Answer:
Dispersion forces
Relative molecular mass
Explanation:
Alkanes experience only dispersion forces. Dispersion forces increase with increasevin the relative molecular mass of the compounds. Hence a higher relative molecular mass implies greater dispersion forces and a greater boiling point.
The answer is heterogeneous mixture<span> because the </span>blood<span> cells are physically separate from the </span>blood<span> plasma.</span>
Answer is: c. It is incorrect because sodium phosphate is a compound that has a single composition.
Sodium phosphate is chemical compound composed of atoms connected with chemical bonds.
Pure substance is made of only one type of atom (element) or only one type of molecule (compound), it has definite and constant composition with distinct chemical properties.
Pure substances can be separated chemically, not physically, that is difference between pure substances and mixtures.
The option which gives the correct mole ratios is H₂S : SO₂ = 2 : 2 and O₂ : H₂O = 3 : 2
<h3 /><h3>What is Mole ratio ?</h3>
It is a conversion factor between compounds in a chemical reaction, that is derived from the coefficients of the compounds in a balanced equation
Molar ratio also known as stoichiometry is the ratio in which the reactants and products are either formed or reacted in the given equation
The balanced equation for given reaction is as follows ;
2H₂S + 3O₂ --> 2SO₂ + 2H₂O
Molar ratio can be determined by the coefficients of the compounds in the balanced reaction
Coefficient is the number in front of the chemical compound and they are as follows
- H₂S - 2
- O₂ - 3
- SO₂ - 2
- H₂O - 2
Therefore, correct option is H₂S : SO₂ = 2 : 2 and O₂ : H₂O = 3 : 2
Learn more about mole ratio here ;
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(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 :)
