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
Answer:
The answer is "152 pm".
Explanation:
The bond length from the values inside the atomic radii is calculated according to the query. This would be the upper limit of a molecule's binding length.
The atomic radius of 
The atomic radius of 
Answer:
The kinetic energy of the translational motion of an ideal gas depends on its temperature.
Explanation:
<span>3 NO2 + H2O -------->. 2 HNO3. + NO
3(46g)------------------------> 2 ( 63g) HNO3
? kg-------------------------5.89 x10^3kg HNO3
Mass of NO2. = 5.89x10^3 x 138/ 2(63) = 6.45 x10^3 kg</span>
Answer:
164.3g of NaCl
Explanation:
Based on the chemical equation:
CaCl2 + 2NaOH → 2NaCl + Ca(OH)2
<em>where 1 mole of CaCl2 reacts with 2 moles of NaOH</em>
To solve this question we must convert the mass of CaCl2 to moles. Using the chemical equation we can find the moles of NaCl and its mass:
<em>Moles CaCl2 -Molar mass: 110.98g/mol-</em>
156.0g CaCl₂ * (1mol / 110.98g) = 1.4057 moles CaCl2
<em>Moles NaCl:</em>
1.4057 moles CaCl2 * (2mol NaCl / 1mol CaCl2) = 2.811 moles NaCl
<em>Mass NaCl -Molar mass: 58.44g/mol-</em>
2.811 moles NaCl * (58.44g / mol) = 164.3g of NaCl
