calculate number of atoms present in 44 gram of nitogen gas?
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Answered by
1
nitrogen mass is 28=1 mole
1 mole=6.023×10²³
28→6.023×10²³
44→?
=44×6.023×10²³÷28
9.46×10²³
1 mole=6.023×10²³
28→6.023×10²³
44→?
=44×6.023×10²³÷28
9.46×10²³
Answered by
2
weight /Molecular wt.=no. of atoms/Na where Na=avagadro number=6.022×10^23. so now=44/28=no.of atom/6.022×10^23. Now...solve yourself!!
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