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3 years ago

Look at the diagram. What number needs to be added to balance the equation?

Chemistry

1 answer:

suter [353]3 years ago

4 0

Answer:

Explanation:

because hydrogen is a diatomic element, so it can't be just H

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3 years ago

Consider the following equilibrium:

Angelina_Jolie [31]

Answer:

1. The equilibrium lies to the left.

$\large \boxed{2. \, \, 5.41 \times 10^{-2} \text{ mol/L}}$

Explanation:

N₂O₄ ⇌2NO₂; Keq = 5.85 × 10⁻³

1. Position of equilibrium

Keq < 1, so the position of equilibrium lies to the left.

2. Equilibrium concentration

(a) Set up an ICE table.

$\begin{array}{rcc}\rm \text{N$_{2}$O}_{4} & \, \rightleftharpoons \, & \text{2NO}_{2} \\\text{E/M}:\qquad \qquad& & 1.78 \times 10^{-2} \\\end{array}$

(b) Calculate the equilibrium concentration

$K_{\text{c}} = \dfrac{\text{[NO$_{2}$]}^{2}}{\text{[N$_{2}$O$_{4}$]}} = \dfrac{(1.78 \times 10^{-2})^{2}}{\text{[N$_{2}$O$_{4}$]}} = 5.85 \times 10^{-3}\\\\\begin{array}{rcl}\\(1.78 \times 10^{-2})^{2}&=&\text{[N$_{2}$O$_{4}$]} \times 5.85 \times 10^{-3}\\\text{[N$_{2}$O$_4$]}&=& \dfrac{(1.78 \times 10^{-2})^{2}}{5.85 \times 10^{-3}}\\\\& = & \mathbf{5.41 \times 10^{-2}}\textbf{ mol/L}\\\end{array}\\$

$\text{The equilibrium concentration of N$_{2}$O$_{4}$ is $\large \boxed{\mathbf{5.41 \times 10^{-2}} \textbf{ mol/L}}$}$

Check:

$\begin{array}{rcl}\dfrac{(1.78 \times 10^{-2})^{2}}{5.41 \times 10^{-2}} & = & 5.85 \times 10^{-3}\\\\5.85 \times 10^{-3} & = & 5.85 \times 10^{-3}\\\end{array}$

It checks.

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3 years ago