2NH3(g) CO2(g) In an experiment carried out at this temperature, a certain amount of NH4OCONH2 is placed in an evacuated rigid container and allowed to come to equilibrium. Calculate the total pressure in the container at equilibrium.

The question is incomplete, here is the complete question:

At 25°C, Kp = 2.9 × 10⁻³ for the reaction:

In an experiment carried out at this temperature, a certain amount of NH₄OCONH₂ is placed in an evacuated rigid container and allowed to come to equilibrium. Calculate the total pressure in the container at equilibrium.

The total pressure in the container at equilibrium is 0.2694 atm

Explanation:

Let the initial concentration of be 'x'

The given chemical equation follows:

Initial:                       x

At eqllm:              x-y                         2y              y

The expression of for above equation follows:

The partial pressure of pure solids and liquids are taken as 1 in equilibrium constant expression. So, the partial pressure of is not seen in the expression.

We are given:

Putting values in above expression, we get:

So, the equilibrium partial pressure of ammonia = 2y = (2 × 0.0898) = 0.1796 atm

The equilibrium partial pressure of carbon dioxide = y = 0.0898 atm

Total pressure inside the container at equilibrium =

Hence, the total pressure in the container at equilibrium is 0.2694 atm

the ozone layer acts as a filter for the shorter wavelength and highly hazardous ultraviolet radiation (uvr) from the sun, protecting life on earth from its potentially harmful effects. when the sky is clear, there is an inverse relationship between stratospheric ozone and solar uvr measured at the earth's surface.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3