Question

calculate no. of electrons present in 9.5 g of PO4 ^-3​

Answer 1

Answer:

Moles of PO43− = Mass in g/molar mass of PO43−                              

= 9.5/31 + 4 × 16                            

= 9.5/95                   

= 0.1 mole

Number of molecules = mole × avogadro number                 

= 0.1 × 6.022 × 10^23                  

= 6.022 × 10^22

1 molecule of PO43− contains 15 + 4 × 8 + 3 = 50 electron

So,

6.022 × 10^22 contains 

= 50 × 6.022 × 10^22        

= 301.1 × 10^22 electrons

Read more on Brainly.in - https://brainly.in/question/5255492#readmore

Answer 2

HELLO ,

mols PO4^3- ion = grams/molar mass = 9.5/approx 95 = approx 0.1 but you need to confirm all of that

There are 6.02E23 PO4^3- ions in 1 mol; therefore, there are 6.02E22 mols in the 9.5 g.

In 1 PO4^3- there are 31 electrons in the P atom and 4*8 = 32 electrons in the 4 O atoms. 31 + 32 = 63 total electrons in each PO4^3- ion. So in 6.02E22 ions there will be ......... electrons.

MAKE

MY

ANSWER

AS

BRAINLIST........ ....

THANK YOU .....