Could you explain the steps? Thanks!
We can use dimensional analysis to convert 4.56 × 10^23 formula units of AlCl3 to moles of AlCl3, to moles of Cl^-We will need the following equalities to set up the conversion factors: 1 mol AlCl3 6.02 × 10^23 formula units AlCl3 1 mol AlCl3 3 mol Cl^- ions [(4.56 × 10^23 formula units AlCl)/1][(1 mol AlCl3)/(6.02 × 10^23 formula units AlCl3)][(3 mol Cl^-)/(1 mol AlCl3)] 2.272425 mol Cl^- ions or 2.27 mol Cl^- ions rounded to three significant figures Answer: About 2.27 mol Cl^- ions are in 4.56 × 10^23 formula units of aluminum chloride.
electrons ALWAYS carry a negative charge protons ALWAYS carry a positive charge neutrons ALWAYS carry NO chargem
