ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 17:40Z
7φ² + 7φ + 6 = 0
AgNO₃ + NaCl → AgCl↓ + NaNO₃
\frac{6}{3} \sum_{i=1}^{n} i^{2}
125K
\binom{n}{k} = \frac{n!}{k!(n-k)!}
91K
9n² + 8n + 4 = 0
\binom{n}{k} = \frac{n!}{k!(n-k)!}
310K
ΔS ≥ 0
det| 5 3 ; 2 6 | = 24
311K
E = mc²
AgNO₃ + NaCl → AgCl↓ + NaNO₃