Rate of reaction is given by the change in concentration of FeSO₄: \[ \frac{-\Delta [FeSO₄]}{\Delta t} \] Substitute the given values: \[ \frac{-10 + 8.8}{30 \times 60} = \frac{1.2}{1800} = 6.67 \times 10^{-4} \] From the given reaction, the rate of production of Fe₂(SO₄)₃ is related to the rate of FeSO₄: \[ \frac{1}{6} \times \frac{-\Delta [FeSO₄]}{\Delta t} \] Substitute the value of \(\frac{-\Delta [FeSO₄]}{\Delta t}\): \[ \text{Rate of production of Fe}_2(SO₄)_3 = \frac{3}{6} \times 6.67 \times 10^{-4} = 333.33 \times 10^{-6} \] Thus, the rate of production of Fe₂(SO₄)₃ is \(333 \times 10^{-6}\) mol L⁻¹ s⁻¹. Hence,
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: