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4 years ago

When heat energy is removed, two changes are observed:

Chemistry

2 answers:

Valentin [98]4 years ago

8 0

Answer : The correct option is, A is condensation and B is freezing.

Explanation :

Melting or fusion : It is a type of process in which the phase changes from solid state to liquid state at constant temperature.

Freezing : It is a type of process in which the phase changes from liquid state to solid state at constant temperature.

Evaporation : It is a type of process in which the phase changes from liquid state to gaseous state at constant temperature.

Condensation : It is a type of process in which the phase changes from gaseous state to liquid state at constant temperature.

Sublimation : It is a type of process in which the phase changes from solid state to gaseous state without passing through the liquid state at constant temperature.

Deposition : It is a type of process in which the phase changes from gaseous state to solid state without passing through the liquid state at constant temperature.

The given observations are:

Change A: Gas changes phases to a liquid. → Condensation

Change B: Liquid changes phases to a solid. → Freezing

Hence, the correct option is, A is condensation and B is freezing.

harkovskaia [24]4 years ago

3 0

Answer: Option (A) is the correct answer.

Explanation:

Condensation is defined as the process in which vapors or gaseous phase changes into liquid phase.

Freezing is the process in which liquid phase changes into solid phase.

Evaporation is the process in which liquid phase changes into vapors or gaseous phase.

Melting is the process in which solid phase changes into liquid phase.

Therefore, we can conclude that names for given phase changes A and B is that A is condensation and B is freezing.

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Calculate the concentration of an aqueous solution of naoh that has a ph of 11.50

Reika [66]

NaOH is a strong base and complete dissociation into Na⁺ and OH⁻ ions.
Therefore [NaOH] = [OH⁻]
To calculate the [OH⁻], we can first find the pOH as NaOH is a basic solution.
pH + pOH = 14
Since pH = 11.50
pOH = 14 - 11.50
pOH = 2.50
We can calculate [OH⁻] by knowing pOH
pOH = -log[OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 3.2 x 10⁻³ M
therefore [NaOH] = 3.2 x 10⁻³ M

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Answer:

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Partial pressure of Helium, P1 = 105 mm Hg

Partial pressure of Nitrogen, P2 = 312 mm Hg

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According to Dalton's law of Partial pressures,

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