In a particular experiment, a 2.00-g sample of cao is reacted with excess water and 2.14 g of ca(oh)2 is recovered. what is the percent yield in this experiment?

81.0% yield First, lookup the atomic weight of all involved elements Atomic weight of Calcium = 40.078 Atomic weight of Hydrogen = 1.00794 Atomic weight of Oxygen = 15.999 Calculate the molar mass of CaO and Ca(OH)2 Molar mass CaO = 40.078 + 15.999 = 56.077 g/mol Molar mass Ca(OH)2 = 40.078 + 2 * 1.00794 + 2 * 15.999 = 74.09188 g/mol Determine how many moles of CaO you started with 2.00 g / 56.077 g/mol = 0.035665246 mol Determine how many moles of Ca(OH)2 you ended up with 2.14 g / 74.09188 = 0.028883057 mol Looking at the formula, you should get 1 Ca(OH)2 molecule per CaO molecule used. So divide the number of moles obtained by the number of moles used to get the percent yield. 0.028883057 / 0.035665246 = 0.809837594 = 81.0%

They are held together by chemical bonds. i am pretty sure.

True, the standard unit for volume is liters which can be represented by either a lower case "l" or a capital "l"