A sample of NH3(g) is placed in a rigid container. When the equilibrium below is established, the partial pressure of hydrogen gas is 0.978 atm. What is the partial pressure of N2(g) at equilibrium?

PpN₂ = 0.326 atm

Explanation:

In this case, let's write the equilibrium reaction taking place here:

2NH₃(g) <> 3H₂(g) + N₂(g)

Now, we know that when the reaction reach the equilibrium the partial pressure of H₂ is 0.978 atm, so, let's write an ICE chart and see what's happening here:

                 2NH₃(g) <> 3H₂(g) + N₂(g)

i)                     y                      0           0

c)                 -2x                   +3x          x

e)                y-2x                    3x          x

At the beggining we just have the sample of NH₃, we don't know how much but we don't need that data to get the partial pressure. In equilibrium, its produced hydrogen and nitrogen, and we know how much of hydrogen was produced, so, with that we can solve for the value of the partial pressure of NO by a simple math operation:

PpH₂ = 0.978 atm = 3x

PpN₂ = x

PpN₂ = PpH₂ / 3

PpN₂ = 0.978 / 3

Hope this helps

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

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8