C C Q = 150
ke = 0.046/hr and V C
notice how
the units are consistent: hr in k
Answer: C
C C Q = 150
ke = 0.046/hr and V C
notice how
the units are consistent: hr in k
Answer: C
C C Q = 300
ke = 0.046/hr and V so C
notice how
the units are consistent: hr in k
Answer: C
C C Q = 450
ke = 0.046/hr and V so C
notice how
the units are consistent: hr in k
Answer: C
f = 1 -e this equation allows calculation of the time (t) need to reach a new steady-state concentration following a constant infusion of drug; f is the fraction of the shift from one steady state to the other. The question here asks for a 50% or 0.5 shift. so: f
= 1 -e or 0.5 = 1 - e or 0.5 = e now take antilogs: -0.693 = -0.046 t -0.693/-0.046 = t
Answer: t = 15 hours;
note that a drug
will always reach 50% of C
Q = ? C C Q = ?
ke = 0.046/hr and V 25 ug/ml = Q / ((0.046/60 min) * 40,000 ml) Answer: Q = 760 ug/min
f = 1 -e this equation allows calculation of the time (t) need to reach a new steady-state concentration following a constant infusion of drug; f is the fraction of the shift from one steady state to the other. The question here asks for a 90% or 0.9 shift. so: f
= 1 -e or 0.9 = 1 - e or 0.1 = e now take antilogs: -2.3 = -0.046 t -2.3/-0.046 = t Answer: t = 50 hours; |