ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:56Z
∑_{k=1}^{n} k = n(n+1)/2
Fe₂O₃ + 3CO → 2Fe + 3CO₂
312K
4β² + 5β + 0 = 0
det| 9 4 ; 1 10 | = 86
91K
d/dφ [φ^6] = 6φ^5
∫₀^∞ e^(-ω²) dω = √π / 2
310K
Fe₂O₃ + 3CO → 2Fe + 3CO₂
311K