). These techniques produce both fresh water and more concentrated salty water (which is one way of getting more sodium ions in your product).

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Oxidation State of Amino Group
What is the oxidation state of nitrogen the following amino acid:

I figured it must be -III:

Hydrogen is always +I
Both electrons of the N-C bond are near the nitrogen (higher electronegativity).

However, I've also heard that the N-C bond sometimes is seen as neutral (+0), which would make the oxidation state of Nitrogen -II. Which one is correct?
Do the zwitterions play a key role here (as the molecule could also be H3N-C-COO)?

I think -III is the correct 'formal' oxidation state of N in your molecule.
The method I was taught is the following. Write down the molecule without any bonds. Sum the number of 'valence' electrons from all atoms. For each covalent bond (counted as single regardless whether it is single, double or triple), put two electrons on the most electronegative atom of the pair (if same type of atom, put one electron on each instead). Distribute all remaining valence electrons in decreasing order of electronegativity (of course without going beyond 8 electrons on any atom). When you have multiple atoms of the same type, put one electron on each alternately. At the end of this process, the formal oxidation state of each atom is its number of valence electrons minus the number of electrons that ended up on it.
For an organic molecule with 'R' moieties, if you can assume the atoms have equal electronegativity (e.g. the C-R bond in your molecule would be a C-C bond), you can just put an electron on C from the beginning and ignore R.
If you try that in your example, you have in total 4*1 (from 4 H) + 1*5 (from 1 N) + 2*4 (from 2 C) + 2*6 (from 2 O) = 4 + 5 + 8 + 12 = 29 valence electrons. Put 1 of them on the C-R carbon. You are left with 28 more electrons to place. Take care of the bonds first. You put 6 electrons on N (3 bonds to less electronegative atoms), 2 electrons on the C=O oxygen (1 bond to a less electronegative atom), 4 electrons on the COH oxygen (2 bonds to less electronegative atoms), 1+2 electrons on the CR carbon (1 bond to another carbon, 1 bond to a less electronegative atom), 1 electron on the carboxyl carbon (1 bond to another carbon). You are left with 12 more electrons to place. 10 of them go on the two oxygen atoms. 2 of them go on the nitrogen. You end up with: 8 electrons on each oxygen (6-8 = -2); 8 electrons on the nitrogen (5-8 = -3); no electrons on any hydrogen (1-0 = +1); 3 electrons on the carbon bound to R (4-4 = 0); 1 electron on the carboxyl carbon (4-1 = +3).
You can also verify that 2*(-2)+1*(-3)+4*(+1)+1*(0)+1*(+3) = - 4 - 3 + 4 + 0 + 3 = 0.
The nice feature of this method is that when you have atoms of the same type but with different oxidation state (e.g. a nitro and an amino group in the same molecule), it should still give a correct result. E.g. if you process $\ce{H2N-CH2-CH2-NO2}$, you should get oxidation state -III for the amino nitrogen, and +III for the nitro nitrogen.
Interesting question about the different charged status; I'll let you figure it out for yourself - the method it the same, except that of course you will have covalents bonds in different places.

Sorry for not accepting your question in so long, I've only revisited Chemistry.SE today and remembered this question.


text:
Is NMR possible using nitrogen as the nucleus?
Is NMR spectroscopy performed using nitrogen nuclei? If yes, then where?

15N is NMR active (spin 1/2) and is commonly used in protein NMR (amongst other uses). Did you see that there is a Wikipedia page on it? https://en.wikipedia.org/wiki/Nitrogen-15_nuclear_magnetic_resonance_spectroscopy

and before you ask  also of many other nuclei, $\ce{^2H, ^{10}B,^{11}B, ^{17}O,^{19}F,^{27}Al,^{29}Si,^{31}P,^{129}Xe}$ among others :)

Of course yes -- because of the spin $\ce{^{15}N}$ and the much lesser used $\ce{^{14}N}$ nuclei posses:

The image is from here, the site provides some useful additional  overview to the source mentioned by @orthocresol.

P.S. If there were no NMR active nitrogen nuclei, 3D NMR would not this widespread in protein research:

(source)

Thanx man! Actually I just started my spectroscopic studies and hence was not aware of the fact.


text:
How does optimizing the molecular orbital coefficients in CASSCF improve its multi-reference capabilties?
The complete active space self-consistent field (CASSCF) method is a special kind of multi-configurational SCF procedure, attempting to combine Hartree-Fock (HF) and configuration interaction (CI). In HF, the molecular orbital (MO) expansion coefficients are varied during the procedure, but the wave function contains just one Slater determinant. In CI, the multielectron determinant expansion coeffcients are varied, but the MO's used to build the excited Slater determinants are taken from HF, and are not varied during the energy-minimization.
During CASSCF, on the other hand, all MO expansion coefficients are optimized during the energy-minimization procedure, in addition to the multielectron expansion coefficients.
CI is able to recover the dynamical correlation of the system (configuration with large weight, many configurations with small weight), whereas (properly selected) CASSCF recovers the static correlation of the system (a few configurations with about the same weight, many configurations with small weight). By having a wave function that have some configurations with about the same weight, means that the wave function has some multi-reference character. I think that it is the MO optimization that leads to this multi-reference capabilities of CASSCF, but I'm not sure I understand why.
If can be shown mathematically, instead of just by concepts, that would be great.

I think you are maybe confusing how dynamical and static correlations are treated with different method. Also CASSCF by itself is not a multi-reference method.
CI in general is able to describe both dynamical and static correlation (FCI does at least). What is treating dynamical correlations (but not static) is the truncation scheme using different degrees of excitation, or example CISD, or in a similar way CCSD. Those are called single-reference methods, because they generate configurations based a single reference wave function.
To deal with static correlations you can for example use the Active Space approach (CASCI), or select configurations by hand (since you might not need to many here).
The term multi-reference refers to doing both, dynamic and static correlation, by generating excitations starting from multiple configurations. One first does a calculation for static correlation (usually MCSCF or CASSCF). Then a second calculation adds dynamical correlations by using the configurations from the first calculation (all or only the most important ones) as multiple reference points (e.g. MRCI-SD or MRCC-SD).
Optimizing the MO coefficients in a MCSCF calculation (e.g. CASSCF) is about optimizing the one-electron basis for your specific CI problem. Since you include more than just one configuration, the HF orbitals are not optimal anymore. In principle you could combine MCSCF with CISD as well, but usually CISD generates to many configurations making such an approach unfeasible.
TL;DR
The optimization of the orbitals in MCSCF does not directly improve the multi-reference capabilities. But it improves the description of static correlation, so you get a better starting point for your multi-reference calculation.
Maybe this explains better the need to optimize both sets of coefficients:

[...] The presence of several important configurations poses a difficult challenge for ab initio electronic-structure theory. The single-configuration Hartree-Fock approximation, by its very construction, is incapable of representing systems dominated by several configurations. By the same argument, methods designed to improve on the Hartree-Fock description by taking into account the effects of dynamical correlation, such as the coupled-cluster and Moller-Plesset methods, are also not suitable for such systems. Furthermore, to carry out a CI calculation 'on top of the Hartree-Fock calculation'is problematic as well since the Hartree-Fock model is inappropriate as an orbital generator: the orbitals generated self-consistently in the field of a single electronic configuration may have little or no relevance to a mutliconfigurational system.
An obvious solution to the multiconfigurational problem is to carry out a CI calculation where the orbitals are variationally optimized simultaneously with the coefficients of the electronic configurations, thereby ensuring that the orbitals employed in the wave function are optimal for the problem at hand and do not introduce a bias (towards a particular configuration) in the calculations. This approach is referred to as the multiconfigurational self-consistent field (MCSCF) method. [...] it may be used either as a wave function in its own right (for a qualitative description of the electronic system) or as an orbital generator for more elaborate treatments of the electronic structure.

Taken from the introduction to the MCSCF chapter in the book T. Helgaker, P. Jorgensen, and J. Olsen. Molecular electronic-structure theory. New York: Wiley, 2000.

To my understanding, static correlation and multi-reference go hand in hand. The more pronounced the multi-reference character of our system becomes  - e.g. breaking a sigma bond - the more important it becomes to use a method capable of giving a "balanced" description of the near-degenerate states. CASSCF mainly recovers the static correlation, but of course a small part of the dynamic correlation is recovered by the fact that the wave function is multi-configurational; we include the dynamic correlation between the active electrons, I suppose.

But a CASSCF calculation is often just qualitatively describing the system; you need to do CASPT2 to get energetically accurate results. Imagine, for some near-degenerate system, making two wave functions made up of the same configurations. Then we optimize only the configuration coefficients in one wave function, but in addition optimize the MO coefficients in the other wave function. My understanding is that the first wave fcn will fail to describe the near-degeneracy properly, and that the second one will succeed (assuming a proper choice of configurations). Is there something wrong here?

Sorry, but your example is not clear to me. "Then we optimize only the configuration coefficients in one wave function, but in addition optimize the MO coefficients in the other wave function." This sentence seems contradicting: First parts says "only in one", second part says "in addition ... in the other". Could you rephrase that? And I assume by "wave function" you mean the electronic state?

Yes, it is the electronic Schrödinger equation we are solving, so the wave function is the electronic wave function. In the truncated CI calculation in my example (i.e. first wave function), we use the MO coefficients from, say, a HF calculation. We therefore don't optimize the MO coefficients for each configuration. The "second" wave function I mentioned referred to a CASSCF calculation, where we minimize the energy with respect to both the MO coefficients and the configuration coefficients.

I added a quote about MCSCF, maybe this does answer your question?

This is a misunderstanding of the term "multi-reference". CASSCF is a multi-reference method as are the MRCI methods. Multi-reference just means that the wavefunction is described using more than one single-particle determinant. HF uses one. MP2 and coupled cluster use one, with some corrections. CASSCF uses more than one, specifically all of the ones generated from placing a set of electrons in all possible occupations of a set of orbitals (hence "complete active space"). MRCI uses (typically) all the ones in CASSCF plus a bunch more. See my answer for more details.

Sorry, but I need to object. "Single-reference" means the method derives configurations (which might be either determinants or configuration state functions) from just one "reference configuration". "Multi-reference" means configurations are derived from multiple configurations. To CAS methods these terms do not really apply, since the configurations are generated in a different way.
Furthermore, coupled cluster uses more than just one configuration! It is basically a different formulation of the CI wave function, which leads to different equations.

HF is a single-determinant method, CISD and CCSD are single-reference methods, which may be extended to their multi-reference variants MR-CI and MR-CC.

CC does not really have more than one configuration because of the way the CC equations are solved. The contributions of excited configs are computed based on the single-determinant ref wf and other integrals, just as in MP2. CAS has many configurations, it is multi-reference. The term MR-CI is because it is doing CI from a multireference wavefunction i.e. the CAS wavefunction.

Just because you don't explicitly write them down, doesn't not mean they are not there. Considering many configurations is still important for thr theory behind CC. I don't see the logic behind calling CAS multi-reference. Also by the same logic you call MR-CI multi-reference, you can call CISD a single configuration method, which would be a very misleading terminology (and is just wrong). Can you provide a reference for your claim?

It's not just that you don't write them down, it's that you don't use them to evaluate the energy, you only approximate their effect on the HF ref, which is a privileged state. Here are David Sherill's notes on multi-ref methods in which he includes CASSCF (http://vergil.chemistry.gatech.edu/research/bondbreaking/multi-ref.html). CASSCF treats all configs on equal footing, there is no special status for the HF state because CAS is not single ref. CISD uses one special ref and so, as you yourself noted, is single ref (though multi-configurational). Can you provide a reference for your claim?

The CC energy depends on the single and double amplitudes (which are coupled to all higher amplitudes). Those amplitudes define the CC wave function, which can be directly related to the CI wave function and therefore configurations (as shown in eq. (13.2.10) in the book of Helgaker, Jorgensen and Olsen).
CISD treats all configs on equal footings as well. The difference to CASCI is just that the configs are selected based on the HF config. But how does this relate to the question why CISD and MRCI are called single/multi-reference?

You stated: "Multi-reference just means that the wavefunction is described using more than one single-particle determinant." But CISD uses more than one single-particle determinant and is single-reference.

CISD does not treat all configs equally. One determinant has all single and double excitations included in wave function, a small number (the singly excited determinants) have all their single excitations and some of their doubles, and a big number (the doubly excited determinants) have some of their singles excitations. The different configurations are not equal; you could look at a CISD determinant list and know which one was the reference.

I agree that my original statement was unclear. Let me clarify it to "Multi-reference just means that the wavefunction is described by using more than one single-particle determinant as the zero-order state." CISD, CC, HF, MP2 all assume that a single, special determinant is a starting point and then improve from there. CAS, MRCI, and MRCC assume several determinants are needed to make a correct zero-order approximation of the wavefunction.

You are mixing something up in the terminology.  In CI a single determinant corresponds to one single configuration! Each single excitation creates one further determinant (doubles etc do as well). Your sentence "one determinant has all single and double excitations included ..." does not make any sense.

Yes, your last comment is what I keep talking about.

I don't think I can be any more clear: In CISD, there exist exactly one determinant $A$ such that the CISD wavefunction determinant list includes all determinant formed from single and double excitations from $A$. Moreover, there are no determinants that are not single or double excitations from $A$. This is true for no other determinant in the CISD wavefunction. Hence $A$ is the special single reference determinant. As you say, the terms "configuration" and "determinant" are interchangeable.

Now that again makes sense, the other stuff did not. Sorry if this seems like nitpicking, but the way you expressed some sentences could give some wrong ideas. Where did I say configuration and determinant are interchangeable? Because strictly speaking they are not (always).

The distinction between static and dynamic correlation is not well-defined.1 The distinction is only sensible with respect to a single-particle picture, i.e. viewing the many-electron wavefunctions as built up from single electrons. Let's start with some notes about configuration interaction (CI). The idea of configuration interaction is to express the wavefunction as a linear combination of Slater determinants (simple wavefunctions with orbitals that are either doubly, singly, or unoccupied by an electron). Generally, most CI methods require that all of the Slater determinants have the same orbitals (although NOCI or non-orthogonal CI expansions are sometimes used) and so these different configurations differ in the occupation of their orbitals. This then gives two sets of parameters to describe the CI wavefunction: CI expansion coefficients and orbitals in the Slater determinant. In methods like CIS, CISD, CASCI, etc. the orbitals are not optimized self-consistently. In so-called MCSCF (multi-configuration self-consistent field) methods, the orbitals are optimized (this is computational relatively expensive and the optimization is much more difficult than with standard SCF methods).
The biggest CI wavefunction one can do (in a given basis set) is to expand the wavefunction in all possible configurations that can be obtained from all possible occupations of all of the orbitals. This is called Full CI and it is exact (in a given basis), that there is no wavefunction lower in energy than the FCI wavefunction. The FCI wavefunction is also invariant to orbital changes; because all possible configurations are considered, changing the orbitals just changes the expansion coefficient weights, but not the total energy (or any density matrices). FCI is too expensive to calculate for almost all molecules and so we make approximations. One such approximation are active-space methods (such as CAS). In CAS, a set of orbitals and electrons are termed "active" and the FCI problem is solved within these active orbitals while the remaining orbitals are solved via mean-field (i.e. HF). These methods account for static correlation, usually described as correlation energy due to low-lying excited states as occurs in bond-breakings and with some transition metal systems. The main hallmark of static correlation is in orbital occupation numbers that differ significantly from 0, 1, or 2. Dynamic correlation is the "other" part of the correlation, which are usually described as correlations due to instantaneous fluctuations of the electrons. MP2 and coupled-cluster are ways of recovering that dynamic correlation for a single reference. Standard multi-configurational methods like CASCI/CASSCF recover static correlation principally, but without having an active-space of all of the electrons, they will not recover all of the dynamic correlation without other tricks. Enter the CASPT2/NEVPT2 method which seek to capture the missing dynamic correlation via perturbation theory on top of the CASSCF wavefunction, or the MR-CI methods which bring in more determinants by considering excitations from some set MR expansion (i.e. a CASSCF procedure).2
On to your question, about what does the orbital optimization step accomplish in CASSCF. The orbitals that are often fed into a CASSCF to define an active-space are based on single-reference orbitals. They represent some mean-field approximation to the wavefunction which may or may not be any good depending on the system. The CAS procedures allows properly fractional electron occupation numbers and so the orbital optimization procedure adjusts the active-space to select the best orbitals in this less approximate wavefunction that has the needed flexibility to correctly describe the system.
Another important note: do not look at the configurations' coefficient weights to determine if the system has a lot of static correlation or not. Changing the orbitals will change the coefficients dramatically. Consider the simplest case, $\ce{H2}$. If the orbitals are the $1s$ on $\ce{H}_a$ and the $1s$ on $\ce{H}_b$, the CAS(2,2) wavefunction will have two configurations of equal weight (where the electrons are on $\ce{H}_a$ or $\ce{H}_b$). If you switch to the orbitals being the $\sigma$ and $\sigma^*$, the CAS(2,2) energy will be identical but the two configurations will now have very unequal weight (it will be almost 100% doubly occupied $\sigma$ and very little of the $\sigma^*$). So you can't just look at the weights.

Ramos-Cordoba, E., Salvador, P., Matito, E. Separation of dynamic and nondynamic correlation. Phys. Chem. Chem. Phys., 2016, 18, 24015-24023. DOI:  10.1039/C6CP03072F
Szalay, P. G., Müller, T., Gidofalvi, G., Lischka, H., Shepard, R. Multiconfiguration Self-Consistent Field and Multireference Configuration Interaction Methods and Applications. Chem. Rev., 2012, 112 (1), 108–181 DOI: 10.1021/cr200137a


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Why is the concentration of pure water is 55.5 mol/L?
Whilst reading about pH, it is given that concentration of water is $\pu{55.5 M}$. Let water be of $\pu{1 mol}$ at $\pu{297 K}$ and $\pu{1 atm}$ pressure. Then the concentration is
$$c =\frac{n}{v} = \frac{P}{RT}\\
  =\frac{1}{8.314\times 297} = 0.004 $$
But it should be $\pu{55.5 mol/L}$. Where I am going wrong?

The ideal gas equation (pV=nRT) isn't going to help you here, you're looking at the concentration of a liquid.

You are calculating the concentration of water vapour (a gas); your question is about liquid water.

First, you want to work out the number of moles in a litre ($\pu{1000 mL}$) of water:
\begin{align}
  n &= \frac{m}{M_r}\\[1ex]
    &= \frac{\pu{1000 g}}{\pu{18 g mol-1}}\\
    &= \pu{55.5 mol}\\
\end{align}
We know that the mass of $\pu{1000 mL}$ of water is $\pu{1000 g}$, and the molecular weight of water is around $\pu{18 g mol-1}$, therefore the calculation gives us an answer of $\pu{55.5 mol}$.
This can then be plugged into the equation for concentration:
\begin{align}
  n &= cV\\[1ex]
  c &= \frac{n}{V}\\
  c &= \frac{\pu{55.5 mol}}{\pu{1 dm3}}\\
    &= \pu{55.5 mol dm-3}\\
\end{align}
Since $\pu{1000 mL}$ of water is precisely $\pu{1 dm3}$ (and the units for concentration are in moles per decimetre), the concentration of water is also $\pu{55.5 mol dm-3}$.


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Why do isomeric complexes often have different colors?
For example, $\ce{[Cr(NH3)2Cl2]+}$
One isomer is green and the other is violet. Why is this so? The ligands are the same, the oxidation state is the same, and so is the coordination number. The only reason I can think of is that if two $\ce{NH3}$ ligands are trans to each other, they repel the 3d orbitals differently than if they were cis. But I don't know why this would occur.
The explanation given in the other question involves anti-bonding orbitals, and I simply have no idea what these are. I'd like an answer which uses highschool chemistry so that I can understand (albeit perhaps in a flawed way ) whats going on.

Sorry for the late followup - if you want an easier answer, please place a bounty on the duplicate question, requesting for an easier answer. When something is marked as a duplicate, that is simply because the question has been asked before; it does not take into account the kind of answer requested. Thanks for understanding.

I can't see the option for adding a bounty..

You seem to have figured it out? :)

Yeah, just read 'on the duplicate question'. Hopefully this is something which is explainable to a highschooler..


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What does "percentage by mass" mean in the context of a compound dissolved in a solution?

Copper can be recovered from low-grade ores by 'leaching' the ore with dilute $\ce{H2SO4}$, which converts the copper compounds in the ore into $\ce{CuSO4(aq)}$. In a $\pu{50 cm^3}$ leach solution, the concentration of copper ions is $\pu{7.8 x 10^-3 mol/dm^3}$. Find the percentage by mass of copper in the leach solution.

Isn't it impossible to find the percentage by mass in this case? We don't know the total mass of the leach solution, and it could be anything. Is the question wrong, or is the definition of percentage by mass different than what I'm using?
Also, how does $\ce{H2SO4}$ react with copper compounds to form $\ce{CuSO4}$? I thought copper couldn't reduce hydrogen ions.

is the definition of percentage by mass different than what I'm using?

Basically, the questioner wants you to calculate mass by volume percentage contrary to your assumption that the question asks for a mass by mass percentage. The existence of a lot of terms referring to the same mass percentage can be a common confusion.
Nevertheless, as a rule of thumb, you'd say, I advise you to go for the calculation of mass by volume percentage unless the questioner explicitly mentions or you've been given enough data.

Find the percentage by mass of copper in the leach solution.

Now that you already have the concentration of $\ce{Cu^2+}$ ions with you, you just need to calculate the number of moles, then the mass and finally divide it with the volume of the mixture.

Also, how does $\ce{H2SO4}$ react with copper compounds to form $\ce{CuSO4}$? I thought copper couldn't reduce hydrogen ions.

Oxidic sources of copper are well known for their amenability to sulfuric acid leaching, for eg. azurite, malachite, tenorite, cuprite, chryosocolla, and brochantite dissolve easily in dilute sulfuric acid. Note that too dilute concentration is refrained from since we don't want hydrous ferric oxides to precipitate (they can adversely effect copper extraction).
And as a side-note, if you remove copper sulfides and iron sulfides, the $\ce{H2SO4}$ requirement would be about five times the weight of the dissolved copper (concentrated $\ce{H2SO4}$ is a good oxidizer).

Interestingly, this question was asked in University of Cambridge international examinations October/November 2010, your question is perhaps a simplification of this:


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DIY Brass Etching methods (safer etchants, waste, and disposal of)
I want to etch brass for making name plates etc. (Also for copper PCBs but brass is probably more complicated given the other metals).
Several etchants commonly used, and I'm sure many more are possible; I'm not fussed what is used. Even vinegar, although something a bit faster would be great.
I want a method easy for someone with minimal chemistry knowledge and equipment to safely perform, and dispose of byproducts etc as household waste, or down a drain.
(Since AFAIK for example copper compounds for example are hazardous to the environment, and etchants can be dangerous/toxic/corrosive), and in the UK for example, getting it disposed of professionally is prohibitively expensive for responsible DIYers on low budgets)
What would be a good choice of etchant? And what would the process be for using and disposing of the waste / bi-products? In general, the less hazardous the better.
PCB resists and techniques would be used, so need an etchant compatible with any of those.

An easy step-up from vinegar would be hydrochloric acid, $\ce{HCl}$. It's known as muriatic acid and is sold in bulk at pool-supply and home-improvement stores. The concentration from such sources is typically ~8.9M, or 32% (by mass). Such is what I measured from home; it's typically advertised as 30%-32%. There are rumors (which I haven't been able to confirm but suspect are true) that $\ce{HCl}$ from these sources often has iron contamination, but that is generally a minor concern for etching processes.
$\ce{HCl}$ has been widely used for etching aluminum hardware and paint-thinning in construction etc. It is highly reactive with (unoxidized) aluminum but can still etch other metals with more effort / more concentrated solution. You probably already know, but it's best practice to clean and polish a surface before etching, to remove impurities like oil and dirt, but also to expose the pure metal by removing protective oxides on the surface. The rougher the better (you can always polish-up later).
Brass is a simple alloy of $\ce{Cu}$ and $\ce{Zn}$. The reaction enthalpy of $\ce{Al + HCl}$ is -1,062.74 kJ, whereas that with $\ce{Zn}$ is only -153.90 kJ. Copper is hardly reactive with $\ce{HCl}$ alone, so it follows that $\ce{HCl}$ is a much stronger etchant for aluminum than for brass, although it is still possible to use it. There are many types of brass -- generally, the more $\ce{Zn}$, the better $\ce{HCl}$ will etch the material:

It is fine to dispose of $\ce{HCl}$ down the drain etc., but only do so after diluting it (carefully, slowly and substantially with water).
You can boost the performance of $\ce{HCl}$ as an enchant by introducing hydrogen peroxide, $\ce{H2O2}$. This combination has very strong oxidative effect and can dissolve metals like copper with ease. If you want to be environmentally friendly, this would be a good stopping point. The most powerful etchant will result from higher concentration of $\ce{H2O2}$. It's typically sold at 3% in stores but food-grade 30% $\ce{H2O2}$ can be bought online or made at home via this method.
I have had great success etching copper (and nickel) using a home-made solution of copper II chloride $\ce{CuCl2}$. Copper (I/II) chloride is very hazardous to aquatic life and (dissolved in solution or not) human life, but with careful handling it's a fine thing to use. It is quite easy to produce and handle, and does not corrode other (more precious) metals, like gold, so you can get selective etching if you have a certain alloy or something like RAM cards from PC parts. (I have isolated gold metal from these using $\ce{CuCl2}$ solution). To make it, simply add copper metal to a solution of concentrated $\ce{HCl}$ and very slowly add concentrated $\ce{H2O2}$. The copper metal will dissolve/oxidize into $\ce{Cu^{+2}}$, and if you dry the resulting blue-green solution (you don't have to) completely, you'll be left with a brown $\ce{CuCl2}$ powder. It's hygroscopic so if you leave it out it will form a blue hydrate. So I keep mine in a tightly closed plastic bottle. To use it as an etchant, just take some of the solid and dissolve it in water (again). The chief downside, as you mentioned, is the toxicity of copper chloride -- I have never thrown mine away. You can re-use it many times.
If I ever had to dispose of it, to do so properly I would use electrolysis. Attaching a DC power supply with the (-) terminal on a piece of copper metal (in dissolved $\ce{CuCl2}$ solution) and the (+) terminal in the solution (or attached to an iron nail) would do the trick, though you'd have to wait some time for the copper in solution to reduce -- and to push equilibrium you'd need a moderate/high voltage. Once you've reduced most or all of the aqueous copper to solid metal, the solution leftover is relatively harmless.

Exactly what I was after - thank you! One question though, are there any fumes / gases to worry about?

HCl fumes can be nasty, and corrosive. If you can, do the etching outside or with good ventilation. I generally wear gloves and a painter breather mask but that's a little thorough. If you get HCl on your skin, just wash with lots of water. Don't neutralize with base.

I have since found (and am still exploring), some resources:

Etching Without Acid: Metal Salt Etching & Electro Etching
Cupric chloride PCB etching
Is the best PCB etchant in every kitchen

My favourite would be the last one, assuming it works!

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Can any other common reagent other than than H3PO4 cause this conversion?

Can any other common reagent other than than $\ce{H3PO4}$ cause this conversion? What about (conc) $\ce{H2SO4}$ or (conc) $\ce{HNO3}$ or (conc) $\ce{HCl}$?

This is an intramolecular Friedel-Crafts acylation.  Other sources call it Haworth reaction.
Concentrated phosphoric acid (in its viscous solution up to $\pu{85 m\%}$) is indicated as it is both strong (to generate the acylium cation), as well as removing the water generated in the course of the reaction (which could trap the acylium cation just generated).  Hydrochloric acid (up to $\pu{37 m\%}$), in comparison, is a strong Bronsted acid, too,  but is lesser hygroscopic than concentrated phosphoric acid.
This does not imply that phosphoric acid will be your solvent of reaction; indeed you may even use polyphosphoric acid (CAS 8017-16-1) -- which is a water free, waxy solid -- to catalyse the reaction.  As alternatives to the above route, I suggest

using strongly acidic ion exchangers (like Amberlyst-15), provided you find a way to remove the water generated (Dean-Stark distillation, trapping in an other reaction).
If you change to anhydrous conditions, a Lewis Acid ($\ce{AlCl3}$ is a classic example of the stronger ones) may be used, too. Along this classical route, of course, you would need to generate the acyl chloride first (typically $\ce{SOCl2}$, or $\ce{PCl5}$), to transform the acid in the acyl chloride. And subsequently, you will perform the Friedel-Crafts acylation.  Perhaps $\ce{AlCl3}$ may serve in a one-pot reaction, too -- yet you would need a bit more.

What do you mean by "catch the water" ? Why can phosphoric acid catch the water but not HCl ?

Concentrated $\ce{H3PO4}$ may be up to 85 m% and is highly hygroscopic.  Concentrated $\ce{HCl}$ is up to 37 m%, lesser hygroscopic than the former (63 m% of water, seen from the other perspective).  In addition, instead of phosphoric acid as solution, you may use polyphosphoric acid (CAS 8017-16-1) which is a water free solid. You need both generation of the acyl cation, and removal of water (because water would react just again on the acyl cation) to drive the reaction forward.  Of course this does need not to imply to run the reaction "just in phosphoric acid" (as sole solvent).

Please edit to add this to your answer so that I can upvote it! Thanks

@blue And till some moments ago, I did not know that instead of the more general "intramolecular Friedel-Crafts reaction" this may be called a "Haworth reaction", too.  The gain here is (at least) for both of us.

On the wiki page they don't show any conditions required for Haworth reaction. Not even H3PO4. Strange! https://en.wikipedia.org/wiki/Friedel%E2%80%93Crafts_reaction#Haworth_reactions

@blue According to Organic Synthesis (http://orgsyn.org/demo.aspx?prep=CV4P0898), once the acyl chloride is formed, even a milder Lewis acid as $\ce{SnCl4}$ (rather than the harsher $\ce{AlCl3}$) may work.  (But $\ce{SnCl4}$ might be not efficient enough to promote opening of the lactone.)  An other related entry, a bit different, although with $\ce{AlCl3}$ is http://orgsyn.org/demo.aspx?prep=CV6P0744. Likely eased because it is an intramolecular reaction, and the aryl does not bear a deactivating substitutent.

Eaton's reagent - $\ce{P2O5}$ in Methanesulfonic Acid - is a good alternative.
JOC 62(1997) 3552
Conc. Nitric acid risks nitrating the aromatic ring. I have never seen conc. Nitric or conc. $\ce{HCl}$ used for this

What about HCl, HNO3....? You answered the question only partly.

Edited to include nitric and HCl


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Will an acyl chloride be attacked by water molecule to form carboxylic acid?

$\ce{H2O}$ molecules can act as a nucleophile. So it should attack the carbonyl carbon. However, I think that that when the $\ce{O-}$ "kicks back" it will once again remove the water molecule keeping the $\ce{Cl}$ intact as water is definitely a better leaving group. But in my book they showed a reaction intermediate, which involved an acyl chloride and they used water molecule as a nucleophile to convert acyl chloride to carboxylic acid. I'm confused as to how that is possible? Isn't water a better leaving group than $\ce{Cl-}$ ?

Don't really remember, but in general proton transfers are very fast. For mechanistic organic chemistry Clayden is a good place to start.

You need to lose a proton, then chloride becomes a better leaving group than hydroxide. In general, proton transfers are incredibly fast; in this scenario there is tons of water around to accept the proton, and the tetrahedral intermediate with a positively charged oxygen is fairly acidic anyway.

Mechanism from Clayden 2nd ed. (p 199).
In the absence of an external base (e.g. pyridine), water will act as the base itself here. Another way of approaching it is to note that hydrolysis of an acyl chloride will release one equivalent of HCl:
$$\ce{RCO\color{red}{Cl} + \color{red}{H}OH -> RCOOH + \color{red}{HCl}}$$
so in the absence of a second base, you will necessarily have to produce $\ce{H3O+}$.


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Carbocation rearrangement with expansion of five-membered ring?
I am confused with this particular problem.

I began the problem with a mechanism as following. In the beginning $\ce{H+} $ protonates the alcohol and leaves as $\ce{H_2O}$. Then everything is about carbocation rearrangement.

But then my professor gave out a different answer saying that ring expansion doesn't happen.

Why doesn't ring expansion take place? Isn't a six-membered ring more stable than a five-membered ring? I am not sure where I am making the mistake.

With your proposed ring expansion you are going from a tertiary cation to a secondary - ain't going to happen

@Waylander. Generally which of the two products are stable?

There's also no real driving force for ring expansion here - many of the common examples involve 3- or 4-membered rings which expand in order to reduce the strain present.

@orthocresol but isn't the 6-membered ring a tad more stable than a 5-membered one?

@PrittBalagopal Just a tad, yes. Not quite the same as 4-ring to 5-ring, though.

@orthocresol  the comparison seems tobe tertiary aliphatic cation with 6 possible hyperconjugative structures and tertiary cation in 6 membred ring  with again 6 hyperconjugative structures .Shouldnt 6 membred ring cation be more stable due to enhanced (tad) stability of 6 membred ring ?

@ChakravarthyKalyan I'm not sure how true that is. It is tertiary to secondary, after all, as has already been pointed out several times. But that is just one half of the issue. The other half is that carbocations do not sit around waiting to see if they get a chance to rearrange. They get intercepted by nucleophiles pretty quickly, and if they want to rearrange before that happens, it has got to be pretty fast. If you don't have a great incentive to rearrange, it doesn't happen very fast.

Your proposed mechanism involves a ring expansion from cyclopentane to cyclohexane. This gives very little driving force in terms of relieving strain in the system. Cyclopentane, in a twisted or "envelope" conformation, has some angle strain ($102-106^o$ vs $109.5^o$ $^{ref~1}$) and it does also has some torsional strain down some of its bonds.

This is mostly alleviated by the move to cyclohexane, if it is in a full chair conformation, but the stabilization of the system is minimal ($25-29kJmol^{-1} $ $^{ref~2}$). Here is a link to a question that deals with ring strain and the calculation of it from $\Delta H$ of combustion.
Along with this, your mechanism proposes this ring opening involving an alkyl shift that moves the carbocation from tertiary to secondary. This involves a destabilizing of the cation due to the reduced hyperconjugation (and inductive effect). Studies have shown that a secondary carbocation is around $67-75kJmol^{-1} $less stable than a tertiary, for simple substituents (and in the gas phase$^{ref~3}$).
In total, the ring opening from a five to six-membered ring in your mechanism does not result in an overall increase in stability. Hence why the second mechanism is suggested to happen instead.

There seems to be something wrong with your ref 2 hyperlink. I can't seem to reach the page it links to.

Sorry. I have fixed the link. It is a secondary source: https://www.masterorganicchemistry.com/2014/04/18/ring-strain-in-cyclopentane-and-cyclohexane/ I will try to get more primary data.

I also edited the post. It was meant to read chair conformation, not boat, for cyclohexane.

Ref 2 updated to umn.edu notes. The data for the overall ring strain of cyclopentane comes from heats of combustion relative to cyclohexane


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What happens to gunpowder when it gets wet?
What happens to the gunpowder and other explosives when they get wet? I know that they stop "working" (won't explode anymore) but for what reason? Is it anyhow similar to the fact, that wet wood is burning worse than the dry one?

In general, putting explosives in water will not prevent them from exploding, because most explosives already contain all of the necessary reactants to produce an explosive chemical reaction.  They do not need to draw oxygen from the air, and the reaction happens much too fast for the water to act as a heat sink.
Traditional "black powder" is an exception; it doesn't explode when wet.  I don't know exactly why this is, but two plausible reasons are:

Unlike most other explosives, black powder is shock-insensitive; it has to be initiated with a flame or spark.  Water can easily soak up all the heat energy from a small flame, like the one you get from a percussion cap.  This is the same as how it's harder to set fire to a living tree than a dry log, but if the fire gets hot enough it will still catch.
Black powder is a "low" explosive, meaning that the chemical reaction happens relatively slowly (for an explosion), which gives the water more time to absorb heat energy and quench the reaction.

Worse, black powder can be permanently ruined by water.  This is because it's a mixture of saltpeter (potassium nitrate, $\ce{KNO3}$), sulfur, and charcoal, and saltpeter is water-soluble.  If you get it wet enough, the saltpeter can be completely washed out of the mixture.  This doesn't always happen; the 19th century industrial manufacturing process for black powder involved deliberately dampening it, both to make it less dangerous during manufacture, and to make saltpeter crystals form inside each particle of charcoal.  But they were careful to use only just enough water.

I suspect an important difference between the water-insensitive explosives and black powder is that it's a low explosive. I.e. the water has time to cool portions of the mixture, before they are themselves ignited by a supersonic shock front. In a high explosive, I'd figure this isn't possible, because in order for the water to take away latent heat it needs to vaporize, thereby expanding. But while the detonation shock front is passing through, the high pressure inhibits this.

@leftaroundabout It may also be relevant that black powder is shock-insensitive, needing a flame to initiate an explosion; water can plausibly absorb the heat energy of a small flame (e.g. from a percussion cap) before the carbon-KNO3 reaction has a chance to get started.

There's some problems with this question.

gunpowder and other explosives when they get wet? ... (won't explode anymore)

"Gunpowder and other explosives" encompasses thousands of substances, all with different properties. And: how wet? And what exactly is meant by "won't explode" -- there's several different types of explosions.
Moreover, OP's premise is wrong. It is not true that explosives, as a general rule become non-explosive when wetted. Most military explosives and many construction/mining explosives are completely waterproof (although their initiation systems may not be.) The product family of commercial emulsion explosives actually need water to be explosive. When completely dried they burn instead of exploding.
The most common explosive in the world -- ANFO -- is rendered (practically) non-explosive by dampening it, but not for the reason you suspect. In this material, the oxidizer (ammonium nitrate) is easily wetted, but the fuel (oil) is repelled by water. Hence when it is dampened the components separate. It is still theoretically possible to detonate it in this state, but it requires an abnormally powerful initiating charge (booster), explodes with greatly reduced power, and produces a lot of toxic fumes. Since the shot-firer probably didn't expect to need an abnormally powerful booster, it will most likely fail to explode.
Note that this occurs strictly because the fuel is repelled by water. It is possible to make a very similar explosive using a hydrophilic fuel. This composition will be able to tolerate much higher levels of dampness. However if enough water is present, the ammonium nitrate will dissolve completely and wash away. To prevent this, a gelling agent may also be added.
The most interesting case is perhaps smokeless powder (gunpowder for modern firearms.) This is a precision application of explosives: the powder needs to burn at a very precisely controlled rate, both to control the pressure that develops inside the firearm, and to give consistent muzzle velocity so you can shoot accurately and hit your target. Allowing small amounts of moisture to contact the powder will alter the burn rate by an unpredictable amount, rendering the powder unfit to use in a firearm, but by no means non-explosive.
This is widely assumed to be due to the cooling effect. That is, each burning granule rises in temperature slower because of its water content, which causes its neighbor to heat slower, and take longer to ignite. However precise measurements show the burning rate change to be larger than can be explained this way, so there must be more to it. However the exact mechanism doesn't seem to be known.
Unlike in ANFO, or even black powder, the water doesn't do any harm to smokeless powder. In powder factories, it is routine to process the powder wet, for safety reasons, and dry it only as one of the last steps before packaging. For several reasons it is infeasible to do this anywhere but in a powder factory.
Although burn rate is affected by even fractions of a percent of water, smokeless powder doesn't become non-explosive until a lot of water is added. Under transport safety regulations, NC based explosives change from "1.3 explosive" to "4.1 flammable solid" at 25% or greater moisture. However, even at 25% water, it is far from safe. It will still burn fiercely, but the burn rate is controlled by the rate of evaporation of water from the surface of the mass. Under some circumstances, this can still be ferocious. In magazines where I have worked, the point at which scrap powder no longer counts against the NEQ limit (Licensed Nett Explosive Quantity) is when it is Under Water: that is, totally soaked in water, and submerged, with at least two inches of standing water above the surface and some sort of lid to prevent evaporation. Even in this state, it wasn't considered to be unable to explode: just unable to be ignited.
Which is why I earlier asked what exactly is meant by "won't explode"? Nitrocellulose explosives are nearly always used as low explosives (propellants), which react by extremely fast burning. However, if suitably compacted (to give good propagation to the shock wave) and given a powerful booster to start it, high-nitrogen nitrocellulose can also be detonated as a high explosive. This reaction can occur even if soaking wet and fully submerged.

For combustion to take place, there needs to be an oxidizer, fuel, and heat. When a substance gets wet, it is surrounded with water and is not in contact with atmospheric oxygen, arguably the most common oxidizer. In the absence of any other oxidizers, combustion stops. Underwater fuses incorporate strong oxidizers like perchlorate salts so the reactions can continue in the absence of atmospheric oxygen. Water is also an incredible heat sink and will steal heat from the reaction and dampen it in that way.

The question was specifically about explosives (which always come with an anærobic oxidiser), wasn't it?


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Why are thioethers not better leaving groups than ethers?

Why are thioethers ($\ce{RSR}$) not better leaving groups than ethers ($\ce{ROR}$)? I learnt that better bases are poorer leaving groups. Here, clearly   the poorer base is thioether, as the lone pair is on a larger atom.

Then you should unlearn it. Being a leaving group is only tangentially related to basicity, as you can see from this table.

What this chart doesn't cover (and would help you with your understanding) is what is meant by "leaving group ability?" Is it kinetics? Is it thermodynamics? You can't rationalize if you don't know what you're trying to rationalize.

@Zhe It's from Wikipedia....

@Mithoron It would be better if you explain why....

@blue Let's assume we are talking about thermodynamics. Then look at it by considering the reverse reaction. Is it easier to S-alkylate RSR or to O-alkylate ROR? Kinetics would be a little trickier to rationalise, as steric hindrance and other effects come into play.

Wikipedia is not a source. Go to the reference they are citing and figure out what it's actually measuring.


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Is a SR2+ substituent really more electron-withdrawing than NH3+?
As the electronegativity of nitrogen and sulfur are around 3.04 and 2.58, respectively (according to the Pauling scale). So according to that, $\ce{NH3+}$ should pull the electrons more towards it than what $\ce{SR2+}$ should. This argument has been put forth by Jan in his answer to a similar question here.
However, my textbook (Organic Chemistry by Subrata Sengupta) states that the exact opposite is true. Who is correct here?

I'm sorry, I'm unable to find sufficient data so as to draw meaningful conclusions at the moment. I'm therefore deleting my answer, if I happen to stumble upon resources that may interest you, I'll add them at a later time lest someone else hasn't provided an answer by then.

Mods decided this wasn't a dupe of the linked question. If anybody would like to discuss it, please make a post on [meta]. // I have reopened this post, which was your original question - in the future please don't repost, instead edit your question, or if you wish, post on meta. Thanks for understanding.

Thanks for your concern @orthocresol.  I will keep these thing in my mind from next time.


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Units on rate constants from transition state theory and DFT
I am using density functional theory (DFT) to compute the energy profile and, by extension, kinetic parameters for a catalytic process. I know that the rate constant from transition state theory (TST) can be found via the Eyring equation:
$$k=\frac{k_{B}T}{h} e^{-(\Delta G^\circ)^\ddagger/RT}$$
I believe that the derivation of this equation from TST implies that it strictly holds for reactions of the form $A\leftrightarrow [AB]^\ddagger\rightarrow P$. It's clear by looking at the units that other molecularities won't work with this equation directly (for example, $A \leftrightarrow A^\ddagger \rightarrow P$).
My concern arises because I am using DFT (specifically Gaussian) to compute the thermochemistry (as discussed here). I am able to generate a free energy landscape for my proposed mechanism, but I am concerned about the accuracy of using the Eyring equation for various molecularities.
It's clear to me how the units work out when looking at this from the perspective of partition functions. From TST, we can say that
$$k=\frac{k_{B}T}{h} \frac{Q^\ddagger}{\prod_{i}Q_{i}} e^{-E/RT}$$
for $i$ reactants, where $E$ is the energy computed from DFT, and
$$Q\equiv \frac{q_{t}}{V}q_{r}q_{v}q_{e}$$
such that the translational partition function is effectively written in units of inverse volume. So, changing the number of reactants $i$ thereby changes the units of $k$ appropriately.
Where is the disconnect between this expression for $k$ and the Eyring equation I originally show? Presumably I can use the above expression for $k$ since the partition functions are used by Gaussian to compute the Gibbs free energies, but it would be much easier to use the Eyring equation directly, especially if I am making a Gibbs free energy profile anyway.
I'm familiar with the related question posted on StackExchange in the past. However, it does not address how to actually approach various molecularities.

The first equation is only applicable to first order reactions $\ce{A \rightarrow B}$ since the rate constant has units $\pu{sec^{-1}}$ , (although it is widely quoted as otherwise).  Using the ratio of partition functions Q is the correct way to go for other reactions.


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Why can the anion NO3- behave as a L-type ligand while BF3 can not?
The fluorine has a lone pair, and the oxygen has a lone pair. Both are donate-able, so they should both act as ligands. I know fluorine is more electronegative, but oxygen is quite electronegative itself, so the reason must lie somewhere else.

Can (...) not behave? Could you correct your grammar to clarify?

Also they both can coordinate.

How in the world could $\ce{BF3}$ act as a ligand? Its the opposite of a Lewis base, a Lewis acid for that matter.

@Mithoron The definition of a ligand given to us is 'a species which tends to form dative bonds with transition metal ions', which is an L-type ligand ( I literally just read this off wikipedia, so not sure ). Can they both act as L-type ligands? The official answer disagrees.

Not bad searching @PrittBalagopal If BF3 coordinates to metal center accepting electrons it is also a ligand but "Z-type".

What’s an L-type ligand? Please edit your question to include that.


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What is litre atm?
"The slope of a given straight line graph with constant temperature is found to be 0.2 l atm at 5 atmospheric pressure." What is l atm here? It is the unit of what?

$\pu{L atm}$ is a unit of (pressure)(volume). As $\pu{atm}$ is a multiple of $\pu{N//m^2}$ and $\pu{L}$ is a multiple of $\pu{m^3}$, then $\pu{L atm}$ is a measure of joules, the unit of energy, which is $\pu{kg m^2//s^2}$ in base SI units.

you might also want to recall the gas law $pV=RT$ and the work done  (from thermodynamics) as $p\Delta V$; both have units of energy.

$\pu{L atm}$ is a unit of pressure ($P$) times volume ($V$). As $\pu{atm}$ is the multiple of $\pu{N m-2}$. $\pu L$ is the multiple of $\pu{m3}$, then $\pu{L atm}$ is a multiple of Joules, the measure of energy, which is $\pu{kg m2/s2}$ in base SI units.


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What makes "Color-Safe Bleach" color safe?
The ability of bleach ($\ce{NaOCl}$) to change the color of garments and introduce "bleach stains" is a well-known issue. In this question:
Does bleach turn a substance white or colorless?, the accepted answer says the mechanism is breaking apart long pigment compounds.
As per Clorox's website, Clorox2 color-safe bleach uses hydrogen peroxide ($\ce{H2O2}$) and:

"[a]t the concentration used in the laundry, hydrogen peroxide is a much less aggressive bleach than sodium hypochlorite."

I'm trying to parse this ad-copy statement into GenChem vocabulary. Is the idea that while $\ce{H2O2}$ is also a powerful oxidizing agent, it isn't strong enough to disrupt the pigment molecules in dyes (like $\ce{NaOCl}$), but still powerful enough for general cleaning / stain removal? Or is it simply a result of having a diluted product?

afaik $\ce{H2O2}$ is not that powerful an oxidizing agent. It can sometimes act as an oxidizer and sometimes even as a reducing agent. I call it double minded. $\ce{NaOCl}$ on the other hand, knows that it'll act like an oxidizing agent.

@BerryHolmes good point, I suppose we aren't filling the washing machine with piranha solution lol


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Why is the melting point of a substance is the same as its freezing point?
Melting point = Freezing Point
Why is the melting point of a substance is the same as its freezing point?

Related: https://chemistry.stackexchange.com/questions/53348/difference-between-freezing-point-and-melting-point

We have a block of 10 g of ice at -5°C. Its melting temperature is ~0°C. If we let it warm up slowly, when it reaches 0°C some ice will melt, forming some (liquid) water. Thermodynamics shows that the temperature must stays constant as long as there are both solid and liquid. So at a given time you could have 5 g ice and 5 g water, necessarily at 0°C. Now you put this in the freezer (say at -5°C). What happens? The ice stops melting, and instead some water starts freezing and turning back into ice. But you still have both solid and liquid, so you must still be at 0°C.

Consider a system that contains only a single substance. At the temperature where the solid and liquid phase of the substance are in thermodynamic equilibrium you cannot tell if the system initially consisted only of solid or only of liquid or of a mixture of both.
Thermodynamic equilibrium means that the substance has the same temperature all over the system.
If a small amount of heat is added to the system, some of the substance melts. So the temperature is the melting point of the substance.
If a small amount of heat is removed from the system, some of the substance freezes. So the temperature is the freezing point of the substance.
This shows that melting point and freezing point are the same.


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Solubility and equilibrium
Is equilibrium between dissolution and recrystallization of a salt in an aqueous solution reached only when the solution is saturated; or can this also be established in an unsaturated solution?
Will Le Chatelier's principle apply in this situation (as it is about chemical equilibria)?
I was actually studying purification of salt by common ion effect where salt crystallizes from saturated aqueous solution on addition of common ion. Will the salt still crystallize if solvent still can dissolve more of it in unsaturated solution on addition of little amount of common ion. Since this crystallization is consequence of Le Chateliers principle, will crystallization still occur if solution taken was unsaturated and the common ion added is also in very small amount?

If the solid salt is present together with the solution of the salt, the solution is saturated and in equilibrium with the solid (provided you have allowed time for the solid to dissolve etc.)

Technically, in an unsaturated solution, you are also at equilibrium.

Le Chatelier's principle will apply

The solubility of a solute in a solvent is determined by the chemical potentials of the solute for the solution and the pure solute. Substances move from high chemical potential to low chemical potential. At the concentration of solubility, these two chemical potentials are equal so there is no net transfer. Below this concentration, the chemical potential of the solute in the solution is lower than that of the pure solute so molecules will transfer from the pure solute to solvent until it is all dissolved or saturation is reached.
When you invoke the common ion effect, you are increasing the chemical potential of the common ion in solution which decreases the solubility. It is possible to have a nearly saturated solution and add enough common ions to reduce the solubility below the original concentration.

Beginning level Thermodynamics is about path independent processes near (or at) equilibrium. Almost no thermodynamic systems are left undisturbed long enough for true equilibrium to occur, so we speak about "near equilibrium" states of isolated, closed, or open systems. What exactly we mean by "near" is not obvious or easy. A solution in equilibrium is one in which the concentration of solute is constant (and stable). Visual observation of a solution to determine if it is stable is not optimum, although it is convenient. The problems with this approach are, I think, beyond the scope of your question. (They involve the actual molecular and atomic scale processes which occur during crystallization/dissolution.) So, to answer your first question: Equilibrium is reached when the concentration is constant. If sufficient solute is present to saturate the system, then it is saturated equilibrium, if less is present, then unsaturated equilibrium. The answer to your second question is a bit more complicated. If I take a saturated salt solution and add more solid salt to it, does the concentration in the solution change? No. So Le Chateliers fails here. You need to be aware of the difference between quantitative chemical relationships (for instance equilibrium constants) and qualitative ones (eg. Le Chats.Principle). I shouldn't have to tell you which one is more trustworthy. (OTOH, quantitative relationships are all specific, while qualitative ones can often be applied without knowing all of the details - they're more general because they're less specific. But along with generality comes exceptions to these "rules"....) Does Le Chatelier's Principle apply here? I suppose it could but I don't see any useful application of it. I guess the common ion effect could be looked at as an example of Le Chat's, but it seems to me easier to explain by looking at the equation for Ksp (quant wins over qual). As to your third question: It isn't coherent. "Will something crystallize out if it isn't at saturation?" is what you're asking. By definition, no. Can you reach saturation by adding a common ion rather that the solute of interest? Yes, as you suggest, this will be (generally) in solutions near saturation (and recall, saturation implies equilibrium.) But keep in mind that adding a common ion is actually adding part of the solute of interest - the [Anion] or [Cation] is increasing, and so the ratio of reactants to products also changes, moving closer to saturation.


text:
Use of significant figures in addition and subtraction
My book mentions:

The proper use of significant figures in addition and subtraction involves a comparison of only the absolute uncertainties of numbers. This means that only as many digits are retained to the right of the decimal point in the answer as the number with the fewest digits to right of the decimal

But I don't understand that why must the answer be rounded off after addition? Is the answer not accurate enough to be retained as it is ?

No. The answer isn't accurate enough. Say, for example I wish to add two numbers: $1.23$ and $2.367$.
I proceed like this, but I've no idea what the third digit after the decimal in $1.23$  is and so I represent it as $?$.
$$\begin{align*} 1 &.23? \\ + 2 &.367 \\ \hline 3 &.59? \\ \hline  \end{align*}$$
This is why the end result is rounded off with the number of digits equal to the number having the fewest number of digits you initially set off with.

But in addition we add the extra digit to the answer like 1kg is added to 1kg and 2g we get 2kg and 1g

@AksaK Ah, in that case it was because the weight "1kg" was accurately known.

Let one rod is measured 1.23m and another rod is measured 2.367m. We combine them into a single rod of the length which is combination of both i.e 3.597m so why do we need to round off this number which gives the accurate length?

As @Barry Holmes has explained, in the 1.23 m rod we do not know if it is 1.231 m, 1.239 m or 1.235 m etc. but we do know the longer one is 2.367 m and so we have to do something about our ignorance of the true length of the shorter rod, and this is rounding.


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Reaction of 3-Chloropyridine
I've given a question , 3-chloropyridine + NaNH2  gives out A . Compound A on further reaction with furan gives compound B.
What I think now , First step is simple elimination addition reaction .
So A should be 3-Aminopyridine . Next step is little bit confusing to me.
I think product B should be nucleophilic substituted at 2-position of Furan by -NH2 group of 2-amniopyridine .

Think you've got this wrong.  I doubt sodamide displaces chloride from the 3 position (2 or 4 much more possible). I think it far more likely it is acting as a strong base.

https://en.wikipedia.org/wiki/Aryne


text:
Hoffmann's rearrangement
In Hoffmann's rearrangement of acid amide...the OH from the basic medium attacked the H ..now my question is...isn't the carbonyl carbon more acidic as it's surrounded by 2 electronegative elements?

I suppose here $\ce{OH-} $ acts as a base and not as nucleophile.

why not a nucleophile then?

$\ce{OH-}$ cannot act as a nucleophile here, since in that case, leaving ground would be $\ce{NH2-}$, which is stronger than hydroxide.

The arrows should be equilibria ones and yes, the hydroxide will probably act as a nucleophile but it's a product dead-end, shuttling around in equilibrium

You are right, the OH- ion can indeed act as a nucleoplhile attacking the unsaturated carbonyl group giving rise to a hemi acetal type of complex (with an amino group instead of an OH group). This is similarly unstable and tends to form back the initial carbonyl compound by the elimination of the OH- group. The NH2- cannot be the leaving group here because it is a bad leaving group as compared to the OH- group. Hence the OH- prefers to attack as a base instead and remove the acidic hydrogem from the amide.


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Reaction of substituted furan with methyl vinyl ketone with subsequent rearomatisation

I am supposed to provide a mechanism for the above transformation. I observed that the order of substituents on the product is same as that on furan, and since furan undergoes Diels–Alder reactions, I came up with the following mechanism:

Is my Diels–Alder product correct? After this, however, I have no idea how aromaticity is restored.

Your Diels–Alder adduct is pretty much correct. Without any further evidence we cannot definitely say whether the exo or the endo adduct is formed (you have drawn the exo adduct, which generally arises via thermodynamic control). However, the stereochemistry of this adduct does not affect the final product, so it does not matter which one is formed.
By the way, you also should think about the regioselectivity of the Diels-Alder reaction, i.e. why do the $\ce{NH2}$ and $\ce{COMe}$ end up ortho to each other and not meta. You need to have a better reason than simply "that's the way it was drawn in the product". Generally, this is rationalised by looking for the largest coefficients in the HOMO of the diene and the LUMO of the dienophile. The HOMO of 2-aminofuran is likely to have a large coefficient on C–5 (opposite the amino group), whereas the LUMO of methyl vinyl ketone has a large coefficient β to the carbonyl group.
The subsequent steps are not anything fancy. Generally, $\mathrm{sp^3}$ carbons that are bonded to two heteroatoms tend to collapse (e.g. tetrahedral intermediates in nucleophilic acyl substitution; hydrates; etc.) In this case the nitrogen kicks out the oxygen. Note that the other option of oxygen kicking out nitrogen is not possible, as that would lead to a bridgehead double bond.

Ordinarily you need some sort of acid or base catalyst, but presumably here, the thermodynamic driving force for the collapse is large (relief of ring strain + product is aromatic), and the high temperature provides the energy required to overcome the activation barrier. The last elimination step probably occurs via an E1 mechanism, since there's no good base present.

On the Diels-Alder reaction of furans, endo/exo addition. https://chemistry.stackexchange.com/questions/12417/why-is-the-exo-product-formed-in-the-diels-alder-reaction-between-furan-and-male

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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How can you make magnesium carbonate?
I am trying to start my own gym company that sells gym chalk however I have not been able to find a good process for making good magnesium carbonate. I have tried mixing equal amounts of magnesium sulfate and sodium bicarbonate into water and pouring it through filter paper but have not ended with a good batch.

Buying is usually better than producing when it comes to chemicals. Have you looked for a supplier?

Strongly related: https://chemistry.stackexchange.com/questions/74370/will-this-reaction-form-magnesium-carbonate/74375

The best way to make magnesium carbonate is to opt the laboratory route (mix magnesium chloride and sodium bicarbonate) as described in the Wikipedia article of magnesium carbonate.
$$\ce{MgCl2(aq) + 2NaHCO3(aq) → MgCO3(s) + 2NaCl(aq) + H2O(l) + CO2(g)}$$
The advantage of this process is that the byproduct do not interfere with the main product (sodium chloride is soluble in water, carbon dioxide bubbles out) leaving behind magnesium carbonate as precipitate.
Alternatively, magnesium carbonate can be prepared from magnesium carbonate by bubbling carbon dioxide in it. The advantage of this process is that there are no byproducts.

Note: If you use sodium carbonate, you will get a  precipitate of basic magnesium carbonate—a hydrated complex of magnesium carbonate and magnesium hydroxide—rather than magnesium carbonate itself is formed.
$$\ce{5MgCl2(aq) + 5Na2CO3(aq) + 5H2O(l) → Mg(OH)2·3MgCO3·3H2O(s) + Mg(HCO3)2(aq) + 10NaCl(aq)}$$
You can see that a basic salt is formed and also magnesium bicarbonate is formed which add to the difficulty of separating them and decomposing them to get magnesium carbonate and thus this process is not feasible.

Bubble carbon dioxide into a milk of magnesia suspension, maybe?

@Oscar that is also possible.


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Why does diesel have a higher energy density than gasoline?
One of the primary reasons why hydrocarbons produce energy is the production of water during the combustion process.
I've been trying to find out why diesel is considered to have a higher energy density compared to gasoline and I can think of a couple of reasons:

Higher molar mass (more carbon/hydrogen to burn)
Easier to burn diesel since it is heavier (weaker bonds)

I think I'm on the right path but I don't have a clear picture of the situation.

If your question were about Diesel and Otto engines, from the perspective of physics I would like to add to your picture drawn that the compression/decompression ratio in a Diesel engine is much higher than a Otto engine; as the mix auto-ignites, there are no spark plugs necessary (glowing plugs are a different story).  In terms of the Carnot cycle, this is one contribution why a Diesel engine may be more efficient than an Otto engine in terms of converting chemical energy into mechanical energy.

related https://chemistry.stackexchange.com/questions/69425/how-does-biodiesel-have-so-much-energy-density

Diesel and gasoline have roughly the same energy per unit mass (lower heating value), about $41~\pu{MJ/kg}$).
The density of diesel is about $833~\pu{kg/m^3}$ compared to $740~\pu{kg/m^3}$ for gasoline. This gives diesel about 13% higher energy density per volume.

The actual structure of the hydrocarbons themselves don't contribute to the energy per unit mass, but their willingness to undergo combustion?

I don't know what you mean by the term "willingness to undergo combustion". But the structure of the hydrocarbons has influence on their heating values. E.g. methane > ethane > propane > butane > pentane (https://en.wikipedia.org/wiki/Heat_of_combustion#Heat_of_combustion_tables)


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Interpretation of 4-benzoyl butyric acid H-NMR Spectrum
I am trying to undertake a first order analysis of the spectrum below:

My issue is that every signal appears to have the same coupling constant (I am aware that the value should be reported in Hz, the spectrum just has some of my rough working on it). This does not make sense to me, as I understand that would imply each hydrogen is equally coupled to each other.  I also am at a loss of what type of multiplicity the signal multiplet at 7.5ppm has.
Any help is really appreciated

With this sort of resolution on a printed spectrum, you're not going to be able to distinguish between similar coupling constants. Even if you had a better spectrum, the coupling constants may just be coincidentally very similar to one another (and that is very likely to be the case in this molecule - that 1:4:6:4:1 quintet doesn't arise from coupling to four equivalent protons - it just happens that the two relevant coupling constants are nearly the same). That aromatic multiplet is probably best called a multiplet and left at that.

You should not assume that just because the coupling constants are the same that this means all the nuclei are coupled. It simply means that two sets of coupled nuclei have the same coupling constants. This is not unexpected when you consider that the nuclei are very similar - they are freely rotating alkyl chain methylene groups. And you will find that freely rotating alkyl chain methylene groups have coupling constants of around 7 Hz. This is because the 3J-coupled nuclei (called vicinal coupling) rotate through a range of conformers that have a range of coupling constant values (described by the Karplus equation) to give an average value. Different types of functional group give characteristic coupling constants. Trans-vinylic coupled protons have a coupling of around 15Hz, for instance. So, given the functionality of these alkyl methylenes, you would expect similar coupling constants.
To work out which nuclei are in fact coupled, you would use other information. A set of homonuclear decoupling experiments, or a COSY experiment would give this information clearly, although there is some very useful information in this spectrum. Your signals at 2ppm, 2.5ppm and 3ppm all show significant 'tenting'; they lean towards their coupled partner. That is, you do not get a true Pascal's triangle intensity, with the peaks closer to their coupled partner having greater intensity, as shown below:

The signals at 2.5 and 3 both lean towards the right, indicating they are coupled to something upfield (the signal at 2ppm). Similarly, the peak at 2ppm is leaning to the left, indicating it is coupled to the peaks at 2.5 and 3ppm. You could use your integrals of 2:2:2 to rationalise a -CH2-CH2-CH2- chain. And this would explain the expected splittings you see of a triplet (3ppm), triplet (2.5ppm) and triplet of triplets (2ppm). Of course, if you draw out your splitting diagram for a triplet of triplets with identical coupling constants, it will give you a pentet with intensity 1:4:6:4:1.
To answer your second question, you should try to predict what nuclei these signals represent. I have drawn out some useful props below to assist in the interpretation.

The signals between 7 and 8ppm come from your mono-substituted benzene ring, so you have three distinct environments: ortho, meta and para to the substituted position. They should integrate 2:2:1. The coupling constants for protons around an aromatic ring aren't important for this exercise, other than to say you will only see splitting due to a proton on an adjacent carbon; that is, 3J coupling. So, you have 5 aromatic nuclei in 3 environments, that integrate 2:2:1, and I can see 2 sets of signals that integrate 2:3. Clearly you have two signals overlapping at 7.5ppm. Which signals?
Let's have a think about the splitting we would expect to see from this ring. The ortho protons would be a doublet, as they are coupled to one other 3J proton only.The meta protons would be a doublet of doublets, as they are coupled to the ortho (doublet) and para (doublet) protons. And the para proton would appear as a triplet (coupled to the two meta protons).
What else can we see? The substituent on the benzene ring is a -(C=O)-R. This is a strongly electron withdrawing group, and so if you draw out your resonance structures for this, you will find that the ortho protons are deshielded compared to the meta protons, and so will be downfield of them. The para protonalso  experiences a bit of shielding, and so will also be downfield of the meta protons. So, you can assign the peak at 8ppm as your ortho proton; furthest downfield fits with your electronics adn doublet fits with your expected splitting. The overlapping peaks at 7.5 are the para proton (the left hand side of the forest); this fits with your electronics and (if you were really keen) you could assign some of these peaks to a triplet. The remaining right hand side of the peaks are your meta signals, and again you could assign this asa doublet of doublets. Of course some of the lines for the para and meta signals are overlapping (actually, possibly just one). You could resolve these signals on a higher field instrument quite easily.


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Balancing simple Redox equation
How would you balance these two redox equations? Would you add a '2' in front of Pt2+ and a '2' in front of H+ ?

Oxidation is loss of electrons and reduction is gain of electrons,so in these equations platinum and cadmium lost 2 electrons(oxidation) which Hydrogen ions used to form Hydrogen gas(reduction).
The redox is as follows
$$\ce{Pt -> Pt^{2+} + 2e-}$$
$$\ce{2H+ +2e- -> H2}$$
So the net equation is
$$\ce{Pt +2H+ -> Pt^{2+} + H2}$$
You can check that out
http://www.chemguide.co.uk/inorganic/redox/equations.html


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What is this Boron Compound called?
What would this aromatic boron compound be called?

Label the atoms.

Is this 1H-Borepin

@blue I used CPK coloring code for atoms. See https://en.wikipedia.org/wiki/CPK_coloring

@PrittBalagopal You should mention that in your question to avoid confusion. I wasn't sure what colouring scheme you were using.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), heteromonocyclic compounds with up to and including ten ring members are named by the extended Hantzsch–Widman system. According to P-68.1.1.3.2, this also applies to heterocyclic parent hydrides containing Group 13 (boron group) atoms.
The Hantzsch-Widman system stem for an unsatured 7-membered ring is “epine”. Therefore, the preferred name for the compound that is given in the question is 1H-borepine.
The name 1H-1-boracyclohepta-2,4,6-triene is not recommended.

Unsaturated or fully conjugated 7-member ring?  Do we use epine for the cycloalkene formulated C7H12?

@OscarLanzi H-W system is for heterocycles, not alkenes.

Then presumably the words "fully conjugated" should be used instead of "unsaturated".

If the black coloured are carbon atoms and the white coloured are hydrogen atoms, and the pink one is Boron. Then it should be 1H-Borepine.

[Source: Chemspider]
See here for why it is aromatic.

There is a standard color code for ball and stick models.  True, carbon = black, hydrogen = white, boron = that pretty pink color. Some others :  oxygen = red, nitrogen = blue, chlorine = yellowish green like the gas.

@Mesentery Thanks for your answer. There is in fact a color code for atoms. See https://en.wikipedia.org/wiki/CPK_coloring


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What is the pKa of iodous acid?
Searching the web readily gives the $\mathrm{p}K_\mathrm{a}$ of hydroiodic acid, hypoiodous acid, and iodic acid, however, extensive searching did not result in any value of the $\mathrm{p}K_\mathrm{a}$ of iodous acid. So I am wondering if this value has been documented before, and if it hasn't, then what progress has been made to determine its $\mathrm{p}K_\mathrm{a}$?

That's because iodous acid is less stable than the others you mention.  Sorry I could not provide a more direct answer, but iodous acid is barely even known.  What is known, now, is what you are up against.  See https://en.m.wikipedia.org/wiki/Iodous_acid

There is a question posted on the Q&A site CheggStudy that reads: "Determine the pH of a 50 mL solution of a 0.15 M iodous acid (HIO2). The Ka of iodous acid is 2.9×10−5. ". Although the answer to the question is behind a paywall, if we trust the question itself then the pKa of iodous acid is 4.5. This is a somewhat dubious source though.

According to Schmitz [1], $\mathrm{p}K_\mathrm{a}(\ce{HIO2})\approx 6$ at $\pu{25 °C}$ (numerical references for the bibliographical entries have been updated):

Stanisavljev [2] has proposed a correlation between thermodynamic values of oxo-halogen ions and has obtained $ΔG_\mathrm{f}^\circ (\ce{IO2-}) = \pu{−51.8 kJ mol−1}$. He has also estimated the dissociation constant of $\ce{IO2H}$, $\mathrm{p}K_\mathrm{a}(\ce{IO2H}) = 4$, and obtained $ΔG_\mathrm{f}^\circ (\ce{IO2-}) = \pu{ −75 kJ mol−1}$. However, this value introduced in the above equation leads to $K_\mathrm{C}^\circ = \pu{2.8e12}$ and an impossibly large value of $k_\mathrm{+C}^\circ$. His correlation supports the idea that $k_\mathrm{+C}^\circ$ is very large but overestimates $ΔG_\mathrm{f}^\circ (\ce{IO2-})$ and probably also the dissociation constant of $\ce{IO2H}$. Figure 1 shows known $\mathrm{p}K_\mathrm{a}$ values of oxy-halogen acids, including the value of $\mathrm{p}K_\mathrm{a}(\ce{BrO2H})$ [3] not considered by Stanisavljev, and suggests that $\mathrm{p}K_\mathrm{a}(\ce{IO2H})$ is closer to $6$ than to $4$.

Figure 1. $\mathrm{p}K_\mathrm{a}$ values of oxy-halogen compounds $\ce{XOH}$ (•) and $\ce{XO2H}$ (◦).

References

Schmitz, G. Inorganic Reactions of Iodine(III) in Acidic Solutions and Free Energy of Iodous Acid Formation. International Journal of Chemical Kinetics 2008, 40 (10), 647–652. https://doi.org/10.1002/kin.20344.
Stanisavljev, D. Consideration of the Thermodynamic Stability of Iodine Species in the Bray-Liebhafsky Reaction. Berichte der Bunsengesellschaft für physikalische Chemie 1997, 101 (7), 1036–1039. https://doi.org/10.1002/bbpc.19971010708.
Faria, R. B.; Epstein, I. R.; Kustin, K. Kinetics of Disproportionation and $\mathrm{p}K_\mathrm{a}$ of Bromous Acid. J. Phys. Chem. 1994, 98 (4), 1363–1367. https://doi.org/10.1021/j100055a051.


text:
Equilibrium constant Kp
If I increase P, to compensate that change the equilibrium will move to the right if the forward reaction produces fewer moles.Then the change in pressure will be undone. Now my question is how is Kp staying constant while at the last pressure is just the same as before but number of moles are changed?Can it be explained with the equation of Kp involving pressures and mole fractions?

Have a look at this answer, its not exactly the same but closely related.https://chemistry.stackexchange.com/questions/73644/how-does-the-partial-pressure-of-gas-c-change-with-the-change-in-volume-of-the-c/73975#73975

If mole fraction changes, partial pressure will change and total pressure will change too.That means the increase of pressure will be undone.Then why are we using the increased pressure in the last mentioned equation instead of using the unchanged pressure?

Mathematically speaking, $$\Delta G_\mathrm{r} = \Delta {G_\mathrm{r}}^{\circ} + RT \ln Q$$
Here, $\Delta G_\mathrm{r}$ is the molar Gibb's free energy of the reaction and $\Delta {G_\mathrm{r}}^{\circ}$ is the standard molar Gibb's free energy of the reaction.
At equilibrium, $Q = K$ and $\Delta G_\mathrm{r} = 0$, thus $\Delta {G_\mathrm{r}}^{\circ} = -RT \ln K$. From this equation, you can clearly see that $K$ is dependent only on temperature and the molar Gibb's free energy of the system. So, adding or removing reactants, products, or changing pressure or volume won't affect $K$ or $K_p$.
For $K_p$ you can replace concentrations of reactants/products in the expression for $K$ by their partial pressures divided by the standard pressure $P^\circ = 1 ~\mathrm{bar}$. And as you know, partial pressure of a gas is equal to the total pressure multiplied by the mole fraction of the particular gas.

Can you please see the edited question?

I've edited the answer accordingly.


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Kinetics of Carbon Nanotubes Growth
I'm reading a paper where the author has come with these consecutive elementary reactions

Where $C_2H_{4\ ads}$ is ethylene adsorved, $C_{ads}$ is carbon adsorved and $C_{CNT}$ is carbon that became nanotubes. The process he used to grow the nanotubes was CVD.
The author has derived this following equation, in which I couldn't derive myself and still doesn't make sense to me
$$\frac{d[C_{ads}]}{dt}=2k_1^+[C_2H_{4\ ads}] - (k_2^+ + k_1^-)[C_{ads}][H_2] $$
In the text he said he used these two approximations: $\frac{d[C_2H_4]}{dt}=0$ and $\frac{d[H_2]}{dt}=0$
First: I don't know why $\frac{d[C_2H_4]}{dt}=0$ is useful, as it never appears in the equation for $C_{ads}$.
Second: Why he didn't put $C_{CNT}$ there? (and even then $k_2^+$ happens to appear in the equation)
Third: Is this correct? Can someone give me some light?

Welcome to Chemistry StackExchange.  Is it not possible that there are some typos?  Taking $\ce{C2H4}$ at steady state seems odd, making this $\ce{d[C2H4_{abs}]/dt=0}$ seems more likely as it is a transient species.  The last step only involves $\ce{C_{ads}}$ so it should only appear as $-k_2^+C_{ads}$ in the rate expression and not multiplied by $\ce{[H2]}$.

Thanks for the answer! Well, he says $d[C_2H_4]/dt=0$ because he is putting ethylene constantly into the system, so I guess it makes sense, and as you said, I agree with your last equation statement, I don't know why he plugged $k_2^+$ in the $d[C_{ads}]$ equation. I have the paper in hands but I think it's unethical to put it here.

No you should not post the paper, seems likely then that the last part is a typo. The hydrogen is then presumably swept away and so $\ce{d[H2]/dt =0}$ also which makes sense.


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Formation of MnO2 vs [Mn(H2O)6]4+ in aqueous solution
I am looking for reasons why $\ce{Mn(IV)}$ wouldn't be stable as a hydrate. I know that from standard reduction potentials it exists as a ppt of $\ce{MnO2}$, but what is stopping it from being a hydrated $\mathrm{4+}$ aqueous cation? Is it that the water ligands would be too polarised and therefore $\ce{H+}$ dissociation would occur very easily? Are there other factors that make it a solid rather than a solvated ion, like lattice enthalpy etc.?

related https://chemistry.stackexchange.com/questions/41038/odd-acidity-trend-in-hexaaqua-metal-complexes

Solid $\ce{MnO2}$ can be hydrated, but it's the solid which precipitates from solutions as a product of redox reactions of other $\ce{Mn}$ compounds. There is also for example crystalline $\ce{[MnF4(py)(H2O)]}$ What about aquo cations you mentioned?
One thing is that $\ce{[Mn(H2O)_n]^4+}$ cations should be indeed quite strongly acidic, but that's no problem as another strong acid in solution would prevent precipitation (similar thing like with $\ce{Fe(OH)3}$). The problem is oxidative power, what may be surprising knowing that Mn has higher oxidation states.
Why do I think so? Well, let's see what happens if we try this trick with acid to make a solution. Concentrated $\ce{H2SO4}$ and $\ce{HCl}$ react with $\ce{MnO2}$, but lets see the reactions:
$$\ce{2 MnO2 + 2 H2SO4 → 2 MnSO4 + O2 + 2 H2O}$$
$$\ce{MnO2 + 4 HCl → MnCl2 + Cl2 + 2 H2O}$$
In first case oxygen gets oxidised and in second chlorine!
Now what if we used $\ce{HF}$? Concentrated solutions are really powerful stuff! Unfortunately the problem persists, as you could see earlier $\ce{Mn^{4+}}$ can oxidise even water if acidity is high enough. With difficulty $\ce{MnF4}$ was synthesised, but guess what, Mn can gradually oxidise fluorine in this compound, making $\ce{F2}$ gas! That was first chemical reaction known to do that.
So, generally not much chance for simple aquo complexes.
Check out these articles on Wikipedia for more info and citations to articles: about $\ce{MnF4}$ and about $\ce{MnO2}$.

, but that's no problem as another strong acid in solution would prevent precipitation  || It's a problem because really strong acid will dissociate in water anyway, a non-water solution would be needed. Monoatomic tetracations are a very rare thing a small cations like Mn definitely are not stable enough to exist in water solution.

@permeakra Well concentrated H2SO4 is still arguably "aqueous"  Yep, I was thinking about adding a lot of acid :D


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Titration - Just need your judgement
A student investigated the volume of sulfuric acid that reacted with 25 cm3 sodium hydroxide. Describe a titration method the student could use in this investigation. {6 marks}

My answer:

Use the pipette and pipette filler to safely add 25 cm3 of $\ce{NaOH}$ to a clean conical flask.
Add a few drops of indicator and put the conical flask on a white tile (so you can see the colour of the indicator more easily).
Fill the burette with $\ce{H2SO4}$ and note the starting volume (record from bottom of the meniscus at eye level).
Slowly add the acid from the burette to the alkali in the conical flask dropwise, swirling to mix.
Stop adding the acid when the end-point is reached (the appropriate colour change in the indicator happens). Note the final volume reading of $\ce{H2SO4}$ (record from the bottom of the meniscus at eye-level).
Repeat steps 1 to 5 until you get consistent readings

Have I missed anything that will prevent me from getting 6 marks?

I guess you haven't missed anything ..yet i would have mentioned an appropriate indicator for the reaction and mention the colour change ex:using phenolphthalein indicator until the colour changes from pink to colourless stable for 30 sec

Do not delete/repost questions

Usually the measurements also need to be stated to the specified precision of the instrument. For example, for my school's pipette, we typically give the measured quantity to 1 d.p. (i.e. 25.0 cm3)

As titration is based on the relationship of molar quantity ($n$), volume  ($V$) and concentration ($c$) of the reagents
$$ c = \frac{n}{V}$$
and as your titration is to determine the concentration of NaOH in a known volume of solution of analyte, you would need to know both the volume as well as the concentration of the $\ce{H2SO4}$ solution deployed by addition via the burette.  The later is missing in your list.
Knowing the stoichiometry of the neutralisation is equally of importance, since each equivalent of $\ce{H2SO4}$ reacts with two equivalents of $\ce{NaOH}$. (This differs from a determination of, say, of / with $\ce{HCl}$, or $\ce{H3PO4}$.)  Your current list neither accounts for a chemical equation, nor this stoichimetric factor.