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tensa zangetsu [6.8K]
2 years ago
6

A. At STP, what is the volume of 708 mol of nitrogen gas? 708 mol = 708 mol X L B. A sample of hydrogen gas occupies 14.1 L at S

TP. Hov many moles of the gas are present? 14.1 L = 14.1 L X 11 mol​
Chemistry
1 answer:
elena-14-01-66 [18.8K]2 years ago
5 0

Answer:

A. 15859.2 L or 15900 L

B. 0.629 mol

Explanation:

At STP, one mole is equal to approximately 22.4 L

L or mL is volume, so you are attempting to solve for L or mL.

A.

708 mol x (22.4 L/1 mol) = 15859.2 L (w/ significant figures included - 15900 L)

B.

(14.1 L) x (1 mole/ 22.4 L) = 0.629 mol.

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Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibriumdescribed by the equation N2O4(g)↔
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Answer : The correct option is, (C) 1.1

Solution :  Given,

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First we have to calculate the concentration N_2O_4.

\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}

\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M

The given equilibrium reaction is,

                           N_2O_4(g)\rightleftharpoons 2NO_2(g)

Initially                      c                 0

At equilibrium   (c-c\alpha)           2c\alpha

The expression of K_c will be,

K_c=\frac{[NO_2]^2}{[N_2O_4]}

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

where,

\alpha = degree of dissociation = 40 % = 0.4

Now put all the given values in the above expression, we get:

K_c=\frac{(2c\alpha)^2}{(c-c\alpha)}

K_c=\frac{(2\times 1\times 0.4)^2}{(1-1\times 0.4)}

K_c=1.066\aprrox 1.1

Therefore, the value of equilibrium constant for this reaction is, 1.1

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