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,
Consider the following compounds. Arrange these compounds in a n increasing order of reactivity with nitrating mixture. The correct order is : 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
Consider the following two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.
Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to