The equilibrium constant for the following reaction is 6.30 at 723K. 2NH3 <=> N2(g) + 3H2(g) If an equilibrium mixture of the three gases in a 10.2 L container at 723 K contains 0.221 mol of NH3(g) and 0.315 mol of N2(g), the equilibrium concentration of H2 is M. (Enter numbers as numbers, no units. For example, 300 minutes would be 300. For letters, enter A, B, or C. Enter numbers in scientific notation using e# format. For example 1.43×10-4 would be 1.43e-4.)

The equilibrium concentration of H2 is 0.4579 M

Explanation:

Step 1: Data given

The equilibrium constant for the following reaction is 6.30 at 723K

Volume = 10.2 L

Number of moles NH3 = 0.221 moles

Number of moles N2 = 0.315 moles

Step 2: The balanced equation

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

Step 3: Calculate concentration

Concentration = moles / volume

Concentration NH3= 0.221 moles / 10.2 L

Concentration NH3 = 0.0217 M

Concentration N2 = 0.315 moles / 10.2 L

Concentration N2 = 0.0309 M

Step 4: Define Kc

Kc = [H2]³[N2] / [NH3]²

6.30 = [H2]³ * (0.0309) / (0.0217)²

[H2]³ = 0.09601 M

[H2] = 0.4579 M

The equilibrium concentration of H2 is 0.4579 M

(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