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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

To learn more about lanthanides click on the link below:

brainly.com/question/12756990

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30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

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Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

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$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

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Convert 16.8 L of nitrogen monoxide gas at STP to grams .

german

Answer:

See explanation

Explanation:

1 mole of a gas occupies 22.4 L

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x = 0.75 moles

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