Chemistry, asked by Jtg, 1 year ago

How many electron r in 3.1 mg of NO3

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Answered by Sanskriti101199
1

heya-------
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Answered by Anonymous
2
NO. OF ELECTIONS =NO. OF MOLES OF NO3 X Avogadro no.
=[3.1x.001/62]x Na =0.00005Na [Na = Avogadro no]
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