Alchemistry - A study on alchemy

Question

(Author Note) A while ago I made some puzzles for a now defunct puzzle hunt project, here's one of them. The answer to this (and most other puzzles of said hunt) is a single word. The puzzle itself is fully contained within the image above. Author notes don't matter.

Answer 1

The symbols are

alchemical symbols.

We can convert to

the atomic number of each element

to transform the statements into

1. 6 (carbon) + 3⦶ = 50 (tin)
2. 82 (lead) + 🜔 = 83 (bismuth)
3. 50 (tin) + ⦶ = 82 (lead)
4. 51 (antimony) + ⦶ - 4🜔 = 79 (gold)
5. 47 (silver) - ⦶ = 29 (copper)
6. 77 (iridium) + 2🜔 = 79 (gold)
7. 79 (gold) - ⦶ = 47 (silver)

We can think of each of these statements as performing alchemy as follows:

Starting with one element, apply some transformations to obtain a different element.

Based on statements 2 and 6, we can infer that

Applying a multiple of 🜔 changes the element's atomic number by that multiple.

Turning our attention to statements 3, 4, and 7, it appears that

Applying a multiple of ⦶ changes the element's atomic number by that multiple of 32.

Finally, from statements 1 and 5, we learn that

Applying ⦶ still results in a large change, but with a smaller magnitude than 32.

These rules could be visualized if

we placed the elements in unequal rows based on atomic number and viewed 🜔 and ⦶ as horizontal and vertical movement, respectively.

Of course, there is a well known way to do this:

The periodic table.

This gives us the proper way to interpret the 7 statements:

1. Carbon (6 C) shifted down by 3 yields tin (50 Sn).
2. Lead (82 Pb) shifted right by 1 yields bismuth (83 Bi).
3. Tin (50 Sn) shifted down by 1 yields lead (82 Pb).
4. Antimony (51 Sb) shifted down by 1 and left by 4 yields gold (79 Au).
5. Silver (47 Ag) shifted up by 1 yields copper (29 Cu).
6. Iridium (77 Ir) shifted right by 2 yields gold (79 Au).
7. Gold (79 Au) shifted up by 1 yields silver (47 Ag).

Turning our attention to the unanswered statements at the end of the puzzle, we can convert and transform them as follows:

1. Bismuth (83 Bi) shifted right by 1 yields polonium (84 Po).
2. Silver (47 Ag) shifted left by 5 and down by 1 yields tungsten (74 W).
3. Samarium (62 Sm) shifted right by 6 yields erbium (68 Er).

Finally, we can

combine the resulting chemical symbols (Po, W, Er)

to obtain the answer:

POWER.