The second-order decomposition of hi has a rate constant of 1.80 x 10−3 m−1 s−1. How much hi remains after 45.6 s if the initial concentration of hi is 3.81 m?

2.9 M

Explanation:

The concentration-time equation for a second order reaction is:

1/[A] = kt + 1/[A°]

Where,

A = concentration remaining at time, t

A° = initial concentration

k = rate constant

1/[A] = (1.80 x 10^-3) * (45.6) + 1/3.81

1/[A] = 0.345

= 1/0.345

= 2.9 M.

answer: the answer is a

part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by