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mestny [16]
1 year ago
14

A 1.0 mole sample of fluorine gas at 25 °C has an average molecular velocity of 415 m/s. What is the total KE of the gas sample?

Report your answer in kilojoules to the nearest tenth.
Chemistry
1 answer:
Mariana [72]1 year ago
4 0

The total kinetic energy of the gas sample is 3.3 KJ

<h3>What is kinetic energy? </h3>

This is the energy possessed by an object in motion. Mathematically, it can be expressed as:

KE = ½mv²

Where

  • KE is the kinetic energy
  • m is the mass
  • v is the velocity

<h3>How to determine the mass of the fluorine gas</h3>
  • Molar mass of fluorine gas = 38 g/mol
  • Mole of fluorine gas = 1 mole
  • Mass of fluorine gas = ?

Mass = mole × molar mass

Mass of fluorine gas = 1 × 38

Mass of fluorine gas = 38 g

<h3>How to determine the KE of the gas sample</h3>
  • Mass (m) = 38 g = 38 / 1000 = 0.038 Kg
  • Velocity (v) = 415 m/s
  • Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.038 × 415²

KE = 3272.275 J

Divide by 1000 to express in kilojoule

KE = 3272.275 / 1000

KE = 3.3 KJ

Learn more about energy:

brainly.com/question/10703928

#SPJ1

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The equilibrium constant is given as :

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The relation between K_p and K_c are :

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K_p = equilibrium constant at constant pressure = ?

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R = gas constant = 0.0821 L⋅atm/(K⋅mol)

T = temperature = 20.0°C =20.0 +273.15 K=293.15 K

\Delta n = change in the number of moles of gas = [(1) - (1 + 2)]=-2

Now put all the given values in the above relation, we get:

K_p=14.45\times (0.0821L.atm/K.mol\times 293.15 K)^{-2}

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