
\(Rate [X]^1[Y]^0\)
\(Rate = k[X]\)
From Exp I and II,
\(\frac{4 \times 10^{-3}}{2 \times 10^{-3}} = \left(\frac{L}{0.1}\right)^1 \left(\frac{0.2}{0.1}\right)^0\)
\(2 = (10 L)^1.\)
Hence L = 0.2 mol/L
From Exp III and IV,
\(\frac{M \times 10^{-3}}{2 \times 10^{-3}} = \left(\frac{0.4}{0.1}\right) \left(\frac{0.4}{0.2}\right)^0\)
\(\frac{M}{2} = 4\)
\(M = 8\)
\(\frac{M}{L} = \frac{8}{0.2}\)
\(\frac{M}{L} = 40\)
So, the answer is 40.
The following data were obtained for the reaction: \[ 2NO(g) + O_2(g) \rightarrow 2N_2O(g) \] at different concentrations: 
The rate law of this reaction is:
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is:
The Order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species taking part in it. In order to obtain the reaction order, the rate equation of the reaction will given in the question.