Abiochemist studying breakdown of the insecticide ddt finds that it decomposes by a first order reaction with a half life of 12.0 years. how long does it take ddt in a soil sample to decompose from 413 ppbm to 10.0 ppbm (parts per billion by mass)?

64.4 years

Explanation:

The half-life (t1/2) for the first order breakdown of DDT is 12.0 years. The rate constant (k) is:

k = ln 2 / t1/2 = ln 2 / 12.0 y = 0.0578 y⁻¹

We can find the time (t) for an initial concentration A₀ = 413 ppbm to decrease to A(t) = 10.0 ppbm, using the following expression.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.