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timama [110]
3 years ago
6

Equal amounts of N2 and O2 are added, under certain conditions, to a closed container. Which changes occur in the reverse reacti

on as the system proceeds toward equilibrium
Chemistry
1 answer:
geniusboy [140]3 years ago
4 0
1) The forward reaction is N2 (g) + O2 (g)  → 2NO

(that reaction requires special contitions because at normal pressures and temperatures N2 and O2 do not react to form another compound.

2) The equiblibrium equation is

  N2 (g) + O2 (g)  ⇄ 2NO

3) Then, the reverse reaction is

2NO → N2(g) + O2(g)

Answer: 2NO → N2(g) + O2(g)
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Where should the neutrons be added to this atomic model?
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with the proton inside the nucleus

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A solution is made by adding 50.0 ml of 0.200 m acetic acid (ka = 1.8 x 10–5) to 50.0 ml of 1.00 x 10–3m hcl. (a) calculate the
Irina18 [472]

Answer:

Final pH of the solution: 2.79.

Explanation:

What's in the solution after mixing?

\displaystyle c = \frac{n}{V},

where

  • c is the concentration of the solute,
  • n is the number of moles of the solute, and
  • V is the volume of the solution.

V(\text{Final}) = 0.050 \;\textbf{L} + 0.050\;\textbf{L} = 0.100\;\textbf{L}.

Acetic (ethanoic) acid:

\displaystyle \begin{aligned}n &= c(\text{Before})\cdot V(\text{Before}) \\&= 0.050\;\text{L} \times 0.200\;\text{mol}\cdot\text{L}^{-1}\\ &= 0.0100\;\text{mol}\end{aligned}.

\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{0.0100\;\text{mol}}{0.100\;\text{L}}\\ &= 0.100\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 0.100\;\text{M}\end{aligned}.

Hydrochloric acid HCl:

\begin{aligned}n &= c(\text{Before})\cdot V(\text{Before})\\ &= 0.050\;\text{L} \times 1.00\times 10^{-3}\;\text{mol}\cdot\text{L}^{-1}\\ &= 5.00\times 10^{-5}\;\text{mol}\end{aligned}.

\displaystyle \begin{aligned}c(\text{After}) &= \frac{n}{V(\text{After})}\\ &= \frac{5.00\times 10^{-5}\;\text{mol}}{0.100\;\text{L}}\\ &= 5.00\times 10^{-4}\;\text{mol}\cdot\textbf{L}^{-1}\\ &= 5.00\times 10^{-4}\;\text{M}\end{aligned}.

HCl is a strong acid. It will completely dissociate in water to produce H⁺. The H⁺ concentration in the solution before acetic acid dissociates shall also be 5.00\times 10^{-4}\;\text{M}.

The Ka value of acetic acid is considerably small. Acetic acid is a weak acid and will dissociate only partially when dissolved. Construct a RICE table to predict the portion of acetic acid that will dissociate. Let the change in acetic acid concentration be -x\;\text{M}. x > 0.

\begin{array}{c|ccccc}\textbf{R}&\text{CH}_3\text{COOH}\;(aq) &\rightleftharpoons &\text{CH}_3\text{COO}^{-}\;(aq) &+& \text{H}^{+}\;(aq)\\\textbf{I}&0.100\;\text{M} & & & & 5.00\times 10^{-4}\;\text{M}\\\textbf{C}&-x\;\text{M} & & +x\;\text{M} & & +x\;\text{M} \\ \textbf{E}&0.100\;\text{M}-x\;\text{M} & & x\;\text{M} & & 5.00\times 10^{-4}\;\text{M} + x\;\text{M}\end{array}.

\displaystyle K_a = \frac{[\text{CH}_3\text{COO}^{-}\;(aq)]\cdot[\text{H}^{+}\;(aq)]}{[\text{CH}_3\text{COOH}\;(aq)]} = \frac{x\cdot(x + 5.00\times 10^{-4})}{0.100 - x}.

Rewrite as a quadratic equation and solve for x:

x\cdot(x + 5.00\times 10^{-4}) = (1.8\times 10^{-5} )\cdot (0.100 - x)

x\approx 0.00111.

The pH of a solution depends on its H⁺ concentration.

At equilibrium

[\text{H}^{+}\;(aq)] = 5.00\times 10^{-4}\;\text{M} + x\;\text{M} = 0.00161\;\text{M}.

\text{pH} = -\log{[\text{H}^{+}]} = 2.79.

5 0
3 years ago
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