1. How many GRAMS of ammonium phosphate, (NH4)3PO4, are present in 2.81×1022 formula units of this compound?
We know that
Moles = no of molecules /Avagadro no
Moles = 2.81x1022 /6.022x1023
Moles = 0.0466
Now moles = weight /molecular weight
But moles, 0.0466= wt/149
Therefore weight = 6.951 gm
1. How many GRAMS of ammonium phosphate, (NH4)3PO4, are present in 2.81×1022 formula units of this...
Part D (NH4)3PO4, ammonium phosphate Part E C17H19NO3. morphine
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