Calculate number of electrons present in 9.5 gram of phosphate (PO4)
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Molar mass of PO4 = 95g/mol
therefore 95g of PO4 contain 6.022*10^23 atoms
1g of PO4 will contain 6.022*10^23/95 atoms
Hence 9.5g of PO4 will contain 6.022*10^22 atoms
Since 1 molecule contain 5+4(6) electrons
therefore total electrons =29*6.022*10^22 e
Answered by
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Number of electrons in 9.5 g of phosphate
According to formula
molar mass=95
9.5/95
=0.1
1 mole contains =6*10^22 ions
but each ion has 50 electrons
15+32+3
Total electrons in 9.5 gram in phosphate is =(6*10^22)*50
=3*10^24 in (PO4)3
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According to formula
molar mass=95
9.5/95
=0.1
1 mole contains =6*10^22 ions
but each ion has 50 electrons
15+32+3
Total electrons in 9.5 gram in phosphate is =(6*10^22)*50
=3*10^24 in (PO4)3
HOPE IT WILL HELP YOU....
PLEASE MARK AS BRAINLIEST .......
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