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Answer : The equilibrium constant for the reaction is, 0.1133
Explanation :
First we have to calculate the concentration of .
Now we have to calculate the dissociated concentration of .
The balanced equilibrium reaction is,
Initial conc. 1.731 M 0
At eqm. conc. (1.731-x) (2x) M
As we are given,
The percent of dissociation of =
= 1.2 %
So, the dissociate concentration of =
The value of x = 0.2077 M
Now we have to calculate the concentration of at equilibrium.
Concentration of = 1.731 - x = 1.731 - 0.2077 = 1.5233 M
Concentration of = 2x = 2 × 0.2077 = 0.4154 M
Now we have to calculate the equilibrium constant for the reaction.
The expression of equilibrium constant for the reaction will be :
Now put all the values in this expression, we get :
Therefore, the equilibrium constant for the reaction is, 0.1133
Answer:
1. 25 moles water.
2. 41.2 grams of sodium hydroxide.
3. 0.25 grams of sugar.
4. 340.6 grams of ammonia.
5. 4.5x10²³ molecules of sulfur dioxide.
Explanation:
Hello!
In this case, since the mole-mass-particles relationships are studied by considering the Avogadro's number for the formula units and the molar mass for the mass of one mole of substance, we proceed as shown below:
1. Here, we use the Avogadro's number to obtain the moles in the given molecules of water:
2. Here, since the molar mass of NaOH is 40.00 g/mol, we obtain:
3. Here, since the molar mass of C6H12O6 is 180.15 g/mol:
4. Here, since the molar mass of ammonia is 17.03 g/mol:
5. Here, since the molar mass of SO2 is 64.06 g/mol:
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