ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 16:25Z
10n² + 5n + 0 = 0
E = mc²
∫₀^∞ e^(-n²) dn = √π / 2
91K
det| 8 7 ; 7 9 | = 23
\frac{6}{3} \sum_{i=1}^{n} i^{2}
17K
d/dt [t^11] = 11t^10
AgNO₃ + NaCl → AgCl↓ + NaNO₃
264K
∫₀^∞ e^(-y²) dy = √π / 2
det| 4 8 ; 1 5 | = 12
\lim_{γ\to 0} \frac{\sin γ}{γ} = 1
364K
\oint_C \vec{F}\cdot d\vec{r} = 0
N₂ + 3H₂ ⇌ 2NH₃
AgNO₃ + NaCl → AgCl↓ + NaNO₃