Answer: This is a typical acid/base equilibrium problem, that involves the use of logarithms.
Explanation:We assume that both nitric acid and hydrochloric acid dissociate to give stoichiometric
H
3
O
+
.
Moles of nitric acid:
26.0
×
10
−
3
⋅
L
×
8.00
⋅
m
o
l
⋅
L
−
1
=
0.208
⋅
m
o
l
H
N
O
3
(
a
q
)
.
And, moles of hydrochloric acid:
88.0
×
10
−
3
⋅
L
×
5.00
⋅
m
o
l
⋅
L
−
1
=
0.440
⋅
m
o
l
H
C
l
(
a
q
)
.
This molar quantity is diluted to
1.00
L
. Concentration in moles/Litre =
(
0.208
+
0.440
)
⋅
m
o
l
1
L
=
0.648
⋅
m
o
l
⋅
L
−
1
.
Now we know that water undergoes autoprotolysis:
H
2
O
(
l
)
⇌
H
+
+
O
H
−
. This is another equilibrium reaction, and the ion product
[
H
+
]
[
O
H
−
]
=
K
w
. This constant,
K
w
=
10
−
14
at
298
K
.
So
[
H
+
]
=
0.648
⋅
m
o
l
⋅
L
−
1
;
[
O
H
−
]
=
K
w
[
H
+
]
=
10
−
14
0.648
=
?
?
p
H
=
−
log
10
[
H
+
]
=
−
log
10
(
0.648
)
=
?
?
Alternatively, we know further that
p
H
+
p
O
H
=
14
. Once you have
p
H
,
p
O
H
is easy to find. Take the antilogarithm of this to get
[
O
H
−
]
.
Answer link
Increase the Surface Area of the Reactants. Increasing the surface area of the reactants increases the rate of the reaction. More surface area means more collisions of the reactant molecules and an increased rate of the reaction. This occurs when reactants are made to react in powdered form
a. The unit cell is the smallest group of atom which have overall symmetry of a crystal, and from which is the entire letters can be buled built up by repetition in 3 dimensions.
b. The volume(v) of the unit cell is equal to the cell edge length (a)cubed.
c. density of polonium is 9.32g/cm3.
Explanation:
If you bring an unopened bag of potato chips aboard an airplane you'll notice that the bag puffs up as the plane ascends to a higher altitude explain why this happens using the information and terminology you learned in this module
Answer:
B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2
Explanation:
Hello,
In this case, we should understand oxidizing agents as those substances able to increase the oxidation state of another substance, therefore, in B. reaction we notice that copper oxidation state at the beginning is zero (no bonds are formed) and once it reacts with nitric acid, its oxidation states raises to +2 in copper (II) nitrate, thus, in B. Cu + 4HNO3 → Cu(NO3)2 + 2H2O + 2NO2 nitritc acid is acting as the oxidizing agent.
Moreover, in the other reactions, copper (A.), sodium (C. and D.) remain with the same initial oxidation state, +2 and +1 respectively.
Regards.
