Dissolution of wax in kerosene

In this nuclear reaction equation, oxygen decays to form nitrogen. Which equation correctly describes

sertanlavr [38]

Answer: Its A

Explanation:

6 0

4 years ago

How many hybrid orbitals do we use to describe each molecule c2h5no (4 c−h bonds and one o−h) key?

yarga [219]

Answer:

see attached

Explanation:

6 0

4 years ago

If 5.00lb of potatoes cost $3.60,how much would 1.30kg of potatoes cost?

Nana76 [90]

Answer:

the correct answer is $2.06

Explanation:

hope it'll help you...

6 0

3 years ago

A reaction in which two ions switch places is which type of reaction?

AURORKA [14]

A reaction in which two ions switch places is called double-displacement reaction.

Explanation:

Double displacement reaction usually takes place between two positive ions or two negative ions. When anions or cations of totally different compounds change places, two totally different compounds are formed.

Syntax for double displacement reaction:

AB + CD --> AD + BC

It can also be defined in simple terms as the reaction in which exchange of ions takes place is called double displacement reaction.

The chemical reaction is as follows:

NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O (l)

6 0

3 years ago

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimen

weqwewe [10]

Answer:

Rate = (0.64 M²/min) [A]¹[B]²

Explanation:

1) Determination of the orders of A & B:

The rate law of the reaction = k [A]ᵃ[B]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:

Reactant [A] has the same concentration in both trials, but [B] has different concentrations and the rate of the reaction changes, so the reaction rate depends on [B].

From trial 1, Rate₁ = k [A₁]ᵃ[B₁]ᵇ, [1.2 × 10⁻² M/min] = k [0.30 M]ᵃ[0.25 M]ᵇ

From trial 2, Rate₂ = k [A₂]ᵃ[B₂]ᵇ, [4.8 × 10⁻² M/min] = k [0.30 M]ᵃ[0.50 M]ᵇ

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:

Reactant [B] has the same concentration in both trials, but [A] has different concentrations and the rate of the reaction changes, so the reaction rate depends on [A].

From trial 2, Rate₂ = k [A₂]ᵃ[B₂]ᵇ, [4.8 × 10⁻² M/min] = k [0.30 M]ᵃ[0.50 M]ᵇ

From trial 2, Rate₃ = k [A₃]ᵃ[B₃]ᵇ, [9.6 × 10⁻² M/min] = k [0.60 M]ᵃ[0.50 M]ᵇ

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.

The rate law of the reaction will be: Rate = k [A]¹[B]²

The overall order of the reaction is third order reaction.

2) Determining the rate law constant:

Taking data of trial 1:

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

3 0

3 years ago