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klio [65]
2 years ago
10

Calculate each of the following quantities: (g) Number of atoms in 0.0015 mol of fluorine gas

Chemistry
1 answer:
Inessa05 [86]2 years ago
8 0

1.81×10^{21} atoms are in 0.0015 mol of fluorine gas

<h3>What is fluorine?</h3>

The chemical element fluorine has an atomic number of 9 and the symbol F. The lightest halogen, it is an extremely poisonous, pale yellow diatomic gas under normal conditions. With the exception of argon, neon, and helium, it reacts with every other element, making it the most electronegative and reactive element.

Fluorine is the 24th most abundant element overall and the 13th most abundant on Earth. When added to metal ores to decrease their melting temperatures for smelting, fluorite, the main mineral source of fluorine that gave the element its name, was first recorded in 1529. The Latin word fluo, which means "flow," gave the mineral its name.

Learn more about fluorine

brainly.com/question/24373411

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what is the volume of the air in a balloon that occupies 0.730 L at 28.0 c if the temperature is lowered to 0.00 C
svetoff [14.1K]

Answer:

The volume of the air is 0.662 L

Explanation:

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:

\frac{V}{T}=k

If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

\frac{V1}{T1}=\frac{V2}{T2}

In this case:

  • V1= 0.730 L
  • T1= 28 °C= 301 °K (0°C= 273°K)
  • V2= ?
  • T2= 0°C= 273 °K

Replacing:

\frac{0.730 L}{301K}=\frac{V2}{273K}

Solving:

V2=273K*\frac{0.730L}{301K}

V2=0.662 L

<u><em>The volume of the air is 0.662 L</em></u>

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