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Lubov Fominskaja [6]
2 years ago
15

The vapor pressure of benzene at 25°C is 95.1 mm Hg and of toluene 28.4 mm Hg. The mass fractions of benzene and toluene in a so

lution are 0.5.
a) Calculate the partial pressure of benzene and toluene above the solution.
b) Calculate the total vapor pressure above the solution.
c) Calculate the vapor composition above the solution.
Chemistry
1 answer:
arsen [322]2 years ago
3 0

Answer:

df  dffdffdfd  dffd  ff d

Explanation:

df fd df f df df dfdf  f fdf  df df df fd

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An equilibrium mixture of CO, O2 and CO2 at a certain temperature contains 0.0010 M CO2 and 0.0025 M O2. At this temperature, Kc
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Answer:

5.35 *10^{-4}M

Explanation:

Equation for the reaction is as follows:

2CO_{(g)}      +      O_{2(g)}        ⇄       2CO_{2(g)}

By Applying the ICE Table; we have

                             2CO_{(g)}      +      O_{2(g)}        ⇄       2CO_{2(g)}

Initial                      x                  0.0025 M              0.0010 M

Change                  0                       0                            0

Equilibrium             x                  0.0025 M              0.0010 M

K_c =\frac{[CO_2]^2}{[CO]^2[O_2]}

Given that K_c = 1.4*10^2 ; Then:

1.4 *10^2 = \frac{(0.001)^2}{(x)^2(0.025)}

1.4 *10^2*0.025 = \frac{(0.001)^2}{(x)^2}

3.5 =( \frac{(0.001)}{(x)})^2

\sqrt {3.5} = \sqrt {( \frac{(0.001)}{(x)} )^2}

1.87=\frac{(0.001)}{(x)}

(x)= \frac{(0.001)}{1.87 }

x = 5.35 *10^{-4}M

∴ The equilibrium concentration of CO = x = 5.35 *10^{-4}M

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