below left is a cylinder containing water. an ball with a mass of 21g and a volume of 15 cm3 is lowered into the water. sketch t
he object and the new water level in the cylinder on the right.
1 answer:
You didn’t show the cylinder containing water, so I created one that you can use as a model (see image).
The water level was originally at 37 mL.
Then you added the ball, and it displaced its volume of water.
The new volume reading is 52 mL, so
Volume of ball = volume of displaced water = 52 mL – 37 mL = 15 mL.
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Answer:
Approximately , assuming that this acid is monoprotic.
Explanation:
Assume that this acid is monoprotic. Let denote this acid.
.
Initial concentration of without any dissociation:
.
After of that was dissociated, the concentration of both
and
(conjugate base of this acid) would become:
.
Concentration of in the solution after dissociation:
.
Let ,
, and
denote the concentration (in
or
) of the corresponding species at equilibrium. Calculate the acid dissociation constant
for
, under the assumption that this acid is monoprotic:
.