Answer
There are $7.45\times10^{28}$ atoms of Helium in the air blimp.
Work Step by Step
First we must convert the units of mass from kg to grams by multiplying by 1000; $495 kg He \times1000 = 495,000 g He$. Now we must convert from grams of Helium to moles of Helium by using the molar mass; $495,000 g He\times\frac{1 mole He}{4.00 g/mol} = 123,750 moles He$. Lastly, we use Avogadro's number to convert from moles to atoms; $123,750 moles He \times\frac{6.023\times10^{23} atoms}{1 mole He} = 7.45\times10^{28} atoms He.$
