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Elis [28]
3 years ago
6

A student has two substances at a lab table. One substance is iron pyrite (fool's gold) and the other is real gold. After placin

g 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
Chemistry
2 answers:
blagie [28]3 years ago
6 0
The answer is A) 1 mL 
skad [1K]3 years ago
5 0

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=d_i=\frac{Mass}{Volume}=\frac{20 g}{4 mL}=5 g/mL

d_g=4\times d_i(Given)

d_g=4\times 5 g/mL=20 g/mL

Mass of gold = 20 g

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

D_g=20 g/mL=\frac{20 g}{V}

V = 1 mL

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

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9.31 g C₂H₂            1 mole               2 moles CO₂          44.0095 g
------------------  x  -------------------  x  ----------------------  x  -------------------  =
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=  31.5 g CO₂

3.8 g O₂             1 mole               2 moles CO₂          44.0095 g
-------------  x  --------------------  x  ----------------------  x  --------------------  =
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=  10.6 g CO₂

10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.

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