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saul85 [17]
3 years ago
8

There are 0.55 moles of carbon dioxide gas in a 15.0 L container. This container is at a temperature of 300 K. What is the press

ure of the gas inside the container? Use 8.31 L*kPa/mol*K for the gas constant.
A.)760 mm Hg\

B.) 271 kPa

C.) 2 atm

D.) 91.4 kPa
Chemistry
1 answer:
sertanlavr [38]3 years ago
6 0

Answer:

\large \boxed{\text{D.) 91 kPa}}

Explanation:

We can use the Ideal Gas Law — pV = nRT

Data:

V = 15.0 L

n = 0.55 mol

T = 300 K

Calculation:

\begin{array}{rcl}pV & =& nRT\\p \times \text{15.0 L} & = & \text{0.55 mol} \times \text{8.31 kPa$\cdot$ L$\cdot$K$^{-1}$mol$^{-1}\times$ 300 K}\\15.0p & = & \text{1370 kPa}\\p & = & \textbf{91 kPa}\end{array}\\\text{The pressure in the container is $\large \boxed{\textbf{91 kPa}}$}

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Answer:

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Answer:

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<u />

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3 years ago
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Answer:

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Explanation:

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

<em>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?</em>

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<em>Remember to use correct significant figures in your answer. Do not include units in your response.</em>

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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The equilibrium constant for this reaction at equilibrium is 0.0119.

You can learn more about equilibrium here: brainly.com/question/4289021

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