text:
Which is not true about Ideal gas

Interaction between atom or molecule is neglected
The transformation to liquid or solid is observed under appropriate condition
The volume of constituent atom or molecule is neglected
The compressibility, PV/RT is independent of pressure

Why is 2 wrong? In formula there is P,V,T which means under certain condition?
Why is 3 true? Then what volume of gas do we calculate if it is not from atom or molecules?
Why are 1 and 4 true? Do atoms in gas interact? Why is the compressibility independent of pressure?

Homework question. You should explain why you think 2+4 are correct, but 1+3 not.

Ahem, you should 1) get a few english lessons, your grammar is highly unorthodox ;-) and 2) I was more thinking about you saying "I think 1. is right,because ...., 2. is wrong, because .... , etc. You've now asked five additional questions, not explained your reasoning!

Interaction between atom or molecule is neglected

True. In an ideal gas the gas particles are unaware of each other's existence.

The transformation to liquid or solid is observed under appropriate condition

False. As mentioned in the answer to (1), the particles don't sense each other. Since there are no attractive interactions there is nothing to prevent the gas particles from flying apart, there is nothing to encourage association with other particles. This can of course be a bad assumption in reality, particularly if the gas is highly compressed, but you should not be applying the ideal gas law under such conditions anyway!

The volume of constituent atom or molecule is neglected

True. This is a central tenet in the definition of an ideal gas. A correction for a definite particle volume is included in other gas laws, such as van der Waals'.

The compressibility, PV/RT is independent of pressure

True. This follows directly from the definition, $PV=nRT$

Elastic collisions are allowed in an ideal gas. Dilute gases approach ideal behaviour long before the mean free path lenght goes towards infinity.

@Karl I agree with you. However it is not particularly more logical to treat particles mathematically as identical elastically colliding "point particles" than to simply ignore collisions, particularly if the result from the observer's point of view is equivalent. Mean free path is a concept brought up in the kinetic theory of gases, which elaborates on these ideas. I think if you consider these as concepts that are superimposed, it is best to start with the simplest possible. There is a conceptual elegance in ignoring particle interactions altogether that is very helpful later.

https://en.wikipedia.org/wiki/Kinetic_theory_of_gases#History

@Karl I think there is something oxymoronic in assuming that particles don't interact, don't have a volume (are infinitesimally small) and yet can collide elastically. They are ultimately a set of mutually inconsistent assumptions, and unnecessary as far as basic thermodynamic application of the ideal gas equation of state is concerned. But still, I admit that these are assumptions made when introducing the kinetic theory of gases, and important for understanding the structure of matter, see eg the wikipedia link above.

I think we're totally on the same side here, just want to mention that collisions in the gas are important from the thermodynamical point of view too: A noncolliding gas would have real trouble to reach thermal equillibrium. Realistic walls would do it, but perfectly hard walls never.

@Karl yes, it is not necessary to invoke particle collisions if you model the container correctly. It's a personal preference to ignore collisions, and as this problem (perhaps an outlier) demonstrates I am not alone :-). In any case, I believe ignoring collisions makes grasping some topics in thermodynamics easier, obviously not the kinetic theory.

Perhaps I should add some of these comments to my answer, but it'd probably do more harm than good.

Well, your point 1 is still wrong. It doesn't matter if your ideal gas has collisions. The "no volume" condition is not very strict, a minimal cross section of the particles immediately gives you a huge amount of hard sphere interactions, but no significant change on the observed $p(V,t)$

@Karl Points (1) and (3) amount to the same thing. Your gas would not behave ideally across all p or T. Mine would (even under conditions that would condense, say, a noble gas). Remember, the question is not, "when do real gases behave ideally".

(correction, they don't amount quite to the same thing, but almost. (3) implies HS interactions; (1) implies all interactions).

Sorry, I give up. Please just remove that wrong, and misleading 1.), before any more undergrad students looking for help read it. It's a corollary from trying to fulfil condition number three, NOT a condition on its own. The correct conditions reads "Particles in an ideal gas must only have hard sphere interactions with themselves or the wall", period.

@Karl The equation $pV=nRT$ does not stipulate the existence of intermolecular interactions. On the contrary, (1) is more important and general than (3), since it also addresses the absence of attractive interactions in the ideal gas model, which (3) does not address. Of course the ideal gas law is an approximation, to be used only when it approximates reality sufficiently under special circumstances. Students should understand that when taught about the difference between the ideal gas model and the behavior of real gases.

https://en.wikipedia.org/wiki/Hard_spheres EOD

@Karl https://en.wikipedia.org/wiki/Ideal_gas_law#Deviations_from_ideal_behavior_of_real_gases I've noticed that the wikipedia page for ideal gas brings up hard spheres and elastic collisions as properties of ideal gases, but the HS condition is strictly incorrect, and the requirement of elastic interparticle collision is unnecessary. Particles in a HS gas have a definite volume, not so in an ideal gas. If you want to model a gas as a collection of hard spheres, I recommend calling it a HS gas, not an ideal gas. A HS gas may behave ideally at low p, high T.

The kinetic theory of matter makes the assumption of colliding elastic point particles, and starting from these and additional axioms can be used to derive an equation of state analogous to the ideal gas equation. But the ideal gas equation does not require such axioms to be defined. On the contrary, for generality it is best to ignore intermolecular interactions altogether.


text:
How can I measure calcium levels in my blood at home?
I need to check my vitamin D levels regularly for my disease. It doesn't seem possible to measure at home, but measuring calcium in the blood may be more achievable since they are directly correlated. But then how can I do it?

You should not use calcium as an indirect means to evaluate your vitamin D status. While both are metabolically connected different mechanisms are involved meaning that there is not a one directional correlation across time, and many other things, unrelated to vitamin D, affect blood calcium levels as well.
In the short run, all else kept equal (which is a culprit), a higher vitamin D status will lead to a greater portion of a dietary dose of calcium to be absorbed. However, this will also lead to increased bone formation, and soft tissue mineralization, which has the opposite effect on blood calcium levels in the short to medium term.
With time and resources you could probably develop a model for a very controlled process to measure your vitamin D status through your blood calcium levels, but it wouldn't be practical for people to use in general, and it would also cost you personally much more time and resources than just getting a vitamin D lab test every 3 months for the next couple of years.
While blood levels of vitamin D may fluctuate in the short term, your overall vitamin D status fluctuates slowly (which is due to the fact that your body fat acts as a storage buffer for it). That and the fact that is has become a very routine test makes it affordable to monitor via routine lab testing.
If you develop a proper habit to expose a greater part of your bare skin to sunlight once every 7-10 days, and have a stable diet, you can just use initial testing to confirm the effectiveness of that habit and then save even more money by not testing for your vitamin D levels at all.
Your body'd capacity to efficiently produce vitamin D will slightly decrease over time as you get older, but that happens only very slowly, so you can consider the execution of whatever habit you have developed to be a good way to predict a good vitamin D status.

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text:
Name of the intermediate in between phosphono and phosphonato group
$\ce{R-PO3^2-}$ is called a phoshonato group (https://www.ebi.ac.uk/chebi/searchId.do?chebiId=30931)
$\ce{R-PO3H2}$ is called a phosphono group (https://www.ebi.ac.uk/chebi/searchId.do?chebiId=48108)
So, what is: $\ce{R-PO3H-}$ called ? I can't find it, it seems to be largely ignored.
Auxiliary question, is one of the two possible enantiomers of this form dominant ? Is there anyway to know this ? Or is it by definition distributed in a 50%/50% manner ?

Edited back to the original format (what was that edit anyway ?)

— Monophosphonic acid

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text:
Ensuring constant ionisation of transition metals using plasma ionisation for accurate mass spec
Mass spectrometry depends on separation of the analyte based on mass to charge (m/z). ICP-MS is used to detect metals, but under the plasma conditions many of these metals could be ionised multiple times, giving them a variety of (m/z) ratios making a difficult to identify peak for any particular metal. Additionally some metals would have m/z ratios that overlap with others. Then it is conceivable that depending on your solution the quantitation of some elements is impracticable. For example if you want to quanitate aluminium (mass ~27) then a solution high in manganese (mass ~54) would make this impracticable, if manganese were twice ionised to give the same mass to charge ratio as a single ionised aluminium ion. How does the ICP-MS ensure atoms are ionised only once, or that only singly ionised species are detected?
This is further complicated if you consider that the elements rarely exist in solution at oxidation states of 0, but rather aqueous ions with varying oxidation states.

In your particular example, Al = 26.981 and Mn = 54.938, so any decent mass spec can easily separate them. But, no, you can't ensure one charge state, but you can look for and correlate where different charge states of one ion would end up. Elements with multiple isotopes will help, of course. A similar issue happens with mass spec of molecules (is m=2 from D or H2?).

Interesting question. Some of the stuff can be justified. As an analytical chemist, we should keep in mind that no technique is perfect or universal, even if the instrument costs thousands or millions of dollars. Mass spectrometer cannot distinguish isobaric species (or ions), for instance, if two ions have identical mass due to a different location of the isotope within the same molecule, the mass spec won't be able to you. However, if there is a small difference in mass at fourth, or even fifth decimal place, modern day mass spectrometers can distinguish them (see e.g., Fourier transform ion cyclotron resonance mass spectrometers). The more money you add, the fancier the mass spec gets, with it comes better and better mass resolution.
Now come to the ion generating source - the plasma, which is at say 10,000 K. Although the name plasma sounds very fancy as the fourth state of matter, and given that its temperature is higher than the surface of the Sun- this temperature is still too low for atoms to ionize like crazy. It is pretty hard to strip off one, then the second and then the third electron from an atom as you are thinking of Al. Therefore in a typical ICP torch, multiple ionization on a single atom is not a major concern. To satisfy yourself, do a Boltzmann distribution calculation for sodium atom at 10,000 K. You will notice that a very small fraction of sodium atoms will be in the excited state.

This is further complicated if you consider that the elements rarely
  exist in solution at oxidation states of 0, but rather aqueous ions
  with varying oxidation states.

Once the ions from solution reach the plasma they get reduced to "zero" oxidation state. Consider the plasma as a rich source of free electrons, which are circulating like crazy following the radiofrequency field of the coil. The emission in ICP mainly comes from neutral atoms.

First ionisation potentials range from Cs (375 kJ mol-1) to Hg (1 007 kJ mol-1). The lowest second ionization potentials are 965 kJ mol-1 and lanthanides are in the 1020 - 1170 kJ mol-1 range. Thus insuring adjusting the temperature in the ICP torch is necessary to get the best compromise between first ionization of the target ion and unwanted second ionization of other metals.


text:
Relationship between Gibbs free energy and potential energy (related to reaction coordinate diagrams)
For a rate law, the rate constant $k$ is given by the equation
$$\ln k = \ln A - \frac{E_\mathrm{a}}{RT}\tag{1}$$
If $k_\mathrm{fwd}$ is the forward rate constant and $k_\mathrm{rev}$ is the reverse rate constant:
$$\ln k_\mathrm{fwd} = \ln A - \frac{E_\mathrm{a,fwd}}{RT}\tag{2}$$
$$\ln k_\mathrm{rev} = \ln A - \frac{E_\mathrm{a,rev}}{RT}\tag{3}$$
Subtracting the two equations yields
$$\ln\frac{k_\mathrm{fwd}}{k_\mathrm{rev}} = \frac{E_\mathrm{a,rev}}{RT} - \frac{E_\mathrm{a,fwd}}{RT} = \frac{1}{RT} (E_\mathrm{a,rev} - E_\mathrm{a,fwd})\tag{4}$$
If $K$ is the equilibrium constant for the reaction, then $\displaystyle K = \frac{k_\mathrm{fwd}}{k_\mathrm{rev}}$:
$$-RT\ln K = \Delta E\tag{5}$$
where $\Delta E$ is the change in potential energy on a reaction diagram, such as the one shown here.
However, we know that
$$-RT\ln K = \Delta_\mathrm{r} G^\circ\tag{6}$$
Does this mean that potential energy is equal to the standard Gibbs free energy of reaction?

Why did you assume that the pre-exponential factor is the same for the forward reaction as for the reverse reaction?

Is it not? I'm not sure if it changes or not.

If that were the case, the equilibrium constant for all reactions would approach 1 at high temperatures.  We know that that doesn't happen.  There is absolutetly no physical reason why it would be the case, however.

@ChetMiller For a single step reaction $K=\frac{k_1}{k_{-1}}$. We also know that $K=e^{\frac{-ΔrG}{RT}}$. But $$\frac{k_1}{k_{-1}}=\frac{A_1\cdot e^{-Ea_1/RT}}{A_{-1}\cdot e^{-Ea_2/RT}}=\frac{A_1}{A_{-1}}\cdot e^{-ΔrG/RT}$$ Shouldn't pre-exponential factors be equal?

The A’s can be absorbed into the exponential

@ChetMiller Oh now I see. When we use $E_a$ we are in an energy diagram where enthalpy is used whereas for an energy diagram with Gibbs free energy we should use $ΔrG^{\ddagger}$ instead of $E_a$ in order to account for the entropic term?

@ChetMiller "The A’s can be absorbed into the exponential". According to transition state theory there is also a pre-exponential factor with some constants and the transmission coefficient. Is the transmission coefficient the same for the forward and reverse rate?

Equation (1) is an empirical one that does not consider entropy. If you substitute it for the Eyring equation (considering free energy of activation $\Delta G^\ddagger$) you end up with equation (6) as expected, not equation (5). Wikipedia has a nice section on the relationship of activation energy and Gibbs energy of activation:

Although the equations look similar, it is important to note that the Gibbs energy contains an entropic term in addition to the enthalpic one. In the Arrhenius equation, this entropic term is accounted for by the pre-exponential factor A. More specifically, we can write the Gibbs free energy of activation in terms of enthalpy and entropy of activation: $ΔG‡ = ΔH‡ – T ΔS‡$. Then, for a unimolecular, one-step reaction, the approximate relationships $E_a = ΔH‡ + R T$ and $A = (k_B T/h) \exp(1 + ΔS‡/R)$ hold. Note, however, that in Arrhenius theory proper, A is temperature independent, while here, there is a linear dependence on T. For a one-step unimolecular process whose half-life at room temperature is about 2 hours, ΔG‡ is approximately 23 kcal/mol. This is also the roughly the magnitude of Ea for a reaction that proceeds over several hours at room temperature. Due to the relatively small magnitude of TΔS‡ and RT at ordinary temperatures for most reactions, in sloppy discourse, $E_a$, ΔG‡, and ΔH‡ are often conflated and all referred to as the "activation energy".

So in a more rigorous treatment, the pre-exponential factor is temperature-dependent (this is almost impossible to measure because the exponential is "very" temperature-dependent) and has an entropy component. That is why equation (5) in the question is not accurate (equation (6) is correct, and can be derived in a rigorous manner from first principles and definitions).
Related question: Y-axis of the reaction co-ordinate graph


text:
Rate law for breakdown of ozone to oxygen
For the following problem:

Example 3: Write the rate law for $\ce{2 O3 -> 3 O2}$ given that the first step is fast and reversible, and the second step is slow.
$$
\begin{align}
&\text{Step 1 (fast, reversible)} &\qquad \ce{O3 &<=>[$k_1$][$k_{-1}$] O2 + O} \\
&\text{Step 2 (slow)} &\qquad \ce{O + O3 &->[$k_2$] O2 + O2}
\end{align}
$$
The rate is determined by the slowest step.
The rate of formation of $\ce{O2}$ is equal to $2$ times the rate of the slow step $(k_2[\ce{O}][\ce{O3}]),$ since two molecules of $\ce{O2}$ are formed.
Thus, rate of formation of $\ce{O2} = 2k_2[\ce{O}][\ce{O3}],$ but $“\ce{O}”$ is an intermediate, solve for $“\ce{O}”$ in
  terms of products and reactants and rate constants.
Since the first step is fast and reversible and the second step is slow, the first step is in equilibrium and we can write
$$\frac{[\ce{O2}][\ce{O}]}{[\ce{O3}]} = \frac{k_1}{k_{-1}} = K_1 \quad\text{or}\quad [\ce{O}] = \frac{k_1[\ce{O3}]}{k_{-1}[\ce{O2}]}$$
Substituting:
$$
\begin{align}
\text{rate} &= \frac{2k_2k_1[\ce{O3}]^2}{k_{-1}[\ce{O2}]} \\
\text{rate} &= k_\mathrm{obs}\frac{[\ce{O3}]^2}{[\ce{O2}]}
\end{align}
$$

The solution indicates that the rate of formation $\ce{O2}$ is $2k_2[\ce{O}][\ce{O3}]$. Why isn't it $2k_2[\ce{O}][\ce{O3}] + k_1[\ce{O3}]$, because the first reaction also produces $\ce{O2}$?

You should write down the rate equations and set [O] as an intermediate so that $\ce{d[O]/dt} = 0$. This is the steady state approach and leads to $\displaystyle \frac{d\mathrm{[O_2]}} {dt}=2\frac{k_1k_2\mathrm{[O_3]}^2}{k_{-1}\mathrm{[O_2]}+k_2\mathrm{[O_3]}}$ which differs from your answer because when you assume a fast equilibrium you also assumed that $k_{-1} \gg k_1$. You should make this assumption after you have the full equation, if necessary, but not before you do the calculation.
To answer your last point you still have the intermediate [O] which is an unknown and so has to be eliminated as shown by using the steady state approach used above.

$$
\begin{align}
&\text{Step 1 (fast, reversible)} &\qquad \ce{O3 &<=>[$k_1$][$k_{-1}$] O2 + O} \\
&\text{Step 2 (slow)} &\qquad \ce{O + O3 &->[$k_2$] O2 + O2}
\end{align}
$$
The rate is determined by the slowest step.

$$r=k_2[\ce{O}][\ce{O3}]$$
If the concentration of  $\ce{ O}$ which  produced from the first step increased, the reaction rate increased in the opposite direction, leading to the arrival of the concentration intermediate material $\ce{ O}$   to a fixed value. This is called Steady State, where the concentration of $\ce{O}$ is constant,
The rate of formation of the intermediate material $\ce{O}$  equal the rate of its consumpution, this material produced from the first step in the forward direction and consumed from the first step in the opposite direction, and in the second step also.
$$\text{ Production rate of}~ \ce{O}  = \text{ Consumption rate of}~ \ce{O}$$
$$k_\mathrm {1}\ce{[O3]}= k_\mathrm{-1}[\ce{O2}][\ce{O}] + k_\mathrm {2} \ce{[O] [O3]}$$
But $k_\mathrm{2} << k_\mathrm{-1}~ \text{and}~ k_\mathrm{1} $ because the second step is the slow step, so the term $k_2\ce{ [O] [O3] }$ can be neglected  at the far right of the last equation.
$$k_\mathrm {1}\ce{[O3]}= k_\mathrm{-1}[\ce{O2}][\ce{O}]$$ $$[\ce{O}] =\frac{ k_1 [\ce{O3}] }{ k_\mathrm{-1}[\ce{O2}]}$$
The value of$ [\ce{O}]$ from the last equation can now be compensated in the reaction rate equation which represents the slow step.
$$\text{r} = {k_\mathrm{2}[\ce{O}][\ce{O2}] = \frac{k_\mathrm{1}k_\mathrm{2}}{k_\mathrm{-1}}\times\frac{[\ce{O3}][\ce{O3}]}{[\ce{O2}]}}
 =\frac{k[\ce{O3}]^2}{[\ce{O2}]}=k[\ce{O3}]^2[\ce{O2}]^{-1}$$
In it we see that $k$ expresses the value of the fractional $\frac{k_\mathrm{1}k_\mathrm{2}}{k_\mathrm{-1}}$ and we note that the concentration of $\ce{O2}$ has appeared in the denominator of the reaction rate law , although it is a substance resulting from the reaction. This means that the order of reaction for oxygen gas $\ce{O2}$ is $(-1)$ that the doubling of the concentration of $\ce{O2}$ leads to a reduction in the rate of the reaction by half.


text:
What will happen if electron as a wave undergo destructive interference while travelling in its Orbit?
Today I came across a justification of Bohr's angular momentum quantization using the De- Broglie's hypothesis.
The justification said that since the electrons are having wave nature also then while travelling in its Orbit there is starting point and ending point must concide in order to avoid destructive interference of wave.
What will happen if electron as a wave undergo destructive interference while travelling in its Orbit ?
If the electron goes under constructive interference then it's amplitude should increase each time. Is it possible ??

It is not possible. That was just a simplistic rationalization of the quantization assumed by Bohr. So no worries!

I am not sure how De Broglie might have come to this conclusion at that time (if indeed he did, I have not read his work), but another interpretation would be that a discontinuity would be inconsistent with a time-independent (stationary) state such as Bohr was postulating.

@Tushar — The thing is that waves are constrained by length/radius... so they are standing waves. In other words, all the wavelengths which are not proportional to the length/radius cancel out (and only standing waves survive)! One way or another, you only get standing waves which, by definition, don't cancel out


text:
Are hydroxides soluble in each other?
Let's say I have molten potassium hydroxide and then I add some Calcium hydroxide. I imagine, since there are hydroxide groups floating around, it would be energetically neutral for the calcium ions to bind to a "new" hydroxide group, therefore they should be floating freely with the hydroxide groups, becoming solvated. I can't find any empirical data; is my reasoning correct and I can dissolve a hydroxide in any other molten hydroxide?

Well it would depend on case. In the one you mention, probably yes.

The most robust way to answer this would probably be to find the corresponding phase equilibria diagram for $\ce{Ca(OH)2}$–$\ce{KOH}$ system.

@andselisk such a phase diagram may require deep digging to get to.  Simply googling for a calcium hydroxide-potassium hydroxide phase diagram fails because other phase diagrams come up instead.  A direct reference would be most welcome, thanks.

Probably not so much calcium hydroxide, but alkali metal hydroxides mixing with each other in the liquid phase.  The pairs $\ce{LiOH-NaOH}$ and $\ce{KOH-NaOH}$ are not only fully miscible, they have been studied as fuel cell electrolytes[1].  The picture below, from the referenced article, shows results for a range of electrolyte compositions that imply the extensive miscibility.  The fuel cell itself is based on carbon oxidation to carbon dioxide.

Reference

A. Kacprzak, R. Kobylecki, Z. Bis (2013)."Influence of temperature and composition of NaOH-KOH and NaOH-LiOH electrolytes on the performance of a direct carbon fuel cell".
Journal of Power Sources
239, 409-414.
ISSN 0378-7753,
https://doi.org/10.1016/j.jpowsour.2013.03.159.

In order for the $\ce{Ca(OH)2}$ to dissolve, it should become solvated, which requires it to bind some of the OH- ions in molten $\ce{KOH},$ leaving some $\ce{K+}$ relatively bare. We already know that $\ce{Ca(OH)2}$ can be dehydrated easily, while $\ce{KOH}$ cannot (it boils at 1325 °C).
The heat of hydration of $\ce{K2O}$ is much greater than the heat of hydration of $\ce{CaO}.$ However, the heat of hydration of $\ce{Ca^2+}$ is greater than the heat of hydration of two $\ce{K+}.$ So, addition of a little water could solvate (hydrate) the calcium ion and make it more soluble.
A relatively easy way to test would be to take a concentrated solution of $\ce{KOH}$ (soluble ~50% in $\ce{H2O}$ @ 15 °C, 64% @ 100 °C) and add some $\ce{Ca(OH)2}.$ More $\ce{Ca(OH)2}$ is likely to dissolve than if no solvating moieties are present other than the $\ce{KOH}.$ Although $\ce{H2O}$ is not powerful enough to bring $\ce{Ca(OH)2}$ into solution, $\ce{OH-}$ ions might do the job — if the $\ce{K+}$ ions are not left bare. It is worth considering whether the water is the actual solvent, with $\ce{KOH}$ as a reactant
$$\ce{Ca(OH)2 + KOH -> Ca(OH)3- + K+},$$
or whether the $\ce{KOH}$ solution is simply a solvent.

Note that $“\ce{Ca^{++}}”$ is an obsolete notation and $“\ce{Ca^2+}”$ should be used instead. As you are already an experienced user, please use proper formatting. (re)Visit this page, this page and this one on how to format your posts better with MathJax and Markdown. Also, I'm not quite following the reasoning in this answer. Probably you can improve it if you add some references your assumptions are based upon.

To the extent that it dissolves, calcium hydroxidevis a string base tending to dissociate the hydroxide ions it has. Certainly not disposed to picking up more. Among Group 2 metals only beryllium forms a hydroxide that's amphoteric in water solution.


text:
$K_c$ vs $K$ in equilibrium reactions
I don’t understand why the following question

The decomposition of nitrosyl bromide $(\ce{NOBr})$ proceeds by the following reaction:
$$\ce{2 NOBr(g) <=> 2 NO(g) + Br2(g)} \qquad K = 0.0142$$
Calculate the $[\ce{NOBr}],$ $[\ce{NO}],$ and $[\ce{Br2}]$ when $\pu{10.0 mol}$ of nitrosyl bromine is placed in a $\pu{5.00 L}$ closed vessel and allowed to decompose.

has the answers that it does:

$$
\begin{align}
[\ce{NOBr}] &= \pu{1.585 M} \\
[\ce{NO}] &= \pu{0.415 M} \\
[\ce{Br2}] &= \pu{0.207 M}
\end{align}
$$

From what I understand
$$K = \frac{\prod a_\mathrm{products}}{\prod a_\mathrm{reactants}},$$
where $a$ is the activity of the products/reactants. The activity of a gas is approximately equal to the partial pressure of the gas divided by a reference pressure (usually 1 atm). This yields unitless quantities with the magnitude of the partial pressures of each gas. Thus
$$K = \frac{P_\ce{Br2}\cdot P_\ce{NO}^2}{P_\ce{NOBr}^2}$$
Converting from a partial pressure to a concentration can be done using the ideal gas equation
$$\frac{P}{RT} = \frac{n}{V} = M,$$
where $M$ is concentration. Thus,
$$K = \frac{[\ce{Br2}][\ce{NO}]^2\cdot RT}{[\ce{NOBr}]^2}$$
However, plugging the concentrations given into the solution of this problem doesn’t yield the given value of $K.$
Am I wrong in the way I am approaching this problem, or is the problem itself wrong? If the equilibrium constant was denoted $K_c$ instead of $K,$ the given solution would be correct, but I don’t think it’s correct to assume that $K_c$ is the same as $K.$

At a first glance I notice that the units don't seem to match - K should be dimensionless, but in your description of K by partial pressures it has a unit of pressure. So I think there is an error somewhere in your derivation.

Does the problem not stipulate the value of T?

It's easier to think of $K$ as dimensionless, particularly if you plan to find the logarithm.

The $K$ you are given is $K_c$ not $K_p$ as the values given do yield the equilibrium const.  If you want to work out these values make up an ICE table with initial concentration $n/V =0.5$. The equilibrium constants are related as $K_C=K_P(RT)^{-\Delta n}$ where $\Delta n$ is the number of moles  of products  minus reactants according to stoichiometry , 1 in your case. The equilib. constants are always dimensionless as they are always divided by 1 M or 1 atm as appropriate.

The thermodynamic equilibrium constant $K$ (different from both $K_p$ and $K_c$) and the reaction quotient $Q$ are defined with reference to a standard state. When all reactants and products are at standard state, $Q = 1$. $K$ and $Q$ are related to the Gibbs energy of reaction via:
$$\Delta_\mathrm{r}G = RT \ln\frac{Q}{K}$$ No matter which standard state you use, this relationship gives the same Gibbs energy of reaction (because $K$ and $Q$ are equally affected). However, depending on which standard state is used, $K$ and $Q$ will have different numeric values.
According to wikipedia,

In chemistry, the standard state of a material (pure substance, mixture or solution) is a reference point used to calculate its properties under different conditions. In principle, the choice of standard state is arbitrary, although the International Union of Pure and Applied Chemistry (IUPAC) recommends a conventional set of standard states for general use.

The recommended standard state for a gas is, again according to wikipedia,

the hypothetical state it would have as a pure substance obeying the ideal gas equation at standard pressure ($\pu{e5 Pa}$, or 1 bar). No real gas has perfectly ideal behavior, but this definition of the standard state allows corrections for non-ideality to be made consistently for all the different gases.

Because neither the temperature nor the pressure is given in the problem the OP asked about, you can't use this standard state. Instead, you can use the standard state for a solute (still wikipedia):

For a substance in solution (solute), the standard state is the hypothetical state it would have at the standard state molality or amount concentration but exhibiting infinite-dilution behavior. The reason for this unusual definition is that the behavior of a solute at the limit of infinite dilution is described by equations which are very similar to the equations for ideal gases. Hence taking infinite-dilution behavior to be the standard state allows corrections for non-ideality to be made consistently for all the different solutes. Standard state molality is 1 mol kg−1, while standard state amount concentration is 1 mol dm−3.

What is the problem with this problem?

Am I wrong in the way I am approaching this problem, or is the problem itself wrong?

The problem is incomplete. If they use a funky standard state, they should say it explicitly.

Well maybe I'm an idiot, but I also get a different result.
First of all, the equilibrium constant is given by
$$
K = \frac{\prod_p a_p^{m_p}}{\prod_r a_r^{n_r}} \,,
$$
where $a$ is the activity of a reactant/product, $p$ is the index for products, $r$ is the index for the reactants, and $m_p$ and $n_r$ are the corresponding stoichiometric constants. So far, so good - you actually used this formula as well, as far as I can see, even though you didn't write it this way.
Now it's important to realize that the activity is dimensionless, and is given by
$$
a_i = \gamma_i x_i = \gamma_i \frac{c_i}{c_\text{tot}} \,,
$$
with the activity coefficient $\gamma$ ($=1$ for ideal gases) and the mole fraction $x_i$, which can be represented by the molar concentration $c_i$ of the reactant or product divided by the total molar concentration $c_\text{tot}$.
Using this and assuming $\gamma = 1$, we get
$$
K = \frac{\prod_p c_p^{m_p}}{\prod_r c_r^{n_r}} \, \frac{\prod_r c_\text{tot}^{n_p}}{\prod_p c_\text{tot}^{m_r}}\,,
$$
which in your case leads to
$$
K = \frac{[\ce{Br2}][\ce{NO}]^2}{[\ce{NOBr}]^2} \, \frac{1}{[\ce{Br2}]+[\ce{NO}]+[\ce{NOBr}]} \,.
$$
This is a dimensionless number, as desired. We know from your initial concentration $[\ce{NOBr}] = 2 \, \text{M}$ how many mol of N, O, and Br are involved. Solving this equation for the concentrations is a little bit annoying, but in the end I get for $K=0.0142$:
$$
[\ce{NOBr}] = 1.48 \, \text{M} \\
[\ce{NO}] = 0.52 \, \text{M} \\
[\ce{Br2}] = 0.26 \, \text{M} \\
$$
I don't see where I could have gone wrong. I can only get the solution that you gave if I set the total concentration to $1 \, \text{M}$, and I don't see how that could be right.
Edit: After some digging, I found another definition of the activity which gives it as
$$
a_i = \gamma_i \frac{c_i}{c^\ominus} \,,
$$
where $c^\ominus$ is the concentration of the reactant or product in its standard state. This concentration is given as $c^\ominus = 1 \, \text{M}$, which then leads to the result given in the OP as the "correct" solution. However, I am not happy with this. It makes no sense and contradicts the other definition of the activity, in which the mole fraction is used. Since $c_i/c^\ominus$ is not a mole fraction, but some other random fraction, I don't think this is correct.

The activity depends on the choice of standard state. If the standard state is the pure substance, the mole fraction is a good approximation in some cases (when Roult's law is a good description). If the standard state is the pure gas at 1bar, you need to know the pressure. If the standard state is 1 M but exhibiting infinite-dilution behavior, the given data allows you to estimate the activity.

@KarstenTheis, yeah. And I think the standard state that is used here is that the concentrations are related to $c^\ominus = 1,\text{M}$, leading to the values given in the solution. But this is not satisfying to me, since I get different values with a different and, as far as I can see, equally valid approach.

Without knowing the standard state, K is a meaningless number.  $$$$ It is like saying the mass of a sample is 5.

@KarstenTheis, I mean, you can do that. As long as you use natural units.

... and say so. Otherwise, someone could always go ahead and say "but I meant milligrams".

@BuckThorn, the activity has to be dimensionless, so you have to have some fraction of two molar concentrations in it. In my opinion the intuitive one is to use the mole fractions, i.e. divide by the total molar concentration. But apparently it's more common in chemistry to use a comparison concentration of 1 mol/l in the denominator. As Karsten Theis said, the question sadly does not specify which definition should be used, so both results should be correct.

$K_c$ is the equilibrium constant for reactions where reactants and products are all in the same phase, which is the case for your reaction. The subscript c implies that $K_c$ is expressed in terms of concentration, which has units of moles per volume. I'm not entirely sure on this, but I don't believe you can use partial pressures in the expression for $K$, because the total number of moles of gas changes during the reaction (2 moles of reactants become 3 moles of products), and the container has constant volume, so the partial pressures would change from the reaction.


text:
Plaster of Paris

One of my Chemistry professors said that the hard mass formed after adding water to Plaster of Paris, cannot be converted back into Plaster of Paris. However, he said that the original Gypsum can be heated to make Plaster of Paris.

What causes the hard mass not to convert back into POP by adding water even though the hard mass and the gypsum share the same chemical formula? If there is a structural change? How is the structure changed here?

Sounds like nonsense afaik: https://en.wikipedia.org/wiki/Plaster#Gypsum_plaster.

If it can be converted once, it can be done again. Are you sure you understood the professor's comments? If the prof seems reasonable, you could easily show this in the lab, using an oven at ~130-180 °C.


text:
How do you calculate a half cell potential Copper(II) and Copper Sulfate?
I have to complete an experiment for school where I have copper and zinc electrodes placed into an electrolyte of copper sulfate and I have measured the potential across it at varying temperatures. For example at $\pu{60°C}$ ($\pu{333K}$) I measured the potential to be $\pu{0.880 volts}$ and at $\pu{75°C}$ ($\pu{348K}$) I measured the potential to be $\pu{0.780 volts}$.
Is there a way I can prove this result? I tried using the Nernst Equation, but my reaction quotient is equal to 1, which disproves my finding. I found out about activity coefficients, but I do not know how to apply it or the Nernst Equation to half cell potentials.
Any advice or help is greatly appreciated!

That's not a Cu(2+)/CuSO4 half cell, it's a Cu/CuSO4 half cell, or Cu/Cu(2+) if you like.

Do you have a single solution vessel containing copper sulfate solution and the two metal electrodes? If so, why not use two vessels and a salt bridge, like in a Daniell cell? Zinc is spontaneously oxidized by copper ions, so a single vessel cell has a short circuit, as it were.

@EdV Yes I do have a single solution vessel containing copper sulfate solution and metals copper and zinc as the electrodes. I essentially wanted to make a Daniel cell with two solution vessels, but instead my chemistry teacher suggested to me to use one solution vessel and the two metal electrodes as it would be easier to set up. However, I did not consider whether the Nernst Equation would change accordingly and I have already collected my data using one solution vessel and need to type a lab report using my collected results. How would I configure the equation for my experiment?

Unfortunately, I do not see a way to salvage this: using a single vessel was a definite problem, but the electrolyte should at least have contained both zinc ions and copper ions (from, say, the sulfate salts). Then, even though the cell was internally shorted, you could have used the Nernst equation, though not accurately at all. Instead, you start with zero zinc ion concentration, so you have an unknown, but spontaneously increasing, zinc ion concentration in solution. And the copper ion concentration is also unknown after a short time and continues to decline. Sorry!

@EdV Thanks for the Advice! I will be changing my experiment then by using two solution vessels insead.

Excellent! You have a great attitude in the face of adversity! Back in the summer of 1969, I was a 'rising sophomore' at RPI. While two of my RPI classmates went to Woodstock, I was 'enjoying' a $600 summer NSF stipend to work like a dog in a physical chemistry research lab at RPI. On the wall of the lab was a sign I have never forgotten: "If you don't have the time to do a job right, where will you find the time to do it over?" So, I am going to post a little 'answer' now, to possibly help you out a bit. It will probably be deleted by the PTB if it does not get an upvote. Best of success!

Here is a figure of the famed Daniell cell:

Note the spelling of the name: it was invented in 1836 by John Frederic Daniell. At $\pu{25 ^\circ C}$, and with $n = \pu{2 equivalents/mole}$, as correct here, $(\ln 10)\frac{RT}{nF} = \pu{0.02958 V} = \pu{29.58 mV}$.


text:
What's greater, enthalpy of combustion or heat of combustion?
What is the difference between the two? In my assignment book there is this question which asks if enthalpy of combustion is greater than, equal to or less than the heat of combustion. The answer given is that heat of combustion is greater and I don't understand why

Have they explained the difference between heat and enthalpy, particularly the condition that leads to them being equivalent? "Heat of combustion" is often used to mean "enthalpy of combustion", so I understand your confusion, but strictly speaking heat and enthalpy are different things, if sometimes equal.

In the question it's also mentioned that it's at constant volume. As far as my understanding goes, enthalpy is the total energy content and heat is basically transfer of energy. Since combustion is exothermic, energy is given to the surrounding but not all of the available energy of the substance is given. So in that case shouldn't the enthalpy of combustion be greater than heat of combustion ?

The enthalpy of combustion (aka the heat of combustion) is the enthalpy change in going stoichiometrically from pure reactants to pure product at constant temperature and pressure.  The change in internal energy between these same two states is equal to $$\Delta U=\Delta H-\Delta(PV)=\Delta H-P\Delta V$$For a reaction in the ideal gas region, this reduces to $$\Delta U=\Delta H-\Delta(PV)=\Delta H-\Delta nRT\tag{1}$$Since the internal energy of an ideal gas is independent of volume, the pressure on the gases can be changed such that the volume of the products is the same as the volume of the reactants with no further change in internal energy.  So, for combustion of ideal gases, Eqn. 1 describes the relationship between the heat effect at constant volume and the heat of combustion at constant pressure.


text:
What is a "ynoate" in the context of alkynes?
In the literature, I have read the term, "ynoate" refer to a sort of alkyne. Searching through the IUPAC gold book and Google has failed to return an explicit definition.

You also hear ynamine, ynone, ynamides, etc.

ynoate refers to a derivative of an alkyne acid. e.g. ethyl-3-butynoate  structure here
The simplest of these is propiolic acid  https://en.wikipedia.org/wiki/Propiolic_acid

How is ynoate pronounce?

Ein.... oate...


text:
What is the oxidation state of Mn in HMn(CO)5?
What is the oxidation state of $\ce{Mn}$ in $\ce{HMn(CO)5}$? If $\ce{H}$ has an oxidation state of $+1$, then $\ce{Mn}$ should have oxidation state of $-1$, which I'm not sure is possible.
On the other hand, if $\ce{H}$ has an oxidation state of $-1$, then $\ce{Mn}$ has an oxidation state of $+1$. However, Wikipedia says:

The pKa of $\ce{HMn(CO)5}$ in water is $7.1$.

This means that the $\ce{Mn(CO)5}^-$ complex exists, implying a $-1$ oxidation state for $\ce{Mn}$. Again, I am not sure that a metal can adopt a negative oxidation state.

As you can see they sure can get neg. ox. state, but because of such cases as this one, this doesn't mean all that much. There are lots of cases where getting clear values of ox. states is virtually impossible. Fun fact: here usual treating H as hydride ligand makes dissociation to be redox reaction :D

It's important to remember that oxidation states are not a fundamental law of chemistry, they're merely a useful fiction for understanding how things behave.

On negative oxidation states, in general
Although it's usually a topic that's covered relatively late in a chemistry education, negative oxidation states for transition metals[1] are actually quite alright. On the Wikipedia list of oxidation states, there are quite a number of negative oxidation states. Some textbooks have tables which only show positive oxidation states, but good authors will be careful to add a disclaimer along the lines of "only 'common' oxidation states are listed".
The key is in the ligand that the metal is paired with. In metal ions with positive oxidation states, the ligands that they pick up tend to be Lewis bases, like $\ce{H2O}$, $\ce{Cl-}$, $\ce{NH3}$, ... This should "make sense" intuitively, because the positively charged metal ion wants to gain electron density, and these ligands are happy to give electron density to it.
In $\ce{Mn(CO)5-}$, the ligand is carbon monoxide: a very poor Lewis base, but it does possess low-lying π* orbitals. In the parlance of coordination chemistry, CO is a poor σ-donor, but a great π-acceptor. This means it tends to seek out metal centres which are relatively electron-rich instead of electron-poor, and thus it makes sense that metals with a negative oxidation state are its preferred partners. [I'll defer a full explanation of the $\ce{M-CO}$ bonding to a proper inorganic textbook.]
It turns out that it's actually quite difficult to find examples of carbonyl complexes with positive oxidation states.[2] Nearly all carbonyl complexes have the metal in oxidation states of zero or lower: see, e.g. $\ce{[Cr(CO)6]}$, $\ce{[Fe(CO)4]^2-}$, $\ce{Ni(CO)4}$, and of course, $\ce{[Mn(CO)5]-}$. Another common feature is that all of these complexes obey the 18-electron rule.

On [HMn(CO)5]
So, back to the question of the oxidation state in $\ce{[HMn(CO)5]}$... IUPAC has the final say on oxidation states;[3,4] see also this writeup by Martin. Quoting from ref. 4,

Oxidation state equals the charge of an atom after its homonuclear bonds have been divided equally and heteronuclear bonds assigned to the bond partners according to Allen electronegativity [...]

Following this, we have to assign the electrons in the $\ce{Mn-H}$ bond to the hydrogen, since hydrogen is more electronegative than manganese on the Allen scale (3.30 and 2.75 respectively). There is also evidence that in metal carbonyl hydrides like these, the hydrogen bears a partial negative charge, although this argument is not without its caveats.[5] Therefore, the oxidation state of $\ce{Mn}$ should be +1. However, IUPAC offers a way out of this:

[...] except when the electronegative atom is bonded reversibly as a Lewis-acid ligand, in which case it does not obtain that bond’s electrons.

According to this, it is possible to have exceptions to the general algorithm for determining oxidation state in the previous quotation, and there is a case to be made for this exception being valid here: the electronegative atom ($\ce{H}$) is indeed bonded reversibly ($\mathrm pK_\mathrm a = 7.1$) as a Lewis acid ($\ce{H+}$). If we accept this line of argument, then $\ce{H}$ does not "obtain" the two electrons in the $\ce{Mn-H}$ bond, and the oxidation state of $\ce{Mn}$ returns to −1, which is perhaps more in line with our chemical intuition.
I leave it to the reader to draw a conclusion for themselves; but this is an excellent illustration of the most important lesson here: oxidation states have their limits, and should not always be taken as meaningful descriptors of the chemical bonding in a species, especially when ambiguity is present, since this makes their assignment essentially arbitrary. If we assign an oxidation state of −1 to hydrogen here, it doesn't magically make the complex any less acidic than it was before. And IUPAC are aware of it (ref. 4):

The applications of OS in chemistry are wide and deal with a cornucopia of chemical compounds and materials. It is therefore not surprising that, for some compounds, one value does not fit all uses, or that dedicated measurements or computations are needed to ascertain the actual OS. In those rare cases when the most convenient OS becomes a matter of choice, this fact must be clearly stated.

Refs. 3 and 4 make for excellent (and fairly accessible) reading on the topic, for those wishing to find out more about these edge cases. Both can be accessed without a subscription.

Notes and references

Negative oxidation states for non-transition metals exist (e.g. alkalides), but these are probably more "exotic" than organometallic complexes.
They do exist, for example $\ce{[Ir(CO)6]^3+}$, although the bonding situation is markedly different (as evidenced by, e.g., IR spectroscopy).
Karen, P.; McArdle, P.; Takats, J. Toward a comprehensive definition of oxidation state (IUPAC Technical Report). Pure Appl. Chem. 2014, 86 (6), 1017–1081. DOI: 10.1515/pac-2013-0505.
Karen, P.; McArdle, P.; Takats, J. Comprehensive definition of oxidation state (IUPAC Recommendations 2016). Pure Appl. Chem. 2016, 88 (8), 831–839. DOI: 10.1515/pac-2015-1204.
Sweany, R. L.; Owens, J. W. Evidence for a negative charge on hydrogen for some metal carbonyl hydrides. J. Organomet. Chem. 1983, 255 (3), 327–334. DOI: 10.1016/S0022-328X(00)99326-4.


text:
Proper (and safe) way to pressurize a vessel to around 70 kPa with gentle (~38* C) heat
I'm trying to force crystallization in a complex mixture by maintaining gentle pressure and heat. I'm thinking of using a Duran pressure plus bottle with a cap that has 2 GL-14 fittings, one attached to a compressed air pump and the other with a pressure gauge with release valve. I'm also considering Ace Glass pressure vessels. I don't have any experience working with pressurizing vessels and want people's input.
Thanks!

Atmospheric pressure is about 100 kPa, so you appear to want a pressure lower than atmospheric. Is this correct ?

Probably gauge I.e. about 0.7 barg, so 1.7 bara (absolute). I've never seen crystallisation induced by pressure though. Cooling by vacuum or vacuum distillation,  yes. Pressure, no.

Possible duplicate of Order internal pressure resistance to expect from laboratory glassware?


text:
Which of the free access molecular modelling program apply the B3LYP/G-31G(d,p) level of theory?
Please, let me know which an open free molecular modelling program will apply
the B3LYP/G-31G(d,p) density functional level of theory? Thanks.
Helen

ORCA will do it. Please consider speaking to somebody experienced before running calculations.

Also, I would recommend checking out this answer on DFT functional selection before choosing a method.

There are several free packages, see here.
Like orthocresol mentioned, you might want to be tutored by someone of experience at first. Quantum calculations are very easy to start (too easy, some might say), but harder to actually interpret correctly. The actual method can also greatly influence the results and needs to be chosen carefully (for example, B3LYP has fallen out of favor after some weird behaviours have been observed (Why the Standard B3LYP/6-31G* Model Chemistry Should Not Be Used in DFT Calculations of Molecular Thermochemistry: Understanding and Correcting the Problem); consider M06 or M062X as good alternatives)


text:
Why does nitrogen molecule have greater ionization energy than nitrogen atom?
I am studying an article in my book that $\ce{N2}$ molecule has higher ionisation than $\ce{N}$ atom and I have no clue it may be because of strong triply bonded atom then why this is not applicable to $\ce{F2}$, $\ce{F}$ and $\ce{O2}$, $\ce{O}$. Also explain and with reason for $\ce{N2}.$

I think you have a good point here, this is unusual.

And here's answer http://www.vias.org/genchem/chembonding_14777_04.html


text:
Steric Inhibition of Resonance in ortho-substituted Benzoic Acids
It is given that due to Steric Inhibition of Resonance (SIR), the acidity of ortho-substituted benzoic acid increases with respect to benzoic acid. However, I'm unable to understand the reason behind this fact.
We know that resonance stabilises a molecule and the absence of which decreases the stability. In case of ortho substituted benzoic acids, resonance is inhibited due to the loss of planarity due to the steric interactions from the ortho substituent.
If this is the case then the conjugate base of the acid will not be stabilized by resonance with the benzene ring. So it seems the ortho-substituted benzoic acid will not readily donate H+ and is less acidic than benzoic acid, contrary to what is being given in various sources.
Then kindly explain why is ortho-substituted benzoic acid is more acidic than benzoic acid.

which ortho substituent are you refering to?

Think of a localized, in plane, benzoate conjugate base. The carbonyl can busy itself by resonance with the ring or resonate with the localized charge-bearing oxygen. The carbonyl is splitting its effort. But in the out-of-plane conformation (say, from o-toluic acid) the ring resonance is lost and the carbonyl can now focus on stabilizing the charge on oxygen. On this basis, the out-of-plane acid would be predicted to be more acidic..

@Waylander  Ortho-substituents like alkyl group(methyl,ethyl,etc.), halogens(F,Cl,Br,I), etc., where the SIR effect depends on the size of the substituent. The substituent I referred does not include small ones like hydroxyl or amine groups.

@user55119  Ok I understood your point. But we know resonance is a stabilizing phenomenon and more the number of resonance structures, more must be the stability right. This one confuses me. Kindly guide me.

Ortho effect is discussed in the following link . https://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases&ved=2ahUKEwj1z9205eHjAhUJL48KHVdoDqsQjjgwG3oECAoQAQ&usg=AOvVaw1YoZ5Nl4BKSodJUJesaS4y

Here resonance is destabilizing as resonance here would mean that two oxygens will have negative charge and that would destabilize the conjugate base.

@ChakravarthyKalyan: Thanks. It was informative but didn't clarify my doubt.

In case of ortho substituted benzoic acids, resonance is inhibited due to the loss of planarity due to the steric interactions from the ortho substituent. This is incorrect for most substituents in 2 position. SIR is a very fuzzy subject and it is based on false premises. All steric effects are essentially a consequence of electronic effects; it is therefore a balance between electronic effects to begin with. It is unfortunately more complicated than what some books make it out to be.

Important information before understanding the reason for steric inhibition of resonance and the ortho effect
In general, resonance stabilizes a molecule. But in the case of carboxylic acids, the conjugate base (anion after losing $\ce{H+}$) there are two equivalent resonance structures.
Take the example of acetic acid, $\ce{CH3COOH}$, and thus acetate ion, $\ce{CH3COO-}$:

The hybrid would look something like this:

It could be easily seen that equivalent resonance structures mostly stabilise the hybrid greatly.
Therefore, equivalent resonance tends to stabilize a molecule/compound more than normal resonance. (in most cases)
Now let's take the case of o-toluic acid.

There are two ways by which resonance can take place:

In (I), it takes place through $\ce{-COOH}$. Its ion is expected to be more stable due to the above result. Also, here $\ce{OH}$ is highly polar and it will tend to lose $\ce{H+}$ to become uncharged and thus be more stable (as compared to $\ce{O-}$).
In (II), it takes place through the ring. Its ion is less stable due to more formal charge and less aromatic character (due to + on the ring). Also, $\ce{OH}$ is less polar than $\ce{H+}$ due to the above point.

Therefore, it will be better to prevent the taking place of resonance through ring.
That is done by $\ce{-CH3}$ group which provides bulky group repulsion and takes the ring out of the plane of $\ce{-COOH}$ and thus providing condition for equivalent resonance in $\ce{-COOH}$.

If you want more explanation on Ortho-Effect. Visit here:- https://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases?noredirect=1&lq=1

I've learnt that more the number of resonance structures more stable the compound is. I understood equivalent resonance structures are possible in all circumstances (whether there is SIR or not). But I don't understand why the molecule gets destabilised if we count the resonance between the carboxylic group and the ring. However, I'm glad that you understood my question properly.

Its not destabilised, its just less stabilised. The compound would prefer to be more stable rather than being less stable.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup, and removed some of the obnoxious markup. You can find more about editing here.

No problem, unfortunately I have to tell you, that what you write is not correct. The 'equivalent resonance' is a widely propagated myth, it doesn't hold up to any scientific rigor. Resonance in itself is only a result from our inability to grasp the actual bonding description.

@Martin-マーチン But it is usually taught to us in this way, even written in books. Equivalent Resonance Str. tend to stabilize the hybrid. What is the real reason then for Carboxylic Acids to be quite acidic?

It is very unfortunate that this has not yet been purged from the syllabus and textbooks. It remains a popular concept, even though it is of little to no use. It is a post-hoc rationalisation based on a few cases, which cannot be generalised; it uses reverse argumentation (i.e. it presents the reasons for something as the conclusion of something else).

@Martin-マーチン If a person has difficulty in understanding a concept, complicating it for them even more is not the step! You needn't bring down every "novice user" by saying "It is very unfortunate that this has not yet been purged from the syllabus and textbooks". Either leave a constructive comment, or give a detailed answer as to WHY it's wrong, it helps imrove morale of such users.

@BongoMan this is an answer and as an answer it should be correct. It isn't. I don't think the author has had any difficulty understanding what they had written. It also is not their fault to be taught a wrong theory. However, it shouldn't be taught to others as the truth. If they wanted to have a detailed explanation, they could ask a question. The comment section is not the place for an elaborate correction. So in my mind, it was reasonable, constructive criticism and an explanation for my down vote. This is so often demanded here.


text:
When diagonalising the overlap matrix, what happens if the square root of the diagonal matrix does not exist?
I was looking at the code required for a simple RHF computation when I came across the section for the diagonalisation of the Fock Matrix.
Based on what I know, the diagonal matrix $\mathbf{s}$ is evaluated as
$$\mathbf{s} = \mathbf{U}^\dagger\mathbf{S}\mathbf{U},$$
where $\mathbf{S}$ is the overlap matrix.
The matrix $\mathbf{X}$ is then evaluated as
$$\mathbf{X} = \mathbf{U}\mathbf{s}^{-1/2}\mathbf{U}^\dagger$$
and the diagonalised Fock operator $\mathbf{F'}$ as
$$\mathbf{F'} = \mathbf{X}^\dagger\mathbf{F}\mathbf{X}.$$
But what happens if the matrix $\mathbf{s}$ has terms that are negative? What happens then? Do we simply take the magnitude and carry on?
Thanks!

The overlap matrix is positive definite (https://en.wikipedia.org/wiki/Definiteness_of_a_matrix). This means all its eigenvalues, the elements of the diagonal matrix you give above, are positive. Thus what you describe can not happen in real arithmetic, though there may be issues on computers where you are using floating point. When I have more time I'll write a proper answer

The proof can be found here

@JanJensen FYI the .dk books.google link appears not to work outside Denmark

@MichaelBanck ah, ok. The book is "Simple Theorems, Proofs, and Derivations in Quantum Chemistry" by Istvan Mayer and the proof appears on page 63

It works in .uk

The matrix $\mathbf{s}$ in your question are the eigenvalues of the overlap matrix $\mathbf{S}$, further the matrix $\mathbf{U}$ are a matrix of the eigenvector of the overlap matrix.
Now looking at the equation that gives you concern:
$$\mathbf{X} = \mathbf{U}\mathbf{s}^{-1/2}\mathbf{U}^\dagger$$
Since $\mathbf{s}$ is a diagonal matrix then taking the power of $\mathbf{s}$ is just taking the power of every element:
$$\mathbf{s}^{-1/2}=\begin{bmatrix}s_1&0&...&0\\0&s_2&...&...\\...&...&...&0\\0&...&0&s_n\end{bmatrix}^{-1/2}=\begin{bmatrix}s_1^{-1/2}&0&...&0\\0&s_2^{-1/2}&...&...\\...&...&...&0\\0&...&0&s_n^{-1/2}\end{bmatrix}$$
The only way as you noted this can cause trouble is if any of the elements of $\mathbf{s}$ are negative.
Now here we have to remember that $\mathbf{s}$ are the eigenvalues of the overlap matrix, it is therefore guareenteed that all of the elements will be equal to or greater than zero!
For the proof of this see my answer to one of your other questions, Basis sets - negative contraction coefficients?


text:
Synthesis of potassium trioxalatocobaltate(III) trihydrate

I've synthesized the compound potassium trioxalatocobaltate(III) trihydrate and for it I have two reaction equations. One of them is:
$$\ce{CoCO3 + 3K2C2O4.2H2O -> K2[Co(C2O4)3] + 2CO2}$$
and the other:
$$\ce{K2[Co(C2O4)3] + PbO2 -> K3[Co(C2O4)3].3H2O}$$
I see that the lead dioxide oxidizes $\ce{Co^2+}$ to $\ce{Co^3+}$ in the second step but I can't figure out what happens in the first. Does cobalt(II) make a ligand complex with the oxalate group?
One of the beginning steps was a also to mix oxalic acid and potassium oxalate. I assume (but I am not sure) that it was done to dissolve the potassium oxalate as other acids might have led to an unwanted reaction that would hinder the reaction following reaction with $\ce{CoCO3}$. I wanted to know that this was in fact the case.

Your first equation is not balanced. Your second equation is not correct? How do you explain $\ce{Pb^4+ -> K+}$ conversion?

@MathewMahindaratne for the first reaction you could also write it as $\ce{Co^2+ + 3C2O4^2- -> [Co(C2O4)3]^4- }$. I took the information from https://bit.ly/2OxsNi1. The second one might be incorrect because of the number of potassium on the product it would have to be K4 meaning the oxidation of Co goes from 2+ to 3+. I'm not 100% sure either so that;s why I asked the question.

Is the iupac naming correct? Shouldn't it be tris(oxalato)?

It appears that Sørensen was the first to publish the synthesis of trioxalatocobaltate(III) trihydrate [1].
A summary written in English (and that should answer both of your questions) including chemical reactions can be found in the very first volume of Inorganic Syntheses1 [2, p. 37]:

C. POTASSIUM TRIOXALATOCOBALTIATE
$$\ce{K3[Co(C2O4)3]·3H2O}$$
$$\ce{H2C2O4 + CoCO3 → CoC2O4 + H2O + CO2↑}$$
$$\ce{2 CoC2O4 + 4 K2C2O4 + PbO2 + 4 HC2H3O2 → 2 K3[Co(C2O4)3] + 2 KC2H3O2 + Pb(C2H3O2)2 + 2 H2O}$$
Twenty-three and eight-tenths grams (0.2 mol) of cobalt carbonate is dissolved in a solution of 25.2 g. of oxalic acid $(\ce{H2C2O4·2H2O})$ and 73.7 g. of potassium oxalate $(\ce{K2CrO4·H2O})$ in 500 cc. of hot water.
When the solution has cooled to 40°C., while it is vigorously stirred, 23.9 g. of lead dioxide (see synthesis 16) is added slowly, followed by 25 ml. of glacial acetic acid added a drop at a time.
The stirring is continued for an hour, during which time the color changes from red to deep green.
After the unused lead dioxide is filtered out, the trioxalatocobaltiate is precipitated by the addition of 500 ml. of alcohol.
The material appears as emerald-green needles, which are sensitive to both light and heat; The yield is 70 g. (71 per cent).

Notes

Please disregard obsolete typography and notations used in this textbook such as full stops after unit symbols, as well as the lack of white space between the numerical value and a degree symbol (as seen in “40°C.”) and use of notations such as “cc.” (cubic centimeter).

References

Sörensen, S. P. L. Kritische Präparatenstudien. III. Darstellung von Einigen Kobaltidverbindungen. Z. Anorg. Chem. 1896, 11 (1), 1–5. https://doi.org/10/fh47nq. (in German)
Inorganic Syntheses; Booth, H. S., Ed.; John Wiley & Sons, Inc.: Hoboken, NJ, USA, 1939; Vol. 1. ISBN:9780470131602, https://doi.org/10.1002/9780470132326.

In the first step, one makes an oxalate complex by decomposing the carbonate salt of cobalt (II) with oxalic acid. I am not sure how an oxalate salt will decompose the carbonate to carbon dioxide. Oxalate ion makes a complex. The second step is lead dioxide oxidation.
See this book chapter for proper synthesis equations of cobalt (III) oxalate complex, search PbO2 in the file.
https://link.springer.com/content/pdf/10.1007%2FBFb0048297.pdf

OP mentioned using oxalic acid too.


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What is Liebermann nitroso test?
From the articles in the internet , I found that there are two reactions which are being called Lieberman nitroso test .
One is the test of 2 degree amine in which a 2 degree amine is converted to N - nitroso amine using nitrous acid  . And this product reacts with phenol to form a red solution which turn blue on addition of alkali . Confirming the presence of 2 degree amine .
And another is the direct reaction of phenol with nitrous acid to form Indo-phenol giving same characteristic red and blue colour .
I am confused and have tried to refer to a lot of sources but with no success .

Yes, it is a test for both phenol and secondary amines. See: https://www.scribd.com/document/369333943/Liebermann-Nitroso-Test-and-Ninhydrin-Test


text:
Further reading on photoconductivity of hemoglobin
I recently ran into this paper which describe experiments that confirm that hemoglobin is photoconductive. Finding this fact as incredible, I wanted to read up more on photoconductive properties of hemoglobin. The paper does not discuss what the conductivity number is, that would give some estimate of charge carrier density created when hemoglobin is illuminated. Do you have any estimates of how conductive hemoglobin gets? Is it close to highly doped semiconductors? Also, if you can provide links to papers that discuss more on photoconductivity of hemoglobin, that would be appreciated, as Google only showed me this sole paper from 1963.

Try searching on Google Scholar. I couldn't find any such paper describing specifically about haemoglobin's photoconductivity except the one by RC Nelson you stated above.


text:
The phlogiston theory
I have looked this theory up on google and found that the theory is that when metal is burnt calx (metal oxide) and phlogiston is created (metal --> calx + phlogiston). Then I found out that phlogiston is a substance supposed by 18th century scientists to exist in all combustible bodies and to be released during combustion. But it later got disproved by Antoine Lavoisier who proved that metal and oxygen when burnt produced calx.
So my questions are:

What exactly is the phlogiston theory and if my understanding of it correct?
What exactly did Georg Ernst Stahl (discovered the phlogiston theory) think phlogiston was?
Why did the 18th century scientists believe that phlogiston existed in all combustible bodies and why not just a product created during the combustion reaction?

1.)"Phlogiston" is, as far as I know, not a scientific theory, because it does not allow (or even attempt) to make any quantitative predictions. A philosophical setup much like Demokrit's atoms https://chemistry.stackexchange.com/questions/74227/democritus-vs-john-dalton-atomic-theory/74230#74230 , only it's actually wrong.

Phlogiston has a negative mass! Metallurgists of the 18th century certainly knew that the calx of a metal weighed more than the base metal. Ironically, Priestley showed that a sealed vessel containing HgO formed Hg but did not change its weight but was, nonetheless,  a proponent of the phlogiston theory.

The phlogiston theory and some of its history have been discussed over at the History of Science and Mathematics stack exchange community.

Don't rely on the random webpages and even Wikipedia. Wikipedia is a good starting point but not the end. Start from Google Scholar which brings up relatively more authentic publications.
Also a quick search of the Journal of Chemical Education brings all the topics you will ever need on phlogiston theory and how it died. https://pubs.acs.org/action/doSearch?AllField=Phlogiston+theory&SeriesKey=jceda8

@Karl No expert on the subject, but unless the wikipedia is wrong, your statement is: https://en.wikipedia.org/wiki/Phlogiston_theory#Challenge_and_demise :-)

I'm voting to close this question as off-topic because it is not about chemistry but instead about a long-discarded historical "theory".

@JonCuster Well, there is a history-of-chemistry tag! I see nothing wrong with this question.

Refer to the book "The Structure of Scientific Revolutions ( fourth edition 2012). It is a book about the history of science by the philosopher Thomas S. Kuhn. To see why and how the scientists abandoned this theory.

It's off topic. And not a valid scientific theory.

@BuckThorn - perhaps more for meta, but at some level I think there is a limit to what on-topic chemical history consists of. History of how currently accepted chemistry came to be, yes. Why any of a myriad of discarded ideas arose or were discarded, not so much. While interesting, the second are more for History of Science and Mathematics to my taste.

@blupotatos It is not off topic. It is about the history of chemistry and a very important part of that history too. A good explanation of why the phlogiston theory was rejected (especially the empirical evidence) would make a good and useful answer here that would provide useful material about how chemistry works and how modern chemistry originated.


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Will a Lewis acid catalyze the reaction of HBr with primary alcohols?
It is well-established that a Lewis acid catalyst (popularly $\ce{ZnCl2})$ is necessary to form an alkyl chloride when reacting $\ce{HCl}$ with a primary alcohol. This actually changes the mechanism: the Lewis acid complexes with the alcohol and the resulting leaving group is $\ce{[Zn(OH)Cl2]-}$ instead of water. For $\ce{HBr}$ and $\ce{HI}$ this is technically not necessary.
However, practically speaking $\ce{HBr}$ alone is also insufficient as its azeotrope is insufficiently concentrated. Generally one adds conc. sulfuric acid to increase the acidity.
I am curious as to whether the addition of a Lewis acid could be used instead. Bromide is a stronger nucleophile (in protic solvents) than chloride, so I expect that it would. Has anyone ever tried this?


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Why CO is a stronger and more common ligand than N2?
Both N2 and CO are considered sigma-donor and pi-acceptor. Their MO diagram is similar, so I wonder why CO binds generally more strongly and it is a more common ligand.

related https://chemistry.stackexchange.com/questions/93797/why-do-nitrogen-molecules-not-act-as-ligands-in-haemoglobin

Both the sigma HOMO and the pi LUMO of CO are polarised towards carbon - so the overlap with metal orbitals is much better

I saw the link and the other question doesn't have an answer jet. I saw they asked about N2 and O2, I think it is a different explanation because N2 and O2 have very different MO instead N2 and CO are very similar

@orthocresol - do you mean that is because CO is polar? (2 different atoms)

 The answer to the above question is attached.Read M.Tarr

Even though it is basically the reversed explanation, it might still be helpful: How to rationalise with MO theory that CO is a two-electron donor through carbon?


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A more efficient method for electrolysis?
If you made the anode and cathode both out of a single piece of zeolite that has been coated inside with graphite (or graphene) wouldn’t that provide a conductive path and greater surface area thus making electrolysis of water more efficient?

You don't really need greater surface.

That's a good question, but keep in mind there is still a diffusion barrier. Zeolites use to trap stuff, and for an effective electrolysis you need the products to be exchanged rapidly.


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Enzyme Kinetics
How come enzymes with a lower $K_\mathrm{m}$ are more easily saturated with substrate than an enzyme with higher $K_\mathrm{m}?$
Is it because enzymes with higher $K_\mathrm{m}$ are able to bind substrate and hold onto it tighter and therefore have a higher chance of becoming a product?
I'm confused because shouldn't a lower $K_\mathrm{m}$ indicate that an enzyme is able to catalyze a reaction quicker?

The Michaelis–Menten kinetics model formula is:
$$ V= \frac{(d¦P¦)}{dt} = \frac{(Vmax)¦S¦}{Km + ¦S¦}$$
Reaction rate V is the rate of formation of product, ¦P¦ is the concentration of product and the ¦S¦ the concentration of a substrate S.
In a saturated regime V = Vmax and this occours only if:
$$ \frac{¦S¦}{Km + ¦S¦} \approx  1 $$
This is nearly equal to 1 only when
$$ ¦S¦ >>  Km $$
In conclusion: If Km is small then a lower concentration of substrate ¦S¦ is required to saturate the enzyme.

ahh, you're right, thank you! the two comments on my question really helped.

Application of the Michaelis-Menten equation generally assumes steady state concentration of the enzymatic complex intermediate. Then $K_M$ allows two interpretations$^1$:

an apparent dissociation constant describing the equilibrium $$\ce{ES<=>E +S}\tag{1}$$
where $E$ and $S$ are the concentrations of enzyme and substrate, respectively
the concentration of substrate at which the rate of formation of products $\nu=\frac{V_{max}}{2}$ where $V_{max}$ is the maximum (limiting) rate at saturation with substrate.

Note that in the case of Michaelis-Menten kinetics, $$V_{max} = k_{cat}[E]_0\tag{2}$$ where $k_{cat}$ is the rate constant for conversion of substrate into product: $$\ce{ES->[k_{cat}]E + P}\tag{3}$$ Equation (2) follows from the general Michaelis-Menten expression at saturation since all of the enzyme is then converted into intermediate ($[ES]=[E]_0$).
In light of these definitions, an enzyme with a lower $K_M$ saturates at a lower concentration of substrate, ie the intermediate complex is more favored.

Is it because enzymes with higher Km are able to bind substrate and hold onto it tighter and therefore have a higher chance of becoming a product?

Beware how you think about this. Larger $K_M$ means enzyme-substrate complex is less favored.
When the intermediate complex $ES$ is more thermodynamically stable than the free enzyme, the equilibrium is shifted toward the intermediate, leading to faster turnover. There is more of the complex to "leak" into product. If this is what you mean by "tighter", then yes, "tighter" leads to more product.

I'm confused because shouldn't a lower Km indicate that an enzyme is able to catalyze a reaction quicker?

All else being equal, and at steady-state, yes.
References

Fersht, Alan. Structure and Mechanism in Protein Science. New York: Freeman, 1999.

this was a good answer! thank you! it helped me look at Km in a different light


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Why does zinc sulfide glow when hit by electrons?
In cathode ray tube experiment in order to check the direction of flow of electrons a hole was made in a note and behind it phosphorescent material zinc sulfide it was coated.
Why does zinc sulfide glow when hit by electrons?
What causes this glow?

Related: Why does zinc sulfide glow?; Why did scientists use a zinc-sulfide coated screen to detect the alpha, beta, and gamma radiation?; If cathode rays travel from the cathode to the anode how do they make zinc sulphide glow?

This is an example of phosphorescence. The Wikipedia page of zinc sulfide and phosphorescence can explain it better than I do, but in short, when zinc sulfide get hit by electrons, electrons transfer some of its energy to zinc sulfide and excites its electron. Zinc sulfide stores the energy for a while, then emit light when the electron goes back to its ground level.

And it is cheaper than diamond ( that also glows).


text:
Stereochemistry of C as non-chiral centre
When a C of a molecule is not a chiral centre, the spatial orientation of its substituent group doesn't affect its stereochemical specification, does it?
Ex:

Also note that hashed or wedged bond should not be used for nonstereogenic atoms. Plain bonds should be used for nonstereogenic atoms, even if the substituents are not physically coplanar.

Actually it depends on the message that must be conveyed. @Loong◆ Geometry can/must be detailed if necessary. Which is not the case for most drawing of reactions, this is true.


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Highly concentrated NaOH solution at high temperature vs borosilicate glassware and cellulose?
I've read here on stack-exchange that NaOH will react with glass through the following reaction:
4 NaOH + SiO2 ⤑ Na4SiO4 + 2 H2O
I would like to briefly heat a highly concentrated NaOH solution to 140-200C over the course of a few minutes just to pass it through a filter.
Beside safety concerns, several questions come up to my mind:
1) Do I need to worry that an NaOH solution at 140-200C could dissolve more than say max 0.5ppm Boron, Aluminium and Na4SiO4 from the glassware to my solution over the course of less than 10 minutes? or is that too short of a time to be a cause of concern?
2) Could it cause problems with my glass-joints over such a short period of time?
3) I think that cellulose doesn't chemically react with NaOH. As for the temperature, wikipedia reads "whereas starch undergoes a crystalline to amorphous transition when heated beyond 60–70 °C in water (as in cooking), cellulose requires a temperature of 320 °C and pressure of 25 MPa to become amorphous in water." In light of that and of the aforementioned reaction between NaOH and glass, am I right to assume a cellulose filter would be better than fritted glass disk or fiber glass filter and suitable for the operation or would the solution risk burning or dissolving my filter?
4) I plan on keeping my filtering apparatus around 90-110C thanks to a steam bath until right before proceeding with the filtration to minimise the incremental dissipation that will occur upon contacting the solution. Wikipedia reads "The temperature differential that borosilicate glass CAN withstand before fracturing is about 165 °C (329 °F)."  From this, is it reasonable to assume my filtration apparatus made of borosilicate glass WILL take the remaining temperature difference without cracking? (there might be up to max 120C difference)
5) I have an A2B2E2K2P3 cartridge for my full face mask. Does the P3 part protect against NaOH fumes?
6) Would you have any advice with respect to this prospective operation? (other than comments such as "don't do it" or the like)

See also: https://chemistry.stackexchange.com/questions/53760/concentration-of-naoh-to-dissolve-glass

@BuckThorn Thanks! That's very good info with respect to the reaction rate between glass and NaOH. This reaction is much less negligible than I thought it was, especially at high temperatures.

You must be filtering under pressure if above about 145 degree C as you'll be above the boiling point. If it's brief enough, you might get away with using stainless steel filtration kit. Otherwise hastelloy /Monel but that may be cost prohibitive

@Beerhunter Thanks for the info! Do you know if maybe NaOH could react with some of the chromium and iron in the steel and contaminate the solution with trace amounts thereof? Or does the alloy somehow prevent this?

@Hans I have a link for a graph that may help https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.bssa.org.uk/topics.php%3Farticle%3D34&ved=2ahUKEwjQ863b_vjjAhUITxUIHbHtCrYQFjAKegQIAhAB&usg=AOvVaw3dC0WzpFUXyEtzUtQHhrKv You can ens up with stress cracking, implying corrosion which is usually an alteration of the metal chemistry itself I would have thought.


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Why does the tautomerism of purine favor 9H-purine?
I tried to explain the equilibrium by comparing the basicity of different nitrogen atoms in a purine anion, but I can’t see any obvious difference. Seems like that the 9-N will be the most basic because it can have two adjacent imine structure while keeping the structure of pyrimidine untouched. Computational results will be appreciated.

P.S. I don’t think the question is a duplicate of the “purine nitrogen basicity” one because that one compares the basicity of nitrogens in a neutral purine molecule whereas my question compares the basicity in a purine anion.

The given purine has two structures as shown below.

In $\ce{9H–Purine}$ (figure 2), lone pairs are far apart from each other (3 and 7). Comparatively in $\ce{7H–Purine}$ (figure 1) lone pairs are very near each other (3 and 9) . Nearness of lone pairs in $\ce{7H–Purine}$ could leed to greater repulsions between them. This could make it energetically unfavorable in comparison to $\ce{9H–Purine}$.
This could be the reason why $\ce{9H–Purine}$ is more favored tautomer compared to $\ce{7H–Purine}$.


text:
Understanding the solubility of Ca(HCO3)2
According to wikipedia $\ce{Ca(HCO3)2}$ has the following solubility values:
$\pu{16.1 g/100 mL} \,(\pu{0^\circ C)}$
$\pu{16.6 g/100 mL} \,(\pu{20^\circ C)}$
$\pu{18.4 g/100 mL} \,(\pu{100^\circ C)}$
So I assume $\ce{Ca(HCO3)2}$ would precipitate beyond this level. However,  the wikipedia page for calcium bicarbonate states that attempts to prepare compounds such as solid calcium bicarbonate by evaporating its solution to dryness invariably yield instead the solid $\ce{CaCO3}$:
$\ce{Ca(HCO3)2(aq) → CO2(g) + H2O(l) + CaCO3(s)}$
which leaves me wondering if, when the solution is saturated beyond ~$\pu{16-18 g/100 mL}$, a precipitate of $\ce{Ca(HCO3)2}$ will exist  in solution? or if it also dissociating beyond that point while still in solution?

What is the reference to solubility at 100 degrees?  This value could very well be the equilibrium value of dissolved calcium bicarbonate which is in equilibrium with solid calcium carbonate at boiling temperatures. Until and unless we don't know how it was measured, it may be futile to discuss this value.

Surprisingly, SciFinder is quiet about its solubility and it is only Wikipedia which quotes these values. No reference sadly!

Yes, all three values are from wikipedia. I edited the text to mention it. But as you write, wikipedia doesn't cite any source... : /

I found one encyclopedia which quoted the same values without any reference. Most likely copied from Wikipedia without any reference. This is the state of affairs these days. There is German work on the solubility of calcium bicarbonate, however it is in sodium chloride solutions.

Left on note on Wikipedia calcium bicarbonate page to add citations.

For low temperatures, the solubility value  may get obtained by mixing  $\ce{2 MHCO3 + CaCl2}$, being formally equivalent to $\ce{Ca(HCO3)2 + 2 MCl}$, where $\ce{M}$ = $\ce{Na}$ or  $\ce{K}$, observing $\ce{CaCO3}$ precipitation.

Calcium bicarbonate $\ce{Ca(HCO3)2}$ simply does not exist. It is impossible to fill up any container with pure $\ce{Ca(HCO3)2}$. Nobody has ever been able to produce a powder containing $\ce{Ca(HCO3)2}$. It is possible and easy to produce a solution containing the ions $\ce{Ca^{2+}}$ and the ions $\ce{HCO3^-}$, by bubbling $\ce{CO2}$ into a solution of calcium hydroxyde $\ce{Ca(OH)2}$. The following reaction will happen:
$$\ce{Ca(OH)2 + 2CO2 -> Ca^{2+} + 2HCO3^-}$$
The result is exactly what could be expected if some $\ce{Ca(HCO3)2}$ had been dissolved in water. And of course, it makes sense to admit that by evaporating the solution, the calcium bicarbonate could be obtained dry. Unfortunately it is not the case. When trying to evaporate this solution, $\ce{Ca(HCO3)2}$ is always and totally decomposed according to the following equation:
$$\ce{Ca^{2+} + 2HCO3^- -> CaCO3 + H2O + CO2}$$
And all you obtain by trying to evaporate a solution containing he ions $\ce{Ca^{2+}}$ and the ions $\ce{HCO3^-}$ is a deposit of calcium carbonate $\ce{CaCO3}$, whatever the method used for carrying out this evaporation.
Of course you can calculate the maximum amount of Calcium and of $\ce{HCO3^-}$ ions that may exist simultaneously in such a solution. You may even calculate the solubility of $\ce{Ca(HCO3)2}$. But this does not prove that $\ce{Ca(HCO3)2}$ exists as a pure compound. $\ce{Ca(HCO3)2}$ is a compound that only exist in solutions.

I would assume that when calcium bicarbonate dissolves in water, the bicarbonate reacts to form water and carbon dioxide gas. With the carbon dioxide gas evaporating, you're inevitably left with only calcium and carbonate ions in the solution, so you can't precipitate calcium bicarbonate out of the solution.

The CRC Handbook does not list solubility numbers for $\ce{Ca(HCO3)2}$ or $\ce{Mg(HCO3)2}$. We know that temporary hard water has Mg and/or Ca salts of carbonic acid that will precipitate in hot water heaters at ~$\pu{60 °C}$, which appears to be due to forced evolution of $\ce{CO2}$ rather than an inverse solubility curve.
The solubility of $\ce{CO2}$ ($\pu{0.058 g/100mL}$ at $\pu{60 °C}$) would appear to be insufficient to dissolve ~$\pu{10 g}$ of $\ce{CaCO3}$, so I question the solubility number of $\pu{18 g}$ at $\pu{100 °C}$.
The likely reason for being unable to isolate solid $\ce{Ca(HCO3)2}$ is that $\ce{HCO3-}$ ion is in a labile equilibrium with $\ce{CO2(aq)}$ and $\ce{OH-(aq)}$,
$$\ce{2HCO3- -> H2CO3 + CO3^2-}$$
At higher temperatures, the $\ce{CO2}$ simply escapes by evaporation rather than because of equilibrium processes. Stalactites form at cold temperatures because the $\ce{Ca(HCO3)2}$ solution is exposed to a large surface where $\ce{H2O}$ and $\ce{CO2}$ evaporate leaving $\ce{CaCO3}$ behind.

"The solubility of CO2 (0.058 g/100mL @ 60 C) would appear to be insufficient to dissolve ~10 g of CaCO3" wrong. One can dissolve way more CO2 in water in presence of CaCO3.  Pressure can also be applied and it is good to know at what pressure the solubility is measured.

@James You are an experienced user. We expect you to learn chemical formatting and typeset so that your posts looks better. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


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Why is p-methoxyphenol less acidic than phenol?
They have slightly different $\mathrm{p}K_\mathrm{a}$ values of 9.9 (phenol) vs 10.2 (p-methoxyphenol).
They both have the same number of resonance structures too.
My guess is that the resonance structure for p-methoxyphenol with the lone pair on the para carbon contributes less to the stabilisation of the negative charge since it's in an electron rich area (near the lone pairs of the methoxy group oxygen).
Is this a valid explanation? If not, what is the true reason?

related https://chemistry.stackexchange.com/questions/27361/why-is-p-methoxyphenol-more-acidic-than-p-methylphenol

Deprotonation creates a negative charge on the phenolic oxygen. The better you can stabilise this charge, the more favored this process is going to be.
You are right in saying that the charge can be stabilised by delocalisation in the aromatic ring. However, the electrons of the methoxy oxygen can also be delocalised in the ring (to a much lesser extend, of course). This has the effect of making the ring richer in electrons. It will thus tend to less receive additional electrons, and thus stabilise less the formal negative charge. (However, keep in mind that the ring does still stabilise the negative charge, making p-methoxyphenol still much more acidic than alcohols with pKa values around 15)


text:
Stritations in a discharge tube
I am reading about A cathode ray discharge tube and I still do not fully understand why that as you keep on lowering the pressure in the chamber it creates striations, Faraday's dark space and Crooke's dark space and why it does not just stay as a constant stream of light? Also why does the gas have to be at a low pressure in the first place for the constant column of light to appear?
To verify my questions are:

Why when you lower the pressure to a certain extent you can see striations, Faraday's dark space and Crooke's dark space and why doesn't it just stay as a constant column of light?
Why does the pressure in the chamber have to be very low for you to be able to see a constant column of light?

The most important question is why do we need low pressures? If you recall kinetic theory of gases, there is a concept of mean free path. This quantity roughly tells us the distance a particle would travel before colliding with another particle. If you want the electrons to travel from one end to the other, you need to provide a large mean free path to them. How can one obtain that? Simply by removing "matter" from the cathode ray tube. Practically this is achieved by vacuum pumps. Although nobody gives importance to vacuum pumps, it was this technology which allowed great progress in the field of cathode ray experiments and hence understanding matter.
Now, the "glow" phenomenon is rather a spectroscopic problem. Electron don't glow green/pink etc, it is the matter which starts to glow when it interacts with electron. The glow can originate from several sources such as recombination phenomenon, fluorescence of the glass, a fluorescent zinc sulfide screen placed inside, or even Xrays. When the pressures are moderate, other phenomena like which you quote as striations, appear. This arises from accelerated electrons exciting the gas particles and then losing energy (dark space). This video explains some of it. Cathode ray striations. However it is not a trivial phenomena. It took ages to understand it.
The observation of striations indicate that there is enough gas (low vacuum) in the discharge tube.


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Will N,N‐diethylbenzamide show geometrical isomerism?

I know for a compound to show geometrical isomerism it should

have site of restricted rotation;
be connected to two different groups on either side of site of restricted rotation;
the distance between the groups on the either side of site of restricted rotation should be different.

I think conditions 1 and 3 hold true here. I am confused if or not 2 is true.

(1) "Restricted rotation" is not a black or white thing; different things have different barriers to rotation. (2) What makes you think that one side of a benzene ring could be different from the other side?

@orthocresol An empty p orbital , on the other side there is p orbital involved in π bonding

Firstly, resonance forms don't exist individually; but that's not the main problem. More crucially, $$\ce{C+-C=C-C=C <-> C=C-C=C-C+}$$

@Orthocresol so that suggests the two group on one side of double bond same ?

https://i.stack.imgur.com/XR9J7.png - there is no way you can treat two sides of a benzene ring differently, ever. And there is no "double bond" there, by the way. It is only there in certain resonance forms, which contribute only a tiny bit to the overall resonance hybrid. Which brings us back to the point that restricted rotation is not something that can be answered with only a yes or no.

@orthocresol Thank you ! Now I understand that there are no two different groups. I understand that the structure is contributing very less to the resonance hybrid but my teacher had said that we consider a molecule to show GI even if any of its resonating structure shows GI

Should really be looking at amide-iminol tautomerism for this compound


text:
Basis sets - negative contraction coefficients?
I was looking through Basis Set Exchange when I noticed quite a few basis sets (STO-3G to STO-6G, for example) have negative contraction coefficients for some primitive Gaussians. As (based on what I know) the overlap between two STO-nG basis functions is equal to the sum of the overlap integrals between the pGTOs used to approximate the STO, this means that the diagonalised overlap matrix may have some negative elements.
If this happens, what would the program do?

It would do whatever it normally does, why do you think negative elements are a problem? The overlap matrix is positive definite, which means its eigenvalues are positive, NOT that all its elements are positive. For instance consider the 2x2 symmetric matrix with 1 down the diagonals and -0.5 in the corners - that has eigenvalues 1.5 and 0.5 and so is positive definite despite the negative elements of the matrix.

I meant the diagonalised overlap matrix, not the inverse of its square root.

The overlap matrix is positive definite. The diagonalised form is one with the eigenvalues down the diagonal, and zeros elsewhere. As it is positive definite all the eigenvalues are positive. There are no negative elements. Please see the proof referred to in a comment to your previous post https://chemistry.stackexchange.com/questions/118761/when-diagonalising-the-overlap-matrix-what-happens-if-the-square-root-of-the-di#comment225774_118761

Oh, and just in case - you add the overlaps of the contracted orbitals together and then diagonalise, NOT diagonalise each contribution in turn and then add the result together. I'm sorry but that's about the only way I can think you are confused over this.

Diagonalization of a matrix means to find the eigenvalues of the matrix and put them into a diagonal matrix:
$$\mathbf{S}^\text{diag}=\mathrm{diag}\left( \lambda_1,...,\lambda_n \right)\tag{1}\label{diagmat}$$
for $\lambda_i$ being the eigenvalues of the matrix $\mathbf{S}$.
When we want to find the eigenvalues of the matrix $\mathbf{S}$ we solve an eigenvalue problem:
$$\mathbf{S}\mathbf{c}=\bar{\lambda}\mathbf{c}\tag{2}$$
for $\mathbf{c}$ being a matrix of the eigenvector, and $\bar{\lambda}$ being a vector of the eigenvalues.
This equation can also be written out in sum-form for the $i$the eigenvalue:
$$\sum_kS_{jk}c_{ik}=\lambda_ic_{ij}\tag{3}\label{eigproblem}$$
Now as you have stated we can define our atomic orbitals as a linear combination of primitive basis functions:
$$\chi_i=\sum_k c_{ik}\phi_k\tag{4}\label{AO}$$
This can be used to evaluate the self-overlap of the atomic orbitals:
$$S_{ii}=\int_\Omega\chi_i^*(\bar{r})\chi_i(\bar{r})\mathrm{d}\bar{r}=\int_\Omega\left|\chi_i(\bar{r})\right|^2\mathrm{d}\bar{r}=\left< \chi_i \left| \chi_i\right.\right>\tag{5}\label{selfoverlap}$$
This can now be expanded out using Eq. (\ref{AO}):
$$S_{ii} = \left< \sum_k c_{ik}\phi_k \left| \sum_j c_{ij}\phi_j\right.\right>\tag{5}$$
Now using the definition of the overlap-integral $S_{jk}=\left< \chi_j \left| \chi_k\right.\right>$ the above equation can be formulated as:
$$S_{ii}=\sum_{j,k}c_{ij}^*S_{jk}c_{ik}\tag{6}$$
Now using Eq. (\ref{eigproblem}) this can be expressed as:
$$S_{ii}=\sum_{j,k}c_{ij}^*\lambda_ic_{ij}=\lambda_i\sum_j\left|c_{ij}\right|^2\tag{7}$$
Now since $S_{ii}$ can be seen to be larger than or equal to zero from Eq. (\ref{selfoverlap}) and $\left|c_{ij}\right|^2$ must also be larger than or equal to zero, it follows that the eigenvalues $\lambda_i$ must all also be larger then or equal to zero.
I.e. from Eq. (\ref{diagmat}) the diagonalized overlap matrix cannot have any negative values.