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A graduated beaker with 375 mL of water is sitting on a scale which measures the weight of the glass and water to be 7.60 N. When a rock is put into the glass, the volume level of the water changes to 450 mL and the scale reading changes to 9.22 N. What is the specific gravity of the rock

Respuesta :

Answer:

Volume of water displaced = 450 - 375 = 75 ml

Vr = volume of rock = 75 ml

Wr = 9.22 - 7.60 = 1.62 N  weight of 75 ml of rock

Density of rock = 1.62 N / 75 ml = .0216 N / ml

Density of water = 1000 g / 1000 ml = 9.8 N / 1000 ml = .0098 N / ml

Density of rock / density of water = .0216 / .0098 = 2.20

onyebuchinnaji

The specific gravity of the rock in the given water volume is 0.2.

The given parameters;

  • initial volume of the water, = 375 ml
  • weight of the water, = 7.6 N
  • final volume of water = 450 ml
  • change in scale reading = 9.22 N

The specific gravity of the rock is calculated as follows;

[tex]S.G = \frac{weight \ in \ air}{Weight \ in \ water} \\\\S.G = \frac{450 - 375}{375} \\\\S.G = 0.2[/tex]

Thus, the specific gravity of the rock in the given water volume is 0.2.

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