Question

avogadro's number of atoms are present in 28g o2

Answer 1

Answer:

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Explanation:

Avagadro's law, 1 Nitrogen mole or 28 g of Nitrogen contains 6.023 x 10^23 no. of molecules. Nitrogen, being a diatomic element, the no. of atoms = 2 x 6.023 x 10^23 atoms.20-Sep-2018

Answer 2

Answer:

As we all know the avogrado number

It's 1 mole = mass of the nitrogen

Here we can say, mass of the nitrogen=1mole

1mole=6.022×10^23

And we can get the mass of nitrogen by

Mass of nitrogen=no. Of protons+neutrons

Mass of nitrogen=7+7

Mass of nitrogen=14u

So we get that,

Mass of the nitrogen=1mole

Then,

14g=1mole

28g=xmole

14:28::1:x

14x=28

X=2

So,

28g=2moles

Or,28g=2×6.022×10^23 no. of atoms

So,the no. Of atoms present in 28g of nitrogen will be 1.204×10^24 no. Of atoms are present

Explanation:

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