A 90-m3 basement in a residence is found to be contaminated with radon coming from the ground through the floor drains. The concentration of radon in the room is 1.5 Bq · L−1 (becquerels per liter) under steadystate conditions. The room behaves as a CMFR, and the decay of radon is a first-order reaction with a decay rate constant of 2.09 × 10−6 s−1. If the source of radon is closed off and the room is vented with radon-free air at a rate of 0.14 m3 · s−1, how long will it take to lower the radon concentration to an acceptable level of 0.15 Bq · L−1?

it will take 24.6467 min to lower the radon concentration to an acceptable level of 0.15Bq.L^-1

Explanation:

To calculate the required time, we make use of the formula

Cout = Coexp [ - ( 1/θ  + k ) t ] lets call this equation 1

where Co is the concentration of radon( 1.5Bq.L^-1 ), Cout id the allowable radon concentration( 0.15Bq.L^-1), k is the radon decay rate constant( 2.09 * 10^-6 s^-1 ), θ is the theoretical detention time and t is the time required to lower the radon concentration.

first we find θ  which is the theoretical detention time using the expression

θ   = _V / Q

_V is the volume of basement (90 m^3) and Q is the rate of free air (0.14 m^3.s^-1)

so we substitute

θ = 90 / 0.14

θ = 642.857 s

Now we substitute all our values into equation 1

Cout = Coexp [ - ( 1/θ  + k ) t ]

0.15 = 1.5exp [ - ( 1/642.857  + (2.09 * 10^-6) ) t ]

divide both LHS and RHS by 1.5

0.15/1.5  = exp [ - ( 0.001555 + 0.00000209)t ]

0.1 = exp[ - ( 0.001557)t ]

next we take the natural log of both sides

In 0.1 = In [ exp( -0.001557t )]

-2.3025 = - 0.001557t

t = 2.3025 / 0.001557

t = 1478.8053 sec

so we convert to minutes

t = 1478.8053 / 60

t = 24.6467 min

therefore it will take 24.6467 min to lower the radon concentration to an acceptable level of 0.15Bq.L^-1

possibly, yes.

explanation:

the amount of sugar you take in is dangerously high, risk of heart disease, high risk of diabetes, and etc.

the answer is definitely c.