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A student has two substances at a lab table. One substance is iron pyrite (fool's gold) and the other is real gold. After placing a 20.0-gram sample of iron pyrite into 40.0 mL of water in a graduated cylinder, it displaces 4.0 mL of water. The density of real gold is almost 4 times larger than iron pyrite. If a 20.0-gram sample of real gold is placed into the same amount of water, approximately how much water will be displaced?
A) 1 mL
B) 4 mL
C) 16 mL
D) 20 mL

Respuesta :

The answer is A) 1 mL 

Answer:The correct answer is option A.

Explanation

Mass of the iron pyrite= 20 grams

Volume water displaced by the iron pyrite = 4 mL

Density of iron pyrite=[tex]d_i=\frac{Mass}{Volume}=\frac{20 g}{4 mL}=5 g/mL[/tex]

[tex]d_g=4\times d_i[/tex](Given)

[tex]d_g=4\times 5 g/mL=20 g/mL[/tex]

Mass of gold = 20 g

Volume of water displaced by the 20 grams of water = V

[tex]D_g=20 g/mL=\frac{20 g}{V}[/tex]

V = 1 mL

20 grams of gold will displace 1 ml of water.So, the correct answer is option A.

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