Im pretty Sure its B Tuesday And Wensday
Answer:
<u>Radiation is the transfer of energy by waves, and conduction is the transfer of heat through contact with air.</u>
Explanation:
Conduction is the transfer of thermal energy through direct contact. Radiation is the transfer of thermal energy through thermal emission.
(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: The value of 
for the half-cell reaction is 0.222 V.
Explanation:
Equation for solubility equilibrium is as follows.
Its solubility product will be as follows.
![K_{sp} = [Ag^{+}][Cl^{-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D)
Cell reaction for this equation is as follows.
Reduction half-reaction: 
, 
Oxidation half-reaction: 
, 
= ?
Cell reaction: 
So, for this cell reaction the number of moles of electrons transferred are n = 1.
Solubility product,
= 
Therefore, according to the Nernst equation
![E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7B0.0592%20V%7D%7Bn%7D%20log%20%5Cfrac%7B%5BAgCl%5D%7D%7B%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D%7D)
At equilibrium, 
= 0.00 V
Putting the given values into the above formula as follows.
![0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}](https://tex.z-dn.net/?f=0.00%20%3D%20E%5E%7Bo%7D_%7Bcell%7D%20-%20%5Cfrac%7B0.0592%20V%7D%7B1%7D%20log%20%5Cfrac%7B1%7D%7B%5BAg%5E%7B%2B%7D%5D%5BCl%5E%7B-%7D%5D%7D)
= 
= 0.577 V
Hence, we will calculate the standard cell potential as follows.
= 0.222 V
Thus, we can conclude that value of for the half-cell reaction is 0.222 V.
