What is the numerical value of Kc for the following reaction if the equilibrium mixture at 750 °C contains 0.20 M CO and 0.052 M CO2?

FeO(s) + CO(g) ⇌ Fe (s) + CO2 (g)

The answer to your question is    Kc = 0.26

Explanation:

Data

Kc = ?

[CO] = 0.20 M

[CO₂] = 0.052 M

For this problem, do not consider the concentration of the reactants and products in a solid-state.

Formula

           Kc =

Substitution

           Kc =

Simplification and result

           Kc = 0.26

The equilibrium constant Kc is 0.26

Explanation:

Step 1: Data given

The equilibrium mixture contains 0.20 M of CO and 0.052 M of CO2

Step 2: The balanced equation

FeO(s) + CO(g) ⇌ Fe (s) + CO2 (g)

Step 3: Calculate Kc

Pure solids and liquids are not included in the equilibrium constant expression. This is because they do not affect the reactant amount at equilibrium in the reaction, so they are disregarded and kept at 1. Remember that the activity, a, of any solid or liquid in a reaction is equal to 1

Kc = [CO2] / [CO]

⇒with [CO] = 0.20 M

⇒with [CO2] = 0.052 M

Kc = 0.052 / 0.20

Kc = 0.26

The equilibrium constant Kc is 0.26

away from the curb

explanation:

the wheels would then hit the curb if it started to rolling, decreasing the chance of an accident.

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied