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Welcome to the article What volume of tex O 2 tex at 0 750 atm and 313 K is generated by the thermolysis of 10 0 g of tex. On this page, you will learn the essential and logical steps to better understand the topic being discussed. We hope the information provided helps you gain valuable insights and is easy to follow. Let’s begin the discussion!

What volume of [tex]O_2[/tex] at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of [tex]HgO[/tex]?

Answer :

0.568 L of O2 is generated by the thermolysis of 10.0 g of HgO at 0.750 atm and 313 K.

The thermolysis of HgO results in the production of mercury and oxygen gas. The balanced chemical equation for the reaction is:

2HgO(s) -> 2Hg(l) + O2(g)

The molar mass of HgO is 216.59 g/mol, so 10.0 g of HgO is equivalent to 0.0461 moles. From the balanced chemical equation, we can see that for every 2 moles of HgO, 1 mole of O2 is produced. Therefore, 0.023 moles of O2 will be produced by the thermolysis of 10.0 g of HgO.

The ideal gas law, PV = nRT, can be used to calculate the volume of the O2 produced. Rearranging the equation to solve for volume, we get:

V = nRT/P

where n is the number of moles of gas, R is the gas constant (0.08206 L atm/mol K), T is the temperature in Kelvin, and P is the pressure in atm.

Plugging in the values, we get:

V = (0.023 mol) x (0.08206 L atm/mol K) x (313 K) / (0.750 atm) = 0.568 L

Therefore, the volume of O2 produced by the thermolysis of 10.0 g of HgO at 0.750 atm and 313 K is 0.568 L.

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