Answer:
A.) ![K_b = \frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D)
Explanation:
The general Kb expression is:
![K_b = \frac{[HA][OH^-]}{[A^-]}](https://tex.z-dn.net/?f=K_b%20%3D%20%5Cfrac%7B%5BHA%5D%5BOH%5E-%5D%7D%7B%5BA%5E-%5D%7D)
In this equation
-----> Kb = equilibrium constant
-----> [HA] = acid
-----> [A⁻] = base
Since liquids are not included in equilibrium expressions, H₂O should not be present. The products are in the numerator while the reactant are in the denominator. In this reaction, CH₃NH₂ is acting as a base and CH₃NH₃⁺ is acting as an acid.
As such, the expression is:
Answer:
It has denatured
Explanation:
When the temperature get high the enzymes tend to change shape and denaturing occurs.
Answer:
A
Explanation:
idk but i do know its right
Answer:
The empirical formula is ZnO2
Explanation:
What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?
Step 1: Data given
Suppose the compound has a mass of 100.0 grams
A compound contains:
67.1 % Zinc = 67.1 grams
100 - 67.1 = 32.9 % oxygen = 32.9 grams
Molar mass of Zinc = 65.38 g/mol
Molar mass of O = 16 g/mol
Step 2: Calculate moles of Zinc
Suppose the compound is 100 grams
Moles Zn = 67. 10 grams / 65.38 g/mol
Moles Zn = 1.026 moles
Step 3: Calculate moles of O
Moles O = 32.90 grams / 16.00 g/mol
Moles O = 2.056 moles
Step 4: Calculate mol ratio
We divide by the smallest amount of moles
Zn: 1.026/1.026 = 1
O: 2.056/1.026 = 2
The empirical formula is ZnO2
To control this we can calculate the % Zinc for 1 mol
65.38 / (65.38+2*16) = 0.67.1 = 67.2 %
(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 :)
