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Answer :
Final answer:
After converting the patient's weight from lb to kg, we multiply it by the dosage of activated charcoal (1 g/kg), getting an answer of 80 grams. Thus, the medical assistant should prepare 80 grams of activated charcoal.
Explanation:
The question inquires how many grams of activated charcoal should be prepared by a medical assistant for a patient weighing 176 lb, given that the dosage is 1 g/kg. To answer this, the weight of the patient originally given in pounds (lb) needs to be converted into kilograms (kg), since the activated charcoal dosage is given in grams per kilogram. As 1 lb equals approximately 0.453592 kg, by multiplying the patient's weight in lb by this conversion factor, the weight becomes approximately 80 kg (176 lb x 0.453592 kg/lb).
Since the dosage of activated charcoal is 1 g/kg, the total grams of activated charcoal required can be found by multiplying the weight in kg by the dosage. This results in 80 g, as 80 kg x 1 g/kg = 80 g. So, the correct answer is (b) 80 grams.
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Answer:
b) 80 g
Step-by-step explanation:
1 g/kg × 176 lb × (0.45359 kg)/(1 lb) = 79.83 g
Answer: 80 g
