A steady-state plasma concentration of 25 mg/L v by IV infusion to healthy volunteers (average weight, 75 kg) at a rate of 7.5 m
1 answer:
Answer:
The correct answer is option B.
Explanation:
Concentration of a steady-state plasma ,= 25 mg/L
Weight of the volunteer =75 kg
Rate of infusion = 7.5 mg/kg hr
Concentration of steady-state plasma in 75 kg weight body:
Total body clearance of this drug:
The total body clearance of this drug is 22.5 L/hr.
You might be interested in
If 0.25mg of atropine is in 1mL
so
0.50mg of atropine is in x
so
0.50mg of atropine is in x
194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.
Explanation:
In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.
It is known that 1 moles of any element has 6.022×10²³ molecules.
Then 1 molecule will have moles.
So
Thus, 1.66 moles are included in BCl₃.
Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.
As it is known as 1 mole contains molecular mass of the compound.
As the molecular mass of BCl₃ will be
Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.
Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.
So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.