1.

The number of oxygen present in 11.2 L of ozone at N.T.P. are______ .

Answer»

`1.20xx10^(24)`
`9.03xx10^(23)`
`6.02 xx10^(23)`
`3.01xx10^(23)`

Solution :`11.2 L " of OZONE " -=(6.02xx10^(23))/(2)" MOLECULES of "O_3`
`-=3xx3.01xx10^(23)" atoms of oxygen "`
`-=9.03xx10^(23)" atoms of oxygen "`


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