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:
Explanation:
The lewis structure (indicating all the atoms and patterns provided as hint in the question) of glycine can be seen in the attachment below. While the chemical structure of glycine can be seen below
H
|
H₂N - C - C =O
| \
H OH
The structure (of glycine) above provides a "fair idea" of how the lewis structure will be.
Answer:
A. 2NO + O2 -> 2NO2
B. 4Co + 3O2 -> 2Co2O3
C. 2Al + 3Cl2 -> Al2Cl6
D. 2C2H6 + 7O2 -> 4CO2 + 6H2O
E. TiCl4 + 4Na -> Ti + 4NaCl
So diffusion is inversely proportional to mass !
so as mass of the particle increases, diffusion decreases !
Answer:
92.04%
Explanation:
Given:
Mass of CO₂ obtained = 53.0 grams
Mass of calcium carbonate heated = 1.31 grams
Now,
the molar mass of the calcium carbonate = 100.08 grams
The number of moles heated in the problem = Mass / Molar mass
= (1.31 grams) / (100.08 grams/moles)
= 0.013088 moles
now,
1 mol of calcium carbonate yields 1 mol of CO₂
thus,
0.013088 moles of calcium carbonate will yield = 0.013088 mol of CO₂
now,
Theoretical mass of 0.013088 moles of CO₂ will be
= Number of moles × Molar mass of CO₂
= 0.013088 × 44 = 0.5758 grams
Thus, the percent yield for this reaction = 
or
the percent yield for this reaction = 
or
the percent yield for this reaction = 92.04%