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Hunter-Best [27]
4 years ago
8

An infant acetaminophen suspension contains 80 mg/0.80 mL suspension. The recommended dose is 15 mg/kg body? weight. How many mL

of this suspension should be given to a child weighing 13 ibs. please help me, im at a total loss at how to approach this problem
Chemistry
2 answers:
jeyben [28]4 years ago
5 0
<span>First, the child's weight must be converted from pounds to kilograms. 1 lb is equal to 0.453592 kg, so a 13 lb child weighs: 13 lb * 0.453592 kg/lb = 5.896696 kg Next, use the child's converted weight to determine the mg dosage. The recommended dose is 15 mg per kg, so the recommended dose is: 15 mg/kg * 5.896696 kg = 88.45044 mg Finally, determine how many mL are needed to provide the calculated mg dosage. One unit of the suspension is 80 mg/0.80 mL. In order to provide 88.45044 mg, you will need 88.45044 mg / 80 mg = 1.1056305 units of the suspension. Multiplying this by the 0.80 mL portion of the unit of the suspension, you get the final mL dosage: 0.80 mL * 1.1056305 = 0.8845044 mL A 13 lb child should receive 0.8845044 mL of the 80mg/.8mL suspension.</span>
lana66690 [7]4 years ago
4 0

\boxed{{\text{0}}{\text{.8845 mL}}} of an infant acetaminophen suspension should be given to a child weighing 13 lbs.

Further Explanation:

The unit is a standard used for the comparison of measurements. Units can be basic or derived units. The units that cannot be further reduced into any other unit are called basic units. For example, mass, time and temperature are basic units. The units that need basic units to express themselves are called derived units. Volume, area, and density are some examples of derived units.

Conversion factors are the ratios that are described in the form of fractions whose multiplication with the original unit yields the desired units.

The weight of the child has to be converted into kg. The conversion factor for this is,

 1{\text{ lbs}} = 0.453592{\text{ kg}}

So the weight of the child can be calculated as follows:

 \begin{aligned}{\text{Weight of child}}&= \left( {13{\text{ lbs}}} \right)\left( {\frac{{0.453592{\text{ kg}}}}{{1{\text{ lbs}}}}} \right)\\&= 5.896696{\text{ kg}}\\\end{aligned}

The recommended dose per kg of the bodyweight is 15 mg. Therefore the recommended dose for the child can be calculated as follows:

 \begin{aligned}{\text{Recommended dose}} &= \left( {\frac{{15{\text{ mg}}}}{{1{\text{ kg}}}}} \right)\left( {5.896696{\text{ kg}}} \right)\\&= 88.45044{\text{ mg}}\\\end{aligned}

An infant acetaminophen suspension contains 80 mg per 0.80 mL suspension. Therefore the quantity of suspension required for the child can be calculated as follows:

\begin{aligned}{\text{Quantity of suspension}}&= \left( {\frac{{0.80\,{\text{mL}}}}{{80\;{\text{mg}}}}} \right)\left( {88.45044{\text{ mg}}} \right)\\&= 0.8845044{\text{ mL}}\\&\approx {\text{0}}{\text{.8845 mL}}\\\end{aligned}  

Learn more:

  1. What is the mass of 1 mole of viruses: brainly.com/question/8353774
  2. Determine the moles of water produced: brainly.com/question/1405182

Answer details:

Grade: Senior School

Chapter: Basic concepts of chemistry

Subject: Chemistry

Keywords: unit, conversion factor, basic, derived units, lbs, kg, 0.8845 mL, 88.45044 mg, 13 lbs, 5.896696 kg, 0.453592 kg, 1 lbs.

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Ca(OH)₂(s) precipitates when a 1.0 g sample of CaC₂(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g sa
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D) Ca(OH)₂ will not precipitate because Q <  Ksp

Explanation:

This is a question in which we will employ the reaction quotient Q to determine whether a precipitate will form.

Here we have first a chemical reaction in which Ca(OH)₂  is produced:

CaC₂(s)  + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂  is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product  constant for Ca(OH)₂  is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

Thus the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M  = 0.002 M  ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q=  [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

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