Number of electrons present in `3.8g` of `F^(-)` ion is (Atomic weight of `F=19`)
(1) `1.8N_(A),`
(2) `2N_(A)`
(3) `2.3N_(A),`
(4) `1.5N_(A)`

The total number of electrons in 4.2 g of N^(3-) ion is ( N_(A) is the Avogadro's number)

The total number of valence electrons in 4. 2g of N_3^- ion are :

In an organic compound of molar mass 108 g mol^(-1) , H and N atoms are present in 9 : 1 : 3.5 by weight. Molecular formula can be: "1.C_(6)H_(8)N_(2) " 2.C_(7)H_(10)N 3.C_(5)H_(6)N_(3) 4.C_(4)H_(18)N_(3)

The number of F^- ions in 2.1 g AlF_3 is (N_A=6xx10^23)

If N_(A) is Avogadro's number then number of valence electrons in 4.2 g of nitride ions (N^(3-))

Mean of the numbers 1,2,3,…,n with respective weights 1^(2) + 1, 2^(2) + 2, 3^(2) + 3,…,n^(2)+n is

If n ^(3)=-8, what is the value of ((n^(2))^(3))/((1)/(n^(2))) ?

Let f(n)=2cosn xAAn in N , then f(1)f(n+1)-f(n) is equal to (a) f(n+3) (b) f(n+2) (c) f(n+1)f(2) (d) f(n+2)f(2)

Let f(n)=2cosn xAAn in N , then f(1)f(n+1)-f(n) is equal to f(n+3) (b) f(n+2) f(n+1)f(2) (d) f(n+2)f(2)

For all n inN,3*5^(2n+1)+2^(3n+1) is divisible by (A) 19 (B) 17 (C) 23 (D) 25