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Find the number of moles of in 571,0 g of tetraphosphorous octoxide.

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Answer:

Think sorry if i did it wrong The molar mass that is the mass of 6.022 X 1023 atoms of helium is 4g. Hence 1 mole of He = 4 g of He = 6.022 X 1023atoms of He. 1 g of He = (6.022 X 1023) / 4 atoms of He. 52 g of He = (6.022 X 1023X 52 ) / 4 atoms of He.

For Helium,

Number of atoms=Number of moles ×6.022×1023

No of Moles = 6.022×10

2312.044×10

23 = 2.

Now, atomic mass of He=4

Weight = 2×4 = 8 g

The molar mass that is the mass of 6.022 X 1023 atoms of helium is 4g. Hence 1 mole of He = 4 g of He = 6.022 X 1023atoms of He. 1 g of He = (6.022 X 1023) / 4 atoms of He. 52 g of He = (6.022 X 1023X 52 ) / 4 atoms of He.

As we know, gram atoms = gram mole

here, no of moles = 23/46 =2 so no of gram atoms are 2.

Explanation:

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