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3 years ago

05. When a gold pebble is placed in a graduated cylinder that contains 12.0 mL of water, the water level rises

Chemistry

1 answer:

OLga [1]3 years ago

6 0

Answer:

142.82 g

Explanation:

The following data were obtained from the question:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Density of gol= 19.3 g/cm³

Mass of gold =.?

Next, we shall determine the volume of the gold. This can be obtained as follow:

Volume of water = 12 mL

Volume of water + gold = 19.4 mL

Volume of gold =.?

Volume of gold = (Volume of water + gold) – (Volume of water)

Volume of gold = 19.4 – 12

Volume of gold = 7.4 mL

Finally, we shall determine the mass of the gold as follow:

Note: 1 mL is equivalent to 1 cm³

Volume of gold = 7.4 mL

Density of gol= 19.3 g/cm³ = 19.3 g/mL

Mass of gold =?

Density = mass /volume

19.3 = mass of gold /7.4

Cross multiply

Mass of gold = 19.3 × 7.4

Mass of gold = 142.82 g

Therefore, the mass of the gold pebble is 142.82 g

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