Natural uranium blankets and thermal neutrons could produce ⁹⁹Mo after all, but this takes a bigger tokamak, and the little molybdenum must be extracted from much uranium. At least, no isotopic separation is needed.

300K neutrons see 450b section for ²³⁵U fission, 67b for ²³⁵U absorption, 2.2b for ²³⁸U absorption. 0.72% and 99.28% abundances give natural uranium mean 5.9b with 55% chances to fission ²³⁵U. Natural C provides 4.9b for collisions (22mm mean free path) and 3.2mb for absorption. Accepting 1/3 neutrons lost in C, 923 moles C pass the neutrons many times through each mole U in a Brownian motion. Mean 510 collisions before absorption let a neutron spread by very roughly 0.5m rms. The uranium load is a bit over 1t.

Neutrons leaving a blanket inwards shall serve at opposite blankets. U absorbs too strongly between 1keV and 3eV so most neutrons must survive in well separated C. I didn't check deuterated polyolefins, light water nor the proliferating heavy water. Liquid oxygen could replace graphite, save a little bit uranium, and bring many drawbacks. Colder moderators bring little and liquid deuterium has big drawbacks. And the tokamak's coils must sit somewhere but will catch some neutrons.

Materials that absorb neutrons little: ethylene and propylene glycol carbonates and their eutectic, oxalates of alcohols or Be, including deuterated variants.

Some 30mg ⁹⁹Mo made in a day must be extracted from >1t U. As the fission destroys the U molecule, maybe the chosen environment could build a new Mo molecule easy to separate.

6.1% of the fissions yield ⁹⁹Mo. Maybe every second neutron misses the blankets or is absorbed elsewhere. 1/3 is lost in C. Among U events, 55% produce Mo. All this takes 100 neutrons per Mo atom, so the tokamak is only 1.7×1.7×1.7× smaller than ITER, alas. But producing no energy and needing no tritium, it's simpler. Adding U blankets at the top and if possible the bottom would reduce the size.

Marc Schaefer, aka Enthalpy

Any modern sophomore organic chemistry book is fine.

For polymer specifics then young and lovell’s polymer textbook is excellent.

You can find stuff on libgen

[...] So 2.1×10¹⁵/s neutrons as well. [...]

Wrong! Every second D+D reaction produces T+p, but T has 1MeV recoil energy while the magnets confine only mean 20keV hydrogen. Tritium escapes the reaction zone.

2.1×10¹⁵n/s need a machine 8×8×8× smaller than Iter rather. Or better magnets, as these progress.

And the neutrons have 2.45MeV as emitted by the D(D, n)T reaction, not 14.1MeV as in T(D, n)⁴He.

Please help me how to calculate...for hydrogen peroxide, I know how strong it is if I use a 3% mixture, so I make one part of 30% H2O2 and 9 parts of water and get a 3% mixture. I want to dilute the ClO2 (chlorine dioxide) that I made from NaClO2 (SODIUM CHLORITE) and otherwise I used 80% powder and distilled water 720 water+280 NaClO2 and thus got about 24% solution. I mixed this solution with 4% HCl (hydrochloric acid) - I mixed 1:1 and got a strong oxidant ClO2- chlorine dioxide...if I put this on the skin it is clear that the skin will burn. So how do you calculate how much distilled water to add to ClO2 to get a comparable solution - comparable to 3% H2O2?
Thanks for the *delete me*

I would plug the definition of r into your second equation.  Then I would consider two cases, x = 1 and x = 2.  What is different about the time courses of these two cases?

With [itex]x = 1[/itex], r increases linearly with the A concentration
[tex] r = \dfrac{-\text{d[A]}}{\mathrm{d}t}  = k_r [\mathrm{A}] [/tex]
With [itex]x = 2[/itex], r will depend on the A concentration square
[tex] r = \dfrac{-\text{d[A]}}{\mathrm{d}t} = k_r [\mathrm{A}]^2 [/tex]
from which it can be deduced that a greater quantity of A will be required to increase r