The rate constant for the second-order reaction !s 0.54 M-1 s-1 at 300°C. How long (in seconds) would 1t take for the concentration of N02 to decrease from 0.62 M to 0.28 M?

3.63s take for the concentration of NO₂ to decrease from 0.62M to 0.28M

Explanation:

The general equation for a second-order reaction is:

1/[A] = kt + 1/[A]₀

Where [A] is actual concentration of the reactant (0.28M)

k is rate constant (0.54M⁻¹s⁻¹)

t is time the reaction takes (Our incognite)

And [A]₀ is initial concentration of reactant (0.62M)

Replacing:

1/0.28M = 0.54M⁻¹s⁻¹t + 1/0.62M

3.5714 = 0.54t + 1.6129

1.9585 = 0.54t

3.63s = t

3.63s take for the concentration of NO₂ to decrease from 0.62M to 0.28M

Ithink this is what you meant. the temperature at which a liquid begins to boil is called the boiling point. the boiling point of water is 212°f or 100°c.  and at that point will start to turn into steam (gas)

The answer is d.potassium