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postnew [5]
4 years ago
6

A catalyst lowers the activation energy for both the forward and the reverse reactions in an equilibrium system, so it has no ef

fect on the equilibrium position of a system. True False

Chemistry
2 answers:
rosijanka [135]4 years ago
5 0

Answer:

True.

Explanation:

  • Catalyst increases the rate of the reaction without affecting the equilibrium position.
  • Catalyst increases the rate via lowering the activation energy of the reaction.
  • This can occur via passing the reaction in alternative pathway (changing the mechanism).
  • The activation energy is the difference in potential energies between the reactants and transition state (for the forward reaction) and it is the difference in potential energies between the products and transition state (for the reverse reaction).
  • in the presence of a catalyst, the activation energy is lowered by lowering the energy of the transition state, which is the rate-determining step, catalysts reduce the required energy of activation to allow a reaction to proceed and, in the case of a reversible reaction, reach equilibrium more rapidly.
  • with adding a catalyst, both the forward and reverse reaction rates will speed up equally, which allowing the system to reach equilibrium faster.
  • Kindly, see the attached image.

stich3 [128]4 years ago
5 0
<h2>Answer: </h2>

<u>The statement is</u><u> true </u>

<h2> Explanation: </h2>

When we add a catalyst to a system, both the forward and reverse reaction rates will speed up equally at the same time thereby allowing the system to reach equilibrium faster. But one thing that is very important to keep in mind that the addition of a catalyst has no effect on the final equilibrium position of the reaction because it only makes the reaction faster and nothing else.

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Masteriza [31]

Answer : The value of pK_a of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}

\text{Moles of }KOH=3.00M\times 10.0mL=30mmol=0.03mol

Now we have to calculate the value of pK_a of the weak acid.

The equilibrium chemical reaction is:

                          HA+KOH\rightleftharpoons HK+H_2O

Initial moles     0.25     0.03        0

At eqm.    (0.25-0.03)   0.03      0.03

                     = 0.22

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[HK]}{[HA]}

Now put all the given values in this expression, we get:

3.85=pK_a+\log (\frac{0.03}{0.22})

pK_a=4.72

Therefore, the value of pK_a of the weak acid is, 4.72

7 0
3 years ago
In the combustion of ethane, shown above, how many moles of carbon dioxide can be produced from 1.00 mole of ethane? A. 0.500 mo
Daniel [21]
The balanced chemical equation for the <span>combustion of ethane is 
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The stoichiometric ratio between C₂H₆(g) and CO₂(g)  is 1 : 2

Hence,
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CO₂(g) produced = moles of reacted C₂H₆(g) x 2
                                                 
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Hence, the correct answer is "C".
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3 years ago
What happens to PH of basic solution as you add water
Arlecino [84]

Answer:

to fall towards 7, making the solution less alkaline as more water is added.

8 0
2 years ago
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The specific gravity of iron is 7.87, and the density of water at 4.00
OlgaM077 [116]

The volume of 6.00g iron is 0.762 cm^3.

<em>Step 1</em>. Calculate the <em>density of Fe</em>

The <em>specific gravity</em> (SG) of a substance is the ratio of its density to the density of water at 4 °C.

SG = density of substance/density of water.

∴ 7.87 = density of iron/1.00 g·cm^(-3)

Density of Fe = 7.87 × 1.00 g·cm^(-3) = 7.87 g·cm^(-3)

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8 0
4 years ago
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Answer:

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Al escribir la ecuación de reacción química para una reacción particular, se utilizan los símbolos químicos de todas las especies involucradas.

Por la reacción. en el cual dos moléculas de Clorato de Potasio en estado sólido, al aplicar calor se descompone en dos moléculas de Cloruro de Potasio en estado sólido y tres moléculas diatómicas de Oxígeno en estado gaseoso, la ecuación de reacción se escribe así;

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La regla para escribir ecuaciones de reacción química balanceada es que el número de átomos de cada elemento en el lado derecho de la ecuación de reacción debe ser el mismo que el número b de átomos del mismo elemento en el lado izquierdo de la ecuación de reacción.

Si realizamos un conteo de átomos simple en ambos lados de la ecuación de reacción, la regla se cumple

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3 years ago
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