Answer:
a) C6H5COOH + H2O ↔ H3O+ + C6H5COO-
b) [ H3O+ ] = 2.517 E-3 M
c) pH = 2.599
Explanation:
a) balanced equation:
C6H5COOH + H2O ↔ H3O+ + C6H5COO-
⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5
mass balance:
0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)
charge balance:
[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant
⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)
b) (2) in (1):
⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]
⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]
⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5
⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0
⇒ [ H3O+ ] = 2.517 E-3 M
c) pH = - log [ H3O+ ]
⇒ pH = - Log ( 2.517 E-3 )
⇒ pH = 2.599
Answer:
A. stored energy
Explanation:
potential energy is stored energy.
kinetic energy is the energy of motion
A saturated solution is one in which no more solute is able to dissolve in a given solvent at a particular temperature. Some amount of the solute is left undissolved in the solution.
Unsaturated solution has solute in lower proportions than required to form a saturated solution.
Supersaturated solution has solute in amounts greater than a saturated solution.
We can take the help of solubility curve in order to find out the amount of a salt required to prepare a saturated solution of that salt at a particular temperature.
The solubility of KI at 10
is 136 g/ 100 mL water
The solubility of
at
is 21 g/100 mL water.
The solubility of
at
is 80 g/100 mL water.
The solubility of NaCl at
is 38 g/ 100 mL water.
So the correct answer will be KI, as it would need 136 g KI / 100 mL water to form a saturated solution at
.So, if we have 80g KI/ 100mL water it would be an unsaturated solution.
Answer:
To tell if something is an acid or a base, you can use a chemical called an indicator. An indicator changes color when it encounters an acid or base. There are many different types of indicators, some that are liquids and others that are concentrated on little strips of "litmus" paper.
Explanation:
A model showing that gases are made from the matter of particles that are too small to see and are moving freely around in space can explain many observations.