Mercury(ii) oxide decomposes to form mercury and oxygen, like this: (s)(l)(g) at a certain temperature, a chemist finds that a reaction vessel containing a mixture of mercury(ii) oxide, mercury, and oxygen at equilibrium has the following composition:

The question is incomplete, here is the complete question:

Mercury(II) oxide decomposes to form mercury and oxygen, like this:

At a certain temperature, a chemist finds that a 9.8 L reaction vessel containing a mixture of mercury(II) oxide, mercury, and oxygen at equilibrium has the following composition:

Compound                   Amount

HgO                              24.0 g

Hg                                 23.6 g

              22.7 g

Calculate the value of the equilibrium constant for this reaction. Round your answer to 2 significant digits.

The value of for the given chemical reaction is

Explanation:

To calculate the molarity or concentration, we use the equation:

For oxygen gas:

Mass of oxygen gas = 22.7 g

Molar mass of oxygen gas = 32 g/mol

Volume of the solution = 9.8 L

Putting values in above equation, we get:

For the given chemical equation:

The expression of for above equation follows:

The concentration of pure solids and liquids are taken as 1

So,

Hence, the value of for the given chemical reaction is

answer: lol

explanation:

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d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong