I think its true but im not 100 percent sure
(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 :)
Answer:
B. a change in their state or substance
Explanation:
A change of state consists of a physical process in which the structure of the substance changes, its appearance. EXAMPLE: in the merger, the solid state becomes liquid, changes the rearrangement of particles from being highly compacted to being less compact.
Answer:
Explanation:
YES BECAUSE YE SIS YES WHEN YES=![\sqrt{x} x^{2} x^{2} \neq \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right]](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%20x%5E%7B2%7D%20x%5E%7B2%7D%20%5Cneq%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D)
GAMER MOMENT FROM LUIGI FOR⇔![\sqrt{x} x^{2} x^{2} \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \alpha \alpha \alpha x_{123} \frac{x}{y} \pi \neq \geq \leq \\ \left \{ {{y=2} \atop {x=2}} \right. \int\limits^a_b {x} \, dx x^{2} \sqrt{x} \sqrt{x} \\](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%20x%5E%7B2%7D%20x%5E%7B2%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%5Calpha%20%5Calpha%20%5Calpha%20x_%7B123%7D%20%5Cfrac%7Bx%7D%7By%7D%20%5Cpi%20%5Cneq%20%5Cgeq%20%5Cleq%20%5C%5C%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20x%5E%7B2%7D%20%5Csqrt%7Bx%7D%20%5Csqrt%7Bx%7D%20%5C%5C)
Answer:
Without taking body measurements, the fabric material can run out of supply if the tailor uses it too much. And more material is required to fix small dresses or suits. So when body measurements are taken accurately, the tailor will know the exact amount of material to be used.
Explanation:
