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sweet-ann [11.9K]
2 years ago
8

How many formula units are in 4.56 moles of LiCl?

Chemistry
1 answer:
pashok25 [27]2 years ago
5 0

Answer:

In 4.56 moles of LiCl 27.46*10^23 formula units is present.    

Explanation:

Formula units is defined as the empirical formula for any ionic, covalent and  network solid compounds which are used as a single unit for stoichiometric calculations. Also It can be defined as the lowest whole number ratio of ions which can be represented in an ionic compound.

We are given

4.65 moles of Lithium chloride.

One mole of Lithium chloride has 6.022*10^23 unit  Formula unit

In 4.56 moles of Lithium Chloride the formula unit will be

              = 4.56*6.022*10^23

Formula units    = 27.46 * 10^23

It is clear that in 4.56 moles of Lithium Chloride will contain 27.46*10^23 Formula unit.

Therefore in 4.56 moles of LiCl 27.46*10^23 formula units is present.    

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

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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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