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1 year ago

Use electron configurations to account for the stability of the lanthanide ions Ce4⁺ and Eu2⁺.

1 answer:

4 0

Eu2+ is stable because of the two electrons in its outermost orbit and after losing 4 electrons. Ce +4 is stable

What are lanthanide ions?

All lanthanide elements form trivalent cations, Ln3+, whose chemistry is largely determined by the ionic radius, which decreases steadily from lanthanum to lutetium. These elements are called lanthanides because the elements in the series are chemically similar to lanthanum.

The Ce metal has the following electronic configuration: [Xe] 4f¹5d¹6s²

It obtains noble gas configuration after losing 4 electrons. Ce +4 is stable, so.

Eu's electronic setup is as follows.

[Xe] 4 f ⁷6s²

[Xe] 4 f ⁷

Eu2+ is stable because of the two electrons in its outermost orbit. Additionally stable is its +3 oxidation state. Ce⁺²

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Use the word bank to complete the charts about the metrics system measurement

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3 years ago

How many chlorine atoms are on the products side of the reaction 2al 6hcl → 2alcl3 3h2? 2 3 6 9

GenaCL600 [577]

The number of chlorine atoms present on the product side of the reaction is 6

<h3>What is a chemical equation? </h3>

Chemical equations are representations of chemical reactions using symbols and formula of the reactants and products.

The balancing of chemical equations follows the law of conservation of matter which states that matter can neither be created nor destroyed during a chemical reaction but can be transferred from one form to another.

<h3>How to determine the number of atoms of Cl</h3>

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Products => AlCl₃ and H₂

Number of Cl atoms = 2 × 3

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3 0

2 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

Answer:

A) 1059 J/mol

B) 17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

$C_p=C_v+R$

We can determine $C_v$ from above if we make $C_v$ the subject of the formula as:

$C_v$ $=C_p-R$

$C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314$

$C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4$

A).

The formula for calculating change in internal energy is given as:

$dU=C_vdT$

If we integrate above data into the equation; it implies that:

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 1059J/mol$

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)$

$U2-U1= 17,920 J/mol$

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4 years ago

put the following in order from LARGEST TO SMALLEST nuclueus,molecule,electron,atom,proton and a teacher

SVETLANKA909090 [29]

Answer:

electron, proton, nucleus, atom, molecule teacher

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3 years ago