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Answer :
Final answer:
To find the mass of manganese in a 196 lb person's body, convert the weight into kilograms and then multiply by the percentage by mass of manganese. Approximately 1.15557 grams of manganese would be found in such a person.
Explanation:
To calculate the amount of manganese in the body of a person weighing 196 lb, we first convert the weight into kilograms by dividing by 2.205. Then, we use the given percentage by mass to find the mass of manganese in the body.
Step 1: Convert pounds to kilograms: 196 lb ÷ 2.205 lb/kg = 88.89 kg
Step 2: Calculate the mass of manganese: (1.3 × 10⁻⁴) × 88.89 kg = (1.3 × 10⁻⁴) × 88,890 g = 1.15557 g of manganese
Therefore, a person weighing 196 lb would have approximately 1.15557 grams of manganese in their body.
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Rewritten by : Barada
Given the mass percentage of Manganese in a normal healthy body is [tex]1.3*10^{-4}%[/tex]
It means that [tex]1.3*10^{-4}[/tex]g Mn is present in 100 g of the body mass.
Given that the person weighs 196 lbs
Converting the person's mass to g: The conversion factors to be used are: 1 kg = 2.205 lb; 1 kg = 1000 g
[tex]196lbs*\frac{1kg}{2.205lbs}*\frac{1000g}{1kg}= 8.89*10^{4}g[/tex]
Calculating the mass of Mn in the person's body:
[tex]8.89*10^{4}g body*\frac{1.3*10^{-4}g Mn}{100 g body}=0.116g Mn[/tex]
Therefore, 0.116 g Mn is present in a person weighing 196 lbs.
