calculate number of atoms of
of
conitituent elemelte in
52g
of Na2, co3
Answers
Answered by
7
Answer:
53 g of Na2CO3 Contain = 53106=12 mol.
No of atoms of Na = 12×2×6.022×1023= 6.022×1023.
No of atoms of C = 12×6.022×1023.
No of atoms of O = 12×3×6.022×1023= 9.033×1023.
Answered by
0
Answer:
Molecular mass of Na2CO3 = ( 2 × 23 ) + 12 + (3 × 16) = 46 + 12 + 48 = 106
Given mass of Na2CO3 = 53 g
106 g of Na2CO3 Contain = 1 mol
53 g of Na2CO3 Contain = 53106=12 mol
No of atoms of Na = 12×2×6.022×1023= 6.022×1023
No of atoms of C = 12×6.022×1023
= 3.011×1023
No of atoms of O = 12×3×6.022×1023= 9.033×1023
:))))
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