The amount of Amoxicillin dose given to the 85.4 lb child daily is determined as 1,743.3 mg.
<h3>
What is the amount of Amoxicillin dose given to the child?</h3>
The amount of Amoxicillin dose given to the child is calculated as follows;
amount of Amoxicillin dose = weight of the child x dosage prescribed
<h3>What is the weight of the child in pounds (lb) </h3>
The weight of the child in pounds (lb) is calculated as follows;
1 lb = 0.453592 kg
85.4 lb = ?
= 85.4 x 0.453592 kg
= 38.74 kg
amount of Amoxicillin dose = 38.74 kg x 45 mg/kg
amount of Amoxicillin dose = 1,743.3 mg
Thus, the amount of Amoxicillin dose given to the 85.4 lb child daily is determined as 1,743.3 mg.
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The complete question is below:
A 85.4 lb child has a Streptococcus infection. Amoxicillin is prescribed at a dosage of 45 mg per kg of body weight per day given b.i.d. Calculate the daily dose of the child.
Keep in mind that when two moles of potassium chlorate are decomposed, 3 moles of oxygen are produced.Think of the coefficients: since the mole ratio is 2 to 3, it means that<span> 2 mol KClO3 will turn into 3 mol O2.
</span>
2kclO3->2kcl+3O2
Explanation:
As we know that sum of only total number of protons are also known as atomic number of an atom. And, the sum of total number of protons and neutrons is known as atomic mass of an atom.
Number of protons help in determining the identity of an atom whereas the number of neutrons can be same for two or more number of atoms. But the number of protons will remain the same for a particular atom and for its isotopes.
For example, tex]^{1}_{1}H[/tex] and 
are isotopes and each of them contains 1 proton.
thus, we can conclude that the number of protons in their nuclei would allow you to determine whether two atoms are of the same or different elements.
The answer is the third one because density is mass divided by volume so 80 divided by 10 is 8
Answer:
0.143L
Explanation:
Molar mass of H2SO4 = 98g/Mol
No of mole = mass/molar mass
No of mole= 49/98 = 0.5 mol
No of mol = concentration × volume
Volume = n/C = 0.5/3.5 = 0.143L
