The no of electrons present in one mole of N3- ion
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Heyya friend.
Here's the answer.
for 1 g-ion. 1 g-ion = (weight in grams of sum of atomic weight of atoms making the ion.)
Here N^3- has 14 amu so 1g-ion 14 gm.
So here we have to find for 14g that is 1mole of nitride ion. 1 ion of nitride has =1.6 ×10^-19 × 3 C
Therefore 6.022 × 10^23 ions has = 6.022 × 10^23 ×1.6 ×10^-19 × 3 C = 2.89 × 10^5 C.
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Here's the answer.
for 1 g-ion. 1 g-ion = (weight in grams of sum of atomic weight of atoms making the ion.)
Here N^3- has 14 amu so 1g-ion 14 gm.
So here we have to find for 14g that is 1mole of nitride ion. 1 ion of nitride has =1.6 ×10^-19 × 3 C
Therefore 6.022 × 10^23 ions has = 6.022 × 10^23 ×1.6 ×10^-19 × 3 C = 2.89 × 10^5 C.
Pls mark as brainest and follow me.
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the no of electrons in azide ion is 22
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