2007-06-16 13:24:51 · 4 answers · asked by archy 1 in Science & Mathematics ➔ Physics
The metals will be lighter in water. The mass of water is subtracted from the mass of the metal. Use these density values: Au = 19.3, Ag = 10.5, water = 1.0, And, since mass of water displaced = volume of water displaced:
Au = 1000g - (1000g / 19.3g) = 948.2g
Ag = 1000g - (1000g / 10.5g) = 904.8g
2007-06-20 05:31:03 · answer #1 · answered by Metallic stuff 7 · 0⤊ 0⤋
Call the density of the gold/platinum D. Immersion in water reduces the apparent mass by the mass of water displaced, which is 1/D liter in volume, 1/D kg in mass. Thus the apparent mass is 1 - 1/D kg.
2007-06-16 15:18:58 · answer #2 · answered by kirchwey 7 · 0⤊ 0⤋
1. multiply the 1kg by the density of gold.
2. mulitply the 1kg by the density of silver.
2007-06-16 13:48:03 · answer #3 · answered by Bob T 1 · 0⤊ 1⤋
Volume of 1kg of gold is 1/ 19,300 =5.181e-5m^3.
Mass of equal volume of water
= 5.181e-5 x 1000
=0.05181kg
Apparent mass = 1-0.05181 = 0.94819 kg.
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Volume of 1kg of silver is 1/ 10,500 = 9.524e-5
Mass of equal volume of water
= 9.524e-5 x 1000
=0.09524
Apparent mass = 0.90476
2007-06-16 19:58:08 · answer #4 · answered by Pearlsawme 7 · 0⤊ 0⤋