The 3 (THREE) particles of the atoms are:
_______________________________________________
protons, neutrons, and electrons .
__________________________________________________________

8 0

2 years ago

The drug labetalol has a dose of 300 mg and is administered in 300 mL of an intravenous infusion at a rate of 2 mg/min. Using an

Evgesh-ka [11]

Answer:

Explanation:

From the given information:

Since 300 ml of intravenous infusion contains 300 mg of drugs;

Then; 1 ml of intravenous infusion contains (x) amount of drugs

(x) amount of drugs = (300 mg × 1 ml)/ 300 ml

(x) amount of drugs = 1 mg.ml

To calculate the drip rate in drops/minutes by applying conversion factors.

$rate (\dfrac{ in \ drops }{minute })= \dfrac{2 \ mg }{min } \times \dfrac{20 \ drops}{1 \ ml} \times \dfrac{1 \ ml }{1 \ mg }$

$rate (\dfrac{ in \ drops }{minute }) = 40 \ drops / min$

4 0

2 years ago

An ideal gas originally at 0.85 atm and 66°C was allowed to expand until its final volume, pressure, and temperature were 94 mL,

svet-max [94.6K]

Answer : The initial volume was, 71.2 mL

Explanation :

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=0.85atm\\V_1=?\\T_1=66^oC=[66+273]K=339K\\P_2=0.60atm\\V_2=94mL\\T_2=43^oC=[43+273]K=316K$

Now put all the given values in above equation, we get:

$\frac{0.85atm\times V_1}{339K}=\frac{0.60atm\times 94mL}{316K}$

$V_1=71.2mL$

Therefore, the initial volume was, 71.2 mL

4 0

3 years ago

An exergonic reaction __________ free energy, and an endergonic reaction __________ free energy.

An exergonic reaction free energy, and an endergonic reaction free energy.