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Calculate the number of electrons, protons and neutrons present in 24 grams of ""_(12)Mg^(24)and ""_(12)Mg^(26). |
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Answer» Solution :24 g of magnesium CONTAINS `6.023 xx 10^(23)` protons For `""_(12)Mg^(24)` `therefore` 24 g of Mg has `12 xx 6.023 xx 10^(23)` protons `=72.2 xx 10^(23)` `therefore 24`g of Mg has `12 xx 6.023 xx 10^(23)` electrons `=72.2 xx 10^(23)` `therefore 24G` of Mg has `12 xx 6.023 xx 10^(23)` neutrons `=72.2 xx 10^(23)` For `""_(12)Mg^(26)` 26 g of Mg has `6.023 xx 10^(23)` atoms 1g of Mg has `(6.023 xx 10^(23))/26 xx 24` atoms `=5.56 xx 10^(23)` atoms `therefore` Totalnumber of electrons in 24 g of `""_(12)Mg^(26) = 12 xx 5.56 xx 10^(23) = 66,72 xx 10^(23` `therefore` Total NUMBER of proton in 24 g of `""_(12)Mg^(26)` `=66.72 xx 10^(23)` `therefore` total number of proton in 24 g of `""_(12)Mg^(26) = 66.72 xx 10^(23)` Total number of neutrons in 24 g of `""_(12)Mg^(26)` `=14 xx 5.56 xx 10^(23) = 77.84 xx 10^(23)` |
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