Can you please show me how you get the answer as well? Thank you.
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Hy there...
to find that follow a series...
1) find the no. of moles of comp. ( here n = weight/ mol.weight i.e. 57/114 = 0.5)
2) now in 1 mole there are 6.023 * 10^23 molecules... i.e. NA ( the avogardo no.) so in 0.5 moles of compound there will be 0.5 * 6.023 * 10^23 molecules of compound = 3.01 * 10^23 or simply 0.5 NA
3)now there are 6 H atoms in one molecule of compound so no. of H atoms in 0.5 NA molecules of compound = 0.5NA * 6 or 3.01*10^23 * 6 = 3NA or 1.8*10^24
thats your answer...
hope this helps :)
to find that follow a series...
1) find the no. of moles of comp. ( here n = weight/ mol.weight i.e. 57/114 = 0.5)
2) now in 1 mole there are 6.023 * 10^23 molecules... i.e. NA ( the avogardo no.) so in 0.5 moles of compound there will be 0.5 * 6.023 * 10^23 molecules of compound = 3.01 * 10^23 or simply 0.5 NA
3)now there are 6 H atoms in one molecule of compound so no. of H atoms in 0.5 NA molecules of compound = 0.5NA * 6 or 3.01*10^23 * 6 = 3NA or 1.8*10^24
thats your answer...
hope this helps :)
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in 1mole C6H5ONH, there are 6moles of H
we have 57g C6H5ONH which is 108g
57g / 108g/mole = 0.528moles
0.528moles C6H5ONH x (6moles H / 1 C6H5ONH) = 3.167moles H present
3.167moles x 6.022x10^23atoms/mole = 1.91x10^24atoms
or
%H in C6H5ONH = 6g / 108g x 100% = 5.55%
0.0555 x 57g = 3.16g H in C6H5ONH
3.16gH / 1g/mole = 3.16moles
3.16moles x 6.022x10^23 atoms/mole = 1.91x10^24atoms
we have 57g C6H5ONH which is 108g
57g / 108g/mole = 0.528moles
0.528moles C6H5ONH x (6moles H / 1 C6H5ONH) = 3.167moles H present
3.167moles x 6.022x10^23atoms/mole = 1.91x10^24atoms
or
%H in C6H5ONH = 6g / 108g x 100% = 5.55%
0.0555 x 57g = 3.16g H in C6H5ONH
3.16gH / 1g/mole = 3.16moles
3.16moles x 6.022x10^23 atoms/mole = 1.91x10^24atoms