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valentinak56 [21]
3 years ago
13

how does the energy required to sublime a substance compare to the energy required to melt the substance

Chemistry
1 answer:
Darya [45]3 years ago
3 0
The energy required to sublime (solid to gas) a substance at 1 ATM pressure is greater than the energy required to melt (solid to liquid) a substance. When you compare the energies in varying pressures, however, this trend is not always the case. 

<span>any 'general phase diagram', you can see that under the triple point, when all phases are in equilibrium, have solid and gas meeting under a certain pressure. In a vacuum, it would require less energy to sublime than to melt.</span>
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Answer:

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Explanation:

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3 years ago
What is the partial pressure of carbon dioxide in a container that contains 3.63 mol of oxygen, 1.49 mol of nitrogen, and 4.49 m
lana66690 [7]

Answer:

Partial pressure of CO₂ is 406.9 mmHg

Explanation:

To solve the question we should apply the concept of the mole fraction.

Mole fraction = Moles of gas / Total moles

We have the total moles of the mixture, if we have the moles for each gas inside. (3.63 moles of O₂, 1.49 moles of N₂ and 4.49 moles of CO₂)

Total moles = 3.63 mol O₂ + 1.49 mol N₂ + 4.49 mol CO₂ = 9.61 moles

To determiine the partial pressure of CO₂ we apply

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Partial pressure of CO₂ = (mol of CO₂ / Total moles) . Total pressure

We replace values: (4.49 moles / 9.61 moles) . 871 mmHg = 406.9 mmHg

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3 years ago
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What can be a catchphrase for cobalt if it were a superhero?
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Answer:lumerman.com › catchphrases › catchphrase

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A fluid occupying has a mass of 4mg. Calculate its density and specific volume in SI, EE, and BG units.
kondaur [170]

The question is incomplete, complete question is:

A fluid occupying 3.2 m^3 of volume has a mass of 4 Mg. Calculate its density and specific volume in SI, EE, and BG units.

Explanation:

1) Mass of liquid = m = 4 Mg = 4 × 1,000 kg = 4,000 kg

(1 Mg = 1000 kg)

Volume of the fluid = V = 3.2 m^3

Density of the fluid = D

D=\frac{m}{V}=\frac{4,000 kg}{3.2 m^3}=1,250 kg/m^3

Specific volume is the reciprocal of the density :

V_{specific}=\frac{1}{Density}

Specific volume of the fluid = S_v

S_v=\frac{1}{D}=\frac{1}{1,250 kg/m^3}=0.0008 m^3/kg

2)

Density of the fluid in English Engineering units  = D (lb/ft^3)

1 kg = 2.20462 lb

1 m = 3.280 ft

D=\frac[1,250\times 2.20462 lb}{(3.280 ft)^3=78.95 lb/ft^3

Specific volume of the fluid :

=\frac{1}{78.95 lb/ft^3}=0.0127 ft^3/lb

3)

Density of the fluid in British Gravitational System units  = D (slug/ft^3)

1 kg = 0.06852 slug

1 m = 3.280 ft

D=\frac[1,250\times 0.0685218 slug}{(3.280 ft)^3=2.43 slug/ft^3

Specific volume of the fluid :

=\frac{1}{2.43 slug/ft^3}=0.412 ft^3/slug

7 0
3 years ago
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