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3 years ago

How many atoms are in 5 mole of water? Steps

Chemistry

1 answer:

Vladimir79 [104]3 years ago

3 0

Answer: 3.01 x 10^24 atoms

Explanation:

Based on Avogadro's law:

1 mole of any substance has 6.02 x 10^23 atoms

So, 1 mole of water = 6.02 x 10^23 atoms

5 moles of water = Z atoms

To get the value of Z, cross multiply

Z x 1 mole = (6.02 x 10^23 atoms x 5 moles)

Z•mole = 30.1 x 10^23 atoms•mole

Divide both sides by 1 mole

Z•mole/1 mole = 30.1 x 10^23 atoms•mole/ 1 mole

Z = 30.1 x 10^23 atoms

[Place the value of Z in standard form]

Z = 3.01 x 10^24 atoms

Thus, there are 3.01 x 10^24 atoms in 5 mole of water

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3 years ago

How can one kg of iron melt more ice than 1 kg lead at 100 °C

Vanyuwa [196]

Answer:

Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)

Explanation:

The given parameters are;

The metals provided to melt the ice and their temperature includes;

One kg (1000 g) of iron;

Specific heat capacity = 0.444 J/(g·°C)

Temperature = 100°C

1 kg (1000 g) of lead

Specific heat capacity = 0.160 J/(g·°C)

Temperature = 100°C

Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;

For iron, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ $_{iron}$ = Heat obtained from the iron by the ice

ΔQ $_{iron}$ = 0.444 m × 1000 × (100 - 0) = 44400 J

Heat absorbed by the ice for melting, H $_l$ = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H $_l$ = Mass of ice × Latent heat of ice

H $_l$ = Mass of ice × 334 J/g = 44400 J

Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g

Mass of ice melted by the iron ≈ 132.9 g

For lead, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ $_{lead}$ = Heat obtained from the iron by the ice

ΔQ $_{lead}$ = 0.160 m × 1000 × (100 - 0) = 16000 J

Heat absorbed by the ice for melting, H $_l$ = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H $_l$ = Mass of ice × Latent heat of ice

H $_l$ = Mass of ice × 334 J/g = 16000 J

Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g

Mass of ice melted by the lead ≈ 47.9 g

Therefore, mass of ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.

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3 years ago