\(Rate [X]^1[Y]^0\)
\(Rate = k[X]\)
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 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:
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.