B. Beta decay
Beta decay occurs when a high-speed electron is lost from the atom, converting one of the atom's neutrons into a proton. The mass number of the atom then remains the same but its atomic number is increased by one.
This is the decay taking place because the transition from carbon-14 to nitrogen-14 involves an increase in atomic number by 1.
<span />
(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 :)
The reaction that results from this is:
H2O + CO2 --> H2CO3
Stoichiometric ratio between water and CO2 is 1:1. So we can say that for every Mole of CO2, we need 1 Mole of water to produce 1 Mole of H2CO3. Thus as n=m/M we can find n = 528/44.01 = 11.997 ~ 12Mol.
therefore, we need 12 moles of water.
An oxide of nitrogen contains 30.45 mass % N, if the molar mass is 90± 5 g/mol the molecular formula is N₂O₄.
<h3>What is molar mass?</h3>
The molar mass of a chemical compound is determined by dividing its mass by the quantity of that compound, expressed as the number of moles in the sample, measured in moles. A substance's molar mass is one of its properties. The compound's molar mass is an average over numerous samples, which frequently have different masses because of isotopes.
<h3>How to find the molecular formula?</h3>
The whole-number multiple is defined as follows.
Whole-number multiple = 
The empirical formula mass is shown below.
Mw of empirical formula = Mw of N+ 2 x (Mw of O)
= 14.01 g/mol + 2 x (16.00 g/mol)
= 46.01 g/mol
With the given molar mass or the molecular formula mass, we can get the whole-number multiple for the compound.
Whole-number multiple = 
≈ 2
Multiplying the subscripts of NO2 by 2, the molecular formula is N(1x2)O(2x2)= N2O4.
To learn more about molar mass visit:
brainly.com/question/12127540
#SPJ4
