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svp [43]
3 years ago
8

An unknown gas is contained in a sealed container. Over time, the gas is gradually cooled until it becomes a solid. Determine wh

ich
statements accurately describe the graphic representation of the cooling process. Choose all of the correct answers.
es )
A)
The freezing point of the gas is 100°C.
B)
It will take 140 minutes for the gas to completely solidify.
C)
D)
As the gas changes state, the intermolecular attraction of the molecules
increases.
Droplets of the gas will begin to condense on the sides of the container at
125°C.
As the temperature of the gas decreases over time, the kinetic energy of
the molecules increase
E)
Chemistry
2 answers:
otez555 [7]3 years ago
7 0

Answer:

BCD

Explanation:

Alecsey [184]3 years ago
5 0

Answer:D molecular attraction increases as temp decreases

Explanation:A,B would require data not available. C is missing. The question is a mess.

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ivolga24 [154]

Answer:

Cobalt Sources

Cobalt-60 is used as a radiation source in many common industrial applications, such as in leveling devices and thickness gauges. It is also used for radiation therapy in hospitals. Accidental exposures may occur as the result of loss or improper disposal of medical and industrial radiation sources.

Explanation:

4 0
3 years ago
Read 2 more answers
1.33 dm3 of water at 70°C are saturated by 2.25
astraxan [27]

Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the

amount amount of solute that will be deposited is 1,927.413 grams.

<h3>How can the amount of solute deposited be found?</h3>

The volume of water 1.33 dm³ of water 70 °C.

The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C  = 2.25 moles

At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles

Therefore;

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;

1.33 dm³ contains 2.25 moles.

Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{2.25}{1.33} \times 4.50 \approx \mathbf{7.613 \, moles}

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;

1.33 dm³ contains 0.53 moles

Number \ of \ moles \ in \ 4.50 \ dm^3 = \dfrac{0.53}{1.33} \times 4.50 \approx \mathbf{1.79 \, moles}

Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles

The number of moles that precipitate out = The amount of solute deposited

Which gives;

Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles

The molar mass of Pb(NO₃)₂ = 207 g + 2 × (14 g + 3 × 16 g) = 331 g

The molar mass of Pb(NO₃)₂ = 331 g/mol

The amount of solute deposited = Number of moles × Molar mass

Which gives;

The amount of solute deposited = 5.823 moles × 331 g/mol =<u> 1,927.413 g </u>

Learn more about saturated solutions here:

brainly.com/question/2624685

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