ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:57Z
∑_{k=1}^{n} k = n(n+1)/2
∫₀^∞ e^(-β²) dβ = √π / 2
302K
4β² + 5β + 0 = 0
d/dφ [φ^6] = 6φ^5
det| 5 2 ; 3 6 | = 24
310K
6n² + 3n + 0 = 0
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
311K
det| 7 4 ; 4 8 | = 40
39K