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3 years ago

Look up and record the boiling point of acetic acid, and explain why only some of it evaporates from the reaction mixture.

Chemistry

1 answer:

REY [17]3 years ago

8 0

Answer:

Heating the mixture to a temperature above the boiling point of acetic acid, but below 100°C (the boiling point of water). The vapours from the acetic acid rise, and go into a tube. They are then condensed within the tube, and run off into a separate storage area. Because water can exist as a gas at pretty much any temperature above 0°C, it will result in an impure mixture, but repeatedly doing this will get the acetic acid to the desired purity.

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Read 2 more answers

Hc2h3o2(aq)+h2o(l)⇌h3o+(aq)+c2h3o−2(aq) kc=1.8×10−5 at 25∘c part a if a solution initially contains 0.260 m hc2h3o2, what is the

Cloud [144]

Answer : The equilibrium concentration of $H_3O^+$ at $25^oC$ is, $2.154\times 10^{-3}m$ .

Solution : Given,

Equilibrium constant, $K_c=1.8\times 10^{-5}$

Initial concentration of $HC_2H_3O_2$ = 0.260 m

Let, the 'x' mol/L of $H_3O^+$ are formed and at same time 'x' mol/L of $HC_2H_3O_2$ are also formed.

The equilibrium reaction is,

$HC_2H_3O_2(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+C_2H_3O_2^-(aq)$

Initially 0.260 m 0 0

At equilibrium (0.260 - x) x x

The expression for equilibrium constant for a given reaction is,

$K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

Now put all the given values in this expression, we get

$1.8\times 10^{-5}=\frac{(x)(x)}{(0.260-x)}$

By rearranging the terms, we get the value of 'x'.

$x=2.154\times 10^{-3}m$

Therefore, the equilibrium concentration of $H_3O^+$ at $25^oC$ is, $2.154\times 10^{-3}m$ .

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3 years ago

Calculate the relative density of carbon monoxide and carbon dioxide with respect to air.

JulsSmile [24]

Density is proportional to molar mass, assuming pressure and temperature remain constant. Therefore, since CO has a molar mass of 28 and CO2 has a molar mass of 44:
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3 years ago