3.34 calculate the following quantities:
(a) mass, in grams, of 5.76x10-3 mol CdS.
the conversion factor from grams -> moles or moles -> grams is the molar mass; calculate it
for cadmium sulfide...
now set the problem up as a simple unit conversion:
(b)number of moles of NH4Cl in 112.6 g of this substance
again, use the molar mass....
(c) number of molecules in 1.305 x 10-2 mol of C6H6
whenever we calculate a number of atoms or molecules, we will use Avogadro's number, since this number gives us the number of things (atoms, molecules, ions, gnus) in 1 mol of those things. Here we're given a mol quantity and asked to convert it to individual molecules:
(d) number of O atoms in 4.88x10-3 mol Al(NO3)3
again, we're given a molar quantity and asked to convert to a numerical quantitiy, The trick here is
to notice that for each aluminum nitrate unit, we have nine O atoms.....
