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KengaRu [80]
3 years ago
9

Think about what you have learned about electrons, energy levels, and electron configurations. In 2 or 3 sentences, explain why

an atom of Argon, Ar is less reactive (more stable) than an atom of Chlorine Cl.
Chemistry
1 answer:
nignag [31]3 years ago
8 0
Alr u see. argon is in group 8 of the periodic table, also known as noble gases. noble gases all have full shells, octect. this means that there is no chance for covalent or ionic bonding, no valence electrons available to bonding.

however, chlorine is in group 7 of the periodic table. this means to obtain a full octect shell, they will have to undergo covalent bonding to form a full octect shell.

hence, argon is more stable compared to chlorine
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A sample of carbon dioxide occupies a 5.13 dm3 container at STP. What is the volume of the gas at a pressure of 286.5 kPa and a
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Considering the ideal gas law and STP conditions, the volume of the gas at a pressure of 286.5 kPa and a temperature of 12.9°C is 1.8987 L.

<h3>Definition of STP condition</h3>

The STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases. And in these conditions 1 mole of any gas occupies an approximate volume of 22.4 liters.

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  • 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.
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<h3>Volume of gas</h3>

In first place, you can apply the following rule of three: if by definition of STP conditions 22.4 L are occupied by 1 mole of carbon dioxide, 5.13 L (5.13 dm³= 5.13 L, being 1 dm³= 1 L) are occupied by how many moles of carbon dioxide?

amount of moles of carbon dioxide=\frac{5.13 Lx1 mole of carbon dioxide}{22.4 L}

<u><em>amount of moles of carbon dioxide= 0.229 moles</em></u>

Then, you know:

  • P= 286.5 kPa= 2.8275352 atm (being 1 kPa= 0.00986923 atm)
  • V= ?
  • T= 12.9 C= 285.9 K (being 0°C= 273 K)
  • R= 0.082 \frac{atmL}{mol K}
  • n= 0.229 moles

Replacing in the ideal gas law:

2.8275352 atm× V = 0.229 moles×0.082 \frac{atmL}{mol K} × 285.9 K

Solving:

V= (0.229 moles×0.082 \frac{atmL}{mol K} × 285.9 K)÷ 2.8275352 atm

<u><em>V= 1.8987 L</em></u>

Finally, the volume of the gas at a pressure of 286.5 kPa and a temperature of 12.9°C is 1.8987 L.

Learn more about

STP conditions:

brainly.com/question/26364483

brainly.com/question/8846039

brainly.com/question/1186356

the ideal gas law:

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