Gaseous water molecules condense when they _. Liquid water molecules evaporate when they _.

1 answer:

3 0

Answer : Gaseous water molecules condense when they cooled. Liquid water molecules evaporate when they heated.

Explanation :

Condensation : It is a process in which a phase changes from gaseous state to liquid state. This process occurs when the water molecules in the air cool down and they join together to form liquid water.

Evaporation : It is a process in which a phase changes from liquid state to gaseous state. This process occurs when the water is heated, the molecules in the water move and vibrate quickly that they escape into the atmosphere as water vapor.

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Determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. group of answer choices

lilavasa [31]

Considering the ideal gas law and the definition of molar mass, the molar mass of the sample of gas is 9.17 $\frac{g}{mol}$ .

<h3>Ideal gas law</h3>

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:

P×V = n×R×T

where:

P is the gas pressure.
V is the volume that occupies.
T is its temperature.
R is the ideal gas constant. The universal constant of ideal gases R has the same value for all gaseous substances.
n is the number of moles of the gas.

<h3>Definition of molar mass</h3>

The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.

<h3>Molar mass of the sample of gas</h3>

In this case you know:

P= 0.980 arm
V= 1.20 L
T= 287 K
R= 0.082 $\frac{atmL}{molK}$
n= ?

Replacing in the ideal gas law:

0.980 atm× 1.20 L= n× 0.082 $\frac{atmL}{molK}$ × 287 K

Solving:

(0.980 atm× 1.20 L)÷ (0.082 $\frac{atmL}{molK}$ × 287 K)= n

<u><em>0.04997 moles= n</em></u>

On the other hand, you know that the<u><em> mass of the sample of gas</em></u> is <u><em>0.458 grams</em></u>. Replacing in the definition of molar mass:

$molar mass=\frac{0.458 grams}{0.04997 moles}$