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Romashka-Z-Leto [24]
3 years ago
9

How does adding oxygen (O2) to this reaction change the equilibrium? 2SO2(g) + O2(g) ⇌ 2SO3(g) A. The equilibrium shifts right t

o produce more SO3 molecules. B. The equilibrium shifts left to produce more O2 molecules. C. The equilibrium shifts right because of decreased collisions between SO2 and O2 molecules. D. The equilibrium shifts left with an increase in SO2 and O2 molecules. E. The equilibrium shifts left because of increased collisions between SO2 and O2 molecules.
Chemistry
1 answer:
liq [111]3 years ago
8 0

The equilibrium law or the Le Chatelier's principle states that when a system encounters a disturbance like a temperature, concentration, or pressure modifications, it will react to restore a new state of equilibrium.  

In the given case, adding reactant, that is, oxygen will shift the reaction in the direction of the product, on the basis of Le Chatelier's principle. Thus, the equilibrium will move toward the product side. So, the correct answer will be option A, that is, the equilibrium shifts right to produce more SO₃ molecules.  


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A gas occupies a volume of 1.00 L at 25.0°C. What volume will the gas occupy at 1.00 x10^2 °C?
Leno4ka [110]

Answer : The volume of gas occupy at 1.00\times 10^2^oC is, 1.25 L

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Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

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\frac{V_1}{T_1}=\frac{V_2}{T_2}

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V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:

V_1=1.00L\\T_1=25.0^oC=(25.0+273)K=298K\\V_2=?\\T_2=1.00\times 10^2^oC=((1.00\times 10^2)+273)K=373K

Putting values in above equation, we get:

\frac{1.00L}{298K}=\frac{V_2}{373K}\\\\V_2=1.25L

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0.27mol × (S/.1800 / 1mol oro) =

<h3>S/.486 es el valor del anillo</h3>
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