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A bullet seized from a crime scene has a composition of lead 11.6 g, tin 0.5 g, and
antimony 0.4 g. What is the percentage of lead in the bullet? Express your answers to
the ones place.

Respuesta :

dsdrajlin

Answer:

92.8%

Explanation:

Step 1: Given data

  • Mass of lead in the bullet (mPb): 11.6 g
  • Mass of tin in the bullet (mSn): 0.5 g
  • Mass of antimony in the bullet (mSb): 0.4 g

Step 2: Calculate the total mass of the bullet

The total mass of the bullet is equal to the sum of the masses of the elements that form it.

m = mPb + mSn + mSb = 11.6 g + 0.5 g + 0.4 g = 12.5 g

Step 3: Calculate the mass percentage of Pb in the bullet

We will use the following expression.

%Pb = mPb / m × 100%

%Pb = 11.6 g / 12.5 g × 100% = 92.8%

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