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3 years ago

A drug is administered at a rate of 13.5 cL/hour. What is this rate in µL/s?

Chemistry

1 answer:

lawyer [7]3 years ago

4 0

Answer:

A drug administered at a rate of 13.5 cL/hour has a rate in μL/s of 37.5 μL/s

Explanation:

The given rate at which the blood is administered = 13.5 cL/hour

The rate at which the question asked to be administered = μL/s

The conversion factor between cL and μL is given as follows;

1 cL = 10,000 μL

Which gives;

13.5 cL/hour × 10,000 μL/cL × 1 hour/(60 × 60 seconds) = 37.5 μL/s

Therefore, a drug administered at a rate of 13.5 cL/hour has an administration rate of 37.5 μL/s.

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Vera_Pavlovna [14]

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3 years ago

1) Look at the graph. What is the independent variable and the dependent variable?

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Answer:

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2 years ago

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4 years ago

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In one sentence, explain how the particles of a substance change as a substance changes from a solid to a liquid to a gas.

mestny [16]

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3 years ago

Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m

Vesna [10]

Answer : The correct option is, (A) -101.37 KJ

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH

Thus, the number of neutralized moles = 0.1392 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $87ml+87ml=174ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 174ml=174g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 174 g

$T_{final}$ = final temperature of water = 317.4 K

$T_{initial}$ = initial temperature of metal = 298 K

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

$q=174g\times 4.18J/g^oC\times (317.4-298)K$

$q=14110.008J=14.11KJ$

Thus, the heat released during the neutralization = -14.11 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -14.11 KJ

n = number of moles used in neutralization = 0.1392 mole

$\Delta H=\frac{-14.11KJ}{0.1392mole}=-101.37KJ/mole$

Therefore, the enthalpy of neutralization is, -101.37 KJ

3 0

3 years ago