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Anni [7]
3 years ago
10

45.6 grams of CH4 is placed into a 275 mL container at 55.00C. What is the pressure inside the container in atm?

Chemistry
1 answer:
xxTIMURxx [149]3 years ago
4 0

<em>Given-</em>

<em>Mass of CH4 - 45.6 grams</em>

<em>Molar mass of CH4 - 16.04 g/mol</em>

<em>volume - 275mL</em>

<em>Temperature- 55°C</em>

<em>Solution,</em>

<em>moles of methane (CH4) n =</em><em> </em><em>45.6/16.04 </em>

<em>= 2.84 moles</em>

<em>Using Formula,</em>

<em>Pv = nRT</em>

<em>P*275 = 2.84*8.314*55</em>

<em>P</em><em> = 1298.64/275</em>

<em>P </em><em>= 4.72 Atm.</em>

<em>Answer - The pressure inside the container in atm is 4.72.</em>

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Answer:

Only changes in temperature will influence the equilibrium constant K_c. The system will shift in response to certain external shocks. At the new equilibrium Q will still be equal to K_c, but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of Q. For most reversible reactions:

  • External changes in the relative concentration of the products and reactants.

For some reversible reactions that involve gases:

  • Changes in pressure due to volume changes.

Catalysts do not influence the value of Q. See explanation.

Explanation:

\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}.

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The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium Q = K_c the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

  • Changes in concentration influence the number of particles per unit space.
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For reactions that involve gases,

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Answer:

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