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garri49 [273]
2 years ago
9

The addition of an inert gas has no effect on the equilibrium position of a gaseous reaction because ______. Multiple choice que

stion. the inert gas has no effect on the equilibrium system at all since it is not involved in the reaction the partial pressures of the reaction components remain the same the reaction quotient compensates for the inert gas the total pressure of the system remain the same
Chemistry
1 answer:
boyakko [2]2 years ago
5 0

Inert gas does not affect the equilibrium position:

It is because the partial pressures of the reaction components remain the same.

What is Inert Gas?

  • Under a given set of conditions, an inert gas is a gas that does not undergo chemical reactions.
  • The noble gases (helium, neon, argon, krypton, xenon, and radon) were previously known as "inert gases" due to their perceived lack of involvement in any biochemical processes.
  • Because inert gases are non-reactive, they do not affect equilibrium partial pressures and thus do not affect volume.
  • An inert gas does not react with the reactants or products; it does not change the concentration of the products and reactants. Furthermore, because the volume is constant, the concentrations are unaffected. As a result, this does not affect equilibrium.

The equilibrium position won't change if an inert gas is added. A volume change won't change the equilibrium position if the total moles of gas in the products and reactants are the same. When the volume is reduced, the process changes to create fewer moles of gas.

Learn more about the inert gas here,

brainly.com/question/15909389

#SPJ4  

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\text{Hg} \text{O} and \text{Hg}_{2} \text{O}.

Assuming complete decomposition of both samples,

  • m(\text{Hg}) = m(\text{residure})
  • m(\text{O}) = m(\text{loss})

First compound:

  • m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g

n = m/M; 0.6498 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.048 \; g  / 16 \; g \cdot mol^{-1}= 0.003 \; mol
  • n(\text{Hg atoms}) = 0.6018 \; g  / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol

Oxygen and mercury atoms seemingly exist in the first compound at a 1:1 ratio; thus the empirical formula for this compound would be \text{Hg} \text{O} where the subscript "1" is omitted.

Similarly, for the second compound

  • m(\text{O}) = m(\text{loss}) = 0.016 \; g
  • m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012  \; g

n = m/M; 0.4172 \; g of the first compound would contain

  • n(\text{O atoms}) = 0.016 \; g  / 16 \; g \cdot mol^{-1}= 0.001 \; mol
  • n(\text{Hg atoms}) = 0.4012 \; g  / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol

n(\text{Hg}) : n(\text{O}) \approx  2:1 and therefore the empirical formula

\text{Hg}_{2} \text{O}.

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3 years ago
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