Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided. a + b yields products trial [a] [b] rate 1 0.30 m 0.25 m 1.2 × 10-2 m/min 2 0.30 m 0.50 m 4.8 × 10-2 m/min 3 0.60 m 0.50 m 9.6 × 10-2 m/min

Rate = (0.64 M²/min) [A]¹[B]²

Explanation:

1) Determination of the orders of A & B:

where, k is the rate law constant,

a is the order of the reaction with respect to reactant A,

b is the order of the reaction with respect to reactant B.

This is initial rate method problem:

From trial 1 & 2:

By dividing Rate₁ / Rate₂:

Rate₁ / Rate₂ = k [A₁]ᵃ[B₁]ᵇ / k [A₂]ᵃ[B₂]ᵇ

[1.2 × 10⁻² M/min] / [4.8 × 10⁻² M/min] = k [0.30 M]ᵃ[0.25]ᵇ / k [0.30 M]ᵃ[0.50 M]ᵇ

0.25 = [0.50]ᵇ

Taking log for both sides; log(0.25) = b log(0.5)

b = log(0.25) / log(0.5) = 2.

The reaction is second order with respect to reactant B.

By the same way for reactant A:

From trial 2 & 3:

By dividing Rate₂ / Rate₃:

Rate₂ / Rate₃ = k [A₂]ᵃ[B₂]ᵇ / k [A₃]ᵃ[B₃]ᵇ

[4.8 × 10⁻² M/min] / [9.6 × 10⁻² M/min] = k [0.30 M]ᵃ[0.50 M]ᵇ / k [0.60 M]ᵃ[0.50 M]ᵇ

0.50 = [0.50]ᵃ

Taking log for both sides; log(0.50) = a log(0.5)

a = log(0.50) / log(0.50) = 1.

The reaction is first order with respect to reactant A.

2) Determining the rate law constant:

Rate = k [A]¹[B]²

[1.2 × 10⁻² M/min] = k [0.30 M]¹[0.25 M]²

k = [1.2 × 10⁻² M/min] / [0.30 M]¹[0.25 M]² = 0.64 M²/min.

So, the rate law will be: Rate = (0.64 M²/min) [A]¹[B]²

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.