Balancing redox reactions:
Oxygen should be balanced by adding 
as needed, while hydrogen should be balanced by adding 
.
What is a redox reaction?
Redox reactions, also known as oxidation-reduction reactions, involve the simultaneous oxidation and reduction of two different reactants.
The Half-Equation Method is one technique used to balance redox processes. The equation is divided into two half-equations using this technique: one for oxidation and one for reduction.
By changing the coefficients and adding , , and 
in that order, each reaction is brought into equilibrium:
- By putting the right number of water () molecules on the other side of the equation, the oxygen atoms are brought into balance.
- By adding ions to the opposing side of the equation, one can balance the hydrogen atoms (including those added in step 2 to balance the oxygen atom).
- Total the fees for each side. Add enough electrons () to the more positive side to make them equal. (As a general rule, and are nearly always on the same side.)
- The on either side must be made equal; if not, they must be multiplied by the lowest common multiple (LCM) in order to make them equal.
- One balanced equation is created by adding the two half-equations and canceling out the electrons. Additionally, common terms should be eliminated.
- Now that the equation has been verified, it can be balanced.
Learn more about redox reaction here,
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Answer:
The correct answer is B.
The 
is samller than 
of the reaction . So,the reaction will shift towards the left i.e. towards the reactant side.
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
For the given chemical reaction:
The expression for is written as:
![Q=\frac{[PCl_3][Cl_2]}{[[PCl_5]^1}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5B%5BPCl_5%5D%5E1%7D)
Given : = 0.0454
Thus as 
, the reaction will shift towards the left i.e. towards the reactant side.
(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 :)
Sucrose is insoluble in dichloromethane because dichloromethane is not a polar solvent.
Sucrose is soluble in water because the molecules of sucrose has the ability to react with the molecules of water and thereby forming hydrogen bond which enhance the dispersion of sucrose in the water.
Sugar can not react in a similar way with dichloromethane because they do not possess chemical species that can react together to form bonds. Thus, sugar is a polar substance which can not dissolve in a non-polar solvent.
