a).
k = 0.693/t1/2
ln(No/N) = kt
No = 1.5mg
N = unknown mg
k = from above.
t = 5.25 hours.
b)
Same equation as above.
No = 1.50 mg
N = 2.50E-6 mg
k from above
solve for t in hours.
Nitrogen Oxides, NOx (mixture of NO and NO2).
The NOx in the atmosphere is slowly broken down to N2 and O2 in a first-order reaction. The average half-life of NOx in the smokestack emissions in a large city during daylight is 3.9 hours.
(a) Starting with 1.50 mg in an experiment, what quantity of NOx remains after 5.25 hours?
(b) How many hours of daylight must have elapsed to decrease 1.50 mg of NOx to 2.50X10^-6 mg?
2 answers
Thank you Bob