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lina2011 [118]
3 years ago
10

Which cellular molecules would you expect to be soluble in a container of benzene?

Chemistry
1 answer:
nalin [4]3 years ago
3 0
The cellular molecules that you would expect to be soluble in a container of benzene would be the triglycerols which ae fats and oils in the cells, the cholesterol and the chains of fatty acids that are present. Benzene is a nonpolar solvent as it does not have any unequal sharing of electrons in its structure. You would expect that molecules that would have equal charges in their structures or the nonpolar molecules would readily dissolve in this solvent since like dissolves like. Along with this, all polar molecules would not be dissolved and be visible in the mixture since these are not miscible in benzene.
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PbO2

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At 920 K, Kp = 0.40 for the following reaction. 2 SO2(g) + O2(g) equilibrium reaction arrow 2 SO3(g) Calculate the equilibrium p
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<u>Answer:</u> The equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.

<u>Explanation:</u>

We are given:

Initial partial pressure of sulfur dioxide = 0.52 atm

Initial partial pressure of oxygen = 0.52 atm

Initial partial pressure of sulfur trioxide = 0 atm

For the given chemical reaction:

                        2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)

<u>Initial:</u>                 0.52      0.52

<u>At eqllm:</u>         0.52-2x    0.52-x        2x

The expression of K_p for above equation follows:

K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times p_{O_2}}

We are given:

K_p=0.40\\\\p_{SO_3}=2x\\\\p_{SO_2}=0.52-2x\\\\p_{O_2}=0.52-x

Putting values in above equation, we get:

0.40=\frac{(2x)^2}{(0.52-2x)^2\times (0.52-x)}\\\\x=-1.13,-0.402,0.077

Neglecting the negative values because partial pressure cannot be negative.

So, x = 0.077

Equilibrium partial pressure of sulfur dioxide = (0.52-2x)=(0.52-(2\times 0.077))=0.366atm

Equilibrium partial pressure of oxygen = (0.52-x)=(0.52-0.077)=0.443atm

Equilibrium partial pressure of sulfur trioxide = 2x=(2\times 0.077)=0.154atm

Hence, the equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.

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