calculate the number of N A 2 c o 3 molecules in its 530 gram
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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=1253106=12 mol
No of atoms of Na = 12×2×6.022×102312×2×6.022×1023= 6.022×10236.022×1023
No of atoms of C = 12×6.022×102312×6.022×1023
= 3.011×10233.011×1023
No of atoms of O = 12×3×6.022×102312×3×6.022×1023= 9.033×10239.033×1023
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Given mass of Na2CO3 = 53 g
106 g of Na2CO3 Contain = 1 mol
53 g of Na2CO3 Contain = 53106=1253106=12 mol
No of atoms of Na = 12×2×6.022×102312×2×6.022×1023= 6.022×10236.022×1023
No of atoms of C = 12×6.022×102312×6.022×1023
= 3.011×10233.011×1023
No of atoms of O = 12×3×6.022×102312×3×6.022×1023= 9.033×10239.033×1023
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Answered by
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Answer:
the answer is 30110×10¹⁹
Explanation:
- na2co3 has a mass of -: na=23g, c=12g and o=16g. mass = 23×2+12×2+16×3=46+12+48=106g. no.of moles = 530/106=5moles. no.of molecules = 5×6.022×10²²=30110×10¹⁹molecules Hope it was helpful
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