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Paraphin [41]
3 years ago
14

What is the overall (X-fold) dilution of 4 serial 2-fold dilutions?

Chemistry
1 answer:
svet-max [94.6K]3 years ago
6 0

Answer:

16-fold dilution.

Explanation:

A serial dilution is any dilution where the concentration decrease by the same quantity in each successive step. So, for a 2-fold dilution, the concentration decrease 1/2, it means that if we have a sample with 10 M of concentration, after a 2-fold dilution it will be 5 M. For the next step it will be 1/2 of 5= 2.5 M, and successively.

Then, we just multiply the factor for each dilution. After 4 serial dilutions:

1/2 x 1/2 x 1/2 x 1/2 = 1/16

So, it would be a 16-fold dilution in the end.

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Answer: Concentration of N₂ is 4.8.10^{9} M.

Explanation: K_{c} is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For

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K_{c} = \frac{[NO2]^{2} }{[N2][O2]^{2} }

From the question concentration of NO2 is twice of O2:

[NO2] = 2[O2]

Substituting this into K_{c}:

K_{c} = \frac{[2O2]^{2} }{[N2][O2]^{2} }

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[N2] = \frac{4}{8.3.10^{-10} }

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The concentration of N2 in the equilibrium is [N2] = 4.8.10^{9}M.

6 0
3 years ago
5.0 g of NaCl is dissolved in water, and the solution is brought to a final volume of 150 mL. What is the molarity of this solut
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