Calculate the no. of atoms of Ca present in 58.4g of CaCo3
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Explanation:
one atom is present in CaCO3
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Answer:
Explanation:
No of atoms = moles *nA* atomicity
= Moles = gievn mass/ molar mass that is = 58.4/100=0.584
Na= avagdros no
Atomicity= total no pf atoms = 5
So no of atoms = 0.584*6.022*10^23*5= 1.758*10^22 atoms
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