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2 years ago

Refer to the following balanced equation in which ammonia reacts with nitrogen monoxide to produce nitrogen and water.

Chemistry

1 answer:

RSB [31]2 years ago

6 0

Answer:4314

Explanation:

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A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci

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Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

$C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}$

The ICE table is then shown as:

$C_3H_6ClCO_2H_{(aq)} \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ + \ \ \ \ H^+_{(aq)}$

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

Since the value for $K_a$ is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

$3.02*10^{-5} *(1.8) = {(x)(x)}$

$5.436*10^{-5}= {(x^2)$

$x = \sqrt{5.436*10^{-5}}$

$x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M$

Dissociated form of 4-chlorobutanoic acid = $C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M$

Percentage dissociated = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)

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