Question

1. Calculate the number of electrons present in:-
(i)1.6g of CH4 and
(ii)2.8g of N2...
Explain it...

Answer 1

(i). 1molecule of CH4 = 6+4(1) =10electrons

=>1mole has 6.022×10^23molecules

=>1mole =16g has( 6.022×10^23)×10electrones

=>1.6g has ( 6.022×10^23)×10×1.6/16

=6.022×10^23 electrones (Answer)

____________________________________

(ii). 1molecule of N2 has 7+7=14electrones

1mole has 6.022×10^23molecules

=>1mole=14×2g=28g

=28g has 6.022×10^23 ×14electrones

=>2.8g will have 6.022×10^23 ×14×2.8/28

=84.308×10^24

=8.4308×10^25 electrons (Answer)

hope it helps you.......
☺✌☺