Question

HEY FRND SOLVE THIS .....

CALCULATE THE NO. OF ELECTON & NEUTRON IN 1.6g OF NAOH.???????

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Answer 1

hope u can see it.......

Answer 2

No. of Mole Of NaOH = 1.6/40 [ Molecular Wt. of NaOH is 40]

Mole = 0.04

No. Of Electron in NaOH = 1(11) + 1(8) + 1(1)..... [ 11,8, 1 are no. of electron present in Na, O and H]

= 20

So No. of Electron = 0.04 * 6.022 * 10$^{23}$ * 20

= 4.8176 * 10$^{23}$

Now For Neutron = 1(12) + 1(8) + 1(0)...... [12,8,0 are neutron in Na, O and H]

= 20

So No. of Neutron Present = 0.04 * 6.022 * 10$^{23}$ * 20

= 4.8176 * 10$^{23}$

Hope this helps you out