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If 500 mL of mineral oil are used to prepare a liter of mineral oil emulsion, how many grams of the oil, having a specific gravity of 0.87, would be used in the preparation of 1 gallon of the emulsion?

Sagot :

680 answers

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We know that  

1 gallon = 3.78541 L  

1000 ml = 1.0 L

Density = mass / volume  

Mass = volume *density

Therefore  

(3.78541 L emulsion) x (500 mL oil / 1 L emulsion) x (0.87 g/mL oil) = 1647 g oil

Density = 0.87 g/ ml  

Because the density of water = 1.0 g/ ml