For the reaction co+2h2=ch3oh at 700 k, equilibrium concentrations are [h2]=0.072 m, [co]= 0.020m, and [ch3oh]= 0.030 m. calculate k

The balanced equation for the reaction is

CO(g) + 2H₂(g) ⇄ CH₃OH(g)

The given concentrations are at equilibrium state. Hence we can use them directly in calculation with the expression for the equilibrium constant, k. expression for k can be written as

   k = [CH₃OH(g)] / [CO(g)] [H₂(g) ]²

[H₂]=0.072 M
[CO]= 0.020M
[CH₃OH]= 0.030 M

From substitution,

   k = 0.030 M / 0.020 M x (0.072 M)²

   k = 289.35 M⁻²

Hence, equilibrium constant for the given reaction at 700 K is 289.35 M⁻².

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

you gotta add the following molecules so we can give u an answer

explanation: