Question

No.of electrons present in 1.4 g of nitride ion

Answer 1

HÈLLØ!!

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Molar mass of dinitrogen (N2) = 28 g mol-1

Thus, 1.4 g of N2

= 1.4/28

= 0.05 mol

= 0.05 × 6.02 × 10²³ no of molecules

= 3.01 × 10²³ no. of molecules

Now, 1 molecule of N2 has 14 electrons.

Therefore, 3.01 × 10²³ molecules of N2 contains,

= 14 × 3.01 × 10²³

= 4.214 × 10²³ electrons.

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THÅÑKẞ!!

Answer 2

Concept:

A nitride ion is a nitrogen compound with a formal oxidation state of -3.

Given:

Mass of nitride ion = 1.4 g

Find:

Calculate the number of electrons present in 1.4 g of nitride ion.

Solution:

In the nitride ion, 8 valence electrons are present.

1.4 g of nitride ion corresponds to mass / molar mass = 1.4 g / 14 g/mol = 0.1 mol

The number of electrons present in 1.4 grams of nitride ion can be calculated as:

Number of electrons = Number of moles * Avogadro's number (NA)

By substituting the given values in the above expression, we can calculate the number of electrons present in the known amount of nitride ion as:

Number of electrons present in 1.4 grams of nitride ion = 0.1 mol * 6.022 * 10²³ * 8 electrons = 4.8176 * 10²³ electrons

Hence, the number of electrons present in 1.4 g of nitride ion is 4.8176 * 10²³ electrons.

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