The mole is a number equal to 602 ,000,000,000,000 ,000,000,000 or 6.02times 10^23 Solve the following problem to give you an idea of how large a number this is: How many dollars would each of the 7 billion (7,000,000 ,000) people of the world get if you distributed 1 mole of pennies to each of them? Converting moles to representative particles: representative particles=molesx(6.02times 10^23reprrssentative particles)/(1mole) Converting representative particles to moles: moles=representative particlesx(1mole)/(6.02times 10^22)representative particles 1. How many moles of magnesium is 1.25times 10^23 atoms of magnesium? 2. Determine the number of atoms that are in 0.58 mol of Se. 3. How many moles of barium nitrate (BaNO_(3)) contain 6.80times 10^24 formula units?

1. To find the number of moles of magnesium, we can use the conversion factor from the given information:<br /><br />$representative\quad particles=molesx\frac {6.02\times 10^{23}reprrssentative\quad particles}{1mole}$<br /><br />Given that there are $1.25\times 10^{23}$ atoms of magnesium, we can set up the equation:<br /><br />$1.25\times 10^{23} atoms \times \frac{1 mole}{6.02\times 10^{23} atoms} = 0.207 moles$<br /><br />Therefore, there are 0.207 moles of magnesium in $1.25\times 10^{23}$ atoms of magnesium.<br /><br />2. To determine the number of atoms in 0.58 mol of Se, we can use the conversion factor from the given information:<br /><br />$moles=representative\quad particlesx\frac {1mole}{6.02\times 10^{22}representative\quad particles}$<br /><br />Given that there are 0.58 mol of Se, we can set up the equation:<br /><br />$0.58 mol \times \frac{6.02\times 10^{22} atoms}{1 mole} = 3.49\times 10^{22} atoms$<br /><br />Therefore, there are $3.49\times 10^{22}$ atoms in 0.58 mol of Se.<br /><br />3. To find the number of moles of barium nitrate $(BaNO_{3})$ that contain $6.80\times 10^{24}$ formula units, we can use the conversion factor from the given information:<br /><br />$representative\quad particles=molesx\frac {6.02\times 10^{23}reprrssentative\quad particles}{1mole}$<br /><br />Given that there are $6.80\times 10^{24}$ formula units of barium nitrate, we can set up the equation:<br /><br />$6.80\times 10^{24} formula units \times \frac{1 mole}{6.02\times 10^{23} formula units} = 11.29 moles$<br /><br />Therefore, there are 11.29 moles of barium nitrate $(BaNO_{3})$ that contain $6.80\times 10^{24}$ formula units.