The concentration of X after [tex]13{\text{ minutes}}[/tex] is [tex]\boxed{0.04359}.[/tex]
Further explanation:
Given:
The rate constant for the decomposition [tex]k = 1.6{M^{ - 1}}{m^{ - 1}}.[/tex]
Explanation:
The second order integrated rate law can be expressed as follows,
[tex]\boxed{\frac{1}{{{N_0}}} = \frac{1}{{{N_1}}} - kt}[/tex]
Here, [tex]{N_0}[/tex] represents the initial concentration, [tex]{N_1}[/tex] represents the concentration after time t and k represents the constant.
The concentration after 13 minutes can be obtained as follows,
[tex]\begin{aligned}\frac{1}{{{N_{13}}}} - kt &= \frac{1}{{{N_0}}}\\ \frac{1}{{{N_{13}}}} - 1.6\left( {13} \right) &= \frac{1}{{0.467}}\\\frac{1}{{{N_{13}}}} &= 2.14 + 20.8\\\frac{1}{{{N_{13}}}} &= 22.94\\{N_{13}} &= \frac{1}{{22.94}}\\\end{aligned}[/tex]
The concentration of X after [tex]13{\text{ minutes}}[/tex] is [tex]\boxed{0.04359}.[/tex]
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Exponential function
Keywords: rate, constant, decomposition, integral, hypothetical, compound, X, part A, 1.60, concentration, 13 minutes.
