Sure, let's solve this problem step-by-step using basic principles of mass, volume, and density:
Given:
- The total mass of the alloy [tex]\( M_{\text{alloy}} \)[/tex] is 25.5 grams.
- The total volume of the alloy [tex]\( V_{\text{alloy}} \)[/tex] is 3 cubic centimeters.
- The density of tin ([tex]\( \rho_{\text{tin}} \)[/tex]) is 7.3 grams per cubic centimeter.
- The density of lead ([tex]\( \rho_{\text{lead}} \)[/tex]) is 11.3 grams per cubic centimeter.
Let's denote:
- The mass of the tin in the alloy as [tex]\( m_{\text{tin}} \)[/tex].
- The mass of the lead in the alloy as [tex]\( m_{\text{lead}} \)[/tex].
From the problem, we can set up two equations based on the given information.
1. The sum of the masses of tin and lead equals the total mass of the alloy:
[tex]\[ m_{\text{tin}} + m_{\text{lead}} = 25.5 \, \text{grams} \][/tex]
2. The sum of the volumes of tin and lead equals the total volume of the alloy. We can express the volumes in terms of mass and density:
[tex]\[ \frac{m_{\text{tin}}}{\rho_{\text{tin}}} + \frac{m_{\text{lead}}}{\rho_{\text{lead}}} = 3 \, \text{cubic centimeters} \][/tex]
Substituting the given densities, we get:
[tex]\[ \frac{m_{\text{tin}}}{7.3} + \frac{m_{\text{lead}}}{11.3} = 3 \][/tex]
Now we have a system of two linear equations:
1. [tex]\( m_{\text{tin}} + m_{\text{lead}} = 25.5 \)[/tex]
2. [tex]\( \frac{m_{\text{tin}}}{7.3} + \frac{m_{\text{lead}}}{11.3} = 3 \)[/tex]
We solve these simultaneous equations step-by-step.
First, solve Equation 1 for [tex]\( m_{\text{lead}} \)[/tex]:
[tex]\[ m_{\text{lead}} = 25.5 - m_{\text{tin}} \][/tex]
Substitute this expression into Equation 2:
[tex]\[ \frac{m_{\text{tin}}}{7.3} + \frac{25.5 - m_{\text{tin}}}{11.3} = 3 \][/tex]
Combine terms:
[tex]\[ \frac{m_{\text{tin}}}{7.3} + \frac{25.5}{11.3} - \frac{m_{\text{tin}}}{11.3} = 3 \][/tex]
Multiply every term by [tex]\(7.3 \times 11.3\)[/tex] to clear the denominators:
[tex]\[ 11.3 (m_{\text{tin}}) + 7.3 (25.5) - 7.3 (m_{\text{tin}}) = 3 \times 7.3 \times 11.3 \][/tex]
Let’s simplify each term:
[tex]\[ 11.3 m_{\text{tin}} - 7.3 m_{\text{tin}} = 3 \times 7.3 \times11.3 - (7.3 \times25.5) \][/tex]
Combine like terms:
[tex]\[ 4 m_{\text{tin}} = 228.57 - 186.15 \][/tex]
[tex]\[ 4 m_{\text{tin}} = 42.42 \][/tex]
[tex]\[ m_{\text{tin}} = \frac{42.42}{4} \][/tex]
[tex]\[ m_{\text{tin}} = 10.605 \][/tex]
Therefore, the mass of the tin in the alloy is approximately [tex]\( 15.33 \)[/tex] grams.
Using the mass of tin [tex]\(m_{\text{tin}}\)[/tex], we can find the mass of lead [tex]\( m_{\text{lead}}\)[/tex]:
[tex]\[ m_{\text{lead}} = 25.5 - 15.33 = 10.17 \][/tex]
Thus, the mass of tin in the alloy is [tex]\(15.33\)[/tex] grams, and the mass of lead in the alloy is [tex]\(10.17\)[/tex] grams.
