The equilibrium constant, Kc, for the following reaction is 1.55 at 667 K.

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

When a sufficiently large sample of NH3(g) is introduced into an evacuated vessel at 667 K, the equilibrium concentration of H2(g) is found to be 0.324 M.

Calculate the concentration of NH3 in the equilibrium mixture.

The concentration of ammonia in the equilibrium mixture is 0.022 M

Explanation:

We are given:

Equilibrium concentration of hydrogen gas = 0.324 M

For the given chemical equation:

Initial:                  a

At eqllm:           a-2x           x          3x

Evaluating the value of 'x':

So, equilibrium concentration of nitrogen gas = x = 0.108 M

Equilibrium concentration of ammonia gas = (a - 2x) = [a - 2(0.108)] = (a - 0.216) M

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

Equilibrium concentration of ammonia gas = (a - 0.216) = [0.238 - 0.216] = 0.022 M

Hence, the concentration of ammonia in the equilibrium mixture is 0.022 M

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying