Answer:
The specific gravity of the mixture is 0.943.
Explanation:
The density of the mixture ([tex]\rho_{M}[/tex]), measured in kilograms per cubic meters, is defined by the following equation:
[tex]\rho_{M} = \frac{m_{A}+m_{B}}{V_{A}+V_{B}}[/tex] (Eq. 1)
Where:
[tex]m_{A}[/tex], [tex]m_{B}[/tex] - Masses of liquids A and B, measured in kilograms.
[tex]V_{A}[/tex], [tex]V_{B}[/tex] - Volumes of liquids A and B, measured in cubic meters.
By applying the definition of density, we expand the density of the mixture as follows:
[tex]\rho_{M} = \frac{m_{A}+m_{B}}{\frac{m_{A}}{\rho_{A}}+\frac{m_{B}}{\rho_{B}} }[/tex]
[tex]\rho_{M} = \frac{m_{A}+m_{B}}{\frac{\rho_{B}\cdot m_{A}+\rho_{A}\cdot m_{B}}{\rho_{A}\cdot \rho_{B}} }[/tex]
[tex]\rho_{M} = \frac{\rho_{A}\cdot \rho_{B}\cdot (m_{A}+m_{B})}{\rho_{B}\cdot m_{A}+\rho_{A}\cdot m_{B}}[/tex] (Eq. 2)
If we know that [tex]\rho_{A} = 850\,\frac{kg}{m^{3}}[/tex], [tex]\rho_{B} = 1060\,\frac{kg}{m^{3}}[/tex] and [tex]m_{A} = m_{B} = 0.075\,kg[/tex], then the density of the mixture:
[tex]\rho_{M} = \frac{\left(850\,\frac{kg}{m^{3}} \right)\cdot \left(1060\,\frac{kg}{m^{3}} \right)\cdot (0.075\,kg+0.075\,kg)}{\left(1060\,\frac{kg}{m^{3}} \right)\cdot \left(0.075\,kg\right)+\left(850\,\frac{kg}{m^{3}} \right)\cdot \left(0.075\,kg\right)}[/tex]
[tex]\rho_{M} = 943.455\,\frac{kg}{m^{3}}[/tex]
And the specific gravity is the ratio of the density of the mixture to the density of water, that is:
[tex]SG = \frac{\rho_{M}}{\rho_{w}}[/tex] (Eq. 3)
If we get that [tex]\rho_{M} = 943.455\,\frac{kg}{m^{3}}[/tex] and [tex]\rho_{w} = 1000\,\frac{kg}{m^{3}}[/tex], then the specific gravity of the mixture is:
[tex]SG = \frac{943.455\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} }[/tex]
[tex]SG = 0.943[/tex]
The specific gravity of the mixture is 0.943.
The specific gravity of the mixture of the liquid A and liquid B is 0.8.
The given parameters;
The volume of each liquid is calculated as follows;
[tex]volume = \frac{mass}{density} \\\\volume_{A} = \frac{0.075}{850} \\\\volume_{A} = 8.82 \times 10^{-5}\\\\volume_{B} = \frac{0.075}{1060} \\\\volume_{B} = 7.08 \times 10^{-5}[/tex]
The specific gravity of the mixture is calculated as;
[tex]S.G = \frac{\rho _A}{\rho_B} \\\\S.G = \frac{m/V_A}{m/V_B} \\\\S.G = \frac{V_B}{V_A} \\\\S.G = \frac{7.08 \times 10^{-5}}{8.82 \times 10^{-5} } \\\\S.G = 0.8[/tex]
Thus, the specific gravity of the mixture of the liquid A and liquid B is 0.8.
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