Answer:
Carbon dioxide reacts with calcium hydroxide solution to produce a white precipitate of calcium carbonate
Explanation:
. Limewater is a solution of calcium hydroxide. If carbon dioxide is bubbled through limewater, the limewater turns milky or cloudy white
The answer is C. They lower the activation energy of an elementary step of a reaction
This makes the reaction rate to increase since less energy is required to make a reaction occur.
Answer:
Individually well, defined identical chemically units such as molecules, ions, atoms, or electrons
Explanation:
The mole is used as the unit of measurement for substance such as molecules, ions, atoms, or electrons. One mole of a substance is equivalent to 6.02×10²³ particles of the substance. The number, 6.02 × 10²³ is known as Avogadro's number.
The particles quantified as moles are individually well, defined identical chemically units such that the mole can be used to describe a part of a substance or the whole substance consisting of several moles of the substance combined.
(a)
pH = 4.77
; (b)
[
H
3
O
+
]
=
1.00
×
10
-4
l
mol/dm
3
; (c)
[
A
-
]
=
0.16 mol⋅dm
-3
Explanation:
(a) pH of aspirin solution
Let's write the chemical equation as
m
m
m
m
m
m
m
m
l
HA
m
+
m
H
2
O
⇌
H
3
O
+
m
+
m
l
A
-
I/mol⋅dm
-3
:
m
m
0.05
m
m
m
m
m
m
m
m
l
0
m
m
m
m
m
l
l
0
C/mol⋅dm
-3
:
m
m
l
-
x
m
m
m
m
m
m
m
m
+
x
m
l
m
m
m
l
+
x
E/mol⋅dm
-3
:
m
0.05 -
l
x
m
m
m
m
m
m
m
l
x
m
m
x
m
m
m
x
K
a
=
[
H
3
O
+
]
[
A
-
]
[
HA
]
=
x
2
0.05 -
l
x
=
3.27
×
10
-4
Check for negligibility
0.05
3.27
×
10
-4
=
153
<
400
∴
x
is not less than 5 % of the initial concentration of
[
HA
]
.
We cannot ignore it in comparison with 0.05, so we must solve a quadratic.
Then
x
2
0.05
−
x
=
3.27
×
10
-4
x
2
=
3.27
×
10
-4
(
0.05
−
x
)
=
1.635
×
10
-5
−
3.27
×
10
-4
x
x
2
+
3.27
×
10
-4
x
−
1.635
×
10
-5
=
0
x
=
1.68
×
10
-5
[
H
3
O
+
]
=
x
l
mol/L
=
1.68
×
10
-5
l
mol/L
pH
=
-log
[
H
3
O
+
]
=
-log
(
1.68
×
10
-5
)
=
4.77
(b)
[
H
3
O
+
]
at pH 4
[
H
3
O
+
]
=
10
-pH
l
mol/L
=
1.00
×
10
-4
l
mol/L
(c) Concentration of
A
-
in the buffer
We can now use the Henderson-Hasselbalch equation to calculate the
[
A
-
]
.
pH
=
p
K
a
+
log
(
[
A
-
]
[
HA
]
)
4.00
=
−
log
(
3.27
×
10
-4
)
+
log
(
[
A
-
]
0.05
)
=
3.49
+
log
(
[
A
-
]
0.05
)
log
(
[
A
-
]
0.05
)
=
4.00 - 3.49
=
0.51
[
A
-
]
0.05
=
10
0.51
=
3.24
[
A
-
]
=
0.05
×
3.24
=
0.16
The concentration of
A
-
in the buffer is 0.16 mol/L.
hope this helps :)
