Michael L. answered • 02/06/16

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Intuitively explains the concepts in Math and Science

Hi Karlee,

To find the specific gravity of a substance, you will need a fluid to submerge the substance in, if we take water as a standard, the density of water is ρ

_{water}= 1.00g/cm^{3}(1) M = 50g in V = 49 mL, 1 mL = 1 cm

^{3}, ρ is the densityρ=M/V = 50 g/ 49 cm

^{3}=**1.02 g/cm**^{3}Specific density = ρ/ρ

_{water}= 1.02g/cm^{3}/1g/cm^{3 }=**1.02****(sinks)**(2) M = 85g in V = 68 mL, 1 mL = 1 cm

ρ=M/V = 85 g/ 68 cm

Specific density = ρ/ρwater = 1.25g/cm

^{3}, ρ is the densityρ=M/V = 85 g/ 68 cm

^{3}=**1.25g/cm**^{3}Specific density = ρ/ρwater = 1.25g/cm

^{3}/1g/cm^{3}=**1.25 (sinks)**(3) M = 20g in V = 30 mL, 1 mL = 1 cm

ρ=M/V = 20g/ 30 cm

Specific density = ρ/ρwater = 0.667/cm

^{3}, ρ is the densityρ=M/V = 20g/ 30 cm

^{3}=**0.667g/cm**^{3}Specific density = ρ/ρwater = 0.667/cm

^{3}/1g/cm^{3}=**0.667 (floats)**(4) M = 15g in V = 60 mL, 1 mL = 1 cm

ρ=M/V = 15g/ 60cm

Specific density = ρ/ρwater = 0.25 g/cm3/1g/cm3 =

^{3}, ρ is the densityρ=M/V = 15g/ 60cm

^{3}=**0.25 g/cm**^{3}Specific density = ρ/ρwater = 0.25 g/cm3/1g/cm3 =

**0.25 (floats)**(5) M = 6g in V = 3 cm

ρ=M/V = 6g/3 cm

Specific density = ρ/ρwater = 2.00 g/cm

^{3}, ρ is the densityρ=M/V = 6g/3 cm

^{3}=**2.00 g/cm**^{3}Specific density = ρ/ρwater = 2.00 g/cm

^{3}/1g/cm^{3}=**2.00 (sinks)**