Areaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by o 3 ( g ) + no ( g ) ⟶ o 2 ( g ) + no 2 ( g ) the rate law for this reaction is rate of reaction = k [ o 3 ] [ no ] given that k = 2.55 × 10 6 m − 1 ⋅ s − 1 at a certain temperature, calculate the initial reaction rate when [ o 3 ] and [ no ] remain essentially constant at the values [ o 3 ] 0 = 7.34 × 10 − 6 m and [ no ] 0 = 6.99 × 10 − 5 m, owing to continuous production from separate sources.

1.31 × 10⁻¹⁵ M/s

Explanation:

Let's consider the following reaction.

O₃(g) + NO(g) ⟶ O₂(g) + NO₂(g)

The rate law is:

rate = 2.55 × 10⁻⁶ M⁻¹s⁻¹ [O₃].[NO]

where,

2.55 × 10⁻⁶ M⁻¹s⁻¹ is the rate constant (k)

[O₃] is the concentration of O₃ raised to the reaction order 1

[NO] is the concentration of NO raised to the reaction order 1

When [O₃] = 7.34 × 10⁻⁶ M and [NO] = 6.99 × 10⁻⁵ M, the rate of the reaction is:

rate = 2.55 × 10⁻⁶ M⁻¹s⁻¹ (7.34 × 10⁻⁶ M).(6.99 × 10⁻⁵ M) = 1.31 × 10⁻¹⁵ M/s

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )

there are two significant figures.