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vladimir1956 [14]
3 years ago
7

The average dosage of oxcarbazepine for an epileptic child between the ages of 4 and 16 is 9.00 mgmg per 1 kgkg of body weight (

9.00 mg/kgmg/kg). Calculate how many milliliters of oxcarbazepine a child who weighs 42.9 pounds should be prescribed considering the medication is served as a suspension of 60.0 mg/mLmg/mL. Keep in mind that 1 pound is equivalent to 0.453 kilograms.​
Chemistry
1 answer:
Helga [31]3 years ago
5 0

Answer:

The answer to your question is 3 ml

Explanation:

Data

Dosage = 9.0 mg/ kg

Child's weight = 42.9 pounds

Suspension = 60 mg/ml

milliliters = ?

Process

1.- Convert the weight to kg

                         1 pound ------------------- 0.453 kg

                     42.9 pounds ---------------  x

                           x = (42.9 x 0.453) / 1

                           x = 19.43 kg

2.- Calculate the milligrams the child needs

                         1 kg of weight ------------ 9 mg

                       19.43 kg ----------------------  x

                            x = (19.43 x 9) / 1

                            x = 174.87 mg of oxcarbazepine

3.- Calculate the milliliters needed

                          60 mg of suspension ------------- 1 milliliters

                        174.87 mg                     --------------  x

                                 x = (174.87 x 1) / 60

                                 x = 2.9 ml ≈ 3 ml

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Determine the number of atoms in 41.0 grams of calcium, Ca. (The mass of one mole of calcium is 40.08 g.)
Dovator [93]

Hi there!

Let me know if you have questions about my answer:

6.15 × 10²⁴

Explanation:

To find the number of atoms, divide the mass of calcium (m) by its molar mass (MM). This will give you the number of moles of calcium (n).

n = m/MM            Start with a formula

=\frac{41.0g}{40.08g/mol}             Substitute values from the question

= \frac{41.0g*1mol}{40.08g}             Simplify

= \frac{41.0mol}{40.08}                 Cancel out the units "g"

= 1.022...mol          Number of moles of calcium, with 4 significant figures

Then, multiply the number of moles of calcium (n) by Avogadro's number (N_{A}) to find the number of calcium atoms.

Number.of.Ca=n*N_{A}      Start with a formula

= 1.022mol*\frac{6.022*10^{23}Ca}{1mol}      Substitute values from the question

= \frac{1.022mol*6.022*10^{23}Ca}{1mol}           Simplify

= 1.022*(6.022*10^{23}Ca)     Cancel out the units "mol"

= 6.154*10^{24} Ca                   Multiply, leave one extra digit to help round. "4" tells you to round down.

= 6.15*10^{24} Ca                     Final answer to 3 significant figures*

*When multiplying and dividing numbers, your final answer will have the same number of significant figures as the number with the least number of significant figures in the question.

40.08 has 4 significant figures. 41.0 has 3 significant figures. So, you want 3 significant figures in your final answer.

I hope this helps!

Learn more about Avogadro's number here:

brainly.com/question/1445383

Check out another answer that uses significant figures:

brainly.com/question/3721164

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