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This gives a = 100 μg/mL if the drug stays in the blood stream only, and thus its volume of distribution is the same as that is = 0.08 L/kg. If the drug distributes into all body water the volume of distribution would increase to approximately V D = {\displaystyle V_{D}=} 0.57 L/kg [ 8 ]
Usually, clearance is measured in L/h or mL/min. [2] The quantity reflects the rate of drug elimination divided by plasma concentration. Excretion, on the other hand, is a measurement of the amount of a substance removed from the body per unit time (e.g., mg/min, μg/min, etc.). While clearance and excretion of a substance are related, they are ...
The procedure is to take the child's weight in pounds, divide by 150 lb, and multiply the fractional result by the adult dose to find the equivalent child dosage.For example, if an adult dose of medication calls for 30 mg and the child weighs 30 lb, divide the weight by 150 (30/150) to obtain 1/5 and multiply 1/5 times 30 mg to get 6 mg.
Another use is in the therapeutic drug monitoring of drugs with a narrow therapeutic index. For example, gentamicin is an antibiotic that can be nephrotoxic (kidney damaging) and ototoxic (hearing damaging); measurement of gentamicin through concentrations in a patient's plasma and calculation of the AUC is used to guide the dosage of this drug ...
mg milligram mg/dL milligrams per deciliter MgSO4 magnesium sulfate: may be confused with "MSO4", spell out "magnesium sulfate" midi at midday min. minimum [or] minim [or] minutum: minimum [or] minim [or] minute mist. mistura: mixture mL millilitre mod. præscript. modo præscripto: in the manner directed MS morphine sulfate or magnesium sulfate
That immediately gets the drug's concentration in the body up to the therapeutically-useful level. First day: 1000 mg; the body clears 100 mg, leaving 900 mg. On the second day, the patient takes 100 mg, bringing the level back to 1000 mg; the body clears 100 mg overnight, still leaving 900 mg, and so forth.
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And since is fraction of the drug that is removed per unit time measured at any particular instant, then if we divide the rate of elimination by the amount of drug in the body at time t, we get; K = d E t d t ÷ A t = ln 2 t 1 / 2 ≈ 0.693 t 1 / 2 {\displaystyle K={dE_{t} \over dt}\div A_{t}={\frac {\ln 2}{t_{1/2}}}\approx {\frac {0.693}{t ...