Respuesta :
Answer:
49,415.43 gram is the mass of a bronze cylinder.
Explanation:
Assuming that the given composition of bronze is mass percentage of copper and tin in bronze.
Let he mass of bronze be x.
Mass percentage of copper in bronze = 79.42%
Mass of copper = 79.42% of x=0.7942 x
Volume of the copper: [tex]V_1=\frac{Mass}{density}=\frac{0.7942 x}{8.94 g/cm^3}[/tex]
Mass percentage of copper in bronze = 20.58%
Mass of copper = 20.58% of x=0.2058 x
Volume of the copper: [tex]V_2=\frac{Mass}{density}=\frac{0.2058 x}{7.31 g/cm^3}[/tex]
Volume of the bronze cylinder = V
Radius of the cylinder = r = 6.44 cm
Height of the cylinder = h = 44.37 cm
[tex]V=\pi r^2h=3.14\times (6.44 cm)^2\times 44.37 cm=5,781.1073 cm^3[/tex]
[tex]V=V_1+V_2[/tex]
[tex]5,781.1073 cm^3=\frac{0.7942 x}{8.94 g/cm^3}+\frac{0.2058 x}{7.31 g/cm^3}[/tex]
x = 49,415.43 g
49,415.43 gram is the mass of a bronze cylinder.
Answer:
starting from binary alloy:
⇒ mass of alloy cylinder is 49430.896 g
Explanation:
be density of binary alloy (da) :
⇒ da = (MCu + MSn) / (VCu + VSn)
∴ VCu = MCu / dCu ∧ VSn = MSn/ dSn
⇒ da = (MCu + MSn) / ((MCu / dCu) + (MSn / dSn))
⇒ 1 / da = ((MCu / dCu) + (MSn/dSn)) / (MCu + MSn)
∴ MCu / (MCu + MSn) = XCu
∴ MSn / ( MSn + MCu ) = XSn
⇒ 1 / da = ( XCu /dCu ) + ( XSn / dSn )
∴ XCu = 0.7942; XSn = 0.2058
⇒ XCu / dCu = 0.7942 / 8.94 = 0.0888
⇒ Xsn / dSn = 0.2058 / 7.31 = 0.02815
⇒ 1 / da = 0.0888 + 0,02815 = 0.116953
⇒ da = 8.55042 g/cm³
Cylinder Volume:
V =[tex]\pi[/tex] * r² * l
∴ r = 6.44cm; ∧ l = 44.37cm
⇒⇒V = [tex]\pi[/tex] * (6.44cm)² * (44.37cm) = 5781.0738 cm³
mass cylinder:
- m = da * V
⇒ m = (8.55042 g/cm³) * (5781.0738 cm³)
⇒ m = 49430.896 g
