text:
What is the mathematical meaning of the plus sign (+) in chemical reaction equations?
Reactants and products are connected with plus signs in a chemical reaction equation, e.g.
$$\ce{CH4 + 2 O2 -> CO2 + 2 H2O}.$$
Is there a strict mathematical interpretation of what looks like sums before and after the reaction arrow? If so, what do the sums represent, and what are the mathematical objects we would use for the chemical species?
Often, when we use a chemical equation to solve a problem, operations other than addition come into play. For an equilibrium constant expression, concentrations of species are multiplied, not added. In stoichiometry, the chemical equation is interpreted as ratio of changes in amounts. In kinetics, we divide by the coefficients to be able to compare rates of changes of different species (and there is an equal sign, not a plus sign between the rate expressions of the different species.
So why are there plus signs between species in a chemical equation, and what is their mathematical interpretation?

Kind of related: https://chemistry.stackexchange.com/q/42075/

$$(c + 4 \cdot h) + 2 \cdot ( 2 \cdot o ) = ( c + 2 \cdot o )  + 2 \cdot ( 2 \cdot h + o )$$

What is the chemical meaning of the  plus sign (+) in mathematical equations? ;-)

I would argue that they don't have a sensible mathematical interpretation at all, they have a chemical interpretation.

Perhaps the free abelian group generated by all possible molecules (or all possible chemical formulas)? By itself, that doesn't pick up the conservation of atoms during a reaction, but additional structure can probably be added without too much hassle.

To approach this from the mathematical side:
It is a common misconception that there is one single meaning of the plus sign (or addition) in mathematics. The additions of numbers, vectors, functions, etc. in mathematics are not the same. For an illustrative example, the addition of a two-dimensional vector and a number is not even defined in common mathematical notation. Very strictly speaking, not even the additions of natural numbers and real numbers are exactly the same. Now, of course, these different kinds of mathematical additions have common properties (commutativity, existence of a zero element, etc.) and some cases can be seen as special cases of the other with the right approach, and that’s why the same symbol is used. The same applies to multiplication and other operations.
The chemical plus sign behaves like mathematical additions in many ways, in particular:

it is commutative: $\ce{CH4 + 2 O2}$ is the same as $\ce{2 O2 + CH4}$.
it is associative: $\ce{CH4 + (2 O2 + H2O)}$ is the same as $\ce{(CH4 + 2 O2) + H2O}$ to the extent that I have never seen parentheses used in this context.
there is a neutral element, in the sense that “chemically adding” nothing doesn’t change anything or corresponds to no reaction. You would immediately understand that $\ce{CH4 + 0 O2}$ is the same as just $\ce{CH4}$, even though it is an exotic notation. This is also why it makes more sense to treat “chemical addition” like an addition and not a multiplication.
there is an inverse element: $\ce{CH4 + 2 O2 - 2O2}$ makes sense and is equivalent to $\ce{CH4}$. It is not without problems and thus barely used, but that’s another issue. However, this becomes very apparent if one considers reactions involving antimatter, where the negative elements can be actual reactants and not just something that is taken away: $\ce{e+ + e- → 0}$ (ignoring that this creates energy in form of two photons).

Mathematically speaking, chemical addition is an Abelian group (on each side of the reaction; the reaction arrow does not correspond to an equals sign).
It is for the above reason that using a plus sign in chemical reactions does not create a cognitive dissonance and I would say that this is why it prevails.
While chemical addition can be treated as a vector addition (as elaborated in these answers), I wouldn’t consider this central, since few people are fluent with vector calculations before learning about chemical notation.
Note how other disciplines use “mathematical” operators in similar manners:

The addition of physical values – which is different to mathematical additions as you cannot add values of a different dimension such as 42 m and 23 s.

The multiplication of physical units – which ironically is very akin to a vector addition.

The usage of mathematical operators for all kinds of things in programming languages. For example, in Python, "abc"+"def" is the same as "abcdef".

For an equilibrium constant expression, concentrations of species are multiplied, not added.

This is similar and actually related to mathematical additions translating to multiplications in exponential space: $e^{x+y} = e^x ·e^y.$
Thus, I don’t see a contradiction here.

It would be better if we first define what our group elements are. Does the left hand constitute one group and the right hand side of the chemical equation another group? It will be Abelian only on one side? Although it indeed sounds fancy, but it seems like an overstretch to extend the group theory treatment to the plus sign in a chemical equation.

@M.Farooq: It’s only one side of the equation (see my edit). I don’t want to apply group theory (or more generally algebraical structures) to chemical reactions beyond using its terminology to retrospectively explain the plus sign (and I mention groups only in a sidenote after all). On the other hand, the nature of the question requires to think about what a plus sign actually signifies and that inevitably brings you to algebraical structures, whether you use mathematical terminology to describe them or not.

As a chemist, I can say that the plus simply signifies chemical combination on the left hand side and a list of products on the right hand side. We could have used a comma on the RHS instead. The key requirement is that the mass is conserved in each balanced equation i.e., mass on the LHS=mass on RHS. If we read the OPs original equation, it can be read as "methane chemically combines with oxygen to produce (the arrow) carbon dioxide and water. If there is an equilibrium sign, it means that the reaction proceeds both ways then we can use the word combine for both sides.

I doubt of the associative property. I can guarantee that following an associative way of leading a reaction is a very effective way to fail. Again, we can interprete chemical equation with linear algebra or whatever one like if we limits their significance to a mass (or atoms) balance. That is all.

@Alchimista: That’s only true if you consider the reaction arrow an equals sign. In terms of describing the reactants or products separately, there clearly is associativity.

@Wrzlprmft I do agree. But all boils down to counting atoms by elements, and loose any other chemical significance.

@Alchimista: Replace atoms with reactants and I agree. And yes, it is not particularly chemically interesting, but I never claimed that – it is what it is.

I think we are over-interpreting the poor $+$ sign. In older books you will also see an $=$ sign instead of an arrow just like what Lavoisier did. We have to get hold of La Traité Élémentaire de la Chimie (1789).
We should look at the historical perspective and I always want to see the earliest usage in chemistry just to see what was the original person thinking. I think we can credit Lavoisier for the "+" sign in chemical equations and let us see what was going on in his mind. I quote from the History of Chemistry, by F.J. Moore 1910 (Full text in Google Books).

The cumulative effect of these researches led to the sudden conversion of most French chemists about $1785,$ and the new ideas were firmly fixed by the publication of Lavoisier's great text-book La Traité Élémentaire de la Chimie in 1789. The change was well called the Chemical Revolution, for it inverted completely the chemical point of view. The mysterious hypothetical substance, phlogiston, which did not obey the law of gravitation, and changed its properties arbitrarily as theoretical considerations dictated, was banished from the science and the law of conservation of mass vindicated once for all. This made possible quantitative analysis and the chemical equation. The latter is a convenience in form of expression which we owe to Lavoisier, and of whose assistance in stating and solving chemical problems he made immediate use. He writes: $^{1}$
If I distill an unknown salt with vitriolic acid and find nitric acid
in the receiver and vitriolated tartar in the residue, I conclude that
the original salt was nitre, and I reach this conclusion by mentally
writing the following equation based upon the supposition that the
total weight is the same before and after the operation.
"If $x$ is the acid of the unknown salt, and $y$ is the unknown base
$I$ write: $x+y+$ vitriolic acid $=$ nitric acid $+$ vitriolated
tartar $=$ nitric acid + vitriolic acid + fixed alkali. Hence I conclude: $x=$ nitric acid, $y=$ fixed alkali, and the original salt was nitre."
Nothing like this had ever appeared before in chemical literature and
what a fog of mystery and superstition it removed!

Update Surprisingly, I cannot find this quote in Lavoisier' book (can't read French) but + sign is obvious! The only equation in his 382-paged book is
moût de raisin = acide carbonique +  alkool.
grape wort = carbonic acid + alcohol
Anyway, plus sign is quite old. In the book titled Historical studies in the language of chemistry by Crosland, Maurice P, a 1744 text is cited showing

Readers should consult this book (Internet Archive, need to make an account for borrowing it) if the want to dig deeper. The author discusses plus sign at several places at length.

Back then, they relied on weights a lot, and that is what is being added up: "by mentally writing the following equation based upon the supposition that the total weight is the same before and after the operation". So you have to make sure you start with pure reactants, and the reaction goes to completion (no reactants left). Not sure how they got the stoichiometry right other than trial and error.

Hence the origin of the + sign in chemical equations. The weight must be conserved.

https://library.si.edu/digital-library/book/traiteyeyleyment1lavo

Here is an interpretation I came up for it, so suppose you have the molecule $\ce{CH4}$, then we can think of this as a vector as $ \langle 1,4 \rangle$ where the $1$ stands for number of carbons and $4$ as the number of hydrogen.
Generally for each 'nice' chemical reaction(*), we can associate a vector length 'n' to the reactant side, where $n$ is the number of 'unique' elements. So, for the example you've posted, I see three unique elements on the reactant side, namely: Carbon, hydrogen and oxygen and hence the vector has length '3'.
For a compound of form $\ce{C_x \! H_y \! O_z}$, I associate a vector $\langle x,y,z \rangle$ where the first entry counts the number of carbons, second entry the number of hydrogens and third entry the number of oxygens.
If we replace each compound with it's vector in the following way:
$$1 \, \langle 1,4,0 \rangle + 2\, \langle 0,0,2 \rangle = \langle 1,0,2 \rangle + 2 \, \langle 0,2,1 \rangle $$
And we can indeed see that the above equation is true (and hence the corresponding chemical equation is balanced).
Here is another interesting idea, say I had a compound $\ce{CH4}$ that is denoted by the vector $ \langle 1,4,0 \rangle$ and suppose I wanted to find the number of hydrogen in it, then easy I just dot it with the 'unit' hydrogen vector $\langle 0,1,0 \rangle$ this gives $4$.
This is exactly analogous to extracting out components between vectors using dot products, like if I have $\vec{v} = 3 \hat{\imath \,} + 4 \hat{\jmath \,}$ then $ \vec{v} \cdot \hat{\imath \,} = 3$ i.e: a way to extract out the number of carbons.
Perhaps there exist meaningful interpretation of the cross product with chemical compounds as well. However, I do not know of them.

Addendum:
I add that though the above idea is not standardly taught, it opens up a more a consistent way of balancing chemical equations. Further, some great discoveries are solely due to what seemed to start off as 'notational hacks', for example see Heaviside (read).
What is 'standard' is subject to change :P and this way of applying the ideas of linear algebra here opens up understanding of both chemical equations in the perspective of linear algebra and the ideas of linear algebra in terms of chemistry

*: I can imagine this needing to be tweaked for nuclear reactions
**: On deeper thought, we can think of analogies between the notions in linear algebra about basis vectors and 'elements' in chemistry.

Sorry about posting a similar answer just after you posted this excellent one. I was in the middle of editing, and did not watch out for your super-fast answer.

Honestly I do not see what this conveys which is not already in a chemical equation. + it is not an arithmetic operation. While your tweaking does somehow give a sense to the + it has the unpleasant results to replace an arrow with a = And I think that I should NOT explain here on this site that products and reactants are not equal.

What is equal is the number of carbon atoms before and after the reaction. @Alchimista

@KarstenTheis uhm well. That is the minimum requirement for a chemical reaction indeed. But still I don't get the point.

I am afraid this is an overinterpretation. Plus sign in chemical equations existed even when molecular formulae did not exist.

What about unbalanced chemical equations? The (+) sign is still valid notation in those cases.

If I understand right, this view is essentially equivalent to Arthur's comment -- you can think of chemical formulae as elements of a free abelian group with the elements as its basis.

Answer is no, there is no strict mathematical interpretation. We can say that 1 mol of methane reacts with 2 moles of oxygen, but that is hardly any kind of mathematical interpretation. Plus sign only denotes what reacts together and what is formed together. We could use multiplication sign without  much difference although I think addition is more intuitive and that is the main reason why is it used.
Also, using plus sign allows easy adding reactants and products on both sides like in an mathematical equation and you can use this to find least number of reactions needed to describe chemical equilibrium in systems where there are many reactions happening by matrix manipulations.

Funny that the straight answer get less upvotes.

@Alchimista First answers gets more votes even if it is plain wrong. And sometimes even if it is "not even wrong".

@fraxinus well I was referring to another moment. Nothing against the current one.

Counting atoms
If you code each species in a vector showing the number of atoms of a given type, you can turn the chemical equation into a meaningful mathematical equation. In these vectors, the first value would be the number of hydrogen atoms, the second those of helium, and so forth. For example, the formula of water ($\ce{H2O}$ in case you forgot) would be:
$$  \newcommand\mycolv[1]{\begin{bmatrix}#1\end{bmatrix}}
\mycolv{2\\0\\0\\0\\0\\0\\0\\1\\0\\\vdots} $$
Now, you can multiply these vectors with the stoichiometric coefficients, and check that the sum for the reactants is equal to the sum of the products.
$$1 \mycolv{4\\0\\0\\0\\0\\1\\0\\0\\0\\\vdots} + 2  \mycolv{0\\0\\0\\0\\0\\0\\0\\2\\0\\\vdots} = 1 \mycolv{0\\0\\0\\0\\0\\1\\0\\2\\0\\\vdots} + 2 \mycolv{2\\0\\0\\0\\0\\0\\0\\1\\0\\\vdots}$$
Adding reactions
When you have two reactions, you can add them, using the rules of addition on both sides of the equation. For example, consider reactions (1) and (2):
$$\ce{CO2(g) + H2O(l) <=> H2CO3(aq)}\tag{1}$$
$$\ce{H2CO3 <=> HCO3-(aq) + H+(aq)}\tag{2}$$
If you add these, you get:
$$\ce{CO2(g) + H2O(l) + H2CO3(aq) <=> H2CO3(aq) + HCO3-(aq) + H+(aq)}\tag{1+2}$$
If you cancel the carbonic acid (i.e. substract on both side), you get the net equation:
$$\ce{CO2(g) + H2O(l) <=> HCO3-(aq) + H+(aq)}\tag{1+2}$$
In a similar manner, you get from
$$\ce{H2 + 1/2 O2 -> H2O}\tag{3}$$
to
$$\ce{2 H2 +  O2 -> 2 H2O}\tag{3+3}$$
Stoichiometry
While the interpretation above is great to check whether an equation is balanced, I would argue that the most common use of the stoichiometric coefficients is to figure out the stoichiometry of how the amounts of different species change in the course of a reaction. If we translate the chemical equation into a mathematical formula for this purpose, it has a very different structure and the species are not added up. Again, we can use vectors but with the different species as the rows of the vector. For example, we could defined the composition of the reaction mix as:
$$\mathrm{mix} = \mycolv{n_\ce{CH4}\\n_\ce{O2}\\n_\ce{CO2}\\n_\ce{H2O}}$$
And we could encode the stoichiometric coefficients into another vector (notice the negative sign for the reactants):
$$\mathrm{reaction} = \mycolv{-1\\-2\\1\\2}$$
Now we can do math with these objects. If we want to show how the mix changes when 0.3 mol of carbon dioxide are formed, we would write:
$$ \mathrm{mix_{after}} = \mathrm{mix_{before}} + \pu{0.3 mol} \cdot\mathrm{reaction}$$
This is essentially the calculation we would do in an ICE table. From this equation, you can also get the following relationship:
$$ \frac{\Delta n_\ce{CH4}}{-1} = \frac{\Delta n_\ce{O2}}{-2} = \frac{\Delta n_\ce{CO2}}{1}= \frac{\Delta n_\ce{H2O}}{2} = \xi$$
So for stoichiometry, thinking of the plus sign as an addition operation is not so useful (and thinking of the stoichiometric coefficient as a multiplier is not so useful either). Instead, seasoned teachers sometimes translate the equation into "for each methane molecule, two oxygen molecules react, and one carbon dioxide and two water molecules are formed". In effect, the chemical equation defines fixed ratios of species being used up and being made.

Same comment as for the answer by Buraian.

Oh I love it!

tl;dr–  It's just normal addition.  Chemical species are basically just vectors of periodic-table-elements (and similar things).  Multiplication/division gets involved when we're talking about the odds of a component being present at a reaction-site, which is multiplicative as that's how statistics works.

It's normal addition.
As other answers have covered, this is normal addition.
Normal addition, such as in the specific case of vector-addition, can add unlike things together.  For example, we can add 1 apple and 2 oranges – that's perfectly normal addition.  Just, obviously, we shouldn't perform an inappropriate reduction, e.g. reducing to 3 apples.  Though we could reduce to 3 fruits.
When dealing with known formulas in balancing stochiometric equations, we can interpret the molecular-species in terms of their contents, then use linear-algebra to balance the equation.  For examples:

"How to Balance Chemical Equations Using Linear Algebra", wikiHow;

"Balancing chemical equations using linear algebraic methods", SE.Math.

Sometimes an equation might be written in unbalanced form.
Say two things react to form a product.  Then a chemist might write
$$
\mathrm{A} + \mathrm{B} \to \mathrm{C}
$$
when they actually mean
$$
\alpha_{\mathrm{A}}\mathrm{A} + \alpha_{\mathrm{B}}\mathrm{B} \to \alpha_{\mathrm{C}}\mathrm{C}
\,.
$$
This is, the coefficients ${\alpha}_{i}$ are omitted.  This is called an unbalanced chemical equation.
When writing, it's generally preferable to avoid the shorthand in favor of more explicit notation, e.g. including the coefficients ${\alpha}_{i}$ even if they're variables.

Why sometimes multiplication/division?

Often, when we use a chemical equation to solve a problem, operations other than addition come into play. For an equilibrium constant expression, concentrations of species are multiplied, not added. In stoichiometry, the chemical equation is interpreted as ratio of changes in amounts. In kinetics, we divide by the coefficients to be able to compare rates of changes of different species (and there is an equal sign, not a plus sign between the rate expressions of the different species.

In short, the multiplication stuff isn't multiplying the chemical-species but rather the concentrations (or activities) of the chemical species.
Regarding kinetics...
In kinetics, rates are like
$$
\frac{\mathrm{d} \xi}{\mathrm{d} t}
~=~
k
\prod_{\forall\text{species}~i}{C_i^{\alpha_i}}
\,,
$$
where:

$\xi$ is the extent of the reaction (how much the reaction's happened);

$t$ is time;

$k$ is the reaction-rate constant;

$C_i$ is the concentration of species $i ;$

$\alpha_i$ is the stochiometric-coefficient of species $i .$

Conceptually, the theory's that we break up the reaction-space into discrete units.. like little boxes or whatever (it doesn't matter).  Then the probability of each component being appropriately there is proportional to its concentration, $C_i .$  And being appropriately there multiple times (this is, $\alpha_i$ times) is multiplicative, much like getting Heads once is 50% likely, then 25% likely for twice, then 12.5% likely for thrice, etc..
And then how large is a reaction-space-unit/box/whatever?  Doesn't matter in this simple formation.  We just measure it and get $k ;$ that's enough.

Regarding equilibrium...
Equilibrium's really the same thing as kinetics, just we're considering the forward and backward reactions together, and then focusing on when they cancel out, i.e.
$$
\frac{\mathrm{d} {\xi_{\text{forward}}}}{\mathrm{d} t}
~=~
\frac{\mathrm{d} {\xi_{\text{backward}}}}{\mathrm{d} t}
\,,
$$
so
$$
k_{\text{forward}}
\prod_{\forall\text{species}~i}{C_i^{\alpha_i^{\text{forward}}}}
=
k_{\text{backward}}
\prod_{\forall\text{species}~i}{C_i^{\alpha_i^{\text{backward}}}}
\,,
$$
allowing us to define
$$
{\begin{array}{rl}
K_{\text{equilibrium}}
& ~ \equiv ~
\frac{k_{\text{forward}}}{k_{\text{backward}}}
\\
& ~=~
\frac{\prod_{\forall\text{species}~i}{C_i^{\alpha_i^{\text{backward}}}}}{\prod_{\forall\text{species}~i}{C_i^{\alpha_i^{\text{forward}}}}}
\\
& ~=~
\prod_{\forall\text{species}~i}{C_i^{\left(\alpha_i^{\text{backward}} - \alpha_i^{\text{forward}}\right)}}
\\
& ~=~
\prod_{\forall\text{species}~i}{C_i^{\alpha_i^{\text{total}}}}
\end{array}}
_{,}
$$
where $\alpha_i^{\text{total}} \equiv \alpha_i^{\text{backward}} - \alpha_i^{\text{forward}} ,$ hence the common convention that reactant-side coefficients be taken as negative.

Summary.
In short, the $`` + "\text{-operator}$ is just the normal mathematical operator for addition, where the added components could be understood as vectors in periodic-table-element-space.
Sometimes multiplication/division's involved because kinetics, and hence equilibrium states, depend on the probabilities of components being appropriately present at a potential reaction-site, which is a multiplicative thing.

The plus sign in chemical reactions manifests the mass conservation. If your reactant mixture has a mass of 1 kg (it is just a random number), then also the product mass will have the same value. Imagine the reactant as money, the double arrows as the shopping to the mall and the products as new shoes and other stuffs you want to buy. Then the total amount of your money must be equal to the total value of the things bought, if you have spent all the money. The only operation that computes the "total amount" is the sum. This motivates the usage of the plus sign.

It manifests elemental amount of substance conservation too.


text:
I have contradictory results in a research paper, how do I proceed with my analysis?
I am performing a simple acid-base titration investigating the efficency of different pure antacid compounds, and in my results, I found out that NaHCO3 neutralized the fastest (using a pH by time graph). Later on, I decided to further prove these results, by calculating pKb values for each base, and out of my three bases (Mg(OH)2, CaCO3, and NaHCO3), NaHCO3 had the highest pKb, which would suggest that it is the weakest base, and wouldn't neutralize the fastest. What should I do in this situation??

You text does not match the title. What research paper?

Look up the pK values. Edit your question, add a table with the values from literature, and those you calculated from your measurements. Show your calculation.

What do you refer to / how do you qualify «fastest»?  This is not a kinetic analysis because (assuming aqueous solution) the neutralization of $\ce{H+}$ by $\ce{OH-}$ to yield $\ce{H2O}$ will be a fast one anyway.  Maybe you refer to «to neutralize amount $x$ of acid, amount $y_1$ of antacid (1) is necessary, which is half the amount of $y_2$ using antacid(2)», or a plot of pH value in function of the volume of antacid added.

As Buttonwood says that you can't really think on kinetics. At most you can relate how long there will be a base depending on availability. It can have different grains size (also overcame by the fast reaction, btw) or it can be in a jelly carrier. But this would mean comparing different drugs / formulation not really the bases.

I think I know what your problem is. Yes, $\ce{NaHCO3}$ is the weakest base of the three, but it is the most soluble in water. Thus, the neutralization can occur throughout the solution. For $\ce{CaCO3}$ and $\ce{Mg(OH)2}$, the rate limiting step is dissolution of the solid. Neutralization only occurs at or near the solid surface.
You also need to be careful mixing kinetic (speed, lower case $k$) and thermodynamic (upper case $K_b$) parameters. The thermodynamic parameters tell you how likely a reaction is, not how fast it will proceed.
However, in this case, we can also see that the reaction with $\ce{NaHCO3}$ is most likely. The acid-base equilibrium is not the only equilibrium to consider. We also need to consider the solubility equilibrium summarized by the solubility product constant $K_{sp}$. For example, for $\ce{CaCO3}$, $K_{sp} = 3.30\times 10^{-9}$ for the following equilibrium.
$$\ce{CaCO3(s) <=> Ca^2+(aq) + CO3^2-(aq)}$$
$$K_{sp}=[\ce{Ca^2+}][\ce{CO3^2-}]$$
The equilibrium equation for neutralization is, with $K_b = 10^{-5}$:
$$\ce{CO3^2-(aq) + H2O(\ell) <=> HCO3^-(aq) + OH-(aq)}$$
$$K_b = \dfrac{[\ce{HCO3^-}][\ce{OH-}]}{[\ce{CO3^2-}]}$$
The overall reaction is then:
$$ \ce{CaCO3(s) + H2O (\ell) <=> Ca^2+(aq) + HCO3^-(aq) + OH-(aq)}$$
$$K = [\ce{Ca^2+}][\ce{HCO3^-}][\ce{OH-}] = K_{sp}K_b$$
Now we have what we need to evaluate the three bases thermodynamically:
$$\begin{array}{l|l|l|l|l}
\text{Formula} & \text{solubility} & K_{sp} & K_b & K = K_{sp}K_b \\ \hline
\ce{NaHCO3} & 96 \text{ g/L} & 1.31 & 2.24 \times 10^{-8} & 2.92\times 10^{-8} \\
\ce{CaCO3} & 1.3\times 10^{-2} \text{ g/L} & 3.30\times 10^{-9} & 1.00 \times 10^{-5} & 3.30\times 10^{-14} \\
\ce{Mg(OH)2} & 6.4 \times 10^{-6} \text{ g/L} & 5.61\times 10^{-12} & 50.1 & 2.81 \times 10^{-10} \\
\end{array}$$
From these data, the overall equilibrium constant is largest for $\ce{NaHCO3}$ because it has the highest solubility despite being the weakest base.

For the interested, lower case $k$, as in IUPAC here, about kinetics' rate constant.  Which is, by convention, different to upper case $K$ about equilibrium constants, as in IUPAC here.

@BenNorris, if it isnt too much to ask, could you please provide me with the source of where you go the values displayed above on the table?

All of these values are from the corresponding Wikipedia articles, (except the $K_{sp}$ of $\ce{NaHCO3}$, which I had to calculate from the solubility.

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text:
Reaction stoichiometry for calculation of Kp

NH$_2$CO$_2$NH$_4$(s)⇌$2$ NH$_3$(g)+ CO$_2$(g)

I want to calculate K$_p$ for the reaction above, given that the total pressure is $0.63$ bar.
My calculations:
$0.63$ = p(CO$_2$) + p(NH$_3$)
,   where p(NH$_3$) = $1/2$ p(CO$_2$) (For every NH$_3$ half a CO$_2$ is needed)
$0.63$ = p(CO$_2$) + ($1/2$ p(CO$_2$) = $3/2$p(CO$_2$) -> p(CO$_2$) = $0.42$ bar , p(NH$_3$) = $0.21$ bar
So, K$_p$ = ($0.21$)$^2$ * ($0.42$) = $0.0185$ bar$^3$
However, the answer key's way of calculating this is:
$0.63$ = p(CO$_2$) + p(NH$_3$)
,   where p(NH$_3$) = $2$ p(CO$_2$)
$0.63$ = $3$p(CO$_2$) , so that  p(CO$_2$) = $0.21$ bar and p(NH$_3$) = $0.42$ bar
How can p(NH$_3$) = $2$ p(CO$_2$)? Isn't that wrong? Or am I thinking wrong? I have contacted my professor but haven't received an answer yet and our exam is tomorrow so hopefully someone here can help me understand. Thank you!

Each time one molecule of salt is decomposed, it produced $\ce{2 NH3}$ for $\ce{1 CO2}$. The number of molecules $\ce{NH3}$ is always double from that of $\ce{CO2}$ in the gas phase. Now the partial pressure of a given compound is proportional to the number of molecules of this compound in the gas phase. As a consequence, the partial pressure of $\ce{NH3}$ is always twice as high as the partial pressure of $\ce{CO2}$

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text:
Is the electric conductivity of chlorine ions influenced by its oxidation state?
I'm a high school student researching the conductivity of electrolyte solutions. In this excerpt of the CRC Handbook, I noticed that different ions containing chlorine have different limiting molar conductivity, and there seems to be a trend between molar conductivity and the oxidation state of chlorine:

Ion
Oxidation state of chlorine
$\Lambda$ in $\pu{e-4 m2 S mol-1}$

$\ce{ClO2-}$
+3
52

$\ce{ClO3-}$
+5
64.6

$\ce{ClO4-}$
+7
67.3

Does the trend really exist? If so, why did it happen? I found previous researches on the conductivity of $\ce{Fe^2+}$ and $\ce{Fe^3+}$, and the difference seemed to be a result of electrons spinning.

It is the similar effect as for molar condutivity of alkali metals.

It is the similar effect as for molar conductivity of cations of alkali metals, which grows from $\ce{Li+}$ toward $\ce{Cs+}$, in spite of increasing ionic radius.
The point is, the effective radius of the hydrated ions has the opposite trend as the "naked" ion radius trend. $\ce{Li+}$ ion with the smallest radius has the strongest electrostatic field and the largest hydration sphere and has therefore the lowest molar conductivity.
The similar effect I expect for chlorine oxyacid anions. In this case, there is also decreasing degree of delocalization of the negative charge in trend $\ce{ClO4-} \gt \ce{ClO3-} \gt \ce{ClO2-}$.

OP is questioning whether there is a trend between oxidation state of center atom and molar conductivity of relevant anion ($\ce{ClO_x^-}$), but gives only three examples. Since given molar conductivity values are from an acceptable source, the trend seems to exist. However, we cannot conclude that it is due to oxidation state of center atom of $\ce{ClO_x^-}$, which are +3, +5, and +7 when $x$ is 2, 3, and 4, respectively. If similar trend is shown by both or either of $\ce{BrO_x^-}$ and $\ce{IO_x^-}$, there is some extra data to consider. Unfortunately, I cannot find the relevant molar conductivity values for these anions.
Nonetheless, I agree with Poutnik that this may be the similar effect as for molar conductivity of cations of alkali metals. In that answer, Poutnik has discussed that the trend is probably caused by the increasing ionic radius. When I was looking for ionic radii of corresponding anions, I found this reference (Ref.1), which described the bond lengths of $\ce{X-O}$ of $\ce{ClO_x^-}$, $\ce{BrO_x^-}$, and $\ce{IO_x^-}$. Out of curiosity, I checked a trend between bond lengths of $\ce{X-O}$ and molar conductivity:
$$
\begin{array}{c|c|c|c}
\text{Oxyhalo anion} & \ce{X-O}\text{ bond length, pm} & \text{molar conductivity, }\Lambda & \Lambda^2 \\
\hline
\ce{ClO2-} & 159.1 & 62 & 3844 \\
\ce{ClO3-} & 150.1 & 64.6 & 4173 \\
\ce{ClO4-} & 145.3 & 67.3 & 4529 \\
\ce{BrO3-} & 167.1 & 55.7 & 3102 \\
\ce{IO3-} & 182.9 & 40.5 & 1640 \\
\hline
\end{array}
$$
The plot of $\ce{X-O}$ versus $\Lambda^2$ is a straight line with good agreement ($R^2 = 0.9769$):

I included the data of $\ce{BrO_3^-}$ and $\ce{IO_3^-}$ for better representation. Unfortunately we don't have other $\ce{BrO_x^-}$ and $\ce{IO_x^-}$ conductivity data. Still, good relationship with five data point is better than three. What I'm saying is it is close to zero chance that to find direct or indirect relationship of molar conductivity versus oxidation state of the center atom as OP is wondering. Yet, there is a good chance to have a relationship between molar conductivity versus $\ce{X-O}$ bond length as depicted in the image. I do not know why is that but it is worth noting.
References:

Lars Eklund, Tomas S. Hofer, Ingmar Persson, "Structure and water exchange dynamics of hydrated oxo halo ions in aqueous solution using QMCF MD simulation, large angle X-ray scattering and EXAFS," Dalton Trans. 2015, 44(4), 1816–1828 (doi: 10.1039/c4dt02580f).


text:
If we put an metal electrode in water with no metal ions present potential should tend to minus infinity?
If we consider an equilibrium between metal electrode and its ions in solution we know that equilibrium potential is proportional to activity of ions in solution from Nernst equation. Since this dependence is logarithmic, what we conclude is that when activity of ions tends towards zero, potential tends to minus infinity. If we do place a metal electrode in pure water, would its potential decrease continously if we disregard any kind of other processes on it like corrosion which create their own potential?

Yes. The potential is in principle infinite if the water is pure. But in reality, when a metallic plate is dipped into pure water, you cannot prevent one or two atoms from being oxidized by air and pass into solution. On the other hand, the same thing can be said about $\ce{O2}$ producing one or two $\ce{OH-}$ ions. As a consequence the potential goes quickly from infinite to a finite value.

Very good answer- this used to puzzle me when I was a student too that the Nernst equation must go to infinity but of course when I started teaching I had to look for an answer lest a student ever ask it.

Are you sure this is not a model issue? Most equations in science have regimes of applicability and it is invalid to extrapolate outside those regimes. My opinion is that the Nernst equation simply does not apply for pure water with an inert metal electrode, e.g., platinum, inserted in it.

Thank you for you answer. Yes, in reality corrosion processes(reduction of oxygen) would determine the potential.

Regarding other comments, Nernst equation is derived from second law of thermodynamics and as such it must hold for any activity of metal ions since this law doesn't put any constraints in terms of activity.

I will wait until the derivation and assumptions are explicitly given. I wonder what Nernst actually did.

Nernst law has no limit at very low concentrations. It may even be applied when there is less than 1 ion in one liter solution. This is the case when determining the solubility product of substances like $\ce{HgS}$ which is $\pu{1.4 10^{-53} mol^2 L^{-2}}$

Frankly, I doubt this. The second law of thermodynamics is ultimately based on quantum mechanics and classical laws are classical limits of quantum mechanics principles and derivations. So I doubt the Nernst equation, though extremely useful in typical situations, is strictly valid down at the bottom. I would be delighted to be proven wrong, but, until then, I see no compelling reason to believe the Nernst equation is not a classical limiting case. I will say nothing more and not vote up or down.

@Ed V Frankly I also doubt it. I cannot imagine of a chemical reaction happening on an electrode with an ion that is practically absent in solution. Less than $1$ ion par cubic kilometer seems absurd. But the potential of this Mercury electrode on a $\ce{Hg/HgS}$ can be measured and it is so different from the standard redox potential that the difference, divided by $\pu{ 0.0296 V}$, it yields a logarithm equal to $53$. As a consequence the mercury concentration is $\pu{10^{-53} M}$ .

I looked at the wikipedia article on the Nernst equation and the short section on limitations notes that the assumption of thermodynamic equilibrium is an issue down at the bottom. I am not trying to give you any grief, it is just my normal skepticism: down in the single digits of the ultra-small, issues of quantum statistics matter. Anyway, this was a thought-provoking question and answer and I am glad the Nernst equation is so reliably robust.

I think the answer to the OP’s question is no. Consider this answer by Ben Norris and assume his posited concentration cell and concentrations. Then, if the two solutions are each 1 L, there is very low probability, via the Poisson distribution with the low concentration as the mean value, of having any mercuric ions in the solution on the low concentration side of the cell. Solution volume can be increased, to have a reasonable probability of having several mercuric ions present, but then they will not just be staying near the electrode.

So then you have concentration shot noise: the cell cannot sense the concentration except in the vicinity of the electrode and, on average, almost nothing but water is there. Increasing the electrode area increases the parasitic stuff, e.g., electrode capacitance, etc. So there is a quantum limit to the concentration cell and its ability to perform at extremely low concentrations.

Well, yes. Wikipedia helped here as you said. In practice, we would never reach potentials predicted by Nernst since kinetics now becomes an issue because at such low activities exchange current density is also very low. Thermodynamics and kinetics are two distinct areas. It is like with diamond/graphite transition, even though thermodynamics says that diamomd should change to graphite, we never see it in real life since kinetics is extremely low in that case. Other reason is that in reality other processes which happen tend to define its potential rather than metal/ion equilibrium.


text:
Why is the dihedral angle of solid hydrogen peroxide less than that of gaseous?
Hydrogen peroxide has an "open book" structure:

The dihedral angle in gaseous H$_2$O$_2$ is close to what we might expect in a tetrahedral arrangement. However, in solid H$_2$O$_2$ it's almost a right angle. My book says that this is due to "hydrogen bonding" but doesn't go into further detail.
Note that this question isn't a duplicate of Dihedral angle of gaseous and crystalline HOOH as the accepted answer is clearly incomplete, and the other answer uses a computer simulation that falls short of an intuitive explanation.

The intuitive explanation is that the crystal structure forces the molecules into this form. Anything more involved will be heavy on math.

Well, hydrogen bonding is surely present in the crystal structure, and that's how it is involved.

Allow this anthropomorphism:  In the gas, the isolated molecule is free to venture out each and any conformation most comfortable for the molecule.  Over time, it may drop into the energetic pit of least energy.  In the solid state, especially the regular crystalline, this optimization about the individual molecule no longer is possible.  The lowest energy is met for the molecule including its interactions with its neighbours; you may say, an optimization of the supramolecule.


text:
Why does hydrogenation of this compound not result in a diastereomeric mixture?
In a question, this compound was given and it was asked:
"Hydrogenation of the above compound in the presence of poisoned Palladium catalyst gives?"

When I was answering it, I thought that the Lindlar catalyst would partially reduce the alkyne to give a cis alkene which would create two compounds that look like this:

Clearly, compound I is optically inactive due to the presence of plane of symmetry but compound II does not have any such plane of symmetry so it must be optically active. Hence, I answered that the resulting product would be a diastereomeric mixture since I and II are diastereomers wrt each other.
But, the solution to the question suggests that only compound I forms, with no explanation given for why compound II does not form. Searching on the net was also of no help. Could someone provide an explanation?
Thanks for your time and apologies for the sloppy drawings.

Is the product of Lindlar hydrogenation of (2Z,4S)-4-methylhept-2-en-5-yne optically active?

Thank you. I have understood the answer. I had not been able to find a solution of this problem when I had searched about it previously.

No prob, let's close this as a duplicate then. Nice drawings by the way, no need to be sorry.


text:
Can I engrave borosilicate glassware without compromising its structural integrity?
Can I use a pen engraver (the kind that's like a tattoo gun with a diamond tip) to engrave my initials on labware without compromising it? I am in a lab space with a shared acid wash, and my glassware keeps getting boosted. I'm worried about e.g. heating a beaker and having it shatter because of the propagation of microfractures potentially caused by the engraver.

Many beakers / Erlenmeyers have a square of etched surface where one may write e.g., with a graphite pen #2 / HB.  It does not come off with organic solvents, you have to use an eraser as in school (or one of the softer rubber stop cocks) to remove these writings prior to bringing the glass ware to the dish washers but leaves the beakers mechanically intact.  Would this be an option suitable for you?

I think folks would balk at having graphite introduced in the acid bath.

What you are experiencing is a common practice occurs in busy and large research groups. To scratch, do not use diamond tips. But sharp steel tips ate okay. It won't initiate glassware breaking (my personal experience).

Not my problem, but they should be really "your beakers".

@Alchimista - Yeah I'm the PI. Ya know, just a tenure track prof at an R1 asking rudimentary lab questions on stack exchange. oh god I'm such a fraud.

You could try reverse engraving: use a Dremel drill and a buffer wheel with mild abrasive powder (toothpaste? stove-top cleaner?), or even a sanding wheel or disc (emery disc?), and make a smooth area on the etched surface already there. This should introduce no new stresses, but be somewhat identifiable. I've never tried it, therefore not suggested this is an answer.

Any scratch weakens glass somewhat. However, a vibratory engraver is very likely to shatter the vessel.
You can label glass permanently by etching with warm concentrated $\ce{NaOH}$ or room-tempeerature $\ce{HF}$, perhaps with a stencil. Use great caution with $\ce{NaOH}$, and even more caution with $\ce{HF}$. Both are very caustic, and $\ce{HF}$ can release toxic fumes and cause deep tissue damage.
There are commercial etching creams with HF that might be less likely to splash.
Note that etching weakens glass slightly, but less so then engraving.

I think you're right that a vibratory tool is the wrong approach. I need either a rotary tool or just hand-scratching. But I don't think etching borosilicate glass with NaOH works. And I am certainly not going to mess with HF for the sake of $1k of glassware.

NaOH works, in my experience, but slowly at room temp. If a concentrated solution is left on borosilicate glass at 25°C for a few weeks, the glass becomes cloudy. It might not be worth weeks to wait, though. There are commercial etching creams with HF that might be less likely to splash, e.g. https://www.walmart.com/ip/Armour-Etch-Etching-Cream-10-oz/12347708


text:
How to identify the concentration of nitric acid?
I have quite a small quantity of concentrated nitric acid I had acquired about two years ago. This was given to me by a friend in a couple of Pyrex volumetric flasks with stopper and some in a borosilicate reagent bottles. The acid does have a whiskey brown color now.
Is there any simple method or visual cue to identify what concentration the nitric acid could be with a simple test? I want to see if the acid is 70% nitric acid. If it's the higher concentration of 90%, then I want to dilute it to 70%. I have attached a picture of one of the bottles for reference to see the color below.

Simply put, if you've had it for years, it can't be 90%.

For starters it would be fuming nitric acid. Why do you even think there could be sth stronger then usual conc. acid?

@IvanNeretin I agree the concentration may have been lost a bit overtime. But it’s still from late 2018.

@Mithoron I actually wanted some 70% acid. The reason I wanted to to know if there is any visual way or simple test to check concentration is because I haven’t  really dealt with concentration nitric much. Only sulfuric and HCl. Become of the change in color from clear I was a bit confused if this is the 70% or 90%+ conc.

$\ce{4 HNO3 -> 2 H2O + 4 NO2 + O2}$ If it's dark, there's plenty of nitrogen dioxide. Concentrated nitric acid should be clear. Btw. This does not look like a safe storage container.

It looks like you have about 40 mL of the solution. You could measure the density by using a small volumetric flask (10 or 25 mL), or even a graduated cylinder (which you could calibrate with water). Add your solution to the mark, weigh reasonably accurately, and you will have your answer (which I predict will lie between 70 and 90%).
At 20C, the density of 70% acid is 1.41; the density of 90% acid is 1.48. The difference is large enough to determine easily, and if there is any deterioration, you should be able to determine the exact concentration fairly easily from density tables, e.g.: https://www.handymath.com/cgi-bin/nitrictble2.cgi?submit=Entry


text:
Oribital comparison
When comparing the orbitals 2s and 2p only in the z axis, do they have the same energy levels and size? I am not so sure just because I know they both have a quantum number of 2. Thank you in advance!

(Take any axis , the result would be same)
Comparision of Energies :
2p  electrons are higher in energy than 2s ones due to screening effects that result from electron-electron interactions. On average, the 2s electrons will be at a slightly greater distance from the nucleus than the 2p electrons.  However, the 2s electrons have a higher probability of being closer to the nucleus due to the inner peak.  As a result, the 2s orbital will lie lower in energy than the 2p orbital in multi-electron atoms.
The letter of the orbital (s, p, d, etc.) is a nameplace for the azimuthal quantum number (orbital angular momentum, ℓ). The s orbital has ℓ=0, p has ℓ=1. The angular momentum being higher, the energy also is.
Comparision of Size :
A p orbital is rather like 2 identical balloons tied together at the nucleus.
The shapes of 2s and 2p can't be compared as they are unique. You can compare all the s orbitals - as these all have the same shape (spherical), equally all the p orbitals (dumb-bell) etc. The only difference there is size - so 1s is a small sphere, 2s is a bigger sphere, 3s even bigger sphere etc.

There is significant difference between s orbitals. ns orbital has n -1 radial nodes.Similarly for p orbitals.


text:
Finding formula of reduced scandium oxide using its mass and mass of obtained metal

An oxide of scandium, with a mass of $\pu{1.423 g},$ is chemically reduced with $\ce{H2}$ to give $\ce{H2O}$ and $\pu{0.929 g}$ of $\ce{Sc}$ metal. What is the formula of scandium oxide? How much water is formed?

I know that the reaction is of the form:
$$\ce{Sc_xO_y + H2 -> H2O + Sc},$$
where $x$ and $y$ are the units to define scandium oxide molecular formula (not empirical).
I don’t know how to get the constants using the info given, and what the info given tells me. I typically use a linear system of equations to solve for the reactants and products. In this case, we’d have:
$$
\begin{align}
\ce{Sc}&: &\quad  Ax - D &= 0;\\
\ce{O}&: &\quad   Ay - C &= 0;\\
\ce{H}&: &\quad   2B - 2C &= 0.
\end{align}
$$
Usually, we would have an equation where $x$ and $y$ aren’t present and you can solve the system of equations. Here, it prevents you from doing the balancing. Instead, you’re just guessing. Any suggestions when linear algebra doesn’t work?

Since scandium oxide is an ionic compound, the method is exactly the same as obtaining the empirical formula, from the ratio of the amounts of elements $n(\ce{Sc})$ and $n(\ce{O}).$
$$n(\ce{Sc})=\frac{m(\ce{Sc})}{M(\ce{Sc})}=\frac{\pu{0.929 g}}{\pu{44.96 g mol-1}}=\pu{2.07E-2 mol}$$
The mass difference comes from the loss of oxygen atoms:
$$n(\ce{O})=\frac{m(\ce{Sc_xO_y})-m(\ce{Sc})}{M(\ce{O})}=\frac{\pu{1.423 g}-\pu{0.929 g}}{\pu{16.00 g mol-1}}=\pu{3.09E-2 mol}$$
The ratio between the amounts of elements is close to $2:3$, therefore the formula is $\ce{Sc2O3}$. You can check that using oxidation states.


text:
A useful way to reuse non-recyclable plastics: Turn them to petrochemicals, shall we?
I read here (https://chemistry.stackexchange.com/a/34016/101483) that strongly acidic catalysts can facilitate dimerization of alkene. With catalysts on zeolite to allow only the isooctane (octane rating 100) to come out as the only product, can non-recyclable plastics be turned into useful petrochemicals.
In this particular case of isooctane, polystyrene (C8H8)n is broken down into styrene (C8H8) the monomer. Styrene is hydrogenated into octenes (C8H16). The hydrogenation on octenes at the crucial third stage with strongly acidic catalysts on zeolite only yields isooctane.
Is the whole process feasible?

The process is scientifically feasible. The problem is always the economics of doing this v the cost of preparing the product from raw ingredients

If you have a cost-effective process for depolymerizing polystyrene back to relatively clean styrene, it's more valuable as styrene than fuel. But so far, such processes,while known, are not economically feasible.

Well, the directions of ideas shall diverge then. Let us take polystyrene (PS) as an example. Firstly, find an alternative and/or replacement of PS. Secondly, pyrolyze PS to simple hydrocarbons and refine them in similar processes existent in petrochemical industries. Thirdly, potential eight carbons monomer -- styrene (S)  -- is made into isooctane (IO) the fuel, instead of polymerization.

Andrew, I think you are suggesting that the cost-effective process -- if exists --- makes S itself valuable than burning S or PS as fuel. Or in other words, the process counters the motives to find a plastic replacement of PS.

And if you found a replacement for PS, then there would be no S. It is produced primarily for making polymers like PS and ABS etc

The broad question here is whether "non-recyclable" plastics can be turned into fuel. The short answer to that is yes. More on that below. However, the specific proposal to convert polystyrene to isooctane is not particularly feasible. I'll address that first.
The proposed process begins with depolymerization of polystyrene back to styrene. Although lab-scale processes have been reported that give as high as 80% yield of styrene monomer, there is not yet an economically feasible process for scale-up. In fact, if styrene could be depolymerized practically, it would not be considered "non-recyclable". Furthermore, if the styrene could be recovered, it would be much more valuable as styrene than as gasoline, so there would be no incentive to spend the money on the subsequent steps described by OP.
Nonetheless, if one wanted to make liquid fuel from styrene, there is no need to produce isooctane specifically. Liquid fuels are generally mixtures of a variety of hydrocarbons, and the exact chemical composition does not especially matter as long as the autocombustion temp, vapor pressure and other relevant properties fall within an acceptable range.
So rather than make isooctane, one could simply hydrogenate styrene to ethylbenzene, an already common anti-knock agent or all the way to ethylcyclohexane (a more likely final product than octene).
Going back to the original premise of getting liquid fuel from plastics, just as there is no need to produce a specific compound from the styrene, there is no need to recover styrene specifically from the polymer, nor even restrict one's self to a specific polymer type such as polystyrene. Instead, bulk plastic mixtures can be pyrolyzed to produce a mixture not all that different from some crude oils and gases. These liquids and gases can be used as fuels themselves or mixed with existing petroleum products and refined.
Many challenges remain to make such a process economically viable, however, not least of which is the cost of collecting and transporting plastic waste (generally quite low density) compared to recovering oil from the ground.

https://www.sasol.com/media-centre/media-releases/sasol-commences-basic-engineering-tetramerization-technology So tetramerizing ethene to 1-octene and specialized hydrogenation of 1-octene to isooctane may be possible. Skip the plastics reusing altogether. Ethene are plenty.


text:
What is the relation of water potential with gibbs free energy?
In my high school textbook, they have stated that diffusion is a process in which water moves along a free energy gradient. I cannot understand how water potential may be related to the Gibbs free energy or the entropy and enthalpy of the system(as Gibbs free energy depends on both of them).

Your two sentences are not mutually consistent, they describe different problems.  What your textbook is referring to is a definition of diffusion within the framework of non-equilibrium thermodynamics. On the other hand the relation between "water chemical potential" and "Gibbs free energy" is very simple: $\mu_i=\left( \frac{\partial G}{\partial n_i} \right)_{p,T}$

It is not very clear, if mechanical or chemical potential is meant. but it does not matter much, as the former is the part of the latter.
At higher position in gravitational potential context, water has higher both mechanical and chemical potentials.
The former is defined as the rate of increase of mechanical potential energy with component mass.  $$V = \frac{\mathrm{d}U}{\mathrm{d}m}$$
The latter as the rate of increase of Gibbs energy with water molar mass.
$$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{T,p,n_j \ne n_i}$$
Note that $V$ is the part of $\mu$, if mutually recalculated to either mass or molar amount.
Water staying at higher gravitational potential has higher Gibbs energy and chemical potential.
Similarly, in solution with solute concentration gradient, water in more diluted solution has higher Gibbs energy and chemical potential. therefore it diffuses ( = has non-zero net flow ) from more to less diluted solution.

Comment feedback:
Adding of solute may both increase or decrease the entropy.
What is important is the change of the Gibbs energy during dissolution, if it is a spontaneous process or not. If the enthalpy of dissolution increases the entropy of neighborhood enough, the Gibbs energy change can be still negative even if the system entropy decreases.
Leveling concentrations decreases the Gibbs energy of the system, what means decrease of the total entropy of system + neighborhood.

So in the physical context of adding solute, we are decreasing the entropy of the system which water tries to regain by going towards a more dilute solute. This makes it a spontaneous process. Am I correct?

See the answer update.

@Poutnik  "Adding of solute may both increase or decrease the entropy." and " if it is a spontaneous process or not"  If I add sugar to water and it spontaneously dissolves, then entropy increases right? but if we keep on adding sugar to the point we have to stir it and heat to make it spontaneous (in other words provide an external agent), is it correct to tell that the entropy has decreased? if not could you tell an example

There is relation of spontaineity and the total entropy increase. Total entropy may increase even if the system entropy decrease. That happens if dissoving decreases thge system entropy, but is exothermic, and the heat increases the neighborhood entropy more, than the system entropy decreased.

dissolving usually increases entropy, but there are exceptions.


text:
Why does green ferrous sulfate solution change color to yellow upon addition of hydrochloric acid?
I was demonstrating the formation of $\ce{Fe(OH)2}$ precipitate at school when the light green $\ce{FeSO4}$ I was using reacted with dil. $\ce{HCl}$ turned yellow. This only happened with dil. $\ce{HCl}$ and not any other acid available at the lab.
[EDIT- I did not wash the reaction and after some time, the ferrous sulphate reacted with nitric acid turning a dark reddish brown, but the colour had vanished the next day.]
I have no idea what this is, but I have a feeling that this is iron being oxidized to $\ce{Fe^3+}$ in solution. Someone please explain what was happening here.

It is $\ce{Fe^3+}$ indeed. You always have it in $\ce{Fe^2+}.$

Complexes with chlorides have frequently different color then aqua complexes. See e.g. blue $\ce{[Cu(H2O)_n]^2+}$ vs yellow $\ce{[CuCl4]^2-} $. So, it would be probably $\ce{[FeCl4]-},$ or $\ce{[FeCl_n]^{2-n}}.$

Dry ferrous sulfate heptahydrate is green(ish) and is expected to make a green(ish) aqueous solution.

Anhydrous ferrous chloride is described as green to yellow, and the dehydrate and tetra hydrate are green to blue-green (CRC Handbook).

Now color is broad and continuous (see the visible spectrum),

not digital, like how many fingers am I holding up. This set of curves relating ferrous to ferric comes from https://www.sciencedirect.com/science/article/abs/pii/S0143720816309263 There's a continuity, not a change that can be described as this color to that color.

There is also an effect called dichromatism (not dichroism) in which the perceived color of a material that has some depth will change depending on the depth/thickness or concentration observed (https://en.wikipedia.org/wiki/Dichromatism).
In addition, ferrous ion picks up oxygen easily and becomes ferric (ferric chloride is deep red in concentrated solution (~20% or more), but quite definitely yellow in more dilute solutions (~5%)). The reaction 2Fe$^2$$^+$ + 0.5 O$_2$ + 2H$^+$ --> 2Fe$^3$$^+$ + H$_2$O is favored with lower pH. So your supposition that the (greenish) ferrous ion in solution is picking up oxygen and producing some yellowish ferric ion is almost certainly true. As ferric is formed green will diminish and yellow will increase; the spectral curve will just slide over from the green to the yellow.
However, because there is no quantification mentioned, everything is ishy. Greenish, yellowish, pHish, and yellow and green are both in the green and in the yellow solutions. You also don't mention specifically what other acids you tried.
To explain this to a class, you need some numbers. What you could do is get the visible absorption curve of fresh ferrous sulfate of known concentration (do it at two concentrations, at least), and then get the spectra after addition of (at least) two levels of known concentration HCl. Versus time, because the oxidation will probably take a few minutes in order to get enough exposure to air/oxygen. That will suggest how much of the color change is due to oxidation.
But you should also take the known concentrations of ferrous sulfate and add sodium chloride at the same molar concentrations as the HCl, to see how much color difference occurs because of chloride complexation while keeping the pH essentially constant. These solutions will yellow in time also, but slower than the acidified solutions.
These are some of the variables in your experiment, but how to express this is a short time will be up to you.


text:
How to calculate molar concentration from equilibrium equations with no initial concentration given?
I have the equations:
\begin{align}
\ce{CO2(g) + H2O(l) &<=> H2CO3(aq)} &\quad K_1 &= 10^{-1.47} \tag{1}\\
\ce{H2CO3(aq) &<=> H+(aq) + HCO3-(aq)} &\quad K_2 &= 10^{-6.35} \tag{2}\\
\ce{HCO3-(aq) &<=> H+(aq) + CO3^{2-}(aq)} &\quad K_3 &= 10^{-10.33} \tag{3}\\
\end{align}
Atmospheric $\ce{CO2}$ is at $\pu{10^{-3.5} atm}$ and the solution is pure water.
So far I have been able to find the equilibrium equations and have got the concentration of $\ce{H2CO3}$ to be $10^{-4.97}.$ However, from this I do not know where to go in order to find the concentration of bicarbonate ($\ce{HCO3-}$).
I have the equilibrium equations as:
\begin{align}
\tag{1}
  \frac{[\ce{H2CO3}]}{[\ce{CO2}][\ce{H2O}]} &=
  K_{\ce{CO2}} = 10^{-1.47}\\
\tag{2}
  \frac{[\ce{H^+}][\ce{HCO3^-}]}{[\ce{H2CO3}]} &=
  K_1 = 10^{-6.35}\\
\tag{3}
  \frac{[\ce{H^+}][\ce{CO3^2-}]}{[\ce{HCO3^-}]} &=
  K_2 = 10^{-10.33}
\end{align}
By multiplying $10^{-1.47}$ by $10^{-3.5}$ I got $10^{-4.97}$ as previously stated.
From there I get $\ce{HCO3-}$ to be $10^{-6.35}+\frac{10^{-4.97}}{\ce{H+}}$.
However I do not know how to find $\ce{H+}$.

Molar concentration of what? It would be ideal if you could provide exact problem and show your calculations. Also, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

Acidity constant 10^-6.35  already implies ALL CO2(aq) is converted to H2CO3(aq).   CO2(aq,tot) + H2O <=> H+ + HCO3-. The truth is, less then 1% of CO2(aq) converts to H2CO3(aq). So it seems to me, by H2CO3(aq) is probably rather consider total CO2(aq) = CO2(aq) + H2CO3(aq)

K(CO2) depends on used units for CO2(g). // the key part is to calculate pH. Relative ratios of carbonic acid / bicarbonate /carbonate depends on that.

Treat the combination of CO2 (aq) and H2CO3 (aq) as any weak acid in pure water, and calculate the pH from the concentration and K1 (you can search on this site or elsewhere for a model calculation). The pH will come out as about 5.7. Then, verify that there will be very little carbonate, and make sure all reactions (including the auto-dissociation of water) are at equilibrium.

Please refrain from asking the same question on different platforms, otherwise it might happen that it ends up here twice and quite a few people will be doing the same work also twice.

I will continue with the data of $[\ce{H2CO3}] = \pu{10^{-4.97} M}.$
Now, as the $K_2$ of $\ce{H2CO3}$ is very small as compared to its $K_1,$ we can assume that all the $\ce{H+}$ will come from the first dissociation of $\ce{H2CO3}.$
$$
\begin{array}{lccc}
& \ce{&H2CO3 &<=> &H+(aq) &+ &HCO3-(aq)} \\
&\text{Initial} & 10^{-4.97} && 0 && 0 \\
&\text{Final} & 10^{-4.97}(1-\alpha) && 10^{-4.97}\alpha && 10^{-4.97}\alpha
\end{array}
$$
$$K_1 = 10^{-6.35}$$
$$10^{-6.35} = \frac{(10^{-4.97}\alpha)^2}{10^{-4.97}(1-\alpha)}$$
As $K_1 \ll 1,$ $(1 - \alpha)\approx 1:$
$$\alpha = \sqrt{\frac{10^{-6.35}}{10^{-4.97}}}$$
$$
\begin{align}
[\ce{HCO3-}] &= \ce{[H+}] = 10^{-4.97}\alpha \\
  &= \sqrt{10^{-6.35}\times 10^{-4.97}} \\
  &= 10^{-5.66} \approx \pu{2.19E-6 M}
\end{align}
$$
Note, that I have not considered self ionization of water because the $\mathrm{pH}$ of this solution will be slightly less than $6$ $(\mathrm{pH} = 5.66).$ Which means even after accounting for self ionization of water, the results won't change much. This is a good approximate of what  we needed. And if you want to calculate $[\ce{CO3^2-}]$ than just calculate it in a similar way to which you calculated $[\ce{H2CO3}],$ because now you have $K_2,$ $[\ce{HCO3-}]$ and $[\ce{H+}].$

Let us continue this discussion in chat.

@NisargBhavsar Please have a look at my edit and feel free to use it as an example. Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.

I just edited the and added the table. Thanks, @andselisk for asking to help!!

@NisargBhavsar No prob, the problem with the new Markdown tables is that you have no control over its width and appearance, so it's a poor choice for aligning chemical reactions and ICE tables.


text:
Interpreting data on molecular orbitals in GaussView 6.0
Introduction
I have been trying to use computational software to gain more insight into how molecular orbitals are formed, as well as their shapes and sizes in accordance with Molecular Orbital Theory.
What I know so far
The molecule I examined was allyl bromide.
What I knew theoretically was that, contrary to VBT, all of the orbitals interact with each other, generating more complex hybrid orbitals. This makes sense, given the fact that GaussView generated 60 molecular orbitals.
Here is a snapshot of the LUMO for this molecule:

Questions
My question is what accounts for the shape of these orbitals?
According to my textbook (which offers a simplified representation for this orbital), the LUMO should behave like an arithmetic combination between the π* orbital of the C-C bond and the σ* orbital of the C-Br bond. But this approach seems localized, as if the influence of other orbitals would be negligible. Since MOT states that all of the orbitals interact to some extent, isn't this approach wrong?

So, how is this LUMO actually supposed to look like? What is the mathematical mechanism behind orbital structure and energy? To what extent is it wrong to just add orbitals together to try and predict their behaviour?
References
Jonathan Clayden, Nick Greeves, Stuart Warren, Peter Wothers - Organic Chemistry-Oxford University Press, USA (2000)

What C-O bond? Do you mean C-Br?

Apologies, I was referring specifically to an SN2' mechanism. That makes use of both the bonds I was talking about.

OK but there is no "O" if by that you mean oxygen.

I edited it. Looking back, I understand why that paragraph seemed like nonsense. Bonds broken: C1=C2 and C-Br. Bonds formed: C2=C3 and C-Nu (random nucleophile).

Another question successfully ignored... :(

You have a lot of questions in your post and it's a bit hard to unwrap it (know where to start explaining). What you see in your textbook is evidently a simplificaction, a schematic, that highlights the most important aspects of the electronic structure for the purpose of explaining the particular mechanism in question. MOs are LCAOs. The coefficients in the expansion determine AO contributions. The important point is that $H\psi=E\psi$ be satisfied, provided $\psi \psi*=1$ for a one electron spin-orbital, since the Schr. equation is linear the coefficients can be any real number 0<c<1.

@BuckThorn, thank you for your response! Can you recommend a textbook or any source that covers AOs and essentially MOT? All of the sources I find either offer extremely simplified descriptions (such as my textbooks, and those simplifications work just fine in those contexts) or stuff that involves a lot of complicated maths that I cannot grasp since I'm a HS Junior. Anything in the middle would be highly appreciated.

You may want to look through https://chemistry.stackexchange.com/questions/37303/resources-for-learning-chemistry My best suggestion is a physical chemistry textbook. An old favorite of mine is Miessler and Tarr which covers MO theory and is not too mathy but might be too advanced for HS.

An alternative is an orgo textbook that introduces MO theory.

BTW How did you get hold of Gaussview as a HS Junior?

Haha, from a friend of mine who studies Chemistry at EPFL, in Switzerland.

Also, some chem SE questions may be useful intros eg: https://chemistry.stackexchange.com/questions/19365/the-rules-for-lcaos-in-molecular-orbital-theory?rq=1 I'm going to vote to close your post since I think that one answers your question.

Let us continue this discussion in chat.


text:
Properties of a macroscopic amount of virus
The mathematics YouTube channel Numberphile just published a video entitled "All the World's Coronavirus Fits in a Coke Can" in which mathematician Kit Yates makes a Fermi estimate that all the SARS-CoV-2 coronavirus particles in the world would fit in a can of Coca-Cola (thereby missing the opportunity to make the pun that they would fit in a can of Corona beer).
What would the physical and chemical properties of such a macroscopic amount of coronavirus -- or even viruses in general other than coronavirus -- particles be? Would the mass as a whole seem solid or liquid-like? Would it be sticky? Would the viral particles attract or repel each other? What would it look like? Would it be transparent or opaque? What color, texture, smell, or other physical and chemical properties would it have?

It is a question rather to be answered by an article, or even a book, instead of a rather limited SE site answer. And the question is rather unfocud, forming a broad topic, that outstretches itself across many domains of science.
It would be look like something between a protein and nucleic acid.

It would depend on method of preparation.

Look for virus crystallisation.


text:
Enthalpy of dissolution of NaOH in small amount of water
is there a way to calculate enthalpy of dissolution of NaOH in small amount of water?
I know that enthalpy of dissolution of sodium hydroxide in water is –44,5 kJ/mol. However it assumes the orders of magnitude higher amount of water than NaOH. (Which is not my case.) Is there any formula which does not omit the volume of solvent? (Or a tool which can answer questions such as: "How much heat is released if you make 60% solution of 80% solution?")
Thanks for help.

The dissolution enthalpy for concentrated final solutions highly depends on the dissociation and hydration degree, that varies a lot with the substance and concentration.
Therefore particular values for concentrated solutions have to be either measured experimentally, either calculated by some empirical formula, derived from experimental data ( if available ).

Ahoj :). Thank you for your answer. Unfortunately, I have no experimental data and no calorimeter in my lab. So I will just have to calculate the worst possible scenario.

@VDVH Those data should be available

If no data are found, you can derive rough values from experimenting with heating up of concentrated solutions being diluted, and from the limit infinite dilution value.


text:
Stability of a carbanion that is attached to three phenyl groups
I had come across a question - which asked me to predict the stability of a carbanion attached to 3 phenyl rings, in opposed to one in which the same 3 phenyl rings that are attached to the carbanion are now attached to each other via sigma bonds at the ortho position.
While comparing the stability, I thought that both of them would be stabilised by the -R effect equally. Hence, I could not choose between either of them.
However, my book says the answer would be the later case without giving an explanation and I couldn't find one anywhere on the net as well.

I think the conceptual issue at play here is the degree of planarity

@Zhe Do you mean the second compound is planar because of the sigma bonds while the first one is non planar and the planar one having more effective overlap hence better resonance and more stable. What I didnt get then is why is the first one non planar?

Try building a model of the first structure. My guess is that 5 member rings will make it a bit hard to build a model in the second structure though...

@zhe thank you for your help I think I have understood now.

the second can be thought of as three cyclopentadiene rings fused together to fulfil some kind of huckle's rule that stabilize the ring to overcome the angle strain. the first has only charge distribution and the hydrogens at 2 and 5 position in the phenyl ring prevent conjugation in at least two phenyl rings.


text:
Magnitude of Lennard-Jones force evaluated with molecular dynamics simulation
The intent is to evaluate the error source of a molecular dynamics (MD) simulation which is only based on the Lennard-Jones interaction between noble gases (helium in this case). What is the magnitude of the force?
For reference: $\varepsilon = \pu{1.49E-22 J}$ and $\sigma = \pu{2.967E-10 m}.$
The equation used to calculate the force is:
$$F = 4\varepsilon\left(\frac{6\sigma^6}{r_\mathrm{abs}^7} - \frac{12\sigma^{12}}{r_\mathrm{abs}^{13}}\right) \sim \pu{E-13 N}$$
Combined with the mass of helium this leads to an acceleration which seems to be too large. Would the above make sense?
Our step size is roughly $\pu{5E-3 s}$ and the velocity has a magnitude of $\pu{E13 m s^-1}.$ This is our problem at the moment because those values are obviously too large. We are using the Verlet algorithm.

Well, the problem might be the step size and not the acceleration (although I haven't checked your calculations). 0.005 s seems a tad long for a simulation of noble gases unless they are very cold (actually even then).

A timestep of  0.005s? That's HUGE! MD timesteps are normally measured in picoseconds. What are you trying to do?

The following plots the force for reasonable distances

Intuitively, if you are implementing the verlet algorithm then the step size will be $\delta x \propto a\delta t^2$ (ignoring velocities). If you want a step size of order ~$f \times \sigma$ (where f is a scaling factor) then you'd like to satisfy a relation such as $\delta t < \sqrt{\frac{f\sigma}{a}}$ (if ignoring velocities). The timestep cutoff $\delta t$ computed according to this equation with a force of $\pu{10^-12 N}$ and f=0.1 is 0.3 psec. But 0.3 psec is still a relatively long timestep. Clearly the problem is not with acceleration (provided the timestep is shorter than the cutoff just calculated), but maybe there is one with the velocities. You can proceed in the same way but solving the complete quadratic verlet equation for the timestep in terms of the step size, and inserting a value for the rms speed based on the desired T and step size, to arrive at a timestep cutoff. Using this reasoning I came up with a cutoff $\delta t < \pu{ 15 fsec}$
One problem with a large step size (and discretization of the trajectory in general) is that the energy is not conserved and can easily blow up (along with the temperature): if you take a step of order $\sigma$ from the potential well minimum toward the maximum in the potential then the atoms will suddenly find themselves in the strongly repulsive high energy region.
You may want to check Ref. 1 which discusses this to some extent, and provides this advice:

One wants to find the largest time step that will maintain the conservation of energy.

Reference

Sangrak Kim. Issues on the Choice of a Proper Time Step in Molecular Dynamics. Physics  Procedia 53 (2014) 60–62.
doi: 10.1016/j.phpro.2014.06.027


text:
Meaning of "Nonyl-phenol 10 mole"
In my workplace we are purchasing "Nonylphenol 10 mole" and "Nonylphenol 30 mole" for use as non-ionic surfactants.
At first I assumed, because of the name, that the substance was simply nonylphenol. But further research shows that commercial non-ionic surfactants are actually ethoxylates of nonylphenol. Thus, I'm assuming that what we are buying is in fact ethoxylated nonylphenol, in two different degrees.
Further research is fruitless though. I simply cannot find any source explaining the meaning of "10 mole" and "30 mole" at the end of the name.
Can anyone tell me the actual real chemical formula of "Nonylphenol 10 (or 30) mole", or at least give me an explanation of the name's meaning?

Nonylphenols are a family of closely related organic compounds composed of phenol bearing a 9 carbon-tail (Wikipedia). Following is an example of nonylphenols:

However, "Nonylphenol $n$ mole" is a commercial name for non-ionic surfactants where $n$ stand for the amount of ethelene oxides (EO) in moles in one surfactant molecule. For example, if the non-ionic surfactant is $\ce{R-O-(CH2CH2-O)n-H}$ where $\ce{R}$ is hydrophobic tail of the surfactant, the commercial name "Nonylphenol 10 mole" means $\ce{R-O = 4-C9H17-C6H4-O}$ and $n = 10$ in the structure. An example for a linear chain non-ionic surfactant is pentaethylene glycol monododecyl ether:

where $\ce{R-O = C12H25-O}$ and $n = 5$ (available in Sigma-Aldrich).


text:
Strength of Pi Donor and Acceptors as Ligands
The spectrochemical series is usually known as the "strength" of the ligands. It makes sense on a basic level as the more willing a ligand is to donating its electrons (via a $\sigma$-bond), the greater the repulsion on the metal's $d$-orbitals. However, I recently learned about $\pi$-donors and $\pi$-acceptors and how they affect a ligand's placement on the spectrochemical series.
$$\ce{CO > CN- > NO2- > NH3 > H2O > OH- > F- > SCN- > Cl- > Br- > I- }$$
So from the very basic spectrochemical series I have shown here, it is said that $\ce{OH-}$ is a weaker-field ligand than $\ce{H2O}$ because it can act as a $\pi$-donor, which lowers the CFSE of the d-orbitals as the $e_g$ orbitals are stabilised by the resulting bond. However, I fail to see how this translates to a lower affinity to complex with the metal. Shouldn't this make $\ce{OH-}$ more stable as a ligand? I know that this isn't the case (proven by the structure of the $\ce{[Cu(H2O)4(OH)2]}$ complex structure, $\ce{OH-}$ takes on the weaker axial position), but I don't know why.
I'll include the molecular orbital diagram I got from chem.libretexts, which supposedly shows how the $\pi$-bonds affect ligand strength.


text:
Understanding Pourbaix diagram
I am recently taking courses in environmental chemistry and I was introduced to Pourbaix diagram. I was taught that the lines in the Pourbaix diagrams are equilibrium lines. Lets take as an example the Pourbaix diagram of the water (diagram is taken from Ibanez):

Lets look at the equilibrium line between $\ce{H2O}$ and $\ce{H2}$. The reaction describing this equilibrium line is (with equilibrium constant $K$):
$$\ce{2H+_{(aq)} + 2e- <=> H2_{(g)}}$$
Now at equilibrium taking the $\log K$ and equilibrium constant we obtaing:
$$\log K = \log p_{_{\ce{H2}}} -2 \, \mathrm{pH} -2 \log {[\ce{e-}]} $$
and by defining $\log {[\ce{e-}]} = \text{p}E$ we can obtain (assuming $p_{_{\ce{H2}}} = 1 \, \text{atm}$):
$$\text{p}E=-\text{pH}$$
So we can plot how $\text{p}E$ varies with $\text{pH}$ at equilibrium conditions. What I can't understand is why if we leave that line and move below it in a point which doesn't satisfy the equation we will have only $\ce{H2}$. How is it possible to change $\text{p}E$ without changing $\text{pH}$ if they are both related through equilibrium constant? Why all the water turns into hydrogen?

Alone, a Pourbaix daigram is of little interest because there are a lot of parameters that come into play: for example, it is not because a reaction is feasible that it will be done because if it is too slow to be done, it will not be done.
For gases, like dissolved species, these are areas of predominance. So, in the example you quote, if we go below the line, the gas $\ce {H_2}$ will be able to form until saturation


text:
In electrolysis, does voltage or amperage produce more gas?
I'm trying to produce a large amount of brown's gas, and I'm wondering if higher voltage or amperage will split the water faster.

What gas do you want formed? If you increase the voltage, you run the risk of electrolysis of other species and therefore of obtaining a mixture of products.
the greater the intensity and the greater the number of electrons, therefore the more products you form

I'm making brown's gas, a mix of h2 and o2. I'm just doing electrolysis of salt water, and I want to produce a lot of gas quickly.

Too high voltage, and/or using concentrated salt solution would produce also the third gas – chlorine.

Note that the voltage and current are not independent. You need a particular voltage for expected current, that is determined by used electrolyte and geometry of electrolytic cell. Too soft source of high voltage cannot provide sufficient current and a hard source of too low voltage cannot provide it either.

The chemical yield in an electrolysis is proportional to the amperage. Increasing the voltage is no use. It will only increase the production of heat. But it is often necessary to increase the voltage to increase the amperage. A good way of increasing the amperage without increasing the voltage is to decrease the distance between the electrodes, and increasing the surfaces of the electrodes.


text:
Can a reaction happen if a product change phase under reaction conditions?
Suppose we have the reaction :
$$\ce{A(s) + B(s) <=> C(g)}$$
The equilibrium constant will be equal to $e^{-\frac{Δ_rG}{RT}}$ and it varies with temperature. Suppose that $\ce{A}$ and $\ce{B}$ come into contact at temperature $T$ and pressure $p$ where the phase diagram for $\ce{C}$ at these conditions correspond to liquid phase. Will the reaction take place at all? I mean if it takes place but all the gas that is produced is converted into liquid then the reaction quotient will be zero at all times. But shouldn't at some time the reaction reach equilibrium with $K=p^{eq}_{c}$? Are equilibrium constants meaningfull only if the conditions allow for the compounds to be in the physical state that the chemical equation indicates?
After reading the comments I felt that I should add some details. If the reaction is happening in a closed vessel then we can combine the reactions ($1$), ($2$) to get ($3$):
\begin{align}
\ce{A(s) + B(s) &<=> C(g)} & (1)\\
\ce{C(g) &<=> C(l)}          & (2)\\
\ce{A(s) + B(s) &<=> C(l)} & (3)
\end{align}
What if the reaction takes place at open vessel so that all gas turns into liquid?

No, the equilibrium constants are always meaningful. See, a liquid has a certain vapor pressure above it.

If A+B is very rapid vs C condensation, an equilibrium may be established rapidly in the vapour phase and so a K(gas) may be measurable. Then as C liquid forms, the reaction proceeds further as the vapour pressure above the liquid C is smaller than the pressure of C as a gas. If A+B reaction is v slow then equilibrium is that between liquid C  and A+B. The end result is the same. Thermodynamics only concerns itself with final equilibrium conditions, i.e A(g)+B(g)=C(liq).

Note that $K = \exp\Big({-\frac{\Delta_rG^\circ}{RT}}\Big)$. Don't forget that you have to use the standard states for calculating $K$. Moreover, it would be more meaningful to measure $K$ for $\ce{A(s) + B(s) <=> C(l)}$ at low temperatures.

Scratch that last. How would you even measure an equilibrium constant for the last one...

Based on your edit I think one of your misconceptions is that you think that C will always be a gas first, but in reality most C particles will just aggregate to a liquid directly at low temperatures.

I bet $\ce{A(s) + B(s) <=> C(l)}$ behaves similar to phase transitions. So there will be a certain $T$ analogous to a melting temperature that dictates on which side the equilibrium lies. $K$ is no longer meaningful, since concentrations of solids and liquids are treated constant.


text:
Preparation of a HCN/NaCN buffer that has a pH of 9.8 and an osmotic pressure of 1.35 atm at 298 K
You are asked to prepare $\pu{2.0 L}$ of a $\ce{HCN/NaCN}$ buffer that has a $\mathrm{pH}$ of $9.8$ and an osmotic pressure of $\pu{1.35 atm}$ at $\pu{298 K}$. What masses of $\ce{HCN}$ and $\ce{NaCN}$ should you use to prepare the buffer? (Assume complete dissociation of $\ce{NaCN}$)
Here is my work for the problem:
$$\mathrm{pH} =-\log K_\mathrm{a} + \log \left(\frac{[\ce{CN-}]}{[\ce{HCN}]} \right) $$
$$9. 8= -\log (4.9 \times 10^{-10}) + \log\left(\frac{[\ce{CN-}]}{[\ce{HCN}]} \right)$$
$$\left(\frac{[\ce{CN-}]}{[\ce{HCN}]} \right) = 3.1$$
Osmotic pressure $= \Pi = cRT$
$\pu{1.35 atm} = c (\pu{0.0821 L atm mol-1 K-1})(\pu{298 K})$
$c = \pu{0.055 M}$
I realized that I can easily solve the problem using a system of equation if I knew which compounds/ions contributed to the osmotic pressure. However, I'm unsure of which ones. Therefore, my question is, which compounds/ions contribute to the osmotic pressure, and why?
Thank you, help is appreciated!

Why would anyone make a buffer with HCN as the conjugate acid? I would be more concerned with the partial pressure of HCN than the osmotic pressure of the solution. It is not a good idea to have a volatile buffer component, especially when the volatile substance is toxic.

@Karsten Theis this was just a problem I was given  for practice. I do not know why one would make HCN /NACN buffer.

Related: https://chemistry.stackexchange.com/q/108783/72973

See also https://en.wikipedia.org/wiki/Van_%27t_Hoff_factor

T respect  Karsten Theis' concern about the buffer in reality. However, I'd rather treated this as an hypothetical practice question for learning.
For all  colligative properties (freezing point and vapor pressure depression, boiling point elevation, and osmotic pressure), ionization of solute(s) must be taken into account (recall the dimensionless van't Hoff factor $i$). For example, if the solution in concern is aqueous $\ce{NaCl}$ for osmotic pressure measurements, then $c = [\ce{Na+}] + [\ce{Cl-}]$ in $\Pi = cRT$ equation (assume $\ce{NaCl}$ dissociate 100% in water).
I assumed your calculation is correct for the $\frac{[\ce{CN-}]}{[\ce{HCN}]}$ ratio using Henderson–Hasselbalch equation. Thus, denote $[\ce{CN-}] = a$ and $[\ce{HCN}] = b$:
$$\frac{[\ce{CN-}]}{[\ce{HCN}]}= \frac{a}{b} = 3.1 \ \Rightarrow \ a = 3.1b \ \tag1$$
Assuming ionization of $\ce{HCN}$ is minimal due to the tiny $K_\mathrm{a}$ the particles in the solution are $[\ce{HCN}], [\ce{Na+}],$ and $[\ce{CN-}]$. Thus,
$$c= 2a + b = 2 \times 3.1b + b = 7.2b \tag2$$
$$\Pi = cRT \ \Rightarrow \ c= 7.2b = \frac{\Pi}{RT} = \frac{\pu{1.35 atm}}{\pu{0.0821 L atm mol-1 K-1} \times \pu{298 K}}\\ = \pu{0.0552 mol L-1} \tag3$$
According to the equation $(3)$, the solution can be diluted enough to be considered as ideal solution, hence $\Pi = cRT$ can be used here.  Solving equation $(3)$ for $b$ gives you:
$$ b = \frac{\pu{0.0552 mol L-1}}{7.2} = \pu{7.66 \times 10^{-3} mol L-1} = [\ce{HCN}]$$
Hence,
$$[\ce{Na+}] = [\ce{CN-}] = a = 3.1b = 3.1 \times \pu{7.66 \times 10^{-3} mol L-1} = \pu{2.38 \times 10^{-2} mol L-1}$$
Now you can calculate the masses of each species to prepare $\pu{2.0 L}$ of appropriate buffer.


text:
Does Pyrite react with HCl?
I was wondering if anyone has some experience regarding my question. The geochemical simulation of mixing a mole of pyrite with a kilogram of acidic solution (pH=1) shows that Pyrite almost remains intact (PHREEQC). Do lab observations confirm this?

I would expect it to react.

I would expect it to react, especially if there is oxygen (e.g. from the atmosphere) present, even in small amounts.  Not sure if PHREEQC assumes this.


text:
Why are the rare Earth elements divided into light and heavy categories?
Rare Earth's up to z=62 (Samarium) are often called 'light' rare Earths, while z=63 (Europium) and beyond are sometimes referred to as 'heavy' ones. Why such distinction? I didn't have come across (or heard about) a reason for? Is this division arbitrary?  Are the light ones often found separately from the heavier ones?   Are they processed and separated differently?
Since they are rarely found in this country and almost never processed here and neither I nor anyone else I've met in person has worked with them, I want to reasoning for this distinction.

Though scandium and yttrium have lower atomic mass, they fall in the category of HREEs. The classification seems to have something to do with the formation of REE ores. Here is an extract of Wikipedia describing this phenomenon.

REE are chemically very similar and have always been difficult to separate, but a gradual decrease in ionic radius from LREE to HREE, called lanthanide contraction, can produce a broad separation between light and heavy REE. The larger ionic radii of LREE make them generally more incompatible than HREE in rock-forming minerals, and will partition more strongly into a melt phase, while HREE may prefer to remain in the crystalline residue, particularly if it contains HREE-compatible minerals like garnet. The result is that all magma formed from partial melting will always have greater concentrations of LREE than HREE, and individual minerals may be dominated by either HREE or LREE, depending on which range of ionic radii best fits the crystal lattice.

The original reference is included below:
References
Winter, John D. (2010). Principles of igneous and metamorphic petrology (2nd ed.). New York: Prentice Hall. ISBN 9780321592576. OCLC 262694332.
Jébrak, Michel; Marcoux, Eric; Laithier, Michelle; Skipwith, Patrick (2014). Geology of mineral resources (2nd ed.). St. John's, NL: Geological Association of Canada.


text:
Best solvent for recrystallization of acetanilide?
How would you recrystallize acetanilide using ethanol and hexanes? I know this compound is not soluble with the hexanes solvent. But, where would I include ethanol in the procedure? Would I heat up the ethanol and then add hexanes dropwise?

Comment: Why ethanol/hexanes?  It may depend on the history of your sample, but why would not do a mixture of water/ethanol do the job which I imagine to find in a reference like the red Vogel book (practical organic chemistry), or in Perrin's about how to purify  lab chemicals?

If you have small amount to crystalize water is the best solvent (Solubility at $\pu{0 ^\circ C}$: $\pu{0.53 g}$/$\pu{100 mL}$; and solubility at $\pu{100 ^\circ C}$: $\pu{5.5 g}$/$\pu{100 mL}$).

There are two ways to do this:
Firstly you could dissolve your acetanilide in hot ethanol and add hexane dropwise (keeping the solution hot) until you detect some cloudiness then leave it to cool.
Secondly suspend your acetanilide in hot hexane and add ethanol dropwise (keeping the solution hot) until it dissolves then leave it to cool.
Scratching the interior surface of the flask can help creating nucleation sites, or the addition of a seed crystal of pure acetanilide if you have one.


text:
On the exact definitions of Frenkel and Schottky defects
The wiki page on vacancy defects says;

It is also known as a Schottky defect, although in ionic crystals the concepts are not identical.

As far as I'm aware, they're the same. Furthermore, the wiki page for Schottky defects contains the term 'vacancies' while describing the phenomenon, which is linked back to the page for vacancy defects. That seems contradictory.
Are they the same? And is the same true for Frenkel and interstitial defects? Or are the named ones subsets?

According to Kittel 'Introduction to Solid State Physics' a lattice vacancy defect is a called a Schottky defect and in the Frenkel defect, also a vacancy defect,  an atom is transferred from a lattice site to an interstitial position.

I thought Frenkel defects weren't vacancy defects, but classified separately under interstitial defects.

@porphyrin: aren't Frenkel defects a seperate class?

Could you provide a link to the wiki page/s you are referring to?

Well, you can say that they are kind of similar...the only major difference is that Schottky defects are only for ionic compounds, and the number of missing cations and anions must be equal. Vacancy and interstitial defects can be shown by non-ionic solids. Ionic solids must always maintain electrical neutrality. Rather than simple vacancy or interstitial defects, they show these defects as Frenkel and Schottky defects.
Vacancy Defect: When some of the lattice sites are vacant, the crystal is said to have a vacancy defect. This results in a decrease in the density of the substance. This defect can also develop when a substance is heated.

Schottky Defect: It is basically a vacancy defect in ionic solids. In order to maintain electrical neutrality, the number of missing cations and anions are equal.
Like a simple vacancy defect, the Schottky defect also decreases the density of the substance. The number of such defects in ionic solids is quite significant. For example, in $\ce{NaCl}$, there are approximately $10^6$ Schottky pairs per $\pu{{cm}3}$ at room temperature. In $\pu{1 {cm}3}$ there are about $10^{22}$ ions. Thus, there is one Schottky defect per $10^{16}$ ions. The Schottky defect is shown by ionic substances in which the cation and anion are of almost similar sizes. For example, $\ce{NaCl, KCl, CsCl}$ and $\ce{AgBr}$. It may be noted that $\ce{AgBr}$ shows both, Frenkel as well as Schottky defects.

Also, with respect to your comment: "I thought Frenkel defects weren't vacancy defects but classified separately under interstitial defects."
Frenkel Defect: This defect is shown by ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location. Frenkel defect is also called dislocation defect. It does not change the density of the solid. Frenkel defect is shown by ionic substances in which there is a large difference in the size of ions, for example, $\ce{ZnS, AgCl, AgBr}$ and $\ce{AgI}$ due to the small size of $\ce{Zn^2+}$ and $\ce{Ag+}$ ions.

Interstitial Defect: When some constituent particles (atoms or molecules) occupy an interstitial site, the crystal is said to have an interstitial defect. This defect increases the density of the substance.

Also, here's why $\ce{AgBr}$ shows both Schottky and Frenkel defects, even though Frenkel defects are shown by ionic substances in which there is a large difference in the size of ions, and Schottky defects are shown by ionic substances in which the
cation and anion are of almost similar sizes.

Schottky defect in $\ce{AgBr}$ is exhibited due to precipitation of both cations and anions. In $\ce{AgBr, Ag+}$ ion is small in size and when removed from lattice point they can occupy interstitial sites and hence show both Frenkel and Schottky defects.

All four defects mentioned are types of stoichiometric point defects....other types of defects include impurity point defects, non-stoichiometric point defects (Metal Excess Defect and Metal Deficiency Defect) and line defects

By "cm3" do you mean cubic centimeter? Because, if so, your calculation about the number of ions is out by a factor exceeding 10^20. If that isn't what you meant you need to define your terms carefully to avoid confusing people.

I am sorry, the text didn't get edited properly, I'll edit my answer rn....i meant 10^22 and 10^16 respectively in one cubic centimeter


text:
Is the Co-ordination number of Cobalt always 6 in every co-ordination compound?
For example in the following  compounds

CoCl3.6H2O
CoCl3.6NH3
CoCl3.5NH3
CoCl3.4NH3

in all of these complexes, we took the
a)oxidation state of Cobalt as +3 ( since there are 3 Cl- ions and NH3 and H2O are neutral ligands )
b) Co-ordination number of Cobalt  as 6
and the answer to writing the compounds was:
1.[Co(H2O)6]Cl3 or [Co(H2O)     5Cl]Cl2.H2O

[Co(NH3)6]Cl3

[Co(NH3)5Cl]Cl2

[Co(NH3)4Cl2]Cl

My question is this: Why do we take the co-ordination number of Cobalt always six?
Can't Co take other co-ordination numbers like 4? or 5?
So for the following compound
CoCl3.4NH3
Can't we write it as [Co(NH3)4]Cl3 instead of
[Co(NH3)4Cl2]Cl
Also how to predetermine the C.N of the central metal atom just by looking at anions and ligands? (cases when no coordination sphere is marked like in my above examples)

This link includes a tetrahedral coordiation complex of cobalt, where the coordination number is 4.

So how do we pre-determine the C.N of the central metal atom ?

@TanyaGupta We need to know the structure (for example, crystal structure). Even then there might be issues with the determination of C.N. — that's why a concept of fractional coordination numbers exists.

Could you give an example that uses the concept of fractional co-ordination number for a compound?

With these compounds, you have coordination number $6$, but like other transition metals cobalt can have different coordination numbers with the right ligands. In this answer the $\ce{CoO4^{3-}}$ ion is discussed as an example of a tetrahedral complex with the rare case of high spin despite the central ion having an oxidation state of $+5$. (The tetrahedral geometry and strongly pi-donating ligands favor the high-spin electron confuguration.) Five-coordination is seen in Co(II) complexes with pi-accepting organic ligands such as $\ce{o-(C6H4)(SCH3)(P(C6H5)2)}$ [1].

Geoffrey Dyer and Devon W. Meek (1967). "Five-coordination. VI. Low-spin cobalt(II) complexes with with bidentate ligands". J. Am. Chem. Soc. 89, 16, 3983–3987.
https://doi.org/10.1021/ja00992a006


text:
I- oxidation to I2 or IO3-
Is there any logical way to determine whether $\ce{I-}$ will oxidize to form $\ce{I2}$ or $\ce{IO3-}$ is a reaction ?
For example,
In reaction of $\ce{KMnO4}$ with $\ce{I-}$ in acidic medium $\ce{I2}$ will be formed,
whereas in the reaction of $\ce{KMnO4}$ with $\ce{I-}$ in alkaline medium $\ce{IO3-}$ would be formed.
$\ce{HNO3}$ will oxidize $\ce{I2}$ to $\ce{IO3-}$.
Is it related to the oxidation power or the reaction conditions and medium or things out of my scope ? PS. I'm currently a high school student.

My question was not really specific to KMnO4 however it does cover the general idea too

See also https://chemistry.stackexchange.com/questions/141624/why-cant-iodide-be-oxidised-to-iodate-in-acidic-medium?noredirect=1&lq=1

In alkaline solution, $\ce{I2}$ does not exist. It is transformed into $\ce{IO3^-}$ and $\ce{I^-}$ according to $$\ce{3I2 + 6OH^- -> IO3^- + 5I- + 3 H2O}$$ In acidic conditions the mixture $\ce{IO3^- + 5I-}$ reacts in the opposite direction : $$\ce{IO3^- + 5I- + 6 H^+ -> 3 I2 + 3 H2O}$$

Oh didnt think of that. This helps a lot but how do you explain this theory with HNO3 + I2 reaction !? Where HNO3 is an acid ?

@Ashish. The two equations I gave you about iodine and its ions are only valid in dilute aqueous solution. $\ce{HNO3}$ is an acid if diluted in water. As a pure substance, it is a strong oxidizing substance. And it reacts with $\ce{I2}$ according to $\ce{3 I2  +  10 HNO3 -> 6 HIO3 + 10 NO + 2 H2O}$


text:
Molecularity of Saponification Reaction
I had a doubt while reading about the saponification reaction of esters in basic medium. What is the molecularity of the reaction and why?
P.S - I think its bimolecular and heres why -
The RDS of the reaction is the formation of the tetrahedral intermediate after the attack of  the OH ion. In this step both the ester and the hydroxide ion are involved (The ester is shifting the carbonyl bond electrons to the oxygen and the hydroxide is attacking the electrophilic carbon). Therefore, the molecularity should be two.

You should know that the molecularity is not relevant in your case and has no significance for multi-step reactions, it is just applicable to elementary reactions. The relevant parameter for multi-step reactions is the, reaction order wich is originally an experimental parameter  indirectly related to the molecularity of all elementary reactions that makeup the reaction mechanism.
So for saponification the proposed mechanism includes 3 steps, the last one of them is irreversible and controls the reaction. The order of the reaction is considered 2, the molecularity for the step that you have mentioned is also 2 but it is just one step in the mechanism.
Also, there are two types of mechanisms through which this reaction takes place: a base catalyzed mechanism (with simple base or alkoxides) and an acid catalyzed mechanism.