Question

My Own Space

Answer 1

The decomposition of \(\mathsf{A^2B^3}\) is second order with a \(\mathsf{k = 6.5x10^{-5} M^{-1} s^{-1}}\) at 25 degrees Celsius. if the initial concentration is 0.50 M, what is the concentration after 3 minutes?

Answer 2

Okay, so it's a second order reaction. The rate law must be,\[
\frac{dx}{dt} = k_2 (a-x)^2\]Where a is the initial concentration or, .50 M.
\[k_2 = \frac{1}{t}\cdot \frac{x}{a(a-x)}\]
\[t k_2 a = \frac{x}{a-x} \implies a^2tk_2 - x tk_2 a = x \implies x = \frac{a^2t k_2}{1 + tk_2 a}\]
\(k_2\) is the rate-constant, and \(t\) is the time.

Answer 3

And x is the final concentration at t=3 minutes.

Answer 4

\[x^4 + ax^3 + bx^2 + cx + 1 = \left(x^2 + \frac a2x\right)^2 + \left(b - \frac{a^2 -c^2}4\right)x^2 + \left(\frac c2 x + 1\right)^2\]

Answer 5

\[s_k \equiv 0 \pmod{m}\]