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zloy xaker [14]
3 years ago
12

How many Moles of NAOH are present in 2.50L of 0.300 M NaOH?

Chemistry
1 answer:
maw [93]3 years ago
8 0
<h3>Answer:</h3>

0.75 moles NaOH

<h3>Explanation:</h3>

We are given;

Volume of NaOH solution = 2.5 Liters

Molarity of NaOH = 0.300 M

We are required to calculate the moles of NaOH

We need to establish the relationship between moles, molarity and volume of a solution.

That would be;

Concentration/molarity = Moles ÷ Volume

Therefore;

Moles = Concentration × Volume

Thus;

Moles of NaOH = 0.300 moles × 2.50 L

                         = 0.75 moles

Therefore, the number of moles of NaOH is 0.75 moles

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where the concentrations are not those of equilibrium, but other given concentrations

Chemical Equilibrium is the state in which the direct and indirect reaction have the same speed and is represented by a constant Kc, which for a generic reaction as shown above, is defined:

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where the concentrations are those of equilibrium.

This constant is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients divided by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.

Comparing Qc with Kc allows to find out the status and evolution of the system:

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  • If the reaction quotient is less than the equilibrium constant, Qc <Kc, the system is not in equilibrium. The concentration of the reagents is higher than it would be at equilibrium, so the direct reaction predominates. Thus, the system will evolve to the right to increase the concentration of products.

In this case:

Q=\frac{[So_{3}] ^{2} }{[SO_{2} ]^{2}* [O_{2}] }

Q=\frac{10^{2} }{0.10^{2} *0.10}

Q=100,000

100,000 < 4,300,000 (4.3*10⁶)

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<u><em> The system is not in equilibrium and will evolve left to right to reach equilibrium.</em></u>

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