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Anna11 [10]
3 years ago
7

The vapor pressure of CCl3F at 300 K is 856 torr. If 11.5 g of CCl F is enclosed in a 1.0-L container, will any liquid be pres-

vap Cs,solid 118 J>mol 3 ent? If so, what mass of liquid?
Chemistry
1 answer:
Lemur [1.5K]3 years ago
7 0

Answer:

Yes, there will be liquid present and the mass is 5.19 g

Explanation:

In order to do this, we need to use the equation of an ideal gas which is:

<em>PV = nRT (1)</em>

<em>Where:</em>

<em>P: Pressure</em>

<em>V: Volume</em>

<em>n: number of moles</em>

<em>R: gas constant</em>

<em>T: Temperature</em>

we know that the pressure is 856 Torr at 300 K. So, if we want to know if there'll be any liquid present, we need to calculate the moles and mass of the CCl3F at this pressure and temperature, and then, compare it to the initial mass of 11.5 g.

From (1), solving for moles we have:

<em>n = PV/RT (2)</em>

Solving for n:

P = 856/760 = 1.13 atm

R = 0.082 L atm / mol K

n = 1.13 * 1 / 0.082 * 300

n = 0.0459 moles

Now, the mass is:

m = n * MM (3)

The molar mass of CCl3F reported is 137.37 g/mol so:

m = 0.0459 * 137.37

m = 6.31 g

Finally, this means that if we put 11.5 g of CCl3F in a container, only 6.31 g will become gaseous, so, this means it will be liquid present, and the mass is:

m = 11.5 - 6.31

m = 5.19 g

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How much energy, in joules, does 150.0 g of water with an initial temperature of 25 C need to absorb be raised to a final temper
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Answer:

31395 J

Explanation:

Given data:

mass of water = 150 g

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Energy absorbed = ?

Solution:

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q = m . c . ΔT

we know that specific heat of water is 4.186 J/g.°C

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2 years ago
A solution is made by dissolving 0.656 mol of nonelectrolyte solute in 869 g of benzene. calculate the freezing point, tf, and b
solmaris [256]
Answer is: the freezing point is 1.63°C and boiling point is 82.01°C.<span>.

1) n(</span><span>nonelectrolyte solute) = 0.656 mol.
</span>m(C₆H₆ - benzene) = 869 g ÷ 1000 g/kg.
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</span>
3) ΔTb = Kb(benzene) · b(solution).
ΔTb = 2.53°C/m · 0.754 m.
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</span>
4 0
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How many grams of oxygen are there in 45.7 grams of Ba(NO2),?
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