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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?

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    $\begingroup$ Kind of related: chemistry.stackexchange.com/q/42075 $\endgroup$ – Nicolau Saker Neto Mar 11 at 13:16
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    $\begingroup$ $$(c + 4 \cdot h) + 2 \cdot ( 2 \cdot o ) = ( c + 2 \cdot o ) + 2 \cdot ( 2 \cdot h + o )$$ $\endgroup$ – Poutnik Mar 11 at 16:01
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    $\begingroup$ What is the chemical meaning of the plus sign (+) in mathematical equations? ;-) $\endgroup$ – Peter - Reinstate Monica Mar 12 at 11:49
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    $\begingroup$ I would argue that they don't have a sensible mathematical interpretation at all, they have a chemical interpretation. $\endgroup$ – pjc50 Mar 12 at 12:54
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    $\begingroup$ 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. $\endgroup$ – Arthur Mar 12 at 19:42
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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.

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    $\begingroup$ 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. $\endgroup$ – M. Farooq Mar 14 at 21:00
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    $\begingroup$ @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. $\endgroup$ – Wrzlprmft Mar 14 at 21:40
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    $\begingroup$ 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. $\endgroup$ – M. Farooq Mar 14 at 21:50
  • $\begingroup$ 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. $\endgroup$ – Alchimista Apr 25 at 15:45
  • $\begingroup$ @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. $\endgroup$ – Wrzlprmft Apr 25 at 16:22
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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

textbook

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.

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  • $\begingroup$ 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. $\endgroup$ – Karsten Theis Mar 11 at 18:11
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    $\begingroup$ Hence the origin of the + sign in chemical equations. The weight must be conserved. $\endgroup$ – M. Farooq Mar 11 at 19:14
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    $\begingroup$ library.si.edu/digital-library/book/traiteyeyleyment1lavo $\endgroup$ – DavePhD Mar 14 at 16:06
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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.

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    $\begingroup$ 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. $\endgroup$ – Karsten Theis Mar 11 at 13:13
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    $\begingroup$ 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. $\endgroup$ – Alchimista Mar 11 at 13:33
  • $\begingroup$ What is equal is the number of carbon atoms before and after the reaction. @Alchimista $\endgroup$ – Karsten Theis Mar 11 at 13:54
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    $\begingroup$ @KarstenTheis uhm well. That is the minimum requirement for a chemical reaction indeed. But still I don't get the point. $\endgroup$ – Alchimista Mar 11 at 14:04
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    $\begingroup$ I am afraid this is an overinterpretation. Plus sign in chemical equations existed even when molecular formulae did not exist. $\endgroup$ – M. Farooq Mar 11 at 15:42
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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.

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    $\begingroup$ Funny that the straight answer get less upvotes. $\endgroup$ – Alchimista Mar 12 at 9:09
  • $\begingroup$ @Alchimista First answers gets more votes even if it is plain wrong. And sometimes even if it is "not even wrong". $\endgroup$ – fraxinus Mar 14 at 9:11
  • $\begingroup$ @fraxinus well I was referring to another moment. Nothing against the current one. $\endgroup$ – Alchimista Mar 14 at 9:40
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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.

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    $\begingroup$ Same comment as for the answer by Buraian. $\endgroup$ – Alchimista Mar 11 at 13:35
  • $\begingroup$ Oh I love it! $\endgroup$ – uhoh Mar 13 at 6:21
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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:


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.

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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.

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    $\begingroup$ It manifests elemental amount of substance conservation too. $\endgroup$ – Alchimista Apr 25 at 15:39

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