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2 years ago

Which is the name of an element that has 3 protons, 4 neutrons, and 2 electrons?

Chemistry

1 answer:

____ [38]2 years ago

5 0

Answer:

LITHIUM-ION.

Explanation:

ATOMIC STRUCTURE -

The smallest unit has the characteristics property of an element.

An atom consists of electron revolving around the nucleus in fixed orbits.

The Nucleus is a centrally located mass formed by the protons and the neutrons.

The atomic mass of an element is given as the sum of the protons and neutrons.

The Atomic number is equal to the number of protons or electrons.

In the given element element,

Atomic mass = 3 + 4 = 7

Atomic number = 3

As the number of protons in the atom is greater than the number of electrons by 1. Therefore, the element is a cation.

The element with atomic number 3 and atomic mas 7 is LITHIUM-ION.

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

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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

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

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

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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

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