The second-order decomposition of NO2 has a rate constant of 0.255 M-1s-1. How much NO2 decomposes in 4.00 s if the initial concentration of NO2 (1.00 L volume) is 1.33 M?

C NO2 = 0.5644 M

Explanation:

The second order decomposition of NO2 (A):

⇒ Kδt = - δCA/CA²

⇒ K∫δt = - ∫δCA/CA²

⇒ K*t = 1/CA - 1/CAo

∴ K = 0.255/M.s

∴ t = 4.00 s

∴ CAo = 1.33 M

⇒ 1/CA = K*t + 1/CAo

⇒ 1/CA = ((0.255/M.s)(4.00 s)) + (1/1.33 M)

⇒ 1/CA = 1.02/M + 0.752/M

⇒ 1/CA = 1.772/M

⇒ CA = 0.5644 M

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.