ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:58Z
∑_{k=1}^{n} k = n(n+1)/2
4β² + 5β + 0 = 0
det| 2 7 ; 7 3 | = -43
91K
d/dφ [φ^6] = 6φ^5
det| 9 4 ; 1 10 | = 86
310K
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
311K
det| 5 6 ; 1 6 | = 24
68K
∑_{k=1}^{n} k = n(n+1)/2
39K
det| 12 9 ; 5 13 | = 111
364K