heating, 33.632 grams, from the mass of crucible, cover and sample after heating, 33.376
grams.
33.632 grams − 33.376 grams = 0.256 grams of water
The 0.256 grams of water is then divided by the 18 g/mole, the molar mass of water, to provide
the moles of water.
65 . C
The sample needs to be heated until all the water is removed from the original hydrate
sample. The student needs to continue heating the sample until the mass after heating is
unchanged from the mass prior to the additional heating, indicating all the water has been
removed.
66 . B
A buffer solution contains both a weak acid and its conjugate base in significant
concentrations. In each case, decide what species will be present in the solution; in each, the
substances will react with each other. You can mark off C and E right away, since they contain
only strong acids and bases; you can’t make a buffer with no weak acid or base present.
A: the reaction will produce 0.1 mol of HSO 3 − and leave 0.1 mol H+. Not a buffer.
B: the NaOH is the limiting reactant; the solution will contain 0.2 mol F− and 0.2 mol leftover
HF, a weak acid. The solution will act as a buffer.
D: NH 3 is the limiting reactant; all the NH 3 will be converted to NH 4 +, a weak acid, leaving 0.2
mol HCl, a strong acid. There is no conjugate base present and therefore no buffer.
67 . D
For a second-order reaction,