ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 17:51Z
7φ² + 7φ + 6 = 0
det| 9 4 ; 1 10 | = 86
312K
AgNO₃ + NaCl → AgCl↓ + NaNO₃
\binom{n}{k} = \frac{n!}{k!(n-k)!}
125K
det| 3 7 ; 2 4 | = -2
91K
9n² + 8n + 4 = 0
\binom{n}{k} = \frac{n!}{k!(n-k)!}
310K
ΔS ≥ 0
det| 9 1 ; 7 10 | = 83
311K
∑_{k=1}^{n} k = n(n+1)/2
264K
E = mc²
AgNO₃ + NaCl → AgCl↓ + NaNO₃