Question

Calculate number of electrons present in 9.5 gram of phosphate (PO4)
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Answer 1

Answer:

28.3*$10^{23}$ electrons..

Explanation:

Number of electrons prensent in one molecule of $PO_{4}$ = 15 + 4*8= 47.

Atomic mass of  $PO_{4}$  =   95 grams.

So, 95 g of $PO_{4}$ contains 6.023* $10^{23}$atoms  i.e., 47*6.023*$10^{23}$ electrons.                            

Therefore,95. g of   $PO_{4}$ contains   47*6.023*$10^{22}$ =  28.3*$10^{23}$ electrons.                                          

Answer 2

Answer:

28.3 * 30

Explanation:

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