PLS HELPPP How many moles of oxygen atoms (not molecules) are present in 6.41×1025 molecules of dinitrogen pentoxide (N2O5)? 1. 532 mol 2. 106 mol 3. 32 × 1025 mol 4. 266 mol

(4) 266 moles

Explanation:

We have Dinitrogen Pentoxide N2O5

6.41*10^25 molecules are given

No of Moles of N2O5 =

                                     =

                                      = 106.5 mol

Now using Unitary Method

2 Mole of Nitrogen pentoxide require 5 mole of Oxygen to form N2O5

1 mole of N =

In 106.5 mole of N  = = 266.25 mole

So, 6.41*10^25 molecules of N2O5 will require 266.25 mole of Oxygen atoms.

c-14 is carbon-14, or radiocarbon.   it is a radioactive isotope of carbon with an atomic nucleus containing 6 protons and 8 neutrons.