1.

Number of electrons in 3.6mg of NH_(4)^(+) are:

Answer»

`1.2XX10^(21)`
`1.2xx10^(20)`
`1.2xx10^(32)`
`2xx10^(-3)`

Solution :Number of electrons in one ION of `NH_(4)^(+)`=10
`=(3.6xx10^(-3))/(18)xx6.023xx10^(23)=1.2xx10^(20)`
`therefore` Number of electrons in 3.6mg `NH_(4)^(+)=1.2xx10^(20)xx10`
`1.2xx10^(21)`


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