A children's pain relief elixir contains \(80 . \mathrm{mg}\) acetaminophen per \(0.50\) teaspoon. The dosage recommended for a child who weighs between 24 and \(35 \mathrm{lb}\) is \(1.5\) teaspoons. What is the range of acetaminophen dosages, expressed in mg acetaminophen/kg body weight, for children who weigh between 24 and \(35 \mathrm{lb}\) ?

Short Answer

Expert verified
The range of acetaminophen dosages for children weighing between 24 and 35 lbs is approximately between 15.10 mg/kg and 22.04 mg/kg.

Step by step solution

01

Calculate the total amount of acetaminophen for the recommended dosage

We know that there are 80 mg of acetaminophen per 0.50 teaspoon. To find the total amount of acetaminophen in the recommended dosage of 1.5 teaspoons, we need to set up a proportion and solve for the unknown: \(\frac{80\:mg}{0.50\:teaspoons} = \frac{x\:mg}{1.5\:teaspoons}\) Cross-multiply and solve for x: \(0.50 * x = 80 * 1.5\) \(x = \frac{80 * 1.5}{0.50} = 240\:mg\) So, the total amount of acetaminophen in the recommended dosage is 240 mg.
02

Convert the weight range from pounds to kilograms

We know that 1 pound is approximately equal to 0.453592 kg. To convert the given weight range (24 lbs to 35 lbs) into kilograms, we multiply each weight by this conversion factor: 24 lbs * 0.453592 = 10.886208 kg (rounded to 10.89 kg) 35 lbs * 0.453592 = 15.875720 kg (rounded to 15.88 kg) So, the weight range in kilograms is approximately 10.89 kg to 15.88 kg.
03

Find the dosages for the lightest and heaviest children in the given range

We need to find the acetaminophen dosages in mg/kg for: 1. A child weighing 10.89 kg 2. A child weighing 15.88 kg We'll use the recommended dosage of 240 mg of acetaminophen that we calculated in Step 1. For the lightest child (10.89 kg): Acetaminophen dosage = \(\frac{240\:mg}{10.89\:kg} = 22.04\:mg/kg\) For the heaviest child (15.88 kg): Acetaminophen dosage = \(\frac{240\:mg}{15.88\:kg} = 15.10\:mg/kg\)
04

Conclusion

The range of acetaminophen dosages for children weighing between 24 and 35 lbs is approximately between 15.10 mg/kg and 22.04 mg/kg.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A rectangular block has dimensions \(2.9 \mathrm{~cm} \times 3.5 \mathrm{~cm} \times 10.0 \mathrm{~cm}\). The mass of the block is \(615.0 \mathrm{~g}\). What are the volume and density of the block?

To determine the volume of a cube, a student measured one of the dimensions of the cube several times. If the true dimension of the cube is \(10.62 \mathrm{~cm}\), give an example of four sets of measurements that would illustrate the following. a. imprecise and inaccurate data b. precise but inaccurate data c. precise and accurate data Give a possible explanation as to why data can be imprecise or inaccurate. What is wrong with saying a set of measurements is imprecise but accurate?

The density of osmium (the densest metal) is \(22.57 \mathrm{~g} / \mathrm{cm}^{3} .\) If a \(1.00-\mathrm{kg}\) rectangular block of osmium has two dimensions of \(4.00 \mathrm{~cm} \times 4.00 \mathrm{~cm}\), calculate the third dimension of the block.

A \(25.00-\mathrm{g}\) sample of a solid is placed in a graduated cylinder and then the cylinder is filled to the \(50.0-\mathrm{mL}\) mark with benzene. The mass of benzene and solid together is \(58.80 \mathrm{~g}\). Assuming that the solid is insoluble in benzene and that the density of benzene is \(0.880 \mathrm{~g} / \mathrm{cm}^{3}\), calculate the density of the solid.

A rule of thumb in designing experiments is to avoid using a result that is the small difference between two large measured quantities. In terms of uncertainties in measurement, why is this good advice?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free