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3 years ago

Write the balanced COMPLETE ionic equation for the reaction when

Chemistry

1 answer:

Vera_Pavlovna [14]3 years ago

3 0

Answer:

Ba²⁺(aq) + 2 NO₃⁻(aq) + 2 Rb⁺(aq) + 2 OH⁻(aq) = Ba(OH)₂(s) + 2 Rb⁺(aq) + 2NO₃⁻(aq)

Explanation:

Let's consider the molecular equation between barium nitrate and rubidium hydroxide to produce barium hydroxide and rubidium nitrate.

Ba(NO₃)₂(aq) + 2 RbOH(aq) = Ba(OH)₂(s) + 2 RbNO₃(aq)

The complete ionic equation includes all the ions and the molecular species.

Ba²⁺(aq) + 2 NO₃⁻(aq) + 2 Rb⁺(aq) + 2 OH⁻(aq) = Ba(OH)₂(s) + 2 Rb⁺(aq) + 2NO₃⁻(aq)

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At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106

lana [24]

Answer:

The new equilibrium concentration of HI: [HI] = 3.589 M

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

Given chemical reaction: H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant ( $K_{c}$ ) for the given chemical reaction, is given by the equation:

$K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}$

At the original equilibrium state:

$K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}$

$K_{c} = \frac {1.6641}{0.002332} = 713.59$

Therefore, at the new equilibrium state:

$K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}$

$\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}$

$\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88$

$\Rightarrow [HI] = \sqrt {12.88} = 3.589 M$

Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M

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2 years ago