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2 years ago

The Kc for the following is 5.0 at 100 degrees Celsius.

Chemistry

1 answer:

murzikaleks [220]2 years ago

5 0

Answer:

1.40M = [N2O4]

Explanation:

Based on the reaction:

2NO2(g) ⇄ N2O4(g)

Kc is defined as:

Kc = 5.0 = [N2O4] / [NO2]²

<em>Where [] are the concentrations in equilibrium of each gas</em>

As the equilibrium concentration of NO2 is 0.53M:

5.0 = [N2O4] / [0.53]²

5.0 = [N2O4] / [0.2809]

5.0*0.2809 = [N2O4]

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Answer:

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2 years ago

The molar mass of NH3 is 17.03 g/mol. The molar mass of H2 is 2.0158 g/mol. In a particular reaction, 0.575 g of NH3 forms. What

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3 years ago

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Luba_88 [7]

Answer : The rate of change of the total pressure of the vessel is, 10.5 torr/min.

Explanation : Given,

$\frac{d[NO]}{dt}$ =21 torr/min

The balanced chemical reaction is,

$2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)$

The rate of disappearance of $NO$ = $-\frac{1}{2}\frac{d[NO]}{dt}$

The rate of disappearance of $Cl_2$ = $-\frac{d[Cl_2]}{dt}$

The rate of formation of $NOCl$ = $\frac{1}{2}\frac{d[NOCl]}{dt}$

As we know that,

$\frac{d[NO]}{dt}$ =21 torr/min

So,

$-\frac{d[Cl_2]}{dt}=-\frac{1}{2}\frac{d[NO]}{dt}$

$\frac{d[Cl_2]}{dt}=\frac{1}{2}\times 21torr/min=10.5torr/min$

And,

$\frac{1}{2}\frac{d[NOCl]}{dt}=\frac{1}{2}\frac{d[NO]}$

$\frac{d[NOCl]}{dt}=\frac{d[NO]}=21torr/min$

Now we have to calculate the rate change.

Rate change = Reactant rate - Product rate

Rate change = (21 + 10.5) - 21 = 10.5 torr/min

Therefore, the rate of change of the total pressure of the vessel is, 10.5 torr/min.

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3 years ago

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elena55 [62]

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2 years ago