How many cations and anions present in 9.5g of mgcl2
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anion=cation - 0.2 NA molecules
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Answer:
No of ions=moles × avagadro number×ions
Explanation:
No of ions =9.5/24×6.022×10^23×2
No of ions =4.75×10^23
Thank you I hope you understand well
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