Find no. Of nitrogen atoms present in 65mg of nan3
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Answer:
At. Mass of NaNo3=23+14+16*3
=85g =8500mg
Given mass =65mg
No. Of moles =65/8500
=0.00764
1 mole =6.022*10^23
0.00764mole =0.00764*6.022*10^23
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