(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 :)
C can be the only correct answer - 6.023 x 10^23 is the amount of molecules in a mol of an element. 4.5 x 6.023 x 10^23 can not equal anything but C.
4.5 x 6.023 x 10^23 = 2.71035 x 10^24
Answer:
Evaporation
Explanation:
Water in the ocean, rivers, lakes, etc. is part of the hydrosphere, and when that water evaporates it enters the atmosphere
Answer:
270g
Explanation:
Given parameters:
Concentration of NaOH = 1.5M
Volume = 4.5L
Unknown
Mass of NaOH added = ?
Solution:
To solve the problem, we need to find the number of moles of the NaOH first;
Number of moles = concentration x volume
Number of moles = 1.5 x 4.5 = 6.75mol
Now;
Mass = Number of moles x molar mass
Molar mass of NaOH = 23 + 16 + 1 = 40g/mol
Mass = 6.75 x 40 = 270g
<u>Answer:</u> The percentage abundance of 
and 
isotopes are 75.77% and 24.23% respectively.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Let the fractional abundance of isotope be 'x'. So, fractional abundance of isotope will be '1 - x'
- <u>For isotope:</u>
Mass of isotope = 34.9689 amu
Fractional abundance of isotope = x
- <u>For isotope:</u>
Mass of isotope = 36.9659 amu
Fractional abundance of isotope = 1 - x
- Average atomic mass of chlorine = 35.4527 amu
Putting values in equation 1, we get:
![35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577](https://tex.z-dn.net/?f=35.4527%3D%5B%2834.9689%5Ctimes%20x%29%2B%2836.9659%5Ctimes%20%281-x%29%29%5D%5C%5C%5C%5Cx%3D0.7577)
Percentage abundance of isotope = 
Percentage abundance of isotope = 
Hence, the percentage abundance of and isotopes are 75.77% and 24.23% respectively.
