Determine the rate law and the value of k for the following reaction using the data provided: 2n2o5(g) → 4 no2(g) + o2(g)0.2241.16 x 10-3 a. rate = 5.6 × 10-2 m-1s-1[n2o5]2 b. rate = 6.0 × 10-1 m-2s-1[n2o5]3 c. rate = 1.6 × 10-3 m1/2s-1[n2o5]1/2 d. rate = 1.7 × 10-2 m-1/2s-1[n2o5]3/2 e. rate = 5.2 × 10⁻3 s-1[n2o5]

The rate law and the value of k is Rate = 5.2 x 10^-3 s-1[N2O5]

The rate law or rate equation for a chemical reaction is an equation that links the reaction rate with concentrations or pressures of reactants and constant parameters (normally rate coefficients and partial reaction orders).

The correct answer between all the choices given is the last choice or letter E. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

(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