How many moles of gas sample are 5.0 L container at 373K and 203kPa

1 answer:

6 0

For the purpose we will here use the ideal gas law:

p×V=n×R×T

V= 5.0 L

T= 373K

p= 203kPa

R is universal gas constant, and its value is 8.314 J/mol×K

Now when we have all necessary date we can calculate the number of moles:

n=p×V/R×T

n= 203 x 5 / 8.314 x 373 = 0.33 mole

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Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the co

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There was a part missing. I think this is the whole question:

Before any reaction occurs, the concentration of A in the reaction below is 0.0510 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0153 M?

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Remember to use correct significant figures in your answer. Do not include units in your response.

First, we have to make an ICE Chart, which stands for initial, change and equilibrium. We will call "x" unknown concentrations.

A (aq) ⇌ 2B (aq) + C (aq)

I 0.0510 0 0

C -x +2x +x

E 0.0510-x 2x x

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$[A]e = 0.0153 M \\[B]e = 2x = 2(0.0357) = 0.0714 M \\[C]e = x = 0.0357 M \\$

The equilibrium constant, Kc, is the ratio of the equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.

$Kc = \frac{[B]^{2} \times [C]}{[A]} = \frac{0.0714^{2} \times 0.0357}{0.0153} = 0.0119$