Chemistry is not my forte so don't laugh to hard when I askthis question... How many moles of each of the following are provided by 0.125moles of Mg(NO3)2? 1. Magnesium ions 2. nitrate ions 3. oxygen atoms Is the answer: 0.125 for Mg 0.25 for N 0.75 for O Thanks! Chemistry is not my forte so don't laugh to hard when I askthis question... How many moles of each of the following are provided by 0.125moles of Mg(NO3)2? 1. Magnesium ions 2. nitrate ions 3. oxygen atoms Is the answer: 0.125 for Mg 0.25 for N 0.75 for O Thanks!
Explanation / Answer
We know that 1.0 mol ofMg(NO3)2 has contained 6.023*1023 Mg ions ,2*6.023*1023 Nitrate ions and 6*6.023*1023 Oxygen atoms. Now we will calculate the number of species present in0.125 mol of Mg(NO3)2 . 1) Magnesiumions: [6.023*1023 Mg ions / 1.0 mol ofMg(NO3)2 ]* 0.125 mol ofMg(NO3)2 = 7.53*1022 Mg ions. 2) Nitrate ions: [2*6.023*1023 Nitrate ions / 1.0 mol ofMg(NO3)2 ]* 0.125 mol ofMg(NO3)2 = 15.1*1022 Nitrate ions. 3) Oxygen atoms : [6*6.023*1023 Oxygen atoms / 1.0 mol ofMg(NO3)2 ]* 0.125 mol ofMg(NO3)2 = 45.2*1022 Oxygen atoms. = 15.1*1022 Nitrate ions. 3) Oxygen atoms : [6*6.023*1023 Oxygen atoms / 1.0 mol ofMg(NO3)2 ]* 0.125 mol ofMg(NO3)2 = 45.2*1022 Oxygen atoms. = 45.2*1022 Oxygen atoms.