the total number of electrons in 4.2 grams of N3 minus ion is
Answers
ANSWER:-
Given mass= 4.2g
Molar mass of N3^- (azide ion) = 14*3= 42
Therefore, No. Of moles of azide ion = given mass/ molecular mass= 4.2/42= 0.1 moles
No. of valence electrons in one nitrogen atom = 5
No. of valence electrons in azide ion (N3^-) ( N=N=N ) is =(5*3)+1=16electrons
Therefore, total no. of valence electrons in 4.2g of azide (N3^-) ion = no. of moles of azide ion * no. of valence electrons in one azide ion * Avogadro's constant
= 0.1* 16* 6.022*10^23
=1.6* 6.022*10^ 23
=9.6352* 10^23 valence electrons.
PLEASE MARK AS BRAINLIEST
Answer:
The number of electrons in 4.2 g of azide ion is 2.2NA.
Explanation:
Given mass of N3= 4.2g
Molar mass of N3= 42 g/mol
No. of moles of N3= 4.2/42 = 0.1 moles
No. of electrons in one mole nitrogen atom N= 7e (hint: same as atomic no)
No. of electrons in one mole of azide N3= 7*3= 21
No. of electrons in one mole of azide ion N3- = 21+1 = 22 (since 1 electron is added in azide ion)
Formula:
No. of electrons = no. of moles x no. of electrons in one mole x NA
No. of electrons = 0.1 x 22 x NA = 2.2 NA